Francis Bonahon – Low Dimensional Geometry

STUDENT MATHEMATICAL LIBRARY

SI IAS/PARK CITY MATHEMATICAL SUBSERIES

Volume 49

Low - Dimensional

Geometry

From Euclidean Surfacesto Hyperbolic Knots

Francis Bonahon

American Mathematical Society,Providence,Rhode Island

Institute for Advanced Study, Princeton, New Jersey

Editorial Board of the Student MathematicalLibrary

Gerald B. Folland Brad G. Osgood(Chair)

Robin Forman Michael Starbird

Series Editor for the Park City Mathematics Institute

John PolkingThe photo on the back coverof the author is reprinted with the

permission of the USC Department of Mathematics.

2000 Mathematics Subject Classification. Primary 51M05,51M10,30F40,

57M25.

For additional information and updates on this book, visit

www.ams.org/bookpages/stml-49

Library of CongressCataloging-in-Publication Data

Bonahon, Francis, 1955-Low-dimensional geometry : from euclidean surfaces to hyperbolic knots /

Francis Bonahon.p. cm. -

(Student mathematical library ; v. 49. IAS/Park City mathematicalsubseries)

Includes bibliographical references and index.ISBN 978-0-8218-4816-6(alk. paper)1. Manifolds (Mathematics) 2. Geometry, Hyperbolic. 3. Geometry, Plane.

4. Knot theory. I. Title.

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\320\276la souns et a I'ecureuil

Table of contents

Table of contents

IAS/Park City Mathematics Institute

Preface

Chapter1. The euclidean plane

\302\2471.1. Euclidean length and distance

\302\2471.2. Shortest curves

\302\2471.3. Metric spaces

\302\2471.4. Isometries

Exercises for Chapter 1

Chapter 2. The hyperbolic plane

\302\2472.1. The hyperbolic plane

\302\2472.2. Some isometries of the hyperbolic plane

\302\2472.3. Shortest curves in the hyperbolic plane

\302\2472.4. All isometries of the hyperbolic plane

\302\2472.5. Linear and antilinear fractional maps

\302\2472.6. The hyperbolic norm

\302\2472.7. The disk model for the hyperbolic plane

viii Table of contents

Exercises for Chapter 2 39

Chapter3. The 2-dimensional sphere 47

\302\2473.1. The 2-dimensional sphere 47

\302\2473.2. Shortest curves 48

\302\2473.3. Isometries 49


Chapter4. Gluing constructions 55

\302\2474.1. Informal examples: the cylinder and the torus 55

\302\2474.2. Mathematical definition of gluings and quotient spaces 58

\302\2474.3. Gluing the edges of a euclidean polygon 61

\302\2474.4. Proofs of Theorems 4.3 and 4.4 67

\302\2474.5. Gluing hyperbolic and spherical polygons 79

Exercisesfor Chapter 4 84

Chapter 5. Gluing examples 89

\302\2475.1. Some euclidean surfaces 89

\302\2475.2. The surface of genus 2 97

\302\2475.3. The projective plane 102

\302\2475.4. The cylinder and the Mobius strip 103

\302\2475.5. The once-punctured torus 114

\302\2475.6. Triangular pillowcases 125


Chapter6. Tessellations 133

\302\2476.1. Tessellations 133

\302\2476.2. Complete metric spaces 134

\302\2476.3. Prom gluing polygon edges to tessellations 135

\302\2476.4. Completeness and compactness properties 147

\302\2476.5. Tessellations by bounded polygons 155

\302\2476.6. Tessellations by unbounded polygons 161

\302\2476.7. Incomplete hyperbolic surfaces 163

Table of contents ix

\302\2476.8. Poincare's polygon theorem 169


Chapter7. Group actions and fundamental domains 185

\302\2477.1. Transformation groups 185

\302\2477.2. Group actions and quotient spaces 187

\302\2477.3. Fundamental domains 192

\302\2477.4. Dirichlet domains 197


Chapter8. The Earey tessellation and circle packing 207

\302\2478.1. The Farey circle packing and tessellation 207

\302\2478.2. The Farey tessellation and the once-punctured torus 212

\302\2478.3. Horocircles and the Farey circle packing 214

\302\2478.4. Shearing the Farey tessellation 217


Chapter 9. The 3-dimensionalhyperbolicspace 227

\302\2479.1. The hyperbolic space 227

\302\2479.2. Shortest curves in the hyperbolic space 230

\302\2479.3. Isometries of the hyperbolic space 231

\302\2479.4. Hyperbolic planes and horospheres 234


Chapter 10. Kleinian groups 241

\302\24710.1. Bending the Farey tessellation 241

\302\24710.2. Kleinian groups and their limit sets 247

\302\24710.3. First rigorous example: fuchsian groups 252

\302\24710.4. Poincare's Polyhedron Theorem 257

\302\24710.5. More examples of kleinian groups 265

\302\24710.6. Poincare, Fuchs and Klein 283


Chapter 11. The figure-eight knot complement 293

\317\207 Table of contents

\302\24711.1. Another crooked Farey tessellation 293

\302\24711.2. Enlarging the group \316\2238 294

\302\24711.3. Limit sets 299

\302\24711.4. The figure-eight knot 303


Chapter 12. Geometrization theorems in dimension 3 315

\302\24712.1. Knots 315

\302\24712.2. The Geometrization Theorem for knot complements 319

\302\24712.3. Mostow's Rigidity Theorem 324

\302\24712.4. Ford domains 326

\302\24712.5. The general Geometrization Theorem 340


Appendix. Tool Kit 355

\302\247T.l. Elementary set theory 355

\302\247T.2. Maximum, minimum, supremum, and infimum 358

\302\247T.3. Limits and continuity. Limits involving infinity 360

\302\247T.4. Complex numbers 361

Supplemental bibliography and references 365

Supplemental bibliography 365

References 369

Index 377

IAS/Park CityMathematics Institute

The IAS /Park City Mathematics Institute (PCMI) was founded in

1991 as part of the \"Regional Geometry Institute\" initiative of theNational ScienceFoundation. In mid-1993 the program found an

institutional home at the Institute for Advanced Study (IAS)inPrinceton,New Jersey. The PCMI continues to hold summer programsin

Park City, Utah.

The IAS/Park City Mathematics Institute encouragesbothresearch and education in mathematics and fosters interaction between

the two. The three-week summer institute offers programs for

researchers and postdoctoral scholars, graduate students,undergraduatestudents, high school teachers, mathematics educationresearchers,and undergraduate faculty. One of PCMI's main goals is to make

all of the participantsaware of the total spectrum of activities that

occur in mathematics education and research: wewish to involve

professional mathematicians in education and to bring modern conceptsin mathematics to the attention of educators. To that end the

summer institute features general sessions designed to encourageinteractionamong the various groups. In-year activities at sites around the

country form an integral part of the High School Teacher Program.

XI

\321\205\320\270 IAS/Park City Mathematics Institute

Each summer a different topic is chosen as the focus of the

Research Program and Graduate Summer School. Activities in the

Undergraduate Program deal with this topic as well. Lecturenotesfrom

the Graduate Summer School are published eachyear in the IAS/ParkCity Mathematics Series. Course materials from the UndergraduateProgram, such as the current volume, are now being publishedaspart of the IAS/Park City Mathematical Subseries in the Student

Mathematical Library. We are happy to make available more of the

excellent resources which have beendeveloped as part of the PCMI.

John Polking, SeriesEditor

April 13, 2009

Preface

About 30 years ago, the field of 3-dimensional topology wasrevolutionized by Thurston's Geometrization Theorem and by theunexpected appearance of hyperbolic geometry in purely topological

problems. This book aims at introducing undergraduate studentsto someof these striking developments. It grew out of notesprepared by the

author for a three-week coursefor undergraduates that he taught atthe Park City Mathematical Institute in June-July 2006. It coversmuch more material than these lectures, but the written version

intends to preserve the overall spirit of the course.Theultimate goal,

attained in the last chapter, is to bring the studentsto a level where

they can understand the statements of Thurston's Geometrization

Theorem for knot complements and, more generally, of the generalGeometrization Theorem for 3-dimensionalmanifolds recently proved

by G. Perelman. Another leading theme is the intrinsic beauty of

some of the mathematical objects involved, not just mathematicallybut visually as well.

The first two-thirds of the book are devoted to 2-dimensionalgeometry. After a brief discussion of the geometry of the euclidean

plane M2, the hyperbolicplaneH2, and the sphere \302\2472,we discuss the

construction of locally homogeneousspacesby gluing the sides of a

polygon. This leadsto the investigation of the tessellations that areassociated to such constructions,with a special focus on one of the

\320\245\320\250

xiv Preface

most beautiful objects of mathematics, the Farey tessellationof the

hyperbolic plane. At this point, the deformations of the Fareytessellation by shearing lead us to jump to one dimension higher, in order

to allow bending. After a few generalities on the 3-dimensionalhyperbolic space H3, we consider the crooked tessellationsobtained by

bending the Farey tessellation, which naturally leads us to discussing

kleinian groups and quasi-fuchsian groups. Pushing the bendingof

the Farey tessellation to the edge of kleinian groups, we reach thefamous example associated to the complement of the figure-eight knot.

At this point, we are readyto explain that this example is amanifestation of a general phenomenon. We state Thurston'sGeometrization

Theorem for knot complements, and illustrate how it has

revolutionizedknot theory in particular through the use of Ford domains. The

book concludes with a discussion of the very recently proved

Geometrization Theorem for 3-dimensional manifolds.

We tried to strikea balancebetween mathematical intuition and

rigor. Much of the material is unapologetically \"picture driven\", aswe intendedto shareour own enthusiasm for the beauty of some of

the mathematical objects involved. However, we did not want to

sacrifice the other foundation of mathematics, namely, the level of

certainty provided by careful mathematical proofs. One drawback

of this compromiseis that the exposition is occasionally interruptedwith a few proofs which are more lengthy than difficult, but can

somewhat break the flow of the discourse. When this occurs, thereaderis encouraged,on a preliminary reading, to first glance at theexecutive summary of the argument that is usually present at its

beginning, and then to grab the remote control dgT and press theV

\"fast forward\" button until the first occurrence of the closing symbol

^0. The readermay later need to return to some of the parts that

have thus been zapped through, for the sake of mathematical rigor orbecausesubsequent parts of the book may refer to specificarguments

or definitions in these sections. For the samereason,the bookis not

intended to be read in a linear way. The reader is strongly advisedtogenerously skip, at first, much of the early material in order to reach

the parts with pretty pictures, such as Chapters 5, 6, 8, 10or 11,as

Preface xv

quickly as possible, and then to backtrackwhen specific definitions

or arguments are needed.

The bookalsohas its idiosyncrasies. Prom a mathematical pointof view, the main one involves quotients of metric spaces. It is

traditional here to focus only on topologicalspaces,to introduce the

quotient topology by fiat, and then to claim that it accurately describes

the intuitive notion of gluing in cut-and-paste constructions; this isnot always very convincing. A slightly less well-trodden road involves

quotient metric spaces,but only in the case of quotients underdiscontinuous group actions. We decided to follow a different strategy,

by discussing quotient (semi-)metrics very early on and in their full

generality. This approach is, in our view, much more intuitive but

it comes with a price: Someproofs become somewhat technical. Onthe one hand, these can serve as a good introduction to thetechniques of rigorous proofs in mathematics. On the other hand, thereaderpressedfor time can also take advantage of the fast-forward

commands where indicated, and zap through these proofs in a first

reading.

Prom a purely technical point of view, the text is written in sucha way that, in theory, it does not require much mathematical

knowledge beyond multivariable calculus. An appendix at the endprovides a \"tool kit\" summarizing some of the main concepts that will

be needed. In practice, however, the mathematical rigor of manyarguments is likely to require a somewhat higher level of mathematical

sophistication. The reader will also notice that the level of difficulty

progressively increases as one proceeds from early to later chapters.Each chapter ends with a selection of exercises, a few of which can

be somewhat challenging. The idea was to provide material suitable

for an independent study by a dedicatedundergraduate student, or

for a topics course. Sucha coursemight cover the main sections of

Chapters 1-7, 9, 12,and whichever parts of the remaining chapterswould be suitable for both the time available and the tastes of theinstructor.

The author is delighted to thank Roger Howefor tricking him into

believing that the PCMI coursewould not require that much work

(which turned out to be wrong), and Ed Dunne for encouraging him

xvi Preface

to turn the original lecture notes into a book and for warning him that

the task would be very labor intensive (which turned out to be right).The general form of the book owesmuch to the feedback received fromthe students and faculty who attended the PCMI lectures, and who

were used as \"guinea pigs\"; this includes Chris Hiatt, who was the

teaching assistant for the course.Dave Puter provided numerous and

invaluable comments on an earlier draft of the manuscript, Roland

van der Veen contributed a few more, and Jennifer Wright Sharp

polished the final version with her excellent copy-editing. Finally,the mathematical content of the book was greatly influencedby theauthor's own research in this area of mathematics, which in recent

years was partially supported by Grants 0103511and 0604866 from

the National Science Foundation.

Chapter 1

The euclidean plane

We are all very familiar with the geometry of the euclidean plane K2.We will encounter a new type of 2-dimensionalgeometry in the next

chapter, that of the hyperbolicplane H2. In this chapter, we first

list a series of well-known properties of the euclidean plane which,

in the next chapter, will enable us to develop the properties of the

hyperbolic plane in very close analogy.Before proceeding,you are advised to briefly consult the Tool

Kit in the appendix for a succinct review of the basic definitions and

notation concerning set theory, infima and suprema of sets of realnumbers, and complex numbers.

1.1. Euclidean length and distance

The euclidean plane is the set

\320\2262={(\321\205,\321\203);\321\205,\321\203\342\202\254\320\251

consisting of all ordered pairs (x,y) of real numbers \317\207and y.

If 7 is a curve in R2, parametrized by the differentiable vector-

valued function

t \320\275->(x(t),y(t)), \316\261<\316\257<6,

1

2 1. The euclidean plane

2/m

7

0<f

Figure 1.1. The euclidean plane

its euclidean length ^euc(7) is the arc length given by

(1.1) = (7)= f V*

J a(t)2+y'(t)*dt.

This length is independent of the parametrization by a well-known

consequenceof the chain rule.

It will be convenient to consider piecewise differentiable curves

7 made up of finitely many differentiable curves 71, 72, ..., 7n such

that the initial point of each 7i+i is equalto the terminal point of 7$.In other words, sucha curve 7 is differentiable everywhere except at

finitely many points, corresponding to the endpoints of the 7*, where

it is allowed to have a \"corner\" (but no discontinuity). In this case,the length ieac (7) of the piecewise differentiablecurve 7 is defined as

the sum of the lengths 4uc(7i) of its differentiable pieces 7*. Thisis equivalent to allowing the integrand in (1.1) to be undefined at

finitely many values of t where, however, it has finite left-hand and

right-hand limits.The euclideandistancedeuc(P,Q) between two points \316\241and

Q is the infimum of the lengths of all piecewise differentiable curves

7 going from \316\241to Q, namely

(1.2) 4uc (P, Q) = inf {4uc (7); 7 goes from \316\241to Q} .

1.3. Metric spaces 3

Seethe Tool Kit in the appendix for basic facts about the infimum

of a set of real numbers. By definition of the infimum, the abovedefinition means that every piecewise differentiable curve 7 going from

\316\241to Q must have length greater than or equal to deac(P,Q),and that

there are curves whose length is arbitrarily close to deac(P, Q).

1.2. Shortest curves

It is well known and easily proved (seeExercise1.2)that the straight

line provides the shortest route between two points.

Proposition 1.1. The distance deuc(P,Q) is equal to the euclidean

length ieac ([P,Q]) of the line segment [P,Q] going from \316\241to Q. In

other words, [P,Q] is the shortest curve going from \316\241to Q. D

In particular, computing the length of a line segment by usingformula (1.1) for arc length (see Exercise 1.1), weobtain the following.

Corollary 1.2. The euclidean distance from Pq = (xo,yo) to P\\ =

(xi,yi) is equal to

(1.3) deuc(P0,Pi)= vO^i-

*o)2 + (2/1-

2/o)2 \342\226\241

1.3. Metric spaces

The euclidean plane R2, with its distance function deUC, is a

fundamental example of a metric space. A metric apace is a pair (X, d)consistingof a set X together with a function d: X x X \342\200\224>R such

that

(1) d(P, Q)^0 and d(P,P) = 0 for every P,QeX;

(2) d(P, Q) = 0 if and only if \316\241= Q;

(3) d(Q, P) = d(P,Q) for every P,QeX;

(4) d(P, R) < d(P,Q) + d(Q, R) for every P,Q,R\302\243 X.

The fourth condition is the Triangle Inequality. The function d

is called the distance function, the metric function,orjust the

metric of the metric space X.A function d that satisfies only conditions (1), (3) and (4) above

is called a semi-distance function or a semi-metric.


Elementary and classicalpropertiesof euclidean geometry show

that (M2,deuc) is a metric space. In particular, this explains the

terminology for the Triangle Inequality. In fact, (M2,deuc) and its

higher-dimensionalanalogsare typical examplesof metric spaces. See

Exercise 1.3 for a proof that deuc is a distance function which, instead

of prior knowledge about euclidean geometry, uses only the definition

of the euclideandistanceby equation (1.2).

The main point of the definition of metric spaces is that thenotions about limits and continuity that one encounters in calculus(seeSection T.3 in the TOOL Kit) immediately extendto the wider

context of a metric space (X, d).For instance,a sequenceof points P1; P2, ..., P\342\200\236,... in .\316\233\316\223

converges to the point P^ if, for every \316\265> 0, there exists an integer\321\211such that d(Pn,Poo) < \316\265for every \316\267^ no- This is equivalent tothe propertythat the sequence (d(P\342\200\236,Poo)) N converges to 0 as a

sequenceof real numbers. The point Poo is the limit of the sequence

(Pn)n\342\202\254N\302\267

Similarly, a function \317\210:X \342\200\224>X' from a metric space (X,d)to a metric space (X',d') is continuous at Pq \342\202\254X if, for everynumber \316\265> 0, there exists a \316\264> 0 such that d'{ip{P), \317\206(\316\241\316\277))

< \316\265for

every \316\241\302\243X with d(P, Pq) < \316\264.The function is continuous if it is

continuous at every Pq G X.

We will make extensive use of the notion of a ball in a metricspace(X,d). The (open) ball with center Pq e X and radius r > 0

in (X, d) is the subset

Bd(P0,r)= {P\342\202\254X;d(P,P0) < r}.

The terminology is motivated by the case where X is the 3-dimensionaleuclideanspaceM3, and where d is the euclidean metric deuc defined

by the property that

4uc(Po, Pi) = V(xl ~\317\207\316\270)2+ \320\253

~\320\243\320\276)2+ (Zl

~Z0)2

when P0 = (zo.2/0,z0)and Px = (xx,yi,zi). In this case, Bdeuc(P0,r)is of course a geometric ball of radius r centeredat Po, without its

boundary.

1.4. Isometries 5

When (X,d) is the euclidean plane (R2,deuc)>a ball Bdeac(Po,r)is an open disk of radius r centeredat Pq. When {X, d) is the realline (R,d) with its visual metric d(x,y) =

\\x,y\\, the ball Bdeuc(xo,r)is just the open interval (xq

\342\200\224r,xo + r).

Incidentally, this may be a goodspot to remind the reader of

a few definitions which are often confused. In the euclidean planeR2, the open disk Bdeuo(P0,r)

= {P \342\202\254R2;deac(P,P0) < r} is nott,he same thing as the circle {P G R2;deUC(P, P0)

= r} with center

Pq and radius r that bounds it. Similarly, in dimension 3, the openball

\316\222^^\316\241\316\277,\316\263)=

{\316\241\342\202\254R3;deuc(P,PQ) < r} should not be confusedwith the sphere {P \342\202\254R3;deuc{P, Pq) = r} with the same center and

radius.

1.4. Isometries

The euclideanplane has many symmetries. In a metric space, thesearecalledisometries.An isometry between two metric spaces (X, d)and {X',d') is a bijection \317\206\302\267.X \342\200\224>X' which respects distances,namely, such that

d>(<p(P)MQ))=d(P,Q)

for every P,Q \342\202\254X.

Recall the statement that \317\210is a bijection means that \317\206is one-to-

one (or injective) and onto (or surjective), so that it has a well-defined

inverse \317\210-1: X' \342\200\224>X. It immediately follows from definitions that

the inverse \317\210-1of an isometry \317\210is also an isometry.

It is also immediate that an isometry is continuous.

When there existsan isometry \317\206between two metric spaces (X, d)and (X', d'), then these two spaces have exactly the sameproperties.Indeed, \317\210can be used to translate any property of (X, d) to the same

property for (X, d').

We are here interested in the case where (X, d) = {X',d') =(M2,deuc)\302\267 Isometries of (R2,deuc) include:

\342\200\242translations along a vector (xo,yo), defined by

\302\245>(s.2/)= (x + xo,y + yo);

6 1. The euclideanplane

\342\200\242rotations of angle \316\270around the origin,

\317\206(\317\207,\321\203)=

(\317\207cos \316\270\342\200\224\321\203sin \316\270,\317\207sin \316\270+ \321\203cos \316\270);

\342\200\242reflections across a line passing through the origin and

making an angle of \316\270with the x-axis,

\317\206(\317\207,\321\203)=

(\317\207cos 2\316\230+ y sin 20, \317\207sin 2\316\230\342\200\224

\321\203cos 20);

\342\200\242more generally, any composition of the above isometries,

namely, any map \317\210of the form

(1.4) \317\210(\317\207,\316\275)=

(\317\207cos0 \342\200\224\321\203\316\262\316\257\316\261\316\270+ xq, x sinO + ycos0 + yo)

or

(1.5) \317\206(\317\207,\316\275)=

(xcos20 + ?/sin20 + a;o,xsin20 \342\200\224\321\203cos 2\316\230+ yo)-

For the last item, recall that the compositionof two maps

\317\206:X \342\200\224>\316\245and -\317\210:\316\245\342\200\224>\316\226is the map \317\210\316\277\317\210:\316\247\342\200\224>\316\226defined by

\321\204\316\277\317\206(\316\241)

=\317\210(\317\206{\316\241))for every Pe X.

The above isometries are better expressed in terms of complex

numbers, identifying the point (x, y) \342\202\254R2 with the complex number\316\266= \317\207+ \\y \342\202\254C. See Section T.4 in the Tool Kit for a brief summary

of the main properties of complexnumbers.

Using Euler's exponential notation (see Section T.4)

e'e = cos\316\270+ i sin \316\230,

the isometries listed in equations (1.4) and (1.5)can then be written

as

\317\210(\316\266)= ewz + z0

and

\317\210(\316\266)= e2i9z + zq,

where Zq= xq + iyo and where \316\266= \317\207\342\200\224

\\y is the complex conjugateof \316\266= \317\207+ iy.

Proposition 1.3. If \317\210is an isometry of(\342\204\226.2,deuc)= (C,deuc), then.

there exists a point zq \342\202\254\320\241and an angle \316\270\342\202\254R such that

\317\210(\316\266)= el9z + zq or \317\206(\316\266)

= e2l9z + z0 )

for every \316\266\342\202\254\320\241.


Proof. SeeExercise2.3(and compare Theorem 2.11) for a proof of

this well-known result in euclidean geometry. D

A fundamental consequence of the abundance of isometriesof the

euclidean plane (R2, deac) is the homogeneity of this metric space. A

metric space (X, d) is homogeneous if, for any two points P,Q \302\243X,

there exists an isometry \317\210:X \342\200\224*X such that \317\210(\316\241)= Q. In other

words, a homogeneous metric space looks the same at every point,sinceany property of (X, d) involving the point \316\241also holds at anyother point Q, by translating this property through the isometry \317\206

sending \316\241to Q.

Actually, the euclidean plane is not just homogeneous, it is

isotropic in the sense that for any two points Pi and P2 \342\202\254R2, and

for any unit vectors v\\ at P\\ and v2 at P2, there is an isometry \317\210of

(M2, deuc) which sends P\\ to P2 and v\\ to $2. Here we are assumingthe statement that \317\210sends the vector \321\211to the vector v2 is intuitively

clear; a more precise definition, using the differential \316\214\317\201\317\207\317\210of \317\210at

Pi, will be given in Section 2.5.2.As a consequence of the isotropy property, not only does the

euclidean plane look the sameat every point, it also looks the samein every direction.


Exercise1.1.Using the expression given in equation (1.1) and a suitable

parametrization of the line segment [\316\241\316\277,\316\241\316\271]going from Po = (rccj/o) toPi =

(rci, 1/1), show that the euclidean length of [P0, Pi] is equal to

4uc([Po,Pi])= \\/(si-So)2+ (2/i-2/o)2.

Exercise 1.2. The goal of this exercise is to rigorously prove that the line

segment [P, Q] is the shortest curve going from \316\241to Q. Namely, considera piecewisedifferentiable curve 7 going from \316\241to Q. We want to show

that the euclidean length ^0(7) dennedby equation (1.1) is greater than

or equal to the length &s\342\200\236c([P,Q]) of the line segment [P,Q].a. First consider the case where \316\241= (xq, j/o) and Q = (xo,2/1) sit on the

same vertical line of equation \317\207= xo- Show that the euclideanlength

4uc(7) is greater than or equalto |t/i\342\200\224

t/o| =4uc([P,Q])\302\267

b. In the general case, let \317\206:{\317\207,\321\203)\302\267-\342\226\272(rccosfl

\342\200\224i/sin0,:rsin0 + t/cos0) be

a rotation such that \317\210{\316\241)and \317\210{0)sit on the same vertical line. Show


that the curve ^(7), going from <p(P)to \317\206(0), has the same euclideanlength as 7.

\321\201Combine parts a and b above to conclude that 4uc(7) > 4uc([P,Q])\302\267

Exercise 1.3. Rigorously prove that the euclidean distance function deuc,

as defined by equation (1.2), is a distancefunction on R2. You may needto use the result of Exercise 1.2 to show that deuc(P, Q) = 0 only when

\316\241= Q. Note that the proof of the Triangle Inequality (for which you mayfind it useful to consult the proof of Lemma 2.1) is greatly simplified by

our use of piecewise differentiable curves in the definition of deuc \302\267

Exercise 1.4. Let (X, d) be a metric space.

a. Show that d(P, Q) - d(P,Q')< d{Q, Q') for every P, Q, Q' e X.b. Conclude that \\d{P, Q) - d{P,Q')\\ < d{Q, Q') for every P, Q,Q' \342\202\254X.

\321\201Use the above inequality to show that for every \316\241\342\202\254X, the function

dp: X \342\200\224\302\273\320\232defined by dp(Q) = d(P, Q) is continuous if we endow thereal line \320\232with the usual metric for which the distance between a and6 \342\202\254\320\232is equal to the absolute value |o

\342\200\2246|.

Exercise 1.5. Let \317\206:X \342\200\224\302\273X' be a map from the metric space(X,d)tothe metric space (X', d'). Show that \317\206is continuous at Po \342\202\254X if and onlyif, for every \316\265> 0, there exists a \316\264> 0 such that the image \317\206{\316\222\316\254{\316\241\316\277,\316\264))of

the ball \316\222\316\254(\316\2410,\316\264)\320\241X is contained in the ball\316\222\316\254>(\317\206(\316\2410),\316\265)\320\241\320\245'.

Exercise 1.\316\230(Product of metric spaces). Let (X, d) and (X',d') be two

metric spaces. On the product \316\247\317\207X' ={(\317\207,\317\207');\317\207\342\202\254\316\247,\317\207'\342\202\254\316\247'},define

D: {\316\247\317\207\316\247')\317\207(\316\247\317\207\316\247')-\302\273\316\232

by the property that D{[x,x'),(y,y')) = max.{d(x,y),d'(x',y')}for every

(re, re'), (y,y') \342\202\254X x X'. Show that D is a metric function on \316\247\317\207\316\247'.

Exercise 1.7. On \316\2322= \316\232\317\207\316\232,consider the metric function D provided by

Exercise 1.6. Namely, D((x,y), (x',y')) =max{|rc

\342\200\224x'\\, \\y

\342\200\224y'\\} for every

\342\226\240

(x,y),(x',y')eR2.

a. Show that ^deuc(P,f) < D(P,P') < deuc(P,P')for every \316\241,\316\2411\342\202\254

K2.

b. Let (Pn)neN be a sequence in K2. Show that {Pn)n& convergesto apoint Poo \342\202\254K2 for the metric D if and only if it converges to \320\240\320\266for

the metric

c. Let \317\206:\316\2322\342\200\224\302\273X be a map from K2 to a metric space (X, d). Show that\342\226\240

\317\206is continuous for the metric D on \320\2502if and only if it is continuousfor the metric deuc

Exercise1.8 (Continuity and sequences). Let \317\206:X \342\200\224\302\273X' be a map from :the metric space(X,d) to the metric space (X', d').


a. Suppose that \317\210is continuous at Po. Show that if Pi, Pi, ..., Pn, ...

is a sequence which converges to Po in (X,d), then \317\210{\316\241\316\271),\317\210{\316\2412),\342\226\240\342\226\240\342\226\240,

\317\210{\316\241\316\267),\342\226\240\342\226\240\342\226\240is a sequence which convergesto <p(Po) in (X', d').

b. Suppose that \317\206is not continuous at Po. Constructa number \316\265> 0

and a sequence Pi, P2, ..., P\342\200\236,... in X such that d(Pn, Po) < ^ and

d(<p{Pn),<p{Po))> \316\265for every \316\267> 1.

c. Combine parts a and b to show that \317\210is continuous at Po if and only if,for every sequence Pi, P2, ..., Pn, \342\226\240\342\226\240\342\226\240converging to Po in {X,d), the

sequence <p(Pi), \317\210{\316\2412),\342\226\240\342\226\240\342\226\240,\317\210{\316\241\316\267),\342\226\240\342\226\240\342\226\240converges to <p(Po) in (X', d').

Exercise 1.9. Let d and d' be two metrics on the same set X. Show that

the identity map Idx: (X, d) \342\200\224>(X, d') is continuous if and only if everysequence (Pn)neN that converges to some P\302\273\342\202\254X for the metric d alsoconvergesto P\302\273for the metric d'. Possible hint: Compare Exercise 1.8.

Exercise 1.10. The euclidean metric of the euclidean plane is an exampleof a path metric, where the distance betweentwo points \316\241and Q is theinfimum of the lengths of all curves joining \316\241to Q. In the plane K2, letU be the U-shaped region enclosed by the polygonal curve with vertices

(0,0), (0,2), (1,2), (1,1),(2,1),(2,2), (3,2), (3,0), (0,0) occurring in this

order. Endow U with the metric du defined by the property that du(P, Q)is the infimum of the euclidean lengths of all piecewisedifferentiable curves

joining \316\241to Q and completely contained in U.

a. Draw a picture of U.b. Show that du{P, Q) > deuc(P,Q) for every P,QeU.

\321\201Show that du is a metric function on U. It may be convenient to use

part b above at somepoint of the proof.

d. If P0 is the point (0,2), give a formula for the distance du{P, Po) asa function of the coordinates of \316\241=

(\317\207,\321\203).This formula will involveseveral casesaccording to where \316\241sits in U.

Exercise 1.11 (Lengths in metric spaces). In an arbitrary metric space

(X, d), the length \302\2430(7)of a curve 7 is defined as

^<j(7)=sup<^2d{Pi-i, Pi);Po, Pi, \342\226\240.\342\226\240,Pn\342\202\2547occur in this order along 7>.

In particular, the length may be infinite. For a differentiable curve 7 in

the euclidean plane (K2,deuc), we want to show that this length ^\320\270\321\201(7)

coincides with the euclidean length 4uc(7) given by equation (1.1). For this,

suppose that 7 is parametrized by the differentiable function

\316\257\316\227^7(\316\257)=

(\316\267;(\316\257),2/(\316\257)),\316\277<\316\257<6.

a. Show that 4jeuc(7) < 4uc(7)\302\267


b. Cut the interval [o, b] into \316\267intervals [U-i, U] of length \316\224\316\257=(b\342\200\224a)/n.

Set \316\241,= 7(ti) = 7(0 + iAt) for i = 0, 1, ..., n. Show that

deuc(Pi-i,Pi) > h'{U-i)\\\\bt-\\K{btY,

where \320\232= max ||7\"(*)|| denotes the maximum length of the second

derivative vector 7\"(i)=

(x'(t),y\"(t)) and where the length ||(\302\253,\302\253)||

of a vector (u, v) is defined by the usual formula ||(w, w)|| = y/u2 + v2.

You may need to use the Taylor formula from multivariable calculus,which says that for every t, h,

7(i + h) = 7(i) + /17'(t)+ h2R!(t,h),where the remainder Ri(t,h) is such that ||Jii(i,ft)|| < \\K.

c. Use part b aboveto show that 4jeuc(7) > 4uc(7).d. Combineparts a and \321\201above to conclude that ^deUC(7)= -4uc(7)\302\267

Exercise 1.12. Consider the length -\302\2430(7)of a curve 7 in a metric space

(X, d) defined as in Exercise 1.11, in the specialcasewhere (-X, d) = (K2, D)is the planeR2=Exl endowed with the product metric D ofExercise1.6,defined by the property that D((x, \321\203),(\321\205',\321\203'))

=\321\202\320\260\321\205{|\320\263\321\201

\342\200\224rc'|, \\y

\342\200\224y'\\} for

every (x,y), (x',y') e K2.

a. Show that ^0(7) > D(P,Q) for every curve 7 going from \316\241to Q.

b. Show that the length (-d{\\P,Q\\) of the line segment [P,Q]is equal to

D(P,Q), so that [P,Q] consequently has minimum length among allcurves going from \316\241to Q.

\321\201Give an example where there is another curve 7 going from \316\241to Qwhich has minimum length ^0(7) \342\200\224

D(P, Q), and which is not the line

segment [P, Q].d. If 7 is differentiably parametrized by 11-> (x(t),y(t)), a < t < b, give

a condition on the derivatives x'(t) and y'(t) which is equivalent to

the property that 7 has minimum length over all curves going from

\316\241=(x(a),y(a)) to Q \342\200\224

(x(b),y(b)). (The answer depends on therelative position of \316\241and Q with respect to eachother.)

Chapter 2

The hyperbolic plane

The hyperbolic plane is a metric space which is much less familiar

than the euclidean plane that we discussed in the previous chapter.We introduce its basic properties, by proceeding in very close analogy

with the euclidean plane.

2.1. The hyperbolic planeThe hyperbolic plane is the metric space consistingof the open

half-plane

H2 = {(x,y) \302\243R2; \321\203> 0} ={\316\266G \320\241;Im(*) > 0}

endowed with a new metric dhyp defined below. Recall that the

imaginary part Im(z) of a complex number \316\266= \317\207+ \\y is just thecoordinate y, while its real part Re(z) is the coordinatex.

Todefine the hyperbolic metric dhyp, we first define thehyperbolic length of a curve 7 parametrized by the differentiable vector-

valued function

t~(x(t),y(t)), a^t^b,

as

(2.1) W7) =ye \320\251

dt.

11

12 2. The hyperbolic plane

Again, an application of the chain rule shows that this hyperbolic

length is independent of the parametrization of 7. The definition

of the hyperbolic length also immediately extends to piecewise dif-ferentiable curves, by taking the sum of the hyperbolic length of thedifferentiablepieces,or by allowing finitely many jump discontinuitiesin the integrand of (2.1).

H2<

\320\243\320\277

P^Q'

Figure 2.1. The hyperbolic plane

The hyperbolic distance between two points \316\241and Q is the

infimum of the hyperboliclengths of all piecewise differentiable curves

7 going from \316\241to Q, namely

(2.2) dhyp(P, Q) = inf {^hyp(7); 7 goes from \316\241to Q}.

Note the analogy with our definition of the euclidean distance in

Chapter 1.

The hyperbolic distance dhypis at first somewhat unintuitive.

For instance, we will see in later sections that the hyperbolic distance

between the points P' and Q' indicated in Figure 2.1 is the sameasthe hyperbolic distance from \316\241to Q. Also, among the curves joining

\316\241to Q, the one with the shortest hyperbolic length is the circlearcrepresented.With practice, we will become more comfortable with

the geometry of the hyperbolic plane and see that it actually shares ;'

many important features with the euclideanplane.

2.1. Thehyperbolicplane 13

But first, let us prove that the hyperbolicdistancedhyp is really

a distance function.

Lemma 2.1. The function

defined by (2.2) is a distancefunction.

Proof. We have to check the four conditions in the definition of adistancefunction. The condition dhyp(P, Q) ^ 0 is immediate, as is

the symmetry condition dhyp(Q,P) =dhyp(P, Q).

To prove the Triangle Inequality, considerthree pointsP,Q, R\342\202\254

H2. Pick an arbitrary \316\265> 0. By definition of the hyperbolicdistanceas an infimum of hyperbolic lengths, there exists a piecewisedifferen-

tiable curve 7 going from \316\241to Q such that ^hyp(7) < dhyP(-P, Q) + \\\316\265,

and a piecewise differentiable curve 7' going from Q to R such that

<Vp(7') < ^hyp(Q, R) + \\s. Chaining together these two curves 7

and 7', one obtains a piecewisedifferentiable curve 7\" joining \316\241to R

whose length is

4yp(7\") =4yp(7) + 4yp(Y) < \320\260\321\212\321\203\321\200(\320\240,Q) + dhyp(Q, R) + e.

As a consequence,

dhyp(P, R) <dhyp(P, Q) + dhyp(Q, R) + \316\265.

Since this property holds for every \316\265> 0, we conclude thatdhyp(-P, R)

< dbyp(P, Q) + dbyP(Q,R) as required.Note that our use of piecewise differentiablecurves, insteadof just

differentiable curves, has greatly simplifiedthis proof of the Triangle

Inequality. (When 7 and 7' are differentiable, the same is usually not

true for 7\" since it may have a \"corner\" at the junction of 7 and 7').

The only condition which requires some serious thought is thefact that dhyP(-P, Q) > 0 if \316\241\316\246Q. Namely, we need to make surethat we cannot go from \316\241to Q by curves whose hyperbolic lengthsare arbitrarily small.

Consider a piecewise differentiable curve 7 goingfrom \316\241to Q,

parametrized by the piecewisedifferentiable function


with \316\241=(x(a),y(a)) and Q = (x(b),y(b)). We will split the

argument into two cases.

If 7 does not go too high, so that y(t) ^ 2y(a) for every t \342\202\254[a,b],

=(7)

\342\200\242^e\\ / \\ \"cut v. J ^ / \342\226\240

2\316\271/(\316\261)

Otherwise, 7 crosses the horizontal line L of equation 1/ = 2y(a).Let \316\257\316\277be the first value of t for which this happens; namely, \321\203(to)

=

2y(a) and y(t) < 2y(a) for every t < tQ. Let 7' denote the part of 7

corresponding to the values of t with \316\261^ \316\257< \316\2570-This curve 7' joins\316\241to the point (x(to),y(to)) G L, so that its euclidean length 4uc(7')is greater than or equal to the euclidean distance from \316\241to the line

L, which itself is equal to \321\203(a). Therefore,

'hyp (7)Ja

Ja

^ 1

VV(*)2+\302\273'(')2

^'(i)2+y'(i)adi_2y(a)

fL..JP.O\\_

1

2y(a)

\302\253hyp(7) ^ \302\253hyp(7 ) =/ ~\316\226\316\273 dt

>Ja 2y(a)M

2y(a)4uc(7)

>\321\210\321\203(\320\260)=*\302\267

In both cases, we found that ^hyP(7) ^ \320\241for a positive constant

C =mm{^deac(P,Q),\302\261}>0,

which depends only on \316\241and Q (remember that \321\203(a) is the y-coor-dinate of P). If follows that dhyp(-P, Q) ^ \320\241> 0 cannot be 0 if

P^Q. D

2.2. Some isometriesof the hyperbolic plane

The hyperbolic plane (H2,dhyp)has many symmetries. Actually, we

will see that it is as symmetric as the euclideanplane.

2.2. Someisometriesofthehyperbolic plane 15

2.2.1. Homotheties and horizontal translations. Someof these

isometries are surprising at first. These include the homothetiesdefined by \317\206(\317\207,\321\203)

=(\316\273\317\207,\316\273\317\207)for some \316\273> 0. Indeed, if the piecewise

differentiablecurve 7 is parametrized by

t~(x(t),y(t)), a^t^b,

its image 1^(7) under \317\206is parametrized by

\320\253(\\x(t),\\y(t)), a^t^b.

Therefore,

4\321\203\321\200(^(7))=

\320\233 xy(t)

*

_ fbV^W\302\261VW,,~L m

=4yp(7)\302\267

Since \317\206establishes a one-to-one correspondence between curvesjoining \316\241to Q and curves joining \317\206(\316\241)to tp(Q), it follows from thedefinition of the hyperbolic metric that dbyP(ip(P), tp(Q)) =

dhyP(P, Q)

for every P, QgH2. This proves that the homothety \317\210is indeed an

isometry of (H2,dhyp)\302\267

The horizontal translations defined by ip(x,y) = (x + xo,y)for some \320\2660G R are more obvious isometries of (H2, dhyP), as is the

reflection \317\206(\317\207,\321\203)=

(\342\200\224\317\207,\321\203)across the y-axis.

2.2.2. The homogeneity property of the hyperbolicplane.The isometries obtained by composing homotheties and horizontal

translations are enough to prove that the hyperbolicplaneishomogeneous. Recall that the composition of two maps \317\206:X \342\200\224>X' and

\342\226\240\317\210:X' -> X\" is the map \317\210\316\277\317\206:X -> X\" defined by ip\302\260tp{P)=

\342\226\240\317\206(\317\206(\316\241))

for every \316\241\302\243X. If, in addition, X, X' and X\"are metricspaces,if\317\206is an isometry from (X, d) to (X', d') and if \317\210is an isometry from

(X', d') to (X\",d\,") then \317\210\316\277

\317\206is an isometry from (X, d) to (X\",d\since

d\"(i>\316\277

\317\206(\316\241),\321\204\320\276<p(Q))

=\316\254\"(\317\210(\317\206(\316\241)),r/>(<p(Q)))

=<\316\244(\317\206(\316\241),\317\206(0))

= d(P,Q).


Proposition 2.2. The hyperbolic plane (H2,dhyP) is homogeneous.Namely, for every P, Q G H2, there exists an isometry \317\206of (\316\2342,dhyp)

such that \317\210(\316\241)= Q.

Proof. If \316\241= (a,b) and Q = (c,d) \342\202\254\320\2502,with b, d > 0, the

homothety \317\206of ratio \316\273=^ sends \316\241to the point R =

(^,\316\254)

with the same y-coordinate d as Q. Then the horizontal translar

tion \321\204(\321\205,\321\203)= (x + \321\201\342\200\224

^,y) sends RtoQ. The composition \317\210\316\277

\317\206

now provides an isometry sending \316\241to Q. D

2.2.3. The standard inversion. We now consider an even lessobvious isometry of (H2,dhyP)- The standard inversion, orinversion across the unit circle, or inversion for short, is defined

by

This map is betterunderstood in polar coordinates, as it sends the

point with polar coordinates [r, \316\230]to the point with polar coordinates

[\\,\316\270].See Figure 2.2.

(-1,0) (0,0) (1,0)Figure2.2.The inversion across the unit circle

Lemma 2.3. The inversion across the unit circle is an isometry ofthe hyperbolic plane (H2,dhyP).

Proof. If 7 is a piecewisedifferentiable curve parametrized by

i>-\302\273(x(t),y(t)), a^t^b,

2.3. Shortest curves in the hyperbolicplane 17

its image \317\210{\316\267)under the inversion \317\210is parametrized by

\316\257\316\271->(xi(t),yi(t)), \316\261^\316\257<6,

with

x(t)*+y(ty\342\204\242\"*1W

*(\316\257)2+ 2/(\316\257)2

'

Then

a/ (i) = (y(t)2-x(t)2)x'(t)-2x(t)y(t)y'(t)

(x(ty + y(t)*)2

und

= (*(*)'-y(0V(*)-2*(*M*V(*)

(\302\273(t)a + \302\273(*)a)a

\320\275\320\276that after simplifications,

*i(*) +\302\273i(*)~

, m2 , ma,a\302\267

\320\230*)2+\302\273(*)2)

It follows that

<-hyp (V(7))=

/'./a

-/ Ja

V^(*)2+2/i(i)2

2/1 (i)di

vMO' + lrt*)2^\302\273(*)

=4yp(7)\302\267

As before, this shows that the inversion \317\210is an isometry of the

hyperbolic plane. D

2.3. Shortest curves in the hyperbolic planeIn euclidean geometry, the shortest curve joining two points is the

line segment with these two points as endpoints. We want to identifyI.he shortest curve betweentwo points in the hyperbolic plane.

We begin with a special case.

Lemma 2.4. If Pq =(\321\205\320\276,\320\243\320\276),Pi = (xo,yi) \342\202\254H2 are located on

the same vertical line, then the Une segment [Pq, \316\241\317\207]has the shortest

hyperbolic length among all piecemse differentiable curves goingfrom


Po to Pi. In addition, the hyperbolic length of any other curve joiningPo to Pi has strictly larger hyperbolic length, and

dbyP(P0, Pi) = 4yP([Po, Pi]) = In

Figure2.3.Vertical lines are shortest

Proof. Assuming yo < yi without loss of generality, let us first com-'

pute the hyperbolic length of [Po, Pi]\302\267Parametrize this line segment

by

i\" (xo,t), \316\271/\316\277< t < yi.

Then,

1\321\212VP ([Po,Pi])= [V1^\302\261^dt= ^.

>yo\320\223 2/0

Now, consider a piecewise differentiable curve 7 going from P0 toPi, which is parametrized by

t^(x(t),y(t)), o<i<6.

Its hyperboliclength is

4yp(7)_ fb^WT7W-J Ja yd) dt>LWdt

fb\\\302\243Ja 2/1

\342\204\226\\

> rbmdt=inM iny\302\261=ebyp{[POiPl]yJa 2/(0 \320\243(\320\260) 2/0

^ u '

2.3. Shortest curvesin the hyperbolic plane 19

In addition, for the first term to be equal to the last one,the above

two inequalities must be equalities. Equality in the first inequality

requires that the function x(t) be constant, while equality in the second

one implies that y(t) is weakly increasing. This shows that the curve

7 is equal to the line segment [.\316\241\316\277,-\316\241\316\271]if ^hyp(7)=

^hyp([-Po,-Pi])\302\267\316\240

For future reference, we note the following estimate, which is

proved by the same argument as the second half of the proof of

Lemma 2.4.

Lemma 2.5. For any two points Pq = (xq,yo), Pi = (x\\,y\\) G H2,

dhyP(Po,Pi) ^ In**2/0

D

In our determination of shortest curves in the hyperbolic plane,the next step is the following.

Lemma 2.6. For any P, Q \302\243\320\2502that are not on the same verticalline, there exists an isometry of the hyperbolic plane (H2,dhyp) such

that \317\206(\316\241)and <p(Q) are on the same vertical line. In addition, the

line segment [\317\210(\316\241),<p(Q)] is the image under \317\206of the unique circle

arc joining \316\241to Q and centered on the x-axis.

Proof. Since\316\241and Q are not on the samevertical line, the

perpendicular bisector line of \316\241and Q intersects the \320\266-axis at some pointR. The point R is equidistant from \316\241and Q for the euclidean metric,so that there is a circle \320\241centered at R and passing through \316\241and

Q. Note that \320\241is the only circle passing through \316\241and Q that is

centered on the rc-axis.

The circle \320\241intersects the \320\266-axis in two points. Let \317\206\316\271be a

horizontal translation sending one of these points to (0,0). Then\320\241= \317\206\316\271(\320\241)

is a circle passing through the origin and centeredatsome point (a, 0).

In particular, the equation of the circle \320\241in polar coordinates is

\316\275= 2a cos \316\230.Its image under the inversion \317\2062is the curve with polar

coordinate equation r = r, namely, the vertical line L whose4 2acos0' ^equation in cartesian coordinates is \317\207= \342\200\224.


\316\2502\302\260\316\250\\{\316\241)

(i.0) (\316\261,\316\237)(0,0) R

Figure 2.4. Circlearcscenteredon the x-axis are shortest.

The composition \317\2102\316\277

\317\210\316\271sends the circle \320\241to the vertical line

L. In particular, it sendsthe points \316\241and Q to two points on thevertical line L. Restricting \317\2102

\316\277\317\2101to points in H2 then provides the

isometry \317\206of (H2,dhyp) that we were looking for. D

Lemma 2.6 can be extended immediately to the case where \316\241and

Q sit on the same vertical line L by interpreting L as a circleof infinite

radius whose center is located at infinity on the rc-axis. Indeed, thevertical line L of equation re = a can be seen as the limit as re tends

to +oo or to \342\200\224oo of the circle of radius \\x\342\200\224

a\\ centered at the point

(re, 0). With this convention, any two P, Q e H2 can be joined by a

unique circle arc centeredon the rc-axis.

Theorem 2.7. Among all curves joining \316\241toQ inM2, the circle arccentered on the x-axis (possibly a vertical line segment) is the unique

one that is shortest for the hyperbolic length ^\321\203\321\200.

Proof. If \316\241and Q are on the same vertical line, this is proved byLemma 2.4.

Otherwise,Lemma 2.6 provides an isometry \317\206sending \316\241and Q

to two points P' and Q' on the samevertical line L. By Lemma 2.4,the shortestcurve from P' to Q' is the line segment [P1,Q'\\. Since an

isometry sends shortest curves to shortestcurves, the shortest curve

from \316\241to Q is the image of the line segment [P\\ Q'] under the inverse

isometry \317\206~\317\207.By the second statement of Lemma 2.6,this imageisthe circlearcjoining \316\241to Q and centered on the rc-axis. D

.\317\21020\317\210\316\220{0\\,

2.3. Shortest curves in the hyperbolic plane 21

In a metric space where the distance function is defined by takingthe infimum of the arc lengths of certain curves, such as the euclideanplane and the hyperbolic plane, there is a technical term for \"shortest

curve\". More precisely, a geodesic is a curve 7 such that for every\316\241G 7 and for every Q G 7 sufficiently close to P, the sectionof 7

joining \316\241to Q is the shortest curve joining \316\241to Q (for the arc lengthconsidered).

For instance, Proposition 1.1 says that geodesies in the euclidean

plane (R2,deuc) are line segments,whereas Theorem 2.7 shows that

geodesies in the hyperbolicplane(H2,dhyp) are circle arcs centered onthe rc-axis.By convention, line segments and circle arcs may include

some, all, or none of their endpoints(in much the same way as aninterval in the number line R may be open, closedor semi-open).

A complete geodesic is a geodesic which cannot be extended to

a larger geodesic.From the above observations, complete geodesiesof the euclideanplaneare straight lines. Complete geodesies of the

hyperbolic plane are open semi-circlescenteredon the re-axis and

delimited by two points of the re-axis (including vertical half-lines

going from a point on the rc-axis to infinity).

For future reference, we now prove the following technical result.

Lemma 2.8. Let Pq = (0,yo) and Pi = (0, y{) be two points of the

upper half L = {(0,y);y > 0} \320\241H2 of the y-axis, with y\\ > yo,

and let g be a complete hyperbolic geodesic passing through Pq. SeeFigure 2.5. Then the following are equivalent:

(1) Pq is the point of g that is closest to Pi for the hyperbolicdistance dhyp/

(2) g is the complete geodesicgo that is orthogonal to L at Pq,namely, it is the euclidean semi-circle of radius yo centered

at (0,0) and joining (yo,0) to (\342\200\224yo,0).

Proof. Lemmas 2.4 and 2.5 show that for every point \316\241=(\320\270,\317\205)on

the geodesic go,

dbyP(Pi, P)^ln^ln^- =dhyP(Pi, Po).

\316\275 2/0


Figure 2.5

As a consequence, the point Po is closestto Pi among all points of

50\302\267

Conversely, if g is another complete hyperbolicgeodesicthat

passes through Pq and makes an angle of \316\270\317\206fwith L at Po, we ,

want to find a point \316\241\342\202\254g with dhyp(-Pi> P) < ^hyp(Pi> ^*o)\302\267

For \316\241= (\320\270,\316\275)\342\202\254g, the standard parametrization of the linesegment[\316\241\316\271,\316\241]gives that its hyperbolic length is equal to

4yp([Pl,P])= /JO

l^u2 + {y_yi)2

2/1 + t(v- t/i)

dt

y/u2 + {v-yi)2 ^yij/i

\342\200\224\317\205 \316\275

We now let the point \316\241=(\320\270,\317\205)vary on the geodesic g near Po\302\267

dvWhen \320\270= 0, we have that \316\275= yo and \342\200\224= cot \316\230.Differentiating

duthe above formula then gives that still at \320\270= 0,

^-4yp([Pl,P])=

-^COt^.du yo

In particular, unless \316\270=^, this derivative is different from 0 and

there existsnear Po= (0,2/o) a point \316\241= (it, v) of 5 such that

<W(Pl, P) < 4yp([Pl, P]) < 4yp([Pl,Po])= dhyp([Pl, Po])\302\267 \316\240

I

2.4. All isometries of the hyperbolic plane 23

2.4. All isometriesof the hyperbolic plane

So far, we have encountered three types of isometries of the

hyperbolic plane (H2,dhyp): homotheties, horizontal translations and theinversion. In this section, we describe all isometriesof (H2,dhyp).

It is convenient to use complexnumbers. In this framework,

H2 ={^eC;Im(2) >0},

where the imaginary part 1\321\202(.\320\263)is the y-coordinate of \316\266= \317\207+ iy.

In complex coordinates, a homothety is of the form \316\266\320\275->Xz for a

real number \316\273> 0, a horizontal translation is of the form \316\266\320\275->\316\266+ \317\207\316\277

with xq \342\202\254R, and the inversion is of the form \316\266\320\275->\321\211\320\267

=|, where

2: = \317\207\342\200\224iy is the complex conjugate of \316\266= \317\207+ iy and where \\z\\

=

sjx2 + y2 =\\fz~i is its modulus (also called absolute value).

We can obtain more examples of isometries by compositionof

isometries of these types. Recall that the composition\317\206\316\277\317\210of two

maps \317\206and \317\210is defined by \317\210\316\277\317\206(\316\241)

=\317\210(\317\206(\316\241)),

and that the

composition of two isometries is an isometry.

Lemma 2.9. All maps of the form

(2.3) \316\266\316\271\342\200\224>with a, b, c, d \302\243R and ad \342\200\224be = 1cz + d

or

(2.4) \316\266\320\275->\342\200\224-\342\200\224-with a,b,c,d \342\202\254R and ad\342\200\224bc= laz + b

are isometries of the hyperbolic plane (H2,dhyp)-

Proof. We will show that every such map isa composition of

horizontaltranslations \316\266\316\271\342\200\224>\316\266+ xq with xq \342\202\254R, of homotheties \320\263\320\275>\320\220\320\263with

\316\273> 0, and of inversions \316\266\320\275->4. Since a composition of isometries isan isometry, this will prove the result.

When \320\260\321\2040, the map of equation (2.4) is the composition of

6 1 1 \321\201\316\266\320\275->\316\266\316\233\342\200\224, \320\263\320\275-, \320\263\320\270\342\200\224\316\266\316\266and \316\266*-*\316\266-\\\342\200\224.

\316\261 \316\266 \316\261* \316\261

In particular, this map is the compositionof several isometries of H2

and is therefore an isometry of H2.

24 2. The hyperbolicplane

Composing once more with \316\266(-\342\226\272A, we obtain the map ofequation (2.3), thereby showing that this map is alsoan isometry of the

hyperbolic plane when \316\261\317\2060.

When a = 0, so that \321\201\321\2040, the map of equation (2.3) is thecomposition of

cz + b + d\316\266\320\275-> \342\200\224,

cz + d

which is an isometry of H2 by the previous case, and of the horizontal

translation \320\263\320\270\320\263-l.It follows that the map of equation (2.3)isalsoan isometry of H2 when a = 0.

Finally, composing with \316\266\320\275->i shows that the map of

equation (2.4) is an isometry of H2 when a = 0. Q

Conversely, we will show that every isometry of the hyperbolic\342\226\240

plane is of one the two types consideredin Lemma 2.9. The proof ofthis fact hinges on the following property.

Lemma 2.10. Let \317\206be an isometry of the hyperbolic plane (H2,dhyp)

such that ip(iy) =iy for every \321\203> 0. Then either \317\206(\316\266)

= \316\266for every-

\316\266\342\202\254\316\2272or \317\206(\316\266)= \342\200\224\316\266for every \316\266.

Proof. Let L = {iy; \321\203> 0} be the upper half of the y-axis. By j

hypothesis, \317\206fixes every point of L. !

For every \\y \342\202\254L, let gy be the unique hyperbolic completegeo-jdesic that passes through \\y and is orthogonal to L. Namely, gy

is \\

the euclidean semi-circle of radius \321\203centered at 0 and contained in ]H2. Since\317\206is an isometry and </?02/)

=*2/> we know that it sends gyi

to a completegeodesicg passing through iy. We will use Lemma 2.8 |to prove that g = gy. j

Indeed, this statement characterizesthe geodesicgy by the prop- J

erty that for any \321\203\320\263> \321\203,the point iy is the point of gy that is closest jto ij/i. As a consequence,since \317\210is an isometry, tp{\\y)

= iy is the ,

point of ip{gy)= g that is closest to <p(iyi)

= ij/i. Lemma 2.8 then I

shows that g =gy,

so that <p(gy)= gy. 1

Now, if \316\241= \320\270+ iv is a point of gy,its image <p(gy) is one of the \\

two points of gy that are at distancedhyP(-P, iy) from iy. One of these |two points is P, the other one is \342\200\224\320\270+ iv by symmetry. 1

2.4. All isometries of the hyperbolic plane 25

We conclude that ip(u+iv) = u+iv or \342\200\224u+'w for every u+iv \342\202\254H2

(since u + iv belongs to somegeodesicgy). Since \317\210is an isometry,it is continous. It follows that either ip(u + iv) = \320\270+ iv for every\320\270+ iv \342\202\254H2 or \317\206(\317\205,+ iv) = -u + iv for every u + iv \342\202\254\320\250\320\2232.This can be

rephrased as either \317\206(\316\266)= \316\266for every \316\266\342\202\254\316\2272or \317\206(\316\266)

= \342\200\224\316\266for every

\320\263\342\202\254\320\2352. D

A minor corollary of Lemma 2.9 is that \317\210(\316\266)= with a,

b, c, d G \320\232and ad \342\200\224be = 1 sends the upper half-space H2 to itself;

this can also be easily checked \"by hand\". This map is not definedat the boundary point \316\266= \342\200\224*. However, if we introduce a point ooat infinity of the real line R (without distinguishing between +oo and

-oo), the same formula defines a map

\317\210:R U {oo} -> R U {oo}

by setting \317\206(\342\200\224*)= oo and <p(oo) = |. This map is specially designed

to be continuous. Indeed,

lim \317\210(\317\207)= oo and lim \317\210(\317\207)

= -

in the \"obvious\" sense, which is made precise in Section T.3 of the

Tool Kit.

The same applies to a map of the form \317\210(\316\266)= \342\200\224\316\266\342\200\224-with \320\276,6,

az + \320\276

c, d \342\202\254R and ad\342\200\224be = 1. These extensions are often convenient, as

in the proof of the following statement.

Theorem 2.11. The isometnes of the hyperbolic plane (H2, dhyp) \320\276.\342\204\242

exactly the maps of the form

\317\210(\316\266)= with a,b,c,d\302\243 R and ad \342\200\224bc=l

cz + aor

\317\206(\316\266)= \342\200\224\316\266\342\200\224-mth a,b,c,d\302\243 R and ad \342\200\224bc\342\200\224\\.

az + b

Proof. We already proved in Lemma 2.9 that all maps of these two

types are isometries of the hyperbolic plane.Conversely, let \317\206be an isometry of H2, and consideragainthe

positive part L = {iy; \321\203> 0} of the y-axis. Since L is a complete


geodesic of H2, its image under the isometry \317\206is also a completegeodesic of H2, namely, a euclidean semi-circleboundedby two distinct

points \317\212,\317\205\316\276\316\234\317\205{oo}\302\267Here \320\270or \320\263;will be oo exactly when <p(L)is avertical half-line. In addition, if we orient L from 0 to oo, we requirewithout loss of generality that the corresponding orientation of y>(L)

goes from \320\270to \320\263;.

First, consider the case where \320\270and \320\263;are both different from oo.The hyperbolicisometry

. . az \342\200\224au

\320\244{*)=

.cz \342\200\224cv

with a and cEl chosenso that ac(u\342\200\224

\320\263;)= 1, sends \317\211to 0 and \320\263;to

oo. It follows that the composition \317\206\316\277\317\210fixes the two points 0 and

oo. As a consequence, the isometry \317\210\316\277\317\210sends the complete geodesic

L to itself, and respects its orientation. In particular, \317\210\316\277\317\206(\316\257)

= it

for some t > 0. Replacing a by \320\260/\\\320\224and \321\201by c\\/i in the definitionof \317\210,we can arrange that \317\210\316\277

\317\206(\317\212)= i. Then, \317\210\316\277

\317\206sends each \\y \342\202\254L

to a point of L that is at the same hyperbolic distance from i as iy;since \317\210\316\277\317\206respects the orientation of L, the only possibility is that

\321\204\320\276</?(\320\270/)

= iy for every \321\203> 0.

Applying Lemma 2.10, we concludethat either \317\210\316\277

\317\206(\316\266)= \316\266for

every z or \317\210\316\277

\317\206(\316\266)= \342\200\224\316\266for every z. In the first case,

where the formula for the inverse function \317\210~1is obtained by solvingthe equation \317\210(\316\266')

= \316\266(compare Exercise 2.10). In the second case,

,\321\207 ,_i, _\321\207 cvz + au

cz + a

In both cases,\317\210(\316\266)is of the type requested.

It remains to considerthe caseswhere \317\211or \320\263;is oo. The argumentis identical, using the isometries

V>(*)=

cz \342\200\224cv

with oc = 1 when \320\270= oo, and

2.5. Linear and antilinear fractionalmaps 27

with ac = 1 when \320\263;= oo. \316\240

The isometries of (H2, dhyp)of the form \317\206(\316\266)

= with a, b,CZ \"T~ u

c, del and ad \342\200\224be = 1 are called linear fractional \321\202\320\260\321\200\320\260with

real coefficients. Those of the form \317\206(\316\266)= \342\200\224\342\200\224-with \320\276,6, \321\201,d \342\202\254R

az + \320\276

and ad \342\200\224bc= 1 are antilinear fractional maps.

2.5. Linear and antilinear fractionalmapsIn this section we establish a few fundamental properties of linear and

antilinear fractional maps. Since later we will need to consider mapsof this type with arbitrary complex (and not just real) coefficients,

we prove these properties in this higher level of generality.

In this context, a linear fractional mapisa nonconstant map

\317\206of the form \317\210(\316\266)= - with complex coefficients a, b, c, d \302\243

cz + d\320\241Elementary algebra shows that \317\206is nonconstant exactly when

ad \342\200\224be \321\2040. Dividing all coefficients by one of the two complex

square roots \302\261y/ad\342\200\224

be, we can consequently arrange that ad\342\200\224bc= 1

without changing the map \317\206.We will systematically require that thecoefficientsa, b, c, d \342\202\254\320\241satisfy this condition ad \342\200\224be = 1.

So far, the map \317\206is not defined at \316\266=^. However, this can easily

be fixed by introducing a point oo at infinity of C. Let the Riemann

sphere be the union \320\241= \320\241U {oo}. Then a linear fractional \317\206(\316\266)=

with a, b, c, d \342\202\254\320\241and ad \342\200\224be = 1 defines a map \317\210:\320\241\342\200\224\342\231\246\320\241cz + d

by setting \317\210{\342\200\224^)= oo and </?(oo) = ^. This map is continuous for

the obvious definition of continuity at infinity because

lim \317\206(\316\266)= oo and lim \317\206\316\257\316\266)

=\342\200\224,

z-*-i\320\263-\320\276\320\276^4

'\320\241

where these limits involving infinity are defined exactly as in

Section T.3 of the Tool Kit, but replacing absolutevalues of real

numbers by moduli (= absolute values) of complexnumbers.

Similarly, a general antilinear fractional map is a map \317\206:

\320\241\342\200\224\342\231\246\320\241of the form \317\206(\316\266)=

\342\200\224\316\266\342\200\224-with \320\276,b, \321\201,d \342\202\254\320\241and ad \342\200\224bc= 1,az + b

with the convention that \317\206(\342\200\224|)= oo and </?(oo)

=|.


SeeExercise 2.8 for an explanation of why the Riemann sphere\320\241= \320\241U {oo} can indeed be considered as a sphere.Seealso

Exercise 2.12 for another interpretation of C, which sheds a different lighton linear fractional maps.

2.5.1. Some special (anti)linear fractional maps. We already

encountered the homotheties

\320\260\320\254+ \320\276\316\266\316\271-\342\226\272Xz =

\316\237\316\266+ \\-\317\204

with positive real ratio \316\273> 0. If we allow complex coefficients, we

can also consider the rotations

w e1^ \316\266+ 0\316\266>-> e \316\266=

0_\320\263+ \320\265~*2

of angle flgl around the origin, and the translationsZ + Zq

for arbitrary complex numbers \320\263\320\276\342\202\254\320\241

We also considered the inversion across the unit circle\316\266

_1 _ Oz + 1

Z\"\\z\\*~~z~ 5+ 0-

Lemma 2.12. Every linear or antilinear fractional map \317\210:\320\241\342\200\224\342\231\246\320\241\320\263$

\320\276composition of homotheties, translations, rotations, and inversionsacrossthe unit circle.

Proof. The proof is identical to the purely algebraicargument that

we already used in the proof of Lemma 2.9 for linear and linear frac- i

tional maps with real coefficients. D i

Actually, there is no reason to prefer the unit circle to any other

circle. If \320\241is the circle of radius R centeredat the point zq \342\202\254\320\241,\\

the inversion across the circle \320\241is the antilinear fractional map \317\206

defined by the property that

\317\210(\316\266)-zo = R2Z Z\302\260

\\z\342\200\224

zo\\z

2.5. Linear and antilinear fractional maps 29

or, equivalently, that

\320\267\320\260?_i_\320\2642~1\320\263\320\27612

**>=

Rl\\.l\342\226\240

Rz R

Namely, \317\210sends \316\266to the point that is on the sameray issued from zqus z, and it is at the euclideandistance i?2/deuc(^, Zq) from Zq. This

inversion fixes every point of the circle C, and exchangesthe inside

iind the outside of C.

There is an interesting limit case of inversions as welet the centerund the radius of the circle go to infinity. For given t, to and 9q \342\202\254R,

set zq = tel9\302\260and R = t \342\200\224to- If we let t tend to +00, the circle\320\241

converges to the line L that passes through the point tQee\302\260and makes

iin angle of \316\2700+ | with the rc-axis. On the other hand, the inversion

\317\206across \320\241converges to the map \316\266\316\271->\342\200\224e2l9\302\260\316\266+ 2\316\257\316\277\316\262\316\2719\302\260,which is justthe reflection across the line L.

In this way, if we interpret the line L as a circleof infinite

radius centered at infinity, we can also considerthe euclideanreflection

across L as an inversion across this circle.Note that every line L canbe obtainedin this way.

2.5.2. Differentials. Recall that if \317\206:U \342\200\224>R2 is a differentiable

function defined on a region U \320\241R2 by \317\206(\317\207,\316\275)=

(f(x,y),g(x,y)),the differential or tangent map of \317\206at a point Pq = (xo,yo) inthe interior of U is the linear map \316\214\317\2010\317\206:R2 \342\200\224>R2 with matrix

\320\250\321\200\320\276)%po)

\320\264\320\271(\320\240\320\276)\302\247\316\276(\316\241\316\277)

Namely,

\317\213\316\241\316\260\317\206{*)= DPofi*, b) =

(a\302\243\302\243(P0)+

\320\254\320\246(\320\2400),aff (P0) +ftg(fl)))

for every vector v = (a, b) \342\202\254R2.

The differential map \316\214\317\2010\317\206also has the following geometricinterpretation.

Lemma 2.13. Let the differentiable map \317\210-.U\342\200\224*\320\2262be defined over

a region U containing the point Pq in its interior. Then, for every

parametrized curve 7 in U which passes through Pq and is tangent to


the vector \316\275there, its image under \317\206is tangent to the vector \316\214\317\2010\317\210(\316\275)

at the point \317\206(\316\241\316\277).

\316\250{\316\212)\317\210{\316\257)

\317\210(\316\241\316\277)

DpMv)

Figure 2.6. The geometry of the differential map

Proof. Supposethat the map \317\206is given by \317\206(\317\207,\321\203)=

(f(x, \321\203),g(x, \321\203)),

'

and that the curve 7 is parametrizedby t \320\275+(rc(i), t/(i)). If the ;

point Pq corresponds to t =to, namely, if Pq = (x(to)~,y(to)), then 1

v =(x'(to),y'(t0)). i

The image of the curve 7 under \317\206is parametrized by

t \302\273<p(x(t),y(t))

=(f(x(t),y(t)),g(x(t),y(t))y

}

Applying the chain rule for functions of several variables, its tangent ;

vector at \317\210(\316\241\316\277)is equal to \\

jt(f(x(t),y(t)),9(mMt)))t=to1

={jtmt)Mt))t=t0,^9(m,y(t))t=t0) ]

=(^(PQ)x'(tQ)

+^(Po)y'(to),^(Po)x'(tQ)

+^(PQ)y'(tQ)^

J

=\317\213\316\241\316\277\317\210(\316\275). \342\226\241;!

6<

An immediate consequence of this geometric interpretation is the I

following property. |

Corollary 2.14.J

\316\267\316\241\316\277(\317\206\316\277\316\250)=

(\317\213\317\206(\316\241\316\277)\317\206)\316\277(\317\213\316\241\316\277\317\206).\316\261

I

The differential maps of linear and antilinear fractional maps have .

a particularly nice expressionin complexcoordinates.j

2.5. Linear and antilinear fractional maps 31

Proposition2.15.Ifthe linear fractional map \317\210is defined by \317\206(\316\266)=

where a, b, c, d \342\202\254\320\241with ad \342\200\224be = 1, its differential maprz + d

0\316\226\316\277\317\210:\320\241\342\200\224>\320\241at zq \342\202\254\320\241with zq \321\204\342\200\224

\\is such that

(czQ + d)2

for every \316\275\302\243\320\241

cz ~f~ d

If the antilinear fractional map \317\210is defined by \317\210(\316\266)= \342\200\224\342\200\224-

az + \320\276

where a, b, c, d \342\202\254\320\241with ad\342\200\224be= 1, its differential map \317\213\316\226\316\277\317\210:\320\241\342\200\224>\320\241

at -\320\263\320\276\342\202\254\320\241with zq \316\246\342\200\224

| is such that

for every \316\275\342\202\254\320\241.

Proof. We will use Lemma 2.13. Given \316\275\342\202\254\320\241interpreted as a vector,consider the line segment \316\267parametrized by t \320\275->z(t)

= zq + tv. Notethat z(0)

= zq and that z'(0) = v.

Lemma 2.13 then implies that

dzmv) =^(*(*))|t=0=\320\271\320\270\320\230*(\320\273))

-^(\302\260)))

'aZQ + ahv + b azo + b\\= lim

h V</i-\302\273oh \\ czq + chv + d czq + d)

= lim -. -\320\2237 rr-h-*0 (CZQ + ChV + d)(CZo + d)

_ 1\"(czo + d)2\"'

using the property that ad\342\200\224bc= 1.

The argument is identical for the antilinear fractional map \317\210. \316\217

For future reference, we note that the samecomputation yields:

Complement 2.16. If<p(z) = where ad\342\200\224beis not necessarily

equal to 1, then

D^{v)=i^Vdyv-D

32 2. Thehyperbolicplane

A consequence of Proposition 2.15 is that the differential map

of a linear fractional map is the composition of a homothety with

a rotation, and the differential map of an antilinear fractional mapis the composition of a homothety with a reflection. This has thefollowing important consequence.

Corollary 2.17. The differential map \316\214\316\226\316\277\317\206of a linear fractional

map \317\210respects angles and orientation in the sense that for any two

nonzero vectors \321\211,V2 \342\202\254C, the oriented angle from \316\214\316\226\316\261\317\210(\316\275{)to

DZoip(v2) is the same as the oriented angle from V\\ to v^,

measuringoriented angles counterclockwise in C.

The differential map DZoip of an antilinear fractional linear map\321\204respects angles and reverses orientation in the sense that for any \342\226\240\342\226\240\342\226\240

two nonzero vectors v\\, #2 \342\202\254C, the oriented angle from \316\214^\317\206^\316\263)to

DZoip(v2) is the opposite of the oriented angle from v*i to V2- \342\226\241;

Incidentally, Corollary 2.17 shows that a linear fractional map

cannot coincide with an antilinear fractional map.

2.5.3. (Anti)linear fractional maps and circles. Another

fundamental property of linear and antilinear fractional mapsis that they j

send circles to circles. For this, we have to include all lines as circlesof infinite radius centered at infinity. More precisely, let a circle inthe Riemann sphere \320\241= \320\241U {oo} be either a euclideancirclein \320\241or ;

the union L U {oo} of a line L \321\201\320\241and the point oo.

Proposition 2.18. A linear or antilinear fractional map \317\210:\320\241\342\200\224>\320\241

sends each circle of \320\241to a circle of \320\241

Proof. By Lemma 2.12, \317\206is a composition of homotheties, rotations,translations and inversions across the unit circle. Since homotheties,rotations and translations clearly sendcirclesto circles,it suffices to

consider the case where \317\206is the inversion across the unit circle.

It is convenient to use polarcoordinates.In polar coordinates r '

and \316\230,the circle \320\241of radius R centered at zq = \320\263\320\276&1\320\262\320\276has equation'\342\226\240

r2-2rr0cos(e~e0) + rl-R2 = 0. I

The inversion \317\206sends the point with polar coordinates [r, \316\230]to !

the point of coordinates[\302\243,0].

The image of the circle \320\241under \317\210is

2.6. The hyperbolic norm 33

therefore the curve of equation

_L_^\302\260cos(0_0o)+r.2-i?2 = o.

If Iro\316\231\316\246R or, equivalently, if the circle \320\241does not contain the

origin 0, simplifying the above equation shows that this curve is thecircleof radius 7-5 =rr; centered at

|\320\2632-\320\2242|

~rl-Rf

If \\zq\\= R, we get the curve of polar equation r =

2r0 cos(0-\316\2700)'

which of course is a line.

Finally, we need to consider the case where \320\241is a line. In polar

coordinates, its equation is of the form r = rr\342\200\224\342\200\224\320\263for some2r0 cos(0 -

\316\2300)

\320\223\320\276and 6q. Then its image under \317\210has equation r = 2rocos(0 \342\200\2246q),

and consequently it is a circle passingthrough the origin. D

2.6. The hyperbolic norm

If \316\275= (a, 6) is a vector in R2, its euclideanmagnitudeor euclidean

norm is its usual length

INeuc= Va*+ V.For instance, if \316\275is the velocity of a particle moving in the euclidean

plane, ||#||\320\265\321\206\321\201describes the speed of this particle.

In the hyperbolicplane, distancesare measured differently

according to where we are in the plane, and consequently so are speeds.If \317\205is a vector based at the point \316\266\342\202\254\316\2272\320\241\320\241,its hyperbolic norm

is

To justify this definition, let 7 be a curve in H2, parametrizedby

11\342\200\224\342\226\272z(t), a < t < 6. In particular, the tangent vector of 7 at the pointz(t) is the derivative z'(t), and must be considered as a vector based

at z(t). Then the euclidean and hyperboliclengths of 7 are given bythe very similar formulas

4uc(7) = /Vwileuc dt

Ja

34 2. The hyperbolic plan

and

4yP(7) = /Vwilhyp dt.Ja

If \317\206is a differentiable map and s is a vector based at P, its

image \316\214\317\201\317\206(\316\275)under the differential map is a vector based at \317\210(\316\241)

Indeed, see the geometric interpretation of the differential \316\214\317\201\317\210given

by Lemma 2.13.

Lemma 2.19. If \317\210is an isometry o/(H2, dhyp), then \\\\DZoip(v)\\\\. =

IKIIhyp for everV vector U based at zq \342\202\254H2.

Proof. We could go back to basic principles about the metric dhypr,

but it is easier to use a straight computation.

Consider the case where \317\210is a linear fractional \317\210(\316\266)= ;s'CZ + d;

with a, b, c, d G R and ad \342\200\224be = 1. By Proposition 2.15, if \317\205is

vector based at zq el2,

\\\\DzMme\\czQ + d\\2

On the other hand,1

MpW) =^ (v(zo)

-<\321\200\320\253)

=

%(

azo + b azo + bs

Zq\342\200\224

Zq

2i \\czQ + d\\2 \\czq + d\\2

czq + d

bn(z0).

czq+ d,

Therefore,

=

ImbNleuc

=l|iT|lhyp\302\267

The argument is essentially identical for an antilinear fractional/

\316\271\\cz + d

\317\200map \317\206(\316\266)

=-7\342\200\2247.

\342\226\241

az + b

2.6.1. The isotropy property of the hyperbolic plane. We now

show that like the euclidean plane, the hyperbolic plane H2 is iso-1

tropic. Recall that this means that not only can we send any point

z\\ \342\202\254H2 to any other point zi e H2 by an isometry \317\210of (H2,dhyP),'

2.6. The hyperbolic norm 35

but we can even arrange that \317\210sends any given direction at Z\\ to

uny arbitrary direction at z^. As a consequence,the hyperbolic plane

looks the same at every point and in every possible direction.

Proposition 2.20. Let V\\ be a vector based at Z\\ \342\202\254H2, and let v*2 be

\316\267vector based at z-i G H2 with||^i||hyp

=H^IL\302\267\342\204\242\302\267Then there is an

biometry \317\206of (\316\2272,dhyp) which sends z\\ to z^ and whose differentialmap \316\214\316\226\316\271\317\206sends v\\ to V2-

Proof. Let 9sKbe the angle from V\\ to v2 measured in the usual(iuclidean way, namely, after moving v\\ to the point z^ by a euclideantranslation of R2.

There exists c, d \342\202\254R such that cz\\ + d = \320\265-,\320\267.Indeed, finding

r. and d amounts to solving a linear system of two equations. If

Zi = X\\ + ij/i, one finds \321\201=-\317\201

sin\302\247

and d = cos | \342\200\224\317\210-

sin f, but the

precise value is really irrelevant.

Then one can find (many) a, b \342\202\254R such that ad \342\200\224be = 1. This

is again a simplelinear equation problem after observing that \321\201and

d cannot be both equal to 0.Let \317\210\\be the linear fractional defined by \317\210\\(\316\266)

=^^. Because

of our choiceof a, b, c, d, Proposition 2.15 shows that \316\217\316\226\316\271\317\206\\is

the complex multiplication by ew, namely, the rotation of angle \316\230.

As a consequence, still comparing angles and directionsin the usual

euclidean way, \320\263?\320\267=

DZl \317\206\316\271(\321\211)is parallel to \320\263\320\2232and points in the samedirection.

Let z%=

\317\206\316\271(\316\266\317\207).Let \317\2102be an isometry of H2 sending z% to

Z2. As in our proof of the homogeneity of H2 in Proposition 2.2, wecan even arrange that \317\2102is the composition of a homothety with a

horizontal translation, so that \316\217\316\2263\317\2102is a homothety. In particular,\316\217\316\2263\317\2102sends each vector to one which is parallel to it.

Then \317\210=

\317\2102\316\277\317\206\316\271sends \316\226\317\207to Zi, and its differential \316\214\316\226\316\271\317\206=

\317\213\316\2263\317\2102\316\277

\316\214\316\226\316\257\317\210\\sends Vi to a vector \320\251=

\317\213\316\226\316\271\317\206(\316\275\316\271)=

\316\214\316\2263\317\2102{\316\2753),

which is based at Z2 and is parallel to \320\263\320\2232.

By Lemma 2.19,

ll*2||hyp=

\\\\DzM*i)hyp=

\320\222\320\224\321\214\321\203\321\200=

Hullhyp\302\267


The vectors tf2 and vi are based at the samepoint z?, they are parallel,

they point in the same direction, and they have the samehyperbolic

norm. Consequently, they must be equal.We therefore have found an isometry \317\206of (H2, dhyp) such that

\317\206(\316\266\316\271)= z% and DZl<p(vi) = v2- \320\236

2.7. The disk model for the hyperbolic plane j

We now describe a new model for the hyperbolic plane, namely, an- j

other metric space (X, d) which is isometricto (H2,dhyp)\302\267This model,j

is sometimes more convenient for performing computations. Another j

side benefit, mathematically less important but not negligible,is that\342\226\240

it often leads to prettier pictures, as we will have the opportunity to )

observe in later chapters. -i

/ % \\ ![ B2 4\302\267\302\267.I ;

Figure 2.7. The disk model for the hyperbolic plaae

Let B2 be the open disk of radius 1 centered at 0 in the com- ;

plex plane C, namely, in the sense of metric spaces introduced in -I

Section 1.3, the ball\320\224\320\263\320\265\320\270\321\201((0,0),1)

in the euclidean plane (M2,deuc). .

For a vector \316\275based at the point \316\266G B2, define its B2-norm as -j

2

1\320\230\320\270==

i_uia ll'?Heuc.

where||\320\263?||\320\265\320\270\321\201

is the euclidean norm of V. Then, as for the euclidean ;

and hyperbolic plane, define the B2-lengthof a piecewise differentiable

curve 7 in B2 parametrized by \316\257\316\271-\302\273\316\266(\316\257),\316\261^ t ^ 6, as \302\243\316\2622(\316\257)

=

2.7. The disk model for the hyperbolicplane 37

\316\231\316\261\316\231\316\231^'\316\257\316\237\316\231\316\231\316\271\302\273^*\302\267FinaUyi given two points P, Q \342\202\254B2, define their

lB2-distance \320\260\320\2502(P, Q) as the infimum of the lengths ^32 as 7 ranges

over all piecewisedifferentiable curves going from \316\241to Q.

\342\200\224z+ iLet \316\246be the fractional linear map defined by \316\246(\316\266)

= \320\263\302\267

Beware that the coefficients of \316\246do not satisfy the visual relation

ad \342\200\224bc= 1. This could be achievedby dividing all the coefficients byone of the complexsquareroots \302\261y/\342\200\2242i,but the resulting expressionwould be clumsy and cumbersome.

\342\200\224z+ iProposition 2.21. The linear fractional map \316\246(\316\266)

= \320\263induces

an isometry from (H2,dhyp) to (\320\2222,\320\271\321\211\320\263).

Proof. Note that \\\316\246(\316\266)\\= 1 when \316\266\342\202\254R, so that \316\246sends R U {\321\201\320\276}

to the unit circle. As a consequence,\316\246sends the upper half-plane H2to either the insideor the outside of the unit circle in \320\241U {00}. Since

\316\246(\317\212)= 0, we conclude that \316\246(\316\2272)is equal to the inside B2 of the unit

circle.

Consider the differential \316\214\316\226\316\246:\320\241-> \320\241of \316\246at \316\266\342\202\254\320\2222. By

Proposition 2.15 and Complement 2.16,

1|\320\224\320\263\320\244^)11\320\2222=

1_|\321\204(\320\263)|2||-\320\240\320\263\320\244(\320\263;)||\320\265\320\270\321\201

2

I |2

2i

(* + i)2

|2 + i|2-|-2 + i|2

(* + i)(*-i)-(-* + i)(-S-i)'

=i(jb)H

=imWM=IMIhyp\302\267

From this computation, we conclude that \316\246sends a curve 7 in

\320\2502to a curve \316\246(7)in B2 such that tBi (\316\246{\316\267))=

\316\257\\^\316\241{\316\267)\302\267

Taking the infimum of the lengths of such curves, it follows that

\320\2713\320\267(\320\244(-\320\240),\320\244(<3))=

dhyp(P,Q) for every P, Q \342\202\254H2. In other words,\316\246defines an isometry from (H2, dhyP)

to (\320\2222,^\320\262\320\267). D


In particular, this proves that dB2 is a metric and not just a semi-

metric (namely that dw(P, Q) = 0 only when \316\241=Q), which is a

property that we had implicitly assumed so far.

Proposition 2.22. The geodesiesof (M2,d&t) ore the arcs containedin euclidean circlesthat are orthogonal to the circle S1 bounding B2,

including straight lines passing through the origin.

Proof. Since \316\246is an isometry from (H2,dhyp) to (B2,^), the geo-S

desics of (B2,^) are just the imagesunder \316\246of the geodesies of:;

(tf.dbyp)\302\267 \\

Because linear fractionals send circles to circles(Proposition2.18)iand respect angles (Corollary 2.17), the result follows immediately*

from the fact that geodesies of (H2, dhyp) are exactly circle arcs inieuclideancirclescenteredon the \320\266-axis or, equivalently, orthogonal^to this \320\266-axis. Di

Proposition 2.23. The isometries o/(B2,d^) are exactly the re-'j

strictions to B2 of all linear and antilinear fractional maps of the\\

form \316\257

\316\273\316\261\316\266+ \316\262 \316\261\316\266+ \316\262 \316\231

\317\210(\316\266)=

\342\226\240% or \317\210(\316\266)=

\342\226\240%\316\216\316\232'\316\262\316\266+ \316\254

^w\316\262\316\266+ \316\254

with |\316\261|2-

\\\316\262\\2= 1.

j

Proof. Since \316\246is an isometry from (H2, dhyp) to (B2, dBa), the isome- 3

tries of (B2, da2) are exactly thosemaps of the form \316\246\316\277^\316\277\317\206-1where-j

\317\206is an isometry of (H2, dhyp). 1, j

If \317\206is a linear fractional map of the form \317\206(\316\266)= ; with \316\261,\317\212

cz + d i

b, c, d \342\202\254R and ad \342\200\224be = 1, then f

\320\266 , \320\273-1/\302\267s (ai-b+c + di)z + (-oi-6-c + di)\302\253

\321\204\320\276\321\221\320\276\321\204I z) = \342\200\224^ * \321\207

(-oi + b + \321\201+ d\\)z + (oi + 6 - \321\201+ di) |

is of the form indicated for j

1 'ia =

-(\320\276+ 6i \342\200\224ci + d) |

and \316\262=-(-\320\260 + \320\253+ a + d). I2 s


(lonversely,writing a + \316\262= \320\253+ d and \316\261\342\200\224

\316\262= a \342\200\224ci with o, 6, c,

CiZ + \316\262d e R, any map \320\263\316\271->-= \342\200\224with |a|2

- |/?|2 = 1 is of the form

\316\246\316\277\317\210\316\277\316\246-1for some a, 6, c, d \342\202\254R with ad \342\200\224be = 1.

The argument is identical for antilinear fractional maps. D


Exercise2.1.Rigorously prove that a horizontal translation \317\206:\320\2302\342\200\224>\320\2302,

defined by the propery that \317\206(\317\207,\321\203)=

(\317\207+ xo, \321\203)for a given xo \342\202\254R, is an

iHometry of the hyperbolic plane (H2,dhyp).Exercise 2.2 (An explicit formula for the hyperbolic distance). The goal

of this exercise is to show that the hyperbolic distance d\\VP(z, z') from \316\266

to z' \342\202\254H2 \320\241\320\241is equal to

\321\210, /x _ W \\Z- Z'| + \\Z~ Z'\\

D(z,z)-loglz__l{_lz_zll.

a. Show that \321\201!\321\214\321\203\320\240(\320\263,\320\263')= Z)(z, \316\266')when \320\263and \320\263'are on the same vertical

line.b. Show that D(<p(z),<p(z'))

= D(z,z') for every z, z' \342\202\254H2 when \317\206:

\320\2302\342\200\224>\320\2302is a horizontal translation, a homothety or the inversionacross the unit circle.

\321\201Use the proof of Lemma 2.6 to show that dhyp(z,z') = D(z,z') for

every z, z' \342\202\254H2.

Exercise 2.3. Adapt the proof of Theorem 2.11 to prove that every isom-etry of the euclidean plane (R2,deUc)is of the form <p(z)

= zq + zeie or\317\210(\316\266)

= zo + zei2e for some z0 \342\202\254\320\241and \316\270e R.

Exercise 2.4 (Perpendicular bisector). The perpendicular bisector of

the two distinct points \316\241and Q \342\202\254H2 is the geodesic bpq defined as follows.

Let \316\234be the midpoint of the geodesic \320\264joining \316\241to Q. Then bpq is thecomplete geodesicthat passes through \316\234and is orthogonal to \320\264.

a. Let \317\201be the inversion across the euclidean circlethat contains bpq.Show that \317\201sends the geodesic \320\264to itself and exchanges P. and Q.

b. Show that dhyp(P, R) = dhyp{Q,R) for every R e bpq. Possible hint:

Use part a.

c. Suppose that \316\241and R are on opposite sidesof gpq, in the sense that

the geodesic fc joining \316\241to R meets bpq in a point S. Combine piecesof fc and p(k) to construct a piecewisedifferentiable curve fc' which

goes from Q to R, which has the same hyperbolic length as fc, and

which is not geodesic. Concludethat dbyP(Q,R) < dhyp(P,R).

40 2. The hyperbolic plan^j

^

d. As a converse to part b, show that dhyp(P, R) \317\206dhyp(Q, R) wheneveifR & bpq. Possiblehint: Use part \321\201 ';

Exercise 2.5 (Orthogonal projection). Let g be a complete geodesic of!

H2, and consider a point \316\241\342\202\254\316\2272. \316\231

\342\226\240i

a. First consider the case where \317\201is a vertical half-line g = {(xo,y) \342\202\254l

R2;y > 0}. Show that there exists a unique complete geodesic\320\233\321\214

containing \316\241and orthogonally cutting g at some point Q (namely, \320\253

and g meet in Q and form an angle of\\ there). j

b. In the case of a generalcomplete geodesic g, show that there exists \321\211

unique complete geodesic h containing \316\241and orthogonally cuttingat some point Q. Possible hint: Use Lemma 2.6.

c. Show that Q is the point of g that is closest to P, in the sense thafejdhyp{P, Q') > dbyP(P,Q)for every Q' \342\202\254g different from Q. Possiblehint: First consider the case where the geodesic h is equal to a verticalhalf-line and where the point \316\241lies above Q on this half-line, then!

apply Lemma 2.6. JExercise 2.6 (Hyperbolic rotation around i). For \316\270\342\202\254R, consider the|fractional linear map defined by

\316\266cos I + sin I

-zsinf +cos|a. Show that \317\210fixes the point i \342\202\254H2, and that its differential \316\214\\\317\210atl

i is just the rotation of angle \316\230.Hint: Use Proposition 2.15 to com-jpute JDi^i.

b. For an arbitrary zq \342\202\254H2, give a similar formula for the hyperbolic'!rotation of angle \316\270around the point zo, namely, for the isometry \317\206:

\316\2272\342\200\224\342\226\272\316\2272for which \317\206(\316\266\316\277)= zq and Dzo \317\210is the rotation of angle \316\230.

iExercise 2.7 (Classificationof hyperbolic isometries).Consideran isom- %

etry \317\206of the hyperbolic plane (H2,dhyP),defined by the linear fractionalij

map <p(z) = ^^ with o, b, c, d \342\202\254R and ad \342\200\224be = 1. I2

a. Show that if (o + d)2 > 4, \317\210has no fixed point in H2 but fixes exactly ij

two points of R U {oo}. Conclude that in this case there exists an 1isometry \317\210of (H2,dhyp) such that \317\210\316\277\317\206\316\277\317\204\317\201-1is a homothety \316\266\316\272+

\316\247\316\266jj

with \316\273> 0. (Hint: Choose ip so that it sends the fixed points to 0 \\

and oo). Find a relationship between \316\273and (o + d)2. A hyperbolic |

isometry of this type is said to be loxodromic. \"

b. Show that if (o + d)2 < 4, \317\210has a unique fixed point in H2. Conclude\302\267;

that in this case there is an isometry V of (\316\2272,dhyp) such that \317\210\316\277\317\206\316\277\317\206-1'

is the linear fractional map of Exercise2.6 for some \316\270\342\202\254R. (Hint: ,

Choose \317\206so that it sends the fixed point to i). Find a relationship


2

between \316\270and (o + d)2. A hyperbolic isometry of this type is said tobe elliptic.

\321\201Show that if (o + d)2 = 4 and if \317\206is not the identity map defined by

<p(z)= z, then \317\210has a unique fixed point in R U {oo}. Conclude that

in this case there is an isometry \317\210of (H2, dhyp) such that \317\210\317\214\317\206\316\277\316\271))-1is

the horizontal translation \316\266\316\271\342\200\224>\316\266+1. (Hint: Choose V 8\302\260that it sendsthe fixed point to oo). A hyperbolic isometry of this type is said to beparabolic.

Exercise2.8 (Stereographic projection). Let S2 be the unit sphere in the.\320\247-dimensional euclidean space R3, consisting of those points (re, y, z) \342\202\254R3

Huch that x2 + y2 + z2 = 1. Consider the map p: S2->I2 U {oo} defined

\320\275\320\260follows. If (\320\266,\321\203,\316\266)\316\246(0,0,1), then

p{x,y,z)=^,Ty-^sR

otherwise, p(0,0,1) = oo.

a. Show that when \316\241= (re, y, z) is not the \"North Pole\" \316\233\316\223= (0,0,1), its

image p(P) is just the point where the line NP crosses the rcj/-plane

inR3.

b. Show that p: S2->R2U {oo}is continuous at every Po \342\202\254S2. When

Po = N \320\262\320\276that \317\206(\316\241\316\277)= oo, this means that for every large \316\267> 0 there

exists a small \316\264> 0 such that deuc(p(P), \316\237)> \316\267for every \316\241\342\202\254S2 with

deac{P,Po) < <5, where \320\236is the origin in K2. (Compare the calculusdefinition of infinite limits, as reviewed in Section T.3 of the Tool

Kit.)\321\201Show that the inverse function p'1: M.2 U {oo} \342\200\224>S2 is continuous at

every Qo \342\202\254R2 U {oo}. When Qo = oo so that p_1(Qo) = N, this

means that for every small \316\265> 0, there exists a large \316\267> 0 such that

deuc(/\302\273-1(Q)>A0 < \316\265for every Q \342\202\254R2 with deuc(Q,0) > \316\267.

In other words, \317\201is a homeomorphism from S2 to R2 U {oo}. (SeeSection 5.1 for a definition of homeomorphisms).

Exercise 2.9. Let \302\253u,z\\ and Zoo be three distinct points in the Riemann

sphere \320\241= \320\241U {oo}. Show that there exist a unique linear fractional map\317\210and a unique antilinear fractional map \317\210such that ^>(0)

= V(0) = ^o,\317\206{\317\212)

=\321\204(1)

= z\\ and p(oo) = V(oo) = \316\266,\302\273.

Exercise 2.10.

a. Show that the linear fractional map \317\210:\320\241\342\200\224*\320\241defined by \317\206(\316\266)=

fjjj,with o, b, c, d \342\202\254\320\241such that ad\342\200\224be= 1, is bijective and that its inverse

\317\206'1is the linear fractional map \317\206~\316\271{\316\266)= _j^7*. Hint: Remember

that \317\206-1(\316\266)is the number \320\270such that \317\210{\316\275)= \316\266.


b. Give a similar formula for the inverse of the antilinear fractional map

#*) = *gfwitb\302\253J-uc=l.

Exercise 2.11.

a. Show that any linear or antilinear fractional map can be written as the '\342\226\240

composition of finitely many inversions across circles. \316\271

b. Show that when a linear fractional map is written as the compositionof finitely many inversions across circles, the number of inversions is \342\226\240

even. (Hint: Corollary 2.17.) Similarly, show that when an antilinear

fractional map is written as the compositionof finitely many inversionsacross circles, the number of inversions is odd.

Exercise 2.12 (Linear fractional maps and projective lines). Let the real

projective line RP1 consist of all 1-dimensionallinear subspaces of the f

vector space R2. Namely, RP1 is the set of all lines L through the origin |

in R2. Since such a line L is determined by its slope \320\260\342\202\254R U {oo}, this \342\226\240

provides an identification RP1 Slu {oo}.

a. Let \316\246\316\261'\302\267R2 \342\200\224\342\226\272R2 be the linear map defined by the matrix A-\320\241\320\255

with determinant ad \342\200\224be equal to 1. Similarly, consider the linear

fractional map ^\320\264:R U {oo} -\302\273HlU {oo} defined by 4>a{s) = -.Show that \316\246\316\261sends the line L \342\202\254RP1 with slope \320\260\342\202\254R U {oo} to theline \320\244\320\260{\320\254)with slope <pa{s)\302\267

b. Use part a to show that \317\210\316\261\316\261'=

\317\210\316\261\302\260\316\250\316\2611,where AA' denotes theproduct of the matrices A and A'.

\321\201Similarly, consider \320\2412= \320\241\321\205\320\241as a vector space over the field C,

and let the complex projective line CP1consist of all 1-dimensionallinear subspaces L \320\241\320\2412.Such a complex line L is determined by its

complex slope s \342\202\254\320\241U {oo} defined by the following property. If

L is not the line {0} \317\207\320\241,it intersects {1} \317\207\320\241at the point (l,a); if

L = {0} \317\207C, its complex slope is s = oo. Let \316\246\316\261-C2 \342\200\224>C2 be the

complex linear map defined by the matrix -4=1 .1 with complex

entries o, b, c, d \342\202\254\320\241and with determinant 1. Show that for the aboveidentification CP1 SCU {oo}, the map CP1 -\342\226\272CP1 induced by \320\244\320\220

corresponds to the linear fractional map ^:CU {oo}\342\200\224>\320\241U {oo}

defined by 4>a{s) = \342\200\224-,.cs + a

Exercise 2.13 (Hyperbolicdisks).Recall from Section 1.3 that in a metric

space (X,d) the ball of radius r centeredat \316\241\342\202\254X is Bd{P, r) = {Q \342\202\254

X;d(P,Q)<r}.


a. Let \320\236be the center of the disk model B2 of Section2.7.Show that the

ball Bd.2 (O, r) in B2 coincides with the euclidean opendiskof radius

tanh j centered at O.b. Show that every hyperbolic ball

Bdbyp (P, r) in the hyperbolic planeH2 is a euclidean open disk. Possible hint: Use part a, the isometry

\320\244:(H2,dhyP) -> (B2,dBa) of Proposition 2.21, Proposition 2.2 and

Proposition 2.18.

\321\201Show that the ball\320\224^ (\316\241,r) centered at \316\241=

(\317\207,\321\203)\342\202\254\316\2272is the

euclidean open disk with euclidean radius 2ysmhr and with euclidean

center (x,2ycoshr). Possible hint: Look at the two points where the

boundary of this disk meets the vertical line passingthrough P.

Exercise 2.14 (Hyperbolic area). If D is a region in H2, define itshyperbolic area as

Areahyp(Z)) = // -^dxdy.

a. Let p: H2 \342\200\224>H2 be the standard inversion. Show that p{D) has thesame hyperbolic area as D. Possible hints: Polar coordinates may beconvenient; alternatively, one can use the change of variablesformula

for double integrals (see part c).b. Show that an isometry of H2 sends eachregion D \320\241\316\2272to a region of

the same hyperbolic area.\321\201Let \316\246be the isometry from (H2,dhyP) to (B2, dBa) provided by

Proposition 2.21. Show that for every region D in H2

It may be convenient to use the changeof variables formula for double

integrals, which in this case says that

// f{u,v)dudv= \320\230/(\320\244(\321\205,\321\203))|detZ)*| dxdy

JJ<HD) JJd

for every function /: $(D) \342\200\224>R, and to borrow computations from the

proof of Proposition 2.21 to evaluate the determinant det Z)* of the

differential map \316\214\317\206.

d. Show that a ballBdbyp (P, r) of radius \320\263in H2 has hyperbolic area

\320\220\320\263\320\265\320\260\320\275\321\203\321\200(Bdbyp (P, r)) = 27r(cosh\320\263- 1) = 4\317\200sinh2^.

Hint: First consider the case where \316\241= (1,0) =\316\246_1(0), and use

part \321\201and the result of Exercise 2.13a.

Exercise2.15(Area of hyperbolic triangles). For every \316\270with 0 < \316\270^ f,let \320\242\320\265be the (infinite) hyperbolic triangle with vertices i, e'e = cos0+i sin \316\230

44 2. The hyperbolic plane:

and oo. Namely, \320\242\320\265is the region of H2 bounded below by the euclidean

circle of radius 1 centeredat the origin, on the left by the y-axis, and on;the right by the line \317\207= cos \316\230.

a. Show that the hyperbolic area Areahyp(Te), as defined in Exercise 2.14,.is finite.

d 7\320\223b. Show that -\320\263:\320\220\320\263\320\265\320\260\321\214\321\203\321\200(\320\242\320\262)

= \342\200\2241.Conclude that \320\220\320\263\320\265\320\260\321\214\321\203\321\200(\320\2429)= \342\200\224\342\200\224\316\270.

c. Let \316\244be a finite triangle in the hyperbolic plane, namely, the regionof H2 bounded by the three geodesies joining any two of three distinctpoints P, Q, R. Let a, /3, 7 \342\202\254[\316\237,\317\200]be the respective angles of \316\244at\342\226\240\342\226\240\302\267

its three vertices. Show that 5

Areahyp(T9)= \317\200- \316\261\342\200\224

\316\262\342\200\224\321\203. 5

Hint: Express this area as a linear combination of the hyperbolic areasjof six suitably chosen infinite hyperbolic triangles, each isometricto al

triangle \320\242\320\265as in parts a and b. ;1

Exercise 2.16 (Crossratio). The crossratio of four distinct points z\\, Z2,\\

Z3, Z4 \342\202\254\320\241= \320\241U {oo} is

\316\232-\316\257,, , _ \321\207_ (gl-

Z3){Z2 -Zj)\316\232(\316\226\316\271,\316\226\302\2672,\316\2263,\316\2264)

=\320\263\321\202

r \342\202\254<L,(Zl

-Z4,){Z2 -Z3)

using the straightforward extension by continuity of this formula when one

of the Zi is equal to 00. For instance, K(z\\, \316\2262,Z3,00)= Z1~*8. Show thatr

\316\232(\317\206{\316\266\316\271),\317\206(\316\2662),\317\206(\316\2663),\317\206(\316\2664,))is equal to K(zi,Z2,Z3,Zi) for every lineari:

fractional map \317\210:\320\241\342\200\224>\320\241,and that it is equal to the complex conjugate?of K(zi,Z2,Z3,Z4,) for every antilinear fractional map \317\206:\320\241\342\200\224*\320\241.Possible\"

hint: First check the property for a few simple (anti)linear fractional maps,'and then apply Lemma 2.12.

Exercise 2.17 (Another formula for the hyperbolic distance). Given two-

distinct points zi, zi of the hyperbolic plane H2 = {z \342\202\254C;Im(z) > 0}, let*

X\\, X2 eR = MU {00} be the two endpoints of the complete geodesic gpassingthrough Z\\ and Z2, in such a way that Xi, z\\, zi and X2 occur in

this order on g. Show that

, , \321\207, (Zl-

X2){Z2~

X\\) ';dbyp(*i. z2) = log f- \342\200\224^-

\342\200\224i. \316\271(Zl

- Xl)(Z2 \342\200\224X2) ;

Possible hint: First consider the casewhere \317\207\316\271=0 and X2 = 00, and theaiuse the invariance property for the crossratio proved in Exercise 2.16.

\\

Exercise 2.18. Show that a crossratioformula similar to that toExercise 2.17 holds in the disk model (B2,dB2)of Section 2.7. Hint: Use theinvariance property for the crossratio proved in Exercise 2.16.


*(*'\302\273)=

(\316\271, V2 \342\226\240\342\200\2362', , -2 , ..2)\302\267

Exercise 2.19 (The projective model for the hyperbolic plane). Let B2bethe open unit disk of Section 2.7. Consider the map \320\244:B2 \342\200\224>B2 defined

by2rc 1y

4 + x2+y2' l + x2 + y2

a. Show that \316\246is bijective.

b. Show that if g is a complete geodesicof (\320\2222,\321\201!\320\262\320\263)which is a circle arccenteredon the rc-axis, its image \316\246(<?)is the euclidean line segmentwith the same endpoints.

\321\201Let pe: B2 \342\200\224>B2 be the euclidean rotation of angle \316\270around the originO. Show that \316\246\316\277\317\201\316\270

=\317\201\316\270\316\277\317\206.

d. Combine parts a and b to show that \316\246sends each complete geodesicg of (B2,dp) to the euclidean line segment with the same endpoints.

For a vector \316\275based at \316\241\342\202\254\320\2222,define its projective norm ||v||Proj =;|\302\243>\321\200\320\244-1(\320\263\320\223)||\320\2222.For every piecewise differentiable curve 7 in B2

parametrizedby t t~* 7(i), a ^ t ^ b, define its projective length as ^Proj(7)=

/ ||/7/,(i)||projcii\302\267 Finally, consider the new metric dpr0j on B2 by the

property that dpIoj{P,Q) =\321\201\320\263\320\2673(\321\204-1(\320\240),\321\204-1(\320\264))

for every P, Q \342\202\254B2. In

particular, \316\246is now an isometry from (\320\2222,\321\201^\320\267)to (B2,dproj).

e. Show that for every P, Q \342\202\254B2, the projective distance dpr0}(P,Q) isequalto the infimum of the projective lengths of all piecewisedifferentiablecurves going from \316\241to Q in B2. Show that this infimum is equalto the projective length of the euclidean line segment from \316\241to Q.

f. Given a vector \316\275based at \316\241\342\202\254\320\2222,draw the line L passing through \316\241

and parallel to tf, and let A and \320\222be the two points where it meets

the unit circle S1 bounding B2. Show that

The computations may be a little easier if one first restricts attentionto the case where \316\241is on the rc-axis, and then use part \321\201to deduce

the general case from this one.

g. For any two distinct P, Q \342\202\254B2, let \320\220,\320\222\342\202\254S1 be the points where theline PQ meets the circle S1 in such a way that A, P, Q, \320\222occur in

this order on the line. Combineparts e and f to show that

(A,Q)deuc(P,B)dproj (\316\233Q)

-2 log ^^ {A p)4uc (p)Q).

(Compare the crossratio formula of Exercise 2.17.)The metric space (B2,dproj), which is isometric to the hyperbolic plane

(H3,dhyp), is called the projective model or the Cayley-Klein model

46 2. The hyperbolic plane;

for the hyperbolic plane. It is closely related to the geometry of the 3- !dimensional projectiveplane \320\250\320\2402,denned in close analogy with the pro- >

jective line MP1 of Exercise 2.12and consisting of all lines passing through;

the origin in R3. The fact that its geodesies are euclidean line segmentsmakes this projective model quite attractive for some problems.

Chapter 3

The 2-dimensional

sphere

The euclidean plane (M2,deuc) and the hyperbolic plane (]HP,dhyp)

have the fundamental property that they are both homogeneousandisotropic. Thereis another well-known 2-dimensional space which

shares this property, namely, the 2-dimensional sphere in R3.

This is a relatively familiar space but, as will become apparent

in later chapters, its geometry is not as fundamental as hyperbolic

geometry or, to a lesserextent,euclideangeometry. For this reason,

its discussion will be somewhat de-emphasized in this book. We only

need a brief description of this spaceand of its main properties.

3.1. The 2-dimensional sphere

The 2-dimenaional sphere is the set

S2 = {(x,y, z) \342\202\254M3; x2 + y2 + z2 = 1}

consisting of those points in the 3-dimensionalspaceR3 which are

at euclidean distance 1 from the origin. Namely, S2 is the euclideansphereof radius 1 centered at the origin \320\236=

(0,0,0).

Given two points P, Q \342\202\254S2, we can consider all piecewisediffer-

entiable curves 7 that are completely contained in S2 and join \316\241to

47

48 3. The 2-dimensional sphere!

1\302\267.(

(

!

\342\226\240I.

\\

Figure 3.1. The 2-dimensional sphere

Q. The spherical distance from \316\241to Q is defined as the infimum [

dsvh{P, Q) = {4uc(7); 7 goes from \316\241to Q in S2}

of their usual euclideanarclengths ^euc(7)\302\267 Here, as in Chapter 1, theeuclideanarclengthof a piecewise differentiable curve 7 parametrized;by

t^ (x(t),y(t),z(t)), o<i<6,isgiven by

4uc(7)=

/ \321\203/*$)* + \342\204\226)*+ *\342\204\226#.J a

The definition immediately shows that this spherical distance

^eph(-P) Q) is greater than or equal to the usual euclidean distancedeuc(-P) Q) from \316\241to Q in R3. In particular, this proves that

dsph(P,Q)= 0 only when \316\241= Q. By the same arguments as in

Lemma 2.1, d8ph also satisfies the Symmetry Condition and theTriangle Inequality. This proves that

d8phis really a metric.

3.2. Shortest curves

A great circle in the sphere S2 is the intersection of S2 with a planepassing through the origin. Equivalently, a great circle is a circleof

.4.3. Isometries 49

nidius 1 contained in \302\2472.A great circle arc is an arc contained in\316\267great circle.

Elementary geometric considerations show that any two P, Q \342\202\254

\342\204\226can be joined by a great circlearc of length < \317\200.In addition,

litis circle arc is unique unless\316\241and Q are antipodal, namely unlessQ =

(\342\200\224x,\342\200\224y,\342\200\224z)if \316\241=(\317\207,\321\203,\316\266).When P and Q are antipodal,

lliere are many great circle arcs of length \317\200going from \316\241to Q.

Theorem 3.1. The geodesieso/(S2,d8ph)are exactly the great circle

arcs. The shortest curves joining two points P, Q e S2 are exactly

the great circle arcs of length < \317\200going from \316\241to Q.

Proof. We sketch a proof of this result in Exercise 3.1. D

Note that we encounter here a new phenomenon, where a geodesicis not necessarily the shortest curve joining its endpoints; this happens

for every great circle arc of length > \317\200.Recall that in the definition

of a geodesic7 the part of 7 that joins \316\241to Q is required to be theshortestcurve joining \316\241to Q only when Q is sufficiently close to P.

Another new phenomenon is that great circles provide closed

yeodesica of \302\2472,namely, geodesies which are closed curves in thesensethat they return to their initial point.

3.3. Isometries

[\316\267\316\2323,a rotation \317\206around a line L respects euclidean distancesand

arc lengths. If, in addition, the rotation axis L passesthrough the

origin O, then \317\206sends the sphere S2 to itself. Consequently,any

rotation \317\210around a line passing through the origin induces an isometryof(S2,deph).

Theseisometries are sufficient to show that the sphere ishomogeneous and isotropic, as indicated by the following statement.

Proposition 3.2. Given two points P, Q \342\202\254S2, \316\261vector \316\275tangent to

S2 at P, and a vector w tangent to S2 at Q such that||\320\263\320\223||\320\265\321\206\321\201

=||\320\263\320\270||\320\265\321\206\321\201,

there exists a rotation \317\210around a line passing through the origin \320\236

such that \317\206(\316\241)= Q and Dp<p{if)

= \316\260\316\220.

50 3. The 2-dimensional sphe:

Proof. One easily finds a rotation \317\206\316\271such that \317\210\316\271(\316\241)= Q, foj

instance a rotation whose axis is orthogonal to the lines OPan<|

OQ. Then, \316\214\317\201\317\210\316\271(if) is a vector tangent to S2at \317\210\316\271(\316\241)= Q whose

cuclidean length is |

H-DpVl(iOlleuc=

Neuc=

IMIeuc\302\267

As a consequence, there exists a unique rotation \317\2062around the lm|OQ whose differential map Dq<P2 sends \316\214\317\201\317\210\316\271(\316\275)to \316\260\316\244.Then th|

composition \317\206=

\317\2102\302\260

\317\210\316\271sends \316\241to Q, and its differential \316\214\317\201\317\210

Dq<P2\320\276Dpipi sends \316\275to w.

By a classical property (seeExercise3.2),the composition \317\210\321\211

\317\210\316\271\316\277(/?\316\271of two rotations is also a rotation around a line passing througthe origin.

We already saw that a rotation \317\210around a line L passing through

the origin provides an isometry of S2. If we compose \317\206with the rej

flection across the plane orthogonal to L at O, we obtain a rotation^reflection. sj

Note that rotations include the identity map, which is a rotad

tion of angle 0. As a consequence, rotation-reflections also includereflectionsacrossa planepassing through the origin.

\"

\342\226\240\302\267!Theorem 3.3. The isometries o/(S , deph)

ore\302\267exactly the above ro-\\

tations and rotation-reflections.i

Proof. This can be proved by an argument which is very close to theproof of Theorem 2.11. See Exercise 3.4. Q

Exercises for Chapter 3Exercise 3.1 (Geodesies of the sphere S2).

a. Let 7 be a piecewise diflferentiable curve in R3 parametrized by t t->

7(i), a < t < b. For each t, let p(t), 0(t) and <p(t) be the sphericalcoordinates of 7(i). Show that the euclidean length of 7 is equalto

4uc(7)= / y/p'it)2+ P(t)2 sinV(t) 0'(*)2 + P(i)2^'(*)2dt.

JaHint: Remember the formulas expressing rectangular coordinates in

terms of spherical coordinates.

Kxercises for Chapter 3 51

t), hi the sphere \302\2472,let \316\241be the point (0,0,1) and let Q be the point

(x, 0, z) with \317\207> 0. Let \316\261be the vertical circle axe going from \316\241to Qin \302\2472,where the spherical coordinate \316\270is constantly equal to 0. Show

that any curve 7 going from \316\241to Q has euclidean length greater than

or equal to that of a.

\321\201Show that if \316\241and Q are two points of the sphere \302\2472,the shortest

curves going from \316\241to Q are the great circle arcs of length < \317\200going

from \316\241to Q. Possible hint: Use a suitable isometry of (S2,rfSpi,) toreduce this to the case of part b.

<l. Show that the geodesies of S2 are the great circle arcs.

Kxercise 3.2. The main goal of this exercise is to show that iu theeuclidean space R3 the composition of two rotations whose axes pass through

(lie origin \320\236= (0,0,0) is also a rotation around a line passing through the

1ii'igiu.

\320\270.In R3, let L be a line contained in a plane \320\237.Let \320\237'be the planeobtained by rotating \316\240around L by an angle of ^0, and let r and r'

be the orthogonal reflections acrossthe planes \316\240and \316\240'.respectively.

Show that the composition \317\204'\316\277\317\204is the rotation of angle \316\270around L,and that \321\202\320\276\321\202'is the rotation of angle \342\200\224\316\230around L. Possible hint:You may find it convenient to considerthe restrictions of \317\204\316\214\317\204and

\321\202\320\276\321\2021to each plane \316\240orthogonal to L.

b. Let L and L' be two lines passing through the origin \320\236= (0,0,0) iu

Ra, and let \317\201and p' be two rotations around L and L', respectively.Show that the composition \317\201\320\276\317\201is a rotation around \320\275line passing

through the origin. Possible hint: Consider the plane \316\240containing L

and L', and use the two properties of part a.

c. Show that iu R3. the composition \320\277\320\276\321\202^\320\276\302\267\302\267

\302\267\316\277\317\2042\316\267of an even number oforthogonal reflections \317\204,\302\267across planes passing through \320\236is a rotation

(possibly the identity).

Exercise 3.3. The main goal of this exercise is to show that if r is anorthogonal reflection across a plane \316\240passing through the origin \320\236=

(0. 0. 0) and if \317\201is a rotation of migle 0 tuound a line L passing through O.then the composition r \316\277\317\201is a rotation-reflection. This means that \321\202\320\276\321\200

is also equal to a compositionr' \316\277\317\201',where \317\201'is a rotation across a line

L' passingthrough \320\236and where r' is the orthogonal reflection across the

plane \320\237'orthogonal to V at O.

Without loss of generality, we can assumethat L is not orthogonal to\320\237.since otherwise we are done. Then, the plane \316\240\316\271orthogonal to \316\240and

containing L is uniquely determined. Let \320\251be the image of \316\240\316\271under

the rotation of angle \342\200\224\\d around L, and let \320\237\320\267be the plane orthogonal

52i

3. The 2-dimensionalspheref

to both \316\240and \316\240\316\271at O. Let \316\267and \321\202\320\263be the orthogonal reflections across :the planes\316\240\316\271and \320\251,respectively. ';

a. Show that \317\204\316\277\317\204\317\207(\316\241)= \342\200\224\316\241for every P e \320\237\320\267(where \342\200\224\316\241denotes the

point (\342\200\224x, \342\200\224y,\342\200\224z)when \316\241=(\317\207,\321\203,\316\266)).Drawing a picture might help. :

b. Show that \317\204\316\277\316\267\316\277ts(-P)= \342\200\224\316\241for every P in the line L' = \316\2402\316\240\316\2403.

c. Let \317\204'be the orthogonal reflection across the plane \316\240'orthogonal to :

L' at O. Show that p' = r'o r \316\277\316\267\316\277\317\204\316\271is a rotation around the line I/. ;;

Hint: First use the result of Exercise 3.2c to show that p' is a rotation :

around some line.

d. Show that r \316\277\317\201= r'o \317\201'

=\317\201'\316\277\317\204',so that \317\204\320\276\321\200is a rotation-reflection. ;

Hint: First use Exercise3.2ato show that \317\201= \317\204\316\257\316\277\316\2442.

e. Show that in R3 the composition \321\202\\\320\276\321\202\320\263\320\276\302\267\302\267

-oTsn+i of an odd number of

orthogonal reflections \316\267across planes passing through \320\236is a rotation-:\302\267!

reflection. Hint: Use Exercise 3.2c.

Exercise 3.4 (Isometries of the sphere (S2,tieph)).

a. Adapt the proof of Theorem 2.11 to show that every isometry of the

sphere (S2,deph)is a composition of reflections across planes passingthrough the origin \320\236= (0,0,0).

b. Show that every isometry of the sphere (S2,dsph) is, either a rotation,or a rotation-reflection.Hint: Use the conclusions of Exercises 3.2cand 3.3d.

Exercise 3.5 (Spherical triangles). A spherical triangle is a region \316\244of

the sphere S3 bounded by three geodesies arcs \316\225\316\271,\302\2432,E3 of S2 meetingonly at their endpoints. We also require that the angle of \316\244at each of itsthree vertices is less than \317\200,and that \316\244is not reduced to a single point.Let q, /3, and 7 be three numbers in the interval (\316\237,\317\200).

a. Suppose that we have found three noncoplanar vectors \320\231,v, vl in R3such that the angle between \320\270and \316\260is equal to \317\200\342\200\224

\316\261,the anglebetween \320\270and w is equal to \317\200\342\200\224

\316\262,and the angle between \316\275and w is

equal to \317\200\342\200\2247. Let U, V and W be the hemispheres of the sphere S2

consisting of all points \316\241\342\202\254S2 with OP-u> \316\237,\316\270\316\241-v^ 0, \316\237\316\241-\316\260^\316\237,

respectively. Show that the intersection \316\244= U DVDW is a sphericaltriangle with angles \316\261,\316\262,\316\267.

b. Consider \320\271= (1,0,0) and \317\205=(\342\200\224cos a, sin a, 0). Show that there

exists a unit vector \302\267\316\260=(\316\261,6, \321\201)with \321\201\317\2060 making an angle of \317\200\342\200\224

\316\262

with \320\270and an angle of \317\200\342\200\2247 with \316\275if and only if

cos(q + \316\262)< \342\200\224cos 7 < cos(q \342\200\224\316\262).


\321\201Show that the double inequality of part b is equivalent to the condition

that

max{a\342\200\224

\316\262,\316\262\342\200\224a}<n\342\200\224

j< min{a + \316\262,2\317\200\342\200\224(a + \316\262)},

which itself is equivalent to the condition that

\317\200<\316\261+ \316\262+ *\316\263<\317\200+ 2 min{a, \316\262,j}.

Hint: Note that -\317\200< \316\261\342\200\224\316\262<\317\200,0 < \316\261+\316\262<2\317\200and 0 < \317\200- j < \317\200.

d. Combine parts a, b and \321\201to show that if

\317\200< q + \316\262+7 < \317\200+2min{a,\316\262,j},

there exists a spherical triangle \316\244\320\241S2 with respective angles \316\261,\316\262

and 7.

e. Show that if two spherical triangles \316\244and T' \320\241S2 have the same

angles q, \316\262,7, there exists an isometry \317\210of (S2, dsph) sending \316\244to T\".

Hint: In part b, there are only two possible unit vectors iZf.

Exercise 3.6 (Area of spherical triangles).a. In the sphere S2, consider two great semi-circlesjoining the \"North

Pole\" (0,0,1) to the \"South Pole\" (0,0, -1), and making an angle ofq with each other at these poles. Show that the surface area of the

digon bounded by these two arcs is equal to 2a. Hint: Use sphericalcoordinates or a proportionality argument.

b. Let \316\244be a spherical triangle with angles \316\261,\316\262,\316\267.Let \320\220,\320\222and \320\241be

the great circles of S2 that contain each of the three edges of T. Show

that these great circles subdivide the sphere S2 into eight spherical

triangles whose angles are all of the form a, /3, 7, \317\200\342\200\224q, \317\200\342\200\224

/3 or \317\200\342\200\224\316\267.

c. Combine parts a and b to show that the area of the triangle \316\244is equal

to q + \316\262+ 7 \342\200\224\317\200.Hint: Solve a system of linear equations.d. Show that necessarily \317\200< \316\261+\316\262+*f < \317\200+2min{a,/3,7}. Hint: First

show that 0 < a + \316\262+ f\342\200\224\317\200<2a.

Chapter 4

Gluing constructions

This chapter and the following one are devoted to the construction of

interesting metric spaces which are locally identical to the euclideanplane, the hyperbolic plane or the sphere, but are globally very

different. We start with the intuitive idea of gluing together pieces of

paper, but then go on with the mathematically rigorousconstructionof spaces obtained by gluing together the edges of euclidean andhyperbolicpolygons.This chapter is concerned with the theoreticalnspectsof the construction, while the next chapter will investigate

various examples.

4.1. Informal examples: the cylinder and thetorus

We first discuss in a very informal way the idea of creating new spacesby gluing. Precise definitions will be rigorously developed in the next

section.

If one takes a rectangular piece of paper and glues the top sideto the bottom side so as to respect the orientations indicated in

Figure 4.1, it is well known that one gets a cylinder.This paper cylinder can be deformed to many positions in 3-

dimensional space but they all have the same metric: As long as we

do not stretch the paper,the euclideanarclength of a curve drawn on

55

56 4. Gluing constructions

Figure 4.1. Creating a cylinder from a piece of paper

the cylinder remains constant under deformations, and it is actually

equal to the arc length of the corresponding pieces of curve in the

original rectangle.

One can also try, in addition to gluing the top sideto the bottom

side, to glue the left side to the right side of the piece of paper,Namely, after gluing the top and bottom sidestogether to obtain

a cylinder, we can glue the left boundary curve of the cylinder tothe right one. This is harder to realize physically in 3-dimensional

space without crumpling the paper but if we are willing to use rubberinstead of paper and to stretch the cylinder in order to put its two

sides in contact, one easily seesthat this creates a torus, namely, aninner tube or the surface of a donut. See Figure 4.2.

3^7\320\267 Q*\"^l 7i Qi

Figure 4.2. Love story on a torus

4.1. Informal examples: the cylinder and the torus 57

Let us try to understand the geometry of this torus from the

point of view of a little bug crawling over it. For instance,suppose

Uiat in order to meet its lover the bug walks from \316\241to Q along the

curve 7 indicatedon the right of Figure 4.2. To measure the distancel.liat it needs to travel, one could considerthe euclideanarc lengthof 7 for a given position of the torus in 3-dimensional space, but this

will depend on the stretching that occurred when moving the torus to

that position. However,if we are interested in prestretching distances,the natural thing to consider is to decompose the curve 7 into pieces

coming from the original piece of paper, and then take the sum of the

lengths of these pieces. For instance, in the situation illustratedonFigure 4.2, the curve 7 comes from three curves 71, 72 and 73 on the.square, in such a way that each 74 goes from a point Pi to a pointQi, and where \316\241= Pi, Q\\ is glued to P%, Q% is glued to P3, andQ3

= Q The distance traveled by our critter friend, as measured

on the original piece of paper, is then the sum of the euclideanarclengths of 71, 72 and 73 on this pieceof paper.

In order to introduce some mathematical rigor to this discussion,let us formalizethis construction.We begin with the rectangle

X = [a,b] \317\207[c,d]=

{(\317\207,\316\257/)\316\27612;\316\261\316\237<\316\234<!/^}\302\267

Let X be the space obtained from X by doing the gluings indicated.

Some points of X correspond to exactly one point of X (located in

the interior of the rectangle),some points of X correspond to two

points of X (locatedon opposite sides of the rectangle), and onepoint correspondsto four points of X (namely, the corners of the

rectangle). In other words, each point of X is described by a subsetof X of one of the following types:

(1) the 1-element set{(\320\266,\321\203)}

with a < \317\207<b and \321\201< \321\203< d;

(2) the 2-element set{(\320\266,\321\201),(\317\207,d)} with a < \317\207< b;

(3) the 2-element set {(a, y), (b, y)} with \321\201< \321\203<d;

(4) the 4-element set {(a, c),(a,d), (b,c), (b, d)}.

These subsets form a partition of X. This means that every point

of X belongs to exactly one such subset.

58 4. Gluingconstructions\342\226\240';

|

We could define the distancebetween points of X by taking the <

infimum of the lengths of curvesjoining them, as in the exampleof our \\

little bug walking on the torus. The next sectiondevelops a definition \"\342\226\240>

that is equivalent to this idea, but is somewhat easierto state and to !

use. ]

Js

4.2. Mathematical definition of gluings andj

quotient spaces \342\226\240

4.2.1. Partitions. Let (X, d) be a metric space,and consider a par- ;tition X of X. As indicated above, a partition of X is a family X of .!

subsets \320\220\320\241X such that each point \316\241\302\243X belongs to one and only ;one suchsubset A.

In particular, every element of the set X is a subset \320\220\320\241X. We

can therefore consider that the set X is obtained from X by deciding'

that for each subset A of the partition all the points in the subset .A now correspond to a single element of X. In other words, all thepoints of A are now glued together to give a single point in X. Sothe formalism of partitions is a good way to rigorously describe the

intuitive idea of gluing points of X together. It takesa while to get

used to it though since a point of X is also a subset of X.The following notation will often be convenient. If \316\241\342\202\254X is a

point of X, let \316\241\342\202\254X denote the corresponding point of X after the

gluing. Namely, in the formalism of partitions, \316\241\320\241X is the element :

of the partition X such that \316\241\302\243P.

4.2.2. The quotient semi-metric. We now introduce a distance

function on the set X, along the lines of the informal discussion of

the previous section.

If \316\241and Q are two points of X corresponding to \316\241and Q \302\243X,

respectively, a discrete walk w from \316\241to Q is a finite sequence\316\241=

\320\240\321\214Qj,P2, Q2, \320\240\320\267,\302\267\302\267\302\267,Qn-i, P\302\273,Qn

= Q of points of Xsuch that Qi = P*+i for every i < n. Namely, such a discrete walk

alternates travels in X from P* to Qi and jumps from Qi to a point

Pj+i that is glued to Qi. Thed-length(or just the length if there isno ambiguity on the metric d considered) of a discretewalk w is the

4.2. Mathematical definition of gluings 59

sum of the travel distances\316\267

*\342\200\236(\302\253;)=

\316\243difl.Qi)\302\267

This is the exact translation of our informal discussion of the little

bug walking on the torus, exceptthat we are now requiring the bugto follow a path that is made up of straight line segmentsin therectangle.Namely, the bug should be a grasshopper instead of a

snail. As it follows a discrete walk, it alternatessteps where it hops

in X from one point P* to another point Qi, and steps where it is

beamed-up from Qi to Pj+i by the gluing process.1

To make the descriptionof a discrete walk w easier to follow, itis convenient to write \316\241~ P' when \316\241is glued to P', namely, when\316\241\342\200\224\316\241'. (Compare also our discussion of equivalence relations inExercise4.2). Then a discrete walk w from \316\241to Q is of the form

\316\241= Pi, Qi ~ Pa, Q2 ~-\320\240\320\267,\302\267\302\267\302\267,Qn-i

~P\302\273,Qn

= Q- This makes it

a little easier to remember that the consecutive pointsQi, \316\241\302\273+\316\271are

glued to each other, while the pairs Pi, Qi correspond to a travel of

length d(Pi,Qi) in X.

We would like to define a distance function d on X by

d(P, Q) =inf{^d(u>); w discrete walk from \316\241to Q}

for any two points P, Q G X.

Lemma 4.1. The above number d(P, Q) is independent of the choice

of the points P, Q \342\202\254X used to represent P, Q \342\202\254X. As a

consequence, this defines a function d: \316\247\317\207X \342\200\224*R.

In addition, this function d is a semi-distanceonX, in the sense

that it satisfies the following three conditions:

(1) d(P,Q)ttOand d(P, P) = 0 for everyP,Q eX (Nonneg-

ativity Condition);

(2) d(P, Q) = d(Q,P) for every P, Q e X (SymmetryCondition);

(3) d(P, R) \316\266d(P, Q) + d(Q,R) for every P,Q, ReX(Triangle Inequality).

The original Park City lectures involved pedagogic sound effects to distinguishbetween these two types of moves.

60 4. Gluing constructions!\302\247

Proof. To prove the first statement, consider two other points PAQ' e X such that P' = \316\241and Q' = Q. We need to show that|d(P',Q') = d(P,Q). I

If w is a discrete walk \316\241\342\200\224\320\240\321\214Qi

~ P2, Q2 ~ P3, ..., Qn-i ~5

\342\226\240Pn,Qn= Q from \316\241to Q, starting from \316\241and ending at Q, we can;

consider another discretewalk w' of the form P' = Po, Q0 ~ Pi,5Qi ~

P2, Q2~

\320\240\320\267,\302\267\302\267\302\267,Qn-i~ Pn, Qn ~ Pn+i, Qn+i = Q'

byj

taking P0 = Qo = P' and Pn+i =Qn+i

=Q'\342\226\240This new discrete

walk|w' starts at P', endsat Q', and has the same length i(w') = i(w) as;w. Taking the infimum over all such discretewalks w, we conclude;

that the \"distance\" d(P',Q'), defined using P' and Q', is lessthan orf

equal to d(P, Q), denned using \316\241and Q. Exchanging the roles of Pf\\

Q and P', Q', we similarly obtain that d(P,Q) < d(P^,Q'), so thaij

d(P,Q) = d(P',Q'). /t

This proves that the function d: \316\247\317\207X \342\200\224>\342\204\226.is well defined.

The Nonnegativity Condition (1) is immediate.To prove the Symmetry Condition (2), note that every discrete!

walk \316\241= Pi, Qi ~ P2, Q2 ~ P3, \302\267\302\267\302\267,Qn-i~ Pn, Qn = Q from \316\241td

Q provides a discrete walk Q= Qn, Pn ~ Qn-i, Pn-i ~ Qn-2,\302\267\302\267\302\267,'

P2 ~ Qi, Pi = \316\241from Q to P. Since these two discrete walks have*

the same length, one immediately concludes that d(P, Q) = d(Q, P).;'Finally, for the Triangle Inequality (3), consider a discretewaffi

10 of the form \316\241=_Plf Qi ~ P2, Q2 ~ P3, \302\267\302\267\302\267,Qn-i

~ Pn, Qn =Q\\

going from \316\241to Q, and a discrete walk w' of the form Q = Qi,Ri ~

<?2i \302\267\342\226\240\302\267>\320\271\321\202-i~ Qmi Rm = R going from Q to P. These

two discrete walks can be chained together to give a discrete walk

w\" of the form \316\241= Plf Qi ~ Q2, \302\267\302\267\302\267.,Qn-i~

P\342\200\236,Qn~ Q[,'

Ri ~ Q'2, \302\267\302\267\302\267,\320\271\321\202-i~ Q'm, Rm = R going from \316\241to P. Since

id(w\") = id(w) + id(w'),taking the infimum over all such discretewalks w and w', we conclude that d(P, R) < d(P,Q) + d(Q, R). \316\217

The only missing property for the semi-distanced to bea distance

function is that d(P, Q) = 0 only when \316\241= Q.

If this property holds, we will say that the partition or gluing

process is proper. The metric space (X, d) then is the quotient space

4.3.Gluing the edges of a euclidean polygon 61

ill' the metric space (X, d) by the partition. The distancefunction d

ih the quotient metric induced by d.In the caseof the torus obtained by gluing the sides of a

rectangle,we will prove in Sections 4.3 and 4.4 that the gluing is proper.'Plie same will hold for many examples that we will consider later

on. However, there also exist partitions which are not proper. SeeKxercise4.1 for an example where each point is glued to at most one

other point.

4.2.3. The quotient map. Forfuture reference, the following

elementary observation will often be convenient.

Let \317\200:X \342\200\224>X be the quotient map defined by \317\200(\316\241)= P.

Lemma 4.2. For everyP, Q G X,

d(P,Q)^d(P,Q).

\316\233\316\261a consequence, the quotient map \317\200:X \342\200\224>X is continuous.

Proof. The inequality is obtainedby consideration of the one-stepdiscrete walk w from \316\241to Q defined by \316\241= Pi, Qi = Q. Bydefinition of d, d(P, Q) < t{w)= d(P,Q).

The continuity of the quotient map \317\200is an immediate

consequence of this inequality. D

4.3. Gluing the edges of a euclideanpolygonThis section is devoted to the special case where X is obtained by

gluing together the edges of a polygon X, as in the exampleof the

torus obtained by gluing together opposite sides of a rectangle.

4.3.1. Polygonsand edgegluing data. Let X be a polygon in

the euclidean plane M2. Namely, X is a regionof the euclidean plane

whose boundary is decomposedinto finitely many line segments, linesand half-lines E\\, E2, \302\267..,En meeting only at their endpoints. We

also impose that at most two Ei can meet at any given point.

The line segments, lines and half-lines Ei bounding X are the

edges of the polygon X. The points where two edges meet are its

vertices.

62 4. Gluingconstructions.)

We require in addition that X and the \302\2437,\302\267are closed, in the sensesthat they contain all the points of K2 that are in their boundary. Inthis specificcase,this is equivalent to the property that the polygonJX contains all its edges and all its vertices. \\

I

However, we allow X to go to infinity, in the sense that it may';

be bounded or unbounded in (]R2,cieuc).A subset of a metric space.]is bounded when it is contained in some ball ; J

\316\220\316\231

Bdruc(P,r)={QeX;dtluc(P,Q)<r} ')

with finite radius r < oo.Using the Triangle Inequality (and changing 1the radius r), one easily sees that this property does not depend ori^

the point \316\241chosen as center of the ball. The subset is unbounded]

when it is not bounded. ..\302\267.*

Figure 4.3. A few bounded euclidean polygons

Figures 4.3 and 4.4 illustrate a lew examples of polygons.

In most casesconsidered,the polygon X will in addition be con-;vex in the sense that, for every P, Q e X the line segment [P, Q] i

joining \316\241to Q is contained in X. We endow such a convexpolygonwith the restriction \316\254\317\207of the euclidean metric r/tuc. defined by the

property that dx(P, Q) =dcuc(P, Q) for every P, Qe X.

Among the polygons \316\247\316\273,\316\2472,\302\267..,X7 described in Figures 4.3and 4.4.Xj is the only one that is not convex.

4.3. Gluing the edgesofa euclideanpolygon 63

Figure 4.4. A few unbounded euclidean polygons

When X is not convex, it is more convenient to consider for P,Q e X the infimum dx(P,Q) of the euclidean length 4uC(7) \302\260fall

piecewise differentiable curves 7 joining \316\241and Q and contained in

X. The fact that the function \316\254\317\207so defined is a metric on X isimmediate, noting that dx(P,Q) > deuc(P,Q). See Exercise1.10for

an explicit example.

We will call this distance function \316\254\317\207the euclidean pathmetric of the polygon X. Note that the path metric \316\254\317\207coincides with

the restriction of the euclideanmetricaeac when X is convex. In the

general case,the path metric coincides locally with aeac in the sense

that every \316\241\342\202\254X is the center of a small ball Bdtuc(P,e)such that

dx(\316\241, Q)= aeuc(P, Q) for every Q &\320\245\320\237Bdeuc(\316\241,\316\265)(because the line

segment [P, Q] is contained in X).

We will occasionally allow the polygonX to be made up of several

disjoint pieces, so that there exist points \316\241and Q \342\202\254X which cannot

be joined by a piecewisedifferentiable curve 7 completely containedin X. In this case, \316\254\317\207(\316\241,Q)

= +00 by convention. This requiresthat

we extend the definition of metrics to allow them to take values in

[0, +00], but the extension is immediate provided we usethe obvious

conventions for inequalities and additions involving infinite numbers

(o < 00 and \320\276+ \321\201\321\216= \321\201\321\216for every \320\276\342\202\254[0, \321\201\321\216],etc ).

64 4. Gluing constructions 1

A polygon for which this does not happen is said to be con- \\

nected. Namely, the polygon X is connectedif any two points \316\241\\

and Q \342\202\254X can be joined by a piecewisedifferentiable curve 7 com- ,

pletely contained in X. Thereaderwho is already familiar with somenotions of topology will notice that the definition we are using here \\

is more reminiscent of that of path connectedness;however, the two

definitions are equivalent for polygons.After these preliminaries about polygons, we now describe how

to glue the edges of a polygon Xcl2 together.For this, we first

group these edges into pairs {Ei,E2}, {E3,Ei}, ..., {E2p-i,E2p},and then for each such pair {E2k-i,E2k} we specify an isometry '

<P2k-i- E2k-i \342\200\224\342\226\272Ew Here </?2fc-i is an isometry for the restrictions

of the metric \316\254\317\207to the edges E2k-\\ and E2k, which also coincide

with the restrictions of the euclideanmetricdeac to these edges. This

is equivalent to the propertythat ip2k-i sends each line segment inthe edgeE2k-ito a line segment of the same length in E2k-

Note that in particular, the edges E2k-i and E2k in a given.pairmust have the same length, possibly infinite. In general,the isometry

V?2fc-i is then uniquely determined once we know how </?2fc-i sends

what orientation of E2k-i to which orientation of E2k- A convenient

way to describethis information is to draw arrows on E2k-i and E2k

corresponding to these matching orientations. It is alsoconvenient to

use a different type of arrow for each pair {E2k-i,E2k}, so that thepairing can be readily identified on the picture. Figures 4.3 and 4.4offer some examples.

There is one case where this arrow information is not sufficient,when E2k-i and E2k are both lines (of infinite length), as in the caseof the infinite strip X5 of Figure 4.4. In this case, \317\206^-\316\271-#2fc-i \342\200\224>E2k

is only defined up to translation in one of these lines, and we will need

to add extra information to describe<p2k-i-

The notation will be somewhat simplified if we introduce the

isometry \317\2062^:E2k \342\200\224>E2k-\\ defined as the inverse \317\2062\316\272=

\317\2062^-\\of

V?2fc-i: E2k-i \342\200\224>E2k- In this way, every edge E% is glued to an edgeEi\302\261iby an isometry ipf. Ei \342\200\224>\302\243?i\302\261i,where \302\2611depends on the parityof i.

4.3. Gluingthe edgesofa euclidean polygon 65

With this data of isometricedge identifications \317\210\316\271:Ei \342\200\224>Ei\302\261x,

we can now describe the gluing of the edges of the polygon X by

Hpecifying a partition X as follows.Recall that if \316\241\342\202\254X, we denote

by \316\241\342\202\254X the corresponding element of the partition X, consisting

of all the points of X that are glued to P. The gluing is then defined

\320\277\320\275follows:

\342\200\242if \316\241is in the interior of the polygon X, then \316\241is glued to

no other point and \316\241={\316\241};

\342\200\242if \316\241is in a edge E^ and is not a vertex, then \316\241consists of

the two points \316\241e E{ and \317\210\316\257(\316\241)\342\202\254\302\243?i\302\261i;

\342\200\242if \316\241is a vertex, then \316\241consists of \316\241and of all the verticesof X of the form ipik \316\277

\317\210\316\257\316\227_\317\207\316\277\302\267\302\267\302\267\316\277

\317\206\316\257\317\207(\316\241),where the indices

\302\253\316\271.\302\2532\316\271\302\267\302\267\302\267.*fc are such that\317\206\316\257\317\214_1

\316\277\302\267\302\267\302\267\320\276<ph (\316\241)\317\202Ei}

for

every j.

The case of vertices may appear a little complicated at first, but

it, becomes much simpler with practice. Indeed, because each vertex

belongs to exactly two edges, there is a simple method for listing

nil elements of \316\241for a vertex P. The key observation is that when

considering a vertex \317\210%\316\227\316\277\317\206\316\257\316\220\316\257_1\316\277\302\267\342\226\240

\302\267\316\277<\317\2014\316\271(\316\241)glued to P, we can alwaysassumethat the range\317\206^(\316\225\316\271})

of the gluing map \317\210^is different from

the region \302\243<\302\273\316\257+1of

\317\206%1+1(since otherwisey>\302\273J+1

=\317\206\316\2261}so that these

two gluing maps cancel out).

This leads to the following algorithm: Start with P\\= P, and let

Eix be one of the two edges containing P\\. SetPi =\317\210\316\2201(\316\241\316\271),and let

Ei2 be the edgecontaining P2 that is different from \317\206^(\316\225^). Iterating

this process provides a sequence of vertices Pj, P2, ..., Pj, ... and

tidges Eix, Ei2, ..., Eia, ... such that Pj \342\202\254Eijf Pj+l=

\302\245>^(\320\240,-)>

and Eij+1 is the edge containing Pj+\\ that is different from\317\206%\316\257(\316\225\316\257\316\257).

Since there are only finitely many vertices, there is an index \320\272for

which Pfe+i = Pj for some j ^ k. If \320\272is the smallest such index, oneeasily checksthat Pfe+i

= Pi, and that \316\241={Pi,P2,.. \302\267,Pfe}.

For instance, in the example of the rectangle X\\ of Figure 4.3,the four corners are glued together to form a single point in \316\247\317\207.For

the hexagon X4, the vertices of X4 project to two points of X4,eachof them corresponding to three vertices of the hexagon.In Figure 4.4,

66 4. Gluing construction^

the infinite strip X5 has no vertex. The quotient space \316\247\316\263hastwoij

elements associated to vertices of \316\247\316\263,one corresponding to exactly!one vertex and another one consisting of two vertices of \316\247\316\263. |

4.3.2. Edge gluings are proper. Let \316\254\317\207be the semi-metric on|the quotient spaceX that is defined by the euclidean path metric \316\254\317\207\\

of the polygon XcR2. $\316\212

Theorem 4.3. If X is obtained from the euclidean polygon X b$

gluing together edge pairs by isometries, then the gluing is proper.*

Namely, the semi-distance \316\254\317\207induced on X by the metric \316\254\317\207of \320\251

is such that \316\254\317\207(\316\241,Q) > 0 when \316\241\317\206Q.

The proof of Theorem 4.3 is somewhat long and is postponed t

Section 4.4. |\"'II

4.3.3. Euclidean surfaces. Let us go backto our informalpaper]

and adhesive tape discussion of the torus X, obtained by gluing op-^

posite sides of a rectangleX. , |_ 4

If \316\241is the point of the torus that corresponds to the four cornersjof the rectangle,it should be intuitively clear (and will be

rigorously

proved in Lemma 4.5) that a point Q \342\202\254X is at distance < \316\265of \316\241\320\251

X exactly when it corresponds to a point Q \342\202\254X which is at distance< \316\265from one of the corners of the rectangle.As a consequence, for

\321\211

small enough, the ball\316\222^\317\207(\316\241,\316\265)is the image in X of four

quarter^

disks in X. We know from experience that if we glue together four!

paper quarter-disks with (invisible) adhesive tape, we obtain an objec|which is undistinguishable from a full disk of the same radius in the

euclidean plane. {

The same property will hold at a point \316\241\342\202\254X that is the image;of a point \316\241\302\243X located on a side of the rectangle,not at a cornergThen the ball Bj(P, \316\265)is obtained by gluing two half-disks along theiidiameters,and again it has the same metric propertiesas a disk. \342\226\240<

As a consequence, if our little bug crawling on the torus X \320\251

in addition, very near sighted, it will not be able to tell that it is

walking on a torus insteadof a plane. This may be compared to th3

(hi)story of other well-known animals who thought for a long time

that they were living on a plane,before progressively discovering that

4.4. Proofs of Theorems 4.3 and 4.4 67

they were actually inhabiting a surface with the rough shape of a verylurge sphere.

When gluing together sectors of paper disks, a crucial propertyfor the result to looklike a full disk in the euclideanplane is that the

iingles of these disk sectors should add up to 2\317\200.As is well known

to anybody who has ever made a birthday hat out of cardboard, theresulting paper construction has a sharp cone point if the sum of

the angles is less than 2\317\200.Similarly, it wrinkles if the angles addup to more than 2\317\200.Not unexpectedly, we will encounter the samecondition when gluing the edges of a euclideanpolygon.

Let us put this informal discussionin a more mathematical

framework.

Two metric spaces (X, d) and (X1,a') are locallyisometricif

for every \316\241\302\243X there exists an isometry between someball Bd(P, \316\265)

centered at \316\241and a ball \316\222^(\316\241',\316\265)in X'.

Theorem 4.4. Let (\316\247,\316\254\317\207)be the quotient metric space obtainedfrom a euclidean polygon (\316\247,\316\254\317\207)by gluing together pairs of edgesof X by isometries. Suppose that the following additional condition

liolds: For every vertex \316\241of X, the angles of X at those vertices P'of X which are glued to \316\241add up to 2\317\200. Then (\316\247,\316\254\317\207)is locally

isometric to the euclidean plane (R2,deuc).

Again, although the general idea is exactly the one suggestedby the above paper-and-adhesive-tape discussion, the proof of

Theorem 4.4 is rather long with several cases to consider. It is given inthe next section.

A metric space (X, d) which is locally isometric to the euclideanplane (R2,deuc) is a euclidean surface. EquivalentJy, the metric \316\254

is then a euclidean metric.

4.4. Proofs of Theorems 4.3 and 4.4This section is devoted to the proofs of Theorems 4.3 and 4.4. These

proofs are not very difficult but are a little long, with many cases

to consider. They may perhaps be skipped on a first reading, which

68 4. Gluing constructions*!^

gives the reader a first opportunity to usethe fast-forward command?;

of the remote control. *

These are the first really complex proofs that we encounter in ]this book. It is important to understand these arguments (and many?!more later) at a level which is higher than a simple manipulation off

symbols. With this goal in mind, the readeris strongly encouraged,,

to read them with a piece of paper and pencil in hand, and to draw \316\271

pictures of the geometric situations involved in order to betterfollow\302\253j

the explanations. ',!

Throughout this section, X will denote a polygon in the euclidean '

space (R2,deUC). We are also given isometries </?2fc-i:E2k-i \342\200\224*\320\2252\320\272]

and v?2fe=

\302\245>2fc-i: ^2fc \342\200\224*E2k-\\ between the edges \316\225\317\207,..., En of(i

X. Then X is the space obtainedfrom X by performing the corre- ;

sponding edge gluings, and dx is the semi-metric inducedon X by;

the euclidean path metric dx introduced in the previous section.

Recall that in the more common case where X is convex, dx is just the

restriction of the euclideanmetric deac to X.

First we need to understandthe balls

BSx(P,e)= {Qe P; dx(P,Q) < \316\265},

at least for \316\265sufficiently small.

4.4.1. Small balls in the quotient space(X,d).Lemma 4.5. For every \316\241\342\202\254X, there exists an \316\2650> 0 such that

for every \316\265< \316\265\316\277omd every Q \342\202\254X, the point Q \342\202\254X is in the ball

Bjx (\316\241,\316\265)if and only if there is a P' \342\202\254\316\241such that dx(P', Q) < \316\265.

We can rephrase this in terms of the quotient map \317\200:X \342\200\224>X

sending \316\241\342\202\254X to \317\200(\316\241)= \316\241G X. Lemma 4.5 states that for \316\265

sufficiently small, the ball\316\222^\317\207(\316\241,\316\265)is exactly the union of the images

under \317\200of the balls Bdx (P\\ \316\265)as P' ranges over all points of P.

For a better understanding of the proof, it may be useful to realizethat the ball Bgx (\316\241,\316\265)can be significantly larger than the union ofthe images\317\200(\316\222\316\254\317\207(\316\241',\316\265))

when \316\265is not small. This is illustrated byFigure 4.5 in the case of the torus. In these pictures,\316\241consists of

the single point P. The shadedareasrepresent the balls Bjx (\316\241,\316\265)


lor various values of \316\265;in each shaded area, the darker area is theimage of the ball Bdx (\316\241,\316\265)under \317\200.

Figure 4.5. Balls Bs(P,e) in X

Proof of Lemma 4.5. Recall from Lemma 4.2 that dx(P,Q) <dx(P',Q) for every P' \302\243P. Therefore, the \"if\" part of the

statement holds without restriction on \316\265.

The \"only if\" part will take more time to prove as we will need

to distinguish cases.

Let \316\241be a point of X. For a number \316\265\316\277which will be specifiedlater on in function of P, we consider a point Q \342\202\254X such that

dx (P, Q) < \316\265< \316\265\316\277\302\267We want to find a point P' \342\202\254P, namely, a pointof X which is glued to P, such that dx(P', Q) < \316\265.

Since dx (P, Q) < \316\265,there exists a discrete walk w from \316\241to Q

of the form \316\241= Pu Qi ~ P2, Q2 ~ ft, \302\267\302\267\302\267,Qn-i~ Pn, Qn = Q

and whose length is such that \302\243(w)=

\316\243\316\223=\316\271dx(Pi, Qi) < \316\265.

We want to prove by induction that for every j < n,

i

(4.1) there exists P' \342\202\254\316\241such that dx{P',Qj) < Y^dx{PuQi) < \316\265.

We can begin the induction with j= 1, in which case (4.1) is trivial

by taking P' = P.Suppose as an induction hypothesis that (4.1) holds for j. We

want to show that it holds for j + 1.

For this, we will distinguish cases according to the type of the

point \316\241\302\243X. We will also specify \316\265\316\277in each case.

18!70 4. Gluing construction

Case 1. \316\241is in the interior of the polygon X.

We first specify the number \316\265\316\277needed in this case. We chooseyit so that the closed disk of radius \316\265\316\277centered at \316\241is

completely^contained in the interior of the polygon.Equivalently, every point oofthe boundary of the polygon is at distance> \316\265\316\277from P. J

In this case, \316\241is the only point of P. I

By the induction hypothesis (4.1) and by choice of \316\265\316\277> \316\265,the!

point Qj is in the interior of the polygon. In particular, it is glued to|

no other point so that Pj+\\= Qj. Combining the TriangleInequality!

Iwith the induction hypothesis, we conclude that

*=1 :i

This proves (4.1) for j + 1. i

Case 2. \316\241is on an edge Ei of the polygon and not at a vertex.

In this case,\316\241consists of \316\241and of exactly one other point \317\210\316\257(\316\241)

on the edge Ei\302\261\\that is glued to E^.

Choose \316\265\316\271> 0 such that \316\241is at distance > \316\265\316\271from the other

edges Ej, with j \317\206\320\263,of the polygon. Similarly, let 62 be such that

\317\210\316\257{\316\241)is at distance > 62 from any edge other than the edge Ei\302\261\\

that contains it. Choose \316\265\316\277as the smaller of \316\265\316\271and 62\302\267

If Qj= Pj+i, combining the induction hypothesis (4.1) with the

Triangle Inequality gives, as in the caseof interior points,

i+i

dx(P',Qj+1) 4dx(P',Qj) + dx(Pj+i,Qj+1) <53dx(Pi,gi) <e,\302\273=i

which proves (4.1) for j + 1 in this case.Otherwise, Qj and Pj+i are distinct but glued together.Because

dx(P',Qj) < \316\265< \316\265\316\277and by choice of \316\265\316\277,these two points cannot

be vertices of the polygon, so that one of them is in the edge E{ andthe other one is in the edge i?j\302\261i glued to E{ by the map \317\210{.In

particular, PJ+1= ip^iQj). Set \316\241\"=

\317\206\316\2631(\316\241').Note that P\" is just

equal to \316\241or to <fii(P); in particular it is in P.

4.4. Proofsof Theorems 4.3 and 4.4 71

We will use the crucial property that the gluing map \317\210\316\271respects

distances. As a consequence, dx(P\",Pj+i) = dx(P',Qj) so that

dx (P\", Qj+i) < dx (P\",Pj+i) + dx(Pj+1,Qj+1)j+i

< dx(P\\ Qj) + dx(Pj+1, Qj+i) \302\253\316\25253dx(Pi' Q*) < \316\265

i=\\

by the induction hypothesis. Again, this proves (4.1) for j +1 in thiscase.

Case 3. \316\241is a vertex of the polygon.In this case, \316\241consists of a certain number of vertices P' of the

polygon. Pick sq such that every point of \316\241is at distance > \316\265\316\277from

the edges that do not contain it.The proof is almost identical to that of Case 2 with only a couple

of minor twists.

If Qj = Pj+i, as in the previous two cases, the combination of

the induction hypothesis (4.1) and the Triangle Inequality shows that

(4.1) holds for j+ 1.If Qj \317\206Pj+i, we distinguish cases according to whether Qj is

a vertex of X or not. If Qj is a vertex, since dx(P', Qj) < \316\265\316\266Eq

by the induction hypothesis, this vertex must be P' by the choice of\316\265\316\277\302\267Then, P\" = Pj+1 is also in \316\241since it is glued to Qj = P', and

therefore we found a P\" \342\202\254\316\241such that

j'+i

dx(P\", Qj+1) \302\253dx(Pj+1, Qj+1) < 53 dx(Pi, Qi) < \316\265

i=l

as required.

Otherwise, Qj is not a vertex and is glued to Pj+1 by a gluing

map \317\210}.:Ef. \342\200\224>Ef.\302\261i\302\267By choice of \316\265\316\277,\316\241'is contained in the edgeEk, so that P\" = 4>k{P') 6 \316\241is defined. Since the gluing map \317\210^

respects distances,

dx(P\", Qj+1) < dx(P\",Pj+1) + dx(Pj+1,Qj+1)

j+i

< dx(P\\ Qj) + dx (Pj+1,Qj+1)< 53 dx (Pi' Q^ < \316\265'

i=\\

using the fact that the induction hypothesis(4.1)holds for j.

72 4. Gluing construction

Therefore, (4.1) now holds for j + 1, as requested.,^

This completes the proof of (4.1) in all three cases,and for all ja

The case j = \316\267proves Lemma 4.5 since Qn = Q. urn

4.4.2. Proof of Theorem 4.3. We now have the tools needed toJs

prove Theorem 4.3, namely that dx (P, Q) > 0 whenever \316\241\317\206Q.

Let \316\265\316\277be associated to \316\241by Lemma 4.5.

Because X is a partition of X, the fact that \316\241\317\206Q implies thatl

these two subsets \316\241and Q of X are disjoint, namely, they have noj

point in common. Since these subsets are finite, it follows that there!exists an \316\265\316\271> 0 such that every point of \316\241is at distance > \316\265\316\271from!

every point of Q. Set \316\265=min{go, \316\265\316\271}> 0.

Then d~x(P, Q) > \316\265.Indeed, Lemma 4.5 would otherwise provide!a point P' \342\202\254\316\241such that dx(P', Q) < \316\265< \316\265\316\271,thereby contradictir

the definition of e\\.

4.4.3. Proof of Theorem4.4.For every point \316\241\342\202\254X, we need \320\246

find an isometry \317\206between a small ball\316\222^\317\207(\316\241,\316\265)\320\241X centered atj

\316\241and a small ball Bdeac(P'',\316\265) in the euclidean plane \320\2502.

Fix an \316\265satisfying the conclusions of Lemma 4.5. In addition,!choose\316\265small enough that each P' \342\202\254\316\241is at a euclidean distanc> 3\316\265from any edge that does not contain it. In particular, thel

balls Bdx {\316\241'\302\267,\316\265)are pairwise disjoint and are disks, half-disksordisk!sectors in R2, according to the type of P' \342\202\254P. Here, a (euclideanUdisk sector is one of the two pieces of a euclidean disk Bdeuc (Po, r]i

in R2 delimited by two half-lines issuedfrom its center Pq, as in a.1

slice of pie. .|In addition, the Triangle Inequality shows that the ballsBdx(\316\241',\316\265|

are at a euclidean distance > \316\265apart, in the sense that deuc(Q', Q\")>'\316\257

\316\265if Qf \342\202\254Bdx (\316\241',\316\265)and Q\" \342\202\254Bdx (\316\241\",\316\265)with \316\241'\317\206\316\241\"e P. j

Lemma 4.5 says that the ball\316\222^\317\207(\316\241,\316\265)

is obtained by gluing!together the ballsBdx (\316\241',\316\265)in X centered at the points P' that are

glued to P. Let ;B= U \316\222\316\254\317\207(\316\241>,\316\265) j

P'\342\202\254P

'

denote the union of these balls. !

4.4. Proofsof Theorems 4.3 and 4.4 73

This subset \320\222\321\201X comes with two natural metrics. The first

one is the restriction of the metric \316\254\317\207.The second one \320\260\320\262is similarly

denned, but by restricting attention to curves that are contained in

B. Namely, de{Q, Q') is the infimum of the euclidean lengths 4uc(7)of all piecewisedifferentiable curves 7 joining Q to Q' and containedin B. In particular, de{Q, Q') = 00 if Q and Q' are in distinct balls

Bdx(P',e)andBdx(P\",e)ofB.

When Q and Q' are in the sameball Bdx (P', e) of B, elementarygeometry shows that de(Q,Q) = dx(Q,Q). Indeed, Bdx(P',e) isa disk, a half-disk or a disk sector. Therefore, the only case which

requires some thought is that of a disk sector of angle > \317\200(since

otherwise de(Q,Q') = dx(Q,Q') =deX]C(Q,Q') by convexity). In

this case, one just needsto checkthat the shortest curve from Qto Q' in the polygon X is either a single line segment completelycontained in B^x (\316\241',\316\265)or the union of two line segmentsmeeting at

the vertex P' in Bdx(P',s); compare Exercise 1.10.Sincethe ball Bjx (\316\241,\316\265)is obtained by performing certain gluings

on \320\222it inherits a quotient semi-metric \320\260\320\262\302\267The advantage of \320\260\320\262is

that it is entirely denned in termsof B, without reference to the restoiX.

Lemma4.6. The metrics \316\254\317\207andde coincide on the ball\316\222^\317\207(\316\241,|\316\265).

The restriction to the ball of radius|\316\265

is used to rule out thepossibilityof a \"shortcut\" through X making Q and Q' closerinX than in

\316\222^\317\207(\316\241,\316\265)= \320\222.The left-hand side of Figure 4.5

provides an example of two such points Q and Q' \302\243Bgx (\316\241,\316\265)such that

dx(Q,Q')<dB(Q,Q').

Proof of Lemma 4.6. By definition of \316\254\317\207and \320\260\320\262,ax(R,R') <

\320\260\320\262(\320\257,\320\257')for every R, R' \342\202\254B. It immediately follows that

a~x{Q,Q') < dB(Q,Q') for every Q, Q' \342\202\254B3x(P,e). Incidentally,

this shows that de is really a metric and not just a semi-metric.To prove the reverse inequality, we need to restrict attention to

Q, Q' \342\202\254Bdx(P, |\316\265).In particular, ax(Q,Q') <

|\316\265by the Triangle

Inequality.

74 4. Gluing constructions\316\220

\320\263\342\200\224 1Let \320\263\320\276be a discrete walk from Q to Q' in X of the form Q = Qi, |

0\316\212~ Q2, Q'2 ~

<9\320\267,\302\267\302\267\302\267,Q'n-i~ Qn. Qn = Q', and whosedx-length J

idx(w) is sufficiently close to dx(Q,Q) that ^dx(\302\2530 <|\316\265.

Then;j

Qi= Qi+i in X and, vising the fact that the quotient map is distance \316\257

nonincreasing (Lemma 4.2), \342\226\240!ii

\320\237 \320\237 ';

Yidx(QuQw)<Yidx(Quqfi)<\\e. )

i=l i=l

A repeated use of the Triangle Inequality then shows that :i

i-\\ ;

dx(P,Qi)^dx(P,Q1) + 4^2dx(QvQj+i)<y+^= e, j

- - iso that all Qi are in Bdx(P,e). Since \316\265satisfies the conclusions\320\276\320\246

Lemma 4.5, we conclude that all Qi and Q'i are in the subset B. 1

If \316\241'\317\206\316\241\"\342\202\254\316\241,then dx(P',P\") > 3\316\265by choice of \316\265,and the,

Triangle Inequality shows that any point of the ball Bdx(P',e)isat;a distance > \316\265from any point of Bdx(P\",e). Sincedx(Qi, QI) <

|\316\265,<

we conclude that Qi and Q,i are in the same ball Bdx(P',e). In;

particular, we observed(right above the statement of Lemma 4.6);that dB(Qi,Q!i)

= dx(Qi,Q'i).

What this shows is that w is also a discrete walk from Q to Q'.in B, whose de-length idB (w) is equal to its dx-lengthidx(w). As a

consequence, 3b(Q,Q') ^ ^dx{w).Sincethis holds for every discrete walk w whose length \302\243dx(w)

is sufficiently close to dx(Q,Q'), we concludethat ds(Q)Q') ^

dx(Q,Q').

Because we have already shown that the reverse inequality holds,this proves that dB(Q, \302\243?')

= dx (Q, Q') for every Q, Q' \342\202\254Bdx (\316\241,|\316\265).

D

We are now ready to prove Theorem 4.4. As in the proof of

Lemma 4.5, we will distinguish cases according to the type of the

point \316\241\302\243X corresponding to \316\241\342\202\254X.


Case 1. \316\241\320\263\320\260in the interior of the polygon X.

In particular, \316\241is glued to no other point so that \316\241consists only

of P. Then \320\222= Bdx (\316\241,\316\265)and, by our choice of \316\265,the ball Bdx (\316\241,\316\265)

is completely contained in the interior of X. In particular, the ball

Bdx (\316\241,\316\265)\321\201X is the same as the euclideanball \320\224\320\263\320\265\321\206\321\201(\316\241,\316\265)\320\241R2, an

open disk in the euclidean plane K2. Also, there are no gluings

between distinct points of \320\222= Bdx (\316\241,\316\265),so that every Q \342\202\254Bdx (P> \316\265)

corresponds to exactly one point Q \342\202\254Bdx (\316\241,\316\265).

Define \317\206:\316\222\316\254\317\207(\316\241,\316\265)-> Bdeuc(P,e) by the property that ip(Q) =

Q for every Q\342\202\254Bdx(P,\316\265).

The map \317\206may not be an isometry over the whole ball, but we

claim that

deMQ),tl>(Q')) = dx(Q,Q')for every Q, Q \302\243

Bix(P, |\316\265). Indeed, d~x(Q,Q') = dB(Q,Q') byLemma 4.6. Sincethere are no gluings in B, one easily sees that

MQ>Q') = deiQ^Q1)(seeExercise4.3). Finally, dB(Q,Q')=

ikociQ, Q')=deuc(i>(Q), \321\204{\320\257'))by convexity of the ball B=Bdx (\316\241,\316\265).

This proves that the restriction of \317\206to the ball Bdx (\316\241,|\316\265)

is an isometry from (Bjx(P, |\316\265),<\316\226\317\207)to the euclidean disk

{Bdmc(P, \320\264\320\265),4\320\270\321\201),as requested.

Having completed the analysis in Case1,we now directly jumpto the most complexcase.

CASE 2. Pisa vertex of the polygonX.Write \316\241= {Pi, P2,...,Pfc} with \316\241=

\320\240\321\212Namely, \320\240\321\212\320\2402,...,

Pfc are the vertices of X that are gluedto P. Lemma 4.5 says that the

ball Bjx (\316\241,\316\265)in X is the image under the quotient map \317\200:X \342\200\224>X

of the union \320\222of the balls Bdx (Pi, \316\265),Bdx (\320\240\320\263,\316\265),..., Bdx (Pfc, \316\265)in

X.

Because of our choice of \316\265,each of the balls Bdx(Pj,e) in themetric space (\316\247,\316\254\317\207)is a disk sector of radius \316\265in R2, and these disksectors are pairwisedisjoint. We now need to work harder than in

the previous case to rearrange these disksectorsinto a full disk.

Each Pj belongs to exactly two edges E^ and Ey. As in ourdescriptionof vertex gluings at the end of Section4.3.1,we can choose

76 4. Gluing constructions^

Pi \320\232

3 ^i3

The polygon X

\320\244\320\267(\320\262*\321\205(\320\240\320\267,*))|

*s(Bdx\342\204\226,e))

*i(B^(Pi,s))

Bdenc(P',e)=Tp(Bdx(P,e))\\

Figure 4.6. Gluing vertices together

the indexings so that for every j with 1 < j < fc, the gluing map

\317\210\316\257}sends the vertex Pj to PJ+i and the edgeEi} to Ey. with the

convention that Pk+i = Pi andi'fc+1

=\320\263'\320\263.

We will construct our isometry \317\206:Bjx (\316\241,\316\265)\342\200\224>

Bd\342\200\236Uc(-\320\240',\316\265)pie

wise from suitable isometries Vj of (R2,deuc)\302\267 For this, we use the

following elementary property, which we list as a lemma for futuref

reference. 1

Lemma 4.7. Let \317\210:g \342\200\224>g' be an isometry between two line segments^

half-lines or full-lines g andg' inM.2. Then, \317\210extends to an isometry]

V?:R2-\302\273K2 \320\276/(\320\2322,4\320\270\321\201). - 1

In addition, if we choose one side of g and another side for g', wecan arrange that \317\210sends the selected side of g to the one selectedforg'. The isometry \317\206is then uniquely determined by these properties*

a

Lemma 4.7 is an immediate consequence of the classification ofisometries of (M.2,deuc) provided by Proposition 1.3.

In particular, for every j, we can extend the gluing map \317\206^:

Eii \342\200\224>Ei>. to an isometry \317\210\316\271,:\320\2502\342\200\224>R2 of (R2, deuc) that sends thepolygon X to the sideof \316\225\316\263that is opposite X.

To define the\317\210$,

we begin with any isometry \317\210\316\271of (K2, deuc),and inductively define

V>j+i=

\316\246,\316\277

\317\210\317\212/= Vi \302\260

\317\210\317\212*\302\260

\316\250\317\212\317\204\302\260\342\226\240\342\226\240\342\226\240\302\260

\321\207>7?\302\267


By induction on j and because Pj+i =<Pj(Pj),

the map \317\206$sends

l.he vertex Pj to the same point P' =\317\206\\(\316\241)for every j. In

particular, the isometry \317\2103-sends the disk sector Bdx(Pj,e) to a diskHector of the disk Bdeuc(P',e). Similarly, the image of the edge

Eii \316\261=

4>ij+i{Ej) under \317\206j+\316\271is equal to the image of\320\225\321\206

under

i/'j. By definition of the extensionof\321\204^

to an isometry of \320\250.2,the

two disk sectors \317\210}(Bdx(Pj,e))and ipj+1 (Bdx(Pj+\316\271,\316\265))

sit on

opposite sides of ipjiEij)= ipj+i(Ei> ). It follows that the disk sectors

i/'j (Bdx (Pj, \316\265))all fit side-by-side and in order of increasingj around

(.heir common vertex P'. See Figure 4.6.It is now time to use the hypothesis that the internal angles of

l.he polygon X at the vertices Pi, P2, ..., Pk \342\202\254\316\241add up to 2\317\200.This

implies that the disk sector\321\204\320\272+1(\320\222\320\260\321\205(\320\240\320\272+1,\320\265))

=\321\204\320\272+1(\320\222\320\260\321\205(\320\240\\,\320\265))

is equal to \321\204\\(Bdx (Pi, \316\265)).In particular, the two isometries V'fc+i and

V'i of (R2, deuc) send Pi =Pk+i to the same point P', send the edge

Eik+X= Ei' to the same line segment or half-line issued from P', and

send a sideofEik+1

=\316\225\316\263to the same side of ipk+i(Eik+1)= ipi(Ei').

By the uniqueness part of Lemma 4.7, it follows that \320\263\321\200\320\272+\316\271=

\316\250\316\271\302\267

Finally note that when Q \342\202\254Ei}

is glued to Q' =\317\210^(Q) \342\202\254\316\225?\316\271,

l.hen ipj(Q)=

V>j+i(Q')\302\267 We can therefore define a map

ir.B3x(P,e)^Bdmia(P',e)

by the property that \317\206(0) is equal to\317\206j(Q)

whenever Q \342\202\254Bd(Pj,\316\265).

The above considerations show that \317\206is well denned.

We will show that \317\206induces an isometry between the

correspondingballs of radius|\316\265.

For this, consider two points Q, Q' \342\202\254Bjx (\316\241,|\316\265).By Lemma 4.6

and by the Triangle Inequality, \320\260\320\262(0,0')= dx(Q,Q') <

|\316\265.Let w

be a discrete walk from Q to Q' in B, of the form Q= Qi, 0\316\212

~ Q2,

Q'2 ~\320\257\320\267,\302\267\342\226\240\302\267,Qn-i

~Qn,_ Q'n = Q, and whose dB-lengthidB(w)

is sufficiently close to \320\271\320\262(\320\257,\320\257')that idB(w) <|\316\265\302\267

^\316\267particular,

each d,B(Qi,Q'i) is finite, so that Qi and Q'i belong to the same ball

Bdx (Pji:e). As a consequence,

deuc(\320\244(0~\320\263),\320\244\320\250)= 4uc (\316\246\317\215(Qi), \316\250\317\215(Qi))

= doac(Qi,Q'i)^dB(Qi,Q'i)

a

re\342\200\2241 i

78 4. Gluing constructions I

since each ipj. is a euclideanisometry. Then, by iterating the Triangle4Inequality and using the fact that Q\302\243

= Qi+\\,*

n-l

^\320\270\321\201(^(<9),^(<9)) <\302\243>\320\265\320\270\321\201\320\234<\320\255\320\263),^(<3<))

i=l

\316\267\342\200\2241

<5>B(Qi,gi)=

W\302\2530- 1

<=\316\271 ':!

Since this holds for every discrete walk w from Q to Q' in \320\222whose \\

length is sufficiently close to 3b(Q, Q'), we conclude that 5

(4.2) ^^(0)^(0)) <dB(Q, Q')\302\267 .;

j

Conversely, let 7 be the oriented line segment from ip{Q) to ip(Q');i,in the disk Bdeac (\316\241',|\316\265).

Recall that Bd,ac (\316\241',|\316\265)is decomposed into}

the disk sectors ipj(Bd(Pj,g^))\302\267Therefore we can split 7 into linei

segments71,72, \302\267\302\267\302\267,7n, in this order, such that each 7\302\273is contained'

in a disk sectoripjt {Bd{Pji, \\\316\255)). J

In the disk sector Bd(Pji, |\316\265)\320\241-X\", consider the oriented lines

segment 7*=

^*(7\302\253) corresponding to 7\302\273.If the endpoints of 7^'are labelledso that 7^ goes from Qi to Q\302\243,we now haye a discrete^walk w from Q to Q' of the form Q

= Q1} Q[ ~ Q2, Q2 ~\320\244\320\267>\302\267\302\267\302\267>'*

Q^_!~ Qn, Q'n = Q, of ds-length ;

tdB w =\316\243dB(Qi> ofi)=\316\243 '\302\253*\320\274

=\316\243\302\253\321\214)

i=l i=l i=l

=4\320\270\321\201(7)

=4\320\270\321\201(^(<9),^(<\320\255)).

It follows that

(4.3) dB(Q,Q')^deUiC(i>(Q),i>(Q)).

Combiningthe inequalities (4.2)and (4.3), we conclude that

dx(Q,Q') = dB(Q,Q')=deuc(i>(Q),i>(Q))

for every Q, Q e Bgx (\316\241,|\316\265).In other words, \317\206induces an isometry

from the ball (Bdx(P, |e), dx) to the ball(\320\224^ (P', \302\247e),\302\253Uc)\302\267

This concludes our discussion of Case 2, where \316\241is a vertex of

X. We have one case left to consider.

4.5. Gluinghyperbolicand spherical polygons 79

Case 3. \316\241is in an edge of the polygon X, but is not a vertex.

The proof is identical to that of Case 2. Actually, it can even be

considered as a specialcaseof Case 2 by viewing \316\241and the point P'

that is glued to it as vertices of X where the internal angle is equalto \317\200.

This concludes the proof of Theorem 4.4. D

4.5. Gluing hyperbolic and spherical polygons

Before applying Theorems 4.3 and 4.4 to specificexamples,letus lookat the key ingredients of their proof in more detail. For the proof ofTheorem 4.3 (and the proof of Lemma 4.5 before), in addition to

.standard properties of metric spaces,we mostly used the fact that

the maps \317\210\316\257gluing one edge of the polygon to another preserved

distances. A critical component of the proof of Theorem 4.4 was

Lemma 4.7.

4.5.1. Hyperbolic polygons. All these properties have

straightforward analogues in the hyperbolic plane (H2,dhyp)> provided we

use the appropriate translation. For instance,the euclideanmetricrieuc just needs to be replaced by the hyperbolicmetric dhyp>

euclidean isometries by hyperbolic isometries, line segments and linesby geodesies, etc Consequently, our results automatically extendto the hyperbolic context.

The only point that requires somethought is the following

property, which replaces Lemma 4.7.

Lemma 4.8. In the hyperbolic plane (IHP^hyp), let \317\210:g \342\200\224*g' be

an isometry between two geodesies g and g'. Then, \317\210extends to an

isometry \317\210:\320\2502-> H2 of (H2, dhyp).

In addition, if We choose one side of g and another sideforg', we

can arrange that \317\210sends the selected side of g to the one selectedfora'. The isometry \317\210is then uniquely determined by these properties.

Proof. Pick a point \316\241\342\202\254g and a nonzero vector \316\275tangent to g at P.In particular, if defines an orientation for g, which we can transportthrough \317\210to obtain an orientation of g'. At the point \316\241'=

\317\210(\316\241)\342\202\254g,

80 4. Gluing constructions \316\231

-J\342\226\240J

let v' be the vector tangent to g' in the direction of this orientation, \302\243

and such that ||tf'||hyp= ||if||hyp.Proposition2.20,which shows that

\\

(iP^hyp) is isotropic, provides a hyperbolic isometry \321\204:\320\2502\342\200\224>\316\2272|

such that \317\206(\316\241)= \316\241'and \316\214\317\201\317\206(\316\275)

= \317\2051'.In particular, \317\206sends g toj

the geodesic which is tangent to Dpip(v)= if' at \317\206(\316\241)= P', namely, \\

9'- \316\257

The restriction of \317\206to the geodesic g is an isometry g \342\200\224>g' \\

which sends \316\241to the same point P' as the isometry \317\210,and sends 1

the orientation of g to the same orientation as \317\210.It follows that \317\206\316\257

coincides with \317\210on g. In other words, the isometry \317\206:H2 \342\200\224>H2 |extends \317\206:g \342\200\224>g'. \\

If \321\204sends the selected side of g to the selectedside of g', we 7

are done. Otherwise, let \317\201be the hyperbolic reflection across g', \\

namely, the isometry of (H2, dhyp) induced by the inversion across !the euclideancirclecontaining g'. Because \317\201fixes every point of

g'-\\

and exchanges its two sides, the hyperbolicisometry \317\210=

\317\201\320\276\321\204now i

has the required properties. .|

The uniquenesseasily follows from Lemma 2.10. , D'{

We are now ready to carry out our automatic translation from;

euclidean to hyperbolic geometry.Let X be a polygon in the hyperbolic plane (H2, dhyp)\302\267Namely,

X is a region in H2 whoseboundary in H2 is decomposed into finitely

many hyperbolic geodesies E\\, E2 En meeting only at theirendpoints. When we consider X as a subset of K2, its boundary ia<

M.2may also include finitely many intervals in the real line \320\232bounding:

the hyperbolic plane H2 in K2 = C; if this is the case, note that Xwill be unbounded for the hyperbolic metric dhyp.

In addition we require that X and the Ei contain all those points

of H2 that are in their boundary. Namely, X and the Ei are closedin H2, although not necessarily in \320\250.2.

The geodesies Ei bounding X are the edges of the polygon X.

The points where two edges meet are its vertices. As in the euclidean,

case, we require that only two edges meet at any given vertex.

4.5. Gluinghyperbolicand spherical polygons 81

Figure 4.7 offers a few examples. In this figure, X\\ is a

hyperbolic octagon, with eight edges and eight vertices; it is bounded for

the hyperbolic metricdhyp\302\267

The hyperbolic polygon X2 is an infinite

Ht.rip, with two edges and no vertex; it touches the line R along two

disjoint intervals, and is unbounded for the hyperbolic metricdhyp

(although it is bounded for the euclidean metric of R2). The

hyperbolic quadrilateral X3 is delimited by our edges, has no vertex in H2,

mid touches R = R \320\270{\320\276\320\276}along four points, one of which is 00. We

will meet these three hyperbolic polygons again in Sections 5.2, 5.4.2mid 5.5, respectively.

H2

Figure 4.7. A few hyperbolic polygons

For such a hyperbolic polygon X, we can then introduce edge

Kluing data by, first, grouping the edges together in pairs {E\\,E2},(/?\320\267,\302\243\321\207},...,{\302\243^2\317\201-\316\271,\302\243^2\317\201}

and then, for each such pair {S2fc-i,\302\243^2fc},

l>y specifying an isometry tp2k-i- #2fc-i \342\200\224\342\226\272i?2fc\302\267Here tfi2k-i is

required to be an isometry for the hyperbolic distancedhyp\302\267

As in the euclidean case, the edgesE2k-\\ and E2k in the same pairmust have the same hyperbolic length, possibly infinite. In general,

the isometry \317\2102\316\220\316\257-\316\271is then uniquely determined once we know how

^ifc-i sends an orientation of E2k-i to an orientation of E2k, and we

will often describe this information by drawing matching arrows on/'/\342\200\2422fc_iand \320\225-2\320\272\302\267The case where drawing arrows is not sufficient to

wpecify v?2fc-i is when Ezk-i and E2k are completegeodesiesof H2,

namely, full euclidean semi-circles centered on the real line.As in the euclidean case, we endow X with the path metric

\316\271\316\220\317\207for which dx(P,Q) is the infimum of the hyperbolic lengths ofall piecewise difFerentiable curves joining \316\241to Q in X. When X is

82 4. Gluing construction\302\256

convex, in the sense that the geodesicarc joining any two P, Q G

is contained in X, the metric \316\254\317\207clearly coincides with the restrictionof the hyperbolicmetricdhyp\302\267

Theorem 4.9. IfX is obtained from the hyperbolicpolygon X by gib

ing pairs of its edges by isometries, then the gluing is proper. \320\230\320\260\321\202\320\265\320\251

the semi-distance \316\254\317\207induced on X by the path metric dx ofXis sucq

that dx(P, Q)>0 when \316\241\317\206\302\247.

Proof. The proof is identical to that of Theorem 4.3. Just folloul

each step of that proof, using the appropriatetranslation.

Theorem 4.10. Let (\316\247,\316\254\317\207)be the quotient metric space obtainefrom a hyperbolic polygon (\316\247,\316\254\317\207)by gluing together pairs of \320\265\320\260\320\264\320\251

of X by hyperbolic isometries. Suppose that the following additiond

condition holds: For every vertex \316\241of X, the angles of X at thos$

vertices P' of X which are glued to \316\241add up to 2\317\200.Then (\316\247,\316\254)

locally isometric to the hyperbolic plane (H^dhyp)\302\267

Proof. The proof is identical to that of Theorem 4.4, provided tha

we replace Lemma 4.7 by Lemma 4.8.

A metric space (X, d) which is locally isometric to the hyperboliiplane(H2,deuc)is a hyperbolic surface. Equivalently, the metric i

is then a hyperbolic metric.

4.5.2. Spherical polygons. The same properties also generalize \320\251

polygons in the sphere (S2,dsph)\302\267

A polygon in the sphere (S2,dsph) is a regionX of S2 who

boundary is decomposed into finitely many geodesies E\\, \316\225\316\271,

En meeting only at their endpoints. TheseEi are the edgesof tlj

polygon X, and the points where they meet are its vertices.

before, we require that X contains all its edges and vertices,and tha

every edge contains its endpoints. Also, exactly two edges meet at;'}

given vertex.

We endow X with the path metric \316\254\317\207for which \316\254\317\207(\316\241,Q) is \317\215

infinitum of the euclidean lengths of all piecewisedifferentiable curve

joining \316\241to Q in X \320\241S2 \320\241R3. When X is convex, in the sense tha

any two P, Q \342\202\254X can be joined by a geodesicarc of S2 of length <;

4.5. Gluing hyperbolic and sphericalpolygons 83

which is completely contained in X, the metric \316\254\317\207clearly coincides

with the restriction of the sphericalmetricdeph\302\267

After grouping the edges together in pairs {E\\,E2}, {\316\225^,\316\225^,

..., {i?2p-i, E2P}, the gluing data used consists of isometries <p2k-i-l''-2k-l\342\200\224*E2k\302\267

As in the hyperbolic case, the key propertyis the following

extension of Lemma 4.7.

Lemma 4.11. In the sphere (S2,deph)> \342\204\226\316\250'\302\2679 ~* 9' be an isometrybetween two geodesies g and g'. Then \317\206extends to an isometry \317\206:

S'^S2 0/(S2,deph).

In addition, if we chooseonesideforg and another side for g1, weam arrange that \317\206sends the selected side for g to the one selectedfor(f. The isometry \317\210is then uniquely determined by these properties.

Proof. Since geodesies of (S2,daph)are great circlearcs (Theorem

!l.l), this is easily proved by elementary arguments in 3-dimensional

(Euclidean geometry. D

As before,weendowthe sphericalpolygon X with the path metric

dx for which \316\254\317\207(\316\241,Q) is the infimum of the sphericallengths of all

piecewise differentiable curves joining \316\241to Q in X.

Then, by replacing Lemma 4.7 by Lemma 4.11, the proofs of

Theorems 4.3 and 4.4 immediately extend to the sphericalcontextund give the following two results.

Theorem 4.12. IfX is obtained from the spherical polygon X by

gluing pairs of its edges by isometries, then the gluing is proper. Namely,

the semi-distance \316\254\317\207induced on X by the path metric \316\254\317\207is really a

metric. \320\236

Theorem 4.13. Let (\316\247,\316\254\317\207)be the quotient metric space obtainedfrom a spherical polygon (\316\247,\316\254\317\207)by gluing together pairs of edgesi>f X by hyperbolic isometries. Suppose that the following additional

conaition holas: For every vertex \316\241of X, the angles of X at thosevertices P' of X which are glued to \316\241add up to 2\317\200.Then (\316\247,\316\254\317\207)is

locally isometric to the sphere (S2,dePh). \342\226\241

84 4. Gluing constructior

As in the euclideanand hyperbolic cases, a metric space (X, \321\201\320\251^

which is locally isometric to the sphere(S2,dsph)is a spherical \320\267\320\270\320\263\320\251

face. Equivalently, the metric d is then a spherical metric.


-P>v

Exercise 4.1. Let X be the closed interval [0,1] in K. Let X be the par^W

tition consisting of all the subsets {^r, %k}where m, \316\267\302\243N axe

integers^,such that m is odd and 1 < m < 2\342\204\242,and of all one-element subsets {Pfflwhere \316\241e [0,1] is not of the form \316\241=

\342\226\240\342\200\224;or\320\251

as above. Let a\\.uc be theft

quotient semi-metric induced on X by the usual metric deuc(P, Q) =|P\342\200\224Q\\

of* =[0,1]. \321\211

a. Show that for every P_e [0,1]and every \316\265> 0, there exists Qi, Q2 \342\202\254$i

[0,1] such that Qi = Qi in -X\\ deuc(0,Qi) < \316\265,and \321\201&\321\210\321\201^\320\244\320\263)< \316\265..'\320\250

b. Show that deuc(0, P) = 0 for every PeX. In particular, the semi\302\273\302\273

metric deuC is not a metric, and the gluing is not proper. A

\321\201Show that deuc(P, Q) = Qfor every P,QeX. JExercise4.2(Equivalence relations and partitions). A relation on a setgf

X is just a subset 31of the product XxY. One way to think of this is tha^\320\257describes a certain property involving two points of X. Namely, \316\241and!

Q e X satisfy this property exactly when the pair (P, Q) is an element of\320\230\320\251

To emphasize this interpretation, we write \316\241~ Q to say that (P, Q) e \320\233.|An equivalence relation is a relation such that 4

(i) \316\241~ \316\241for every \316\241e X (Reflexivity Property); 1(ii) if \316\241~ Q, then Q ~ \316\241(Symmetry Property); |

(iii) if \316\241~ Q and Q ~\320\233,then \316\241~ Q (Transitivity Property). |

a. Given an equivalence relation, define the equivalenceclass of \316\241e X\302\253

as .J

P={QeX;P~Q}. 4

Show that as \316\241ranges over all points of X, the family of the equiva*jlence classes\316\241is a partition of X. \342\226\240\342\200\236

b. Conversely, let X be a partition of the set X and, as usual, let \316\241\302\243X\\

denote the subset that contains \316\241\316\266.X. Define a relation on X by the-

property that \316\241~ Q exactly when \316\241and Q belong to the samesubset\316\241= Q of the partition. Show that ~ is an equivalencerelation.

Exercise4.3 (Trivial gluing). Let (X, d) be a metric space, and consider

the trivial gluing where the partition X consists only of the one-element

subsets \316\241~ {P}. In other words, no two distinct elements of X are gluedtogether in X. Let (X, d) be the resulting quotient semi-metric space.


I Ising the definition of the quotient semi-metric d in terms of discrete walks,

rigorously prove that the quotient map \317\200:X \342\200\224>X defined by 7r(P) = \316\241is

mi isometry from (X, d) to (X,d).Exercise4.4(Iterated gluings). Let X be a partition of the metric space(X, d), and let X be a partition of the_quotient space X. Let d be theHomi-metric induced by d on X, and let d be the semi-metric induced by dou X.

a. If \316\241e X, the element \316\241= \316\241\316\266.X is a family of subsets of X, andwe can considertheir union \316\241\320\241X. Show that the subsets \316\241form a

partition X of X.

b. Let \317\206:\316\247\342\200\224\342\226\272X be the map defined by the property that \317\206(\316\241)= \316\241for

every \316\241e X. Show that ^> is bijective.

\321\201Let w be a discrete walk \316\241=PiLQi

~ P2, ..., Qn-\\~ Pi, Qn =_Q

from Pto Q in X. Show that \316\241=Pi4 Qi ~ P2,=..., Qn-i ~ Pi,

Qn= Q forms a discrete walk w from \316\241to Q in X, whose length is

such that la{w) < ld{w). (Beware that the same symbol ~ is usedtorefer to gluing with respect to the partition X in the first case, andwith respect to X in the second instance.) Conclude that if d is the

quotient semi-metric induced by d on X, then d(P, Q) < d(P,Q) for

every P,Q eX.

d. Given a small \316\265> 0, let w be adiscrete_walk \316\241= P\\, Qi ~ P2, \302\267\302\267\302\267,

Qn-i ~_Pi, Qn = Q from Pto Q in X whose length is sufficiently

close to J(P, Q) that la{w) ^ d(P, Q) + f \302\267Similarly, for every i, choosea discretewalk Wj from \316\241to Qi, consisting of \316\241=

\320\240\320\264,Qi,i ~ P,2,\342\200\242\302\267\302\267,Qi,ki-i

~ P.fci, Qi,kt = Q<whoselength is sufficiently close to

d(P,Qi) that ld{v>i) ^ d(P,Qi) + ^. Show that the w\302\273can be

chained together to form a discretewalk w from \316\241_to_Q in X suchthat id{w) \316\266ia{w) + f. Conclude that d{P,Q) *\316\266d(P,Q)+s.

e. Show that \317\210is an isometry from (X, d) to (X,d).In other words, a two-step gluing construction yields the same quotient

semi-metric space as gluing everything together in one single action.

Exercise 4.5. Let X be the interval [0,2\317\200]\320\241\316\234,and let X be the partition

consisting of the two-element subset {0,2\317\200\302\267}and of all the one-elementsubsets{P}with \316\241e (0,2\317\200). Let d be the quotient semi-metric induced

on. X by the usual metric d(P, Q) =\\Q

- P\\ of X =[0,2\317\200]. In other words,

(X,ct) is obtained by gluing together the two endpoints of the interval

X =(0,2\317\200) endowed with the metric d.

Let S1={(re, y) e M2; x2 + y2

= 1} be the unit circle in the euclideanplane, endowedwith the metric d$i for which dsi(P,Q) is the infimum

86 4. Gluing constructions\302\267

of the euclidean axelengths \302\243euc(7) of all piecewise differentiable curves 7Jgoing from \316\241to Q in S1. Consider the map \317\210:\316\247~

[0,2\317\200]-+ S1 defined|

by tp(t) = (cost,sint). I_ \316\257

a. Show that if \317\200:X \342\200\224\342\226\272X denotes the quotient map, thereexistsa uniques

map \317\206:X \342\200\224\302\273\342\226\240S1 such that \317\206=

\317\206\316\277\317\200.Show that ^ is bijective. ;i_ _ _ $

b. Show that for every discrete walk w from \316\241to Q in X, there ex-'ists a piecewise differentiable curve 7 going from <p(P)to \317\210{0)in S1

whose length 4uc(7)_is equal to the length id(w) of w. Conclude that

dsi (<p(P),<p{Q)) \316\266d(P, Q) for every P, Q e X.c. Conversely, show that for every P, Q e X, there exists a discrete walk

w from \316\241to Q in X involving at most four points of X, whose length

id{w) is equal to\316\254$\316\271(\317\206(\316\241),\317\206((3)).(Hint: Consider a shortest curve

from\317\210(\316\241)_\316\247\316\277\317\210(0)

in S1.) Conclude that d{P,Q) < dgi (<p(P),tp(Q))for every P,QeX.

d. Combinethese results to show that \317\210is an isometry from (X, d) to

Exercise 4.6. In the euclidean plane \320\2502,let D\\, D2 and D3 be three

disjoint euclidean disk sectors of radius \320\263and respective angles \316\270\316\271,\316\222\316\271and

\316\2303with \316\270\316\271+ 02 = 03 ^ \317\200.Let X be the quotient space obtained fromX = Z?i U Z?2 by isometrically gluing one edge of Z?i to an edge of D2,sending the vertex of Di to the vertex of D2. Show that if dx and dD3

are the euclidean path metrics defined as in Section 4.3.1, the quotientspace (X,dx) is isometric to (D3,dD3)\302\267 Hint: Copy parts of the proof ofTheorem 4.4.

Exercise 4.7 (Euclidean cones). Let D\\, D2, ..., Dn be \316\267disjoint

euclidean disk sectors with radius \320\263and with angles \316\270\316\271,02, \342\226\240\342\226\240\342\226\240,\316\270\342\200\236,respec-;

tively, and let Ei and E[ denote the two edges of Di. Isometrically glue\302\267

each edge Ei to E'i+1,sending the vertex of Di to the vertex of Di+\316\271and

counting indices modulo \316\267(so that \316\225'\316\267+\316\257= E[). Show that the resulting

quotient space (X, dx) depends only on the radius \320\263and on the angle sum\316\270=

\316\243\"=\316\271\316\270*\302\267Namely, if D[, D'2, ..-, D'n, is another family of n' disjoint

euclidean disk sectors of the sameradius \320\263and with respective angles \316\270[,

#21 \302\267\302\267\302\267,\316\270'\316\267\316\271with \316\243?=\316\271% \342\200\224\316\243\316\223=\316\271^\302\267\316\271an(^ if these disk sectors are glued:

together as above, then the resulting quotient space (\316\247',\316\254\317\207\316\271)is isometric

to (X,d~x). Hint: Use the results of Exercises 4.4 and 4.6 to reduce theproblem to the case where \316\267= \320\263\320\260'and \316\270\316\257=

\316\270\\for every t, and to make surethat the order of the gluings does not matter.

The space (X,d~x) of Exercise 4.7 is a euclideanconewith radius \320\263

and cone angle \316\230.Figure 4.8 represents a few examples. Note the differentshape accordingto whether the cone angle \316\270is less than, equal to, or


more than 2\317\200.When \316\270is equal to 2\317\200,the cone is of course isometric to atmclidean disk.

cone angle < 2\317\200 cone angle = 2\317\200 cone angle > 2\317\200

Figure 4.8. Three euclidean cones

Exercise 4.8 (Surfaces with cone singularities). Let (X,d~x) be the

quotient metric space obtained from a euclideanpolygon (X, dx) by isomet-rically gluing together its edges. Show that for every \316\241\316\266.X, there existsii. radius \320\263such that the ball B^x (P, r) is isometric to a euclidean cone,defined as in Exercise 4.7.

A space (X, dx) satisfying the conclusions of Exercise4.8 is a

euclidean surface with cone singularities. One can similarly define

hyperbolic and spherical surfaces with cone singularities.

Exercise 4.9.

a. Let (C, d) be a euclidean conewith center Po, radius \320\263and cone angle\316\270as in Exercise 4.7. Show that for every r' < r, the \"circle\"

Sd(Po, r) = {PeC; d(P, Po)= r'}is a closed curve, whose length \302\243d(Sd(Po,r')) in the sense ofExercise 1.11 is equal to Or'.

b. Conclude that the angle condition of Theorem 4.4is necessary for its

conclusion to hold. Namely, if, when isometrically gluing together thesidesof a euclidean polygon X, there is a vertex \316\241such that the anglesof X at those vertices P' which are glued to \316\241do not add up to 2\317\200,

then the quotient space (X, d) is not locally isometric to the euclideanplane (K2,deuc)

Chapter 5

Gluing examples

After suffering through the long proofs of Section4.4, we can now

harvest the fruit of our labors, and apply the technology that we

have built in Chapter 4 to a few examples.

5.1. Some euclidean surfaces

We begin by revisiting, in a more rigoroussetting,the exampleof the

torus that we had informally discussed in Section 4.1.

5.1.1. Euclidean tori fromrectanglesand parallelograms. Let

X\\ be the rectangle [a,b] \317\207[c,d], consisting of those (x, y) \342\202\254R2

such that a < \317\207< b and \321\201*\316\266\321\203\316\266d. Glue the bottom edge\316\225\316\271

= [o, b] \317\207{c} to the top edge E%=

[a, b] \317\207{d} by the isome-

try \317\210\317\207\\[a, b] \317\207{c}\342\200\224>

[a, b] x {d} defined by </?i(x,c) = (x,d), and glue

the left edge E3 ={\316\261}\317\207[\321\201,d] to the right edge E4 = {b}\317\207[\321\201,d] by

\317\2063\302\267{a} x [c,d] -> {bj x [c,d]defined by <\302\243>\320\267(\320\262,2/)= (b,y). Namely,

we consider the edgegluing that already appeared in Section 4.1 andwhich we reproduce in Figure 5.1.

With these edge identifications, the four vertices of the rectangleare glued together to form a single point \316\241of the quotient space X\\.

The sum of the angles of \320\245\320\263at these vertices is

7\320\2237\320\2237\320\2237\320\223_

2+

2+

2+

2=2\317\200\302\267

89

90 5. Gluing examples ,

;;*\320\267

\342\200\224>\342\200\224

\342\226\240Z?2

\316\247\316\271

\320\225\320\263\342\200\224>\342\200\224

\320\257,\342\200\236

Figure 5.1. Gluing opposite sidesof a rectangle

tWe can therefore apply Theorems4.3and 4.4. Note that X\\ is:'<

convex, so that the path metric \316\254\317\2071coincides with the restriciton offthe euclideanmetric dexic of R2. Then Theorems 4.3 and 4.4 showj

that the metric space (\316\247\316\271,<\317\212\317\2071)is locally isometric to the euclidean*metricof the euclidean plane (R2, deuc). \320\246

IThis is our first rigorous example of a euclidean surface. , |In our informal discussion of this example in Section 4.1, we exrj

plained how Xi can be identified to the torus illustrated on the right!of Figure 5.1 if we are willing to stretch the metric. The mathemalM

ically rigorous way to express this property is to usethe languageofj

topology and to say that the space X\\ is homeomorphic to the torusA

A homeomorphism from a metric space (X, d) to another met^

ric space (X', d') is a bijection\317\206:X \342\200\224>X' such that both \317\206and its^

inverse \317\206-1are continuous. (See Section T.l in the Tool Kit foi!

the definition ofbijectionsand inverse maps.) The homeomorphism \317\206|

can be used as a dictionary between X and X' to translate back an4,forth every property involving limits and continuity. For instance, al

sequence Pi, P2, ..., P\342\200\236,... converges to the point P\302\273in X if \320\260\320\277\320\251

only if its image \317\206{\316\241\\),\317\206(\316\2412),\302\267\302\267\302\267,\317\210(\316\241\316\267),\342\226\240\342\226\240\342\226\240converges to\317\210(\316\241\316\277\316\277);\316\257

in X'. As a consequence, if the metric spaces(X,d) and (X', d') aref

homeomorphic, in the sense that there exists a homeomorphismj]

(X, d) \342\200\224>(X', d') , then (X, d) and (X', d') share exactly the sameV

limit and continuity properties. ;?

An example of a homeomorphism is providedby an isometry from|

(X, d) to (X',d'). However, a general homeomorphism \317\206:X \342\200\224>X'|

is much more general in the sense that the distance\316\254'[\317\206(\316\241),\317\210{0)^

may be very different from d(P,Q). The only requirement is thai

5.1. Some euclidean surfaces 91

d'(ip(P),ip(Q))is small exactly when d(P,Q) is small (in a sense

quantified with the appropriate \316\265and \316\264).

In general, we will keep our discussion of homeomorphisms at a

very informal level. However, we should perhaps go through at leastone examplein detail.

Let the 2-dimenaional torus be the surface T2 of the 3-dimen-

sional space R3 obtained by revolving about the 2-axis the circlein

the xz-plane that is centered at the point (Ft,0,0) and has radius

r < R, as in the right-hand side of Figure 5.1. We consider T2 as a

metric space by endowing it with the restriction of the 3-dimensionalmetric deuc of R3. Different choices of r and R give different subsetsof R3, but these are easily seento be homeomorphic.

Lemma 5.1. Let (\316\247\316\271,\316\254\317\207^be the quotient metric space obtainedfrom

the rectangle \320\245\320\263= [a, b] x [c, d] by gluing together opposite edges by

euclidean translations. Then X\\ is homeomorphic to the 2-dimensionaltorus T2.

Proof. To simplify the notation, we restrict attention to the case of

the square X\\=

[\342\200\224\317\200,\317\200]\317\207[\342\200\224\317\200,\317\200].However, the argument

straightforwardlyextends to general rectangles [a, b] \317\207[\321\201,d] by suitable rescalingof the variables.

Let \317\201:\316\247\316\271\342\200\224>T2 be the map defined by

\317\201(\316\270,\317\206)=

((R + r cos \317\206)cos 9,(R + r cos \317\206)sin \316\230,r sin\317\206).

(!eometrically, \317\201(\316\270,\317\206)is obtained by rotating by an angle of \316\270around

I. lie 2-axis the point (R + r cos\317\210,0, r sin \317\210)of the circle in the xz-plane with center (R, 0,0) and radius r. From this description it is

immediate that \317\201(\316\270,\317\206)=

\317\201(\316\270',\317\206')exactly when (\316\270,\317\206)and (\316\270',\317\206')are

Klued together to form a singlepoint of X\\. It follows that \317\201induces

\320\270bijection p: X\\ \342\200\224>T2 defined by the property that p(P) = p(P) for

any \316\241\316\266P.

The map \317\201is continuous by the usual calculus arguments. UsingI lie property that d(P, Q) < d(P,Q) (Lemma 4.2), it easily followsI hat \317\201is continuous.

92 5. Gluing examples

To prove that the inverse function p_1: T2 \342\200\224>\320\245\320\263is continuous,!

at the point Qo \342\202\254T2, we need to distinguish cases accordingto theJtype of the point Qq.

Consider the most complexcase,where Qq =(\342\200\224i?+ r, 0,0) is the j

image under \317\201of the point Pq \342\202\254X\\ corresponding to the four vertices i

(\302\261\317\200,\302\261\317\200)of \316\247\316\271.If Q = (x,y,z) \342\202\254T2 is near Qo, we can explicitly^compute all (\316\270,\317\206)such that \317\201(\316\270,\317\206)

= Q. Indeed, \317\210= \317\200

\342\200\224arcsin f \320\251

\316\266> 0, \317\206= \342\200\224\317\200\342\200\224arcsin ^ if \316\266< 0, and \317\206

= \302\261\317\200if \320\263= 0. Similarly,!\316\262= \317\200- arcsin

\320\264+/\321\201\320\276\320\260\321\203if \321\203> 0, \316\262= -\302\273- arcsin

\320\264+/\321\20108\321\203if \321\203<

\316\270|and 0 = \302\261\317\200if \316\271/

= 0. By continuity of the function arcsin, it follow?!that (\316\270,\317\206)will be arbitrarily close to one of the corners (\302\261\317\200,\302\261\317\200)\317\212$

Q~

(\317\207,\321\203,\316\266)is sufficiently close to Qo = (~R + r, 0,0).For \316\265> 0 small enough, Lemma 4.5 showsthat the ball B^(Pq, ej

is just the imageunder the quotient map X\\ \342\200\224>\316\247\316\271of the four quarterdisks of radius \316\265centered at the four vertices (\302\261\317\200,\302\261\317\200)of X\\. By the

observations above, thereexistsa \316\264> 0 such that whenever Q \342\202\254T2 \320\251

such that deac(Q, Qo) < \316\264,any (\316\270,\317\206)\302\243X\\ with \317\201(\316\270,\317\206)= Q is withE

a distance < \316\265of one of the vertices (\302\261\317\200,\302\261\317\200).Since p~1(Q) is the

image in X\\ of any (0, \317\210)\342\202\254\316\247\316\271with p(0, \317\210)= Q, we conclude tha

d(r4Q).-Po)<\302\243.Since Po

= p~1(Qo), this proves that /\316\2231is continuous at Qo

(~R + r,0,0).Thecontinuity at the other Qq \342\202\254T2 is proved by a similar case

by-case analysis.

Figure 5.2. Gluing opposite sides of a parallelogram

We can consider a variation of this example by replacing th|rectangleby a parallelogram X2 and gluing again opposite sides

translations. As in the case of the rectangle, the four vertices

5.1. Some euclideansurfaces 93

the parallelogram are glued to a single point. Becausethe angles

of a euclidean parallelogram add up to 2\317\200,the angle condition of

Theorem 4.4 is satisfied, and we conclude that the quotient metricspace(J?2) d) is a euclidean surface.

The parallelogram X2 can clearly be stretched to assumetheshape of a rectangle in such a way that the gluing data for \316\247-\317\207gets

transposed to the gluing data for X\\. See Exercise 5.1. It follows

that the quotient surface X2 is again a torus.

5.1.2. EuclideanKleinBottles.Given a rectangle X3 = [a,b] \317\207

[c,d], we can also glue its sidestogetherusing different gluing maps.

For instance, wecan still glue the bottom edgeE\\ = [a, b] \317\207{\321\201}to the

top edge E2 = [a,b] x {d} by the isometry \317\206^.[\320\276,b] \317\207{\321\201}\342\200\224>

[\320\276,b] x {d}

defined by \317\210\316\271(\317\207,\316\277)= (x,d), but glue the left edge E3 = {a}x [c,d]

1.0 the right edge E4 = {b}\317\207[c,d] by \317\206^:{\316\261}\317\207[c,d] \342\200\224>{6} \317\207[c,d]

defined by <\321\200\320\267(\320\260,\321\203)= (b, d \342\200\224

y). Namely, the gluing map flips the left

(dge upside down before sending it to the right edge by a translation.

Again, the four vertices of X3 are glued together to form a single

point of the quotient spaceX3. Sincethe angles of X3 at these fourverticesadd up to 2\317\200,the combination of Theorems 4.3 and 4.4shows

Lhat (X3, \316\254\317\2073)is locally isometric to the euclidean plane (R2, \302\2534uc)\302\267

>

E2

\316\271\\\320\225\320\267X3 Ei

\316\225\317\207I >.

Figure 5.3. Another way of gluing opposite sidesofa rectangle

To understand the global shape of X3, we first glue the bottom

and top sides together,to form a cylinder as in the case of the torus.

Wie then need to glue the left side of the cylinder to the right side by

\321\217translation followed by a flip. This time, the difficulty of physically

realizing this in 3-dimensional space goeswell beyond the need for


stretching the paper. It can actually be shown to be impossible to

realize, in the sense that there is no subset, of R3 which is homeomorphictoX3-

The right-hand side of Figure 5.3 offers an approximation, where

the surface crosses itself along a closedcurve. Each point of this

self-intersection curve has to be understood as corresponding to two

points of the surface X3. Introducing an additional space dimension,this picturecanalsobe used to represent an object in 4-dimensional

space. This is similar to the way a figure eight 00 in the plane can

be deformed to a curve 00 with no self-intersection in 3-dimensional

Figure 5.4. A physical Klein bottle

5.1. Some euclidean surfaces 95

space,by pushing parts of the figure eight up and down near thepoint where it crosses itself. In the same way, the object represented

on the right-hand side of Figure 5.3 can be deformed to a subset of

the 4-dimensional space R4 that is homeomorphicto X3.The surface X3 is a Klein bottle.

The Klein bottle was introduced in 1882 by Felix Klein (1849-1925),as an example of pathological surface. The \"bottle\"

terminology is usually understood to reflect the fact that a Klein bottle

can be obtainedfrom a regular wine bottle by stretching its neck andconnecting it to the base after passing inside of the bottle. Another

interpretation (unverified, and not incompatible with the previous one)claims that it comes from a bad pun, or a bad translation from the

German, in which the Kleinsche Flache (Klein surface) becametheKleinscheFlasche(Kleinbottle). Thelatter version probably

provides a better story, whereas the first one makes for better pictures.This secondpoint is well illustrated by the physical glassmodelof

Figure 5.4, borrowed from the web site www.kleinbottle.comof the

company Acme Klein Bottle, which offers many Klein bottle-shaped

products for sale.

5.1.3. Gluing oppositesidesofa hexagon. Let us now go

beyond quadrilaterals and consider a hexagon X4 whereweglueoppositeedgestogether, as indicated on Figure 5.5; compare alsoFigures4.3and 4.6. The vertices of X4 project to two points of the quotient spaceX4, each corresponding to three vertices of X4. More precisely,if we

label the vertices Pi, P2, ..., Pe in this order as one goes around thehexagon, the odd vertices \316\2411;P3 and P5 are glued together to form

one point of X4, and the even vertices P2, P4 and Pq form anotherpoint of X4. Consequently, we need the hexagon X4 to satisfy the

following two conditions:

(1) opposite edgeshave the same length;

(2) the angles of X4 at the odd vertices Pj, P3, P5 add up to

2\317\200.

Recall that the sum of the anglesof a euclidean hexagon is alwaysequal to 4\317\200,so that condition (2) is equivalent to the property that

the angles of X4 at its even vertices P2, P4, Pe add up to 2\317\200.A little


exercisein elementary euclidean geometry shows that in a hexagonsatisfying conditions (1) and (2) above, opposite edgesare necessarily

parallel; see Exercise 5.7.

If the hexagon X4 satisfiesconditions (1) and (2), we can againapply Theorems4.3and 4.4 to show that the quotient metric space(X4, \316\254\317\2074)is a euclidean surface.

Pi P2

\320\240\321\214 Pa

Figure 5.5. Gluing opposite sides of a hexagon 1

To understand the global shape of X4 up to homeomorphism, i

we can consider the diagonals P1P5 and P2P4 of the hexagon X4,as in Figure 5.5. These two diagonals cut the hexagon into three \342\226\240

pieces, the parallelogram P1P2P4P5 and the two triangles P2P3P4 !

and \316\241$\316\241\316\262\316\241\\\302\267Gluing the edges P1P2 and P4P5 of the parallelogram

P1P2P4P5 provides a cylinder, whose boundary consistsof the two \316\257

images of the diagonals P\\P$ and P2P4. Similarly, gluing the two i

triangles P2P3P4 and PsPePi together by identifying the edge P2P3 I

to the edge P5P6 and the edgeP3P4 to \316\241^\316\241\317\207gives another cylinder, >

whose boundary again corresponds to the imagesof the diagonals i

P1P5 and P2P4. See the right-hand side of Figure 5.5. This proves j

that splitting the quotient space X4 along the imagesof the diagonals >

P1P5 and P2P4 gives two cylinders. In particular, X4 can be recovered

from these two cylinders by gluing them back together according to \342\226\240\342\226\240]

the pattern described on the right of Figure 5.5. It easily followsthat :

X4 is homeomorphic to the torus. i

As announced in an earlier disclaimer, this discussion of the

construction of a homeomorphism from the quotient space X4 to the \\

torus is somewhat informal. However, with a little reflection, youshould be able to convince yourself that this description could be '-\342\226\240

\302\2433

\320\250

5.2. The surface of genus2 97

made completely rigorous if needed. The same will apply to other

informal descriptions of homeomorphisms later on.

5.2. The surface of genus 2

We saw in the last section that if we start with a euclidean rectangleor parallelogram and if we glue opposite edges by a translation, weobtain a euclideantorus.Similarly, for a euclidean hexagon satisfying

appropriate conditions on its edgelengths and angles, we found outthat gluing opposite edges again yields a euclidean torus.

We can go one step further and glue opposite sides of an octagonX, as on the left-hand side of Figure 5.6.

x

Figure 5.6. Gluing opposite edges of \320\260\320\273octagon

We claim that the quotient spaceX is homeomorphic to the

surface of genus 2 represented on the right of Figure 5.6, namely, a

sphere with 2 handles. (A torus is a spherewith one handle added.)

To see this, one can cut out a smaller octagon from X asindicated in Figure 5.6. This smaller octagon X\\ can be seen as a

rectangle whose corners have been cut off. Gluing oppositeedgesof this

rectangle, we see that the image \316\247\317\207in X is just a torus from which

a square (corresponding to the triangles removed from the rectangle)has been removed. SeeFigure5.7.

It remains to consider the four strips forming the complement

X2 of \316\247\316\271in X. Gluing these four strips together along their shortedgesgives a big square minus a smaller square,namely, some kind

of square annulus, as in the left-hand side of Figure 5.8. Flippingthis annulus inside out, as in the middle picture of Figure 5.8, and

Q\302\251

98 5. Gluingexamples

\316\230

Figure 5.7. One half of Figure 5.6

Figure 5.8. The other half of Figure 5.6

then gluing the outside sides, we see that the image X2 of X2 in the ;

quotient space X is againa torus minus a square.

Finally, the quotient space X is obtained by gluing the two

surfaces \316\247\316\271and X2 along their boundaries, which gives the surface of i

genus 2 of Figure 5.6. ,

Let us now try to put a euclidean metric on the quotient space !

X. The eight vertices of the octagonX are glued together to form a-|single point of X. Consequently, if we want to apply Theorems 4.3 \\

and 4.4, we need to use a euclideanoctagonwhere opposite edgesl

have the same length, and where the sum of the angles at the vertices;is equal to 2\317\200.Unfortunately, in euclidean geometry, the angles of an |

octagon add up to 6\317\200! (

Therefore, it seems impossible to put a euclideanmetricon X.t.

(It can be proved that this is indeed the case, and that the surfaced

of genus 2 admitsno euclideanmetric;seeExercise5.16).However,

hyperbolic geometry will provide us with a suitable octagon. S

The first step is the following. .;

Lemma 5.2. In the hyperbolic plane H2, there exists a triangle T\\with angles f, f and ^. ^

5.2. The surface of genus 2 99

SeeProposition5.13and Exercise 5.15 for a more generalconstruction of hyperbolic triangles with prescribed angles.

Proof. We will actually use euclidean geometry to construct thishyperbolic triangle.

We begin with the hyperbolicgeodesicg with endpoints 0 and \321\201\321\216.

Namely g is the vertical half-line beginning at 0. Then consider the

complete geodesic ft that is orthogonal to g at the point i. Namely,

ft is a euclidean semi-circle of radius 1 centeredat 0. We are looking

for a third geodesic \320\272which makes an angle of f with both g and ft.

For every j/ < 1, let ky be the complete geodesic that passesthrough the point \\y and makes an angle of | with g. Namely, ky is

a euclidean semi-circleof radius \321\203esc | centeredat the point \321\203cot f,and consequently meets g as long as (sinf) / (1+ cos

f) < J/ < 1.

See Figure 5.9.

0 ycotf

Figure 5.9. A hyperbolic triangle with angles f, j and -|

The angleay betweenky

and g at their intersection point dependscontinuously on y. It is equal to ^ when \321\203

= 1, and approaches 0 as\321\203tends to (sin|)/(l + cos|). Consequently,by the IntermediateValue Theorem there exists a value of \321\203for which ay = f. Byapplying the Cosine Formula to the triangle formed by the intersection

point of g andky

and by the centers of these euclideansemi-circles,one could actually find an explicit formula for this y, but this is not

necessary.

100 5. Gluing examples |

For this value of y, the hyperbolic geodesies g, h and ky delimit Jthe hyperbolic triangle \316\244required. D I

Lemma 5.3. In the hyperbolicplane \320\2502,there exists an octagon X |where all edges have the same length and where all angles are equal Ito \342\200\224 1to 4. j

Proof. Let \316\244be the triangle provided by Lemma 5.2. Listits vertices 1

as P0, P\\ and P2 in such a way that Pq is the vertex with angle \\.i

We start with 16 isometric copies Tj, T2, ..., \316\244\317\207\316\262of T. Namely, |the Ti are hyperbolic triangles for which there exists isometries</?\302\253:I

\320\2302-> H2 such that Tt =\317\210\316\257{\316\244). }

Figure 5.10. A hyperbolic octagon with all angles equal to\\

Pick an arbitrary point Q \342\202\254H2. We can choose the isometriesf

\317\210\316\257so that \302\245>\302\273(\320\240\320\263)= Q for every i. In addition, using Lemma 4.8, we ii

can arrange that: |

(1) if \320\263is even, \317\206\316\257and tp,-_i send the edge P0P2 to the same|geodesic, so that T,- and T,_! have this edge ^\302\253(-\320\240\320\276\320\240\320\263)=|

tpi-i(PoP2) in common; I

(2) if \320\263> 1 is odd, \317\210{and </?i_i send the edge P1P2to the same;1

geodesic, so that T\302\273and \316\244\302\273_\316\271have this edge \317\210\317\212{\316\241\\\316\2412)=|

tpi-i(PiP2) \321\210common; 1

(3) for every i > 1, the triangles T\302\273and Tj-j sit on opposite\302\273!

sides of their common edge. \316\252

5.2. The surface of genus 2 101

Since16f =2\317\200,the T\302\273fit nicely together around the point Q,

and the last edgeof Tjg comes back to match the first edge of \316\244\316\212.In

particular, \317\206\316\271\316\262\316\231\316\241\316\271\316\241\316\266)=

\317\206\\(\316\241\\\316\2412)\302\267See Figure 5.10.

When \320\263is even, the two geodesic arcs ^(PqPj) =\317\206\316\257-\316\271(\316\2410\316\241\317\207)

meet at \317\210\316\257(\316\241\316\277)=

\317\210\316\257-\316\271(\316\241\316\277)and make an angle of ^ + ^ = \317\200at that

point. It follows that the union of these two geodesic arcs forms asinglegeodesicarc.

Therefore, the union X of the 16triangles T{ is an octagon in the

hyperbolic plane H2. Its angles are all equal to 2^ = j. Itsedges all

have the same length, namely, twice the length of the edge PqPi ofthe original triangle \320\242. \320\236

The symmetries of this hyperbolic octagon are more apparent inFigure 5.11,which represents its image in the disk model for H2,

introduced in Section 2.7, if we arrange that the centerQofX corresponds

to the center \320\236of!2.

Figure 5.11. The hyperbolic octagon of Figure 5.10 in the

disk model

Let X be the hyperbolicoctagon provided by Lemma 5.3,and let

X be the quotient space obtainedby gluing together opposite sidesof X, as in Figure5.6. Then Theorems 4.9 and 4.10 assert that themetric dx induces a quotient metric dx on this quotient space X, and

that the metric space (\316\247,\316\254\317\207)is locally isometric to the hyperbolicplane.

102 5. Gluingexamples )

In particular, we have constructed a hyperbolicsurface (\316\247,\316\254\317\207)i

which is homeomorphic to the surface of genus 2. \\

5.3. The projective plane

We now construct a spherical surface, which is different from the

sphere S2.

Let X be a hemisphere in S2. lb turn X into a polygon,picktwo

antipodal points Pi and Pi =\342\200\224P\\on the great circle \320\241delimiting

X, which will be the vertices of the polygon. These two vertices split\320\241into two edges E\\ and Ei. We now have a sphericalpolygon X,

which we endow with the path metric dx. Note that X is convex, so

that \316\254\317\207is just the restriction of the sphericalmetricdeph\302\267

Now, glue \316\225\317\207to E2 by the antipode map \317\210\\:E\\ \342\200\224>E2 defined

by \317\210\317\207{\316\241)= \342\200\224\316\241-This gluing data defines a quotient space(X,dx)

The polygon X and its gluing data are represented in Figure 5.12.

Figure 6.12. The projective plane

The angles of X at P\\ and P2 are both equal to \317\200,and

consequently add up to 2\317\200.Therefore, we can apply Theorems 4.12 and4.13, and show that the quotient space (X, dx) is locally isometric

to the sphere (S2,dePh)\302\267 This quotient space (\316\247,\316\254\317\207)is called the

projective plane.In Exercise 5.10,we show that the projective plane can alsobe

interpreted as the space of lines passing through the origin in the

3-dimensional space R3.

5.4. The cylinderand the Mobius strip 103

5.4. The cylinder and the MobiusstripWe now consider unbounded polygons and the surfaces obtainedby

gluing their edges together.

The simplest caseis that of an infinite strip where the two edges

are glued together which provides a cylinderor a Mobiusstrip.These

examples may appear somewhat trivial at first, but they already

display many features that we will encounter in more complicatedsurfaces.

\320\225\320\273

Xi

Ei

Figure 5.13. A euclidean cylinder

5.4.1. Euclidean cylinders and Mobius strips. We can begin

with an infinite strip \316\247\317\207in the euclidean plane R2, bounded by two

parallel lines \316\225\317\207and E2- Orient \316\225\317\207and E2 in the same direction, andglue them by an isometry \317\210\317\207:E\\ \342\200\224>E2 respecting these orientations.

Because there are no verticeson E\\ and E2, this is a situation where

the gluing map \317\210\317\207is not uniquely determined by these properties.Indeed,thereare many possible choices for \317\210\317\207,all differing from each

other by composition with a translation of R2 parallel to \316\225\317\207and E2.

Pick any such gluing map \317\206\316\271:\316\225\317\207\342\200\224>E2, and consider the

corresponding quotient space (Xi, dxj. Since \316\247\316\271has no vertex, the anglehypothesis of Theorem 4.4 is automatically satisfied. Therefore,Theorems 4.3 and 4.4 show that (\316\247\317\207,\316\254\317\207\317\207)is a euclidean surface.

104 5. Gluing examples |

This euclidean surface is easily seen to be homeomorphic to the

cylinder.

For another vertical strip X2 in R2 bounded by parallel lines E\\

and E2, we can orient E\\ and E2 so that they now point in opposite

directions and choose a gluing map \317\206:E\\ \342\200\224>E2 respecting these

orientations. Again, there are many different choices for this gluing

map, differing by composition with translations parallel to E\\ and

Ei.

For each choice of such a gluing map \317\206\\,another application of

Theorems 4.3 and 4.4 shows that the corresponding quotient space(X2, \316\254\317\2072)is a euclidean surface.

The topology of the quotient space (X2, \316\254\317\2072)is now very different.

Indeed, this space is homeomorphic to the famous \320\234\320\264\320\253\320\270\320\260strip.

Xi

I

I

Figure 5.14. A euclidean Mobius strip ]

The Mobius strip is named after August Mobius (1790-1868),j

who conceived of this nonorientable surface in 1858while working J

on geometric properties of polyhedra.He never published this work, \316\257- \320\276- i-jr \321\217\320\263~ \302\253 - --jr- \316\273~

which was only discovered after his death. Credit for the discovery I

of the Mobius strip shouldprobably go to Johann Benedict Listing J(1808-1882)instead, who independently described the Mobius strip 1in 1858. Incidentally, Listing made another important contribution |to the themes of this monograph. He coined the word \"topology\" |

(or Topologiein the German original) in an 1836 letter, and the first |

printed occurrence of this term appears in his book Vorstudien zur'

Topologie,published in 1847.

5.4. The cylinder and the Mobius strip 105

5.4.2.Hyperboliccylinders.We can make completely analogousconstructions in hyperbolic geometry by replacing the euclidean stripwith a strip X3 in the hyperbolic plane H2, bounded by two disjoint

complete geodesies E\\ and E2. However, there are essentially two

different shapes for such an infinite strip in H2.

Figure 6.15. A hyperbolic cylinder

First, the endpoints of \316\225\316\273and E2 on Ru {\321\201\320\276}may all be distinct.

By an easyalgebraicexercisewith linear fractional maps, there existsan isometry of H2 sending E\\ to the completegeodesicwith endpoints

\302\2611and sending E2 to the complete geodesicwith endpoints \302\2610,for

some a < 1. Consequently,we can assume,without loss of generality,that Ei goes from \342\200\2241to +1 and that E2 goes from \342\200\224\320\276to +a.

To glue E\\ to E2 we need an isometry \317\210\\sending E\\ to E2.

Among such isometriesof H2, the simplest one is the homothety

defined by \317\210\317\207{\316\266)= \316\261\316\266.Choose this specific isometry as a gluing map,

and let (\320\245\320\267,\320\260\320\245\320\267)be the corresponding quotient metric space. Thecombination of Theorems 4.9 and 4.10 shows that (X3, dXa) is locallyisometric to the hyperbolicplane (H2,dhyp); namely, it is a hyperbolic.surface.We are now in the situation of Figure 5.15.

This surface is easily shown to be homeomorphic to the

cylinder. However, its geometry is very different from that of a euclidean

cylinder.


Indeed,in the euclidean case, assume that Xi is the vertical

strip {(x,y) \342\202\254R2;0 < \317\207< a} and that we glue E\\=

{(\320\266,y) \342\202\254

R2; \317\207= 0} to \320\225\321\212=

{(x, y) \342\202\254R2; \317\207= a} by the horizontal translation\317\210\317\207'\302\267(\320\266!\321\203)*-*(\321\205+ \320\276>,\321\203)-(See Exercise 7.9 in Chapter 7 for an analysisof the other possible gluing maps, which shows that the general caseis essentiallyequivalent to the one considered here.) The quotient

space (\316\247\316\271,dXl) can then be decomposed as a union of closed curves

7t where, for each t, */t is the image in \316\247\316\271of the horizontal line

segment {(x,y) \342\202\254\316\2332;0 < \316\257< o, j/ = i}. These closed curves all have

length o, and the set of points at distance \316\264from the central curve 70consists exactly of the two curves 7,5 U 7_\320\263.In particular, the curves

7t show that the \"width\" of the euclidean cylinder \316\247\317\207is the same at

every point.In the hyperbolic case, where X3 is the strip in H2 delimited by

the geodesies \316\225\316\271and Ei going from \342\200\2241to +1 and from \342\200\224\320\276to +a,

respectively, we glued E\\ to E2 by the homothety \317\210\317\207(\316\266)= \316\261\316\266.For

every \316\270with \342\200\224f < \316\270< ^, we now have a closedcurve 79 in the

quotient space X3, which is the image of the euclidean line segmentconsistingof all \316\266= re1^-9) with a < r < 1.

The closed curve 70 is geodesic because, for each \316\241\342\202\25470, there

exists an isometry between a small ball Bj (\316\241,\316\265)and a ball of

radius \316\265in (H2,<ihyp) sending 70 \316\240Bjh (\316\241,\316\265)to a geodesic arc of

H2. The only point \316\241where this requires a little checking is when

\316\241corresponds to the two points i and oi \342\202\254X3, in which case thelocal isometry provided by the proof of Theorem 4.10 is easily seento satisfy this property. The curves 79 with \316\270jiO are never geodesic.

A few rather immediate computations of hyperbolic lengths showthe following:

f 1(1) The curve 79 has hyperbolic length / -zdr=log a sec \316\230.

,/j r cos a

(2) Every point \316\266= rel(-^~9^ of 79 is at distance

-dt = log(sec \316\270+ tan |0|) from the curve 70.Im r

0 r cos t'

Compare Exercise2.5for the proof of (2).

5.4. The cylinder and the Mobiusstrip 107

Therefore, for every \316\264> 0, the set of points of X3 that

are at distance \316\264from 70 consists of the two curves 7\302\261ewith

log(sec0 + tan|0|) = \316\264.By elementary trigonometry, this isequivalent to the property that sec0 = coshi. Therefore, by (1) above,

the length of each of these two curves is equal to log \320\276cosh \316\264.In

particular, the width of the cylinder grows exponentially with the

distance \316\264from 70, so that 70 forms some kind of \"narrow waist\" for

the hyperbolic cylinder X3.The picture on the right-hand side of Figure 5.15 attempts to

convey a sense of this exponential growth. This pictureis necessarily

imperfect. Indeed, as one goes toward one of the ends of the

hyperbolic cylinder, its width grows faster than that of any surface of

revolution in the 3-dimensional euclidean space R3.

00

X,

0 1

Figure 5.16. Another hyperbolic cylinder

In the hyperbolic plane H2,there is another type of infinite stripbounded by two complete geodesies\316\225\317\207and E2 which occurs when

Ei and Ei have one endpoint in common in R U {00}. Applying an

isometry of H2, we can assume without loss of generality that this

common point is \321\201\320\276,namely that E\\ and Ei are both vertical half-

lines. By a horizontal translation followed by a homothety, we caneven arrange that E\\ is the vertical half-line with endpoints 0 and

\321\201\320\276,while E2 goes from 1 to 00. Thesetwo geodesies now delimit the

strip X4 = {ze H2;0 sj Re(^)sj 1}in H2.


To glue the edgesE\\ and Ei together, the simplest gluing map

\317\206\316\271:\316\225\316\271\342\200\224\342\226\272E2 is the horizontal translation defined by \317\206\316\271(\316\266)= \316\266+

1. Let {X4, dhyp)be the quotient space obtained by performing this

gluing operation on X4. This quotient space is a hyperbolic surface

by Theorems 4.9 and 4.10, and is easily seen to be homeomorphic tothe cylinder.We should note that our choice of the gluing map \317\206\316\271is

critical here. Indeed, we will see in Section 6.7.1 that other choicesleadto hyperbolic cylinders with very different geometric properties.

For every i\342\202\254i,let 7t be the closed curve in the quotient space

X4 that is the image of the horizontal line segment consistingof those

\316\266\342\202\254X4 such that \\m(z) = e*. By definition of the hyperbolic metric,it is immediate that 7t has hyperbolic length e-i. Also, everypoint of

7t is at hyperbolic distance \\t\\ from the central curve 70. Therefore,the width of X4 grows exponentially toward one end of the cylinder

and decreases exponentially toward the other end.

Again, the right-hand side of Figure 5.16 attempts to illustrate

this behavior. As in the case of X3,this picture is necessarily

imperfectfor the end with exponential growth. Surprisinglyenough,theend with exponential decay can be exactly represented as a surface

of revolution in the ^-dimensional space M3.

More precisely, let Xf denote the upper part of X4 consisting of

those \316\266G X4 such that Im(z) > 7^, and let Xf be its image in X4.

In the rcj/-plane, consider the tractrix parametrizedby

it-\342\200\224>(t\342\200\224

tanhi,sechi), 0 < t < 00,

and let the paeudosphere S be the surface of revolution in M3

obtained by revolving the tractrix about the \320\266-axis. This surface is

represented in Figure 5.17.

Endow the pseudosphereS with the metric ds defined by the

property that for every P, Q \342\202\254S, ds(P,Q) is the infimum of theeuclideanlengths ^euc(7) of all piecewise differentiable curves 7contained in S and joining \316\241to Q.

As usual, we endow X% with the metric dx+ for which the

distance between two points is the infimum of the hyperbolic lengthsof all curves in Xf joining these two points. Becausehyperbolic

geodesies are euclidean semi-circles, it is relatively immediate that


Xf is convex, so that this metric dx+ actually coincides with the

restriction of the hyperbolic distance. However, it is convenient to keep

a distinct notation because of (minor) subtleties with quotient

metrics; compare Lemma 5.8. Let dx+ be the quotient metric induced

by dx+ on the quotient space X\302\243.

Proposition 5.4. The metric space (X^,dx+) is isometricto the

surface (S,ds).

Proof. The proof will take several steps, in part because of thedefinition of the quotient metric. It may perhaps be skipped on a first

reading. However, you should at least have a glance atLemma 5.5,which contains the key geometricidea of the proof.

Let p:Xj-^S be the map which to \316\266\342\202\254X^ associates the point page 114

p(z) = (u, v, w) with

. = log(2.1\320\250\320\234

+^\320\234*)\320\260-1)

\"^tS)\"1'lm(z)

\316\275=-\342\200\224\317\202\342\200\2247-rcos (2irRe(z)),2\321\202\320\2631\321\202(,\320\263)

v v \">

w =-\342\200\224;\342\200\224\320\223\320\242sin (2\317\200Re(z)).2\321\202\320\2631\321\202(,\320\263)

v v \"

Geometrically, p(z) is better understood by setting

t = arccosh(27rIm(,z)) = log (2nlm(z) + y/4n21m(z)2-l) .

Then p(z)= (t

\342\200\224tanhi,sechi cos(2\317\200Ile(\320\263)),sech\316\257 sm(2nBe(z)))

is

obtained by rotating the point of the tractrix corresponding to the

parameter t by an angle of 2\317\200Be(z) about the rc-axis.

In particular, two points z, z' e X/ have the same image under

\317\201if and only if z' \342\200\224\316\266is an integer, namely, if and only if \316\266and z'

are glued together to form a single point of the quotient spaceXf.As a consequence, \317\201induces an injective map p: Xf \342\200\224>S defined bythe property that p(P) is equalto p(P) for every \316\241\342\202\254X\302\243and any

element \316\241\342\202\254\316\241,namely, for any point \316\241G Xf that corresponds to\316\241in the quotient space Xf.

The map \317\201is surjective by definition of the pseudosphereS. Itfollows that p: X\302\243

\342\200\224>S is surjective and is therefore bijective.


Figure 5.17\302\267The pseudosphere

We will prove that \317\201is an isometry from (Xf,dx+) to (S,ds).The key step is the following computation.

Lemma 5.5. The map \317\201sends every curve 7 in the half-strip X% toa curve in the surface S whose euclidean length 4uc(p(7)) is equal to

the hyperbolic length ^hypM of \316\267.

Proof. Let us compute the differential map Dzp. Remember that

Dzp sends the vector \316\275=(a, b) to

d.pw =(i\302\253+$*, i\302\253+S6\302\267 %a+wb)

if we write \316\266= x + iy.

With t =arccosh^i/) as before, we find that \342\200\224

dx \302\260.s

2\317\200\316\262\316\262\316\233\316\257tanhi, |^= -2\317\200\316\262\316\262\316\233\316\2578\316\220\316\267(2\317\200:\317\204),J|

= 2\317\200\316\262\316\262\316\2332\316\257\321\201\320\260\320\260(2\320\266\321\205),

fjjf= 2ttSechi cos (2ira;), and f| = 2\317\2008\316\265\316\2771\316\2712t 8\316\252\316\267(2\317\200:\317\204).

After simplifications,

ll^P^lleuc = 2*sechi ||tf||euc= -||iT||euc

=||i;||hyp.


If7 is a curve parametrized by s t-* z(s), Si ^ s ^ 82, its image\317\201{\316\267)is parametrized by s 1-*p(z(s)), si < s < \302\2532-Therefore,

4uc(\302\245>(7))= ra||(po2ry(*)||eucd*= \320\223||\320\257\320\263(\320\262)\320\230*))||\320\265\321\210>

=\320\223\342\204\226)\320\235\321\214\321\203\321\200^

=4\320\243\321\200(7). D

Lemma 5.6.

ds(p(P),P(Q))^dxt(P,Q)

for every P, Q \342\202\254X4\".

Proof. Let the points \316\241=\316\241\317\207,Qi

~ P2, \302\267\302\267\302\267,Qn-i~ Pi, Qn = Q

form a discrete walk w from \316\241to Q. The length of the discretewalk

w is i(w) =\316\243)?=\316\271<**+ (\320\233,Q\302\273)\302\2674

By convexity of Xf, there is a geodesicarc 7* joining P\302\273to Qj

whose hyperbolic length is equal to dx+(Pi,Qi). Lemma 5.5 shows

that the image 7^=

\317\201(7\302\273)is a curve joining p(Pi) to p\\Qi)= p(Pi+i)in S whose euclidean length is equal to the hyperboliclength of 7\302\273.

Chaining together these 7^ provides a curve 7' joiningp(P) =p(P)

to p(Q)= p(Q) in S. Its euclideanlength is

\316\267 \316\267 \316\267

i=l i=l i=l

By definition of the metric ds, this shows that ds(p(P),p(Q)) < i(w).Since this holdsfor every discrete walk w from \316\241to Q, we conclude

that ds(p(P), p(Q)) < dx+ (P, Q). D

Lemma 5.7.

djc+(^Q)<ds(p(P),p(Q))

/or every P, Q \342\202\254X4\".

Proof. For a given \316\265> 0, there exists a piecewisedifferentiable curve

7 going from p(P) to p(Q) in 5 whose length is sufficiently close to

ds(p(P),p(Q)) that

ds(p(P),p(Q)) < 4uc(7) < ds(p(P),p(Q))+e.

We want to decompose 7 into pieces coming from Xf. For this, we

use the following estimate.


In the surface S, consider the tractrix \316\244parametrized by t t-*

(t \342\200\224tanhi, sechi, 0). If \316\261is a curve in S whose endpointsP' and Q'

are both in T, and if \316\262is the portion of \316\244going from P' to Q', weclaim that 4uc(/?) < 4uc(a)\302\267

To see this, parametrize aljysi-\302\273 (x(s),y(s),z(s)),a < s < b,

with 'x(s) = t(s)- tanhi(s)

< y(s)= sechi(s) cos0(s)

k z(s)= sechi(s) sin0(s)

for some functions s t-\302\273i(s) and s \316\271-\302\2730(s). The curve \316\262has a similar .

parametrization, where 6(s) is a constant equal to 0. An immediate '

computation then yields\320\263\321\214

4uc(a)=

/ \\\316\233'(\316\262)2+ \316\257/'(\302\253)2+ ,\316\265'(\302\253)2\317\216\316\257 :J a s

=J \\/t'(s)2

tanh2 \316\257(\316\262)+ 0'(s)2 sech2 t(s) ds \\Ja '.

>/ i'(s)tanhi(e) ds = eeuctf). 1

Jaj

As a consequence, we can arrange that the intersection of 7 with I

\316\244consists of a single curve contained in T, possibly empty or reduced J

to a single point. Indeed, if P' and Q' are the first and last points 1

where 7 meets T, we can replace the part \316\261of 7 going from P' to Q' 3

by the part \316\262of \316\244joining P' to Q' without increasing its length. j

We are now ready to conclude. The key observation is that \316\244I

is also the image under \317\201of the \"suture\" of Xf consisting of thosej

points which correspond to two points of X\302\261(located on the vertical ]

part of the boundary of the half-strip X4\.") |

If 7 is disjoint from T, then 7 is the image under \317\201of a curve\\

7' contained in Xf and going from a point \316\241to a point Q. By \\

Lemma 5.5, the hyperbolic length of 7' is equal to the euclidean length >

of 7. It follows that \316\257

dx+ (P,Q) < dx}(P,Q) < 4\321\203\320\240(\320\243)=

4uc(7) < ds(p(P), p{Q)) + \316\265.j

IIf 7 meets the tractrix T, wecan split it into a first piece 71 going

from p(P) to a point of T, a secondcurve 72 contained in \316\244(possibly<

5.4. The cylinder and the Mobius strip 113

reduced to a single point), and a third piece73 going from \316\244to the

point p(Q). Then 71, 72 and 73 are the respective images under \317\201of

curves 7i, 72 and 73 in Xf. Thereare two possible choices for 72, onein each of the two vertical sides of Xf, and we just pick one of them.

We have observed that two points of X\302\261have the same image under

\317\201if and only if they are glued together in the quotient space Xf.

Consequently, if P\302\273and Qi denote the initial and terminal points of

each 7i, we have that \316\241~ Pi, Qi ~ P2, Q%~ P3 and Q3 ~ Q. In

particular, the P\302\273and Qi form a discrete walk w from \316\241to Q whose

length is

3 33

i=l i=l i=l

since the euclidean length of 7\302\273is equal to the hyperbolic length of 7^

by Lemma 5.5. Therefore,

dx} (P, Q) < dxt (P, Q) <\302\243\316\254\317\207}

(w) < 4uc(7) *Sds{p(P),p(Q)) + \316\265.

We have now proved that dx+(P, Q) < ds(p(P),p(Q))+ \316\265in

both cases, and this for every \316\265> 0. Therefore, dx+(P,Q) <

ds(p(P),p(Q)) as requested. D

Thecombination of Lemmas 5.6 and 5.7 showsthat ds(p(P), p(Q))= dx+(P,Q) for every P, Q \342\202\254X4\", namely that \317\201is an isometry

from (Xf,dx+) to (S,ds). This completes the proof of

Proposition5.4.*

D

We conclude our discussionof Proposition 5.4 by addressing alittle subtlety. By convexity of X4 and Xf, dx+ (P, Q) = dx (P, Q)

=

e?hyp(P, Q) for every \316\241and Q \342\202\254X%. However, in the quotient spaceX4\", there might conceivably be a difference betweenthe quotient

metric dx+ and the restriction of the quotient metric dx of X4.

Indeed, the definition of the quotient metric dx involves discretewalks valued in X4, whereas dx+ is denned using discretewalks that

4

are constrained to X\302\261. Compare our discussion of Lemma 5.12 inthe next section.


In the specific case under consideration, it turns out that we do

not need to worry about this distinction:

Lemma 5.8. On the subspace Xf of X4, the two metrics dx+ and

dx coincide.T4

Proof. Because Xf is contained in X4 and because the metrics dxand dx+ coincideon X4, every discrete walk w valued in X% is also

valued in X4, and the two lengths id + (w) and id (w) coincide.It__X4 _ *4 _

follows that dx (P, Q) sj dx+(P,Q) for every P, Q \302\243Xf.

Conversely, let \316\241=\320\240\321\214Qi

~ P2, ..., Qn-i ~Pn, Qn

= Q be

a discrete walk w valued in X4 where \316\241and Q both belong to theupperhalf-strip X^. For each Pi, let P/ be equal to P* if P\302\273is in X\302\261,

and otherwise let P/ = Re(Pj) + -^i be the point of the boundary of

Xf that sits right above P\302\273.Similarly, define Qi to be Qi if Q \342\202\254X4

and Re(Qi) + \302\2431if Q \302\243X%.

We claim thatdhyP(-Pj> Qi) ^ ^hyp(Pi, Qi)\302\267Indeed, if 7 is a curve

joining Pi to Qi in X4, let 7' be obtained from 7 by replacing eachpiecethat lies below the line of equation Im(z) = ^ with the line

segment that sits on the line right above that piece. From thedefinition of hyperbolic length, it is immediate that 4yP(7') ^ ^hyp(7)\302\267

Considering all such curves 7 and 7', it follows that<2\321\214\321\203\321\200(\320\240\320\263',\320\241?'\302\273)^

dhyp(Pi,Qi)\302\267

Since P[ =P\\ and P'n

= Pn, we now have a discretewalk \316\241= P{,

Q\\ ~ P'2, ..., Qn_x~

P\302\243,Q'n= Q which is valued in X/. Since

dhyp(-P/, Qi) < dhyp(P\302\273, Qi), this new walk w' has length \302\243d (u/) sj

^d (\320\263\320\276).As a consequence, id (w) ^ dx+(P,Q).

Considering all such walks w, we conclude that dx+ (P, Q) ^

dx (P,Q). Therefore, dx+(P,Q)=

dx (P,Q) for every P, Q \342\202\254

x/.

4 4

\342\226\241

5.5. The once-punctured torus

The once-punctured torus is obtained by removing one point fromthe torus. Toexplainthe terminology, think of what happens to the

5.5. The once-puncturedtorus 115

inner tube of a tire as one drives over a nail. If we describe the torusas a square with opposite edges glued together, we can assumethat

the point removed is the point correspondingto the four vertices of

the square.

This surface of course admits a euclidean metric by restriction of

a euclidean metric on the torus. However, we will see in Chapter 6that such a metric is not complete (seethe definition in Section 6.2),and that complete metrics are moredesirable.Our goal is to constructa hyperbolic metric on the once-puncturedtorus, which we will later

on prove to be complete.This example will turn out to be very important. In particular,

it will accompany much of our discussionin Chapters8, 10and 11.

Consider the hyperbolic polygon X describedin Figure 5.18.Namely, X is the region in the hyperbolic planeH2 bounded by the

four complete geodesies E\\, E2, E3 and E4 where E\\ joins \342\200\2241to 00,

E2 joins 0 to 1, E3joins 1to 00,and E4 joins 0 to \342\200\2241.

00

\320\225\320\263

1

X

;

I ^4 \\/ ^2 \320\233

\320\225\320\267

\316\271\316\271

-1 0 1

Figure 5.18. A hyperbolic square

As such, X is a \"quadrilateral\" except that its vertices are in\316\234U {00}, namely, at infinity of H2. As a subset of \316\2272,\316\247is therefore

a quadrilateral with its four vertices removed.

In a hyperbolic polygon of this type where the vertices are at

infinity of H2 in R U {oo} (and consequently are not really verticesof


the polygon in H2),we say that the vertices are ideal. If the verticesof the polygon are all ideal, and if the polygon touches \316\234U {00} onlyat these vertices, we say that it is an ideal polygon.

With this terminology, X is now an ideal quadrilateral in H2. It

is even an ideal square,in the sensethat it has all the symmetries ofa square;seeExercise5.13.

Glue together opposite edges of X, while respecting theorientations indicated in Figure 5.18. Because these geodesiesdo not have

any endpoints in H2, this is a situation where there are manypossibleisometric gluings. In particular, the gluing data is not completelydetermined by the picture. Consequently, we need to be more specific.

To glue the edge \316\225\317\207to E2, we need a hyperbolic isometry \317\206\316\271

sending \342\200\2241to 0, and 00 to 1. The simplestoneis the linear fractional

map

Similarly, we can glue E3 to E4 by the hyperbolic isometry

As usual, define \317\210%=

\317\210\317\2121and \317\2104=

\317\210^1.

Let (\316\247,\316\254\317\207)be the quotient metric space obtained from (\316\247,\316\254\317\207)

by performing these edge gluings. Note that X is convex so that themetric \316\254\317\207is just the restriction of the hyperbolicmetricdhyp-

Since X has no vertices in H2, there is nothing to be checked and

Theorem 4.10 shows that (\316\247,\316\254\317\207)is a hyperbolic surface. From thedescriptionof X as a quadrilateral with its vertices removed, we seethat X is (homeomorphic to) a once-punctured torus.

We want to better understand the metric \316\254\317\207near the puncture.

For a > 1, considerthe horizontal line La of equation lm(z) = o.

By Proposition 2.18, linear fractional maps send circlesto circlesin \320\241= \320\241U {oo}. Consequently, \317\206\316\271sends La U {00} to a circleCipassing through <^i(cx>)

= 1. Since \317\206\316\271also sends the half-plane H2 toitself, this circlemust be tangent to the real line at 1. The circle Cialso contains the image \317\206\\(\342\200\2242+ai)

=1+^i of the point \342\200\2242+ai\342\202\254La.

5.5. The once-punctured torus 117

It follows that C\\ is the euclidean circleof radius ^ centered at 1+î,and ^(La) = Ci-{l}.

Similarly, 1\321\200\320\267(\320\254\320\260)is contained in a circle C_i tangent to the real

line at </?\320\267(\302\260\302\260)= \342\200\2241and containing \317\2063(2+ oi)

= \342\200\2241+ î. Namely,

\320\241-\317\207is the circle of radius ^ centeredat \342\200\2241 + î, and tpz(La) =

C_,-{-l}.The map \317\206\317\207also sends \342\200\2241to 0. Consequently, it sends the circle

C-i to a circle Cq tangent to the real line at \317\206\316\271(\342\200\2241)= 0 and passing

through \317\210\\\320\276</?\320\267(3+ oi)

=\\i (since (/?\320\267(3+ ai) \342\202\254<\321\200\320\267\342\204\226\320\260)\320\241C_i).

Namely, Co is the circle of radius ^ centered at î.Finally, \317\2103sends C\\ to a circle C'0 tangent to the real line at

</?\320\267(1)= 0 and passing through \317\2103

\316\277\317\206\316\271(\342\200\2243+oi)

=î. It follows that

this circle C'0=

\317\206\316\266{0{)is exactly equal to the circle Co=\317\210\$0-\$.

Let Uoa be the set of points of X that are on or above the lineLa, and let \320\251,U\\ and U-\\ consist of the points of X that are on orinside the circlesCo, C\\ and C_i, respectively. In addition, because\320\276> 1, the Oi are disjoint. See Figure 5.19.

What the above discussion shows is that when \316\241belongs to some

of these Ui and is glued to some P* \342\202\254X, then P' must be in some

other Uj. See Figure 5.19.

1;

vv i

II

V2(t/l)

1

'\342\226\240u00 = v00 ;

\"n. /^~~~\\ /~

II

*>\302\253(U-i)

11

Vo

II

\302\245\342\200\24240^2(^0)

La

-2-1012 4

Figure 5.1\316\230


Let U denotethe union Uoo UUqUUiUU-i in X, and let U beits imagein the quotient X. Namely, U is obtainedfrom U by gluingits eight sides though the restrictionsof the maps \317\210\\,\317\2102

=\317\210\317\2121,\317\2103

and \317\2104=

\316\2503\302\267

Consider on U the metric du defined by the property that du (P, Q)is the infimum of the hyperbolic length of all curves joining \316\241to Q

in U. In particular, du(P, Q) = 00when \316\241and Q are in different Ui,since they cannot be joined by any curve that is completelycontainedmU.

Each Ui is convex. This is fairly clear for the vertical half-stripUoo,sinceevery hyperbolic geodesic is a circle arc centeredon the x-axis.The property is slightly less obvious for the other Ui but follows

from the fact, provedbelow, that Ui is isometric to a vertical half-strip

Vi. A consequence of this convexity is that du(P,Q)=

^thyp(P,Q)when \316\241and Q are in the same Ui.

Themetricdu induces a quotient semi-metric du on the quotient

space U.

We begin by showing that the metric space(U,du) is isometricto a spacethat we have already encountered. Let Sa be the surface

of revolution obtained by revolving about the rc-axis the portion of

the pseudosphere parametrized by

\316\257\316\271\342\200\224>\342\226\240(i\342\200\224tanh t, sech t), arccosh ^ < t < 00.

Equivalently, Sa consists of those points (\320\266,\321\203,z) of the pseudosphere

S such that \317\2072 log (<f + yj^f-\316\233-

y/l-J^.As in Proposition 5.4, endow Sa with the metric dsa defined by

the property that dsa (P, Q) is equal to the infimum of the eucfidean

lengths of all curves joining \316\241to Q in Sa.

Proposition 5.9. The metric space (U,d~u) \317\216isometric to the space

(Sa,dSa).

Proof. The proof is very similar to that of Proposition 5.4, exceptthat we now have to glue four vertical half-strips instead of a single

one.


Figure 5.20. The portion Sa of the pseudosphere

In the hyperbolic plane H2, considerthe subsets VOo = Uoo,

V_i =\317\2062(\317\205-1)

=tp^iU-i), Vi =

\317\2104{\317\205\316\273)=

\317\206\317\212\316\227\317\213!)and V0 =

\317\2104\302\260lf2(Uo)=

\316\25031\316\277

\317\206\317\2121^).

Remember that VOo= Uoo is the vertical half-strip delimitedby

the horizontal line La = {z;lm(z) = a} and by the vertical lines of

equation Re(z)= \302\2611.

We already observed that \317\206\316\271sends the line La to the circleC\\.

It also sends the points 00, \342\200\2242and \342\200\2241to 1, 00 and 0, respectively.SinceU\\ is delimited by C\\ and by the geodesies joining 1 to 00and 1to 0,it follows that V\\ =

\317\206\317\212x(Ui) is delimited by La and by the two

geodesies joining \342\200\2241to 00 and \342\200\2242to 00. These hyperbolic geodesiesare alsothe vertical lines of equations Re(z) = \342\200\2242and Re(z)

=\342\200\2241,

so that Vi is a vertical half-strip.

Similarly, since 4>2{La) = C-i, V_i= yJJ1(i7_i) is the vertical

half-strip delimitedby La and by the two vertical lines of equations

Re(z) = 1 and Re(z) = 2.Finally, because Co =

\317\206\\(0-\316\271)=

\317\210\\\316\277(\321\200\320\267(\320\254\320\260),the same type of

arguments show that Vo=

\317\206^1\302\260f^Wo) is delimited by La and by

the vertical half-lines of equations Re(z) = 2 and Re(z)= 4.

This situation is illustratedin Figure5.19.Let V be the vertical half-strip union of V\\, VOo, V-i and Vo. Let

V be the quotient space obtained from V by gluing its left-hand side

{z \342\202\254V; Re(z)= -2} to its right-hand side {z \342\202\254V; Re(z)

= 4} by


the horizontal translation \317\210:\316\266\316\271-\302\273\316\266+ 6. Since V is convex,endow it

with the restriction dy of the hyperbolicmetric dhyp>a^d let dy be

the quotient metric inducedby dy on V.

We now split the argument into two steps.

Lemma 5.10. The quotient space (V, dy) is isometric to the subset(Sa,dsa) of the pseudosphere.

Proof. This is essentiallyProposition5.4.Indeed,let V =

{z \342\202\254H2;0 sj Re(*) \302\253\316\252l,Im(z) > \302\247}.Let V'

be the quotient space obtainedfrom V by gluing its left-hand side{z \342\202\254V;Bs(z)

= 0} to its right-hand side{z \342\202\254V;Re(z)= 1} by the

horizontal translation \317\210':\316\266\316\271-\302\273\316\266+ 1. Endow V with the restriction

dy/ of the hyperbolic metricdhyp> and endow V' with the induced

quotient metric dy.

The hyperbolic isometry \317\206:\316\266\316\271-\302\273|\316\266+ | sends V to V, and it

sends the gluing map \317\206of V to the gluing map \317\206'of V. Consequently,

\342\200\242\317\210induces an isometry \317\210from (V, dy) to (V', dy).Note that V is a subset of the half-strip Xf considered in

Proposition 5.4, so that V \320\241X. The surface Sa is exactly the image of V'

under the isometry \316\262constructed in the proof of Proposition 5.4.Therefore, \317\201restricts to an isometry from (V',dv<) to (Sa,dsa).

The composition \317\201\316\277\317\210consequently provides an isometry from

(V,dv) to (Sa,dSa). \316\217

Lemma 5.11. The quotient spaces (U,du) and (V, dy) are

isometric.

Proof. Let \321\204:U \342\200\224>V be the map defined by the property that

\321\204coincides with \317\2102=

\317\210\317\2071on U\\, with \317\2104=

\317\206^1on U-\\, with

\316\250\316\261\302\260\317\2102=

\316\25031\302\260\317\210\317\2121\316\277\316\267Uq, and with the identity map on \317\205,\317\207.

A case-by-case inspection of the eight sides of U that are glued

together shows that P, Q \342\202\254U are glued together in U if and only

if \321\204(\320\240)and \317\206(0) are glued together in V. For instance,if \316\241is in

the intersection of Ui with the vertical line of equation Re(z) = 1,it is glued to the point Q =

\317\210\316\266{\316\241)contained in the intersection

of Uq with the geodesic joining 0 to \342\200\2241.Then\317\206(\316\241)

=\317\2102(\316\241)and


%I>(Q)=

\317\210^\316\277\317\2062{0)differ by the map \317\210\316\221\316\270\317\2102\316\277\317\210\316\206\316\277\317\21021\342\226\240A computation

shows that </?4 \302\260\317\2102\302\260\316\2503\302\260V3^1^)

= \316\266+ 6, so that V'(-P) and VKQ) axe

indeed glued together in V. The other casesare similar, and actually

easier since no gluing in V is needed.

It followsthat \317\210induces a map \317\210:U \342\200\224>V, defined by the

property that \317\210(\316\241)is equal to the point of V corresponding to \321\204{\320\240)\342\202\254

V, for an arbitrary point \316\241\302\243Ucorresponding to \316\241G U. Let us showthat \317\210is an isometry.

If \316\241= Pi, Qi ~ P2, ..., Qn-i ~Pn, Qn

= Q is a discretewalk \320\263\320\276from \316\241to Q in tf, then \317\210(\316\241)

=\317\210(\316\241\316\271),V(Q0

~\317\210{\317\2012),

..., V'(Qn-i) ~ 'i/'i-Pn))\321\204(\320\257\320\277)=

\321\204(\320\257)is a discrete walk from \321\204(\320\240)

to \321\204{0) in \316\220^,which has the same length as w. It follows that

dv${P)MQ)) < du(P, Q) for every P,QeU.Conversely, let \321\204(\320\240)

= P{, Q\\ ~P^..._, QU ~ Pn, Q'n

=

V>(Q) be a discrete walk w' from \317\210(\316\241)to VKQ) m \320\243\302\267Consider the

decomposition of V into the four vertical half-strips V(^i)>VK^oo))

tp(U-i), ip(Uo). If P/ and Q\302\243are not in the same half-strip %l>{Uj),

draw the geodesic g joining P/ to Q\302\243,consider the points Rlt ...,i?fc (with \320\272< 3) where 5 meets the vertical lines Re(z)

=\342\200\2241,1, 2

separating these half-strips, and replacethe part Qi_! ~ P/, QJ~

P/+1 of the walk w' by Q^ ~ P/, \320\2241=

\320\224\321\214..., Rk = i?fc, QJ ~

P/+1.By performing finitely many such modifications wecan arrange,without changing the length of \317\205/,that any two consecutive P/, Q\\

belong to the same half-strip ip(Uj). As a consequence,there existsP{,Qi \342\202\254Uj such that \321\204{\320\240^

=\316\241/,V(Q\302\273)

= Qi, and du(Pi,Qi) =

dhyP(Pi,Qi) =<WP/,Qi)\302\267

ThenP = Pi, Q}~ P2,..., Qn_i ~ Pn,

Qn= Q is a discrete walk from \316\241to Q in U, whose length is equal

to the length of \317\205/.This proves that d~u(P, Q) sj dv(ip(P),ip(Q))for

every P,Q \342\202\254U.

This completes the proof that \317\210:(U, d~u) \342\200\224*(V,dv) is an

isometry. . D

The combination of Lemmas 5.10 and 5.11completesthe proof

of Proposition 5.9. D

Without any assumption on a, the metric d~u on U may be

different from the restriction of the metric \316\254\317\207of X. For instance, when


\320\276is very close to 1, the images \316\241and Q \342\202\254\316\214corresponding to

\316\241= oi and Q = ^i are very close with respect to the metric \316\254\317\207,

since dx(P,Q) < dx(P,Q) = loga2, but they are quite far from

each other with respect to the metric du since there clearly is a ball

Bdbyp (\316\241,\316\265)which contains no point that is glued to another one, so

that any discrete walk from \316\241to Q in U has length ^ \316\265in U.

Lemma 5.12. If a is chosenlargeenough that \316\261log \316\261> \\, the

metrics \316\254\317\207and du coincide on U.

Proof. Becauseevery discrete walk valued in U is also valued in

X, and because dx(Pi,Qi) < du(PuQi) for every P{, Q\302\273\342\202\254U, it is \\

immediate that du(P, Q) < d~x(P,Q) for every P, Q \342\202\254U. \\

To prove the reverse inequality, pick another number a' > 1

sufficiently close to 1 that logo' < logo \342\200\224^, and \316\265> 0 small enough \302\273

that \316\265< log ^7\342\200\224

^. Let U' and U' be associatedto a' in the same J

way as U and U were associatedto o. Becauseo' < o,the portion 1

Sa' of the pseudosphere S contains Sa. j

Let P, g \342\202\2540, and let \316\241=Pi_, Qi _~ Pa, \302\267\302\267\342\226\240,Qn-i

~P\342\200\236,

|Qn = Q be a discretewalk w from \316\241to Q in X, whose dx length \\

is such that idx{w) < d~x(P,Q) + \316\265.Without loss of generality, we ican assumethat whenever the hyperbolic geodesic 7*joining P\302\273to Qi |meets one of the circles Co, Ci, C_i and La delimiting U in X, it does

j

so only at its endpoints Pj, Qi\\ indeed, if this property does not hold, Iwe can just add to the discrete walk w the intersection points of 7\302\273J

with these circles, which will not change the dx-length of w. Adding 1

a few more points if necessary,wecan arrange that the sameproperty ;

holds for the circles C0, C[, C'_x and La' similarly associated to a'. |i

We will show that because of our choice of a' and \316\265,the discrete \\

walk w stays in U'. Namely, all the Pi and Qi as well as the geodesic 3

arcs 7j joining them, are contained in U'. 1

Indeed, suppose that the property does not hold. Let i\\ be the \\

smallest index for which 7^ <(_ U and let i2 be the largestindex for\302\267

which 7i2 \317\210.U. By the condition that we imposedon the geodesies7\302\273,\342\226\240\"

the points P^ and Qi2 are both contained in the union of the circles ;

Co, Ci, C_i and La. Similarly, let i\\ be the smallest index for which

7i' <jL U' and let i'2 be the largest index for which 7\302\273/<jL U', so that ;


Pi' and Qi> belong to the union of the circlesC'Q, C[, C'_x and La>\302\267

Note that i\\ < i[ < i'2 < i2.

In the quotient space U', the geodesies7* with \320\263\320\263< i < i\\ project

to a continuous curve 7 joining \316\241\302\273!to Qj/ iaU'. Consider the isometry\317\206:(U',du')

\342\200\224>(Sa',dsa,) provided by Proposition 5.9. Then \317\210(\316\267)

is a piecewise diiferentiable curve joining \317\206{\316\241%\316\271)to\317\206(0\316\257>1)

in Sa>,

Note that because P^ and Q^ are in the boundary of U and \302\243/',

respectively, \317\206(\316\241%1)and <p(Qi') respectively belong to the circlesdSaand dSa' deUmiting SO and SO' in the pseudosphereS. We then

use an estimate which already appeared in the proof of Lemma 5.7.

Parametrize the curve \317\210(\316\267)by

8 1-\302\273(i(s)-tanhi(s),sechi(s) cos0(s),sechi(s) sin#(s)), 0 < \316\262< 1,

for some functions s i->t(s) and s i-> 9(s), so that \317\206(\316\241\316\2571)\342\202\254dSa

and\317\206(0\316\257'1)

\342\202\254dSa< correspond to s = 1 and s = 0, respectively. In

particular, i(0) = arccosh ^ and i(l)= arccosh ^ by definition of

Sa and Sa<. Then,

\316\243d*(p*. <w =\316\243\320\265^\321\213)

=^(\321\202)

1=11

= / V*'(s)2+2/'(s)2+z'(\302\253)2dsJO

= /\"^\316\257'(\316\262)2

tanh2 i(s) + 0'(s)2 sech2 t(s) ds

^ / i'(s)tanhi(s)dsJo

= log cosh i(l) - logcoshi(0)= log \342\200\224.

a

\302\253=\302\2531 \302\253=\302\2531

Similarly, } dx(Pi,Qi) ^ log \342\200\224.As a consequence,7\342\200\224f, a,\320\263=\320\263'2

\316\267

dx(P>Q)>^x(w)-e=

J2dx(pi'Qi)-\302\243i=\\

ii\342\200\2241 \316\267

>\316\243d*(Pi- GO + 2loS \"7

+\316\243 dx(Pi> GO

\" \316\265\302\267

i=l i=ia+l


On the other hand, the boundary circledSa has length |.Consequently, it is possible to join the two points \317\210(\316\241\316\2571)and <p(Qi3) bya curve of euclideanlength ^ | in Sa so that

du(Pix,Qia) = ds.Mfli).v(Qia)) < f \342\226\240

It follows that there is a discretewalk P^= P[,Q\\ ~ Pi, \342\226\240\342\226\240\302\267,\316\241'\316\267\316\271-1

~

Ki'i Q'n' = Q*2 \302\260flength < | +e. Chaining this discrete walk with

the beginning and the end of w, we obtain a discrete walk \316\241= P\\,

Qi ~ Pa, ...Qh-i ~\316\241*\302\273,P<t

=\316\241{,Qi

~ Pi, \302\267\302\267\302\267,K'-i _~ P\302\253>

Qn'= Qia, Qii ~ Pia+i, ..., Qn_i

~ Pn, Qn = Q from \316\241to Q in X.

Therefore,

dx(P,Q)<Y/dx(Pi,Qi) + l+e+\302\243 dxifl.Q,)

i=l i=ia+l

Combining this with our earlier estimate for \316\254\317\207(\316\241,Q), we

conclude that 21og^-\302\243<|+\302\243, which is impossible by our choices of

\320\276,\320\276'and \316\265.

This contradiction shows that our initial hypothesis was false.

Namely, the geodesies7* are all in U'. In particular, the Pi and Qiform a discrete walk from \316\241to Q in U', so that

\316\267 \316\267

\302\253to\302\273(\316\241,Q) <\316\243

dy' (\317\201\"<?i)=

\316\243\317\201(7\316\257)= tdx \320\230< \316\257\317\207(\316\241,<5) + \320\265.

i=l t=l

Since this holds for every \316\265that is small enough, we conclude that

dul(P,Q)^dx(P,Q).

Finally, the two metrics du and du> coincide on V'. This canbe seen by applying the proof of Lemma 5.8 to the space (V, dy)

of Lemma 5.11, and to (V1, dy1) similarly associated to a'. One canalso show that the metrics dsa and dsa, coincideon Sa by a simple

estimate of euclidean lengths of curves.

Therefore, du(P,Q) < du>(P,Q) < d~x(P,Q)for every P, Q \342\202\254

U. Since we had already proved the reverse inequality, this shows

that the metrics du and \316\254\317\207coincide on 0'. \316\240

In Lemma 5.12, the condition that a loga > \302\247is not quite sharp.

With a better hyperbolic distance estimate to improve the inequality

5.6. Triangular pillowcases 125

d~u (Pii) Qi2) ^ |) OQecan show that the conclusions of Lemma 5.12 fb\302\273\" \320\277\320\260\321\201\320\26512

still hold for \316\261^ 2.This second estimate is sharp, in the sensethat

Lemma 5.12 fails for a < 2.

5.6. Triangular pillowcasesWe conclude with an example where the angle condition of

Theorems 4.4, 4.10 or 4.13 fails, so that we obtain surfaces with cone

singularities, as in Exercise4.8.

Proposition5.13.Let \316\261,\316\262and 7 be three numbers in the interval

(\316\237,\317\200).Then:

(1) if a + \316\262+ 7 =\317\200,there exists a triangle \316\244of area 1 in the

euclidean plane (R2,deuc) whose angles are equal to \316\261,\316\262

and 7;

(2) if a + \316\262+ 7 < \317\200,there exists a triangle \316\244in the hyperbolic

plane (H2, dhyp)whose angles are equal to \316\261,\316\262and 7;

(3) if \317\200< \316\261+ \316\262+\316\267< \317\200+ 2\317\200\317\213\316\267{\316\261,/?,7},there exists a triangle\316\244in the sphere (S2, daph) whose angles are equal to \316\261,\316\262and

7\302\267

In addition, in each case, the triangle \316\244is unique up to isometry of(R2,deuc),(H2,dhyp) o/(S2,daph), respectively.

Proof. The euclidean case is well known. See Exercises 5.15 and 3.5for a proof in the hyperbolic and sphericalcases. D

SeealsoExercises2.15and 3.6 for a proof that the conditions of

Proposition 5.13 are necessary.

Given \316\261,\316\262and 7 e (\316\237,\317\200),let \316\244be the euclidean, hyperbolic or

spherical triangle provided by Proposition 5.13.

Choose an isometry \317\210of (R2,deUC), (H2,dhyp) or (S2, daPh),

according to the case, such that \317\210(\316\244)is disjoint from T. Theexistence of \317\206is immediate in the euclidean and hyperboliccase. Forthe spherical case, we observe that the proof of Proposition 5.13(3)

in Exercise 3.5 shows that \316\244is always contained in the interior of ahemisphere.Then, we can use for \317\210the reflection across the greatcircle delimiting this hemisphere.


Let X be the union of \316\244and \317\206(\316\244).We can consider X as anonconnectedpolygon, whoseedgesare the threesidesE\\, E3, E$ of

\316\244and the corresponding three sides E% =\317\210(\316\225\316\271),\316\225= \317\210{\316\225\316\266)and

Eq =<\321\200{\320\225\321\212)of \317\206(\316\244).We can then glue these edges by the gluing

maps \317\210\316\271:\316\225\316\271\342\200\224>E2, \317\2103'.\320\225\320\267\342\200\224>E^ \317\206$:E$ \342\200\224>\320\225\320\262defined by the

restrictions of \317\206to the edges indicated.

By Theorems 4.3, 4.9 or 4.12,this gluing data provides a

quotient metric space (\316\247,\316\254\317\207).This metric space is locally isometricto (R2,deuc),(H2,dhyp)

or (S2,daph) everywhere, except at the threepoints that are the images of the vertices under the quotient map.

The metric has cone singularities at thesethree points, with

respectivecone angles 2a, 2/3, 27 < 2\317\200,in the sense of Exercise 4.8.

Note that X is obtained by gluing two triangles \316\244and \317\206(\316\244)along

their edges. In the euclideancase,this is the familiar construction of apillowcaseobtainedby sewing together two triangular pieces ofcloth.Figure5.21attempts to describe the geometry of X in all cases.

\316\261+/3 + 7 = \317\200 \316\261+ /? + 7<\317\200 \316\261+ \316\262+7 > \317\200

Figure 5.21. Triangular pillowcases


Exercise5.1.We want to rigorously prove that the metric spaces (\316\247\316\271,dxl)

and (X2, <\302\243y3)of Section 5.1.1 are homeomorphic. Without loss of

generality,we can assume that the lower left corners of the rectangle Xi and theparallelogram X2 are both equal to the origin (0,0) \342\202\254R2.

a. Show that there is a unique linear map R2 \342\200\224>R2 that sends the bottomedgeof Xi to the bottom edgeofX2, and the left edge of Xi to the left

edge of X2\302\267Show that this linear map restrictsto a homeomorphism

\317\206:X\\ \342\200\224*X2\302\267

b. Show that two points \316\241and Q \342\202\254Xi are glued together to form a singlepoint of Xi if and only if their images <p(P) and \317\206{0)are glued together


in X2. Conclude that \317\206:\316\247\316\271\342\200\224\302\273X2 induces a map \317\206:\316\247\316\271\342\200\224*X2, defined

by the property that \317\206(\316\241)=

\317\210{\316\241)for every \316\241\342\202\254\316\247\316\271.

c. Show that \317\206:\316\247\316\271\342\200\224*X2 is a homeomorphism.

Exercise 5.2. Let X be the torus obtained by gluing opposite sides ofa euclideanparallelogram X. Let \317\210:X \342\200\224>X be the rotation of angle \317\200

around the center of the parallelogram, namely, around the point where thetwo diagonals meet. Show that there existsa unique isometry \317\206:\316\247\342\200\224\342\226\272X of

the quotient metric space(X,dx)such that \317\206(\316\241)=

\317\210{\316\241)for every \316\241\342\202\254X.

Exercise 5.3. Let (X, dx) be the torus obtained by gluing opposite edgesof a euclidean rectangle X = [a,b] \317\207[\321\201,d] \320\241\320\2322by translations, where Xis endowed with the euclidean metric dx = deuc\302\267Recall that a geodesicof (X, dx) is a curve 7 in X such that for every \316\241\342\202\2547 and every Q \342\202\2547

sufficiently close to P, there is a piece of 7 joining \316\241to Q which is theshortest curve going from \316\241to Q; here the length of a curve in (X, dx) isdefined as in Exercise 1.11.

a. Let 7 be a curve in X and let \317\200:X \342\200\224>X denote the quotient map.Show that if 7 is geodesic in (\316\247,\316\254\317\207),then its preimagep\"1 (7) consistsof parallel line segments in the rectangle X. You may need to use themain result ofExercise1.11,which says that the length of a curve in the

metric space (R2,deuc) coincides with its usual euclidean length 4oc

b. Consider the case where X is the squareXi = [0,1]\317\207[0,1]. Show

that every closed geodesiccurve in Xi has length > 1. Possiblehint:

Consider the projection of the preimage p_1(7) \321\201Xi \320\241R2 to each ofthe coordinate axes.

\321\201Consider the case where X is the square Xi =[0, 5] x [0,2]. Show that

(X, dx) contains a closed geodesicof length \\.

d. Conclude that the euclidean tori (\316\247\316\271,\316\254\317\207^and (X2, dx2) are not

isometric.

Exercise 5.4. Let (X, dx) be the torus obtained by gluing togetheropposite sides of the square X = [0,1]x [0,1].Let \317\200:X \342\200\224\302\273X denote the

quotient map.

a. Given \320\276\342\202\254[0,1], let \317\210\316\261'\342\226\240X \342\200\224*X be the (discontinuous) function

defined by the property that \317\206\316\261(\317\207,\321\203)=

(\317\207+ \320\260,\321\203)if 0 ^ \317\207^ 1 \342\200\224a and

<fa(x,y) = (x + o, \342\200\2241)2/)if 1 \342\200\224\316\261< \317\207^ 1. Show that <pa induces a map\317\206\316\261:X \342\200\224\342\226\272X, uniquely determined by the property that \321\204\320\260on =

\317\200\316\277\317\206\316\261.

b. Show that if w is a discrete walk \316\241\342\200\224Pi, Qi ~ P2, ..., Qn-i ~Pn,

Qn = Q is a discretewalk from \316\241to Q in X, there exists a discretewalk w' from \321\204\320\260(\320\240)to \317\206\316\261(0)whieh has the same dx-length \302\243dx(\302\253/)

=

tdx (w) as w. Hint: When Pi and Qi are on oppositesidesof the line


\317\207= 1 \342\200\224a, add to w the point where the line segment [PjQi] meetsthis

line.

\321\201Use part b to show that dx(<pa(P),<pa(Q)) ^ dx(P,Q) for every \316\241

Q \342\202\254X. Then show that \317\206is an isometry of (X, dx).d. Use the above construction to show that the metric space (X, dx)

is homogeneous, namely that, for every P, Q \342\202\254X, there exists an

isometry of (X, dx) which sends \316\241to Q.

e. More generally, let \320\243be a parallelogram in K2 and let (\316\253,\316\254\316\263)be

the quotient metric space obtained by gluing opposite sides of \320\243by

translations as in Section 5.1.1. Show that (\316\253,\316\254\316\263)is homogeneous.

Exercise 5.5. Let (X, dx) be the Klein bottle obtained by gluing togetherthe sides of the squareX = [0,1]x [0,1] as in Section 5.1. Let \317\200:X \342\200\224*X

denote the quotient map.

a. Recall that a geodesic of (X, dx) is a curve 7 in J? such that for every\316\241\342\202\2547 and every Q \342\202\2547 sufficiently close to P, there is a pieceof 7

joining \316\241to Q which is the shortest curve going from \316\241to Q, wherethe length of a curve in (X, dx) is defined as in Exercise 1.11. Showthat the image 71 =

\317\200([0,1]x {0}) is a closed geodesicof length 1,

and that there exists at least one closed geodesic of length 2 which is

disjoint from 71. Show that 72 =\317\200([0,1]

\317\207{i}) satisfies the same two

properties, namely, it is a closedgeodesicof length 1 which is disjoint

from a closed geodesic of length 2.

b. Show that 71 and 72 are the only two closed geodesies in X satisfyingthe properties of part a. Hint: For a closedgeodesic 7 in X note, asin Exercise 5.3, that the preimage \317\204\"1(7) must consist of parallel linesegments in the square X.

\321\201Conclude that the Klein bottle X is not homogeneous.

Exercise 5.6. In the euclidean plane, let Xi be the square with vertices

(0,0), (1,0), (0,1), (1,1); let X2 be the parallelogram with vertices (0,0),(1,1), (0,1)and (1,2); and let X3 be the parallelogram with vertices (0,0),

(1,1), (1,2) and (2,3). Let X\\, X2 and X3 be the euclideantori obtainedby

gluing opposite sides of these parallelograms.Show that the euclidean tori\316\247\316\271,X.2 and X3 are all isometric. Hint: Show that each of these euclideantori is obtained by gluing the sides of two suitably chosen triangles. (Youmay need to use the result of Exercise 4.4 to justify the fact that the orderof the gluings does not matter.)

Exercise 5.7. Given a euclidean hexagon X, index its vertices as Pi, P2,

..., Pq in this order as one goes around the hexagon. Suppose that opposite

edges have the same length, and that the sum of the anglesof X at its oddvertices Pi, P3 and P5 is equal to 2\317\200.Use elementary euclidean geometryto show that opposite edges of X are parallel.


Exercise 5.8. A surface is nonorientable if it contains a subset home-omorphic to the Mobius strip. For instance, the Klein bottle is anonorientable euclidean surface. Construct a nonorientable hyperbolic surface.

Exercise 5.\316\230.The surface of genus g is the immediate generalization ofthe caseg = 2 and consists of a sphere with g handles added. Construct ahyperbolic surfaceof genus g for g = 3, 4 and then for any g ^ 2. Possiblehint: Glue opposite sides of a hyperbolic polygon with suitably chosen

angles; Proposition 5.13may be convenient to construct this polygon.

Exercise 5.10 (The projective plane). Let X be the spherical polygon of

Section 5.3, with the gluing data indicated there, and let (X, dx) be the

corresponding quotient space. Let \320\256\320\2402be the set of lines passing through

the origin \320\236in R3. For any two L, V \342\202\254\320\256\320\2402,let 9{L,L') \342\202\254[0, \302\247]be the

angle between these two lines in R3.

a. Show that \316\270defines a metric on RP2.b. Consider the map \317\210:X \342\200\224>RP2 which to \316\241\342\202\254X associates the line

OP. Show that \317\206induces a bisection \317\206:X \342\200\224*RP1.

c. Show that this bijection \317\206is an isometry between the metric spaces(X,d~x)and(RP2,0).

d. Show that the metric space (RP2, \316\230)(and consequently (X, dx) as well)

is homogeneous.

Exercise 5.11. Let (X,dx) be the projective plane of Section 5.3 (orExercise5.10above). Show that for every great circle \320\241of S2, the image ofX C\\C under the quotient map \317\200:X \342\200\224>X is a closed geodesicof length \317\200.

Conclude that the spherical surface (X, dx) is not isometric to the sphere(S2,dsPh).Exercise 5.12.Let (X, dx) be the projective plane constructed in

Section 5.3, by gluing onto itself the boundary of a hemisphere X \320\241S2. Let

Po be the center of this hemisphere, namely, the unique point such that

X = BdspbiPo,|). Show that X \342\200\224Po is homeomorphic to the Mobius stripof Section5.4.Exercise5.13.Show that the \"hyperbolic square\" of Figure 5.18 reallyhas the symmetries of a square, in the sense that there exists an isometry

of \320\2302which sends 0 to 1, 1 to oo, oo to \342\200\2241and \342\200\2241to 0.

Exercise 5.14. Let X be the hyperbolic polygon of Figure 5.18, namely,

the infinite square in H2 with vertices at infinity 0, 1, oo and \342\200\2241.We

will glue its edges in a different way from the construction of Section 5.5.

Namely, we glue the edge loo to the edge 01 by the map \316\266\342\226\240-+ -, and\342\200\224\316\266+ 2

\"1 - -

the edge (-l)oo to the edge 0(-l) \320\254\321\203\320\263\320\275\302\273-. Let (\316\247,\316\254\317\207)be the

quotient metric space provided by this gluing construction.


a. Show that (\316\247,\316\254\317\207)is locally isometric to (H2,dhyp).b. Show that X can be decomposed as X = Xo U U\\ U \302\243/2U \302\243/3where

each (Ui, \316\254\317\207)is isometric to a subset (S\342\200\236,dsa) of the pseudosphere asin Proposiiton 5.9, and where Xo is the image of a bounded subset

Xo \320\241X under the quotient map X \342\200\224\302\273\342\226\240X. Hint: Adapt the analysis of

Section 5.5.

c. Show that (\316\247,\316\254\317\207)is homeomorphic to the complement of three points

in the sphere S2.

Exercise 5.15 (Hyperbolic triangles). Let \316\261,\316\262and 7 be three positivenumbers with \316\261+\316\262+f< \317\200.We want to show that there exists a hyperbolictriangle in H2 whose angles are equalto \316\261,\316\262and 7, and that this triangle is

uniqueuptoisometryofH2. For this, we just adapt the proofof Lemma 5.2.

Let g be the completehyperbolic geodesic of H2 with endpoints 0 and00. Let h be the complete hyperbolic geodesic passingthrough the pointi and such that the angle from g to h at i, measured counterclockwise, is

equal to +\316\262.For \321\203< 1, let ky be the complete geodesic passing through

\\y and such that the counterclockwiseangle from g to ky at \\y is equal to

\342\200\2247.Compare Figure 5.9.

a. Show that the set of those \321\203< 1 for which ky meets h is an interval

(2/0,1).

b. When \321\203is in the above interval (1/0,1), let ay be the counterclockwiseanglefrom h to ky at their intersection point. Show that

lim ay = \317\200\342\200\224\316\262

\342\200\224~f and lim ay = 0.

\321\203^*~ v-*vt

Conclude that there exists a value \321\203\342\202\254(yo, 1) for which ay = a.

\321\201Show that the above \321\203\342\202\254(yo, 1) with ay = a is unique.

d. Let \316\244and T' be two hyperbolic triangles with the same angles \316\261,\316\262and

7. Show that there is an isometry \317\210of the hyperbolic plane (\320\2302,dhyp)

such that \317\210(\316\244)= \316\2441.Hint: For g and h as above, apply suitable

hyperbolic isometries to send \316\244and T' to triangles with one edge in gand another edge in h, and use part \321\201

Exercise 5.16 (The Gauss-Bonnet formula). LetX bea bounded polygon

in K2, H2 or S2 consisting of finitely many disjoint convex polygons Xi,X2, . \302\267\302\267,Xm- Let X be obtained by gluing together pairs of edges of X.As usual we assume that for every vertex \316\241of X, the_ angles of X atthe vertices that are glued to \316\241add up to 2\317\200,so that X is a euclidean,hyperbolic or spherical surface.

The quotient space X is now decomposed into m images of the convexpolygons Xj, \316\267images of the In edges of X, and \317\201points images of the


vertices of X. The Euler characteristic ofX is the integer

\317\207(\316\247)= m \342\200\224n + p.

A deep result, which we cannot prove here, asserts that the Euler

characteristic \317\207(\316\247)is independent of the way the space X is obtained by gluing

edges of a polygon in the sense that if two such surfaces X and X' arehomeomorphic, they have the same Euler characteristic \317\207(\316\247)

= x(X')-

See, for instance, [Massey, Chap.IX, \302\2474]or [Hatcher2, \302\2472.2]

a. Compute the Euler characteristic of the torus, the Klein bottle, thesurface of genus 2, and the projective plane.

b. Show that for each of the convex polygons Xi,

\316\2430,=

(\321\211-

2)\317\200+ KArea,(Xi),3=1

where: Xi has n* vertices, and \316\270\\,#2, \302\267\302\267\302\267,0Ui are the angles of Xi at

these vertices; if Xi \320\241\320\2322is a euclidean polygon, Area(Xj) denotesits

usual area and \320\232= 0; if Xi \320\241\320\2302is a hyperbolic polygon, Area(Xj)denotesits hyperbolic area Areahyp(Xj) as defined in Exercise 2.14, and\320\232=

\342\200\2241;if Xj \320\241S2 is a spherical polygon, Area(Xj) denotes its surfacearea in S2 and \320\232= +1. Hint: Use the convexity of Xi to decomposeit into (n.j

\342\200\2242) triangles, and apply the results of Exercises2.15and

3.6 to these triangles.c. Usepart b to show that

2\317\200\317\207(\316\247)=

\320\232\320\220\321\202\320\265\320\260(\320\245).

In particular, if a surface X is obtained by gluing together the sides ofa euclidean,hyperbolic, or spherical polygon X as above,its Euler

characteristic \317\207(\316\247)must be zero, negative or positive, respectively.The equation

of part \321\201is a special case of a more general statement known as the Ganes-Bonnet formula.

Chapter 6

Tessellations

We are all very familiar with the tiling of a kitchen floor. The floor

is divided into polygons (the tiles) with disjoint interiors. When onlyone type of tile is used, any two such polygons are isometricin thesensethat there is an isometry of the euclideanplane sending the first

polygons to the secondone.The standard patterns used to tile a kitchen floor can usually be

extended to providea tiling of the whole euclidean plane. In this case,there is a morehighfalutin word for tiling, namely \"tessellation\". Inthis chapter, we will see how the edge gluings of euclidean orhyperbolic polygons that we considered in Chapters 4 and 5 can be usedtoobtain interesting tessellations of the euclidean plane, the hyperbolicplane or the sphere.

6.1. Tessellations

We first give a formal definition of tessellations.

A tessellation of the euclidean plane, the hyperbolicplane or

the sphere is a family of tiles Xn, \316\267\342\202\254\316\235,such that

(1) each tile Xm is a connected polygon in the euclidean plane,

the hyperbolic plane or the sphere;

(2) any two Xm, Xn are isometric;

133

134 6. Tessellations

(3) the Xm cover the whole euclideanplane,hyperbolic plane or i

sphere, in the sense that their union is equal to this space;

(4) the intersection of any two distinct Xm, Xn consists only ofvertices and edgesof Xm, which are also vertices and edgesof Xn\\

(5) (Local Finiteness) for every point \316\241in the plane, there existsan \316\265such that the ball of radius \316\265centered at \316\241meets only

finitely many tiles Xn.

Figure 6.1 provides an example of a tessellation of the euclideanplaneby isometric hexagons.

Figure 6.1. A tessellation of the euclidean plane (R2,deuc) by hexagons

Exactly like a kitchen can be tiled with tiles with different shapes,one can slightly extend the definition of a tessellation to allow several

types of tiles. Namely, condition (2) can be relaxedby asking only

that every tile Xn is isometric to one of finitely many model polygons

\316\244\316\271,\316\2442,..., Tp. We will restrict our discussion to the original

definition, but will later indicate how to extend the arguments to this more

general context.

6.2. Complete metric spaces

In Section 6.3, we will use edge gluings of polygons to construct

tessellations of the euclidean plane, the hyperbolic planeor the sphere.

6.3. From gluing polygon edges to tessellations 135

This construction will use a new ingredient, the completeness of a

metric space.

In a metric space(X,d), define the length of a sequence Pi, P2,..., Pn, ... as the infinite sum

\316\243\342\204\242=1d(Pn,Pn+i). This length maybe finite or infinite.

The metric space (X, d) is completeif every sequence of points

Pi, P2, \302\267\342\226\240\302\267,Pn, \302\267\302\267\302\267in X whose length is finite converges to some

Poo eX.

In another mathematics course you may have encountered adifferent definition of complete metric spaces involving Cauchy sequences.

Exercise 6.1 shows that the two definitions are equivalent. The aboveformulation has the advantage of being more geometric because of itsanalogy with the length of a discrete walk. Intuitively, a metric space

(X, d) is completeif one cannot escape from it by walking a finite

distance. Indeed, a finite length walk is the same as a finite length

sequence, and such a sequence (\316\241\316\267)\316\267\316\265\316\235must converge to some pointPooG X if the space is complete.

In Section6.4,we will establish several general properties which

can be used to prove that a metric spaceis complete.However, we

first see how this property is relevant for constructing tessellations of

the euclidean and hyperbolicplanes.

6.3. From gluingpolygonedges to tessellations

Let X be a polygonin the euclideanplane (R2, deuc), in the hyperbolicplane (H2,dhyP),or in the sphere (S2,d8ph) with edges E\\, E2, \302\267\302\267\302\267,

E2p. As in Chapter 4, group theseedgesinto pairs {E2k-i, #2fc} and

specify isometries </?2fc-i: E2k-\\ \342\200\224>E2k and </?2fc=

\317\2102^-\316\271'\342\226\240\320\2252\320\272\342\200\224>

E2k-i\302\267 As in Sections 4.4 and 4.5, extendeach \317\210\316\257to an isometry of

the euclidean plane,the hyperbolic plane or the sphere in such a way

that\317\210\316\257(\316\247)is on the side of the edge(fi(Ei)that is opposite to X.

Let \316\223consist of all isometries \317\210of (R2, deuc), (H2 dhyp)or (S2, deph)

that can be written as a composition

\317\210=

\317\210\316\257\316\271\316\277\317\210\316\257\316\271_1\320\276\302\267\302\267\302\267\316\277

\317\210\316\257\316\271


of finitely many such gluing maps. By convention this includestheidentity map, which can be written as a compositionof 0 gluing maps.

We will refer to \316\223as the tiling group associated to the polygonX and to the edge gluing isometries \317\210\316\257\302\267.Ei \342\200\224\302\273Ei\302\261\\. This is our first

encounter with a transformation group, a notion which will be further

investigated in Chapter 7.

As usual, endow X with the metric \316\254\317\207for which \316\254\317\207(\316\241,Q) is the

infimum of the euclidean,hyperbolic or spherical length of all curvesjoining \316\241to Q in X. Let (\316\247,\316\254\317\207)be the quotient metric space of

(\316\247,\316\254\317\207)defined by the above gluing data.

Theorem 6.1 (Tessellation Theorem). Let X be a euclidean,

hyperbolic or spherical connected polygon with gluing data as above. In

addition, supposethat for each vertex ofX, the angles of X at allvertices that are glued to \316\241add up to

^\302\243for some integer \316\267> 0

depending on the vertex; in particular, by Theorems 4.4 or 4.10, the

quotient space (\316\247,\316\254\317\207)is a euclidean, hyperbolic or sphericalsurface with cone singularities. Finally, assume that the quotient space

(\316\247,\316\254\317\207)is complete.

Then, as \317\206ranges over all elements of the tiling group \316\223,the

family of polygons \317\210(\316\247)forms a tessellation of the euclidean plane,the hyperbolic plane or the sphere.

The proof will take a while.

By construction, the tiles \317\210(\316\247)are all isometric. Therefore, theissue is to show that they cover the whole plane and that they have

disjoint interiors.

Our strategy will be that of the tile-layer, beginning at X and

progressively setting one tile after the other. In this approach, there

are two potential snags that need to be ruled out. The first one is

that the tiles \317\206(\316\247)might not necessarily cover the wholeplane.Thesecondoneis that, as we are progressively laying the tiles beginningat X, two distinct tiles might end up overlappingas in the tile-layer's

nightmare of Figure 6.2 (well known to weekend do-it-yourselfers,

including the author). Ruling out these potential problemswill use the

completeness of (X, a) in a crucial way; see in particular Lemma 6.4.

6.3. Fromgluing polygon edges to tessellations 137

Figure 6.2. The tile-layer's nightmare

Before we proceed, it is convenient to establish some terminology.If the points \316\241and Q of the boundary of X are glued together, then,

by definition of the gluing process, Q =v?\302\273i

\302\260\302\245>*i-i

\302\260\"\"\"\302\260\302\245>\302\273\316\271(\316\241)where

the\317\210\316\257$

are restricted by the condition that\317\210\316\271$_\316\271

\320\276\302\267\302\267\302\267oy>fl(\316\241)belongs

to Eij for every j \316\266I. In this situation, we observed in the proof of

Theorem 4.4 that the tiles\317\210~^

\316\277\321\203\302\273\"1

\316\277\302\267\302\267\302\267\316\277\317\2107\316\271{\316\247)fit nicely side-by-

side around the point P. We will say that these tiles are adjacenttoX at the point P.

More precisely, a tile \317\206(\316\247)is adjacent to X at \316\241if there exists a

sequence of gluing maps y>fl, \317\206\316\2572,..., ipit such that\317\206^^

\316\277\302\267\302\267\302\267\316\277\317\210^(\316\241)

belongs to\316\225^

for every j < I (including the fact that \316\241\302\243E^) and

\317\210=\317\21071\316\277\317\206\316\244^\316\277-'-\316\277\317\206-1.

Note that \317\206is an element of the tiling group \316\223,since each \317\206^1is

equal to some gluing map </?\302\273^\302\2611\302\267However, the condition that\317\210\316\2571_1

\316\277

\342\200\242\302\267\302\267\302\260\317\210\316\220!(\316\241)belongs to E^ is quite restrictive, so that not every tile

\317\210(\316\247)with \321\204\342\202\254\316\223is adjacent to X.

By convention, we allow the family of gluing maps \317\210^to be

empty, in which case \317\206is the identity map. In particular, the tile Xis always adjacent to itself at \316\241

More generally, we will say that the tiles \317\206(\316\247)and \317\210(\316\247)are

adjacent at the point \316\241\342\202\254\317\210(\316\247)\316\240\317\210(\316\247)if \317\210-1\316\277

\317\210(\316\247)is adjacent to

X at \321\204~1(\320\240),in the above sense.

Lemma 6.2. There are only finitely many Hies \317\210(\316\247)that are

adjacent to X at the point P.


Proof.If\316\241is in the interior of X, then X is the only tile adjacent toX at P. If \316\241is in an edge Ei but is not a vertex, then Ei is the only

edge containing \316\241and \317\206{(\316\225{)=

Ei\302\261\\is the only edge containing theunique point \317\210\316\257(\316\241)that is glued to P\\ it follows that X and

\317\206\317\212\317\207(\316\247)

are the only tiles adjacent to X at P.Consequently, we only need to focus on the case where \316\241is a

vertex. In this case, the proof will use in a critical way the hypothesis

that the angles of X at \316\241and at all the points that are glued to \316\241

add up to an angle sum of the form 2\302\243,for some integer \316\267> 0. When

\316\267= 1, so that the angle sum is equal to 2\317\200,the result is a propertythat we have already encountered in the proof of Theorem 4.4. Thegeneralargument is just a mild extension of this case.

As in the end of Section 4.3.1 when considering a sequence of

gluing maps <Pi1,<Pia,...,<Pii,... such that Pj =<Pij_1o\302\267\302\267-\316\277\317\206^(\316\241)e Ei}

for every j < I, we can always require that the edge Ei} isdifferentfrom the image \316\275^-\316\271\316\257-^-\316\271))

since otherwise\317\210\316\257\316\257_1

and\317\210

are

the inverse of each other and cancelout in the composition.As a

consequence, the sequence of indices ii, %2, ..., ij, ... is uniquely

determined once we have chosen an edgeE^ containing \316\241= Pi, byinduction and because ij is the unique index for which the edge Ei}contains

Pi:l=

\317\210\316\257}_\317\207\316\277\302\267\302\267\302\267\320\276

<\321\200\321\214(\320\240)and is not

\317\206^.^\316\225^^).

Consider the sequence of gluing maps \317\210^,\317\206\316\2572,..., \317\206^,... as

above, associated to the choiceof the edge E^ containing P. As in

the end of Section 4.3.1, the sequenceof points Pj eventually returns

to \316\241= Pi after visiting all the points of P, and there exists a number\320\272^ 1 such that Pj+k = Pj and ij+k = ij for every j. In particular,the set of points that are glued to \316\241is \316\241=

{\316\241\317\207,\316\2412,...,P^}.

As in the proof of Theorem 4.4,choose\316\265small enough that each

ball Bdx(Pj,e) in X is a disk sector for each j = 1,2, ..., \320\272.For

every j' ^ 1, consider the euclidean,hyperbolic or spherical isometry

'\320\244\320\267=

^i\302\260

\316\250\316\2522\302\260'\"' \302\260

f7j1.1 (with the convention that \317\210\316\271is the

identity map). Still as in the proof of Theorem 4.4, the disksectorsil>j(Bdx{Pj,e))fit side-by-side around the vertex \316\241= Pi. However,

as we return to Pfc+i= Pi, the disk sector ^k+i(Bdx(Pi,\316\265)) is not

equal to Bdx(Pi,\316\265) any more and is instead obtained by rotating thisdisksectoraround Pi by an angle of 2e.


If we keep going, %l>2k+\\{Bdx{P\\,\302\243))is obtained by rotating

Bdx(P1,e) by an angle of 2^, ip3k+i(Bdx(Pi,s)) by an angle of 3^,

etc..., so that eventually we reach \321\204\320\277\320\272+\316\271(Bdx (Pi,\316\265))= Bdx (Pi, \316\265).

The isometry \321\204\320\277\320\272+\\fixes the point Pi and the two geodesies

delimiting the disk sector \316\222^(\316\241\316\271,\316\265).It follows that \321\204\320\277\320\272+\316\271is the identity.

(Use, for instance, Lemma 2.6to prove this in the hyperbolic case).As a consequence,\321\2043-+\320\277\320\272

=\316\250\316\257for every j ^ 1, and there are only

finitely many (in fact exactly nk) tiles ipj(X).Similarly if, instead of E^, we had selectedthe other edgeE^

adjacentto P, the resulting family of gluing maps \317\206?, \317\206?2,..., \316\275\316\275,\316\271

... and maps \317\210^,=

\317\206\316\244,1\316\277

\317\206\316\244,1\316\277\342\226\240\302\267\302\267\316\277

\317\206~* is such that\321\204'\321\203+\320\277\320\272

=\320\244'\321\203

for every j' ^ 1. So this other choicealsogives only finitely manytiles <p'j/(X) adjacent to X at P.

Since E^ and \320\225#are the only edges containing the vertex P, this

proves that there are only finitely many tiles adjacent to X at P. D

We can make Lemma6.2a little moreprecise.When \316\241is a vertex

of X and with the notation of the above proof, note that Ei> is the

edge containing \316\241=\320\240\320\263

= Pk+i that is different from E^ =Ek+i-

Therefore, E^ is the image of the gluing map ipils: Eik \342\200\224></?\302\273*\342\204\226*)

=

Ei>. As a consequence, \317\206\316\275is the inverse of ipik. Similarly, E^ is

the edge containing \317\206\316\275(\316\241)=

tp^iPk+i)= Pt that is different from

\317\210\316\257^(E^)= E{k so that the gluing map \317\206^

is equal to the inverse of

4>ik_1. Iterating this argument, we conclude that</?\302\273',

=\317\206\316\244\\.,

for

every j' with 1 < j' < k.Using the property that ij+k =

i3-and

ij,+fc= %'\302\267,for every j,

f ^ 1, one concludesthat \317\206\316\275t=

\317\210\316\2441_.,for every j' with 1 < f <

nk. As a consequence,

*j>=

Vv/\302\260

\316\250\316\275*\302\260\302\267\302\267\302\267\302\260

\316\250^_\302\261

= \317\206\316\244\302\260\317\2107\316\277\302\267\302\267\302\267\316\277w7

=\320\244\320\277\320\272-j'+2


for every / with 1 < / < nk, since^o^o- \302\267-o^J;

=\321\204\320\277\320\272+\\

=\317\206\316\271

is the identity map.

The point of this is that we do not need to considerthe tiles

\342\226\240\321\204'\321\203(\320\245),since they were already included among the tilesipj(X).

In particular, this proves the following.

Complement 6.3. Under the hypotheses of Theorem 6.1, for everypoint \316\241\302\243X, there is a small \316\265such that the tiles \317\210(\316\247)that are

adjacent to X at \316\241decompose the disk \316\222^(\316\241,\316\265)\320\241R2, \316\2272or S2

into finitely many euclidean or hyperbolic disk sectors Bd(P,e) with

disjoint interiors, whered =deuC, ^hyp or dsph according to whether

X is a euclidean, hyperbolic or spherical polygon. \320\236

We can rephrase Complement 6.3 by saying that the tiles \317\206(\316\247)

that are adjacent to X at \316\241fit nicely side-by-side near P. However,atthis point, we have no guarantee that they do not overlap away from

P. This is particularly conceivableif X has a funny shape instead ofbeingconvex as in most pictures that we have seenso far. We will

need the full force of all the hypotheses of Theorem 6.1 to rule outthis possibility. Similarly, it will turn out that the adjacent tiles arethe only ones meeting X, which is far from obvious at this point.

6.3.1. Hyperbolictilings.After these preliminary observations,

we now begin the proof of Theorem 6.1. Having to systematicallywrite \"the euclidean plane, the hyperbolic plane or the sphere\"

becomes clumsy after a while. Consequently, we will restrict our

attention to the less familiar case, for practice, and assumethat the plane

considered is the hyperbolic plane (lP,dhyp)\302\267All the arguments will

almost automatically extend to the euclideanand spherical context.

See Sections 6.3.2 and 6.3.3.Our first goal is to show that every \316\241\342\202\254\316\2272is covered by a tile

\317\206(\316\247)with \317\210\302\243\320\223.For this, pick a base point P0 in the interior of X,consider another point \316\241\342\202\254\320\250\320\2232,and let g be the hyperbolic geodesicjoining Pq to P. We will progressively set tiles over the geodesicg.

Look at the first point P\\ where g leaves X. At Pi, g enters one

of the finitely many tiles that are adjacent to X at P\\. Let \317\210\316\271(\316\247)

be this tile. Then let P2 be the first point where g leaves \317\210\316\271(\316\247)and


enters a tile ip2(X)\302\267 Repeating this process, one inductively definesa sequenceof points Pn \342\202\254g and of tiles \317\210\316\267{\316\247)such that g leaves

\317\210\316\267-\316\271(\316\247)at Pn to enter \317\210\316\267{\316\247),and \317\210\316\267(\316\247)is adjacent to \317\206\316\267-\317\207(\316\247)

at Pn. See Figure 6.3.

Note that the tile ipn(S) is not always uniquely determined. This

happens when, right after passing Pn, the geodesic g followsan edgeseparating two of the tiles that are adjacent to \317\210\316\267\316\271(\316\247)at Pn.

This process will terminate when the geodesic g enters a tile

\317\210\316\267(\316\247)and never leaves it, namely, when the point \316\241is in the tile

\317\210\316\267(\316\247)as requested.

Figure 6.3. Laying tiles over a geodesic

Lemma 6.4. The above tiling process must terminate after finitely

many steps.

Proof. The hypothesis that the quotient space (\316\247,\316\254\317\207)is complete

is here crucial.

Suppose that the tiling process continues forever and provides aninfinite sequence of points Pn and tiles \317\206\316\267(\316\247).

By construction, Pn is in both the tiles \317\210\316\267(\316\247)and \317\210\316\267-\316\271(\316\247).

The two points ip^iPn) and \316\250\316\267-\316\271(\316\241\316\267)\342\202\254X are glued together and

consequentlyprojectto the same point Pn in the quotient space X.

Becausethe quotient map X \342\200\224>X is distance nonincreasing byLemma 4.2,

dx(Pn,Pn+i) < \316\254\317\207(\317\206-1(\316\241\316\267),\317\210-1(\316\241\316\267+1)).

In addition, the two points Pn and Pn+i are joined in \317\206\316\267{\316\247)by a

geodesic curve contained in the geodesicg. Therefore,

\316\254\317\207(\317\210-\\\316\241\316\267),\317\206-\\\316\241\316\267+\316\273))=

\320\260\321\212\321\203\321\200(<\321\204-\\\320\240\320\277),\321\204-\\\320\240\320\277+1)),


which is alsoequal to dbyp(Pn, Pn+i) since \317\210\316\267is a hyperbolic isometry.As a consequence,\316\254\317\207(\316\241\316\267,\316\241\316\267+\316\271)< dhyp(Pn,Pn+i).

The sequence (Pn)n\342\202\254Nhas length

\316\243dx(Pn,Pn+1) sj ^2dhyp(Pn,Pn+1) sj 4yp(ff)< oo

n=l n=l

and thereforeconvergesto someP\302\273\342\202\254X in the complete metric space(X,dx).

Let P^,, P\302\243,,..., P\302\243,\342\202\254X be the points of \316\241\316\214\316\277,namely, the

points of the polygon X that are glued together to form the pointPoo in the quotient space X. By Lemma 4.5, for \316\265> 0 sufficiently

small, the ball Bjx (\316\241^,\316\265)in X is exactly the image of the union of

all the balls Bdx(P^,\316\265) in X. In addition, the balls Bdx(Pl0,e)aredisjoint for \316\265sufficiently small.

Fix such an \316\265.Since (Pn)n\342\202\254Nconverges to \320\240\320\266,there exists

an ra0 such that \316\254\317\207(\316\241\316\267,\316\241^) < \316\265/2for every \320\263\320\260> rao. By

convergence of the sum ]C^Li ^\321\214\320\243\320\240(\320\240\320\277,Pn+i), we can also choose rao so that

dhyP(Pn, Pn+i) < \316\265/2for every ra > ra0.

Since ^(^,\316\241\316\277\316\277)< \316\265/2,the point \317\210^\316\257\316\241\316\267)\342\202\254X must be in someball Bd(P\302\243s > \316\265/2)by another application of Lemma 4.5. Usingtheproperty that

\316\254\317\207{\317\210-\\\316\241\316\267),\317\210-\\\316\241\316\267+1))=

dhyp(Pn,Pn+i) < \316\265/2,

we observe that \342\226\240\317\206~\316\273(\316\241\316\267+\\)is at distance < \316\265from P%\302\243in X. The

gluing map Vn+i\302\260^n, which sends ^(Pn+i) \342\202\254X to Vn+i^n+i) \342\202\254

X, must therefore send P^ to somepoint \316\241\342\202\254P^ and which is atdistance< \316\265from Vv+i^n+i)\302\267 By choice of \316\265,\316\241\302\243?+1is the only

such \316\241\342\202\254\316\241\302\273,so that \317\210\316\267{\316\241&)=

\317\210\316\267+\316\271(\316\241\302\243\316\266+1)-In particular, the

tiles \317\210\316\267(\316\247)and \317\210\316\267+\316\271(\316\247)are also adjacent at the point \317\210\316\267(\316\241\302\243\316\266)=

\316\250\316\267+\316\220(\316\241&+1).

If we set Poo =ipno(Px\302\260), this proves that \317\206\316\267{\316\241^)

\342\200\224Poo for

every ra ^ rao and that the tiles \317\210\316\267(\316\247)are all adjacent to \317\210\316\267\316\277(\316\247)at

Poo- In particular, there are only finitely many such tiles.

By construction, either there is an edge\316\225of the tile \317\210\316\267{\316\247)such

that Pn is the unique intersection point of \316\225with the geodesic g, or

g coincides with an edge E' of \317\210\316\267(\316\247)for a while and Pn is one of


the vertices of \317\210\316\267(\316\247).Since there are only finitely many tiles \317\210\316\267(\316\247)

that are adjacent to\317\210\316\267\316\277(\316\247)

at P^ and since each has only finitely

many edges and vertices, we concludethat there are only finitely

many points Pn on g.But this contradicts our original assumption. Therefore, the tiling

process of the geodesic g must terminate after finitely many steps. D

An immediate corollary of Lemma 6.4 is that every point PsH2

is contained in at least one tile.

More precisely, if \317\210\316\267(\316\247)is the last tile needed to cover the

geodesic g joining Pq to \316\241as in Lemma 6.4, we will say that a tile \317\206(\316\247)

is a canonical tile for the point \316\241(with respect to the base pointPo) if it is adjacent to

\317\210\316\267(\316\247)at P. In particular, if \316\241is contained

in the interior of \317\210\316\267(\316\247),then \317\210\316\267(\316\247)is the only canonical tile forP. (This requires a little thought when g follows an edgeseparatingtwo tiles, in which case there are several possibilities for the tiles X,\316\250\316\271(\316\247),ih(X), \342\226\240\342\226\240\302\267,\317\210\316\267{\316\247)covering g.)

Lemma 6.5. For every \316\241\342\202\254\316\2272,there exists an \316\265> 0 such that foreveryP' \342\202\254Bdhyp (\316\241,\316\265)\320\241\320\250\320\240the canonical tiles for P' are exactly the

canonical tiles for \316\241that contain P'.

Proof. Let us use the notation of the proof of Lemma 6.4. Inparticular, let \317\210\316\267(\316\247)be the last tile in the coveringprocess,namely, the

one containing the endpoint \316\241of the geodesic g.

Let 7 be the finite collection of tiles consisting of X, of all the

tiles \317\210\316\257(\316\247)and, in addition, of any tile that is adjacentto some\317\210\316\257(\316\247)

at Pi.

A case-by-case analysis shows that for every Q in the geodesicgthere is a small ball Bdhyp (\316\241,\316\265)in H2 which is contained in the union

of the tiles \316\245\302\2437. Indeed, there are three types of suchpoints Q \342\202\254g:

those in the interior of sometile ipi(X),the points Pi, and the pointswhere g followsan edgeof \317\210\316\257(\316\247).

As one slightly moves the point \316\241to a nearby point P', thegeodesicg moves to the geodesic g' joining Pq to P' which stays very

close to g. If we investigate what happens near the points Pi, we seethat the tiling process of g' will only involve tiles of 7, and that the


final tile will be adjacent to \317\210\316\267(\316\247)at P. The result easily follows.SeeFigure6.3. D

Lemma 6.6. Let \316\241and Q be in the interior of the same tile \317\206(\316\247)-

If \317\210(\316\247)is canonical for P, it is also canonical for Q. In addition,

\317\210(\316\247)is the only canonical tile for \316\241and Q.

Proof. This is a relatively simple consequence of Lemma 6.5.

Because of our assumption that X is connected, there is a curve

in \317\210(\316\247),parametrized by \316\257\321\214->z(t), a < \316\257< b, and joining \316\241=

z(a)

to Q = z(b). After pushing it in the interior of \317\210(\316\247),we can even

assume that this curve is completely contained in this interior. We

can then consider

\316\257\316\277= sup{i; \317\206(\316\247)is a canonical tile for z(t)}.

By definition of i0, there are points z(t) arbitrarily close to z(to)such that the tile \317\210(\316\247)is canonical for z(t). By Lemma 6.5,it follows

that \317\210(\316\247)is canonical for z(to). Because z(tQ) is in the interior of

\317\210(\316\247),this tile is the only canonical tile for z(i0) by definition of

canonicity. Another application of Lemma 6.5 then shows that \317\210(\316\247)

is canonical for z(t) for every t sufficiently close to to. If \316\257\316\277was less

than b, this would contradict the definition of \316\257\316\277as a supremum.

Therefore, \316\257\316\277= b, and \317\210(\316\247)is canonical for z(t0) = z(b) = Q, as

requested.Since \316\241and Q are in the interior of \317\210(\316\247),no other tile can be

canonical for them. D

Lemma 6.7. Every tile \317\210(\316\247)is canonical for some \316\241in its interior.

Proof. Let us first prove that if \317\210(\316\247)is canonical for some \316\241in

its interior, then for every gluing map \317\210\316\271the tile \317\210\316\277

\317\210\316\257(\316\247)is also

canonical for some P' in its own interior. Later, we will show how

this proves Lemma 6.7.Thetiles \317\206(\316\247)and \317\206\316\277\317\206^\316\247)meet along the edge \317\206\316\277\317\206^\316\225\316\271).Let

Q be a point of this edge,not a vertex.Let P\" be a point of the interior of \317\210(\316\247)that is close to Q. By

Lemma 6.6, \317\210(\316\247)is the unique canonical tile for P\". ChoosingP\"sufficiently close to Q and applying Lemma 6.5, we concludethat


\317\210(\316\247)is also canonical for Q. Therefore, every tile that is canonicalfor Q is adjacentto \317\210(\316\247)at Q, namely, it is either \317\210(\316\247)or \317\210\316\277\317\210^\316\247).

If P' is a point of the interior of \317\206\316\277\317\210\316\257(\316\247)that is sufficiently close

to Q, Lemma 6.5 again shows that each canonical tile for P' will be

either \317\210(\316\247)or \317\206\316\277

\317\206\316\271(\316\247).This canonical tile cannot be \317\210(\316\247)since

P' is not in \317\210(\316\247).Therefore, \317\210\316\277

\317\210\316\257(\316\247)is canonical for P'.

This concludesthe proof that if \317\206(\316\247)is canonical for some \316\241in

its interior, then \317\206\316\277\317\206\316\271{\316\247)is canonical for some P' in its interior. Bydefinition of \316\223,every \317\206\342\202\254\316\223is obtained from the identity bysuccessivemultiplication on the right by gluing maps \317\210\316\257.By an induction

starting with the fact that X is canonical for Pq, it follows that everytile \317\210(\316\247)is canonical for some \316\241\342\202\254\316\2272. D

Proof of Theorem 6.1. Putting all these steps together, we cannow conclude the proof of Theorem 6.1, namely that the tiles \317\210(\316\247)

with \317\206\342\202\254\316\223form a tessellation of the hyperbolic plane H2. We have

to check that:

(1) each tile \317\206(\316\247)is a polygon, and any two tiles \317\206(\316\247)and

\317\210'(\316\247)are isometric;

(2) the plane \320\250\320\2232is equal to the union of the tiles \317\210(\316\247);

(3) any two distinct tiles \317\210(\316\247)and \317\210'(\316\247)have disjoint interiors;

(4) every point \316\241is the center of a ballBdhyp (\316\241,\316\265)which meets

only finitely many tiles \317\206(\316\247).

The first condition is trivial. '

Lemma 6.4 shows that every point \316\241\342\202\254\316\2272is contained in a tile

\317\210(\316\247).This proves (2).

If the interiors of the tiles \317\206(\316\247)and \317\206'(\316\247)contain the same point

P, the combinations of Lemmas 6.6 and 6.7 show that \317\210(\316\247)and

\317\206'(\316\247)each are the unique canonical tile for P. Therefore, \317\206(\316\247)=

\317\210'(\316\247).This proves (3).

Finally, a point \316\241\342\202\254\316\2272has only finitely many canonical tiles,and there exists a ball

Bdhyp (\316\241,\316\265)contained in the union of thesecanonical tiles. No other tile \317\210(\316\247)can meet this ball; indeed, itsinterior would otherwise meet the interior of one of the canonical tiles

146 6. Tessellations ]

of \316\241,contradicting (3). Therefore, the ball Bdhyp(P,e)meets only ]

finitely many tiles, namely, the tiles that are canonical for P. \320\236j

6.3.2. Euclidean tilings. The proof of Theorem 6.1 in the eu- j

clidean setup is identical to that for the hyperbolicplane. j

6.3.3. Spherical tilings. When X is a polygon in the sphere S2, j

the only difference is that the shortest geodesicg from P0 to \316\241may

not be unique. We need to prove that the set of canonical tiles for \316\241'

is independent of the choice of this geodesic g. Several such shortest \\

geodesies exist only when \316\241is the antipodal point \342\200\224P0 of Po. In this ;

case, any two geodesiesg and g' connecting Po to \316\241=\342\200\224Pqare great j

semi-circles and can be moved from one to the other by a rotation !

about the line OPq. We then have to checkthat the set of canonical '.

tiles for \316\241does not change as one rotates from g to g\\ which is easily ,

done by the arguments used in the proofs of Lemmas 6.5 and 6.6. \\

Once this is proved, the rest of the proof of Theorem 6.1 is identical.'

6.3.4. Slight generalizations.We briefly indicate two ways in

which the hypotheses of Theorem 6.1 can be somewhat relaxed.

Self-gluing of edges. So far, the edgesof the polygon X were paired'

by gluing maps ipf. Ei \342\200\224>Eji with ij \342\200\224i\302\261land \317\206^

=\317\206\316\244\316\271.However,

we never used the fact that these two edges Ei and E^ were distinct.I

As a consequence, we can allow the possibilitythat ji\342\200\224i. In this'

case, because we still need that \317\206-\317\207=

\317\206^=

\317\206~[, the isometry \317\206\316\271of I

Ei must be the identity or a flip reflecting Ex across its midpoint.

The casewhere \317\206\316\271is a flip can actually be reducedto the previous\342\226\240

setup, by converting the midpoint of Ei into a new vertex and splitting;Ei into two new edges.

Beware that when (/?,:\302\243\342\226\240,\342\200\224>Ei is the identity, the isometric?extension\317\210\316\220to all of R2, H2 or S2occurringin the tiling group will

be the reflectionacrossthe completegeodesiccontaining Ei because

it must send X to the oppositesideof Ei.

The statement and proof of Theorem 6.1 immediately extends to

this context.

6.4. Completeness and compactnessproperties 147

Allowing nonconnected polygons. We can also relax the hypothesisthat X is connected and allow it to be the union of finitely many

disjoint connected polygons X\\, X2, ..., Xp- Then, the tiling

process win provide a tessellation where each tile is isometricto one of

the model tiles \316\247\316\271,X2; ..., Xp. This again followsthe intuition of

practical tile setting.

More precisely,we just need to take into account some additionalinformation, namely, which of these connected polygons Xi contains

each edge Ei. We then introduce the tiling groupoid \316\223consisting

of all isometries of the form

\317\206=

\317\210\316\257\316\271\316\277\317\210\316\257^\316\277..\302\267\316\277\317\210\316\257\316\271,

where, for each \320\272= 1, 2, ..., \316\231- 1, the edge Eik is in the same

connected polygonXi as the edge'Pik_1(Eik_1)that is glued to Eik_1

by <\316\241\316\260_\316\271\302\267The need for the barbaric terminology \"groupoid\" arises

from the fact that \316\223is not a group in the sense that we will encounter

in Chapter 7; indeed, the compositionof two elements of \316\223is not

always in \316\223.

The tiles of the tessellation are then all the tilesof the form \317\210{\316\247\317\212),

where \317\206\342\202\254\316\223is decomposed as \317\210=

\317\206^\316\277

\317\210\316\257\316\271_1\320\276\302\267\302\267\302\267\316\277

\317\210\316\257\317\207as above and

where Xi is the connectedpolygon of X that contains the startingedgeEh.

The fact that these tiles form a tessellationis proved by exactly

the same arguments as Theorem 6.1.

6.4. Completeness and compactnesspropertiesTheorem 6.1, and in particular Lemma 6.4, pointsto the importance

of the completeness property for our geometric endeavors.We will

need a few criteria to guarantee that a space is complete. This sectionis devoted to results of this type.

6.4.1. The euclideanand hyperbolicplaneare complete.Theorem 6.8. The euclidean plane (R2,deuc) *s complete.

Proof. This is an immediate consequenceof the following deep

property of real numbers.


Fact 6.9. The real line R, endowed with the usual metric d(x, y) =

\\x\342\200\224

y\\, is complete. \316\217

We refer to any of the many textbooks on real analysis for a proofof Fact 6.9, which requires a deep understanding of the definition of

real numbers. Historically, real numbers were introducedprecisely

for this property to hold true. (Of course,it took many centuries to

state the property in this way).

To prove that (R2, deuc) is complete,considera sequenceof points

Pi, P2, \342\226\240\342\226\240\342\226\240,Pn, \342\226\240\342\226\240\342\226\240in R2 with finite length \316\243\316\223=\316\271deuC(Pn, \316\241\316\267+1)<

00. We want to show that the sequence (Pn)n\342\202\254Nconverges to some

point \316\241\317\207e R2.

If Pn =(\321\205\320\277,\320\243\320\277),note that

deuc\\Pn>Pn+l) ^ Fn+1 ~~xn\\

= d(Xni xn+l)

by equation (1.3). As a consequence,the sequenceof real numbers

x\\, x2, ..., xn, \302\267\302\267\302\267has length

00 00

^d(xn,Xn+1) < ]JPdeuc(-Pn,-Pn+l) <00\302\267.

n=l n=l

By completeness of (R,d) (Fact 6.9), it follows that the sequence

(xn)n\342\202\254Nconverges to some number \317\207,\317\207,in R.

The same argument shows that the sequence (yn)n\342\202\254Nconverges

to some number j/oq.

Now, consider the point P*, = (x00,\321\20300)in R2\302\267For every \316\265> 0,

by definition of the convergence of sequences, there exists a number

\317\200\316\271such that |xn\342\200\224

x^l < -% for every \316\267^ \317\204\302\273\316\271,and there exists

another number n2 such that |yn\342\200\224

3/<x>| <77s

f\302\260revery \316\267^ n2\302\267If no

is the larger of n\\ and n2, we conclude that

deuc(Pn, Poo)= \320\273/(\320\245\320\277-

Xoo)2 + (\320\243\320\277~

3/oo)2 <\\J\316\244

+ \316\244= \316\265

for every \316\267^ ra0.

This proves that the sequence {Pn)n& converges to the point

Poo in (R2,deuc)> and completesthe proof that this metric space is

complete. \316\240

6.4. Completeness and compactness properties 149

We now use the completeness of the euclideanplaneto prove that

the hyperbolic plane is also complete.This is one more property that

these two spaces have in common.

Theorem 6.10. The hyperbolicplane (H2,dhyp) is a complete metric

space.

Proof. Let Pi, P2,..., Pn, ... be a sequence of points in H2 with

finite lengthoo

L = ]JPdhyP(Pn,Pn+i)< oo.n=l

We need to show that the sequence (Pn)ne^ converges to some P*,

for the hyperbolic metric dhyP\302\267

Let 7\342\200\236be the hyperbolic geodesic arc joining Pn to Pn+i. Iterateduses of the Triangle Inequality give that for every \316\241\302\243\316\267\316\267,

n-l

dhyP(P, pi) <\316\243dbyp(ft\302\273 P\302\245+^

+ d(P\">P)fc=ln-l

<\316\243 \321\203\320\240(\320\233\321\201,Pk+l) + d(Pn, Pn+i) < L.fc=l

The estimate provided by Lemma 2.5 showsthat dhyp(P, Pi) ^| In -*-1if \316\241=

(\317\207,\321\203)and Pi =(\317\207\316\271,3/1). Combining these two inequalities, we

concludethat

Vie~L < \320\243< yieL

for every \316\241=(\317\207,\321\203)\342\202\254~fn\302\267To ease the notation, set c\\

\342\200\224yie~L and

C2= !/i eL so that the y-coordinate of \316\241is between ci and c2 for every

\316\241\342\202\2547\342\200\236.

Comparing the formulas for the euclideanand hyperboliclengths

4uc and ^hyP) we conclude that

\"hyp(-'n) *n+l)= thyp^TnJ ^ *-\320\265\\1\320\241\\\320\243\320\277)

=\302\253euc(-'nj *n+l)\302\267

c2 c2

In particular, the sum ]Cnt=i^euc(Pn>Pn+i)is finite. Since (R2,deuc)

is complete (Theorem 6.8), we conclude that Pn converges to somepoint Poo \342\202\254M2 for the metric deuc.


We showed that the y-coordinate yn of Pn is such that yn > Ci

for every n. It followsthat the y-coordinate of the limit P*, is also^ ci > 0,sothat Pqo is in the upper half-space \320\250\320\240.Comparing the

hyperbolic and euclidean lengths of the line segment [Pn, P*,] alsogives

1 1Uhypl-fni \342\226\240'ooj\316\266*\302\267hyp ll/n\302\273Px>J) ^ *\320\265\320\270\321\201([-'\320\237!Px)JJ

=\302\253euc(-'n)-\"ooj\302\267

C\\ Ci

Therefore, since the sequence (Pn)n\342\202\254Nconverges to P^ for theeuclidean metric deuc, it also converges to P^ for the hyperbolic metric

dhyp\302\267\316\240

6.4.2. Compactness properties. For metric spaces, the notion of

completeness is closely related to another property, namely,

compactness.

This compactness property is based on subsequences.Given a

sequence of points Pi, P2, ..., Pn, ... in a space X, any increasing

sequence of integers n\\ < \320\263\320\260\320\263< \302\267\302\267\302\267< \320\237\320\272< \302\267\302\267\342\226\240provides a new

sequence Pni, Pna, ..., Pnk, ... in X. This new sequence(PnJfceN

is a subsequence of the original sequence(Pn)n\342\202\254N\302\267In other words,

one goes from a sequenceto a subsequence by forgetting elements of

the original sequence,while keeping infinitely many of them so that

they still form a sequence.

A metric space (X, d) is compact if every sequence (Pn)n\342\202\254Nin

X admits a converging subsequence,namely, a subsequence (Pnfc)fc\342\202\254N

such that

lim \320\240\320\237\320\272=

\316\241\302\273\320\272\342\200\224>oo

for some Poo \342\202\254X.

The connection between compactness and completenessisprovided by the following result.

Proposition 6.11. Every compact metric space (X, d) is complete.

Proof. Let Pi, P2, ..., Pn, ... be a sequence with finite length

\316\243^=\316\271d(Pn, Pn+i) < oo. We want to show that the sequence (Pn)n\302\243N

converges.


By compactness, it admits a converging subsequence (Pnk)keN

such that

lim Pnk = Poo-k\342\200\224*oo

For an arbitrary \316\265> 0, there exists a number k0 such that

d(Pnk,Poa) < | for every \320\272^ k0 by definition of the limit of a

sequence.

Similarly, by convergence of the series \316\243)^ d(Pi,Pi+i), there

exists a number no such that Y^ln d{Pi, Pi+i) <\302\247

for every \316\267^ no-

As a consequence,the Triangle Inequality shows that

n' \342\200\2241 oo

d(Pn,Pn>) <\316\243d(pi>pi+i) <

\316\243^\342\204\226,\316\241\316\257+\316\271)< f

\320\263=\320\277 \320\263=\320\277

for every \316\267,\316\267'^ n0 with \316\267< n'.

Then, for every \316\267^ max{n0, \316\267^0}after picking any \320\272such that

\320\237\320\272\316\257\317\212\316\267,we conclude that

d(Pn,Poo) < d(Pn,Pnk) + d{Pnk,Poa)<\302\247+ \302\247=\302\243.

Therefore, for every \316\265> 0, we found an n'0= max{n0,ftfc0} such

that d(Pn, Poo) < \316\265for every \316\267^ n'0. This proves that the sequence(-Pn)neN converges to Poo. \342\226\241

The advantage of compactness over completenessis that it is often

easier to check. For instance, the following criterion is particularlyuseful.

Proposition 6.12. Let \317\206: X \342\200\224\302\273X' be a continuous map from ametric space(X,d) to a metric space {X',d'). If, in addition, X iscompact, then its image \317\210(\316\247)\320\241X' is compact for the restriction of

the metric d'.

Proof. Let P{, \320\240\320\267,..., P\342\200\236,\342\226\240\342\226\240\302\267be a sequence in \317\206(\316\247)\320\241X'. By

definition of the image, each P'n is the image of some P\342\200\236G X under

\317\210,namely, P'n =\317\206{\316\241\316\267).

Since X is compact, there is a subsequencePni, P\342\200\2362,..., \320\240\320\237\320\272,

... of the sequence (P\342\200\236)Nwhich converges to some point Poo G X-

By continuity of \317\206,\320\240'\320\237\320\272=

<\321\200{\320\240\320\237\320\272)converges to \317\206(\316\241\317\207)G \317\210{\316\247)as \320\272

tends to oo.


Therefore, for every sequence (\320\240\320\277)\320\277\320\265\321\206m

\317\210(\316\247)>we can nnd a

subsequence (Pnk)keNw^<^1 converges in \317\210{\316\247).This proves that

\317\210(\316\247)is compact. D

6.4.3. Compactness properties in the euclideanplane,thehyperbolic plane and the sphere. A subset \320\243of a metric space

(X, d) is bounded if, for an arbitrary point Pq \342\202\254X, there exists a

number \320\232such that d(P, Pq) < \320\232for every \316\241\342\202\254\316\245.This is

equivalent to the property that \316\245is contained in a large ball Bd(Po,K).Using the Triangle Inequality, one easily verifiesthat this property is

independent of the choiceof the point Pq.

A subset \316\245of the metric space (X, d) determines three types of

points in X. The interior points \316\241are those for which there existsan \316\265> 0 such that the ball Bd(P,\316\265)is completely contained in Y. An

exterior point is a point \316\241\342\202\254X such that for some \316\265> 0, the ball

Bd(P,e) is disjoint from Y. A boundary point is a point \316\241\342\202\254X

which is neither interior nor exterior,namely, such that every ball

\316\222\317\215(\316\241,\316\265)centered at \316\241contains points that are in \316\245and points that

are not in Y.

A subset \316\245of the metric space (X, d) is closedif it contains all

its boundary points.

Theorem 6.13. Every closed bounded subset of the euclidean plane(R2,deuc)is compact.

Proof. Again, the key ingredient is a deepproperty of real numbers,

whose proof can be found in most textbooks on real analysis.

Fact 6.14. In the real line R endowed with the usual metric d(x, y) =

\\x\342\200\224

y\\, every closed interval [a, b] is compact. D

Let X be a closedbounded subset of R2. To show that X iscompact,considera sequence(Pn)n\342\202\254Nof points Pn = (xn, yn) e X.

Because X is bounded, there is a closedinterval [a, b] \342\202\254R which

contains all the xn. Therefore, by Fact 6.14, there is a subsequence(xnk)k\342\202\254Nwhich converges to some number \317\207,*,\302\243[\320\276,b].

Turning our attention to the y-coordinates,the boundedness of \316\245

again implies that the \321\203\320\237\320\272are all contained in some interval [c,d]. We


can therefore extract from the subsequence (ynk)ke^ a subsequence(Vnki)ieN

which converges to some \321\203\320\266G [c,d\\.

Note that the sequence (Pnki)i\342\202\254Nis a subsequence of (Pn)neN\302\267

We now have

lim xnk,=

\317\207\317\207and lim \321\203\320\237\320\272.=

\321\203\320\266-\320\263\342\200\224>\320\276\321\201\320\263 \320\263\342\200\224\302\273\320\276\320\276\320\263

The argument concluding the proof of Theorem 6.8can again be used

to show that the subsequence(Pnfci)ieN converges to \316\241\317\207,=

(\321\217<\321\205>,\320\243\320\260\320\260)\302\267

The point Poo cannot be an exteriorpoint of X since there are

points \320\240\320\237\320\272that are arbitrarily close to it. Therefore, P*, is either an

interior point or a boundary point of X. Since X is closed, it follows

that \316\241\302\273is in X. Therefore, every sequence (Pn)n\342\202\254Nin X admits

a subsequence which converges to some point \316\241\317\207,\342\202\254X. Namely,

(X, deuc) is compact. D

Theorem 6.15.Every closed bounded subset of the hyperbolic plane(IH2,dhyp)

is compact.

Proof. Let X be a boundedsubset of the hyperbolic plane (H2, dhyp)\302\267

In particular, the subset X is contained in a large ball\320\224\320\263\320\254\321\203\321\200(\320\240\320\276!K)\302\267

Let (Pn)n\342\202\254Nbe a sequence in X, and write Pn=

(xn, yn). Since

rfhyp (Pm Po) < \320\232,the same argument as in the proof of Theorem 6.10

shows that ci < yn < c-i with c\\= yoe~K and c2 = yoeK,and that

deuc(Pn, Po) < C2dhyp(Pn, \316\241))< C2K.

In particular, the sequence (Pn) gNis contained in a large euclidean

ball Bdeuc(Po,c2K).Applying Theorem 6.13 to the closed ball with

the same radius and the same center, there consequently exists a

subsequence (Pnfc)fceN which converges to some point \316\241*,=

(\317\207,\317\207,,yoo)

in(R2,deuc).

Since ynk > c\\ for every k, the coordinate j/oo is greater than or

equal to c\\ > 0, so that \316\241\317\207,belongs to the hyperbolic plane H2.

Borrowing another argument from the proof of Theorem 6.10,

^\320\254\321\203\321\200^-'\320\277,-^ooJ ^ \302\243*euc(.Xn>-ioojCl

for every n. It follows that Pnk also converges to P\302\273for the hyperbolic

metric dhyp) as k tends to oo.


Since X is closed, it again followsthat P^ is an element of X.This proves that every sequence in X has a converging

subsequence. In other words, (X, dbyp)is compact. D

For subsets of the sphereS2,the boundedness hypothesis is

irrelevant since dsPh(-P, Q) < \317\204for every P, Q \302\243S2.

Theorem 6.16. Every closed bounded subset of the sphere (\302\2472,dsPh)

is compact. I

Proof. Let X be a closed subset of S2, and let (Pn)neN be a se- \\

quence in X. By the immediate generalization of Theorem 6.13 to i

three dimensions, the sphere S2 is compactfor the restriction of the i

3-dimensional euclidean metric deuc- Therefore, there existsa sub- :

sequence (Pnk)keN converging to some \316\241\317\207\342\202\254S2 for the euclidean i

metric deuc. !

By an elementary euclidean geometry argument comparing the j

length of a circle arc to the length of its chord, the euclidean andj

spherical distances are related by the property that \316\257

dsph(Pnk, \320\240\320\276\321\201)= 2 arcsin

deuc{Pn2*'Poo).I

It follows that dBph(Pnk,P<x>) converges to 0 as \320\272tends to oo, namelythat the subsequence (Pnk)keN converges to Poo in (S2, daph)\302\267

\316\240

6.4.4. A few convenient properties. For future reference, we

mention here several easy propertieswhich will be convenient in the

future.

Lemma 6.17. In a metricspace(X,d), let (Pn)neN be a sequencewith finite length. If there exists a subsequence(\320\240\320\237\320\272)feeN

which

converges to some point \316\241\317\207,then the whole sequence converges to \316\241\317\207.

Proof. This property was the main step in the proof of

Proposition6.11. D

Lemma 6.18. The length of a sequence (Pn)n\342\202\254Nis greater \320\233\320\260\320\277or

equal to the length of any of its subsequences (Pnk)keN-

6.5. Tessellations by bounded polygons 155

Proof. By an iterateduse of the Triangle Inequality,

d(Pnk,Pnk+1)^ \316\243 d(Pn,Pn+i).

n=nk

It follows thatoo oo

53d(Pnfc,Pnfc+1)<53d(Pn,Pn+i). Dfc=l n=l

Lemma 6.19. Suppose that the metric space (X,d) splits as the

union X =\316\247\317\207U \316\247\316\271of two subsets X\\ and \320\245\320\263-If, for the

restrictions of the metric d, the two metric spaces (Xi,d) and (X2,d) arecomplete, then (X, d) is complete.

Proof. Let (Pn)n\342\202\254Nbe a sequence with finite lengthoo

Y^d(Pn,Pn+x) <oo.n=l

We want to show that this sequence convergesto someP*, \342\202\254X.

At least one of X\\ and \320\245\320\263must contain Pn for infinitely many

\316\267\342\202\254N. Without loss of generality, let us assumethat this is X\\.

Consequently, we have a subsequence(Pnfc)feeNwhich is completely

contained in X\\.

This subsequence has finite length by Lemma 6.18.Since(X\\, d)

is complete, this implies that(Pnk)k\342\202\254Nconverges to some limit P^ \342\202\254

\316\247\317\207.By Lemma 6.17, it follows that the whole sequence (Pn)n\342\202\254N

converges to \316\241\317\207,G \316\247. \316\237

6.5. Tessellations by bounded polygons

We now apply the results of the previous sections to construct various

examples of tessellationsof the eucfidean plane, the hyperbolic planeor the sphere.

The simplest examples come from bounded polygons. Indeed,becauseof our convention that a polygon X contains all of its edges,

it is always closedin(R2, deuc), (H2, dhyp) or (S2, dsph)\302\267If, in addition,


the polygon X is bounded, it is therefore compact by Theorems 6.13,6.15or 6.16.Recall that a spherical polygon is alwaysbounded.

If (\316\247,\316\254\317\207)is the quotient metric space obtained by gluing

together the edges of X, recall from Lemma 4.2 that the quotient mapX \342\200\224>X is continuous. It follows that (X, \316\254\317\207)is compact byProposition 6.12, and therefore complete by Proposition 6.11. This proves:

Proposition 6.20. Let X be a bounded polygon in the euclideanspace(R2,cieuc), the hyperbolic plane (H2,dhyp) or the sphere (\302\2472,d9ph)\302\267

Then, if we glue together the edgesofX, the resulting quotient metric

space (\316\247,\316\254\317\207)is compact, and therefore complete. D

In particular, we can now apply Theorem 6.1 to several of theexamplesthat we considered in Chapter 5.

6.5.1. Tessellations of the euclideanplane.Let us look at the

euclidean polygon gluings that we examined in Section 5.1. All these

polygons were bounded so that we can apply Proposition 6.20 andTheorem 6.1to createtessellations of the euclidean plane.

Figure 6.4. A tessellation of the euclidean plane by squares

First, we considered the casewhere X is a square, and where weglue oppositesidesby translations. In this case, the tiling group \316\223

consists entirely of translations, and the correspondingtessellation of

the euclidean plane by squares is illustratedin Figure6.4.If we consider the same square, but glue its sideswith a twist to

obtain a Kleinbottle,the tessellation of the euclidean plane associatedto this polygon gluing is the same as for the torus. The difference here


is that the tiling group \316\223is different. In addition to translations, itcontains glidereflections, namely, compositions of a reflection with a

nontrivial translation parallel to the axisof the reflection.

Figure \316\262.5.A tessellation of the euclidean plane by parallelograms

We also considered the case where X is a parallelogram, andwhere oppositesidesare gluedby translations. Since the

parallelogramis bounded, this provides a tessellation of the euclidean plane

by parallelograms. The tiling group \316\223in this case again consists

entirely of translations, and a tessellation of this type is illustrated in

Figure 6.5.

A different, and perhapsmoreinteresting,tessellationof the plane

by parallelograms is associated to the gluing of edges of the

parallelogram that give the Klein bottle. See Exercise6.4.Finally, we saw that the torus can also be obtained by gluing

opposite sides of a hexagon.This leads to the tessellation of theeuclidean plane by hexagons that we already encountered in Figure 6.1.

6.5.2. Tessellations of the hyperbolic plane by bounded

polygons. Let us now switch to hyperbolic polygons. The hyperbolic

octagon of Section 5.2 is bounded in the hyperbolicplane. The

hyperbolic surface of genus 2 that we obtain by gluing opposite edges of

this octagon consequentlygives rise to a tessellation of the hyperbolicplane H2. This tessellation is represented in Figure 6.6.

Thetilesof this tessellation may look very different from eachother, but they are actually all isometric by isometries of thehyperbolic plane. Some of these isometries are a little more apparentin


Figure 6.6. A tessellation of the hyperbolic plane by octagons

Figure 6.7. A tessellation of the disk model by hyperbolic octagons

Figure 6.7, which represents the image of the same tessellationinthe disk model B2 for the hyperbolic plane. Note the nice rotational

symmetry. Compare Figure 5.11.

More hyperbolic rotational symmetries of the tessellation arerevealed if one transports it by an isometry of the disk model that

sends to the euclideancenterof the disk any of the points where eight

octagons meet. See Figure 6.8.


Figure 6.8. Another view of the tessellation of the disk model

by hyperbolic octagons

Figure \316\262.9.A tessellation of the euclidean plane R2 by

triangles of angles f, f, f

6.5.3. Tessellations by triangles.

Theorem 6.21. For any three integers a, b, \321\201> 2,

(1) if \316\276+ | + \316\276=

\317\200,there exists a tessellation of the euclidean

plane R2 by euclidean triangles of angles \316\276,|, \316\276;


Figure \316\262.10. A tessellation of the disk model B2 of the

hyperbolic plane by triangles of angles f, f, \321\203

(2) if \316\276+1 + \316\276< \317\200,there exists a tessellation of the hyperbolic

plane \320\2502by hyperbolic triangles of angles *, f, \316\276;

(3) if \316\276+ f + \316\276

> \317\200,there exists a tessellation of the sphereS2by spherical triangles of angles \316\276,f, \316\276.

Proof. Proposition 5-13 provides a triangle X with the angles

indicated. As gluing data, let us take for each edge the self-gluingof Section 6.3.4so that the tiling group \316\223is generated by the

reflections across the complete geodesies containing each side of the

triangle. Since the triangle X is bounded, Theorem 6.1 (generalized

6.6. Tessellations by unbounded polygons 161

as in Section 6.3.4) \320\260\320\273\320\260Proposition 6.20 then provide the requestedtessellations. D

Figure6.11.A tessellation of the sphere S2 by triangles of

angles f, f, }

Examplesof these tessellations are illustrated in Figures 6.9, 6.10and 6.11.

Note that the spherical case can only occur for {a, b, c} = {2,2,c},{2,3,3},{2,3,4}or {2,3,5}, with \321\201^ 2 arbitrary in the first case.The euclideancasearisesonly when {a,b,c}

= {2,3,6}, {2,4,4} or

{3,3,3}.

6.6. Tessellations by unboundedpolygonsThe punctured torus discussed in Section 5.5 offers an interestingchallengebecauseit is less obviously complete.

Let X be the hyperbolic square of Figure 5.18. Consider U =Uaa U U\\ U Uq U U-i as in Section 5.5, and let \316\245be the union of the

complement X \342\200\224U and of all its boundary points. Let \316\214and \316\245be

(heir respective images in the quotient space (\316\247,\316\254\317\207).

The subset \316\245is closed and bounded in the hyperbolic planeH2.[f follows that \316\245is compact by Theorem 6.15, and therefore that \316\253

is also compact by Proposition 6.12. As a consequence,(\316\253,\316\254\317\207)is

complete by Proposition 6.11.


In Proposition 5.9 and Lemma 5.12 we showed that (f/,d~x) is

isometric to the portion (Sa,dsa) of the pseudosphere S in R3. Letus show that (Sa, dsa) is complete.

Let \316\241\317\207,\316\2412,..., Pn, \302\267\302\267\302\267be a sequence of finite length in (Sa, dsa)-Recall that the distancedsa (P, Q) is defined as the infimum of the eu-

clidean lengths of all curves joining PtoQ and contained in Sa\\ it

follows that deuc(P, Q) < ds\342\200\236(P, Q) for every P, Q \342\202\254Sa. In particular,the sequence (Pn)n\342\202\254Nhas finite length in (R3, deuc),and consequentlyit converges to some point \316\241\317\207in the complete space (R3,deUc) (bythe immediate generalization of Theorem 6.8 to three dimensions).Sincethe surface Sa is defined by finitely many equations and weak

inequalities involving continuous functions, the limit \316\241\317\207is in Sa,and one easily checksthat the sequence (-Pn)neN converges to \316\241\317\207in

(Sa,dsa)\302\267

This proves that (Sa, dsa) is complete;therefore, so is the

isometricmetric space (U,d).

The space (X,d) is the union of the two subsets \316\253and f/, each;of which is complete for the metric d. Lemma 6.19 shows that this;

implies that (X, d) is complete.We will see another proof that (X, d)is completein Theorem6.25.

As before, an application of Theorem 6.1 provides a tessellationof the hyperbolic plane by tiles isometric to the hyperbolicsquare X.'

This tessellation is illustrated in Figure 6.12.

00

-3 -| -2 -| -1 -\\ 0\\

1\302\247

2 \302\247 3

Figure 6.12. A tessellation of the hyperbolic plane comingfrom the once-punctured torus

6.7. Incomplete hyperbolic surfaces 163

6.7. Incomplete hyperbolic surfacesWe now look at a few examples where the quotient space (\316\247,\316\254\317\207)is

incomplete.

6.7.1. Incomplete hyperbolic cylinders. Let us revisit the

example of Figure 5.16. Namely, X is the vertical strip delimited in the

upper half-space H2 by the two vertical half-lines E\\ and E2 of therespective equationsBe(z)= 0 and Re(z)

= 1, and we glue the two

sides by an isometry \317\210\317\207:\316\225\316\271\342\200\224>E2- As usual, extend \317\210\317\207to an isome-

try of \316\211.2that sends X to the side of E2 opposite X. An elementaryalgebraic computation shows that \317\210\\is of the form \317\210\317\207(\316\266)

= \316\261\316\266+ 1

for a real number a > 0.Since we only have two edges, the tiling group \316\223consists of all

possible compositions of \317\210\\and \317\2102=

\317\210^1\302\267As a consequence, everyclement of the tiling group is of the form \317\206\342\204\242

=\317\210\317\207\316\277

\317\2061\316\277\302\267\302\267\302\267\316\277

\317\210\316\271or

\316\250\316\252\316\267=

\317\206\342\204\242=

\317\210\317\207\316\247\316\277

\317\210\317\212\317\207\316\277\302\267\302\267\302\267\316\277

\317\206^\317\207,where \316\267^ 0 is the number of mapsoccurringin the composition.

In Section5.4.2,we considered the case where a = 1. In thiscase,an immediate computation shows that all elements of the tiling

group \316\223are horizontal translations of the form \317\206\342\204\242(\316\266)= \316\266+ \316\267with

11 G Z. In particular, each tile \317\210(\316\247)is a vertical strip of the form

{z G H2; \316\267< 1\321\202(,\320\263)< \316\267+ 1}, and it is immediate that these tilesform a tessellation of the hyperbolic plane. As in the case of the

punctured torus that we just examinedin Section 6.6, the quotient

\342\200\242space(\316\247,\316\254\317\207)is complete.

The situation is completely different when \320\260\321\204\\. In this case,(he tiles \317\210\317\207(\316\266)are still vertical strips, delimited by the vertical half-

lines 1\321\202(,\320\263)= an and lm(z) =

an+\\, where an =\317\210\317\207(0).However, by

induction on n, we see that for every \316\267^ 0

1 _ nn

\316\250\316\271[\316\266)= \316\261\316\266+ a + \316\261 +\302\267\302\267\302\267+\316\261+ \316\220= \316\261\316\266+ ,

while

,.\342\200\224n/_\\ \342\200\236\342\200\224n_\342\200\236\342\200\224n\342\200\236\342\200\224(n\342\200\2241) \342\200\236\342\200\2242\342\200\236\342\200\2241\316\250\316\220(\316\266)

= o z \342\200\224a \342\200\224\320\276v ' \342\200\224\302\267\302\267\302\267\342\200\224a \342\200\224a

1n~\316\267

-\316\271 \342\200\224\316\267

= a~nz - \316\261\316\2231- \320\223= a~nz -\\ .

1 - a~l 1 \342\200\224a


As a consequence,an=

\317\210\317\207(0)

l-anfor every \316\267e

OO

-2 o\342\200\224i 0 1 \316\2702\320\260\320\267\316\261\317\207\316\261^

Figure 6.13. A partial tessellation coming from anincompletehyperbolic cylinder

Consider the case where a < 1. Then an converges to \316\261<\317\207,=

jr^ j

as \316\267tends to +oo and, more importantly, an < \316\261<\317\207,for every \316\267Gj

Z. This implies that for every element \317\206\342\204\242of the tiling group, the |corresponding tile \317\210\317\207(\316\247)stays to the left of the vertical half-line of !

equation Be(z) =\316\261<\317\207,.In particular, the tiles \317\210\317\207(\316\247)cannot cover the jj

whole hyperbolic plane H2. The case where a = 0.75is represented )

in Figure 6.13. j

A symmetric phenomenon occurswhen a > 1, in which case thej

tiles \317\210\317\207(\316\247)are all to the right of the vertical half-line of equation jlm(z) =

a,\302\273,where \316\277,\317\207,=

j^= limn^-ooOn. Again, this prevents ]

the tiles from coveringthe whole plane H2.

If we combine these observations with Theorem 6.1, we conclude I

that the quotient metric space (\316\247,\316\254\317\207)is not complete if \316\261\316\2461. The]accumulation of the tiles along the line 1\321\202(,\320\263)

= ax explains why the jproof of Lemma 6.4 fails in this case. \316\257

IThis abstract argument showing that (\316\247,\316\254\317\207)

is incomplete is jfine mathematically, but it will be more gratifying to exhibit a finite- J

length sequence (Fn)n\342\202\254Nin (\316\247,\316\254\317\207)that does not converge. We re-3strict our attention to the case where a < 1. SetPn

= \\a~n for everyjinteger \316\267> 1. Note that Pn is glued to the point Qn = 1+ io_n+1,:and that the points Pn and Qn+i can be joinedby a horizontal line!

segment whose hyperbolic length is equal to an. Therefore, the length]

6.7. Incomplete hyperbolicsurfaces 165

of the sequence (Pn)n\342\202\254Nis

oo \320\276\321\201 \320\276\321\201

53 dx(pn, pn+1) = 53 dx(pn>Qn+1)< 53a\"< \302\260\302\260>

n=l n=l n=l

where the geometricseriesconverges since a < 1.

We claim that this finite length sequence (Pn)neN cannot converge

to any point \320\224\320\276in (X, dx). Indeed, the proof of Theorem 4.10 shows

that there is a small ball Bjx(Poc,e)such that the set of points\316\241G Ei U E2 projecting to some\316\241G Bgx (.Poo \316\265)under the quotient

map is either empty if P\302\273corresponds to a point Poo in the interior of

X or two geodesicarcs of hyperbolic length 2\316\265when P*, correspondsto a point \316\241\302\273G E\\ glued to another point Qoo G #2\302\267In the first

case, \320\222\320\264\321\205(Poo, \316\265)does not contain any Pn. In the secondcase,eachof

the two geodesic arcs can contain at most finitely many points Pn and

Qn, since Pn and Qn converge to oo as \316\267tends to oo. In both cases,this shows that

\316\262^\317\207{\316\241\316\277\316\277,\316\265)contains no Pn with \316\267large enough, so

that (Pn)neN cannot converge to any \316\241\317\207,G X.

Figure 6.14. The incomplete cylinder of Figure 6.13

It is interesting to understanda little better the geometry of

(\316\247,\316\254\317\207).Again, we focus our attention on the casewhere a < 1. The

case where a > 1issimilar.In this case where a < 1, the tiles \317\206\342\204\242(\316\247),with \316\267G Z, tessellate

the open quadrant \316\227consisting of those \316\266G HP with Re(^) < \316\262\316\277\316\277\342\200\224


\321\2034^\302\267On \316\227,consider the partition \316\227consisting of the subsets of \320\257of

the form {\317\206\342\204\242(\316\241):\316\267\316\276\316\226}with \316\241\302\243\316\227.Since \317\206\317\207is the gluing map usedto gluethe two edges of X together, there is a natural map \317\206:X \342\200\224+\316\227,

uniquely determined by the property that it associates \316\241\302\243\316\227to

\316\241\342\202\254J for every \316\241\302\243X. Because the \317\206\317\207(\316\247)tessellate H, one easilysees that \316\227is bijective. Anticipating results from Chapter 7, we will

prove in Theorem 7.12 that \317\210is actually an isometry from (\316\247,\316\254\317\207)

to (tf, dH).

It turns out that there is another natural way to tessellate the

quadrant H. Let \316\245be the quarter annulus delimited in \316\227by the two

euclidean quarter circles \316\237\317\207and \320\241\320\263centered at \316\261=

\316\263^and passing

through the points 0 and 1, respectively. Note that \317\206\317\207(\316\266)= az + 1

can also be written as \317\206\317\207(\316\266)=

a(z\342\200\224

\316\261^)+ \316\261<\317\207,,so that \317\206\316\271sends

the larger quarter circle to the smaller one. It easily follows that the ,

\317\206\317\207(\316\245)also tessellate H. By the same argument as above, (H,\320\260\320\275)

is therefore isometric to the quotient space (\316\245,\316\254\316\263)obtained from \302\267

(Y, \316\254\316\263)by gluing together its sides \320\241\320\263and C2 by \317\206\317\207.

As a consequence, the two metric spaces (\316\247,\316\254\317\207)and (\320\243,\316\254\316\263)are

isometric.

It turns out that we have already encounteredthe quotient space

(\316\245,\316\254\316\263)in Section 5.4.2. More precisely, it corresponds to one half ofthe hyperboliccylinder X3 of Figure 5.15 because \316\245is one half of the

hyperbolic strip X3. Comparing the left-hand side of Figure 5.15 tothe left-hand side of Figure 6.14, we concludethat (\316\245,\316\254\316\263)is isometric

to any one of the two halves of the hyperbolic annulus X3 delimitedby the waist 70. The correspondenceis rigorously described by the

hyperbolic isometry \317\206(\316\266)=

\316\261^\316\266+ \316\237\317\207,and is illustrated by the right-hand sidesof Figures 5.15 and 6.14. In particular, the open annulus

(\316\253,\316\254\316\263)has one end that flares out with exponential growth, while

the other one gets very close to the closed curve 70.It is interesting to consider the image \320\201of the edges \316\225\317\207and

E2 of X under the isometriccorrespondencebetween (\316\247,\316\254\317\207)and

(\316\245,\316\254\316\263).By careful inspection of Figure 6.14, one easilyseesthat

this image \320\201consists of an infinite curve which, in one direction,goesto infinity in the end of \316\245with exponential growth whereas,in the other direction,it spirals around the closed curve 70. This is

6.7. Incompletehyperbolicsurfaces 167

illustrated on the right-hand side of Figure 6.14. A more detailed

description of this construction can also be found in [Weeks4, \302\2472].

6.7.2. Incomplete punctured tori. We can similarly deform the

hyperbolic once-punctured torus of Section5.5. We keep the same

polygon X, namely, in this case the hyperbolic square delimitedby

the complete geodesies \316\225\317\207,E2, E3, \302\2434,where \316\225\317\207goes from \342\200\2241to 00.

E2 from 0 to 1, E3 from 1 to \321\201\321\216and \302\2434from 0 to \342\200\2241.

To glue \316\225\317\207to E2, we need a hyperbolic isometry \317\210\\sending\342\200\2241

to 0, 00 to 1, and X to the side of E2 opposite X. These isometriesare the linear fractional maps of the form

,\316\273

z + 1

\316\266+ a

with a > 1. Similarly, the E3 is glued to E4 by a map of the form

with b > 1. In Section5.5,we considered the case where a = b = 2,

but we now consider the general case.As in Section 5.5, chop off little pieces U-\\, Uq, U\\ and t/,\302\273

near the four corners \342\200\2241,0, 1 and \321\201\321\216of the square X, in such away that the intersections Ui \316\240Ej are glued together by the gluing

maps \317\210\321\203However, in this general case, we cannot always arrange

that the Ui are delimited by euclidean circlestangent to R as in

Section 5.5. To construct the Ui, first select Pj and Q\\ e \316\225\317\207with

Im(Pi) > Im(Qi), P3 and Q3 G E3 with Im(P3) > Im(Q3), and setP2

=\302\245>i(Pi), Q2 =

\302\245>i(Qi), ^4 =^\320\267(\320\240\320\267)and Q4 = <p3(Q3). Then

pick disjoint curves 7_1( 70, 71, 7<\317\207,in X connecting Q\\ to P4, Q4 to

Q2, P2 to Q3, and P3 to Pj, respectively. Finally, define Ui as the

portion of X that is delimitedby 7; and is adjacent to the corner i.As in Section 5.5, define Voo

= Ugo, V\\ = 4>2{U\\),V-\\= ipi(U-\\)

and Vo =\316\250\316\261

\316\277\317\2102(\317\2050),aad let U = t/_j U U0 U Ux U t^, and V =

V 1 U Vo U Vj U Voo- As before, V and the subsets Vi are vertical

half-strips delimited on the sides by vertical lines, and from below

by curves that are the images of the curves 74. SeeFigure6.15,and

compare Figure 5.19 in Section 5.5.


-1 0 1Figure 6.15. An incomplete punctured torus

The left side of V is glued to its right side by the map

\317\210\316\257\302\260\317\2102\316\270\317\210\316\206\302\260\317\210\\{\316\266)

-miz + (ab - l)(a+ b -

2)

(a -1)2

which, in general, is not a horizontal translation any more (unlessa = b). So we are essentially in the same situation as for the

incomplete cylinder of Section 6.7.1, and, when \316\261\317\206b, the tiles

(\317\2104\302\260\317\2102\302\260\316\250\316\254\302\260

\317\210\316\271)\316\267(\316\247)accumulate along the vertical half-line ofequation 1\321\202(,\320\263)

=\316\261^,where a^ =

^f^ is the only real number such

that \317\2104\316\277\317\2102\316\277\317\2103\316\277\317\210\316\271(\316\261\316\261\316\277)

= \316\261\317\207. '\342\226\240

Looking a little more closely into the tiling process, we can easilyconvince ourselves that the tiles \317\206(\316\247)do not overlap. The simplest:way to see this is to observe that the tiling construction provides anatural correspondencebetween the tiles constructed here and thetiles coming from the complete punctured torus of Section6.6. Inparticular, if an edge of one tiling separates two tiles of the same \342\226\240

tiling, then in the other tiling the corresponding edge separates the ;

corresponding tiles. Since there is no overlap between the tiles of the j

example of Section 6.6, we concludethat there cannot be any overlap'

between the tiles in the exampleof this section either.

This implies that all the tiles stay on the same the side of theline 1\321\202(,\320\263)

=\320\260^.More precisely, the tiles \317\206(\316\247)are all disjoint from

6.8. Poincare's polygon theorem 169

the hyperbolic half-plane1 \316\227={\316\266e lf;Im(z) ^ \316\261,\317\207,}if a < b, and

\316\227={\316\266G \320\250\320\240;Im(z) < a<x>} if a > b.

**J?>i**J(/~^fcbJl.-1\320\273\320\273\320\253I\342\200\224

\320\257

-10 1 \316\277,\317\207,

Figure 6.16. The tessellation associatedto the incomplete

punctured torus of Figure 6.15

The tessellation (in a case where a < b) is illustrated in

Figure 6.16.

Note that the tiles \317\210(\316\247)are also disjoint from every hyperbolichalf-plane of the form \317\210(\316\227),where \317\206is an element of the tilinggroup \316\223.Indeed, if

\317\210(\316\247)met \317\210(\316\227),then \316\271/\302\273-1\316\277\317\210(\316\247)would meet ff;

however, it immediately followsfrom the definition of the tiling groupthat \317\206~\316\271\316\277\317\210is also an element of \316\223,contradicting the fact that alltiles are disjoint from H. Therefore, the tiles are disjoint from all the

hyperbolic half-planes \317\210(\316\227)as \317\210ranges over all the elements of thetiling group \316\223.Some of these half-planes are visible in Figure 6.16where they appear as white euclidean half-disks.

6.8. Poincare's polygon theorem

Let (\316\247,\316\254\317\207)be the quotient space obtained by gluing together the

edges of a polygon (\316\247,\316\254\317\207)in the hyperbolic plane. We will provide a

relatively simple criterion to checkwhether or not the quotient metric

space (\316\247,\316\254\317\207)is complete.

The fundamental ideas involved have already appeared in ouranalysis of the complete example of Section 6.6 and in the incomplete

examples of Section 6.7. The completenesscriterionessentially says

that there is no new phenomenon.For this reason,this section may

be skipped on a first reading.

1Not to be confused with H2, which ia a euclidean half-plane but ia the wholehyperbolic plane.


We already observed in Proposition 6.20 that when X is bounded

the quotient space (X, dx) is compactand therefore complete. Thus,

the only relevant case here is the one where X is unbounded and, in

particular, when some of its edges are infinite.

Edges \316\225of the polygon X can be of three types. A bounded

edge goes from one vertex of the polygon to another one. A singly

infinite edge goes from a vertex to a point of RU {\321\201\321\216},which we will

call a vertex at infinity or an ideal vertex. A doubly infinite

edge joins two points of R U {\321\201\321\216},which consequently are two idealvertices of the polygon.

To simplify the exposition, we will restrict attention to the casewhere a vertex at infinity is the endpoint of at most two edges of

X. Note that we had already imposeda similar, and equally natural,

condition for the vertices of X that are in H2.

When we glue the edgesE{ and -Z?i\302\261i,the gluing map \317\210\316\271:\316\225\316\271\342\200\224*

Ei\302\261\\sends each ideal vertex \316\276G R U {\321\201\321\216}adjacent to \302\243?,-to an ideal

vertex adjacent to Ei\302\261\\. In this way, we can extend the gluing of

edges to gluings between the ideal vertices of X, as in the case of

nonideal vertices. An element of the correspondingquotient space

will be an ideal vertex of the quotient space X of X under thegluing operation. Note that exactly as ideal vertices of the polygon

X are not elements of X, idealvertices of the quotient space X arenot elements of X.

Let \316\276=

{\316\276\317\207,\302\243\320\263,\302\267\302\267\302\267,\302\243fc}be an ideal vertex of X. Namely, \316\276is the

set of ideal vertices of X that are glued to a given ideal vertex \316\276.The

gluing maps identifying the\302\243j

can be organized in a nice manner,as in the caseof nonideal vertices in Section 4.3.1, but with a minor

twist because an ideal vertex is allowed to be adjacent to only one

edge of X. We state this as a lemma for future reference.

Lemma 6.22. The indexing of the ideal vertices in \316\276=

{\302\2431,62, \342\226\240\302\267\302\267,\302\243fc}

can be chosen so that there exists gluing maps \317\206^:Ei}\342\200\224*Ey. =

\316\225\317\212\317\214\302\261\\,with j =1, 2, ..., k\342\200\2241, such that

(1) \302\243jis an endpoint of E^, \302\243J+i

is an endpoint of \316\225\317\207., and

the gluing map \317\210^sends

\302\243jto

\302\243j+i/


(2) for every j with 1 < j < \320\272\342\200\2241, the two edges E{. and

Ei'. adjacent to\316\276jare distinct, so that

\317\210^is different from

(3) exactly one of the following holds:

(i) either there existsa gluing map \317\210^:Eik\342\200\224*Ey such

that Eik is an edgeadjacent to \302\243fcand different from

the range E^ of \317\210\316\271\316\271^_1,such that \316\225\316\263is an edgeadjacent to \316\276\316\271and different from the domain E^ of \317\210^,

and such that \317\210sends \316\276^to \316\276\\;or

(ii) each of \316\276\316\271and \302\243&is adjacent to a unique edge of X,

namely, E^ and E^ , respectively.

Proof. As in Section 4.3.1, start with any element \316\276\316\271=

\316\276G \316\276.Let

\316\225\316\257\317\207be one of the < 2 edges containing \316\276\317\207.Set \316\2762=

\317\210\316\2571(\316\276\316\271)\302\267If

\317\206%\316\271(Eh) is the only edge that is adjacent to \316\2762>stop here. Otherwise,

apply the same processto the other edge Ei2 leading to \316\2762,define

\302\243\320\267=

\320\243\320\2633(\302\2432))and iterate this construction.

If this process goeson forever, we eventually reach an index \320\272

such that \302\243fc+i=

\302\243jfor some j ^ k. If \320\272is the smallest such index,

one easilychecksthat \316\276\316\271,,+\316\271=

\316\276\316\271and that we are in the situation of

alternative 3(i).

Otherwise, we reach an index \320\272such that \316\276\316\271,is adjacent to onlyone edge, namely, \317\210\316\271\316\231\316\271_1(\316\225\316\257\316\220\316\271_\316\271)

= Ey . In this case, we restart theprocessbeginning at \316\276\316\271,but going backward. Let \316\225\316\263be the edge that

is adjacent to \316\276\316\271and is different from E^ (if any). Let Eio be the

edge that is glued to Ey by the gluing map \317\210\316\2710:\316\225\316\2710\342\200\224*E^ , and set

\302\2430=

\316\250\317\212^\316\257\316\266\316\271)\302\267If Ei0 is the only edge that is adjacentto \316\2760,stop here.

Otherwise, let E^ be the other edgeadjacentto \302\243o>consider the edge

Ei_z that is glued to Ei0 by the gluing map \317\210\316\257_\316\271:\316\225\316\271_\316\273\342\200\224*\316\225\316\263,set

\316\276-\316\271=

\317\206\317\212}(\316\2760),and iterate the construction. In this case, the process

must eventually conclude by the discussion of the previous case, and

we reach an index \342\200\224I ^ 0 such that \316\276-\316\271is adjacent to only one edgeEi_t of X. Shifting all indices by I \342\200\2241 to make them positive, we arenow in the situation of alternative 3(ii).

In both cases,conditions (1) and (2) are satisfied by construction.D


In the case of alternative 3(i), we will say that we have an edge 3

cycle around the ideal vertex \316\276=

{\316\276\316\271,\302\243\320\263,\302\267\302\267\302\267,\342\202\254fc} of X. In this case, \316\257

it is convenient to consider the indicesj modulo k, namely, in such a ,way that j is identified with j + k. This convention has the advantage j

that each gluing map \317\206^then goes from Eij to \316\225\317\207.. Note that

'

the composition \317\206^\316\277

\317\210\316\257\316\231\316\271_\316\271\316\277\342\226\240\342\226\240\342\226\240\320\276

\317\206\316\2572\316\277

\317\210\316\257\316\263sends \316\276\316\271to itself. Our ]completeness criterion will be expressed in terms of this composition\\

of gluing maps, considered for all edge cyclesaround ideal vertices.

A horocircle or horocycle centeredat the point \316\276G RU {\321\201\321\216}is

a curve C\342\200\224{\302\243}in H2 where \320\241is a euclidean circle tangent to RU {\321\201\321\216}

at \316\276(and lying \"above\" R U {\321\201\321\216}).When \316\276=

\321\201\321\216,this means that ;\320\241-

{\316\276}is a horizontal line. Note that any isometry of H2 must send 1

any horocycle centered at \316\276to a horocycle centered at\317\206(\316\276),

since \317\206I

sends a euclidean circle to a euclideancircle.SeeExercises 6.10-6.12 \316\212

for an interpretation of a horocircle as a hyperbolic circle of infinite

radius centered at \316\276.

An isometry \317\206of the hyperbolic plane H2 is horocyclic at \316\276if it

respects some horocircle centered at \316\276.When \316\276= \321\201\321\216,this just means t

that \317\206is a horizontal translation \316\266>-^ \316\266+ b (possibly the identity j

map) or a reflection \316\266\316\271\342\200\224>\342\200\224f+ \320\252across a vertical line. I

In particular, a hyperbolicisometry which is horocyclic at \316\276re- j

spects every horocircle centered at \316\276.When \316\276\342\200\224

\321\201\321\216,this immediately!

follows from the above observation. In the generalcase,it suffices!

to apply this special case to \302\267\317\206\316\277\317\206\320\276\321\204~\320\263,where \317\210is any hyperbolic

isometry sending \316\276to \321\201\321\216,for instance \316\266\316\271->1/(\316\266

\342\200\224\316\276).

As usual, extend each gluing map iff. Ei -+ Ei\302\261\\to a hyperbolic

isometry \317\210\316\257:\316\2272\342\200\224>\316\2272that near Ei, sends X to the side of\302\243?i\302\261i'j

opposite X. |.1

Proposition 6.23. The following two conditions are equivalent: jI

(1) (Horocircle Condition) At each ideal vertex \316\276of X, one cam

choose a horocircle \316\237\316\276centered at \316\276such that whenever thd

gluing map \317\206^:\316\225\316\271\342\200\224>Ei\302\261x sends \316\276to another ideal\321\203\320\265\320\235\320\265\321\211

\316\276',its extension \317\210\317\212.\316\2272\342\200\224>\316\2272sends\316\237\316\276

to\320\241?; .j

(2) (Edge Cycle Condition) for every edge cycle around anideafy

vertex \316\276\342\200\224

{\316\276\316\271,&, \302\267\302\267\302\267> \302\243fc} of X with gluing maps \317\206^.:E^ \342\200\2241|


\320\251+1=

Eij\302\2611 sending \302\243jto \316\276,\302\267+1for j = 1, 2, ..., \320\272as in

alternative 3(i) of Lemma 6.22 (counting indices modulo \320\272

so that j +k is consideredthe same as j), the correspondingcomposition\317\206^

\316\277y>ifc \320\263

\320\276\302\267\302\267\302\267\316\277\317\206\316\2572

\316\277\317\210is horocyclic at \316\276\317\207.

Proof. Suppose that the first condition (1) holds. Then, for an edgecycle around an ideal vertex consisting of gluing maps \317\206^:E^

\342\200\224>

Ei'.+l\342\200\224

Eij\302\261x,with j \342\200\224

1, 2, ..., k, the composition\317\206\316\257\316\220\316\271\316\277\317\206\316\257\316\220\316\271_1\316\277\302\267\302\267\302\267\316\277

\317\210\316\2572\316\277

\317\210\316\257\316\257sends the ideal vertex \316\276\316\271to itself, and consequently sendsthe horocircle

\316\237\316\276\316\271to itself. It follows that \317\206\316\257\316\220\316\271\320\276y>ifc_1 \320\276\342\226\240\302\267\302\267\316\277

\317\206\316\2572\316\277

\317\210^

is horocyclic. This proves that the second condition (2) holds.

Conversely, suppose that the second condition (2) holds.Foreachideal vertex \316\276

={\316\276\316\271,\316\2762,.\302\267\302\267,\302\243fc}of X, organize the gluing data as in

Lemma 6.22. Pick an arbitrary horocircleC^ centeredat \316\276\316\271,and

define

Ci;(=

\317\210\316\257}_\317\207\316\277

\317\210^_2\316\277\302\267\342\226\240\302\267\320\276

\317\210\316\257\316\271(\316\237\316\266\316\271)

for j < \320\272.Note that\317\210\316\271\316\257_\316\271

\316\277\317\210\316\257-_2

\316\277\302\267\302\267\302\267\320\276\317\206\316\257\316\263sends \316\276\316\271to

\316\276^\302\267,so that

\316\237\316\276\316\257is really a horocircle at \302\243j.

If we are in the case of alternative 3(ii) of Lemma 6.22, there isnothing to prove. Otherwise, we have to make sure that the gluing

map <pik sends the horocircle C(k to\316\237\317\2021.

This is where the hypothesisthat tpik

\320\276<pik_1

\320\276\302\267\302\267\302\267\316\277\317\210\316\2572\316\277\317\206^is horocylic is needed, since it shows

that

<#*(%) =\316\250\316\257*\302\260Vik-i

\302\260' ' ' \302\260\316\250\316\257\316\271\302\2604>ix(Cix)

=\302\260\316\276\316\271-

Performing this construction for every ideal vertex \316\276=

{\316\276\316\271,\316\2762,\302\267\342\226\240\302\267> \316\276&}

of X provides the conclusions of condition (1). D

Complement 6.24. If the conditions of Proposition 6.23 hold, we

can choose the horocircles \316\237\316\276in condition (1) so that they are

arbitrarily small. (When \316\276\342\200\224oo, this means that \316\237\316\276

is a horizontal line

which is arbitrarily high.)

Proof. In the above proof that condition(2) implies condition (1),

we have a degree of freedom in the choice of the initial horocircle\316\237\316\2761

lor each ideal vertex \316\276=

{\316\276\316\271,\316\2762,\302\267\302\267\302\267>\316\276*;}of X. In particular, thisliorocirclecan be chosensmall enough that all other horocircles C^.are alsoarbitrarily small. \316\240


In practice, the Edge Cycle Condition (2) of Proposition 6.23 isoften easier to check than the Horocircle Condition(1). However,

the Horocircle Condition (l) will be better suited for the proof of our

main Theorem 6.25 below.

Exercise 6.14 shows that in the specific case of the compositionmap <pik \320\276

\317\210\316\257\316\231\316\271_\316\271\316\277-\317\214^\316\277^ of gluing maps that occurs in the EdgeCycle Condition (2) of Proposition 6.23, this isometry is horocyclic

if and only if it is parabolic in the sense defined in Exercise 2.7.

However, we will not need this property.

Theorem 6.25 (Poincare'sPolygon Theorem). Let (\316\247,\316\254\317\207)be the

quotient space obtained by gluing together the edges of a polygon

(\316\247,\316\254\317\207)in H2, using edge gluing maps <pi\\ Ei \342\200\224*Ei\302\261\\. The quotient

space (\316\247,\316\254\317\207)is complete if and only if the (equivalent) conditions ofProposition 6.23are satisfied.

We will split the proof of equivalence into two parts.

Proof of the \"only if\" part of Theorem 6.25. We need to showthat the conditionsof Proposition 6.23 hold if (\316\247,\316\254\317\207)is complete.

We will actually prove the contrapositive of this statement. More

precisely, we will assume that the Edge CycleCondition (2) of

Proposition6.23 does not hold, and then show that (\316\247,\316\254\317\207)is not complete.

By our assumption, there is an edge cycle \316\276=

{\302\243\316\271,&2,\302\267\342\226\240\302\267,\302\243fc},

with data as in alternative 3(i) of Lemma 6.22, such that thecomposition

is not horocyclic.

Modifying X and all gluing data by a hyperbolic isometry sending\316\276\316\271to oo, we can arrange that \302\243j

= \321\201\321\216without loss of generality. Then

\317\206sends \321\201\321\216to \321\201\321\216,and is consequently of the form \317\206(\316\266)\342\200\224az + 6 or

\342\200\224az+ b with a > 0, 6 \342\202\254R. The fact that \317\206is not horocyclic is

equivalent to the propertythat \320\260\321\2041.

The argument is essentially the one that we already encountered,

in Sections 6.7.1 and 6.7.2.First considerthe casewhere a > 1.

Near each ideal vertex\302\243j

of X, chop off an infinite half-strip:

Uj delimited by a curve joining the two edges E^ and E^. that are


adjacent to \316\276;.Then i/b \317\210^{112), \317\210^\316\277

tpi2\\U3), ..., \317\206^1\316\277\317\206.*\320\276

\342\226\240\"\302\260\316\250\316\252l(^fc) are vertical half-strips which are adjacent to each other,

and their union forms a vertical half-strip V. In addition,\317\210~\317\207sends

the vertical side of V that is contained in E^ to the other vertical

side of V.

Pick a point \316\241\317\207\342\202\254E^ so that the horizontal line L\\ passing

through P\\ lies above the curves delimiting V from below. Let Q2be the point where this line L\\ hits the other vertical side of V. We

already observed that this other vertical sideis contained in \317\210~\316\216{\316\225^),

so that we can consider the point P2 \342\200\224f(Q2) \302\243Eh\302\267Repeating the

process, we can thus define a sequenceof points Pn \342\202\254E^ such that

Pn is the image under \317\206of the point Qn of the other vertical side of

V that is at the sameeuclideanheight as \316\241\316\267_\316\271\302\267

We now consider the sequence (Pn)n\342\202\254Nin the quotient space X.

Let yi be the y-coordinate of P\\ (and Q2). By construction,the ^-coordinates of Pn and Qn+\\ are both equal to any\\, and Pn

and Qn+\\ have the same x-coordinate as P\\ and Q2, respectively.

Considering the hyperbolic length of the horizontal line segment sngoing from Pn to Qn+i, we conclude that

dx(Pn,Pn+1) = dx(Pn,Qn+1)< dx{Pn,Qn+x)

As a consequence, the total length00 00

\316\243dx(Pn,Pn+1) <\302\273\316\223^\302\253\316\271\316\265(\316\262\316\220)\316\243\316\262_\316\267

< \302\260\302\260

n=1 n=1

of the sequence (-Pn)neN is finite, since \320\276> 1. However, by the

argument that weusedin Section6.7.1,this sequence cannot converge1.0any point of the quotient space X. In particular, the metric space(\316\247,\316\254\317\207)is not complete.

The argument is very similar when a < 1. This time, we will

\316\271lioose the points Pn on the vertical side of V that is not contained in

/'/,!. Namely, we start with P\\ in that side, consider the point Q2 \342\202\254

/'/ij that is at the same euclideanheight as Pi, set P2 \342\200\224\317\210~\317\207((^2),

:\302\253tditerate the process. This again gives a sequence(Pn)n\342\202\254Nin X


that has finite length (this time because a < 1) but cannot converge

in X. Therefore, (\316\247,\316\254\317\207)is not complete in this case as well. D

Proof of the \"if\" part of Theorem 6.25. Let us assumethat the

Horocircle Condition (1) of Proposition 6.23holds.We will prove that

the quotient space (\316\247,\316\254\317\207)is complete.

For this, consider in X a sequence(Pn)n\342\202\254Nwith finite length. We

want to show that this sequence converges. The problemis that of

course, there is no reason for the corresponding sequence (Pn)n\342\202\254Nto

have finite length in X.

We will split the proof into four distinct cases, only two of which ;

involve significant geometric ideas. The first case is relatively easy todeal with.

CASE 1: There exists a hyperbolic ball Bdhyp(Po,r) that contains in- \342\226\240

finitely many Pn.

The Pn that are contained in Bdbyp(Po,r)form a subsequence

(Pnk)keN valued in that ball. Since the closed hyperbolic ball \320\222of;

center Po and radius r is compactby Theorem 6.15, this subsequence \342\226\240

itself admits a subsequence (Pn/, ),gNwhich converges to some point.;

\342\226\240Poo\316\265\316\222for the hyperbolic metric dhypithe point Poo must be in X \342\226\240

since all polygons are closed. It easilyfollows that Pnit also converges,to Poo \342\202\254X for the path metric dx, as / tends to \321\201\321\216.(Hint: For;

every \316\241s X, there exists a small ballBdhyp (\316\241,\316\265)in H2 such that j

dx(P,Q)= dbyp(P,Q) for every Q e \320\245\320\237

Bdhyp(P,e)). \\

Since the quotient map X i\342\200\224>X is continuous by Lemma 4.2, the1(sub)subsequence(Pnfci);eN converges to P*, in (\316\247,\316\254\317\207).

Now the sequence (\316\241\316\267)\316\267\342\202\254%has finite length and admits a

subsequence converging to \316\241\317\207,\342\202\254X. The whole sequence consequently;convergesto P^ in (\316\247,\316\254\317\207)by Lemma 6.17.

We now consider the casewhere the Pn stay away from the edgesof the polygon X.

CASE 2: There existsan \316\265> 0 such that every Pn is at hyperbolic

aistance > \316\265from every edge of X.

6.8. Poincare's polygontheorem 177

In particular, no two distinct points of the ball Bdx (Pn,\316\265) aie

glued together.

Since the total length ]T^\302\260=1dx(Pn, Pn+i) of the sequence (\320\240\320\277)\320\277\320\265\320\277

is finite, dx(Pn,Pn+i) tends to 0 as \316\267tends to \321\201\321\216,and in particularthere exists an no such that \316\254\317\207(\316\241\316\267,\316\241\316\267+\316\271)< \316\265for every \316\267^ no\302\267

Consider a discrete walk w from Pn to Pn+i in X, whose length

ldx (w) is closeenough to the infimum dx(Pn. Pn+i) that idx (w) < \316\265.

Let Pn - Qx, Rx~ Q2, ..., Rk-\\

~ Qk, Rk =Pn+\\ be the steps

of this discrete walk. By induction and using the Triangle Inequality,the points Qi stay in the ball \316\222\316\261\317\207(\316\241\316\267,\316\265)and are equal to the points\320\224\320\263+\316\271,since there is no nontrivial gluing in this ball. Therefore, bythe Triangle Inequality,

fc

dx(PniPn+i)^^2dx(Qk,Rk)='tdxM\302\267

Since this holds for every discrete walk w whose length is sufficiently

close to dx(Pn,Pn+i), it follows that dx(Pn,Pn+1) ^ dx(Pn,Pn+i)-As the reverse inequality always holds by Lemma 4.2,this inequality

is actually an equality, and dx(Pn,Pn+i) \342\200\224dx(Pn,Pn+i) for every

\316\267> ra0.

In particular, the sequence (Pn)n\342\202\254Nbas finite length in X. Since

(H2,dhyP)is complete,the finite length sequence (Pn)n\342\202\254Nconverges

to some point P^ \342\202\254H2. This point \320\240\320\266must be in the interior ofX sincethe Pn stay at distance > \316\265from its boundary. Since the

quotient map X \342\200\224>X is continuous (Lemma 4.2), it followsthat thesequence(Pn)ngN converges to the point P*, in the quotient space

(\316\247,\316\254\317\207),which concludes the proof in this case.

Next, we consider the casethat is completely opposite to Case 2.This is the crucialcase,which will strongly use our hypothesis thatthe HorocircleCondition(1)of Proposition 6.23 holds.

('ase 3: All the Pn are contained in edges of X.The HorocircleCondition (1) of Proposition 6.23 asserts that

there exists a horocircle \316\237\316\276centered at each ideal vertex \316\276such that

whenever the gluing map \317\206^:\316\225\316\271\342\200\224\342\226\272Ei\302\261\\sends \316\276to another ideal

vertex \316\276',its extension \317\210\316\271:\316\2272\342\200\224>\316\2272sends \316\237\317\202to\316\237\317\202>.


Let \316\222\316\276be the \"inside\" of

\316\237\316\276in H2. Namely, when \316\276\317\206\321\201\321\216,\316\222\317\202

is the closed euclidean disk bounded by \316\237\316\276minus the point \316\276.When

\316\276\342\200\224

\321\201\321\216,\316\222\316\276is the euclidean half-plane bounded by \316\237\316\276

from below.

Such a\316\222\317\202

is called a horodisk centered at \316\276.

By Complement 6.24, we can assume without loss of generalitythat the horocirclesare small enough that they are disjoint from eachother, and that the only edges of X that are met by \316\222\316\276

are those

which are adjacent to the idealvertex \316\276.

Let \320\222\342\200\224(L \316\222\316\276denote the union of all the\316\262\317\202,

as \316\276ranges over all

ideal vertices of X. Thereisa convenient function ft: \320\222\342\200\224>R defined

as follows. For \316\241e\316\222\316\276,

let Q be the point of \316\237\316\276that is closest to

P; then ft(-P)=

dhyp(-P, Q)\302\267The point Q can be easily constructedfrom the property that it is the intersection of

\316\237\316\276with the complete

geodesic g passing through \316\276and P. This property is easily checkedwhen \316\276

\342\200\224\321\201\321\216,for instance using Lemma 2.5; for the general case,just

transport everything by an isometry of H2 sending\316\276to \321\201\321\216.

The function ft is called the Busemann function. SeeExercises 6.10, 6.11 and 6.12 for another geometric interpretation of ft.:

This function has the following two important properties:

(1) if \316\241and Q e \320\222are glued together, then h(P) = ft(Q);

(2) for any two P, Q in the same horodisk\316\222\316\276,dhyp(P,Q) ^

\\h(P)-h(Q)\\-

The first property is an immediate consequence of the fact thatthe gluing maps are isometries and send each

\316\237\316\276to some

\316\237\316\276>.Thei

second property is a consequence of Lemma 2.5 in the case where\316\276

\342\200\224\321\201\321\216.The general case follows from this specialcaseby modifying!

the geometric setup through a hyperbolic isometry sending\316\276to \321\201\321\216.\\

After bringing up all this machinery, we now return to our se-s

quence of points Pn in the edges of X. |Fixan arbitrary \316\265> 0, whose precise value will not be important.!

By the same argument as in Case 2, there is a number rii such that;

dx(Pn, Pn+i) < \316\265for every \316\267> \321\211. \\

In addition, we can assume that we are not in the situation o|

Case 1, since otherwiseweare done. This guarantees that there exists


an ra0 ^ rii such that for every \316\267^ ra0, the point Pn is sufficiently far

away on the edge \316\225containing it that it belongs to the horodisk\316\222\316\276\316\267

associated to some ideal vertex \302\243n,and that it is even at distance> \316\265

from the point where the edge \316\225meets the horocircle \316\237\316\276\316\267delimiting

For \316\267^ ra0, consider a discrete walk w from Pn to Pn+i in X,

whose length idx(w) is close enough to the infimum \316\254\317\207(\316\241\316\267,\316\241\316\267+\316\271)

that idx (w) < \316\265and whose steps are Pn \342\200\224Q\\, R\\ ~ Qi, ..., Rk-i ~

Qk, Rk \342\200\224Pn+i- We wiH prove by induction that all the Qi and Ri-i

belong to \320\222\342\200\224(J? \316\222\316\276,

and that

\320\263-1 i-l

(6.1) h(Pn) -J^dxiQ^Rj) < h{Qi) < h(Pn)+ J2dx(Qj>Rj)-

Of course, the Qi and Ri-\\ may jump from one horodisk \316\222\316\276to

another.

We start the induction with the case where i \342\200\2241, in which case

(he property trivially holds since Q\\ \342\200\224Pn.

Suppose that the induction hypothesis holdsfor i. If Qi is in thehorodisk

\316\222\316\276,the distance from Qi to the horocycle\320\241\316\276bounding \316\222\317\202

is h(Qi). By the induction hypothesis,

ih(Qi)- dx(Qi,Ri)> h(Pn)

-\302\243dxiQj, Rj)

> h(Pn)- tdx \320\230> h(Pn)

- \316\265> 0

.since idx (w) ^ \316\265and since Pn is at distance > \316\265from the horocycle

hounding the horodisk\316\222\316\276\316\267

that contains it. In particular, thedistance from Qi e \316\222\316\276

to Ri is less than the distance from Qi to the

boundary horocircle\316\237\316\276bounding \316\222\316\276.

It follows that \320\224\302\273is also in\316\222\316\276.

Since Qi+\\ is glued to \320\224\302\273,and since the gluing maps sendpointsof \320\222to points of B, we conclude that Qi+i is alsoin B.

Also, combining (6.1) with the two fundamental properties of ft,

i

KQi+i) = h(Ri) < h(Qi)+ dx(QuRi) ^ h(Pn) + Yjdx{Qj,Rj)


and\320\263

h(Ql+l)= h(Ri) > h(Qi) - dxiQuRi)> h(Pn) -^dxiQ^Rj).

This proves that (6.1) holds for \320\263+ 1.

This completes our proof by induction that all Qi and \320\257*_\316\271with

1 < \320\263'^/c belong to \320\222and satisfy (6.1).

One more step in the same proof gives that

\320\272

h(Pn+1)= h(Rk) < h(Qk) + dx(Qk,Rk) < h{Pn)+Yjdx{Qk,Rk)

j=iand

\320\272

h(Pn^) = h(Rk) > h(Qk)-dx(Qk,Rk) > h(Pn)-Y^dxiQ^Rk),

so that

\320\272

]h(Pn+i)-

h(Pn)\\ < Y^dxiQ^Rk)= idx{w).

Sincethis holds for every discrete walk w from P\342\200\236to Pn+\\ whose

length is sufficiently close to the infiinum \316\254\317\207(\316\241\316\267,Pn+i)i we conclude

that

HPn+u-hiP^^dxiPn.P^!).

A consequence of this inequality is that the length of the sequence

(h(Pn))n Nin R is bounded by the length of the sequence {Pn)neNin

X, and in particular is finite. Since (R, <ie\342\200\236c)is complete (Fact 6.9),if follows that the sequence(ft(Pn)) converges in R. As a

consequence, it is bounded by some number M.

However, this shows that for \316\267> no, the P\342\200\236stay at hyperbolic

distance < \316\234from the finitely many points where the edgesof X

meet the horocircles C^. In particular, these Pn are all contained in a

very large hyperbolic ball. So we are in the situation of Case 1 after

all. which proves that the sequence (\316\241\316\267)\316\267\316\266:>converges in A\".

We are now ready to conclude.


Case 4: General case.

We will use the three previous cases to extract from (Pn)n& a

converging subsequence and then apply Lemma 6.17.If we are in the situation of Case2, namely, if the Pn stay away

from the edges of X, we are done. Consequently, we can assumethat

t.his is not the case.

We then construct by induction a subsequence (Pnk)ke^ such

(hat for every k, there exists a point Qnk in an edge of X such that

dx{P\342\200\236k.Qnk) < 2_fc. (We could replace 2~k by the kth term of any

convergent series.) Indeed,supposing the first \320\272terms of the

subsequence have been constructed, the fact that the hypothesis of Case 2

does not hold implies that there exists an index n^-i > \316\267\320\272such thai

\320\240\320\237\320\272+1is at distance < 2~(fc+1' from the edges of X. This means that

(here exists a point Qn^+x m the boundary of X which is at distance< 2~(fc+1) from Pnk+1- This inductive process clearly provides the

subsequence (\320\240\320\237\320\272)\320\272eNand the sequence (Qnk)keK requested.

We claim that the sequence (Qnk.)kGNhas finite length. Indeed,

dx{Qn,,Qn*_J < d{Qnk,Pnk) + dx(Pnk,Pnk+J)+ dx(Pn,.^,Q\342\200\236,_J

^2-k + dx(Pnk,Pnk+l) + 2-(k^

by the Triangle Inequality, so that

\320\276\321\201 \320\276\321\201 \320\276\321\201 \320\276\321\201

Y,d*(Qnk,Qnk+i) <\316\2432~\" +Ed*(p\302\253*'p\302\253*+1)

+E2~(fc+1)

fc=l fc=l fc=l fc=l

iti finite since the length of(Pnk)k(Z^

is finite.

Therefore, we can apply Case3 to the sequence(Qnfc)fceN\302\267Our

;malysis of this case shows that this sequence converges. Since

<lx(Pnk;Qrn.) < 2~k tends to 0 as A: tends to oo. it follows that

l.he sequence (Pnk)ke\342\204\242 converges to the same limit.

Thus, our original finite length sequence (Pn)neN has a converging\320\240*\302\260\321\202

subsequence (P,u )kGS- It therefore convergesby Lemma 6.17. >\\

This concludes the proof of the \"if\" part of Theorem 6.25, andtherefore completes the proof of this statement. D ^


Exercisesfor Chapter 6

Exercise 6.1. A Cauchy sequence in a metric space(X,d) is a sequenceof points Pi, P2, ..., Pn, \342\226\240\342\226\240\342\226\240in X such that for every \316\265> 0, there exists anumber no such that d(Pn, Pn') < \316\265for every \316\267,\316\267^ no-

a. Show that a sequence (Pn)ne^ that has finite length is Cauchy.

b. Conversely,let (Pn)ncN be a Cauchy sequence. Show, by induction on

k, that it contains a subsequence (Pnk)ksf; such that d(Pnk, \320\240\320\237\320\272_\320\263)\316\266

1~k for every k. Show that this subsequence (Pnk )A..Nhas finite length.

\321\201Show that if a Cauchy sequence admits a converging subsequence, thenthe whole sequence is convergent.

d. Combine the previous steps to show that a metric space (X, d) iscomplete in the sense of Section 6.2 if and only if every Cauchy sequencein (X, d) is convergent.

Exercise 6.2. Let (X, d) and (X', d') be two metric spaces for which thereexistsa homeomorphism \317\206:X -+ X'. Prove, or disproveby a

counterexample,each of the following two statements:

a. If (X, d) is compact, then (X1, d') is also compact.b. If (X, d) is complete, then (X',d') is also complete.

Exercise 6.3. Show that a tessellation of the sphere S2 has only finitely

many tiles. Hint: The sphere is compact.

Exercise 6.4. In the euclidean plane, let X be the parallelogram with

vertices (0,0), (2,1), (2,2), (0,1). Glue the left edge of X to the right

edge by the reflection-translation (x, y)'\342\200\224>

(\320\266-t- 2,\342\200\224

\321\203+ 2), and the bottomedgeto the top edge by the translation (x,y) >\342\200\224>[x,y + 1); we saw in

Section 5.1 that the corresponding quotient space X is homeomorphic tothe Klein bottle. Draw the tessellationassociatedby Theorem 6.1 to this

gluing data.

Exercise 6.5. Construct a tessellationof the euclidean or hyperbolic planeby tiles which are not convex. Extra credit: Find one which is particularlypretty (solution not unique).

Exercise 6.6. For three integers a, b, \321\201> 2 with\302\243+ | + \\

> 1, consider

a tessellation of the sphere S2 by spherical triangles of angles f, f, fas in Theorem 6.21. Show that this tessellation consists of ob4b^?^._ obc

triangles. Hint: Use the area formula of Exercise 3.6.

Exercise 6.7.

a. Show that for every euclidean triangle TcR2, there exists atessellationof the euclidean plane \320\250.2whose tiles are all isometric to T.


b. Give an example of a spherical triangle TcS2 for which there existsno tessellationof the sphere S2 by triangles isometric to T. Hint: Use

the area formula of Exercise 3.6or the hint for part \321\201below.

\321\201Give an example of a hyperbolic triangle \316\244\320\241\320\2352for which there existsno tessellation of the hyperbolic plane H2 by triangles isometric to T.Hint: Consider the angles of the tiles of the tessellation around one

vertex.

Exercise 6.8. Show that for every integer p, q ^ 2 with | + | < \\,there

exists a tessellation of the hyperbolic plane H2 by p-gons such that exactlyq of these p-gons meet at each vertex. Possiblehint: Use appropriate

triangles.

Exercise 6.9. Let X be a polygon in the euclidean plane R2, and let

(X,dx) be a quotient space obtained by isometrically gluing its edgestogether. Show that (X,dx) is always complete. Hint: Adapt the proof ofPoincare'sPolygon Theorem 6.25, noting that Case 3 of this proof becomesmuch simpler in the euclidean case.

Exercise 6.10. Fix a point Po \302\243H2 in the hyperbolic plane and a point

\316\276e RU{oo} at infinity of E2. Show that as Q e H2 tends to \316\276inR2U{oo},

the limit

h(P) = lim dhyp(Q, P)-

dhyp(Q, P0)

exists for every Pel2, Possiblehint: Use the explicit formula for the

hyperbolic metric provided by Exercise 2.2.

Exercise 6.11. In Exercise 6.10, consider the case where \316\276= oo. If

Po = (zo, 3/o) and \316\241=(\317\207,\321\203),show that h(P) = log f.

Exercise6.12.In Exercise 6.10, show that the set ofPel2 with h(P)=

0 is exactly the horocirclecentered at \316\276and passing through Po\302\267Possible

hint: Use Exercise 6.11 and a hyperbolic isometry sending \316\276to oo.

Exercise 6.13. In Figure 6.15,Vq is delimited from below by a straightline segment with negative slope, whereas Uo =

\317\206J1 \302\260\317\206\\'(Vo) is delimited

from above by a curve \320\241which appears to be a euclideancircle arc.

a. Show that \320\241is indeed a circle arc containedin a circle passing throughthe point 0.

b. This circle appears to be tangent to the z-axis at 0. Is this actually

the case? Explain.

Exercise 6.14. Consideran edge cycle as in condition (2) ofProposition6.23, and use the same notation as in that statement.

a. Show that i^jfc\302\260V</t -i \302\260'''\302\260f'2 \302\260f*\\ cannot be an antilinear fractional

map and is different from the identity. Possible hint: Look at the way

the tiles\317\206~^

\316\277\317\206~^

\316\277\342\226\240\342\226\240\342\226\240\316\277\317\206^1

\316\277\317\206~^(\316\247)

sit side-by-side near \316\276\316\271.


b. Show that </?ifc \320\276(^>ifcl \320\276\342\226\240\302\267\302\267\316\277\317\206\316\2572\316\277\317\206\316\257\316\273is horocyclic if and only if it is

parabolic in the sense of Exercise2.7.Exercise6.15.All the tessellations constructed in this chapter areinvariant under a transformation group \316\223,as defined in Chapter 7. Do an

Internet search for examples of tessellations which are invariant under nonontrivial transformation group. Suggested key word: Penrose.

Chapter 7

Group actions andfundamental domains

The tiling groups we encountered in Chapter 6 are examplesof

isometric group actions. The current chapter is devoted to definitions

and basic properties of group actions.

7.1. Transformation groupsA transformation group on a set X is a family \316\223of bijections

7: X \342\200\224\342\231\246X such that

(1) if 7 and 7' are in \316\223,their composition 707' is also in \316\223;

(2) the group \316\223contains the identity map Idx;

(3) if 7 is in \316\223,its inverse map 7_1 is also in \316\223.

Recall that the identity map \316\231\316\254\317\207:X \342\200\224>X is defined by the

property that Idx(x) = \317\207for every \317\207G X. Also, the inverse of abijection\317\206:X \342\200\224>\316\245is the map \317\206-1:\316\245\342\200\224\342\231\246X such that \317\206-1(\321\203)is the

number \317\207such that \317\206(\317\207)=

\321\203.This is equivalent to the propertiesthat \317\206\316\277

\317\206-1=

\316\231\316\254\316\263or \317\206\"1\316\277\317\206

=\316\231\316\254\317\207.

A transformation group \316\223naturally gives rise to a map \316\223\317\207\316\247\342\200\224>

X which to (7, x) associates y(x). This map is traditionally defined

as the group action of \316\223over X.

185

186 7. Group actions and fundamentaldomains

We are particularly interested in the situation whereX isa metricspace(X, d) and where all the elements of \316\223are isometrics of (X, d).In this case we say that \316\223is a group of isometries of (X,d) or

that \316\223acts by isometries on (X, d) or that we have an isometric

action of \316\223on (X, d).

It is time for a few examples.

Example 7.1. The trivial group \316\223= {Idx}, consisting of the singlebijection Idx, satisfies all three conditions since Idx \316\277\316\227\317\207

=\316\231\316\254\317\207and

Id^1= Idx. It is the smallestpossibletransformation group for X,

and clearly acts by isometries.

Example 7.2. The set \316\223of all isometries \317\206:\316\247\342\200\224*X of (X, d) is atransformation group, obviously acting by isometries. By definition,this is the isometry groupof (X, d). Clearly, it is the largest groupof isometries acting on (X, d).

Example 7.3. In the hyperbolic plane (H2, dbyp), the set of linear

fractional maps \316\266\320\275->^^

with integer coefficients a, b, c, d \302\243\316\226and

with ad \342\200\224be = 1 forms a group of isometries. Indeed, a computationshows that the compositionof two such linear fractionals also hasinteger coefficients. The identity map can be written as \316\266\320\275->

<jf+f'

and the inverse of \316\266~ ^ is \320\263~-\302\243\302\243.

The set of antilinear fractional maps with integer coefficients is

not a transformation group.Indeed,the composition of two antilinear

fractionals is linear fractional, and not antilinear fractional. Also, theidentity is not antilinear fractional.

Example 7.4. Let \317\210\\,\317\2102,\342\226\240\342\226\240\342\226\240,\317\210\316\267be a family of bijections of a setX. Let \316\223be the set of all bijections \317\206of X which can be writtenas a compositionof finitely many \317\210\316\271and their inverses. Namely,\316\223is the set of all \317\206of the form \317\210

=\317\206^1

\316\277\317\206^\317\207

\316\277\342\226\240\342\226\240\342\226\240\316\277\317\206^1

where

the indices ij G {1,2,... ,n} are not necessarily distinct. Then \316\223is

a transformation group. It is actually the smallesttransformation

group containing \317\206\316\271,\317\2102,. \342\226\240.,\317\210\316\267\302\267By definition, the transformation

group \316\223is then generated by \317\206\316\271,\317\2062,..., \317\210\316\267\302\267

As a fundamental example, all the tiling groups constructed in

Chapter 6 are transformation groups of the euclidean plane, thehyperbolic plane or the sphere, acting by isometries.

7.2. Group actions and quotient spaces 187

7.2. Group actionsand quotient spaces

Let \316\223be a group of isometries of the metric space(X,d). We will use

the action of \316\223to create a new type of quotient space X.

For every P, let \316\241consist of those Q G X which are of the form

Q = j(P) for some 7 \302\243\316\223.This subset \316\241\321\201X is the orbit of thepoint \316\241under the action of \316\223.The orbit is also denoted as \316\241= \316\223(\316\241).

Lemma 7.5. As \316\241ranges over all points of X, the orbits \316\241form a

partition of X.

Proof. The proof uses all three conditions in the definition oftransformation groups at the beginning of Section 7.1.

SinceIdx must be in \316\223,the point \316\241= Idx(P) is an element of

P. Therefore, the sets \316\241cover all of X.

It remains to show that distinct \316\241and Q must be disjoint. Forthis, supposethat \316\241and Q share a common point R. We want to

show that \316\241= Q. By definition, there exists \316\261,\316\262\302\243\316\223such that

R = a{P) and R =/3(Q). As a consequence, Q =

\316\262~\317\207\316\277\316\261(\316\241).

For every S \302\243Q, there exists an element 7 \302\243\316\223such that S =

7(<5). Therefore, S =\316\267\316\277\316\262~\316\271\316\277\316\261(\316\241).The conditions in the definition

of a transformation group imply that 7 \316\277\316\262-1\316\277\316\261is an element of \316\223.

Therefore, 5 is an element of the orbit P. Sincethis holds for every

S \302\243Q, we conclude that Q is contained in P.

Conversely, \316\241is contained in Q by symmetry so that \316\241= Q

whenever \316\241and Q intersect. D

Let X be the partition defined by the orbits P. Lemma4.1 shows

that the metric d induces a quotient semi-metric d on X.

When the semi-metricd is a metric, the metric space (X, d) is

the quotient space of X under the action of \316\223,or the orbit spaceof the action of \316\223on X. It is also denoted by X = X/Y.

In this situation, the semi-metricd is much simpler than in the

general settingof Chapter 4.

Proposition 7.6. Let the group \316\223act by isometries on the metric

space (X,d), and consider the quotient space (X, d). Then, for every


P,QeX,d(P,Q)= iui{d(P',Q');I* eP,Q'e Q}

= inf{d(P,7(Q));7Gr}.

In other words,the (semi-)distancefrom \316\241to Q is equal to theinfimum of the distances from points in the orbit of \316\241to points in the

orbit of Q, which is also equal to the infimum of the distances from

the point \316\241itself to points in the orbit of Q.

Proof. Define

d'(P,Q) = inf{d(P',Q'); P' \342\202\254P,Q'e Q}.

We want to show that d = d'.

Let us first show that d! satisfies the Triangle Inequality, namely

that d'(P,R) < d'(P,Q) + d'(Q,R) for every P,Q, Re X.

By definition of the infimum there exists, for every \316\265> 0, pointsP' eP and Q' e Q such that

d(P',Q')^d'(P,Q) + e.

Similarly, there existsQ\" G Q and R\" G R such that

d{Q\",R\^d'(Q,R)") + e.

The fact that both Q' and Q\" are in the orbit Q means that there

exists 7', 7\" G \316\223such that Q' = -y'(Q) and Q\"= l\"{Q). As a

consequence, Q\" = y(Q') for 7 = 7\" \320\276\321\203'-1g \320\223.Consider the pointR' =

7_1(\320\224\,") which is also in the orbit R of R. Then,

d\\P, R) < d(P', R') < d(P',Q')+ d{Q\\ Rf)

<d(P',Q') + db(Q')n(R'))

^d(P',Q') + d(Q\",R\

^d'(P,Q)+d'(Q,R) + 2e

using the fact that 7 is an isometry of (X, d).

Since this folds for every \316\265> 0, it follows that d'(P, R) < d'(P,Q)+

d'(Q,R). Namely, d! satisfies the Triangle Inequality.

Also, note that the definition of d! immediately implies that

d'{P, Q) iS d{P, Q) for every P,QeX.


After these two preliminary observations, we are now ready to

complete the proof. By definition, d(P, Q) is equal to the infimum of

the lengths idiw) of all discrete walks w from \316\241to Q, namely, walks

of the form \316\241= PY, QY ~ P2, ..., Qn-Y~ Pn, Qn = Q. Then,

\316\267 \316\267

*dH =\316\243d(p\" Qi) >

\316\243d'(p\302\273 U) > J'(-Pb 0\342\200\236)= d'(P, Q)

i=l i=l

by repeated use of the Triangle Inequality for d'. Sincethis holds for

every discrete walk w, we conclude that d(P, Q) > d'(P, Q).

Conversely, if P' e \316\241and Q' G Q, d(P,Q) = d(P',Q') <d(P',Q') by Lemma 4.2. Since d'(P,Q) is defined as the infimum

of those distances d(P', Q'), we concludethat d(P, Q) < d'(P, Q).The combination of the two inequalities shows that

d(P, Q) = S{P,Q)=

inf{d(p't \317\201');\317\201'e \317\201,Q' e Q}\302\267

It remains to show that d'(P, Q) is equal to

d\"(P,Q) = mf{d(P,Q');Q' e 0} = inf{d(P,7(<2));7e \316\223}.

This is immediate, once we realize the following equalities between

subsets of \316\234

{d(P',Q'):P' eP.Q'eQ} = {d(a(P),/3(Q));a,/3 e \316\223}

= {d(P,a-1o/3(Q));a,/?er}

=\320\264\320\2607((?));7\320\265\320\263}

using the fact that every \316\261G \320\223is an isometry of (X, d). \320\236

The isometric action of \320\223on the metric space (X, d) isdiscontinuous if, for every \316\241\302\243X, there exists a ball \316\222\316\261(\316\241,\316\265)centered at

\316\241such that there are only finitely many 7 e \316\223with j(P) \302\243Bd(P, \316\265).

Theorem 7.7. //\316\257/ie transformation group \316\223ocis 61/ isometries and

discontinuously on the metric space (X,d). the semi-metric d inducedon the quotient space X =

\316\247/\316\223is a metric. In particular, the quotient

space (X, d) is a metric space.

Proof. We have to show that d(P, Q) \317\2060 whenever \316\241\317\206Q.

190 7. Group actions and fundamental domains

Becausethe action is by isometries, Proposition 7.6 assertsthat

d(P, Q) is equal to inf {d(P, Q'); Q' e Q}. Since the action isdiscontinuous, there exists an \316\265> 0 for which there are only finitely many

7 G \316\223such that -\316\263(\316\241)e Bd(P,e).

This implies that the ball Bd(P,\302\247)can contain only finitely many

points Q' of the orbit Q of Q. Indeed, if j(Q) and y'(Q) G Q areboth in the ball Bd(P,f), then

d(P,7O7-1(P)) =\316\254(\316\212'-1(\316\241),\316\212-\\\316\241))

^d(1'-1(P),Q)+d(Q,1-1(P))

< d(P,y'(Q)) + d(y(Q), P) < f + f=

\316\265,

using the fact that 7 and 7' are both isometries. By discontinuity of

the action, 7\316\2147\"1 can take only finitely many values in \316\223,so there

can be only finitely many such 7. This proves that the intersectionof the orbit Q with the ball Bd(P,|) isa finite set {Qi, Q2,. \302\267.,Qn},

possibly empty.

If the intersection is nonempty, note that no Qi can be equal to\316\241since \316\241^Q, Therefore,

d(P, Q) = mm{d(P,Qx),d(P;Q2),...,d(P, Qn)} > 0

since the numbers d(P,Qi) areall different from 0 and since there areonly finitely many of them.

If Bd(P, |) contains no Q' G Q, then d(P, Q) ^ \302\247> 0. Therefore,

d(P, Q) \317\2060 in both cases. D

The stabilizer of the point \316\241G X for the action of \316\223on X is

\316\223\317\201= {7 G \320\223;7(\320\240)

=\316\241}.The stabilizer \316\223\317\201is easily seen to be a

transformation group acting on X.

If the action of \316\223on the metric space (X, d) is by isometries, note

that every element of the stabilizer\316\223\317\201sends each ball \316\222\316\261(\316\241,\316\265)to

itself. In other words, \316\223\317\201acts on \316\222^{\316\241,\316\265)and we can consider the

quotient space\316\222^(\316\241,\316\265)/\316\223\317\201with the quotient metric induced by d.

Theorem 7.8. // the transformation group \316\223acts by isometries and

discontinuously on the metric space(X,d), then, for every \316\241in the

quotient metric space (X,d), there exists a ball \316\222^(\316\241,\316\265)which is

isometric to the quotient space \316\222\316\261{\316\241,\316\265)/\316\245\317\201.


Proof. By hypothesis, there exists a ball \316\222^\316\241,\316\265\316\271)which contains

7(P) for only finitely many 7 G \316\223.As a consequence, there exists an\316\265> 0 such that d(P, 7(F)) ^ 5\316\265for every 7(F) G \316\241different from P.

Let \316\222\316\261(\316\241,\316\265)denote the quotient space \316\222\316\261(\316\241,\316\265)/\316\223\317\201,endowed

with the quotient metric d induced by the restrictionof d to B^{P, \316\265).

Also, let \316\241denote the point of B,j(P, \316\265)corresponding to \316\241G Bd(P, \316\265).

Since \316\223\317\201is contained in \320\223,any orbit under the action of \316\223\317\201is

contained in a unique orbit of \320\223.It follows that there is a well-defined

map \317\206:\316\222\316\261(\316\241,\316\265)\342\200\224>

Bj(P,e) defined by the property that \317\210{0)= Q

for every Q G \316\222^\316\241,\316\265).Note that the point Q G X is indeedin theball B$(P, \316\265)since the quotient map X \342\200\224>X is distance nonincreasing

by Lemma 4.2. We will show that \317\210is an isometry.

Let Q, Q' G \316\222\316\261(\316\247,\316\265).Proposition 7.6 shows that

<%(Q), \317\206{$))= d{Q, Q') = inT{d(Q, 7(Q')),7 \342\202\254\316\223}< d{Q, Q').

If d(Q, j(Q1)) is sufficiently close to the infimum that d(Q, 7(Q')) ^J(Q,Q')+ \316\265,a repeated use of the Triangle Inequality shows that

d{P, 7(P)) < d(P, Q)+ d(Q,7(Q')) + d(y(CT), 7(P))

< d(P, Q) + d(Q,0') + \316\265+ d(Q\\ P)

< d{P, Q) + d{Q,P) + d(P,Q')+ \316\265+ d{Q', \316\241)< 5\316\265.

By choice of \316\265,this implies that j(P) = P. What this proves is that if

d(Q, j(Q')) is very close to the infimum d(Q, Q'), then 7 is in \320\223\321\200\321\201\320\223.

As a consequence, the two infimums

d(g,Q') = mf{d(Q,7(Q'));7er}and

d(Q,Q')=

inf{d(QMQ'))-n e \320\223\321\200}

are equal. This proves that d(ip(Q), ip(Q')) = d(Q,Q')=d(Q, Q') for

every Q, Q' \342\202\254\316\222\316\254(\316\241,\316\265).

In particular, \317\206is injective.

If_Qis a point of Bg{P,\316\265),then inf {d(P', Q'); P' G P, Q' G Q}

=

d(P, Q) < \316\265,so that there exists a Q' G Q such that d{P, Q') < \316\265.

As a consequence, Q = <p(Q')is in the imageof \317\206.This proves that

\342\200\242\317\201:\316\222\316\254{\316\241,\316\265)\342\200\224*

\316\222^(\316\241,\316\265)is surjective, and concludes the proof that \317\206

i.s an isometry. D


The action of \316\223on X is free if, for every 7 \342\202\254\316\223\342\200\224{Idx}, j(P) \317\206\316\241

for every \316\241\302\243X. In other words, the action is free if the stabilizer

\316\223\317\201of every point \316\241\316\225\316\247is the trivial group {Idx}.

Corollary 7.9. If the transformation group \316\223acts by isornetnes, dis-

contimiously, and freely on the metric space (X,d), then the quotient

metric space (X,d) is locally isometric to (X,d).

Proof. If \316\223\317\201= {Idx}, then Bd(P, \316\265)is equal to \316\222^(\316\241,\316\265),and the

quotient metric d coincides with d (compare Exercise 4.3). Therefore,Theorem 7.8 shows that every \316\241\316\225\316\247is the center of a ball Bg,{P,\316\265)

which is isometric to a ball Ba(P,\316\265)in \316\247, \316\237

When (X,d) is the hyperbolic plane (H2;\321\2012\321\214\321\203\321\200),Exercises 7.12

and 7.14 discuss the possibletypes for the quotient space Bd{P, \316\265)/\316\223\317\201

when the stabilizer is nontrivial. It is isometric,either to ahyperbolic cone with cone angle ^\317\212,

or to a hyperbolic disk sector of angle

^, for some integer \316\267^ 1. By Theorem 7.8, this consequentlydescribes the local geometry of the quotient of the hyperbolic plane by

a discontinuous isometric group action.Thesameresults(andproofs) hold without modifications for

quotients of the euclidean plane (M2,deuc)or of the sphere (\302\2472,efcph) under

a discontinuous isometric group action.

7.3. Fundamental domains

Let the group \316\223act by isometries on the euclidean plane (M2. deuc),

the hyperbolic plane (H2, e?hyp) or the sphere (S2, \320\265?\320\262\321\200\321\214).For simplicity,

write (X, d) = (R2,deuc),(K2,dhyp) or (S2, dsph), according to thecaseconsidered.

A fundamental domain for the action of \320\223on X is a connected

polygon \316\224\320\241X such that as 7 ranges over all elements of \316\223,the

polygons 7(\316\224) are all distinct and form a tessellationof X.

The tessellations that we constructed in Chapter6 provide many

examples of fundamental domains. For instance, Figures6.4 and

6.5 provide two different fundamental domains for the same group \316\223

acting isometrically on the euclidean plane (\320\2342,\321\201?\320\265\320\270\321\201))and consisting

7.3. Fundamental domains 193

of all integral translations (x, \321\203)\320\275->

(x + m,y + n) with \317\200\316\271,\316\267G \316\226;see

Exercise 7.7.

We will see that fundamental domains can be very useful for

proving that an action is discontinuous and for determining the geometryof the quotient spaceX/Y.

7.3.1. Fundamental domains and discontinuity.

Proposition 7.10. If the isometric action of \316\245on (X, d) = (M.2,deac),(\320\2352,\320\265?\321\214\321\203\321\200)or (S2,dBph) admits a fundamental domain A, the action

of \320\223is discontinuous.

Proof. Consider \316\241& X. By the Local Finiteness Condition in thedefinition of tessellations, there exists an \316\265> 0 and a finite subset

{7it72t \342\226\240\342\226\240\342\226\240,7n} \320\241\320\223such that the 7,- are the only 7 \302\243\320\223for which

7(\316\224) meets the ball \316\222\316\261(\316\241,\316\265).Without loss of generality, \316\241is in the

polygon 7! (\316\224).

If 7 G \316\223is such that 7(F) G \316\222\316\254(\316\241,\316\265),then 7 \320\27671 (\316\224)contains

7(P), and consequently meets the ball Bd(P,\316\265).Therefore, 707! = 7$and 7 =

7,- \320\2767^~* for some i = 1, 2, ..., n. It follows that there are

only finitely many such 7 G \320\223. \342\226\241

In particular, by Proposition 7.6, there is a well-defined quotient

metric space (X, d) = (X/Y,d) when \320\223admits a fundamental domain

\316\224.The next section describes this quotient space in terms of edge

gluings of the polygon \316\224.

7.3.2. Fundamental domains and quotient spaces. Let \316\224be

;i fundamental domain for the isometric action of \316\223on (X, d) =

(M2,deuc), (H2,dhyp)or (S2,c4ph).Let Eu E2, ..., En be its edges.In the tiling of X by the images of \316\224under the elements of \316\223,each

odge Ej separates \316\224from some other tile 7^(\316\224) and coincides with

l he image jj (\302\243^)of an edge Ei}, possibly with EiJ

= Ej. Note

(hat Ei. separates\316\224from 7~1(\316\224), and is equal to ^~l(Ej), so that

~}ji=

771\302\267As a consequence, the rule j 1\342\200\224>\320\263defines a bijection of

\\{.2.,...,n}.

194 7. Group actionsand fundamental domains

In this situation, it is more convenient to write j =j,- and \317\206\316\271

=

\316\212\316\267=

7\302\273\"\"1\302\267^n particular, \317\206\316\257is an element of \316\223,and sends the edge

Ei to the edgeEjt.

Theorem7.11.Let A be a fundamental domain for the isometric

action of the group \316\223over (X, d) = (M2,deuc),(H2,dhyp) or (S2, dBph).As above, consider for each edge Ei of A the element \317\210\316\271G \316\223sending

Ei to some other edgeEji (possibly equal to Ei). Then the group \316\223

is generated by the \317\206^.

Proof. Let 7 G \316\223.Let g be an oriented geodesicarc in (X,d) going

from a point \316\241in the interior of \316\224to a point Q in the interior of

7(\316\224).

We claim that g can meet only finitely may vertices and finitely

many edges of the tessellation of X by the images of \316\224under the

elements of \316\223.This follows from the fact that as a boundedclosedsubset of X = M2, H2 or \302\2472,the geodesic g is compact (seeTheorems 6.13, 6.15 or 6.16). If g met infinitely many edges, looking at

the intersection points of g with these edges and extracting a

converging subsequence, the limit of that subsequence would contradictthe LocalFinitenessCondition in the definition of tessellations.

In particular, g meets only finitely many vertices of thetessellation. By slightly moving g (and its endpoints\316\241and Q), we can

arrange that it actually meetsno vertex of the tessellation.

Let \316\224,7\316\271(\316\224),72 (\316\224),..., 7m-i(A), 7\321\202(\320\224)= 7(A) be the tiles of

the tessellationtraversed by g, in this order. It is quite possiblethat

two 7fc are equal when the fundamental domain \316\224is not convex. Let

Eik and Ejk be the edges such that the geodesic g enters the tile jk (\316\224)

by the edge jk{Eik) and exits it by the edge jk(Ejk). In particular,lk{Eik) =

7fc_i(\302\243j/,_1),with the convention that 70 = Idx.

By construction, \317\210is the unique element of \316\223that sends \316\224to

a tile (fik(A) adjacent to \316\224in such a way that ipik{Eik) coincideswith another edge of \316\224.It follows that \317\210^

=Jk-i \302\2607*\302\273namely>

Ik =7fc-i 0iPik-

By induction,

7 =7m =\317\210\316\267

\302\260\317\210\316\2572

\302\260\342\226\240\342\226\240\342\226\240\302\260<Pim-

7.3. Fundamental domains 195

Since this holds for every 7 G \316\223,this proves that \316\223is generated by

the \317\206\316\271. \320\236

The isometries ipi, sending an edge Ei of the polygon \316\224to another

edge Ejt, define a gluing data of the type considered in Sections 4.3

and 4.5. However, note that it is quite possible that ji = i, so that

the edge Ei may actually be glued to itself as in Section6.3.4.Let (\316\224,\320\265?\320\264)be the quotient space obtained from the polygon \316\224

by performing the edge gluings specified by the isometries<p,-. Here,

we are using hats\"\"

instead of bars\"

to distinguish objects in (\316\224,\320\265?\320\264)

from elements of the other quotient metric space(X,d)= (X/Y, d),

the quotient of (X,d) by the action of the group \320\223.In particular, a

point \316\241\316\265\316\224defines points \316\241G A and \316\241\302\243X m. each of these two

quotient spaces.

Theorem 7.12. Let the group \316\223act by isometries and discontinu-

ously on (X,d) = (R2,deuc),(\320\2502,dhyp) or (S2, dsph), and let A be a

fundamental domain for the action \316\277/\316\223.Then, for the above

definitions, the space (\316\224,\320\265?\320\264)obtained by gluing edges of A is isometric to

the quotient space (\316\247/\316\223,d) of (X, d) by the action of \316\223.

Proof. By definition of the gluing process, if two points \316\241and Q G \316\224

give the same point \316\241= Q in the quotient space \316\224,then Q is the

image of \316\241under a composition of gluing maps \317\206\316\271(none if \316\241is an

interior point of \316\224,exactly one \317\206\316\271if \316\241and Q belong to edges butare not vertices, and possibly several \317\210\\when \316\241and Q are vertices).As a consequence,when \316\241and Q define the same point in \316\224,there

exists an element 7 \302\243\316\223such that Q = 7(P), so that \316\241and Q also

define the same element \316\241= Q in X/Y.

We can therefore define a map p: A \342\200\224+X/Y by associating \316\241G

X/Y to \316\241G \316\224.Indeed, the above observation shows that \316\241G X/Y

does not depend of the point \316\241that we used to represent \316\241G \316\224.

To show that \317\201is surjective, consider an element \316\241G X/Y,

represented by \316\241G X. Because the images 7(\316\224) of the fundamental

domain \316\224under the elements 7 of \316\223cover all of X, there exists apoint P' e A and a group element 7 G \316\223such that \316\241= j(P')- Then,


\316\241is equal to P' in the quotient space\316\247/\316\223,which itself is the imageunder \317\201of the element P' G \316\224.This proves that \317\201is surjective.

To prove that \317\201is injective, suppose that P, Q G \316\224are such

that p(P) = p(Q), namely, such that \316\241= Q in \316\247/\316\223.This property

means that there exists 7 G \316\223such that \316\241= j(Q). In particular,the two tiles \316\224and 7(\316\224) meet at P. If \316\241is in interior point of \316\224,

this is possible only if 7= Idx, so that \316\241is equal to Q. If \316\241is in

an edge and is not a vertex, then by definition 7 is one of the gluing

maps \317\210\316\257,so that \316\241and Q =\317\210\316\257(\316\241)define the same point \316\241= Q

in \316\224.Finally, if \316\241is a vertex, let \316\224=7\316\277(\316\224),7\316\271(\316\224),72(\316\224), ...,

7*;(\316\224)= y(A) be a sequence of tiles goingfrom \316\224to 7(\316\224) around

P. There are two possible such sequences,accordingto the direction

in which one turns around P. By construction, each tile jj(A) meets7^+1(\316\224) along an edge 7J+i(\302\243'ij)

near P, so that7J\"1

\320\276jj+1 is equal

to the gluing map \317\206^.In particular, the two vertices j~1(P) and

7~\320\224(\316\241)are glued together by the gluing map \317\210^.As a consequence,

the element \316\241G \316\224contains the vertices \316\241=7\316\277\"1(\316\241),\316\212\316\263\316\273{\316\241),...,

lk\\ (\320\237 \320\252\320\247\320\240)=

\316\244-1(\316\241)= Q of \316\224,so that \316\241= Q.

In all cases we therefore showed that if two points \316\241and Q \302\243\316\224

define the same point \316\241= Q G \316\247/\316\223,then they also correspond tothe samepoint \316\241= Q in the quotient space \316\224.This proves that \317\201is

injective, and therefore bijective.

Since \316\224is a subset of X, everydiscretewalk w from \316\241to Q \342\202\254\316\224in

\316\224is also a discrete walk from \316\241to Q G X/T in X, and \302\243d(w)< id& (w)since d(P',Q') < db{P',Q') for every P', Q' G \316\224.It follows that

d(P,Q)^dA(P,Q).

Conversely, consider P, Q G \316\224\321\201X. For \316\265> 0, Proposition 7.6

provides a 7 G \316\223such that d(P, j(Q)) < d(P, Q)+ \316\265.As in the proofof Theorem 7.11,let g be a geodesicarc going from \316\241to j(Q) in

X, and meeting the tiles \316\224=7\316\277(\316\224),7\316\271(\316\224),72(\316\224), ..., 7\342\204\242-\316\271(\316\224),

7m (\316\224)= y(A) in this order. This time, g is allowed to cross the

vertices of the tiling.For every i, let Pi and Qi G \316\224be such that the piece of g crossing

7,(\316\224) goes from ^(P;) to 7\302\273(<?\302\273)\342\226\240Note that dA(Pi,Qi) = d(Pi.Qi)since these two points are joined by a geodesiccontained in \316\224.Also

7.4. Dirichlet domains 197

7i+i(Pi+i) = 7i(Qi),so that Pi+1= Qi in \316\247/\316\223;it follows that Pi+i

and Qi are glued together in \316\224since we just proved that \317\201is injec-

tive. Similarly, the last point Qm is glued to Q in \316\224.Therefore, the

sequence \316\241= P0, QQ ~ Pb Qi ~^P2, u-, Qm-i~Pm, Qm ~ Q

forms a discretewalk w from \316\241to Q in \316\224,whose e^-length is

m mm^H = Y,dA(P,Qi)= Y^diPuQi)

=^<7i(Pi),7i(Qi))

i=0 i=0 i=0= i{g)= d{Pn{Q))<d{P,Q)+e,

where\302\243(g)is the euclidean, hyperbolic or spherical arc length

according to whether X =\320\2502,\320\2502or S2. It follows that dA(P,Q) <

d(P, Q)+ \316\265for every \316\265,and therefore that \320\260\320\220(\320\240,Q) < d(P, Q).

Since we had already proved the reverse inequality, this showsthat dA(P,Q) = d(P,Q) for every P, Q G \316\224,and completes the

proof that \317\201:\316\224\342\200\224+X/T, which associates \316\241G X/Y to \316\241G \316\224,is an

isometry from (\316\224,\321\201/\320\264)to (\320\245/\320\223,d). D

7.4. Dirichlet domains

We now prove a converse to Proposition 7.10by showing that for any

discontinuous group \320\223of isometries of the euclidean plane (R2, denc),

the hyperbolic plane (H2, d^yp)or the sphere (S2, dsph)) there always

exists a fundamental domain for the action of \320\223.We will not use this

property until Section 12.4, where a variation of Dirichlet domains,calledFord domains, will play an important role.

This conversewill hold provided we slightly extend our definition

of polygons in (X,d) =(M2, c4uc); (H2,dhyP) or (S2,deph). Namely,

we will allow a polygon \316\224to have infinitely many edges and vertices,provided that the families of edges and vertices are locally finite in

the following sense: For every \316\241\302\243\316\224,there is a ball Bd(P, \316\265)centered

at \316\241which meets only finitely many edges and vertices of \316\224.Such

a polygon will be called a locallyfinite polygon to distinguish this

notion from the finite polygons considered so far. By compactness of

S2, one easily sees that a locally finite polygon in the sphere S2 canhave only finitely many edges and vertices, and consequentlyis finite.


If the group \316\223of isometries of the metric space (X,d) acts dis-

continuously on X, the Dirichlet domain of \316\223centered at the pointPo G X is the subset

\316\224\316\223(\316\2410)= {PeX; d(P, P0) < d[P,7(P0))for every 7 G \316\223}.

In other words, the Dirichlet domain centeredat Po consists of those

points that are at least as close to Po as to any other points of its

orbit. It is named after Gustav Lejeune Dirichlet (1805-1859), who

introduced a similar construction to study the number-theoreticproperties of quadratic forms with integer coefficients.Dirichlet domainsand their variations are ubiquitous in mathematics, where they ariseunder many different names. For instance, if we replace the orbit of

Po by an arbitrary locally finite subset \320\220\320\241X, the similarly defined

subset of X is calledthe Voronoi domain of A centered at Po G A.

See also the Ford domains that we will consider in Section 12.4.

Theorem 7.13. Let the group \320\223act by isometries and discontinu-

ously on (X, d) = (R2,deuc), (\320\2302,dhyp) or (S2, d8ph). Then, for every

Pq G X, the Dirichlet domain \316\224\317\201(\316\241\316\277)is a locally finite polygon and,as 7 rangesover all the elements of the group \316\223,the

7(\316\224\317\201(\316\241\316\277))form

a tessellation of X.

If, in addition, the point Pq is fixed by no element of \316\223except

for the identity, then the Dirichlet domain \316\224\317\201(\316\241\316\277)is a fundamentaldomain for the action \316\277/\316\223.

We will split the proof of Theorem 7.13into several lemmas. The

following elementary construction is the key geometric ingredient.

Lemma 7.14. Let \316\241and Q be two distinct points in (X,d) =

(R2,dexic), (H2,dhyP) or (S2,\302\24328ph). Then the set of R e X such that

d(P,R)= d(Q,R) is a complete geodesic Ppq of X. In addition, the

set of R G X such that d(P, R) < d(Q, R) is the closed half-space

delimited by Ppq in X and containing P.

Proof. This is immediate by elementary geometry in the euclideanand spherical case. See Exercise 2.4 for the hyperbolic case. D

By analogy with the euclidean case, the geodesicPpq is theperpendicular bisector of the points \316\241and Q.


Lemma 7.15. if \316\223acts by isometry and discontinuously on (X,d) =(M2, deuc), (H2, dhyp) or (S2, dgph) then, for every \316\241e X and r > 0,there are only finitely many 7 e \316\223such that d{P, 7(Po)) ^ r-

Proof. Suppose,in search of a contradiction, that there are infinitely

many elements 71, 72\302\273\302\267\302\267\302\267,7n, \342\226\240\342\226\240\342\226\240*\320\277\320\223such that all 7n(Po) belongto the closedball \320\222of radius r centered at P, consisting of all Q \302\243X

such that d(P, Q) < r.One easily checks that \320\222is a closed subset of X. Sinceit is

clearly bounded, it follows from Theorems 6.13, 6.15and 6.16 that \320\222

is compact. As a consequence, there existsa subsequence (7nfc)fceN

such that jnk (Pq) convergesto someP*,G \320\222as \320\272tends to 00.

This implies that for every \316\265,there exists \320\260\320\272\320\276such that

d(lnk {Pq), Poo) < I for every \320\272^ \320\272\320\276-In particular, for every \320\272^ \320\272\320\276,

d(Po, lnl0\302\260lnk (Po))

=d(jnk0 (Po), 7n, (Po))

< d(7n*0(Po),Poo)+<\316\241\316\277\316\277,7\316\267,(\316\241\316\270))

So, for every \316\265> 0, we found infinitely many 7 G \316\223for which 7(Po)is in the ball \316\222,\316\271(\316\241\316\277,\316\265),contradicting the fact that the action of \316\223is

discontinuous.

This proves that our original assumptionwas false and therefore

that \320\222contains only finitely many 7(P). \316\240

Lemma 7.16. The Dirichlet domain \316\224\317\201(\316\241\316\277)is a locally finitepolygon.

Proof. For every 7 G \316\223,let ffpo7(p0) be the set of points \316\241\302\243X

such that d(P,P0) < d(P,7(P0)). Lemma 7.14 shows that if 7(P0) \317\206

Po, i/po7(p0) is a half-space delimited by the perpendicular bisector

Pi'o-y(Po) of po and 7(Po)\302\267 If 7(Po) = Po, \316\267\317\201\316\277\316\257(\317\2010)is of coursc thc

whole space X and we set \316\262\317\2010\317\2010to be the empty set in this case.

By definition, the Dirichlet domain \316\224\317\201(\316\241\316\277)is the intersection of

t.he half-spaces HPol(Po)as 7 ranges over all elements of \320\223.To prove

that \316\224\317\201(\316\241\316\277)is a locally finite polygon, it consequently suffices to

show that the family of geodesies /3po7(p0) bounding the\320\224\"\321\20007(\321\2000)

is


locally finite. Namely, for every \316\241G X, we need to find an \316\265such

that the ball \316\222\316\261(\316\241,\316\265)meets only finitely many \316\262\316\241\316\277\316\212^\316\241\316\277\316\263

Actually, any \316\265> 0 will do. If/\320\227\321\200\320\2767(\321\2000)

\320\237Bd(P, \316\265)is nonempty,

pick a point Q in this intersection. By the Triangle Inequality, the

point Q is at distance< d(P.Po)+ \316\265from Pq. Since Q G /J7, it is

also at the same distance from 7(Po) as from Po. Anotherapplicationof the Triangle Inequality then shows that j(Pq) is at distance

< 2d(P, Pq) + 2\316\265from Po\302\267By Lemma 7.15, this can happen for only

finitely many 7 G \316\223.

This proves that the family of the perpendicularbisectors/3p07(p0)

is finite, and therefore that the Dirichlet domain \316\224\317\201(\316\241)is a locally

finite polygon. D

Before proving that the images of the Dirichlet domain \316\224\317\201(\316\241\316\277)

under the elements of \316\223tessellate X, let us first observe that each of

these images is also a Dirichlet domain.

Lemma 7.17. For every 7 G \316\223,7(\316\224\316\223(\316\241\316\277))=

\316\224\316\223(7(\316\2410)).

Proof. The Dirichlet domain\316\224\317\201(7(\316\241\316\277))

consists of those \316\241G X

which are at least as closeto 7(Po)asto any other point of the orbit

\316\223(7(\316\241\316\277))=

\316\223(-\316\241\316\277)\302\267Since 7 is an isometry of X, this property is

equivalent to the fact that 7_1 (P) is at least as closeto Po as to any

other point of its orbit. Therefore, \316\241is in\316\224\317\201(7(\316\241\316\277))

if and only if

7_1(P) is in \316\224\316\223(\316\241\316\277),namely, if and only if \316\241is in7(\316\224\316\263(\316\241\316\277))\302\267

\342\226\241

Lemma 7.18. As 7 ranges over all the elements of \316\223,the Dirichlet

domains\316\224\316\263(7(\316\241\316\277))form a tessellation of X.

Proof. A point \316\241G X belongs to \316\224\316\223(7(\316\241\316\277))if and only if

d(P, 7(Po))is equal to

\320\253{\320\260(\320\240,\320\243(\320\240\320\276\320\243\320\277'eT} = d(P,P0),

where the equality comes from Proposition 7.6. Lemma 7.15 showsthat there are only finitely many 7' G \320\223such that d(P, y'(Po)) <

d(P,Po) + 1. This has two consequences.

The first one is that there exists at least one 7 such that

d(P,7(P0))= d(P,P0). Indeed, the above observation shows that

it suffices to considerthe infimum over finitely many elements, so


that the infimum is actually a minimum. As a consequence, there

exists 7 G \316\223such that \316\241G\316\224(7(\316\2410))\302\267

The second consequence is that the set of 7 G \316\223such that \316\241G

\316\224(7(\316\241\316\277))*s a finite set {71,72, \302\267\302\267\302\267,7n}. In particular, \316\241belongs to

only finitely many Dirichlet domains\320\224\321\200(7(-\320\240\320\276))and, near P, these

Ar(7i(-Po)) arc delimitedby the perpendicularbisectors\316\262\316\212\316\257(\317\2010)\316\212\316\257{\316\2410)\302\267

These two properties show that the union of the domains

\316\224\316\263(\316\257{\316\241\316\277))is equal to X and that when two distinct

\316\224\317\201(7(\316\241\316\277))and

\316\224\317\201(7'(\316\241\316\277))meet, they meet along edges and/or vertices.Sinceany

two of these Dirichlet domains are isometric by Lemma 7.17, this

completes the proof. D

The combination of Lemmas 7.17 and 7.18 shows that the

images of the Dirichlet domain \316\224\317\201(-\316\241\316\277)under the action of \316\223form a

tessellation of X. This was the main statement of Theorem 7.13.

To prove the secondpart of this theorem, assume that thestabilizer \316\223\317\2010consists only of the identity map. Then, whenever 7 \317\206\320\243,

the two points j(Pq) and j'{Pq) are distinct so that the Dirichlet

domains AT(y(P0))=

7(\316\224\316\223(\316\241\316\277))and AT(y'{Po))

=Y(Ar(P0)) are

distinct. Thisis exactly the additional condition needed to prove that

\316\224\316\263(-\316\241\316\277)is a fundamental domain.

This concludesthe proof of Theorem 7.13. D

For Theorem 7.13 to provide a (locally finite) fundamental

domain for the action of \316\223,we need to find a point Pq whose stabilizer

Tp0 consists only of the identity. This of course is automatic when theaction is free. In the generalcase,almost every Pq G X will have the

required property. Indeed, the analysis of all possible stabilizers in

Exercise 7.12 (suitably extendedto the euclideanand spherical

contexts) shows that every point of X can be approximated by a point Flom

page 197with a trivial stabilizer.

We will let you check that the results of Sections 4.3, 4.4, 4.5, 6.3and 7.3on edgegluings of polygons immediately extend from finite

polygonsto locally finite polygons \316\224in X = M2 or H2, provided that

we impose the following additional condition.

Finite Gluing Condition.Each point of the polygon \316\224is glued to

only finitely many other points of \316\224.



Exercise 7.1. For every rational number \302\243,let 72: \320\232\342\200\224\342\226\272\320\226be the trans-

lation defined by 72 (\320\266)\342\200\224\317\207+ f.

a. Show that the set \316\223= {jz; f \342\202\254Q} of all such rational translations is

a goup of isometries of the metric space (\320\226,deUc).

b. Show that the quotient semi-metric d induced by d = deuc on the

quotient space X =\316\247/\316\223is not a metric. Actually, compute d(P, Q)

for every P, Q \302\243X.

Exercise 7.2. This exercise is devoted to a few classical transformation

groups that arise in linear algebra.

a. Let the general linear group GL\342\200\236(M)consist of all the linear maps\320\226\"\342\200\224\342\226\272Mn whose associated matrix \316\234has nonzero determinant

det(M) \317\2060. Show that GL\342\200\236(M)is a transformation group of Kn.Hint: It may be useful to remember that any \316\267x \316\267-matrix \316\234with

det(M) \317\2060 admits an inverse, and that det(MJV)= det(M)det(JV).

b. Let the special linear group SLn(M) consist of all the linear mapsMn \342\200\224\342\226\272Mn whose associated matrix \316\234has determinant det(iW) equalto 1. Show that SLn(M) is a transformation group of Mn.

\321\201Let the projective apace \320\256\320\240\"-1consist of all the lines passing through

the origin in Mn (compare Exercise 2.12). Every \317\206\342\202\254GL\342\200\236(K)induces

a map \317\210:]RPn\"' -\302\273MP\342\204\242\"1,which associates to each line L G MP\"\"1

the line tp(L) G MP\342\204\242\"1.Show that as \317\210ranges over all elements ofGLn(M), the correspondingmaps \317\210form a transformation group ofKPn_1. This group is called the projective general linear groupand is traditionally denoted by PGLn(K)\302\267

d. Let the projective special linear group PSLn(K)consistof all the

maps \317\206:MP\342\204\242\"1\342\200\224\342\226\272MPn_1 induced as in part \321\201by linear maps \317\206G

SL\342\200\236(R).Show that PSLn(M) is a transformation group of MPn_1.

Exercise 7.3 (Abstract groups). A group law on a set \316\223is a map \316\223\317\207\316\223\342\200\224>

\316\223,denoted by (7,7') \320\275-+7\342\226\240

7' G \316\223for every 7, 7' G \316\223,such that:

a- 7 \302\267(V

\302\2677\")

= (7 \302\2677')

\302\2677\" for every 7, 7', 7' G \316\223;

b. there exists an element t G \316\223such that 7 \302\267t \342\200\224l \342\226\2407 \342\200\2247 for every 7 G \316\223;

c. for every \316\223G \316\223,there exists an element 7' G \316\223such that 7 \302\2677'

=

7' \342\226\2407

= L.

Show that if \316\223is a transformation group on a set X, the map \316\223\317\207\316\223\342\200\224\342\226\272\316\223

defined by (7.7') \316\271-\302\2737 \320\2767' is a group law on \320\223.

An abstract group is the data of a set \320\223and of a group law on \320\223.


Exercise 7.4. Let the group \316\223act discontinuously and by isometries onthe metric space(X,d). Show that if X is compact, \316\223is necessarily finite.

Exercise 7.5. Let the group \316\223act by isometries on (X,d) \342\200\224(R2,deuc),

(H2,dhyP) or (S2,dsph)\302\267 Show that if \316\223acts discontinuously at some point

Po \342\202\254X, in the sense that there exists an e > 0 such that *f(P) G B<j(Po,e)for only finitely many 7 \342\202\254\316\223,then it acts discontinuously at all points\316\241G X. Possible hint: Borrow ideas from the proof of Lemma 7.15.

Exercise 7.6. Let the group \316\223act discontinuously on the complete metricspace(X,d).We want to show that the quotient space (X/T, d) is complete.For this, let (Pn)neN be a sequencein this quotient space X = X/T with

finite length \302\243~=1d(Pn,Pn+1) < 00.

a. By induction on n, construct a sequence (Pn)neN in X such that for

every n,P'n = Pn and d{P'n,P'n+\\) < d(Pn,Pn+i) + 2~n. Hint:

Proposition 7.6.

b. Show that the sequence (i^)nern convergesto some point Poo \342\202\254X.

c. Conclude that (Pn)neN converges to P*, in (X/F,d).

Exercise 7.7. Let \316\223be the group of integral translations of the euclidean

plane (.R2,deuc),of the form (x,y) >-<\342\226\240(x + m,y + n) with (m,n) G Z2.

a. Let \316\245be the parallelogram with vertices (0,0), (1,1), (2,1)and (1,0),

as in Figure 6.5. Provethat \320\243is a fundamental domain for the action

of \320\223.Possible hint: Glue opposite sides of Y, apply Theorem 6.1, and

check that the corresponding tiling group is equal to \320\223.

b. Do the same for the parallelogram whose vertices are (0,0), (1,1),(3,2)and (2,1).

Exercise 7.8. Let \320\223be the group of isometries of (S2,dsPh) consisting of

the identity map IdS2 and of the antipode map 7 defined by 7(P) = \342\200\224P.

a. Check that \320\223is really a transformation group acting discontinuously

onS2.

b. Show that the quotient space (S2/T, dsph) is isometric to the projectiveplaneof Section 5.3. Hint: Find a fundamental domain for the actionof \320\223and apply Theorem 7.12.

Exercise 7.9. In the euclidean plane, consider the infinite strip Xw =

[0, \317\211]\317\207(\342\200\224oo,+oo) of width w > 0 and, for t G K, let Xw,t be the cylinder

obtained from Xw by gluing the left-hand side {0} x (\342\200\22400,+00) to the

right-hand side {w} \317\207(\342\200\224oo,+00) by the translation \317\206:{0} x (\342\200\22400,+00) \342\200\224\342\226\272

{w} x (-00,+00) defined by <p(0,y)= (w,y + t). Endow Xw,t with the

quotient metric deuc induced by the euclidean metric deuc of Xw \342\226\240

a. Show that (Xw,t,deac) is isometric to a quotient space (R2/TWtt,dcac),where rWtt is the group of isometries of the euclideanplane (K2,dCUc)


generated by a translation along a certain vector of length \\/w2 +12.Hint: Theorems 6.1 and 7.12.

b. Supposethat w\\ + if =w\\ +1\\. Show that the cylinders {Xw-L,t1, deuc)

and (-Xw2,t2>(ieuc) are isometric. Possiblehint: First find a euclideanisometry \317\206:(R2,de\342\200\236c)\342\200\224\342\226\272(K2,deuc) such that \320\223\320\2502,*2

={\317\210\316\277\316\267\316\277\317\210^-,\316\267

\302\243

\316\223^\316\271,\316\257\316\271},and then use \317\206to construct and isometry ip:(M.2/Fwl,tl, deuc)\342\200\224*

(\320\226/\320\223\321\2102,*2!\"eucj-

\321\201Show that the cylinders (Xw1,o,deac)and (Xw2,o, dmc) are notisometric if \302\253/-\316\271\317\206Wi. Possible hint: Look at closed geodesiesin these

cylinders.

d. Show that the cylinders (-Xtui.tudeuc) and (-Xu>2,t2>deur.) are isometric

if and only if w\\ + t2 =w\\ + t\\.

Exercise 7.10. Let the group \316\223act isometrically and discontinuously onthe metric space (X,d). Let \316\224be the Dirichlet domain of \316\223at P. Show

that \316\224is invariant under the stabilizer \316\223\317\201of P, namely that 7(\316\224)= \316\224

for every 7 \342\202\254\316\223with j(P) = P.

Exercise 7.11. For a \302\243\320\226,let \316\223be the group of isometries of the euclideanplane (M2,deuc) generated by the translations \317\206\316\271:(\317\207,\321\203)

\316\271\342\200\224>(\317\207+ l,y) and

\317\2102'-(\321\205,\321\203)*-* (\317\207+ a,y + 1). Determine the Dirichlet domain \316\224\316\223\316\261(\316\237)of \316\223\316\261

centered at the origin \320\236= (0,0). The answer will depend on a.

Exercise 7.12 (2-dimensionalstabilizers). Let \320\223act by isometries anddiscontinuously on the hyperbolic plane. The goal of the exercise is todetermine the possibletypes for a stabilizer \316\223\317\201.It will be convenient touse the ball model (B2, dB2). Using a hyperbolic isometry \317\210sending \316\241to \320\236

and replacing \320\223by the group {\317\210\316\277\316\267\316\277\317\206~\316\271;\316\2676 \316\223},we can assume without

loss of generality that \316\241is the euclidean center \320\236of the disk B2.

a. Show that every element of the stabilizer \320\223\320\276is the restriction to B2 ofeither a euclidean rotation around \320\236or a euclidean reflection across aline passing through O.

b. Suppose that every element of \320\223\320\276is a rotation. Show that there exists

an integer n > 1 so that \320\223\320\276consists of the \316\267rotations around \320\236of

angles 252E, vvith fc = 0, 1, ..., \316\267\342\200\2241. Hint: Consider an element of\320\223\320\276

\342\200\224{IdH2} whose rotation angle is smallest.

\321\201Suppose that \320\223\320\276contains at least one reflection. Show that there

exists an integer \316\267> 1 and a line L passing through the origin such

that \320\223\320\276consists of the n rotations around \320\236of angles ~-, with \320\272= 0,

1, ..., \316\267\342\200\2241, and the \316\267reflections across the lines obtained by rotating

L around \320\236by angles ^, with fc = 0, 1, ..., \316\267\342\200\2241. Hint: First

consider the elements of \320\223\320\276that are rotations.


Exercise7.13.Let the group \316\223act by isometries and discontinuously

on the hyperbolic plane (E2,dhyp). Let \316\241be a point whose stabilizer \316\223\317\201

contains more elements than just the identity map.

a. Show that there is an angle sector A, delimited by two infinite geodesiesissued from P, which is a fundamental domain for the action of \316\223\317\201.It

may be convenient to use Exercise7.12.b. Let \316\224\316\263(\316\241)be the Dirichlet domain of \316\223at P. Show that the

intersection\316\221\316\240\316\224\316\263(\316\241)is a fundamental domain for the action of \316\223.

Exercise 7.14. Let \316\223act by isometries and discontinuously on the

hyperbolic plane (H2,dhyp).

a. Let \316\223\317\201be the stabilizer of a point \316\241\302\243H2, and choose an arbitrary e >

0. Show that the quotient space (Bdhyp(P,\320\265)/\320\223\321\200,<4\320\243\320\240)is isometric,

either to a hyperbolic cone of radius e and cone angle 2zl for some

integer \316\267^ 1 (as defined in Exercise 4.7 for the euclidean context)or

to a hyperbolic disk sector of radius e with angle ^ for some integer\316\267> 1. Hint: Use Exercises 7.12and 7.13.

b. Assume, in addition, that every element of \316\223is a linear fractional map.Show that the quotient space (\320\2532/\320\223,dhyp) is a hyperbolic surface with

cone singularities, as defined in Exercise 4.8. Hint: Use Theorem 7.8.

Exercises7.12,7.13and 7.14 have immediate generalizations to theeuclidean and sphericalcontext.

Chapter 8

The Farey tessellation

and circle packing

The Farey tessellation and circle packing rank among some of the

most beautiful objects in mathematics. It turns out that they are

closely related to the punctured torus that we encountered in

Section 5.5, and to the tiling group of the corresponding tessellationconstructed in Section6.6. Some of the exercises at the end of this

chapter explore additional connections between the Farey tessellationand various number-theoretic and combinatorial problems.

8.1. The Farey circle packingandtessellation

oo=i

Figure 8.1. The Farey circle packing

207

208 8. The Farey tessellationand circlepacking

For every rational number | G Q with p, q coprime and q > 0,draw in the plane R2 the circle Ce of diameter X that is tangentto the \320\266-axis at

(|,0)and lies above this axis. Thesecircles\320\241\320\265fit

together to form the pattern illustratedin Figure8.1.We can get a better view of this circlepattern by zooming in, as

in Figures 8.2 and 8.3.

\302\273*rrrt\"

516

13

5 414 11

300

513

25 12 UJs

Figure 8.2. Zooming in on the Farey circle packing

413631 26907968 57

21 3746 81

16 4327 3835 9459 83

1124

392845 17 40 23856198 37 87 50

29 3563 76

Figure 8.3. Zooming in once more on the Farey circle packing

8.1. The Farey circle packing and tessellation 209

By inspection, we can make the following experimental

observations:

\342\200\242the circles Ce have disjoint interiors;\342\200\242two circles Ce and CV are tangent exactly when

pq' -p'q =\302\261l;

\342\200\242three circles Ce , Cp> and CP\" with 2 < ^77 < \320\252are tangent9

'fr \302\24777

\320\257 \320\257 9 ^

to each other exactly when ^77 is the Farey sum | \316\270\320\252of

Iand ^7, namely, when

2l \342\200\224\316\225\320\266e! \342\200\224p+p'

\321\217\"~

\321\217w

q' q+q''

The same properties hold if we consider, in addition, the infinite

rational number 00 =g

=^ and introduce \320\241\320\266as the horizontal

Figure 8.4. TheFarey circle packing in the disk model B2

210 8. The Farey tessellation and circlepacking

line of equation j/= 1, while the interior of C^ is dennedas the

half-space consisting of those points (x,y) with \321\203> 1.

The family of the circlesOr is the Farey circlepacking.

Some of the beauty of this collection of circles is also visible in

Figure 8.4, which represents the image of the Farey circlepacking

in the disk model B2 for the hyperbolicplane, under the linear

fractional map \316\246(\316\266)= \342\200\224~

^r sending the upper half-plane to the diskB2.The numbers in that picture label some of the images\316\246(2) of the

corresponding points | G Q-

Quite remarkably, this Farey circle packing is relatedto thehyperbolic punctured torus that we considered in Section 5.5.

We can get a hint at this relationship if we erase the circles Cs.,

and if we connect the two points (^;0) and (^7,0) by a semi-circle

centered on the \320\266-axis exactly when the circles Cs. and C,/ are tan-

gent. The resulting collection of hyperbolic geodesiesis illustrated in

Figure 8.5, which bears a strong analogy with the tessellation of

Figure 6.12 that we associated to our hyperbolic puncturedtorus. This

connection is made precise in Section 8.2.

Figures 8.6 and 8.7 are obtainedby zooming in on this collectionofsemi-circles,calledthe Farey tessellationof the hyperbolic plane.

Figure 8.5. TheFarey tessellation

The Farey circle packing and the Farey tessellation are named

after John Farey (1766 1826), a geologistwho experimentally observed

(without proof) the following elementary property. For a fixednumber N > 0, consider all rational numbers \302\243with 0 < \317\201< q < N

8.1. The Farey circlepackingand tessellation 211

* i \316\234 !A ! & f I \302\243M I

Figure 8.6. Zooming in on the Farcy tessellation

Figure 8.7. Zooming in once more on the Farey tessellation

coprime, and list these numbers by order of increasing size

1 Pi-i Pi Pi+i TV\342\200\2241

TV qi-l Qi Qi+i TV

Then, Pi+iqi\342\200\224

Piqi+i= 1 for any two consecutive \302\243*\342\226\240< \316\2661^, and

\317\212\317\213_\316\241\316\257^\317\206\316\241\316\257\317\213for any three consecutive ^^ < %\342\226\240< \317\210*1.The same

<li \320\257\320\263-1QUI J \320\257\320\263-1\320\257\320\2639\316\271+ 1

property had actually been discovered,with partial proofs, earlier in

L802by C. Haros. However, the name of Farey series of order TV

became attached to the above sequence when Augustin Cauchy, who

was only aware of Farey's note [Farey], provided a completeproof of


these two statements in 1816. A good reference for the Farey seriesis [Hardy & Wright].

Of course, Farey did not know anything about hyperbolic

geometry.The first published account of what we call herethe Farey circle

packing is due to Lester R. Ford (1886-1967) [Ford2]. In particular,these circlesare also often called Ford circles.

Figure 8.8. TheFarey tessellation in the disk modelfor H2

8.2. The Farcy tessellation and theonce-punctured torus

Let us go back to the tessellation of the hyperbolic plane H2

discussed in Section 6.6. We started with the hyperbolic square X with

vertices 0, +1, -1 and ooelU {oo}at infinity of H2. We alsoconsidered the transformation group \316\223generated by the transformations

\317\210\316\271(\316\266)=

f^i and \317\206^{\316\266)=

_z~\\- We then showed that the polygons

8.2. The Fareytessellationand the 1-puncture torus 213

\317\210(\316\247),as \317\210ranges over all the elements of \316\223,form a tessellation of

H2.

Let us split the squareX along the diagonal Ooo into the two

triangles T+ and T~, where T+ has vertices 0, +1 and oo and T~

has vertices 0,-1 and oo. Note that the hyperbolic reflection \320\263\320\275-\320\263

across the vertical half-line Ooc exchangesT+ and T~, so that these

two triangles are isometric. (Actually, Lemma 8.4 below will show

that any two triangles with vertices at infinity are isometric.)

It follows that the collection of the triangles \317\210{\316\244+)and \317\206(\316\244~),

with \317\210e \316\223,forms a tessellation of H2 by ideal triangles.Let \316\244denote

this tessellation.

We will prove that this tessellation is exactly the Farey

tessellation,in the sense that its edges are exactly the hyperbolicgeodesiesjoining | to |j when pq'

\342\200\224p'q = \302\2611.

For this it is convenient to consider,as in Example7.3,the set

PSL2(Z) consisting of all linear fractional maps

, . az + b\317\210(\316\266)

=\320\263-j.cz + a

with a, b, c, d \302\243\316\226and ad \342\200\224be = 1. We already observed inExample 7.3 that PSL2(Z) is a transformation group of H2. See

Exercises 2.12 and 7.2 for an explanation of the notation.

Since the gluing maps \317\206\316\271and </?\320\267generating \320\223are elements of

PSL2(Z), we conclude that \320\223is contained in PSL2(Z).

When we consider a rational number | \317\202Qu{oo}, we will use

the convention that \317\201and q are coprime and that q > 0, with the

exception of oo =\302\243

==\302\243.

We say that E, ^ eQU {oo}form a

Farey pair if pq' \342\200\224p'q = \302\2611.

Lemma 8.1. If \317\206e PSL2(Z) and if 2, 4 e Q U {oc} form a Farey

pair, then\302\245>(\302\247)

an(^ fib) form a Farey pair.

Proof. Immediate computation. D

Note that the pairs {0,oo},{1,oo}and {\342\200\2241,oo} are Farey pairs.From Lemma 8.1 and the fact that \320\223\321\201PSL2(Z), we conclude that

I he endpoints of each edge of the tessellation\316\244form a Farey pair.

214 8. The E^irey tessellation and circle packing

Conversely, let g be a geodesicof H2 whose endpoints form a

Farey pair.

Lemma 8.2. Let gi and gi be two distinct geodesies of H2 whose

endpoints each form a Farey pair. Then gi and #2 ane disjoint.

Proof. If the endpoints of g\\ are\317\210-

and ~h and if we choose the

indexing so that p[qi - piq[ = +1,the map \317\206e PSL2(Z) denned by

\317\210{\316\266)\342\200\224

1\316\271\317\212^+1'sends g to the geodesic with endpoints 0 and 00. Let

\342\200\224and ^ be the endpoints of <p(g2)- If g and g' meet each other,

then 4>{g') must meet ip(g) so that 22. an(j Ea have opposite signs.

But this is incompatible with the fact that by Lemma 8.1, they must

satisfy the Farey relation ^2^2 ~ ^2^2 - \302\2611\302\267 \316\240

Lemma 8.2 shows that a geodesicg whoseendpointsform a Farey

pair must be an edgeof the tessellation T. Indeed, the tiles of the

tessellation \316\244are ideal triangles, and their interiors consequentlycannot

contain any complete geodesic. Therefore, g must meet an edge g' of

T, and must be equal to g' by Lemma 8.2.

This shows that the edges of the tessellation \316\244are exactly the

complete geodesies of H2 whose endpointsin \316\234U {00} form a Fareypair. As a consequence, the tessellation 7 coincides with the Farey

tessellation. In particular, this proves that the Farey tessellation is

indeed a tessellation,something that we had taken for granted so far.

8.3. Horocircles and the Farey circle packing

If \316\276\302\243\316\234U {\320\276\321\201}is a point at infinity of the hyperbolic plane H2,recall from Section 6.8 that a horocircle centered at \316\276is a curve

\316\227= \320\241-{\316\276}\320\241\320\2302where \320\241is a euclidean circle passing through \316\276

and tangent to M. In particular, when \316\276= oo, a horocircle is just a

horizontal line contained in H2.

Note that since linear and antilinear fractional maps send circles

to circles, every isometry of H2 sends horocircle to horocircle.We had already encountered horocircles when analyzing the

punctured torus of Section 5.5. For a given o, we indeed used horocircles

Coo, C_i, Co, C\\ centered at 00, \342\200\2241,0, 1, respectively, to cut out

8.3. Horocircles and the Fareycirclepacking 215

oc

Elj

(

\316\271

^

\320\225\321\217. 1

& )

Figure 8.9. \316\233hyperbolic square with horocircles

pieces \320\270\320\266,U-i, Uq, U\\ from the hyperbolic squareX. If we take the

value \320\276to be equal to 1 (which was not allowed in Section 5.5), thesehorocirclesarc tangent to each other. Sec Figure 8.9.

This is the secretbehind the Farey circle packing. Indeed, weobservedin Section5.5that the gluing maps \317\210\\,\317\2102

=\317\206^1,\302\245>\320\267,\316\250\316\261

\342\200\224

\317\206^1respect this set of four horocircles, in the sensethat\317\210\316\257(\316\237\317\202)

= \320\241\317\202/

whenever \317\206\316\257(\316\276)\342\200\224

\316\276'\342\226\240Therefore, as we tessellate the hyperbolicplane'HI2 by the tiles \317\206{\316\247)with \317\210\302\243\316\223,the images of these four horocirclesunder the transformations \317\206\302\243\316\223form a family of horocircles, allcentered at points

E eQU {oc}\302\267Looking at their intersection with

a given tile, it is immediate that two of these horocirclesmeet only

when they are tangent to each other, and that this happens exactlywhen their centers are the ends of an edge of the Farey tessellation

T, namely, when their centers form a Farey pair.

The family of these horocircles,combined with the Fareytessellation T, is illustrated in Figure 8.10 for the upper half-space H2, andin Figure 8.11 for the disk model B2 for the hyperbolic plane.

To identify this family of horocircles to the Farcy circle packing,

it suffices to combine the following lemma with the fact that \316\223is

contained in PSL2(Z).

Lemma 8.3. Considerthe horocircle C^ = {z e \320\2352,1\321\202(,\320\263)= 1}

centered at oc. For every \317\206\302\243PSL2(Z) of the form

216 8. The Farey tessellationand circle packing

with a, b, c, d \316\266\316\226and ad\342\200\224be = I, the image \317\210(0\316\277\316\277)is equal to

Cs, \342\200\224{f},

where Cs, is the euclidean circle of diameter ^ tangent to

at*.

Figure 8.10.The Farey tessellation and circle packing

Figure 8.11. The Farey circle packing and tessellation in the

disk model B2

8.4. Shearing the Farey tessellation 217

Proof.Since \317\206sends horocircle to horocircle and since it sends oo to

\", the only issue is to compute the (euclidean)diameter of the circle

Cs.. This can be easily checked by decomposing \317\210as a composition of

horizontal translations, homotheties and inversions as in Lemma 2.12.

D

8.4. Shearing the Farey tessellation

We now revisit the complete and incomplete punctured tori of

Sections 5.5 and 6.7.2, and their associated tessellationsor partial

tessellations of the hyperbolic plane.

8.4.1. Revisiting the partial tessellationsassociatedtoincomplete punctured tori. The key property is the following.

Lemma 8.4. Given a triple of distinct points \316\276\316\271,\316\2762,\302\243\320\267\316\225\316\234U {\316\277\316\277}

and another such triple of distinct points \316\276[,\316\276'2,^elU {\320\276\321\201},there

is a unique isometry \317\210of the hyperbolic plane (H2, dhyp) sending each

\316\276\316\271to the corresponding \316\276[.

Also, at each \302\243;,there is a unique horocircle Ci centered at &

such that any two Ci and Cj are tangent to each other and meet at a

point of the complete hyperbolic geodesic going from \316\276\316\271to \302\243j.

Proof. The first part is a simplealgebraiccomputation, using the

fact that every isometry of the hyperbolic plane is a linear or an-

tilinear fractional map; compare Exercise 2.9. Observethat \317\206is

orientation-preserving, namely, it is linear fractional, preciselywhen

(\316\257\316\271-\316\2572)(6-\316\2573)(\316\2573-\317\211and (ii-U)(U-ft)(i3-ii) have the samesign (suitably interpretedwhen one of these points is equal to oo).

To prove the second statement, the first part shows that we canrestrict attention to one specific example, such as the case where

the points are 0, 1 and oo. The result then follows from elementary

geometric considerations. CompareFigure8.9. \316\240

A consequence of Lemma 8.4 is that everyedgeof an ideal triangle

has a preferredbasepoint, namely, the point where the two horocirclescenteredat its endpoints and singled out by Lemma 8.4 touch each

other.


The hyperbolic punctured tori constructed in Sections 5.5 and

6.7.2 were obtained from the ideal squareX with vertices \342\200\2241, 0, 1

and oo by gluing the edge E\\ going from \342\200\2241 to oo to the edge E2going from 0 to 1 by the map

, \316\275z+1

\316\266+ a

and by gluing the edge E3 going from 1 to oo to the edgeE\\ going

from 0 to \342\200\2241 by the map

In the first case that we considered, in Section 5.5, the constantswere

chosen so that \320\276= b = 2 and the construction provided a completepunctured torus.

Split the square X along its diagonal Ooo, namely, along the

geodesic going from 0 to oo. This gives two ideal triangles, one with

vertices \342\200\2241, 0, oo, and another one with vertices 0, 1 and oo.

ApplyingLemma 8.4 to these two triangles now provides a base point oneach of the edges of X, namely, \320\240\320\263

= \342\200\2241 + i e \320\225\320\263,P2 =\\ + ^ \316\225E2,

\320\240\320\267= 1 + i e E3 and P4

= -\\ + | e i\302\2434.Compare Figure 8.9.

When a = b = 2, the gluing maps \317\210\\\302\267.E\\\342\200\224*E2 and \317\2063:\316\225\302\2673\342\200\224>E4

exactly send base point to base point. This is why when we tessel-late the hyperbolicplane by the images of X under the elementsof

the tiling group generated by \317\210\316\271and \317\2103,the horocircles fit nicely

together, as in Figure 8.10.

However, this is not so in the generalcase. We now give a

geometricinterpretation to the constants \320\276and b occurring in the definitionof \317\210\316\271and \317\2103.First of all, recall the definition of the other gluing

maps defined by \317\2062=

\316\250\316\2201and \317\206\316\221=

\317\206^1.

In the edge E\\=

\317\2062(\316\2252),the base point determined by thehorocircles of \317\2102(\316\247)is the image \317\2062(\316\2412)

= \342\200\2241+ i(a\342\200\224

1) of the base pointP2 = | + ^.

In particular, the hyperbolic distance from this base

point to the base point \316\241\317\207= \342\200\2241 + i determined by the horocircles of

X is equal to |log(o - 1)|. More precisely, as seen from the interiorof \316\247,\317\2062(\316\2412)is at signed distance S] = \342\200\224

log(o\342\200\224

1) to the left of

Pi, where we count the distance as negative when the point is to the

right.

8.4. Shearing the E^irey tessellation 219

S&ai fSXi

-1 0 1

Figure 8.12. Horocirclearcsin the tessellation of Section 6.7.2

Note that this signed distanceconvention is symmetric. If wc

look at the edgeE\\ =\317\2062(\316\2252)from the interior of \317\2062(\316\247),the base

point \320\240\320\263marked on \316\225\317\207=

\317\2102(\316\2252)by the horocircles of the \"outside\"

polygon X is also at signed distance S\\ to the left of the base pointdeterminedby the horocircles of \317\2102(\316\247).

Transporting the geometric situation to the edgeE2=

\317\210\316\271(\316\225\316\271)by

\317\206\316\271=

\317\2102\316\271\316\271we see that, as seen from the interior of X, the base pointdeterminedby \317\210\316\271(\316\247)is at signed distance s\\ to the left of the basepoint P2 determined by X.

Similarly, on the edgeE3=

\317\206\316\221(\316\225\316\271),the base point \316\250\316\261{\316\241\316\261)

determined by \316\250\316\261(\316\247)is at signed distance s3 = log(6-1) from the base

point P3 determined by the horocirclesof X as seen from the interiorof X. Also, on the edge E4 =

\302\245>\320\267(\302\243?\320\267),the base point \321\203>\320\267(\320\240\320\267)is to the

left of the base point P4 as seen from inside X.

Actually, we have the same situation on each edgeof the partial

tessellation of the hyperbolic plane associatedto the square X and

to the gluing maps \317\210\316\271and ^3. Each such edge \316\225is of the form \316\225=

\317\210(\316\225\316\271)or \317\206(\316\225\316\266)for some element \317\206of the tiling group \316\223generated

by \317\210\316\271and \317\2063,and separates the polygons \317\210(\316\247)from the polygon

\317\210(\316\247)associated to another \302\267\317\210\302\243\320\223.As seen from inside \317\210(\316\247),the

base point determined by \321\204(\320\245)is at signed distance si if \316\225~\317\206(\316\225\316\271),

and at signed distance S3 if \316\225=\317\206(\316\2253),to the left of the base point

determinedby \317\206(\316\247).This immediately follows by transporting to\317\210(\316\247)by \317\210our analysis of the partial tessellation near X.

220 8. The Farey tessellation and circle packing

This is illustrated in Figure 8.12 in the casewhere si = 0.25

and S3 = \342\200\2241. Compare this figure to Figures 8.9 and 8.10, and to

Figures 6.12 and 6.16.In particular, every tile of this partial tessellation \316\244\316\2621\316\2623

corresponds to a tile of the tessellation \316\244associated to the case where

\302\253i= S3 = 0 (and corresponding to a completepunctured torus).

Actually one goes from \316\244to TSl S3 by progressively sliding all the tilestothe left along the edges, and by a signeddistanceof s\\ or S3 accordingto whether the edgeconsideredis associatedto E\\ or E3 by the tiling

group. We say that TSlS3is obtainedby shearing 7 according to theshear parameters Si and .S3.

8.4.2. Shearing the Farey tessellation. We can generalize the

above construction by introducing an additional edge, namely, the

diagonal of X formed by the geodesicjoining0 to 00. This geodesic

splits X into two ideal triangles: the triangle T4 with vertices 0, 1and 00; and the triangle T~ with vertices \342\200\2241,0 and oc. Given anadditional shear parameter S5, we can then deform X by slidingT~along \302\2435by a distance of S5 to the left of T+. Namely, we can replaceT~ by its image under the hyperbolic isometry \316\266\320\275->c~Sr,z.

00

1

\316\225,

\317\204-

,

~-\\ ^

\320\224\320\273

\317\204+

\320\225\320\252\316\2711

fk\\

\320\225\321\212

\316\271

0 1

Figure 8.13

The corresponding new hyperbolic quadrilateral (not a squareany more) is illustrated in Figure 8.13 in the case where S5 = \342\200\224.25.

Starting with this sheared quadrilateral, we can then constructa partial tessellationof the hyperbolic plane as before, using shear

8.4. Shearingthe Farey tessellation 221

parameters si and S3. In practice,one begins by gluing the sides ofthe new quadrilateral X using the gluing maps \317\206\316\271:E\\ \342\200\224>E2 and

\317\2063:E3 \342\200\224>E4 denned by

, \321\207 eS5z + lu ' es^z+ e~Si+1

and

\317\2063(\316\266)= \316\262-**

\316\226~\316\233.\316\216\316\275' -z + e1*+ 1

Theseformulas are easily obtained from the formulas for the original

gluing maps, using the fact that \320\276= c-Sl +1 and b = e\"3 + 1.

One can then considerthe partial tessellation of the hyperbolic planeassociated to theseedgegluings.

Figure 8.14 illustrates the case where sx = 0.25,S3= \342\200\2240.75and

-e-85 0 1

Figure 8.14. Shearing the Farey tessellation

Lemma 8.5. The imagesof X under the tiling group \320\223generated by

the gluing maps \317\210\316\271and \317\206$is the whole hyperbolic plane \320\2502exactly

when Si + S3 + S5= 0.

Proof. We will use Poincare's Polygon Theorem 6.25. The gluing

maps \317\210\317\207\\\316\225\316\273-> E3, \317\2103:\316\2253-> E4, \317\2062:E2 -\302\273\320\225\320\263and \317\2064-\316\2254-\302\273E2

form an edge cycle around the ideal vertex 00 = {00,1,0,\342\200\224e-\"5}.

This is the only edge cycle in the gluing data.

Theorem 6.25 asserts that the quotient space (X,dx) iscompleteif and only if the composition map \317\2064\316\277\317\2062\316\277\317\2103

\316\277\317\2061is horocyclic at

222 8. The Farey tessellationand circlepacking

oo. Remembering that </?2=

\316\250\317\2121and \316\250\316\261=

\320\243\320\267\"1!an immediate

computation yields

\316\250\316\261\316\277\317\2102\316\277\317\2063\316\277\316\250\316\271(\316\266)= c2\302\253i-+^\320\267+\320\263^+ \316\271+ eS3 + e-\"3+Sl

In particular, this map is horocyclic if and only if Si + S3 + S5= 0.If si + S3 + S5 = 0, Theorem 6.25 guarantees that the quotient

space (\316\247,\316\254\317\207)is complete. Then Theorem 6.1 shows that the images

of X under \316\223tessellate H2. Note that X has no finite vertex, so that

the completeness of the quotient space is the only hypothesis that wehave to checkwhen applying Theorem 6.1.

Conversely, suppose that the images of X under \316\223tessellate H2.

Then, Theorem 7.12 shows that (\316\247,\316\254\317\207)is isometric to the quotient(H2/r.(Jr) of H2 under the action of \316\223,and Exercise 7.6 impliesthat (\316\2272/\316\223,dr) is complete. The converse part of Theorem 6.25 now

proves that si + S3 + S5 must be equal to 0. (We could also have usedthe analysis of Section 6.7.2.) D

In Chapter 10, we will generalize this construction by allowing,in addition to shearing, bending into a third dimension. For this,we need to considerthe 3-dimensional hyperbolic space, which we

introduce in the next chapter.


Exercise8.1. Rigorously show that if the endpoints of an edge of theFarey tessellation are ?, ^ g Q \320\270{oo}, this edge separates two tiles of the

tessellation which are ideal triangles with respective vertices (at infinity) |,

2jTand \302\24377with \302\24377= | \317\206^ and ^7

= | \317\206^ (abandonning the conventionthat all fractions have positive denominator, by keeping numerators anddenominators coprime), where \317\206denotes the Farey sum.

Exercise 8.2. Consider the hyperbolic square X of Figure 8.9, with

vertices \342\200\2241, 0, 1, 00. In Section 8.4, we split it along the geodesic Oooto

obtain two ideal triangles, and applying Lemma 8.4to these two triangles

provided us with base points on each of the edges of X. Compute the basepoints that one would have obtained if we had instead split X along theother diagonal of the square, namely, along the hyperbolic geodesic going

from 1 to \342\200\2241.


Exercise 8.3 (The Farey Property). For every N > 0. consider all rational

numbers | G Q whose denominator is such that 0 < q < N. Sincethere

are only finitely many such rational numbers between any two consecutive

integers, we can order all these rationals and list them as

\342\200\242\342\200\242\342\200\242<fi=i<fi<r\302\261i<\302\267\302\267\302\267

with \320\263G Z, where qi > 0 and p\302\273and 5\302\273are coprime. This bi-infinitesequence is called the Farey series of order N.

a. Show that if | andjjr

G Q are such that p'q\342\200\224

pq' = 1 and g, q' > 0,then |g\"| ^ q + q' for every ^7 G Q with f < f^ < ^j and equality

holds only for\302\243

=^\302\261\302\243.

Hint: If \316\265< f^ <f\302\261\302\243,

then

_J^ < Eli _ \316\225^ P+P' _ \302\243= 1

b. Prove by induction on N that that any two consecutive terms in the

Farey series form a Farey pair, namely that pi+iqi\342\200\224

\316\241\302\273?\302\273+\316\271= 1 for

every \320\263G Z. Hint: Use part a.

c. Show that any three consecutive terms of the Farey seriesare such

that 2i = \302\243lri\302\251\302\243ixinamely that ^ = ^i\"1^<Tl for every i G \316\226

(although pi_i +pi-i-i and 5,-1 +qn-1 axe not necessarily coprime).

Exercise 8.4 (Pythagoreantriples).A Pythagorean triple is a triple(a, 6,c) of three coprime integers a, 6, \321\201> 0 such that a2 + b2 = c2. For

instance, (1,0,1) and (3,4,5) are well-known Pythagorean triples, while

(387, 884,965) is probably a lessfamiliar one.

a. Let S1 be the circleof radius 1 and center (0,0) in the plane K2. Showthat the map \316\246:(\316\261,b,c) 1\342\200\224>

(\302\247,\302\243)defines a one-to-one correspondence

between Pythagoreantriples (a, b, c) and rational points of S1 locatedin the first quadrant, namely, points (x,y) G S1 whose coordinates x,\321\203are rational and nonnegative.

b. Considerour usual isometry \316\246(\316\266)= \342\200\224

fizf from the upper half-space(H2,dhyp) to the disk model (B2,dB2). Show that \316\246sends each rational

point J G Q \316\240[0,1] to a rational point of S1 in the first quadrant.

\321\201Conversely, for (x,y) G S1 with x, \321\203rational and nonnegative, showthat \320\244~1(\321\205,\321\203)is a rational number in the interval [0,1].

As a consequence, the composition\316\246-1\316\277\316\246provides a one-to-one

correspondencebetween rational numbers in the interval [0,1] and all PythagoreanIriples. For instance,\316\246_1\316\277\316\246(^)

= (3,4, 5), and\316\250-1\316\277\316\246(^)

= (387,884,965).


Figure 8.15. Traveling in the Farey tessellation

Exercise 8.5 (Travelling in the Farey tessellation). Consider a Farey

triangle \316\244with nonnegative vertices |, E*P7, S ^ 0. This triangle can be

connected to the base triangle To with vertices 0, 1, oo by a sequence of

Farey triangles To, \316\244\316\271,\316\2232,..., T\342\200\236_1, T\342\200\236= \316\244in such a way that each \316\244\316\257+\316\271

is adjacent to \316\244\316\257and that the \316\244\316\257are all distinct. As one enters \316\244\316\257from

Ti_i, since we are not allowed to backtrack to \316\244\316\257_\316\271,there are exactly two

possibilities for \316\244\316\257_\316\271,which are respectively to the left or to the right ofTi as seen from \316\223,_\316\271.The above sequence of Farey triangles To, \316\244\316\212,T2,

..., Tn -i, Tn = \316\244can therefore be described by a sequence of symbolsS1S2 \342\226\240..Sn, where each Si is the symbol L if 7\316\257+\316\271is to the left of T\302\273,and

Si is the symbol R if it is to the right of \316\223,.By convention, To is

consideredto have been entered through the edge Ooo, so that Si = L if 7\316\212is the

triangle 12oo, and Si = R if \316\244\316\212is the triangle 0|l.For instance, Figure 8.15 illustrates the case of the triangle \316\244with

vertices ^f, \320\251-and |, associated to the symbol sequence S1S2S3S4S5S6 =

LLRLRL. (The vertex\320\251-

is not labeled due to lack of space.)Finally, consider the matrices \316\273=

(\317\214\\) and \317\201=

(\\ \302\260),and let the

matrix \317\203be associated to the symbol sequenceS1S2...Sn by the propertythat \317\203= \317\203\\\317\2032\342\226\240\342\226\240\317\203\316\267where \317\203,= \316\273when Si = L and to \317\203,=

\317\201when

Si = R.

Show that the vertices 2^\316\206,

^ of \316\223can be directly read from the

matrix \317\203,by the property that \317\203~(p, p). Hints: Proof by induction on

n. It may also be convenient to consider, as in Exercise 2.12, the linear

fractional map \317\206\342\200\236(\316\266)=

^\320\231 associated to the matrix a\342\200\224(\\ bd), and notethat \317\210\\sends To to the triangle 12oo and that <pp sends To to the triangleoil.

Exercise 8.6 (The Farey tesseUation and continued fractions). A

continuedfraction is an expression of the form


[\316\237\316\271,\316\2372,. . . , On] =

\316\261\317\212\316\227

\316\2612+1

\316\223

\302\267+\342\200\224.

For instance, [2,1,1,1] = f and [2,1,1,1,1] =\317\210.With the notation of

Exercise 8.5,write the symbol sequence S1S2 ...Sn as LmiRniL\"12Rn*...LmkRnk, where Lmi denotes m, > 0 copiesof the symbol L, and similarly\320\257\"'denotes n\302\273> 0 copies of R.

Show that

I= [mi, \316\240\316\271,m2, n2,..., mk,nk]

and

*r = [mi,ni,m2,ra2,... ,nk-i,mk],using the conventions that 5 = 00, ^ = 0 and 0+00 = 00. Hint: Induction

on k.

Exercise 8.7 (Domino diagrams). In the diagram

!\302\267\342\200\224\342\226\272\302\267\342\200\224\342\226\272\302\267\342\200\224\342\226\272\302\267\342\200\224\342\226\272\302\267\342\200\224\342\226\272\302\267\342\200\224\342\226\272\302\267!

(8.1) \\ \\ / \\ / \\2\302\267\342\200\224+\302\267\342\200\224>\302\267\342\200\224+\302\267\342\200\224+\302\267\342\200\224+\302\267\342\200\224+\302\2672

consider, for i, j G {1,2}, the paths that jump from bullet to bullet byfollowing the arrows, and that go from the bullet marked \320\263on the left tothe bullet marked j on the right; let riij be the number of such paths. Forinstanceone relatively easily sees that among the paths starting from the

bullet marked 2 on the left, three of them end at the bullet marked 1 while(ive of them end at the bullet marked 2, so that \320\23721= 3 and \320\23722= 5. Alittle more perseverance shows that \320\237\321\206= 8 and \320\23712= 13.

The above diagram can also be describedby chaining together two

l.ypes of \"dominos\", namely

\\ and /\342\200\242\342\200\224+\302\267 \302\267\342\200\224+\302\267

If we denote the domino on the left by L and the domino on the right by R,l.he above diagram correspondsto the chain LLRLRL. Comparing with the

example discussedin Exercise 8.5, we may find it surprising that \302\24311=

f,.Hid 2ia =

\317\210.Namely, in the Farey tessellation, the path described by the

siime symbol LLRLRLleadsto the ideal triangle \316\244with vertices 211 ( \321\210\320\264(

iitid \"\302\273+\"\". We want to show that this is no coincidence.n21+n22

226 8. The Farey tessellation and circle packing

Given a symbol sequence S1S2 . --Sn, where each Si is L or R, considerthe domino diagram obtained by chaining together the dorninos L and R

associated to the Si. Let \317\203be also the matrix associated to S1S2\342\226\240\342\226\240- Sny as

in Exercise 8.5. We will use the notation of that exercise.

a. Write \317\203=(*jj ^2) \316\206\316\261^\317\203*=

(\"(i) *\") )\302\267Remember that \317\203=

^s21 S22 '

\317\203\316\271\317\2032- - \302\267\317\203\316\267.Show that

... _ V J1)J2) J3) >)\302\273l,<2,--,\302\253n-l6{l,2}

It may be useful to first do a few examples with small n.

b. In our case, \317\203,\342\226\240=({

\302\260) or (J \\) according to whether the symbol Si is

L or \320\257.Show that each term sL s,-?LsinL \302\267\302\267\302\267s* \342\226\240>is equal to 0 or 1.

c. In the domino diagram associated to S1S2\342\226\240\342\226\240\342\226\240Sn, label each bullet onthe top row by 2 and each bullet on the lower row by I. Show that

Sill s) {\342\200\236\320\265\320\271... Si' . = 1 if and only if there is a path in the domino\320\2630\320\2631\320\2631\320\2632*2*3 \320\263\320\277\342\200\2241 *n \" *

diagram where, for each fc=l,2)...,n+l) the fcth bullet is labeled

byifc-i \342\202\254{1,2}.

d. Conclude that stj is equal to the number ny of paths from \320\263to j inthe domino diagram.

e. Let \316\244be the Farey triangle associated to the symbol sequence S1S2 \342\226\240\342\226\240\342\226\240

S\342\200\236as in Exercise 8.5. Show that the vertices of \316\223are 211, ^ and\302\25321' n22

\"11+\"12\"21+\320\23722

\"

Chapter 9

The 3-dimensional

hyperbolic space

We now jump to one dimension higher, to the 3-dimensionalhyperbolic space H3. This space is defined in completeanalogy with the

hyperbolic plane. Most of the proofs are identical to those that we

already used in dimension 2.

9.1. The hyperbolic spaceThe 3-dimensional hyperbolic space is the metric spaceconsisting

of the upper half-space

H3 ={(x,y,u)\342\202\254R3;u>0},

endowed with the hyperbolic metric e?hypdefined below. Here we are

using the letter \320\270for the last coordinate in order to reserve \316\266for the

complex number \316\266= \317\207+ \\y.

If 7 is a piecewisedifferentiable curve in H3 parametrized by thevector-valued function

\316\257\316\227\302\273(\317\212(\316\257),\316\271/(\316\257),!\316\220(\316\257)),a^t^b,

its hyperbolic length is

4\320\243\320\240(7)=

I W)dt.

227

228 9. The 3-dimensionalhyperbolic space

Figure 9.1. The hyperbolic space H3

The hyperbolic distance from \316\241to Q \342\202\254H3 is then defined as

dhyP(P, Q)= inf

{4\321\203\321\200\320\253;7 goes from \316\241to Q} .

The proof that(\320\2353,\320\265?\321\214\321\203\321\200)is a metric space is identical to the

proof of the same statement for the hyperbolicplane in Lemma 2.1.

As in dimension 2, the hyperbolic norm of a vector \317\213based at

the point \316\241=(\317\207,\321\203,\320\270)\342\202\254\320\2303is defined as

1

\320\270ll^llhyp= -

\"\"\320\237\320\265\320\271\321\201

where ||#||euc= \320\273/\320\2602+ 1? + \321\2012is the usual euclidean norm of \316\275=

(\320\276,b, \321\201).

The hyperbolic space admits several \"obvious\" isometries \317\206;

H3 \342\200\224>H3, which are immediate extensions of their 2-dimensionalcounterparts.

Theseinclude the horizontal translations defined by

\317\210(\317\207,\321\203,u)= (x + x0, \321\203+ \316\257/\316\277,\320\270)

for xo, \320\263/\320\276\342\202\254R, and the homotheties

ip(x,y,u) = (Xx,Xy,Xu)

for \316\273> 0.

9.1. The hyperbolic space 229

A new, but not very different, type of symmetry consists of the

rotations around the u-axis,defined by

\317\210(\317\207,\321\203,\320\270)=

(\317\207cos \316\270\342\200\224\321\203sin \316\270,\317\207sin \316\270+ \321\203cos \316\270,\320\270)

for \316\270\342\202\254R.

Finally, the 3-dimensional analogue of the inversion across the

unit circle is the inversion across the unit sphere. This is the

map \317\206:R3 U {00} -> R3 U {00} defined by

/ , = ( * \320\243 \320\270

W,V>U)\\^\317\2072+ y2 + U2< \317\2072+ y2 + u2

>\317\2072+ y2 + u2

This inversionclearly sendsthe upper half-space H3 to itself.

Lemma 9.1. If \317\206:\316\2343\342\200\224*H3 is the inversion across the unit sphere,

and if \316\275is a vector based at \316\241\316\266\316\2343,then its image \316\214\317\201\317\210(\316\275)under

the differential of \317\206,which is a vector based at\317\206(\316\241),

is such that

As a consequence,\317\206is an isometry of (H3,dhyp).

Proof. This is an immediate computation, identical to the one usedin the proof of Lemma 2.3 to show that the inversion across the unit

circle is an isometry of the hyperbolic plane. D

Consider the vertical half-plane

II ={(\317\207,\316\237,\317\205)\342\202\254R3;u > 0} \320\241\316\2273.

lleplacing the letter \320\270by the letter y, this half-plane has a natural

identification with the hyperbolic plane H2. For this identification,

the 3-dimensional hyperbolic length of a curve in // \320\241\316\2273is the same

as the 2-dimensionalhyperboliclength of a curve in H2, The sameholds for the hyperbolic norm |M|hyp of a vector \316\275tangent to \316\227at

\316\241\342\202\254\320\257\320\241\320\2353.

We can therefore identify the hyperbolic plane (H2,dhyp) to the

half-plane \320\257\321\201!3,endowed with the metric d# for which \320\260\320\275(\320\240,Q)

is the infimum of the hyperboliclengths of all curves that join \316\241

l.o Q in H. Theorem 9.4 below, which identifies the shortest curves

joining \316\241to Q in H3, shows that \320\260\321\206is just the restriction to \316\227of

I.he hyperbolic metricdhyp

of H3.

230 9. The 3-dimensional hyperbolic space

Theorem9.2. The hyperbolic space (H3,dhyp) is homogeneous andisotropic.Namely, for every vector \317\213at the point \316\241and every iSatQ

with\\\\v\\\\hyp

=||^||hyp, there exists an isometry \317\206of (H3,dhyp) such

that \317\210{\316\241)= Q and \316\243>\316\241\317\210(\316\275)

= vS.

Proof. Modifying \316\275,\316\241,w and Q by suitable horizontal translationsand rotations about the it-axis (which are isometrics of H3), we canassumewithout loss of generality that \316\241and Q are both in the half-

plane H2 \320\241\316\2273,and that \316\275and w are both tangent to that half-plane.

Then, because the hyperbolic plane H2 is isotropic

(Proposition2.20), there exists an isometry \317\206:H2 \342\200\224>H2 which sends \316\241

to Q and whose differential map sends \316\275to w. This isometry is a

composition of translations along the \320\266-axis, homotheties, and

inversions across the unit circle in the plane. Extending the inversions by

inversions across the unit sphere, all of these factors extend to isome-tries of H3, so that \317\206extends to an isometry \317\210:H3 \342\200\224>H3 which has

the desired properties. D

Theorem 9.3. The hyperbolic space (H3,dhyp) is complete.

Proof. The proof is identical to that of Theorem 6.10. \316\240

9.2. Shortest curves in the hyperbolic space

Theorem9.4. The shortest curve from \316\241to Q \342\202\254H3 is the circle arc

joining \316\241to Q that is contained in the vertical circlepassing through

\316\241and Q and centered on the xy-plane (possiblya vertical line).

Proof. The proof is identical to the one we used to prove the sameresult in dimension 2, namely, Theorem 2.7. Namely, first consider

the case where \316\241and Q lie on the same vertical line as in Lemma 2.4.

Then, for the general case, use a suitable compositionof horizontal,

translations, homotheties and inversions to send\316\241and Q to the samevertical line, and then use the previous case. We leave the details asan exercise. D

9.3. Isometries of the hyperbolic space 231

9.3. Isometries of the hyperbolicspaceTo list all isometries of the hyperbolic spaceH3, it is convenient to

identify the xy-pla,neto the complexplaneC, in the usual manner.

Lemma 9.5. Every linear or antilinear fractional map of \320\241= \320\241U

{\320\276\321\201}continuously extends to a map \317\206:\316\2273U \320\241\342\200\224>H3 U \320\241whose

restriction to H3 is an isometry of (H3, dhyp)\302\267

Proof. By Lemma 2.12, the linear or antilinear map \317\206:\320\241\342\200\224>\320\241is

a composition of translations, rotations, homotheties, and inversions

across the unit circle. We observed that all of these factors extend tocontinuous transformations of H3UC inducing isometries of (H3,dhyp)\302\267

D

A priori, the extension \317\206\316\277\316\257\317\210provided by the proof of Lemma 9.5might depend on the choice of the decompositionof \317\206as a compositionof translations, rotations, homotheties, and inversions across the unit

circle. To show that this is not the case, we use a simple geometricobservation.

Lemma 9.6. The inversion across the unit sphere sends any sphere Scenteredon the xy-plane to a sphere centeredon the xy-plane, possibly

a vertical plane.

Proof. The inversion and the sphere S are both symmetric with

respect to rotations around the line joining the origin \320\236to the

center of the sphere S. The property consequently follows from the 2-

dimensional fact that the inversion across the unit circle is an isometryof H2 and therefore sends a circle centeredon the \320\266-axis to another

circle centered on the \320\266-axis (possibly a vertical line). \342\226\241

Lemma 9.7. The extension \317\210:H3 \342\200\224>H3 of the linear or antilinearmap y>: \320\241\342\200\224>\320\241provided by the proof of Lemma9.5 is independent of

choices.

Proof. If \316\241\342\202\254H3, arbitrarily pick three spheres Si, S2 and S3

centered on the xy-plane, passing through P, and sufficiently genericthat \316\241is the only point of the intersection \320\2303\320\237Si \320\237S2 \320\237S3. There

;ire of course many spheres like this.

232 9. The 3-dimcnsional hyperbolicspace

The spheres Si, 52, 5\320\267intersect the xy-plane \320\241in three circles

Cj, C2, C3. By Proposition2.18,\317\206sends the circles \320\241\321\214C2, \320\2413to

circles \320\241(,C2, \320\241'\321\212in C. Then there exist unique spheresS[, \302\273S2,

S'3 centered on the xy-plane which intersect this xy-plaae along C[,C2, C3. By Lemma 9.6 and similar (and obvious) statements for

horizontal translations, rotations about the z-axis and homotheties,the extensionof \317\206must send the sphere Si to S[, 52 to S'2 and S3

to \302\2433.As a consequence, \317\206(\316\241)is the unique point of the intersection

In particular, \317\206(\316\241)is independent of the decomposition of \317\210as

a composition of translations, rotations, homotheties, and inversions

that was used in the proof of Lemma 9.5. D

Theorem 9.8. Every linear or antuinear fractional map \317\210:\320\241\342\200\224>

\320\241has a unique continuous extension \317\206:tfuC \342\200\224>tfuC whose

restriction to H3 is an isometry of (H3, dhyP)\302\267

Conversely, every isometry o/(H3,dhyp) is obtained in this way.

Proof. We just proved the existenceof the extension in Lemma 9.5.

Conversely,let \317\210be an isometry of (H3,dhyP)\302\267 We want to find

a linear or antilinear fractional map \317\210whose isometric extension to

H3 coincideswith \317\210.

Since \321\204is an isometry, it sends the oriented geodesicOoc to

another complete geodesic of H3, going from zi to \320\2632\342\202\254\320\241.By

elementaryalgebra, there is a linear fractional map \317\206sending 0 to Z\\ and

00 to \320\263\320\263.If \317\210is its isometric extension to H3, the isometry \321\204\320\276\317\210~\316\271

now sends the oriented geodesic Oooto itself.

Replacing \317\206by \317\206\316\277\317\210~\316\273if necessary, we can therefore assume

that \317\206sends the geodesic Ooo to itself, without loss of generality.

Composing \317\206with a homothety if necessary, we can even assumethat it fixes somepoint of this geodesic. Since \317\206is an isometry, it

now fixes every point of the geodesic Ooo.

Let g be a completegeodesiccontained in the vertical euclidean

half-plane \320\2302\320\241H3 and crossing Ooo at somepoint Pq. Then,\321\204(\320\264)is a

complete geodesic passing through Pq. Note that H2 is the union of allcompletegeodesiesthat meet both Ooo and g\342\200\224{Pq}- Therefore, \321\204(\320\250?)

9.3. Isometrics of the hyperbolic space 233

is the union of all complete geodesies meeting Oooand \317\210{\316\261)\342\200\224

{\316\250(\316\241\316\277)},

and consequently is the vertical euclidean half-plane containing Ooo

and \321\204(\320\264).Composing \317\206with a rotation around Ooo if necessary, wecan therefore assume that \317\210(\316\211?)

= K2.

In particular, \317\206now restricts to an isometry ofH2. By the

classification of isometries of (H2, dhyp)in Theorem 2.11, there consequently

exists a linear or antilinear fractional map \317\210,with real coefficients,whose isometric extension \317\206coincides with \302\267\317\206\316\277\316\267\316\2342.Then, \317\210

\316\277\317\206-1is

a hyperbolic isometry of H3 which fixes every point of H2. The sameargument as in the proof of Lemma 2.10then shows that \317\210

\316\277\317\206-1is

either the identity map or the euclideanreflection across H2.

This concludes the proof that every isometry of (H3, dhyp)

coincides with the isometric extension \317\206of some linear or antilinearfractional map \317\210.

Lemma 9.7 shows that the extension is unique. D

It is possible to provide explicitformulas for the extension of a

given linear or antilinear fractional map \316\266\316\271\342\200\224>^jj^

\320\276\320\263\320\263\320\270^\302\261^

to

an isometry of H3. SeeExcercise9.2.However, these expressions are

not very illuminating.

In general, we will use the same letter to denotea linear or anti-

linear fractional map \317\210:\320\241\342\200\224>\320\241,the associated hyperbolic isometry

\317\210:H3 -> H3, and the continuous map ^:H3UC^i3UC made up

of these two maps. We will say that \317\206:H3 \342\200\224+H3 is the isometric

extension of \317\210:\320\241\342\200\224>\320\241to an isometry of (H3, dhyP) which, although

stricto sensu mathematically incorrect (sincethe actualextensionis\317\206:H3 U \320\241\342\200\224\342\226\272H3 U \320\241),is a convenient short-hand terminology.

For future reference, we indicate here the following property.

Lemma 9.9. Let Bdhyp(Po,r) and Bdhyp(Po,r) be two balls with thesame radius r > 0 in

(H3,dhyp).Endow each of these two balls with

the restriction of the hyperbolicmetric rfhyp\302\267 Then every isometry

\317\210:Bdhyp(Po,r)\342\200\224>

\320\222\320\260\320\254\321\203\320\240(\320\240\320\276>\320\263)has a unique extension to an isometry

\317\210:\316\2273-* \316\2273.

Proof. We just follow the lines of the proof of Theorem 9.8.

234 9. The 3-dimensionalhyperbolic space

Transporting everything by \320\260\320\273isometry of H3 if necessary, we

can assume without loss of generality that Pq is on the geodesicOoo. Then, by the arguments used in the proof of Theorem 9.8,we can progressivelycompose\317\206with various isometries of (H3, dhyp)

to reduce the problem to the case where \317\206sends Pq to Pq, andthen sends the intersection of Bdhyp(Po,r) with Ooo to itself, andthen sends Bdhyp(Po,r) \316\240H2 to itself, and then fixes every point of

\320\222<1\321\212\321\203\320\240(\320\240\320\276,\320\263)\320\237\320\250\320\240,and then fixes every point of Bdhyp(Po,r).At this point, the existence of the extensionto a global isometry

(namely, the identity map) of H3 is immediate.Theuniqueness of the

extension is immediate once we notice that it must send each geodesicemanating from Pq to itself. D

9.4. Hyperbolic planes and horospheres9.4.1.Hyperbolic planes. A hyperbolic plane in the hyperbolic

space H3 is the intersection // of H3 with a euclidean sphere centeredon the xy-plaxve or with a vertical euclidean plane. In particular,

a special case of \320\276hyperbolic plane is that of the hyperbolic plane\320\2302\320\241\316\2273.

Proposition 9.10. Every isometry of (H3,dhyp) sends hyperbolic

plane to hyperbolic plane.

Proof. The propertyholds for horizontal translations, homotheties,rotations around vertical axes, and inversions across spheres centeredon the xy-plane(including vertical planes) by Lemma 9.6 andelementary observations. The result then follows from Theorem 9.8 and

Lemma 2.12, which prove that every isometry is a compositionof

transformations of this type. D

Conversely,using an appropriatecomposition of horizontal

translations and homotheties, one easily checks that everyhyperbolicplane\316\227\320\241\316\2273is the image of H2 under an isometry of (H3, dhyp)\302\267

9.4.2. Horospheres. A horosphere centered at \316\266G \320\241is the

intersection with H3 of a euclidean sphere S which is tangent to the

a;i/-plane \320\241\320\241\316\2343at \316\266,and lies above this xy-plane. A horosphere


Figure 9.2. A hyperbolic plane \320\257and a horosphere S

centered at the point \321\201\320\276is just a horizontal euclidean plane containedinH3.

Proposition9.11.Every isometry of (H3, dhyp) sends horosphere to

horosphere.

Proof. As in Lemma 9.6, the inversion across the unit sphere must

send a horosphere to a sphere,which must be tangent to \320\241since the

inversion respects \320\241Therefore, the standard inversion sends horo-spheresto horospheres.Since the same property clearly holds forhorizontal translations, homotheties and rotations around vertical axes,the result immediately follows from Theorem 9.8 and Lemma 2.12.

D

Exercises for Chapter 9Exercise 9.1 (Dihedral angles). A (hyperbolic) dihedron isthe region D

of H3 which is delimited by two hyperbolic half planes \316\240\316\271and \316\2402meeting

along a complete geodesicg = \316\240\316\271\316\2231\316\2402,where a hyperbolic half-plane is oneof the two regions of a hyperbolic plane delimited by a complete geodesic. If\316\241is a point of this geodesic g, let Up be the unique hyperbolic plane whichis orthogonal to g at P. Then, D \316\267Up is a angular sector in Up delimited

by the two semi-infinite geodesies Up \316\240\316\240\316\271and Up \316\240\320\251issued from P.

Show that the angle of this angular sector D \316\240Up at \316\241is independent of

the choice of \316\241\302\243\320\264.Possible hint: Use an isometry of (H3,dhyp) sending gto the geodesic going from 0 to 00.

This angle is the dihedral angle of the dihedron D along its edge g.Dihedral angles will play an important role when we consider hyperbolicpolyhedra in Chapter 10.


Exercise9.2(Explicit formulas for hyperbolic isometries).

a. Show that the extension of the linear fractional map \317\206(\316\266)=

^^, with

ad - be = 1, to an isometry of \320\2303\320\241R3 = \320\241\321\205R is given by

\316\271\\ _ (az + ^ |cu|2 \320\270 \\

\317\210[\316\226'\316\267'~

VcT+d+

c{cz + d){\\cz + d\\2 + \\cu\\2)' \\cz + d\\2 + \\cu\\2)

for every \316\266\302\243\320\241and \320\270> 0. Hint: Decompose \317\210as a compositionof horizontal translations, rotations, homotheties and inversions, andthen take the composition of the isometric extensionsof these factors.

b. Give a similar formula for the isometric extension of the antilinear

fractional map \317\206(\316\266)=

jjfif.

Exercise 9.3 (The ball model for the hyperbolic space). Let

B3= {(x,y, u) \302\243R3; x2 + y2 + u2 < 1}

be the open unit ball in R3. For every piecewisedifferentiable curve 7 in

B3, parametrized by t 1-+ (x(t),y(t),u(t)), a ^ t ^ b, define its B3-lengthas

4\320\267(7)=

2/J a

by/x'{ty+y'{ty+u'{ty

l-x(ty-y{ty-u{ty\342\226\240

Finally, let the B3-distance from \316\241to Q \302\243B3 be

dBz(P,Q) = inf{43(7);7 goesfrom \316\241to Q},

where the infimum is taken over all piecewise differentiable curves goingfrom \316\241to Q in B3.

a. Let \316\246:R3 \342\200\224>R3 be the inversion across the euclideansphereof radius

-y/2 centered at the point (0,0,-1). Show that it sends the upperhalf-space E3 toB3.

b. Show that the restriction of \316\246to H3 defines an isometry from (\316\2273,dhyp)

to (B3,dB3). Possible hint: If if is a vector based at \316\241\302\243\320\2223,compute

11^\321\204_1(^)11\321\214\321\203\321\200.

c. Let S be a euclideanspherewhich orthogonally meets the unit sphereS2 bounding B3. Show that the inversion \317\201across S respects B3, andrestricts to an isometry of (B3, dB3). Possible hint: For \316\246as in part a,it may be easier to prove that \316\246-1\316\277\317\201\316\277\316\246is an isometry of (H3,dhyp).

d. Show that every composition of inversions across euclidean spheresorthogonal to S2restricts to an isometry of (\320\2223,\320\271\320\262\320\267).Hint: part \321\201

e. Conversely, show that every isometry of (\320\2223,\320\271\320\262\320\267)is a composition ofinversions across euclideansphereswhich orthogonally meet S2. Hint:First consider isometrieswhich fix the origin O.


f. Show that every geodesic of (B3,dB3) is a circle arc contained in a

euclidean circle which orthogonally meets the sphere S2 in two points.Hint: Use the isometry \316\246of part b.

Exercise 9.4. Let S =<S,dhyp(-Po>''\")

be a hyperbolic sphere of radius \320\263in

H3, consisting of those \316\241\342\202\254\320\2303such that dhyp(P, Pq) = r. Endow S with

the metric ds for which ds(P,Q) is the infimum of the hyperbolic lengthsof all piecewise differentiable curves joining \316\241to Q in S. Show that the

metric space (S,ds) is isometricto (S2, (sinhr-)dsph) where (sinhr-)dsph isthe metric on the standardeuclidean sphere \302\2472obtained by multiplying thespherical metric dsPh by sinhr. Possible hint: It may be useful to consider

the ball model (\320\2223,\320\260\320\262\320\267)of Exercise 9.3, and hyperbolic spheres centeredat

the origin \320\236in this ball model.

Exercise 9.5 (The Mobius group). Let the Mdbius group \320\234\320\267be the

group of transformations of R3 = R3 U {oo} generated by all inversions

across euclidean spheres (including reflections across euclidean planes).The elementsof \320\234\320\267are Mobius transformations. Show that every

Mobius transformation \317\210\302\243\320\234\320\267sends every euclidean sphere to a euclideansphere (including planes among spheres), and every euclideancircle to a

euclidean circle (including lines among circles). Hint: Compare the proofof Lemma 9.6, and note that every circle is the intersection of two circles.

Exercise 9.6. Show that if \317\210G \320\234\320\267is a Mobius transformation and if

\316\275and w are vectors based at the same point \316\241\316\276\316\2313,the angle betweenthe vectors \316\214\317\201\317\210{\316\262)and \316\214\317\201\317\210{\316\277)is equal to the angle between \316\275and v}.

In particular, isometries of (H3,dhyp) are angle-preserving, in this sense.Hint: First consider the case of inversions.

Exercise 9.7 (A characterization of Mobius transformations). The goalis to prove the converse of the statement of Exercise 9.5. Let \317\210be a

homeomorphism of R3 which sends sphere to sphere.

a. Show that there is a Mobius transformation \317\206\302\243\320\234\320\267such that for

the usual identification of K3 with \320\241\321\205\320\250and the standard inclusionsR \320\241\320\241= \320\241\317\207{0} \320\241\320\241\321\205R, \317\206(0)

=\317\206(0),\321\204(1)

=<\317\201(\317\212),\317\206(\316\277\316\277)

=<\317\201(\316\277\317\214),

\321\204(\320\226)=

<\321\200(\320\226),\321\204(\320\241)=

<\321\200(\320\241),\321\204{\320\2503)=

\317\206(\316\2343),\321\204sends each half-planeof \320\241delimited by R to the same half-planeas \317\206.

b. Set \317\206=

\317\210\320\276\321\204-1. Show that \317\206'sends sphere to sphere, fixes thepoints 0, 1,oo, and respects C, R, H3 and each of the two half-planes

delimited by R in C.

\321\201Show that, for any x, \321\203\302\243R, the midpoint m = ^{x+ y) is uniquelydetermined by the fact that there exists two circles Ci, C2 and two

lines L\\, Li in \320\241with the following properties: The points m and \317\207

belong to C\\\\ the points m, \321\203belong to C2; the lines L\\, la are each

tangent to both of C\\ and Ci\\ the three lines L\\, Li and R are disjoint.

238 9. The 3-dimensionalhyperbolicspace

(Draw a picture). Conclude that \317\210'(\317\204\316\267)is the midpoint of \317\210(\317\207)and

v'(y).

d. Use part \321\201to show, by induction on \\p\\, that \317\206'(\317\201)=

\317\201for every integerpel Thenshow, by induction on n, that

\317\210'{-^)=

-%rfor every \317\201\302\243\316\226

and \316\267\302\243N. Conclude that \317\210'{\317\207)= a; for every i\302\243R,

e. Show that \317\210'sends every circle in \320\241with center in R to another circle

centered on R. Hint: Such a circle \320\241is characterized by the propertythat the lines tangent to \320\241at each of the two points of \320\241\320\237R are

disjoint.

f. Use part e to show that \317\206'fixes every point of \320\241Hint: Compare the

proof of Lemma 9.7.

g. Show that \317\206'sends each sphere of R3 centered on \320\241to a sphere centeredon C. Hint: Look at the planes tangent to S at the points of S \316\240\320\241.

h. Use part g to show that \317\206'is the identity map of R3. Hint: Compare

the proof of Lemma 9.7.

i. Conclude that every homeomorphism of R3 which sends sphere tosphere is a Mobius transformation.

Exercise 9.8. Let \317\206G \320\234\320\267be a Mobius transformation of R3 which

respects the upper half-space H3. Show that the restriction of \317\206to H3 is an

isometry of (H3,dhyp). Hint: Adapt the steps of Exercise 9.7; more

precisely show that, in part a of that exercise, we can take \321\204so that it inducesan isometry of H3, and then follow the steps in parts b-i to concludethat

Exercise 9.9. Remember that if \317\210is a function of three variables valued

in R3, its jacobian at the point \316\241is the determinant detDp^i. Show

that if \317\206:HI3 \342\200\224>H3 is the isometric extension of a linear or antilinear

fractional map \320\241\342\200\224>\320\241,the jacobian det Dp<p is positive at every \316\241\302\243\316\2273

for a linear fractional map \320\241\342\200\224>\320\241,and it is negative at every Pel3 for an

antilinear fractional map. Hint: First consider an inversion, and rememberthat Dp (\317\206

\316\277\317\210)

=:\317\213\317\210(\317\201)\317\210

\316\277\316\214\317\201\317\210.

In other words, the extension of a linear fractional map to an isometryof (H3,cihyp) is orientation-preserving, and the isometric extensionof

an antilinear fractional map to H3 is orientation-reversing. See our

discussion of 3-dimensional orientation in Section 12.1.1.

Exercise 9.10 (Classificationof orientation-preserving isometriesof \320\2303).

Let \317\206:\320\250\320\273\342\200\224>\320\2303be the isometric extension of a linear fractional map

\317\206:\320\241\342\200\224>\320\241.Assume in addition that \317\206is not the identity map.

a. Show that \317\206:\320\241\342\200\224>\320\241fixes exactly one or two points of \320\241


b. If \317\210fixes only one point \316\266\302\243\320\241,let ^bea hyperbolic isometry sending

\316\266to oo. Show that \317\206\316\277\317\210\316\277\317\206-1is a horizontal translation; in other

words, transporting all points of H3 by \317\206replaces ^bya horizontaltranslation. In this case, \317\206is said to be parabolic.

c. Now consider the case where \317\210fixes two distinct points z\\ and zi of

C. Let \321\204be a hyperbolic isometry sending z\\ to 0 and zi to oo. Show

that \317\210\316\277\317\206\320\276\321\204~1is a homothety-rotation \316\266\316\271\342\200\224>az with a \302\243\320\241-{0}.

Conclude that \317\206respects the complete hyperbolic geodesic g going

from z\\ to Z2, and acts on g by a translation of constant hyperbolicdistance | log \\a\\ |. In this case, \317\210is said to be elliptic if it fixes everypoint of g (namely, if \\a\\

= 1), and loxodromic otherwise.d. Show that a parabolic or loxodromic isometry of (H3,dhyp) fixes no

point of E3. Show that the fixed points of an elliptic isometry form a

complete geodesic in H3.

Exercise 9.11 (Classification of orientation-reversing isometries of H3).Let \317\206:HI3 \342\200\224>H3 be the isometric extension of an antilinear fractional map

\317\206:\320\241\342\200\224>\320\241.Consider its square \317\2062=

\317\210\316\277\317\206.

a. Show that \317\2062:\320\241\342\200\224>\320\241is a linear fractional map, and that \317\210sends each

fixed point of \317\2101to a fixed point of \317\2062.

b. With the terminology of Exercise 9.10,supposethat \317\2102is parabolicand is not the identity. Show that there existsa hyperbolic isometry

\321\204such that \321\204\316\277\317\210\320\276\321\204~\320\263is the composition of the euclidean reflection

across a vertical euclideanplane with the horizontal translation alonga nonzero vectorparallel to that plane. In this case, \317\206is orientation-

reversing parabolic.

\321\201Suppose that \317\2062fixes exactly two points z\\, zi \302\243C, and suppose that

\317\210fixes both z\\ and zi (compare part a). Show that there exists ahyperbolic isometry \321\204such that \317\206\316\277\317\210\316\277\317\206~'is the composition of ahomothety with the reflection across a vertical euclideanplane passing

through the point 0. Show that \317\206respects the complete geodesic ggoing from z\\ to \316\2262,and acts on g by a nontrivial hyperbolic translation.In this case, \317\210is a orientation-reversing loxodromic.

d. Supposethat \317\2062fixes exactly two points z\\, zi \302\243C, and suppose that\317\210exchanges z\\ and zi (compare part a). Show that there exists ahyperbolic isometry \321\204such that \317\206\316\277\317\210\316\277\317\206~\316\273is the composition of theinversion across a sphere centered at the origin 0 with the rotationaround the vertical line passing through 0 (namely, our it-axis) by

an angle \316\270which is not an integer multiple of \317\200.In this case, \317\210is

orientation-reversing elliptic.

e. Finally, suppose that \317\2062is the identity map. Choose an arbitrary

point Pq \302\243H3 which is not fixed by \317\210(which must exist since \317\210is


not the identity), and let jbea unique complete hyperbolic geodesicpassing through Po and <p(Po). Let \317\206be an isometry of H3 sending

g to the vertical half-line going from 0 to oo. Show that \317\206\316\277\317\210\316\277\317\210~\316\273

is either the inversion across a euclidean sphere centered at 0 or the

composition of the inversion across a sphere centered at the origin 0

with the rotation of angle \317\200around the vertical line passing through

0. Then \317\210is a hyperbolic reflection in the first case, and it is again

orientation-reversing elliptic in the second case.

f. Show that an orientation-reversing parabolic or loxodromic isometry

of (H3, dhyp) fixes no point of H3, that an orientation-reversing ellipticisometry fixes exactly one point of \316\2273,and that the fixed points of ahyperbolic reflection form a hyperbolic plane \316\240\320\241\320\2353.

Exercise 9.12 (Hyperbolic volume). If D is a region in the hyperbolicspace H3 = {(x,y,u) \302\243\320\2323;\320\270> \320\236},let its hyperbolic volume be defined as

volhyp(O)=

/// -^dxdydue [0,oo].

Note that this volume may be infinite if D is unbounded in H3.

a. Let \317\206:\320\2303\342\200\224\342\226\272\320\2303be a horizontal translation, a rotation around the

it-axis, a homothety or an inversion across the unit sphere. Show that

volhyp(^>(0))= volhyp(O). As for the hyperbolic area in Exercise 2.14,

you may need to remember the formula for changes of variables in triple

integrals.

b. Conclude that volhyp(^>(0)) =volhyp(0) for every isometry of (H3, dhyp)-

Exercise 9.13 (Hyperbolic volume in the ball model). Let B3 be the ball

model of Exercise 9.3. If D is a region of B3, define its hyperbolic volume as

volB3(D)=

volhyp(p(D)),

where \317\206:\320\2223\342\200\224\342\226\272\316\2273is an arbitrary isometry from (\320\2223,\320\260\321\212\320\267)to (H3, dhyp).

a. Show that this volume is independent of the choiceof the isometry \317\206.

Hint: Exercise 9.12.

b. Show that

vol33(0) =fffD {1_x^8y2_z2)3d*dyd*.

Hint: Remember the change of variable formula for triple integral, anduse a convenient isometry \317\206.

Exercise 9.14. Show that the hyperbolic volume of a hyperbolic ball

i?dhyp(P, \320\263)\320\241\320\2303of radius \320\263is equal to \317\2008\316\220\316\2671\316\2712?-\342\200\2242nr. Hint: It might beconvenient to use the ball model B3, and to considerthe case where theball is centered at the origin O, as in Exercise 2.13a.

Chapter 10

Kleinian groups

This chapter is devoted to kleinian groups, which are certain groups of

isometries of the hyperbolicspaceH3. We begin with an experimental

investigation, in Section 10.1,of the tiling groups that one obtains bysuitably bendingthe Farey tessellation into the 3-dimensionalhyperbolic space H3. The following sections develop the basic definitions

and properties of kleinian groups. We then return to the examplesof Section10.1by rigorously proving, at least in some cases, that

l.hc phenomena we had experimentally observed correspondto realmathematical facts. This includes a few surprising properties, suchus the natural occurrence of continuous curves in the Riemann sphere

\320\241which are differentiable at very few points.

10.1. Bending the Farey tessellation

10.1.1.Crooked Farey tessellations. In Section 8.4.2, weassociatedto real numbers Si, S3, S5SM with Si + S3 +s5 = 0 a tessellationof the hyperbolic plane H2, defined as follows. We started with the

hl\302\253altriangle T+ with vertices 0, 1 and 00, and the hyperbolictriangleT~ with vertices \342\200\224

e-\"5, 0 and 00. We also consideredthe linear

fractional maps

/\\ es*z + l _w -,,.-(e~Sl +1)^+1

241

242 10. Kleinian groups

\"s5. z~\\\302\245>4(*)

=\302\245>3(*)

=\320\265.5\320\263+1

-\302\245*(*) \"-, + \320\265^\320\267+ \320\223

\342\200\224e^ +

The tessellation then consistedof all triangles of the form \317\206(\316\244+)

or \317\210{\316\244~)as \317\210ranges over all elements of the transformation group

\316\223generated by \317\210\\and \317\206$.Recall that \316\223consists of all the linearfractional maps of the form

\317\206-\317\210\316\257,\316\277\317\206\316\2572\316\277\342\226\240

>4>ik

where each\317\206^

is equal to \317\206\316\271,\317\206\316\271,\317\2063or \317\206\302\261.The standard Fareytessellation corresponded to the casewhere Si = s2 = s3 = 0.

\316\2502(\316\244+)

\316\225\316\271

T-

E4

&21\321\202\320\273.

H-e-s*\\

T+

\320\225\321\212

\316\225\316\271

(\316\275\316\233\316\244-)\316\257

Vi(T )

\316\2253

1'

Figure 10.1

We now make the key observation that this setup makessenseeven when Si, S3, s5 \342\202\254\320\241are complex numbers, provided that weconsider ideal triangles in the hyperbolic space H3. See SectionT.4in the Tool Kit for the definition of the complexexponential e*

when s \342\202\254\320\241is a complex number.

Indeed, given complex numbers .si, s3, s^ \342\202\254C, let T+ and \316\244

be the ideal triangles in the hyperbolic space H3 with vertices 0, 1,;

oo\342\202\254C= CU {00}, and \342\200\224e~\"5, 0 and oogC, respectively. Let E\\,

\302\2432,\302\2433,\302\2434,\302\2435be their edges, where \302\2431goes from \342\200\224e-\"5 to \321\201\321\216,Eq

from 0 to 1, \302\2433from 1 to \321\201\321\216,\302\2434from 0 to \342\200\224e-\"5, and \302\2435from 0 toi

00. See Figure 10.1. \342\226\240

10.1. Bending the Farey tessellation 243

The linear fractional maps \317\210\\{\316\266)\342\200\224

e,5ezl^X\\+1and <\321\200\320\267{\320\263)

=

\320\265~\"\321\212-\320\263+\321\221*\320\267+\\

define isometries of the hyperbolic space (H3,dhyp),which we denote by the same letter. We can then consider the group\320\223of isometries on H3 that is generated by \317\206\316\271,\317\2102,\317\2103and \317\206\302\261,and the

family \316\244of all ideal triangles \317\210^^) as \317\206ranges over all elements of\316\223.

Of course, because the triangles \317\210^\316\223^)are 2-dimensional, there is

no way they can cover the whole 3-dimensionalspaceH3. However,

we will see that these triangles still lead to some very interesting

geometric objects. We will call the family \316\244of the triangles \317\210{\316\244^)a

(2-dimensional) crooked Farey tessellation in H3.

Figure 10.2.Side view of the Farey tessellation

10.1.2. Experimental examples. Let us lookat a few examples

before going any further. In the real case,the situation where s\\ +

\342\226\240\321\207\320\267+ \302\2535=0 was better behaved, as it provided a tessellation of thewholehyperbolicplaneH2.Forthe same reasons (both mathematicaland aesthetic), we will restrict our attention to complex s\\, S3, S5 \342\202\254\320\241

with S\\ + s3 + s5 = 0. In particular, this guarantees that the isometry

244 10. Kleiniangroups

of H3 defined by

\316\2504\302\260

\316\2502\302\260

\316\2503\302\260

\317\210\316\271(\316\266)= \320\265281+28\320\267+28*\320\263+1 + \320\2658\320\267+eS3+Sl

\316\271\320\265\302\253\320\267+\302\2731+\302\2735

\316\271e2s3+si+s5 \316\271e2s3+2si+s5

= z + 2 + 2eS3 + 2eS3+Sl

of \316\223is a horizontal translation. Note that we already encounteredthis element \317\2104

\316\277\317\2102

\316\277\317\206\316\206

\316\277^ of \320\223in Sections 5.5 and 6.7.2, where itwas usedto gluetogetherthe two sides of the vertical half-strip V.

Since \317\210\\\316\277

\317\2102\316\277

\317\2063\320\276(^! is an element of \320\223,the crooked tessellation

\316\244is invariant under this horizontal translation. Indeed,it sends the

triangle \317\206{\316\244^)of \316\244to \317\2104\316\277\317\2062\316\277\317\2063\316\277\317\206\317\207

\316\277\317\206\316\266\316\244^),which is another

triangle of T.

First considerthe casewhere si = s$ = s^ = 0. In this situation,

there is no shearing or bending,and we just get the standard Fareytessellationin the hyperbolic plane \320\2302\320\241\320\2353.See Figure 10.2 for a3-dimensionalview.

Figure 10.3. Shearing and bending the Farey tessellation a little

The situation becomesmore interesting if we slightly move the s\302\273j

away from 0. Figure 10.3 represents the casewhere Si \320\270\342\200\2240.19+0.55iJ

s3 \320\2700.15 + 0.42i and s5 = 0.04- 0.97i.Here, we have rotated the!

10.1. Bending the Farey tessellation 245

picture, so that the translation \317\2104\316\277

\317\2102\316\277

\317\2103\316\277\317\210\316\273\302\243\320\223,which leaves the

crooked Farey tessellation \316\244invariant, appears to be parallel to the\317\204-axis in the standard coordinate frame for M3.

Note the bendingalongthe edgesgoingto \321\201\321\216.The bending along

the other edges is a little harder to see, but results in the wiggly

shape of the crooked tessellation as it approachesthe Riemann sphere

C = CU {\321\201\320\276}bounding the hyperbolic space H3.

Figure 10.4.Shearing and bending the Farey tessellation

much more

Figure 10.4 offers another example where si = \342\200\2240.5+ 1.4i, s3 =

0.3 \342\200\2241.4i and s5 = 0.2. Note that there is no bending between T+and T~ because s$ is real. On the other hand, the picture clearly

appears more intricate than Figure 10.3,which is consistent with the

fact that Si and S3 are \"less real\" in this case, in the sensethat theirimaginary parts are larger.

L0.1.3. Footprints. In both Figures10.3and 10.4, the ideal

triangles of the crooked tessellation appear to draw a relatively complex

curve in the sphere C = Cu {\321\201\321\216}bounding the hyperbolic space H3.More precisely,let us plot the vertices of all ideal triangles. For thecasesof Figures 10.2, 10.3 and 10.4, these \"footprints\" of the crooked


tessellationin \320\241are represented in Figures 10.5, 10.6 and 10.7,respectively.

Figure 10.5. The vertices of the ideal triangles of Figure 10.2(a boring straight line)

Figure 10.6. The vertices of the ideal triangles of Figure 10.3

Figure 10.7. The vertices of the ideal triangles of Figure 10.4

One can construct many examples in this way. We cannot helpincluding one more example in Figure 10.8, correspondingto Si =1.831\342\200\224

2.16355i, S3 = \342\200\224s\\ and S5 = 0, which is particularly pretty.

The picturesof Figures 10.5-10.8 (and Figure 11.2 in Chapter 11)were drawn using the wonderful piece of software OPTi, developed

by Masaaki Wada [Wada]. Figures10.2-10.4(and many other

complex pictures in this book) were drawn using Mathematica\302\256 programs

written by the author.

What is very clearly apparent in Figures 10.5, 10.6and 10.7 is

that these footprints appear to form a continuous curve, going fromthe left-hand side of the picture to the right-hand side, and invariant

10.2. Kleinian groups and their limitsets 247

Figure 10.8. Another pretty example

under a horizontal translation. Upon closer inspection, the sameactually holds for Figure 10.8, where the intricacy of the curve is muchmore amazing. Except for the straight line of Figure 10.5,thesecontinuous curves do not appear to have a very well-defined tangent line

at mast points.

Experimentally, as one moves the parameters 8\\, S3 and S5 awayfrom 0. one observesthe following phenomenon. For a while, thecrookedtessellationand the limit set look very similar to the examplesthat we have seen so far, except that the picturesare gettingmorecomplex.Then, suddenly, complete chaos occurs: The triangles of

the crooked tessellation are everywhere in the space H3 and in all

directions, and its footprints cover the whole Riemann sphere C.

10.2. Kleinian groups and their limitsetsThe tiling groups \316\223of the crooked tessellations of the previous

section are examples of kleinian groups, and the curves drawn by their


footprints in \320\241are the limit sets of these kleinian groups.Let us give,

the precise definition of these concepts.A kleinian group is a group \320\223of isometries of the hyperbolic

space (H3,dhyp) whose action on H3 is discontinuous.

Arbitrarily choose a point Pq G H3, and considerall limit points

of its orbit in M3 U {\321\201\321\216}.More precisely, a limit point of the orbit

\320\223(\320\240\320\276)is a point Pel3U {\321\201\320\276}for which there exists a sequence!(ln)neNof elements of \320\223such that 7n(Po) \316\246\316\241for every \316\267and, more

importantly, such that

\316\241= lim 7\342\200\236(P0)n\342\200\224*oo

in M3 U {\321\201\320\276}and for the euclidean metric deiic.This means that

lim deuc(P,7n(P0)) = 0 \\n\342\200\224*<x>

when \316\241\317\206\321\201\320\276,and that

lim 4uc(Pi,7n(Po)) = +00n\342\200\224*oo

for an arbitrary base point Pi when \316\241= 00.

The limit set of the kleinian group \316\223is the set \316\233\317\201of all limit;

points of the orbit \320\223(\320\2400)in M3 U {00}. /

Lemma 10.1. The limit set \320\233\321\200of a kleinian group \320\223is contained in

the Riemann sphere \320\241= \320\241U {\321\201\320\276}bounding the hyperbolic space H3and is independent of the point Pq e H3 chosen. ;

i

Proof. Clearly, a point \316\241=(\317\207,\321\203,\320\270)\320\265\316\2343with \320\270< 0 cannot be

a limit point, since all points of the orbit \320\223(\320\240\320\276)\320\241\320\2303havepositiveij

\316\220\316\257-coordinate. !

Suppose that a point \316\241=(\317\207,\321\203,\320\270)\320\265\320\2303,with \320\270> 0, is in]

the limit set. Then limn-*oodeuc(P,jn(Po)) = 0 for some sequence!

(7n)neN in \320\223with 7n(Po) \317\206\316\241for every n. A simple comparisonsof the euclidean and hyperbolic metrics, similar to the one used ati

the end of our proof of Theorem 6.10or to the estimatesthat we

use below in this proof, shows that linin-^oo dhyp(P,7n(Po))= 0 as]

well. In particular, for every \316\265,there are infinitely many jn suchj

that dhyP(P,7n(Po)) < \316\265,contradicting our hypothesis that \316\223actsj

10.2. Kleinian groups and their limit sets 249

discontinuously on H3 (compare Lemma 7.15). Therefore, noPei3

can be a limit point of the orbit \316\223(\316\241\316\277)either.

The combination of these two arguments shows that the limit set

is contained in the set of (x, y, u) e M3 with \320\270\342\200\2240 plus the point \321\201\321\216,

namely, that the limit set is contained in C.To show that \316\233\317\201is independent of the choice of base point,

consider another point Pq G H3. Let \316\276=

\316\231\317\212\317\200\316\220\316\267-\317\207\317\207,7n(Po) G \320\241be a pointon the limit set of \320\223(\320\240\320\276).

First consider the case where \316\276\317\206\320\276\321\201,so that \316\276=

(\317\207,j/,0) e

\320\241. Write 7n(P0) = (xn,yn,un), dn =deuc(\302\243>7n(Po))

and D =

dhyp(Po,Po) to simplify the notation. Then, for every point P' =(\320\266',\321\2031',u') of the hyperbolic geodesic g joining 7n(Po) to 7n(Po)>

log\342\200\224

< 4yP(7n(Po),P') < dhyp(7n(Po),7n(PO)) =dhyP(Po,PO) = D

by Lemma 2.5, and since jn is a hyperbolicisometry. As a

consequence, for every point (\320\266',\321\203',\320\270')of the geodesic g going from 7n(Po)to 7n(-Fo)i i*s third coordinate is such that

u' < eDun < eDdeuc(\302\243,7n(P0))= eDdn.

From the formula for the hyperbolic length, we conclude that

deuc(7n(-Po),7n(fo)) < 4\320\270\321\201\320\253< eDdn\302\243byp(g)

< eDdndbyP('ln(Po),ln(Po))

^eDdndhyp(P0,Pi)=eDdnD.

Since liirin-^oo dn = 0, this proves that dexiC (7n(Po), 7n(Po)) converges

to 0 as \316\267tends to oo, so that

lim 7n(Jft = lim 7n(P0)=

\316\276\316\267\342\200\224+oo n\342\200\224*oo

for the metric deuc. As a consequence,\316\276is also a limit point of theorbit \316\223(\316\241\302\243).

The argument is similar when \316\276= oo. Then

lim deuc(O,7n(P0)) = oo

for the origin \320\236= (0,0,0). If the coordinates un admit an upper

bound U, namely, if un < U for every ra, then the same argument as


before shows that

deuc(7\302\273(fi>),7\302\273(^)) <eDUD,

so that by the Triangle Inequality,

<\316\234\316\270,7\302\273(\316\257\316\277))> deuc(0)7n(fb)) -deuc(7n(Po),7n(i8))

^douc(0)7n(^))-eDi70

tends to oo as ra tends to infinity. Otherwise, there is asubsequence

(7nfc)fceNsuch that lim*-,,\302\273 unk

= oo. Then, anotherapplication of Lemma 2.5 shows that the third coordinate of 7\316\267*(-\316\241\317\214)

=

(x'nk. \320\243'\320\237\320\272,u'nk)is such that

\320\270'\320\237\320\272^ \320\270\320\237\320\272\320\265~0,so that deuc (O, -ynk (\316\241\316\262)^

\320\270\320\237\320\272tends to oo as \320\272tends to oo. Therefore, oo is a limit point of the

orbit\320\223(\320\240,5)in both cases.

This proves that every limit point of \320\223(\320\240\320\276)is also a limit point of\316\223(\316\241\317\214)\302\267By symmetry, it follows that \316\223(\316\241\316\277)and \316\223(\316\241\317\214)

have the same

limit points, so that \316\233\317\201does not depend on the choice of the base

point P0 e H3. D

Lemma 10.2. The limit set \320\233\321\200of a kleinian group \320\223is closed in theRiemann sphere C, and is invariant under the action of \320\223on \320\241in

the sense that it is respected by every 7 e \320\223.

Proof. If ^ is the limit of a sequence of points \316\276\316\267e \320\233\320\263,then,

picking for each \316\267a point 7n(Po) of the orbit of the base point Pq e H3that is sufficiently close to \316\276\316\267,we see that \302\243<,<,is also the limit of asequenceof points 7n(Po) of the orbit \316\223(\316\241\316\277).(Exercise: Write a

complete proof of this statement with the appropriate e's, and distinguishthe casesaccordingto whether \316\276

= oo or not.) As a consequence,\302\243<,<,

is in the limit set \316\233\317\201.Since this holds for any such convergingsequence in Ap, this proves that the limit set \320\233\320\263is closed.

If \316\276G \320\233\320\263and 7 G \320\223,consider a sequence of points 7n(-Po)of the

orbit\320\223(\320\2400)converging to \316\276.Then the points 7 \316\277

7\316\267(\316\241\316\277)are also in

\320\223(\320\240\320\276),and converge to j(\302\243)by continuity of 7. As a consequence,7(\316\276)is also in the limit set \316\233\317\201.Since this holds for every \316\276e \320\233\320\263and

7 G \316\223,this proves that \320\233\320\263is invariant under the action of \320\223. D

10.2. Kleinian groups and their limit sets 251

The following result shows that except in a few degenerate cases,

the limit set \316\233\317\201is the smallest subset of \320\241satisfying the conclusions

of Lemma 10.2.

Proposition 10.3. Let \320\232be a closed subset of the Riemann sphere\320\241which is invariant under the action of a kleinian group \320\223and which

has at least two points. Then \320\232contains the limit set Ap.

Proof. Since\320\232has at least two points, there existsa completegeodesic g of H3 whose endpoints are in \320\232\320\241\320\241.Pick a base point Pqon this geodesicg.

If \316\276is a point of the limit set \316\233\317\201,first consider the case where \316\276

is different from oo. Then, for every \316\265,the euclidean ball Bdeuc(\302\243,e)

contains an element y(Po) of the orbit T(Pq),with 7 G \316\223.In

particular, the geodesic 7(5) meets this ball Bd,H<.(\342\202\254,\316\265).Because g is

also a euclidean semi-circleorthogonal to M2 in M3, elementaryeuclidean geometry shows that at least one of the endpoints \316\276'of 7(5)

is contained in Bde\342\200\236c(\302\243,e).Note that \316\276'is in the subset K, since theendpointsof g are in \320\232and since this set is invariant under the

action of \320\223.Therefore, for every \316\265,we found a point \316\276'\302\243\320\232such that

^euc(\302\243,\302\243')< \316\265\302\267Since \320\232is closed, this shows that \316\276belongs to K.

The argument is similar in the casewhere \316\276\342\200\224oo. The

combinationof both cases proves that every point of \316\2337also belongs to K, sothat \320\233\321\200is contained in K. D

SeeExercise10.1for an example showing that in Proposition 10.3the hypothesisthat \320\232has at least two elements is necessary.

Proposition 10.4. Suppose that the quotient (\320\2353/\320\223,dhyp) of the

hyperbolic space H3 by the kleinian group \320\223is compact. Then the limit

set \320\233\321\200is the whole Riemann sphere C.

Proof. Picka base point Pq G H3.

We claim that there is a number D such that dbyP(P, Pq) < D for

every \316\241e \316\2273/\316\223.Indeed, we would otherwise be able to construct,by induction, a sequence (Pn)neN in \320\2353/\320\223such that diiyp(Pn, Pq) ^ \316\267

for every n. By the compactness hypothesison \320\2353/\320\223,there exists a

252 10. Kleinian groups ):j

- - ^

subsequence (Pnk)keN converging to some P^, so that i

_ \342\200\224 \342\200\224\342\200\224 \316\257

lim dhyp (Pnk, Po) =dhyp (\316\241<\317\207>,Pq ) j

k\342\200\224+oo '

(for instance, use the inequality of Exercise 1.4). But this would \\

contradict the fact that \\\\\321\202\320\277-+00<1\321\212\321\203\321\202>{\320\240\320\237\320\272,\320\240\320\276)= \321\201\321\216by construction j

of the sequence (Pn)neN- This proves the existenceof a constant Dj

as required.1 ;

By Proposition 7.6, \320\260\321\212\321\203\321\200(\320\240,\320\240\320\276)is the infimum of the distances ]

dhyp(-P, 7(Po)) as 7 ranges over all elementsof \320\223.Therefore, for every\316\241G H3, there exists a point j(Pq) of the orbit \316\223(\316\241\316\277)such that j

d(Pn(P0)) < D.\\

Given a point \316\276of the sphere at infinity \320\241which is not oo, namely, .

given \316\276e C, apply the above property to any point Pel3 which

is at euclideandistance< \316\265from \316\276.This provides a point j(Pq) G ;

\316\223(\316\241\316\277)such that dhyp(-P,7(-Po)) < -D\302\267The same estimates as in the ;proof of Lemma 10.1 then show that dew:(P, j(Pq)) < eDeD,so that \302\267!

^euc(^) l(Po)) < e(eDD + 1). Taking \316\265sufficiently small, we conclude \\

that there are points of the orbit \316\223(\316\241\316\277)which are at arbitrarily small \342\226\240

euclidean distance from \316\276.Namely, this shows that \316\276G \320\241is in the ;

limit set \320\233\321\200of \320\223.

For the point oo, pick a point \316\266e \320\250that is very close to oo, '

namely, very far from \320\236= (0.0,0) for the euclidean metric. By the ;

previous case, the orbit \320\223(\320\240\320\276)contains points which are very close :to z, again for the euclidean metric. It follows that there are points .of \320\223(\320\240\320\276)which are arbitrarily close to oo for the euclidean metric. ;

Namely, oo is in the limit set \320\233\321\200. \342\226\241

i

10.3. First rigorous example: fuchsian groups

In Chapter 6, our examples of various tessellations of the hyperbolic j

plane H2 provided us with many examples of groups acting by isom- ,etry and discontinuously on H2, arising as the tiling groups of these .

tessellations (Proposition 7.10 guarantees discontinuity). Let us see .how these also lead to examples of kleinian groups.

If you have taken a course in analysis or topology, you may recognize here the \342\226\240

proof that every continuous function on a compact space is bounded, as applied to the :

function f(P) = d(P, Po).

10.3. First rigorous example: fuchsian groups 253

An isometry \317\210of (H2, dhyp) is determined by its extensionto thecircleat infinity l=KU {oo}, which is a linear or antilinear map

V(x) =\320\250+l

offf+\320\221

wi*11 a,b,c,deR and ad-bc= 1. ReplacingieRbyzeC = CU {\321\201\321\216}in the above formula yields a linear

or antilinear transformation \317\206:\320\241\342\200\224>\320\241defined by \317\206(\316\266)=

|^ in

the first case, and ^1^ in the second case. This linear or antilinear

transformation of \320\241uniquely extends to an isometry \317\206:\316\2273\342\200\224+\316\2273of

(H3,dhyp).

Since \317\206:\320\241\342\200\224\342\231\246\320\241respects the circle RcC, its extension \317\206:\316\2273\342\200\224>

\316\2273respects the hyperbolic plane that is bounded by this circle, which

is

H2 = {(re, u) e R2; \320\270> 0} ={(\317\207,0, \320\270)G \320\2322;\320\270> 0} \320\241\316\2273.

The restriction of \317\206:\316\2273\342\200\224\342\231\246\316\2273to H2 then is exactly the original

isometry \317\210:\316\2272\342\200\224\342\231\246\316\2272.Indeed, the property is immediate when \317\206is

a homothety, a horizontal translation of an inversion across a circlecentered on \320\241(in which case the 3-dimensional extension is theinversion across the sphere with the same center and the sameradius),and

the general case follows from these specialexamplessinceLemma 2.12

(or, more precisely, the proof of Lemma 2.9) provides a decompositionof every isometry of H2 as a composition of homotheties, translations

and inversions.

In this way, each isometry \317\210of the hyperbolic plane H2 has anatural extension to the hyperbolic space H3.

We can give a less algebraic and more geometric descriptionof

this extension in the following way. There is a naturalorthogonalprojection \317\201:\320\2303\342\200\224\342\231\246\316\2272defined as follows (compareExercise 2.5). For \316\241G H3, there is a unique complete geodesicg that

passes through \316\241and is orthogonal to H2; then p(P) is the point

where g meets H2. Elementary geometry shows that in cartesian

coordinates, p(P)= (x, 0, yjy2 + u2) if \316\241=

(\317\207,\321\203,\320\270).This construction

also provides a signed distance function q: H3 \342\200\224\342\231\246R defined by

<l{P) =\302\261dhyP(P,p(P)) where \302\261is the sign of the y-coordinate of

\316\241=(\317\207,\321\203,\320\270).See Figure 10.9.

Lemma 10.5. If \317\206is an isometry of the hyperbolic plane (\320\2502,\320\260^\321\203\321\200),

Us extension to an isometry of the hyperbolic space (H3, dhyp) sends


Figure 10.9. The orthogonal projection \317\201

the point \316\241G H3 to the point of H3 that is at the same signeddistance q(il>{P))

= q(P) from H2 as P, and that projects to the point

\317\201(\317\210(\316\241))=

\317\206(\317\201(\316\241))G \316\2272image of p(P) under \317\206.

Proof. This immediately follows from the fact that the extension

of \317\206to H3 is an isometry and respectseach of the two hyperbolic

half-spaces delimited by H2 in H3. D

Lemma 10.6. Let the group \316\223act by isometries and discontinuouslyon the hyperbolic plane (\320\2352,\320\265?\321\214\321\203\321\200)\302\267

Extend this action to an isometricaction on the hyperbolic space (H3,dhyp) as above. Then, the action

of \320\223on H3 is discontinuous.

Proof. Consider a point Pel3 and its projection p(P) G H2. Byhypothesis, the action of \320\223on H2 is discontinuous at p(P). This .means that there exists a small radius \316\265> 0 such that there are only

finitely many 7 G \316\223for which j(p(P)) is in the ballBdhyp (\317\201(\316\241),\316\265).

By continuity of the orthogonal projection p: H3 \342\200\224>H2, there

exists an \316\267such that dhyP(p(P),p(Q)) < \316\265whenever dhyp(P, Q) < \316\264.'

In other words, the image of the ballBdhyp (\316\241,\316\264)under \317\201is contained

intheballBdhyp(p(P),e).

10.3. First rigorous example: fuchsiangroups 255

If j(P) is in the ball Bdhyp(P,S),then p(j(p)) is in the bal1

Bdhyp(p(P),e). Lemma 10.5 shows thatp(-y(P))

=-y(p(P)).

Therefore, 7(P) G Bdhyp(P,S) only when\316\212(\317\201(\316\241))

GBdhyp(\317\201(\316\241),\316\265).By

choice of \316\265,this occurs only for finitely many 7 G \316\223.

This proves that the action of \316\223is discontinuous at every \316\241G

at3. d

By Lemma 10.6, every discontinuous isometric group action onthe hyperbolicplane H2 extends to a group \316\223of isometries of H3

whose action is also discontinuous on H3. Namely, this group \316\223is

a kleinian group. A kleinian group obtained in this way is called a

fuchsian group.

The limit set \320\233\320\263of a fuchsian group \320\223is clearly contained in thecircle l = Ru {00}.Indeed, if we choose the base point Pq in H2,

its orbit \320\223(\320\240\320\276)is completely contained in H2 so that its limit points

must be in R.

If the quotient space\320\2302/\320\223is compact, Proposition 10.4 (adaptedto one dimension lower) shows that \320\233\320\263is the whole circle R. Forinstance,this will occur when \320\223is the tiling group of a tessellationassociatedto the gluing of edges of a bounded polygon in H2,as inthe hyperbolic examples of Section 6.5.

The following lemma enables us to generalize this to more tiling

groups.

Lemma 10.7. In the hyperbolic plane E2, let X be a polygon with

edge gluing data satisfying the conditions of Theorem6.1, so that the

images of X under the element of the tiling group \320\223generated by the

gluing maps form a tessellation0/H2. Suppose in addition that X

touches the circle at infinity R at only finitely many points (possibly

none). Then, if we extend \320\223to a fuchsian group, the limit set \320\233\320\263is

equal to the whole circle R.

Proof. Let (\316\247,\316\254\317\207)be the quotient space obtained by performingthe prescribed gluings on the edges of X. By Theorem 6.25, we can

choose a horocircle\316\237\316\276at each ideal vertex \316\276of X so that whenever;i gluing map sends \316\276to \316\276',it also sends \316\237\316\276to \316\237\316\276>.

In addition, by

(complement 6.24, these horocirclescan be chosen small enough that


they are disjoint and that the only edgesmet by \316\237\316\276are the two edges

leading to \316\276.For each ideal vertex \316\276,let \316\222\316\276be the horodisk bounded

by the horocircle C!j.Then the imagesof \316\247\316\240(L \316\222\316\276

under all elements of the tiling

group \316\223is the union \320\222of a family of (usually infinitely many) disjoint

horodisks in H3. Also, the hypothesisthat X touches the circle at

infinity only at the ideal vertices \316\276implies that if we clip the\316\222\316\276

off X, what is left is bounded in H2. Namely, for an arbitrary base

point PQ e X, the complement X \342\200\224(J?

\316\222\316\276

is contained in a large ball

Bdb\342\200\236(Po,D).

The argument is then very similar to that of Proposition 10.4.

Let re e \320\232be a point of the circleat infinity \316\234which is different from

oo. Pick a point \316\241G H2 which is at euclidcan distance< \316\265from

x. If \316\241happens to be in one of the horodisksof B, we can move it

out of \320\222while keeping it at euclidean distance < \316\265from \317\207(since \320\222

consists of disjoint euclidean disks tangent to R). As a consequence,

we can always choose the point \316\241outside of B. Since the imagesof X under the elements of \316\223tessellate H2, there exists 7 G \316\223be

such that \316\241is in the tile y(X). Then, \316\241is at hyperbolic distance

< D from the point y(Po) of the orbit \316\223(\316\241\316\277)and, as in the proofsof Lemma 10.1and Proposition 10.4, at euclidean distance < eDeDfrom this point 7(^0)\302\267 Taking \316\265sufficiently small, it follows that theorbit \316\223(\316\241\316\277)contains points which are at arbitrarily small euclideandistancefrom x. In other words, \317\207G \316\234is contained in the limit set

Ar.For the point \321\201\321\216,pick a point \317\207\342\202\254\320\232which is very close to 00,

namely, very far from the origin \320\236for the euclidean metric. By theprevious case, there are points of T{Pq)that are very close to x, againfor the euclidean metric. It follows that there are points of \320\223(\320\240\320\276)that

are arbitrarily close to 00 for the euclidean metric. Namely, 00 is inthe limit set Ar. D

According to a traditional terminology, a fuchsian group is saidto be of the first type if its limit set is the whole circle at infinity

R; otherwise, it is of the second type. Lemma 10.7and Chapter 5

provide many examples of fuchsian groups of the first type.

10.4. Poincare's PolyhedronTheorem 257

Simple examples of fuchsian groups of the secondtype include

(he following: the group \316\223\316\271generated by the inversion \316\266\320\275->|, in

which case the limit set \316\233\316\263\316\257is empty (\316\223\316\271is finite with only two

elements); the group \316\2232generated by the translation \316\266\320\275->\316\266+ 1, for

which \316\233\317\2012consists of the single point \321\201\321\216;the group \320\2233generated by

l.he homothety \316\266\320\275->2z, for which Ap3 = {0, \321\201\321\216}has two elements. See

Rxercise 10.11for a fuchsian group of the secondtype whose limit set

in not finite.

Fuchsian groups are convenient examples of kleinian groups, but

not very exciting because of their intrinsically 2-dimensionalnature.Theresultsof the next section will enable us to construct moreinteresting examples.

L0.4. Poincare's Polyhedron Theorem

I 'oincare's Polyhedron Theorem is the natural generalization to three

dimensions of the Tessellation Theorem 6.1 and of Poincare's

Polygon Theorem 6.25. It will enable us to construct tessellationsof the

hyperbolic space H3 by hyperbolic polyhedra, and will provide many

examples of kleinian groups.

10.4.1. Gluing the faces of a hyperbolic polyhedron. We de-line polyhedra in the 3-dimensional hyperbolic space H3. In ordertodo this, we first need to fix conventions about polygons.

A polygon in H3 is a subsetF of a hyperbolic plane \316\240\320\241\320\2353

which is delimited in \316\240by finitely many geodesies called its edges.Recall that the hyperbolic plane \316\240is isometric to the standardhyperbolic plane H2. Under any such isometry, the polygon contained in \316\240

Mien corresponds to a polygon in H2, as defined in Section 4.5.1. As

in the case of H2, we require that edges can only meet at their end-points, calledvertices, and that each vertex is adjacent to exactlyl wo edges. Some of these edges are allowed to be infinite and lead to.1 point of the sphere at infinity \320\241U {00}; such a point is a vertexat infinity or an ideal vertex of the polygon. Also, the polygon F

is required to contain all its edgesand vertices, so that F is a closedsubset ofH3.


A polyhedron in the hyperbolic space H3 is a regionX in H3

delimited by finitely many polygons, called its faces. By convention,

faces can only meet along someof their edges and vertices, and anedgeis adjacent to exactly two faces. As for polygons,werequire that

a polygon contains all of its edgesand vertices, so that it is a closedsubsetof H3.

The edges and vertices of a polyhedron are the edgesand

vertices of its faces. \316\233\316\267ideal vertex or vertex at infinity is a point

of the sphere at infinity \320\241which touches two distinct faces of thepolyhedron.Thereidealvertices can be of two types. The most com-,

mon one is that of an idealvertex of a face, also corresponding to anendpoint of an infinite edge. However, it is also possible for two faces

Fi and F2 to touch at a point \316\276G \320\241without \316\276being the endpointof an edge; this happenswhen Fi and F2 touch \320\241along two circle

arcs that are tangent to each other at \316\276.To avoid unnecessary

complications, we assume that the polygon X approachesan idealvertex

\316\276from only one direction, in the sense that there exists a small eu-:clideanball Bdeuc (\316\276,\316\265)in R3 such that the intersection \316\247\316\240Bdeuc (\316\276,\316\265):

is connected.

We now introduce face gluing data in complete analogy with,

the case of polygons in Sections 4.3 and 4.5. Namely, we group the)faces of the polyhedron X into disjoint pairs {\316\257\\,F2}, {F3, F4}, ... \342\226\240

{F2P-\\, F2P}. For each such pair {F2k-\\,F2k},we are given an isom-

etry \317\206-ik-i'- F2k-\\ \342\200\224*F2k with respect to the restriction of the hyper-;

bolic metric dhyp to these faces, and we define tp2k- F^k \342\200\224>F2k-i to

be the inverse ip2k=

\317\210^-\316\275

Let X be the partition of X defined by the property that for

every \316\241e X the corresponding element \316\241G X containing \316\241consists!

of all points Q =<fik

\320\276</?\320\263\320\272_,

\302\260\302\267\302\267\302\267\302\260<Pu (P), where the indices ii, %2i

..., ik are such that\317\210\316\257\316\257_\316\271

\316\277\302\267\342\226\240\342\226\240\316\277\317\210\316\257\316\271(\316\241)g Ftj for every j.

Endow X with the metric dx defined by the property thai

dx(P,Q) is the infimum of the hyperbolic lengths of all piecewise

differentiable curves joining \316\241to Q in X. We can then considerthe

quotient semi-metric \316\254\317\207an. X.

The proof of Theorem 4.3 automatically extends to this context;to prove:

10.4. Poincare's PolyhedronTheorem 259

Lemma 10.8. The semi-metric \316\254\317\207is a metric, in the sense that

dx(P, Q) > 0 whenever P^Q. \316\240

LO.4.2. Poincare's Polyhedron Theorem. We now consider a 3-dimensional version of the Tessellation Theorem 6.1 and of Poincare's

Polygon Theorem 6.25. It is convenient to combine the two results

into a single statement.

Using the homogeneity and isotropy of H3 (Theorem 9.2), Lemma

4.8 immediately extends to the 3-dimensionalcontext,and provides

;i unique extension of each gluing map \317\210^.Fi \342\200\224>F,\302\261jto a hyperbolic

isometry ipt: \316\2273\342\200\224>\316\2273that along Fi, sends X to the sideof ii\302\261i that

is opposite X.

As before, let the tiling group associatedto the above gluing

data be the group \316\223of isometries of (M3,dhyp) generated by these

extended gluing maps \317\210^.\316\2273\342\200\224>\316\2273.

As in dimension 2, the images of the polyhedron X under theelements of \316\223form a tessellation of the hyperbolic spaceH3 if:

(1) as 7 ranges over all elements of the tiling group \316\223,the tiles

j(X) cover the wholespaceH3,in the sense that their union

is equal to H3;

(2) the intersection of any two distinct tiles j(X) and \316\257'(\316\247)

consists only of vertices, edges and faces of j(X), which are

also vertices, edges or faces of \316\216(\316\247);

(3) (Local Finiteness) for every point \316\241\342\202\254\316\2273,there exists a

ball Bdhyp(P,e) which meets only finitely many tiles \"f(X)\302\267

If, in addition, distinct 7 e \316\223give distinct tiles j(X), the polyhedronV is a fundamental domain for the action of \316\223on H3.

The bending of the boundary of a hyperbolic polyhedron X alongan edge \316\225is measured by its dihedral angle along E, which is defined

as in euclidean geometry. More precisely, for a point \316\241\302\243\316\225which

is not a vertex and for \316\265> 0 sufficiently small, the hyperbolic planeI [ orthogonal to \316\225at \316\241cuts the intersection \316\247\316\240

Bdhyp (\316\241,\316\265)along

a disk sector of radius \316\265and of angle \316\230.The dihedral angle of the

polyhedron X along the edge \316\225is exactly this angle \316\230.One easily

(-decks that it does not depend on the point \316\241\342\202\254\316\225(see Exercise 9.1).


Theorem 10.9(Poincare's Polyhedron Theorem). For a connected

polyhedronIci3 with face gluing data as in Section 10.4.1, supposein addition that the following three conditions hold:

(1) (Dihedral Angle Condition) for every edge \316\225of the

polyhedronX, the dihedral angles of X along the edges that are

glued to \316\225add up to ^- for some integer \320\277\320\265> 1 depending

onE;

(2) (Edge Orientation Condition) the edges of the polyhedron Xcan be oriented in such a way that whenever a gluing map\317\206\317\212.Fi \342\200\224*Fi\302\261i sends an edge \316\225to an edge E', it sends the

orientation of \316\225to the orientation of E';

(3) (HorosphereCondition) for every ideal vertex \316\276of X, we

can select a horosphere \316\236\316\276such that whenever the gluing

map \317\210\317\212.Fi \342\200\224*Fi\302\261isends the ideal vertex to the ideal vertex

\316\276',it also sends\302\273%

to\302\273%'.

Then, as 7 ranges over all the elements of the tiling group \316\223generated

by the extended gluing maps \317\206\316\257:\316\2273\342\200\224>\316\2273,the tiles y(X) form atessellation of the hyperbolic space H3.

In addition, the tiling group \316\223acts discontinuously \320\276\320\277\320\2503,the two

quotient spaces (\320\2353/\320\223,(Jhyp) and (\316\247,\316\254\317\207)are isometric, and these two

metric spaces are complete.

In the Edge Orientation Condition (2), an orientation for the edge\316\225is the choice of a traveling direction along E, usually indicated byan arrow. This is easilyensured by adding a few vertices, providedthe HorosphereCondition (3) holds. Indeed, for a semi-infiniteedge\316\225going from a finite vertex \316\241\342\202\254I3 to an ideal vertex \316\276G C, we

can orient \316\225in the direction from \316\241to \316\276.For a finite edge \316\225joining

two finite vertices, we can introduce a new vertex \316\234at its midpoint,

split \316\225accordingly, and orient the resulting two edges E' and E\"

away from M. For a bi-infinite edge \316\225joining two ideal vertices \316\276'

and \316\276\",we can again introduce a new vertex \316\234at the midpoint of

the two points \316\225\316\2405\317\202\302\273and \316\225\316\240

\316\236\316\276\302\273determined by the horospheres

provided by the HorosphereCondition (3), and orient the resultingtwo new semi-infinite edges E' and E\" away from M. The setup is

10.4. Poincare's Polyhedron Theorem 261

then speciallydesignedso that the gluing maps <\302\243\302\273:Fj \342\200\224>F{\302\261irespect

these new vertices, and also respectthe edgeorientations so defined.

As in the 2-dimensional case, the DihedralAngle Condition (1)

and the Horosphere Condition (3)are much more critical. The EdgeOrientation Condition (2) was essentially introduced to simplify thestatement of the Dihedral Angle Condition (1).

Proof of Theorem 10.9.The proof is essentially identical to the 2-dimensional case, with only a minor twist indicated in Lemma 10.10

l>elow. Consequently, we only give a sketch of the arguments.Our proof of Poincare's Polygon Theorem 6.25 immediately ex-

Lends to the 3-dimensional context to show that because of the Horo-

.sphere Condition (3) (to be compared with condition (1) of

Proposition6.23), the quotient space (\316\247,\316\254\317\207)is complete.

To prove Theorem 10.9, we follow the strategy of the proof ofi,he Tessellation Theorem 6.1, by setting one tile after the other. In

particular, a tile \317\206(\316\247)is adjacent to X at \316\241if there exists a sequenceof gluing maps tptl, \317\206\316\2572,..., \317\206^such that

\317\210\316\257^1\302\260\342\226\240\342\226\240\342\226\240\302\260

\317\210\316\2571(\316\241)belongs

l.o the face Fii for every j < I (including the fact that \316\241\342\202\254F^) and

\316\250=\316\250\317\212*\302\260

\316\250\317\212\316\266\302\260\302\267\302\267\302\267\302\260

\316\250\316\252?'\302\267

More generally, the tiles \317\206(\316\247)and \317\210(\316\247)are adjacent at the point\316\241G \317\206(\316\247)\320\237\321\204(\320\245)if \317\210-1

\316\277\317\210(\316\247)is adjacent to X at \321\204~1(\320\240)in the

above sense.

The only point that really requires some thought is the following

iinalogue of Lemma 6.2 and Complement 6.3. We need some

terminology. Let a polyhedral ball sector with center \316\241and radius \316\265in

IH3 be a region in the ball Bdhyp (\316\241,\316\265)delimited on its sides by finitely

many hyperbolic planes passing through the point P; we allow thecase where the polyhedral ball sector is the whole ball

Bdhyp(P,\316\265).

This is the 3-dimensional analogue of disksectors,specially designed

\320\275\320\276that every point of X is the center of some ball Bdhyp {\316\241,\316\265)such

l hat \316\247\316\240Bdhyp (-Pi\316\265)>3 \320\260polyhedral ball sector.


Lemma 10.10. There are only finitely many tiles \317\210(\316\247)that are\\

adjacent to X at the point P. In addition, these adjacent tiles decom-lposea small ball Bdhyp(P,e) into finitely polyhedral ball sectors with .

disjoint interiors.

Proof. First, we observe that there are only finitely many points:

\316\241= Pi, P2, ..., Pn of X that are glued to P. This is automatic':

when \316\241is an interior point, since it is only glued to itself. Similarly,

a face point, namely, a point of a face which is not contained in;an edge, is only glued to one other point, and a vertex can onlyS

be glued to some of the finitely many vertices. The case of edge a

points, namely, points of edges that are not vertices, will require 1

more thought and critically relies on the HorosphereCondition (3) of i

Theorem 10.9. jWe claim that no two distinct edge points P' and P\" of the same \342\226\240\342\226\240

edge \316\225can be glued together. Indeed, if two such points P' and P\" = )

\317\210(\316\241')are glued together by the gluing map \317\206=

\317\206^\316\277

\317\206\316\2572\316\277\302\267\342\226\240\342\226\240\316\277

\317\210\316\257\316\271_,

then \317\210sends the edge \316\225to itself, and respects its orientation by the

Edge Orientation Condition (2). In addition, it must send at least

one point of \316\225to itself: If both endpoints of \316\225are finite vertices, j

\317\206must fix these two endpoints since it respects the orientation of'!

E; if exactly one endpoint of \316\225is a finite vertex, \317\206must clearly fix :j

this endpoint; finally, when \316\225is a bi-infinite edge joining two ideal \342\226\240

vertices \316\276and \316\276',\317\206must fix the intersection points \316\225\317\200\316\236\316\276and

\316\225\317\200\316\262\317\202'?

determined by the horospheres S$ and5\316\276/

associated to \316\276and \316\276'by the i

Horosphere Condition (3). As a consequence,the restrictionof \317\206to;

\316\225is an orientation-preserving isomctry fixing at least one point, and \302\267

is consequently the identity. But this would contradict our hypothesis ,

that the edge points P' and \316\241\"=\317\206(\316\241')

are distinct.

This proves that the set of points that are glued to an edgepoint!\316\241can include at most one point of each edge, and is consequently ]

finite. I

Therefore, in all cases, there are only finitely many points \316\241\342\200\224P\\, j

P2, ..., \316\241\316\267\316\277\316\271X that are glued to P. \\

For \316\265small enough, the intersection of X with each of the balls \342\226\240

Bdliyp(P,,e)is a polyhedral ball sector. Let

<Sdhyp(-Ppe)denote the

10.4. Poincare's Polyhedron Theorem 263

hyperbolic sphere consisting of the points that are at distance\316\265from

/';, and let Sj be the intersectionX\320\237Sdhyp[Pj,s).

As usual, endow Sjwith the metric dsj where dsj {Q,R) is the infimum of the hyperbolic

lengths of all curves joining Q to R in Sj.

If we rescale the metric d<,v by a factor sinhe, Exercise 9.4 showsl.hat the metric space (Sj, ^jjjfe^j) ^ isometricto a polygon in the

sphere (\302\2472,\321\2014\321\200\321\214)>whose angle at each vertex \316\241is equal to the

dihedral angle of X along the edge containing P. The restriction of

(.liegluing maps \317\210\316\271to the edges of the Sj then defines isometric

gluing data for these spherical polygons. Because of the Dihedral Angle

Condition (1) of Theorem 10.9and because no two distinct points ofl.lie same edge are glued together, the TessellationTheorem6.1 (as

extended to disconnected polygons in Section 6.3.4, and usingProposition 6.20 to guarantee completeness of the quotient space) shows

\316\271hat this gluing data provides a tessellationof the sphere \302\2472.

In particular, if we start from the polygon 5\316\212\320\241<Sdhyp (\316\241,\316\265)and

proceed with the tiling procedure using this gluing data, we obtain;i tessellationof the sphere <Sdhyp(P, \316\265).It immediately follows from

definitions that the tiles of this tessellation are exactly the polygons\342\200\242p(Sj),where \317\206G \316\223is such that the polyhedron \317\210(\316\247)is adjacent to

\320\232at \316\241and where\317\206(\316\241,)

= \316\241.

By compactness of S2, this tessellation ofSdhyp(P, \316\265)has finitely

many tiles (compare Exercise 6.3). Lemma 10.10 now immediately

follows from these observations. D

Substituting Lemma 10.10 for Lemma 6.2 and Complement 6.3,ihe proof of Theorem 6.1 now immediately extends to show that the

polyhedra j(X) with 7 G \316\223tessellate the hyperbolic space H3. Inother words,X is a fundamental domain for the action of the tiling

\321\206\320\263\320\276\320\270\321\200\320\223on I3.

Once we have shown that X is a fundamental domain for the

action of \316\223,the proofs of Proposition 7.10 and Theorem 7.12immediately extend to three dimensions to show that the action of \316\223is

discontinuous, and that the quotient space (\320\2353/\320\223,dhyp) is isometric

i*(X,dx). D

Similarly, we have the following analogue of Theorem 4.10.


Theorem 10.11. Under the hypotheses of Theorem 10.9, suppose in\\

addition that:

(1) the Dihedral Angle Condition (1) of Theorem 10.9 is strength*

ened so that for every edge\316\225of X, the sum of the dihedral

angles of X along the edges that are glued to \316\225is equal to

(2-3) Conditions (2) and (3) are unchanged;

(4) the 3-dimensional extensions (/?,: \316\2273\342\200\224>\316\2273of the gluing

maps are orientation-preserving.

Then, the quotient space (\316\247,\316\254\317\207)is locally isometric to the hyperbolic

space (H3,dhyp)\302\267

Proof. Wc will prove that the action of the tiling group \316\223is free,

namely, that \317\210{\316\241)\317\206\316\241for every \316\241G H3 and every \317\206G \316\223different

from the identity.

Suppose, in search of a contradiction, that some \317\210\342\202\254\316\223\342\200\224{\320\253\321\206\320\267};

fixes a point \316\241of H3. Then Exercise 9.10 shows that because \317\206is1

orientation-preserving by condition (4), it fixes each point of a whole*

complete geodesic g of H3.

The geodesicg cannot meet the interior of a tile \317\215(\316\247),with 7 \320\261

\320\223,of the tessellation provided by Theorem 10.9. Indeed, \317\210would;

otherwise send this tile to itself, which is excluded by the fact thai

X is a fimdamental domain for the action of \316\223.Therefore, g contai:

an edge \316\225of some tile j(X),

Replacing \321\203>\320\254\321\2037_1\320\276\321\203>\320\2767(which fixes each point of the edg<

7_1(\302\243J) of X) if necessary, we can assumewithout loss of generalitjthat 7 is the identity. In particular, \317\206now fixes every point of ari

edge \316\225of X.

Let \316\247\316\273=X,X2 = 12(X), \320\245\320\267=

7\320\267\320\2400,\302\267\302\267\302\267,Xn =

7\342\204\242P0be th

tiles sitting around the edgeE, and let \316\270\\be the dihedral angle of tpolyhedron Xi alongthis edge E. Then 5\316\226?=\316\271^\316\257

= 2\317\200\302\267

By the second half of Theorem 10.9,the inclusion map X \342\200\224

induces an isometry X \342\200\224>\320\2353/\320\223.In particular, the fact that this maj

is bijectiveimpliesthat two points P, Q \302\243X are glued together if anc

only if there exists an element 7 e \320\223such that Q \342\200\224j(P). Therefore

10.5. More examples of kleiniangroups 265

the edges of X that axe glued to \316\225are the edges \316\225,7^\"1(\302\243), \316\267^\316\271{\316\225),

\342\226\240..,7~1 (E). Note that the dihedral angle of X along the edge 7\320\263~1(\320\225)

is also equal to #,.

However,there are duplicationsin this list. Indeed, the tile \317\210(\316\247)

is one of these JQ, so that \317\206= ji and jJ'1(E) =

\317\206~1(\316\225)= \316\225since \317\206

fixes every point of E. Therefore, the sum of the dihedral angles ofX along the edgesthat are glued to \316\225is strictly less than \302\243)7=\316\271^\302\267

=

2\317\200,contradicting the stronger Dihedral Angle Condition (1) in the

hypotheses of Theorem 10.11.

This proves that the action of \316\223on H3 is free. Since this action isdiscontinuous by Theorem 10.9, Corollary 7.9 showsthat the quotient

space (\320\2353/\320\223,Jhyp) is locally isometric to (H3,dhyP). \320\237

It can actually be shown using Lemma 10.10that conditions (2-

3) of Theorem 10.9 are not really necessary for Theorem 10.11 tohold. Similarly, condition (4) of Theorem 10.11 can be significantly

relaxed, but at the expense of making the precise result somewhat

more cumbersometo state.Exercise10.8shows that Theorem 10.11 fails when only its

conditions (1-3) are realized, without condition (4) or any similar

additional hypothesis.

10.5. More examples of kleinian groups

We return to the examples of crooked Farey tessellations of

Section 10.1. We now have the tools to rigorously justify some of our

observations.

10.5.1. Kleinian groups associatedto crookedFarey

tessellations. We will prove that when the shear-bend parameterss\\, S3,

s5 are sufficiently small, the tiling group \316\223of the crooked Fareytessellation associated to these parameters act discontinuously on thehyperbolic spaceH3.We will restrict our attention to the casewhere

S5 is real, as the arguments are a little simpler with this condition.

First consider our original punctured torus exampleof Sections

5.5 and 6.6, corresponding to S\\= S3 = s$ = 0. Then, the square

T+ U T~ is a fundamental domain for the action of \320\223on i2. If we


extend the action of \316\223on H2 to a fuchsian action on H3,Lemma 10.5

shows that this action is well behaved with respect to the

orthogonalprojection p: \320\2503\342\200\224>H2. As a consequence, the preimage X =p~1(T+U T~) is a fundamental domain for the action of \320\223on H3.

Let us analyze this fundamental domain a little better. As usual,let \316\225\316\271,Eh, E3 and E4 be the edgesof the square \320\223+U T~, where E\\

goes from \342\200\2241to \321\201\320\276,\316\225\316\271from 0 to 1, E$ from 1 to oo, and E4 from 0 to

\342\200\2241. Then X is bounded by the preimages\320\251= p~1(Ei), and each Hi

is a hyperbolicplane in H3. More precisely, #1 and II3 are vertical

euclidean half-planes, while H2 and ff4 are euclideanhalf-spheres of

radius I centered at\302\261|.

In addition, when we extend the gluing

maps to hyperbolic isometries \317\206^.\316\2273\342\200\224>H3, \317\210\316\271sends Hi to Hi and

\317\2103sends \320\2573to Hi.

We will now create a very similar fundamental domain in the moregeneral casewhere S5 is real and where Si and S3 have an imaginary

part which is not too large (and where, as usual, the relation Si +

S3 + s5 = 0 is satisfied).Recallthat the tiling group \316\223of the corresponding crooked Farey

tessellation is generatedby

. . eS5z +1 , . . _Ss z-\\<Pnz)

=^:\342\200\224;\342\200\224_,,, \317\204

and <\320\240\320\267(\320\263)= \320\265

\320\265\320\2575\320\263+ \320\265-\321\2171+ 1^ w

-\320\263+ \320\2655\320\267+ 1

Consider the four circles C\\, C2, C$, C4 in \320\241defined as follows: C3 is

the euclidean line passingthrough the point 1 and making an angle of

\316\270=|(\317\200

\342\200\224Im(si)) with the \320\266-axis, counting angles counterclockwise;

C\\ is the line parallel to C3 passingthrough \342\200\224e_Ss;C-i is the circle

passing through the points 0 and 1 and tangent to the line C3 at 1;C4 is the circle passing through 0 and \342\200\224e\"S5 and tangent to the lineCi at -e_S5.SeeFigure 10.10.

Lemma 10.12. If S5 is real and iflm(si) is sufficiently small that

cos(Im(si)) >\320\264|1\320\264|~|,any two distinct Ct and Cj meet only at 0 or 1

of the points {0,1,\342\200\224

e~S5,00} and are tangent to each other at their

intersection point when they meet.

In addition, \317\210\316\271sends Ci to C2, and \317\206$sends C3 to C4.


Figure 10.10. A fundamental group for the tiling group of a

crooked Farey tessellation

Proof. By elementary euclidean geometry,the two circles Ci and \320\2414

have the same tangent line at 0, which makes an angle of \317\200\342\200\224\316\270with

the z-axis. Therefore, for the first statement, we only need to checkthat the circle C2 is disjoint from the line C\\ and that the circleC4 is disjoint from the line C3. This is equivalent to the propertythat the distance between the two lines C\\ and C3 is greater than

the diameters of both \320\241\320\263and C4. Computing these quantities byelementary trigonometry in Figure 10.10 gives the condition stated.

To prove the second statement, note that the differential of </?\320\267

at the point 1 is the complex multiplication by e~S5~S3, by a

computation using Proposition 2.15. It follows that the line tangent to

^\320\267(\320\241\320\267)at 0 =9?\320\267(1)makes an angle of \316\270+ Im(\342\200\224s5

\342\200\224S3) with the

x-axis. By choice of \316\270=|(\317\200

\342\200\224Im(si)) and because of our hypothesis

that \342\200\224s-, is real, this angle is equal to \317\200\342\200\224\316\270.Since a linear fractional

map sends circleto circle(Proposition 2.18), it follows that ^\320\267(\320\241\320\267)

is a circle passing through the points 0 =^\320\267(1) and \342\200\224

e_S5= <Pz(og)

and whose tangent line at 0 makes an angle of \317\200\342\200\224\316\270with the :r-axis.

There is only one circle with these properties, namely, C\\. Therefore,^\320\267(\320\241\320\267)

= C4-

The proof that \317\210\317\207{\316\237\317\207)\342\200\224Ci follows from very similar

considerations. D


Proposition 10.13. Let \316\223be the tiling group of the crooked Farey

tessellation associated to the parameters \316\262],S3, 55 with s$ real and

cos(Im(si))> eils'i+i (and Si + s3 + s5 = 0). Then, the action of \316\223

on H3 is discontinuous. In particular, \316\223is a kleinian group.

Proof. For \320\263= 1, 2, 3, 4, let \320\251be the hyperbolic plane in H3 that

touches the sphere at infinity \320\241along the circle C{. Let X be thepolyhedron bounded by these four hyperbolic planes. Namely, X

consists of those points in the upper half-space H3 that are between

the euclidean vertical half-planes H\\ and \320\251and above the euclidean

half-spheres H2 and ff4. Note that X is a hyperbolicpolyhedron with

four faces Hi, ff2, #\320\267,#4, with no edge or finite vertex, and with

four ideal vertices 0, 1, \321\201\321\216,\342\200\224e\"5.

By Lemma 10.12, the isometry \317\2061:\316\2273\342\200\224>H3 sends Hi to H2-By consideringthe image of an additional point, it is also immediatethat it sends X to the side of \320\251that is opposite X. Similarly,\317\2103:H3 \342\200\224>H3 sends H3 to \320\251and sends X to the side of H3 that

is opposite X. Consequently, we are in the situation of Poincare's

Polyhedron Theorem 10.9.

The polyhedron X has no edge. Therefore, in order to applyTheorem 10.9, we only need to check the Horosphere Condition (3)of that statement at the ideal vertices of X. As in the 2-dimensional

setup of Section 8.4, the hypothesis that s\\ + s3 + s5 =0 will be

critical here.

We need to find horospheres SO, S\\, Sx, Sl_e-\302\2735 centered at 0,

1, 00, \342\200\224e_S5,respectively, such that \317\206\316\271(3\316\277\316\261)= Si, ifi(S_e-,5) =

So, tpziSac) =<S_e-\302\2735 and <pz(Si) = So- We could construct these

horospheres\"by hand\", but it will be easier to usean argument similar

to that of Proposition 6.23.

Start with an arbitrary horosphere S<x, centered at 00, namely,

with an arbitrary horizontal euclidean plane. Then set 5\316\212=

v?i(<Soo)

and \302\243Le-s5= <P3(Sao). We will then be done if we can choose SO to be

simultaneously equal to </?3(5\316\212)=

\317\206-s\316\277\317\210\316\271(3(\316\247>)and to ^\316\271(5_\316\262-.\316\222)

=

\317\210\316\271\320\276</?\320\267(\302\243\320\276\320\276),namely, if these two horospheres centered at 0 are equal.

The condition that </?\320\267\302\260

<Pi(Soc)=

\317\210\316\271\302\260\317\206-iiSoo) is equivalent to the

property that \317\210^\316\271

\302\260\317\206^1

\302\260</?\320\267

\302\260\317\210\316\271

=\317\2104

\302\260\317\210\316\271\302\260</?\320\267\302\260

\317\210\316\271respects the

10.5. More examples of kleinian groups 269

Figure 10.11. The images of the fundamental domain X\302\260

under the tiling group \316\223\302\260of the standard Farey tessellation

Figure 10.12. The images of the fundamental domain Xunder the tiling group \320\223of a crooked Farey tessellation

liorosphere Sao- We already computed that because si + .S3+ S5 = 0,

^40^0^30 \316\250\316\271(\316\266)= e2si+2s3+2s*z + 1 + eS3+ eS3+Sl

I gS3 + Sl+S5_|_ g2S3+Sl+S5 I g2S3+2si+S5

= z + 2 + 2e\"3 +2eS3+Sl

in a translation of C, so that its isometric extension to H3 is ahorizontal translation. In particular, this horizontal translation respectsl he horizontal plane SOo, thus we are done.


We can therefore apply Poincare's Polyhedron Theorem 10.9,and

conclude that X is a fundamental domain for the action of \316\223on H3

and that this action is discontinuous. D

The images of the fundamental domain X under the tiling groups

are illustrated in Figure 10.11 for the standard case where Si \342\200\224s3

=

s5 = 0, and in Figure 10.12for a more general example satisfying thehypothesesof Proposition 10.13, in this case Si = \342\200\2240.5+ 1.4i, S3 =

0.3 \342\200\2241.4i and s5 = 0.2. Moreprecisely,thesefigures represent in \320\241the

images of the circles C\\, C2, C3, C4 (bounding the hyperbolic planesH\\, #2, #3, #4 delimiting the polyhedron X) under the elements of

the tiling group \320\223.

We already encountered in Figures 10.4 and 10.7the crookedFarey tessellation corresponding to the kleinian group of Figure 10.12.

When comparing these figures, it is important to remember that

Figures 10.4 and 10.7 have been rotated soas to be invariant by a

translation parallel to the a;-axis.

10.5.2. Limit sets. We now consider the limit sets \320\233\321\200of the kleinian

groups \320\223provided by Proposition 10.13.

The set of ideal vertices of the standard Farey triangulation T\302\260

is the set <Q>= Q U {00} of all rational points in the circle at infinity

R. Given a crooked Farey tessellation T, we indicated in Section 10.1

that there is a one-to-onecorrespondencebetween the faces, edges

and vertices of \316\244and the faces, edges and vertices of the standard

Farey tessellation T\302\260.In particular, each ideal vertex \317\207G Q of T\302\260is

associated to an ideal vertex \316\273(\316\271)\342\202\254\320\241of \320\242.

This correspondence defines a map on the set of all rational points

<Q> in R. We will show that under the hypothesesof Proposition 10.13

(so that the tiling group \320\223of \316\244is a kleinian group), it has a continuous

extension to all real numbers.

Proposition 10.14.Let \316\244be the crooked Farey tessellationassociated to parameters s1; s2, *\320\267such that s5 is real and cos(Im(si))>e|\342\200\236j)|~^(and Si + s3 + s5 = 0). The above map \317\207\316\271\342\200\224>X(x) extends to

a homeomorphism\316\247:\316\232-> \320\233\320\263


between the circle at infinity R and the limit set \316\233\317\201of the tiling group

\320\223\320\276/\320\242.

The proof will take a while and is split into several partial steps.Our strategy is reasonably well illustrated by the comparison ofFigures 10.11 and 10.12. We will use the family of circles j{Ci), with

7 G \320\223to capture the structure of the limit set, and compare this datato the data similarly associated to the standard Farey tessellation.

With this strategy in mind, it will be convenient to denote by\320\223\302\260the tiling group corresponding to si =

s^= s5 = 0, whereas \320\223

will denote the tiling group of the crookedFarey tessellation \316\244

associated to arbitrary Si, s3 and s5 with si + s3 + s5 = 0, s5 real,and cos(Im(si)) >

\302\260|.\320\264|~|\302\267Proposition 10.13 guarantees that \320\223is

a kleinian group, namely, it acts discontinuously on H3. Rememberfrom Section 10.1 that there is a natural correspondence between

elements of \320\223and elements of \320\223\302\260,and between faces, edges and verticesof the crooked Farey tessellation \316\244and faces, edges and vertices ofthe standard Farey tessellation T\302\260of \320\2302\320\241H3. Given any objectassociated to \320\223or T, we will try as much as possible to denote the

corresponding object of \320\223\302\260or T\302\260by adding a superscript \302\260.

Let \320\241denote the collection of the circles of Figure 10.12, namely,the images of C\\, \320\241\320\263,\320\241\320\267,C4 under the elements of the tiling group

\320\223.Following the above convention, let G\302\260denote the

correspondingcollection of circles in the case where Si =s3

= s5 = 0, as in

Figure 10.11.To fix some terminology, let us say that a point is surrounded

by an actual circle \320\241(namely, not a line) if it is located on the insideof \320\241or on \320\241itself. More generally, a subset ofthe planeis surrounded

by the circle \320\241if every point in this subset is surrounded by C.

Lemma 10.15.For every \316\265> 0 and every \320\241\342\202\254G that is an actualcircle (namely, not a line), there are only finitely many circles \320\241G \320\241

that are surrounded by \320\241and whose euclidean diameter is greaterthan \316\265.

Proof. For each such circle C, considerthe hyperbolic plane H' in

H3 bounded by this circle.


Pick an arbitrary base point Pq = (xq, yo, uq) G H3 whose

projectionxq + ij/o G \320\241is surrounded by C. We claim that the hyperbolic

distance from Pq to H' is uniformly bounded in terms of the euclideandiameter D of C, of the coordinate u0 and of \316\265.Indeed, if \316\241is the

apex of the euclidean half-sphere H', namely, the point of H' with

the largest third coordinate, we can join \316\241to Pq by a curve made upof a vertical line segment of hyperbolic length < log ^f and of ahorizontal line segment whose hyperbolic length is < ^-. It follows that

the hyperbolic distance from Pq to H' is bounded by r = log ^ + ^-.By local finiteness of the tessellation ofH3 by the polyhedra\317\210(\316\247)

with \317\210G \316\223,the closed ball Bdhyp(Po; r) can only meet finitely many

\317\210(\316\247),and therefore only finitely many such H'. (Comparethe proof

of Lemma 7.15). This concludes the proof. D

We can now describe the image of a point \317\207\342\202\254\320\226under the map

\316\273:I -> \320\233\320\263.

If \317\207= oo, we just set \316\273(\316\277\316\277)= \321\201\321\216.Otherwise, let G\302\260be the set of

those circles in 6\302\260that surround x. Let Qx \320\241\320\241be associated to 6\302\260

by the correspondence between circles in \320\241and circles in G\302\260.

The circles of 6\302\260are also the boundaries of the hyperbolicplanes

p~x(E), where \316\225ranges over the edges of the tessellationof H2 by

hyperbolic squares constructed in Section 6.6, and associated to our

standard punctured torus. In particular, all these edges \316\225are also

edges of the Farey tessellation.By inspection of the Farey tessellation,we consequentlysee two different patterns according to whether \317\207is

rational or not.

If \317\207is irrational, it cannot belong to any circle of 6\302\260.Therefore,

the circles of 6\302\260are nested together and can be listedas\320\251,K%, ...,

ET\302\260,... in such a way that each ET\302\260is surrounded by \320\232^_\320\263.

If \317\207is rational, the elements of 6\302\260can be listed into three families.First thereis a common stalk (possibly empty) consisting of finitely

many circles K%, K%, ..., K\302\260which do not contain the point x,with each A\"\302\260surrounded by K^_x. Then, 6\302\260splits into two infinite

familiesK\302\260p+1,K\302\260p+2,..., K\302\260,... and L\302\260+1, L\302\260+2, ...,L'\302\260, ... of

circles which do contain x, with A\"\302\260to the right of \317\207and L\302\260to the

left; again, each AT\302\260is surrounded by \320\232^_\320\263and L\302\260nis surrounded by


L\302\260n_jfor every \316\267> \317\201+ 1. In addition, K\302\260surrounds all A\"\302\260and L\302\260

with \317\204\316\271> \317\201+ 1.

Let \316\225\316\223\316\267and Ln be the circles of Qx associated to ET\302\260and L\302\260,

respectively, by the correspondence between Q% and Qx.

Lemma 10.16. If \317\207\302\243KC \320\233\320\263\302\260,there is a unique point \316\276G \320\233\320\263that

is surrounded by all the circles in Qx.

Proof. Let \316\232\317\207,K2, ..., Kn, ... be the circlesof Qx defined above.

Apply Lemma 10.15 to construct, by induction, an increasing

sequence (nk)keN such that the diameter of each \320\232\320\237\320\272is less than\317\210.

If

we pick an arbitrary point \316\276*,surrounded by \320\232\320\237\320\272,the convergence of

l.he seriesY^L-\317\207\317\210implies that the sequence (\302\243fc)fcew has finite length

in C, and therefore converges to some \316\276by completeness of \320\241= R2

(Theorem 6.8). Every Kn surrounds \316\276\316\271^for every \320\272large enough that

n-k > n, and therefore surrounds \316\276by passing to the limit.

By Lemma 10.15,the diameter of Kn tends to 0 as \316\267tends to oo.

It easily follows that \316\276is the only point that is surrounded by all the

\316\233'\316\267-

When \317\207is irrational, this proves that \316\276is surrounded by everycircle of Gx. When \317\207is rational we have to worry about the remainingcircles Ln. However,in this case, note that the Kn all surround thevertex of the crooked Farey tessellation \316\244that corresponds to the

vertex \317\207of the standard Farey tessellation 7\302\260.Therefore, \316\276is equal

l.o this vertex of T, which also belongs to all the Ln. As a consequence,

\302\243is surrounded by all circles of Gx when \317\207is rational as well.

To concludethe proof, we need to show that \316\276is in the limit set

\316\233\317\201.Let PQ G X be an arbitrary base point. For every circle Kn G Qxlot \316\240\316\267be the hyperbolic plane that is bounded by Kn and, among

l.he tiles of the tessellationof H3 by images of X under elements7 G \320\223,let 7nP0 be the tile that is just below \320\237\342\200\236.The euclidean

distance \316\254\316\265\316\2670(\316\276,jn{Po)) is bounded by the euclidean diameter of Kn,

;md consequently converges to 0 as \316\267tends to +oo by Lemma 10.15.Therefore, \316\276

=limn_\302\273oo \316\212\316\267(\316\241\316\277)is in the limit set \316\233\317\201\302\267 \342\226\241

Define \316\273:R \342\200\224*\320\233\320\263by the property that \316\273(\316\277\316\277)= oo and that the

image X(x) of \317\207G R is the point \316\276provided by Lemma 10.16. Notel hat this is consistent with our earlier definition when \317\207is rational.


Lemma 10.17. The map \316\273:R \342\200\224*Ay is continuous.

Proof. We will use the notation of Lemma 10.16and of its proof.

For \317\212\302\243\316\232irrational, take an arbitrary \316\265> 0. By Lemma 10.15,there existsal. least otie (and in fact infinitely many) \316\267such (hat the

circle Kn G 6.,. has diameter < \316\265.Since \317\207is irrational, it does not

belong to the correspondingcircleK\302\260G C\302\260.Therefore, there exists

an \316\267> 0 such that every \321\203G R with (ijiicf1, v) < V is ^lso surrounded

by K]). The image X(y) of such a point \321\203is surrounded by A\"\342\200\236by

construction of \316\273,so that deut:(\\(x),X(y)) < \316\265.Therefore, for every\316\265> 0, we found an \316\267> 0 such that deuc(X(x).\\(y)) < \316\265whenever

dvuci-r-'V) < \316\240-Namely, \316\273is continuous at x.

The argument is very similar for \317\207rational, except that we haveto also use the circlesLn. For every \316\265> 0, Lemma 10.15 againprovides an n such that Kn and Ln both have diameter < \316\265.Now,

there is an \316\267such thai, every \321\203G R with rfeuc(^i y) < V is surrounded

by K^ if \321\203^ \317\207and by L^ if \321\203^ a\302\267.In both cases this guarantees that

^\320\265\320\270\321\201(\320\220(\320\260;),\320\220(\320\263/))< \316\265.This proves the continuity at every rational x.

For the continuity at \317\207= oo. it will be convenient to consider the

translation r =\317\2104

\316\277<\317\2012\302\260\316\250\316\254

\302\260<Pi G \316\223,and the corresponding element

\317\2040G \316\2230.We already computed that \317\204{\316\266)= \316\266+ 2 + eS3 + 2eS3+Sl

and that \317\204\302\260(\316\266)= \316\266-f 6. Note that \316\273is well behaved with respect

to r and r\302\260,in the sense that\316\273(\317\204\302\260(.\317\204))

=r(A(a;)) for every a; G R:

If a; G R is very close to oc, then, for the euclidean line C\302\260G G\302\260

passing through 1, it is outside of the vertical strip delimited by the

lines (r\302\260)n(C10) and (\321\202\302\260)~\320\277(\320\241?)for n > 0 very large; it follows that

X(x) is outside of the strip delimited by the lines rn(Ci) and \321\202~\320\277{\320\241\\)

and is consequently close to oo. Formalizing this reasoning with the

appropriate quantifiers proves that \\(x) tends to oo = A(oc) as \317\207

tends to oc. Namely, \316\273is continuous at oc. Q

We are now ready to complete the proof of Proposition 10.14{

which we restate here as:\\

Lemma 10.18. The map \316\273:R \342\200\224>\320\233\320\263is a homeomorphism.

Proof. We have to show that \316\273is injective, surjective, and that itsj

inverse \316\273-1:\320\233\320\263\342\200\224>R is continuous. ,11 !


If \317\207\317\206\321\203G R, we can find a circle C'J G <?.\302\260which surrounds \317\207

but not y. Then, by construction, the correspondingcircleC'eEsurrounds X(x) but not X(y), so that \316\273(\317\207)\317\206X(y). Also, X(x) \317\206\321\201\320\276

when \317\207\317\206oo. This proves that \316\273is injective.

To prove that \316\273is surjecl.ive, consider a point \316\276\317\206oo in the limitset \320\233\320\263.Pick a base point Pq in the triangle T\"- of the crooked Fareytessellation T. Then for every \316\265> 0 there exists a 7 \316\265\316\223such that

''<4ic(7(-Po);0 < \316\265\302\267We claim that the image of at least one of the

vertices of T~ under \316\267is at euclidean distance < \316\265from \316\276.Indeed,

I he part of H3 that lies outsideof the euclidean ball \320\224/\342\200\236,^(\316\276,\316\265)is a

hyperbolic half-space bounded by a hyperbolicplane,and is therefore

convex in (he hyperbolic sense. It follows that if the three verticesof an ideal triangle are outside of \316\262^\316\220\316\223(\316\276,\316\265),the whole triangle is

outside of that ball. In our case, since the ideal triangle j(T+) meets'^iru,-(\302\243j\302\243)m \320\224)!at least one of its vertices must be in \316\222,\316\271.\342\200\236,,(\316\276,\316\265).

As a consequence, for every n. tlieie is a vertex \316\276\342\200\236of the crooked

Farey tessellation \316\244with \316\254\316\262{]0(\316\276\316\267,\316\276)< ^. The definition of the map\316\273shows that \316\276\316\267

= X(xn), where :r\342\200\236G Q U {00} is the vertex of the

standard Farey tessellation \316\244corresponding to \316\276\316\275.If the sequence

(\342\200\242''iJneN*s bounded in R, it admits a converging subsequence by

Theorem 6.13; otherwise, one can extract from (xn)neN a subsequence

converging to the point oo in R. Therefore, we can always find a

.subsequence {xn,,)keN which converges to somex~^ G R. Then, \316\276is

1 he limit of \316\276\342\200\236\316\221.= X(xnh) as \320\272tends to 00, which is equal to A(a;oo)

hy continuity of \316\273.

For every \316\276G \316\233\342\200\224{oo}, we consequently found an \317\207xG \316\233\317\201\302\260such

l hat \316\273(^\316\277\316\277)\342\200\224

\316\276\302\267For \316\276= 00, note that \316\273(\316\277\316\277)

= oo. This proves that

\316\233is surjective. In particular, the inverse map \316\273-1:\316\233\316\223\342\200\224>R is now

well defined.

To show that \316\273-1is continuous, we use a proof by contradiction..Suppose that \316\273\"\"1is not continuous at \316\276G \320\233\320\263different from 00. This

means that there exists an \316\265> 0 such that for every 77 > 0, there

exists \316\276'G \320\233\320\263such that \316\254\316\262\316\2670(\316\276,\316\276')< \316\267but deuc(X 1(\316\276),\316\273~1(\316\257'))>

. Applying this to each \316\267= ~ provides a sequence (\316\276\316\227)\316\267\316\265\316\235which

converges to \316\276but. such that the points xn= \316\273-1(\316\276\316\267)stay at distance

\342\200\242\316\265from \317\207=

\316\247~1{\316\276)in R. As above, extract a subsequence(xnk)keN


which converges to some \321\205\320\266G R. Then, by continuity, A(a;00)=

\316\276=

X(x) but dexic(x,Xqo) ^ \316\265,contradicting the fact that \316\273is injective.

Therefore, \316\273-1is continuous at every \316\276G \320\233\320\263different from \321\201\321\216.The

continuity of \316\273-1at \321\201\321\216is proved by an almost identical argument.

This concludes the proof that \316\273is a homeomorphism. D

10.5.3. Nondifferentiability of the limitset. Proposition 10.14

justifies our observation that many of the limit sets of Section 10.1

form a closedcontinuous curve with no self-intersection points in the

Riemann sphere \320\241We also noted that the curves in Section 10.1 do

not appear very differentiable. We now prove that this is indeed thecase,at least under the hypotheses of Proposition 10.14.

A subset \320\233of \320\241has a tangent line L at the point \316\276G \316\233if

for every \316\267> 0 there exists an \316\265> 0 such that for every \316\276'G \316\233\316\240

-Bdeuc(\302\243i\302\243)>the line \316\276\316\276'makes an angle < \316\267with L.

Proposition 10.19. Under the hypotheses of Proposition 10.14, sothat the homeomorphism X: R \342\200\224>\320\233\320\263is well defined, suppose in

addition that the parameter sj is not real. Then, for every irrational

point \317\207G R \342\200\224Q, the limit set \320\233\320\263admits no tangent line at X(x).

Remember that \"most\" real numbers are irrational, in a sensewhich can be made precise in many mathematical ways; for instance,

there are uncountably many irrational numbers but only countablymany rationals. See Exercise 10.12 for a proof that \320\233\320\263does admit

a tangent line at each point X(x) with \317\207\342\202\254Q rational. When si is

real, it is immediate from definitions that the limit set \320\233\320\263is equal to

R, arid consequently it is everywhere tangent to the real line R (in a

very strong sense!)

Proof. As in the proof of Proposition 10.14, let K\\, K^, ..., Kn,

... be the circlesof \320\241that surround X(x), with each Kn surroundingthe next circleKn+\\. Because \317\207is irrational, X(x) belongs to nocircleKn.

We will first consider the simpler case where, for infinitely many

\316\240\316\271,\316\2672,..., \316\267*,,..., the circle \320\232\320\237\320\272+\320\263is disjoint from \320\232\320\237\320\272.By

definitionof G, the two circles \320\232\320\237\320\272and K\342\200\236k+ \\ bound two faces of some tile

jk(X)\302\267In other words, there exists 7fc G \320\223such that 7^\"1(^nfc) and


7^\" (\320\232\320\237\320\272)are two of the original circles C\\, C2, C3, C\\. By throwing

away some of the n^ (but keeping infinitely many of them). we canassume that we always get the same two circles, namely, that there

exists i, j \342\202\254{1,2,3,4} such that ^1{\320\232\320\237\320\272)=

\320\241,-and^1{\320\232\320\237\320\272)

=Cj

for every \320\272G N. Note that C{ and Cj are disjoint, and consequently

bound two disjoint disks Di and Dj in the Riemann sphere \320\241(One

of the two circles is actually a line, in which case the diskit bounds

is a euclidean half-plane).

Suppose that the limit set \320\233\320\263has a tangent line L at \\{x). Then,for every \316\267> 0, there exists an \316\265> 0 such that \320\233\320\263\320\237-Bdeuc(A(:r),\302\243)

is located between the lines Lv and L.v that make an angle of \302\261\316\267

with L at X(x), or more preciselyit is contained in the two disksectors of angle 2\316\267and radius \316\265delimited by these two lines L\302\261n.By

Lemma 10.15, the circles \320\232\320\237\320\272and \320\232\320\237\320\272+\\are contained in the ball

.\320\222^\320\265\321\206\321\201(\320\220(:\320\263),\302\243)for \320\272large enough. We will use the corresponding 7*,to \"zoom\" over the region near x. Note that jk{Dj) and 7(C

\342\200\224Di)

are contained in Bdeac(X(x),\316\265).

Consider the two circles J+ = 7^1(L7))and J~ = 7^\"1(L_7)) in

C. These two circles have the following properties:

(1) They crosseachother in one point 7^ (\316\273(a;))& Dj and one

point 7^\"1(oo)e D{,and make an angle of 2\316\267with each

other at these points. (Remember from Proposition 2.18

and Corollary 2.17 that linear fractional maps send circles

to circles and respectangles).(2) The part of the limit set \320\233\320\263that is outside of D{ is located

between J+ and J~. More precisely, \320\233\320\263\342\200\224

Di is contained

in the two \"moons\" that are delimited by the intersectingcirclesJ^ and whose angles at their vertices are equal to

2\316\267.

We now vary \316\267.Take a sequence (jjm)meN of positive numbersconverging to 0. For each of these \321\206\321\202,the corresponding circle

J%m

is completely determined by the four distinct points in which it

intersects the two disjoint circles d and Cj. Replacing (r7m)meN by a

subsequence if necessary, wecan assumeby compactness of the circles

Ci and Cj that these intersection points converge to four points in


Figure 10.13

Ci U Cj. As a consequence, the circle J,+ converges to a circle Jq\"

as m tends to oo; this limit circle Jq may be tangent to the circleCi or Cj, which will happen exactly when the two intersection points

ofJ7~n

with tin's circle converge to a single point. Further passing to

a subsequence if necessary, we can similarly assume that in addition,the circle

3\320\237\321\202converges to a circle J0~~.

The circles Jq and Jq have one point in common in Di andanother one in Dj. Indeed, if Jq\" \316\240Jq \316\240-Djor JqC\\JqC\\Dj was empty,the same property would hold for J+ \316\240,/,\" \316\240\320\224or J^ \320\237J~ \316\240Dj by

continuity, contradicting the fact that JJ\" and J^ meet at 7^\"1(oo)GDi and ijT^-Mz)) G Dj. Similarly, because intersection angles vary

continuously, the circles Jq make an angle of hmm_\302\273oo r\\m = 0 at

these two intersection points. Thesetwo properties imply that the

circles actually coincide,namely, that J0~= Jq. (Here it is crucial

that the circles C, and Cj be disjoint, since otherwise Jq and J0~could be two distinct circles passing through the point d \316\240Cj and

tangent to each other.)In particular, the two moons delimited by the circles J* and

containing \320\233\320\263\342\200\224

Di limit to the circle Jq , so that the part of the limit

set that is outsideDi is completely contained in the circle Jq .

10.5. More examplesof kleinian groups 279

However, we saw that the limit set \320\233\320\263contains all the vertices of(he crookedFarey tessellation T, and so this is clearly impossible if s\\

(and S3 = \342\200\224sj

\342\200\224S5) is not real. For instance, the circleJq must then

contain the vertices 0, 1,00 and e \320\2335of T, and consequently must beR. However, the vertices \317\210{(0). ^,(1) and </?i(\302\260o)are all outside of

I),. and cannot be all real by inspection of the formulas for the gluing

maps \317\206\316\271,\317\2062,<\302\243\320\267>\317\210\316\257\302\267

This concludes the proof of Proposition 10.19under the

additional assumption that the circles Kr, and A',1+i G Gx are disjoint for

infinitely many n.

The proof is very similar in the remaining case where the circles

A'\342\200\236and A'\342\200\236+imeet for every \316\267large enough. Here it is convenient\316\231\316\277look a I. the corresponding data for the standard Farey tessellation.Therethe circlesAT\302\260and

A'\302\260+1surround the point i\302\243l and touch

each other at a rational point. This intersection point A'\302\260\316\240\316\221\316\244\302\260+1

cannot be equal to \317\207since \317\204is irrational, so it will be either to the

light or to the left of x. The intersectionpoint AT\302\260\316\223\316\232\302\260^cannot

be systematically to the left or systematically to the right for \316\267

sufficiently large; indeed the point K\302\260\316\223K\302\260+1

would then be independentof n, and Kn would eventually stop surrounding \317\207since its diameter

converges to 0. Therefore, the side changes for infinitely many values

of n. Namely, we can find a sequence (rifc)fceN sucn *batK\302\260k

\320\237AT\302\260fcfl

is to the left of \317\207and K^k+1 \320\237K^k ,.2 is to its right, or conversely. A

consequence of this is that \320\232\320\237\320\272is now disjoint from \320\232\320\237\320\272+2for every

AgN.

Now, the circles \320\232\320\237\320\272and \320\232\320\237\320\272,2 bound faces of two adjacenttiles of the tessellation of H3, which arc of the form jk(X) and \320\273/,\320\276

\317\206\316\271,(X) with 7it G \316\223and \316\266*.G {1,2,3,4}. This again means that

t here are only finitely many possibilitiesfor the circles7^1(AT\342\200\236lj)

and

\321\203\320\2721(\320\232\320\237\320\272+2).Therefore, replacing the subsequence (ATnA.)fceN by a

subsiibsequence if necessary, we can assumewithout loss of generality

(hat there exists fixed \320\263,j and I G {1,2,3,4} such that7jT1(A'\342\200\236/,)

\342\200\224

(\342\226\240iand7\321\2141(\320\232\320\237\320\272+2)

=4>i{Cj) for every \320\272G N. The argument is

then identical to that used in the first case, replacing Cj by <pi(Cj)

everywhere. D


10.5.4. The parameter space. Our shearing and bending of theFarey tessellation depends on three complex parameters sj, S3, S5 G \320\241

such that si + S3 + s5 = 0. Actually, by inspection of the formulas atthe beginning of this chapter, these depend only on the exponentialseSi. (Remember that by definition of the complex exponential,e* =ez if and only if \316\266\342\200\224\316\2661is an integer multiple of 2\316\271\317\204\\.)

We saw that for some values of these parameters, the tiling group\316\223of the corresponding crooked Farey tessellation acts freely and dis-

continuously on the hyperbolic space \320\2303. We might be interestedin plotting all the values of the parameters for which this happens.

However, even if we take into account the fact that s\302\247= -.si \342\200\224

S3,

this leaves us with two free complex parameters eAl and \320\265\320\257\320\267,so that

a crooked Farey tessellation is determinedby the point (cSl,e*3) G

C2 = R4. Most of us are not very comfortable with 4-dimensional

pictures. We can try to decrease the dimension by imposing anadditional condition. The simplest such condition is that s5 = 0, sothat S3 =

\342\200\224si.The corresponding set of crooked Farey tessellationis known as the Earle slice in the set of all crooked Farey

tessellations. An element of the Earle slice is completely determined by the

complex number \317\205= eSl G C, since e\"3 = u_1 and e\"5 = 1.

Figure 10.14 represents the space of values of \317\205G \320\241for which

the tiling group \320\223of the crooked Farey tessellation associated to \317\205

is a kleinian group acting freely on H3. More precisely, it uses a

change a variables and, in the complex plane C, the lighter shaded

and cauliflower-shaped area of Figure 10.14is the set of values of

_ 2v~ 1 _ 2eSl- 1U~

\320\252\320\223+\316\212

~

2e\"i + 1

for which the tiling group \316\223acts discontinuously and freely on H3.Note that \316\275= eSl is easily recovered from u. This picture was drawn

using the software OPTi.The large circle in Figure 10.14, which is the circle of radius 1

centeredat the origin, encloses the set D of values of \320\270for which the

hypotheses of Proposition 10.13are satisfied. See Exercise 10.13.

The horizontal line segment corresponds to real values of u, forwhich the tiling group \320\223is fuchsian. These fuchsian groups are exactlythe tiling groups associated to the complete hyperbolic tori that we

10.5. More examples of \320\272lei ni an groups 2H1

Figure 10.14. The Earlo slice

consideredin Section G.7.2 (corresponding to the case a = b G (1, +oc)in the notation of that section).

We have also indicated the values u^ =^ corresponding to the

standard Fare.y tessellation, and it2 = \342\200\2241.205-1- 0.714i associated to

the example of Figure 10.8.Chapter 11 will be devoted to the tiling

group corresponding to the point \321\211\342\200\224\\ x

-7=1,located exactly on

the boundary of (he lighter shaded area. The point u2 is near the

boundary but not quite on it.

In the general case,let \316\251\320\241\320\2412be the set of values of theparameters (e,Sl .e''3) for which the tiling group of the associated crooked

Farey tessellationacts discontinuously and freely on the hyperbolicspace \320\2303. It can be shown that the properties of the limit set \320\233\320\263

that we proved under the hypotheses of Proposition10.13hold for all

parameters (ea,-,e*3) located in the interior of \316\251.More precisely, for

these parameters, the limit set \320\233\320\263is homeomorphic to the circle, andhas no well-defined tangent line at most points unlessthe parameters

all real.

For points (eSl,cS3) on the boundary of the parameter set \316\251,

the tiling group \320\223of the corresponding crooked Farcy tessellation isstill a kleinian group acting freely and discontinuously on HI3, but

its geometry is much more complex. \320\22311particular, the limit set is


not homcomorphic to the circle any more. We will investigate such a

group in Chapter 11.

A result of Yair Minsky [Minskyx] implies that the space \316\251is

path-connected, in the sense that any two points can bejoinedtoeachother by a continuous curve contained in \316\251,However. Figure 10.14

suggests that its geometry is quite intricate near its boundary. It isactually much more intricate than what is visible on that picture.Ken Brombergrecently showed in [Bromberg] that there are pointswhere \316\251is not even locally connected. \316\223\316\267particular, at such a point\316\241G \316\251,we can find a radius \316\265> 0 and points Q, R \342\202\254\316\251which are

arbitrarily close to \316\241and such that any continuous curve from Q to

R in \316\251must leave the euclklean ball \316\222,\316\271^{\316\241.\316\265)\320\241\320\2412= R4.

Figure 10.15

This is illustrated in Figure 10.15, which represents a different 2-dimensionalcross-sectionof \316\251.Each of the three squares is obtainedby zooming in on a piece of the squareimmediately to the left. The

little \"islands\" that converge toward the center \316\241of the right-hand

square can be connectedto each other by a continuous curve in \316\251

(leaving the cross-section represented and using the additional two

dimensions), but not by a curve which remains within \316\265of P. The

pictures of Figure 10.15were created by David Dumas, using his

software Bear [Dumas].What we observed for crooked Farey tessellationsis part of a more

general phenomenon. Any fuchsian group \316\223\302\260can be deformed to a

family of kleinian groups depending on complexparametersbelonging

to a certain domain \316\251in C\342\204\242,where the dimension \316\267depends on the

topology of the quotient space \316\2272/\316\223\302\260.For any such kleinian group \320\223

10.6. Poincare, Puchs and Klein 283

correspondingto an interior point of \316\251,the limit set \320\233\320\263is homeomor-

phic to the circle,and admits no tangent line at most points unless thegroup \320\223is essentially fuchsian and the limit set is a circle in \320\241If the

quotient space \320\2352/\320\223\302\260is compact, for instance for the tiling group \320\223\302\260

of the tessellation of H2 by hyperbolic octagons that we constructed

in Section 6.5.2, the limit set of the deformed kleinian group \320\223even

admits no tangent line at any point, at all. The kleinian groupscorresponding to points on the boundary of \316\251are much more complex: it is

only very recently that a reasonable understanding of their geometry

has been reached [Minsky2, Brock & Canary &Minsky].

10.6. Poincare, Fuchs and Klein

The terminology of kleinian and fuchsian groups is due to Henri

Poincare (1854-1912), one of the mathematical giants of his time.

Poincare was led to hyperbolicgeometry through an unexpected route.

He was studying the linear differential equation

(10.1) x\"(t)+p(t)x(t) = 0,

where the the function p(t) is given, where the function x(t) is the

unknown, and where, more importantly, all quantities x, p, t are

complex-valued. Compared to the setup usually taught in calculus

or in a differential equations course, the fact that t is complex maybe somewhat surprising,but the basic definitions and properties ofdifferential equations immediately extend to this complex context.

When investigating a specific family of examples, wherep(t) was

a rational function with real coefficients,Poincarefound that the

space of local solutions to (10.1)gave rise to a certain group \316\223of

linear fractional maps with real coefficients. Using (and proving) thePoincare Polygon Theorem 6.25, he then showed that \316\223acts discon-

tinuously on the upper half-space \320\2302.In addition, the construction

provided a preferred homeomorphism f:X\342\200\224+\316\2272/\316\223from the domain

X of the given function p(t) to the quotient space \320\2352/\320\223.The groups

\320\223considered by Poincare in these examples are very closely related

to the tiling groups associated to complete punctured tori that we

investigated in Sections 5.5, 6.6 and 6.7.2.


In a subsequent work [Poincare5], Poincare started perturbing

equation 10.1 by allowing the coefficients of the rational fraction p(t)to be complex,although close to real numbers. This process is very

similar to the deformations of the standard Farey tessellation to the

crooked Farey tessellations that we investigated in Sections 10.1 and10.5. In particular, it yields a group \316\223of linear fractional maps with

complex coefficients, now acting on the Riemann sphereC. Poincare

proved that these linear fractional maps extendto isometries of the

hyperbolic space \320\2303and, using (and proving) the PoincarePolyhedron Theorem 10.9, showed that the action of \316\223on H3 is

discontinuous.He then used this technique to producea natural homcomor-

phism f-.X\342\200\224+\316\251/\316\223where, again, X is the domain of p(t) and \316\251is a

subset of the Riemann sphere\320\241invariant under the action of \320\223.This

subset \316\251is not the upper half-plane \320\2302any more. Rather, the limit

set \316\233\317\201is homeomorphic to the circle, and \316\251is one of two pieces ofthe complement \320\241\342\200\224

\316\233\317\201.See Exercise 10.4 for a proof that \316\223acts

discontinuously on \320\241-\320\233\320\263,so that the quotient space \316\251/\316\223makes

sense.

In the real coefficientcase, the homeomorphism /: X \342\200\224\342\231\246\316\231\316\2572/\316\223

and its derivatives are essentially equivalent to the data of certain

functions which are well behaved with respect to the action of \316\223.In

the papers [Poincarei, \320\240\320\276\321\202\321\201\320\260\320\263\321\221\320\267,Poincare4], Poincare decided to

call these functions fuchsian functions in honor of Lazarus Fuchs,whosearticles[Fuchsi,Fuchs2]had inspired him. He also gave thename of fuchsian groups to the groups of isometries of H2 occurringin this way. Felix Klein then complained that he had alreadyconsidereda notion equivalent to these fuchsian functions and groups acting

discontinuously on IE2, and consequently deservedmorecredit than

Fuchs. Poincare did not change his terminology but, when moving

to the case with complex coefficients (which Klein had neverconsideredat that time, although he did later on),Poincaregave the name

of kleinian groups to the discontinuousgroupsof isometries of H3

occurring in this context. See the note added by Klein, acting aseditor of the journal, to Poincare's announcement [Poincarei] andPoincare'sresponsein [Poincare2].Theupshot of the story is that

fuchsian groups have almost no connection to Fuchs, and that Klein

had little to do with kleinian groups.

10.6. Poincare, Puchs and Klein 285

There is another story about this. The late nineteenth century

was a period of turmoil in mathematics, as the endeavor to setmathematics on a solid foundation would sometimes lead to ratherunintuitive results. This included the emergence of pathological examples,such as functions which are continuous but nowhere differentiable.

Poincare had mixed feelings about this trend, and clearly thought

that \"real life\" mathematical objects should be well behaved. The

following quote from [\320\240\320\276\321\202\321\201\320\260\320\263\321\221\320\261](translated by the author)summarizeshis thinking rather well.

Logic sometimesgeneratesmonsters.The past half

century has seen the emergence of a multitude of

bizarre functions which try to resemble as littleas possiblethe honest functions which have some

applications. No more continuity, or else

continuitybut no derivatives, etc... Even more, from a

logical point of view, it is these strange functions

which are the most general, and those that one

encounters without looking for them only appear as aspecialcase In the old days, when one inventeda new function, it was with the goal of some

practical application; today, they are invented for thesinglepurposeof exhibiting flaws in the reasoningsof our fathers, and this is the only thing that one

will ever extract out of them.

Little did he realize that these \"monsters\" already appeared in a

natural way in his work. For instance, the article [Poincare5]contains

many examples which are similar to the ones that we considered in

this chapter. Poincare was aware that in these examples,the partial

tiling of the Riemann sphere that he was constructing would

accumulate on a continuous curve \320\233\321\200,and thought that this curve wasunlikely to be very differentiable. However, it seems that the worst

he envisioned was that this curve might not have second derivatives;

see [Poincarei, page 559]. It is only the limited computing power

available to him which prevented Poincare from observing that this

closed curve was nowhere differentiable, namely, it would produce one


of these pathological curves that he thought would never occur in a

natural way.


Exercise 10.1. Let \317\206be the isometry of H3 who.se extension to the Rie-mann sphere \320\241is the homothety \320\263\316\271\342\200\224\342\226\2722\320\263.and let \320\223=

{\317\206\316\267\\\316\267G \316\226}be the

transformation group of H3 generated by \317\206.

a. Show that \316\223is a kleinian group.b. Determinethe limit set of \316\223.

c. Show that there exists a nonempty closed subset \320\232of \320\241which is

invariant under \320\223but which does not contain the limit set \320\233\320\263.(Compare

Proposition 10.3.)

Exercise 10.2. Show that the limit set of a kleinian group \320\223is nonemptyif and only if \320\223is infinite.

Exercise 10.3. Consider a kleinian group \320\223which fixes the point oo \342\202\254\320\241

a. Show that every element of \320\223is the isometric extension of a linear or

antilinear fractional map of the form ipa,b(z) = az + b or ipa,b(z) =

az + b, with \316\261\316\2460.

b. Show that if \316\223contains an clement 70 of the form \317\206\316\261\316\233or va.b with

\\a\\ \316\2461, then it also fixes a point of \320\241in addition to 00. Hint: For

7 G \320\223,consider the elements 7_1 \320\27670 \"07070 \320\265\320\223with \\n\\ large, and

remember that \320\223acts discontinuously on \316\210.3.

\321\201Conclude that either \320\223consists entirely of euclidean isometries of

(K3,deuc)respectingH3 or it respects a unique complete geodesic gofH3.

Exercise 10.4 (Discontinuity domain). If a group \320\223acts on a metric space(X, d), not necessarily by isometries, the action is discontinuous atPeXif there exists a ball \316\222\316\261(\316\241,\316\265)which meeth its images y(Bd(P,s)) for only

finitely many 7 G \316\223.When the action is by isometries, one readily checksthat this definition is equivalent to the one given in Section 7.2.

We want to show that the action of a kleinian group \316\223on \320\241is

discontinuous at every point \316\266of the complement \320\241\342\200\224\320\233\320\263of the limit set \320\233\320\263,for

the euclidean metric deuc. Since the limit set is closed, there exists a small

euclidean ball Ddeuc(z,2s) which is disjoint from \320\233\320\263.Suppose, in searchof a contradiction, that there exists an infinite sequence (7n)neN such that

\"fn(Bdi.nv(z,\316\265)) meets Bd^u(z,e) for every \316\267G N. By Proposition 9.10,each 7\342\200\236(\320\222\320\271\320\265\320\270\321\201(\320\263,\320\265))is delimited in K3 by a euclidean sphere centered on

C; let rn be the euclidean radius of this sphere. Pick a base point Pq G H3

on the euclidean spheredelimiting Bde\342\200\236c(z,\316\265).


a. Suppose,in addition, that Km\342\200\236.,sr\342\200\236= 0. Show that there exists asubsequence (7,,,. )fc v such that {lnk(Po))kes convergesto some point

zx & \320\233\320\263\302\267.Show that necessarily dt.uc(z,zx) \316\266\316\265,which contradicts the

choice of \316\265.

h. Al the other extreme, supposethat the r\342\200\236are bounded from below.namely, that there exists r(1 > 0 such that rn ^ \316\271\316\267for even- \316\267\302\243N.

Show that there exists a constant \320\241> 0 such that for every r> \302\243N,

there exists a point P\342\200\236\302\243-\302\267\342\226\240\342\200\236(\320\224/.,\320\2461.(\320\263.=)) with dhYi,(Pa, Pn) s? C. Show

thai there exists a subsequence (~\302\253\320\264)Ar-for which

(\"\316\271\316\2317,1(^\316\257\316\271))\316\221.\316\271=.

converges to some point ;v \316\266\316\233\317\201.I'se this property \316\271\316\277show thai

necessarily(1,,,\317\210,:\317\207) \316\265,which contradicts the choice \316\277\316\257\316\265.Hint: Comparethe euclidean and hyperbolic distances from 7,7L(Po) to 7\342\200\236l{Pn) \302\243

BH .(\320\263.\316\265).

c. Combine parts a and b to reach a contradiction in all eases, Concludethat the action of \320\223at every z \302\243\320\241\342\200\224

\320\233\320\263is discontinuous.

d. Show that for every z \302\243\320\2337\320\237\320\241,the action of \320\223is not discontinuousat z.

A similar argument shows that for the obvious extension of definitions,

the action of \320\223is discontinuous at the point 00 if and only if 00 is not in

the limit set \320\233\320\263\302\267.For this reason, the complement \320\241\342\200\224\320\233\320\263is called the

discontinuity domain of the kleinian group \320\223.

Exercise 10.5. Let p: H3 \342\200\224\342\226\272H2 be the orthogonal projection constructedin Section 10.3.

a. Let \320\271be a vector based at \316\241\302\243E\\ Show that its image under the

differential of \317\201is such that ||Dpp(?)||euc \316\266||v||eUc- and ||\302\243>pp(v)||hyP ^

||v||hyP coshd, where d is the hyperbolic distance from \316\241to p(P).

b. Let 7 be a piecewise differentiable curve in H3 which stays at distanceat least D > 0 from \320\250\320\2232.Show that 4yP(p(7)) ^ thyP(7)coshD.

In particular, when projecting from far away, the hyperbolic orthogonalprojection p: H3 \342\200\224\342\226\272H2 decreases lengths much more that the euclidean

orthogonal projection X3 \342\200\224\342\226\272K2.

Exercise 10.6. Let \317\201:\320\2303-> \320\2302and q : \320\2503-\302\273IK be the orthogonalprojection and the signed distance function introduced for Lemma 10.5.

Endow the product \316\2272\317\207\316\232with the product dhyp x dcuc of the hyperbolicmetric dhyp of \316\2112and of the euclidean metric deuc of \320\226,as defined in

Exercise 1.6.

a. Show that the product function \317\201\317\207q: \320\2303\342\200\224\302\273\316\2272\317\207\316\232,defined by

\317\201\317\207q(P)= (p(P)- q(P)) for every \316\241\302\243\316\2273,is a homeomorphisrn.


b. Let \316\223be a discontinuous group of isometries of \320\2302,extended to a fuch-sian group which we also denote by \316\223.Endow the quotient spaces \320\2302/\320\223

and \320\2353/\320\223with the quotient metrics dhyp defined by the hyperbolicmetrics dhyp of H2 and H3 (and denoted by the same symbols). Finally,

endow the product \316\2272/\316\223\317\207\317\213with the product metric dhyP x deuc- Showthat there is a homeomorphism

px q: \320\2353/\320\223\342\200\224\320\2302/\320\223\321\205R

defined by the property that \317\201\317\207q(P) = (p(P), q(P)) for every \316\241G H3.

Exercise 10.7 (Twisted fuchsian groups). Let \316\223be a fuchsian group,acting on H3, and supposethat we are given a map \317\201:\316\223\342\200\224\342\231\246

{\342\200\2241,+1} such

that /0(7 \320\2767')= p(l)p(l') (namely, if you know what this is, \317\201is a group

homomorphism from \316\223to the group Z2 ={\342\200\2241,+1}, where the group law

is defined by multiplication). For every 7 G \316\223,we define a new isometry \316\267\317\201

of H3 as follows: If /0(7)= 1, then \316\267\317\201= 7; if \317\201{\316\267)

=\342\200\2241,then 7P = \317\204\316\2777,

where r: H3 \342\200\224\342\226\272H3 is the euclidean reflection across the vertical half-planeH2 \320\241\316\2273.Then consider the set

\320\263\321\200={/\320\273\320\265\320\263}

of all 7P obtained in this way.

a. Show that \320\223\321\200is a group of isometries of (H3,dhyp).

b. Show that the action of \320\223\321\200on H3 is discontinuous. Hint: Compare

Lemma 10.6.

A kleinian group \320\223\321\200obtained in this way is called a twisted fuchsiangroup.

\321\201Show that the limit set of a twisted fuchsian group is contained in

\320\232\321\201\320\241.

d. Let \320\223'be a kleinian group whose limit set is contained in a euclideancircleand has at least three points. Show that there exists a hyperbolicisometry \317\210and a twisted fuchsian group \316\223\317\201as above such that

\316\244'={\317\206-\317\207\316\277\316\212\317\201\316\277\317\206\302\267\316\267\317\201\302\243TP}.

Hint: Choose \317\206so that it sends the circle \320\241to \320\232\321\201\320\241

Exercise 10.8. Let X be the hyperbolic half-spaceX =

{(x,y,u) \302\243\320\2503;\321\203> 0,ii > 0}.

Consider X as a hyperbolic polyhedron by decomposing its boundary into

one vertex \316\241= (0,0,1), two edges

El = {(x,y, u) G \316\2323;\317\207= 0, \321\203= 0, \320\270> 1}

and E2 = {(x,y,u) \320\265\320\2323;\320\266=0,y = 0,0 < it < 1},


and two faces

F, = {(x, y, u) \302\243\316\2323;\317\207> \320\236,\321\203= 0, \320\270> 0}

and F2 = {(x, y, u) \342\202\254\316\2323;\317\207< 0, \321\203= 0, \320\270> 0}.

Glue the faces Fl and F2 together by the restriction \317\206\316\271:Fi \342\200\224*F2 of the

hyperbolic isometry \317\206\316\271:\316\2273\342\200\224\342\226\272\316\2273extending the antilinear fractional map\316\266

a. Show that hypotheses (1-3) of Theorem10.11are satisfied, but that

hypothesis (4) does not hold.

b. For any \320\263> 0, endow the sphereS =

5*yp(A>.r)= {Pe X;dhyp(P,P0) = r}

with the path metric ds, where ds(P,Q) is defined as the infimum

of the hyperbolic lengths of all curves joining \316\241to Q in S. Letainh r ds be the metric obtainedby multiplying ds by the factor

Bi^h r

(compare Exercise 9.4, and the proof of Lemma 10.10). Show that

(Sjhyp(Po,r), ^^ds) is isometric to the projectiveplane (KP2,dsph)

of Section 5.3.

c. Show that the quotient space (\316\247,\316\254\317\207)is not locally isometric to

(H3,dhyp). It may be convenient to use the results of Exercise 5.11.

Exercise10.9.Let \316\244be a crooked Farey tessellation associated toparameterssi, S3 and S5with si+s3\342\200\224.S5 = 0, and let \316\223be the corresponding tilinggroup. Consider the hyperbolic isometry defined by the linear fractional

map p(z) = \342\200\224e~S5^.

a. Show that \317\201exchanges the two ideal triangles \316\223\"1and T~.

b. Show that \317\201\316\277\317\206\316\273=:

\317\206~*\316\277\317\201,\317\201\316\277\317\206~\316\271=:

\317\206\316\271\316\277\317\201,\317\201\316\277\317\206\316\271=

\317\206%l

\302\260p and

\316\241\316\277\316\250\316\220'=

\317\2102\316\277\316\241-

\321\201Conclude that \317\201respects the crooked Farey tessellation \316\244and that

when the tiling group \316\223acts discontinuously on \316\210?,\317\201also respects thelimit set \320\233\320\263-

d. Let r =\317\201\316\277\317\2063\316\277\317\206\\.Show that r is a horizontal translation, whose

translation vector is equal to half the translation vector of the translation

\317\206\316\271\302\260\317\2102\320\276'-\320\240\320\267\302\260\317\206\316\271\302\267

e. Show that \317\204respects the crooked Farey tessellation \316\244and, when the

tiling group \316\223acts discontinuously, that r respects the limit set \320\233\320\263-

Exercise 10.10 (Shear-bend parameters). We want to give a geometricinterpretation of the parameters s\302\273\302\243\320\241defining a crooked Farey tessellation7. Do not let the long definitions intimidate you.

Let \316\212\\and \320\2232be two adjacent triangles of T, meeting along an edgeE. Orient \316\225to the left as seen from \316\244\316\212.Namely, first orient, in the Farey


tessellation T\302\260,the edge E\302\260corresponding to \320\232to the left as seen from

the face T\\ corresponding to T\\, and then transport this orientation to an

orientation of E. (We need to do this because in dimension 3 the notionof right and left depends on which way we arc standing on the triangle

T\\.) Let Pi and Pi be the base points determined on \316\225by the standard

horocircles of T\\ and Ti, respectively, as defined in Lemma 8.4. The shear

parameter betweenthe triangles T\\ and Ti of \316\244is the signed distancefrom P\\ to Pi in this oriented edgeE. Compare the shear parameters ofSection 8.4.

Let \316\270G [0,2\317\200)be the angle by which one needs to rotate Ti along \316\225in

order to bring it in the hyperbolic plane containing \316\244\316\257,but on the side of \316\225

opposite \316\244\316\212;here we measure the angles counterclockwise,as one looks in

the direction of the orientation of E. This angle \316\270is the external dihedralangle \316\270between \316\212\\and \316\2232.

Suppose that \320\232is the image of one of the standard edges \316\225\317\207,with

\320\263G {1,3, 5}, under an element 7 of the tiling group \320\223.Show that the shear

parameter t is equal to Im(s,), and that the external dihedral angle \316\270is

equal to Re(si) up to an integer multiple of 2\317\200.Hint: First consider thecase where 7 is the identity map.

Exercise 10.11 (Schottky groups). Let B\\, B2, B3 and B4 be four eu-

clidean balls in K3, centered on the xy-pla&e \320\241,and far apart enough fromeach other that the corresponding closed balls are disjoint. Let X be the

complement H3 \342\200\224Bi U \320\2572U B3 U Ih- In particular, X is a hyperbolicpolyhedron in E3 delimited by four disjoint hyperbolic planes \316\240\316\271,\316\240.2,\320\237\320\267,

\320\223\320\246,with no vertex at infinity. Choose isometries \317\206\316\271,\317\206\316\271,\317\206$and \317\206^of

(\320\2303,cihyp) such that the following holds: \317\206\316\271sends \316\240\316\271to \316\2402,and sends Xto the side of \316\2402that is opposite \316\247;\317\206\316\271

=\317\206~\316\231\317\207\\\317\2063sends \320\237\320\267to \320\2374,and

sends X to the side of \320\2374that is opposite X; and 1^4=

\316\250\316\2201\342\226\240Let \316\223be the

group generated by \317\206\316\271,\317\206\302\2672,\317\206\316\271,and \317\206\\.

a. Show that \316\223acts discontinuously on H3. Hint: Use Poincare's

Polyhedron Theorem 10.9.

b. Show that every element 7 \302\243\316\223can be written in a unique way as

7= \317\206\317\212\316\273\316\277\317\206\316\2572\316\277\302\267-\302\267\316\277\317\2061\316\267,

where, for each j, ij is an element of {1,2,3,4} and the set {ij,ij+i}is different from {1, 2} and {3,4} (which is a fancy way of saying thatthere is no obvious simplification); by convention, n= 0 when 7 is theidentity map. Hint: To prove the uniqueness, look at the tiles \"/'(X)that are crossed by an arbitrary geodesic going from a point of X to a

point of \"f(X).


c. Let (in)nifi be a sequence valued in {1,2,3,4}, such that the set{in,in+i} is different from {1,2} and {3,4} for every \316\267G N. Set

Show that the sequence (^/n(Po))nen converges in (M3,deuc) to some

point \316\266of the limit set \320\233\320\263.Hint. Adapt the proof of Lemma 10.16.d. Show that for every \316\266G \320\233\320\263,there exists a unique sequence (\320\263\320\277)\320\237\302\243\321\206

such that \316\266= \316\231\316\257\316\267\316\220\316\267-,\316\277\316\2777\316\267(\302\267\316\241\316\277),with the setup of part \321\201In particular,this proves that there is a one-to-one correspondence betweenpoints

of the limit set \320\233\320\263and sequences (in)neN as in part \321\201

e. Choose a specific example and draw as many of the (euclidean circlesdelimiting the) hyperbolic planes \316\271^(\316\240\316\257),w'th \317\206G \316\223and \320\263G {1,2, 3,4}as you can. Similarly sketch the limit set \320\233\320\263-

Exercise 10.12. Let \320\223be one of the kleinian groups considered in

Section 10.5, associated to a crooked Farey tessellation \316\244whose parameters

si, S3, S5 satisfy the hypotheses of Proposition 10.13. Let \316\273:\316\232\342\200\224\342\226\272\320\233\320\263be

the homeomorphism constructed in Proposition 10.14.

a. Show that there is a strip in C, delimited by two parallel lines, which

contains all the limit set \320\233\320\263(minus oo, of course).

b. Show that for every rational point ? G Q, there exists two disjoint open

disks Dl and D2 in C, delimited by two circles tangent to each other

at\316\273(|),

such that \320\233\320\263is contained in \320\241- Di U D2. Hint: Consider an

element 7 G \320\223such that 7(00) = ?, and use part a.

\321\201Conclude that the limit set \316\233\317\201admits a tangent line at each \316\273(2)with

f eQ.

Exercise 10.13. Let \316\244be a crooked Farey tessellation associated toparameters si, S3 and ss such that s6 = 0 and si + s3 = 0. Let \316\223be the

corresponding tiling group, which is contained in the Earle slice of

Section 10.5.4. Show that these parameters satisfy the conditions of

Proposition10.13, in the sense that cos(Tm(s!)) > 0, if and only if fifrr < -1\302\267

(Compare with the circle represented in Figure 10.14.)

Exercise 10.14 (Domino diagrams revisited). This is a continuation of

Bxercise 8.7. Let \316\244be the crooked Farey tessellation determined by the

parameters \316\262\316\271,s2, S3 G \320\241with si + 82 + s3= 0. Let \320\223be an ideal triangle(if T, associated to the Farey triangle T\302\260in the standard Farey tessellationJ\302\260.We want to give an explicit formula for the vertices of T. We will

restrict our attention to the case where T\302\260is to the right of 0oo, namely,

where all its vertices are nonnegative.


Let S1S2\302\267\302\267\302\267Sn, with each Sk equal to L of R, be the symbol sequence

describing the Farey triangle T\302\260as in Exercise 8.5, and consider theassociated domino diagram. By construction, each domino is associated to oneof the edges \316\225of the Farey tessellation T\302\260that are traversed as one travelsfrom the triangle Oloo to the triangle T\302\260,as in Exercise 8.5. The edge\316\225is

of the form 'y(Ei), where 7 is an element of the tiling group \320\223\320\276,and where

Ei is one of the standardedges\316\225\316\212=(\342\200\224l)oo,E3 = loo and \320\225\321\212= Ooo. For

such a domino associated to an edge ~t{Ei),label its upper right bullet byeSi/2 and its lower right bullet by e~s*/2. In particular, the two leftmost

bullets of the domino diagram receiveno labels.

For instance, the domino diagram (8.1) of Exercise 8.7 is labelled asfollows.

e\302\2733/2 e\302\273i/2 es3/2 efi/2 e.-3/2 gi.j/21 \302\267\342\200\224+\302\267\342\200\224\302\273\302\267\342\200\224>\302\267\342\200\224>\302\267\342\200\224>\302\267\342\200\224\302\273\302\2671

\\ \\ / \\ / \\2\302\267\342\200\224+\302\267\342\200\224+\302\267\342\200\224+\302\267\342\200\224+\302\267\342\200\224+\302\267\342\200\224+\302\2672

e-\302\2733/2 e-\316\212/2 e-33/2 e-al/2 \320\265-\"\320\267/2\320\265~\320\237/2

For each path \317\201in the domino diagram, let

W(p) = e\302\261sH/2e\302\261s*2/2...e\302\261^/2

be the product of the labels thus associated to the bullets traversed by p.

Finally, for every i, j G {1, 2}, let

Sij =\316\243 W(P)\302\267

\317\201goes from \320\263to j

a. Show that the triangle \316\244of the crooked Farey tessellation \316\244has vertices!XL \"11+\320\2372 gj^J \302\24312.S21 '

S21+S22 ' \302\25322'

b. Devise a similar formula when \316\244is associated to a standard Fareytriangle T\302\260which is on the left of Ooo, namely, with all vertices non-

positive.

We are not giving any hint, with the idea that this problem can be turned

into a more challenging research project.

Chapter 11

The figure-eight knot

complement

This chapter is devoted to the detailed analysis of one more exampleof a crooked Farcy tessellation whose tiling group is a kleinian group.

We will sec that its features are quite different from those that we

encountered in Chapter 10. This examplealso has an unexpected

connection with the figure-eight knot that one can tie in a piece of

string:.

11.1. Another crooked Farey tessellation

Consider the crooked Farey tessellation corresponding to the shear-bend parameters si =

'^fi, S3 = \342\200\2244^i

and .95= 0. Let ns call this

crooked tessellation Tg, for reasons that we just hinted at.

In this case, the tiling group Tg is generated by

Z+\\ 2+1

2 + 1 + e 1' \316\266+ uj

and

\320\263-1 \320\263-1

-\316\266+ 1 + \321\201~~\" -\316\266+ \317\211

if we set \317\211= e^'. Note that u;3 = \342\200\2241.so that \317\2112\342\200\224\317\211+ 1 =

and 1 + \317\211~2= 1 \342\200\224\317\211= \342\200\224\317\2112= \317\211~\316\271.

\321\2103+1

293

294 11. The figure-eight knot complement

This crooked tessellation is illustrated in Figure 11.1. However,

this picture is necessarily imperfect, because one can only print a

finite number of triangles (and becausethis finite number is further

limited by the poor programming skills of the author).

Figure 11.1. An approximation of the crooked Farey

tessellation 7g

Similarly, an approximation of the limit set \320\233\320\263of \320\223,namely, of

the footprints of the crookedtessellation,isrepresentedin Figure 11.2.

Again, the picture cannot be completely accurate, since it includes

only finitely many points.

In the examples of Section10.1,the imperfections of the figureswere unimportant because they were sufficiently close to the actualgeometric object that the difference was essentially undetectable. Inthis case, however, the differences are very noticeable. Indeed, wewill see in Section 11.3 that the limit set \320\233\320\263is the whole Riemann

sphere \320\241In particular, Figure 11.2 should just be a (mathematically

correct albeit aesthetically less pleasing) solid black rectangle.

11.2. Enlarging the group \320\2238

One can observe that although mathematically incomplete, Figure11.2exhibits some interesting symmetries under translation.

11.2. Enlarging the group\320\223\320\271 295

Figure 11.2. An approximation of the limit sot of Fs

One of (.hem, which is an actual symmetry of the picture, is by

horizontal translation. You should remember that the figure has been

rotated so as to make this symmetry horizontal: in reality, it

corresponds to the clement \317\206^\316\277

\317\206~\316\277

\317\206^\316\277

\317\2063which is a translation by

1 2\317\211with, as before \317\211- \320\265\320\2671.Since this translation is an elementof the tiling group V$, it respects the crooked tessellationT\302\253in H2,

and consequently respects its \"footprints\". As in lOxercise 10.9, there

actually is a translation symmetry by half this distance, namely, by

\316\212\342\200\224\317\211.

However, we can also discern in the figure an imperfect symmetryby a vertical translation. The incompleteness of the computeroutput prevents this symmetry from being completely accurate on the

picture, although the same incompleteness has the advantage of

making the symmetry somewhat visible. When rotated back to its actual

position, this symmetry corresponds to the translation \317\204defined by

r(z) = \316\266+ \317\211.

296 11. The figure-eight knot complement jj

Let \316\223\316\262be the transformation group generated by \317\210\317\207,\317\206-sand by *

this new translation r. We will consider \316\2238 as a group of isometriesofH3.

\\

It turns out that discontinuity is easier to show for the action for j

\316\223\316\262than for Tg. Indeed, we will exhibit in the next section a relatively j

simple fundamental domain for \316\223\302\273.It can be proven that \316\2238 admitsj

no fundamental domain with finitely many faces. This is a special !case of a theorem of Al Marden [Mardeni], but the proof of this

j

property is far beyond the scope of this book. :

Another direct proof that the actions of \316\223\302\273and \316\2238on H3 are 1

discontinuous is provided by Exercise 11.2. jii

\342\226\240j

11.2.1. A fundamental domain for the action of \320\223\302\273.In the hy- ,

perbolic space H3,let \320\220\320\263be the tetrahedron with vertices at infinity ,

0, 1, oo and \317\211= \320\265%\\and let \316\2242be the tetrahedron with vertices ;

at infinity 0. 1. 00 and \317\211~\316\273= 1 \342\200\224\317\211.The union \316\224of the two ideal ;

tetrahedra \316\224\316\271and \316\2242is a polyhedron with five ideal vertices, nine j

edges and six faces. We want to show that \316\224is a fundamental domain

for the action of \316\2238 on H3. In particular, this will prove that rs acts \317\212

discontinuously on H3. [

Figure 11.3 offers a top view of \316\224\316\271and \316\2242,namely, describes the \">\342\226\240

vertical projection of these objects from the hyperbolic space H3 to \316\257

the plane \320\241In this projection, an ideal tetrahedron with one vertexj

equal to 00 appears as a euclideantriangle in the plane. Similarly, anj

ideal triangle with a vertex at oc projectsto a lino segment, whereas \\

any other ideal triangle projects to a euclideantriangle. jNote that \316\224\316\271and \316\2242meet along the ideal triangle T~ of the \316\257

crooked tessellation 7\321\206.Consequently, every time an element \317\206\342\202\254:;

I?8 sends T4 to an ideal triangle with one vertex equal to do. I hej

ideal tetrahedra \317\206(\316\221\\)and ^(\316\224^) 1\321\202\320\273_\320\265one vertex at oc. Since thej

combinatorics of the crooked tessellation T\302\253is controlled by that of i

the standard Farey tessellation T\302\260,we can find many examples of j

such \317\206.Some of these are illustrated in Figure 11.3. To save space. I

we have omitted the symbol \320\276when writing a composition of maps; j

yon should also remember that \317\2062=

\317\210^1and \317\206\302\261=

\317\206$\316\273\342\226\240

'

11.2. Enlarging the group \316\223\302\273 297

\317\211\342\200\2242 \317\211\342\200\2241 \317\211 \317\211+ 1

\\ / \\ /\316\264\316\233

\\ / \316\2752(\316\224\316\271)\\0/ \\1/

\\\302\267\302\267\316\2413(\316\2242)/ \\ \316\224-/ \\

\342\200\224\317\211 \316\231\342\200\224\317\2112\342\200\224\317\211 3 \342\200\224\317\211

Figure 11.3. A fundamental domain for \316\223\316\262

In particular, this analysis enables us to find several elements \317\206

of the extended group Tg for which \316\224and \317\206(\316\221)meet along along a

face of \316\224.More precisely, consider the elements \321\204\\=

\317\204,\321\204\320\267=

\317\210\316\257\316\270\317\2062

and \317\2105= \317\204\316\277\317\2063of \316\223$.If [\316\261,b, c] denotes the ideal triangle with ideal

vertices a, b. \321\201e C, we see that \317\206\317\207sends the face [0,1 \342\200\224\317\211.\316\277\316\277]of \316\224

to the face [\317\211,1, \320\276\321\201.,that \321\204\320\267sends j0,1, u>] to [oc, 1 \342\200\224u>, 1], and that

V'r, sends [0,1,1\342\200\224\317\211;to [0.u>,oc].

Proposition 11.1. 77ie group \316\2238 acts discontinuously and freely onthe hyperbolic space H3, and the ideal polyhedron A is a fundamentaldomain for this action.

Proof. We will apply Poincare's PolyhedronTheorem 10.9.Gluethefaces of the polyhedron \316\224by the isometries \317\210\316\257.More precisely, glue

the face Fi = [0,1- \317\211,\321\201\320\276]to F2 =[\317\211,1, \321\201\320\276]by \317\210\316\271,F3

=[0,1,\317\211] to

Fi =[\321\201\320\276,1 \342\200\224

\317\211,1] by \321\204\320\267and F5 = [0,1,1\342\200\224

\317\211]to Fe =[0,\317\211,\321\201\320\276]by

t',-,. As usual, set \317\2102~

\320\244^1\302\267%'\321\207=

\320\263''\320\267and \321\211=

\321\204^1.

Let (\316\224,rfhyp) be the quotient space so obtained.Theedges[0,1],[oc, 1 \342\200\224

\317\211],[oc, 1; and [0, u>] are glued together to

form an edge of \316\224,as are the edges [0, \321\201\320\276],[\317\211,\321\201\320\276],[1,1 \342\200\224\317\211]and [\317\211,1].

Also, if we orient each of the above edges [a, b] by the direction from

a to b, we see by inspection that these orientations are respectedby

the gluing maps \317\210^.Therefore, the Edge Orientation Condition (2)of Theorem 10.9 holds.

The consideration of Figure 11.3immediately shows that the

dihedral angles of \316\224along the edges [0, \321\201\320\276],[1, \321\201\320\276],[\317\211,\321\201\320\276]and [1\342\200\224\317\211,\321\201\320\276]


are\317\210,\317\210,f, \302\247,respectively. \316\216\316\277determine the other dihedral

angles, observe that the inversion across the sphereof radius 1 centered ;

at 0 defines an isometry \317\201of H3 that exchanges 0 and oo, and fixes

each of 1, \317\211and 1 \342\200\224\317\211.It follows that \317\201sends \316\224to itself and

consequently that for each a = 1, \317\211or 1 \342\200\224\317\211,the dihedral angle of \316\224along

the edge [0, a] is equal to the dihedralangle along [\321\201\321\216,a].

From these considerations, it follows that the sum of the dihedral '

angles of \316\224along the edges that, are glued to [0,1]is equal to\320\251-+

\\ + ^ + \\ = 2\317\200.Similarly, the dihedral angles along the edgesthat

are glued to [0, oo] add up to\317\210+ | + \302\247+ \317\210

= 2\317\200.Therefore, '.

the Dihedral Angle Condition (1) in the hypotheses of Poincare's \\

Polyhedron Theorem 10.9 is satisfied. ,:

For the Horosphere Condition (3), let So, Si, 5\317\211and Si \321\210be ;

the horosphere centered at 0, 1, \317\211and 1 \342\200\224\317\211,respectively, that have ;

euclidean diameter 1, and let Soo be the horizontal plane of equation ',

\320\270= 1. Note that for any face [a,b, c] of \316\224,the three horospheres Sa,

Si, Sc are tangent to each other. Therefore, if [a,b, c] is glued to a

face [a1,b', c'}by a gluing map \317\206,this map \317\206sends Sa, Sb and Sc to

horospheres centeredat a', b' and d and tangent to each other. By

elementary euclidean geometry (compare Lemma 8.4), there are only j

three such horospheres, namely, Sa', SV and 5^. Therefore, if thej

gluing map \317\206sends an ideal vertex \316\261of \316\224to \317\206(\316\261),it also sends the |horosphereSato

Sv^a).This shows that the Horosphere Condition (3) \\

in the hypotheses of Poincare's Polyhedron Theorem 10.9is satisfied, j

Poincare's Polyhedron Theorem 10.9 then asserts that the group \\

Tg generated by \317\210\\,\321\207\321\200\320\267,\317\2105acts discontinuously on H3 and admits \316\224\\

as a fundamental domain. \316\271

Similarly, Theorem 10.11 shows that the action of \320\2238is free.

It remains to show that T's=

\320\223\302\273.Every element of F's can beexpressedas a composition of terms \317\206\316\263\317\207.Since ip\\

=\317\204,\317\206^

=\316\250\316\261\302\260\317\2102\\

and \321\204\321\212= \317\204\302\260

\316\250\316\254-,it can be expressed as a compositionof terms rpml,'

ipf1 and ipf1- Therefore, every element of T's is also an element of ;

IV Conversely, we can manipulate the above equations and find that :

r =\317\210\316\271,1\321\200\320\267

=\317\210^1\316\277\317\2105,\317\206\316\271

=\321\204^\320\263

\320\276\317\210^1

\320\276\321\2041. Therefore, every

element of \320\2238can be written as a composition of terms \317\210^1,and is'

consequently an element of \316\2238.This proves that \320\2238=

\320\223\302\273. \320\236

11.3. Limit sets 299

Proposition 11.2. The quotient space (\320\250.3/\320\223\302\273,\320\260^\321\203\321\200)is locally

isometric to the hyperbolic space H3.

Proof. ThisisanimmediatcconseqiienceofTheoremlO.il. D

Since Tg is contained in Tg, an immediate consequenceis thefollowing statement.

Corollary 11.3. The tiling group Fg is a kleinian group, namely,acts discontinuously on the hyperbolic space H3. D

11.3. Limit sets

We now determine the limit sets of the kleinian groups T& and \320\223\302\247-

Lemma 11.4. The limit set of the enlarged group Tg is the whole

Riemann sphere \320\241

Proof. The proof is identical to that of Lemma 10.7. The key

property is that if, for each ideal vertex \316\276of \316\224,5\316\276is the horosphere

introduced in the proof of Proposition 11.1 and \316\222\316\266

is the ball bounded

by 5\316\276,then X -(BqUBiU \320\222\321\210U Bx. \321\210)is bounded in H3. D

It turns out that the two groups T% and Tg have the same limit

set. This follows from a relatively simplealgebraicfact. We begin

with a definition.

A normal subgroup of a transformation group \320\223is a

transformation group \320\223'contained in \320\223and such that for every 7 G \320\223and

7' e \320\223',the composition 7 \316\277\316\267'

\316\2777-1 is also an element of \320\223.Normal

subgroups play an important role in algebra becausethey are well

behaved with respect to quotient spaces; see Exercise 11.1.

In our case,we are more interested in the following property.

Proposition 11.5. //\320\223' is a normal subgroup of the kleinian group\320\223and if the limit set of V has at least two elements, then the limit

sets \320\233\321\200and \320\233\320\263'coincide.

Proof. The key point in the proof is that \320\233\320\263<is invariant under the

action of \320\223.To prove this, consider \316\276\302\243\320\233\320\263'and 7 G \320\223.Fix a

base point Pq e H3, and considerits orbit T'(Pq) under the action

300 11. The figure-eightknot complement

of \316\223\".Then there is a sequence of elementsy'n \342\202\254\316\223\"such that \316\276=

\320\235\320\263\320\277\320\277-\320\273\321\215\320\276j'n(Po) in R3 U {oo} for the euclidean metric. Now,

7(0= \316\231\317\212\317\204\316\261\316\212\316\277\316\212'\316\267(\316\2410)=Urn\316\267'\316\267(\316\241\302\243)

\316\267\342\200\224*\316\277\316\277\316\267\342\200\224*\316\277\316\277

if we set Pq = j(Pq) and \316\267'\316\267=

\316\267\316\277\316\267'\316\267\316\277\316\267~\316\273G \316\223'.Since the limit set is

independent of the choice of base point (Lemma 10.1), we conclude

that 7(\316\276)is in \320\233\320\263'for every \316\276G \320\233\320\263'and 7 G \320\223.In other words, the

limit set \320\233\320\263'of the normal subgroup \320\223\"is invariant under the actionofT.

Now, \320\233\320\223'is closed in \320\241by Lemma 10.2, it is invariant under

the action of \320\223,and it has at least two points by hypothesis. By

Proposition 10.3, it follows that \320\233\320\263'contains \320\233\320\263-On the other hand,

\320\233\320\263'is clearly contained in \320\233\320\263since \320\223'is contained in \320\223. \320\237

We now prove that \316\223\302\273is a normal subgroup of IV This will

follow from the following computation.

Lemma 11.6.

(11.1)(11.2)(11.3)(11.4)

\317\204'1\316\277\317\206\316\271

\316\277\317\204=\317\206'3\316\233.

\317\204-1\316\277\317\2063\320\276\321\202=\317\206\316\220\316\277\317\206\316\273

\316\277\317\2063)

\317\204\316\277\317\2061\316\277\317\204-1= \317\210\\\316\277

\317\2063\316\277

\317\206\317\207,

\317\204\316\277\317\2063\316\277\317\204~1=

\317\206\317\2121.

Proof. We could use brute force computations of linear fractional \\

maps but we prefer a more conceptual argument, which in addition \\

should hint at the way these relations were discovered. \\

The gluing map \317\210\316\271sends the edge [0,00] of \316\224to the edge [\317\211,oo], j

which is sent by \316\246\316\262=

\316\246^1to [1,1\342\200\224

\317\211],which is sent by \316\2464=

\317\2063\316\271to !

[\317\211,1], which is sent by \316\2502=

\317\206^1to [0,1\342\200\224

\317\211],which is sent back to j

[0,oo] by \321\204\321\212. j

Therefore, the element \321\204\321\214\320\276\317\206~\316\273\316\277\317\2063\316\273

\316\277\317\210^1

\316\277\317\206\316\273\302\243\316\2238sends the

oriented edge [0,00] to itself, and in particular must fix the point j

So \316\240[0, oo] provided by the horosphere Sq. Sincethe actionof \316\223\302\273is j

free by Proposition 11.1, this proves that jj

\320\244\321\212\302\260\320\244\320\242\320\245\302\260\320\244'\320\267\320\245\302\260^51 \302\260\316\250\316\271=\320\230\320\275\320\267. 1

11.3. Limit sets 301

Substituting \317\206\316\271=

\317\204,\321\204\320\267=

\317\206$\316\277\317\2062

=\317\2063

\316\273\316\277

\317\210\316\2631

and \321\204\321\212= \317\204\302\260

\302\245>\320\267,this

gives

\317\204\316\277\317\2063or-1 \320\276

<\321\200!\320\276

<\321\200\320\267\320\276(\321\200^-1\320\276\321\202-1\320\276\321\202=

\320\253\321\211\320\267,

and

\321\202~\320\276\317\206\316\273\320\276\321\202=

\317\206^1

after simplification. This proves (111).

Considering a similar edge cycle beginning with the edge [0,1],the sameargument provides the relation

\320\244\321\2141\302\260

^\320\267*

\302\260\316\250\316\271\302\260^\320\267=

\320\230\320\2753

and

\317\206^1or-1 o<p! \316\277

\317\210\316\206\320\276\321\202\320\276<\321\200\320\267

*\302\260

^i\"1=

\320\230\321\206\321\200>

which simplifies to

\321\202~\320\276\317\206\316\271

\316\277\317\206\316\266

\316\277\317\204=\317\206$\320\276<pj \320\276

\317\206%.

Using (11.1) (which we just proved)on the left-hand side of the

equation and simplifying gives

\317\204'1\316\277<\317\2013

\316\277\317\204=v?3\302\260\302\245>i\302\260<A3,

which proves (11.2).

An algebraic manipulation of (11.2) gives

\316\2503=

{\317\204\316\277\317\206\316\266\316\277\317\204~\317\207)\316\277

(\317\204\316\277

\317\2063\320\276\321\202~\320\263)\316\277(\317\204

\316\277\317\206\316\271\320\276\321\202-1)\320\276

(\317\204\316\277\317\2063or-1)

^\317\206\317\2122

\316\277(\317\204

\316\277\317\206\316\273

\316\277\317\204'1)

\316\277\317\206~\316\233,

using (11.1), from which (11.3) easily follows.

Finally, (11.4) is an immediate consequenceof (111). \316\240

Lemma 11.7. The group \316\2238 is a normal subgroup of the enlarged

group \316\2238.

Proof. We need to show that for every 7 G \316\223\302\247and every 7' e \320\223\321\217,

the element 707\316\2147\"1 \302\243Tg belongs to the subgroup \320\223\302\273.

This is immediate for 7 =<p, and 7 =

\317\206, since these elements

belong to \316\223\316\262,and for 7 = \321\202\302\261\320\263by Lemma 11.6 since Tg is generatedby \317\210\\and \317\210^-The general result then follows, since every 7 G Tg can

be written as a compositionof terms of the form \317\206^1,\317\206\316\206\317\207or\317\204\316\257\316\271.D


The combination of Lemma 11.4, Proposition 11.5, and Lemma

11.7 provides the following result.

Corollary 11.8. The limit set of the kleinian group \316\2238 is equal to

the whole Riemann sphere C. D

As such, the tiling group \320\223&of the crooked tessellation 7$ is very

different from the tiling groups \320\223of the crooked tessellations that we

analyzed in Section10.5.Indeed,we had seen that the limit sets ofthose tiling groups were homeomorphic to the circle \320\250.

Actually, Jim Cannon and Bill Thurston proved a truly amazing

property in [Cannon & Thurston].For the kleinian groups \320\223considered in Section 10.5, we

constructed a homeomorphism \316\273:\320\250\342\200\224+\320\233\320\263between \320\250= Ru{oo} and

the limit set \320\233\320\263of \320\223.By continuity, the map \316\273is uniquely

determined by the property that for every rational point ieQU {oo}corresponding to a vertex of the standard Farey tessellation 7\302\260,its

image \\(x) G \320\241is the corresponding vertex of the crookedFarey

tessellation 7.

In the case of the crookedtessellation 7s, we still have a well-defined map \316\273:Qu{oo}

\342\200\224\302\273\320\233\320\2638,which to each vertex of 7\302\260associates

the corresponding vertex of 7^\302\267Cannon and Thurston prove:

Theorem 11.9 (Cannon-Thurston).The above map \316\273has a unique

extension X: \320\250\342\200\224\302\273\320\233\320\2638

= \320\241which is both continuous and surjective.D

In particular, this natural map \316\273provides a continuous curve

which passes through every point of the Riemann sphere \320\241Poincare

would undoubtedly have been again very surprised to find out that his

theory of kleinian groups would lead to such a mathematical

\"monster\".

In contrast to what happened in Section 10.5,the map \316\273is very

far from being injective.The proof of Theorem 11.9 is widely beyond the scope of this

text, and can be found in [Cannon & Thurston].

11.4.Thefigure-eight knot 303

11.4. The figure-eight knot

Theenlarged kleinian group V$ of the previous sectionhas anunexpected connection with the figure-eight knot familiar to sailorsand

rock climbers, and represented in Figure 11.4. Here,instead of a long

rope, we should think of the knot as an infinite curve going from oo

to oo in R3.

Figure 11.4. The figure-eight knot in R3 U {oo}

If, in \320\2503U {oo}, we move the knot away from oo and deform it

to make the picture esthetically more pleasing,we obtain the closed

curve \320\232represented in Figure 11.5. We are particularly interested

in the complement X = \320\2313- \320\232of this curve \320\232in \320\2313= \320\2503U {oo}.

This is why the knot is drawn here as a \"hollow\" curve, to indicate

that the knot is not really there.

Figure 11.5. The figure-eight knot in R3

Theorem 11.10. The complement X = \320\2503\342\200\224\320\232of the figure-eightknot is homeomorphicto the quotient space H3/IV

Proof. The proof of Theorem 11.10 will be a typical exampleof a

\"proof by pictures\". This type of argument is standard fare for topol-ogists, although sometimes regarded with a certain levelof perplexity


(orworse) by mathematicians from other fields. The generaljustificationfor such a proof is that it would be too time consuming to write

(and read) it with all the details rigorously expounded. The sheerbulk of these details would even obscure the key ideasunderlying the

arguments. Consequently, the basic principle of the exposition of the

proof is to explainthesegeometricideasin such a way that, at each

step, you should be able to convince yourself that you could rigorouslywrite down the complete details if absolutely necessary. Of course,it takes some experience (and it can be a matter of appreciation) to

decide which details are legitimate to skip and which ones must

absolutely be included. However, with practice, one reachesa level of

proficiency that makes this type of proof as mathematically valid as,

say, a long algebraic computation.Our proof of Theorem 11.10 will be a good introduction to this

type of mathematical reasoning. As a warm-up, you can begin bytrying to figure out a way to make rigorous the statement \"let \320\232be

the closed curve in R3 represented in Figure 11.5\". Next, you should

convince yourself that one can prove the existence of a homeomor-

phism \317\210:R3 \342\200\224\342\226\272R3 sending the closed curve \320\232\321\201R3 of Figure 11.5 tothe closedcurve \320\232'\321\201R3 of Figure 11.4. For this, first consider the

physical process of moving the string \320\232to K' in R3; then mentally

define \317\210:R3 \342\200\224>R3 by the property that this process sends eachparticle \316\241\342\202\254I3 of the universe to the particle \317\206{\316\241)\342\202\254R3; and finally,convince yourself that this homeomorphism can be mimicked by a

composition of homeomorphisms rigorously defined by explicit

equations, although this exposition would be extraordinarily cumbersome.We now begin the proof of Theorem 11.10.We will first show that

the complementX = R3 \342\200\224\320\232is homeomorphic to the space obtainedby gluing two polyhedra X+ and X~ along their faces. For this, wefirst arrange by a homeomorphism of R3 that most of the knot \320\232lies

in the xy-plane, except near the crossings,where one strand slightlyrises above the plane while the other strand slightly dips under it.Roughly speaking, X+ will consist of the part of X that lies above

the xy-plaiic, while X~ will include those points which are below the

plane. At least, this will be true away from the crossings.

11.4. The figure-eight knot 305

More precisely, we cut X along a certain number of (curved)

faces, which we now describe. Away from the crossings, these faces

just coincidewith the xy-pl&ne. Understanding the arrangement ofthe faces near a crossingrequires a little more thought. Figure 11.6describesthe situation near such a crossing.

We have here four faces meeting along an edge EF, joining apoint \316\225in the upper strand to a point F in the lower strand. Theendpoints\316\225and F belong to the knot K. In particular, they are not,

in its complement X, and consequentlyare not part of the edge EF.

Near the rectanglePQRS drawn in the a;i/-plane, the first face

coincides with a warped pentagon AEFBQ, where the edges AE and

BF follow the knot and consequently are part of neither X nor the

face. Similarly, the secondface is a warped pentagon BFECR with

the edges BF and EC removed,the third face is CEFDS minus the

edges \320\241\316\225and FD, and the last face is DFEAP minus DF and \316\225A.

SCR

Figure 11.\316\262.Splitting the knot complement near a crossing

Having described these four faces,we now need to understand how

l.hey decompose the knot complement X near the rectangle PQRS.The upper polyhedron X+ of the decomposition touches the edgeEFalong two opposite dihedra, one delimited by AEFBQ and BFECR,and the other delimited by CEFDS and DFEAP. If we separate

X+ along the edge EF, this edge splits into two edges EF' and EF\",and X+ can now be deformed to the part of R3 that lies above the

diagram on the left-hand side of Figure 11.7, namely, in the picture,the part that lies in front of the sheet of paper.

Symmetrically, the lower polyhedron X~ touchesthe edgealongthe remaining two dihedra, namely, the one delimitedby AEFBQ

and DFEAP, and the other one delimited by BFECR and CEFDS.

306 11. The figure-eight knot complement I

\320\241 R

DF'

\316\225

F\"\320\222

\316\241 A Q \316\241 A Q

Top polyhedron Bottom polyhedron j

Figure 11.7

Separating X~ along the edgeEF yields two new edges E'F and\\

E\"F. Then X+ can be deformedto the part of R3 that lies below the

diagram on the right-hand side of Figure 11.7, namely, behind the?sheet of paper. \316\257

Let the faces of the decompositionbe defined in such a way that ]they coincide with the xy-pl&ne away from the crossings, and that]they are of the type described above near the crossings.If we split j

X = R3 \342\200\224\320\232along these faces, we then obtain two pieces Xf and:x-

1

Before we describe these two pieces, it is convenient to introduce!

some notation. Let R+ = R2 \317\207[0, +oo) consist of those points of R3;

that lie on or above the xy-plane R2, and let R-l = R2 \317\207(\342\200\224\316\277\316\277,\316\237]!

consist of those points that are located on or below R2. Then define

R3 = R3 U {\321\201\320\276},considered as a subset of R3 = R3 U {oo}. !

As in Figure 11.7, the upperpieceX+ can be deformed (and is;consequently homeomorphic) to the complement, in the upper part\320\251_

of R3, of the four arcs that are indicatedas hollow curves on the,

left-hand side of Figure 11.8.Thesefour arcs correspond to the trace1of the knot \320\232on the boundary of X+. In this picture,we also use:

the standard convention thatR3^

consists of those points of R3 which

lie in front of the sheet of paper. The boundary of X+ is decomposedinto six faces and eight edges as indicatedon the left-hand side of

Figure 11.8, with two edges associated to each crossing of the picture

in Figure 11.5.

11.4. The figure-eightknot 307

Figure 11.8. The top pieceX+ of the decomposition of the

figure-eight knot complement.

By shrinking each of the missingarcsto a point, X+ can also be

deformed to R+ minus the four points indicated on the right-hand

side of Figure 11.8. The edges of the decomposition of its boundaryare l-hen as indicated in that picture. To clarify this deformation we

should probably note that it does not provide a homeomorphism of

E^ , since it crushes each of the four arcs to a single point (andtherefore is far from being injective). However,it can indeedbe chosen

so that it is injective outside of these arcs, and induces a

homeomorphismbetween\320\250+

minus the four arcs on the left and R+ minus the

four points on the right.

Figure 11.\316\230.The bottom piece X of the decomposition of

the figure-eight knot complement.


Similarly, the lower piece X~ can be deformed to the lower half

R3 of R3 minus the four arcs drawn on the left-hand side of

Figure 11.9, or minus the four points indicated on the right-hand side.

In particular, X~ now lies behind the sheet of paper in Figure 11.9.

Again, the boundary of X~ is decomposed into six faces and eight

edges, as drawn in these pictures.The right-hand sides of Figures 11.8 and 11.9beginto make the

X* look like ideal polyhedra, namely, like polyhedra with their

vertices deleted. However, X+ and X~ each have two faces which are

\"digons\", namely, faces with only two edges and two vertices (at

infinity). This is clearly not possible for ideal polyhedra in the hyperbolicspaceH3, since two hyperbolic geodesies with the same endpointsat infinity must coincide. Consequently, we need to perform a last

modification on the X*.

Before After

Figure 11.10. Collapsing a digon to one edge

The two digons of X+ and the two digons of X~ are gluedtogether to give two digons in the decompositionof X. We will modifythis decomposition by collapsingeach of these two digons of X to asingleedge.Under the following collapse, the two polyhedra X\302\261are

transformed into two other polyhedra Y\302\261.

This process is illustrated by Figure 11.10. We have indicated

the cross section with a plane \316\240to clarify the picture.

In the collapseof the first digon corresponding to the faces Fjf \320\241

X+ and Fx \320\241X~, the edges marked \342\200\224*\342\200\224and <i get combined

to form a single edge\342\200\224*\342\200\224in the new decomposition. Similarly, in

11.4. The figure-eightknot 309

the second digon corresponding to F% \321\201X^, the edges\302\253\316\271are replaced by a single edge \302\273 .

and

Figure 11.11. The tetrahedra \320\243\302\261

After collapsing these two digons, X+ gets replacedby another

region Y+ of K3, homeomorphicto\320\251_

minus the four points indicatedon the left-hand side of Figure 11.11. In addition, Y+ inherits from

the nondigon faces of X+ a decompositionof its boundary into four

faces and six edges,as indicated on the left half of Figure 11.11.Compare Figures 11.8 and 11.11, and remember that K+

lies in front

of the sheet of paper in these figures.

oo

oo

Figure 11.12.Unfolding an ideal tetrahedron

Similarly, after collapsingthe digonsand flipping the picture over,X~ is replacedby a region Y~ homeomorphic to K3 minus four

points, and with four faces and six edgesas indicated on the right-hand side of Figure 11.11.The flip is used to exchange front and half,

since X~ was in the backof the sheet of paper in Figure 11.9.


With a little more thought, we can convince ourselves that Y+

and Y~ areeachhomeomorphic to an ideal tetrahedron, by a homeo-morphism sendingedgesto edgesand faces to faces. It may be easierto go backward to see this, and start with an ideal tetrahedron in

H3 with ideal vertices a, b, c, oo\342\202\254l2. Unfolding its faces onto the

xy-plane deforms the ideal tetrahedron to the upper part R+ of K3

minus {a,b, c, oo}, and the four faces then are as indicated in theright-hand side of Figure 11.12. That picture is then easily deformed

in K2 = K2 U {\321\201\321\216}to those of Figure 11.11, after moving the fourth

vertex away from \321\201\321\216.

The upshot of this discussion is that up to homeomorphism, the

figure-eight knot complement K3 \342\200\224\320\232is obtained from Y+ and Y~by gluing each face F4+ of Y+ to the correspondingface F/ of Y~,in a manner compatiblewith the edge identifications represented on

Figure 11.11.Let us now return to the fundamental domain \316\224for Tg that we

considered in Section 11.2.1.This \316\224was the union of the two tetra-hedra \316\221\317\207

=[0,1,\317\211, \321\201\321\216]and \316\2242

=[0,1,1 \342\200\224\317\211,\321\201\321\216],meeting along their

common face [0,1, \321\201\321\216].Namely, we can consider that \316\224is abstractly

obtained from the tetrahedra \316\224\316\271and \316\2242 by gluing the face [0,1, \321\201\321\216]

of \316\224\316\271to the face [0,1, \321\201\321\216]of \316\2242by the identity map. If we unfold

these tetrahedra as in Figure 11.12, we then obtain the tetrahedrarepresentedon Figure11.13,where \316\224\316\271is on the left, \316\2242is on the

right, and both tetrahedraare in front of the sheet of paper.

00 0

Figure 11.13.Gluing the tetrahedra \316\224\316\271and \316\2242.

The gluing data of the faces of \316\224is such that the face Fj =

[0,1\342\200\224\317\211,\321\201\321\216]of \316\224-2is glued to the face F2 =

[\316\231,\317\211,\321\201\321\216]of \316\224\316\271by the

11.4. The figure-eight knot 311

gluing map \317\206\317\207,the face F3 = [0,1, \317\211]of Ax is glued to the face

FA = [1,1 \342\200\224\317\211,\321\201\321\216]of \316\2242by \317\2063,the face F5 = [0,1,1 \342\200\224\317\211]of \316\2242is glued

to the face F$=

[\342\200\224,\317\211,\321\201\321\216]of \316\224\316\271by \317\2065,and the face F7 = [0,1, \321\201\321\216]of

\316\224\316\271is glued to the face Fg=

[0,1,\317\211]of \316\2242by the identity map. In

addition, the edges[0,1],[\321\201\321\216,1 \342\200\224

\317\211],[\321\201\321\216,1], [0, \317\211]are glued together,and marked on Figure 11.13as \342\200\224\302\273\302\273\342\200\224; similarly, the edges [0, \321\201\321\216],

[\317\211,\320\276\321\201],[1,1\342\200\224

\317\211]and [\317\211,1] are glued together and marked as \342\200\224\302\253\342\200\224.

Comparing Figures 11.11 and 11.13, it is now immediate that the

gluing data are the same.

This showsthat the figure-eight knot complement \320\2323\342\200\224\320\232is home-

omorphic to the quotient space\316\224obtained by gluing the ideal tetra-hedra \316\224\316\271and \316\2242according to the gluing maps described in

Figure 11.11.

Theorem 10.9 shows that \316\224is homeomorphic to H3/IV

Therefore, the quotient space \316\2273/\316\223\316\262is homeomorphic to K3 \342\200\224K, which

concludes the proof of Theorem 11.10. \316\240

We can rephrase Theorem 11.10 in a slightly different way. We

say that two metrics d and dl on the same set X are topologicallyequivalent,or induce the same topology, if the identity map\316\231\316\254\317\207: {X, d) \342\200\224>(X, d') is a homeomorphism. This is equivalent to

the property that for every sequence (Pn)ngN in X, the sequence

converges to the point \316\241\317\207for the metric d if and only if it convergesfor to Poo for the metric d'\\ see Exercise 1.9.

In our case,the euclidean metric deuc of K3 does not quite give a

metric on X = K3 \342\200\224K, because dcnc{P, Q) is undefined when Q

= \321\201\321\216.

Wc will say that a metric d is topologically equivalent to, or induces

the same topology as, the metric deut. if a sequence (P\342\200\236)\342\200\236gNin X

converges to the point Poo (with possibly Poo =\321\201\321\216)for the metric d

if and only if it converges to Poo for the metric deuc\302\267

Theorem 11.11. The complement X = K3 \342\200\224\320\232of the figure-eightknot admits a metric d which is topologically equivalent to the

euclidean metric and which satisfies the following two properties:

(1) (X, d) is complete;

(2) (X, d) is locally isometric to the hyperbolic plane H3.

312 11.Thefigure-eight knot complement

Proof. Let \317\210:X \342\200\224>\316\2273/\316\223\316\262be the homeomorphism provided by \316\271

Theorem 11.10. If dhyp is the quotient metric inducedon \316\2273/\316\2238by the

hyperbolic metric dhypof H3, define a metric d on X by the property

'

that d{P,Q) =d~hyp{<p{P),<p{Q))\302\267 Namely, d is uniquely determined

by the propertythat \317\206is an isometry from (X, d) to (\316\2273/\316\2238,dhyp)\302\267:

The metric dhyp is complete by Theorem 10.9,and locally

isometricto (H3,dhyp) by Theorem 10.11. Since \317\210:(X,d) -> (H3/f8,dhyp) .

is an isometry, the sameproperties hold for d. D

Theorem 11.10was discoveredand first proved by Bob Riley in

[Rileyi]. This property originally appeared to set the figure-eightknot apart from the other knots until, a few years later, Bill Thurston .'showed that this is actually a manifestation of a more generalresult. Thurston's Geometrization Theorem completely revolutionized :3-dimensionaltopology. The next chapter is devoted to these results.

Exercises for Chapter 11Exercise 11.1. Let \316\223be a transformation group acting on the set X, andlet \316\223'be a normal subgroup of \316\223.Let \316\247/\316\223'denote the quotient \320\262\321\200\320\260\320\273\320\265of X

under the action of \320\223',and let \320\223/\320\223'be the quotient of \320\223under the action of\320\223'by left composition. Namely, 7'(7) =

7\316\2147 for every 7' G \320\223'and 7 G I\\and the image of 7 G \320\223in the quotient space \320\223/\320\223'is 7 ~ {7' \320\2767; 7' G \320\223'}.

a. For every 7 G \320\223/\320\223'and \316\241G \316\247/\316\223,consider the element 7(F) G \316\247/\316\223'

represented by y{P) G X. Show that f(P) is independent of the choiceof 7 G 7 and of \316\241G P.

b. Show that as 7 ranges over all the elements of \316\223/\316\223',the corresponding

maps \316\241\316\275-*\321\203(\316\241)form a transformation group \316\270of X/y'.

\321\201Consider the map \320\223/\320\223'\342\200\224\342\226\272\316\270which to 7 G \316\223/\316\223'associates the

transformation \316\241f-\302\273j(P). Show that when X = \316\223and when the action ofthe group \316\223on the set X = \316\223is by left multiplication, the above map\316\223/\316\223'

\342\200\224\342\231\246\316\270is a bijection, so that \316\270has a natural identification with

\316\223/\316\223'.

Exercise 11.2. For \317\211= e^' as usual in this chapter, consider the subset\316\226[\317\211]

= {m + \316\267\317\211;\317\200\316\271,\316\267G \320\251of C, and let PSL2(Z[o;])be the set of linear

fractional maps \316\266\316\271\342\200\224>^\320\271

with a, b, c, d G \316\226[\317\211].Extend each of theselinear fractional maps to an isometry of (H3,dhyp)\302\267

a. Show that PSL2(Z[<j]) is a group of isometries of HI3. Hint: Rememberthat \317\2112= \317\211+ 1.


Our goal is to show that PSLi2(Z[a;]) acts discontinuously on H3. Assume,

in search of a contradiction, that PSL2(Z[a/|) does not act discontinuously

at some Po G H3. Namely assume that for every e > 0, there are infinitely

many 7 G PSL2(Z[o;]) such that y{P0) G Bdtlyp{P0,e).

b. Show that there is a sequence (\321\203\320\277)\320\277\320\265\320\243of distinct elements of the group

PSL2(Z[a;]) such that lim,,-,,\302\273 7\342\200\236(P0)= Po in (E3,dhyp).

c. Show that there exists a subsequence (\321\203\320\237\320\272) Nsuch that the sequences

\320\253\320\233\302\260))\320\272\320\265\320\232> (7\302\273*(l))fceN \302\267(7nfc(\302\260\302\260))fceNconverge to points \302\253,,zu

Zoo G C, respectively, for the euclidean metric deuc\302\267

d. Show that the limits \320\263\320\276,z\\ and Zoo are pairwise distinct. Hint:

Consider the complete geodesic g going from 0 to oo, and the geodesic h

joining the point Pi = (0,0,1) G H3 to the point 1 G C; then observethat the images fnk{Pi) must remain at bounded hyperbolic distancefrom Po.

e. Show that the coefficients ank, bnk, Cnk, dnk G \316\226[\317\211]of fnk{z) =

c\"ftz7d\"'1can be chosen so that they each converge in \320\241as \320\272\342\200\224>oo.

Hint: If f(z) = ^f^j with ad \342\200\224be = 1, express o, b, c, d in termsof 7(0), 7(1) and 7(00), and note that these coefficientsare uniquely

determined up to multiplication by \342\200\2241.

f. Show that for every R > 0, there are only finitely many a G \316\226[\317\211]such

that \\a\\ \316\266R.

g. Conclude that we have reached the contradiction we were looking for.

As a consequence, PSL2(Z[o;]) acts discontinuously on H3. In particular,so do the groups \320\223\320\262\320\241Tg \320\241PSL2(Z[o;]) considered earlier in this chapter.

Chapter 12

Geometrization

theorems in dimension 3

This chapter generalizes to a wider framework the hyperbolicmetricon the figure-eight knot complement that we encountered in

Chapter11. The goal is to show the dramatic impact of techniques of

hyperbolic geometry on very classical problems in 3-dimensionaltopology. The first half of the discussionis focusedon knot theory, where

results are easier to state. After developing the necessary material, wethen considerthe generalGeometrization Theorem for 3-dimensional

manifolds, whose proof was completedvery recently.

The proofs of many results in this chapterrequirea mathematical

expertise which is much higher than what we have neededso far. We

will not even attempt to explainthe ideas behind these arguments.

We will be content with providing the background necessaryto

understand the statements and illustrate these with a few applications.

12.1. Knots

Mathematical knot theory aims at analyzing the many different ways

in which a piece of string can be tied into a knot. As for the figure-

eight knot that we already encounteredin Section11.4,it is more

convenient to consider strings where the two ends have been joined

together.

315

316 12. Geometrizationtheoremsin dimension3

Therefore, from a mathematical point of view, a knot in K3 is a

regular simple closed curve \320\232in K3. We need to makeall these terms

precise.

The fact that the curve is regular means that A\" can be

parametrized by a differentiable function 7: \320\232\342\200\224*\320\2323such that for each

value t e \320\232of the parameter, the derivative Y(t) is different from the

zero vector. Recall from multivariable calculus that this hypothesisguarantees that the curve \320\232has a well-defined tangent line at eachpoint.

The fact that the curve \320\232is closed means that the above pa-rametrization can be chosen so that it wraps onto itself, namely, so

that there exists a T > 0 such that j(t + T) = j(t) for every i\302\243i.

Finally, the closed curve \320\232is simple if it does not cross itself,namely, if j(t) \316\246j(t') for every t, t' G [0.\316\223)with t \317\206t'.

Given two knots \320\232and K' in K3, a problem of interest to sailors

and mathematicians alike is to determine whether it is possible to

deform \320\232to K' in K3 in such a way that the knot never crossesitself throughout the deformation. To express this in a mathematical

way, we want to know if there is a family of homeomorphisms \317\210\316\257:

\320\2323\342\200\224*\320\2323,depending continuously on a parameter t e [0,1] such

that the following holds: at the beginning, when t = 0, \317\210\316\277is just the

identity map; at the end, when I = 1, \317\210\317\207sends the first knot \320\232to the

second one K'. To seehow this is a reasonable model for the physicalprocessthat we are trying to analyze, think about what happens to

each molecule of our world as one moves the string \320\232to the string

K'\\ for each value of the time parameter t, define y>t: K3 \342\200\224>K3 by the

property that the moleculethat was at the point \316\241at the beginning

is now at the point <pt(P) at time t.

When there is such a family of homeomorphisms<pt, we will saythat the knots \320\232and K' are isotopic. The family of

homeomorphisms</?t is an isotopy from \320\232to K'.

Knot theory is essentiallyconcernedwith the problem of

classifyingall possible knots up to isotopy. Ideally,one would like to have a

complete catalogue of knots so that every possible knot is isotopic toone and only one entry of the catalogue. Seethe classic[Ashley] for

12.1. Knots 317

a nonmathematical version of this catalogue; note that knotpractitioners also care about additional properties, such as friction, so that

isotopic knots appear severaltimes in that book, under configurationsthat have different physical characteristics.

The figure-eight knot of Chapter 11 would be one entry in the

catalogue. Another famous knot is the unknot, represented by a

round circle in the plane \320\2322\320\241\320\2323.See Exercise 12.5 for a proof that

the figure-eight knot and the unknot really represent distinct entriesin the catalogue, namely that, in agreement with our own experience

handling strings and ropes,it is not possible to unknot a string tiedas a figure-eight knot.

12.1.1. Isotopy versus isomorphism. When \320\232and K' are

isotopic, we can just focus on the final homeomorphism \317\210\317\207:\320\2323\342\200\224>\320\2323,

which sends \320\232to \320\232'.Note that we can extend \317\210\317\207to a

homeomorphismof K3 = K3 U {\321\201\321\216}by setting \317\210\\(\321\201\321\216)= \321\201\321\216.We will say that

two knots \320\232and K' are isomorphic if there is a homeomorphism

\317\206:\320\2323\342\200\224>\320\2323which sends \320\232to \320\232'(but we do not necessarily require\317\210to send oc to \321\201\321\216).

Clearly, isotopic knots are also isomorphic. However, the property

of being isomorphic appears much weaker than being isotopic, sincewe only have to worry about finding one homeomorphism, as opposedto a whole family. In fact, it is only half as weak, as we will now

explain.

A homeomorphism \317\206of K3 can be, either

orientation-preservingor orientation-reversing. These properties are difficult to

define for general homeomorphisms, but are easily explainedwhen \317\210

is differentiable at some point \316\241G M3. Recall that if the coordinatefunctions of \317\210are \317\206\316\271,\317\2062,<\302\243\320\267(so that '~p(Q)

= {ipi(Q),ip-2(Q), <\302\243\320\267(<2))G

M3), the jacobian of \317\210at \316\241is the determinant

Jacp(</?)

of the differential map \316\214\317\201\317\210:\316\2343\342\200\224>\316\2343.If, additionally, Jacp(</?) \321\204\320\236,

then \317\210is orientation-preserving if Jacp(tp) > 0 and \317\210is orientation-

&(P)

ifr(P)

&(P)

%?(P)

*ft(P)%?(P)

&(P)*&(P)&(P)

318 12. Geometrization theorems in dimension 3

reversing if Jacp(</?) < 0. Compare the 2-dimensional case that \342\226\240,;

we encountered in Section 2.5.2. Using the branch of mathemat- ;

ics called algebraic topology, it can be shown that this property isindependentof the point \316\241with Jacp(</?) \317\2060 chosen, and it can '

even be defined when no such \316\241exists. See for instance [Hatcher 1; .:Sect.2.2],[Massey, Chap. VIII, \302\2472],or any other textbook where the r\\

notion of degree of a continuous map is defined; a homeomorphism is <

orientation-preserving exactly when it has degree +1. \320\267

k

Suppose that we are given an isotopy i/?t: M3 \342\200\224>M3, 0 < \316\257< 1, ]

beginning with the identity map \317\210\316\277=

\320\230\321\206\320\267.The identity map is l-

clearly orientation-preserving. The same algebraic topology tech- jniquesused to define the orientation-preserving property show that jthe final homeomorphism \317\210\\is also orientation-preserving. As a con- j

sequence, if the two knots \320\232and K' are isotopic, they are alsoisomor- >

phic by an orientation-preserving homeomorphism. Namely, there ex-j

ists an orientation-preserving homeomorphism \317\210:\320\2503\342\200\224>\320\2323such that j

\317\210{\316\232)=\316\232>. \\

It turns out that by a deep result of topology, the converse is also )

true.j

Theorem 12.1. The knots \320\232and K'are isotopic if and only if they

are isomorphic by an orientation-preserving homeomorphism o/K3. j

D ':

We will not prove this result. The key to its traditional proof >.

is a relatively simple construction of J. W. Alexander [Alexander], ;

known as \"Alexander's trick\". However, this proof is somewhat unsat- \302\267

isfactory, because the isotopy that it provides is very badly behaved i

at at least one point. This would be considered very unrealistic bymost nonmathematicians, who would require all maps involved to bedifferentiable. Fortunately, the result still holds in the differentiable

context but its proof is much more difficult. This differentiable result :

is due to JeanCerf [Cerf], and was later extended by Allen Hatcher

[Hatcher i].

In spite of Theorem 12.1,there is a difference between isotopyand isomorphism. For instance, the two trefoil knots represented in

Figure 12.1 are isomorphic, since one can sendoneto the other by

a reflection across a plane. However, it can be shown that there is

12.2. Geometrizationof knot complements 319

no isotopy moving the left-handed trefoil knot to the right-handedtrefoil. This fact is mathematically quite subtle, and cannot be easily

proved within the framework of this book; see for instance [Rolfsen,

\302\2478.E.14].

Left-handed trefoil Right-handed trefoil

Figure 12.1. The two trefoil knots

Note that the jacobian of a reflection is equal to \342\200\2241at every

point, so that the reflectionexchangingthe two trefoils is orientation-

reversing.

12.2. The Geometrization Theorem for knotcomplements

This section is devoted to the statement ofThurston'sGeometrization

Theorem for knot complements. We first begin with a discussion ofthe two exceptions of this result, torus and satelliteknots.

12.2.1. Torusknots.Among all knots, torus knots are among thesimplestonesto describe.Given a rational number ^ G Q with p,

q G \316\226coprime, the R-torus knot is the knot \320\232formed by a curve

contained in the standard torus \316\244in M3, and which turns q timesaround the hole of the torus and \317\201times counterclockwise around its

core C. Moreprecisely, for R > r > 0, thicken the horizontal circle

\320\241of radius R centered at the origin to a tube of width 2r. Then,

on the torus \316\244bounding this tube, the 2-torus knot \320\232is the curve

parametrized by the map 7: \320\232\342\200\224>\316\234.3defined by

7(\302\243)=

((R + r cospi) cosqt, (R + r cos pt) sin qt, r sinpt).

320 12. Geometrizationtheoremsin dimension 3

Compare our parametrization of the torus in Section 5.1. It is

immediate that up to isotopy, \320\232is independent of the choice of the radiiR and r.

Figure12.2describesthe ^-torus knot as seen from above in \320\250.3.

Namely, it represents the projection of this knot to the xy-pl&ne.

Figure 12.2. The -jp-torus knot

We have already encountered the |- and ^-torus knots. Indeed,it is relatively immediate that they are respectively isotopicto theleft-handed and right-handed trefoil knots of Figure 12.1. Also, the

-j?j-torus knot is isotopic to the trivial knot, as is the \342\200\224-torus knot.

It can be shown that when \\p\\, \\q\\, \\p'\\, \\q'\\ are all greater than 1,the ^-torus knot is not isomorphic to the unknot, and it is isotopic

to the ^-torus knot if and only if ^ is equal to 2 or to 2. Seefor

instance [Rolfsen, \302\2477D.10] for a proof, and compare Exercise 12.1.

12.2.2. Satelliteknots.The satellite knot construction is a methodfor building complex knots from simpler ones.

Supposethat we are given two objects. The first one is a knot

\320\232in M3. The second one is a nontrivial knot L in the standard solidtorus V. This requiressomeexplanation.

The standard solidtorusV is the inside of the standard torusin \320\250.3.Namely, supposing two radii R and r with R > r > 0 aregiven, V consists of those points of the form

((R + pcosi;)cosii, (R + \317\201cos v) sin u, \317\201sin v)

12.2. Geometrization of knot complements 321

with 0 < \317\201< r and u, \316\275G R. In other words, V consists of those

points which are at distance at most r from the horizontal circle \320\241of

radius R centered at the origin. A nontrivial knot in V is a regularsimpleclosedcurve L in the interior of V such that

(1) L is not isotopic to the central circle\320\241in the solid torus V;

(2) L is not isotopic in V to a knot U which is disjoint from one

of the cross-sectiondiskswhere the coordinate \320\270is constant

(namely, L is really spreadaround V).

The companion \320\232 The solid torus V The satellite K'

Figure12.3.The construction of a satellite knot

Given a knot \320\232in R3 and a nontrivial knot L in the standard

solid torus V, we can then tie V around the knot K, and considerthe image of L. More precisely, choose an injective continuous map

\317\210:V \342\200\224*R3 which sends the central circle \320\241to the knot K. Assumein addition that \317\210is differentiable and that its jacobian is everywhere

different from 0, so that the image K' =<p(L) is now a new knot in

R3.

Any knot K' obtained in this way is said to be a satellite of the

knot K. Conversely, \320\232is a companion of K'.

Figure 12.3offers an example.

A knot admits a unique factorization into companion knots, which

ia analogous to the factorization of an integer as a product of prime

numbers. This factorization of knotswas initiated by Horst Schubert

[Schuberti, Schubert2] in the 1950s, generalized by Klaus Johann-\316\272\316\277\316\267[Johannson], Bus Jaco and Peter Shalen [Jaco & Shalen],and


finally improved by Larry Siebcnmann [Bonahon & Siebenmann,\302\2472]in the 1970s. This unique factorization essentially reducestheanalysis of knots to studying knots that are not satellites.1

12.2.3. The Geometrization Theorem for knot complements.We are now ready to state Thurston's Geometrization Theorem for

knot complements.

Theorem 12.2 (Geometrization Theorem for knot complements).Let \320\232be a knot in M3. Then exactly one of the following holds.

(1) \320\232is a \302\243-torus knot with q > 2;

(2) \320\232is a satellite of a nontrivial knot;

(3) the complement K3 \342\200\224\320\232admits a metric d which is:

(i) complete;

(ii) topologically equivalent to the euclideanmetric;(iii) locally isometric to the hyperbolic metric

dhyp of the

hyperbolic space H3.

A knot \320\232which satisfies conclusion (3) of Theorem 12.2is said

to be hyperbolic.

Theorem 12.2 was proved by Bill Thurston in the late 1970s. Itcompletelyrevolutionized the world of knot theory (and 3-dimensionaltopology),for reasons we hope to make apparent in the next sections.

This astounding breakthrough was one of the main reasons for which

Thurston was awarded the Fields Medal (the mathematical

equivalentof a Nobel Prize) in 1983. A key step in Thurston's proof was

subsequently simplified by Curt McMullen [McMullen]. This part

of McMullen's work was, again, one of the major contributions cited

when he himself receivedthe FieldsMedal in 1998.

In addition to the announcement [Thurstom], Thurston gave,numerous lectures detailing key stepsof the proof of his

GeometrizationTheorem. However, he never wrote a completeexpositionbeyond

several influential preprints which never made it to final publication

To be completely accurate, one needs to study nonsatellite links, consisting of

finitely many disjoint knots in R3. The analysis of links in K3 ia not much more complexthan that of knots.

12.2. Gcometrization of knot complements 323

(with the exception of [Thurstone]). Details are now available in

[Otal1? Otal2, Kapovich], for instance.

Although stating this as a precise mathematical statement is not

completely clear, it is an experimental fact that a knot randomly

drawn in R3 is neither a torus knot nor a satelliteknot. Therefore, in

\"most\" cases, the Geometrization Theorem 12.2provides a complete

hyperbolic metric on the knot complement X = R3 \342\200\224K, as in its

conclusion (3).Theorem 12.2is an abstract existence theorem, whose proof does

not provide a convenient method for finding the hyperbolic metric

whose existence is asserted.Theremarkable piece of software Snap-

Pea, written by Jeff Weeks [Weeksi] and improved over the years

with various collaborators, usually succeedsin describing such a

metric. Namely, given a knot \320\232in R3, SnapPea attempts to find a group

\320\223acting isometrically, freely and discontinuously on the hyperbolicspaceH3,such that the quotient space \320\2353/\320\223,endowed with the

quotient metric dhyp induced by the hyperbolic metricdhyp

of H3, is

homeomorphic to X = R3 \342\200\224\320\232by a homeomorphism \317\206:X \342\200\224\342\226\272\316\2273/\316\223.

When it succeeds, the metric d defined on X by the property that

d(P, Q) =JhyP(v3(-P)> <f(Q)) satisfies conclusion (3) of Theorem 12.2.

To find such a group \320\223,SnapPea uses a method very similarto the one that we employed for the figure-eight knot in Chapter 11.

N'amely, it decomposes X into finitely many topological ideal tetrahe-dra, and then tries to identify them with geometric ideal tetrahedra in

E3 in such a way that the corresponding gluing data satisfies the

hypotheses of Poincare's Polyhedron Theorem 10.9. This often requires

changing the original decomposition ofX into ideal tetrahedra, which

the program does by (educated)trial and error. See [Weeks4] for amore detaileddescription.

In practiceand as long as the knot \320\232is not too intricate (so thatone does not encounter issues of computational complexity),SnapPea always finds the hyperbolic metric on X = R3 \342\200\224\320\232that it is

looking for, unless there is no such complete hyperbolic metric

because the knot is a torus knot or a satelliteknot. Thereis a curious

mathematical situation here. The Geometrization Theorem 12.2hasbeen rigorously proved, but its abstract proof doesnot provide any


method for explicitly finding the hyperbolic metric whose existenceis usually asserted by this statement. Conversely, SnapPea almostalways finds this hyperbolic metric, but at this point in time thereis no rigorous proof that the algorithm used by SnapPea will always

succeed in doing so.

Since it first became available, SnapPea has been a wonderful

tool, widely used by researchers in topology. It is also an equallywonderful toy, and you are strongly encouraged to play with it and

explore its many features.

We conclude this section with a complement,which sharpens

Theorem 12.2.

Let (X,d) be a metric spacewhich is locally isometric to (H3, dhyp).

In particular, every point of X belongsto a ball Bd(P, r) which is

isometric to a ballBdhyp (f, r) in H3. We will say that (X, d) has finite

volume if there exists a sequence of such balls{Bd{Pn,Tn)}neN,each

isometric to a ball Bdhyp(Pn,fn) in H3, such that

(1) the set X is the union of all balls Bd(Pn, fn)\302\267Namely, every

\316\241& X belongs to some ball Bd{Pn^n)'i(2) if

volhyp(Bdhyp(Pn!rin))=

7rsinh2r-\342\200\236- 2nrn denotes the

hyperbolic volume of the ball Bdbyp{Pn, rn) in H3 (compare

Exercise 9.14), the seriesoo

$>olhyp(Bdhyp(P>n))n=l

converges.

This definition is somewhat ad hoc, but will suffice for our pur-.poses. SeeExercise12.6for a more precise definition of the volume

of (X, d).

Complement 12.3. Whenever Theorem 12.2 provides a completehyperbolicmetric on the complement M3 \342\200\224\320\232of a knot K, this

hyperbolic metric has finite volume unless \320\232is isotopic to the unknot.

12.3. Mostow's Rigidity Theorem

What makes the Geometrization Theorem so powerful is that when it

provides a complete metric locally isometric to the hyperbolicspace,

12.3.Mostow'sRigidity Theorem 325

this metric is actually unique. This is a consequenceof a

fundamentalresult which, about 10 years earlier, had been proved by George

Mostow [Mostow!] and later improved by Gopal Prasad [Prasad]

(see also [Mostow2,Benedetti & Petronio]for a proof).

Theorem 12.4 (Mostow's Rigidity Theorem). Let (X, d) and (X', d')be two complete metric spaces which are locally isometric \316\257\316\277\316\2273and

which have finite volume. If there exists a homeomorphism between

(X,d) and (X,d'), then there exists an isometry \317\210:(X,d) \342\200\224>(X',d').

\316\240

In the case that is currently of interest to us, where X = R3 \342\200\224\320\232

is the complement of a knot \320\232which is neither a torus knot nora satellite,the Geometrization Theorem 12.2 and Complement 12.3

provide a completefinite volume metric d which is locally isometricto the hyperbolicmetric of the hyperbolic space H3. We can then

rephrase Theorem 12.4 as follows.

Theorem 12.5. Let \320\232be a knot in R3 which is neither a torusknot nor a satellite knot, and let d and d' be two hyperbolic metrics

on the knot complement X = R3 \342\200\224\320\232as in conclusion (3) of theGeometrization Theorem 12.2. Namely, these two metrics are locally

isometric to the hyperbolic metric of \320\2503,are complete, have finitevolume, and are topologically equivalent to the euclidean metric. Thenthe metric spaces (X, d) and (X, d') are isometric.

Proof. The identity map Idx defines a homeomorphism between(X,d) and (X,d') because d and d' are both topologicallyequivalentto the euclidean metric. It then suffices to apply Theorem 12.4to thesetwo spaces. D

In other words, the hyperbolicmetric d provided by Theorem 12.2is unique up to isometry. As a consequence,given two hyperbolic

knots \320\232and K', if we succeed in proving that the correspondinghyperbolic metrics are not isometric, then we are guaranteed that thel.wo knots are not isomorphic. Here is a typical application.

Corollary12.6.Let \320\232and K' be two knots for which Theorem 12.2

provides complete hyperbolic metrics d and d' on X = R3 \342\200\224\320\232and


X' = R3 \342\200\224K', respectively. If (X, d) and (X,d') have different

hyperbolic volumes (as defined in Exercise 12.6), then the two knots are

not isomorphic. \320\236

In practice, this simple test is remarkably effective.

Mostow's Rigidity Theorem holds in a much more generalframework than the one stated in Theorem 12.5. In particular, for anydimension n, one can define an \321\202\320\265-dimensional hyperbolic space \320\250\320\277

by a somewhat straightforward extension of the formulas we used for

H2 and H3. Mostow's Rigidity Theorem 12.5 extends to alldimensions \316\267> 3 by systematically replacing H3 by H\342\204\242everywhere.

Surprisingly enough, the same statement is false for \316\267= 2.

Actually, we have already encountered this phenomenon in Section 8.4.2,when we associated to each triple of shear parameters s\\, S3, s$ G \320\232

with s\\ + S3 + S5= 0 a completehyperbolic metric on the once

punctured torus. The area of such a metric is twice the hyperbolic area of

an ideal triangle in H2, namely, it is equal to 2\317\200;in particular, it is

finite. It can be shown that if we slightly change the shearparameters

defining a hyperbolic metric d, the metric d' associatedto the new

shear parameters is not isometric to d.

12.4. Ford domains

A Ford domain is a variation of the Dirichlet domains that we

encountered in Section 7.4. We first introduce some preparatory material.

12.4.1. Hyperbolic metrics and kleiniangroups.In Section 7.2,

we saw that if \316\223acts by isometries, freely and discontinuouslyon the

hyperbolic space H3, then the quotient space\316\2273/\316\223is locally

isometricto H3. In addition, Exercise 7.6 shows that \320\2353/\320\223is complete, so

that \320\2353/\320\223is a complete hyperbolic 3-dimensional manifold.

Conversely, every connected complete hyperbolic 3-dimensionalmanifold can be obtained in this way, as a quotient of the hyperbolic

space H3 by a kleinian group \320\223acting freely. This is a general fact

but, in the case we are interestedin, we prefer to state this as acomplement to Theorem 12.2, which will enable us to include a few

properties that will be needed later on.

12.4. Ford domains 327

As in Exercise 9.10, an isometry of the hyperbolic space H3 is

parabolic if, by definition, it fixes exactly one point of the sphere at

infinity C. A point \316\276G \320\241is parabolic for a group \320\223of isometries of

H3 if \316\276is fixed by some parabolic 7 G \316\223.

Complement 12.7. Whenever Theorem 12.2 and Complement 12.3

provide a finite volume complete hyperbolic metric d on the

complementX = R3 \342\200\224\320\232of a knot K, there exists a group \320\223of isometries

ofM3 such that

(1) \320\223acts freely and discontinuously on \320\2503,and (X,d) is

isometric to the quotient space (\320\2353/\320\223,4\320\243\320\240);

(2) the set \316\223,\317\207,of those 7 G \316\223with 7(00)= \321\201\321\216is generated

by two horizontal translations 71 and 72 along two linearly

independent horizontal vectors \321\211and \321\211,respectively; in

particular, 00 is a parabolicpoint;

(3) every parabolic point of 7 is of the form 7(00) for some

(4) there existsa euclidean half-space

Boo = {(x, \320\243,\317\211)G H3; \320\270> u0},

defined by some constant \321\211> 0, which is disjoint from all

its images 7(600) with 7 e \320\223\342\200\224\320\223^.

For (3), observe that conversely, every 7(00)e \320\241with 7 G \320\223is

parabolic. In contrast to (4), note that (2) implies that 7(^00) = #oofor every 7\320\265\320\223\320\266.

Proposition 12.8. Under the hypotheses of Complement 12.7, let \320\223

and \320\223\"satisfy the conclusions of this statement. Then there exists an

isometry \317\210of (H3,dhyP) such that

\316\223\"\342\200\224

{^070 \317\206~\316\271;\316\212G \316\223}.

In addition, one can arrange that \317\210{\316\277\317\214)= \321\201\321\216.

Proof. This is a relatively simple consequenceof Mostow's Rigidity

Theorem. Indeed, the version given in Theorem 12.5 provides an

isometry \321\204:\320\2303/\320\223->

\320\2303/\320\223'between (\320\2303/\320\223,4\320\243\320\240)and

(\320\2303/\320\223',4\321\203\320\240)\302\267

Start with an arbitrary P0 G H3. Considerits image Q0 =\317\200(\316\2410)


under the quotient map \317\200:\316\2273\342\200\224>\316\2273/\316\223,the image Q'0

= <p(Qo) G

\316\2273/\316\223\"of this point Qo under the isometry -\317\206,and finally a point

Pq G \317\200'_1(<5\317\214)G H3 of the preimage of Q'0 under the quotient mapn': \320\2303-> \320\2303/\320\223',namely, such that, \321\202\320\263'(\320\240^)

=Q'0. The following

diagram may be useful to keep track of all these maps and points.

P0 G \320\2303 \316\233\316\241,;e H3

-I F

Qo G \320\2303/\320\223\320\233

Q'o G \320\2353/\320\223'

By the proof of Theorem 7.8 and Corollary 7.9,the quotient map

\317\200is a local isometry. As a consequence,for \316\265> 0 sufficiently small,the restriction

\321\211\320\262\320\233\321\212\321\203\321\200{\320\240\321\214^)'\302\267\316\222\316\254^\317\201(\316\241\316\277,\316\265)->

Bahyp(Qo,e)is an

isometry; in particular, it is bijective. Similarly, we can assume that

\"tahypCO.*)1^\321\214\321\203\321\200\320\241-\320\240\320\276'.\320\265)

->#dhyp(<2c\302\273e)

is also an isometry, by

taking \316\265small enough. Consider the composition

\317\206=

(\"\"^\342\200\236(pg,\302\273))-10^ \302\260

\316\267\316\262^\317\201(\316\241\316\277,\316\265):^\316\227\316\252\316\241(^\316\277,\316\265)-\342\226\272

Bdhyp{P^e).

This map \317\206:Bdhyv,(Po,z)\342\200\224>

\320\222\320\262,\320\252\321\203\321\200{\320\240\320\276,\302\243)is an isometry between

two hyperbolic balls of H3. Therefore, it extends to an isometry

\317\206:\316\2273-> \316\2273by Lemma 9.9.

Consider an arbitrary 7 G \316\223.By definition of the quotient map\317\200:\316\2273\342\200\224>\316\2273/\316\223,\317\200\320\2767

= \321\202\321\202.Since \317\200'\320\276<\321\200(\320\240)

=\317\210

\320\276\321\202\320\263(\320\240)for every

\316\241G5dhyp(-Poj\302\243) by construction of </?, we conclude that

\317\200'\316\277\317\206

\316\2777(-\316\241)

=\321\204on \320\276

7(\320\240)=

\317\210\316\277

\317\200(\316\241)= -\316\271\317\204'\316\277

\316\257/?(\316\241)

for every P GBdhyp(Po,\302\243).

In the particular case where \316\241= Pq,

this implies that \317\206\316\277

7(\316\241\316\277)and \321\203>(\320\233\320\267)\320\266\320\265glued together when one

takes the quotient under the action of \320\223\",namely, that there existsan element 7' G \320\223\"such that 7' \316\277

\317\206(\316\241\316\277)\342\200\224

\317\210\316\277j(Po).

Now, for every \316\241GBdhyp(P0,\316\265),

\342\226\240\320\272'\320\276(\342\226\240y')~1oipo 7(P)

= -\316\272'\316\277\317\206\316\2777(\316\241)= ir'o <p(P).

Note that (7')_1 \302\260\317\206\302\260j(Po)

=\317\210(\316\241\316\277)

=^0 by construction of 7' so

that (\320\243)-1\320\276\317\210\316\277

7(\316\241) and <p(P) are both in the ballBdhyp(PO,\302\243).

Since the restriction of \320\266'to this ball is injective, we concludethat

{\316\212')-\316\271\316\277\317\210\316\277\316\212{\316\241)=

\317\210{\316\241).


This proves that the two isometries {^')~\316\273\316\277\317\210\316\277\316\267and \317\210of (H3, dhyp)coincide on the ball

Bdhyp(Po,^)\302\267 Consequently, they coincide

everywhere on H3 by the uniqueness statement in Lemma 9.9. Composing

both of them with 7' on one side and \317\206-1on the other side, weconclude that \317\206

\316\277j \316\277\317\206_1

= 7' g \316\223'.

This proves that {\317\210\320\276\321\203\316\277\317\206-1;\316\267G \316\223}is contained in \316\223'.

Conversely, given 7' G \316\223\",the same argument but replacing \317\206

by \317\206~\316\271provides an element 7 G \316\223such that \317\210-1\316\2777' \316\277\317\210=

\321\203or,

equivalently, 7' =\316\271\317\201\316\277\316\263\316\277\317\201\"1.This proves that \316\223'is contained in

{\317\210\316\2777 \316\277

\317\206~\316\273;7 G \316\223}.Since we just proved the reverse inclusion,thesetwo sets are equal.

We need to prove the last statement of Proposition 12.8, namely

that we can choose the isometry \317\210so that f^C00)= \302\260\302\260\302\267

For the isometry \317\210that we have constructed so far, we only know

that \317\210\316\233{\316\277\317\214)is a parabolic point of V. Indeed, if 7' is a parabolic

element of \316\223\"fixing oc, then \317\210~\316\273\320\276\321\203'

\316\277\317\206is a parabolic element of \316\223

which fixes \317\206\"1^).

Therefore, since \316\223satisfies conclusion (3) of Complement 12.7,there exists an element 70 G \316\223such that 70(00) =

(/'\"'(\321\201\321\216).Set

\317\206'=

\317\206\316\277jQ. Note that \317\206'is an isometry of H3 and sends\321\201\321\216to 00. In

addition

{\302\245>'o7o(\302\245>')-1;7G \320\223}=

{\321\203?\320\27670 \320\2767 \316\277

7\316\277-1\316\277\317\206~1;\316\212\302\243\316\223}

\342\200\224{\317\210

\316\277\316\261\316\277\317\206~\316\271;\316\261G \316\223}\342\200\224

\316\223'

as required. (To justify the second equality, note that Q = 70 \302\2607 \302\2607\316\277-1

belongs to the group \316\223for every 7 G \316\223and that conversely, everyq G \316\223can be written as q =

70 \320\2767 \320\276y^1 for some 7 G \320\223.) D

12.4.2. Ford domains. As indicated before, Ford domains are avariation of the Dirichlet domains of Section 7.4. For this reason, it

may be useful to first repeat the definition of Dirichlet domains. Let\320\223be a group of isometries of the metric space(X,d) whose action is

discontinuous. The Dirichlet domain of \320\223at \316\241G X is the subset

\316\224\316\223(\316\241)= {QeX; d(P, Q) sj \316\254(\316\212(\316\241),Q) for every 7 \342\202\254\316\223},


consisting of those points Q G X which are at least as close to \316\241as

to any other point of its orbit \316\223(\316\241).

In Section 7.4, we considered the casewhere X is the hyperbolic

plane H2 (as well as the euclidean plane, but this is irrelevant here).In particular, Theorem 7.13 showed that in this case the Dirichletdomain \316\224\317\201(\316\241)is a locally finite polygon in H2 and that, as 7 ranges

over all elements of 7, the polygons 7(\316\224\316\223(\316\241))form a tessellation of

H2. In addition, the two tiles7(\316\224\316\223(\316\241))

and \316\224\316\223(\316\241)are distinct if

and only if 7(P) \317\206P.

The natural generalization of these results holds when X is the

hyperbolic space H3. Namely, \316\224\316\223(\316\241)is then a locally finite

polyhedron in H3. To prove this, one just needsto replaceLemma 7.14 by

the statement that for any P, Q G H3, the set of points R that are at

the same hyperbolic distance from \316\241and Q is a hyperbolic plane \316\240,

while the set of R with d^yp{P, R) < dhyp(Q, R) is a hyperbolichalf-

space \316\227delimited by this perpendicular bisector plane \316\240.The proof

of Theorem 7.13 then immediately generalizes to show that \316\224\317\201(\316\241)

is a locally finite polyhedron and that the polyhedra 7(\316\224\316\263(\316\241))with

7 G \316\223form a tessellation of H3. Again, 7(\316\224\316\263(\316\241))is distinct from

\316\224\316\223(\316\241)unless 7(P) \317\206P.

For a hyperbolic knot \320\232in R3, let \320\223be a kleinian group satisfyingthe conclusionsof Complement 12.7. In the definition of the Dirichletdomain \316\224\316\263(\316\241),we now replace the point \316\241G H3 by the point 00.

Namely, we want to consider the set \316\224\316\263(\316\277\316\277)of those points Q G H3which are at least as close to 00 as to any other 7(00) with 7 G \316\223.

There is of course a problem here, becausea point Q G H3 is atinfinite hyperbolic distance from 00, and from any 7(00)! However,

some infinities are larger than others, and it turns out that we can

really make senseof this statement.

For this, let us pursue the sameformal reasoning. Since 7 G \316\223is a

hyperbolic isometry, the distance from Q to 7(00) should be the sameas the distancefrom j~1(Q) to 00. Therefore, if we neglect the fact

that we are talking about infinite distances, the statement that Q iscloserto 00than to 7(00) should be equivalent to the statement that

00 is closer to Q than to 7~1(Q). If we remember our discussion of

the Busemann function in Section 6.8 (see also Exercises6.10-6.11),

12.4. Forddomains 331

we can now make sense of this property. Indeed,if Qi= (xi, j/i, Ui)

and Q2 =(z2> i/2; U2) are two points of H3,

lim dhyp(P, Qi)- dhyp(P, Q2) = log\342\200\224,

P\342\200\224>oo U2

where the limit is taken for the euclidean metric in R3 = R3 U {00}.

As a consequence, we can decidethat Q\\= (xi,yi,u-[) is closer

to 00 than Q2 = (x2, i/2,1^2) if \317\211\316\271> ^2- To avoid any ambiguity, wewill use the euclidean terminology and say that in this case, Q\\ is

higher than Q2. Similarly, Qi is at least as high as Q2 if \320\270\320\263^ \320\270\320\263-

This leads us to define the Ford domain of the kleinian group\320\223\320\260\320\267

\316\224\316\223(\316\277\316\277)= {Q e H3; Q is at leastas high as -y~1(Q) for every 7 G \316\223}.

We will show that \316\224\316\263(\316\277\316\277)is a locally finite hyperbolicpolyhedron.

Lemma 12.9. Let 7 be an isometry of \320\2503such that 7(00) \317\206\317\203\316\277.

Then the set of those Qei3 which are at least as high as 7\"\316\273(Q) is

a hyperbolic half-space \316\227\316\212,bounded by a hyperbolic plane 1I7.

Proof. For \320\270> 0, let Su be the horizontal plane passing through the

point (0,0, \317\211).In hyperbolic terms, S(u) is a horosphere centeredat00. By Proposition 9.11, its image j(Su) is therefore a horospherecenteredat the point 7(00) e C, namely, a euclidean spheretouching

\320\241at that point.

<7\"1(Q) p s-\320\237) \320\233\320\260\320\264

/f'Q

\316\240\317\215/ \\y(Su0)

\320\236 7(\320\276\320\276)

Figure 12.4. The proof of Lemma12.9

When \320\270is near +00, the euclidean radius of j(Su) is very smalland the horospherej(Su) is disjoint from Su. On the other hand,


when \320\270is very close to 0, the euclideanradius of j(Su) is very largeand Su is near the rcj/-plane, so that the two horospheres y(Su) and Sumeeteachother. (To check these two facts, note that y(Su) containsthe point 7(0,0, \317\211),which is on the hyperbolic geodesic g joining7(0,0,0) to 7(00).)Therefore, there exists a value Uq > 0 for which

SUa and 7(<SUo) are exactly tangent to each other, at a point Pq.Let

\316\2407be the hyperbolic plane that passes through P0 and isorthogonal to the geodesic joining 7(00) to 00. Namely, \316\2407is the

intersection with H3 of the euclideanspherecenteredat 7(00) and

passing through Pq.

Let \317\201be the hyperbolic reflection across\316\2407,

which is also the

restriction to H3 of the inversion across the euclidean sphere centeredat 7(00)and containing Pq. In particular, \317\201exchanges 00 and 7(00),and fixesthe point Pq. It follows that \317\201also exchanges the horospheres

SUo and y(SUo). In particular, the hyperbolic isometry \316\261=\317\201

\316\2777

sends 00 to 00, and respectsthe horosphereSUo. As a consequence,a respects every horospherecenteredat 00, and it sends each pointof H3 to a point which is at the same height.

If Q is a point of \320\2377,then y~i(Q) = \316\261-1\316\277p~1(Q)

= a_1(Q) is

at the sameheight as Q by the above observation on a.If Q is a point of the open hyperbolic half-space delimited by \320\2377

and adjacent to 00, namely, if Q e H3 is located outside of the closed

euclidean ball centered at 7(00) and containing Pq in its boundary,then p(Q) is lower than Q by elementary geometry. Noting that

p~l = p, the same propertyof a shows that y~1(Q) = \316\261-1\316\277p~1(Q)

is lower than Q. In other words,Q is higher than y~1(Q).

Similarly, if Q is in the other open hyperbolic half-space delimitedby \320\2377,the same argument shows that Q is lower than y~1(Q).

This completes the proof of Lemma 12.9. \316\240

By definition, the Ford domain \316\224\316\223(\316\277\316\277)is equal to the intersectionof all thesehalf-spaces ff7 as 7 ranges over all elementsof \316\223such that

7(00) \317\206oo. Namely,

\316\224\316\223(\316\277\316\277)=f| tf77er-r00


using in the notation the stabilizer \320\223= {7 G \320\223;7(00)

= 00} of \321\201\321\216.

To show that this is a locally finite polyhedron, we need the following

fact.

Lemma 12.10. The family of hyperbolic planes {\320\2377;7G \320\223\342\200\224

\320\223^}

defined by Lemma 12.9 is locallyfinite.

Proof. We need to show that for every Pel3, thereexistsan \316\265> 0

such that the ballBdhyp(P,\316\265)

meets only finitely many \316\2407.Actually,

any \316\265> 0 will do.

Note that we are not saying that there are only finitely many

7 G \316\223such that\316\2407

meets Bdhyp (\316\241,\316\265).Indeed, if \316\261is an element of

the stabilizer \316\223,(namely, if \316\261G \316\223and \316\261(\316\277\316\277)=

\321\201\321\216),conclusion (2) of

Complement 12.7 shows that \316\261sends every point of H3 to a point atthe sameheight.It then follows from our definitions that

\320\23770(1=

\316\2407.

In particular, there are infinitely many 7 G \316\223with the same associated

plane IT7.After these preliminary comments, let us begin the proof of

Lemma 12.10. Suppose that the ballBdhyp(P,\316\265)

meets infinitely

many distinct\316\2407\316\267,with \316\267G N. For each n, pick a point Qn G

Bdhyp(P,e)nUyri.

By Complement 12.7, the stabilizer \316\223of 00 is generated by two

horizontal translations, with respective translation vectors \321\211and \321\211.

In particular, an arbitrary parallelogram 7 in the plane with sides

respectively parallel to v\\ and V2 is a fundamental domain for the

action of \316\223\316\267on the horizontal plane R2. As a consequence, there

exists an \316\225\320\242\320\266such that the first two coordinates (xn, \320\243\320\277)of a~x \320\276

\316\212\316\2671(Qn)= (xn, \320\243\320\277,\320\270\320\277)are located in \320\243.

Note that the last coordinate un of \316\261\"1\316\277j~1(Qn) is also the

last coordinate of 7f^1(Qn)i which is equal t,o the last coordinate

of Qn since Qn G\320\2377\342\200\236.

As a consequence, since dhyp(-P, Qn) < \316\265,

the usual comparison between euclidean and hyperbolicmetrics (see

Lemma 2.5) shows that ue~\302\243< un < \317\211\316\262\316\265,where \320\270is the third

coordinate of \316\241=(\317\207,\321\203,\320\270).

It follows that all points a\"1 \302\260~/n1(Qn) are in the parallelepipedicbox 7 \317\207[ue~e,uee] \321\201\316\2273\320\241R3. By compactness, or by another

explicit comparisonbetween hyperbolic and euclidean metrics, there


is a large radius R such that this box \320\243\321\205[ue \316\265,\317\211\316\262\316\265]is contained in

the ball Bdhyp(P,R). Then,

dhyp{P,\316\212\316\267\320\276

\320\260\320\277{\320\240))< dhyp(P, Qn) + dhyp(Qn,7\302\253\302\260

\302\253\302\253(-P))

< dbyp(P,Qn) + \302\253W^1 \302\260In^Qn),P)

\316\266\316\265+ R.

Since the action of \320\223is discontinuous, Lemma 7.15 shows thatthe ball

Bdhyp (\316\241,\316\265+ R) contains only finitely many points of the

orbit. In particular, there existstwo indices \316\267\317\207\317\206\317\204\317\205\317\207such that ynL \320\276

am (\316\241)=

7\317\2002\302\260an2 (\316\241)\302\267Additionally, because the action of \316\223is free,

7\316\267\316\271\316\277

\316\261\316\267\316\271=

-\316\263\342\200\2362\316\277

\316\261\316\267!.However, by the observation that we made atthe beginning of the proof,

contradicting our originalassumptionthat the hyperbolic planes \316\2407\316\267

are all distinct. D

Proposition 12.11. The Ford domain \316\224\316\263(\316\277\316\277)is a locally finitepolyhedron in the hyperbolic space \320\2503.

Proof. We already saw that by definition, \316\224\316\263(\316\277\316\277)is equal to the

intersection of all the half-spaces\316\227\316\212,defined in Lemma 12.9, as 7

rangesover all elements of \316\223\342\200\224\316\223^.

By the local finiteness result of Lemma 12.10, every \316\241e \316\224\316\263(\316\277\316\277)

is the center of a ballBdhyp(P,\316\265)

which meets only finitely many ofthe hyperbolicplanes \316\2407bounding the

\316\227\316\212.Let us write these planes

as\316\24071,\316\24072,..., \316\2407\316\267.

As a consequence, the intersection of the ball

Bdhyp(P,\316\265)with \316\224\316\263(\316\277\316\277)is also the intersection of

Bdltyp(P,\316\265)with

\316\24071\316\240\302\267\302\267\302\267\316\240

\316\2407\316\267.It follows that the boundary of \316\224\316\263(\316\277\316\277)is the union

of all sets \320\2377\320\237\316\224\316\223(\316\277\316\277)with 7 G \316\223-

\316\223^.

By construction, the set F7 =\316\2407

\316\240\316\224\316\263(\316\277\316\277)is the intersection of

the hyperbolic plane \320\223\320\246with all half-spaces \320\235\321\203

with 7' \302\243\320\223\342\200\224\316\223\316\277\316\277\302\267

Since the family of the planes \320\237\321\203bounding these half-spaces is locallyfinite (Lemma 12.10),F7 is either the empty set, a single point, a

geodesicof H3, or a locally finite polygon in H3.

When F7is a polygon, we will say that it is a face of \320\224\320\263(\320\276\321\201).

Prom the local picture of the Ford domain \316\224\316\263(\316\277\316\277)near each of its


points P, it is immediate that two such faces can only meet alongoneedgeor onevertex.

This concludes the proof that the Ford domain \316\224\316\263(\316\277\316\277)is a locally

finite polyhedron. D

Remark 12.12. Here,let us collect a few simple properties of the

Ford domain \316\224\316\223(\316\277\316\277).

(1) The intersection of \316\224\316\263(\316\277\316\277)with a vertical line is a half-line. Indeed, if \316\241= (\317\207,\321\203,\320\270)is in the half-space \316\227\316\212

of

Lemma 12.9, every point P' =(\320\266,\321\203,\320\270')above it (namely,

with u' > u) is alsoin #7.(2) For the vertical projection of R3 to the rcy-plane, every face

of \316\224\316\223(\316\277\316\277)projects to a euclidean polygon in R2.

(3) The Ford domain \316\224\316\223(\316\277\316\277)is invariant under the stabilizer

Too. Indeed, if \316\241e\316\227\316\212

and \316\261G Too, a simple algebraicmanipulation shows that a(P) belongs to Haay.

(4) A consequence of (2) and (3) above is that the faces of

\316\224\317\201(\321\201\321\216)are polygons with finitely many edges.

Figures 12.6and 12.8 below provide a few examples of the topview of Ford domains, namely, of their projectionsto the xy-plane.

Note the symmetry of these pictures under the action of the stabilizer

Too. This stabilizer acts by translations on the plane,and its action is

represented here by indicating a parallelogramwhich is a fundamental

domain for the action; in particular, \320\223 is generated by the two

translations along the sidesof the parallelogram.

These pictures were drawn using the software SnapPea [\\Veeks2] \342\226\240

Indeed, once we are given the kleinian group \320\223,the definition of the

Ford domain \316\224\317\201(\316\277\316\277)makes its computation quite amenable tocomputer implementation.

These examples are fairly typical, as projections of Ford domainsprovide (often quite intricate) tessellations of the euclidean planeR2

by euclidean polygons (usually triangles), which are invariant under

the action of a group acting by translations.


12.4.3.Uniqueness and examples.

Proposition 12.13. Let \320\232be a knot in R3, to which Theorem 12.2,

Complement 12.7 and Section 12.4.2associatea kleinian group \320\223and

a Ford domain \316\224\316\223(\316\277\316\277).If \316\223'and \316\224\316\263'(\316\277\316\277)are similarly associated

to K, then there is an isometry \317\210:\316\2273\342\200\224>\316\2273fixing oo and sending

\316\224\316\223(\316\277\316\277)to \316\224\316\223'(\316\277\316\277).

Proof. This is an immediate consequenceof Proposition 12.8. D

To apply Proposition 12.13,note that an isometry \317\210of H3 that

fixes oo is just the composition of a homothety and of a euclideanisometry of R3 respecting H3. (Look at the linear or antilinear

fractional map induced on C.) In addition, the possibilitiesfor \317\210are

further limited by the fact that Proposition 12.8 implies that it mustsend Too to \316\223^,in the sense that \316\223'^

={\317\206

\320\276\321\203\320\276\317\206-1;j e \320\223^}.

Figure 12.5. Two very similar knots

Because of the wealth of information encoded in them,Proposition12.13 makes Ford domains a very powerful tool to prove that two

knots are not isomorphic. For instance, considerthe two knots

represented in Figure 12.5. Their respective Ford domains are representedin Figure12.6.By inspection, there is no similitude (= compositionof

a homothety with a euclidean isometry) of the plane sending the

projection of one of these Ford domains to the other. For instance, some

vertices of the second tessellationare adjacent to 20 edges, whereas

none have this property in the first tessellation. One can also look atthe (finitely many) shapes of the tiles of each tessellation.

12.4. Forddomains 337

Figure 12.6. The Ford domains of the knots of Figure 12.5

It follows that the knots \320\232and K' of Figure 12.5 are not

isomorphic. These two knots were not completely chosen at random. The

fact that they differ only by an exchange of two twisted parts (thetechnical term is \"mutation\") make them very hard to distinguishby the methodsof algebraic topology which were standard prior tothe introduction of hyperbolic geometry. This example illustrates thepower of these new hyperbolic techniques.

As an aside, these two knots are still very close from the point

of view of hyperbolic geometry. For instance, the hyperbolic metricson their complements have exactly the same volume. It should alsobe clear from the parallelograms in each picture of Figure12.6(and

it can be rigorously proved) that the stabilizers\316\223\316\214\316\277and Tx coincide

up to isometry of K3. Consequently, one needs the full force of Ford

domains to distinguish these two knots.

As a final example, consider the two knots of Figure 12.7.Snap-Peaprovides the same Ford domain for each of them, as indicated in

Figure 12.8. In particular, we cannot use Ford domains to show that

these two knots are not isomorphic.

There is a good reason for this, because the two knots represented\321\217\320\263\320\265actually isotopic...! It took a long time to realize this fact. Thesel,wo knots were listed as distinct in the early knot tables established

by Charles Little [Little] in the nineteenth century, and remained so

lor over 90 years;seefor instance [Rolfsen], where they appear as

338 12. Geometrizationtheoremsin dimension 3

Figure 12.7. Two more knots

Figure 12.8. The Ford domain of the knots of Figure 12.7

the knots numbered 10i6iand IO162, respectively. It is only in 1974

that Kenneth Perko, an amateur mathematician, noticed that theywere isotopic [Perko]. SeeExercise12.8.

We are mentioning this example because it illustrates very well

the power of hyperbolic geometric methodsin knot theory. Indeed,

the isotopy which took 90 years to discover is now a simple

consequence of the computation of Ford domains, provided we use a little

more information than we have indicatedso far.

More precisely, the faces of a Ford domain \316\224\317\201(\316\277\316\277)are paired

together by the same principle that we already encountered for Dirich-

let domains. Namely, using the notation of the proof of

Proposition12.11, if f7 is a face of the Ford domain \316\224\316\263(\316\277\316\277),then F -1 is

another face of \316\224\316\263(\316\277\316\277)and 7 G \316\223sends F7-i to F7. In particular,the faces F7 and F7-1 are glued together under the quotient mapH3 ->

\316\2273/\316\223.


As in the case of Dirichlet domains, we can therefore reconstructthe quotient space \320\2353/\320\223from the following data: the Ford domain

\316\224\316\263(\316\277\316\277);the action of the translation group \316\223\316\214\316\277on \316\224\316\263(\316\277\316\277);the gluing

data between the faces \316\224\316\263(\316\277\316\277).Indeed, H3/7 is homeomorphic to thequotient space obtained by, first, taking the quotient \316\224\316\263(\316\277\316\277)/\316\223\316\277\316\277and

then gluing the faces of \316\224\316\223(\316\277\316\277)/\316\223\316\277\316\277(namely, the images of the facesof \316\224\317\201(\316\277\316\277)under the quotient map) according to the gluing data.

If we are only interested in reconstructing \316\2273/\316\223up to homeomor-

phism as opposed to up to isometry, we do not need to know exactly

the gluing maps F7 \342\200\224*F7-i, but only their combinatorics.

More precisely,let \320\223and \320\223\"be two kleinian groups as in

Complement 12.7, so that they have well-definedFord domains\316\2247(\316\277\316\277)

and \316\224\316\263'(\316\277\316\277).We will say that the Ford domains \316\224\316\263(\316\277\316\277)and \316\224\316\263'(\316\277\316\277)

and their gluing data have the same combinatoricsif there is a

one-to-one correspondence between the faces, edgesand vertices of

\316\224\316\263(\316\277\316\277)and the faces, edges and vertices of \316\224\316\271-'(\321\201\321\216)such that

(1) an edge of \316\224\317\201(\316\277\316\277)is contained in a given face if and only if

the corresponding edge of \316\224\316\263-'(\316\277\316\277)is contained in the

corresponding face; similarly, a vertex of \316\224\317\201(\316\277\316\277)is contained

in a given edge if and only if the corresponding vertex of

\316\224\317\201'(\321\201\320\276)is contained in the corresponding edge of \316\224\317\201\302\273(\321\201\320\276);

(2) two faces, edges or vertices of \316\224\317\201(\316\277\316\277)differ by the action

of an element of \316\223\316\214\316\277if and only if the corresponding faces,edgesor vertices of \316\224\317\201'(\321\201\320\276)differ by the action of an elementofrj\302\273;

(3) two faces, edges or vertices of \316\224\317\201(\316\277\316\277)are glued togetherif and only if the corresponding faces, edges or vertices of

\316\224\316\223/(\316\277\316\277)are glued together.

SnapPea easily computes the combinatorial data thus associated

to the Ford domain of a hyperbolic knot.

Proposition 12.14. Let \320\232and K' be two knots, and let \320\223and \320\223'be

two kleinian groups such that \320\2353/\320\223and \320\2353/\320\223'are homeomorphic to

R3 \342\200\224\320\232and R3 \342\200\224\316\232',respectively, as in Complement 12.7. Suppose

that the Ford domains \316\224\317\201(\316\277\316\277)and \316\224\317\201'(\316\277\316\277)and their gluing data are

combinatorially equivalent.


Then, the knots \320\232and K' are isomorphic.

Proof. The definition of combinatorial equivalence is designed sothat when it holds the quotient spaces \320\2353/\320\223and \320\2353/\320\223\"are home-

omorphic. Therefore, there exists a homeomorphism \317\206:\320\2313\342\200\224\320\232\342\200\224*

R3 \342\200\224\320\232'between the corresponding knot complements. Then, a deeptheoremof Cameron Gordon and John Luecke [Gordon & Luecke]shows that \317\206can be chosen so that it extends to a homeomorphism

\317\206:S3 -> S3, such that \317\210(\316\232)= \320\232'.This proves that \320\232and K' are

isomorphic. D

Ford domains were originally introduced, in the context of thehyperbolic plane ~H?, by Lester Ford in 1935 [Fordi]. BobRiley was

probably the first one to realize that 3-dimensional Ford domains

could be a powerful tool to distinguish knots, and one which was very

amenable to computer implementation [Riley2,RUeys, Rileyi]. A

conceptually different description of Ford domains is due to David

Epstein and Bob Penner [Epstein & Penner].

12.5. The general GeometrizationTheoremThe Geometrization Theorem 12.2 for knot complements is a specialcaseof a more general result for 3-dimensional manifolds. To explainthis statement, we need a few definitions about manifolds.

Consider the \316\267-dimensional euclidean space M\342\204\242,consisting

of all \316\267-tuples (xi,x2, \342\226\240\342\226\240\342\226\240,\317\207\316\267)with each xt e R. We endow Kn with

the euclidean metric dcuc defined by

deuc(P, \320\237=

\\/\316\243?=\316\220(*<-*\316\257)2

when \316\241=(\317\207\316\271,\317\2072,\342\226\240\342\226\240\342\226\240,\317\207\316\267)and \316\241'=

(\317\207\\,\317\207'2,\342\226\240\342\226\240\342\226\240,\317\207'\316\267)-This is of course

the natural generalizationof the cases where \316\267= 2 and \316\267= 3 that we

have consideredso far. When \317\201< \316\267,we identify the space Mp to the

subset of IK\342\204\242consisting of those points whose last (n \342\200\224p) coordinates

are all equal to 0.An n-dimensional manifold is a metric space (X,d) which is

locally homeomorphic to the euclidean space (\320\234\",\321\201?\320\265\320\270\321\201)\302\267Namely, for

every \316\241G X, there exists a small ball Bd(P,e) \320\241X and a

homeomorphism \317\210:Bd(P, \316\265)\342\200\224*U between Bd(P, \316\265)and a subset U of M\342\204\242

12.5. The general Geometrization Theorem 341

which contains a small euclidean ball\316\222\316\2541\316\271\316\263\317\201(\317\210(\316\241),\316\265')\321\201U centered

at \317\206(\316\241).

A manifold X is connected if, for every \316\241and Q G X, thereexists a continuous curve 7: [a, b] \342\200\224>X going from \316\241= 7(a) to

Q =\320\263\321\203(\320\252)(where [a, b] is an arbitrary closed interval in R).

Manifolds naturally appear as in the mathematical modelling of

various physical phenomena and, consequently, they occur in manydifferent branches of mathematics.

12.5.1. Geometrizationofsurfaces.A surface is the same thing

as a 2-dimensional manifold. In Chapter 5, we constructedvarious

euclidean, hyperbolic and spherical surfaces, namely, metric spaceslocally isometric to the euclidean plane (\320\2502,deuc), the hyperbolic plane

(H2,dhyp) or the sphere(S2,deph)\302\267

This is part of a more general phenomenon. The following

theorem is the culmination of work by a long line of mathematicians in the

nineteenth century; see for instance [Bonahon, \302\2471.1]for a discussion.

Theorem 12.15 (GeometrizationTheorem for surfaces). Let (X,d)be a connectedsurface. Then X can be endowed with another metricd' such that

(1) d' is topologically equivalent to the original metric d, in the

sense that the identity map \316\231\316\254\317\207provides a homeomorphism

between (X,d) and (X,d');

(2) the metric space (X, a\") is complete;

(3) (X,a\") is locally isometric to the euclidean plane (M.2,deac),the hyperbolic plane (EC2,dhyp) or the sphere (S2,deph)\302\267

In addition:

(1) when (X,d') is locally isometric to the sphere, it is homeo-morphic to the sphere or to the projective plane;

(2) when (X, d!) is locally isometric to the euclidean plane, it ishomeomorphic to the plane, the cylinder, the Mobius strip,

the torus, or the Klein bottle;


(3) when (X,d') is locally isometric to the hyperbolic plane, it

is not homeomorphic to the sphere, the projective plane, the

torus, or the Klein bottle. D

In particular, outside of four exceptions, every connected surface

can be endowedwith a complete hyperbolic metric. This includes therelatively nice surfaces that we encountered in Chapter 5, but alsowilder examples such as the surface of infinite genus represented in

Figure 12.9 (also known as the \"infinite Loch Ness Monster\,")or thecomplement, of a Cantor set in the plane (if you know what this is).

Figure 12.9. The infinite Loch Ness Monster

As indicated earlier, Mostow'sRigidity Theorem 12.4 is not validin dimension 2, so that this hyperbolic metric is not unique up to

isometry. However, there are still relatively few such metrics. Indeed,

for a given surface (X, d), we can consider the set of all complete eu-

clidean, hyperbolic or sphericalmetrics d' as in Theorem 12.15, andidentify two such metrics when they are isometric. There is a way

to turn the corresponding quotient space into a metric space2 called

the moduli space MeUcPO, 3Vthyp(^) or Msph(X), of euclidean,hyperbolic or spherical metrics on the surface X. When X is compact,

or more generally when it is obtained from a compact surface by

removing finitely many points, this moduli space is almost a manifold

in the sense that it is locally isometric to the quotient of a manifold

by the action of a finite group of isometries; such an object is calledan orbifold. In particular, it is locally a manifold at most of its

points, but can have singularities similar to the cone singularities ofExercise 7.14at those points that are fixed by nontrivial elements ofthe finite group.

For instance, when X is homeomorphicto the sphereor to theprojective plane, the moduli space Msph(^0 consists of a single point

'with several classical choices possiblefor the metric


(and consequently it is a manifold of dimension 0), which is another

way to say that any two spherical metrics on X areisometric.For euclidean metrics, the moduli space MeUc(^0is a 3-dimen-

sional orbifold when X is homeomorphic to the torus or the Kleinbottle, and is a 1-dimensional orbifold (homeomorphic to a semi-openinterval) when X is a cylinder or a Mobiusstrip.

When X is obtained by removing \317\201points from the compact ori-entable surface of genus g, then \320\234\321\212\321\203\321\200(\320\245)

is an orbifold of dimension

3(2g+ p\342\200\224

2). The fact that this dimension is finite can be interpreted

as the property that a complete hyperbolic metric on X is essentiallycontrolled by finitely many parameters (such as the shearparametersof Section 6.7.2) or, as a physicistwould say, that there are only

finitely many degrees of freedom in the choice of such a metric. Forinfinite surfaces such as the infinite Loch Ness Monster, the moduli

space Mhyp(-X') is infinite-dimensional.

12.5.2. Essential surfaces in 3-dimensional manifolds.Let

(S3, dsph) be the 3-dimensional sphere dennedby straightforward

generalization of the 2-dimensional sphere (\302\2472,dSph) to three

dimensions. Namely,

\302\2473={(x1,x2,X3,Xi) GM4;Xi +x22 + x\\ + xl = 1},

and the spherical distance dev^{P,q)is denned as the infimum of

the euclidean lengths of all piecewise differentiable curves joining \316\241

toQ in\302\2473.

One might optimistically hope that a version of Theorem 12.15

also holds for 3-dimensional manifolds,and that every 3-dimensional

manifold admits a metric which is locally isometric to the euclideanspace(\320\2263,dcuc), the hyperbolic space (H3. dhyp)

\302\260rthe 3-dimensional

sphere (\302\2473,rfsph)\302\267This is not quite the case.

Let X be a 3-dimensional manifold.

A typical example of 3-dimensionalmanifold is the complementX = K3 - \320\232of a knot \320\232in M3. In that case, Theorem 12.2as-nerted the existence of a complete hyperbolic metric on X provided

\320\232is neither a satellite nor a torus knot. Fora general3-dimensional


manifold X, the nonsatellite condition is replaced by a condition on

a certain type of surfaces in X.

A two-sided surface in the 3-dimensionalmanifold X is a

subset S \321\201X for which there exists a 2-dimensionalmanifold \316\245and a

homeomorphism \317\210:\316\245\317\207(\342\200\224\316\265.\316\265)\342\200\224*U to a subset U \320\241X such that

\317\206(\316\245\317\207{0})

= S. Here, we are endowingthe product \316\245\317\207(-\316\265,\316\265)with

the product of the metric of \320\243and of the euclidean metric of theinterval (\342\200\224\316\265,\316\265),as denned in Exercise 1.6. The sameconvention will

hold for all products in this section.

One can always find two-sided surfaces in a 3-dimensionalmanifold X by using relatively trivial constructions. For instance, by

definition of manifolds, X contains a subset U which is homeomor-

phic to an open ball V in M3; the preimages of small euclideanspheresin V then provide many two-sided spheres in X. If we start from a

simple closed curve \320\232in X, we can also try to thicken it to a smalltube, and then take the boundary of this tube; in most cases, the

boundary of the tube will be a two-sided torus, but it can also be atwo-sided Klein bottle.

Such constructions are clearly too simplefor the corresponding

surfaces to have much significance. We will focus attention on

surfaces which are not obtained in this way. Such surfaces will be called

essential, where this word is denned on a case-by-casebasis.An essential sphere in the 3-dimensional manifold X is a two-

sided surface S which is homeomorphicto the sphere,and which is

contained in no subset \320\222\321\201X where \320\222is homeomorphic to a ballinM3.

An essential projective plane in X is simply a two-sidedsurface which is homeomorphic to the projective plane, with no further

condition.

The definition of an essential torus involves more properties.

Namely, an essential torus in the 3-dimensional manifold X is atwo-sided surface \316\244such that

(1) \316\223is homeomorphic to the torus T2;

(2) \316\244is not contained in a subset \320\222\320\241X which is

homeomorphicto a ball in M3;

12.5. The general GeometrizationTheorem 345

(3) \316\244is not the boundary of a solid torus V \320\241X; namely,

there is no homeomorphism\317\210:S1 \317\207\320\256>2\342\200\224*V \320\241X such that

\316\244=\317\206(51\317\207S1), where S1 and D2 denote the unit circle and

the closed unit disk in M2, respectively;

(4) \316\244is not the boundary of a collar W going to infinity in X;

namely, there is no homeomorphism \317\206:\316\2442\317\207[0, \320\276\320\276)\342\200\224>W \320\241

X such that T =\317\206(\316\2442

\317\207{0})

and such that\317\206(\316\2442

\317\207{\316\257})

goes to infinity in X as t tends to +oo, in the sense that

lim ini\\d{P0,P);Pe \317\206(\316\2122\317\207{t})}= too

for an arbitrary base point Pq \302\243X.

The definition of essential Klein bottles will involve twisted (alsocalled nonorientable) solidtori. Recallthat the circle S1 is homeomor-

phic to the quotient space obtained from the interval [0,1] by gluing

the point 0 to the point 1 (see Exercise 4.5). As a consequence,thesolid torus S1 \317\207D2 is homeomorphic to the quotient spaceobtained

from [0,1] \317\207D2 by gluing each point (0, P) with \316\241e D2 to the point

(1,P).Let the twisted solid torus S1\317\207D2 be the quotient space

obtained from [0,1] \317\207D2 by gluing each point (0,P) to (1,P),where \316\241

denotes the complex conjugate of \316\241G D2 \320\241\320\2322= \320\241.The

boundary Klein bottle of this twisted solid torus is the image \302\247l\317\207S1 of

[0,1] \317\207S1 under the quotient map.An essential Klein bottle in the 3-dimensionalmanifold X is

a two-sided surface \320\232such that

(1) A\" is homeomorphic to the Klein bottle K2;

(2) \320\232is not the boundary of a twisted solid torus V \320\241X;

namely, there is no homeomorphism \317\210:S1 \317\207D2 \342\200\224*V \320\241X

such that \320\232=\317\206{81xS1);

(3) \320\232is not the boundary of a collar W going to infinity in X;

namely, there is no homeomorphism \317\210:\316\2322\317\207[0, \320\276\320\276)\342\200\224>W \320\241

X such that \316\232=\317\206(\316\2122

\317\207{0}) and such that

lim inf [d(P0, \316\241):\316\241G \317\206(\316\2322\317\207{\316\257})}= +oo.

t\342\200\224*+ OC


Proposition 12.16.Let [X, d) be a 3-dimensional manifold. If dis topologically equivalent to a complete hyperbolic metric d', then X

contains no essential sphere, projectiveplane, torus or Klein bottle.

D

See [Scott, Bonahon} for a proof and further references.

12.5.3. Seifert fibrations. Recall that the standard solid torus is

the subset V of M3 consisting of all points of the form

((R + \317\201cos v) cos u, (R + \317\201cos v) sin u, \317\201sin v)

with 0 ^ \317\201^ r and u, \317\205\302\243\316\234,where the two radii R and r with

R > r > 0 are given, For any rational - \302\243Q, we can consider in V

all the torus knots parametrized by

11\342\200\224\342\226\272((i? + \317\201cos qi) cos (pt + i0), (R + \317\201cos qt) sin(pi + \316\257\316\277),\342\200\224/osingi)

with 0 < \317\201< R and \316\257\316\277G K. When \317\201- 0, the corresponding curve

is the corecircleof V. When \317\201> 0, the corresponding curves aredisjoint --torus knots wrapping around this core.

There is another possiblepresentation of this partition of thesolid torus. Indeed,we already saw that V is homeomorphic to thequotient space S1 \317\207D2 obtained from [0,1] \317\207D2 by gluing each point

(0. P) to (1,P). However, if \317\203:D2 \342\200\224>D2 denotes the rotation of angle\342\200\2242\317\2002\316\263is aiso homeomorphic to the other quotient space obtained

from [0,1] \317\207\316\2172by gluing each (0, P) to (1,\317\203{\316\241)).(To check this, first

use your personal experiencewith playdough, and then try to builda more rigorous argument.) Then the partition of [0, lj x D2 into linesegments[0,1]\317\207{\316\241}projects to a partition of this second quotient

space into disjoint simple closed curves, which is homeomorphic to

the partition of V by ^-torus knots as above.

This secondpresentation immediately extends to the twisted solidtorus W = S1 x32. Recall that W is the quotient space obtained

from [0, 1] \317\207D2 by gluing (0, P) to (1,P), where \316\241is the complex

conjugate image of \316\241\302\243ID2 \321\201\316\2342- \320\241under the reflection across thex-axis. Again, the partition of [0,1] \317\207D2 into line segments [0,1] \317\207

{\316\241}

projects to a partition of W into disjoint simple closed curves. Notethat the curves corresponding to \316\241in the x-axis (so that \316\241= P) wrap


once around S1 xD2, but that all other curves wrap twice around this

twisted solid torus.A Seifert fibration of a 3-dimensional manifold X is a partition

of X into disjoint simple closedcurves, calledthe fibers of the

fibration, which locally looks like the above decompositionsof the twisted

and untwisted solid torus. More precisely, for every fiber K, thereis a subset U \320\241X containing K, a rational number

jje Q and a

homeomorphism \317\210:U \342\200\224*S1 \317\207D2 or S1 \317\207D2 between U and the solidtorus S1 \317\207D2 or the twisted solid torus S'xD2, such that

(1) U is a union of fibers;

(2) \317\210sends \320\232to the core circle of the twisted or untwisted solid

torus;

(3) \317\210sends any other fiber contained in U to a E-torus knot

in S1 \317\207D2 or to one of the curves of the decomposition ofS1 \317\207D2, as above.

For most fibers, the solidtorus U is untwisted and we can take ^ =\316\263;

all other fibers are called exceptional.A 3-dimensional manifold that admits a Seifert fibration is a

Seifert manifold.

Since a Seifert fibration of X is a partition, it gives rise to a

quotient space X. It easily follows from the local description of theSeifert fibration that this quotient space is a surface with boundary.

To a large extent, X behaves very much like the product X xS1 of X

with the circle S1, and we should think of X as a certain thickeningof the surface X.

This intuition is expressed in the early work [Seifert] of

Herbert Seifert, who classified Seifert fibrations up to homeomorphism

respecting the fibration, in terms of the quotient surface X and of thelocal type of the exceptional fibers. A little over thirty years later,

Friedhelm Waldhausen [Waldhausen] and Peter Orlik, Elmar Vogt

and Heiner Zieschang [Orlik, Vogt & Zieschang]showed that the

Seifert fibration of a Seifertmanifold is unique up to homeomorphism,outside of a few well-understood exceptions, so that the classification

of Seifert manifolds up to homeomorphismis equivalent to Seifert's

348 12. Geometrization theorems in dimension3

classification up to homeomorphism respecting the fibration. Themain consequence of these results is that Seifert manifolds form a

class of 3-dimensional manifolds which is very well understood.

12.5.4. The general GeometrizationTheorem.We are now

almost ready state the Geometrization Theorem for 3-dimensionalmanifolds. However, we need a last couple of definitions.

Let R\" denote the closed half-space Kn_1 \317\207[0, \320\276\321\201)\320\241\316\206\316\267

consisting of those points of the euclidean spaceR\342\204\242whose last

coordinate is nonnegative. An \316\267-dimensional manifold-with-boundary

(in one word) is a metric space (X, d) which is locally homeomor-

phic to (R\",cieuc).Namely, for every \316\241e X, there exists a smallball Bd{P,e) \320\241X and a homeomorphism \317\210:Bd{P,e)

\342\200\224>U be-

twen Bd.(P,e) and a subset U \320\241R\" which contains a small ball

R+nBdou<;(P, \316\265')of the metric space (R\302\243,detlc). The boundary dX ofX consistsof those \316\241as above which are sent to a point of Rn_1 \317\207{0}

by the homeomorphism. Its interior is the complement X \342\200\224dX.

A manifold is said to have finite topological type if it is home-

omorphic to the interior of a compact manifold-with-boundary. For

instance, a compact manifold has finite topological type. The infinite

Loch Ness Monster surface of Figure 12.9is a typical example of a

manifold of infinite topological type.

Theorem 12.17 (The Geometrization Theorem for 3-dimensional

manifolds). Let (X,d) be a connected3-dimensionalmanifold with

finite topological type. Then at least one of the following holds:

(1) X admits a complete hyperbolic metric d! which is

topological equivalent to d;

(2) X contains an essential sphere, projective plane, torus, orKlein bottle;

(3) X is a Seifert manifold. D

In addition, there is relatively little overlap with these threepossibilities. We already saw in Proposition 12.16 that (1) and (2) areincompatible. With respect to (1) and (3) thereare,up to

homeomorphism,only six exceptional connected Seifert manifolds that admit

complete hyperbolic metrics, such as the interior of the solid torus,

12.5. The general GeometrizationTheorem \320\250

or the product of the torus T2 with the line R. Similarly, only two

connected Seifert manifolds contain an essential sphereor projective

plane. However, most Seifert manifolds contain many essential tori

or Klein bottles.

Complement 12.18. Under the hypothesis of Theorem 12.17,suppose in addition that X is the union of a compact subset \320\241and of

finitely many subsets homeomorphicto \316\2442\317\207[0,\320\276\321\201)andK2 \317\207[0,\320\276\321\201),

where \316\2442and \320\2322denote the torus and the Klein bottle, respectively.

Then, we can additionally require in conclusion (1) of Theorem 12.17that the hyperbolic metric d' has finite volume. D

It then follows from Mostow's Rigidity Theorem 12.4 that thehyperbolicmetricd! is unique up to isometry under the hypothesesand conclusions of Complement 12.18.

Theorem 12.17may not quite look like the GeometrizationTheorem 12.15 for surfaces because it involves only one geometry, the

hyperbolic geometry. However, Seifert manifolds can be endowed

with other geometric structures, namely, complete metricswhich

are locally isometric to some homogeneousmodel space.Among homogeneous spaces of dimension 3, we have already

encountered the hyperbolic space (H3,cihyp),the euclidcan space(R3, cieuc) and the 3-dimensional sphere (S3,dsph).Thesethreespacesarealsoisotropic.However, in dimension 3, there also existhomogeneous spaces which are not isotropic, such as the products i2xl

and \302\2472\317\207R. There also exist twisted versions of these products.

Altogether, if we identify two geometric models when they only differ byrescaling, and if we require each of them to appear for at least one 3-

dimensional manifold with finite volume, there are only eight possiblegeometriesin dimension 3. This observation is due to Bill Thurston,and is explainedin detail in [Scott, Bonahon], for instance.

It turns out that every Seifert manifold can be endowed with a

geometricstructuremodeledafter one of six of these eight geometries.Therefore, only essential spheres, projective planes, tori and Kleinbottlescan createproblems for endowing a 3-dimensional manifoldwith a geometric structure.


It turns out that earlier theories provide a kind of unique

factorization of 3-dimcnsional manifolds of finite topological type into

pieces which either contain no such essential surfacesor can be Seifert

fibered. To some extent, these factorizations are analogous to the

unique factorization of an integer as a product of primes. The first

such facLorizationis the theory of connected sums initiated by Hel-muth Kneser [Kneser] and completed by Wolfgang Haken [Haken]and John Milnor [Milnor] (see also the graduate textbook[Hempel]),which essentially reduces the analysis of 3-dimensionalmanifolds to

that of manifolds containing no essentialsphereor projective planes.

The second such factorization is provided by the CharacteristicToricDecomposition, originally suggested by Friedhelm Waldhausen andfully developed by Klaus Johannson [Johannson] and BusJacoand

Peter Shalen [Jaco & Shalen].Combined with these factorizations, Theorem 12.17 then asserts

that every conneci.ed 3-dimensional manifold of finite topological type

has a natural splitting into pieces, each of which admits a geometricstructure. In particular, this explainswhy this statement is called aGeometrization Theorem, and not just a Hyperbolization Theorem.

Theorem 12.17and Complement 12.18were proved by Bill

Thurston in the late 1970s in the case of unbounded manifolds, or for

bounded manifolds which contain essential two-sided surfaces (in asensewhich we will not define here). See [Thurstoni], and

expositions of the proof in [Otali, Otal2, Kapovich]. The Geometrization

Theorem for knot complements that we presented as Theorem12.2isa specialcaseof this result.

The final cases of the proof of Theorem 12.17 were completed

by Grisha Perelman shortly after 2000, following a program startedin the early 1980s by Richard Hamilton. Perelman was offered the

Fields Medal for this work in 2006, but declined the award. Followinga definitepattern in this area of mathematics, the completedetailsof

the proof have not been written down by Perelman, who only released

a few unpublished preprints [Perelmani, \320\240\320\265\320\263\320\2651\321\202\320\260\320\277\320\263,\320\240\320\265\320\263\320\2651\321\202\320\260\320\277\320\267].

However, detailed expositions are now beginning to become available

[Chow & Lu & Ni, Chow & al., Cao& Zhu, Kleiner & Lott,

Morgan & Tianl5 Morgan & Tian2].

Exercisesfor Chapter 12 351

An important corollary of Theorem 12.17 is that it solves a

century-old problem, the Poincare Conjecture-We saw that a manifold X is connected if any two points P,

Q G X can be joined by a continuous curve 7: [0\320\233]\342\200\224>X with

7(0) = \316\241and 7(1)= Q. The manifold X is simply connected

if, in addition, the space of curves going from \316\241to Q is connected;

namely, if, for any two curves 70: [0,1] \342\200\224>X and 71: [0,1] \342\200\224*X with

endpoints 70(0) = 71(0) = \316\241to 70(1)= 7i(l) = Q, there existsa

family of curves 7S: 11\342\200\224>7S(i) depending continuously on a parameters e [0,1],all going from 7S(0) = \316\241to js(l) = Q and coincidingwith

the original curves 70 and 71 when s = 0, 1. The fact that the curves

7S depend continuously on s means that the map H: [0.1] \317\207[0,1] \342\200\224*X

defined by H(s, t) = js(t) is continuous.

Examples of simply connected 3-dimensional manifolds includethe euclideanspace\320\250\320\273and the 3-dimensional sphere

\302\2473=-{{x, y, z, t) G M4; x2 + y2 + z2 + t2 = 1}.

In 1904,Henri Poincare proposed the following conjecture, which

became a theorem 100 years later after many unsuccessful attempts in

the intervening years.

Theorem 12.19(The Poincare Conjecture). Every simply connectedbounded3-dimensionalmanifold is homeomorphic to the

3-dimensionalsphere S3. D

The Poincare Conjecture comesas a corollary of the Geometriza-

tion Theorem 12.17by using relatively simple arguments on

hyperbolic manifolds, by exploiting the Kneser-Haken-Milnor theory of

connected sums to reduce the problem to manifolds without

essentialspheres or projective planes, and by applying an earlier result of

Herbert Seifert [Seifert] that S3 is the only simply connected Seifert

manifold.


Exercise12.1.a. Show by a series of pictures that the | -torus knot is isotopic to the

I-torus knot.


b. More generally, show that the. |-torus knot is isotopic to the\302\243

-torus

knot for every ? \302\243Q.

Exercise 12.2. Let \320\232be the figure-eight knot represented in Figure 11.5,and let K' be its mirror image, namely, let K' be the image of \320\232under a

reflection across a planeof R3. Show by a series of picturesthat \320\232and K'

are isotopic.

Exercise 12.3. Consider the unknot, formed by the unit circle S1 in \320\2322\320\241

\320\2323.Show that there exists a kleinian group \320\223\320\276,generated by a singleelement and acting freely on H3, such that the quotient space \320\2353/\320\223is

homeomorphic to the complement R3 \342\200\224S1. Possible hints: First show that

R3 \342\200\224S1 is homeomorphic to K3 \342\200\224\316\232\317\207{(0, 0)}, perhaps by using a suitableMobius transformation; then consider the map H3 \342\200\224\342\226\272R'1 \342\200\224\320\226\321\205{(0,0)}

defined by (x,y,u) \320\275-*(\320\266,\320\270cos \321\203,\320\270sin \321\203).

Exercise 12.4. Let \320\223and \320\223'be two groups acting freely and discontinu-

ously on the hyperbolic space H3. Supposethat the corresponding quotient

spaces are homeomorphic by a homeomorphism \317\206:\320\2303/\320\223'\342\200\224>

\320\2303/\320\223.

a. Let \316\261:[\316\261,b] \342\200\224>\316\2273/\316\223be a parametrized continuous curve in (\320\2303/\320\223,diiyp),

beginning at Pa = q(o) with Pa \302\243H3. Show that there exists aparametrized continuous curve S: [o, b] \342\200\224\342\226\272H3 such that 5(a) = Pa and

q = \317\200\316\277\316\261,where \317\200:\316\2273\342\200\224>\316\2323/\316\223is the quotient map. Hint: Consider

the supremum of the set of those t \302\243[o, b] for which such a map a can

be defined on [\316\277,\316\257],and remember that \317\200is a local isometry by (the

proof of) Theorem 7.8.

b. Show that in part a, the map S is unique. Hint: Given another suchmap q', consider the infimum of those t G [o, b] such that a'(t) \321\204a(t).

\321\201Suppose that we are now given a family of continuous curves au :

[a, b] \342\200\224\342\231\246\320\2533/\320\223,depending continuously on a parameter \320\270\302\243[\321\201,d\\ in the

sense that the map [o,b] \317\207\\c, d\\ \342\200\224\302\273\342\226\240\316\2273/\316\223defined by (i, \320\270)\320\275-*au(t) is

continuous. Assume in addition that au(a) \342\200\224-Pq for every \320\270\302\243[\321\201,dj,

and let au'- [a, b] \342\200\224\342\226\272HI3 be associated to qu as in parts a and b. Show

that the map [a, b] \317\207[\321\201,d\\ \342\200\224\342\226\272HI3 defined by (t, u) >-* din {i) is continuous.Hint: If the map was not continuous at some (\316\257\316\277,\316\220\316\257\316\277),consider the

infimum of those t \302\243[o, b] such that the map (t, u) \316\271\342\200\224>au (t) is notcontinuous at (t,uo)-

d. Under the hypotheses and conclusionsof part c, suppose in addition

that the map \320\270>-> au(b) \302\243\320\2353/\320\223is constant. Show that the map\320\270>\342\200\224>au(b) \302\243HI3 is also constant.

e. Fix two base points P0, \316\241\317\214\302\243\316\2273such that \317\206(\316\2410)= Pq. For \316\241\302\243HI3,

choose a continuous curve S: [o, b]\342\200\224\342\226\272HI3 with a(a) = Po and 5(6) = P,

and lift the curve a' =\317\210\316\277\317\200\320\276\320\260:[\320\276,\320\254]-* \320\2353/\320\223'to a continuous curve


5': [o, b]\342\200\224\342\226\272H3 with a'(a) = Pq as in part a. Use parts a-d to show-

that its endpoint P' = a'(b) dependsonly on P, and not on the choiceof the curve a. In particular, there is a well-definedmap \317\206:\316\2273\342\200\224\342\226\272\316\2273

associating to each PgH3 the point \317\206(\316\241)= \316\241'as above.

f. Show that the above map \317\206:\320\2303\342\200\224\302\273\342\226\240\320\2303is a homeomorphism.

g. Show that for every 7 G \320\223,the composition \317\210\316\277\316\267\316\277\317\206~\316\271is an elementof \316\223'.

h. Show that the map \317\201:\316\223\342\200\224\302\273\342\226\240\316\223'defined by p(j) \342\200\224\317\201\316\263\316\277^\"1is a

bijection, and that \317\201{\316\267\\\316\277

\302\267\316\2632)= p{y\\) \320\276p{y2) for every 71, 72 \302\243\320\223.

(Namely, \317\201is a group isomorphism between the groups \316\223and \316\223',if

you know what this is.)

Exercise 12.5. Let \320\223\320\262be the kleinian group of Section 11.2,such that

E3/Ts is homeomorphic to the complement R3 \342\200\224\320\257\"of the figure-eight knot,and let \320\223\320\276be a kleinian group as in Exercise 12.3, generated by a single

element and for which \320\2303/\320\223\320\276is homeomorphic to the complement \320\2503\342\200\224S1

of the unknot.

a. Show that \320\223\320\262contains elements 71, 72 such that 71 \302\26072 # 72 \302\26071 \342\226\240

b. Show that 71 \302\26072= 72 \302\26071 for every 71, 72 \302\243\320\223\320\276.

\320\241.Use part h of Exercise 12.4to conclude that R3 - \320\232and M3 - S1are not homeomorphic, and therefore that the figure-eight knot is not

isomorphic to the unknot.

Exercise 12.6. Let the kleinian group \320\223act freely on H3, so that the

quotient space \320\2353/\320\223is locally isometric to E3 by Theorem 7.8. We wantto define the hyperbolic volume of \320\2303/\320\223.

a. Let A be a subset of \320\2303/\320\223which is contained in a ball 5jh (\316\241,\316\265)\320\241

\320\2303/\320\223isometric to a ball of H3 by an isometry \317\206: Bjh (\316\241,\316\265)\342\200\224<\342\226\240

Bdhyp(Q,s) \320\241\316\2273.Define the hyperbolic volume volhyp(A) to be the

hyperbolic volumeof \317\210{\320\220)\320\241\320\2303,as introduced in Exercise 9.12. (Here,we are implicitly assuming that the subset A is sufficiently \"nice\" thatthe triple integral involved in the definition of the hyperbolic volumeof \317\206(\316\221)makes sense; all subsets involved in this exercise will satisfy

this property, so that we do not have to worry about the deepermathematical issues that could arise in the general case.) Show that this

hyperbolic volume volhyp(^4) is independent of the choice of the ball

Bjh (\302\267\316\241>\316\265)an(3 of the isometry \317\206.Hint: Use Exercise 9.12.

b. Now consider a subset A which is contained in the union of finitely

many balls B\\, B2, \342\226\240..,B\342\200\236in \320\2303/\320\223,each of which is isometric to aball in E3. For every subset / of {1, 2...., n}, let Bt be the subset of


those \316\241G \316\2273/\316\223such that {i;P G Bij= I. Define

volhyp(A)=

\316\243 volbyP(A \316\240\316\222\316\271),

ic{\\ n}

where each volhyp(A \316\240\316\222\316\271)is defined by part a. Show that volhyp(A)is independent of the choice of the balls Bi.

\321\201Fix a base point Pq G H3. Show that for every R > 0, there existser > 0 such that dbyp(P,f(P)) > \316\265\321\217for every \316\241

\342\202\254Bdhyp(Po,R)and

every 7 G \316\223\342\200\224{Idr}\302\267Conclude that the ball Bjh (Pq,R) is contained

in the union of finitely many balls \316\222\316\271,\316\2222,\342\226\240\342\226\240\342\226\240,B\342\200\236in \320\2303/\320\223,each of

which is isometric to a ball in H3.

As a consequence, the hyperbolic volume volhyp(5jh (Pq, R)) is well

defined.

d. Show that the limit

volhyp(H3/r)= lim vohyp(Bd (P0,R))\302\261t\342\200\224\302\273+00j *

exists (possibly infinite), and is independent of the choiceof the base

point Po-

Exercise 12.7.

a. Consider the perpendicular bisector of two points \316\241and Q e H3,defined as

TipQ={R\302\243H3; dhyp(R, P) =

dhyp(R, Q)}.

Show that Hpq is a hyperbolic plane. Hint: Compare Exercise 2.4.

b. Let 7 be an isometry of H3 such that 7(00) \317\206\316\277\316\277,and let (Pn)nen bea sequencein H3 converging to the point 00 for the euclidean metric.Show that the perpendicular bisector plane \316\240\317\201\316\2677(\317\201\316\267)converges to the

hyperbolic plane \316\2407defined in Lemma 12.9, in the sense that thecenter and radius of \316\220\316\257\317\201\342\200\236\316\212(\317\201\342\200\236),considered as a euclidean hemisphere,convergesto the center and radius of \316\2407as \316\267tends to infinity.

Exercise 12.8. Show by a series a pictures that the two knots ofFigure 12.7 are indeed isotopic. Tying the knots in shoe laces or another typeof string may (or may not) be helpful.

Exercise 12.9. Let \320\232be the |-torus knot in K3, drawn on a torus \316\244

consisting of those points that are at distance r from a central circle \320\241

of radius R > r. Show that the complement X = \320\2503\342\200\224\320\232of \320\232in R3 =

K3 U {oo} admits a Seifert fibration where one fiber is the union of the

z-axis and of the point 00, where another fiber is the central circleC, and

where all other fibers are isotopicto the E-torus knot.

Tool Kit

This appendix, referred to as the Tool Kit in the text, is a quicklist of notation and basic mathematical definitions used in the book.

T.l. Elementary set theory

For us, a set X is a collection of objects called its elements. In

theory, we need to be morecareful in the definition of sets in order to

avoid logical inconsistencies. However,thesedeep and subtle issues

do not arise at the level of the mathematics described in this book.

Consequently, we will be content with the above intuitive definition.

When the object \320\266is an element of the set X, we say that \317\207

belongs to X and we write \317\207\302\243X.

In practice, a set can be describedby listing all of its elementsbetween curly brackets or by describing a property that characterizes

the elements of the set. For instance,the set of all even integers thatare strictly between\342\200\2243and 8 is denoted by {\342\200\2242,0,2,4,6}, or by

{x; there exists an integer \316\267such that \317\207= 2n and \342\200\2243 < \317\207< 8}.

A subset of a set X is a set \316\245such that every element of \316\245is

also an element of X. We then write \316\245\320\241\320\245.

A particularly useful set is the empty set 0 ={ }, which

contains no element.

355

356 Tool Kit

A diagonal bar across a symbol indicates that the corresponding

property does not hold. For instance,\317\207\302\243X means that \317\207does not

belong to the set X.Hereis a list of classical sets of numbers, with the notation used

in this book:

N ={1,2,3,...} is the set of all positive integers;in

particular, 0 is not an element of N, a convention which is not

universal.

\316\226= {..., -3, -2,-1,0,1,2,3,...} is the set of all integers.

Q is the set of all rational numbers, namely, of all numbersthat can be written as a quotient 2 where p, q e \316\226are

integers with q \317\2060.

R is the set of all real numbers.

\320\241is the set of all complexnumbers.(SeeSection T.4 later

in this Tool Kit.)

Given two sets X and Y, their intersection X(lY consists of all

elements that are in both X and Y. Their union X U \316\245consists of

those objects that are in X or in \316\245(or in both). The complementX \342\200\224\316\245consists of those elements of X which do not belong to \316\245.For

instance, if X = {1,2,3,4} and \316\245= {3,4,5}, then

\320\245\320\237\320\243= {3,4},

XUK = {1,2,3,4,5}and X - \316\245= {1, 2}.

Two sets are disjoint when their intersection is empty.

More generally,is X is a set of sets, namely, a set whose elements

are themselves sets, the union of these JeXis the set

[J I={a;;a;eIfor some X 6 X}xex

of those elements \317\207that belong to at least one X e X. Similarly,

their intersection is the set

\316\241)\316\247=

{\317\207;\317\207e X for all X e X}

of those elements \317\207that belong to all X \302\243X.

T.l. Elementary set theory 357

We can also consider the product of X and Y, which is the set\316\247\317\207\316\245consisting of all ordered pairs (\320\266,y) where \317\207\302\243X and \321\203G Y.

More generally, the product \316\247\316\271\317\207X2 x \302\267\302\267\302\267x Xn of \316\267sets \320\245\320\263,\320\2452,

..., Xn consists of all ordered \316\267-tuples (\320\266\321\214\320\2662,\342\226\240..,xn) where each

coordinate x\\ is an element of Xi-

In particular,

R2=lxR= {{\321\205,\321\203);\321\205\320\265\342\204\226.,\321\203\320\265\320\251

is naturally identified to the plane through cartesian coordinates. The

same holds for the 3-dimensional space

R3 = R \317\207R \317\207R = {(x, y, \316\266);\317\207e R, \321\203G R, \316\266G R}.

A map or function \317\206:X -> \316\245is a rule \317\206which to each \317\207e X

associates an element \317\206(\317\207)G Y. We also express this by saying that

the map \317\210is defined bym \317\206(\317\207).Note the slightly different arrow

shape.

When X = Y, there is a special map, calledthe identity map\316\231\316\254\317\207: X \342\200\224>X which to \317\207e X associates itself, namely, such that

Id^(i) = \317\207for every \317\207G X.

The composition of two maps \317\206:X \342\200\224>\316\245and \317\210:\316\245\342\200\224>\316\226is the

map \321\204\316\277\317\206:\316\247\342\200\224>\316\226defined by the property that \317\210\316\277\317\206(\317\207)

=\317\210(\317\206(\317\207))

for every \317\207G X. In particular, \317\210=

\316\231\316\254\316\263\316\277\317\210=

\317\210\316\277\317\212\316\254\317\207for any function

\317\206:\316\247-\302\273\316\245.

The map \317\206is injective or one-to-one if \321\203>(\320\266)^ y>(a;') for everyx, x' \342\202\254X with \320\2667^ \317\207'.It is surjective or onio if every i/ G \316\245is

the image \316\271/=

\317\210(\317\207)of some a; G X. The map y> is bijective if it

is both injective and surjective, namely, if every \321\203G \316\245is the image

\320\263/=\317\210(\317\207)

of a unique \320\266G X. In this case, there is a well-defined

inverse map \317\210~1:\316\245\342\200\224*\316\247,for which \302\245>_1(2/) is tne unique \320\266G X

such that \316\271/=

\317\206(\317\207).In particular, y> \316\277\317\210\"1

= Idy and \317\206~\316\273\316\277\317\206=

\316\231\316\254\317\207.

When \317\210:\316\247\342\200\224>\316\245is bijective, we also say that y> is a bijection, or

that it defines a one-to-one correspondence between elements of

X and elements of Y.

The image of a subset \320\220\320\241X under the map \317\206:X \342\200\224>\316\245is the

subset

V?(A) = {y G \320\243;\316\271/=

\317\206(\317\207)for some \320\266G X}

358 Tool Kit

of Y. The preimageoiBcY under \317\206:\316\247\342\200\224>\316\245is the subset

\321\207>-1{\320\222)=

{\321\205\320\265\320\245;\321\207>{\321\205)\320\265\320\222}

of X. Note that the preimage \317\206~\316\273is defined even when \317\206is not

bijective, in which case the inverse map \317\206~\316\271may not be defined andthe preimageof a point may be empty or consistof many points.

The map \317\206:X \342\200\224>X preserves or respects a subset \320\220\320\241X if

\317\210(\316\221)is contained in A. A fixed point for \317\206is an element \317\207G X

such that \317\206(\317\207)=

\317\207;equivalently, we then say that \317\206fixes x.

If we have a map \317\206:X \342\200\224*\316\245and a subset \320\220\320\241X, the restriction

of \317\206to A is the function\317\210^\316\221'-

\316\221\342\200\224>\316\247defined by restricting attention

to elements of A, namely, defined by the property that\316\250\\\316\261{\316\277)

=\317\210{\316\277)

for every a e A.

When the map \317\206:\316\235\342\200\224>X is defined on the set N of all positive

integers, it is called a sequence. In this case, it is customary towrite \317\210(\316\267)

\342\200\224Pn (with the integer \316\267as a subscript) and to denote thesequenceby a list Pi, P2, \342\226\240\302\267\302\267,Pn, ..., or by (Pn)n\342\202\254Nfor short.

T.2. Maximum, minimum, supremum, andinfimum

If a set A ={\317\207\317\207,\317\2072,\342\226\240\342\226\240\302\267,xn}consists of finitely many real numbers,

there is always one of these numbers which is larger than all the otherones and another one which is smaller than the other ones. Thesearethe maximummax A and the minimum min A of A, respectively.

However, the same does not hold for infinite subsets of R. Forinstance,the set A = {2n; \316\267e Z} does not have a maximum, because

it contains elements that are arbitrarily large. It has no minimum

either because there is no a e A such that a < 2\342\204\242for every \316\267G Z.

We can fix this problem by doing two things. First, we introducea point \302\26100at each end of the number line R =

(\342\200\22400,+00), so as to

get a new set [\342\200\22400,+00]= R U {\342\200\224\321\201\321\216,+\320\276\320\276}.Then, we will say that

an element \316\234e [\342\200\224\321\201\321\216,+\321\201\321\216]is a supremum for the subset \320\220\320\241R if:

(1) a < \316\234for every a e A;

(2) \316\234is the smallest number with this property, in the sensethat there is no \316\234'< \316\234such that a < M' for every a \302\243A.

\320\242.2. Maximum, minimum, supremum, and infimum 359

The secondcondition is equivalent to the property that we can find

elements of A that are arbitrarily closeto M.Similarly, an infimum for \320\220\321\201R is an element m e [\342\200\224\321\201\321\216,+\320\276\320\276]

such that

(1) \316\261> m for every a e A;

(2) m is the largest number with this property, in the sense thatthere is no m' > m such that \316\261> m' for every a e A.

It is a deep result of real analysis that any subset \320\220\320\241R admits

a unique supremum \316\234= sup \320\233,and a unique infimum m = inf A.

The proof of this statement requires a deepunderstanding of the

nature of real numbers. To a large extent,realnumbers were precisely

introduced for this property to hold true, and somepeopleeven use

it as an axiom in the construction of real numbers. We refer to any

undergraduate textbook on real analysis for a discussion of this

statement.

For instance,

sup{2n;ra6 Z} = +oo

and inf{2\";\316\2676 \316\226}= 0.

It may happen that sup A is an element of A, in which case we

say that the supremum is alsoa maximum and we write sup \320\233=

max A; otherwise, the maximum of A does not exist. Similarly, theminimum min \320\233of \320\233is equal to inf A if this infimum belongs toA, and does not exist otherwise. In particular, the maximum and

the minimum are elements of A when they exist. The supremum andinfimum always exist, but are not necessarily in A.

For instance,min{2n;nG N}= 2, but min{2n;n e Z} does not

exist since inf{2n; \316\267e Z} = 0 0 {2n;\316\267e Z}.

Be aware of the behavior of suprema and infima under arithmeticoperations.For instance,if we are given two sequences (xn)neN aQd

(\320\243\320\277)\320\277\320\265\320\277\302\260fre&l numbers, it is relatively easy to checkthat

sup{a;n +t/\342\200\236;neN}< sup{a;n;\316\267 G N} + sup{yn;\320\277\320\265\320\251

and inf {xn +i/\342\200\236;neN}J inf{a;n; neN} + inf{i/n; \316\267\302\243\316\235}.

360 Ibol Kit

However, these inequalities will be strict in most cases. Similarly,

sup{-a;n; \316\2676 \316\235}= - inf {xn; \316\2676 \316\235}

and inf{\342\200\224xn;\316\2676 \316\235}= \342\200\224

sup{a;n; \316\267G \316\235}.

Finally, it may be enjoyable to considerthe caseof the empty set

0, and to justify the fact that sup0 = \342\200\224\321\201\321\216and inf 0 = +oo.

T.3. Limits and continuity. Limits involvinginfinity

In Section 1.3, we define limits and continuity in metric spaces by

analogy with the corresponding notions that one encounters in

calculus. It may be useful to review these calculus definitions.

Let /: \316\246-> R be a function with domain \316\257CR. The function

/ is continuous at xq e D if f(x) is arbitrary closeto f(xo) when

\317\207G D is sufficiently close xq- This intuitive statement is made

rigorous by quantifying the adverbs \"arbitrarily\" and \"sufficiently\" with

appropriate numbers \316\265and \316\264.In this precise definition of continuity,

the function / is continuous at \317\207\316\277if, for every \316\265> 0, there exists

a number \316\264> 0 such that \\f(x) \342\200\224f(xo)\\ < \316\265for every \317\207e D with

\\x\342\200\224

xo\\ < \316\264.This property is more relevant when \316\265and \316\264are both

small, and this is the situation that we should keep in mind to better

understand the meaning of the definition.

We can reinforce the analogy with the metric space definition

given in Section 1.3 by using the notation d(x,y)=

\\x\342\200\224

y\\, namely, by

considering the usual metric d of the real line R. The above definition

can then be rephrased by saying that / is continuous at xq 6 D if,

for every \316\265> 0, there exists a \316\264> 0 such that d(f(x), f(xo)) < \316\265for

every \317\2076 D with d(x, xq) < \316\264.

Also, a sequence of real numbers x\\, x2, \342\226\240\342\226\240\342\226\240,xn, \342\226\240\342\226\240\342\226\240converges

to Zoo \316\265R if xn is arbitrarily close to Zoo when the index \316\267is

sufficiently large. More precisely, the sequence {xn)n& converges to \320\226\320\276\320\276

if, for every \316\265> 0, there exists an no such that \\xn\342\200\224

Zool < \316\265f\302\260r

every \316\267> no. Again, if we replace the statement \\xn\342\200\224

x^ < \316\265by

d(xn>Xoo) < \316\265,we recognize here the definition of limits in metric

spaces that is given in Section 1.3.

T.4. Complex numbers 361

In calculus,we also encounter infinite limits, and limits as the

variable tends to \302\26100. Recall that f(x) has a limit L 6 R as \317\207tends

to +00 if, for every \316\265> 0, there exists a number \316\267> 0 such that

\\f(x) \342\200\224L\\ < \316\265for every \317\207with \317\207> \316\267.Similarly, f(x) converges to

L as \317\207tends to \342\200\224\321\201\321\216if, for every \316\265> 0, there exists a number \316\267> 0

such that \\f(x) \342\200\224L\\ < \316\265for every \317\207with \317\207< \342\200\224\316\267.In both cases, the

more relevant situation is that where \316\265is small and \316\267is large.

In the book, we combine +00 and \342\200\224\321\201\321\216into a single infinity \321\201\321\216.

Then, by definition, f(x) converges to L as \317\207tends to \321\201\321\216if, for

every \316\265> 0, there exists a number \316\267> 0 such that \\f(x) \342\200\224L\\ < \316\265for

every \317\207with \\x\\ > \316\267.

Beware that the symbols 00 and +00 representdifferent

mathematical objects in these statements. In particular, lim f(x)= L

X \342\200\224\320\256\320\236

exactly when the properties that lim f(x) = L and lim fix) = Lx\342\200\224>+ oo x\342\200\224>\342\200\22400

both hold.

Similarly, f(x) converges to 00 as \320\266tends to xq if, for everynumber \316\267> 0, there exists a \316\264> 0 such that \\f(x)\\ > \316\267for every

\317\207with 0 < \\x\342\200\224

xq\\ < \316\264.In particular, lim f(x) = \321\201\321\216if eitherx\342\200\224>xo

lim f(x) = +00 or lim f(x)= \342\200\224\321\201\321\216.However, the converse is not

X\342\200\224*Xo X\342\200\224>X0

necessarily true, as illustrated by the fact that lim \342\200\224= \321\201\321\216but thatJ J x^O X

neither lim - = +00 nor lim - = \342\200\224\321\201\321\216hold.\317\207\342\200\224>0\316\247 \317\207\342\200\224>0\316\247

Immediate generalizations of these limits involving infinity occur

in the book when we consider the Riemann sphere \320\241= \320\241U {00}=

R2 U {oo} in Chapter 9,13UC=H3UCU {\321\201\321\216}in Chapter 9, or

R3 = R3 U {\321\201\321\216}in Chapter 11.

T.4. Complex numbers

In the plane R2, we can consider the \320\266-axis Rx {0} as a copy of the reallineR, by identifying the point (x, 0) G R \317\207{0} to the number \317\207G R.

If we set i = (0.1), then every point of the plane can be written as a

linear combination (\320\266,\321\203)= \317\207+ iy. When using this notation, we will

consider \317\207+ iy as a generalized number, called a complex number.

362 Ibol Kit

It is most likely that you already have somefamiliarity with complex

numbers, and we will just review a few of their properties.Complex numbers can be added in the obvious manner

(x + iy) + (a/ + V) = (x + x') + i(y + y'),

and multiplied according to the rule that i2 =\342\200\2241,namely,

(x + iy)(x' + iy') = (xx'- yy') + i(x'y + xy').

These additions and multiplications behave according to the standardrules of algebra. For instance, given three complexnumbers \316\266= x+iy,

z' = x' + iy' and z\" = x\" + iy\", we have that z(z' + z\")= zz' + zz\"

and z(z'z\") = (zz')z\".Fora complexnumber \316\266= \317\207+ iy, the \320\266-coordinate is called the

real part Re(z) = \317\207of z, and the y-coordinate is its imaginary partIm(z)

= y. The complex conjugate of \316\266is the complex number

\316\266= \317\207\342\200\224iy

and the modulus, or absolute value of \316\266is

\\z\\= \\/x2 + y2 = yfzH.

In particular,

1 1 \316\266 \316\266 \317\207 . \321\203

x + iy \316\266 zz \\\316\266\\2 \317\2072+y2 \317\2072+ y2'

Also,

zz' = (xx' \342\200\224yy')

\342\200\224i(xy' + \321\203\321\205')

=(\317\207

\342\200\224iy)(x'

\342\200\224iy')

= zz!

and

\\zz'\\= Vzz'zz' = i/zlVz7? =

\320\230\320\230

for every \316\266\342\200\224\317\207+ iy and z' = x' + iy' G \320\241

In the book, we make extensiveuse of Euler's exponential

notation, where

cos \316\270+ i sin \316\270- ew

for every \316\270e R. In particular, any complex number \316\266= \317\207+ iy can

be written as \316\266= re10, where [r, \316\230]are polar coordinates describingthe same point \316\266as the cartesian coordinates (x, y) in the plane R2.

T.4. Complex numbers 363

Thereare many ways to justify this exponential notation. Forinstance,we can remember the Taylor expansions

sl^ =E(-1)fe(2FTI)!=0-3!

+ 5!-7!+-

\321\201V4 iMb\320\2622\320\272

\316\271\316\2702 \316\270\316\233\316\270*

co^ =E(-1)fe(2fc)T

=1-^+4!-6!+\302\267\302\267\302\267

\342\200\236~0\" 02 03 Q4 \316\2705 06 076

=\316\243^= 1+ *+^ +

\316\231\316\223+

4!+

5!+

6! +7!+\302\267\302\267\302\267

\316\267=0

valid for every 0 e R. If, symbolically, we replace 0 by i0 in the last

equation and remember that i2 = \342\200\2241,

02 03 04 05 06 07= 1 +

i0-2!-i3!+

4!+i5!\"6!-i7!+\302\267\302\267

,, 02 04 06 \321\207./\320\273 03 05 07 \321\207=

(1-2!+

4!-6!+\302\267\302\267)+1(0-3!+

5!-7!+\302\267\302\267)

= cos0 + isin0.

There is actually a way to justify this symbolic manipulation byprovingthe absolute convergence of this infinite series of complex numbers.

In the same vein, vising the addition formulas for trigonometricfunctions,

e'V = (cos0+ isin0)(cos0'4 isin0')=

(cos 0 cos 0' \342\200\224sin 0 sin 0') + i(cos0sin0' + sin 0 cos 0')

= cos(0 + 0') +isin(0+ 0')=

\320\265\320\232\302\2534\320\262')1

which is again consistent with the exponential notation.

Note the special case

\320\265\321\213= -1,

known as Euler's Formula, which combines two of the most famousmathematical constants (three if one includes the number 1 among

famous constants).

364 Tool Kit

More generally, for an arbitrary complex number \316\266= \317\207+ iy G C,we can define

ez = ex+iy = exely = ex(cosy + isiny).It immediately follows from the above observations that this

complex exponential satisfies many of the standard propertiesof real

exponentials, and in particular that ez+z = e2e2 for every \316\266,\316\266e C.

Supplemental

bibliography

and references

In this section, we list the references that were mentioned in the

text, but begin with a bibliography suggesting additional materialfor further reading. Someentriesappear in both lists.

Supplemental bibliography

This supplemental material is rated with bullets, according to itsdifficulty. One bullet \302\267indicates a textbook which is roughly at the

same mathematical level as the presentmonograph or easier. Two

bullets \302\267\302\267are used for more advanced textbooks, at the graduatelevel. Three bullets\302\267*\302\267denote research-level material.

We have included software [Heath, Wada, Weeks2, Weeks3]that currently is freely available on the Internet, and which the reader

is strongly encouraged to explore. Of course, these electronic

references are quite likely to become unstable with time. The beautiful

(and mathematically challenging) movie [NotKnot] is also highlyrecommended.

365

366 Supplemental bibliographyand references

Some of this suggested reading, such as the excellent[Stillwelli]and [Mumford, Series & Wright], has a significant overlap with

the current text but offersa different emphasis and viewpoint.

More elementary background on hyperbolicgeometry can be

found in [Anderson], [Greenberg], [Henle] or [Stillwelli].Advanced textbooks include the very influential [Thurston.^], as well as

[Benedetti & Petronio], [Marden2], [Maskit], [Ratcliffe],

[Mumford, Series & Wright]. See also [Stillwell2]for easy

access and an introduction to some historical references.The topologicalclassification of surfaces, which is not discussed

in this book, is nicely presented in [Massey, Chapter 1].

With respect to the topicsdiscussed towards the end of the book,[Adams], [Lickorish]and [Livingston] provide nice introductions

to knot theory, and more advanced material can be found in [Rolfsen]

and [Burde & Zieschang]. See also [Flapan]for applications of

knot theory.

A topological approach to 3-dimensional manifolds can be found

in [Hempel], [Jaco] or [Rolfsen]. The book [Weeksx] is a lively

introduction to a more geometric approach to 3-dimensionalmanifolds,

while the articles [Scott] and [Bonahon] survey more advanced

topics.

You should also consider learning some topology and algebraictopology. There are numerous textbooks in these areas, such as

[Armstrong], [Gamelin & Greene], [Hatcher2],[Massey] and

[McCleary].

[Adams] \302\267Colin C. Adams, The knot book. An elementary

introductionto the mathematical theory of knots, American

MathematicalSociety, Providence, RI, 2004.

[Anderson]\302\267 James W. Anderson, Hyperbolic geometry, SpringerUndergraduateMathematics Series, Springer-Verlag London

Ltd., London, 1999.

[Armstrong] \302\267M. Anthony Armstrong, Basic Topology,UndergraduateTexts in Mathematics, Springer-Verlag, New York-Berlin,1983.

Supplemental bibliography 367

[Benedetti & Petronio]\302\267\302\267 Riccardo Benedetti, Carlo Petronio,Lectures on hyperbolicgeometry, Universitext, Springer-Verlag,

Berlin, 1992.

[Bonahon]\302\273\302\267\302\273Francis Bonahon, \"Geometric structures on

3-manifolds\", in: Handbook of geometric topology (R. J. Daverman and

R. B. Sher, eds.), 93-164,North-Holland, Amsterdam, 2002.

[Burde & Zieschang]A Gerhardt Burde, Heiner Zieschang,

Knots, (Second edition), de Gruyter Studies in Mathematics 5,

Walter de Gruyter GmbH, Berlin,Germany, 2003.

[Flapan] \302\267\302\267Erica Flapan, When topology meets chemistry. A

topological look at molecular chirality, Cambridge University Press,Cambridge, United Kingdom\302\267, Mathematical Association of

America, Washington, DC, 2000.

[Gamelin& Greene]\302\267Theodore W. Gamelin, Robert E. Greene,Introduction to topology, (Second Edition), Dover Publications,Mineola,NY, 1999.

[Greenberg]\302\267 Marvin J. Greenberg, Euclidean and non-Euclideangeometries:development and history, W. H. Freeman and Co.,SanFrancisco,Calif.,1973.

[Hatcher2]\302\267\302\267Allen E. Hatcher, Algebraic topology, Cambridge

University Press, 2001.

[Heath] Daniel J. Heath, Geometry playground, mathematical

software, freely available at www.plu.edu/-heathdj/java/.

[Hempel]\302\267\302\267 John Hempel, 3-Manifolds, Annals of MathematicsStudies86,PrincetonUniversity Press, Princeton, NJ, 1976.

[Henle] \302\267Michael Henle, Modern geometries. The analytic approach,

Prentice Hall, Inc., Upper Saddle River, NJ, 1997.

[Jaco]\302\267\302\267William Jaco, Lectures on three-manifold topology, CBMS.Regional Conference Series in Mathematics 43, American Math-eamtical Society, Providence, RI, 1980.

[Lickorish] \302\267\302\267W. B. Raymond Lickorish, An introduction to knot

theory, Graduate Texts in Mathematics 175, Springer-Verlag,New York, NY, 1997

368 Supplemental bibliography and references

[Livingston]\302\267Charles Livinston, Knot theory, MathematicalAssociation of America, Washington, DC, 1993.

[\320\234\320\260\320\263\320\260\320\265\320\277\320\263]\302\267\302\267Albert Marden, Outer circles: an introduction to

hyperbolic 3-manifolds, Cambridge University Press, Cambridge,United Kingdom, 2007.

[Massey]\302\267\302\267 William S. Massey, A basic course in algebraic topology,

Graduate Texts in Mathematics 127, Springer-Verlag,Berlin,1997.

[Maskit]\302\267\302\267 Bernard Maskit, Kleinian groups, Grundlehren derMathematischen Wissenschaften 287, Springer-Verlag, Berlin,1988.

[McCleary] \302\267A first course in topology, Student MathematicalLibrary 31, American Mathematical Society, Providence, RI, 2006.

[Mumford,Series& Wright]\302\267 David Mumford, Caroline Series,David Wright, Indra's pearls. The vision of Felix Klein,

Cambridge University Press, New York, 2002.

[NotKnot] \302\267Not Knot, video by the Geometry Centerof the

University of Minnesota, A.K. Peters Ltd, Wellesley,MA, 1994.

[Ratcliffe] \302\267\302\267John Ratcliffe, Foundations of hyperbolic manifolds,GraduateTextsin Mathematics 149, Springer-Verlag, New York,1994.

[Rolfsen]\302\267\302\267Dale Rolfsen, Knots and links, Publish or Perish,Berkeley,1976.

[Scott] \320\233G. Peter Scott, \"The geometries of 3-manifolds\", Bulletin

of the London Mathematical Society 15(1983),401-487.[Stillwelli]

\302\267John C. Stillwell, Geometry of surfaces, Universitext,Springer-Verlag,New York, 1992.

[Stillwell2] \302\267\302\267John C. Stillwell, Sources of hyperbolicgeometry,

History of Mathematics 10, American Mathematical Society,Providence, RI, 1996.

[Thurston4] \302\267\302\267William P. Thurston, Three-dimensional geometryand topology, Vol. 1 (Edited by Silvio Levy), PrincetonMathematical Series 35, Princeton University Press, Princeton, NJ,1997.

References 369

[Wada] Masaaki Wada, OPTi, mathematical software, freelyavailable at http://vivaldi.ics.nara-wu.ac.jp/-wada/OPTi.

[Weeksi]\302\267 Jeffrey Weeks, The shape of space (Secondedition),Monographs and Textbooks in Pure and Applied Mathematics249, Marcel Dekker,Inc.,New York, 2002.

[\\Vieeks2] Jeffrey Weeks, SnapPea, mathematical software, freely

available at www. geometrygames. org.

[Weeks3] Jeffrey Weeks, Curved Spaces, mathematical software,

freely available at www. geometrygames. org.

References

Note that some of the referencesbelow correspondto books,while

others point to articles in academic journals.

[Alexander] James W. Alexander, \"On the deformation of an n-

cell\",Proceedingsof the National Academy of Sciences of the

United States of America 9 (1923),406-407.

[Ashley] Clifford Ashley, The Ashley book of knots, Doubleday,New

York, 1944.

[Bonahon & Siebenmann] Francis Bonahon, LaurenceC. Sieben-

mann, New geometric splittings of classicalknots, and the

classification and symmetries of arborescent knots, to appear,Geometry &; Topology Monographs, Coventry, UK.

[Brock & Canary & Minsky] Jeffrey F. Brock, Richard D. Car-

nary, Yair N. Minsky, \"The classification of Kleinian surface

groups, II: The Ending Lamination Conjecture\",preprint,available at http://arxiv.org/abs/math/0412006.

[Bromberg] Kenneth W. Bromberg, \"The space of Kleinian

punctured torus groups is not locally connected\",preprint,available

at http://arxiv.org/abs/0901.4306.

[Cannon & Thurston] James W. Cannon, William P. Thurston,

\"Group invariant Peano curves\", Geometry & Topology 11

(2007),1315-1355.

[Cao& Zhu] Huai-Dong Cao, Xi-Ping Zhu, \"A complete proof of

the Poincare and geometrization conjectures\342\200\224application of the


Hamilton-Perelman theory of the Ricci flow\", Asian Journal ofMathematics 10 (2006), 165-492;\"Erratum\", Asian Journal ofMathematics 10 (2006),663.

[Cerf]Jean Cerf, Sur les diffeomorphismes de la sphere dedimension trois (\316\2234

= 0), Lecture Notes in Mathematics 53, Springer-

Vcrlag, Berlin-New York, 1968.

[Chow & Knopf] Bennett Chow, Dan Knopf, The Ricci flow:an introduction, Mathematical Surveys and Monographs110,American Mathematical Society, Providence, RI, 2005.

[Chow & Lu & Ni] Bennett Chow. Peng Lu, Lei Ni, Hamilton's

Ricci flow, Graduate Studies in Mathematics 77, American

Mathematical Society, Providence, RI; SciencePress,New York,

2006.

[Chow & al.] Bennett Chow, Sun-ChinChu, David Glickenstein,

Christine Guenther, James Isenberg, Tom Ivey, Dan Knopf,

Peng Lu, Feng Luo, LeiNi, The Ricci flow: techniques and

applications, Part I. Geometric aspects, Mathematical SurveysandMonographs135,American Mathematical Society, Providence,

RI, 2007; Part II. Analytic aspects, Mathematical Surveys and

Monographs 144, American Mathematical Society, Providence,

RI, 2008.

[Dirichlet] Gustav Lejeune Dirichlet, \"Uber die Reduktion der pos-itiven quadratischen Formen mit drei unbestimmten ganzenZahlen\", Journal fur die Reine und Angewandte Mathematik 40(1850),209-227.

[Dumas]David Dumas, Bear, a tool for studying Bers slices

of punctured tori, mathematical software, freely available at

bear.sourceforge.net.

[Epstein & Penner] David B. A. Epstein, Robert \320\241Penner,

\"Euclidean decompositions of noncompact hyperbolic manifolds\",Journal of Differential Geometry 27 (1988), 67-80.

[Farey] John Farey, \"On a curious property of vulgar fractions\",

Philosophical Magazine 47 (1816), 385-386.

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[Fordi]Lester R. Ford, \"The fundamental region for a Fuchsian

group\", Bulletin of the American Mathematical Society 31(1935),531-539.

[Ford2] Lester R. Ford. \"Fractions\", American MathematicalMonthly 45 (1938), 586-601.

[Puchsi] Lazarus Fuchs, \"Sur quelques proprietes des integrales des

equations differentielles, auxquelles satisfont les modules de

periodicitedesintegrales elliptiques des deux premieres especes\",Journal fur die Reine und Angewandte Mathematik 83 (1877),13-57.

[Fuchs2] Lazarus Fuchs, \"Uber eine Klasse von Functionen mehrererVariabeln, welche durch Umkehrung der Integrale von LosungendeslinearenDifferentialgleichungen mit rationalen Coefficienten

entstehen\", Journal fur die Reine und Angewandte Mathematik

89 (1880), 151-169.

[Gordon & Luecke] CameronMcA. Gordon, John Luecke, \"Knots

are determined by their complements\", Journal of the AmericanMathematical Society2 (1989),371-415.

[Haken] Wolfgang Haken, \"Ein Verfahren zur Aufspaltung einer 3-Mannigfaltigkeit in irreduzible 3-Mannigfaltigkeiten\", Mathema-

tische Zeitschrift 76 (1961),427-467.

[Hardy& Wright] Godfrey H. Hardy, Edward M. Wright, An

introduction to the theory of numbers, Fifth edition, The

ClarendonPress, Oxford University Press, New York, 1979.

[Hatcheri] Allen E. Hatcher, \"A proof of a Smale conjectureDiff(53)

~0(4)\", Annals of Mathematics 117 (1983),553-607.

[Hatches]Allen E. Hatcher, Algebraic topology, CambridgeUniversityPress, 2001.

[Jaco & Shalen] William H. Jaco, Peter B. Shalen, Seifert fibered

spaces in 3-manifolds, Memoirs of the American Mathematical

Society 220, American Mathematical Society,Providence, 1979.

[Johannson]Klaus Johannson, Homotopy equivalences of 3-

manifolds with boundary, Lecture Notes in Mathematics 761,Springer-Verlag,Berlin-NewYork, 1979.


[Kapovich] Michael Kapovich, Hyperbolic manifolds and discrete

groups, Progress in Mathematics 183, Birkhauser Boston,

Boston, MA, 2001.

[Kneser] Hellmuth Kneser, \"Geschlossene Flachen in dreidimen-

sionalen Mannigfaltigkeiten\", Jahresbericht der Deutschen

Mathematiker-Vereinigung 38 (1929),248-260.

[Kleiner& Lott] Bruce Kleiner, John Lott, \"Notes on Perelman'spapers\",Geometry & Topology 12 (2008), 2587-2855.

[Little] CharlesN. Little, \"Non-alternate \302\261knots\", Transactions ofthe Royal Society of Edinburgh 39 (1898-9), 771-778.

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Index

absolute value, 23, 362action of a group, 185. 185-205,

248adjacent tilea, 137, 137-146, 261Alexander, Jamea Waddell II

(1888 1971), 318antilinear fractional map, 27,

27-33, 38, 186,231-235antipodal points, 49, 102arc length, see length

area, 131hyperbolic area, 43, 43-44

spherical area, 53,182

B2, the disk model for the

hyperbolic plane, 36, 36-39,43,44, 101, 158-160, 209, 212,216,223

Bd{P,r),a ball in a metric space,4ball

ball in R3, 5, 344ball in a metric apace, 4, 42-43,

68-79, 240, 262

ball model for the hyperbolic

apace, 236, 240

belong to, 355bijection, 5, 90, 185, 357bijective, 357

Bonahon, Francis, iii

Bonnet, Pierre Oaaian (1819-1892),131

boundary

boundary of a manifold, 348boundary point, 152

bounded, 62, 81, 152,152 161

Brock, Jeffrey F., 283Bromberg, Kenneth W., 282

BusemannBusemann function, 178, 183

Buaemann, Herbert (1905-1994),178

C, the set of all complex numbera,

356, 361-364

Canary, Richard D., 283Cannon, Jamea W., 302canonical tile, 143, 143-146

CauchyCauchy aequence, 135, 182

Cauchy, Augustin (1789-1857),182,211

Cayley, Arthur (1821-1895), 45center

center of a horocircle, 172, 214center of a horosphere, 234

Cerf,Jean, 318circle, 5, 17-22, 32-33, 38, 48,

85-87,172,207-212,230, 288,

317

377

378 Index

great circle, 48, 129

great circlearc, 49,50-51closed

closed curve, 49, 316closedgeodesic,49,127-129,204

closed subset, 62, 80, 152,152-154,250-251, 257-258,

286

compact metric space,150,15& 154, 182, 203, 251, 348

companion knot, 321

complement, 356knot complement, 293-313,

319-340, 352-354complete

complete geodesic, 21

complete metric space,135,147-155,174, 182, 183, 203,230, 260,311,322, 325, 341,346, 348

complex number, 6, 1 1, 23, 361,361-364

complex conjugate, 6, 23, 44,362

complex exponential, 242, 364,362 364

composition, 6, 15, 28, 42, 51,52,185, 186, 357

cone, 86surface with cone singularities,

87, 125-126,136,192, 205, 342

conjugate, see complex conjugate

connected, 64, 341continued fraction, 224

continuous function, 4, 8, 9, 61, 90,151,302, 360

convergence, 4, 135, 147-155,360limit at infinity, 361, 360-361toward infinity, 361, 360-361

convexeuclidean polygon, 62

hyperbolic polygon, 82spherical polygon, 82

crossratio, 44-45

differential map, 29, 29-38, 40,49-50,229-230,317

dihedral angle, 235, 259, 259-265,290

dihedron, 235

Dirichlet

Dirichlet domain, 198, 197-201Lejeune Dirichlet, Guslav

(1805-1859), 198discontinuity domain, 286-287

discontinuous group action, 189,

189-197, 203-205, 248, 254,260,268, 280-283, 286, 313

discrete walk, 58, 58-60, 69-79,85-86

disjoint sets, 356

diskdisk in the plane, 5disk model for the hyperbolic

plane, 36, 36-39,43,44, 101,

158-160, 209, 212, 216,223disk sector, 72, 86, 205

distance

distance function, 3euclidean distance, 2, 1-3, 8, 340hyperbolic distance, 12, 12-14,

37, 39, 44,45,228signed distance, 218-220, 253,

253-254spherical distance, 48, 343

domino diagram, 225, 291

Dumas, David, 282Dunne, Edward G., xv

Rarle, Clifford J., 280

edgeedge cycle, 172, 172-181of a hyperbolic polyhedron, 258,

258-265of a polygon, 61, 61-83, 257

clement of a set, 355elliptic isometry, 41, 239, 240empty set, 355

Epstein, David \316\222.\316\221.,340

equivalence class, 84equivalence relation, 84

essential surface, 343-346essential Klein bottle, 345essential projective plane, 344

essential sphere, 344

Index 379

essential torus, 344

euclideaneuclidean diatance, 2, 3, 340euclidean geodesic,3, 21euclidean iaometriea, 5-7, 39euclidean length, 2, 48

euclidean metric, 67,340euclidean norm, 33

euclidean plane, 1-10euclidean polygon, 61, 61-66,

183euclidean apace, 340, 349euclidean aurface, 67, 89-97,

103-104, 341 343euclidean triangle, 125

EulerEuler characteristic, 131

Euler exponential notation, 6,362

Euler, Leonhard (1707-1783),131, 362

exceptional fiber, 347

exponentialcomplex exponential, 242, 364,

362-364exterior point, 152

face of a polyhedron, 258, 257-265,334-340

Farey

crooked Farey teaaellation, 243,241-247,265-279, 291-297

Farey circle packing, 210,207-226

Farey aeries, 223

Farey aum, 209Farey teaaellation, 210, 207-226

Farey, John (1766-1826), 210

fiber, 347fibration, see Seifert fibrationFielda Medal, 322, 350

figure-eight knot, 303, 293-313,352, 353

finite volume, 324fix, 358

fixed point, 358Ford

Ford circlea, 212

Ford domain, 331, 326-340

Ford, Lester Randolph

(1886-1967), 212, 340free group action, 192, 264,

280-282, 297,352Fuchs, Lazarus Immanuel

(1833-1902), 284fuchaian group, 255, 252-257, 280,

283-284fuchsian group of the firat type,

256

fuchsian group of the second

type, 256twiated fuchsian group, 288

function, 357fundamental domain, 192,

185-205, 259, 259-269,296-298

Gauas, Johann Carl Friedrich

(1777-1855), 131Gauaa-Bonnet formula, 131

generatetranaformation group generated

by bijectiona, 135-136, 186,194,260

genua, 97, 129

geodeaic, 21closedgeodesic, 49, 127-129, 204

complete geodesic,21euclidean geodesic, 3, 21

hyperbolic geodeaic,17 22,230

spherical geodesic, 49-51

geometric atructure, 349

Gordon, Cameron McA., 340great circle, 48, 129

great circle arc, 49group, 185, 185-205

abstract group, 202fuchsian group, 255, 252-257,

280, 283-284group action, 185, 185-205

group of iaometriea, 186

isometry group, 186kleinian group, 248, 241-302,

326-340tiling group, 136, 135- 184,

212-222,259

380 Index

transformation group, 185,185-205

H2, the hyperbolic plane, 11, 11-46H3, the 3-dimenaional hyperbolic

apace, 227, 227-240Haken, Wolfgang, 350

Hamilton, Richard S., 350Haros, C, 211

Hatcher, Allen E., 318Hiatt, Christopher, xvi

homeomorphic, 90homeomorphiam, 41, 90, 91, 96-98,

126,129, 130, 182, 270, 288,303-311, 317 319,325

homogeneous, 7, 15-16, 34-36,49-50,128,129,230, 349

homothety, 15, 28, 40, 228,239,

286

horizontal translation, 15, 28,41,228, 239

horocircle, 172, 172-181, 214-222,290

horocycle, 172

horocyclic iaometry, 172, 172-176,221-222

horodiak, 178, 178-180, 256

horoaphere, 234, 234-235,260-265,268, 298, 331-332

Howe, Roger E., xv

hyperbolic

hyperbolic area, 43, 43-44hyperbolic diak, 42

hyperbolic diatance, 12, 12-14,37, 39, 44, 45, 228

hyperbolic geodeaic, 17 22, 230hyperbolic isometry, 14 -17, 25,

23-27, 40, 232,231- 234, 238,239

hyperbolic knot, 322, 322-340

hyperbolic length, 11, 17-22,33-34, 227, 230

hyperbolic metric, 12, 12-14,37,

39, 44, 45, 82, 228hyperbolic norm, 33, 33-36,

228, 229hyperbolic plane, 11, 11-46, 234hyperbolic polygon, 80

hyperbolic reflection, 15,240hyperbolic rotation, 40

hyperbolic apace, 227, 227-240,349

hyperbolic aurface, 82, 102,341-343

hyperbolic triangle, 43, 98, 125,130

hyperbolic volume, 240, 353

Idx, the identity map of X, 357

idealideal polygon, 116

ideal vertex, 116, 170, 170-180,257,258,257-265

identity map, 185, 357image, 357imaginary part, 11, 362infimum {pi. infima), 359, 358-360infinite limit, 361, 360-361

infinity

limit at infinity, 361, 360-361vertex at infinity, 170, 170-180,

257, 258,257-265

injective, 357

interiorinterior of a manifold, 348

interior point, 152intersection, 356

inverae map, 185, 357inveraion

acroaa a circle, 16, 16-17,28,28,42

across a aphere, 229, 237iaometric

iaometric extenaion, 233,231-233, 236

iaometric group action, 186

locally iaometric, 67

iaometry, 5elliptic isometry, 41, 239euclidean isometriea, 5 7, 39

group of iaomelriea, 186

horocyclic iaometry, 172,172-176,221-222

hyperbolic isometry, 14-17, 25,23-27, 40,232,231-234, 238,

239

Index 381

iaometry group, 186

loxodromic isometry, 40, 239parabolic isometry, 41, 239, 327spherical isometriea, 50

isomorphic knots, 317isotopicknots, 316

iaotopy, 316

isotropic, 7, 34-36,49-50,230, 349

Jaco, William H-, 321,350jacobian, 238, 317

Johannaon, Klaus, 321,350

Klein

Klein bottle, 95, 93-95, 128,341eaaential Klein bottle, 345

Klein, Felix (1849-1925), 45, 95,284

kleinian group, 248, 241-302,326-340

Kneaer, Hellmuth (1898-1973), 350

knot, 316companion knot, 321

figure-eight knot, 303hyperbolic knot, 322, 322-340knot complement, 293-313,

319-340,352-354

satellite knot, 321torus knot, 319

trefoil knot, 319unknot, 317

lengtheuclidean length, 2, 48hyperbolic length, 11, 17-22,

33-34, 227, 230length in a metric apace, 3, 9, 87,

127length of a discrete walk, 58length of a aequence, 135,

147-155limit, 4, 135, 147-155

limit at infinity, 361, 36&361limit point, 248

limit aet, 248, 247-252,255-256,

270-279, 299-302

linear fractional map, 27, 27 33,

38, 186, 231-235linear groupa, 202

Listing, Johann Benedict(1808-1882),104

Little, Charlea N., 337

locally finite, 134, 145, 197, 259,261,333, 334

locally iaometric, 67, 82,83loxodromic isometry, 40, 239Luecke, John, 340

manifold, 340, 340-351map, 357Marden, Albert, 296maximum (pi. maxima),358,

358-360

McMullen, Curtia \320\242.,322

metric, 3euclidean metric, 67, 340hyperbolic metric, 12, 12-14, 37,

39,44,45

metric function, 3metric apace,3, 3-5,58-61,

134-135,150-155path metric, 9, 63, 81, 82product metric, 8, 344quotient metric, 61, 5861,

141-143,187-189spherical metric, 48, 343

Milnor, John W., 350

minimum (pi. minima), 358,

358-360

Minaky, Yair N-, 282, 283Mobiua

Mobiua group, 237

Mobiua atrip, 104, 103-104Mobiua tranaformation, 237

Mobius, August (1790-1868),104Mobiua atrip, 341

moduli apace, 342modulus (pi. moduli), 23, 362Moatow

Moatow'a Rigidity Theorem,325, 324-326,336

Mostow, George Daniel, 325

N, the aet of all positive inlegera,356

nonorientable surface, 104, 129

norm

382 Index

euclidean norm, 33

hyperbolic norm, 33, 33-36,228,229

normal subgroup, 299, 299-302

once-punctured torua, 114,

114-125, 161-162, 167-169,212-214

one-to-one correspondence, 357

one-to-one map, 357onto, 357

OPTi, 246, 280orbifold, 342

orbit, 187, 187-192, 198,248orbit space, 187, 187-192

orientation-preserving, 238,317orienlalion-reveraing, 238, 317

Orlik, Peter, 347orthogonal projection, 40, 253

parabolic isometry, 41, 239, 327

parabolic point, 327

partition, 57, 57-87, 187-192proper partition, 60

path metric, 9, 63,81,82

Penner, Robert Clark, 340Perelman, Grigory Y., 350

Perko, Kenneth \316\221.,338

perpendicular bisector, 39, 198piecewiae differentiable curve, 2

planeeuclidean plane, 1-10

hyperbolic plane, 11, 11-46,234projective plane, 102, 129, 341

essential projective plane, 344Poincare

Poincare Conjecture, 351Poincare'sPolygon Theorem,

174, 169-181Poincare'a Polyhedron Theorem,

260, 257-265Poincare, Jules Henri

(1854-1912), 283, 351

polygon, 61-87,89-131,133-184,192 -201, 335

euclidean polygon, 61, 61-66,183

hyperbolic polygon, 80, 257

ideal polygon, 116

locally finite polygon, 197apherical polygon, 82

polyhedral ball aector, 261polyhedron (pi. polyhedra), 258,

257-265Prasad, Gopal, 325preimage, 358preaerve, 358

product, 8, 357

product metric, 8, 344projective

projective line, 42

projective linear groupa, 202

projective model for the

hyperbolic plane, 45projective plane, 102, 129, 341

eaaential projective plane, 344

projective apace,202proper

proper gluing, 60

proper partition, 60

pseudosphere, 108, 108-114,118-125

Pythagorean triple, 223

Q, the aet of all rational numbers,356

quotient

quotient map, 61

quotient metric, 61, 58-61,141-143,187-189

quotient apace, 60, 58-61, 187,187-197

K. the set of all real numbers, 356K2, the euclidean plane, 1-10, 357R3, the 3-dimensional euclidean

apace, 4, 357real part, 11, 362

reflectioneuclidean reflection, 6, 50 52

hyperbolic reflection, 15, 240regular curve, 316

reapect, 358reatriction, 358

Index 383

RiemannRiemann sphere, 27, 41,

241-252, 276-279

Riemann, Georg FriedrichBernhard (1826-1866),27

Riley, Robert F. (1935-2000), 312,340

rotation

euclidean rotation, 6, 49-52hyperbolic rotation, 40

rotation-reflection, 50, 51 52

52, the 2-dimenaional sphere, 47

53, the 3-dimensional sphere, 343,351

satellite knot, 321

Schubert, Horat (1919-2001),321Seifert

Seifert fibration, 347, 354Seifert manifold, 347

Seifert, Herbert Karl Johannes

(1907-1996), 347semi-distance function. 3

semi-metric, 3, 58-61, 84, 187,189,202

sequence, 358

set, 355Shalen, Peter \320\222.,321, 350shear parameter, 220, 290shear-bend parameter, 265, 289

ahearing, 220Siebenmann, Laurence C, 322

signed distance, 218-220, 253,253-254

simple curve, 316

SnapPea, 323, 335, 337,339sphere, 5, 47, 341

3-dimensional sphere, 343, 349,351

essential sphere, 344

sphericalspherical area, 53, 182spherical distance, 48, 343

spherical geodesic, 19-51spherical isometries, 50

spherical metric, 48. 84, 343

spherical polygon, 82

spherical surface, 84, 341 343

spherical triangle, 52, 125

stabilizer, 190, 190-192,204-205stereographic projection, 41

subsequence, 150, 150-155subset, 355

supremum (pi. suprema), 358,358-360

surface, 341

euclidean surface, 67, 89-97,103-104,341-343

hyperbolic surface, 82, 102,341-343

in a 3-dimensional manifold, 344nonorientable surface, 104, 129spherical surface, 84, 341-343surface of genus 2. 97, 97-102surface of genus g, 129

two-sided surface, 344surjective, 357

tangent line, 276tangent map, 29

tessellation, 133, 133-184,207-226,259,259-265

Thurston, William P., 302, 312,322,350

tile, 133, 133-184tiling group, 136, 133-184,

212 -222,259tiling groupoid, 147

topologically equivalent, 311, 322,341,348

topology

induce the same topology, 311,

322, 341, 348torus (pi. tori), 56, 91

essential torus, 344

once-punctured torus, 114,114-125,161-162,167-169,212214

torus knot, 319

tractrix, 108transformation group, 185,

185 205translation

euclidean translation, 5horizontal translation, 15, 28,

41, 228, 239

384 Index

trefoil knot, 319

triangleeuclidean triangle, 125

hyperbolic triangle, 43, 98, 125,130

spherical triangle, 52, 125

Triangle Inequality, 3

trivial group, 186twisted fuchsian group, 288two-sided surface, 344

unbounded, 62, 81, 103,161-162union, 356

unknot, 317

van der Veen, Roland, xvivertex (pi. vertices)

of a polyhedron, 258ideal vertex, 116, 170, 170-180,

257, 258,257-265

of a polygon, 61, 80,82,257vertex at infinity, 170, 170-180,

257, 258,257-265

Vogt, Elmar, 347volume

finite volume, 324

hyperbolic volume, 240, 353

Wada, Masaaki, 246

Waldhausen, Priedhelm, 347,350walk, see discrete walk

Weeks, Jeffrey R., 323

Wright Sharp, Jennifer, xvi

Z, the set of all integers, 356

Zieschang, Heiner (1936-2004),347

Francis Bonahon – Low Dimensional Geometry

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