Riemann Surfaces and Complex Dynamicskochsc/old.pdfdiscuss Thurston’s theorem, ... The bifurcation...

Riemann Surfaces and Complex Dynamics

Course Notes

Class 1: January 9, 2014

Consider the family of real quadratic polynomials fc

: R ! R, given by fc

: x 7! x2 + c. Thenth iterate of f

c

is a polynomial of degree 2n; depending on the parameter c 2 R, there are manydi↵erent kinds of behaviors one can observe if the polynomial is iterated.

Figure 1. Here is a gallery of iterates for four di↵erent maps fc

; the c-value isspecified in the lower left corner. For each of the four maps, the first iterate isplotted, the fifth iterate is plotted, and the 15th iterate is plotted.

Definition. A dynamical system is a topological space X with a map f : X ! X. Associated toany point x0 2 X is an orbit sequence; that is the sequence n 7! fn(x0).

Fundamental questions: For the dynamical system f : X ! X, what do the orbit sequencesn 7! fn(x0) do as n!1? How does this behavior change as x0 is varied?

We will focus on iterating rational maps F : P1 ! P1, and polynomials p : C ! C. When wediscuss Thurston’s theorem, we will iterate ramified covers f : S2 ! S2, where S2 is an orientedtopological 2-sphere.

Fact. We will see that the forward orbits of the critical points of the maps that we iterate willplay a crucial role in these dynamical systems.

Parameter Space.

Consider the following bifurcation diagram which is drawn as follows: for c 2 [�2, 0), let fc

begiven by f

c

: x 7! x2 + c. consider the orbit sequence of x0 := 0, the critical point of fc

. Take theset

Sc

:= {x1000, . . . , x2000} ✓ [�2, 2].Over each value of c in the horizontal axis, draw the interval [�2, 2] vertically and plot the set S

c

.The following picture results. This picture is fantastically complicated, and there are windows of

Figure 2. The bifurcation picture for the family fc

: x 7! x2 + c.

calm inside. It is self-similar in the sense that if we zoom into one of these windows, we will seethe SAME picture. This is a parameter space picture for the family f

c

: x 7! x2 + c; it encodesinformation about how the dynamical system f

c

changes as the parameter c is varied.

Working over the complex numbers yields more information. Now consider the complex familyfc

: C ! C given by fc

: z 7! z2 + c, where c 2 C. Taking the fact for granted that orbit of thecritical point plays a key role, we again examine the orbit sequence associated to the critical pointz0 := 0:

{z0, z1, z2, . . .}.Given a parameter value c 2 C, we would like to come up with a coloring scheme that is math-ematically meaningful for the dynamical system f

c

: z 7! z2 + c. The point at 1 is naturally adistinguished point for a polynomial, so it is natural to ask if the orbit sequence of z0 = 0 divergesto 1.

Coloring Scheme. Color the parameter c 2 C black i↵ the orbit sequence of z0 = 0 is bounded.

Definition. The Mandelbrot Set is

M := {c 2 C | the orbit sequence n 7! fn

c

(0) is bounded}.The Mandelbrot Set is a compact subset of parameter space.

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Dynamical Space. There is a natural compact set in the dynamical plane associated to a poly-nomial p : C! C.

Definition. The filled Julia Set of the polynomial p is

Kc

:= {z 2 C | the orbit sequence n 7! fn

c

(z) is bounded}.

Exercise. Find all parameters c for which the critical point z0 = 0 is periodic of period 3.

Figure 3. The center picture is in parameter space for the family fc

: z 7! z2 + c.In black, we see the Mandelbrot set. Each of the four outer pictures are picturesin the dynamical plane. The critical point z0 = 0 is in the center of each of thesepictures. In pictures A,C, and D, the critical point is periodic of period 3, and inpicture B, the critical point is periodic of period 2.

The constraints impose the following condition on the parameter c:

(c2 + c)2 + c = 0,

which has three “valid” roots, and one extraneous root at c = 0. Of the three valid roots, there is areal one, and a pair of complex conjugate roots: ↵ 2 R, and �,� /2 R; suppose that the imaginarypart of � > 0.

Cast of Characters. The polynomial z 7! z2�1 (see picture B) is called the ‘basilica’ polynomial.The polynomial in picture A is the polynomial z 7! z2+↵; it is called the ‘airplane’. The polynomialin picture D is the polynomial z 7! z2 + �; it is called the ‘Rabbit’, and the polynomial in pictureC is z 7! z2 + �; it is called the ‘CoRabbit.’

Real and Complex. The bifurcation picture and the Mandelbrot Set go hand-in-hand.


Thurston’s Theorem is about associating finite combinatorial data to dynamical systems. Thurston’stheorem contains an existence and uniqueness statement. It arose first in the real setting; Milnor

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Figure 4. The structure in the bifurcation picture is mirrored in the MandelbrotSet. The areas of calm live over baby Mandelbrot Sets; one can see the “self-similar”structure in M as well.

and Thurston were studying entropy of real quadratic polynomials; specifically the logistic fam-ily

f�

: [0, 1]! [0, 1] and � 2 (0, 4].

As � increases, the graphs of the parabolas get steeper. Note that the critical point of this familyis always at x0 = 1/2.

Given a compact metric space X, and a continuous map f : X ! X, we can define assign areal number to the dynamical system f : X ! X. This quantity measures how complicated thedynamics can be.

Fact. The topological entropy of f : X ! X, denoted by h(f) is a conjugacy invariant; that forall homeomorphisms g : X ! X, we have

h(f) = h(g � f � g�1).

Dynamical systems with entropy 0 are very simple. In our particular case, the definition simplifiesto the following.

Definition. The topological entropy of f�

: [0, 1]! [0, 1] is given by

h(f�

) := limn!1

1

nlog(|fixed points of fn

�

|).

The function h : (0, 4] ! R given by h : � 7! h(f�

) has been extensively studied. Douady provedthat it is continuous. Milnor-Thurston proved that it is monotonic. They required a rigidityresult.

Change variables. We will work with the more familiar form fc

: [a, b] ! [a, b], and c 2[�2, 0].

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We have already seen that the orbits of critical points are important in studying dynamics. If weinsist that all critical points have finite forward orbits, this imposes a lot of constraints on thedynamics and motivates the following definition.

Definition. The polynomial fc

: x 7! x2 + c is postcritically finite if the orbit of x0 := 0 is finite,or equivalently, if the orbit of 0 is eventually periodic.

For every postcritically finite quadratic polynomial fc

: x 7! x2 + c, we wish to associate somefinite combinatorial data such that if x2 + c1 and x2 + c2 have ‘the same’ combinatorial data, thenthey are ‘the same’ from the point of view of dynamics. That is, we need to find the right kind ofinvariant to get a rigidity result.

Definition. The postcritical set of fc

is

P :=[

n>0

fn

c

(0);

that is, it is the union of the points in the forward orbit of the critical point x0 = 0. Note that x0 mayor may not be contained in P . Observe that f

c

is postcritically finite if and only if |P | <1.

Kneading data. Let fc

(x) = x2+c be a postcriticlaly finite quadratic polynomial with postcriticalset P . Label the elements of P as follows:

p1 := c, and pi

:= f i(0), 1 i n.

Define the set Q := f�1(P ). Note that P ✓ Q, and that Q ✓ R. Write the set Q as

Q = {q�n

< q�n+1 < . . . , q�1 = 0 = q1 < . . . , qn�1 < q

n

}.Definition. The kneading sequence k(f

c

) of fc

is the map

k : {1, . . . , n}! {�n,�n+ 1, . . . ,�2, 1, 2, . . . , n� 1, n} given by pi

= qk(i) for 1 i n.

We encode this map k with a kneading sequence

< k(1), . . . , k(n) > .

Exercise. Find a real quadratic polynomial of the form x2 + c for which the critical point x0 = 0is periodic of period 7. Compute the kneading sequence.

Theorem. (Thurston) The kneading data of a postcritically finite polynomial x2+ c is a completeinvariant.

Fact. Each hyperbolic component of the Mandelbrot Set contains a unique parameter c for whichthe polynomial x2 + c is postcritically finite.

Corollary Each hyperbolic component of the Mandelbrot Set can be ‘labeled’ with kneadingdata.

Milnor and Thurston used these facts to ultimately keep track of how the entropy changes as cmoves from �2 to 0 along the real axis; this gave them the tools they needed to prove that theentropy function is monotonic.

Immediate Question. Given a map k : {1, . . . , n} ! {�n,�n + 1, . . . ,�2, 1, 2, . . . , n � 1, n}, isit a kneading sequence for a quadratic polynomial of the form x2 + c?

Observations. There are some obvious necessary conditions.

• The first is that k(1) must be equal to �n.

• The second is that k must be injective.5

• The third is that k(n) must either be equal to 1, or else there exists a unique index 1 j < nfor which k(n) = �k(j).

• The last is that for all i, j satisfying k(i) < k(j) < 0, we have k(i + 1) > k(j + 1), and forall i, j satisfying 0 < k(i) < k(j), we have k(i+ 1) < k(j + 1).

Definition. A map k : {1, . . . , n}! {�n,�n+1, . . . ,�2, 1, 2, . . . , n�1, n} is an abstract kneadingsequence if it satisfies the properties above.

Do all abstract kneading sequences come from polynomials?

Answer. NO. It is rather complicated. Thurston came up with an algorithm, that can be easilyimplemented on the computer, to test if a given kneading sequence comes from a polynomial. Itis a beautiful recipe that involves iterating a contracting map on a certain space and looking for afixed point. We will see this in detail.

Thurston realized that kneading data could actually be generalized to other kinds of maps, andthat one could ask similar questions.

Thurston’s topological characterization of rational maps (First version). Let S2 be anoriented topological 2-sphere, and let f : S2 ! S2 be an orientation-preserving ramified coveringmap of degree d. Such a map has 2d� 2 critical points. Define the postcritical set of f to be

P :=[

n>0

fn(critical points).

The map f is postcritically finite if |P | <1. Thurston’s theorem says that the floppy topologicalmap f : S2 ! S2 “comes from” a rigid analytic object F : P1 ! P1 i↵ there are no obstruc-tions.

Huh? Think of f : S2 ! S2 as defining topological kneading data. That kneading data comes froman analytic map if and only if some purely topological condition is satisfied. We will explore thisa lot.

Class 3: January 16, 2014 - thanks Greg!

References for today: Milnor, ch1 and ch2, and Hubbard, ch2.

We will cover some essential background from complex analysis in this class. One of the mostimportant theorems guarantees that Riemann surfaces come in one of three di↵erent kinds.

Uniformization theorem. Let X be a simply connected Riemann surface. Then X is isomorphicto either

• the Riemann sphere P1, or

• the complex plane C, or

• the open unit disk D (and therefore also to the upper-half plane H )

Corollary. All Riemann surfaces can be classified according to their universal cover.

Examples. The torus C/⇤ is Euclidean. So is the surface C⇤ := C � {0}. The surface P1 �{at least three points} is hyperbolic.

We will make use of the following result from complex analysis:6

Schwarz Lemma. If f : D ! D is a holomorphic map such that f(0) = 0, then |f 0(0)| 1. If|f 0(0)| = 1, then f is a rotation about the origin. If |f 0(0)| < 1, then |f(z)| < |z| for all z 6= 0.

Idea of Proof. Apply the Maximum Modulus Principle to the map f(z)/z.

Automorphisms. Here are some useful facts:

Aut(D) =⇢z 7! az + b

bz + a: |a|2 > |b|2

�,

Aut(C) = {z 7! ↵z + �},

Aut(P1) =

⇢z 7! az + b

cz + d

�⇡ PSL2(C), the Mobius transformations

Aut(H) =

⇢z 7! az + b

cz + d: a, b, c, d 2 R and ad� bc > 0

�⇡ PSL2(R).

Recall that D ⇡ H by the following isomorphisms

H! D given by w 7! i� w

i+ w, and

D! H given by z 7! i(1� z)

1 + z.

More fun facts.

(1) The action of Aut(P1) is simply-three transitive on P1; that is, given any six points x1, x2, x3, y1, y2, y3,there exists a unique Mobius transformation µ 2 Aut(P1) so that µ(x

i

) = yi

.

(2) cross ratios are invariant by Mobius transformations, and

(3) elements of Aut(H) come in three flavors

(a) elliptic: has a fixed point in H,

(b) parabolic: conjugate to either w 7! w + 1 or w 7! w � 1 (has only one fixed point onthe “boundary”), and

(c) hyperbolic: conjugate to w 7! �w (has two fixed points on the “boundary”)

Proposition. Every holomorphic map from a Euclidean Riemann surface to a hyperbolic Riemannsurface is constant. Similarly, every holomorphic map from P1 to a hyperbolic Riemann surface isconstant.

Proof. Lift the holomorphic map f : X ! Y to a map on the universal covers. And note that anyholomorphic map P1 ! C or C! D must be constant.

Related: Picard’s Theorem. Every holomorphic map f : C! C which omits 2 di↵erent valuesis constant.

Poincare Metric There is a natural metric on the unit disk, called the Poincare metric, givenby

⇢D(z) =2|dz|

1� |z|2 .

This metric is the UNIQUE metric (up to scale) on D for which Aut(D) acts by isometries.

In general, the conformal metric '(z)|dz| is invariant under a conformal automorphism f : X ! Xi↵

'(f(z)) ='(z)

|f 0(z)| .7

An element f 2 Aut(X) satisfying the property above is called an isometry.

We can use the isomorphism between H and D to transport the Poincare metric on D to a metricon H

⇢H(w) =|dw|Im(w)

.

Similarly, every Riemann surface whose universal cover is isomorphic to D will inherit a Poincare(or hyperbolic) metric from the Poincare metric on D.

Exercise. Recall that the map H! D� {0} given by z 7! exp(iz) is a universal cover. Show thatthe Poincare metric on D� {0} is

⇢D�{0}(w) =|dw|

|w · log |w|| .

Proposition. Every hyperbolic surface S is complete with respect to its Poincare metric, and anytwo points are joined by at least one minimal geodesic. (This geodesic is unique if S is simplyconnected).

Proof. See Milnor, chapter 2.

The following result is CRUCIAL and will be used over and over again in this course.

Theorem. (Pick) Let S and S0 be hyperbolic Riemann surfaces, and let f : S ! S0 be a holomor-phic map between them. Then exactly one of the following holds:

(1) f is a conformal isomorphism S ! S0, and f maps S with its Poincare metric ⇢S

to S0 withits Poincare metric ⇢

S

0 , or

(2) f is a covering map, but it is not injective. If this happens, then f is a local isometry forthe infinitesimal metric, and we have

for all p, q 2 S, distS

0(f(p), f(q)) distS

(p, q).

(3) In all other cases, f strictly decreases all nonzero distances.

Motto: “A holomorphic map between hyperbolic Riemann surfaces is distance non-increasing.”

Proof. See Milnor, ch 2.

Exercise. Consider the map D! D given by z 7! z2. It is not an automorphism, it is not a cover,so by Pick’s theorem, it must be a contraction, which can be explicitly verified.

Companion Exercise. Now consider the map D� {0}! D� {0} given by z 7! z2. It is not anautomorphism, but it IS a covering map. It is not injective, so we see that it is a local isometry,by Pick’s theorem.

Interesting Remark: If we are doing dynamics, and we iterate a holomorphic map f : S ! Sfrom a hyperbolic Riemann surface to itself, we know from Pick that f is distance nonincreasing. Ifwe further know that the map f is a UNIFORM contraction, then we conclude by the ContractionMapping Fixed Point Theorem that there is a unique fixed point of f (this follows from uniformcontraction, and from the fact that the Riemann surface is complete). If, however, we are unlucky,and we don’t get uniform contraction, we may not have a fixed point at all :(

In our setting, we will iterate a holomorphic map, on a hyperbolic space. That map will be acontraction, but not a uniform contraction, which means that we cannot expect that this map willalways have a fixed point, and in fact, it often has no fixed point.

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Interesting Sub-Remark. Uniform contraction guarantees the existence of a fixed point, but justcontraction on our path-connected space implies that this fixed point is unique. We will leveragethis very simple fact to great reward in this course.

Dynamical Application. Suppose that S0 is a hyperbolic Riemann surface, and that S ✓ S0 is aconnected open subset. If S 6= S0, then

distS

0(p, q) < distS

(p, q) for all p 6= q in S.

Tagline: “distances measured relative to a bigger Riemann surface are always smaller”


The Uniformization Theorem tells us that Riemann Surfaces come in three di↵erent types. We canask about iterating holomorphic maps on all three di↵erent types of surfaces. Because of Pick’stheorem, we can imagine that iterating holomorphic maps on hyperbolic surfaces may be “boring”in the sense that there will not be too many things that can happen.

We will begin by iterating holomorphic maps on P1. It is a hw exercise to prove that a holomorphicmap on P1 is rational.

Dynamics on the Riemann Sphere. First, let f : P1 ! P1 be a rational map of degree d. Sucha map has 2d� 2 critical points counted with multiplicity (this is a Riemann-Hurwitz argument).Recall some results from Complex Analysis:

Weierstrass Uniform Convergence Theorem: If a sequence of holomorphic functions fn

: U !C converges uniformly on compact subsets to a limit function f : U ! C, then f is holomorphic,and the sequence of derivatives n 7! f 0

n

converges uniformly on compact subsets of U , to f 0.

Normal Families. A family F of holomorphic maps of a domain U ✓ P1 is called normal if forall sequences n 7! f

n

2 F , and for all p 2 U , there is a neighborhood of p, and a subsequencem 7! f

nm which converges uniformly on N .

Montel’s Theorem. Any family F of maps U ! P1 � {0, 1,1} is necessarily normal.

Arzela-Ascoli Theorem. Let U ✓ P1. Any family of continuous maps U ! P1 is normal i↵ it isequicontinuous.

There are some natural dynamical subsets of the Riemann sphere.

Definion. Let f : P1 ! P1 be a rational map. Consider the family of iterates

{f, f2, f3, . . .}The largest domain of normality of this family is called the Fatou Set of f , denoted ⌦

f

.

Observe that the Fatou set is automatically open. It does NOT follow from the definition that itis nonempty; in fact, there are examples for which the Fatou set is empty!

Definition. The Julia Set of f is Jf

:= P1 � ⌦f

.

Observe that the Julia set is compact. The Fatou set is a region of stability for the family ofiterates... the Julia set is the chaotic region.

Example. Let f : P1 ! P1 be given by z 7! z2. Compute ⌦f

and Jf

. The family of iterates canbe explicitly written down:

{z2, z4, z8, . . . , z2n , . . .}9

Figure 5. This is a gallery of dynamical pictures associated to iterating a rationalmap P1 ! P1. In each picture, the colorful part is the Fatou set, and the boundarybetween the colors is the Julia set, which is “fractal.”

Guess: ⌦f

= P1 � S1, and Jf

= S1. Let z 2 D. Then there is a neighborhood N around z so thatN ✓ D; on this neighborhood, the sequence of iterates will converge to the constant function z 7! 0.Similarly, if z 2 P1 � D, then there will be a neighborhood of z on which the sequence of iteratesconverges to the constant function z 7! 1. So we can see that the Fatou set contains P1� S1. Letz 2 S1. Any neighborhood of this point will contain points inside D and outside D; there is no waythat any subsequence of the family could converge on any neighborhood since the limit would haveto be holomorphic, and evidently, there would be a jump discontinuity there.

Easy Consequence of Definiton. Both the Fatou and Julia sets are fully invariant; that is,

z 2 Jf

() f(z) 2 Jf

.

Recall that periodic points are distinguished for a dynamical system. According to Poincare,periodic cycles, like torches, enlighten the properties of a dynamical system. Let

z0 7! z1 7! z2 7! . . . 7! zm

= z0

be a periodic cycle of period m. We define the multiplier of the cycle as follows:

Definition. The multiplier � of the cycle above is

� := (fm)0(zi

) where zi

is any point in the cycle.

Exercise. Use the chain rule to prove that this is well-defined.

Exercise. Prove that this is really an “intrinsic” quantity in the sense that if we conjugate by anautomorphism µ 2 Aut(P1), then the multiplier in the new coordinates is equal to the multiplierin the old coordinates.

Classification of periodic cycles. The multiplier gives geometric information about the periodcycle. As should be no surprise, the map should behave like its derivative in a neighborhood of,say a fixed point. More generally, the mth iterate of f should behave like (fm)0(z

i

), near any zi

inthe cycle.

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• If |�| < 1, then the cycle is attracting,

• if |�| > 1, then the cycle is repelling, and

• if |�| = 1, the cycle is called indi↵erent.

The last case above can be excruciatingly complicated, and it breaks up into a bunch of sub-cases.

Subcase 1: The multiplier � is a root of unity, but no iterate of f is the identity map. (Note thatif the degree of f is at least 2, then it is impossible for some iterate of f to be the identity. Funexercise: Try to think of an example of a rational map f : P1 ! P1 which is not the identity, suchthat some iterate of the map is the identity.) Anyway, back to business: if � is a root of unity, andno iterate of f is the identity, then the cycle is called parabolic.

Subcase 2: If |�| = 1, and � is not a root of unity, then ?????? This case splits up into two pieces.This is hard. This is really a question of linearizability.

(1) If fm(z0) = z0, and if there is a neighborhood U containing z0 for which fm is conjugate tow 7! �w, (that is, if fm is conformally conjugate to an ‘irrational rotation’), or equivalently,fm is linearizable), then U is called a Siegel disk.

(2) fm(z0) = z0, and if there is NO neighborhood U containing z0 for which fm is conjugateto w 7! �w, that is, fm is not linearizable, then the point z0 is called a Cremer point.

If a rational map is linearizable at a fixed point, then this means that the map looks like w 7! �w,where � is the multiplier of the fixed point. We should be able to picture what a linearizablemap looks like near its fixed point. If, on the other hand, the map is NOT linearizable, then it ismuch less clear what is going on in a neighborhood of the fixed point - Cremer points are highlymysterious, and very strange. Almost by definition, we do not understand the range of the di↵erentkinds of behavior that rational maps can exhibit in a neighborhood of a Cremer point.

A new parameterization. Instead of using the standard fc

: z 7! z2 + c form for quadraticpolynomials, we will change coordinates and use

fa

: z 7! z(z + a).

Note that the critical point of this family is z0 = �a/2. Further note that there is a fixed pointat p = 0, and the multiplier at the fixed point is � = a. For each value of a 2 C, we get a newquadratic polynomial f

a

, with a fixed point at p = 0, with multiplier a. So we can use FractalStreamto draw the picture in the a-plane. Here is the script (we have to change a to c for FractalStream’sbenefit).

par set z to �c/2.iterate z2 + cz until z escapes.

This will yield the following picture. Note that the parameterization z 7! z(z + a) does not give amoduli space for the family of quadratic polynomials, since for any a 2 (a-plane) � {1} there is aunique a0 2 (a-plane)� {1} for which the polynomials

z 7! z(z + a) and z 7! z(z + a0)

are conjugate by an automorphism of C. In fact, this a-plane is a ramified double cover of the c-planefor the family z 7! z2 + c, and the c-plane IS a moduli space for quadratic polynomials.

Notice that there is some interesting structure in this picture. There are bulbs growing o↵ of theunit circle in this picture; they are attached at all of the roots of unity. If we click on one of

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these points and look in the dynamical planes, we will see pictures of filled-Julia sets for quadraticpolynomials, each of which has a parabolic fixed point at p = 0.

Figure 6. Notice the two prominent disks in this picture. The one on the left isthe unit disk. These two disks are tangent to each other at a = 1. This a-plane forthe family f

a

: z 7! z(z + a) is a ramified double cover of the c-plane for the familyfc

: z 7! z2 + c.

Figure 7. This is a picture of the dynamical plane for z(z + a), where a = 1. Youcan see the forward orbit of a point close to the parabolic fixed point at p = 0. Ithas a spiraling behavior. It comes from the parameter labeled ‘a’ in the a-plane.

What can we say about these di↵erent kinds of periodic cycles?

Definition. Let z0 7! . . . 7! zm

= z0 be an attracting periodic cycle for a rational map f : P1 ! P1.The set of points z 2 P1 for which the sequence

z, fm(z), fm(fm(z)), fm(fm(fm(z))), . . .

converges to some zi

in the periodic cycle, is called the basin of attraction of the cycle. (We haveto take the m-th iterate in the sequence above since the length of our cycle is m).

Note that a basin of attraction is automatically an open set.

Proposition. Every attracting periodic orbit is in the Fatou set, and hence every basin of attractionis in the Fatou set also. All repelling periodic points are in the Julia set.

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Figure 8. Here is the dynamical plane for z(z+a), where a = i. It is the parameterlabeled ‘c’ in the a-plane. There is a parabolic fixed point at p = 0. You can seewhat happens to orbits near this fixed point.

Figure 9. Here is the dynamical plane for z(z + a), where a = exp(2⇡i↵), where↵ is the golden ratio. Note that the fixed point at p = 0 is NOT parabolic. Tofigure out if it is a Cremer point, we need to determine if f is linearizable in aneighborhood of the fixed point. This is usually very hard to do. It turns out thatfor this example, f is conjugate to w 7! �w in a neighborhood of 0. You can seethis in the picture - you can see the orbit of a point under f - look at the rotation!NOTE: if p were a Cremer point, we could not draw a picture :(

Proof. Let’s specialize to the case of a fixed point: f(z0) = z0, and set � := f 0(z0). Since z0 isan attracting fixed point, |�| < 1. By Taylor’s theorem, choose c < 1 so that |�| < c < 1, andwe have |f(z) � f(z0)| = |f(z) � z0| c|z � z0| for z close enough to z0. This implies that thesuccessive iterates converge uniformly to the constant function z 7! z0. This implies that there isa neighborhood around z0 for which the family of iterates is normal.

If z0 is a repelling fixed point, then there can be no neighborhood around z0 on which the familyof iterates is normal. Indeed, if there were such a neighborhood, the sequence of derivatives

f 0(z0), (f2)0(z0), (f

3)0(z0), . . . or �,�2,�3, . . .13

will converge (uniformly on compact subsets) to the derivative of the limit function. But thissequence n 7! �n diverges to 1 since |�| > 1. Hence the repelling periodic points are in the Juliaset.

Proposition. The parabolic cycles are contained in the Julia set.

Proof. Suppose that z0 = 0 is a parabolic fixed point of f . There is an iterate of f , say fm, forwhich there is a neighborhood of 0 where fm has the following power series expansion

w 7! w + aq

wq + aq+1w

q+1 + . . . , q � 2 and aq

6= 0.

This means that the k-th iterate of fm has the series

w 7! w + kaq

wq,

so the q-th derivative of fmk at 0 is equal to q|kaq

, which diverges to 1 as k ! 1. So by theWeierstrass Uniform Convergence Theorem, there is no subsequence of the sequence k 7! (fm)k

which converges in a neighborhood of z0 = 0, so the parabolic fixed point z0 must be in the Juliaset.

What can we say about these interesting sets?

Proposition. Let f : P1 ! P1 be a rational map of degree d � 2. Then the Julia set Jf

6= ;.

Proof (extracted from Milnor, p46) If the Julia set is empty, then ⌦f

= P1; that is, the entireRiemann sphere is in the Fatou set. This means that the family of iterates is normal on all of P1.So some subsequence of iterates j 7! fnj would converge, uniformly over all of P1, to a holomorphiclimit g : P1 ! P1. The map g is necessarily rational. A topological argument implies that thedegree of fnj should equal the degree of g for j large enough. Indeed, for j large enough, for allz 2 P1, we can deform fnj (z) to g(z) along the shortest geodesic between the points, giving ahomotopy between fnj and g. Maps P1 ! P1 which are homotopic have equal degrees. But thesequence of degrees j 7! fnj diverges to 1, so there is no way that the sequence j 7! fnj canconverge to g. Note that this argument is using the topology of P1 in a crucial way.

Lattes Examples. While the Julia set of a rational map f : P1 ! P1, deg(f) � 2 is neverempty, it is possible for the Fatou set to be empty. This happens for rational maps which are“double-covered by a torus endomorphism.” Consider the torus T = C/(Z � iZ), equipped withthe map F : T ! T given by z 7! (1 + i)z. Note that all periodic cycles of this map are repelling.Furthermore, note that the periodic cycles of this map are dense in the torus. Let p : T ! P1 bea ramified double cover of degree 2 (think of the Weierstrass } function). In a fundamental squarefor T , the function p is ramified at the lower left vertex, halfway up the vertical edge, halfway downthe horizontal edge, and right in the middle of the square. The map p is a ramified cover of degree2. One can show that the torus map F : T ! T descends to P1 to yield a map f : P1 ! P1 so thatthe following diagram commutes

TF

//

p

✏✏

T

p

✏✏

P1 f

// P1

and f is holomorphic, hence rational. Any map that is covered by a torus map in this way is calleda Lattes map. As you can see, there is something Euclidean about them. They will come back tohaunt us in future discussions.

In this particular example, the repelling periodic cycles of the map F : T ! T descend to giverepelling periodic cycles of the map f : P1 ! P1. Since the cycles are dense upstairs, they are dense

14

downstairs, and the closure of the repelling cycles is the whole sphere P1. As we have seen,

{repelling periodic cycles} ✓ Jf

,

and since Jf

is closed,

{repelling periodic cycles} ✓ Jf

=) Jf

= P1.

There are other maps which do not come from this construction, for which the Julia set is all ofP1. As a note of independent interest, Lattes maps are postcritically finite! (This is why we willsee them again).

Class 5: January 23, 2014 - thanks Paul!

We will see today that for purposes of identifying the Fatou and Julia sets of a rational mapf : P1 ! P1, it is hard to use the definition. For instance, if we want to draw these sets on thecomputer, how do you tell your computer to test if a family of functions is normal? hmm... Itturns out that there are some nice theorems which make these things much easier.

Theorem. Let f : P1 ! P1 be a rational map of degree d � 2. Let z1 2 Jf

, and let N be anyneighborhood of z1. Then

Jf

✓[

n>0

fn(N);

in fact,

P1 � [

n>0

fn(N)

!

contains at most 2 points.

This theorem says that the Julia set really “mixes things up.”

Proof. Consider P1 � [fn(N). This can contain at most 2 points. If it contained more, then thefamily of iterates

{f, f2, f3, . . .}would be normal on N since every map in this family takes values in P1 � {3 points}. But thatwould mean that z1 2 ⌦

f

, contradicting the fact that z1 2 Jf

, so we see that P1�[fn(N) containsat most 2 points. Note that this set of 2 points must be TOTALLY invariant - that is, both pointsmust be critical, and both must map forward with degree d (the points cannot have extra inverseimages). This means that the points are both critical, and periodic. This means that they are bothin ‘superattracting cycles” for the map f ; hence, the 2 points are in the Fatou Set, and we havethat

Jf

✓[

n>0

fn(N).

Corollary. If Jf

contains an interior point, then Jf

= P1.

Corollary. If A is the basin of attraction for an attracting periodic orbit, then Jf

= @A, thetopological boundary of A.

Upshot: In order to determine the Julia set (and hence the Fatou set), all we have to do isfind ONE basin of attraction for an attracting periodic orbit! Then we just draw this basin, andfigure out what the boundary is. This will tell us what our Julia set is, and then we just takethe complement to get the Fatou set. Excellent! So in order to tell FractalStream how to drawpictures, all we have to do is to find an attracting periodic cycle.

15

Application: polynomials. Every polynomial has an attracting fixed point at 1. So in Frac-talStream, we compute the basin of 1 (this is the colorful part), and the boundary of this basin isthe Julia set. The complement of the basin is the filled-Julia set.

Problem. In general, how do we find attracting periodic cycles?

Solution. Critical points!

Phenomenal Fact: Let f : P1 ! P1 be a rational map of degree d � 2. Let z0 7! . . . 7! zm

= z0be an attracting periodic cycle of period m. Then there is a critical point p of f for which thesequence

p, (fm)(p), (fm)2(p), (fm)3(p) . . .

converges to zi

for some 0 i m � 1. That is, p is contained in the basin of attraction of thecycle.

Corollary. There can be AT MOST 2d�2 attracting periodic cycles for a rational map f : P1 ! P1

of degree d.

Corollary. In order to determine if the rational map has any attracting cycles, all we have to do isfollow around all of the critical points to see what they do under iteration; if there is an attractingcycle, the critical point will definitely find it!

Application: quadratic polynomials. This is one reason that quadratic polynomials are soincredibly nice. When viewed on the Riemann sphere, a quadratic polynomial p : P1 ! P1 is arational map of degree 2, with one fixed critical point (the point at1), and one other critical point,which does... whatever. It is the behavior, or the fate of this critical point that is responsible forthe WIDE array of di↵erent behavior that we see. So if we want to know if a quadratic polynomialhas an attracting cycle, all we have to do is to follow around the extra critical point and see whereit goes under iteration.


The short-term goal is to get a local picture in our minds about what a rational map f : P1 ! P1

looks like in a neighborhood of a fixed point, say f(0) = 0, depending on whether the fixed pointis attracting, repelling, indi↵erent. In the attracting and the repelling cases, the map is conjugatein a neighborhood of the fixed point to w 7! �w, where � = f 0(0) is the multiplier. The followingtheorem makes this precise. We ollow the statement in Milnor, ch8.

Theorem. (Koenig’s Linearization) Suppose f(0) = 0. If the multiplier of the fixed point at 0,� satisfies |�| 6= 0, 1, then there exists a local holomorphic change of coordinate w = �(z), with�(0) = 0, so that ��f ��1 is the linear map w 7! �w for all w in some neighborhood of the origin.Furthermore, � is unique up to multiplication by a nonzero constant.

That is, there is a neighborhood U containing 0 so that f(U) ✓ U , and the following diagramcommutes:

(U, 0)

f

✏✏

�

// (C, 0)

w 7!�w

✏✏

(U, 0)�

// (C, 0)

Sketch of proof. The idea here is that when |�| < 1, we can find a neighborhood of the origin sothat |f(z)| < c|z| where c < 1, and z is in some small disk D

r

, centered at 0. If we want to build16

a map �, which conjugates our map f to w 7! �w in a neighborhood of 0, here is how we shouldstart:

(U, 0)

f

✏✏

�k// (C, 0)

w 7!�w

✏✏

(U, 0)�k�1

//

f

✏✏

(C, 0)

w 7!�w

✏✏

...

f

✏✏

...

w 7!�w

✏✏

(U, 0)�1//

f

✏✏

(C, 0)

w 7!�w

✏✏

(U, 0)id// (C, 0)

Start with the identity map on the bottom, and lift it to a map �1. Continue this way, definingthe map

�k

(z) :=fk(z)

�k.

By the way the map was constructed, we will have �k�1(f(z)) = � ·�

k

(z). So we want to constructour map � as the limit of the sequence of �

k

as k ! 1. We need to check that this sequenceconverges uniformly on an open set, namely the disk D

r

. It is a quick complex analysis proof (usingTaylor’s theorem) to prove that the sequence of maps k 7! �

k

converges uniformly on Dr

to aholomorphic limit � : D

r

! C. This map � is what we want. We will see this ladder constructionlater in the course as well.

NOTE: it follows from the way that we constructed �, that it is a biholomorphism; that is, there issome disk D

✏

✓ C, which contains 0, where there is a holomorphic map ✏

: (D✏

, 0)! (Dr

, 0) suchthat �( (w)) = w.

So the map � constructed above is local. But in fact, we can extend it to the ENTIRE basinof attraction of the attracting fixed point at 0. Here is how: let the basin of attraction of theattracting fixed point at 0 be denoted at A. Let p 2 A. Note that there is a k large enough sothat fk(p) will be inside the domain of �; that is, there is a k so large so that fk(p) 2 D

r

. Wedefine

�(p) := limn!1

fn(p)

�n.

This map � will be holomorphic, and it will be defined on the entire basin � : A! C. Moreover,it is an extension of �, which remember is a biholomorphism in a neighborhood of 0. So in fact, wehave the following commutative diagram:

(A, 0)

f

✏✏

�// (C, 0)

w 7!�·w✏✏

(A, 0)�// (C, 0)

and (A0, 0)

f

✏✏

�// (C, 0)

w 7!�·w✏✏

(A0, 0)�// (C, 0)

where A0 is the immediate basin of the attracting fixed point at 0. Recall that the immediate basinof an attracting fixed point is just the connected component that contains the fixed point.

17

The following lemma is INCREDIBLY IMPORTANT. It is a regular old complex analysis lemma,but it has huge dynamical consequences.

Lemma. (Finding a critical point.) The local inverse ✏

: D✏

! A0 extends by analytic contin-uation to some maximal disk D

R

, giving a holomorphic map : DR

! A0 so that (0) = 0 and�( (w)) is the identity map on D

R

. Furthermore, extends continuously over @DR

, and (@DR

)necessarily contains a critical point of f .

The statement in is what we are after. This tells us that there is a critical point of the map f inthe immediate basin A0.

Proof. First, extend ✏

to a map through radial lines in D✏

. There must be an obstruction tothis extension, else we could extend

✏

to a holomorphic map : C ! A0 having the propertythat �( (w)) is the identity map on C. Note that this would mean that (C) ✓ A0 in other words : C ! A0 is a holomorphic map. What can we say about (C)? For instance, let’s examineP1 � (C). Note that

f | (C) : (C)! A0

is injective (using the commutative diagram). The set P1� (C) must in fact be infinite; if it werefinite, then f would extend to an injective holomorphic map P1 ! P1 that is, f would be a Mobiustransformation. But f : P1 ! P1 is a rational map of degree at least 2, so P1 � (C) is infinite.This means that

(C) ✓ P1 � {at least 3 points}therefore we have a holomorphic map : C ! a hyperbolic space which implies that must beconstant (by Picard’s theorem, or by lifting to universal covers and using Liouville). We thereforeMUST have an obstruction to extending the map

✏

to the whole complex plane. So there is somemaximal disk D

R

on which one can extend ✏

to a map : DR

! A0 so that �( (w)) is theidentity map on D

R

.

Now, here is the idea behind the statement in red. If (@DR

) did not contain a critical pointof f , then we could take a local inverse of f at any point in (@D

R

), and use those inverses viathe commutative diagram, to extend the map to a strictly larger disk. This contradicts themaximality of the disk D

R

, and we have bumped into a critical point. We will see this again.

Takeaway. To determine if a rational map f : P1 ! P1 of degree at least 2 has an attractingcycle, follow around the 2d� 2 critical points to see if they are attracted anywhere. If there is anattracting cycle, one of the critical points will find it.

Challenge Exercise. Use FractalStream to draw the Julia Set of f : z 7! 1� 1/z2. This is NOTa polynomial, so we don’t have a good way to draw this set yet.

Remark. We are specializing to the case of a fixed point f(z0) = z0, or even f(0) = 0 to simplifyour presentation. To extend the results to the more general case of periodic cycles, one can justlook at the appropriate iterate of the map; for instance, given the periodic cycle

z0 7! z1 7! · · · 7! zm

= z0,

our analysis of the local theory of fixed points applies to the fixed points of fm; note that fm(zi

) = zi

for all i = 0..m.

18

Figure 10. The original picture is a picture of Milnor. This is a picture of the filledJulia set for the quadratic polynomial z 7! z2 + 0.7z. The origin is an attractingfixed point with multiplier � = 0.7. There is a local change of coordinate � definedon some neighborhood of 0 which conjugates f on this neighborhood to the mapw 7! �w in a neighborhood of the origin in C. That map can be extended to a map� : A ! C as discussed previously. Note that the interior of the filled Julia setis the basin A for the attracting fixed point at 0. The curves that are drawn are��1(circles in C, centered at 0). Notice that there is a figure ‘8’ that passes throughthe critical point of f .

Class 7, February 4, 2014

We now strive to get a local picture for the parabolic fixed points. Let’s suppose that f(0) = 0, andthat � = f 0(0) = 1. The more general case is when � is a root of unity, but we begin by focusingon the case where it is equal to 1. Locally, for z near 0, we can write

f(z) = z(1 + azn + higher order terms), n � 1, and a 6= 0.

The number n + 1 is called the multiplicity of the fixed point. Why this terminology? Parabolicpoints fixed points are strange; they correspond to a lack of transversality or a failure of the implicitfunction theorem. A parabolic fixed point with multiplier � = 1, is a repeated root of the fixedpoint equation

f(z)� z.

So the “multiplicity” n + 1 is the algebraic multiplicity of the fixed point z0 = 0 as a root of thefixed point equation f(z)� z, which you can see by using the form above.

Definition. A complex number v is a repulsion vector for f at 0 if the product navn = 1. A complexnumber v is an attracting vector if navn = �1. Think of v as a tangent vector anchored at theparabolic fixed point. Note that these abstract and repulsion vectors are immediately determined

19

by the normal form of f above - all you need is the quantity n and the quantity a. Once you havethem, you can draw a “cartoon” which gives a local picture of the dynamics of f near the fixedpoint at the origin.

There will be n equally spaced attracting vectors and n equally spaced repelling vectors all anchoredat the fixed point. There will be an attracting vector between any two repelling vectors and viceversa.

Question. Are the quantities n and a intrinsic?

Answer. The quantity n is intrinsic - meaning that in any local model (in any choice of coordinates)we will be able to recover this same quantity. This means that it is really a geometric quantity -in fact, it is equal to the number of attracting petals (or equivalently, it is equal to the number ofrepelling petals). The quantity a gives the length of the repelling vectors (and attracting vectors).This number gives the “size” of the petals. It is NOT intrinsic; you can conjugate by Mobiustransformations and change this number, whereas you will not be able to change the number n byconjugating.

Figure 11. This is a picture from an early version of Milnor’s book. In this ex-ample, n = 3. There are three attracting vectors (red), and three repelling vectors(blue). Each vector has a “petal” associated to it. If P is a petal surrounding anattracting vector v, then points in P converge (under iteration of f) to the fixedpoint, asymptotically to the direction v. If Q is a petal surrounding a repellingvector u, then points in Q converge (under iteration of f�1 [which is well-defined ina neighborhood of our fixed point!] to the fixed point asymptotically to u.

Notice that the repelling petals and the attracting petals overlap; this is not inconsistent. If apoint is in the intersection of a repelling petal and an attracting petal, then it will map away from

20

the repelling vector in the repelling petal, and map toward the attracting vector in the attractingpetal.

Exercise. Consider the filled Julia Set of f(z) = z(z + 1) (alternatively, consider the filled JuliaSet of p(z) = z2 + 1/4; these polynomials are conjugate by an a�ne transformation, so the filledJulia sets will be related by the a�ne map that conjugates the polynomials. The filled Julia set off is called the cauliflower. Make a cartoon for the attracting and repelling directions; sketch somepetals, and use FractalStream to study the dynamics near the parabolic fixed point.

Figure 12. This is a picture of the filled Julia Set for the map f(z) = z(z + 1).There is a parabolic fixed point with multiplier 1 at z = 0. You can use the curvesin the picture to study the dynamics of f in its filled Julia set near a parabolicpoint. Try to see where the petals should be. There is one attracting petal, and onerepelling petal.

Definition. As we discussed in class, there is a petal associated to attracting vectors and associatedto repelling vectors. Milnor (chapter 10) defines a petal for the attracting vector v as follows: LetN be a neighborhood of the parabolic fixed point, and suppose that f restricted to N is injective. Aset P ✓ N is a petal associated to v if P is simply-connected, if f(P ) ✓ P , and if for all z 2 P , theforward orbit of z under f converges to the parabolic fixed point, asymptotically in the directionof v. Think about how you would define a repelling petal associated to a repelling vector u.

As we have seen, if f : P1 ! P1 is a rational map of degree at least 2, and if f has a parabolicfixed point f(z0) = z0 (or more generally, a parabolic cycle), then there is a nonempty set of pointsthat are attracted to z0 under iteration. For example, there are points in the attracting petals thatare attracted (asymptotically along the attracting directions) to the parabolic fixed point. Thatmotivates the following definition

Definition. The set of points that are attracted to the parabolic cycle under iteration is calledthe basin of the cycle.

21

Fact. The basin of a parabolic cycle of f is contained in the Fatou set of f .

Fact. The boundary of the parabolic basin is equal to the Julia Set. (The proof of this imitatesthe analogous statement about the boundary of the basin of attraction for an attracting cycle. Wehave seen this before).

Fact. (We have already seen). All parabolic cycles of f are contained in the Julia set of f .

FACT! Just as in the attracting case, the basin of every parabolic cycle of f necessarily containsa critical point of f .

Application! To draw a picture of the Julia set of a rational map, all we have to do is find aparabolic cycle, ask the computer to color the basin of this cycle, and the boundary of that basinwill be the Julia set!

Question. How do you tell if f has a parabolic cycle?

Answer! Follow around the critical points! If f has a parabolic cycle, then one of the 2d � 2critical points will necessarily find it!

Application: Quadratic Polynomials. Consider the family fc

: z 7! z2 + c. This is a rationalmap of degree 2, and as such, it has 2 critical points. HOWEVER, one of them, namely 1, hasbeen spoken for; it is fixed. There is only one ‘free’ critical point, or active critical point thatmight 1) find an attracting cycle, or 2) find a parabolic cycle, or 3) do something else! It is thisactivity of this free critical point that gives us the beautifully rich parameter space picture that isthe Mandelbrot Set. We now know that quadratic polynomials are ‘simpler’; not only because theyare only of degree 2, but because there is only ONE critical point that can get into trouble. Thismeans that there is only room for an attracting cycle OR a parabolic cycle OR possibly somethingelse. But quadratic polynomials are simpler because only one of these things can happen for agiven polynomial.

Example. Consider the cubic polynomial f(z) = z2(z+2i). This polynomial has a superattractingfixed point at p = 0, and it has a parabolic fixed point at q = �i. The critical points of f are x = 0and y = �4i/3.

How many cycles? Let f : P1 ! P1 be a rational map of degree d � 2. How many periodiccycles does f have? To find cycles of length n, we just have to solve the equation fn(z) � z = 0;the map fn(z) is a rational map of degree dn, so we write

fn(z) =r(z)

s(z)and fn(z)� z = 0 () r(z)� zs(z) = 0

where r(z) and s(z) are polynomials of degree dn, so the equation fn(z) � z can be rewritten asa polynomial equation of degree dn + 1. So as n ! 1, the degree of this equation goes to 1, sothere should be infinitely many periodic cycles, right? This requires an argument! It is a prioripossible that the SAME complex numbers are repeated roots of all of these equations, giving usjust some finite number of cycles. Seems unlikely, but we need to check something. This is a goodquestion.

One thing we do know at this point is

(number of attracting cycles) + (number of parabolic cycles) 2d� 2.

One neat observation that came up, is this number 2d�2, which is of course, the number of criticalpoints that f has. But it is ALSO something else!

22

Figure 13. Here is a dynamical picture for the polynomial f(z) = z2(z+2i). Thereis a superattracting fixed point at 0, and the basin of this point is colored red. Thereis a superattracting fixed point at 1; the basin is blue. There is also a parabolicfixed point at �i, and the basin of this point is colored light blue. Look at theshading in this picture - you can use it to see the dynamics! Compare this to theprevious picture. The critical point at 0 is attracting to the attracting fixed pointat 0. The critical point at 1 is attracted to the attracting fixed point at 1. Thecritical point at �4i/3 is attracted to the parabolic fixed point at �i.

Moduli space of rational maps. Consider the space of rational maps of degree d; write f(z) =p(z)/q(z), where p and q are polynomials of degree d. Given a list of 2d+ 2 complex numbers, wecan write down a rational map, just using the coe�cients of the numerator and denominator. Thisis naturally a point in P2d+1 because these coe�cients are determined up to scaling. So we mayidentify the space of rational maps of degree d with an open subset of P2d+1. Since we are studyingdynamical systems, we want to study rational maps up to conjugation by Mobius transformations.Define the space

Md

:= Ratd

/(conjugation by Mobius transformations)

This space is a complex orbifold and an a�ne algebraic variety of dimension 2d � 2. Interesting!We get a dimension for every critical point!

This is consistent with what happens for quadratic polynomials fc

: z 7! z2 + c. The parameterspace, or moduli space of quadratic polynomials is given by the c-plane; it is clearly 1-dimensional.How many critical points do maps in this family have? Well, a quadratic polynomial has 2 criticalpoints. But we are looking at polynomials - one critical point (namely, the point at1), is fixed forall of the maps f

c

: z 7! z2 + c, and there is only ONE ‘free’ critical point, namely z0 = 0.

We will continue with the last picture to keep in mind; that is, suppose f(0) = 0, and that� = f 0(0) = 0. What does f look like in a su�ciently small neighborhood of 0?

23

Class 8 - February 6, 2014

Today we talk about the model of a rational map f : P1 ! P1 in a neighborhood of a superattractingfixed point. We may suppose that f(0) = 0, and that there is a neighborhood around 0 on whichf is of the form:

f(z) = an

zn + an+1z

n+1 + · · ·where n � 2, and a

n

6= 0. We guess that f is conjugate to the map z 7! zn in a neighborhood of0. How do we find the conjugating map? Here is a guess (use the ladder diagram idea from classto find this guess):

�k

(z) = nkqfk(z)

but this is a little fishy - we have to know what root to take. We want to prove that this sequenceof maps (once we know which root to take), converges on an open set containing 0.

Theorem. [Bottcher, 1904] Let f(z) = an

zn + an+1z

n+1 + · · · n � 2, and an

6= 0. There existsa local change of coordinate �, with �(0) = 0, conjugating f to w 7! wn in a neighborhood of 0.Furthermore, � is unique up to multiplicating by an (n� 1)st root of unity.

The proof is hardcore complex analysis - see Milnor ch 9. Note the Milnor defines which roots totake.

Taking what happened from the linearizing coordinate as inspiration, we can ask: “does � extendthroughout the entire basin of 1?” The answer is: not always. Note that there will be trouble ifthe basin is not simply connected (since there is a choice of a root involved), or if there is a criticalpoint in the basin, there could be trouble.

We discussed that this theorem is HUGE in polynomial dynamics, and we looked at checkerboardpictures using polynomials.

Figure 14. The model map, z 7! z2 is on the right. The checkerboard pattern isgiven by polar coordinates in the basin of 1. The Bottcher coordinate � extendsthroughout the entire basin of 1, conjugating the basilica map on its basin of 1 tothe squaring map on its basin of 1.

There are many applications of this to polynomial dynamics. We will see this next time.


We discuss extending the Bottcher coordinate �; as previously mentioned, the map does not nec-essarily extend throughout the entire basin of attraction of the superattracting fixed point. Therewill be trouble if there are critical points in the basin (not including the superattracting fixed point

24

itself), or if the basin is not simply connected as there will be monodromy obstructions. Eventhough the Bottcher coordinate does not necessarily extend, the absolute value of it DOES. Thisfollows from the construction of the map �.

Fact. Let f : P1 ! P1 be a rational map of degree at least 2. Suppose that f(0) = 0, and that� = f 0(0) = 0. Let � : (U, 0)! (D, 0) be the Bottcher coordinate defined on a neighborhood U ofthe superattracting fixed point at 0, and let A be the basin of attraction of 0. Then the map

|�| : A! [0, 1) given by p 7! |�(fk(p))|1nk for large k,

is well-defined and continuous on A.

Note that when FractalStream draws pictures of our fractals, we see level curves of this function|�| in the basin of attraction of the superattracting fixed point.

Figure 15. Here is the filled Julia set for the basilica polynomial; in green is thebasin of 1. Note that 1 is a superattracting fixed point for a polynomial, so inparticular, there is a Bottcher coordinate defined in a neighborhood of1, which con-jugates the basilica map to z 7! z2 near 0. In this example, the Bottcher coordinate� extends to the ENTIRE basin of attraction of 1.

The following theorem makes this precise (it is theorem 9.3 in Milnor).

Theorem (Critical points in the basin). Let f : P1 ! P1 be a rational map of degree d atleast 2. Suppose that f(p) = p, and that � = f 0(p) = 0. Let � : (U, p) ! (D, 0) be the Bottchercoordinate which conjugates f : (U, p) ! (U, p) to the map w 7! wn on a neighborhood of 0 inD. Note that there is a holomorphic inverse

✏

, defined on some small neighborhood of 0 so that�(

✏

(w)) is the identity map on this neighborhood. There is a unique disk Dr

, 0 < r 1 such that ✏

extends holomorphically to a map from Dr

to A0, the immediate basin of p. If r = 1, then maps the unit circle (ie, the boundary of D) biholomorphically onto A0, and p is the ONLY criticalpoint in A0. However, if r < 1, then there is at least one other critical point in A0, lying on theboundary of (D

r

).25

Figure 16. This is the filled Julia set for z 7! z3 + z2. There is a superattractingfixed point at z = 0, and of course there is also a superattracting fixed point at1. The Bottcher coordinate in the basin of 1 extends throughout the entire basin(grey). However, the Bottcher coordinate defined near z = 0 DOES NOT extendthroughout the entire basin of 0; it only extends until it “bumps” into a criticalpoint. Note that there is an extra critical point of the polynomial in the basin of 0.So the Bottcher coordinate � cannot possibly extend throughout the entire yellowregion. However, |�| IS defined in the entire yellow region.

Phew!

Proof. Following parallel arguments to those used when we tried to extend the linearizing map, wefirst extend

✏

to a larger disk by analytic continuation; we can either extend it to the whole unitdisk D, or there is some r < 1 on which we can extend it to D

r

. In either case, the plan is to showthat maps D

r

biholomorphically onto its image. Note that can have NO critical points (thinkabout why??). So is locally injective, and we will show that it is globally injective. Recall that|�( (w))| = |w| for w close to 0, and thus for all w 2 D

r

by analytic continuation. Suppose that (w1) = (w2) with w1 6= w2. Then applying |�| yields that |w1| = |w2|. Choose a pair (w1, w2)so that (w1) = (w2), w1 6= w2, so that |w1| is minimal. (Why can we do this? Hint: considerthe set of all (w1, w2) 2 D

r

⇥ Dr

so that (w1) = (w2)). Anyway, since is an open map, if wechoose any w0

1 su�ciently close to w1. Find w02 su�ciently close to w2 so that (w0

1) = (w02) (why

can we do that?). If you choose w01 so that |w0

1| < |w1|, we get a contradiction. So we must havethat is injective on D

r

.

If r = 1; that is, if ✏

extends to a holomorphic map : D ! A0, then in fact (D) = A0; thatis, the Bottcher coordinate � extends to a map on the ENTIRE immediate basin A0 ! D, and itextends as a biholomorphism which conjugates f : (A0, p) ! (A0, p) to the map (D, 0) ! (D, 0)given by w 7! wn. To see this, if (D) is only a proper subset of A0, there would be a point z0 2 A0

in the boundary of (D). Approximate z0 by points (wi

), where wi

2 D, and examine

|�( (wi

))| = |wi

|.26

Taking a limit as |wi

| ! 1, we see that |�(z0)| = 1, which is impossible since z0 2 A0, and|� : A0 ! [0, 1).

Now if r < 1, the proof that there must be a critical point of f in the boundary of (Dr

) followsfrom the fact that if this were not the case, we could locally invert f in a neighborhood of everypoint in the boundary of (D

r

), so we could actually extend the map to a larger domain. Thisis a parallel argument to that used when extending the linearizing coordinate in the case of anattracting fixed point (or repelling fixed point).

NOTE: Even if the Bottcher coordinate extends throughout the entire immediate basin; or equiv-alently, if there is a map : (D, 0) ! (A0, p) which conjugates f : A0 ! A0 to the map(D, 0) ! (D, 0) given by w 7! wn, we may NOT get a continuous extension of to the bound-ary of D; moreover, even if we do get a continuous extension, we may not get a homeomorphism.Sometimes we will, sometimes we won’t. This is where things get interesting.

Okay, now how are Bottcher coordinates useful??

Polynomials. Let f : C! C be a polynomial of degree d; so we may write

f : z 7! ad

zd + · · · a1z + a0, where d � 2, ad

6= 0.

We may conjugate f by an a�ne transformation to make it monic, so we might as well have assumedthat

f(z) = zd + ad�1z

d�1 + · · · a1z + a0.

Recall that associated to any polynomial is the filled Julia set; that is

Kf

:= {z 2 C|n 7! fn(z) is bounded as n!1}.

Lemma. For any polynomial of degree at least 2, the set Kf

is compact, full, the interior of Kf

isequal to the union of all bounded Fatou components of f .

Proof. The ratio f(z)/zd tends to 1 as z ! 1, so there is some r0 > 0 so large, so that for all|z| � r0, ��

f(z)

zd� 1

�� <1

2=) 8 |z| � r0, |f(z)| > 2|z|.

This implies that for all |z| > r0, z 2 B1, where B1 is the basin of 1. This means that Kf

is bounded. Moreover, Kf

= P1 � B1, and since B1 is open, Kf

is closed, and therefore Kf

iscompact.

To prove that Kf

is full, we must show that its complement, B1, is connected. Let U be anybounded component of the Fatou set C � J

f

(note: there may not be any!). Then we must havefor all z 2 U , and for all n � 1, we have |fn(z)| r0 because otherwise, by the Maximum ModulusPrinciple, there would be some z 2 @U for which |fn(z)| � r0. But this means that z 2 B1,however, @U ✓ J

f

, so z 2 Jf

, which contradicts the fact that z 2 B1.

So every bounded component of the Fatou set is in Kf

, and the unique unbounded component isB1. Clearly, @B1 = J

f

. Let U be a bounded component of Kf

, and let � be a simple closed curvein U . Let V be the bounded component of C� �. It follows from the maximum modulus principlethat V ✓ K

f

. Note that in fact, V ✓ U since V \U 6= ;, and if there is z 2 V �U , then V \Jf

6= ;.But this would mean that V \B1 6= ;, but this means that V cannot be a subset of K

f

, which isa contradiction.

So every connected component of the filled Julia set is simply connected. This is very cool.27


Let f(z) = zn + an�1z

n�1 + · · ·+ a1z + a0. We will explore the connections of the topology of thefilled Julia set K

f

, and the orbits of the critical points. In fact, there is a lot of topology that willbecome relevant to the subsequent discussions.

Theorem. (Connected Kf

() all critical points in Kf

)

Let f(z) = zn + an�1z

n�1 + · · · + a1z + a0, n � 2. If all critical points of f are contained in Kf

,then K

f

and Jf

are connected, and the Bottcher map defined in a neighborhood of 1 extendsthroughout the entire basin of 1, and it conjugates

f |C�Kf : C�Kf

�! C�Kf

to w 7�! wn on C� D.

However, if there is at least one critical point c 2 B1 = C�Kf

, then Kf

and Jf

are disconnected(in fact, they have uncountably many connected components).

Proof. (Following Milnor, ch9).

First suppose that all critical points are in Kf

. It follows from the previous discussion that �extends to a map throughout the entire basin of 1 and conjugates f on its basin to the mapw 7! wn on the basin of 1. Moreover, the inverse of �, called , is defined on C � D. Let ✏ > 0,and define

A1+✏ := {z 2 C|1 < |z| < 1 + ✏}be the open annulus in C� D. Consider

(A1+✏) ✓ C.

Note that Jf

✓ (A1+✏), so we have a nested family of connected, compact sets

(A1+✏)

each of which contains the Julia set. In fact,

Jf

=\

✏>0

(A1+✏).

Note that this implies that Jf

is connected. To conclude that Kf

is also connected, we recall thatK

f

is actually simply connected, and since Jf

= @Kf

, we can conclude that Kf

is connected.

Consider the opposite direction. We know from our previous discussion, the Bottcher map �,naturally defined in a neighborhood of 1, can be extended throughout the entire basin until ithits a critical point. To be more precise, consider the inverse of �,

r

: C � Dr

! C �Kf

, wherer > 1 (here we are conjugating f in a neighborhood of 1 to the map w 7! wn in a neighborhoodof 1). Since we are formulating our question this way, r > 1 is the smallest such radius so that : C� D

r

! C�Kf

is defined. It follows that @ (Dr

) ✓ C�Kf

MUST contain a critical pointof f ; call this critical point c, and set v := f(c). Consider the ray t ·�(v) for t � 1 in C�D

r

(makesure you see why �(v) 2 C� D

r

).

Consider the ray Rv

:= (t · �(v)) in C�Kf

, and consider the inverse image

f�1(Rv

) = R1, . . . , Rn

.

Each of the Ri

will be a ray emanating from a point zi

2 C �Kf

such that f(zi

) = v. Note thatsince c 2 f�1(v), and c is a critical point, we must have that there is i 6= j so that z

i

= zj

= c, sothe rays R

i

and Rj

emanate out of c.28

CLAIM: Ri

[Rj

cuts C into two components: V0 and V1. Make sure you see why this is true, andin particular, why R

i

6= Rj

(hint: use the Bottcher coordinate �).

To finish the argument, set J0 := V0 \ Jf

and J1 := V1 \ Jf

are two open sets (open in Jf

) whichseparate J

f

; that is, they are disjoint, and their union is equal to Jf

. We therefore conclude thatJf

is disconnected. It then follows that Kf

is disconnected.

But how do we know that J0 and J1 are nonempty? To this end, note that f(V0) and f(V1) containALL points in C except for possibly the points in the ray R

v

. Think about why this is true.Moreover, f(J0) = f(J1) = J

f

.

To obtain that Jf

and Kf

have uncountably many connected components, one can continue thisprocedure by defining:

Jk0 := J

k

\ f�1(J0), Jk1 := J

k

\ f�1(J1)

to obtain J00, J01, J10, J11, four nonempty compact sets whose union is the whole Julia set. Note

Figure 17. Here is the filled Julia set of the map z 7! z3+2z2 .Note that this maphas a superattracting fixed point at z = 0. The other critical point is in the basinof infinity. Note the figure 8 curve that this critical point is sitting on. You can seethe relevant rays in this picture (compare with the discussion in the proof above).

that it follows form this discussion that if the filled Julia set Kf

is not connected, then the basinof 1 is not simply connected. So we have encountered Fatou components which are not simplyconnected. In class, I showed you an example of a rational map of degree 3 with a Herman ring.This is a Fatou component on which the rational map is conjugate to an irrational rotation of anannulus. By the maximum modulus principle, polynomials cannot have Herman rings.

Remark. In fact, if ALL critical points of the polynomial f are in the basin of 1, then Kf

= Jf

is a Cantor set, andf |

Jf : Jf

! Jf

29

Figure 18. This is a dynamical picture for the rational map f(z) =(�0.7493909542 � 0.6621277805i)z2(z � 4)/(1 � 4z). You see the orbits of a pointin white. There is a Herman ring for this rational map, and you can see it! On thisHerman ring, the map f is conjugate to an irrational rotation of an annulus. Thismap also has a superattracting fixed point at 1 and one at 0, the basins of whichare colored red and green in the picture. Note that the Herman ring is not simplyconnected.

is conjugate to the shift map on X, where X is the set of infinite strings of n symbols.

Corollary. If f is a quadratic polynomial, then Kf

is connected, or it is a Cantor set.


We continue with the discussion of topology during this class. Suppose that f : C ! C is apolynomial with connected K

f

. Then from the previous discussion, � : C � Kf

! C � D is aconformal isomorphism that conjugates f to w 7! wn. Call the inverse of this map : C �D !C�K

f

. We would like to ask

• can be defined at any point of @D?

• if yes, can it be defined to yield a continuous map � : @D! Jf

?

• if yes, is it a homeomorphism?

This turns out to be a question of topology.

Recall. A topological space X is locally connected if for all x 2 X, and for all open neighborhoodsU ✓ X, which contain x, there is a connected open neighborhood V ✓ U , which contains x. Anexample of a space which is connected but not locally connected is the “comb” that we discussedin class.

There is a nice discussion of what the issues are in chapter 17 of Milnor. Caratheodory worked outmuch of the underlying mathematics in 1913. The issue is actually NOT a dynamical one - this ispurely a result about extending a Riemann map to the boundary of a simply connected subset ofthe plane.

30

Theorem. (Caratheodory) A conformal isomorphism : D ! U ✓ P1 extends to a continuousmap from the closed disk D onto U i↵ the boundary @U is locally connected or i↵ P1�U is locallyconnected.

Proof. See Milnor, for example.

Why is this relevant to the study of dynamics? It allows us to introduce external rays. We havealready encountered them in the course, but here we lay the groundwork.

Let t 2 R/Z, and consider the ray r · exp(2⇡it) for r � 1. Note here that our angle t is measuredin ‘turns.’ Using the map , we can transport this ray over to the basin of 1 for the map f .Define

Rt

:= (r · exp(2⇡it)) for r � 1.

Definition. The ‘ray’ Rt

in C�Kf

is called the external ray at angle t.

If the map : C � D ! C � Kf

extends continuously to a map @D ! Jf

, then this provides away to LABEL points of the Julia set with angles t 2 R/Z. Moreover, we can use these angles tounderstand the map f restricted to its Julia set.

Note that f(Rt

) = Rnt

, and if the angle t 2 R/Z is periodic under multiplication by m (mod Z),then the ray R

t

will be periodic of period m as well.

Consider the limit�(t) := lim

r!1 (r exp(2⇡it)) as r decreases to 1.

If this limit exists, then we say that the ray at angle t, Rt

, lands at �(t) 2 Jf

.

The following lemma is immediate.

Lemma. If Rt

lands at �(t) 2 Jf

, then Rnt

lands at f(�(t)) = �(nt). Moreover, each of the nrays

Rt/n

, R(t+1)/n, . . . , R(t+(n�1))/n

lands at one point in f�1(�(t)), and every point in f�1(�(t)) is the landing point of at least onesuch ray.

Observe that if Rt

is periodic of period m, then �(t) 2 Jf

is periodic of period dividing m.

Question. How many rays can land at a given point? More than 1? Infinitely many?

Good questions!

The following theorem is a consequence of work of Fatou and the Riesz brothers.

Theorem. (Most rays land) For all t 2 R/Z outside a set of measure 0, the ray Rt

lands at�(t) 2 J

f

. Furthermore, for each fixed z0 2 Jf

, the number of rays Rs

for which �(s) = z0 hasmeasure 0.

We have the following criterion for landing.

Theorem. (Landing Criterion.) For any polynomial f with connected Julia set, the following areequivalent

(1) Every external ray Rt

lands at a point �(t) which depends continuously on the angle t.

(2) The Julia set Jf

is locally connected.

(3) The filled Julia set Kf

is locally connected.31

(4) The map : C�D! C�Kf

extends continuously over the boundary @D, and the extendedmap takes exp(2⇡it) to �(t) 2 J

f

.

And if any (hence all) of the conditions is satisfied, the map � : R/Z! Jf

satisfies the semiconju-gacy

�(nt) = f(�(t))

and maps the circle R/Z onto the Julia set Jf

.

Proof. Suppose that (1) holds. Then �(R/Z) ✓ Jf

is a nonempty compact subset. Choosez0 2 �(R/Z), and we note that for all n � 1,

f�n(z0) ✓ �(R/Z).

By earlier work we know that[

n>0

f�n(z0)

is everywhere dense in Jf

. This implies that

�(R/Z) = Jf

=) �(R/Z) = Jf

.

From here, we obtain (4), and from Caratheodory’s theroem, we obtain that Jf

is locally connected,which implies that K

f

is locally connected.

It is perfectly possible for the map � to be defined at ALL t 2 R/Z, but Jf

may not be locallyconnected. This happens for the Cantor bouquet Julia set.

Figure 19. This is a dynamical picture for z 7! c · exp(z). There is a unique implyconnected Faout component in black, and the Julia set is in color. It looks like itis made of fat fingers; in fact, this is not true. These fingers consist of Cantor setsof Cantor sets of Cantor sets.... worth of hairs, and the hairs are packed so tightlytogether, so that the Julia set has Lebesgue measure 0 but Hausdor↵ dimension2. There is a conformal isomorphism from D to the Fatou component; it can beextended to all boundary points @D, but it is NOWHERE continuous.

32


We continue with a monic polynomial f(z) = zn+an�1z

n�1+ · · ·+a1z+a0 of degree n � 2. Recallthat we would like our polynomial to be monic because we want f to be conjugate to w 7! wn near1 (opposed to conjugate to w 7! ↵wn near 1). We will work under the assumption that the Juliaset (and thus the filled Julia set) is connected, which means our Bottcher coordinate

� : C�Kf

! C� D

extends to the entire basin of 1, so it has an inverse : C� D! C�Kf

. Recall the ray

Rt

:= (re2⇡it) for r > 1

which lives in C�Kf

; it is called the external ray at angle t. Consider the radial limit

�(t) := limr&1

(re2⇡it)

which may or may not exist; if this limit exists, then Rt

lands at t. The following result is an easyconsequence of the Landing Criterion Theorem.

Proposition. The Julia set Jf

is a simple closed curve if and only if � : D ! Jf

is a homeomor-phism.

Theorem. (Lyubich, Douady, Sullivan). If the Julia set of f is locally connected, then everyperiodic point in J

f

is either repelling or parabolic.

This is interesting because this means that if f is a polynomial with a fixed point which is ofCremer type (this means that the fixed point is indi↵erent, and there is NO neighborhood aroundthe fixed point on which f is conjugate to w 7! �w), then the Julia set is not locally connected, so : C� D! C�K

f

does not extend continuously over @D.

Question. But are there polynomials that admit fixed points of Cremer type?

Answer. Yes! In fact, Bu↵ and Cheritat proved that there are quadratic polynomials f withCremer fixed points, such that J

f

has positive Lebesgue measure.

As a consequence, we know that there are quadratic polynomials such that Jf

is not locally con-nected - remember this.

Here are some facts that comprise the proof of the theorem above.

Let f be a monic polynomial with connected and locally connected Julia set. We will first showthat if z0 2 J

f

is a fixed point, then there are finitely many rays that can land at z0. Note thatthis will imply that if z1, . . . , zm 2 J

f

is a periodic cycle, then at most finitely many rays land ateach of the points z

i

.

Since Jf

is locally connected, the map � : R/Z ! Jf

is continuous and onto. Set X := ��1(z0) ✓R/Z; this is a nonempty compact set. We now require the following lemma.

Lemma. Let n � 2, and let X ✓ R/Z be a compact set. Suppose that the map t 7! nt mapsX ! X homeomorphically. Then X is finite.

Proof. This is a homework exercise.

Armed with the lemma and the discussion in the paragraph above, we conclude that the numberof rays that land at z0 is finite. Moreover, since z0 is not z critical point (why?), the map f sends asmall neighborhood of z0 di↵eomorphically to itself; this means that the little pieces of the finitely

33

many rays landing at z0 inside this neighborhood get permuted by f , so in particular, every ray Rt

landing at z0 is periodic under f .

Taking some iterate of f , we reduce to the case where there is a fixed ray Rt

which lands at z0.Now we study how f maps R

t

to itself.

Lemma. If a fixed ray Rt

lands at z0, then z0 is either a repelling fixed point, or z0 is a parabolicfixed point.

It is clear that z0 will be a fixed point (since Rt

is a fixed ray). One can prove this lemma byparameterizing the ray R

t

(by arc length for instance), and by studying what f does to Rt

. Thelocal picture is that f will push points of R

t

away from z0; this local picture forces z0 to be eitherrepelling or parabolic (see Milnor ch 18 for details).

The Sullivan, Douady, Lyubich theorem then follows: if Jf

is locally connected, then necessarilyall periodic cycles of f that live in J

f

are repelling or parabolic.

Remark. Note that it is a corollary of the previous discussion that if Jf

is locally connected, andif z0 is a fixed point, then finitely many rays land at z0.

Periodic points. The ray Rt

is obviously periodic if t = mt mod Z. Note that Rt

is eventuallyperiodic if and only if t = p/q is rational.

Recall that we are working in the situation where f is a monic polynomial of degree n, and Jf

isconnected but not necessarily locally connected.

Theorem. (Rational Rays Land) Every periodic external ray lands at a periodic point which iseither repelling or parabolic. If t is rational but not periodic, then R

t

lands at a point which iseventually periodic but not periodic.

Note that the ray at angle 0 ALWAYS lands at a fixed point, and the ray at angle 1/n always landsat an inverse image of this fixed point.

Topological models. If Jf

is connected and locally connected, then the fact that � : D ! Jf

iscontinuous and semiconjugates

f |Jf : J

f

! Jf

to the map t 7! nt on R/Z

is very nice. In fact, one can use this map � to obtain a topological model for Jf

.

Definition. A lamination is an equivalence relation on the circle S1 such that the convex hulls ofdistinct equivalence classes are disjoint.

Here is what this means: suppose ⇠ is the equivalence relation on S1; if x ⇠ y, then connect xto y with a hyperbolic geodesic in D; take the convex hull of the union of geodesics correspondingto points in the same equivalence class. This is “one leaf” of the lamination. The leaves will bedisjoint.

A natural equivalence relation. Say that t ⇠ t0 if and only if Rt

and Rt

0 land at the same pointin J

f

.

Theorem. Let f be a monic polynomial with connected and locally connected Julia set. Let ⇠be the induced equivalence relation on D. Then the quotient D/ ⇠ is homeomorphic to the filledJulia set of f .

Fact. The basilica and rabbit Julia sets are locally connected. This is NOT obvious.

Here is another nice picture.34

Figure 20. This is a picture of the filled Julia set for the cubic polynomial z 7!z3 � iz2 + z. Exercise: Label a few of the angles in the picture above.

Figure 21. On the left is the lamination associated to the basilica polynomial.On the right, is the lamination associated to the rabbit polynomial. In the basilicapicture, if we collapse the geodesics to points, we obtain a set which is homeomorphicto the filled Julia set of the basilica. In the rabbit picture, we see convex hullswhich are triangles, and if we collapse them to points, we will obtain a set which ishomeomorphic to the filled Julia set of the rabbit.

We will now try to transport as much of this discussion as we can over to the world of parameterspace. For instance, can we define angles in the complement of the Mandelbrot Set? There shouldbe some immediate questions that arise if we try to do this.

35

Figure 22. Here you see the filled Julia set (with a few rays drawn in), and thecorresponding lamination.


Our topological discussions up until this point provide an alternative definition (and in fact, his-torically accurate) definition for the Mandelbrot set M .

Alternative Definition. The Mandelbrot set M is the set of all parameters c 2 C for which thefilled Julia set of z 7! z2 + c is connected.

Since there is only one critical point to keep track of, there is a dichotomy (in the case of quadraticpolynomials:

Either:

• the critical point escapes to 1 under iteration (that is, it is in the basin of 1) OR

• it does not, in which case it is contained in the filled Julia set.

If the critical point is in the basin of 1, then the Julia set is a Cantor set. So either, the filledJulia set is connected, or it is a Cantor Set. We will use all of our complex dynamics background

Figure 23. On the left, we see a connect filled Julia set for a quadratic polynomial,and the orbit of the critical point is marked. On the right, we see a disconnectedfilled Julia set for a quadratic polynomial; in fact, this Julia set is a Cantor set, andthe orbit of the critical point is marked.

36

this far and focus now on quadratic polynomials. Here are some things we can say (because thereis only ONE critical point to worry about!)

Bottcher coordinates in this case. We only have one critical point, so if the critical point isin the filled Julia set, then the Bottcher coordinate extends throughout the ENTIRE basin of 1,and it has an inverse : C� D! C�K

f

.CAUTION: the biholomorphism may not extend continuously over @D; we know from the LandingCriterion that extends continuously over @D to yield a continuous map

� : R/Z! Jf

if and only if the Julia set (and therefore the filled Julia set) is locally connected.

We also know exactly how far the Bottcher coordinate will extend through the basin of 1: it willextend until we hit the ‘figure 8’ curve that contains the critical point (see Figure 10. How do we

Figure 24. This is the Julia set of a quadratic polynomial. The critical point isin the basin of 1, so the filled Julia set (=Julia set) is a Cantor set. The Bottchercoordinate �, which is defined naturally in a neighborhood of 1 can be extendedto a biholomorphism on the unique unbounded component (in this picture) in thecomplement of the ‘figure 8’ curve that contains the critical point. Note that eventhough the Bottcher coordinate � does not extend throughout the entire basin of1, the function |�| does.

know which quadratic polynomials have locally connected Julia sets? A su�cient condition is thatof hyperbolicity.

Hyperbolic rational maps. A rational map f is dynamically hyperbolic if it is expanding on itsJulia set; that is, if there exists a conformal metric µ, defined on some neighborhood of J

f

, suchthat for all z 2 J

f

, we have Df |z

satisfies

kDf |z

(v)kµ

> kvkµ

for every nonzero v 2 TP1z

.

Since Jf

is compact, there is an expansion constant k > 1 so that

kDf |z

kµ

� k37

for all points z in a neighborhood of Jf

. In fact, for all z 2 Jf

, there is a neighborhood N containingz so that for all x, y 2 N , we have

distµ

(f(x), f(y)) � k · distµ

(x, y).

Recall that the postcritical set of a rational map is the union of the forward orbits of the criticalpoints; that is, let Crit(f) be the set of critical points of f ; define

Pf

:=[

n>0

fn(Crit(f)).

Note that the inequality is strict, so the critical points may or may not be contained in the post-critical set.

Hyperbolic rational maps are “well-behaved.” The following theorem is incredibly useful.

Theorem. (Hyperbolic maps) A rational map f : P1 ! P1 of dgree d � 2 is dynamically hyperbolicif and only if P

f

is disjoint from Jf

, or if and only if every critical point is attracted to an attractingcycle under iteration. If f is hyperbolic, then all points of the Fatou set ⌦

f

are attracted toattracting periodic cycles under iteration.

Note that the last condition does not imply hyperbolicity; consider the map z 7! z2+ i; the criticalpoint z0 = 0 is not attracted to an attracting cycle under iteration, so this map is not hyperbolic.However, the unique Fatou component is the basin of 1, so we can see that every point of ⌦

f

isattracted to the superattracting fixed point under iteration.

Very nice facts about hyperbolic rational maps: If f is hyperbolic, then

• the Lebesgue measure of Jf

is 0

• every periodic cycle for f is either attracting or repelling

• if g : P1 ! P1 is a su�ciently small perturbation of f , then g is also hyperbolic, and theJulia set J

f

deforms continuously to Jg

through hyperbolic maps (this is a famous result ofMane, Sad, and Sullivan).

The last point above is one to think about. Let’s take a map that is not hyperbolic; like thequadratic polynomial z 7! z2 + 1/4. any neighborhood of c = 1/4 in parameter space will containc-values for which the Julia set is a Cantor set, and those for which the Julia set is connected; wecan see that the Julia set does not move continuously as we perturb.

Theorem. (Hyperbolic rational maps and local connectivity.) If f : P1 ! P1 is a hyperbolicrational map, then if J

f

is connected, then it is locally connected.

In fact, in the polynomial case, we can get away with just the following theorem.

Theorem. If U is a simply connected Fatou component for a hyperbolic map, then @U us locallyconnected.

The proof of this theorem is fairly di�cult, and I refer you to Milnor ch 19 for it.

Note that if f is hyperbolic, then Crit(f)\Jf

= ;. There is a wider class of maps, the subhyperbolicmaps which are also helpful to study.Definition. The rational map f : P1 ! P1 is subhyperbolic if and only if every critical point iseither finite or converges to an attracting periodic orbit.

Theorem. If f is a subhyperbolic rational map with Jf

connected, then Jf

is locally con-nected.

38

Examples. Note that z 7! z2+ i is not hyperbolic, but it is subhyperbolic. The map z 7! z2+1/4is neither hyperbolic or subhyperbolic. Recall that the parameter c0 = 1/4 is at the cusp of themain cardioid in the Mandelbrot set; in particular, it is on the boundary of M . Therefore, asc moves in any open neighborhood containing c0 = 1/4, the Julia set will transition from beingconnected to totally disconnected; the Julia set does not vary continuously with the parameter cas c moves in ANY neighborhood containing c0 = 1/4.

Corollary. The boundary @M contains no parameter c for which the map z 7! z2 + c is hyper-bolic.

Topology of M . The Mandelbrot set is not completely understood, and this is a serious issue asthe set is in a very rigorous sense, universal.

Proposition. The set M ✓ C is full (the complement is connected), and it is compact.

Proof. In fact, M ✓ D2, where D2 is the closed disk of radius 2. Note that 2 is sharp, in the sensethat c = �2 belongs to M since the critical point z0 = 0 of the map z 7! z2 � 2 has boundedforward orbit.

Suppose that c 2 C satisfies |c| > 2. We will show that c cannot be in M . Associated to c, is thesequence

c1 := c, c2 := c21 + c, c3 := c22 + c, . . . .

We will prove that n 7! |cn

| is an increasing sequence by math induction. The inductive hypothesisgives for some fixed n, |c

n

| � |c| > 2. Examine:

|cn+1| = |c2

n

+ c| � |cn

|2 � |c| � |cn

|2 � |cn

| = |cn

|(|cn

|� 1) � |cn

|(|c|� 1).

This implies that |cn+1| � |c|(|c|�1), so the sequence n 7! |c

n

| diverges to1 as n!1. Thereforeif |c| > 2, then c /2M , and thus M ✓ D2.

To prove that M is closed, we note that

M =\

n>0

{c 2 C : |cn

| 2}.

Therefore, M is closed and hence compact.

To prove that M is full, we suppose the opposite. If C � M is not connected, then there is abounded component of the complement - call it U . For all c 2 @U ✓ M , the associated orbitsequence satisfies

8 n > 0, |cn

| 2.

Let c 2 U . There is N > 0 so that |cN

| > 2. The map c 7! |cN

| attains its maximum on @U by themaximum modulus principle. But for all c 2 @D, |c

N

| 2, and this is a contradiction.

Remarkable Property of M . The Mandelbrot Set parameterizes all quadratic polynomialsz 7! z2 + c for which the filled Julia set is connected. In this way, we may call it the connectednesslocus of the family z 7! z2 + c. Wouldn’t it be miraculous if the set itself also has this property?That is, if M is connected?? This turns out to be true, and the proof is a beautiful application ofcomplex analysis.

Theorem. (Douady-Hubbard). The Mandelbrot Set M is connected.

We will prove this next time. The way to get our foot in the door is to write down the Riemannmap from the complement of M to the complement of the closed unit disk.

39


We have a big goal for today. We will prove that the Mandelbrot set M is connected. The plan isas follows:

• Define a map � : C�M ! C�D; show this map is a conformal isomorphism from whenceit follows that M is connected.

To show that � is a conformal isomorphism:

(1) Prove that � is holomorphic.

(2) Prove that � is proper, from whence it follows that � is surjective.

(3) A proper map between oriented manifolds has a well-defined degree; we will show thatthe degree of � is equal to 1, which proves that � is injective.

Before we get started, let’s review some key features of proper maps and locally compact Hausdor↵topological spaces.

Definition. Let X and Y be topological spaces. The continuous map f : X ! Y is proper if forall compact sets K ✓ Y , the set f�1(K) ✓ X is compact.

Definition. The Hausdor↵ space X is locally compact if every point x 2 X has a compactneighborhood W ✓ X.

Proposition. A proper map f : X ! Y is closed if Y is a locally compact Hausdor↵ space.

Proof. Let W ✓ X be a closed set, and consider f(W ) ✓ Y . Let u 2 Y � f(W ). Since Y is locallycompact, there is a compact neighborhood of u; call it K

u

. Consider f�1(Ku

); this is a compactsubset of X since f is proper, and the set

W \ f�1(Ku

)

is compact, so f(W \ f�1(Ku

)) is compact. Since Y is Hausdor↵, this set is also closed. Con-sider

Ku

� f(W \ f�1(Ku

));

the interior of this set is an open neighborhood of u, which does not intersect f(W ).

The following proposition will be required eventually; it turns out to be incredibly useful.

Proposition. Let f : X ! Y be a proper map between locally compact Hausdor↵ spaces. If f isa local homeomorphism, then f is a (finite) covering map.

Proof. Let y 2 Y ; we need to find a neighborhood Vy

of y so that Vy

is “evenly covered”. Considerf�1(y) ✓ X. This set is compact since f is proper map. Moreover, since f is a local homeomor-phism, this set is also discrete, so it is finite:

f�1(y) = {x1, . . . , xn}.Since X is Hausdor↵, there are disjoint open neighborhoods U

i

3 xi

, and by taking them smaller ifnecessary, we may assume that

f |Ui : Ui

! f(Ui

)

is a homeomorphism. We would like to take

Vy

:=n\

i=1

f(Ui

),

40

but the problem is that

f�1

n\

i=1

f(Ui

)

!

may not be contained in the union of the Ui

(there may be extra stu↵).

Since Y is locally compact, there is a compact neighborhood Ky

✓ \f(Ui

), which contains thepoint y. Consider the set

Vy

:= int(Ky

)� f�f�1(K

y

)� [Ui

�;

this is an open neighborhood of y 2 Y which is evenly covered.

First note that f�1(Ky

) is compact since f is proper, and f�1(Ky

) � [Ui

is a closed subset off�1(K

y

), so it is compact, and therefore

f�f�1(K

y

)� [Ui

�

is a compact subset of Y , so it is also closed since Y is Hausdor↵. So the set

Vy

:= int(Ky

)� f�f�1(K

y

)� [Ui

�

is an open neighborhood containing y. The set f�1(Vy

) is equal to the union of

f�1(Vy

) \ U1, . . . , f�1(V

y

) \ Un

,

which are disjoint open sets that map homeomorphically onto Vy

by f . It helps to draw somewiggly pictures here!

Okay, back to the complex analysis.

The first step is to define the map � : C�M ! C�D. To this end, we will remember that we areworking with a parameter space for the family z 7! z2+c, and we will take our cues from dynamicalspace. It would be great if we could define a Bottcher coordinate. But we have NO dynamics onthe complement of the Mandelbrot set. But that is not going to stop us.

Let c 2 C�M . Draw the dynamical plane of the polynomial fc

(z) := z2+c. Because c is not in theMandelbrot set, the filled Julia set is equal to the Julia set, which is a Cantor set. Because 1 is asuperattracting fixed point, there is a neighborhood of 1 on which there is a Bottcher coordinate�c

that conjugates fc

on this neighborhood to the map w 7! w2 on a neighborhood of1. We knowthat we can extend �

c

throughout the basin of 1 UNTIL we crash into the critical point. There isa figure ‘8’ curve in the basin of 1 that contains the critical point, and we may extend �

c

to theunique unbounded component of the complement of this figure ‘8’ curve. Moreover, the extendedBottcher coordinate will conjugate the polynomial f

c

to w 7! w2 on this region.

In the dynamical plane of fc

, note that there are a few distinguished points. There is a uniquecritical point, and that critical point has a critical value. The critical point is NOT in the domainof �

c

, BUT the critical value c IS in this domain. This strongly suggests that we consider themap

� : C�M ! C� D given by c 7! �c

(c).

We have our candidate map. We must now check all of the required properties.

The map � is holomorphic because it is the composition of the maps

c 7!✓

cc

◆and

✓zc

◆7! �

c

(z).

Make sure you see why this composition is well-defined.41

Figure 25. This is the same Julia set as in Figure 24; the critical point is markedwith a star and the critical value is marked with a ‘c’. Notice that the critical valuec IS in the domain of the Bottcher coordinate.

We will now try to see that the map is proper. One tool that will be useful is the following functiondefined in the dynamical plane

Gc

: C! [0,1) given by z 7! log+ |�c

(z)|where log+ |z| := max{log |z|, 0}. Note that

Gc

(z) := limn!1

1

2nlog+ |fn

c

(z)|.

Recall that we may write

�c

(z) = limn!1

"z

fc

(z)1/2

z

! f2c

(z)1/4

fc

(z)1/2

!· · ·

fn

c

(z)1/2n

(fn�1c

(z))1/2n�1

!#

where a general term in the product is

fn

c

(z)1/2n

(fn�1c

(z))1/2n�1 =

h(fn�1

c

(z))2⇣1 + c

(fn�1c (z))2

⌘i1/2n

�fn�1c

(z)�1/2n�1 =

✓1 +

c

(fn�1c

(z))2

◆�1/2n.

So we have

Gc

(z) = log+ |z|+1X

n=1

1

2nlog+

��1 +c

(fn

c

(z))2

�� .

This is relevant because

log+ |�(c)| = Gc

(c) = log+ |c|+1X

n=1

log+��1 +

c

(fn

c

(c))2

�� .

Note that if |c| > 2, then ��1 +c

(fn

c

(c))2

�� 2,

so Gc

(c)� log+ |c| is bounded on the region in C�M where |c| > 2. Consider the map

G : C! [0,1) given by c 7! Gc

(c).

This is the Green function associated with M . Since

G(c) = log+ |�(c)|we note that G is continuous on C. We will first show that G : C ! [0,1) is proper. LetK ✓ [0,1) be a compact set (it is closed and bounded), and consider G�1(K) ✓ C. This set isclosed since G is continuous. We must show that this set is also bounded. If not, there is a sequencem 7! z

m

2 G�1(K), and m 7! |zm

| tends to 1. But since G(c) � log+ |c| = Gc

(c) � log+ |c| is42

bounded for |c| > 2, and log+ |zi

| will diverge to 1 as |zi

|!1, we must have G(zi

) tends to 1 aswell. However, the set K ✓ [0,1) is bounded, so we must have that G�1(K) is bounded as well.It follows that G : C! [0,1) is proper.

We will use this to show that � is also proper. Note that � : C �M ! C � D. Let K ✓ C � Dbe a compact set. Note that there are numbers r, s > 1 so that K is a closed subset of the closedannulus

{z 2 C� D : r |z| s}.

Consider ��1(K) ✓ C � M . Note that it follows from the argument above about the Greenfunction, that this set is bounded. We must show that there is no sequence m 7! c

m

2 ��1(K)which converges to some point c0 2 M . If there were such a sequence, then by continuity of theGreen function G

limm!1

G(cm

) = G(c0) = 0.

But note that 0 < log |r| G(��1(K)) log |s|, and we have a contradiction. So this implies that� : C�M ! C� D is proper.

By one of the warm up propositions, we know that this means that the image of � in C�D is closed.But because � is holomorphic, the image is also open. This means that the map � : C�M ! C�Dis surjective since C� D is connected.

We will now show that the map is injective by studying what happens near 1. A proper mapbetween oriented manifolds has a degree, and we will conclude that the degree is 1.

Since Gc

(c) � log+ |c| is bounded near 1, this implies that the map c 7! �(c)/c has a removablesingularity at 1, and so it can be extended there. It follows that the map c 7! �(c) extends to aholomorphic map P1 �M ! P1 �D, where � :1 7! 1. Note that the unique inverse image of 1is 1, so this means that the map � has degree 1.

We conclude that � is a conformal isomorphism, and we conclude that M is connected.

Since we have a Riemann Map � : C �M ! C � D, we can transport the discussion of externalangles in C� D to C�M .

Let : C � D ! C �M be the inverse of �, and consider the ray r · e2⇡is, where r > 1. We canlook at the ray �(r · e2⇡is) in C�M and ask does the following limit exist?

limr&1

(re2⇡is)

exist? Does this ray land at a well-defined point of @M? If yes, then we can label that pointof the boundary of M with the angle s 2 R/Z. From Caratheodory’s work, we know that themap will extend continuously over @D if and only if @M (and equivalently M itself) is locallyconnected.

Open Problem. Is the Mandelbrot set locally connected?

If it is, we can form the lamination equivalence relation on D whereby the point ✓ is equivalent to✓0 if and only if the ray at angle ✓ and the ray at angle ✓0 land at the same point in the boundaryof M . If we take the quotient by this equivalence relation, then we will get a topological model forthe Mandelbrot set.

43

Figure 26. On the left is the Mandelbrot set drawn with a bunch of external rays.If M is locally connected, then all rays have well-defined landing points, and themap : C � D ! C �M extends continuously over the boundary @D. If M islocally connected, we have a topological model for it, given as a quotient of D bythe lamination coming from the ray equivalence relation. The topological model isshown on the right.

Class 15 - March 11, 2014

Today we will explore the Mandelbrot Set a bit.

Hyperbolic Components. Recall that a rational map is hyperbolic if all critical points areattracted to attracting cycles under iteration. Hyperbolic rational maps are particularly nice tostudy. Here is a basic question: what do the hyperbolic parameters in the Mandelbrot set M looklike? A hyperbolic component of M is a connected component of the interior of M ; note that itis therefore simply connected. If c belongs to the hyperbolic component U ✓ M , then the criticalpoint z0 = 0 of z 7! z2 + c is attracted to some attracting periodic cycle ⇣0, . . . , ⇣n�1 of period n.In fact, for any c0 2 U , the critical point z0 = 0 of z 7! z2 + c0 is attracted to a periodic cycle ofperiod n; that is, there is a positive integer associated to every hyperbolic component; that integercorresponds to the period of the attracting cycle that attracts the critical point. We make thisprecise below.

Theorem. Let U be a component of the interior of M such that for some c 2 U , the polynomialpc

: z 7! z2 + c has an attracting cycle of period n. Then

(1) There are analytic functions

Z0, · · · , Zn�1 : U ! C

such that

pc

(Zi

(c)) = Zi+1(c) and p

c

(Zn�1(c)) = Z0(c).

(2) The Multiplier Map µU

: U ! D given by

c 7! (pnc

)0 (Zi

(c))

is an isomorphism, and44

(3) the map µU

extends continuously over @U ! R/Z as an injective map.

We will prove the first assertion. The proofs of the other two require similar kinds of argumentsand a bit more advanced material.

Proof. This is basically an argument involving the implicit function theorem, applied to the rightmap. Consider the locus

Xn

:= {(z, c) 2 C2 | pnc

(z)� z = 0}.

This gives a curve in C2, with possibly many components. Note that the point (z, c) 2 C2 lieson X

n

if and only if z is periodic for z 7! z2 + c, of period dividing n. There is an irreduciblecomponent of X

n

corresponding to pairs (z, c) for which z is periodic for the map pc

: z 7! z2+ c ofexact period n. This is called the dynatomic curve of period n. It is smooth, and irreducible (thisis NOT obvious).

Consider the map ⇡ : Xn

! C given by ⇡ : (z, c) 7! c. This is evidently continuous. We will provethat it is proper; since the map is continuous, the inverse image of a closed set is closed. It remainsto show that the inverse image of a bounded set is bounded. Let K ✓ C be a bounded set, andconsider ⇡�1(K) ✓ X

n

. There is some R > 0 so that for any (z, c) 2 ⇡�1(K), the filled Julia setof p

c

: z 7! z2 + c is strictly contained in the disk of radius R. The filled Julia set contains allperiodic cycles of z 7! z2+ c (except for the super-attracting fixed point at1). Therefore, ⇡�1(K)is bounded in X

n

, and the map ⇡ : Xn

! C is proper.

Consider the restriction ⇡ : ⇡�1(U) ! U ; this is also proper. We will prove that it is a localhomeomorphism. This is just an application of the implicit function theorem. We can solve for zas an analytic function of c as long as

@

@z(pn

c

(z)� z)��(z,c) 6= 0.

But note that if the derivative above is 0, then pc

has a parabolic cycle of period n. This isimpossible since c 2 U , which means that p

c

has an attracting cycle of period n; a quadraticpolynomial is entitled to either a parabolic cycle OR an attracting cycle, not both. This meansthat the map

⇡ : ⇡�1(U)! U

is proper, and it is a local homeomorphism. It is therefore a finite covering map. The degree of thiscover is evidently equal to N :=the number of cycles of z 7! z2 + c of length n; there are N simply-connected components of ⇡�1(U) living above U , and we can define the functions Z0, . . . , Zn�1.Note that I am using the fact that U is simply-connected in this argument.

Take-Away: Every hyperbolic component has a positive integer associated to it.

The Multiplier Map. It follows from the fact that M is full that U and D are isomorphic. Butwe have a canonical isomorphism handed to us by the dynamics. This is a very natural map toconsider. Let U be a hyperbolic component of M which has the positive integer n associated to it(that means from (1), that for all c 2 U , the polynomial p

c

: z 7! z2 + c has an attracting cycle ofperiod n).

The multiplier map takes the parameter c and returns the multiplier of the unique attracting cycle;it is denoted µ

U

: U ! D. The multiplier map µU

associated to the hyperbolic component Uprovides a canonical parameter c0 := µ�1

U

(0). This is called the center of the hyperbolic componentU .

45

Part (3) of the proposition provides another canonical point in @U . Let’s abuse notation and callthe extended map

µU

: U ! D;

the other canonical point is µ�1U

(1). This is necessarily a point of @U . This point is called the rootof the hyperbolic component.

The center. Dynamically, the center c0 of U is a special point. The multiplier of the periodiccycle is 0, so the cycle is super-attracting. This of course means that the critical point z0 = 0 ispart of the cycle. Aha! The map p

c0 is postcritically finite!!!!!

The root. What is so special about the parameter c1 := µ�1U

(1)? The polynomial pc1(z) = z2+ c1

has a parabolic cycle, with multiplier 1. BUT CAUTION! The length of the parabolic cycle maybe SMALLER than n, the period associated to the hyperbolic component U .

Two kinds of hyperbolic components. There are hyperbolic components which have cusps at aunique point on the boundary (like the main cardiod), and there are hyperbolic components whichdo not (like the basilica component). The components with cusps are called primitive components,and the components without cusps are called satellites.

Primitive roots. The root point of all primitive components is the cusp.

Satellite roots. The root of the satellite components is a certain boundary point which is sharedwith another satellite component. We say that a hyperbolic component U of M is attached to thehyperbolic component U 0 of M if their boundaries intersect. In this case, the boundaries intersectat a unique point. This point of intersection is the root of U if the period of U is greater than theperiod of U 0.

Figure 27. Every hyperbolic component has a period associated to it. A few of theperiods are drawn on the components. There are two di↵erent shapes of hyperboliccomponents: satellite and primitive. The main cardiod is a primitive component(it has a cusp), and the basilica component is a satellite component. The root of aprimitive component is at the cusp, and the root of the satellite components will beat a certain parabolic parameter on a common boundary point between two satellitecomponents.

46

The root of the basilica component is at the parameter c = �3/4. At this parameter, pc

(z) =z2 + c has a parabolic fixed point and NO cycle of period 2, even though the period of the basilicacomponent is 2. This is an example where the period dropped.

Confusing point. If you compute the multiplier of the fixed point of z2 � 3/4, you will get(p

c

)0(at the fixed point) = �1. So how could c = �3/4 be the root of the basilica component? Themultiplier of the fixed point is not 1. BUT the multiplier map for the basilica component is themap that evaluates

(p2c

)0(at the 2-cycle).

For the polynomial p(z) = z2 � 3/4, the 2-cycle collapsed into a fixed point, so when we evalu-ate

(p2c

)0(at the fixed point, which is a degenerate 2-cycle) = 1,

which is consistent with the definition of the root.

Flash-forward. We will want to tell these hyperbolic components apart. At a very coarse level, wealready have a filter for doing this; we can label every hyperbolic component with its period. But forall periods greater than 2, there is more than one hyperbolic component. Because every hyperboliccomponent has a unique center, we can use the postcritically finite polynomial corresponding tothe center to tell the hyperbolic components apart. BUT we want to extract some finite data toencode the postcritically finite polynomial. What will it be? It will be something that generalizesthe idea of a ‘kneading sequence.’ Thurston’s theorem will tell us that this data is enough to tellthe components apart; that is, it provides a rigidity result that says if the data is di↵erent, thenthe components are di↵erent, and conversely, if the data is the ‘same,’ then the components arethe same.

Classes 16 & 17 - March 13 & 18, 2014

We continue exploring the Mandelbrot set today. Here is an open problem:

Open Problem. What does the complement of the union of the hyperbolic components look likein int(M)? Is the complement empty?

Finding hyperbolic components. We can use the fact that every hyperbolic component ofperiod n has a unique center, where the critical point is periodic of period n. To find thesecomponents in parameter space, we set up the equation

pnc

(0) = 0.

This is a polynomial equation in the parameter c; moreover, it is a monic polynomial with integercoe�cients. Factoring this polynomial over Q, we see that there is a factor corresponding to alldivisors m|n.

Define Q1(c) := c, and Qn+1(c) := (Q

n

(c))2 + c.

Definition. The Gleason polynomial of period n is the polynomial

Gn

(c) :=Q

n

(c)Qm<n : m|n

Gm

(c).

Open Problem. Are the Gleason polynomials irreducible over Q?

These are called Gleason polynomials because of the following proposition and its remarkable proof(due to A. Gleason).

47

Proposition. (A. Gleason) For any n 2 N, Gn

(c) has simple roots.

Proof. We will show that the polynomial Qn

(c) has simple roots from whence it follows that Gn

(c)has simple roots. Define P

n

(c) := Qn

(c) mod 2. Note that

Pn+1(c) = (P

n

(c))2 + c =) P 0n+1(c) = 2P

n

(c) · P 0n

(c) + 1

which is equal to 1 mod 2. In particular, the derivative of Pn

(c) is nonzero mod 2, which meansthat the discriminant of Q

n

(c) cannot be 0 mod 2, which means that the discriminant of Qn

(c)cannot be 0, so Q

n

(c) has simple roots.

This is really a transversality result. We will mention this later.

Postcritically finite quadratic polynomials. Recall that a rational map F : P1 ! P1 ispostcritically finite if all critical points have finite forward orbit under F ; or equivalently, if allcritical points are eventually periodic under F .

Fact. Postcritically finite maps are entitled to have ONLY superattracting cycles (if a critical pointis periodic), or repelling cycles, and that’s IT.

Note that if pc

(z) = z2 + c is postcritically finite, then necessarily z0 = 0 is periodic (in whichcase the parameter c is a center of a hyperbolic component), or z0 = 0 is strictly preperiodic,which means that z0 = 0 eventually maps into a repelling cycle. Note that the postcritically finiteparametnecessarily

It is clear from the discussion above that the centers of all hyperbolic components are algebraicintegers. What about the parameters c which correspond to critical points which are strictlypreperiodic?

Definition. Let c 2 M be a parameter for which the critical point z0 = 0 of pc

: z 7! z2 + c isstrictly preperiodic. The parameter c is called a Misiurewicz parameter.

The Misiurewicz parameters are necessarily in the boundary of M . In fact, they are dense in theboundary. This is a point of interest since the boundary of M is not well-understood. Recall thatM is locally connected if and only if the Riemann Map

M

: C�D! C�M extends continuouslyover the boundary.

What can we say about the boundary of M? It is precisely the locus in the c-parameter plane forwhich the critical point z0 = 0 for p

c

: z 7! z2 + c is “active.” This means that the boundary of Mis precisely the locus where the family of maps C! C

c 7! c, c 7! c2 + c, c 7! (c2 + c)2 + c, . . . , c 7! pnc

(0), . . .

is not normal.

Lemma 1. Let X be a complex manifold, and let s1, s2, s3 : X ! P1 be three holomorphic mapswhose graphs are disjoint. Then the set of all holomorphic maps f : X ! P1 whose graphs aredisjoint from those of s1, s2, and s3 is a normal family.

Proof. See homework.

Lemma 2. Suppose that f : D ⇥ C ! C is a holomorphic family of polynomials of degree d � 2,parameterized by D. Let a : D! C be an active marked point. Then there is a sequence of distinctparameters t

n

2 D for which a(tn

) is preperiodic for ftn for all n 2 N. Furthermore, we may choose

tn

so that a(tn

) lands on a repelling cycle of ftn for each n.

Proof. To say that a is an active marked point means precisely that the family of maps D! Ct 7! a(t), t 7! f

t

(a(t)), t 7! f2t

(a(t)), . . . , t 7! fn

t

(a(t)), . . .48

is not normal on D.

Let U be the largest open set in D on which the family above is normal. Note that U may beempty. But since a is active, we know that D�U is nonempty. Let t0 be an “activity parameter”;that is choose t0 2 D� U , and consider the map f

t0 : C! C. Let

z1(t0), . . . , zr(t0)

be a repelling cycle for the map ft0 of period r � 2.

By the Implicit Function Theorem, we can follow this cycle in a small neighborhood of t0. Let pi(t)denote the ith point in the corresponding repelling cycle for the map f

t

: C! C.

Since r > 1, we have p1(t) 6= p2(t) for all t close enough to t0. Failure of normality and Montel’stheorem imply that there is a parameter t1 2 D and an integer k > 1 such that

fk

t1(a(t1)) 2 {p1(t1), p2(t1)}.

If not, then there would be a neighborhood around t0 so that for all t in this neighborhood, for allk 2 N,

fk

t

(a(t)) is distinct from {p1(t), p2(t)},but by Lemma 1, this means that the family

t 7! a(t), t 7! ft

(a(t)), t 7! f2t

(a(t)), . . . , t 7! fn

t

(a(t)), . . .

is normal at t0 2 D, which is a contradiction.

Corollary. The postcritically finite parameters are dense in the boundary of M .

Proof. Take a disk Dr

large enough so that M ✓ Dr

. Use this disk in Lemma 2 above. The resultthen follows from the fact that @M is characterized by the fact that it is the ‘activity’ locus for thecritical point z0 = 0.

Figure 28. The centers of 983 hyperbolic components of M are plotted above.

We will present a structure theorem for the Mandelbrot Set. All rational rays in C �M land atwell-defined points of @M . The question is where?

49

Theorem. (Douady-Hubbard; Schleicher)

(1) Every parameter ray at a periodic angle ✓ lands at a parabolic parameter c✓

such that inthe dynamical plane of z 7! z2 + c

✓

, the ray at angle ✓ lands at a “root” of the componentof the filled Julia set which contains the critical value.

(2) Every parabolic parameter c is the landing point of exactly 2 rays at periodic angles (notethat at the cusp of the main cardiod, the 2 rays that land there are at angle ✓ = 0 and✓ = 1.

(3) Every parameter ray at a preperiodic angle ✓ lands at a parameter c✓

, which is a Misiurewiczparameter. In the dynamical plane z 7! z2 + c

✓

, the external ray at angle ✓ lands at thecritical value, which is in the Julia set.

(4) Every Misiurewicz parameter c is the landing point of a finite nonzero number of parameterrays at preperiodic angles. These are precisely the angles that land at the critical value inthe dynamical plane.

The two periodic angles that land at the root of

• a satellite component are in the same cycle under angle-doubling, and

• a primitive component are NOT in the same cycle under angle-doubling.

Figure 29. Some rational angles landing on the boundary of M . Compare theangles which land at roots of satellite components to those landing at roots ofprimitive components to those landing at Misiurewicz parameters.

We will proceed to the Thurston theorem part of the class. We will review kneading data, thendefine spiders, and then discuss Thurston’s theorem for the remainder of the course.


We will begin by addressing a question that came up last time, namely, why are all the Misiurewiczparameters in @M?

50

Here is the idea: Let c0 be a Misiurewicz parameter for which the critical point of z 7! z2 + c0eventually maps to a fixed point. We would like to show that the family of maps

c 7! c, c 7! c2 + c, c 7! (c2 + c)2 + c, . . . , c 7! fn

c

(0), . . .

is not normal on ANY neighborhood of this point. Indeed, if the family were normal on a neigh-borhood, then there would be a convergent subsequence which converges locally uniformly to alimit function, and the corresponding sequence of derivatives would converge as well. We want toanalyze the sequence of derivatives and prove that it does converge on any neighborhood of c0. Fornotation, set P

n

(c) := pnc

(0). Write down the first few derivatives:

c 7! 1, c 7! 2c+ 1, c 7! 2(c2 + c)(2c+ 1) + c, . . .

or D0(c) = 1, and setD

n

(c) = 2Pn

(c)Dn�1(c) + 1.

Let’s examine this sequence at c = c0. Note that c0 is the critical value, and it will EVENTUALLYmap onto a fixed point ⇢ of the polynomial z 7! z2 + c0. So for n large enough, P

n

(c0) = ⇢, andthe sequence of derivatives becomes:

Dn

(c0) = �Dn�1(c0) + 1,

where � is the multiplier of the fixed point ⇢; that is, pc0(⇢) = ⇢, and � = p0

c0(⇢). Since this fixed

point is necessarily repelling, |⇢| > 1. In other words, if we set an

:= Dn

(c0), we have a recursivesequence defined by

a0 := 1, an

:= �an�1 + 1

which will eventually agree with the sequence of derivatives n 7! Dn

(c0) for n large enough. Showthat this sequence n 7! a

n

must diverge to 1. There are some missing details. Some of this willprobably be a bit subtle. You might also think about the Implicit Function Theorem.

Kneading Data

Recall that if we begin with a real polynomial fc

: x 7! x2+c, for c 2 [�2, 0], and x 2 [�2, 2], we candefine kneading data associated to f

c

in the case where the polynomial is postcritically finite.

Let fc

(x) = x2+ c be a postcritically finite quadratic polynomial with postcritical set P . Label theelements of P as follows:

p1 := c, and pi

:= f i(0), 1 i n.

Define the set Q := f�1(P ). Note that P ✓ Q, and that Q ✓ R. Write the set Q as

Q = {q�n

< q�n+1 < . . . , q�1 = 0 = q1 < . . . , qn�1 < q

n

}.Definition. The kneading sequence k(f

c

) of fc

is the map

k : {1, . . . , n}! {�n,�n+ 1, . . . ,�2, 1, 2, . . . , n� 1, n} given by pi

= qk(i) for 1 i n.

We encode this map k with a kneading sequence

< k(1), . . . , k(n) > .

This is really a vector of information that keeps track of the relative position of the postcriticalpoints.

Main Question. Given a piecewise linear map of degree 2, is there a postcritically finite quadraticpolynomial which has the same kneading sequence? That is, can the postcritical points be wiggledenough to fit into the graph of a parabola x2 + c?

This is the first context in which we will study Thurston’s theorem.51

Figure 30. This is an example of a piecewise linear topological map of degree 2.There is a critical point at c0 = c7, and it is periodic of period 7. We can writesome kneading data associated to it (that is, record the relative positions of thepostcritical points on the real line).

Let’s make some immediate observations about what kneading sequences should be allowed. Forinstance, if a kneading sequence is going to have a prayer of arising from an honest polynomial,there are some immediate constraints. Here they are:

• The map k : {indices of P}! {indices of Q} should be injective.

• We must have k(1) = �n (or it should be the minimum value of k). The last entry is either1 (in the case where the critical point is periodic), or there is a unique index 1 j < nfor which the last entry k(n) = �k(j). This is saying that there is a unique point in thepostcritical set; in this case p

j+1, which has TWO inverse images pj

and pn

contained inthe postcritical set. This corresponds to the preperiodic case.

• The kneading sequence satisfies

k(i) < k(j) < 0 =) k(i1) > k(j + 1) and 0 < k(i) < k(j) =) k(i+ 1) < k(j + 1).

Definition. We say that a kneading sequence is admissible if it satisfies the criteria above.

Proposition. Let k : {1, . . . , n} ! {�n, . . . ,�2, 1, 2, . . . , n} be an admissible kneading sequence.Then there exists a piecewise linear map f : [a, b]! [a, b] of degree 2 which is postcritically finite,such that f has kneading sequence k.

Immediate Question. What does it mean if two kneading sequences are the same? For us,this will mean that the relative positions of the points is the same on the real axis. Technically,a kneading sequence is a map k : {1, . . . , n} ! {�n, . . . ,�2, 1, 2, . . . , n}, but two of them are thesame if the positions of the points can be wiggled on the real axis (WITHOUT COLLISIONS), sothat one kneading sequence can be transformed into the other (keeping the signs the same! That is,you must remember where the critical point is, even if it is not part of the postcritical set!).

52

Proof. Here is a recipe for building the piecewise linear map. Plot all ordered pairs:

(1, k(1)), (k(1), k(2)), . . . , (k(n� 1), k(n))

and connect the points with lines, from left to right. If k(n) = 1; that is, if the critical point isperiodic, then STOP. If k(n) 6= 1, then the critical point is preperiodic, and there are is some j forwhich k(j) = �k(n) (that is, the postcritical points p

n

and pj

are symmetric with respect to theorigin and should have the same image, namely p

j+1). In this case, add the point (�k(j), k(j+1)) =(k(n), k(j + 1)). Then connect the points from left to right.

The map will have degree 2, and it was cooked up so that it will realize the kneading sequence thatwe started with.

GREAT!

So given an admissible kneading sequence, we can obtain a piecewise linear map of degree 2 thatrealizes it. Now our important question resurfaces: can we do better? Is there a polynomial thatrealizes it? How would you answer this question?

Exercise. Find a piecewise linear map that realizes the following kneading data< �4, 2,�3,�2 >.

One answer. From the dynamics of the map f , (take the one constructed in the example above),write down the ramification portrait. In this case, it is:

02// p1 // p2 // p3 // p4

^^

Choose coordinates, so that the map has the form fc

(x) = x2 + c. The portrait above forces thecondition that

f3c

(0) = f5c

(0),

and this gives a polynomial equation in the parameter c. We are interested in real solutions. UseMaple to find them, and test the critical orbit to see if the kneading sequence “is” < �4, 2,�3,�2 >.You should find that there is NO quadratic polynomial with this kneading data. Hmmm....

This example proves that not all admissible kneading data is realizable. Thurston provides an algo-rithm which will determine if an admissible kneading sequence is equivalent to a polynomial.

Consider the a subset T ✓ [�2, 2] ⇥ · · · ⇥ [�2, 2] which consists of n copies of [�2, 2]. We areinterested in the open subset which is consistent with the kneading data (note that the kneadingdata tells us the order of the postcritical points on R; this defines an open set in this productspace). The closure of this set T is compact and convex, so a fixed point of �

k

: T ! T exists bythe Brouwer fixed point theorem. HOWEVER, the fixed point may be in the locus where at leasttwo coordinates coincide. And a priori, there could be more than one fixed point. Given kneadingdata k, we will define a map �

k

: T ! T in the following way:

�k

: (x1, . . . , xm) 7! (y1, . . . , yn)

where yi

:= ✏i

pxi+1 � x1, where ✏

i

:= sgn(k(i)). Define yn

:= 0 if k is ‘periodic’, and defineyn

:= ✏n

pxj+1 � x1 where j is the unique index so that k(n) = �k(j). This map is well-defined.

Note that a fixed point (t1, . . . , tn) 2 T corresponds PRECISELY to a postcritically finite quadraticpolynomial with kneading data equal to k. Which polynomial is it? It will be the polynomialx 7! x2 + t1. If there is no fixed point of �

k

in T , then there is a topological obstruction. We willexplore this next time.

Theorem. The kneading data k is realized by a postcritically finite quadratic polynomial if andonly if �

k

has a fixed point in T . Moreover, the fixed point is unique.53

However, recovering the uniqueness of the fixed point from this point of view is not easy; complexmethods are required. For instance, when we get to Thurston’s theorem, the space T will be acomplex manifold, which will carry a natural metric, the Kobayashi metric, with respect to whichall holomorphic maps are contractions. Moreover, the space T is complete with respect to thismetric, and so a fixed point, if it exists, is necessarily unique.


We will begin with some good questions from last time. Namely, it appears that the map �k

depends ONLY on the sequence of ± that occurs in the critical orbit, and NOT on the relativeorder of the postcritical points. Let’s test this.

Figure 31. Here we have a topological map of degree 2, where the unique criticalpoint is periodic of period 7. We encode the kneading data (the relative order ofthe postcritical set on the real axis) with the vector < �7,�5,�2,�6, 3,�4, 1 >.Note that we could have also encoded the relative data of the postcritical set withthe vector < �7, 6,�3,�5, 2,�4, 1 >. It is not the exact integers that show up inthe vector that are important, rather it is the induced order of the points in thereal line from the vector. So there are choices. In fact, we don’t have to restrictourselves to using integers {�7, . . . ,�2, 1, 2, . . . , 7}; this is just a convention. Wecan use and subset of R that contains enough points. We need to keep track of therelative positions of the points, and where the critical point is; this is key.

Given the relative positions of the points above (and the location of the critical point), we candefine the Thurston pullback map associated to this kneading data:

�k

: T ! T �k

: (x1, . . . , x7) 7! (y1, . . . , y7)

wherey1 = �

px2 � x1, y2 = �

px3 � x1, y3 = �

px4 � x1,

y4 = �px5 � x1, y5 = �

px6 � x1, y7 = 0.

54

For right now, take T to be the product of 7 copies of the interval [�2, 2] with itself. This is notactually what we want to do, but for now, it is okay.

So we start with some initial guess of x0 := (x1, . . . , x7) = (�1.5, 1.3,�0.6,�1.4, 1,�1.1, 0) accord-ing to the relative order of the postcritical set on the real axis (and using the fact that 0 shouldbe the critical point of our map x 7! x2 + c). We iterate the map �

k

, generating a sequencexn

:= �k

(xn�1). Using Maple, we see that this sequence converges. As mentioned before, it is the

Figure 32. Here is the maple script for iterating the pullback map �k

: T ! T .Note that our sequence n 7! x

n

converged to a fixed point. This list of 7 numbersIS the postcritical set of the polynomial p(z) = z2 � p1 = x2 � 1.674066092 whichwill necessarily have the kneading data k. Pretty cool.

relative positions of the points that is important. But we seem to only need to use the signs of thepoints in our definition of �

k

. It turns out that given an admissible kneading sequence, there isonly ONE order of postcritical points on the real axis which is compatible with it.

Exercise. Switch p3 and p6 in Figure 31 and see what happens. You have just changed the relativeorder of the points, so you have changed the kneading data. BUT you have not changed the signsof the postcritical points, so you will get the SAME map �

k

: T ! T that you got before. Ifyou iterate this map, you will get a fixed point, and in fact, it is the SAME fixed point as before.WHAT! This means that you will get the same polynomial x2 � 1.674066092 back. How can thisbe? Prove that if you switch p3 and p6, the kneading data is no longer admissible. In fact, showthat the recipe defined on p53 fails in the sense that you will get a piecewise linear topological map,but it will not have degree 2!!!

Let’s define the pullback map �k

: T ! T for the kneading data k =< �4, 2,�3,�2 >. First, weshould check that this data is admissible. Okay, here, we will take the space T to be the productof 4 copies of [�2, 2] with itself. The pullback map �

k

is given by

�k

(x1, x2, x3, x4) 7! (�px2 � x1,

px3 � x1,�

px4 � x1,�

px3 � x1).

In the example k =< �4, 2,�3,�2 >, we know that there is no real polynomial x 7! x2 + c withthis kneading data. When we run Thurston’s algorithm using this data, we find a fixed point of�k

, but it is a bad fixed point. The last two coordinates coincide, which means that the kneadingdata for this polynomial will be ‘degenerate’ and it will not match the kneading data k. This iswhat will happen in general. The model for Thurston’s theorem is the following

55

Figure 33. The Maple script for iterating the pullback map associated tothe kneading data k =< �4, 2,�3,�2 >. Notice that there IS a fixedpoint, but it is a weird fixed point. The fixed point has coordinates(1.543763410, 0.8388917040,�0.8389153369,�0.8388917040). If we define the poly-nomial p(x) = x2+p1 = x2+1.54376341, we expect this polynomial to have the samekneading data as k =< �4, 2,�3,�2 >, but it cannot possibly have this kneadingdata. For one thing, the last two coordinates of the fixed point are equal. Thismeans that for the polynomial p we just defined, p3 = p4, so this cannot possiblywhat we want.

• Given admissible kneading data, or more generally, admissible ‘topological combinatorics’

• forge a topological model with this combinatorics; call it f ,

• define a map �k

: T ! T where T is some space that depends on the map f (so it changesfor every problem)

• iterate the map �k

on T and look for a fixed point; BUT you want to find a GOOD fixedpoint.

• Either there will be a good fixed point, or POINTS COLLIDE.

In the examples we have been able to do by hand, we had a map �k

which was defined on theproduct of n copies of the interval [�2, 2], and we are interested in when �

k

has a fixed point.HOWEVER, it might be a bad fixed point in the sense that some of its coordinates may coincide.We are only interested in good fixed points.

Question. Why are there just two behaviors? Either we will get a good fixed point, or points willcollide. We will see this later.

56

Another Question. What is it about the admissible kneading data k =< �4, 2,�3,�2 > that isbad? Let’s forge the topological model f associated to this data. That is an exercise. There is atopological obstruction that is preventing f from being ‘equivalent’ to a quadratic polynomial.

Definition. The admissible kneading sequence k is obstructed if there are pairwise disjoint intervalsI1, . . . , I

k

, whose endpoints are all elements of P , such that for all j 2 [1, k], we have

f |Ij : Ij ! I

j+1

is a bijection.

This is important because of the following theorem.

Theorem. (Thurston; warm-up version) Let k be admissible kneading data. Then there is apostcritically finite quadratic polynomial p(x) = x2 + c realizing kneading data k if and only if k isnot obstructed.

Idea of proof. Suppose that k is obstructed. Then one should be able to wiggle the map f aroundto manufacture an attracting cycle in one of the intervals I

j

from the obstruction. This is NOTobvious. If f is equivalent to a polynomial, then this polynomial would also have an attractingcycle in this interval. But then the attracting cycle for the polynomial would have to attract acritical point. But since the polynomial is postcritically finite, then this means that the criticalpoint would have to be IN the cycle. However, that would mean that the intervals I

j

7! Ij+1

could not map to each other bijectively as one of them would contain the critical point. This is acontradiction. In summary, if these intervals exist, then the topological map cannot possible ‘comefrom’ a polynomial. If these intervals don’t exist, then one can prove that there must be a fixedpoint of �

k

which is good. This will give us the polynomial in question.

There are lots of details missing. We will fill them in when we discuss the full statement ofThurston’s theorem. The idea is that the topological map f is ‘equivalent’ to an analytic mapp(x) = x2 + c if and only if there are no obstructions in the sense of these intervals.

Classes 20 & 21 - March 27, & April 1, 2014

We will talk about how to generalize the version of Thurston’s theorem above, first to the caseof quadratic polynomials which are complex (not necessarily real). That is, we want to definekneading data for a postcritically finite map p

c

(z) = z2 + c and understand what properties it has.Then we want to turn the problem around: starting with kneading data of a certain type (oncewe figure out what that is), is there a systematic algorithm or way to check to see if this kneadingdata comes from an honest polynomial? If it does not come from a polynomial, then why? Whatis the obstruction? In the case of real polynomials, we used the relative order of the points on thereal line to help us define a kneading sequence. Working over the complex numbers, we don’t havean ‘order’ per se. What do we have? We’ll use angles.

Let ✓ 2 Q/Z, and consider the two halves of ✓, ✓/2, and (✓ + 1)/2. The chord connecting ✓/2 to(✓ + 1)/2 cuts the circle into two sectors. Label the sector containing ✓ with ‘A0, and label theother sector ‘B0 as in Figure 34.

Definition. For any ↵ 2 R/Z, define the ✓-kneading sequence of ↵ to be

k✓

(↵) := a1a2a3 . . . where aj

=

8><

>:

A if e2⇡i✓2j�1 2 A

B if e2⇡i✓2j�1 2 B

⇤ if e2⇡i✓2j�1 2 {✓/2, (✓ + 1)/2}

57

Figure 34. For ↵ 2 R/Z, use the circle above to define the kneading sequencek✓

(↵) = a1a2a3 . . ..

Think of K(✓) := k✓

(✓) as a generalization of the sequence of signs +1,�1 we used when discussingpostcritically finite real quadratic polynomials.

Exercise. Show that K(✓) is periodic if and only if ✓ = p/q has odd denominator (when writtenin lowest terms).

Exercise. Using our treatment for real polynomials, consider the airplane p(x) = x2 + cA

. Writedown the sequences of +1,�1, 0 that arises from the critical orbit. Now consider the angle ✓ = 3/7.Compute K(3/7). What do you notice?

Just like for the real case, we don’t only want to consider the sequence of signs of the postcriticalpoints, but we want to take into account the relative order of the postcritical points. We will needa way to do this in the complex plane. We can no longer exploit the natural order of R. For thispurpose, we will use spiders. These are motivated by external angles.

Definition. Given ✓ 2 Q/Z, define the standard ✓-spider as follows

S✓

:= {1} [[

j>0

{re2⇡i✓2j�1 |r � 1},

and denote xj

:= e2⇡i✓2j�1 2 S1. Observe that it is possible to recover the kneading sequence

K(✓) from the ✓-spider. The standard ✓-spider is a template that we will use to define our spiderspace.

Definition. Consider the space

S0✓

:= {' : S✓

,! bC|' is injective, continuous, '(1) =1,'(x1) = 0,

and ' preserves the circular order of the legs at 1.}

You should image that an element of this space ' : S✓

,! bC is a list of points zj

:= '(xj

), andattached at each point z

j

is a “leg”, called �j

, which stretches out to 1. The space S0✓

consistsof these objects. Ideally, we would like to start with the data of ' 2 S0

✓

, and ask: “does it comefrom an honest postcritically finite polynomial?” What does that even mean? Here is an exampleof spider-data that is associated to an honest polynomial. The plan is to define a map �

✓

on aspace that we have yet to determine, such that �

✓

is a contraction on our path-connected, completemetric space (so that if there is a fixed point, that fixed point is unique), and if there is a fixedpoint of �

✓

, then there is an honest postcritically finite quadratic polynomial p(z) = z2 + c which58

Figure 35. The standard ✓-spider for ✓ = 9/56.

Figure 36. Here are three elements of S0✓

, for some ✓ 2 Q/Z. Two of them areequivalent in S0

✓

, and the third is not.

has the “same” spider data as that of the fixed point. The problem is that the space S0✓

is TOOBIG. We will want to consider '0 and '1 as giving the same spider data if

• there is a homotopy 't

2 S0✓

between '0 and '1, and

• for all t 2 [0, 1], the vectors

['0(x2) : · · · : '0(xn)] and ['t

(x2) : · · · : 't

(xn

)]

represent the same point in Pn�2.

Definition. Let ⇠ be the equivalence relation on S0✓

generated by the two points above. Definethe ✓ spider space to be S

✓

:= S0✓

/ ⇠.59

Proposition. The spider space S✓

is a complex manifold of dimension n� 2.

Idea of the proof. Consider the map S0✓

,! Pn�2 given by

' 7! ['(x2) : · · · : '(xn)];the image of this map is precisely Pn�2��, where � is the forbidden locus, the locus where any ofthe n � 2 coordinates is 0 or any of the two coordinates coincide. In fact, the image of this spaceis isomorphic to M0,n+1, the moduli space of genus 0 curves with n + 1 marked points. This mapinduces a map

⇡✓

: S✓

! Pn ��which is a universal covering map. It will then follow that S

✓

is a complex manifold of dimensionn� 2. One can first (directly) show that the map is a covering map, and then it remains to provethat the space S

✓

is simply connected. Here is the argument. Take a loop in S✓

, represented by't

: S✓

,! bC, as t runs over [0, 1). Find a disk DR

✓ C of large enough radius R so that

8 t 2 [0, 1), 8 2 j n, zj

(t) 2 DR

.

For each fixed t 2 [0, 1) pull the leg �j

(t) until the endpoint zj

(t) is on the circle. The points arein the right circular order. The idea is that for any t 2 [0, 1), there is a path in S

✓

from ['t

] to thislast configuration. Hence in the loop we started with in S

✓

can be contracted to the point in S✓

given by the class of this last configuration of points and legs.

The map. Our spider space is a complex manifold. We now have to define the map � : S✓

! S✓

.To this end, we change coordinates. It will be cleaner to work with the normal form

P�

: z 7! �(1 + z/2)2.

This is preferable because the critical value is at 0. Note that the critical point is at �2. Thecritical orbit is

�2 7! 0 7! � 7! . . .

Let ' 2 S0✓

. The data of ' is a list of points

z1 = '(x1) = 0, z2 = '(x2), · · · , zn = '(xn

)

and the legs �j

, which is the leg emanating from zj

. We wish to take this data, and using thepolynomial P

�

, we will manufacture a point e' 2 S0✓

.

First, consider the points. Given z1 = 0, z2, . . . , zn 2 C, we define the polynomial Pz2(z) =

z2(1 + z/2)2. The leg �1 connects the critical value z1 = 0 to 1. Take the inverse image P�1z2

(�1)in C. This curve has two components which meet at the critical point �2. The union of thesecomponents cuts the plane into two parts. Call the part that contains z1 = 0 part A, and call theother part B. For all 1 j n�1, the point z

j�1 is either in part A or it is in part B. If zn

6= �2,then it is also in either side A or side B. If however, z

n

= �2, then it is on the boundary of thetwo regions.

Defineezj

2 P�1z2

(zj+1)

in the following way. The inverse images of P�1z2

(zj+1) are separated by P�1

z2(�1), so one inverse

image is in A or one inverse image is in B (almost... the inverse image might be the critical point).The point z

j

is either in part A or part B. Choose the inverse image of P�1z2

(zj+1) which is in the

same part of the plane as zj

. This is how to define the point ezj

.

What about the legs? Define e�j

2 P�1z2

(�j+1); choose the component of the inverse image which

lands at ezj

. If ezj

= �2, then choose e�j

to be the inverse image that maintains the circular order ofthe legs at 1.

60

Figure 37. A pictorial representation of the map �✓

. Observe the critical channelpassing through the critical point z = �2 on the left-hand side determined byP�1x2

(�1).

The discussion above provides a way to build e' 2 S0✓

, given ' 2 S0✓

. This construction descends toyield a well-defined map

�✓

: S✓

! S✓

, given by �✓

: ['] 7! [e'].What does it mean if this map has a fixed point? What does it mean if it doesn’t?

Class 22 - April 3, 2014

Given ✓ 2 Q/Z, we define the following objects:

(1) The standard ✓-spider, S✓

,

(2) the space S0✓

, and the quotient

(3) S✓

:= S0✓

/ ⇠, where ⇠ is generated by scaling the points zi

, and moving the legs �i

up tohomotopy, keeping the points fixed.

(4) A map �✓

: S✓

! S✓

; this is called the spider map.

The map �✓

is a holomorphic map defined on S✓

. We already know that S✓

is a complex manifoldof dimension n� 2, but in fact, it has a hyperbolic structure.

Exercise. Let ✓ 2 Q/Z so that n = 3.Subexercise. Find all such angles ✓.Back to the exercise: identify S

✓

in this example, and use your answer to draw a (very important!)conclusion about the holomorphic map �

✓

: S✓

! S✓

.

Fixed points. Just as before, a fixed point, '✓

of the map �✓

should correspond to an honestquadratic polynomial p

✓

, so that the spider of p✓

is “the same” as the spider given by the fixedpoint '✓. We now make this precise.

Let ✓ 2 Q/Z, and suppose that �✓

: S✓

! S✓

has a fixed point; call the fixed point '✓

2 S✓

, set� := '

✓

(x2), and consider the polynomial

P�

(z) = �⇣1 +

z

2

⌘2.

Proposition. If ✓ is preperiodic but not periodic, then for all ✓j

:= 2j�1✓, the external ray of the

polynomial P�

at angle ✓j

lands at the point zi

:= P j�1�

(0). The union of these rays is equivalentto the fixed spider '

✓

.61

Figure 38. Start with ✓ = 1/6, and build the spider space and spider map �✓

:S✓

! S✓

. There will be a unique fixed point '✓

2 S✓

which is equivalent to thespider data given by the polynomial P (z) = z2 + i, which is the postcritical setz1 = i, z2 = �1 + i, z3 = �i, and the (homotopy classes of )rays �1 at angle 1/6, �2at angle 1/3, and �3 at angle 2/3.

Proposition. Suppose that ✓ is periodic of period n under angle-doubling. Then the polynomialP�

has a superattracting cycle of period n, and the external ray of P�

at angle ✓j

lands at a root

of the Fatou component containing the postcritical point zj

:= P j�1�

(0). The jth leg of the fixedspider '

✓

is homotopic relative to the postcritical points {z1, . . . , zn} to the union of the externalray at angle ✓

j

and the “internal ray” which joins zj

to the root. See the figure below for moredetails.

The proofs of these propositions follow from the definition of the objects in question, in particular,the definition of the map �

✓

: S✓

! S✓

.

But when does the spider map have a fixed point? What can we say if it does not??? It turns out,that in this special case, there is something to say. Remember what the motto was for the realcase: wither �

k

has a fixed point in T , or points collide. Something similar happens here.

Theorem. (Hubbard, Schleicher) For ✓ 2 Q/Z, let ' 2 S✓

, and denote the forward orbit of 'under �

✓

as

'(1) := �✓

('), '(2) := �✓

('1), . . . ,'(n) := �✓('(n�1)), . . .

For the corresponding points, set

z(n)j

:= '(n)(xj

).

Then for all 1 j n, the sequence

n 7! z(n)j

converges. Moreover, the ith and jth sequences will have the same limit if and only if the knead-ing sequence of ✓ from the ith position is EQUAL to the kneading sequence of ✓ from the jthposition.

62

Figure 39. For ✓ = 1/7, there is a fixed point '✓

of �✓

: S✓

! S✓

. The spider dataof the fixed point should be “equivalent” to the spider data of the rabbit above. Notethat we need to take care to connect the legs up to the postcritical points, whichare pink. We do this with a small segment in the interior of the Fatou componentcontaining each of the postcritical points z1, z3, z3. Then we wiggle the legs justenough to make sure that they do not cross each other to get the spider data forthe rabbit.

NOTE: The sequence of spiders may not converge! What happens is that if �✓

has no fixed point,then the points collide, and in the limit, we will NOT get a spider (since we will lose injectivity).However, the sequence of points will converge.

Here is one way to look at it. We have the following diagram where Pn�2 � � ,! Pn�2 is theinclusion.

S✓

�✓//

⇡✓

✏✏

S✓

⇡✓

✏✏

Pn�2 �� _

✏✏

Pn�2 �� _

✏✏

Pn Pn

The orbit sequence n 7! '(n) lives in S✓

, and it may or may not converge. Recall that S✓

is NOTcompact. We can look at the projection of the orbit in Pn�2 � �, and in fact, we can study the

limit in the compact space Pn�2. This theorem is saying that in this space, the sequences n 7! z(n)j

all have a limit. BUT two of these sequences may possible have the same limit (ie, POINTSCOLLIDE).

Example. Let’s consider the case ✓ = 9/56, which we have seen before in Figure 35. What is thekneading sequence of ✓ = 9/56? Usinge Figure 35, we see that it is

K(9/56) = AABAAA.

But isn’t this just the same as the kneading sequence AABA? What’s going on? Does the map�✓

: S✓

! S✓

have a fixed point? If it does, then there is a postcritically finite quadratic polynomial63

p�

(z) = �(1 + z/2)2 such that the kneading sequence of the “critical value rays” is equal toAABAAA. Look in parameter space at where 9/56 lands on the boundary of M .

Figure 40. The Julia set of the quadratic polynomial which (in parameter space)is the landing point of the ray 9/56 on the boundary of M . Note that the kneadingsequence associated to this quadratic polynomial is AABA, and NOT AABAAA.What’s the di↵erence?

Exercise. What is the “spider data” of the polynomial p(z) = z2 � 0.101096 + 0.956287i. Whathappens if you try to assign points and legs?

Notice that for ✓ = 9/56, the kneading sequence K(9/56) = AABAAA is the SAME from the

• 4th position,

• 5th position, and

• 6th position.

Therefore, by the theorem, the sequences

n 7! z(n)4 , n 7! z

(n)5 , and n 7! z

(n)6

all have the same limit, and all of the other sequences n 7! z(n)i

converge to distinct limits (fromeach other, and from the common limit of the sequences above). Look at Figure 40. What do younotice about the postcritical set? If you define ⇣

i

:= pi(0) for the polynomial p(z) = z2�0.101096+0.956287i in Figure 40, what do you notice about the points ⇣

i

?

Conclusion. There is NO quadratic polynomial with the kneading sequence AABAAA. However,there IS a quadratic polynomial with kneading sequence AABA.

The criterion given in the theorem might look a little weird. We will discuss the topologicalobstruction that is preventing �

✓

from having a fixed point when ✓ = 9/56. It has everything to dowith the following picture.

64

Figure 41. There is a simple closed curve which surrounds the points X4, X5,and X6. This curve has a certain invariance property which makes it a Thurstonobstruction. An honest polynomial P

�

(z) = �(1 + z/2)2 cannot have a Thurstonobstruction. Hence, if the kneading sequence of ✓ admits an obstruction, there canbe NO fixed point of �

✓

in S✓

.

Question. Is the kneading sequence of the polynomial enough? Recall that in the real case,we found that the sequence of +1 and �1 was enough to determine the relative order of thepostcritical points on the real axis, that is, just knowing these symbols was enough to determine allof the relevant data. In fact, the pullback map �

k

was defined using ONLY the sequence of +1 and�1. We can ask the same question about the complex case. Given “spider data” of a polynomial,we can determine a sequence of 2 symbols coming from the critical orbit.

Exercise. Find the spider data of the rabbit, and then find the spider data of the corabbit.Compare the two kneading sequences you get when you look at the critical orbit. The will be theSAME (?!). Are the spiders di↵erent? They had better be.


This Hubbard-Schleicher spider theorem says that either there is a fixed point of �✓

: S✓

! S✓

, or“points collide.” Reconsider the example of ✓ = 9/56. The sequences

n 7! z(n)4 , n 7! z

(n)5 , and n 7! z

(n)6

have the same limit. We will use the fact that these points collide to build a topological obstructionthat will prevent �

✓

from having a fixed point.

Let ' 2 S✓

be given by the configuration on the right of Figure 42, and consider the curve �drawn there. It separates the points z4, z5, z6 from the other marked points. Moreover, the curveintersects the legs �4, �5, �6 transversally, and � does not intersect any other leg of the spider '. Thepoint z2 determines the polynomial P

�

(z) = �(1 + z/2)2, where � = z2. The inverse image of thespider given by ', under the polynomial P

�

is given on the left. The points {ez1, . . . , ez6, ew1, . . . , ew6}comprise the inverse image of {z1, . . . , z6} under P

�

. Note that ew3 = ez6.

The inverse image of � under P�

consists of two components, one on each side of the critical channel;call them e�1 and e�2. As displayed in Figure 42, the component e�1 contains the points ez4, ez5, andez6, while the curve e�2 contains the points ez3, ew4, and ew5. Note that

P�

|e�1 : e�1 ! �65

Figure 42. The curve � is drawn on the right, and its inverse images (underthe polynomial P

�

) are on the left: e�1 and e�2. Note that P�1�

({z1, . . . , z6}) ={ez1, . . . , ez6, ew1, . . . , ew6}, where � = z2, and ew3 = ez6. The configuration on the rightis given by ' 2 S

✓

, and e' 2 S✓

.

has degree 1. We will use hyperbolic geometry to show that a fixed point of �✓

cannot possiblyexist in the setting where such a e�1 and � exist.

Hyperbolic lengths. Consider the two maps

P�

: C�P�1�

({z1, . . . , z6})! C�{z1, . . . , z6}, and ◆ : C�P�1�

({z1, . . . , z6}) ,! C�{ez1, . . . , ez6}.

Notation:

• Let lP

�1�

(e�1) denote the hyperbolic length of e�1 in the hyperbolic metric of C�P�1�

({z1, . . . , z6}),

• let le'(e�1) denote the hyperbolic length of e�1 in the hyperbolic metric of C � {ez1, . . . , ez6},and

• let l'

(�) denote the hyperbolic length of � in the hyperbolic metric of C� {z1, . . . , z6}.

Note that sinceP�

: C� P�1�

({z1, . . . , z6})! C� {z1, . . . , z6}is a covering map, it is a local isometry for the hyperbolic metrics in the domain and in the range.Moreover, since

P�

|e�1 : e�1 ! �

has degree 1, we havelP

�1�

(e�1) = l'

(�).

Since the map ◆ : C � P�1�

({z1, . . . , z6}) ,! C � {ez1, . . . , ez6} is a holomorphic inclusion, and sincethe inclusion is proper, we necessarily have that

lP

�1�

(e�1) > le'(e�1).

Putting these inequalities together we see that

l'

(�) > le'(e�1).

Now if �✓

has a fixed point, then ' = e', so we have

l'

(�) > l'

(e�1).66

Moreover, note that � and e�1 are homotopic relative to the marked points z1 = ez1, . . . , z6 = ez6. Let�0 be the unique geodesic in the homotopy class of �. Then the inequality above says

l'

(�0) > l'

(e�01), where P�

|e�01 : e�01 ! �0,

but e�01 is homotopic to �0, and �0 is the unique geodesic in its homotopy class, so we must have

l'

(�0) l'

(e�01)which is a contradiction. Therefore, �

✓

can have NO fixed point if this curve exists.

Curves. This is a generalization of something we have seen before. There is a topological obstruc-tion preventing �

✓

from having a fixed point. This is a special case of a Thurston obstruction.


We will state Thurston’s theorem in its full generality.

Definition. Let S2 be an oriented topological 2-sphere, and let f : S2 ! S2 be an orientation-preserving branched cover of degree d � 2. By Riemann-Hurwitz, there are 2d� 2 critical points off (counted with multiplicity). The map f : S2 ! S2 is called a Thurston map if |P

f

| <1.

Definition. Let f : (S2, Pf

) ! (S2, Pf

) and g : (S2, Pg

) ! (S2, Pg

) be Thurston maps. We saythat f and g are combinatorially equivalent if there exist orientation-preserving homeomorphisms�0,�1 : (S2, P

f

) ! (S2, Pg

) so that 1) �0 � f = g � �1, and 2) the maps �0 and �1 are isotopicrelative to P

f

.

Given all Thurston maps f : (S2, Pf

)! (S2, Pf

) of degree d � 2, we can consider the combinatorialequivalence classes. Thurston’s theorem provides a purely topological criterion to test whetherthere is a rational map F : P1 ! P1 which is combinatorially equivalent to a given Thurston mapf : S2 ! S2. In fact, the theorem says that as long as f has hyperbolic orbifold, then there isat most one rational map F : P1 ! P1 (up to conjugation by Mobius transformations) which iscombinatorially equivalent to a given Thurston map f : S2 ! S2.

Theorem. (Thurston’s topological characterization of rational maps)Let f : (S2, P

f

)! (S2, Pf

) be a Thurston map with hyperbolic orbifold. Then f is combinatoriallyequivalent to a rational map F if and only if f admits no Thurston obstructions. If f is equivalentto a rational map F , then F is unique up to conjugation by Mobius transformations.

The blue sentence above is called Thurston rigidity. The hyperbolic orbifold condition is a smalltechnical point; maps which do not satisfy this are z 7! z2, Chebyshev polynomials, and Lattesmaps. If |P

f

| > 4, then f automatically has hyperbolic orbifold. If |Pf

| = 4, then if f is not Lattes,it will have hyperbolic orbifold. Moreover, if f is a Lattes map, then this is completely detectableby using the ramification portrait of f as stated in the following proposition.

Proposition. Let f : (S2, Pf

)! (S2, Pf

) be a Thurston map with four postcritical points. Thenf is Lattes if and only if every critical point is simple, and no critical point is postcritical.

Curve Systems. A Thurston obstruction is a dynamical object. It is some kind of invariant curvesystem � = {�1, . . . , �

k

} on S2 � Pf

.

Definition. The curves �1, . . . , �k

comprise a multicurve � := {�1, . . . , �k

} on S2 � Pf

if

• no two of them are homotopic

• they are pairwise disjoint67

• they are all essential (that is, none of them is nullhomotopic)

• they are all nonperipheral (that is, none of them bounds a once-punctured disk

Definition. The multicurve � = {�1, . . . , �k

} is f -stable or f -invariant if for all 1 i k, for all↵ 2 f�1(�

i

), the curve ↵ is either

• “erased” (in the sense that it is either nullhomotopic or peripheral), or

• there exists j so that ↵ is homotopic in S2 � Pf

to �j

.

Associated to any f -stable multicurve � is a linear transformation

f� : R� ! R�

defined as follows. Let �i,j,↵

be the components of f�1(�j

) homotopic to �i

in S2 � Pf

, and di,j,↵

be the degree of the mapf |�i,j,↵ : �

i,j,↵

! �j

.

Definition. The Thurston Linear Transformation associated to � is

f�([�j ]) :=X

i,↵

1

di,j,↵

[�i

].

Note that the entries in this matrix are non-negative rational numbers.

Definition. The Thurston eigenvalue �(f�) of the f -stable multicurve � is the Perron-Frobeniuseigenvalue of f�.

Definition. The f -stable multicurve � is a Thurston obstruction if �(f�) � 1.

Not all f -stable multicurves are “bad.”

Exercise. Consider the basilica polynomial p(z) = z2 � 1. Build a ramified cover f : S2 ! S2

as follows. Choose a homeomorphism h1 from C to the upper hemisphere of S2, and choose ahomeomorphism h2 from C to the lower hemisphere of S2 in such a way that h1(R

✓

) hits theequator of S2 in the same point that h2(S�✓) where R

✓

is the ray at angle ✓ in the basin of 1for one basilica polynomial, and S

✓

is the ray at angle ✓ in the basin of 1 for the other basilicapolynomial. In this way, we have built a ramified cover of degree 2 from S2 ! S2, which extendscontinuously across the equator like z 7! z2. Check that the equator curve � = {�} forms anf -stable multicurve with Thurston matrix [1/2]. Prove that the multicurve � = {�} which is theconcatenation of

h1(R1/3), h1(R2/3), h2(S1/3), and h2(S�1/3),

is an f -stable multicurve with Thurston matrix [1].

This mating of the basilica with itself is a special case of a more general construction which we nowdescribe.

Formal mating of two polynomials. We add to the complex plane C the circle at infinity whichis symbolically denoted {1 · e2i⇡✓ ; ✓ 2 R/Z}. We define

eC = C [ {1 · e2i⇡✓ | ✓ 2 R/Z},

which is homeomorphic to a closed disk with the standard topology. As we have seen, a monicpolynomial P : C! C of degree d extends to eC by setting

P (1 · e2i⇡✓) =1 · e2i⇡d✓.68

Figure 43. The curve � above forms a Thurston obstruction for the map f : S2 !S2, which is the topological mating of the basilica with itself.

Let eCP

and eCQ

be two copies of eC and assume P : eCP

! eCP

and Q : eCQ

! eCQ

are monicpostcritically finite polynomials with common degree d � 2. (Recall that the Julia set and filledJulia set of a postcritically finite polynomial are both locally connected). Let ⇠ be the equivalencerelation in eC

P

t eCQ

(disjoint union) which, for ✓ 2 R/Z, identifies 1 · e2i⇡✓ in eCP

to 1 · e�2i⇡✓ ineCQ

. The quotient by ⇠ is topologically an oriented 2-sphere. The formal mating P tQ of P and

Q is the orientation-preserving branched self-covering which is given by P on the image of eCP

andby Q on the image of eC

Q

.

Here is an alternative definition. Let S be the unit sphere in C⇥ R. If P : C! C and Q : C! Care two monic polynomials of the same degree d � 2, the formal mating of P and Q is the ramifiedcovering f = P tQ : S ! S obtained as follows.

We identify the dynamical plane of P to the upper hemisphere H+ of S and the dynamical planeof Q to the lower hemisphere H� of S via the gnomonic projections:

⌫P

: C! H+ and ⌫Q

: C! H�

given by

⌫P

(z) =(z, 1)��(z, 1)

�� =(z, 1)p|z|2 + 1

and ⌫Q

(z) =(z,�1)��(z,�1)

�� =(z,�1)p|z|2 + 1

.

Since P and Q are monic polynomials of degree d, the map ⌫P

� P � ⌫�1P

defined on the upperhemisphere and ⌫

Q

� Q � ⌫�1Q

defined in the lower hemisphere extend continuously to the equatorof S by

(e2i⇡✓, 0) 7! (e2i⇡d✓, 0).

The two maps fit together so as to yield a ramified covering map f : S ! S, which is called theformal mating P tQ of P and Q.

Geometric mating. Let us now consider the smallest equivalence relation ⇠ray on S such thatfor all ✓ 2 R/Z,

• points in the closure of ⌫P

�R

P

(✓)�are in the same equivalence class, and

• points in the closure of ⌫Q

�R

Q

(✓)�are in the same equivalence class.

69

In particular, for all ✓ 2 R/Z, ⌫P

�R

P

(✓)�and ⌫

Q

�R

Q

(�✓)�are in the same equivalence class since

the closures of these sets intersect at the point (e2i⇡✓, 0) on the equator of S.

We say that a rational map F : P1 ! P1 is a geometric mating of P and Q if

(1) the quotient space S/⇠ray is homeomorphic to S (which will have a natural orientation),and

(2) the formal mating P t Q induces a map S/⇠ray ! S/⇠ray which is topologically conjugateto F : P1 ! P1 via an orientation preserving homeomorphism.

Figure 44. The rational map F : z 7! (z2 � e�2i⇡/3)/(z2 � 1) is a geometric matingof the basilica polynomial and the rabbit polynomial (see Figure 1). The rationalmap F has a superattracting cycle of period 3 (the basin of which is colored inmagenta), and a superattracting cycle of period 2 (the basin of which is colored inyellow).

Theorem. (Rees) Assume P : C ! C and Q : C ! C are two postcritically finite hyperbolicpolynomials and F : P ! P is a rational map. The formal mating P t Q is combinatoriallyequivalent to F if and only if F is a geometric mating of P and Q.

A similar result also holds in the case P and Q are postcritically finite polynomials, not necessarilyhyperbolic.

Question. Which polynomials are mateable?

The know the answer to this question for hyperbolic quadratic polynomials.

Theorem. (Tan Lei, Rees, Shishikura) Given two hyperbolic quadratic polynomials z 7! z2 + c1and z 7! z2 + c2, the polynomials are mateable if and only if c1 and c2 do not belong to conjugatelimbs of the Mandelbrot set.

Definition. The limbs of the Mandelbrot set are precisely the parts that are growing o↵ of themain cardiod.

No dynamics. More generally, given two connected, compact, full subsets K1 ✓ C and K2 ✓ C,which are locally connected, we can form the mating of the two sets by using the Riemann mapsfrom �1 : C�K1 ! C�D and �2 : C�K2 ! C�D. Why is it important that the sets are locallyconnected?

70

Figure 45. Every limb of M is connected to the main cardiod at a unique parabolicparameter on @M . All points of M that can be disconnected from the main cardiodby removing this one point are in the same limb. By the theorem above, the rabbitand the basilica are mateable, but the rabbit and cokokopellli are not. The basilicaand the airplane are not mateable, but the airplane and a 7! z2 + i are mateable.


Testing the criterion in Thurston’s theorem requires an infinite search to find all f -stable multic-urves. This is HARD. To prove Thurston’s theorem, we transform the question of whether or notf is equivalent to a rational map into the question of whether or not a pullback map has a fixedpoint.

Teichmuller theory. Recall that a Riemann surface is a connected, oriented topological surfacetogether with a complex structure: a maximal atlas of charts � : U ! C with holomorphic overlapmaps. For a given oriented, compact topological surface X, we denote the set of all complexstructures on X by C(X). It is easily verified that an orientation-preserving branched covering mapf : X ! Y induces a map f⇤ : C(Y )! C(X) (there is something to check at the branch points off ! But by a removable singularities argument, we can define f⇤). Note that for any orientation-preserving homeomorphism : X ! X, there is an induced map ⇤ : C(X)! C(X).

The set of all complex structures is too large for our purposes. After marking a finite set of pointson our topological surface, we define an equivalence relation on the set of complex structures inthe following way. Let A ⇢ X be finite (note that A may be empty). The Teichmuller space of thepair (X,A) is

Teich(X,A) = C(X)/⇠A

where c1 ⇠A

c2 if and only if c1 = ⇤(c2) for some orientation-preserving homeomorphism : X !X which is isotopic to the identity relative to A.

In view of the homotopy-lifting property, if

• B ⇢ Y is finite and contains the critical value set Vf

of f , and

• A ✓ f�1(B),71

then f⇤ : C(Y )! C(X) descends to a well-defined map �f

between the corresponding Teichmullerspaces:

C(Y )

✏✏

f

⇤// C(X)

✏✏

Teich(Y,B)�f// Teich(X,A).

This map is known as the pullback map induced by f .

Genus 0; X = S2. For the special case of the topological 2-sphere, we have the following descriptionvia the Uniformization Theorem. Note that a complex structure on S2 is an orientation-preservinghomeomorphism h : S2 ! P1. For h1, h2 2 C(S2), there is a Mobius transformation µ : P1 ! P1 sothat h1 = µ � h2, or the following diagram commutes

P1

µ

✏✏

S2

h2

>>

h1

P1

For A ✓ S2 finite, the equivalence relation ⇠A

on C(S2) implies that the complex structuresh1 : (S2, A) ! (P1, h1(A)) and h2 : (S2, A) ! (P1, h2(A)) are equivalent if there is an orientation-preserving homeomorphism : (S2, A) ! (S2, A) such that h2 = ⇤h1, and is isotopic to theidentity relative to A. In other words, the following diagram commutes.

(S2, A)h2//

✏✏

(P1, h2(A))

µ

✏✏

(S2, A)h1// (P1, h1(A))

In genus 0, we will use the notation TA

:= Teich(S2, A). In summary, we have

TA

:= {orientation-preserving homeomorphisms h : (S2, A)! (P1, h(A)) such that h1 ⇠A

h2

() 9 µ 2 Aut(P1) where h1 = µ � h2 on A and h1 is isotopic to µ � h2 relative to A}.Note that an element [�] 2 T

A

records two pieces of data: i) information about the points in set A;that is �|

A

: A ,! P1, and ii) the homotopy class of � : (S2, A)! (P1,�(A)).

The Teichmuller space TA

has a natural metric called the Teichmuller metric, and it is a complexmanifold of dimension |A|� 3.

Moduli space. Given an orientation homeomorphism � : (S2, A)! (P1,�(A)), we can get rid of thehomotopy information and just record �|

A

: A ,! P1. This gives us a map

⇡ : TA

!MA

where MA

is the moduli space of (S2, A). By definition, this is the space

MA

:= {' : A ,! P1 up to postcomposition by Mobius transformations}.72

Let A = {a1, . . . , an}, and suppose that n � 3. By choosing coordinates, we may suppose that' 2 M

A

satisfies '(a1) = 0,'(a2) = 1, and '(a3) = 1. In this way, the class ['] 2 MA

isdetermined by the complex numbers

('(a4), . . . ,'(an)) 2 Cn�3,

In this way, we identify MA

with an open subset of Cn�3. The map

⇡A

: TA

!MA

given by [�] 7! [�|A

]

is a universal covering map.

The pullback map. In the case of genus 0, we have a concrete description of the pullback map

�f

: TB

! TA

.

Let f : (S2, A) ! (S2, B) be an orientation-preserving branched cover, and let B ✓ S2 be a finiteset which contains the critical values of f . Let A ✓ f�1(B), and finally suppose that |A|, |B| � 3.There is a holomorphic map �

f

: TB

! TA

described explicitly below.

Write B = {b1, . . . , bn}, and write A = {a1, . . . , am}. Let � : (S2, B) ! (P1,�(B)) be anorientation-preserving homeomorphism, and suppose that �(b1) = 0,�(b2) = 1, and �(b3) = 1.Then there are

• a UNIQUE orientation-preserving homeomorphism : (S2, A)! (P1, (A)) so that (a1) =0, (a2) = 1, and (a3) =1, and

• a UNIQUE rational map F : (P1, (A)) ! (P1,�(B)) so that the following diagram com-mutes.

(S2, A)

f

✏✏

// (P1, (A))

F

✏✏

(S2, B)�

// (P1,�(B))

The map �f

: TB

! TA

is given by �f

: [�] 7! [ ]. This map is holomorphic and contracting forthe corresponding Teichmuller metrics.

Dynamics. We will apply the map above to a Thurston map. Note that the pullback map isactually a nondynamical object. Given a Thurston map f : (S2, P )! (S2, P ) with postcritical setP , we will refer to the associated pullback map �

f

: TP

! TP

.

Exercise. Show that the map �f

: TP

! TP

has a fixed point if and only if f is combinatoriallyequivalent to a rational map.

The hardest part of Thurston’s theorem is to prove that �f

has a fixed point in TP

if and only iff : (S2, P )! (S2, P ) admits no Thurston obstructions.


In our last class, we will wrap up the discussion of Thurston’s theorem. Understanding the pullbackmap is not easy, and part of the di�culty is that understanding points of T

P

is not easy. The moduli73

space MP

is somewhat more user-friendly. An immediate question is: does the pullback map�f

: TP

! TP

descend to yield a map MP

!MP

so that the following diagram commutes?

TP

�f//

⇡P

✏✏

TP

⇡P

✏✏

MP

//MP

This almost NEVER happens. I know of only one class of examples where �f

descends to themoduli space, and for those examples, �

f

is actually constant (so of course it extends!). However,sometimes an inverse of �

f

descend, yielding a map MP

MP

so that the following diagramcommutes.

TP

�f//

⇡P

✏✏

TP

⇡P

✏✏

MP

MP

oo

We’ll call this map gf

: MP

! MP

. Let’s compute it a familiar example. Let f : (S2, P ) !(S2, P ) be a Thurston map with P = {a, b, c, d}, and suppose that f has the following ramificationportrait.

a2// b // c

aa

d 2ee

Let � : (S2, P ) ! (P1,�(P )) be an orientation-preserving homeomorphism normalized so that�(a) = 0,�(b) = 1, and �(d) =1. For notation, set y := �(c). Note that the point [�|

P

] 2MP

iscompletely determined by the complex number y 2 P1 = {0, 1,1}. Think of y as a moduli spacevariable.

There is:

• a unique orientation-preserving homeomorphism : (S2, P ) ! (P1, (P )), so that (a) =0, (b) = 1, and (d) =1, and

• a unique rational map F : (P1, (P ))! (P1,�(P )) so that the following diagram commutes

(S2, P )

f

✏✏

// (P1, (P ))

F

✏✏

(S2, P )�

// (P1,�(P ))

For notation, set x := (c); again, think of x 2 P1 � {0, 1,1} as a moduli space variable. Let’sdetermine as much as we can about the rational map F :

• F is a quadratic polynomial (why?)

• F has a critical point at 0, and the corresponding critical value is 1; that is, F (0) = 1

• F (0) = x, and

• F (1) = y.

The previous conclusions are based on the fact that the diagram above commutes, and the coordi-nates we chose. The first two points above imply that a normal form for F is given by

F : t 7! At2 + 1, where A is a complex parameter.74

Imposing the third condition above implies that A = �1/x2, so we really have

F : t 7! � t2

x2+ 1.

Imposing the last condition above implies that y = �1/x2 + 1. We just did something REMARK-ABLE! We found a relation between the moduli space variables x and y. In fact, we did this:

��f//

⇡P

✏✏

⇡P

✏✏

y xgf

oo

where gf

: P1 ! P1 is gf

: x 7! y, y = �1/x2 + 1. This map gf

is rather interesting; note that it isalso postcritically finite!!!! In fact, the ramification portrait for g

f

is

02//1 2

// 1bb

This is not a coincidence. The map gf

has three fixed points: ⇣r

, ⇣c

, and ⇣a

. Note that each ofthese fixed points is in P1� {0, 1,1}, or each of the fixed points is in the moduli space M

P

.

Exercise. Prove that each of these fixed points corresponds to a quadratic polynomial of the formt 7! At2 + 1, such that the critical point t0 = 0 is in a cycle of period 3. Then draw the Julia setof g

f

and find these fixed points.

Figure 46. This is a superposition of two parameter spaces. One of them, themoduli space M

P

⇡ P1 � {0, 1,1} is a nondynamical parameter space, and it hasa dynamical system on it, the map g

f

. The Julia set of gf

is visible in this pic-ture. There are three repelling fixed points of the map g

f

: ⇣r

, ⇣c

, ⇣a

. The otherparameter space is a dynamical parameter space; that is, it parameterizes quadraticpolynomials as dynamical systems, and it contains the Mandelbrot set M . In thecoordinates chosen, the three fixed points of g

f

line up EXACTLY with the cor-responding parameters of M : the rabbit, corabbit, airplane. This is my absolutefavorite picture.

We can lift the Julia set of gf

up to the Teichmuller space TP

to help understand the dynamics ofthe map �

f

: TP

! TP

.75

Figure 47. On the left is the Julia set of gf

: x 7! �1/x2 + 1, and on the rightis a picture of ⇡�1

P

(this Julia set) ✓ TP

. The Teichmuller space in this exampleis isomorphic to the open disk, and if f is the rabbit polynomial, then �

f

has aunique fixed point ⇣ 2 T

P

. The disk on the right is centered at this fixed point. Thedynamics of �

f

is visible in this picture. The three large components comprise onecycle of period 3. The rabbit fixed point in the center is attracting, and one can usethe map g

f

to compute the derivative of �f

at the rabbit fixed point. Just computethe multiplier of g

f

at the repelling fixed point ⇣r

and invert it.

Note that the only thing that I used to compute the map gf

was the ramification portrait of f .I did not use the Thurston class at all. This is another reason why the map g

f

is very helpful;it can be computed from some finite combinatorial data. This is somewhat misleading; the mapgf

depends on just more than the portrait. It actually depends on some nondynamical data; thecovering combinatorics of f , or the Hurwitz class of f . It turns out that if f is a Thurston map ofdegree 2, then there is a unique Hurwitz class, so the portrait is enough to determine g

f

.

Small wrinkle. Unfortunately, the map gf

does not always exist. What does exist is a Hurwitzcorrespondence W

f

which is sometimes the graph of a map. Regardless of whether it is or not, itis an algebraic object that can be used to understand the dynamics of �

f

.

And that’s it! Thanks for attending Math 636!! I would love to talk to any

of you more about any of this stu↵!

76

FractalStream Scripts.

(1) Draw the Mandelbrot set:

iterate z2 + c until z escapes.

(2) Draw a Julia set of a polynomial:

iterate z2 � 1 until z escapes.

(3) Draw a Julia set of a rational map:

iterate 1� 1/z2 until z escapes or z vanishes or (z � 1) vanishes.

(4) Draw a Herman ring:

iterate (�0.7493909542� 0.6621277805i)z2(z� 4)/(1� 4z) until z escapes or z vanishes. Ifz vanishes then [green]. If z escapes then [red].

(5) Draw the Julia set of the second iterate of the basilica, and color the superattracting basinsin di↵erent colors.

iterate (z2�1)2�1 until z escapes or z vanishes or (z+1) vanishes. If z escapes then [blue].If z vanishes then [green]. If (z + 1) vanishes then [yellow].

77

Riemann Surfaces and Complex Dynamicskochsc/old.pdfdiscuss Thurston’s theorem, ... The bifurcation...

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