Explorations in Topology: Map Coloring, Surfaces and Knots

EXPLORATIONS IN TOPOLOGY:MAP COLORING,

SURFACES, AND KNOTS

DAVID GAY University of Arizona

Amsterdam Boston London New York Oxford ParisSan Diego San Francisco Singapore Sydney Tokyo

Academic Press is an imprint of Elsevier

Academic Press is an imprint of Elsevier30 Corporate Drive, Suite 400, Burlington, MA 01803, USA525 B Street, Suite 1900, San Diego, California 92101-4495, USA84 Theobald’s Road, London WC1X 8RR, UK

This book is printed on acid-free paper.

Copyright © 2007, Elsevier Inc. All rights reserved.

No part of this publication may be reproduced or transmitted in any form or by anymeans, electronic or mechanical, including photocopy, recording, or any informationstorage and retrieval system, without permission in writing from the publisher.

Permissions may be sought directly from Elsevier’s Science & Technology Rights Department in Oxford, UK: phone: (+44) 1865 843830, fax: (+44) 1865 853333, e-mail:[email protected]. You may also complete your request on-line via the Elsevier homepage(http://elsevier.com), by selecting “Customer Support” and then “Obtaining Permissions.”

Library of Congress Cataloging-in-Publication DataGay, David.

Explorations in topology / David Gay. -- 1st ed.p. cm.

ISBN 0-12-370858-31. Topology. I. Title.QA611.G39 2006514--dc22

2006017400

British Library Cataloguing in Publication Data A catalogue record for this book is available from the British Library

ISBN 13: 978-0-12-370858-8ISBN 10: 0-12-370858-3

Cover Art: The cover art is based on a photograph of the Ungar-Leech coloring of the torus.Norton Starr of Amherst College made the model and took the photograph. We thank him for hispermission to use them.

For all information on all Elsevier Academic Press publicationsvisit our Web site at www.books.elsevier.com

Printed in the United States of America07 08 09 10 9 8 7 6 5 4 3 2 1

To the memory of Dana Clyman and to all my students: you made

this project worthwhile

This page intentionally left blank

Contents

v

Preface viiAcknowledgments xi

Chapter 1: Acme Does Maps and Considers Coloring Them 1Chapter 2: Acme Adds Tours 27Chapter 3: Acme Collects Data from Maps 49Chapter 4: Acme Collects More Data, Proves a Theorem, and Returns

to Coloring Maps 73Chapter 5: Acme’s Solicitor Proves a Theorem: the Four-Color Conjecture 89Chapter 6: Acme Adds Doughnuts to Its Repertoire 103Chapter 7: Acme Considers the Möbius Strip 125Chapter 8: Acme Creates New Worlds: Klein Bottles and Other Surfaces 149Chapter 9: Acme Makes Order Out of Chaos: Surface Sums and

Euler Numbers 177Chapter 10: Acme Classifies Surfaces 205Chapter 11: Acme Encounters the Fourth Dimension 225Chapter 12: Acme Colors Maps on Surfaces: Heawood’s Estimate 253Chapter 13: Acme Gets All Tied Up with Knots 271Chapter 14: Where to Go from Here: Projects 313

Index 331

Preface

vii

This book will (I hope!) give you a rich experience with low-dimensional topology, get you really hooked on solving mathematical problems, and launchyou into an adventure of creating ideas, solutions, and techniques—one that givesyou a “research experience” with topology. Of course, I also hope you get excitedabout this adventure, talk about it with your friends, and have a good time.

Topology is a geometric way of looking at the world and the ideas you willencounter in this book emerged from a long evolutionary process. I hope that thisbook engages you in this process while it introduces you to the topological pointof view. Along the way, you will encounter many of the concrete, simply-statedproblems for which much of modern topology was created to solve—problemsassociated with maps, networks, surfaces, and knots.

The book is more than just an introduction to topological thinking. It wasdesigned to provide you with the opportunity to

● experience solving open-ended problems by guessing, drawing pictures,constructing models, looking for patterns, formulating conjectures, findingcounter-examples, making arguments, asking questions, using analogies, andmaking generalizations;

● observe new mathematical techniques springing from solutions to problems;

● see the connectivity of mathematical ideas—the solution to one problemhelps solve other problems; the answer to one problem leads to new questions.

Your participation in all this will help you place mathematical connections into bold relief and will elicit from you gasps of surprise! This participation will havefrustrations, but it will also be full of pleasure, drama, and beauty. It will enhance your geometrical and topological intuition, empower you with new approaches to

solving problems, and provide you with tools to help you on your next mathematicaljourney.

Finally, the book is set up so that—as much as possible—you are in charge of the development of the mathematical ideas that emerge. You ask the questionsthat propel the unfolding of your own topological knowledge and understandingand you initiate the process of answering them. The structure of the book maximizes the possibilities for all of this to happen. It poses key questions to begin the discussions and give them shape and direction. It models the problem-solving process and provides you with ways to communicate and explain solu-tions. And then it lets you loose, as a former student of mine put it, “to write ourown book.”

Format of the Chapters

Each chapter (excluding chapter 14) poses a big problem and takes a stab at itssolution. This takes place in the context of a story whose characters are employ-ees of Acme Maps. The big problem and smaller related problems come up natu-rally as the characters carry out their work. The jobs Acme takes on expandnaturally as the book unfolds. The characters fiddle around with the problemsmuch as real problem-solvers would. They make mistakes. They find themselvesin dead ends and work their way out. They create their own definitions and termi-nology. Sometimes they don’t solve the big problem, but they work on it and comeup with partial solutions. These solutions are sometimes what a “mainstream” textwould call theorems. As you follow the story, paper-and-pencil icons involve youby pointing out explicit tasks for you to carry out. These tasks are called YourTurn. A set of Investigations, Questions, Puzzles, and More replaces the tradi-tional problem set at the end of the chapter. An investigation is a non-routine,open-ended problem: Look at such-and-such. What do you see? What can yousay? Can you explain it? In an investigation you make observations, look forpatterns, make a conjecture, and prove the conjecture (or come up with a counter-example and refine the conjecture). When I teach a course that goes with thisbook I ask students to carry out many of these investigations. They form the heartof the course.

To relate what goes on in the chapter to the rest of the world of mathematicseach chapter ends with a section entitled Notes. It places the problems, concepts,and results of the chapter in their historical context. It introduces the standardterminology used by other mathematicians. Occasionally it will summarize theresults of the chapter and set these results out in the form of theorems. A list ofappropriate References follows the Notes section.

The format of chapter 14 differs from that of chapters 1-13. This chapter is anannotated list of possible projects that extend the concepts of the book, but whichmay involve research beyond the materials of the book. It includes an extensive list

viii Preface

of helpful resource materials and a guide for carrying out a project and commu-nicating its results.

Overview of the Chapters

● Chapter 1. This is an introduction to coloring maps on an island and on thesphere. Particular attention is paid to maps requiring two colors.

● Chapter 2. This is an introduction to networks (graphs)—to taking particu-larly “nice” trips on networks, and identifying those networks where such tripsare possible.

● Chapter 3. This is an introduction to collecting data about maps and observ-ing relationships among these data.

● Chapter 4. This continuation of chapter 3 applies relationships among mapdata to map coloring.

● Chapter 5. This final chapter on coloring maps on the sphere is an expositionof the first “proof” of the four color theorem.

● Chapter 6. This chapter considers the torus, a new surface on which to drawmaps and networks and to solve (or not) problems which came up for the sphere.

● Chapter 7. This chapter looks at twisted strips, cuts them, and adds theMöbius strip to the repertoire of surfaces.

● Chapter 8. This chapter takes the idea of a “pattern” for a surface introducedin chapters 6 and 7 and uses it to create new, unusual surfaces (Klein bottle,crosscap) and to observe relationships among them.

● Chapter 9. This chapter introduces the Euler number for a surface anddescribes a technique for creating new surfaces out of old.

● Chapter 10. The study of surfaces becomes algebraic as symbol strings replacepatterns and rules are developed for arriving at equivalent, recognizablesymbols. The consequence: a classification theorem for surfaces.

● Chapter 11. This chapter tidies up the classification of surfaces by looking atboundaries (lakes). It also considers the “existence” of surfaces and offers thepossibility of “assembling” them in four-space.

● Chapter 12. This chapter relates the number of colors needed to color a mapon a surface (not the sphere) to its Euler number.

● Chapter 13. This introduction to knot theory and one of its invariants ismotivated by problems in chemistry and biology.

● Chapter 14. As mentioned above, this chapter is devoted to projects.

Preface ix

Success in the Course this Book is based on

To get the goodies a course using this book has to offer, we all have to dig in andtake risks—by offering a solution that may turn out to be incorrect, by making aguess that might be wrong, or by using an approach to a problem that might notpan out. We will laugh at our mistakes. We will listen to others’ ideas and possiblybuild on them. We will learn to accept a solution only when we understand it,when it feels good in our tummies. It’ll be a hoot!

Instructor’s Manual

This supplement describes how the author uses this book with a class. It highlightseach chapter’s important aspects and points out investigations important for thedevelopment of the main ideas and it includes solutions to many of problems andhints for others. The manual contains large-size versions of diagrams and patternsthat appear in the text so that the instructor could have copies made for use inclass.

x Preface

xi

More than thirty years ago, I taught an experimental, low-dimensional topologycourse with an incredible group of talented and enthusiastic students. I wrote upnotes as the course progressed. The course was so successful that I have taught itseveral times since, revising and supplementing my original notes each time. Thenotes eventually became this book.

Several individuals played parts in the development and evolution of the courseand book. Initial credit goes to my college teachers Albert Tucker and Ralph Foxwho, when I was a mathematical tad, first fed me these topological ideas and gotme hooked. Over the years, Philip Straffin and David T. Gay taught the courseusing the notes and Noah Snavely and Adam Spiegler assisted me in teaching it;I appreciate their pedagogical insights and enthusiasm for the project. I thank theThe University of Arizona’s Department of Mathematics for supporting curriculumdevelopment and for valuing innovation in teaching. My colleague Olga Yparakiwas particularly excited with the story format and encouraged me to continuedeveloping it. I am especially grateful to Susan Lowell, good friend and author,for suggestions on making the story’s characters come alive.

I have borrowed freely from several authors whose books have given me math-ematical ideas, problems, and approaches. Most of these appear as references inthe text. One author that deserves special mention is Martin Gardner. He has theuncanny ability to come up with problems that are accessible to students, thatgrab them, and that lead them nicely and naturally to deeper mathematicalinvolvement.

Of course, the lion’s share of thanks for this project’s development goes to thestudents who took the course, who struggled with the problems and projects andtaught me what worked and what got them excited. It was their enthusiasm forthese ideas and for discovery that kept me on the project all these years.

I also need to thank Deborah Yoklic, who typed the first version of the manu-script thirty years ago and who has been an enthusiastic supporter of the project

Acknowledgments

ever since. To those who reviewed the manuscript, I am grateful for the many help-ful criticisms and encouraging comments during the phases of its development.

Finally, let me thank my editor Tom Singer at Academic Press who believed inthe project from the beginning, Michael Troy at Graphic World PublishingServices who kept me on task and let me know what was happening duringproduction, Fritz Simon who turned my scribbles into effective illustrations andhelped me make the manuscript “look like a book,” and Tulley Straub who helpedrender three-dimensional objects realistically. It was a pleasure and a privilege towork with all of you.

xii Acknowledgements

Acme Does Maps andConsiders Coloring Them

1

Scene 1

It is early afternoon in a small storefront company. The owner, Boss, 50-somethingand portly, wears a rumpled shirt with bola tie and cowboy hat. Joe, 30-somethingand the assistant, wears jeans and a t-shirt. The telephone rings.

BOSS: Howdy. Acme Maps. We can put your country on the map!... Yes....Uh, huh ....Okee, dokee. We'll see what we can do ....Call you back when we have something ....'Bye, and thanks for calling Acme.(Hangs up and turns to Joe.) Well, Joe, we've got a new project.Some guy looks at a map and can't tell one country from another. Strange feller. Sees Switzerland and Austria merging intoeach other. He wants a map of Western Europe where he can tell the countries apart, and he wants it cheap. Can we give him something?

C H A P T E R

1

JOE: Hmm. Interesting. The G7 model with bold black borders oughta workfor anybody ....Here's an idea; let's color each country a different color.Then there'd be no mistake. Something like this. (Joe shows Boss amap.)

BOSS: Not bad. It's purty...too purty. Nope, it won't do.

JOE: Why not?

BOSS: This guy wants it cheap. You know how much it costs to print 11 colors. Think of something better. (Boss goes back to whatever hewas doing before the telephone rang.)

JOE: (Looking at map he colored.) Hmm...You can tell the countries apart,but there are too many colors. I need fewer colors. (Looks at Boss.)Would it hurt, Boss, if Portugal and Holland were the samecolor...maybe pink? And Austria could be pink too.

BOSS: Sure....Hey, wait! If you keep goin' like that, we'll be right back towhere every country is colored the same color, pink. Are you loony?Who wants a pink map?

JOE: Nooo, nooo, Boss. All you have to do is color a country a color differ-ent from its neighbors. So Portugal can be colored anything exceptthe color of Spain.

BOSS: Well, then whaddaya color Spain?

JOE: Huh?

BOSS: To know what to color Portugal, you gotta know what to color Spain,and then to color Spain you gotta know–

2 Chapter 1 Acme Does Maps and Considers Coloring Them

JOE: Stop. You're making it too complicated. Start with some country, likeSwitzerland, and color it red. Now you know none of its neighborscan be colored red.

BOSS: So whaddaya color them?

JOE: Something else.

BOSS: Like?

JOE: OK. France can be blue. Then Germany is green 'cause it's got to bedifferent from red and blue. Then comes Austria; it can't be red, can'tbe green. OK, color it blue, just like France. And then color Italygreen! Look, let me show you.

Boss, we've colored five countries already and used only three colors.

BOSS: Yeah. Nice. But ya haven't finished yet.

JOE: Hey, takes time, Boss. Look, you can color Spain red and thenPortugal blue. Two more countries; no new colors!

BOSS: Big deal.

JOE: (Notices something on the paper on which he's been doodling.) Hey!Neat! Look, Boss, really cheap. Only three colors! Luxembourg is red,Belgium...hmm...it can't be blue, or red, or green. Hmm.

BOSS: So?

JOE: OK. Color it yellow for the moment. Only four colors so far. ThenHolland can be red, and Denmark blue. Done! Four colors. Look.

Scene 1 3

BOSS: Thought you said only three.

JOE: Well, wait a minute, Boss. Luxembourg is red. It's got France,Germany and Belgium as neighbors so they've got to becolored....Hey, Boss, you're not listening.

BOSS: (Looks at latest version on Joe's paper.) Not bad. I'm a genius! Whydidn't I think of this before? Color maps with three colors. Cheap. I'llget a patent on it. Wait'll the guys down at El Grande Maps see this!Joe, make up a few more like that. South America. Africa. Keep itdown to three colors, will ya?

JOE: But Boss! (Turns back to his paper, thinks, doodles, and then writes some rules.)


Rules and Terminology for Coloring a MapTo color a map properly:● Each country must be completely colored by one color.● Two countries sharing a common border must be colored different colors.

1. Your TurnHelp Joe. Can he color the map of Western Europe in three colors? There are twochoices:

1. Color the map properly in three colors.

2. Give an argument for Boss that explains why the map cannot be colored properly in three colors.

2. Your TurnJoe still needs help. Color the following map of South America properly and mini-mally. Boss will want to know how the coloring is minimal so provide Joe with anexplanation. (Joe will also need help with the map of Africa, at the end of the chapter.)

Scene 2

JOE: (Has dealt successfully with coloring the maps of Western Europe,South America, and Africa. He thinks to himself.) It's only a matter oftime before Boss will want me to color all maps properly and mini-mally. (He makes a list.)

Scene 2 5

Problem: Color a map properly with the fewest number of colors; that is, a minimal coloring.

Question: Do this for the map of Western Europe. Will three colors work?

Hmm. The United States. That should be fun!

3. Your TurnHelp Joe. Color the map of the 48 lower states properly and minimally. How manycolors do you need? Do you encounter anything unusual?

Joe started coloring the 48 states from the lower left with California, then Arizona,and then Nevada and Utah. He stopped and stared at the Four Corners, where Utah,Arizona, New Mexico, and Colorado meet.


List of Maps● Western Europe● South America● Africa● Provinces of Australia● Provinces of Canada● The 48 “lower” states of the United States

JOE: How should I color that? The four states all touch at the Four Cornersso each one should be a different color. But four colors could reallypush the number of colors needed way up. What if a map had a ''FiveCorners'' or a ''Six Corners''?

I'd need six colors just for those six countries. Why if there were a''Seventeen Corners, '' I'd need 17 colors! Boss wouldn't like that.Noooo, nooo. (He writes.)

With this new rule I can color the Four Corners area this way.

Nice! (He takes a new page and starts a new list.)

New Rule for Coloring Maps ProperlyA border between two countries must consist of a positive length; it cannotconsist of a single point or even two or three points or any finite number ofpoints.

Scene 2 7

Maps● The countries of Western Europe● The countries of South America● The countries of Africa

Whew. The list goes on and on and I haven't even touched Asia! Then there'sa new map created by a new configuration of countries in Eastern Europeand the former Soviet Union. There're maps that haven't been thoughtabout, that haven't happened yet! (He quickly sketches the following.)

I'd have to recolor it. How many colors would I need? Hmm. Depends.Is there anything I could predict about the number of colors? Boss seemsto think I'd never need more than three, but I know better. What if I hada very complicated map, maybe one with thousands of countries orsubdivisions? Wouldn't that need a lot of colors? (He takes out anothersheet and writes.)


● Provinces of Australia● Provinces of Canada● The 48 ''lower'' states of the USA● Counties of England● Departments of France● The countries of Central America

Big ProblemGiven any map of countries (one that exists or one that could exist in thefuture), how many colors would I need to color it properly and minimally? Is there a fixed number of colors that would be sufficient for all these maps?Or could I find more and more complicated maps that would need more andmore colors?

4. Your TurnWhat do you think? Make a guess. Write it down and save it!

Scene 3

JOE: (Looking at maps in Acme's files.) So many possibilities and so littletime. Infinitely many possibilities! This is a big problem that needssimplification. (He finds a map and pulls it out.)

How would I color this one? Well, that's a silly question; just color it asbefore. The countries have been distorted, some have been stretched, andsome shrunk. But the bordering ''relationships'' are the same. (He pullsanother map out of the file.) Oh, my, have a look at this one.

Scene 3 9

Same thing. Color all of these the same. It's as if the ''original'' map ofWestern Europe had been printed on a thin piece of rubber and thenstretched-pulled-shrunk into the modern version. Hey, I have an idea! Whystop with ''modern''? (He draws.)

Cool. Would this sell? Western Europe as a rectangle where the countrieshave borders made up of straight lines! I could do that to any map.(He takes the outline of South America, draws a ''map,'' and turns it into a map of the future.)

5. Your TurnHelp Joe, again: Transform the ''real'' map of South America (from Exercise 2)into a map of the future. Label countries of the transformed map with the appropriate country names.

JOE: Progress! (He writes.)


6. Your TurnIt's now easy for Joe to draw maps and practice coloring them. Here are ones hedrew. Help him color them properly and minimally. For each one, write down thenumber of colors you used. Make up a couple of maps of your own, and do thesame thing with them.

Scene 4

Joe is drawing maps and talking aloud to himself.

JOE: OK. Things are a little simpler. Maps on a rectangle. Still a lot of possi-bilities. How about starting really simply:

Hmm. I could just keep adding lines from one side of the rectangle toanother. Let's see what happens.

Big Problem, Simple VersionAny map can be transformed to a map on a rectangle where borders can bemade up of straight lines. Solving the coloring problem for any map is reducedto solving the coloring problem for the transformed map.

Scene 4 11

Whoa! Look at that! Two colors. Amazing. I wonder if that alwaysworks. Add another line ...

Hmm. Works again. Now why does it do that? Now let's take the newmap and put it next to the old one, but before the line was added.Then color them.

Wow. I see. Keep old coloring on one side of the added line, reverseold coloring on other side. I've got to show this to Boss! (Boss enters.)

BOSS: Heard m' name.

JOE: Boss, remember we were coloring maps? Well, I found a whole bunchwhere you only need two colors.

BOSS: Ya don't say.

JOE: Take a rectangle. Draw lines from one side to another, as many lines asyou want. You always get a map you can color in two colors. Like this.


BOSS: I like the idea of two colors. But, those ''maps'' don't look likeanything I've seen before.

JOE: Think of them as maps that have been distorted.

BOSS: Distorted?

JOE: Yeah. Let me show you why you can color this particular kind of mapin two colors. The main idea works like this. Think of such a map asbeing made one line at a time. Here it is colored at one stage, justbefore adding another line that we'll call L.

You can see that the map (with line L added) is colored properly onboth sides of L. The only problems are pairs of countries that have aborder that's a piece of L. The solution is to reverse the colors on oneside of L. You get a properly colored map. That tells you what to dowhen you go from one stage to another. This is ''the transition step.''It's the general method. Build the original map one line at time, andyou get a sequence of increasingly complex maps:

The first map in the sequence has one line. Easy to color in two colors.Then add a line to get the next map in the sequence, and use the tran-sition step to color this. Add yet another line, and use the transitionstep again. Keep doing this until all the lines have been added and theoriginal map is colored in two colors.

Scene 4 13

BOSS: Humph. Lot of coloring and recoloring. That's too much work. Theremust be an easier way. (He walks away, leaving Joe a little crestfallen.)

JOE: (Slowly turns back to his desk and talks to himself.) Boss is hard toplease. Anyway, I think we're making progress—one step at a time.(He takes a new piece of paper and writes.)

Solving the special case didn't solve the whole problem, but it's stillpretty powerful. One of these maps–formed by millions of lines—canbe colored properly with only two colors! Solving the special case waskind of an investigation. I didn't really know ahead of time what theoutcome would be. It's worth documenting what I did. (He takesanother sheet and carefully writes.)

Progress on Map Color ProblemSpecial case solved:Map on a rectangle created by drawing lines from one side to another can becolored in two colors. Moreover, I can describe exactly how to carry out thecoloring.


How To Carry Out an InvestigationCreate a bunch of examples of items satisfying certain property. (Drew lots of

the special type of map; colored them.)Look for patterns in the examples. (Noticed that all examples considered

could be colored in two colors.)Make a conjecture about all items satisfying certain property. (''All the special

type maps–even ones I didn't draw—can be colored in two colors.'')Do one of two things:

1. Give an argument explaining why the conjecture is true (I gave Boss anargument explaining my two-color conjecture), or

It had been quite a day, and it was time to go home. Before leaving, Joe wrote downsome investigations and some questions, which he included in the following, for youto carry out and answer.

Investigations, Questions, Puzzles, and More

1. InvestigationJoe has considered the special case of maps on rectangles formed by drawing linesfrom one edge to another. As a new case for investigation, he alters this prescriptionto allow, in addition to lines, circles drawn inside the rectangle as well as arcs of circleswith each end touching a side of the rectangle. Here is an example of such a map.

Again, the countries are the regions or areas bounded by pieces of the lines andthe circles (the pieces of lines and circles thus form the borders of the countries).

Joe has outlined steps to carry out the investigation:

1. Draw a few such maps. Color them properly and minimally.

2. Look for a pattern emerging from step 1. If I don't see anything, draw andcolor some more until I find one.

3. Make a conjecture.

4. Justify the conjecture or find a counterexample to it. (If I find a counter-example, go find a conjecture that I can justify!)

Investigations, Questions, Puzzles, and More 15

2. Find a counterexample to the conjecture. (There is no counterexample tomy two-color conjecture. However, Boss made the following conjecture: ''Allmaps can be colored in three colors.'' But I found that the map of WesternEurope needed four colors. So the map of Western Europe is a counter-example to Boss' conjecture.)

5. Try to generalize the outcomes of this investigation. That is, alter the rules forforming these maps in some simple way so that my conjecture still holds andcan be justified by an argument similar to the one I gave in step 4.

Carry out the investigation and write a report about it to Boss.

2. InvestigationJoe thinks of another way to alter his basic map-on-a-rectangle: Form a map on a rectangle by drawing any number of Y-shaped figures such that each of the threeends of every Y touches a side of the rectangle.

A brief outline of the investigation:

● Draw some maps of this kind. Color them properly and minimally.

● Look for a pattern.

● Make a conjecture.

● Justify the conjecture.

● Tell Boss about my results.

Carry out the investigation. Write a dialog between Joe and Boss in which Joetells Boss what he has discovered and convinces Boss of his conjecture.

3. InvestigationJoe thought of the following variation to Investigation 2: In addition to Y-shapedfigures, allow any number of circles with one chord to be drawn in the rectangle.


● Of course, he's interested to know how many colors such a map would need, whatargument would justify that fact, and how to report to the boss on this special case.

● Joe also wonders whether or not the results of the investigation can be gener-alized; that is, can the rules for creating maps in this investigation be alteredeven further so that the number of colors needed stays the same and the argu-ments justifying it remain pretty much the same? (For example, in addition toY's and circles with a single chord, what would happen if he were also to allowlines from one side of the rectangle to another?)

4. InvestigationIn the real world, there are oceans, lakes, and seas, and many real maps reflect that.So it might be worth considering maps on a rectangle with a rectangular lake inthe middle, a sort of island with lake.

Joe figures that a simple case of this type to investigate would be a map on thisisland with lake created by drawing lines from one side of the island to another(but not allowing a line to pass through the lake) or drawing lines from one sideof the island to the lake. Like this.

He outlines the investigation:

● Draw a few maps like this. Color them properly and minimally. (Assume thelake is not a region to be colored.)

● Observe how many colors I need, and look for patterns.


● Make a conjecture.

● Justify the conjecture or find a counterexample.

● If I find a counterexample, revise my original conjecture to account for this newevidence and try to justify the new conjecture. I want to arrive at a conjecture Ican justify although I may have to revise my conjecture several times until I'mhappy with it. Then it will be the best result possible and I can justify it.

Help Joe out. Carry out the investigation and write a report to Boss.

5. QuestionJoe tells himself: ''I can draw a lot of maps that can be colored properly in twocolors.'' Then he asks:

● Can I draw a map on a rectangle that requires three colors? Can I draw a mapon a rectangle that requires four colors?

● Can I draw a map on a rectangle that requires five colors?

● Of course, none of these maps would be created by drawing lines from oneside to the other. But what else could I say about these maps other than theyrequire three, four, or five colors?

Help Joe answer these questions.

6. QuestionSuppose you have a map in a rectangle that can be colored properly in two colors.What else can you say about this map?

7. QuestionJoe showed Boss the method of coloring a map on a rectangle gotten by drawinglines from one side to another, and Boss thought there must be an easier method.What do you think? Do you have a more efficient method of coloring such a mapin two colors? Could you explain (to Boss) why your method works? What Joe didhas two parts: (1) a proof that, given such a map, there exists a way to color itusing two colors, and (2) a method for coloring such maps that always works. Youcould assume part 1 and argue from there.

8. InvestigationDraw a bunch of circles on a piece of paper. These create regions, including theregion outside of all the circles. How many colors will you need?


Next, draw a bunch of distorted circles. Make sure they intersect like circlesintersect.

Now color the regions. How many colors will you need?


Carry out this investigation.

9. InvestigationYou get a distorted circle if you take a pencil, and starting at a point on a piece ofpaper, you draw a curve without letting the pencil leave the paper and continueuntil you end up at the point where you started. Don't allow the curve to intersectitself. (Such a curve is called a simple closed curve.)

Now allow such a curve to intersect itself. (Such a curve is called a closed curve.)Make sure the curve intersects itself only in isolated points. Replace the simpleclosed curves of Investigation 6 with closed curves. What would happen? Howmany colors will you need? Carry out this investigation.

10. QuestionSuppose you have a map in a rectangle that can be colored properly in threecolors. What else can you say about this map?

11. PuzzleJoe drew the following maps for Boss. Each one is a distortion of some well-known map. What is the well-known map in each case?


(The map on the left comes from the book by Farmer and Stanford, p. 19.)

12. QuestionA problem analogous to coloring maps on the plane or sphere would be to color the''countries'' of a three-dimensional world. (Joe figures that working on this problemmight shed some light on the ''real problem.'') What exactly would this analogousproblem be? What would a three-dimensional map look like? What would the coun-tries be? What would the rules for coloring such maps be? Do the following three-dimensional configurations satisfy your definition of a three-dimensional map?How would they be colored? How many colors would they need? Is there anythingyou can say in general about coloring three-dimensional maps?


13. Gathering EvidenceHere are more maps to color properly and minimally.


14. PuzzleTake a checkerboard and a bunch of dominos, each the size of two squares of thecheckerboard. The checkerboard has 64 little squares.

● Can you cover the checkerboard with 32 dominos so that all squares of thecheckerboard are covered and there are no overlaps?

● Remove a little square from one corner of the board and another from theopposite corner. Can you cover the resulting board with 31 dominos?

● From a complete board, remove a square from each of a pair of adjacentcorners. Can you cover the resulting board with 31 dominos?

15. SummarizingJoe knows that Boss needs to see things written down so Joe usually writes Boss areport of his activities. Help Joe out by writing a summary of what he accom-plished this afternoon. Include goals, statements of problems, terminology,results, and arguments. Make sure it's well organized and convincing because Joelikes map coloring and wants to continue working on it.

Notes

''Suppose there's a brown calf and a big brown dog, and an artist is makinga picture of them ....He has got to paint them so you can tell them apart theminute you look at them, hain't he? Of course. Well, then, do you want himto go and paint both of them brown? Certainly you don't. He paints one ofthem blue, and then you can't make no mistake. It's just the same withmaps. That's why they make every state a different color ....''

–from Tom Sawyer Abroad, Mark Twain

The problem of coloring maps goes back to October 23, 1852, when FrancisGuthrie (then a graduate student at the University of London) posed it to histeacher Augustus de Morgan (of de Morgan's Law fame), who in turn wrote to theIrish mathematician William Rowan Hamilton:

Notes 23

“A student of mine asked me today to give him a reason for a fact which I didnot know was a fact, and do not yet. He says that if a figure be anyhow divided,and the compartments differently colored, so that figures with any portion ofcommon boundary line are differently colored—four colors may be wanted, butno more ...What do you say? And has it, if true, been noticed? My pupil says heguessed it in coloring a map of [the counties] of England. The more I think of it,the more evident it seems.”

We will see more of this problem in this book.Here is some language that will help in discussing maps and that you will see

used in sources outside this book. The ingredients of a map—on sphere, a rectan-gle, or an island—are countries, borders, and vertices. A country is the interior of apolygon or distorted polygon. The countries of a map do not overlap. A border ofthe map is an edge of one or more of the country polygons. Two countries maymeet along one of these borders. A vertex of the map is where two or more bordersmeet. The map is the union of all these elements.

A major idea of this chapter is the equivalency of two maps, one a stretched-shrunk version of the other. This is an instance of ''rubber sheet'' geometry. Printyour map on a thin sheet of rubber, and then distort the rubber sheet to get otherequivalent maps. The distortion preserves the countries, borders, vertices, andtheir relationships to each other. However, the distortion does not preserve exactdistances or angles. The investigation of geometric properties preserved underthis notion of geometric equivalence was first proposed by Leibnitz (see Kline,p. 1163f). He called the study of such properties, geometria situs. The modernword is topology.

In this chapter Joe proved the following theorem.

Theorem. On a rectangle, form a map by drawing lines from one side of the rectangleto another. Then this map can be colored properly with two colors.

You may have noticed that Joe's argument could be formalized into a proof bymathematical induction on the number of lines forming the map.

In a map, the order of a vertex is the number of borders that meet there.In papers of 1879 and 1889, Kempe and Tait (respectively) remarked that any map


with vertices that all have even order can be colored with two colors. Neitherprovided a proof (see Biggs et al. and Stein).

References

Biggs, N.L., Lloyd, E.K., and Wilson, R.J. Graph theory: 1736–1936. Oxford: Oxford University Press,1976.

Farmer, D.W., and Stanford, T.B. Knots and surfaces: a guide to discovering mathematics. Providence:American Mathematical Society, 1996.

Kline, M. Mathematical thought from ancient to modern times. New York: Oxford University Press,1990.

Stein, S.K. Mathematics: the man-made universe. San Francisco: W. H. Freeman, 1969.

References 25

Acme Adds Tours

27

Scene 1

A few days later at Acme Maps. Millie, the gum-chewing, middle-age company secre-tary, receptionist, and bookkeeper has returned from vacation. The phone rings.

MILLIE: Hi ya! Acme Maps. We can put you on the spot!

BOSS: Millie!

MILLIE: ‘Scuse me. I mean we can spot you on the map.

BOSS: Millieee!!

MILLIE: What can I do for ya?…Yeah? …Map…Tour…Where? …Königsberg? Never heard of…Bridges?…OK. I’ll see what we’vegot. Call you back.

BOSS: What was that about?

C H A P T E R

2

MILLIE: It’s a possible job, Boss, for Walking Tours, Ltd. They sell pack-age tours to interesting places. Fly you there, put you in a hotel,provide you a kit for exploring the city. You know.

BOSS: So what did they want?

MILLIE: They want us to help set up a tour of Königsberg.

BOSS: KÖNIGSBERG?

MILLIE: Yeah. Never heard of it neither. He said it’s somewhere inRussia, near the Polish border. Where Emmanuel Kant lived.Known for its beautiful bridges.

BOSS: Kant? Bridges?

MILLIE: First, he’d like to know if we’ve got a map of downtownKönigsberg with the bridges.

BOSS: Doubt it.

MILLIE: And, second, could we design a walking tour of the city. He saidhe’d give us a good deal.

BOSS: Hmm. Walking tours would be a new line for us. Hey, Joe,look in the “city” file for a map of Königsberg! It’s a city in…uh…

MILLIE: Russia.

BOSS: (Turns to Joe.) Russia, Joe. (Turns to Millie.) Did he say howmuch he’d offer us?

MILLIE: No, but he said something else. He wants the walking tour toinclude all the bridges.

BOSS: Bridges?

JOE: Boss! I found something but I’m not sure it’s what you want.

BOSS: Whaddaya mean?

JOE: Map of Königsberg dated 1650. It’s the only one we’ve got.Here. (Hands Boss the map.)

BOSS: From 1650? Does it have the bridges on it? (Looks at map.) Yes,it’ll do. City can’t have changed that much. (Turns from map to look at Joe and Millie.) Joe, Millie! Sketch a tour of the city withthe bridges!

JOE, MILLIE: Right, Boss!

JOE: (Turns to Millie.) Well, this should be easy. Where should westart the tour?

28 Chapter 2 Acme Adds Tours

MILLIE: That big church looks like a good place to start. It’s right smack inthe middle.

JOE: OK. First walk to that bridge there, and then cross it. Traverse the island, and go through the big gate and over the bridge. Head tothe hospital and then back. Another bridge along the quay. Bridge.Whew! Lotta bridges. Island again. Walls. Bridge again….No, no good.We went over that bridge already. OK…no…cross …yes…uh…no!

MILLIE: What’s the mumbling about? You’ve lost me already. And what’s allthat scribbling?

Scene 1 29

JOE: Look, aren’t we trying to set up a tour?

MILLIE: So?

JOE: I can’t do it without crossing some bridges more than once.

MILLIE: Well, try again. We want it to be a nice tour.

JOE: I’M TRYING!

MILLIE: OK, Hon. Just try it a little slower, please. Leave out the towers andchurches and things for the moment. Just do the bridges. That’s lessconfusing, calmer too.

JOE: All right. Let’s start here: bridge, bridge. So then you have to crossthis bridge twice, see?

I must be going out of my mind. A simple city tour. Boy, it’s justimpossible!

MILLIE: Impossible. Impossible? Let me see the map. Hmm. (She blows abubble in her gum. It pops.) That’s it, Joe! Can’t do it.

JOE: Huh? Whaddaya mean? You mean you can’t do it either? And whatmakes you so excited?

MILLIE: Nobody can take a tour of Königsberg going over all the bridgesunless they cross some bridge twice. There ain’t no nice tour!

JOE: But that’s impossible!

MILLIE: Just what I been a tellin’ ya! Impossible to have a tour crossing eachbridge exactly once. Here, I’ll show you. Draw a plain picture of theriver, the island, and bridges. Label the land masses A, B, C, and D.Like this.


Scene 1 31

JOE: What are the A, B, C, and D for?

MILLIE: Makes things easy. You’ll see. Anyway, this is a lot less complicatedthan the map of the city. And, we can throw out even more stuff.(Millie is intent now.) The shape of each hunk of land isn’t impor-tant so make each hunk a dot. Then, instead of a bridge connectingtwo hunks, draw a line to join the dots.

JOE: What’s this mumbo jumbo? Sounds like a meeting of mathematicians.

MILLIE: Luv, we’re doing topology, aren’t we?

JOE: Huh?

MILLIE: Here’s what you get if you join the dots, a connecting line for eachbridge.

Do you agree that this diagram is all you need for talking about tours?Do you agree that starting with a pencil at a dot on the diagram, trac-ing over each line exactly once and, without ever lifting your pencilfrom the paper, returning to the dot you started from is exactly thesame as drawing–on the map of Königsberg–a tour crossing eachbridge exactly once and returning to the starting point?

JOE: Well, not exactly, but I see what–

MILLIE: It means that the city and rivers disappear, leaving just the dots andconnecting lines. That’s all we need for our problem.

JOE: Holy smoke! You’re right! This is neat!

MILLIE: Slicker’n a soapy rhombicosadodecahedron in a bubble bath.

JOE: What? Wait’ll Boss sees this. Only the dots and bridges are left.

MILLIE: Ssshhh! We’re not ready for him yet. OK, look at what we’ve got.Dot A is odd.

JOE: Odd?

MILLIE: Yep. There are exactly three lines attached to dot A, and three is anodd number.

JOE: So?

MILLIE: That’s the key. Suppose you’ve designed a tour that crosses eachbridge exactly once and returns to the starting point.

JOE: But you said you couldn’t!

MILLIE: I didn’t say you could. I said suppose you could. Just pretend.

JOE: OK.

MILLIE: Take a typical dot with some bridges attached.

Then put an arrow on each bridge pointing in the direction of thetrip as the bridge is crossed.

MILLIE: Can you conclude anything?

JOE: A lot of arrows pointing in different directions?

MILLIE: No! I mean, can you say anything about the number of arrowspointing toward the dot or the number of arrows pointing awayfrom the dot?

JOE: Hmm. The numbers have to be the same!

MILLIE: Tell me why.

JOE: Take an arrow pointing toward the dot. That’s a bridge you arriveon. Then there’s got to be a bridge you leave on. That means anarrow pointing away from the dot. So a “toward’’ arrow has to bepaired with an “away from’’ arrow.

MILLIE: What about the dot where you start and the bridge you first leaveon, the first “away from’’ arrow? What “toward’’ arrow does that pairup with?


Scene 2 33

JOE: Well, that… huh…I thought this was your argument. (Millie shrugs.)OK. At the starting dot, the bridge you first leave on is paired with thebridge on which you last arrive at the starting dot for the last time.Yes, the first “away from” arrow is paired with the last “toward” arrow!

MILLIE: Now you’re cookin’ with hot gas! What is the total number ofbridges connected to each dot?

JOE: Huh? Well, that depends….Oh, I see. Supposing we can take a nicetour, at any dot there are a certain number n of “toward” arrowsand the same number n of “away from” arrows. The total numberof arrows is n + n = 2n. It’s always an even number.

MILLIE: That’s odd.

JOE: No, even.

MILLIE: So back to Königsberg. Look at dot A. The number of arrowsattached to dot A is not even. Not at all even.

JOE: Wow! (Boss walks in.)

BOSS: What’s got you guys so hopped up? You finished designing that tourof…uh…

JOE: Königsberg.

MILLIE: Well, Boss, we got news for ya.Music rises and lights dim as Joe and Millie animatedly explain to Boss what theyhave discovered. Boss does not appear happy.

Scene 2

Later that day. Millie and Joe are writing up what they discovered about tours. Joewrites.

The Essential Ingredients of a Tour Goal:● Reduce the map of city, state, or park (whatever) to its essential ingredients

Assumptions:● Exact sizes and shapes of land masses and bridges are not important.● Connections between land masses are important.

Method:● Replace land masses by dots.● Join two dots by a line exactly when the two land masses are connected by

a bridge.● Obtain dot-line diagram on which to design tour.

Millie writes.

1. Your Turn.Joe and Millie thought they should try out their tour ideas on some famousplaces. First they sketched the following maps of New York City and Paris show-ing the bridges. (For New York City, some of the bridges are really tunnels.) Helpthem with the next steps: turn each map into a dot-line diagram, and design a nicetour for each, if possible.

A Nice Tour on a Dot-Line DiagramIngredients of a Nice Tour:With a pencil, trace a path that● Has the pencil touching the paper at all times● Covers each line of the diagram exactly once● Returns to the dot where the path started

When is a Nice Tour Possible?If a nice tour of a dot-line diagram is possible, then the number of lines attachedto every dot must be even. In other words, if the diagram has a dot such thatthe number of lines connected to it is odd, then no nice tour is possible.


Scene 3

Later that week. Joe and Millie are talking.

JOE: Golly. The first time we’re asked to design a tour, we mess up. Nowwe’re getting all these orders.

MILLIE: We didn’t really mess up. We showed that a nice tour could never bedesigned in that case. Maybe customers like our honesty and ourfine minds.

JOE: Huh? (Boss walks in.)

BOSS: That’s cool beans! We’re gettin’ all this work. After you guys failedto come up with a tour of the bridges of…What’s that place again?

JOE: Königsberg.

BOSS: Whatever.

JOE: You know, one of these days we’re going to come up with a tour wecan’t design.

BOSS: I been tellin’ ‘em to send us only situations with even numbers ofbridges. Everybody knows that you mess up if there is a dot with anodd number of bridges.

Scene 3 35

JOE: So far we’ve been lucky. What if one of those turns out to be impos-sible? Now that would really be embarrassing.

BOSS: Waddaya mean “turn out to be impossible”? Never say can’t! Nothere. Acme is a can-do eee-stab-lish-mint. We been able to do allright so far, haven’t we? Besides, you and Millie said if you couldtake a nice trip (start from one dot, traverse each connecting lineexactly once, and return to starting dot), then there must be an evennumber of connecting lines attached to each dot.

JOE: Yes, yes. But I think you’re concluding that if every dot has an evennumber of connecting lines attached, then you can take a nice trip.

BOSS: Huh? I sure daggum am!

JOE: Remember, Boss, we’ve got only a conjecture, not a theorem.

BOSS: Look, folks. We’re now in the tour business. Figure it out!Music rises. Lights dim. Boss stomps out, leaving Millie and Joe with their jaws open.

2. Your Turn.Arizona Air, a new company, plans to offer the flights between Arizona towns andcities shown on the map below. As part of setting up a schedule, Arizona Air hashired ace pilot Piper Cub to fly all the routes and establish reasonable flying times.What Piper wants to do is get in his plane at Tucson in the early morning, fly allthe routes in one day, and get back to Tucson at night. Clearly he doesn’t want todouble up on any route; he wants to fly each route exactly once. Piper has engagedAcme, the local expert on taking efficient (i.e., nice) trips, to figure out such aflight plan. Can Acme really do the job? Help Joe and Millie out.


Scene 4

Later that day. Joe and Millie are working. Boss is nowhere to be seen.

MILLIE: You’re right. We need some new tour terminology. Make things alittle clearer. Ya know, these dot-and-line diagrams look a lot likemaps. Why don’t we use that terminology? (Millie shows Joe whatshe has written.)

MILLIE: Here are examples of networks with labels. (She draws.)

JOE: Neat! I can use this language to state what we know about nicetours. And the conjecture, too. (He writes.)

Tour Terminology● Dot → vertex● Line → edge● Dot-line diagram → network● The order of a vertex is the number of edges attached to it● A vertex is even if its order is even● A vertex is odd if its order is odd● Two edges are connected if they share a common vertex● A path in a network is a list of connected edges, with the first connected to

the second, the second to the third, and so on ● A network is connected if, given any pair of vertices in the network, there is

a path from one to the other (i.e., starting with an edge connected to thefirst vertex and ending with an edge connected to the second)

● A nice tour on a network is a path beginning and ending at the same vertexand containing each edge exactly once

Scene 4 37

3. Your Turn.To enhance its services, City Parks and Recreation would like to offer each visitorto Central Park a guide showing the visitor how to traverse its paths by beginningat the park entrance, walking each path exactly once, and returning to the parkentrance. City Parks has engaged Acme to do this. Central Park’s paths are shownbelow. Joe realizes he can think of this as a network: the intersections are thevertices, and the paths joining them are the edges. He also realizes that thenetwork is connected, that all the vertices are even, and that what City Parks wantsis for Acme to design a nice tour. Since all the vertices are even and the network isfairly complicated, finding a nice tour would be a test of Joe’s conjecture. Help Joeout. See if you can find a nice tour.

MILLIE: Boss thinks that just because we’ve proved the theorem, the conjec-ture must also be true. I suppose he thinks that because everysquare is a rectangle, then every rectangle must also be a square?(Phone rings. Millie answers it. Boss walks in. Joe has becomeengrossed in a problem.)

MILLIE: (Hangs up and turns to Boss) Boss, I just talked with a lady on thephone. She said something about inspecting doors. She had faxedus these floor plans.

Nice Tours on a NetworkTheorem: If there is a nice tour on a network, then the network must be

connected and every vertex of the network must be even.Conjecture: If all the vertices of a connected network are even, then, given any

vertex, there is a nice tour beginning and ending at that vertex.


BOSS: We don’t do doors! We do maps!

MILLIE: Well, I told her that. But she heard about our work with tours andthought we could help her.

BOSS: Huh? What do doors have to do with tours?

MILLIE: The woman inspects doors. Checks hinges, locks, whether or notthe doors stick or slam too fast—that sort of thing. Sometimes when she does a door inspection, she winds up passing through thesame door more than once. She wants to avoid that so she wants usto design a tour of the building that goes through every doorexactly once.

BOSS: Hmm. Buy gum, sounds like that job we had with the city of…

MILLIE: Königsberg.

BOSS: Yeah. I thought that was impossible! (Stomps off. Joe and Millie turnto the floor plans.)

4. Your Turn.Help Millie and Joe with the problem of designing a tour for the door inspectorfor each of the buildings whose floor plans are shown above.

MILLIE: This Königsberg thing has really gotten outta hand. Door inspec-tors, your conjecture, lotta work. Let’s get to it!

Lights fade. Music comes up.

Scene 4 39


1. InvestigationJoe’s conjecture is that if every vertex in a connected network is even, then there is anice trip starting at a vertex and returning to the same vertex. Part of the investiga-tion has already taken place: Joe has taken nice trips on a lot of networks; he has a lotof examples for which he knows it can be done. Boss is convinced that it can alwaysbe done, but Joe wants to be sure. What Joe wants is a sure-fire, guaranteed methodfor taking a nice tour on any network of the prescribed kind. This would prove hisconjecture. Right now, Acme’s method seems pretty hit or miss. Of course, it’s possi-ble that something really nasty could happen: the presence of a counterexample tothe conjecture—a connected network having every vertex even for which there is nonice tour. Of course, then there would be no sure-fire, guaranteed method. So whatis it? Can you show that Joe’s conjecture is true? Can you find a sure-fire method? Or can you find a counterexample? Acme needs your help in this matter!

2. InvestigationConsidering that the door inspector may return to Acme for more help with tourdesign (and perhaps other door inspectors may come by), Millie would like somegeneral rules to help her. Give her all the advice you can! (If, as Boss suggests, it’sreally like the Königsberg bridge problem, it might be helpful if you could trans-form the problem into a nice tour on a network. If you do that, you have todescribe the network pretty explicitly.) Write a report to Millie on your findings.

3. InvestigationJoe, Millie, and Boss are convinced that no nice tour of the bridges of Königsbergexists. So they’re thinking of what they might come up with as an alternative. Milliesuggests they design a tour with a drop-off location and a pick-up location. A vanwould drop off the tourists and pick them up later after they have crossed each bridgeexactly once. She calls this an OK tour. Is an OK tour possible in Königsberg and whatwould the tour be—the drop-off and pick-up spots and the path between the two? Of course, she is not just interested in Königsberg, she is also interested in other toursituations. Help her to investigate the possibility of an OK tour for any connectednetwork. Of course, you will need to write a report to Boss about your investigation.

4. PuzzleAcme is having an open house for all its customers. As an ice-breaker, Boss has theguests shake hands with each other and asks each individual to keep track of the


number of times he or she shakes hands. After people shake hands a bit, Boss askshow many people shook hands an odd number of times. (Boy, Boss is weird!)What kind of response do you think he got? Of course, you also want to explainwhy he got what he got.

5. InvestigationWord has gotten to the postal service in Tucson about Acme’s expertise withdesigning tours. The service wants Acme to design a tour for the mail deliveryperson for the downtown area included in the map below. The delivery persondelivers mail to buildings on both sides of all the streets in the map. The path ofthe tour should start at the corner of Franklin and Main, traverse each blockexactly twice (once for each side of the street), and return to the starting point.Help Acme design the tour (if it’s possible!). Generalize what you get to tours onany connected network. (Describe the new kind of tour and the possibilities forfinding one.)

6. InvestigationWith all the business Acme has been getting with tours, Joe is trying to anticipatethe next problem. Acme has been asked to look at networks where the edges arestreets and the vertices are intersections. Joe thinks, “What if those streets were allone-way? What if a road inspector wanted to take a tour of all the streets, travel-ing each street exactly once (obeying the traffic laws!) and returning to his start-ing point? What conditions on the network have to be satisfied so I will be able todesign such a tour for her?’’ Help Joe carry out this investigation. Boss will expectjustifications for Joe’s answers in the report.


7. QuestionMillie has been working on door inspector tours for some time now. She has aparticular fondness for houses whose rooms all have an even number of doors.She has noticed something curious about the number of outside entrances such ahouse has. What could this be? (Of course, Millie would like an explanation forthis phenomenon.)

8. InvestigationSouthwestern Amalgamated Circuits has the following problem. It wants todesign a circuit board with the following properties. Six isolated terminals A, B, C,G, L, and W will be printed on the board. Electrical connections will be madefrom each of A, B, and C to each of G, L, and W. Of course, connections (printed“wires” on the board) cannot cross. Thinking that the “world expert in networkdesign” should be able to help them with this, Southwestern Amalgamated hasengaged Acme to carry out the basic research for this problem. As a member ofAcme’s staff, you have been asked to investigate and report your findings to Boss.(To clarify things, the board is a rectangular piece of plastic. Furthermore, to avoidweakening the structure of the board, no holes can be punched in it.)

9. InvestigationSouthwestern Amalgamated has additional requests for help in circuit board design.This time the company has several problems, which are all related. The simplest oneis the following. There are three isolated terminals: A, B, and C. Electrical connectionsare to be made between each pair of terminals. Of course, this has an easy solution:


The other problems are similar to the one with the easy solution: There are tobe a bunch (more than three) of isolated terminals (A, B, etc.); electrical connec-tions are to be made between each pair of terminals. Again, SouthwesternAmalgamated has hired Acme to investigate. Because you are the staff person incharge, your task is to carry out the appropriate research and report your resultsto Southwestern Amalgamated.

10. InvestigationThe National Association of Museums (NAM) wants to help its member muse-ums design tours for its visitors. One issue that has arisen is the possibility ofdesigning a tour through a museum that begins at the museum entrance, passesthrough each room exactly once, and ends at the museum exit. NAM hascontracted with Acme to resolve this issue. You, an Acme employee, have beengiven the task of working on this problem for rectangular museums with thefollowing floor plans. (Each plan shows the entrance, exit, rooms, and doorsbetween rooms.)

You are to investigate the possibility of such tours for these museums (andmuseums similar to them) and report your findings to Boss.

11. PuzzleAt the open house mentioned in investigation 3, Millie set up a puzzle. On a tableshe lays out all the pieces of a standard domino set, 28 of them from double zeroto double six—all possible pairs from 0 through 6.


The puzzle has two tasks:

● Arrange the dominos in a straight line so that the dots of adjacent piecesmatch.

● Arrange the dominos in a closed circle so that the dots of adjacent piecesmatch.

A task may be impossible. If it is, you’ve got to convince Millie it can’t be done.

12. PuzzlePinkley Smith, Acme’s legal advisor, was watching the other guests play the puzzleMillie set out in puzzle 11. He began to think of variations: “Suppose I limitmyself to just those dominos with 0, 1, 2, 3, or 4 dots on them. There are 15 suchdominos. Could I solve either of the tasks above for this restricted set of dominos?(Hmm. This suggests a whole bunch of puzzles!) Call the set of 15 a double-4 setand call the original a double-6 set. Then there’s a double-3 set and, hey! a double-nset! Can I solve the tasks for them all, just some, or none? Which, I wonder?’’ HelpPinkley!

13. QuestionThe citizens of Königsberg are a bit dismayed that a nice tour of the city’s bridgesis impossible. They are considering building new bridges to remedy the situationand have asked Acme for advice on how many new bridges they would need andwhere to build them. What would you tell them?

14. SummarizingHelp Joe and Millie write a report to Boss describing what they accomplished inthis chapter. Include problems that came up, clarifications, terminology, results,and connections. Of course, Boss will want the report to be well organized andconvincing.


15. ConnectingIt is a stroke of genius that Millie noticed that the dot-line diagram looked a lot likea map and decided to adopt the terminology they had used for maps. Not thatanything will come of it, but it sets the stage for making connections between thetwo problems that Acme has been considering: coloring maps and designing tours.You are a problem solver, and you look for new ways to think about your problem.Making a connection like this gives you a new way. Here are some questions thatmight get you started in exploiting the connection. Does working on map coloringgive you any insight on tour design, and vice versa? In addition to the overall visualsimilarity between networks and maps, are there other similar features that havearisen in the two problem areas? If Joe were able to prove his conjecture, would thathelp him solve any of the map coloring problems? Conversely, is there a map color-ing problem that, were you able to solve it, would it help you with tour design? Ifthe answer to any of these questions, or one like them, were yes, you’d be excited.You’d receive new energy and new insight for solving these problems. So take timeto reflect on the connection and report back to us.

Notes

In modern terminology, a collection of points in space (called vertices) and lines(called edges) joining selected pairs of those points is called a graph, and the studyof graphs is called graph theory. A graph that can be drawn on the plane so that thejoining lines intersect only at vertices is called a planar graph. In this chapter theterm network is interchangeable with planar graph.

In a paper of 1736 (see Biggs et al., p. 3f), the Swiss mathematician Leonhard Euler(1707–1783) considered the problem of devising a tour of the bridges of Königsberg.Much as we did in this chapter, he showed that a tour was impossible. Moreover, hecame up with a general method for other problems of the same type. In the languageof graph theory, here are the theorems he proved:

Theorem 1. If a tour on a connected graph returning to one’s starting position ispossible traversing every edge exactly once (such a tour is now called an Euleriancircuit), then every vertex is even.

Theorem 2. If a tour on a connected graph is possible without returning to one’sstarting position and traversing every edge exactly once (such a tour is now calledan Eulerian path), then all vertices but two are even. Moreover, in taking the tour,one must begin on one odd vertex and end on the other.

Euler saw these results as part of a new kind of geometry. Here is how he beginsthe paper:

“The branch of geometry that deals with magnitudes has been zealously studiedthroughout the past, but there is another branch that has been almost unknown up

Notes 45

to now; Liebnitz spoke of it first, calling it the “geometry of position’’ (geometricasitus). This branch of geometry deals with relations dependent on position alone,and investigates the properties of position; it does not take magnitudes into consid-eration nor does it involve calculation with quantities. But as yet no satisfactory defi-nition has been given of the problems that belong to this geometry of position or ofthe method to be used in solving them. Recently there was announced a problemthat, while it certainly seemed to belong to geometry was nevertheless so designedthat it did not call for the determination of a magnitude, nor could it be solved byquantitative calculation; consequently I did not hesitate to assign it to the geometry of position, especially since the solution required only the considera-tion of position, calculation being of no use.”

The problem Euler refers to in this quote is the bridges of Königsberg problem.Later geometry of position becomes more commonly known as topology (from theGreek topos, place, + logos, word).

In his paper, Euler replaced the map of the city by a simple picture similar to theone Millie and Joe created. Below is a picture of another rivers-bridges-land massessituation that Euler analyzed in addition to the Königsberg bridges problem.

Euler did not take the additional step of replacing the picture by the network of vertices and edges. The latter seems to have been done first by Listing. In apaper of 1847 (see Biggs et al., p. 16f), Listing includes the following diagram,which, he says, can “be drawn in a single stroke, since it has only two points ofodd type….’’

The converses to Euler’s theorems were first proved in 1873 by Hierholzer (see Biggs et al., p. 11f).


Several of the problems from this chapter have been around for a while. In 1849Terquem proposed the domino problem (Puzzle 11) and related it to the work ofEuler (see Biggs et al., p. 16). Investigation 8 first appeared in a book of puzzles byDudeney in 1917 (see References). The problem is usually referred to as TheUtilities Problem: “The puzzle is to lay on water, gas, and electricity, from W, G,and E, to each of the three houses, A. B, and C, without any pipe crossing another.Take your pencil and draw lines showing how this should be done.’’

The graph associated with n-terminals in Investigation 9 is called the completegraph on n points and is denoted as Kn.

The ideas in this chapter can be used to solve many old problems, such as howto escape from a maze or labyrinth (see Biggs et al., p. 18f), and many new ones,such as municipal vehicle routing (see Tannenbaum and Arnold, p. 183).

References

Biggs, N.L., Lloyd, E.K., and Wilson, R.J. Graph theory 1736–1936. Oxford: Oxford University Press,1976.

Dudeney, H.E. Amusements in mathematics. London, Nelson: 1917.Farmer, D.W., and Stanford T.B. Knots and surfaces: a guide to discovering mathematics. Providence:

American Mathematical Society, 1991. See chapter 1.Mr. Simplex saves the Aspidistra. Film. Washington, DC: Mathematical Association of America, 1965.

(Investigation 10 is considered in this movie short.)Newman, J. The world of mathematics. New York: Simon and Schuster, 1956.Steinhaus, H. Mathematical snapshots. New York: Oxford University Press, 1950.Tannenbaum, P., and Arnold, R. Excursions in modern mathematics. Upper Saddle River, NJ: Prentice

Hall, 1998. (See Chapter 5.)

References 47

Acme Collects Datafrom Maps

49

Scene 1

Early that summer. Acme has hired Brandon, its first summer intern. Brandon isa math major at the state university. Boss and Joe are discussing what Brandonshould do.

JOE: Boss, you hired this kid from the university as a summer intern, butour customers are taking their vacations and things are slow. What’rewe going to have him do?

BOSS: There must be somethin’. Have him collect data.

JOE: Data?

BOSS: Dunno. We got all those maps. Isn’t there some kinda data we couldgather from ‘em? Think of somethin’.

C H A P T E R

3

JOE: OK, Boss. (Joe mutters to himself.) What’ll we do with a lot ofdata? Sounds like a lotta busywork to me.

BOSS: What? Yer mumblin’ agin, Joe.

JOE: Huh? I’ll get on it, Boss.

(Brandon enters. Joe turns to greet him.)

JOE: Hello, Brandon, I’m Joe. Welcome to Acme! I’ve got somethingfor you to do. See that file cabinet? (Joe points.) That’s where wekeep our maps. I’d like you to go through the maps and get mesome data on them.

BRANDON : You mean, like count how many maps there are?(eager)

JOE: Noooo. Data on the maps themselves. Let’s pull out a map andhave a look. (Goes to file cabinet, opens it, and pulls out map.)Every map has three ingredients at least: countries, bordersbetween countries—we call those edges, places where three ormore borders meet—we call those vertices.

BRANDON: Why don’t I count countries, edges, and vertices for each map?

JOE: Hey, good idea! I’ll leave ya to it! That desk over there will beyours.

Brandon sits down and gets to work. He takes out a piece of paper andconstructs the following table.

1. Your TurnHere are some maps Brandon found in the file. Help him out. For each one,count the number of vertices (V), the number of edges (E), and the number ofcountries (C). Enter the data in the table.

Data on MapsItem V E C

50 Chapter 3 Acme Collects Data from Maps

Scene 2

A few hours later. Brandon is working at his desk. Joe walks up to him.

JOE: How’s data gathering going?

BRANDON: Data gathering’s fine. After I did a lot of the maps in the cabi-net, I decided to draw some diagrams to see if I could drawsome conclusions from the data.

JOE: Conclusions? Waddya mean?

BRANDON: Well, I was looking at the table of data and wondering if someof those numbers were related.

JOE: Related?

BRANDON: Like suppose you knew that a map had 20 countries and 40vertices. Could you tell how many edges it had?

JOE: You mean like some sort of formula?

BRANDON: Something like that. Anyway, I was looking for more maps, likesome simple ones or maybe even made-up ones.

JOE: Wait a minute, Acme doesn’t deal in made-up maps. This is theREAL world!

BRANDON: Hold on. Let me explain. If there’s some sort of relationship—maybe a formula—between those numbers for any map, then itshould work for made-up maps, too. Look, here’s what I mean.

Scene 2 51

Draw an imaginary continent or an island. Acme’s maps arereally maps of some kind of continent, right?

Then add imaginary countries:

There, doesn’t that look like a real map-wanna-be? It’s even gotthe V, E, and C data like a real map! (Brandon writes the data forLa-La Land in his table.)

JOE: Jeepers! We could sell tours to that place. Come one, come all!Take a tour to a world with 20 vertices, 29 borders, and 10 coun-tries! Let’s call it something exotic!

BRANDON: How about “20-29-10’’?


La-La Land 20 29 10


JOE: That would really make me want to take a vacation there.

BRANDON: OK. Now you understand what I mean. With those diagrams Iwas trying to come up with a whole bunch of maps at once.Something that would give us a lot of data—in fact, an infiniteamount—but without much work.

JOE: Infinite? You some kinda math major or somethin’?

BRANDON: Look, these diagrams are really maps.

Imagine that the one on the left is an n × m rectangle. Then V = (n + 1)(m + 1), C = nm, and E = n(m + 1) + m(n + 1). Forevery pair n and m of whole numbers, there’s a map and datathat goes with it. We can add all that data to my table.

For the second diagram, you can vary the number of spokes andget another infinite family; for the third, change the sides of thelarge equilateral triangle and get yet another infinite family. So, twomore infinite sets of data.

JOE: Infinite families? Boy, you must really be a math major.

2. Your TurnHelp Joe and Brandon. Figure out what V, C, and E are for the two additional infi-nite families of maps. Put the corresponding data you get into your Data on Mapstable. Look for patterns; see if you can find a relationship among V, E, and C forall maps considered so far—by you, Brandon, and Joe.

Scene 3

An hour later. Brandon is working alone. Millie walks in.


La-La Land 20 29 10n × m rectangle (n + 1)(m + 1) n(m + 1) + m(n + 1) nm

Scene 3 53

MILLIE: Hi, ya! You must be Brandon! We haven’t met. I’m Millie. I heardyou found out some weird stuff about maps.

BRANDON: I’m pleased to meet you, Millie. Yes, I think I’ve found somethingreally interesting. If V is the number of vertices, E the number ofedges (borders), and C the number of countries on a continent orisland map, then it looks like you’ll always have V − E + C = 1. I’vebeen trying to come up with an explanation for it.

MILLIE: Explanation? All ya have to do is check all the maps in the cabi-net. You’ve done that, haven’t ya?

BRANDON: I’ve checked a lot of them. It’s true that the formula works for allthe maps I looked at and even some imaginary maps I’ve drawn.What I’d like to do is explain why that formula is true for allmaps, including any maps that haven’t been drawn or eventhought of yet. Of course, I might not be able to do it. In fact,there might be some yet-unthought-of map for which theformula doesn’t work. But, if I can show it’s true for all maps,then the formula will express some basic and immutable princi-ple of maps.

MILLIE: Well, I declare! Sounds deep, like a Newton’s Law of Motion.Maybe I kin hep ya with the explanation?

BRANDON: Sure. Let me show you an approach I was thinking about.Suppose I take a typical map on an island. We’ll assume that allour maps are maps on an island, OK? The map is made up ofcountries, borders (edges), and vertices. I’ve been looking atsome really simple maps, and I think we have to be carefulabout what we’re going to allow in order to draw a usefulconclusion.

MILLIE: Allow? Useful?

BRANDON: Here’s a really simple map on an island. One country, one edge,zero vertices. So V − E + C = 0. Doesn’t fit the pattern.

MILLIE: Isn’t that what you were looking for, a map where the formuladoesn’t work?


BRANDON: Wait. Look what happens if you add one or two vertices:

MILLIE: But those vertices weren’t there before.

BRANDON: Yes. But it’s also OK to have them, isn’t it? But in one case we’vegot V − E + C = 0; in two other cases we’ve got V − E + C = 1.It would be neat to have our formula be true! We’ve just got topin down when it’s true. Maybe it’s true for maps where everycountry has at least one vertex on its boundary?

MILLIE: What about a map where you’ve got a country that doesn’t?

BRANDON: Well, our theorem wouldn’t be true, so we won’t allow that. But itwouldn’t hurt if we add a vertex to every country that doesn’t haveone. The map would be pretty much the same. Then we could saythat for every map on an island where every country has at leastone vertex on its boundary, it’s true that V − E + C = 1.

MILLIE: Sounds a little awkward.

BRANDON: OK. We’ll keep working on it. I found another simple map withanother problem:

Every country has a vertex on its boundary, but in this caseV − E + C = 2. Most of our countries are like this.

All of these are distorted versions of a polygon or a disc.

MILLIE: Distorted?

Scene 3 55

BRANDON: Yeah. Imagine a disc made of rubber that you can stretch tomake into any one of these countries.

.

MILLIE: Then you could just prick a hole with a pin in one of those andstretch it into the ring.

BRANDON: No, no. You can’t cut or make holes.

MILLIE: OK, OK! Picky, picky!

BRANDON: We’ve got to have rules….We’ll call a map OK if every countryhas a vertex and is a distortion of a disc. Here’s my idea. You’vegot this complicated OK map on an island. Maybe it’s got 5283countries so you sure don’t want to count V, E, and C. Thenyou’ve got this one-country map on an island.

We know V − E + C = 1 for the one-country map. Is there a waywe could change the 5283-country map little by little, step bystep, into the one-country map? I don’t mean really do itbecause then we might just as well have counted V, E, and C inthe first place. But we describe what steps we could take to do it.Suppose we could do it so that V − E + C doesn’t change?

MILLIE: V − E + C doesn’t change?

BRANDON: Yeah. Here’s a pic.


We’d want V − E + C = V' − E' + C'. I guess at each step we’dwant to change what we have into another OK map. Do youhave any ideas?

MILLIE: Hmm. Here’s a map.

Then V' = V, E' = E − 1, C' = C − 1. Thus

V' − V +C' = V − (E − 1) + (C − 1) = V − E + C.

They’re the same again!

BRANDON: That’s a great idea! Will that always work?

MILLIE: Waddya mean? It worked for me!

BRANDON: I mean, if you take away any country from the island’s edge, willthings always work out so well?

MILLIE: Why not?

BRANDON: OK. Look at your map. What if we had decided to remove adifferent country from the island’s edge. Any problems?

MILLIE: Uh, oh. I think I see one. Lookit that country:

Scene 3 57

Then V' = V, E' = E − 2, C' = C − 1. So

V' − E' + C' = V − (E − 2) + (C − 1) ≠ V − E + C.

They’re not the same!

BRANDON: But look what happens. It’s not an island; it’s two islands! I thinkwe’d want V − E + C = 2 for two islands.

MILLIE: Wait. I know what. The “bad’’ country in my map has two of itsedges on the island’s shoreline.

We’ll remove a country when just one of its edges lies on theisland’s shore.

BRANDON: Hmm. I think that works pretty well. V − E + C doesn’t change.Let’s go back and look at your original map and the change wemade in it.

What we could do next would be to remove the shaded country.Trouble is, there are those vertices left over.

MILLIE: No problem, Hon. Those are vertices of order two. If you removeone of those and replace the two edges that meet there with a single edge, then both V and E would go down by one, and V − E + C wouldn’t be affected. Here


Then V' = V − 1, E' = E − 1, C' = C. Thus

V' − E' + C' = (V − 1) + (E − 1) + C = V − E + C.

They’re the same!

BRANDON: That sounds good. One thing though, you’d have to make surethat the vertex you remove isn’t the last vertex on the country’sboundary.

MILLIE: I think we’ve got it! Start with the original. Remove a countrythat shares just one edge with the island’s shoreline. Thenremove vertices of order two. So far no change in V − E + C.Then remove another country that borders the “new’’ islandwith just one edge. Keep doing this until you get an island withjust one country. Along the way there are no changes in V − E + C.At the end (with one country), V − E + C = 1. So V − E + C = 1for the original!

BRANDON: You’re right. Dude! (Brandon and Millie do high fives.) Let’s goback over the argument and make sure it’s OK….Hmm, Millie,after you remove vertices of order two, how do you know youcan always find a country that borders the island with just oneedge?

MILLIE: It’s obvious, isn’t it?

BRANDON: I don’t know.

Millie goes to her desk and starts painting her nails. Brandon sits at his deskand writes:

Scene 3 59

3. Your TurnHelp Brandon solve the problem. Millie claims that, given an OK map on anisland, you can (possibly after removing some vertices of order two) always find acountry that shares exactly one edge with the “shoreline’’ of the island. Find anargument that justifies her claim or find a counterexample that disproves it.

Scene 4

Brandon is working at his desk. Joe approaches.

JOE: Way to go, Brandon! I hear you and Millie found a formula formaps. Say, what’s that gizmo you’ve got?

BRANDON: (Looks at Joe.) Thanks. (Turns to object on desk.) This “gizmo’’ isa calendar. My uncle gave it to me. It’s got 12 faces, one for eachmonth.

It’s neat and I look at it a lot. I’d been thinking about maps andthe formula we found and I thought,“The calendar is like a mapon the surface of the earth. The month-faces are the countries.The borders are the edges of the faces, and so on.’’ That makesense to you?

JOE: Boy, I don’t see that. You some kind of math major or somethin’?Thing is, it ain’t flat like the maps in the cabinet.

BRANDON: You’re right. I thought of that. You’d have to cut one of the facesoff and then stretch and flatten out the rest.

JOE: C-c-cut a face off? F-f-flatten out the rest?


Map Data Theorem (almost)A map on an island is OK if every country has at least one vertex and if everycountry is the distortion of a disk.

Theorem. If V, E, and C are the number of vertices, edges, and countries(respectively) of an OK map on an island, then V − E + C = 1.

BRANDON: Yeah. Let me show you what I mean with a cubical box. Sameidea, but simpler than the calendar. Cut off one of the faces.

Now imagine the box-without-lid is made of rubber. Stretch itand flatten it out. You’d get something like this.

JOE: Gosh darn! I see what you mean. There’s the map. Except for theface you cut off, the faces of the cube are now the countries of themap. The edges of the cube are the edges. The sharp corners ofthe cube have become the vertices of the map. What’s your calen-dar look like when you flatten it out?

BRANDON: That’s more complicated, but I’ve got it right here.

4. Your TurnBrandon has other three dimensional shapes on his desk. To get some insight intoBrandon’s idea, do the same for each one as Brandon did for the cube and thegizmo: remove a face and draw the island map you’d get if you were able to stretchand flatten it out.

Scene 4 61

JOE: What are you going to do with these flattened-out maps?

BRANDON: Well, the formula Millie and I found has consequences for theseshapes and for maps on the surface of a sphere.

JOE: Maps on a sphere?

BRANDON: Let me get to the sphere later. Let’s talk about the shapes first.When you flatten out the shape, its faces become the countries,its edges become the edges, and its vertices become vertices of anisland map. So, if the shape has F faces, E edges, and V vertices,then V − E + F = 1.

JOE: What about the face you cut off?

BRANDON: Yes, V − E + F = 1 is not quite right for the original shape. If theshape has F faces, then the flattened-out part has F − 1 faces. Soit really should be V − E + (F − 1) = 1 for the flattened-out part.But then we would have this formula for the entire shape:

V − E + F = 2.

Not bad, eh? Let’s check it out with the cube.

Brandon writes:


Data for Cube

● V = 8 (4 vertices on top, 4 on the bottom)

● E = 12 (4 on the top, 4 on the bottom, 4 around the sides)

● F = 6 (1 on the top, 1 on the bottom, 4 around the sides)

● V − E + F = 8 − 12 + 6 = 2 (just as predicted!)

Formula for Three-Dimensional ShapesIf a three-dimensional shape has V vertices, E edges, and F faces, then

V − E + F = 2

JOE: Let me do it for your calendar gizmo. Because there are 12months of the year, there are 12 faces. You can also see them: 1on top, then 5 surrounding that in a belt. That makes 6 so far.Then 1 on the bottom and 5 surrounding that in a belt. Thatmakes 6 more.

And that’s all: the top plus the top belt meshes nicely togetherwith the bottom plus bottom belt….Neat. Now to countedges….Hmm, it’s complicated to keep track of everythingbecause you want to make sure you don’t count anything twice.Maybe there’s an easier way. Let’s see. Every face is a pentagon.The number of pentagons is 12. Five is the number of edges of apentagon. Multiply 5 by 12 to give you the total number of edges;except every edge would be counted twice: once for each of thetwo faces that meet along the edge. So E = (12)(5)/2 = 30. Countvertices the same way. Five is the number of vertices of a penta-gon. Multiply 5 by 12 to give you the total number of vertices;except that every vertex would be counted three times: once foreach face that the vertex sits on. So V = (12)(5)/3 = 20. Also V − E + F = 20 − 30 + 12 = 2. Hey, it works again! That’s amazing!

Joe writes:

5. Your TurnCheck out the formula for the shapes in Exercise 4. Describe how you count theelements.

Data for Gizmo

F = 12

E = 30

V = 20

V − E + F = 20 − 30 + 12 = 2

Scene 4 63

6. Your TurnA certain mystery three-dimensional shape has the following properties: all itsfaces are triangles and every vertex is of order five. Of course the formula V − E +F = 2 holds for this shape. Use what you know to find out more about this shape,such as the values of V, E, and F.

BRANDON: Now let me tell you about maps on the sphere. All the three-dimensional shapes we’ve been looking at correspond to maps onthe surface of the sphere. Take the surface of a cube, and imagineit’s made of rubber and that you can inflate it like a balloon. You’llget a sphere. The imprint of the cube’s faces, edges, and verticesforms the countries, edges, and vertices of a map on the sphere.

You can do the same with all the other shapes.

JOE: You know, if you take one of the original three-dimensionalshapes, you can remove a face and flatten it out to make a mapon an island. You can do the same thing with the correspondingspherical shape: remove a “face,’’ flatten, and get a map on


an island. That face is really a country for a map on the sphere.I think you could take any map on a sphere, cut out a country,flatten it out, and make a map on an island. Our argument forthe shapes says that V − E + C = 2 for any map on a sphere!

BRANDON: We could do the whole thing in reverse. Take a map on an island,make this an island on the sphere, and then make the part of thesphere that isn’t the island (the sea?) into a country.

That means if you take a map on an island and think of thesurrounding water as a country, then you’re really thinking of amap on the sphere. Neat stuff!

Brandon writes.

7. Your TurnThe calendar gizmo corresponds to a map on the sphere such that every countryhas five edges and every vertex is of order three. Something similar is true of thecube: the corresponding map on the sphere is such that every country has fouredges and every vertex is of order three. Investigate maps on the sphere of thefollowing kind: every country is a hexagon (the number of edges is six), and everyvertex is of order 3 or more.

Scene 5

The next day. Brandon and Joe are working at their desks. Millie walks up toBrandon’s desk.

Formula for Maps on a SphereIf V is the number of vertices, E the number of edges, and C the number ofcountries of an OK map on a sphere, then

V − E + C = 2.

Scene 5 65

MILLIE: Still thinkin’ about our formula?

BRANDON: Yeah. When I was hired, Boss told me about a problem you guyswere working on. You were supposed to design a network on acircuit board starting with five terminals and connecting each ofthe five terminals once to each of the other four, without lettingthe connections cross.

MILLIE: I remember. We couldn’t solve it.

BRANDON: Well, no wonder. I don’t think it can be solved.

MILLIE: You mean, you can’t solve it either?

BRANDON: No. I mean nobody can solve it, ever. Want me to show you why,using our formula?

MILLIE: Sure. But who ever heard of a problem that couldn’t be solved ifyou had enough smarts.

JOE: (Overhears the conversation.) I don’t suppose anyone has heardof a nice tour of the bridges of Königsberg.

BRANDON: Huh? OK, suppose you had a solution to the problem on thesphere.

MILLIE: I feel this abstract rash coming on—

BRANDON: Just concentrate. You have each of the five terminals connectedto the other four without any of the connections crossing. Thewhole network would form a map on the sphere; the partoutside all the connections would be a country. The terminalswould be the vertices: V = 5. The connections would be theedges: E = 5 × 4/2 = 10 because each of 5 terminals is connectedto each of the 4 other terminals, giving you 20 edges; but eachedge would be counted twice, once for each endpoint. We don’tknow much about the countries yet. Since V − E + C = 2, thenumber of countries, including the part outside all the connec-tions, would be 2 − 5 + 10 = 7. Seven countries. Let’s pick a coun-try and see if we can say anything about it, maybe the number ofedges it can have. The boundary of the country would be formed


by a trip along edges from one terminal back to itself. You cansee that no country could have exactly two edges because wearen’t connecting a terminal to itself. So every country musthave at least three edges. Let Ck denote the number of countrieswith exactly k edges. Then C = C3 + C4 + C5 + … We can use thisto count edges:

3C3 + 4C4 + 5C5 + … = 2E

Since 3C = 3C3 + 3C4 + 3C5 + …, we have that 3C < 2E. Pluggingthe values we have for C and E into this inequality, we get 3 × 7 <2 × 10, or 21 < 20. That can’t be! Assuming the problem could besolved led us into this horrible mess! It must be that the problemcan’t be solved! What do you think of that?

MILLIE: Eeek!

Millie puts her hands over her eyes and runs out of the room. Lights fade.


1. InvestigationBrandon wonders if his technique for resolving the five terminals problem might be helpful in investigating the utilities problem, in which each of threehouses is to be connected to three utilities, without the connections crossing (See Investigation 8 of Chapter 2). Carry out the investigation for Brandon.Write a dialog between you and the Acme employees in which you present yourresults.

2. Gathering DataThe maps shown below suggest two more infinite families of maps. Figure outwhat V, E, and C are for the two families and add the corresponding data to yourData on Maps table. Do they confirm what you know?


3. InvestigationEach of the maps on the sphere corresponding to the cube and Brandon’s calen-dar gizmo has the following properties:

● All countries have the same number of edges.

● All vertices have the same order.

Acme wants to know if there are other maps on the sphere having these two properties. Carry out the investigation for them. Write a report ofthis to Boss. Suggest to him how you might advertise a trip to one of these locations.

4. InvestigationBrandon is also interested in maps on an island with a lake. He wants to know ifthere is anything he can say about V − E + C for such a map. Help him. Of coursehe’ll be interested in generalizing the problem to maps on an island with n lakes.Help him with that, too.

5. Gathering DataCount V, E, and F for the surface of the following solid figure and calculate V − E + F.

(The square hole goes all the way through; the bottom looks just like the top.)What happens?


6. QuestionSuppose you know that any map on an island can be colored with 132 colors orless. What can you say about the number of colors needed to color any map on anisland with a lake?

7. QuestionSuppose you know that any map on an island with a lake can be colored in 5,283 colors or less. What can you say about the number of colors needed to colorany map on an island?

8. QuestionSuppose your world is a cylinder—the outside surface of a tin can with both endsremoved.

Can you say anything about V − E + C for maps on your world?

9. QuestionActually two questions!

1. Suppose you know that a whole number N has the following properties:

● Any map on the sphere can be colored properly with N or fewer colors.

● There is a map on the sphere that requires N colors to be colored properly.

What can you say about the number of colors sufficient to color any map on anisland?

2. Suppose you know that a whole number M has the following properties:

● Any map on an island can be colored properly with M or fewer colors.

● There is a map on the island that requires M colors to be colored properly..

What can you say about the number of colors sufficient to color any map on a sphere?

10. QuestionOne way to classify OK maps on a sphere would be by number of countries. Forwhich whole numbers n is there a map with C = n?


11. InvestigationConsider OK maps on a sphere such that every vertex is of order three or more.A way to classify such maps might be by the triple of numbers (V, E, C) such thatV − E + C =2. (Compare this classification method with those of Investigations 10 and 12.) To test the value of such a classification, you come up with two questions for investigation:

● For any three positive whole numbers L, N, M such that L − N + M = 2,can you always find a map with V vertices, E edges, and C countries such thatV = L, E = N, and C = M?

● Is it possible to find a pair of maps that have the same V, E, C data but are other-wise significantly different (i.e., you can’t transform the network of one into thenetwork of the other by shrinking or stretching).

12. InvestigationAcme wants to maintain a catalog of different types of maps. One type of map isone on a sphere having 12 or fewer countries, each of which is a quadrilateral, i.e.,a 4-edged country. Investigate all maps of this type.

13. InvestigationAnother type of map Acme is interested in for its catalog is one on a sphere havingeight or fewer countries, each of which is a triangle, i.e., a three-edged country.Investigate all maps of this type.

14. SummarizingThis time the job of summarizing Acme’s investigations has been given toBrandon. Help him out by writing a summary to Boss of what was accomplishedthese last few days. Be sure to include goals, statements of problems, new termi-nology introduced, results, and arguments. Brandon would like good letters ofrecommendation to come out of his internship with Acme, so be sure to make thereport well organized and convincing.

Notes

The three-dimensional shapes that appear in this chapter—dodecahedron,cube, and others—are called polyhedra. More about these shapes can befound in many of the references listed below. The formula relating vertices,


edges, and countries of a map is called Euler’s formula and was originally aformula relating the edges, vertices, and faces of a polyhedron.

Theorem. If V, E, F denote the number of vertices, edges, and faces (respectively) ofa polyhedron, then V − E + F = 2.

Leonard Euler’s discovery of his famous formula was first conveyed in a letter to Christian Goldbach in 1750 (Biggs et al., p. 76). The first proof of theformula appears to be due to Adrian-Marie Legendre (1752–1833). Descartes(1596–1650) studied polyhedra and obtained an expression for the sum of theangles of all the faces of a polyhedron. An easy consequence of this expression isEuler’s formula, but Descartes never made the connection. In a paper of 1813,Augustin.-Louis Cauchy proved the formula for a network of what we would call a map, although Cauchy was not interested in maps per se (Biggs et al., p. 81).The first use of this formula for studying maps occurred in 1879 (see the Notessection in Chapter 5).

The search for a description of maps on the sphere for which Euler’s formula istrue deserves a bit of discussion. As we have seen, there are some very simple mapsthat do not satisfy Euler’s formula—a map consisting of two countries, each adisc, is a good example: V = 0, E = 2, C = 2, V − E+ C ≠ 2. Thus, if we want anEuler’s formula theorem for maps, we need to specify those maps for which thetheorem applies. Then, if we want to use the theorem, we need to make sure thatthe map satisfies the conditions of the theorem. So much for restrictions.Allowing vertices of order two in a map is another matter. This seems artificialbecause in the real world these maps have no use! However, this was useful for usas it enabled us to prove the theorem, which is a small price to pay. If it seemed wewere changing the rules as we went along, we were not. Instead, we were creatingthe definitions appropriately in order to get the mathematics to do what wewanted. Mathematics is a human endeavor, and we were exercising our humanity;watch for this in subsequent chapters. Incidentally, “OK map’’ is not standardterminology; there is no standard term for the concept.

References

Beck, A., Bleicher, M.N., and Crowe, D.W. Excursions into mathematics. New York: Worth Publishers,1970.


Gay, D. Geometry by discovery. New York: John Wiley & Sons, 1998: Chapter 3.Holden, A. Shapes, space and symmetry. New York: Columbia University Press, 1971.Kappraff, J. Connections: the geometric bridge between art and science. New York: McGraw-Hill, 1991.O’Daffer, P., and Clemens, S.R. Geometry: an investigative approach. Menlo Park, CA: Addison-Wesley,

1997.Pearce, P., and Pearce, S. Polyhedra primer. New York: Van Nostrand Reinhold, 1978.Wenninger, M.J. Polyhedron models. Cambridge: Cambridge University Press, 1971.

References 71

Acme Collects More Data,Proves a Theorem, andReturns to Coloring Maps

73

Scene 1

A week later. A couple of days ago, Acme hired Tiffany, another math major from the otherstate university. She is working at her desk. Millie enters the shop and walks up to her.

MILLIE: What’s cookin’, Tiff?

TIFFANY: Like, I’m so totally absorbed. Ever since he came up with hisformula, Brandon has got everybody collecting data on maps.

MILLIE: Our formula.

TIFFANY: Whatever. Anyway, Brandon kept track of the number of vertices,edges, and countries, but I’ve been zooming in on just the coun-tries and counting the number of edges they have. I’m trying to seeif there is some kind of pattern.

MILLIE: Any luck?

TIFFANY: I don’t know. I’ve found something curious. Look at my data:

C H A P T E R

4

74 Chapter 4 Acme Collects More Data, Proves a Theorem

Map Number of countries with this many edges:

2 3 4 5 6 7 8 9 10 11 12 13

Western Europe 2 2 5 1 1

Lower 48 2 6 8 16 13 1

South America 5 4 1 2 1

La-La Land 3 3 3 1 1

What do you think?

MILLIE: Looks like every map has a lot of dinky countries, ones with not a lot of edges.

TIFFANY: That’s what I thought. I wonder if we could find something different,maybe a map where every country has a lot of borders. Think whatan attraction that would be for our customers! “Take your vacationon an island where every country has more than six borders!”

MILLIE: Why not for the moon and let every country have 17 or moreborders! The formula Brandon and I found relates V, E, and F.May be you and I could find a formula that relates the number ofcountries of different sizes (i.e., different number of borders) withother stuff.

TIFFANY: That would be, like, so totally awesome. Here’s a thought.Brandon was telling me about the counting arguments he and Joe were using, one was for his calendar gizmo and the other for theutility problem. The big idea was to let Ck indicate the number ofcountries with k edges. Then C = C2 + C3 + C4 + … The next thingwas to count edges: the guys got

2C2 + 3C3 + 4C4 + … = 2E.

Isn’t that a neat connection? We could do the same thing with vertices:

V = V2 + V3 + V4 + …

Of course, there are no vertices of order two in the maps we’re looking at, so

V = V3 + V4 + V5 + …

MILLIE: Who cares about vertices? We’re looking at countries.

TIFFANY: Don’t get too squirrelly. We’re lookin’ for connections, right? By using the same idea as the guys used for countries, we cancount edges from vertices:

3V3 + 4V4 + 5V5 + … = 2E.

As you said, we don’t care too much about the fine points of vertices.This last equation gives us a cruder inequality:

3V ≤ 2E.

(I think Joe and Brandon got something like this, too.) Now we canput all this into yours and Brandon’s formula, V − E + C = 2.

MILLIE: That’s Brandon’s formula for the sphere. Why ya using that?

TIFFANY: Not sure. Let’s see where it leads us. So

(2/3)E − E + C ≥ E + C = 2 or −(2/3)E + C ≥ 2

Hey, it’s working! Got rid of one variable! Multiply both sides of theinequality by three:

−E + 3C ≥ 6

Whoa! Remember, way back when? We had this equation:

2C2 + 3C3 + 4C4 + … = 2E.

Rewrite our inequality as 3C ≥ 6 + E and multiply both sides by two:

6C ≥ 12 + 2E.

Then substitute for 2E from our equation:

6C ≥ 12 + 2C2 + 3C3 + 4C4 + …

Then, since C = 2C2 + 3C3 + 4C4 + …, we have

6(C2 + C3 + C4 + …) ≥ 12 + 2C2 + 3C3 + 4C4 + …

or

6C2 + 6C3 + 6C4 + … ≥ 12 +2C2 + 3C3 + 4C4 + …

Now, with a little subtracting of the same thing (2C2 + 3C3 + 4C4 +5C5 + 6C6 + 6C7 + 6C8 + …) from both sides, we’d have

4C2 + 3C3 + 2C4 + C5 ≥ 12 + C7 + 2C8 + 3C9 + …

MILLIE: By gum! You’ve got something that tells you just about the Ck’s,and nothing else. Amazin’! But you know, Tiff, it’s not possible.

TIFFANY: Like there’s something the matter with my inequality?

MILLIE: No, no. It’s impossible to do what we wanted to do when westarted with this. We can’t find a map where every country hasmore than six borders. We can’t do that.

Scene 1 75

TIFFANY: Why not?

MILLIE: Look at your last inequality. If you’ve got a map where all thecountries have seven or more edges, then everything on the leftside of your inequality would be zero and everything on the rightside would add up to something positive. I mean, you’d have 0 >12 + some positive number. That’s not possible!

TIFFANY: Hmm. You’re right! Every country having more than five edgeswouldn’t work, either.

Tiffany writes:

1. Your TurnSuppose a map on the sphere (all vertices of order three or more) has no coun-tries with exactly three borders and none with exactly four borders. What can yousay about the number of countries with exactly five borders?

Scene 2

A little later. Millie is alone in the shop. Joe enters.

JOE: Millie! I hear you and Tiffany just made a new breakthrough onmaps. Every map on a sphere—all vertices of order three or more—must have a country with five or fewer edges. That’s pretty wild! I never would have guessed that. But you look a little disappointed.

MILLIE: Well, we thought we had a great advertising scheme: “Take a vaca-tion where every country has more than six borders!”

JOE: Chin up. I’ve been thinking about what you and Tiffany were able to show. Remember when we were coloring maps? What youand Tiffany did gives me a way to color any map on the sphere withsix or few colors. Guaranteed!

Possible “Sizes” of Countries in a MapIf Ck = the number of countries with exactly k edges, then

2E = 2C2 + 3C3 + 4C4 + …

Also,

4C2 + 3C3 + 2C4 + C5 ≥ 12 + C7 + 2C8 + 3C9 + …

Theorem. If a nice map has every vertex of order three or more, then theremust be at least one country whose number of edges is five or less.


Voilà! You’ve colored the map in six colors!

MILLIE: OK. But what’s your method for coloring the altered map? Andwhat do Tiffany and I have do to do with it?

JOE: Patience! I’m getting to it. You start with a map—on the sphere, thistime every vertex has order exactly three. You and Tiffany showedthat the map has a country with five or fewer edges, so you find one.What you want to do with that country is remove an edge to get a map with one fewer countries. Like this.

Scene 2 77

Order-Three Reduction for Map Coloring

● Replace every vertex of order greater than three by a little country. Everyvertex of the altered map is now of order three!

● Color the altered map in six colors using my method

● Keep this same coloring of the map while shrinking the little countries youadded.

MILLIE: Six colors. We never found a map that needed more than four.

JOE: Doggone it, you’re right, Millie. But I said guaranteed. We’ve hadour suspicions, but never a guarantee. Let me show you how itworks. I’m going to show you how to color, in six colors, any mapon the sphere that has all its vertices of order three. But, beforedoing that, here’s how to use that method for any map on thesphere with vertices of order three or greater:

Joe wrote.

You have to be careful. The map you wind up with has to be an OK map. You have to remove any vertices of order two that may havecome up because the new map should have the same properties as the original. All vertices should be of order three. For one thing,you want the Millie–Tiffany result to hold for the new map.These pictures show you the possibilities when you’re dealing with a four- or five-edged country and what I want to do with them.(I’ll let you figure out what to do when you have a two- or three-edged country.)

MILLIE: So what do you do with the new map?

JOE: You keep altering it, decreasing the number of countries one by one until you get a map with six countries. Leave the dotted lines in to show where you’ve gotten rid of edges along the way, and keeptrack, somehow, of the order in which you have eliminated edges,vertices, and countries. Then you color the last map you get—theone with six countries—properly with six or fewer colors. You cando that, right?

MILLIE: Well, I can.

JOE: Next, you restore the countries, vertices, and edges you’ve elimi-nated one country at a time, in reverse order, coloring the restoredcountries as you go. Leaving the rest of the map colored as it is,color the last country you eliminated with a color not used by itsneighbors. A color will be available. The last country has five orfewer edges so there are at most five countries surrounding it; atmost five colors are used to color those neighbors so there is a colorleft over (from the six colors) to color the restored country. Themap with the last country eliminated restored is colored properly!


Repeat this process: restore a removed country (along with an edgeand two vertices) one at a time until you arrive at the original map.You’ll get a sequence of maps that are each colored properly with sixcolors, because at each stage you color the restored country a colordifferent from its neighbors in the coloring of the previous stage map.Here’s a map I colored using this method.

Scene 2 79

MILLIE: That’s a real, rootin’, tootin’ explanation if ever I seen one!

Millie leaves. Joe smiles, goes to his desk, and writes.

2. Your TurnTry out Joe’s method on the following map.

Scene 3

Two hours later. Joe is alone in the shop. Tiffany walks in.

TIFFANY: Hola, Joe! Millie told me what you’d done. A method that colors amap in six or fewer colors guaranteed to work every time!

JOE: Yep. Whaddaya think of it?

TIFFANY: It’s rad! I’ve been thinking about it, and I think I have an idea forimproving on your method.

JOE: Lotta thinkin’ goin’ on….Wait. Improving? You mean a fastermethod?

TIFFANY: No, I mean fewer colors, guaranteed.

JOE: Wow! Like four colors?

TIFFANY: No, five. Let me tell you my idea. Your method revolves around thefact that, when you restore a country, it’s surrounded by five orfewer other already colored countries, so that there’s a sixth colorleft to color the restored country.

JOE: Huh?

TIFFANY: Let me show you.

Coloring Maps on a Sphere

Theorem. Every map on a sphere can be properly colored in six or fewer colors.


JOE: So, what’s your point?

TIFFANY: What if the five or fewer already colored countries only use upfour colors? Like this.

JOE: That would be guaranteed if there were always a country with fouror fewer edges, but that’s not always the case. Check out Brandon’sgizmo: all the countries have five edges!

TIFFANY: You’re right, they do. But it’s not the number of bordering countries that count. It’s the number of colors those countries useup. A restored country with four or fewer edges has four or fewer countries bordering it and uses at most four colors, so the restoredcountry can be colored a fifth color. The problem is when the country has exactly five edges. Look, here are the pictures youdrew for Millie.

In both cases you can remove two edges and get a map with two fewer countries.

You have to be careful, though. The two edges you remove can’t be contiguous.

JOE: Contiguous?

TIFFANY: Touching. You don’t want to do this.

Scene 3 81

And, you have to make sure the end result is an OK map. For exam-ple, you wouldn’t want to do this either.

Anyway, the key is that, when you restore the country, at most fourcolors have been used for its bordering countries. Look.

JOE: Neat! Removing two edges makes three bordering countries intopart of the same country in the new map.

TIFFANY: Absolutely. So, I just plug this new idea into your method and,tada, a method of coloring that uses only five colors: You keep altering the map until you get a map with five or fewer countries,and then color it properly with five or fewer colors. After that,you can begin restoring the countries and edges just as in your method.

JOE: Hey, this sounds awfully familiar.

TIFFANY: I told you it was yours. As you restore, one by one, each countryand edge, you leave the rest of the map colored as it is and colorthe restored country with a color not used by its neighbors.A color will be available for the restored country because thecountries surrounding the restored one use four or fewercolors. The resulting map with the country restored is coloredproperly! Repeat this process and you’ll get a sequence of mapsthat are each colored properly with five colors, because at eachstage you color the restored country a color different from itsneighbors in the coloring of the previous stage map. Here’s howI colored your map using this method.


JOE: Well, if that isn’t a gas!!

Joe walks off. Tiffany writes.

Lights fade. Music rises.


1. InvestigationBecause Tiffany and Millie can’t find vacation hideouts where all the countrieshave a large number of borders (six or more), maybe they could take on a new

Coloring Maps on a Sphere, Part Deux

Theorem. Every map on a sphere can be colored in five or fewer colors.


search: find vacation hideouts where all the orders of the vertices are large. Helpthem out. See what you can find out.

2. InvestigationThe folks at Acme’s rival, El Grande Maps, have been looking at OK maps on thesphere with these properties:

● Every vertex is of order three.

● Every country has an even number of edges.

They claim that such maps can be colored properly in three colors, but theyhaven’t been able to explain why. Investigate their claim. You’ll want to draw a fewsuch maps and color them. Also, you’ll want either to find a counterexample or toexplain why their claim is true. Of course, you will want to write up your findingsin the form of a report.

3. InvestigationMillie and Tiffany know that a map in which every country has more than fiveedges is impossible. The next best thing would be maps with some five-edgedcountries and the rest six-edged, seven-edged, eight-edged, and so on. As a start atinvestigating these, help them look at OK maps such that every vertex is of orderthree and every country has either five or six edges.

4. InvestigationIn a map with vertices that are all of order three or more, countries with two,three, four, or five edges are ubiquitous. Consequently, Millie and Tiffany thinkthat it would be good to have a list of all OK maps in which every vertex is of orderthree and every country has five or fewer edges. Help them out. Of course, you willwant to write a report of your findings.

5. InvestigationConsider OK maps on a sphere such that every vertex is of order three or more.Acme is thinking about classifying these maps by combinatorial type. (Comparethis classification with those of Investigations 12 and 13 in Chapter 3.) Two mapsare of the same combinatorial type if they have the same network of edges andvertices; that is, you can get one from the other by shrinking or stretching. Hereare four shapes, each corresponding to maps of the same combinatorial type.


Here are three shapes, each corresponding to maps with eight countries but ofa different combinatorial type.

Help Acme find all combinatorial types for maps with five or fewer countries.Of course, you will want to write a report to the Acme staff justifying the resultsof your investigation.

6. QuestionThey say if a map on the sphere (all vertices of order three or more) has exactlyfive countries, then at least two of the countries must have exactly three edges.Is this an old wives’ tale, or what?

7. QuestionYou have an OK map on the sphere such that


● The map is properly colored in three colors.

What else can you say about this map?

8. QuestionYou want to color a bunch of non-overlapping circles drawn in the plane. Circlestangent to one another must be colored different colors. What can you say aboutthe numbers of colors necessary to color all the circles?


9. PuzzlePinkley Smith, Acme’s solicitor, has been caught up in the map coloring mania.He’s interested in maps that can be colored properly in four colors. He claims thatan OK map on the sphere of the following kind can be colored in four colors:


● You can take a trip traversing edges of the map’s network that passes once(and only once) through each vertex and returns to your starting point.

Pinkley Smith gave out the following hints to help show the Acme team why hisclaim is true:

● The trip he mentions forms a simple closed curve on the sphere. This cuts thesphere into two pieces each of which can be distorted into a disc.

The part of the map on each disc has the following properties:

● There are an even number of vertices on the border of the disc.

● There are no vertices interior to each disc.

● A line on a disc joins a pair of vertices on the border.

● The lines on each disc do not intersect.

Are Pinkley’s hints valid? Could you use them to show that four colors willcolor the map properly?

10. SummarizingBoss has asked Tiffany to summarize the day’s investigations. Help her out bywriting a report to Boss about the accomplishments of Tiffany, Joe, and Millie.


Be sure organize your summary report carefully; write clearly; mention goals,problems, results, and arguments; and tie in today’s activities with those carriedout in previous days.

Notes

[Huck Finn to Tom Sawyer in their flying boat] “We’re right over Illinois yet.And you can see for yourself that Indiana ain’t in sight….Illinois is green,Indiana is pink. You show me any pink down here, if you can. No sir; it’sgreen”“Indiana pink? Why, what a lie!”“It ain’t no lie. I’ve seen in on the map, and it’s pink.”

-from Tom Sawyer Abroad, Mark Twain

In this chapter we have proved the following

Theorem. Any map on the sphere can be properly colored in five colors.

The proof that we have given of this is essentially that given in 1890 by PercyJohn Heawood (see Biggs et al., p. 105f). Tiffany’s proof can be neatened up byusing mathematical induction on the number of countries of the map (see theproofs in Stein and Beck et al.).

The idea of reducing the problem of coloring maps to that of coloring maps allof whose vertices are of order three is due to Cayley and Kempe in separate papersof 1879 (see Biggs et al., p. 93f.) In his paper Kempe also showed that in a map—all vertices of which are of order three or greater—there is always a country withfive or fewer borders. In addition, he seemed to know that, a map, all of whosevertices are of order three and every country having an even number of borders,must be able to be colored with three colors (see Investigation 2).

References

Beck, A., Bleicher, M.N., and Crowe, D.W. Excursions into mathematics. New York: Worth Publishers,1970.


Gay, D. Geometry by discovery. New York: John Wiley and Sons, 1998.Stein, S.K. Mathematics: the man-made universe. San Francisco: W. H. Freeman, 1969.

References 87

Acme’s Solicitor Proves aTheorem: the Four-ColorConjecture

89

Scene 1

A couple of days later. Tiffany, Millie, and Brandon are in the shop, working at theirdesks. Joe enters and calls to the others.

JOE: Guess what? Boss says Pinkley Smith will be here any minute.Pinkley wants to show us a guaranteed way to color maps usingfour colors. Pinkley says he based it on our methods, Tiffany.

BRANDON: Wow! Who’s Pinkley Smith? How did he find out what we’vebeen doing?

MILLIE: Pinkley Smith is the company’s lawyer. I saw him the other dayand told him about our stuff.

BRANDON: I thought we were going to keep quiet about it. We were thinkingabout publishing.

MILLIE: Well, we shouldn’t keep our lawyer in the dark. Anyway, he’s agood cat.

C H A P T E R

5

JOE: Cat? (Pinkley Smith walks in. He is wearing a seersuckerjacket and polka dot bow tie.)

MILLIE: Hey, here he is!

PINKLEY SMITH: Hello, fellow map enthusiasts!

JOE, MILLIE,TIFFANY, BRANDON: Hello, Pinkley Smith!

JOE: We hear you have something to tell us.

PINKLEY SMITH: Have I! (Approaches whiteboard, a new Acme acquisition.)Gather ’round and I shall tell you of a mighty map color-ing breakthrough!

ALL: Groan...

PINKLEY SMITH: I’ve been following what you’ve been doing. A lawyer must keep on top of his clients’ goings-on, I always say.I know about your two-color maps and your “nice trips”and the sometime impossibility of finding them. I’ve alsoheard about the immutable fact that every map (all verticesof order three or more) has to have a country with five orfewer borders. Just yesterday Millie filled me in on thelatest. You have a method for coloring a map with fivecolors that will work every time. Well, I asked myself,why can’t I find a method with four colors that worksevery time? After all, none of us—yes, my friends, Iconsider myself one of you—none of us has been able tofind a map that requires more than four colors. So, guesswhat, I found a method.

MILLIE: Get on with it, Pinkley.

PINKLEY SMITH: Hold on to your hats, friends. My method uses thefollowing idea. Suppose you color a map properly using(among others, possibly) colors α and β.

TIFFANY: Greek letters?

BRANDON: Shhh.

PINKLEY SMITH: (Ignores interruption.) In that map, there may be asequence of contiguous countries colored, sequentially,α—β—α—β—α—β. Like this.

If the sequence begins with country X and ends withcountry Y, then I call it an a-b chain from country X to

90 Chapter 5 Acme’s Solicitor Proves a Theorem

country Y. I can reverse the coloring of the chain and getthe following.

JOE: When you reverse the coloring do you get a β-α chain?

PINKLEY SMITH: No. I mean, yes. A β-α chain is also an α-β chain. Is that clear?

JOE: As mud...

BRANDON: No, JOE. What Pinkley means is that the colors alternate.It doesn’t matter which color starts the chain. An α-β chainand its reversal are both α-β chains.

PINKLEY SMITH: There’s a second idea related to this that I’ll use in my method.Remember we have a map that is colored properly. Suppose Xis one of the countries of the map and it’s colored α or β. Thenconsider the set S of all countries Y connected to X with an α-β chain, that is, S = {Y: there is an α-β chain from X to Y}.The set S might look something like the following.

OK? Now I’m ready to describe my method. It really startswhere you left off, Tiffany.

MILLIE: You mean Brandon, not Tiffany.

BRANDON: I think it’s Joe. Or you, Millie.

PINKLEY SMITH: Whatever. You start with a map, all the vertices of which areof order three. You know that there must be a country withfive or fewer edges. You choose one of these and, by erasingedges, create a map (with vertices that are all of order three)with fewer countries. You keep doing this until the alteredmap has four or fewer countries, which you naturally colorin four or fewer colors. Then you reverse your steps, putback erased edges, and color reinstated countries.

Scene 1 91

BRANDON: Sounds just like our method. How do you show that a fifthcolor won’t be needed?

PINKLEY SMITH: I was just getting to that. As you say, so far everything’s thesame as your method. Let me tell you what I do differently.First, when I am erasing edges and have chosen a countrywith exactly four edges, I erase two edges as follows.

Thus, if the altered map is colored properly with four orfewer colors, then—when you restore the erased edges—there is a fourth color to color the reinstated country. Inyour method, you only erased one edge.

TIFFANY: How do you know you can color the altered map in fourcolors?

MILLIE: I think Pink’s referrin’ back to our method when we have torestore a four-edged country. He’s assumin’ he’s been ableto color with four up to then.

BRANDON: I think Pinkley Smith has more to do before we can believethat.

PINKLEY SMITH: Yes. I said there were two differences. I’ve just described one ofthem. The second change in your method occurs when I rein-state a five-edged country, which is denoted by X. We assumethat the rest of the map—all the countries except X—is coloredproperly in four colors, including the countries that border X.If only three colors are used for these bordering countries, thena fourth color is left for X itself, and so we would have a propercoloring for the map with the reinstated country.

A worse-case scenario occurs when four colors are used forthe countries bordering X. Like this.


I want to show you how to reduce the number of colorsused (by the countries bordering X) to three. Doing thiswill involve altering the colors of the remainder of the map.I’ll illustrate the method referring to the diagram above andusing the following three steps.

Pinkley Smith turns to the whiteboard and writes:

Three Steps to Color Entire Map

1. Does a red-yellow chain exist from countries 1 to 4? If yes, go to step 2.If no, reverse colors in set {Y: a red-yellow chain exists from 1 to Y}. At thispoint, countries 1 and 4 are both colored yellow. Except for X, the entiremap is colored properly in four colors. Furthermore, only yellow, blue,and green are used to color countries 1 through 5. Color country X redand the whole map will be colored properly in four colors. We’re done!

2. Does a red-green chain exist from countries 1 to 3? If yes, go to step 3.If no, reverse colors in set {Z: a red-green chain exists from 1 to Z}. At thispoint, countries 1 and 3 are both colored green. Except for X, the entiremap is colored properly in four colors. Furthermore, only yellow, blue,and green are used to color countries 1 through 5. Color country X redand the whole map will be colored properly in four colors. We’re done!

3. If the process has gotten to this step, it must be the case that there is anred-yellow chain from countries 1 to 4 and an red-green chain from coun-tries 1 to 3. Thus the following situation—or something similar—holds.

Hence, there is no blue-green chain from countries 5 to 2 nor is there a blue-yellow chain from countries 2 to 5. Reversing colors in the sets {U: a blue-green

Scene 1 93

I rest my case.

TIFFANY: All right!

JOE: Way to go, Pinkley Smith!

BRANDON: Sweeeet!

MILLIE: Didn’t know ya had it in ya!

High fives all around. Lights fade. Music rises.


1. Gathering EvidenceUse Pinkley Smith’s method to recolor the map and color country X so that theentire map is colored properly in four colors.


chain exists from country 5 to country U} and {V: a blue-yellow chain existsfrom country 2 to country V} leaves country 5 colored green and country 2 colored yellow. Thus we would have

Except for country X, the map is colored properly in four colors. Furthermore,only red, green, and yellow are used to color countries 1 through 5.

Color country X blue and whole map is colored properly in four colors.We're done!

2. Gathering EvidenceUse Pinkley Smith’s method to recolor the following map and color the blankcountry so that the entire map is colored properly in four colors.

3. GameThis is a map coloring game called Tetrachrome (see McConvill, p. 54). At thebeginning of the game, the players subdivide the rings into various numbers ofregions, as in the example.

Several players may play, each taking a turn coloring any region of his or herchoice with any one of four colors in such a manner that adjacent regions becolored differently. The first player unable to play is the loser. (If two players play,is there is winning strategy for one of them? Which one, the first or second player?Could the game end in a tie?)



4. GameThis is called Brams’s map coloring game (see Gardner, p. 253f). Game equipment consists of a map on an island and a supply of crayons of n differentcolors. The first player, the Minimizer, selects a crayon and colors a region of themap. The second player, the Maximizer, then colors another region using one ofthe crayons. Players continue in this way, alternately coloring a region with anyone of the n colors and always following the rules for the proper coloring ofa map. The Minimizer tries to avoid the need for the use of an n +1 - st color tocomplete the coloring of the map. The Maximizer tries to force the use of it.The Maximizer wins if either player is unable to play, using one of the n colors,before the map is fully colored. The Minimizer wins if the entire map is coloredwith n colors.

1. To get started, play a few games with the following map and four colors.

According to Martin Gardner, the Maximizer can win this game; that is,she can always force the Minimizer to use a fifth color. Can you explain this? Martin Gardner also claims that if you use this map with five colors, the Minimizer can always win. Do you believe this? Can youexplain it?

2. Martin Gardner makes yet another claim: There is a map such that, if you playwith five colors, the Maximizer can always win. Can you find such a map andgive the Maximizer’s winning strategy?

3. Consider the smallest value of n such that when the game is played on any mapthe Minimizer can, if both players play optimally, always win. Brams conjec-tures that this smallest value of n is six. What do you think?

5. InvestigationWhen Joe was coloring the map of the United States, he noticed that the state of Michigan is not connected—it consists of two pieces. This did not create an unusual coloring problem because the two pieces are separated by a lake.


This gave Joe an idea: “Suppose I have a map of empires. Each empire wouldconsist of one or more connected regions that don’t share common borders.I could think of one of the regions in an empire as the ‘mother’ country and theothers as colonies. To color a map of empires, I would color all the regions ofa single empire the same color. Otherwise the rules would be the same as before:two different empires that border each other must be colored different colors.”Here is an example.

Joe wants to know how many colors such a map would need. Help Joe with theinvestigation. He thought the following questions might help:

● Is there an empire map that needs five colors to be colored properly?

● Is there an empire map that needs six colors?

● Is there an empire map that needs seven or more colors?


6. InvestigationYou have a bunch of pennies in the plane. Some are tangent, and none overlap.You want to color them so that touching pennies are colored different colors. Givea simple argument that shows four colors will always be sufficient.

7. Looking BackGo back and have a look at what you know about maps on the sphere or on anisland that can be colored with three colors. In particular, have a look atInvestigations 2 and 3 of Chapter 1 and Investigation 2 of Chapter 4. CanInvestigation 3 of Chapter 1 be generalized to the sphere? (Compare the general-izations of two-colorable maps on an island to a sphere.) Given a map, could youtell if it is one of these? Are there easier methods for coloring such maps thanthose methods suggested by your earlier investigations? (Compare different meth-ods for coloring two-colorable maps.) Suppose you have an OK map in whichevery vertex is of order three and the map can be colored properly in three colors.Is there anything else you can say about the map? Is there anything you can say ingeneral about maps on the sphere that are colorable in three colors?

Notes

[Scott Parris, chief of police, is trying to sell Charlie Moon, Southern Utedetective, on the latest in pick-up trucks.] “It’s got a GPS navigation system,with nine-color map display.”“I thought four colors was enough.”“A real red-white-and-blue American consumer don’t settle for enough.”

–from The Witch’s Tongue by James D. Doss

Even the moonlit track ahead of him faded from his consciousness, for intohis head had come a theorem which might be true or might be false, and hismind darted hither and thither seeking proofs to establish its truth andcounter-examples to show that it could not possibly be true.

–from Kandelman’s Krim, J. L. Synge

Notes 99

In the notes to Chapter 1, we mentioned that in 1852 the English graduate student Francis Guthrie first posed the problem of coloring maps andthought they could all be colored in four colors but did not have a proof ofthe fact.

Then, in 1879 Sir Alfred Kempe, British lawyer and amateur mathematician,published an article entitled “How to Colour a Map with Four Colours” in the firstissue of The American Journal of Mathematics that contained the followingremarks:

Some inkling of the nature of the difficulty of the question, unless its weak point bediscovered and attacked, may be derived from the fact that a very small alteration inone part of a map may render it necessary to recolor it throughout. After a some-what arduous search, I have succeeded, suddenly, as might be expected, in hittingupon the weak point, which proved an easy one to attack. The result is, that the expe-rience of the map makers has not deceived them, the maps they had to deal with,viz.: those drawn on a [sphere] can in every case be painted with four colors.

Kempe’s arguments were accepted by the mathematical community for sometime. Even Felix Klein, of the university at Göttingen and one of the foremostmathematicians of the time, remarked when asked about the map color problem,“Oh, that. An Englishman proved that four will work.” (Klein did not have a highregard for the English mathematicians of the time.) In this chapter we putKempe’s argument into the mouth of Pinkley Smith. Kempe’s paper appears in thebook by Biggs et al. (p. 96f).

Not much was done with map coloring until 1890, when P. J. Heawoodpublished a paper that began:

The Descriptive-Geometry Theorem that any map whatsoever can have its divisionsproperly distinguished by the use of but four colors, from its generality and intangi-bility, seems to have aroused a good deal of interest a few years ago when the rigor-ous proof of it appeared to be difficult if not impossible, though no case of failurecould be found. The present article does not profess to give a proof of this originalTheorem; in fact its aims are so far rather destructive than constructive, for it will beshown that there is a defect in the now apparently recognized proof [of Kempe’s]....

In this paper Headwood produced a counterexample to the technique thatKempe had used in his proof. This is the map of Gathering Evidence 2. Heawood’spaper also appears in Biggs et al. (p. 105f).

So Heawood found that Kempe’s “proof” had flaws. However, in the samepaper, Heawood showed that Kempe’s method could be used to prove the five-color theorem. This was the proof we saw in Chapter 4. For many years afterHeawood’s paper appeared, no one was able to patch up Kempe’s “proof” of thefour-color theorem nor was anyone able to produce a map on the sphere thatneeded five colors.

The four-color problem became famous but remained unsolved until 1976when Professors Kenneth Appel and Wolfgang Haken, from the University ofIllinois, announced a proof. The Times (of London) reported the following on July 23, 1976:


Their proof, published today, runs to 100 pages of summary, 100 pages of detail anda further 700 pages of back-up work. It took each of them about 40 hour research aweek [for four years] and 1,000 hours of computer time. Their proof contains10,000 diagrams, and the computer printout stands four feet high on the floor.

‘It was terribly tedious, with no intellectual stimulation,’ Appel said. ‘There is nosimple elegant answer, and we had to make an absolutely horrendous case analysisof every possibility. I hope it will encourage mathematicians to realize that there aresome problems still to be solved, where there is no simple God-given answer, andwhich can only be solved by this kind of detailed work. Some people might thinkthey’re best left unsolved.’

Contrast this description of Appel and Haken’s proof with the followingcomments by René Thom, a leading 20th-century French geometer:

Bertrand Russell has said that ’mathematics is the subject in which we never knowwhat we are talking about nor whether what we are saying is true.’ Unfortunately, thepurely formal view is difficult to uphold, paradoxically for formal reasons. We knowthe difficulties presented by the formalization of arithmetic associated with Gödel’sTheorem....For myself, I am content with the following illustration: Let us supposethat we have been able to construct for a formal theory S an electronic machine Mcapable of carrying out at a terrifying speed all the elementary steps in S. We wish toverify the correctness of one formula F of the theory. After a process totaling 1030

elementary operations, completed in a few seconds, the machine M gives us a positive reply. Now what mathematician would accept without hesitation the validityof such a ‘proof,’ given the impossibility of verifying all its steps?” (See Thom’s articlein The American Scientist.)

In an article that appeared in the September/October 1976 issue of ScientificAmerican, Haken and Appel were reportedly “not troubled by the unwieldiness oftheir proof”:

Indeed, they believe it is intrinsic to problems such as the four-color theorem. SaidAppel: ’While we are excited to solve a problem as well known as the four-colortheorem, we think that the real significance of our work lies in the area of mathe-matical proof. ’

The Illinois mathematicians believe a concise, elegant proof of the four-color theo-rem does not exist. They think they have found the first example of a type of prob-lem that cannot be solved by the usual mathematical techniques. ’We believe theremust be many true statements in mathematics that can only be proved by suchmethods,’ said Haken. ‘We hope our work on this problem will encourage othermathematicians to investigate these techniques.’

Expositions of the ideas used by Appel and Haken to prove the four-color theoremcan be found in the books by Barnette, by Fritsch and Fritsch, and by Wilson.

Let’s have a brief look at one difficulty with resolving the four-color problem.One approach to solving it might be the following: If a map needs five colors, thenit seems reasonable to assume that there will be five countries that each border onall of the others, thus forcing the map to be colored in five colors. So suppose thereare five such countries. Put a point inside each of these—call it the capital. Join two

Notes 101

capitals with an edge if the two countries share common borders. (This is anexample of the construction of the network dual to the network of a map.)

In this way we are able to connect the five capitals on the sphere without any ofthe connections crossing. But this is impossible! (See Investigation 9 of Chapter 2and Scene 5 of Chapter 3.) Therefore no map needs five colors.

The catch here is the phrase “it seems reasonable to assume that.” The assump-tion that needing five colors means five countries that each border on the otherfour is a kind of “local” assumption; that is, the difficulty with coloring a map withfour colors can be localized to a few neighboring countries. The problem with thisassumption is illustrated as follows. Consider map A in the figure below. Threecountries are colored.

What color should be used for the fourth blank country? We must color it eitherred or a fourth color. Suppose we take the second alternative and color it yellow, asin map B. Then we add another region (as in map B). Now it is impossible to colorthe map without using a fifth color. If we return to map A and color the blank regionred, we are in trouble if a situation such as map C occurs in which two regionsborder the first four. Fourth and fifth colors are needed for the two blank countriesin map C. Of course, the countries in map C can by colored with four colors (try it).But it means going back and changing the coloring already in place in map C.

Empire Maps. In the same paper in which he provided a counterexample toKempe’s proof and in which he proved the five-color theorem, Heawood alsoconsidered the empire maps described in Investigation 6, in which countriescan have colonies. If you allow only one colony per country, then Heawoodshowed that 12 colors will suffice. He also produced an empire map thatrequires 12 colors (see Beck et al., p. 70f. and Gardner, p. 85f.).


Computational Complexity. Given a map on the sphere, it is easy to decide whetheror not it can be colored in two colors. All you do is determine the orders of allthe vertices. This can be carried out in polynomial time, that is, the time it takesis approximately equal to some polynomial function f(n) of the number n of“elements” of the map. The elements in this case would be vertices, edges, coun-tries, and adjacencies. Some problems, however, cannot be carried out in poly-nomial time, for example, the problem of listing all the arrangements of seatingn persons at a table. Every solution takes exponential time. Consider anotherproblem very close to the two-color problem. Given a map on the sphere, decidewhether or not it can be colored in three colors. It appears that this is not an easyproblem; every known algorithm for solving it takes exponential time. It turnsout that the problem is also computationally equivalent to every problem in aclass of problems called NP complete. Two problems are computationallyequivalent if one can be transformed to the other in a time that is polynomialin the number of elements. Many NP-complete problems have been identifiedthat are very difficult and are probably exponential. For more information onNP-complete problems, see the article by Cipra and the books by Garey andJohnson and by Neapolitan and Naimipour.

References

Appel, K., and Haken W. The solution of the four-color-map problem. Scientific American, October1977, p. 108f.

Barnette, D. Map coloring, polyhedra and the four color problem. Washington, DC: MathematicalAssociation of America, 1984.

Beck, A., Bleicher, M.N., and Crowe D.W. Excursions into mathematics: the millennium edition. Natick,MA: A. K. Peters, 2000.


Cipra, B. Advances in map coloring: complexity and simplicity. SIAM News, December 1996, p. 20f.Fritsch, R., and Fritsch, G. The four-color theorem: history, topological foundations, and idea of proof.

Springer-Verlag: New York, 1998.Four colors do it, Scientific American, September/October 1976.Gardner, M. The last recreations. New York: Copernicus, 1997.Garey, M.R., and Johnson, D.S. Computers and intractability: a guide to the theory of NP-completeness.

New York: W. H. Freeman, 1979.McConville, R. The history of board games. Palo Alto: Creative Publications, 1974.Neapolitan, R., and Naimipour, K. Foundations of algorithms. Lexington, MA: D. C. Heath, 1996.Saaty, T.L. “Thirteen colorful variations on Guthrie’s four-color conjecture.” American Mathematical

Monthly, January 1972, pp. 2–43.Thom, R. “Modern” mathematics: an educational and philosophical error? The American Scientist,

November/December 1971, pp. 695–699.Thomas, R. An update on the four-color theorem. Notices of the American Mathematical Society,

August 1998, pp. 848–859.Wilson, R. Four colors suffice. Princeton: Princeton University Press, 2002.

103

Acme Adds Doughnuts toIts Repertoire

Scene 1

A week or so later. Boss and Brandon are alone in the Acme shop working. Each isworking at his desk. The phone rings.

BOSS: Howdy. Acme Maps and Tours. We can give you the trip of yourlife!...Oh, yes. Amalgamated...Yes, I remember....But we reallydon’t....Hole?...OK. I’ll see what we can do. Adios. (Hangs up.)Those fellers are really barkin’ at a knot. Next thing ya knowthey’ll be asking us to do their laundry and having us makedoughnuts. (Turns to Brandon.) You remember those problemsSouthwestern Amalgamated Circuits sent us awhile back, theones about designing circuit boards?

BRANDON: Yes, Boss. We told them we couldn’t solve some of their problems.

BOSS: What? We couldn’t solve their problems?

BRANDON: We told them nobody could. Ever.

C H A P T E R

6

BOSS: Huh? Well, they’re kickin’ up a row about this new circuit boardthey’ve got. One with a hole in it. I guess the old one didn’t havea hole. They asked if we could solve their problems using thenew board.

BRANDON: Hmm...Hole...That’s interesting. Let me...

BOSS: Well, don’t spend too much time on it. (Turns from Brandon. Mutters.) Circuit boards! Can’t they read our sign?(Walks out.)

BRANDON: (Turns to his desk, opens a drawer, pulls out a sheet of paper, andspreads it out on his desk:

104 Chapter 6 Acme Adds Doughnuts To Its Repertoire

Five-terminal problemProblem: Take five points. Connect each pair of points with an edge so that the

edges don’t cross.If this could be done on the sphere, then we’d have a map with

● V = 5

● E = 10

● 2 = V − E + C = 5 − 10 + C

● C = 7

● C = C3 + C4 + C5

● 2E = 3C3 + 4C4 + 5C5

● 2E > 3C

● 2 × 10 > 3 × 7 No! 20 can’t be bigger than or equal to 21!

Impossible. Can’t do it on the sphere.

He talks to himself.) Now, what could a hole do to make theconnections possible? I guess you could allow the connections to pass through the hole. Hmm. How would that change my argument? The surface of original circuit board is sort of like thesurface of a sphere if you allow distortions. Cutting a holethrough the circuit board makes the surface more like the surfaceof...a washer or, hey, like that box-with-a-hole puzzle we hadaround a while ago!

Now what do I remember about the box-with-a-hole? Yes.We got V − E + F = 0. We thought something was the matter.(Joe and Tiffany walk in.)

TIFFANY: Que pasa, Brandon? We brought back some awesome doughnutsfrom Krusty Krums. Want one?

BRANDON: Yeah, sure. (Takes one. Looks at it.) Hey, doughnuts! That’sfunny; I’m working on a doughnut problem!

JOE: In your dreams?

BRANDON: Remember Amalgamated’s circuit board problems? (Shows Joeand Tiffany the paper.) Now they want us to solve the problem ona doughnut.

JOE: Here, eat this quick before you faint. Dreams is right. I thinkyour brain needs a little food.

BRANDON: No, listen. They have a new board. It has a hole in it; with a littledistortion, you get the surface of a doughnut. Look.

TIFFANY: Awesome. Like, aren’t these Krusty Krums so the best?

JOE: I thought that problem couldn’t be solved.

BRANDON: Well, yes. But I think things are different. That was on the sphere.Now we’re on a doughnut. Remember the box-with-a-hole

Scene 1 105

puzzle? That’s really the surface of a doughnut, and we got V − E + F = 0. We thought that was weird. But maybe some-thing’s going on. I mean, maybe that’s what it should be for amap on a doughnut.

JOE: Ooooh. Slow down! This is getting heavy. You need to get out ofthe shop more. Have another doughnut!

BRANDON: Hey, thanks! These Krusty Krums are really inspiring! Let’sconsider my impossibility argument in case we’ve got V − E +F = 0. (He writes:)


Five-Terminal Problem on a DoughnutProblem: Take five points. Connect each pair of points with an edge so that the

edges don’t cross.If this could be done on the doughnut, then we’d have a map with

● V = 5

● E = 10

● 0 = V − E + C = 5 – 10 + C

● C = 5

● C = C3 + C4 + C5

● 2E = 3C3 + 4C4 + 5C5

● 2E > 3C

● 2 × 10 > 3 × 5

But 20 is bigger than 15!

JOE: That settles it! You can solve the five-terminal problem on adoughnut. Good job! Call up Amalgamated!

TIFFANY: Not so fast. You need to prove that V − E + C = 0 on a doughnut.All you have is one item of evidence.

BRANDON: Tiff, you’re right. And, there’s more. Joe, all I’ve done is show that ifI can solve the problem on a doughnut, then there is no contradic-tion. I haven’t shown anything for sure, only that my old argumentisn’t an impossibility argument. Even that evidence of progressdepends on the assumption V − E + C = 0. The invisible emperor iswearing invisible clothes! But I think we’ve got some ideas to

work on. Joe, why don’t you work on seeing if you can actuallydraw the connections for the five terminals on the surface of adoughnut. Tiff, let’s see if we can show that V − E + C = 0 for anymap on the surface of a doughnut.

Joe walks back to his desk and begins to work on the problem.

1. Your TurnJoe needs help. Place five points on the surface of a doughnut. See if you can draw aconnection between every pair of points without having any of the connections cross.

Tiffany pulls up a chair to Brandon’s desk and writes:Conjecture: For any OK map on a doughnut, V − E + C = 0.

TIFFANY: Invisible emperor. That’s bad! OK. I think I’d like to do someexamples to get a feel for the problem. We’ve only actuallycounted V, E, and C for one case. But the box-with-a-hole givesme an idea for more maps, a whole family of them.

2. Your TurnHelp Tiffany figure out V, E, and C for the infinite family of maps on a doughnutsuggested by the diagrams above. Then see if you can conclude anything about V − E + C for these maps.

BRANDON: Hey, those box maps give me an idea. Imagine those shapes aremade of cardboard so they aren’t solid. We can cut them openlike this.

Scene 1 107

Hmm. I’d really like to flatten the whole thing out. Then maybewe could use what we know about sphere or island maps.

TIFFANY: Cut it again. Like this.

Hmm. Need to do some stretch-shrinking to get it to flatten out.But, look, it’s an island! V − E + C = 1. Oh, that’s not quite right. Wemust be doing something wrong. We got V − E + C = 0 when it wasput together.

BRANDON: You’re right. I get the same thing. Where did we mess up?

TIFFANY: I see something. Look at the vertices on the corners of therectangle. Those are all the same.

BRANDON: Same?

TIFFANY: They come from the same vertex on the original map. When wecounted V on the original, that vertex only counted for one, butwhen we count V on the square, we count four. One has becomefour!

BRANDON: So we’d have to subtract three from the vertex count on theisland to get the correct count for the map on the doughnut!

TIFFANY: Yes, but there’s more. Anything along the cut of the box will beduplicated on the island’s border. So we have to subtract thoseduplications in the island count to get the correct doughnut count.

BRANDON: Whew. This is getting complicated. How do we keep track ofeverything? That one vertex on the box turns into four. Otheredges and vertices turn into two.


TIFFANY: Chill. We can handle it. Let’s be careful with our notation. Let’s letV, E, C denote the doughnut-map data and V ', E', C' the island-map data. First, it’s easy to see that C = C'. Right? That’s easy.No duplications there. Now, suppose there are k edges along theisland’s boundary. Then there are also k vertices along the island’sboundary. All of these are duplicated. So E = E' − k and V = V' − k.

BRANDON: Whoa! Not so fast. What about the vertex that’s been “dupli-cated” four times!

TIFFANY: You’re right. In the count V ' − k, there’s a “vertex” that’s beencounted twice. We really should write V = V ' − k − 1. The finallist should look like this.

C = C', E = E' − k, V = V' − k − 1.

Now, we know that V ' − E'+ C' = 1 because V ', E', C' are data fora map on an island. Thus,

V − E + C = V ' − k − 1 − (E' − k) + C' = V ' − k − 1 − E' + k + C' = V ' − E' + C' − 1 = 0.

Hey, it works!

Lights fade. Tiffany and Brandon shake hands.

3. Your TurnThe surface of a doughnut is a new world for the Acme team. It’s only a matter oftime before the team has a look at related worlds. For example, consider thesurface of a doughnut with a lake.

Suppose you have a map on a doughnut-with-lake and you count V, E, and C.Can you say anything about V − E + C? What can you say about V − E + C for amap on a doughnut with n lakes?

Scene 2

A little later. Brandon, Tiffany, and Joe are working at their desks. Suddenly, Joejumps up and leaps around.

Scene 2 109

JOE: Eureka! I’ve got it!

BRANDON: OK, Archimedes, what have you got?

JOE: I’ve solved the five-terminals problem on the doughnut! (Hopsback to his desk.) Let me show you! (Turns to paper on desk.)Here’s the doughnut before the hole has been put in. So it’sreally a sphere at this point. I’m going to add the hole later. Putfive terminals right in the middle, and number them 1, 2, 3, 4,and 5. Make all the connections for terminals 1, 2, 3, and 4.That’s easy. Now connect 5 to 1, 2, and 3.

You can’t connect 5 to 4 because 4 is inside the large “triangle”formed by the connections 1➔2, 2➔3, 3➔1; 5 is on the outside.What to do? Put the doughnut hole inside one of the smaller“triangles”; then make the connection from 5 to 4 going aroundand under the doughnut and up through the hole:

BRANDON,TIFFANY: Way to go, Joe! (High fives all around.)

4. Your TurnBrandon and Tiffany have a way of looking at the surface of a doughnut that Joedoes not know about, namely, a square with opposite edges identified. Now thatthey know the five-terminal problem can be solved on the surface of a doughnut,they figure that there ought to be a solution “on the square.” Help them out. Hereis a start. (In these diagrams, a single-arrow edge gets glued to the other single-arrow edge. The method is similar for the two double-arrow edges. Edges to beglued must have arrows pointing in the same direction.)


Scene 3

A half hour later. Brandon and Tiffany are alone in the office writing up their solu-tion to the five-terminal problem on the surface of a doughnut. Brandon is annoyedthat Tiffany has just pointed out a possible glitch in their reasoning.

BRANDON: Whaddaya mean we need to prove that V − E + C = 0 on adoughnut? I thought we did that!

TIFFANY: Well, we explained why we were getting V − E + C = 0 for thatparticular kind of map. It might be that V − E + C = 0 is only truefor some maps. We should see if we can show V − E + C = 0 forevery map.

BRANDON: V − E + C = 2 for every map on a sphere. We know V − E + C = 0for some maps on the doughnut. Doesn’t it follow that any othermap would have V − E + C = 0?

TIFFANY: A doughnut isn’t a sphere. Look, I think I have an argument.Want to see it?

BRANDON: If it makes you happy.

TIFFANY: What we did before would probably give us an argument if wewere able to cut the surface open along edges of the map. I don’tthink we can always do that and get something nice, like a squarewith opposite sides identified. Here’s an example of what I mean.

BRANDON: That’s pretty wild.

Scene 3 111

5. Your TurnCut out the doughnut along edges and flatten it out. What do you get?

TIFFANY: OK. Here’s my idea. Start with a map on the doughnut. Supposethat V, E, and C are its data. Look at a circle that goes around thedoughnut’s “waist.”

The circle might go through a vertex or two, and it might coincidewith large segments of the map’s edges at certain instances. However,if you “nudge” the circle a little at those spots, you can get it to avoidthe vertices and intersect edges transversely—at single points—so thata narrow “belt” surrounding the “circle” would look like this.

The idea is to add this closed curve (circle) to the map, along with thenew vertices, edges, and countries it creates. All the new stuff can bedetermined by what happens in the belt. Suppose there are k edgesfrom the original map that the circle crosses. Then as you travel alongthe circle and it crosses an edge, it creates a new vertex, two new edges,and a new country.


So, as you travel along the entire circle, the circle creates k newvertices, 2k new edges, and k new countries. With all of thisadded to the original map, the new data are V ' = V + k, E' = E + 2k,C'= C + k. Moreover, V' − E' + C' = V − E + C. Isn’t that great?

BRANDON: So what’s V − E + C?

TIFFANY: I’m getting to that. Now think of the surface of the doughnut asan inner tube. Cut it open along the added curve-circle.

BRANDON: Isn’t that what we did before?

TIFFANY: This time we’re only going to cut once. So, as I said, cut along theadded curve and open up the inner tube.

We get a cylinder, which we stretch and flatten out.

An island with a lake!

BRANDON: Hey, an island with a lake has V − E + C = 0. Not bad!

TIFFANY: Not so fast! There’s some duplication going on. On the innertube, the vertices and edges along the shore of the island wouldmatch up with the vertices and edges along the shore of the lake.There are k edges and k vertices along the shore of the island.So if the data for the island-with-lake were V ', E', C' and the datafor the inner tube were V′′, E′′, C′′, then V ' = V ′′ − k, E' = E′′ − k,and C' = C′′. Consequently, V ' − E' + C' = V ′′ − k −(E′′ − k) + C′′ =V ′′ − E′′ + C′′ = 0. There you have it!

Tiffany goes to her desk and writes:

Scene 3 113

Theorem about Data from Maps on a Doughnut

Theorem: If V, E, and C denote the number of vertices, edges, and countriesrespectively of an OK map on the surface of a doughnut, then

V − E + C = 0.

Brandon, overwhelmed, puts his head down on his desk. Tiffany has a satisfiedlook on her face. Lights fade. Music rises.

Scene 4

Millie eats the last Krusty Krum while listening to Joe, Brandon, and Tiffany recountthe day’s triumphs.

MILLIE: You mean you solved the five-terminal problem on the surface of a doughnut? (Looks suspiciously at what’s left of herdoughnut.)

JOE: Yeah, we did. Cool, eh? Then, while I was doin’ that, Bran and Tiffwere workin’ on an Euler’s formula for maps on the doughnut.(Boss walks in.)

BOSS: Maps on a doughnut? What’s goin’ on? Too much sugar from those Krusty Krums? I pay good money, and this is how you repay me? (Boss grabs the last Krusty Krum from Millie, takes a bite, and tosses the rest in the trash. Joe is horrified.)

BRANDON: I think I can explain. Do you remember the five-terminal problem Amalgamated wanted us to solve? Well, we did it.It turns out that the surface of a circuit board with a hole in it isjust like the surface of a doughnut. We thought it would be easierto think about doughnuts.

BOSS: (Regaining composure, trying–not quite successfully–to cover uphis outburst.) You solved their problem! The next thing is thesix-terminals problem and then the seven-terminals problem!Go to it, gang! Whaddaya waitin’ fer? And keep away from thosedoughnuts, would ya? (Walks off.)

JOE: Whew! Grrr–oow–chee.

MILLIE: Ya know, there’s a map on the doughnut that needs five colors.

JOE: What? Where did that come from?

MILLIE: I can take your solution to five-terminals and turn it in to a map,and it’ll need five colors.

BRANDON: Do you mean that the map formed by the solution needs five colors?

MILLIE: Nope. You have to change it. Take Joe’s solution and enlarge eachterminal to make a country, then label the terminal countries 1,2, 3, 4, and 5.


Now you enlarge each of the five countries even more along theedges that connect pairs of terminals.

You keep growing them that way until each of the five countriestouches each of the others.

So. Five countries, each country touching the other four. That’sa map that requires five colors.

TIFFANY: What do you do with the rest of the doughnut?

MILLIE: Huh?

BRANDON: (Turns to Tiffany.) As long as you don’t mess with the five coun-tries, you can fill the rest in with countries any way you like. (Turnsto Millie.) That’s amazing! Neat idea!

JOE: And you know what? We could do the same thing if we solved thesix-terminal problem: We could make a map on the doughnut thatrequires six colors! Maybe Boss is onto something. Let’s get to work!

High fives around. Music rises. Lights fade.

Scene 4 115


1. InvestigationNow that Acme has solved the five-terminals problem on the surface of a doughnut,Amalgamated is going to want them to solve the utilities problem. Help the Acmeteam investigate this.

2. InvestigationElated (if that is a possible emotion for Boss) by Acme’s success with the five-terminals problem, Boss thinks that Acme should also solve the six-terminalsproblem and the seven-terminals problem on the surface of the doughnut. Helpthe Acme team investigate these problems.

3. QuestionMillie thinks, “Boss wants us to solve the six-terminals problem on a doughnut.If we could, what would that say about map coloring on the doughnut?” HelpMillie answer her question.

4. InvestigationThe solution to the five-terminals problem has opened up all kinds of possibilities.Joe thinks he ought to revisit the door inspector problem and see if some of thefloor plans that couldn’t be handled before—on the sphere—can now be done on a doughnut. Help Joe investigate the door inspector problem on a doughnut.

5. InvestigationBrandon remembers that any OK map on the sphere having every vertex of orderthree or more must have a country with five or fewer edges. Because that conclu-sion seemed to depend on V − E + C = 2, he wonders what kind of effect V − E +C = 0 would have for the same kind of map on the surface of a doughnut. Is therea restriction on country “sizes” for maps on the doughnut similar to that for mapson the sphere? Help Brandon investigate this issue.

6. InvestigationThe “polyhedron” block-with-the-square hole corresponds to a map on thesurface of a doughnut.


This map has the property that every country has four sides and every vertex isof order four. Investigate all maps on the surface of a doughnut such that everycountry has n edges and every vertex is of order m.

7. QuestionSuppose a map on the surface of a doughnut can be colored in two colors. Whatelse can you say about the map?

8. QuestionSuppose a network is drawn on the surface of a doughnut on which you can takea nice trip starting and returning to a single vertex. What else can you say aboutthe network?

9. QuestionIf a connected network on the sphere has exactly two of its vertices odd and therest even, then you can take a trip from one odd vertex to the other passing overeach edge exactly once. What could you say if such a network were to be drawn onthe surface of a doughnut?

10. QuestionOn the sphere a map made up of a bunch of closed curves, each intersecting withitself or with another in a finite number of points, can be colored in two colors.Is the same true of such a map on the surface of a doughnut?

A

A

E

D

D

C

C

G

G

F

F

H

HI

B

B


11. QuestionTake two distinct points on a simple closed curve and join the two points withanother curve (a chord) that doesn’t intersect the simple closed curve.

A map on the sphere made up of such simple closed curves with a chord can be colored in three colors. Is the same true of such a map on the surface of adoughnut?

12. GameWhen Brandon cut open the inner tube and flattened it out to make a square withopposite sides identified, he got an idea for a new game called inner tube tic-tac-toe (IT4). The board is a three-by-three square with opposite sides “glued,” andthe rules are the same as ordinary tic-tac-toe. The gluing of opposite sides changesthe look of winning configurations. For example, the three X’s on the playingboard below make a winning three-in-a-row.

Find a friend and play several IT4 games. How many really different openingmoves does the first player have? How many really different responses does thesecond player have? Is there a way the first player can play to guarantee that shewins? (What about the second player?)

13. PuzzleHere are two simple closed curves on an inner tube.


When you cut along each one of them, you get different things. In the first case,you will get a single piece, a cylinder. In the second case, you get two pieces.Investigate other closed curves on the surface of an inner tube. Could you find asimple closed curve on an inner tube so that when you cut along it the result willbe a knotted band? Could you find two simple, non-intersecting closed curves onan inner tube so that when you cut along them you get two bands interlocked likethe links of a chain?

14. Something to MakeCut out the patterns below and glue them together to make the three-dimensional object called Csaszár’s polyhedron. The following directions willhelp:

● Cut out two copies of each of the figures from light cardboard about the weight of a manila folder. Fold back on all solid lines, and fold forward(toward you) on all dotted lines. Label one copy as in the illustration and the other with A', B', etc. Cut off the flaps on A' and C' and the dotted flap on D'.

● Turn all pieces over.

● Glue A to A' by the flap 1 on A.

● Glue flap 2 on D' to the base of F.

● Glue the long flap 3 of E to G'.

● Pull triangles C and C' together and glue by the flap on C. (This is not impossible, just tricky!)

● Fold triangles D and D' over C and C' and glue by the flap on D.

● Glue the flap of E to C and the flap of G' to A'.

● Glue the flap of D to F' and the flap of E' to C'.

● Finally, glue the flap of G to A and the flap of E' to G.

Spend time looking over what you have made. What is it? What is its significance?


15. InvestigationAmalgamated is interested in the following general problem in circuit boarddesign. Take two sets of terminals {A1, A2, ..., An} and {B1, B2, ..., Bm}. Connectevery terminal in the first set with every terminal in the second set without havingthe connection cross. The utilities problem is the case where n = m = 3. Investigatethe problem in two cases, with each case on the surface of a doughnut: (1) n = 3and m = 4 and (2) n = m = 4.

16. SummarizingWrite a report to Boss summarizing the Acme team’s accomplishments in this chapter. Include the main ideas, problems, solutions, arguments, and future directions for investigation. Make sure your report is clear and convincing.

17. ConnectingSolving the five-terminals problem and obtaining the formula V − E + C = 0,all on the surface of a doughnut, suggests a range of issues to investigatecomparing two worlds: the surface of the sphere and the surface of the dough-nut, especially with respect to networks and maps. A general line of investiga-tion might be this: If a certain problem can be solved on the sphere, can it besolved on the doughnut? If another problem cannot be solved on the sphere,could it be solved on the doughnut? What is it about a doughnut and a spherethat makes a problem solvable on one and not the other? A specific issue mightbe this: although every map on the sphere can be colored properly in four orfewer colors, Millie showed that the same is not true of maps on the doughnut.Is there a number that will work? If so, what is that number? Trying to answersuch questions gives you ways of getting acquainted with the new world; it alsoprovides new insight on the old one. Given what we all have done, make a listof some general lines of investigation and of questions to ask comparing spherewith doughnut.

Notes

In an earlier chapter, the Acme team showed that the five-terminals problem could not be solved on the surface of the sphere. The argument used wasbased on Euler’s formula for the sphere. An alternative argument goes as follows.Start with five points on the sphere. If you make all the connections among fourof the points, you will get something like the following.

Notes 121

(It may not look exactly like this. But I claim that your connections will bea distortion of this. This claim is easy to see if you start with one connectionand then add the others one at a time.) The connections divide the sphereinto four regions. Each region is a “triangle” with vertices that are three ofthe four points. The fifth point, 5, will lie in one of these four regions.

If point 5 lies in the interior of the triangle whose vertices are points 1, 2,and 3, then point 5 cannot be connected to point 4 without crossing a sideof the triangle. Similarly, if point 5 lies in any of the other three regions,there is a point that cannot be connected to point 5 without crossing thetriangle in whose interior point 5 sits. This argument seems straightforwardenough, but what lies behind it is the following theorem, known as theJordan curve theorem.

Theorem. A simple closed curve in the plane divides the plane into the union ofthree disjoint sets: the curve, the interior of the curve, and the outside of thecurve. Moreover, any path joining a point in the interior with a point in theoutside must intersect the curve. Two points in the interior set can be joined by apath lying entirely in the interior set, similarly for two points in the outside set.


(A similar theorem for the surfaces of the sphere follows easily.) This theoremseems almost self-evident, but a remarkable thing about it is that a proof of it isnot trivial. In fact, many false proofs were given by outstanding mathematiciansbefore an acceptable proof appeared in 1913. The theorem is named after theFrench mathematician Camille Jordan (1838–1922), who stated the theorem andclaimed to have proved it in 1887. One of the reasons for the difficulty of its prooflies in the meaning of simple closed curve and the fact that the proof must coverall possible simple closed curves. For example, it must include the followingsimple closed curve of infinite length.

For a proof of the Jordan curve theorem for simple closed polygonal curves, seeBoltyanskiı and Efremovich. For a general proof, see Newman.

Of course, the Jordan curve theorem is not true on the surface of a doughnut.The following simple closed curves are obvious examples that do not satisfy theconclusions of the theorem.

In a paper of 1811, the Swiss mathematician Simon-Antoine-Jean L’Huilier wasthe first person to note that the formula V − E + C = 2 is not true for a ring-shapedpolyhedron, that is,“it has but a single surface, [but] there is an opening passing rightthrough it.” Moreover, in the paper L’Huilier provides a proof that V − E + C = 0 for such a polyhedron (see Biggs et al., pp. 83–86).

The standard term for the surface of a doughnut, or inner tube, is torus. Thegraphs suggested by Investigation 15 are called the complete bipartite graphs Kn,m

(see Biggs et al., pp. 30 and 142).

12π

{(x,y) : y = x sin 1/x, 0 < x < 1/π} U {(x,y): y = x(1/π − x), 0 < x < 1/π}

1π

Notes 123

References

Beck, A., Bleicher, M.N., and Crowe, D.W. Excursions into mathematics. New York: Worth Publishers, 1970.Biggs, N.L., Lloyd, E.K., and Wilson, R.J., Graph theory: 1736–1936. Oxford: Oxford University Press,

1976.Boltyanskiı, V.G., and Efremovich, V.A. Intuitive combinatorial topology. New York: Springer-Verlag,

2001.Farmer, D.W., and Stanford, T.B. Knots and surfaces: a guide to discovering mathematics. Providence:

American Mathematical Society, 1991: pp. 31–39.Newman, M.H.A. Topology of plane sets. Cambridge: Cambridge University Press, 1951.Weeks, J.R. Exploring the shape of space. Emeryville, CA: Key Curriculum Press, 2001: Chapters 4 and 5.Weeks, J.R. The shape of space, 2nd ed. New York: Marcel Decker, 2002: pp. 13–24.


125

Acme Considers theMöbius Strip

Scene 1

Tiffany, Brandon, Millie, and Joe are having lunch at a local restaurant,Café Cosita. They are having an animated discussion. The small restaurant hasupbeat waitress-proprietors, sisters Sandra and Marcella, who dart back and forth among the tables, greeting clients, taking orders, and giving hugs and kisses.

JOE: Tell me again what Boss said about the airport conveyor belt?

BRANDON: Chill, dude! No shop talk. It’s lunch.

MILLIE: Boss nearly went postal: “Who do they think we are? Some kindof fix-it shop?” Then he stomped around the shop with smokecoming out of his ears.

SANDRA: (Comes up to table.) Hola, chicos. What would you like to drink?(To Brandon) Jamaica. (To Millie.) Horchata. (To Joe.) Iced tea,three lemons. (To Tiffany.) And?

C H A P T E R

7

TIFFANY: Lemonade. (Turns to Joe.) So what set him off?

MILLIE: They got a replacement belt. It’s for luggage so it’s pretty long.A Möbius strip. Has one side so it gets longer use and avoids oneside getting used up when the other is good as new. Well, thereplacement is twice as wide as it should be. The airport thoughtwe could advise them what to do. They were thinking aboutcutting it in half down the middle, but they thought we shouldso some experiments before they mess things up.

SANDRA: (Brings drinks.) There’s a special today, Pescado Ranchero (fishin a special sauce).

BRANDON: Thank goodness. Food. I’ll take the chicken with mole sauce.

JOE: Barbacoa.

MILLIE: Chile relleno.

TIFFANY: Combinación. (To Millie.) What is this Merbious thing?

JOE: Yeah. Waddaya mean by “one side”?

MILLIE: Thought you’d never ask. I brought paper and tape to makesome while we’re waitn’ fer the victuals.

BRANDON: Groan.

MILLIE: (Passes out paper strips and tape to everybody.) You’re going totape the ends of a strip of together so you have a loop.

OK. Now put a twist in the strip before taping. You’ll get this.

That’s a Möbius strip!

BRANDON: A burr-lee-Q dancer, a pip / Named Virginia, could peel in a zip./ But she read science fiction / And died of constriction /Attempting a Möbius strip!

126 Chapter 7 Acme Considers the Möbius Strip

JOE: Huh? Get with it, Brandon. I wanna see the one-sided stuff.

MILLIE: O ye of little faith. Let’s try cutting them in half first. (Hands outscissors to everybody.) Cut along the dotted line!

Lights fade. Music rises.

1. Your TurnMake a Möbius strip and cut it as Millie suggests. What happens? Make sure youdo this before Scene 2 begins.

Scene 2

A few minutes later. Lights up. Music down. Sandra brings the food. The table iscovered with the results of Möbius strip cutting. Sandra has to find a place to put theplates.

SANDRA: What a mess! You sure you wanna eat? ‘Cause we’ve gotcustomers outside waitin’ ta get in.

BRANDON: Yes, we want to eat! Pardon my friends; they always have to bringtheir work with them.

MILLIE: I think someone needs a nap.

TIFFANY: Look what I got! I didn’t get two Möbius strips. I got one! Halfas wide, twice as long!

BRANDON: That’s not what I got. I got two interlocking loops!

Scene 2 127

JOE: You weren’t paying attention, Brandon. You put two twists in thestrip before you taped it together.

BRANDON: No, I didn’t. I twisted it all the way around, once. Isn’t that whatI was supposed to do?

TIFFANY: OK, OK. Calm down everybody. Actually, it’s not supposed to betwisted all the way around. Maybe Millie should have called it ahalf-twist. Here.

Hey, that gives me an idea! What would happen if we madestrips with different numbers of twists—or rather half-twists—in them and then cut them down the middle? Maybe we’d learnsomething.

BRANDON: (Lighting up for the first time.) I have another idea! We’ve beencutting them down the middle, in half. Why don’t we try cuttingthem in thirds?

MILLIE: We could do both! We could make any number of half-twists,and cut the strip in half, in thirds, or in fourths. Let’s make atable and keep track.

THIS NOT THIS


Twisted Strips

No. of Half-Twists No. of Divisions Results

1 2 (halves) 1 strip twice as long

2 2 2 strips, same size, linked

1 3 (thirds) ?

2 3

1 4 (fourths)

3 2

3 3

4 2

TIFFANY: Let’s each take a different one and try it. I’ll start with three half-twists cut in half.

BRANDON: I’ll do a half-twist cut in thirds!

JOE: Make sure it’s a half-twist, pal.

BRANDON: Blah, blah, blah. (Everybody makes strips and starts cutting.Sandra comes up to the table with a mixed expression: amused,appalled, and intrigued.)

SANDRA: I can’t believe this!

MILLIE: You can do it, too! Grab some scissors! Here’s a strip! Hey, getyour sister, too.

(Sandra starts cutting and joins them.)

SANDRA: Marcela, come on out here. (Marcela joins them. The group cutsaway. Sounds of “Wow!” and “Lookit that!” break the concentra-tion. Other customers are getting agitated.)

CUSTOMER 1: Hey, miss. Could we have our check?

CUSTOMER 2: Sandra, where’s our food?

CUSTOMER 3: Marcela, when do you think we could get a table?

Sandra and Marcela, lost in cutting twisted strips, ignore the questions. Acme’steam is also oblivious to the outraged customers as the small café overflows with thechaos of paper loops and twisted strips. Lights fade. Music comes up.

2. Your TurnCarry out the investigation suggested by the table above. Fill in the blanks. Lookfor patterns. Extend the table by considering cases not listed. Make predictions.(e.g., if I made a strip with five half-twists and three divisions, then such and suchwould happen.)

Scene 3

Twenty minutes later. Millie, Joe, Tiffany and Brandon are walking into the shop,still animated from lunch.

BRANDON: Boy, that was a real bomb! I’d like to understand what’s goingon. You cut a Möbius strip in half down the middle and you get a single strip, twice as long. (Gestures in the air with invisiblescissors and paper.)

Scene 3 129

TIFFANY: Do you suppose it has anything to do with being one-sided?

JOE: I don’t know, but I think it’s time for somebody to explain thisone-sided bit. Millie?

MILLIE: OK, everybody. Make a strip. (They all go their desks and tape astrip together.) Joe, you use my nail polish. Brandon and Tiffany,you get a felt-tip pens. Ready? Start somewhere and paint thestrip, lengthwise, until you meet up with where you started.

3. Your TurnMake a Möbius strip and “paint” it as Millie suggests. What happens? Do thisbefore continuing with the story!

JOE: Wow. You’re right. You paint along the strip until you get towhere you started. Without turning it over, you’ve painted theother side! What a blast!

MILLIE: There ain’t no other side. It’s all one side. That’s why the airport usesit for a conveyor belt. The belt gets used up evenly and lasts longer.

JOE: Like burning a candle at both ends?

MILLIE: Slowly. That way you get to use the whole candle. No stubs.

BRANDON: You know something else? Not only does the Möbius strip haveone side, it’s got only one edge.

JOE, MILLIE,TIFFANY: One edge?

BRANDON: Yeah. Look at the strip with zero half-twists. (Don’t we alwaysforget to consider special cases?) It’s a cylinder. It’s got two edges.Remember? It’s really an island with a lake. There’s the edge of theisland itself, and then there’s the edge of the lake. Two edges.


MILLIE: So what about the one-edged Möbius strip?

BRANDON: Just like you had us paint it, Millie. Take a pen. Start at a pointon the edge and run the pen along the edge until you get back towhere you started. (Everybody tries this.)

4. Your TurnJoin the crowd. Make a Möbius strip and follow Brandon’s instructions. Whathappens? Do this before returning to the story.

While the others are running their pens around the Möbius strip, Brandon makesanother strip with two half-twists and uses the pen on it just as he did on the one half-twist strip.

5. Your TurnMake a two-half-twist strip and do what Brandon is doing: counting edges.What happens? Do this before returning to the story.

BRANDON: Hey, gang, know what? A two-half-twist strip has two edges. Youknow, maybe we ought to check the number of edges of thatdouble-long strip we got when we cut the Möbius strip down themiddle. Maybe we’ll learn something. If it’s got more than oneedge, it ain’t a Möbius. (Brandon writes.)

TIFFANY: I’m going to add a line to our table. (She writes.)

Facts about Twisted Strips

● A Möbius strip (half-twist strip) has one side and one edge.

● A cylinder (two-half-twist strip) has two sides and two edges.

Scene 3 131

Twisted Strips

No. of Half-Twists No. Divisions Results

0 2 2 strips, same size, unlinked;each one like the original

1 2 (halves) 1 strip twice as long

2 2 2 strips, same size, linked

1 3 (thirds) ?

BRANDON: Good idea, Tiff. Let’s count edges for the items in the Resultscolumn and see what we get. I’ll take the second line—the singlehalf-twist cut in half. Joe, you take the third line. Millie, you…

The Acme team zeros in on the new task as lights dim and music comes up.

6. Your TurnCarry out the investigation begun by the Acme team. Look at the items describedin the right column of your table. For each line, there is an original strip and stripsresulting from cutting the original. Figure out how many edges the original striphas and how many edges each resulting strip possesses. Add this new informationto the right column of your table. Look for patterns, and make predictions.

7. Your TurnThe Acme team has come across the following items in its investigations.

For each figure, out how many sides it has (one or two) and how many edges it has.

Scene 4

Three hours later. Things have calmed down since lunch. Brandon and Tiffany areworking at their desks. Millie is painting on eye shadow at the reception desk. Joe isstaring out the window. Tiffany turns to Brandon and breaks the silence.

TIFFANY: You remember the dressmaker’s pattern for an inner tube?

BRANDON: Dressmaker’s pattern?

TIFFANY: Here.


You glue the top to the bottom to form a cylinder. Then stretch toglue the two ends of the cylinder together to make an inner tube.

BRANDON: I remember. Why?

TIFFANY: You can use the same idea for a Möbius strip, as well as for othertwisted strips. A dressmaker’s pattern for a Möbius strip is

When you glue, you have to make the arrows match so go in thesame direction. That forces you to make the half-twist.

BRANDON: What about the top and bottom of the rectangle? There aren’tany arrows.

TIFFANY: They don’t get glued together like they were for the inner tube.

BRANDON: Hey, if you draw the arrows in the same direction—both goingup—then you get a dressmaker’s pattern for a plain, old cylinder!

TIFFANY: Totally. Well, I thought the dressmaker’s pattern might help explainwhat’s going on when we cut the Möbius strip down the middle.

Scene 4 133

BRANDON: How?

TIFFANY: Instead of assembling it from the pattern and then cutting itdown the middle, you could cut the pattern first and thenassemble.

BRANDON: But the results might be different!

TIFFANY: True. But we might learn something. Let me show you what I’vefound out. Start with a pattern. Cut it.

Next, glue. But how? We’ve got to be careful to put the two piecestogether as they were supposed to be originally. I came up withthis scheme.

The part labeled A on one side gets glued to the part labeled A onthe other side. And B gets glued to B. (I’ve numbered the two piecesjust to keep track of things.) And when you glue, the arrows have tobe going in the same direction. OK.You cut and get these two strips.

Now we’re ready. Use the labels and arrows to glue themtogether. Wait. Can’t glue yet. Got to flip piece 2 over so thearrows match. Now we can glue!


BRANDON: It’s twice as long as the original; it’s half as wide! And, it’s apattern for a cylinder! I told ya, that’s not what we got when wedid things in the right order.

TIFFANY: I thought the same thing. But what’s a pattern for a two-half-twiststrip? And what’s a pattern for a four-half-twist strip?

BRANDON: Huh? They’re the same! And they’re the same as the pattern fora cylinder. All strips with an even number of half-twists have thesame pattern!

TIFFANY: Exactly. Cutting the pattern and then assembling doesn’t tell youeverything, but it tells you something. (She writes.)

BRANDON: You know, the pattern for a Möbius strip is the same as thepattern for a three-half-twist strip and a five-half-twist strip; anystrip with an odd number of half-twists. So your little trick tellsyou more. (Brandon writes.)

Cutting a Strip with an Odd Number of Half-Twists in HalfCut a strip with an odd number of half-twists down the middle. The result isa strip half as wide and twice as long with an even number of half-twists.

Cutting a Möbius Strip in HalfIf you cut a Möbius strip in half down the middle, then the result is a strip halfas wide and twice as long with an even number of half-twists.

Scene 4 135

Tiff, you’re a genius! I think we can use this idea to explain a lot ofwhat we were seeing with twisted strips. (Brandon writes again.)


Patterns for Twisted Strips

● A pattern for a strip with an odd number of half-twists is

● A pattern for a strip with an even number of half-twists is

TIFFANY: You know, it might be obvious, but I think we can use this ideato tell us definitively how many edges a twisted strip will have.The edges of the strip come from the top and bottom edges ofthe pattern, which then get glued together.

In each case, point P gets glued to point P, point Q to point Q.Here’s what happens to each.

(Tiffany writes.)

Scene 4 137

Number of Edges of a Twisted Strip

● A strip with an even number of half-twists has two edges.

● A strip with an odd number of half-twists has one edge.

BRANDON: That’s baaad! Wait’ll Joe and Millie hear about all of this! We’llneed their help to check things out when we cut twisted strips inthirds, fourths, and...

Lights dim. Music comes up. The two approach Joe and Millie. Curtain fallswhile the four are engaged in animated conversation. Brandon andTiffany gesture wildly while Joe and Millie alternate between eyesscrunched up and eyes open wide as they move back and forth betweenincomprehension and understanding.


1. Gathering EvidenceBrandon and Tiffany have come up with ideas suggesting similarities—at least asfar as cutting goes—among all strips with an even number of twists and amongall strips with an odd number of twists. These ideas are new, and the Acme teamwould like confirming evidence. Help them out by completing the following table.Use the evidence you already have from cutting twisted strips.

What do you observe?

2. InvestigationBrandon and Tiffany have found out what happens when they cut a strip with anodd number of half-twists in half. Use their new ideas to continue the investiga-tion. Some first steps would be to figure out what happens in the following cases:

● Cut a strip with an even number of half-twists in half.

● Cut a strip with an odd number of half-twists in thirds.

● Cut a strip with an even number of half-twists in thirds.

● Cut a strip with an odd number of half-twists in fourths.

● Cut a strip with an even number of half-twists in fourths.

● Cut a strip with an odd number of half-twists in fifths.

Cutting a Twisted Strip: How Many Pieces Do You Get?

No. Half-Twists in No. Pieces Resulting When Cut In

Original Halves Thirds fourths

0 2 3 4

1 1

2

3

4


● Cut a strip with an even number of half-twists in fifths.

In each case you will want say how many pieces result and describe each one:its size (compared with the original) and whether it has an odd or an even numberof half-twists.

3. GeneralizingHelp the Acme team generalize the results of Investigation 2. Figure out whathappens in the following two cases.

● Cut a strip with an even number of half-twists into nths.

● Cut a strip with an odd number of half-twists into nths.

In both cases you will want say how many pieces result and describe each piece:its size (compared to the original) and whether it has an odd or an even numberof half-twists.

4. QuestionThe Acme team suspects that every strip with an even number of half-twists istwo-sided and that every strip with an odd number of half-twists is one-sided.Can you provide them with a simple argument showing that this is always so?

5. Gathering EvidenceHere are more surfaces. For each one, decide how many edges it has and whetherit is one- or two-sided.


6. InvestigationTo test Acme’s new methods for analyzing twisted strips using patterns, considerthe following dressmaker’s patterns.

● As they stand, the patterns are incomplete. Each can be made complete by placingarrows on the labeled edges. This can be done in different ways. What are they?(Of course, you want the different ways to be really different.) For each really different labeling, decide the following for the glued-together

object:

● How many sides it has

● How many edges it has

● Cut each of the objects along the dotted lines shown. Do this in two ways:(1) after gluing, and (2) before gluing. Describe the results.

7. InvestigationBrandon likes Tiffany’s simple explanation for the fact that strips with an oddnumber of half-twists have a single edge and strips with an even number of half-twists have two edges. But Brandon would like to know what the edge (or edges)would look like for a specific strip. He begins his investigation with Tiffany’s idea,taking the rectangle that will become the twisted strip and looking at the twoedges that will become the edge (or edges) of the twisted strip.


Then, without gluing, he puts twists in the strip and looks just at these twoedges. Here is what he gets in the first few cases.

Then he glues the ends together. In the cases above, here is what happened tothe original edges of the rectangle.

Brandon called these edge diagrams. Based on these edge diagrams, Brandonwrote the following.

A D B C A DBC

ZERO TWISTS ONE TWISTS TWO TWISTS THREE TWISTS


Edges of Twisted Strips

● The edges of a strip with zero half-twists are two unlinked, unknotted closedcurves.

● The edge of a strip with one half-twist is an unknotted closed curve.

● The edges of strip with two half-twists are two unknotted closed curves,linked.

● The edge of a strip with three half-twists is a knotted closed curve.

Help Brandon out with this investigation by drawing the edge diagrams for stripswith four, five, and six half-twists. Describe their features. Generalize by drawing a picture of the edge diagram for a strip with n half-twists and describe its features.

8. InvestigationBrandon, Tiffany, and you (see Investigations 2 and 3) were able to gain and explaina lot of information about the results of cutting twisted strips by ignoring the exactnumber of twists in the original strip, while retaining only that the strip had an evenor an odd number of half-twists. This simplification enables you to count the result-ing number of strips, determine the size of each, and say whether each has an evenor odd number of half-twists. However, a lot more seems to be going on. Sometimesa strip may be knotted, and other strips may be linked in some way. Brandon thinksthe edge diagrams defined in Investigation 7 can be used to provide information ofthis kind when a twisted strip is cut in half down the middle.

You can put in as many twists as you like. The result will look similar to one inthe first set of diagrams in Investigation 7, and when you glue the ends together,you will get something similar to one in the second set of diagrams. Thus, youcould say that the result of cutting a strip with three half-twists down the middleis one knotted strip. Use this idea to complete the following table and, from thetable, make predictions about what would happen with more than six half-twists.


Cutting Twisted Strips Down the MiddleNo. Half-Twists Knotting/Linking of Resulting Strip/Strips

0

1

2

3 One knotted strip

4

5

6

9. InvestigationInvestigate the results of cutting a strip with n half-twists in thirds. Use the tech-nique of Investigation 8 to describe how the resulting strips are knotted andlinked. Make a table similar to that in Investigation 8 to display your observations.

10. Things To MakeTake two strips of paper of the same size and place one on top of the other.

Put one twist in both strips simultaneously. Tape the ends together, A to A andB to B, to get a kind of double Möbius strip. Look carefully at the result. What canyou say about this? Do the same with three strips: Lay them on top of one another,twist them simultaneously, and tape the ends together. What happens?

Generalize: Take n strips, place on top of each other, put m twists in the wholebunch simultaneously, and then tape. What happens? Does any of this haveanything to do with cutting twisted strips?

11. Something To Make: Möbius ShortsTake a T-shaped piece of paper with a long stem. Bend the top of the T to make aring (not twisted) and glue A to B. Pass C upward through the ring, turn C down(without twisting), and glue C to the outside of the ring at the place where A andB are glued.


How many sides does the surface have? How many edges? Guess what happensif you cut both the ring and what was originally the stem along their midlines.Then go ahead and cut both the ring and the original stem and see if you can putthe result back together again (see Boas).

12. GameLooking at a pattern for a Möbius strip reminded Brandon of the game inner tubetic-tac-toe which is played on a pattern for an inner tube. He thought: Why notMöbius tic-tac-toe (MT3)? The board is a three-by-three square with one pair ofopposite sides “glued” (arrows in opposite directions), and the rules are otherwisethe same as ordinary tic-tac-toe. As with inner tube tic-tac-toe, the gluing of thepair of opposite sides changes the look of winning configurations. On the playingboard below, what would be winning moves for X?

Find a friend and play a several MT3 games. Is there a way the first player canplay to guarantee that he wins? (What about the second player?)

13. InvestigationImagine the Möbius strip has no thickness and that you are a creature that lives“in” the surface. Investigate the possibility of solving, on the Möbius strip, prob-lems that Acme considered for the surface of the sphere and torus: the utilitiesproblem and the five cities problem. One way of visualizing these problems is toassume that, when you draw a dot or a line, the ink soaks through to the other“side.” Another way is to assume the strip is made of transparent material so thatwhat you draw on one side shows up on the other.

14. QuestionA Möbius strip is a surface with one side and one edge. An island is a surface withtwo sides and two edges. Can you find a surface that has one side and two edges?


15. QuestionThe original question for the Acme team was the following: Can you cut theMöbius strip down the middle to get a Möbius strip half the width? You know theanswer is no. However, is there a way to cut it so that the result is a Möbius striphalf its original width? There might be many ways. What’s the best? You can throwsome of the original material away, but you do not want to make any cuts that youwill have to glue back together. Such cutting and regluing would weaken theresulting strip.

16. Something To MakeMake a photocopy of the design below. Cut it out, fold along the dotted line withthe design on the outside, and glue the two halves together on the inside to forma long strip of double thickness paper with the writing on the outside. Put a half-twist in the strip and tape the ends together. Pull the printed strip through yourfingers to read the message. Enjoy!

17. SummarizingBrandon, Tiffany, Joe, and Millie have given Tiffany the job of producing a summaryof the day’s investigations to be delivered to Boss. Help her out. Include statementsof problems, terminology, results, explanations, and directions for future investiga-tions. Of course, Boss will expect the summary to be clear, organized, and under-standable. He would also like to know that their time was well spent.

Notes

A mathematician confidedThat a Möbius strip is one-sided.You’ll get quite a laughIf you cut it in half,For it stays in one piece when divided.

ONCE UPON A TIME THEREWAS A STORY THAT BEGAN

Notes 145

Image reprinted from Lost in the Funhouse by John Barth.

In 1858 two mathematicians, Augustus Ferdinand Möbius (German,1790–1868) and Johann Benedict Listing (Czech, 1808–1882), independentlydiscovered the Möbius strip and the fact that it is a one-sided surface. We encoun-tered Listing in Chapter 2 (see the Notes) in reference to taking nice trips onnetworks. We will encounter both again as our story unfolds. Both mathemati-cians were students of Gauss at Göttingen. In 1816, Möbius was appointedExtraordinary Professor of Astronomy in Leipzig, where he stayed for the rest ofhis life. In an 1840 lecture, Möbius presented the following puzzle:

There was once a king with five sons. In his will he stated that after his deaththe sons should divide the kingdom into five regions in such a way that theboundary of each one should have frontier line in common with each of the otherfour. Can the terms of the will be satisfied?

Because of this occurrence, many mathematicians apparently and erroneouslycredited Möbius with the four-color conjecture 12 years before Guthrie proposedit. For more on Möbius and the German mathematical world of his time, see thebook edited by Fauvel et al. (see also Kline, p. 1165).

Möbius strips have entered our culture in many ways. They appear as objects ofart, as in Maurits Escher’s woodcut, Möbius Band I, showing a Möbius strip cutdown the middle (See the book by Escher, picture 41). Möbius bands occur inliterature, as in the short stories “No-Sided Professor,” “A. Botts and the MoebiusStrip,” and “A Subway Named Moebius” collected by Clifton Fadiman in hisFantasia Mathematica. See also above Something to Make 16.


M.C. Escher’s “Möbius Strip I” © 2006 The M.C. Escher Company-Holland. All rights reserved.www.mcescher.com

Magicians have exploited properties of twisted strips in a cloth-tearing trickcalled “Afghan Bands” and described in Martin Gardner’s Mathematics, Magic andMystery (p. 70f).

A number of patents have been obtained for industrial devices whose underly-ing idea is the Möbius strip: a tape recorder with a twisted tape recorded on both“sides” that runs twice as long as it would otherwise, a conveyer belt designed towear equally on “both” sides, an abrasive belt, a self-cleaning filter belt for a dry-cleaning machine (the dirt can be easily washed from both “sides” as the belt goesround), a conveyer of hot material, and a nonreactive resistor.

For more about Möbius strips, see chapter 9 of Martin Gardner’s MathematicalMagic Show from which much of the material presented here has been drawn.Also see the book by Pickover.

A recent interest in Möbius strips and other twisted strips comes from chemistryand molecular biology. This is partly because a DNA molecule often resembles astrip with an even number of half-twists. Sometimes such molecules are knotted,and some are linked with others. Scientists would like to identify the possible waysthis can happen and determine how each possibility affects chemical and biolog-ical processes of DNA, such as recombination (see the book by Sumners, as wellas Chapter 13 and Investigation 8 of the present chapter).

Chemists have been studying Möbius ladders, molecules similar to the DNAmolecules above. In a Möbius ladder, the surface of the strip is replaced by rungsof carbon-carbon double bonds and the edge is a polyether chain of carbon andoxygen molecules (see the book by Flapan).

We have used the word “edge” in describing twisted strips and other surfaces,such as in “the edge of an island.” The commonly used word is “boundary.” Thusan island with lake would have two boundaries, one the shore line of the islanditself and the other the shore line of the lake.

The limerick Brandon quotes (“A burlesque dancer...”) is by Cyril Kornbluth andis titled “The Unfortunate Topologist” (see Fadiman, p. 266.)

References

Barth, J. Lost in the funhouse. New York: Doubleday, 1968.Boas Jr., R.P. Möbius shorts. Mathematics Magazine, April 1995, p. 127.Escher, M. The graphic work of M. C. Escher. New York: Hawthorne Books, 1960.Fadiman, C. Fantasia mathematica. New York: Simon and Schuster, 1958.Fauvel, J., Flood, R., and Wilson, R., editors. Möbius and his band. Oxford: Oxford University Press, 1993.Flapan, E. When topology meets chemistry: a topological look at molecular chirality. Washington,

DC: Mathematical Association of America, 2000.Gardner, M. Mathematical magic show. New York: Alfred A Knopf, 1977.Gardner, M. Mathematical puzzles and diversions. New York: Simon and Schuster, 1959.Gardner, M. Mathematics, magic and mystery. New York: Dover, 1956.Kline, M. Mathematical thought from ancient to modern times. New York: Oxford University Press, 1972.Pickover, C.A. The Möbius strip: Dr. August Möbius’s marvelous band in mathematics, games, literature,

art, technology, and cosmology. New York: Thunder’s Mountain Press, 2006.Sumners, D.L., et al., editors. New scientific applications of geometry and topology. Proceedings of

Symposia in Applied Mathematics, vol. 45. Providence, RI: American Mathematical Society, 1992.

References 147

Acme Creates New Worlds:Klein Bottles and OtherSurfaces

149

Scene 1

Several days later. The team is busy at work. Boss is just getting off the telephone.He looks a little perplexed.

BOSS: Well, this is a new one. It was someone from Play More; they makevideo games. They heard that Acme has been doing things with innertubes and “Merr-bee-us’’ strips (whatever they are) that might turninto video games. They want us to come up with new stuff….Videogames? (Boss rolls his eyes then looks around to room to see if anyone is paying attention. They all ignore Boss and work more intently onwhatever it was they were working on.) OK, anyone wanta tell mewhat’s goin’ on?

JOE: I’ll try, Boss. Didja ever play airplane videos?

C H A P T E R

8

BOSS: Airplane videos?

JOE: You shoot down enemy planes. The planes move in straight linesacross the screen. When a plane reaches one side, it doesn’t just disap-pear but comes out the other side.

BOSS: Other side?

JOE: Let me show you. Here’s an airplane. It gets to an edge of the screenand appears on the other side.

BOSS: OK. I get it. What’s this have to do with inner tubes?

JOE: Look at the screen here. It’s like a point here on the right edge ofthe screen is the same as the corresponding point on the left edge ofthe screen.

BOSS: The same?

JOE: For the airplanes.

BOSS: Oh.

JOE: So the whole right edge of the screen is really the same as the wholeleft edge, at least for the airplanes.

And the same is true of the top and bottom of the screen. The leftedge is the right edge, and the top is the bottom.

150 Chapter 8 Acme Creates New Worlds

So the world of the airplane is really the surface of an inner tube.

BOSS: Whoa! Not so fast. Where do you get the inner tube? And “worldof the airplane’’ is a little spacey.

JOE: The airplane really lives in the screen. Since the left edge of thescreen is really the right edge, you can pick the edges up and gluethem together.

BOSS: Wait a minute. You can’t do that. The airplane would bend!

JOE: Imagine that the airplane is flexible like a flat decal. OK? Now itsworld is a cylinder. But the top is really the bottom, so you gluethem together.

BOSS: No you don’t. You’ve gone too far. You’re distorting things. Howcan you do that?

JOE: Boss, this is not the real world. Imagine the cylinder is made ofsomething you can stretch and shrink, like rubber. You glue theends together and get an inner tube!

BOSS: Oh.

JOE: Where have you been, Boss? We’ve been distorting things all along.

BOSS: Yes, distorting my brain. Maybe you guys can figure out what PlayMore wants. Don’t bother me. Right now, my head hurts. (Walks offholding his head. Once Boss leaves, the others come out of “hiding.’’)

Scene 1 151

BRANDON: Poor Boss. Good job, Joe. Wow! Video games. What do yousuppose they mean by “new stuff?’’

TIFFANY: I think we ought to have a look at what we have already.“Worlds’’ for airplanes to fly in.

There. Inner tube, cylinder, and Möbius strip. Hmm. Howwould the Möbius strip work?

MILLIE: Plane comes to the right edge, flips over, comes out the left edge.

JOE: What happens when the plane hits the top of the Möbius strip?

MILLIE: Disappears. Goes right off the edge of the world.

JOE: Neat. Hey, same with the cylinder. Wonder if Play More would likethat. What if we combined Möbius world with inner tube world?

BRANDON: Huh?

JOE: Plane goes to right edge, appears on left edge flipped over. Planegoes to top, appears on bottom flipped over. Planes wouldn’tdisappear. Here’s the scheme.

BRANDON: Wait. There’s another combination; it’s kinda mixed. Plane arrivesat left edge, then appears on right flipped over. Plane arrives attop, appears at bottom just as in inner tube world.

JOE: Cool.


MILLIE: I sorta like planes goin’ off the edge of the world. That’s wherethe monsters are.

BRANDON: Here’s the pattern for my combination.

MILLIE: OK, smarty pants, what does it look like when you put it together?

BRANDON: Huh?

TIFFANY: Yeah, like the inner tube for the original video game. It would giveus a model of the plane’s world. Tell us what you get, Brandon.

BRANDON: Uhh…

MILLIE: Glue top to bottom, get a cylinder…just like we did for an innertube. Then glue one end of the cylinder to the other.

BRANDON: Yeah, that’s right. Put a twist in the cylinder before you glue.

MILLIE: OK, smart one, would you be so kind as to tell me just how youexecute that twist? If you line up the two ends of the cylinderthis way, then the arrows match.

But if you put the two ends together, just about to be glued, thearrows don’t match.

Scene 1 153

Words of wisdom, Mighty One?

BRANDON: Unnhh…

JOE: I have an idea, but it’s way out. You guys might not like it.

MILLIE: Out with it. Brandon seems to have lost his voice.

JOE: Hold on to your hats. It involves pushing the surface throughitself.

MILLIE: Ouch.

JOE: If you accept that, it’s really pretty simple. You start out withMillie’s first picture of the cylinder. Grab the top end and pushit through the side near the other end.

MILLIE: Ouch, again.

JOE: Now the arrows of the two ends of the cylinder are lined up.Glue them together.

Brandon, are you there? (Brandon has his head on his desk withhis eyes closed.)


TIFFANY: Awesome! It looks like some kind of bottle. You know, I don’tthink we have a choice. We have to do something strange, justlike Joe did. Whatever it is, the thing has to be one-sided.

MILLIE: One-sided?

TIFFANY: Totally. When we started, we glued the top to the bottom tomake a cylinder. But there was another choice. If we glue left toright first, we get a Möbius strip and that has only one side.

If we were able to put the thing together like an inner tube or asphere, without any boundaries and without some trick likeJoe’s, we’d get something with an inside and an outside. A two-sided thing. And that would be impossible: a thing one-sidedand two-sided. No good.

BRANDON: (Revives. Runs around the room holding his head. The others arestunned, not so much at Brandon, but at Tiffany’s statement.)Ahhhggghhh…


1. Your TurnIf Tiffany is right, then you ought to be able to show the path of a bug starting “inside’’the bottle, crawling around, and winding up on the bottle’s “outside.’’ Can you?

2. Your TurnBefore they started being creative, the Acme team began with three video screens,ones that could be assembled, respectively, into an inner tube, a cylinder, and aMöbius strip. Right now the team is looking at other patterns in hopes of comingup with new video games and of understanding their new “bottle.’’ Here are someof the patterns. Help them assemble them.

Scene 1 155

Scene 2

An hour later. Calm has returned to the Acme shop. Tiffany, Millie, Joe, andBrandon—especially Brandon—are at work “assembling’’ other patterns. Brandonlooks up from his desk, glances around the room, and smiles a big smile.

BRANDON: Hey, gang. I think I’ve got something new. I took Joe’s originalidea, the one where the planes flip over going vertically andhorizontally.

I started by gluing left to right. Got a Möbius strip.

Couldn’t figure out what to do next, even with Tiff ’s new trick.So I thought of another approach. I pushed the interior of thesquare out and stretched it into a hemisphere. The boundary ofthe square becomes the hemisphere’s equator. (I’m using lettersand arrows to label sides.)

Now I’ve got to do some gluing. I want to put A and A togetherwith arrows matching. First, fold up the hemisphere a bit, like aclam shell. Then pinch it together at points labeled X, the pointsat the ends of the A arrows, and the beginnings of the B arrow.Those two points are really supposed to be glued together even-tually anyway.


You can see how the arrows for the B’s are going in the samedirection. Glue them together.

MILLIE: Know what? You’ve got yerself a Möbius strip! The two A’s makeup its edge. The edge is a circle in a plane.

JOE: Oh-ma-gosh! That’s right. Nice goin’. The edge wasn’t flatbefore.

BRANDON: OK. Next, we have to glue the two B’s together. The arrows arepointing in the same direction, but we have to do somethingalong the lines of Joe’s trick: we have to make the surface passthrough itself. Here it is. Ta da!

MILLIE: It still hurts.

TIFFANY: That’s really weird. Looks like a beehive. Doesn’t even resemblethe weird bottle.

The team stares at the diagram in disbelief. They try different angles ofviewing. They shrug their shoulders and scratch their heads. They go back totheir desks. Music comes up. Lights fade.

X

X

X

XA B

AAB

AB

BB

AA

XX XA

B

B

A

A

B

B

A

X B

X X

X

A

AB

X X

XX X

Scene 2 157

3. Your TurnThe Acme team decided it might give them some insight if they didn’t limit them-selves to a square screen. They thought looking at other polygonal screens—evena di-gon!—would be useful. Help the team assemble the following patterns. (Theteam has decided to label every edge, with the convention that only labels thatoccur twice get glued.)

Scene 3

A half hour later. Members of the team are back at their desks. This time it’s Joe wholooks up with the big smile.

JOE: This stuff gets curiouser and curiouser. Let me show you whatI found. I was working on this triangular pattern.


I remembered that when we were cutting up a Möbius strip wedecided to see what would happen if we cut the strip first and thenglued. We found out a lot when we did that. Well, I thought,“Whynot try the same thing here?’’ I cut the triangle in half down themiddle. I figured I’d eventually have to glue the new edges whereI cut back together. So I did some labeling and relabeling.

Then I glued A to A and look what I got!

MILLIE: Well, I’ll be hornswoggled: a Möbius strip!

TIFFANY: Know what else? I was looking at Joe’s pattern and thinking ofdoing what Brandon did: push the insides back, stretch, andturn it into a hemisphere with the triangle’s boundary becomingthe equator.

Split each A side up into two halves: first X and then Y.

AA

B

Scene 3 159

Now, except for B, the part that doesn’t get glued, you have justwhat Brandon had.

BRANDON: Yeah, and since the two ends of B are identical, you can glue themtogether. Then you’ll have something really close to what I had.

TIFFANY: Yes. Now put the thing together, just like you did, Brandon.

There’s the beehive again. And this time it’s got a lake! You knowwhat that means? Joe’s triangular pattern was really a pattern-in-disguise for a Möbius strip. It’s also a pattern for a beehive-with-lake. Sooo…in some sense, a beehive-with-lake is the sameas a Möbius strip.

BRANDON: In the same sense that a strip with three half-twists is the sameas a strip with one half-twist?

TIFFANY: Totally.

MILLIE: Tiff, I tried to do the same thing you did with Joe’s trianglepattern, but this time I started with the bottle pattern. I wantedto see if I’d get something else. I cut it down the middle andadded new labels and glued A to A. Here’s what happened.

YX

B

X

Y


BRANDON: Wow. Stretch that a bit, you get a square. But what the heck is it?

TIFFANY: Hey! Cut the square down the diagonal. Look what you get.

MILLIE: Hell’s-a-poppin’! Two beehives-with-lake!

JOE: Or two Möbius strips!

BRANDON: Know what that means, gang? Take two beehives-with-lake. Gluethem together along the shorelines of their lakes. Whaddaya get?Our “bottle!’’ We may not know much about the bottle or aboutthe beehive, but we do know they’re related. All the weird stufflives in the same house. But, wait, we do know more—and this’llknock your socks off—our bottle is two Möbius strips gluedtogether, the edge of one glued to the edge of the other. Can youbelieve that? (He writes.)

C C

CC

C

B

B

BB

BB

A A A

A

A

Scene 3 161

While Brandon writes, Joe starts to clap. Millie stamps her feet. The others pick itup and dance about. They “do-si-do their partners.’’ They try an “allemand left’’around the room as square-dance music comes up and lights fade.


1. PuzzleMake a Möbius strip, the simple one with one half-twist. Figure out how youwould cut it out so that it would flatten out into the following pattern.

2. PuzzleThe following hexagon is a pattern for some surface.


Strange Facts about Patterns and SurfacesPattern Surface

One-sided bottle

Beehive

Alternate pattern for Möbius strip (a.k.a. beehive-with-lake)

Alternate pattern for two-sided bottle (a.k.a. twobeehives-with-lake glued together along the shorelines of their lakes; a.k.a. two Möbius bands glued together along their edges)

We’re going to try to figure out what it is by cutting it up, labeling along thenewly cut edges, and then gluing the pieces back together another way. Cut alongthe dotted lines A and B. Label the freshly cut lines with labels A, B, and arrows asshown. You’ll get three pieces. Glue the two D’s together, then the two E’s, andfinally the two C’s. What do you get? What do you conclude?

3. InvestigationHere is an interesting surface.

We’re going to cut it open, label cuts, flatten it out, stretch, and bend it to see ifit has a pattern that is the same as some familiar object.

C D

D C

EE

B

A

Investigations, Questions, Puzzles and More 163

It’s an inner tube with a lake. Analyze the following surfaces in the same way.For each you should obtain a pattern. Examine the pattern carefully to see if itcan be stretched and bent into a pattern for something you have seen before.If this is not successful, cut up the pattern, label the cuts, and tape together atsome other place.

untwist

untwist once

untwistagain

stretchit out

glue Bback

together

A

A

A

A

AA

A

A

A

A A

A A

A

A

BB

B B

B

B

B

B

B

B


4. InvestigationConsider the surfaces in Gathering Evidence 5 in Chapter 7. For each one, find apattern, just as you did for the surfaces in Investigation 3 above: cut the patternout, label cuts, flatten it, stretch, and bend it to get something you recognize.

5. Gathering Evidence: Möbius ShortsThis is the surface you made for Something to Make 11 in Chapter 7. What is it?

6. Gathering EvidenceMake a double strip in the form of a cross as shown below. The idea is to glue theends of the cross as indicated. However, the arrows have been left out. Find allthe ways to put arrows on the labels so that you get really different surfaces.


Then make them. (Did you have any trouble putting them together?) Find namesfor these surfaces based on what you know.

7. Pattern RecognitionHere are patterns for surfaces. Cut, label, tape, bend, and stretch each one to see ifit is a pattern for a familiar object.

8. InvestigationHere are patterns for two surfaces. Describe each one as best you can.

A

A

C C

E

ED

D F

FB

B

C C

A

B

A

C C

A

B

A

B C

B

A

CB

C

C

B

A

A

B

C

C

B

A A

C

D

D

C A

B

B

A

C D

A

B B

A

F E

A

C D

A

B B

A

F E


We’ve heard a rumor that these two patterns are patterns for the “same’’ surface.Is it true?

9. Something To MakeHere is a pattern for something that can actually be put together! Here are thesteps.

● Use a copy machine to enlarge (about 150%) the diagram on heavy paper.

● Score along all the fold lines, and cut the enlarged figure out carefully.

● Fold away from you along the solid lines.

● Fold toward you along the broken lines.

● Glue the four tabs to join edges: A and A; B and B; C and C; D and D.

When you get done, look it over and see if you can find a familiar name forthe object.

10. Something To MakeHere are patterns to assemble to make a surface. The steps for each are

● Use a copy machine to enlarge (about 150%) the pattern on heavy paper.

● Score along all the fold lines, and cut the enlarged figure out carefully.



● Glue together the central tube using the two smaller patterns.

● Fold the large cube into shape (do not glue yet); put the central tube in posi-tion and glue it into place.

● Glue sides flaps of the cube. (Do not glue the lid down; it is more interestingto be able to look inside.)

GLU

E GLU

E

GLU

E

GLU

E

A B

D

CD

C

A B


When you get done, look it over and see if you can find a familiar name forthe object. (This model is based on ones from the books by Barr and by Jenkinsand Wild.)

GLU

EG

LUE

GLU

EG

LUE

GLU

EG

LUE

GLU

EG

LUE

GLUEGLUE


11. A GameLooking at a pattern for a Klein bottle reminded Brandon of the games inner tubetic-tac-toe and Möbius tic-tac-toe and led him to think of Klein bottle tic-tac-toe(KBT3). The board is a three-by-three square with pairs of opposite sides identi-fied (arrows in same direction from left to right, in opposite directions from topto bottom), and the rules are the same as those in ordinary tic-tac-toe. As with theother games of tic-tac-toe, gluing of pairs of opposite sides changes the look of

GLUE

GLUE

GLUE

GLUE

CUTCUT

CUT

LID


winning configurations. On each of the playing boards below for KBT3, whatwould be winning moves for X?

Find a friend and play several KBT3 games. Is there a way the first player canplay to guarantee that she wins? What about the second player? (This gameappears in the book by Weeks.)

12. InvestigationAn airplane on the “one-sided’’ bottle video screen moves off the left edge ofthe screen and appears on the right, flipped over. It moves off the top edge of thescreen and appears on the bottom, not flipped over. In both cases it is as if theairplane lives “in’’ the surface of the bottle so that the surface has no thickness.Imagine that you live “in’’ the surface of the bottle, and investigate the possibilityof solving problems there that Acme considered for the surface of the sphere andtorus: the utilities problem and the five-cities problem. Use the video screen withidentified edges as your model, much in the way you did with the torus. See YourTurn 4 in Chapter 6.

13. ExperimentImagine that you are a creature living “in’’ a Möbius strip. (For ways of thinkingabout this, look at Investigation 12 above and Investigation 13 in Chapter 7.) Youare setting up street signs in your world. Streets on “the strip’’ consist of MainStreet—going right down the middle—and cross streets (First, Second, Third, …,Tenth) perpendicular to Main Street.

Starting at First and Main and heading toward Second Street, you put up thesign West First on the left side and East First on the right side. You continue toSecond and do the same thing: on the left side the sign is West Second, and on the

2nd A

v.

10th A

v.

MAIN ST.

1st A

v.

XX XX XX

O

O O

O

OO


right it is East Second. You keep going up the street, installing signs at each inter-section. Eventually you pass the corner of Tenth and Main and arrive again at thecorner of First and Main. What happens? Carry out this experiment in two ways,one way on a paper Möbius strip, the other on a video screen Möbius strip. Whatdo you conclude?

14. QuestionHere is a picture of the Klein bottle as “put together’’ by Joe. Draw a closed path onthis Klein bottle so that if you cut along it you would get two Möbius strips.

15. QuestionHere is another way to “put together’’ the Klein bottle. It involves having thesurface intersect itself in another way.

Draw a closed path on this version Klein bottle so that if you cut along it youwould get two Möbius strips.

16. Something To Make● Use a copy machine to enlarge the following pattern (at least 150%) on heavy

paper.

● Score along all the fold lines, and cut out the enlarged figure carefully.




● Glue (or tape) tabs A, B, E′, G, A′, B′, E, G′.

● Glue tabs J, J′, C, D, C′, D′.

● Finally, glue tabs F, F′, H, H′. (These last couple of steps may take some care-ful maneuvering. Have patience!)

What familiar surface does this model represent?

A

A'

B

B'

C

C'

E

E' D

D'

F

G

J

H

F'

G'

H'

J'

CUT

CUTCUT

CUT


17. SummarizingThis time Joe has the job of producing a summary of the day’s investigations to bedelivered to Boss. Help him out. Include relevant diagrams and statements ofproblems, definitions, results, explanations, and directions for future investiga-tions. Of course, Boss will expect the summary to be clear, organized, and under-standable. He would also like to know that the team’s time was well-spent.

Notes

Three jolly sailors from Blaydon-on-TyneThey went to sea in a bottle by Klein.Since the sea was entirely inside the hullThe scenery seen was exceedingly dull.

– Frederick Winsor, from The Space Child’s Mother Goose

A mathematician named KleinThought the Möbius strip was divine.Said he: “If you glueThe edges of two,You’ll get a weird bottle like mine.”

–Anonymous

The bee-hive surface, which the Acme team discovered, is usually called across-cap, or the projective plane. The latter needs a little explanation. As theEuclidean plane is the place where the points and lines of Euclidean geometrylive, so the projective plane is the place where the points and lines of projectivegeometry live. In projective geometry, the usual axioms of geometry hold exceptfor the parallel axiom, in which two lines always intersect. A model for projectivegeometry might be the surface of a sphere in which the lines are great circles. Buta problem with this model is that a pair of great circles intersect in two antipodalpoints. To eliminate this difficulty, you take the sphere and identify (glue!)antipodal points. Consider this new object. A pair of antipodal points of thesphere becomes one point of the new object. A great circle of the sphere becomesa “line’’ of the new object, and the new object becomes a good model for a projec-tive plane. To find out what this object is, start with a sphere and identify thepoints of the northern hemisphere with their antipodal counterparts in thesouthern hemisphere. (Save the equator for later.) One way to see how to do thisis to slice the sphere along the equator, and then turn the northern hemisphereinside out so that it now sits snugly inside the southern hemisphere. Now, keep-ing the southern hemisphere rigid, rotate the northern hemisphere inside it 180°.Antipodal points are now aligned, and the northern hemisphere can be glued tothe southern hemisphere.

Notes 173

To complete the job, all we need to do is glue diametrically opposed points onthe equator. That is just what the Acme team did when it put together the cross-cap/beehive.

Experiment 13 above can be captured by the following “trip.’’ Place a circle withan arrow on a point on the Möbius strip. This specifies what is meant by clock-wise at that point; it defines an “orientation’’ there.


Travel in the strip—lengthwise—accompanied by the little circle-with-arrowuntil you get to the point where you started. Remember you are “in’’ the strip. Thelittle circle-with-arrow now indicates counterclockwise instead of clockwise as atthe start. Your original orientation has reversed itself. As a creature living on theMöbius strip, you can have no notion of clockwise/counterclockwise or ofleft/right. However, if you were a creature on a torus or sphere and you took yourclockwise circle-with-arrow on a trip along a closed path, every time you returnto the starting point, the arrow at the end points in the same direction as thearrow at the beginning of the trip. Thus the sphere and the torus are calledorientable. Because there are closed paths on the Möbius strip and Klein bottlealong which the orientation changes, these surfaces are called nonorientable.

In 1874 German mathematician Felix Klein (1849–1925) and Swiss mathemati-cian Ludwig Schläfli were the first to investigate the projective plane as a surfaceand to show that it is nonorientable (see Pont, p. 123). Klein was a professor ofmathematics at Göttingen, where Möbius had been a student of Gauss.He presented the world with the one-sided bottle, usally called the Klein bottle,that appeared in this chapter in 1882 (see Kline, pp. 1167–1168).

The cross-cap has two-fold rotational symmetry when viewed from the top. In1903 the mathematician W. Boy assembled the pattern into an object, called Boy’ssurface, having three-fold symmetry (see Hilbert and Cohn-Vossen, pp. 317–321and Lietzmann, pp. 151–153). (For help in visualizing the Klein bottle and crosscap,visit the website www.math.arizona.edu/~ ura/013/bethard.steven/crosscap.mov)

References

Barr, S. Experiments in topology. New York: Thomas Y. Crowell Company, 1964.Hilbert, D., and Cohn-Vossen, S. Geometry and the imagination. New York: Chelsea, 1956.Jenkins, G., and Wild, A. Mathematical curiosities 1. Norfolk, England: Tarquin Publications, 1989.Kline, M. Mathematical thought from ancient to modern times. New York: Oxford University Press,

1972.Lietzmann, W. Visual topology. New York: American Elsevier, 1969.Pont, J.C. La Topologie Algébrique des Origines à Poincaré. Paris: Presses Universitaires de France, 1974.Weeks, J.R. Exploring the shape of space. Emeryville, CA: Key Curriculum Press, 2001.

References 175

Acme Makes Order Out ofChaos: Surface Sums andEuler Numbers

177

Scene 1

A few days later. The Acme team is at the University’s Robotics Lab. Everybody,including Boss, is dressed to the nines. They might get a new assignment! ProfessorGomfrina Demasiadas Palabras has just finished giving the group a tour of the labo-ratory. Now she is explaining how Acme might be of use to her.

PROFESSOR: As you can see, we are designing robots to carry out specifictasks. Of course, we are looking for efficient and effective meth-ods of controlling the robots. We would like to develop sometheoretical tools for dealing with this in general. Consequently,we have decided to focus some of our attention on some verysimple robots in the hopes that they might shed some light onour problems. We are studying robots that operate in a singleplane. They’re made up of rods, connector pivots that link up

C H A P T E R

9

E E

E

the rods, and fixed pivots. A fixed pivot attaches one end of arod to the plane but otherwise allows it to move freely in theplane. A connector pivot also allows free movement of the rodsit connects. We call such a system a linkage. Before I go anyfurther, let me show you a very basic example. This may giveyou some idea of where all of you might come in. Are there anyquestions at this point? Yes, Mr. Boss?

BOSS: I’d like to say that my staff is very accomplished and versatile.They can color maps and design tours. They are experts on theMer-bee-us strips and other video games, such as the Kleinbottle.

PROFESSOR: Yes, thank you. Actually, I have heard of your work on videogames. I think the example I’m about to show you will explainwhy I brought you here.

BOSS: We’re thinking of changing our name to Acme Video and Maps.

PROFESSOR: I see. (Gives a long squint at Boss.) Well, here’s my example. Thislinkage has just two rods. At the end of the first rod, there is afixed pivot. Between the first and second rods, there is aconnecting pivot.

As you can see, the entire linkage can rotate freely about thefixed pivot, and the second rod can rotate freely about theconnecting pivot.

BRANDON: Can the second rod pass over the first?

PROFESSOR: I was just going to mention that. These are kind of idealizedrods. Infinitely thin and thick.

Linkage #1

FREE PIVOT

FIXED PIVOT

178 Chapter 9 Acme Makes Order Out of Chaos

BOSS: Infinitely…?

TIFFANY: Shhh…

1. Your TurnCut out pieces of a manila folder in the sizes shown below. Use a paper punch topunch holes in the spots indicated, and assemble the linkage using brass brads.Keeping the end of the long piece of paper fixed, manipulate the linkage to get afeeling for the set of possible positions of the linkage.

PROFESSOR: We can control the movement of the linkage by moving acursor on a computer screen. We call it a control screen. I thinkit behaves in a fashion similar to one of the video games I’veheard you’ve worked on. Here’s how the control screen works.(She plops a transparency on the overhead.)

5''

BRADS

3 4''

Scene 1 179

Each rod makes an angle to the horizontal. We impose a coor-dinate system on the screen so that the lower left-hand corneris the origin, the bottom of the screen is the positive x-axis, andthe left edge of the screen is the positive y-axis. When the cursoris at the point (s, t), the angle of the first rod to the horizontalis s and the angle of the second is t. Thus, as the cursor movesfrom left to right along the bottom of the screen, the angle thefirst rod makes to the horizontal increases smoothly to 360°.The first coordinate parameterizes the angle of the first rod tothe horizontal.

BOSS: Pair-am-uh-tries…?

MILLIE: (In a stage whisper.) Quiet, Boss.

PROFESSOR: (Squints again at Boss. Tiffany raises her hand.) Yes, señorita.

TIFFANY: What happens to the second rod?

PROFESSOR: You mean, as the cursor is moved to the right, no? The secondrod remains parallel to the horizontal. Similarly, if the cursormoves from the lower left corner up the left side of the screen,the angle the second rod makes to the horizontal increasessmoothly to 360°. At the same time, the first rod lies parallel to

Control ScreenLinkage

Linkage/Control Screen Coordination

A

B

A

360

360

270

270

180

180

90

900

0

B

ANGLE A (DEGREES)

AN

GLE

B (

DE

GR

EE

S)


the horizontal line. Consequently, the second coordinate para-meterizes the angle of the second rod. So, for example, if thecursor is at (90, 30), then the first rod is perpendicular to thehorizontal and the second makes an angle of 30° to the hori-zontal. (She writes on a transparency.)

BOSS: I get it! You move the cursor to (s, t) and the linkage moves overs and up t. Ingenious!

PROFESSOR: (Squints again, but otherwise ignores the comment from Boss.)

BRANDON: Suppose you move the cursor to the top of the screen and thesecond rod has rotated 360°. What if you want to keep rotating it?

PROFESSOR: I think that’s where you fit in. It’s like your video game. If thecursor is at the top of the screen, the second rod has rotated360°. It’s back where it started. The point on the top of thescreen produces the same position of the linkage as the corre-sponding point at the bottom.

MILLIE: So if you want to do negative angles, you start at the bottom ofthe screen and come out the top going down.

BOSS: Huh?

PROFESSOR: Similarly, if the cursor hits the right of the screen, then the firstrod has rotated 360° from the horizontal. It’s back where it started.

t

360˚

(t, 360)

(t, 0)

Control Screen

(90, 30)

Linkage

30˚

Scene 1 181

The point at the right edge and the point directly to its left on theleft edge both correspond to the same position of the linkage.

JOE: If you look at it backwards and spun the first rod, keepingthe second rod at the same angle to the horizontal, you’d see thecursor going from left to right across the screen. Each timethe cursor hit the right edge, it would reappear at the left again.Hey! That’s just like our video game with airplanes.

PROFESSOR: !Si! The right edge is identified with the left edge. The top isidentified with the bottom.

MILLIE: It’s a doughnut!

PROFESSOR: I believe the correct terminology is torus.

MILLIE: Whatever.

PROFESSOR: What’s important is that the torus smoothly parameterizes thepositions of the linkage: to every point on the torus, therecorresponds a position of the linkage, and conversely, to everyposition of the linkage, there corresponds a point on the torus.

BOSS: Con-verse-lee? You mean the linkage is a torus?

PROFESSOR: In a sense, yes. The torus captures most of what we want toknow about the linkage. There is something more. I used theword “smoothly.’’ What I meant is, moving the cursor smoothlyaround the screen produces a smooth motion of the linkage.There are no lurches or bumps.

2. Your TurnImagine that the cursor makes a path on the screen’s diagonal from lower left toupper right. What would the motion of Linkage 1 be in order to create this pathof the cursor? What would the motion of Linkage 1 be in order for the cursor tomake a path along the other diagonal, from upper left to lower right?

PROFESSOR: Well, that’s my first example. Now, I’d like to show you a linkagethat’s a little bit more complicated. We’ve been able to unravelsome information about a control screen, but not everything.


We think you might be able to help us with this. This time thereare four arms, three free pivots, and two fixed pivots.

Let me describe to you what we know about the control screen.We thought that we might try to control the angles as we didwith the first linkage. But that would give us four numbers,and we wanted something two-dimensional. Besides, wethought there might be complicated relationships between theangles. We want something simple, no? The goal is to describea set of points corresponding to the positions (or states) of thelinkage. We chose to look at the central free pivot p and describeits possible positions. Instead of telling you what we came upwith, I thought I’d let you find out for yourselves. This willgive you a feeling for the linkage and what I’ll have to say later.You’ll find paper versions in your packet to help in yourexploration.

Lights fade. Music comes up as Professor Gomfrina Demasiadas Palabrasleaves the podium and Acme team members play with the linkages.

2. Your TurnJoin the Acme team in exploring Linkage 2. Cut out pieces of a manila folder inthe sizes shown below. Use a paper punch to punch holes in the spots indicated,assemble the linkage using brass brads. Holding the long bar fixed on your table,manipulate the linkage to see if you can get some idea of the set of points thatcorrespond to possible positions of the pivot p. (As with Linkage 1, assume thatrods can pass over one another easily.)

10''

6''

3''

3''6''

Linkage #2

Scene 1 183

Scene 2

Fifteen minutes later. Professor Demasiadas Palabras has returned to the podium.The Acme team is still investigating the linkage.

PROFESSOR: (Looks around at the group.) Does anyone have any insight toshare? Yes. Please come up. (Tiffany comes up to the podium.)

TIFFANY: First let me label the linkage: P is the central free pivot, Q andR are the other free pivots, S and T are the fixed pivots.

I started with the rods set up like an isosceles triangle with baseof 10 inches and two sides of 9 inches each. So nine inches is thefarthest P can be from T and from S. The position of P has tolie inside two circles, each of radius 9 inches, one with center Sand the other with center T.

P

Q R

TS

10''

9'' 9''

TS 10''

9'' 9''

P

Q

R

TS

7'' 4'' 11''

8 1 2''


I thought of these circles as “farthest’’ circles. So I asked myself:How close could P be to T (or S)? Instead of extending PRT asbefore, I decided to bend it as much as I could, so P wouldactually lie on top of the rod RT. This is the closest P could beto T. So P must lie outside a circle with a radius 3 of inches withcenter T. Similarly, P would have to lie outside the same sizecircle with center S. These are the “closest’’ circles. Now youknow P is inside both of the farthest circles and outside both ofthe closest circles. Other than these four circles, I don’t thinkthere are any more constraints. Here’s the picture:

The curvy hexagon is the set of positions of the point P.

PROFESSOR: ¡Muy bien! (Tiffany returns to her seat.) Thank you very much.It doesn’t quite look like a video screen, but it’s two-dimensional.And every point of the “hexagon’’ corresponds to a position ofpoint P. But there’s more to the linkage than the position ofpoint P, no? In fact, every position of P corresponds to fourdifferent states of the linkage, depending on how the secondaryelbows are bent. The left elbow could be bent up or down; theright elbow could be bent up or down. Here’s an example.

P

Q

R

R

TS

QP

Q

R

TS

10''

9''

3''

FARTHEST CIRCLES

POSSIBLEPOSITIONS FOR P

CLOSESTCIRCLE

CLOSESTCIRCLE

Scene 2 185

To cover these possibilities, make four copies of the hexagonand label them UU, UD, DU, and DD. The first letter describesthe state of the left elbow (up or down), the second the state ofthe right one. So the state of the linkage in the example justshown corresponds to a point in hexagon UD with the leftelbow up and the right elbow down.

JOE: So you wind up with four screens.

PROFESSOR: Yes, but that’s not the full story. It’s more interesting than that.The four hexagons are connected. Each edge of a hexagon getsidentified with the edge of another hexagon—a sort of wildvariation of identifying opposite edges of the single screen forour first linkage.

MILLIE: You mean you glue the four screens together.

PROFESSOR: That’s sort of the idea. Let me explain the gluing exactly.To start, a point along the top right edge (TR) of hexagon UU isthe same as the corresponding point on the top right edge ofhexagon DU. These are points where the left elbow is fullyextended and not bent so that “crossing’’ this edge from UU toDU (or DU to UU) smoothly switches the left elbow from upto down (or down to up, respectively).

Similarly, a point along the top left edge (TL) of UU is the same as the corresponding point on the top left edge of UD.

UUDU

UU UD DD DU


Crossing edges TR, BR, and ML amounts to switching the left elbowsmoothly from up to down (and vice versa). Thus, when two hexa-gons are labeled with the same second letter, the correspondingedges labeled TR, BR, and ML get identified. Call them TR

U, BRU, and

MLU when the second letter is U and TR

D, BRD, and ML

D when thesecond letter is D. Crossing the remaining three sides of the hexagon—TL, BL, and MR—amounts to switching the right elbowfrom up to down and back. Combining all this information, weobtain four hexagons having their edges labeled like this:

BRANDON: Wow. So when you glue them, that will be your control screen.What do you get?

PROFESSOR: That was what I was hoping you could answer. Are you willingto take on the task?

MILLIE: Whaddaya say, Boss? (Nudges Boss, who has been dozing.)

BOSS: Huh? Where are we?

MILLIE: (Whispers.) Shh. The Professor has a job for us. Should we take it?

BOSS: (Straightens up.) Of course we’ll do the job! Very interestingproblem! Right up our alley! We’ll get on it right away.

Lights fade. Music up. Professor winces as Boss shakes her hand. The Acmeteams says its good-byes and leaves the room. Professor puts away her notes.

4. Your TurnHelp the Acme team. Put together the four hexagons as best you can. See if youcan figure out what the assembled object should be.

Scene 3

Back at Acme headquarters, Brandon, Millie, Joe, and Tiffany are having a strategysession on how to attack the Professor’s problem. Boss is in the corner figuring outhow much money this is going to bring in.

TLU

BRU

MRU

MLU

BLU

UU

TLD TR

UTR

U

BRU

MRD

MLU

BLD

DU

TLD TR

D

BRD

MRD

MLD

BLD

DD

TLU TR

D

BRD

MRU

MLD

BLU

UD

Scene 3 187

JOE: I glued the hexagons together side by side. That gave me asingle shape, with 18 edges left unglued. Then I stretchedthe whole thing into an 18-gon, like an 18-sided video screen!

BRANDON: That’s an answer to the Professor’s question all right. But Idoubt that’s what she wants. Does anybody have a clue aboutwhat it is? When you put it together completely?

TIFFANY: Looks like something we’ve seen. Sort of a doughnut, a Kleinbottle. But so much bigger! Those were rectangles, and this isan 18-gon.

MILLIE: We did look at hexagons and octagons.

BRANDON: Do we have a clue as to what we’re looking for?

JOE: Yeah. Something we recognize.

TIFFANY: How can we recognize it when we haven’t done any 18-gons?

MILLIE: We could cut it up and reglue ‘til we see something familiar.

TIFFANY: As if! Look. I think we need to be systematic. Let’s start with adefinition. What are all these things we’ve been looking at lately?

BRANDON: They’re all surfaces. They can all be put together from patterns.

TIFFANY: Hold on. Let’s start with surface. What’s a surface?

JOE: How about something you can draw a map on? We do maps,right?

TIFFANY: OK. Good start.

Now, can we connect this with something that comes from apattern?

MILLIE: Well, you can always draw a map on a pattern. Include theborder of the pattern as part of the map’s edges.

Definition of SurfaceA surface S is an object on which you can draw a map; that is, you can expressthe set S as a union of vertices, edges, and countries of an OK map.

TLU TR

U

BRU

MRU

MLU

MLU

BLU

UU

TRD TL

U

BLU

MRD

BRD

UD

TLD

TRD

BRD

MLD

BLD

DD

TRU

TLD

BLDBR

U

DU

TLU

TRU

BRU

MLU ML

U

BLU

TRD

TLU

BLU

BRD

TLD

TRD

BRD

BLD

TRU

TLD

BLD

BRU


BRANDON: What about the other way?

MILLIE: What other way?

BRANDON: Start with a surface. Something you can draw a map on. Doesit come from a pattern?

JOE: You know, the Professor’s four hexagons, they’re just countriesin some map on some surface, far, far away. In another galaxy!We just don’t know what it is.

TIFFANY: Hey, great idea! Work it backward. Take a surface. It’s made upof the edges, vertices, and countries of some map. Cut out allthe countries. (Don’t forget to label all the edges with lettersand arrows ahead of time.) Flatten them out and straighten outthe edges. You’ll get a bunch of polygons.

JOE: Cool. Then start gluing them back together. But not completely.Like we did with the hexagons. Keep everything in the plane,and you’ll wind up with a pattern!

A

B

C

D

E

I

F

F

G

G

II

H

D

III

E

IV

G

E

F

F

G

H

D

D

E

A

B

C

DE

I

F C

G

II A

F

H

DIIIE G

B

IV

A B

B A

CC

Scene 3 189

Joe writes.

BRANDON: I’ve been thinking about what we were doing with the Kleinbottle and the cross-cap. Remember when we found out thatthe Klein bottle was two Möbius strips glued together alongtheir boundaries? Well, we also found out that a Möbius stripis a cross-cap-with-lake. So a Klein bottle is just two cross-caps-with-lake glued together along the shorelines of the lakes.

Why not do the same thing with any two surfaces? Add a laketo each one, and then glue together along shorelines.

MILLIE: That could make a rabbit hug a hound! We can call it the“Brandon’’ of the two surfaces!

BRANDON: Let’s just call it the “sum.’’

MILLIE: That’s about as exciting as watching grass grow. (Brandonwrites.)

Theorem Connecting Surfaces with PatternsGiven a surface, you can cut it open and flatten it out to get a pattern.Conversely, given a pattern, it can be assembled into a surface.


TIFFANY: This is awesome. I think we might be able to detect a sumsometimes just by looking at a pattern. In fact, we did that withthe Klein bottle and the cross-caps. One pattern for the Kleinbottle we cut up into two cross-caps-with-lake.

That’s how we discovered the connection between Klein andcross-cap. Let’s go backward again. Take a pattern for surface S.Then it’s easy to make a pattern for S-with-lake.

F

G

HS

E

F

G

H

X

E

G H

X

E

F

A

A

B

B

C A

A

CB

B

C

Surface SumLet S and T be surfaces. Let S′ = S-with-lake and T′ = T-with-lake.Glue S′ and T′ together along the shorelines of the two lakes. The resultingsurface is called the sum of S and T and is denoted S#T.

Scene 3 191

If you have a pattern for S and one for T, it’s easy to find apattern for S#T. (She writes.)

JOE: OK. Nice. So how does all this help with our problem?

BRANDON: Two things. First, we can build more complicated surfaces. Likewe can put two doughnuts together.

E E

E


Pattern for Surface SumSuppose patterns for surfaces S and T are as follows:

So patterns for S-with-lake and T-with-lake are

Then a pattern for S#T is

C Y

A

G

BZ

GF W

D

EX

CB

A

G

Y Z

D XW

E F

C

A

BZ

Y F

D

EX

W

JOE: A two-holed doughnut!

MILLIE: We can add another doughnut to that.

A three-holed doughnut!

BRANDON: See what I mean? It enlarges our repertoire of surfaces. It alsomeans we can go in the other direction and “break down’’ asurface. The Klein bottle is the sum of two crosscaps: K = C#C.This breaks the Klein bottle down into simpler surfaces. Maybewe could do that with the four-hexagon surface: break it down.(Brandon writes.)

JOE: Well, let’s go! Let’s try this idea out on some of the patternswe’ve seen.

Fade.

5. Your TurnGo back to the investigations, questions, etc., of Chapter 8, look at the patternsthere, and see if you recognize them as sums of two or more surfaces.

Scene 4

An hour later. The four are working intently at their desks. Tiffany leans back, smiles,gives a thumbs-up, and turns to address the others.

TIFFANY: I have a great idea. We’re map people, right? A surface issomething you can draw a map on. That gives us some data.We can count V, E, and C like we did before. We can then

Strategy for Analyzing a SurfaceTo understand a surface S, see if you can make it the sum of two simplersurfaces. You can try this with the pattern: see if you can recognize thepattern as being a pattern for the sum of two surfaces.

Scene 4 193

calculate V − E + C. This is a number. For the sphere it’s alwaystwo, and for the torus it’s always zero. If V − E + C isn’t two orzero, then we know it’s not a sphere or a torus. That would bea start.

MILLIE: I don’t think we know what V − E + C is for a Klein bottle, oreven for a Möbius strip.

BRANDON: We don’t even know that the Klein bottle has a number.

MILLIE: Huh?

BRANDON: We know that V − E + C = 2 for any OK map on the sphere. Fora Klein bottle, V − E + C might change if you change maps.It could happen!

MILLIE: Show me!

TIFFANY: Calm down, you guys! We’re new at this stuff. Let’s back off abit and start small. We can build from there using the sum idea.If you started with surfaces—like the sphere and torus—forwhich V − E + C is the same no matter the map, maybe thesurface sum would have the same property. I think we need adefinition.

Now, suppose S and T are two surfaces with Euler numbersN(S) and N(T).

MILLIE: What if a surface doesn’t have an Euler number?

TIFFANY: Well, then it doesn’t. Right now I’m only considering surfacesthat do. S and T do. So, take a map on S#T. If we were to countV, E, and C, would there be some way we could predict whatV −E + C is if we knew N(S) and N(T)?

Euler NumberSuppose that S is a surface and that, for every OK map on the surface, V − E + Cis always the same. Then call that common number the Euler number of thesurface. Denote the Euler number by N(S). So N(sphere) = 2 and N(torus) = 0.


JOE: I think we did something like this for the torus. We cut it openalong its “waist,’’ flattened it out into an island-with-lake, andthen used the Euler number for that.

MILLIE: Well, S#T has a cute little waist.

BRANDON: I remember we drew a circle around the waist and added tothe map the new vertices, edges, and countries it created.We showed that doing that didn’t change V − E + C. We couldtry the same thing with S#T.

TIFFANY: So we add the circle. The same argument as before says V − E + Cstays the same. Now cut along the circle. We get S-with-lake and

T

S

T

S

Scene 4 195

T-with-lake with maps having data VS, ES, CS and VT, ET, CT,respectively.

Also, VS − ES + CS = N(S) − 1, and VT − ET + CT = N(T) − 1.Now, V − E + C = VS − ES + CS + VT − ET + CT, except that thevertices and edges along the added circle have been countedtwice. We need to subtract those off from the V and E counts.

But, look! The circle has the same number of vertices asedges. We’d be subtracting n (vertices) and adding n (edges) toV − E + C. They cancel each other out! So our equation holds:

V − E + C = VS − ES + CS + VT − ET + CT.

(Tiffany writes.)

Euler Number of a Surface SumIf S and T are two surfaces having Euler numbers N(S) and N(T), then thesurface sum has an Euler number given by the formula

N(S#T) = N(S) + N(T) − 2.

S

S

T


MILLIE: So if S is a torus and T is a torus, S#T is a two-holed doughnutand it has Euler number −2. If that don’t beat all.

BRANDON: Well, we have our work cut out for us, figuring out the Eulernumbers for a bunch of things: a Klein bottle (if it has one), ann-holed doughnut, and many other things we can put togetherwith the surface sum.

JOE: Before we do anything, I was thinking. That four-hexagonthing that goes with the linkage? It’s really a map on whateversurface it is. Each of the hexagons is a country. So C = 4. Eachedge of a hexagon is an edge of the map, but each gets gluedto another edge. So E = (6 × 4)/2 = 12. Vertices are trickier.Each hexagon has six vertices, so there are at least six vertices.But I claim there’re only six. Corresponding vertices are thesame. Stack the four hexagons on top of each other so thatrunning a pin through the four of them would correspond tothe same position of the central pivot. Four vertices that lie ontop of each other are really just one vertex—think about it!Those four vertices from the four hexagons correspond to thesame position of the linkage. That means that the surface hasonly six vertices. So, V − E + C = 6 − 12 + 4 = −2. That’s newinformation! That number is also the Euler number of a two-holed doughnut. Now that’s the cat’s meow! Could we assem-ble that four-hexagon thing into a two-holed doughnut?We’ve made an amazing start. But there’s work ahead so let’sget going!

High fives all around. Lights fade. Music comes up.


1. Gathering EvidenceThe Acme team would like to begin recognizing surfaces, as well as how they arebuilt up from simpler surfaces by the surface sum, just by looking at theirpatterns. Given that T is a torus and C a cross-cap, help them find patterns for thefollowing surfaces: T#T#T, C#C#C, T#T#C, and T#C.

2. QuestionMembers of the Acme team wonder what can be said about R#Sphere where R isa surface. Help them out!


3. QuestionSuppose that S is a surface and that S′ is gotten by removing a disc from S. (ThusS′ is S-with-lake.) The Acme team wants to know the answers to the followingquestions:

● If S has an Euler number, does S′? If the answer is yes, are the two numbersrelated? How?

● If S′ has an Euler number, does S? If the answer is yes, are the two numbersrelated? How?

● What about S-with-n-lakes?

4. InvestigationIf T is a torus, members of the Acme team know that N(T) = 0 and N(T#T) = −2.They’d like to know what N(Tn) is where Tn = T#…#T (n times). Help them.Of course, whatever you find out, they would like an explanation.

5. InvestigationThe Acme team knows that the torus and sphere have Euler numbers and thatN(sphere) = 2 and N(torus) = 0. The team would like to know the answers to thefollowing questions.

● Does the Möbius strip have an Euler number? If it does, what is it?

● Does the cross-cap have an Euler number? If it does, what is it?

● Does the Klein bottle have Euler number? If it does, what is it?

● If Cn = C#…#C (n times), does Cn have an Euler number? If it does, what is it?

Help them out by investigating these problems. Of course, they will wantexplanations for everything.

6. QuestionIf M is the Möbius strip, how would you describe M#M? Does M#M have an Eulernumber? If so, what is N(M#M)? (See Investigation 5.)

7. QuestionSuppose that K is a Klein bottle, C a cross-cap, and T a torus. Rumor has it thatK#C is the same as T#C (see Investigation 8 from Chapter 8). If K and C have


Euler numbers, then knowing that the rumor is true and knowing the formulafor the Euler number of a surface sum ought to enable you to find N(C). Whatwould N(C) be? Does your answer jibe with what other investigations haverevealed?

8. QuestionThe Acme team came up with a method for taking a pattern for surfaces A and Band coming up with a pattern for A#B. This method is flawed in some cases. Canyou see when it works and when it does not?

9. InvestigationHere are the first few items in an infinite sequence of surfaces. Your task is to iden-tify as many of these as you can. Ultimately you would like to identify all of them.

10. QuestionThe Acme team feels that the Euler number of a surface might be useful inidentifying it—given that no other information about the surface is available.Suppose you have a mystery surface P about which you have two clues: you knowit has an Euler number, and you know N(P). Furthermore, there is a surface Qyou do know completely, and it turns out that Q has an Euler number and thatN(P) = N(Q). Are P and Q the same surface?

11. InvestigationAlthough we cannot assemble the cross-cap C and the Klein bottle K withouthaving the surface intersect itself, we can assemble C-with-lake and K-with-lake.What about (C#C#C)-with-lake? What about Cn-with-lake? (Cn = C#…#C[n times].)


12. InvestigationProfessor Demasiadas Palabras has given the members of the team the followinglinkage for them to analyze. It is very similar to the second linkage the Professordiscussed in her lecture. However, the lengths of the rods have been changed. Helpthe team figure out a control screen.

13. InvestigationProfessor Demasiadas Palabras gave the team another linkage to investigate.Although the linkage has three fixed pivots, she feels that it can be analyzed in afashion similar to the one with two fixed pivots. Help the team in its investigation.

A

B C

O

4''

3''2''

5''


14. InvestigationHere is a another linkage from Professor Demasiadas Palabras. It has one fixed pivotand is similar to the first one she presented. Find a control screen for this linkage.

15. SummarizingWrite a summary of the Acme team’s activities for the day. Include statements ofproblems, definitions, results, explanations, relevant diagrams, and directions forfuture investigations. Be sure to make the summary clear, organized, and under-standable. Convince Boss that the time spent was worthwhile.

Notes

The greatest impetus to the study of surfaces came from within mathematics itselfthrough the work of German mathematician Bernhard Riemann (1826–1866).In his inaugural dissertation of 1851, Riemann, a student of Gauss, saw the studyof surfaces as a major ingredient in understanding certain complex functions ofone variable (for a description of the Riemann surfaces of certain complex func-tions, see Blackett, p. 85f; see also Lietzmann, p. 157+). Riemann classified surfaces(without boundary) by the notion of connectivity, the smallest number (plus one)of simple closed curves that can be drawn on the surface so that when you cut thesurface along these curves, you obtain something that can be flattened andstretched/shrunk into a disc. A special case is the sphere, defined to have connec-tivity 1 (see Kline, p. 1166+).

In 1861 Möbius was the first to undertake a general study of surfaces. Althoughhe and Listing are responsible for the creation of the Möbius band, his studyincluded only orientable (two-sided) surfaces without boundaries. In 1874 FelixKlein added possible boundaries to the surfaces and began the study of nonori-entable surfaces. In the 1880s two of Klein’s students, Guido Weichold andWalter von Dyck, carried out more systematic studies of nonorientable surfaces(see Huggett and Jordan, p. 135 and Pont, pp. 91, 121, 131, 137).

The Euler number of a surface is usually referred to as the Euler characteristicof a surface and is frequently denoted by χ (S). In 1813 Simon-Antoine-Jean

Notes 201

Lhuilier found a formula for the Euler characteristic of an n-holed doughnut (seeBiggs, et al., p. 84+). The Euler characteristic of a cross-cap is an easy consequenceof the work of von Dyck (see references above and discussion below).

The sum of two surfaces S#T is usually referred to as the connected sum of thetwo surfaces. The notion seems to be a 20th-century one. Klein and his studentsused other ideas to build surfaces. They started with the surface of a sphere onwhich several round holes were cut, and then to a pair of holes, they could glue a “handle.’’

Or to a hole they could glue a cross-cap. Von Dyck observed that gluing a cross-cap to each hole did not change the Euler characteristic of the surface. (Thisparaphrases von Dyck using our terminology.)

The study of linkages has a long history, in part associated with the real prob-lem of turning linear motion into circular motion. The construction of a straightline by means of a linkage was a famous problem solved by Peaucellier in 1864 (fora description of this and other linkages, such as those that construct a plane, seeHilbert and Cohn-Vossen, pp. 272–275). For more on linkages similar to the onesappearing in the story of this chapter, see the article by Thurston and Weeks.

A recent reason for studying surfaces comes from astronomy. Astronomers ask“What is the size of the universe? What is the nature of the three-dimensionalworld in which we live?’’ To answer these questions, it would be useful to knowthe answer to a related mathematical question: “What are all the possible three-dimensional worlds?’’ To answer the latter, it might be good first to answer ananalogous question: “What are the possible two-dimensional worlds, that is, whatare the possible surfaces?’’ For more on possible three-dimensional worlds, see thebook by Weeks.

References

Biggs, N.L., Lloyd, E.K., and Wilson, R.J. Graph theory: 1736–1936. Oxford: Oxford University Press,1976.

Blackett, D.W. Elementary topology: a combinatorial and algebraic approach. San Diego: AcademicPress, 1982.

Hilbert, D., and Cohn-Vossen, S. Geometry and the imagination. New York: Chelsea PublishingCompany, 1956.


Huggett, S., and Jordan, D. A topological aperitif. London: Springer, 2001.Kline, M. Mathematical thought from ancient to modern times. New York: Oxford University Press,

1972.Lietzmann, W. Visual topology. New York: American Elsevier, 1969.Pont, J.C. La topologie algébrique des origines à Poincaré. Paris: Presses Universitaires de France, 1974.Thurston, W.P. and Weeks, J.R. The mathematics of three-dimensional manifolds. Scientific American,

July 1984, pp. 108–120.Weeks, R. The shape of space, 2nd ed. New York: Marcel Dekker, 2002.

References 203

Acme Classifies Surfaces

205

Scene 1

Two days later. Tiffany, Millie, Joe and Brandon are having a meeting to discussprogress on their problem.

BRANDON: I know we haven’t figured out what the 18-sided polygon is. Wehave a clue, but nothing definitive. So let’s go around the groupand see what we do have. No details needed, just rough ideas.Tiff, you start.

TIFFANY: I came up with this algebra of surfaces. You turn each polygonpattern into an algebraic expression. I think it might make iteasier to identify what you’re looking at, like when a surface isthe connected sum of simpler surfaces.

BRANDON: Sounds great. Hold on to that. We’ll come back and hear moreabout it. Joe?

C H A P T E R

10

JOE: Well, I’ve been looking at patterns. I’ve developed some rules forcutting and regluing patterns that might help us figure out whatthe surfaces are.

BRANDON: Neat. Hold on to that. Millie?

MILLIE: I don’t have much, but I’ve got a plan we could carry out. (Shewrites.)

BRANDON: Wow. That’s pretty ambitious. I like it! What do the rest of youthink?

TIFFANY: I agree: it’s ambitious. But do we really want a list of all surfaces?I mean, do we want one-half-twist strip, two-half-twist strip,three-half-twist strip, etc., to all be on the list?

JOE: Yeah. Then there’s all these different things you can put togetherwith the pattern for a doughnut-with-lake.

Should all those things be on the list? The possibilities areendless. How could you list them all?

MILLIE: Keep your britches on, guys. We’ve already lumped the odd-twisted strips together, and we lumped the even-twisted stripstogether. We lumped things together if they had the same pattern.

BRANDON: You’re right, Millie. We did do some lumping. I guess what we needto do is decide, once and for all, when two surfaces are the “same.’’

MILLIE: Well, I say it’s when they have the same pattern.

TIFFANY: I have another idea that might turn out to be equivalent. Wedefined a surface as something made up of a map—a union ofvertices, edges, and countries—with certain properties satisfied.We could say that two surfaces are the same if both have thesame maps.

A

B B

A

How To Figure Out What a Certain Surface Might Be● Make a list of all possible surfaces.● Find a method that will take any surface and figure out which thing on the

list it is.

206 Chapter 10 Acme Classifies Surfaces

JOE: What does “same map’’ mean?

TIFFANY: Hmm. Good question. I guess I mean that there is a one-to-onecorrespondence between the countries of the two maps, a one-to-one correspondence between the two sets of vertices, andanother between the two sets of edges—all of these so that allthe “relationships’’ match. For example, if two countries of onesurface border, then the corresponding countries of the othermust also border. If two edges meet at a vertex of one surface,then the corresponding edges meet at the corresponding vertexof the other. Hmm. This may be complicated to write downcompletely.

JOE: I think I see what you mean. But I think your definition wouldlead to Millie’s. Take the two maps, and cut out the countries.Then reassemble each set of them like we did before to makepatterns. You could reassemble them in the “same’’ way. Get thesame patterns!

MILLIE: What do you mean by the “same’’ way?

JOE: Well…

TIFFANY: It’s not so bad. I think I’ve got it. Anyway, here’s a start. (Shewrites.)

What do you think?

JOE: I think it’s great, and I think that answers Millie’s question.“Reassembling in the same way’’ means that, when you gluecountry c and c′ of surface S1 together along edge e, you also gluecountry F(c) and F(c′) of surface S2 together along edge F(e).

TIFFANY: OK. Here’s where we are. (She writes.)

Definition of “Two Maps Are the Same”Start with maps M1 and M2. Denote the sets of vertices, edges, and countriesof Mi by Vi, E1, and C1, respectively for i = 1, 2. Suppose F is a 1-1 onto func-tion F: V1 � E1 � C1 → V2 � E2 � C2 such that● F takes vertices to vertices, edges to edges, and countries to countries.● If v and v′ are two vertices in M1 joined by edge e, then F(v) and F(v′) are

two vertices in M2 joined by edge F(e).● If e and e′ are two edges in M1 that meet in vertex v, then F(e) and F(e′)

are two edges of M2 that meet in vertex F(v).● If c and c′ are two countries in M1 that border along edge e, then F(c) and

F(c′) are two countries of M2 that border along edge F(e).Then maps M1 and M2 are the same.

Scene 1 207

I think the other direction is pretty clear. If two patterns are thesame, then the two surfaces have the same map. You draw a mapon one pattern and copy it onto the other. (She writes.)

MILLIE: What if a surface has more than one pattern? Like the Kleinbottle. We took one pattern, cut it up, labeled the cuts, and gluedit in another way. What would you call that? An equivalentpattern? A map on one would carry over to a map on the other.(She writes.)

ALL: Way to go, Millie!


1. Your TurnDoes Tiffany’s definition really capture what one would mean by “same maps’’?Here are some properties you would want “same maps’’ to have:

● If vertex v is attached to edges e1, … , ek, then vertex F(v) is attached to edgesF(e1), … , F(ek). In particular, v and F(v) must be vertices of the same order.

● If country c has as borders the edges e1, … , em, then country F(c) has asborders the edges F(e1), … , F(em). In particular, c and F(c) must have thesame number of edges.

Equivalent Patterns and SurfacesPatterns P1 and P2 are equivalent if you can cut up P1, label the cuts, reassem-ble the pieces, and glue them to get pattern P2.

Theorem. Two surfaces are the same if and only if they have equivalent patterns.

Theorem. If surfaces S1 and S2 have the same pattern, then they are same surface.

Two Surfaces Are the SameDefinition: Suppose that S1 and S2 are two surfaces such that S1 has map M1

on it and S2 has map M2 on it. If maps M1 and M2 are the same, then surfacesS1 and S2 are the same.

Theorem. If surfaces S1 and S2 are the same, then they have the same pattern.


Do these properties follow from Tiffany’s definition? Are there other criteriayou’d want “same maps’’ to satisfy? Do they follow from Tiffany’s definition?Should her definition be modified?

Scene 2

The group is still huddled around Brandon, discussing its progress.

BRANDON: Tiff, you said you’d come up with an “algebra’’ of surfaces. Tellus about it.

TIFFANY: I propose that we replace a pattern for a surface by an algebraicsymbol. It will be more efficient, because we wouldn’t have todraw the pattern, and it might be easier to identify familiarsurfaces. Here’s how to create the symbol. (She writes.)

It shouldn’t matter at which vertex you start. If you start at vertexQ on the pattern, then you’d get the symbol A−1D−1ABC−1B. Thisis different from the one I got before. But they’re for the samesurface! So I call them equivalent and write ABC−1BA−1D−1 ~ A−1D−1ABC−1B. There are other symbols I’d call equivalent.Like the symbols I’d get by using different letters. If insteadof using A, B, C, and D, I used W, X, Y, Z (respectively), I’d getWXY−1XW−1Z−1. So I’d want to write

Scene 2 209

Surface SymbolTake a pattern for the surface. Pick a vertex of the polygon and travel clockwisearound its perimeter, writing down the edge labels as you go, one after theother. If an edge is labeled A and the arrow on the edge is also going clockwise,write A; if the arrow is going counterclockwise, write down A−1. Do this untilyou arrive back at the vertex where you started. The result is the surface symbol.Example:Start with vertex P on the following pattern.

Result: symbol ABC−1BA−1D−1.

A

P Q

C

D

B

A

B

ABC−1BA−1D−1 ~ WXY−1XW−1Z−1.

JOE: What if you went counterclockwise instead of clockwise?Starting at P you’d get DAB−1CB−1A−1. The “−1’’ means the arrowis going “against the grain,’’ or pointing clockwise, the oppositeof the direction of your trip.

MILLIE: And you’d get another symbol if you changed the direction of allthe arrows. But it’s really the same pattern. So

ABC−1BA−1D−1 ~ A−1B−1CB−1AD.

TIFFANY: Absolutely. I agree with you both. You’d want all of thosesymbols to be equivalent. In fact, you’d want the symbols youget from equivalent patterns to be equivalent themselves. Forexample, the Klein bottle has patterns

with symbols ABAB−1 and CCDD, respectively. You’d wantABAB−1 ~ CCDD. Maybe we can make a list of rules to tell whentwo symbols are equivalent.

A

B

A

B

C D

C

D


Equivalent Symbols

ABC…Z ~ BC…ZA

(Cyclically permute letters.)

AB…YZ ~ ZY…BA

(Write letters in the opposite direction.)Assume a Greek letter denotes a string of letters.

αAβAγ ~ αA−1βA−1γ

αBβB−1γ ~ αB−1βBγ

(Reverse arrows on all occurrences of a single letter.)

JOE: This is neat. I think we should make a list of symbols for someof our favorite surfaces.

BRANDON: And what about the symbol for the sum of two surfaces?

MILLIE: Well, back to the drawing board.


2. Your TurnHelp the team. Here is a list of surfaces you and they have worked with before.Find a surface symbol for each one. You may know more than one pattern for asurface; in that case, there would more than one symbol. Write down these extrasymbols, too.

Scene 2 211

Symbols for Familiar Surfaces

Surface Pattern Symbol

Sphere XX−1

Cylinder

AA

B

C

X

X

Equivalent SymbolsαAβAγ ~ αXβXγ

αBβB−1γ ~ αYβY−1γ

(Replace all occurrences of a letter by another letter not already being used.)

ABAB−1 ~ CCDD

(Different symbols for a Klein bottle.)

αEE−1β ~ αβ

Continued

3. Your TurnWrite down a symbol for the connected sum of three tori. Write down the symbolfor the connected sum of two tori and three cross-caps. What is the general situ-ation? That is, suppose you know symbols for surfaces S and T. Is it easy to find asymbol for S#T? If you were to be presented with a surface symbol, when couldyou recognize it as the symbol for the connected sum of two surfaces?

Scene 3

Half an hour later. The group is back together.

JOE: Remember those rules I told you about at the beginning oftoday’s session? Well, I’ve been translating my rules for patternsinto rules for symbols. Let me tell you how I got the idea forthe rules. First, you all agree that in a pattern—and now, in asymbol—a given letter can appear one time or two times, butnot more. If a letter happened three times, you’d have threeedges to glue together and that certainly wouldn’t be a surface.

TIFFANY: Good point. We need to write it down. (She writes.)


Symbols for Familiar Surfaces

Möbius band

Torus

Klein bottle

Cross-cap

Two-holed doughnut

Connected sum ofthree cross-caps

F

H H

JG

E E

I just had to add that other bit.

JOE: Well, I noticed that, in patterns I recognized, a letter occurringtwice shows up close together. Let me show you.

One of the things about the linkage surface pattern we’re tryingto decode is that those pairs of letters are scattered all over.

I wanted to figure out some way to bring such a pair of letterscloser together. Here is the first thing I came up with. (He writes.)

Rule 1 for Equivalent Surface SymbolsαXβγXδ ~ αXγXβ−1δ

“A string of symbols to the right of an X can be moved to the right of theother X after inverting.’’

TLU

TRU

BRU

MLU ML

U

BLU

TRD

TLU

BLU

BRD

TLD

TRD

BRD

BLD

TRU

TLD

BLD

BRU

D

E

G

D F

G

E

F

CONNECTEDSUM OF

TWO TORIA

B

C

B

C

A

CONNECTEDSUM OFTHREE

CROSSCAPS

Occurrence of Letters in Surface SymbolIn a surface symbol, a letter can occur at most twice.A letter that occurs only once must be the edge of a lake or part of the edgeof a lake.

Scene 3 213

I used Tiffany’s convention: a Greek letter stands for a string ofsymbols. And, if β = ABCD, then I also mean that β−1 = D−1C−1

B−1A−1. (Notice that [β−1]−1 = β.) OK now, let me show you whyit works. Take a pattern that goes with the symbol on the left.Then cut it, label cuts, reassemble, and glue.

The pattern you wind up with has symbol αYγYβ−1∆. So, usingTiffany’s list of equivalent symbols, you’d get

αXβγXδ ~ αYγYβ−1δ ~ αXγXβ−1δ

That’s my Rule 1. Notice how that brings the pair of X’s closertogether.

BRANDON: Wow. If your pattern were just αXβXδ, then you’d have αXβXδ~ αXXβ−1δ. And then αXXβ−1δ ~ XXβ−1δα from Tiffany’s list ofequivalent symbols. The connected sum of a cross-cap withsomething else! All you need to do is figure out what that some-thing else is. Hey, we’re getting somewhere!

MILLIE: And that “something else’’ is shorter than what you started with.

TIFFANY: That “something else’’ is also a string of letters. If it contains a pairof letters that are the same and that appear as Z and Z–and not asZ and Z−1, then you can use Joe’s rule again. Let’s check it out.Suppose XXβ−1∆α = XXπZλZµ. Then, using Rule 1, you’d get

XXπZλZµ ~ XXπZZλ−1µ.

Hmm. How to get rid of that little “π.’’

JOE: I have another rule that might help. (He writes.)

I’ll let you guys figure out why the rule works.

Rule 2 for Equivalent Surface SymbolsαβXγXδ ~ αXγβ−1Xδ

“A string of symbols to the left of an X can be moved to the left of the otherX after inverting.’’

Cut GlueY

X

X

β

δ

α γ

Y

Xδ

α γ

β

Y

Y

X

δ

α γβ

XY


4. Your Turn.Help Millie, Tiffany, and Brandon. Explain why Rule 2 works.

TIFFANY: Rule 2 is just what we need to move the little “π’’:

XXπZλZµ ~ XXZλπ−1Zµ

Hmm. The “λπ−1’’ is to the right of the leftmost Z…Using Rule 1,

XXZλπ−1Zµ ~ XXZZ(λπ−1) −1µ

Another cross-cap. We’ve got the sum of two cross-caps andsomething else.

MILLIE: You could keep doing that until you get a “something else’’where there aren’t any pairs of letters occurring as A and A or asA−1 and A−1. (She writes.)

JOE: Good deal, Mill! I just had a look at our linkage surface. Everyletter occurs twice but none as…A…A…or…A−1…A−1.…I havea couple rules that relate to that situation. I wonder if they’llhelp? (He writes.)

Rule 3 for Equivalent Surface SymbolsαXβγX−1δ ~ αXγβX−1δ

“In a surface symbol containing X and X−1, a string of symbols to the rightof X can be moved to the left of X−1.’’

Rule 4 for Equivalent Surface SymbolsαβXγX−1δ ~ αXγX−1βδ

“In a surface symbol containing X and X−1, a string of symbols to the left ofX can be moved to the right of X−1.’’

Assembling the Cross-caps

Theorem. If a surface symbol S has a pair of letters occurring as…A…A…or…A−1…A−1…, then S ~ XXYY…ZZσ, where σ is a symbol stringhaving the property that, if a letter B occurs twice in it, it occurs as…B…B−1…or…B−1…B.…So S is the sum of several cross-caps and something else.

Scene 3 215

TIFFANY: They don’t really help you get X and X−1 closer together.

BRANDON: Maybe you don’t want them to get too close. If they’re right nextto each other, then poof, they’re gone. They cancel each other out.

MILLIE: Darn tootin’. In the symbol for a doughnut, ABA−1B−1, there aretwo pairs of symbols: A and A−1, B and B−1. If you got bothtogether as AA−1 and BB−1, you’d have a sphere. Not good.

JOE: I wonder if we could work with surface symbols where two pairs of letters occurring twice are meshed, like this.αXβYγX−1δY−1π.

TIFFANY: Boy, that looks like a doughnut just waitin’ to happen!

BRANDON: Hmm. I wonder. If you use Rule 4 for Y, then you’d have

αXβYγX−1δY−1π ~ αXYγX−1δY−1βπ

Then Rule 3 for Y would get you

αXYγX−1δY−1βπ ~ αXYX−1δγY−1βπ

JOE: Hey, we haven’t used the rules for X yet.

MILLIE: We can use Rule 4 for X backward to get

αXYX−1δγY−1βπ ~ αδγXYX−1Y−1βπ

Sure looks like a doughnut in that thar thing. Whatcha think?

TIFFANY: I think it’s a theorem! (She writes.)

BRANDON: Wow. This is really turning into something. Not bad, not bad.

TIFFANY: We get the sum of a torus and something else. The next thingwould be to work on the “something else.’’ We could “assemblethe tori.’’

BRANDON: We’d have to have another pair of meshing letters.

XYX−1Y−1κUλVµU−1υV−1θ.

MILLIE: Take what we did before, including my last step. You’d have

XYX−1Y−1κUλVµU−1υV−1θ ~ XYX−1Y−1κυµUVU−1V−1λθ

Theorem. If two pairs of like letters in a surface symbol are meshed as…X…Y…X−1…Y−1…, then the surface is the sum of a torus and something else.In fact,

αXβYγX−1δY−1π ~ XYX−1Y−1βπαδγ.


This time we want to get the “κυµ’’ out of there without mess-ing up the XYX−1Y−1. I think we can do it with a lot of“left-right’’ maneuvers:

(Left of the U to right of the U−1.) (Left of the V−1 to right of the V.)

XYX−1Y−1κυµUVU−1V−1λθ ~ XYX−1Y−1UVU−1κυµV−1λθ

(Left of the U−1 to right of the U.)

~ XYX−1Y−1UVκυµU−1V−1λθ

(Left of the V to right of the V−1.)

~ XYX−1Y−1UκυµVU−1V−1λθ

~ XYX−1Y−1UVU−1V−1κυµλθ

BRANDON: The sum of two tori and something else. Then keep doing itwith the “something else.’’ Millie, you’re a genius! Tiffany, timeto chalk up another theorem. (Tiffany writes.)

BRANDON: This calls for a little celebration. And a little break.

Tired high fives all around. A little bit of “allemand-left’’ and “do-si-do-ing’’ musiccomes up as the gang slowly dances about. Lights fade.

Theorem: Assembling the ToriPart A. If S is a surface symbol such that every letter A that occurs twice occursas…A…A−1…or…A−1…A…, then

S ~ X1Y1X1−1Y1

−1…XnYnXn−1Yn−1σ,

where σ is a string of letters in which every letter occurring twice occursas…A…A−1…or…A−1…A.…So the corresponding surface is the sum of n toriand something else. In addition, no pairs of letters X, Y occurring twice in σalso mesh; that is,…X…Y…X−1…Y−1 will not occur in σ.

Part B. Moreover, if S is any surface symbol, then

S ~ U1U1…UmUmX1Y1X1−1Y1

−1…XnYnXn−1Yn−1σ,

where σ satisfies the same properties as σ in part A. Thus any surface is thesum of m cross-caps, n tori, and something else, where the something else hassymbol σ.

Scene 3 217


1. InvestigationThe surface symbol for the linkage “control screen’’ the Acme team is concernedwith is

ABCD−1B−1EDC−1E−1A−1F−1G−1HFJ−1H−1GJ.

Use Rules 3 and 4 to figure out what it is.

2. InvestigationSuppose a surface symbol S has these properties

● Every letter in the symbol occurs twice.

● Every letter occurs as…A…A−1…or…A−1…A….

● No pair of letters X, Y mesh; that is,…X…Y…X−1…Y−1 does not occur in S.

An example is S = AB−1CC−1BD−1DA−1. Investigate all such surface symbols. Makeup more examples of your own. What can you say about the surface that corre-sponds to such a symbol? (Of course, you will want to justify your conclusions.)

3. InvestigationThis is very much like Investigation 2. In this case, you are to investigate thesurfaces that correspond to symbol S with the following properties:

● There may be letters occurring only once.

● Every letter occurring twice occurs as…A…A−1…or…A−1…A….

● No pair of letters X, Y–each occurring twice–mesh; that is,…X…Y…X−1…Y−1 does not occur in S.

(The only difference between the symbols here and those in Investigation 2 isthat, here, there may be letters occurring only once.) Investigate all such surfacesymbols. Make up several examples. In general, what can you say about the surfacethat corresponds to such a symbol? (Of course, you will want to provide goodexplanations for your conclusions.)

4. InvestigationCombine Part B of Assembling the Tori Theorem with the results of Investigations2 and 3 to come up with a theorem that describes all possible surfaces.


5. InvestigationMillie claims that she can use Rules 1 through 4 to show that XXABA−1B−1 ~

CCDDEE. Is she right? (If she’s right, then the sum of a cross-cap and a toruswould be the same as the sum of three cross-caps. What would this do to the theo-rem you came up with in Investigation 4?)

6. QuestionEarlier, there was a question: does every surface have an Euler number? Givenwhat you know from Investigation 4 (and all the work leading up to it), what isthe status of this question?

7. QuestionJoe claims that, if S is any surface, then its Euler number has the following prop-erty: N(S) ≤ 2. Is he correct?

8. QuestionBrandon has been thinking about the algebraic properties of the sum of surfaces.He has observed that the sphere acts as a kind of “identity’’ because, if S thesphere, then S#T = T for any surface T. He wonders if there is such a thing as an“inverse.’’ That is, given any surface T, can you find a surface U such that T#U = S,the sphere? He also wonders if a “cancellation law’’ holds. That is, if S1#S2 = S1#S3,can you conclude that S2 = S3?

9. QuestionDo surfaces S1 and S2 exist such that the following are satisfied?

● Neither S1 nor S2 is equivalent to the torus.

● S1#S2 is equivalent to the torus.

10. QuestionDo surfaces S1 and S2 exist such that the following are satisfied?

● Neither S1 nor S2 is equivalent to the cross-cap.

● S1#S2 is equivalent to the cross-cap.


11. QuestionBrandon has defined a “prime’’ surface T as one in which T = U#V is impossibleunless U is the sphere and V is T itself (or the other way around). What are theprime surfaces? Brandon wonders about “prime factorization.’’ Can every surface(not the sphere) be written as the sum of prime surfaces? If so, is the prime factor-ization unique?

12. InvestigationInvestigate the surface having the following symbol:

A1A2…An−1AnA1−1A2

−1…An−1−1An.

What can you say about this surface? What is a pattern for this surface? Tryusing Rules 1 through 4 to identify this surface exactly. What do you conclude?

13. InvestigationInvestigate the surface having the following symbol:

A1A2…An−1AnA1−1A2

−1…An−1−1An

−1.

What can you say about this surface? What is a pattern for this surface? Tryusing Rules 1 through 4 to identify this surface exactly. What do you conclude?

14. InvestigationAssembling the Tori (part B) and Investigations 4 and 5 provide us with acomplete list of surfaces. Identify the items in that list that can be obtained usingthe following two steps:

1. From a disc in the plane remove a finite number of smaller, nonintersectingdiscs from its interior.

2. Join the boundaries of certain pairs of these discs by tubes. Each tube may beattached in two different ways:


15. InvestigationAssembling the Tori (part B) and Investigations 4 and 5 provide us with acomplete list of surfaces. Which items on this list can be obtained by using soapbubble solution? (Recall that to make a soap bubble surface, you take a wire, puta knot in it, dunk it in soap solution, and poke holes where necessary in the soapfilm so that you get a surface with the wire as the only edge.)

16. SummarizingWrite a summary of the Acme team’s activities for the day. Include statements ofproblems, questions, definitions, results, explanations, relevant diagrams andpictures, and directions for future investigations. Make the summary so clear,organized, and understandable that Boss will be convinced that the time spent wasworthwhile.


Notes

This chapter contains two big accomplishments. The first is a definition of equiv-alent surfaces. This may seem like a triviality. But deciding when two surfaces arethe “same’’ is an important step. We had been headed for this definition sincewe looked at twisted strips. The definition clarifies and focuses the discussion.It makes explicit now what may have been only implicit before. But it is not thelast word, it is just a beginning. We may want to adjust the definition later.

The second accomplishment is a complete classification of surfaces. The resultsof this chapter plus those of Investigations 4 and 5 show that any surface must beequivalent to one of the items on the following list.

● Sphere

● Sphere with r lakes

● Sum of s tori

● Sum of s tori with r lakes

● Sum of s cross-caps

● Sum of s cross-caps with r lakes

The results are stronger than just a classification. Given a surface (or rather apattern for a surface), the results of the chapter (and Investigations 4 and 5) notonly tell you that the surface is one of the items on the list but also give a methodfor deciding exactly which item on the list it corresponds to.

Möbius gets the credit for the first classification of orientable surfaces in 1863(Pont, p. 91f), and Guido Weichhold gets the credit for the first classification ofnonorientable surfaces in 1883 (Pont, p. 129f). Möbius also found it useful tothink of a surface as being constructed by gluing together flat polygonal pieces,that is, as something made up of a map (see Fauvel et al., p. 108).

The standard 20th-century proof of the classification theorem for surfaces isthat of Lehrbuch der Topologie, by Seifert and Threlfall. Their book became thestandard topology textbook soon after it appeared in 1934. In it, they think of asurface as a polygon with certain edges identified, that is, as a pattern, and usesurface symbols just as they are used by the Acme team (see the English transla-tion of their book below, p. 134f). Acme’s development is a variation due to A. W. Tucker (see article by James).

Other books that treat surfaces by using some variation of the Seifert-Threlfallmodel are those by Blackett, Fréchet/Fan, Carlson, and Firby/Gardiner (listedbelow).

There are several more 20th-century proofs of the classification theorem. Oneuses a method called “surgery’’ created by Zeeman. A treatment of this can be foundin the book by Stewart. A second method is due to Conway (see article by Francesand Weeks), and a fourth can be found in the book by Boltyanskii and Efremovich.


References

Blackett, D.W. Elementary topology: a combinatorial and algebraic approach. London: Academic PressLimited, 1982.

Boltyanskii, V.G., and Efremovich, V.A. Intuitive combinatorial topology. New York: Springer-Verlag,2001.

Carlson, S.C. Topology of surfaces, knots, and manifolds. New York: John Wiley & Sons, 2001.Fauvel, J., Flood, R., and Wilson, R., editors. Möbius and his band. New York: Oxford University Press,

1993.Firby, P.A., and Gardiner, C.F. Surface topology. West Sussex, England: Ellis Horwood Limited, 1982.Francis, G.K., and Weeks, J.R. Conway’s ZIP proof. American Mathematical Monthly, May 1999,

pp. 393–399.Fréchet, M., and Fan, K. Initiation to combinatorial topology. Boston: Prindle, Weber & Schmidt, 1967.James, R.C. Combinatorial topology of surfaces, Mathematics Magazine, September/October 1955,

pp. 1–1139.Pont, J.C. La Topologie Algébrique des Origines à Poincaré. Paris: Presses Universitaires de France, 1974.Seifert, H., and Threlfall, W. A textbook of topology. New York: Academic Press, 1980.Stewart, I. Concepts of modern mathematics. New York: Dover, 1995.

References 223

Acme Encounters theFourth Dimension

225

Scene 1

The next day. Millie, Tiffany, Brandon, and Joe are recapping the previous day’ssuccesses.

BRANDON: Well, Millie, we accomplished everything you suggested.We’ve got a list of all possible surfaces. And, if we are presentedwith a surface, we have a method for deciding which item on thelist it is.

MILLIE: Thanks to Tiffany for coming up with the idea of surfacesymbols and to Joe for his rules for equivalent symbols.

TIFFANY: We have all of us to thank. We worked pretty well as a team.

JOE: I think we need to write down that list and have a look at it.(He writes.)

C H A P T E R

11

Opening image from Senechal, M., and Fleck, G. Shaping Space: A Polyhedral Approach, p. 122. Boston:Birkhäuser, 1988. Reprinted with kind permission of Springer Science and Business Media.

Isn’t that neat? Let’s gloat for a while. (Everybody stares at list.)

TIFFANY: Hmm. I don’t want to rain on our parade, but I’ve got a question.How do we know all of those are different?

ALL: Different?

TIFFANY: Could there be any duplications on that list?

MILLIE: Waddaya mean, “duplications’’?

JOE: No way!

BRANDON: Do we really care? (They all turn to stare at Brandon.) Well,everything on that list is something we “know.’’ Those are allthings we could have visualized a long time ago. And, if you giveme a surface, I know it’s one of those and I can figure out whichone it is. Who cares if the “one’’ is actually two, or more?

TIFFANY: Well, I do. Wouldn’t it be nice if a surface had a unique identity?

JOE: Yeah. I care, too. Just look at that list. How could there be anyduplications?

MILLIE: Yeah. How could the sum of seven cross-caps be the same as asphere with three lakes?

TIFFANY: We were pretty excited when we discovered that the sum of across-cap and a torus was actually the same as the sum of threecross-caps. That duplication meant that all those “hybrid’’surfaces—such as the sum of s cross-caps and t tori—were reallynot hybrid at all. The sum of s cross-caps and t tori is really thesame as the sum of s + 2t cross-caps when s is positive. That revelation enabled us to whittle our list down to what we havenow. The question is, can we whittle it down further? That reve-lation wasn’t at all obvious. Is there another obscure revelationlurking in there?

BRANDON: OK, OK. I get your point. Any ideas, gang? (Silence.)

Complete List of Surfaces● Sphere● Sphere with r lakes● Sum of s tori● Sum of s tori with r lakes● Sum of s cross-caps● Sum of s cross-caps with r lakes

226 Chapter 11 Acme Encounters the Fourth Dimension

JOE: I’ll stick my neck out. One thing we’ve never used is the Eulernumber N(S).

TIFFANY: Could we use it?

JOE: Well, what do we know about it? We know that two surfaces thatare the same have maps on them that are the same.

MILLIE: We know that every surface on our list has an Euler number.So every surface has an Euler number. Hey, that’s something new!(She writes.)

BRANDON: And, we know how to calculate the number for everything onour list. Let’s do it. (He writes.)

JOE: We also know that if two surfaces have different Euler numbers,then the surfaces are different.

MILLIE: That’s good. Everything in the second row of the table is different.Same with the third row and with the fifth row. Now, if twosurfaces have the same Euler number...

JOE: Not so fast. The Euler numbers of the torus and the Klein bottleare both zero. And, look, the number of a sphere with two lakesis also zero. Those three surfaces can’t all be the same!

TIFFANY: Nobody here thinks they are. We just need to do more work.Let’s start with the torus and the Klein bottle. Why do we thinkthey’re different? For one thing, one is orientable, the other isn’t.Can we show from the definition that an orientable surface can’t bethe same as a nonorientable surface? We’d accomplish a lot if wecould do that. Look at the table. It would mean that everythingin the third and fourth rows of our table is different from every-thing else. Then there’s the torus and the sphere with two lakes. Mygut feeling is that something with lakes can’t be the same as some-thing without. Let’s take a break and think about those things.

Members of Team Acme return to their desks. Lights dim.

Euler Numbers of Surfaces

Surface Euler Number

1. Sphere 2

2. Sphere with r lakes 2 − r

3. Sum of s cross-caps 2 − s

4. Sum of s cross-caps with r lakes 2 − s − r

5. Sum of t tori 2 − 2t

6. Sum of t tori with r lakes 2 − 2t − r

Every surface has an Euler number.

Scene 1 227

1. Your TurnHelp the group make more distinctions between the items on their list. Let C(s, r)denote the sum of s cross-caps with r lakes, T(s, r) the sum of t tori with r lakes,and S(r) the sphere with r lakes. Use what you know about the Euler number ofsurfaces to show the following pairs are not the same (assume s is positive in each case):

● S(r) and T(s, r) for any r

● S(r) and C(s, r) for any r

● T(s, r) and T(s, t) where r π t

● C(s, r) and T(s, t) where r π t

● C(s, r) and C(t, r) where s π t

● T(s, r) and T(t, r) where s π t

2. Your TurnBelow are patterns for a couple of surfaces. Figure out how many lakes each one hasjust by looking at the patterns.

Scene 2

A half hour later.

MILLIE: Gather ‘round, luvs. I think I’ve got it. (The group surroundsMillie at her desk.)

BRANDON: What do you have, Millie?

MILLIE: You all remember that if you were a creature living “in’’ theMöbius band, you could take a trip around the band in such away that at the end of the trip your orientation was reversedfrom what it was at the beginning of the trip. Of course, thesame thing can happen on other surfaces. In fact, it can happen

A B

P

C A

B

C Q

P Q

B

A A

R

S B


on any surface with a symbol that has a letter X occurring twiceas ...X...X... or ...X−1...X−1..... Here is its pattern:

If you take a trip along the dotted path, your orientationreverses—just as it did on the Möbius band. The converse is alsotrue: if you can take an orientation-reversing trip on the surface,mark a little slit at the beginning and end of the trip. Label the slit X. Cut the surface up starting with X. You’ll get a bunch ofpieces. Glue them back together, but don’t glue the two labelsmarked with an X! You’ll eventually get a pattern whose symbollooks like ...X...X... or ...X−1...X−1.... So here’s what we know.(She writes.)

Of course that means that all the surfaces in Rows 3 and 4 of ourtable are nonorientable.

BRANDON: Where are you going with this, Millie? How do you know that asurface in another row can’t have a pattern that would make itnonorientable?

TIFFANY: If you follow our algorithm, that can’t happen.

BRANDON: Maybe you can get there some other way.

MILLIE: Hold yer horses. I’m gittin’ there. Suppose you start with asurface S that has a symbol where every letter that occurs twiceoccurs as ...X...X−1.... Certainly, no amount of stretching orshrinking of the pattern that goes with the symbol would changethat fact. No amount of relabeling would change it either. Theonly thing that could change it would be to cut up the patternand reglue it. What’s a typical cut and paste operation amount to?You pick a letter that occurs twice (X), cut the pattern along somepath (Y) that separates the two X’s, and then paste the X’s together.Here’s what it looks like.

Nonorientable SurfaceA surface is nonorientable if and only if it has a pattern whose symbol is of theform ...X...X....

X X

Scene 2 229

The symbol for S we started with was αXβγX−1δ. Every letter Z inthe symbol occurring twice occurs as ...Z...Z−1.... The symbol for thenew pattern is αδXγβX−1. It’s clear that this new symbol also has theproperty that a letter Z occurring twice occurs as ...Z...Z−1.... Thatmeans that the surface S can’t have a symbol where, when a letterW occurs twice, it occurs as ...W...W.... The surface can’t be anonorientable surface. It must be orientable. We’ve got a theorem!(She writes.)

JOE: Hey, that’s great! Tiff, whaddaya think?

TIFFANY: This is just what we needed: turn our gut feelings into goodthinking. Make our facts really tight.

BRANDON: There’s the fact that the sphere with two lakes and the torus havethe same Euler number. Shall we tackle lakes next? My gut feelingis that if two surfaces are the same, they’ve got to have the samenumber of lakes. Anybody have any ideas?

JOE: Well I wasn’t making any headway on any of this stuff. So I gotto thinking about the Euler number and how neat it was that wecould use it to tell some surfaces apart. I began to wonder ifthere was some way to calculate the Euler number straight fromthe symbol, without knowing ahead of time which item on ourlist it was. I thought knowing that might be pretty useful.Like, suppose you knew the surface had no lakes and wasnonorientable. From what Millie has done, we know how todefinitively decide what is nonorientable. Then it seems prettyclear to me that it’s the sum of a bunch of cross-caps. Then, ifyou knew the Euler number, you’d know exactly how many inthe bunch—without having to use my rules!

Theorem Relating Orientability and SymbolsIf S is a nonorientable surface, then every symbol has the property that if aletter X occurs twice, it occurs as ...X...X....In particular, if surface S has a symbol with the property that every letteroccurring twice occurs as ...X...X−1..., then that surface can’t be nonorientable;that is, it must be orientable.Corollary: Every surface in Rows 3 and 4 of the table must be different from asurface in any other row.

X

X β

δ

α γ βδ

α

γY

X

Y

X

X

βδ

α

γ

X

Y

δ

Y

X β

α

γ


BRANDON: Hey, Joe, you’re right. Might not tell us about the number oflakes, but it’ll give us something useful. Who knows? Maybe it’llbe useful. Fire away!

JOE: Anyway, it still needs a little work, but here’s my idea for calculatingthe Euler number from the pattern. My big assumption is thatcounting the vertices and edges of the pattern and counting thepolygon as one country would give me all the data I need tocalculate the Euler number. You don’t exactly count the edges ofthe pattern: your count has to account for duplications, a pair ofedges to be glued really counts for just one edge. You can get theadjusted count of the number of edges right from the symbolitself: count the number of distinct letters that appear there.That’s E. Also, C = 1, the polygon itself. That’s easy.

TIFFANY: So all you need to do is calculate V.

JOE: Yup. Again, you have to account for duplications. But it’s a littletrickier than counting edges. Here’s my method. It involveslabeling all the vertices. (Joe writes.)

Labeling the Vertices of a Surface SymbolWe use letters to label edges, numerals to label vertices. If edges (letters) A and B share vertex 1 and they appear consecutively in the symbol, then we’ll write...A1B... .Step 1. Start with a pair of consecutive letters, A and B, in the symbol: ...AB...(this assumes there are at least two letters in the symbol; if there is only oneletter, then I think you know what to do) and write A1B. From here, there are two choices: work with A or work with B. Let’s work with B first. There are twopossibilities: B is unmatched or B is matched in the symbol. In case B isunmatched, stop. If B is matched, go to the other occurrence of B in the symboland write ...C1B... or ...B−11C.... Now work with the “new’’ letter C. Again, thereare two possibilities: C is unmatched (in which case you stop) or C is matched(in which case you go to the other occurrence of C in the symbol and write...D1C... or ...C−11D... if you identified 1 as the initial vertex of C or write...C1D... or D1C−1 if you identified 1 as the terminal vertex of C). Keep doingthis until you reach an unmatched letter or you repeat a pair. (This has got tohappen because there are only a finite number of letters.) If you reach anunmatched letter, go back and work with A (just as you did with B). This timeyou’ll also reach an unmatched letter and you’ll be done. If you repeated a pair,the repeated pair will be ...AB... and you’ll be done. This process identifies all thevertices of the pattern polygon that should all be labeled “1.’’Step 2. Choose a pair of consecutive letters E and F whose shared vertex hasn’tyet been labeled and follow the procedure in Step 1, using a new label. Keeprepeating this process until no such pairs exist. This process has got to endbecause there are only a finite number of letters. That does it!

Scene 2 231

MILLIE: Gracious! Give us an example, Joe.

JOE: OK. Here’s one: XYXAZ−1YWBZW−1. Start with the first pairXY... and label the shared vertex:

X1YXAZ−1YWBZW−1

Work on the Y–

X1YXAZ−11YWBZW−1, and

then work on the Z−1–

X1YXAZ−11YWB1ZW−1.

OK B is unmatched.Go back to the initial pair and work on the X:

X1YX1AZ−11YWB1ZW−1.

This time A is unmatched. You’re done with vertex “1.’’ This is anexample where the sequence begins and ends with unmatched letters.

Pick another pair ...YX... where the shared vertex hasn’t yet beenlabeled. Label it

X1Y2X1AZ−11YWB1ZW−1.

Proceed as before,

X1Y2X1AZ−11YWB1ZW−1

→ X1Y2X1AZ−11Y2WB1ZW−1 → X1Y2X1AZ−11Y2WB1ZW−12.

Now, you’ve got to remember that the end of the symbol is really“hooked up’’ to its beginning:

2X1Y2X1AZ−11Y2WB1ZW−12.

So the next step would be exactly where we got started in label-ing vertex “2.’’ This is an example where the sequence “loops back’’on itself. So we’re done with vertex “2.’’

Keep going. Pick another pair ...AZ−1... where the shared vertexhasn’t yet been labeled. Label it

2X1Y2X1A3Z−11Y2WB1ZW−12.

Then

2X1Y2X1A3Z−11Y2WB1Z3W−12→2X1Y2X1A3Z−11Y2W3B1Z3W−12

Hey, you’ve labeled all the vertices. You’re done! (If youweren’t paying attention and just followed the rules, you’dnotice that you had reached an unmatched letter, B. Therules then say to stop, go back, work on the A in ...AZ−1..., butA is also unmatched. The rules say you’re now done, which we


knew since everything was labeled!) This is another examplewhere the sequence begins and ends with unmatched letters.

BRANDON: Wow. I think it’s time for a break to practice this technique on afew of our favorites.

The group disperses. Each goes back to his or her desk, on the way congratulatingJoe on the new idea. Lights fade. Music comes up.

4. Your TurnTry out Joe’s technique. Label the vertices in the following symbols:

LMNPEEFFGGABA−1B−1CDC−1D−1

XYXZ−1YWZW−1

AXBYCWXDW−1YABA−1CDC−1E

5. Your TurnA certain surface has symbol XWYZX−1WZY−1. By using Joe’s technique, we gotthe following labeling of vertices: 2X1W2Y2Z1X−12Z1W2Y−12. Now use Joe’s ideafor calculating the Euler number to figure out directly which surface on the list hethinks this one should be.

Scene 3

A half hour later. Members of the ACME team are busy at their desks.

TIFFANY: Hey. I think I can use Joe’s method of labeling vertices to count lakes. (Everybody looks up and listens to Tiffany.) When Joelabels a single vertex, two things happen: You can get a loop ofpairs of letters that returns back to the beginning pair. Or, youcan get a string of pairs that doesn’t connect but that begins witha pair having an unmatched letter and ends with a pair havingan unmatched letter.

MILLIE: What’s this have to do with lakes?

TIFFANY: Unmatched letters! Those form pieces of the shorelines of thelakes. A single shoreline consists of a list of unmatched letters,one after another. Consecutive letters in the list will share asingle vertex. The list should form a loop: If you start with an

Scene 3 233

unmatched letter and move along the list, eventually youwind up back at the initial unmatched letter. Here’s a picture.

Here’s the list of letters and the vertices they share 4A1B2C3D4 .

BRANDON: So how do you come up with that “list’’ from the symbol?

TIFFANY: Glad you asked. Let me show you. (She writes.)

Identifying Lakes from a Surface SymbolA. Finding consecutive pairs of shoreline pieces. Start with an unmatched

letter A. If it’s the only letter in the symbol, you’re done. Typically, it appearsin the symbol like this: ...AX.... Label the vertex between A and X: ...A1X....The vertex 1 is the terminal vertex of A. Use Joe’s method to label all theother positions for vertex 1. Eventually, you’ll arrive at another, uniqueunmatched symbol B with its initial vertex 1. (It might be that B = A. Butthat’s OK.) So you’ve found the first two pieces of the shoreline of a lake.The list starts with A1B....

B. Forming a Lake. Do the same thing with B that you did with A. Theunmatched letter will appear in the symbol as ...B2Y... or ...Y2B−1.... Thevertex “2’’ is the terminal vertex of B and can’t be the same as “1.’’ Followstep I with A1... replaced by B2.... Keep repeating Step A until you duplicatean unmatched letter. (This has got to happen since the number of unmatchedletters is finite.) I claim this duplication first occurs with the initial letter A.(If you duplicate before you get back to A—at C, say—then some thing likethis would happen:

A1B2C3D4E5F6C

That would mean that starting with C−16, and using Joe’s method, you’darrive at both F and B. Can’t happen.)

C. Collecting All the Lakes. Once you’ve formed your first lake, you look for an unmatched letter that hasn’t been used. If you don’t find one, stop.If you do, repeat Step B with that letter. Keep doing this until you use up all the unmatched letters.

D. Counting the lakes. Of course, once you have created all the lake shorelines,it’s easy to count them. You just do it!

D3

C

2

B

1A

4


BRANDON: That’s impressive. You can count the lakes just from the symbolitself. But how do you know that the number you get will be thesame if you have another, equivalent symbol?

TIFFANY: Well, I thought of that. The way to really change a symbol/patternis to do a cut-and-paste operation. The basic one goes like this.

What happens is that vertices of both patterns are the same, andthe edges with unmatched letters are the same. Moreover, foreach unmatched letter, the initial and terminal vertices are thesame. So if one pattern has ...A1B..., then the terminal vertex ofA is 1 and the initial vertex of B is 1. The same will be for the otherpattern, so the other pattern also has ...A1B.... The unmatchedletters will line up to form lakes in exactly the same way for bothpatterns. So the number of lakes will be the same for bothpatterns. (She writes.)

JOE: Hey, that does it! Everything on our list is really different!

MILLIE: This has been hard work. Haarrrddd work! We deserve anotherbig theorem. (She writes.)

Theorem about Same SurfacesTwo surfaces are the same if and only if the following three conditions are

satisfied:● The two surfaces are either both orientable or both nonorientable● The two surfaces have the same Euler number● The two surfaces have the same number of lakes

Theorem Relating Number of Lakes and SymbolsIf a surface has two symbols S and S′, then the number of lakes counted for S equals the number of lakes counted for S′.

C

C

D

A C

D

B

1

E

1

2

2

2

1

A

D

D

B

1

1

1

E

E

2

2

2

Scene 3 235

Whew! Time for a little nap.

The group, dazed, does a few high fives and a little dance before leaving theAcme shop for the day. Sounds of “nite’’ and “see ya’’ are heard. Musiccomes up. Lights fade.

6. Your TurnConsider surfaces that have the following symbols:

AXBYCWXDW−1YAXBYCZ−1DYEZFUGXHUFour things to do with these:

1. Use Joe’s technique to label the vertices.

2. Use Tiffany’s technique to find the lake shores for each one.

3. Use Joe’s idea to find the Euler number of each surface.

4. Use Millie’s theorem to pinpoint exactly what each surface is.

7. Your TurnTiffany showed that different symbols for the same surface should yield the samenumber of lakes. Her argument involved a cut-and-paste operation in which thecut went from one vertex of the pattern to another. She did not consider the possi-bility of the cut going from the midpoint of one edge to another vertex or even tothe midpoint of another edge. Her argument might be complete if she had shownthat adding the midpoint of an edge changes the symbol but does not change the number of lakes (two cases: the edge is unmatched; the edge is matched).Show this, and show how knowing this would make her argument complete.

Scene 4

Early Monday morning. The Acme team has just returned from the weekend.

JOE: What a weekend! Didn’t do a thing. Sat around, read, watched TV,looked at the sunset.

TIFFANY: Brandon and I went hiking. We couldn’t keep from talking aboutthe rad stuff we did last week. A complete classification of surfaces!And a way to identify a surface without using the rules to put it in“canonical’’ form.

MILLIE: Took your work home, huh?


BRANDON: After our hike, I started thinking again about those surfaces wecan’t really put together.

MILLIE: You mean the Klein bottle and the cross-cap? That’s had yourknickers in a twist for a long time.

BRANDON: Well, I think I have an idea. I’d like to run it by you all. It involvesputting them together in a four-dimensional space.

TIFFANY: What? You’re kidding! Have you gone off the deep end?

JOE: That’s what we need: a little time travel.

MILLIE: Poor Brandon. You haven’t been feeling well; you must have a fever.Let me feel your forehead.

BRANDON: OK, OK. It’s nothing weird. Just a new way of looking atsurfaces. Think of the “problem’’ with the Klein bottle. Here’s theway we usually put it together.

The surface crosses itself along a circle. Every point on the circlerepresents two points on the surface: one for each of the two“branches’’ of the surface that meet there. That doesn’t happenwith the sphere or the torus or a bunch of other surfaces. I got myidea for getting around the problem with the Klein bottle bythinking about a figure-eight path in the plane:

In this case, it’s the path that crosses itself. If you take a trip along thepath from beginning to end, there’s that one point that really repre-sents two instances on the trip. If you’re a train, you can’t turn cornersso there’s really only one trip you can take. So how can you avoid thatdouble point? It might even be that the train is long enough so thatthe double point would make the train crash into itself!

Scene 4 237

MILLIE: Well, duh. Build a bridge!

BRANDON: Exactly. But you need to go out of the plane to do that. You needthree dimensions.

JOE: So your idea with the Klein bottle is to “build a bridge’’ wherethe surface crosses itself, but do it in four dimensions?

BRANDON: That’s exactly right!

TIFFANY: I think we’re going to need a little help with just how you “builda bridge’’ in four dimensions.

BRANDON: Thought you’d never ask! To show you how to build a bridge infour-space, I need to go back and show you a way to build thebridge in three-space for the figure-eight path. Very carefully.Also, I’m going to use coordinates to tell you what I mean bythree-space and four-space. Three-space is the set of points ofthe form (x, y, z), where x, y, and z are real numbers. The valuesx, y, and z usually represent the distance of a point along perpen-dicular axes to some fixed point called the origin. Four-space isthe set of objects of the form (x, y, z, w), where x, y, z, and w arereal numbers.

JOE: What’s “w’’? Is that time?

BRANDON: Not necessarily. Depends on your interpretation. I just wantyou to think of four-space as something just like three-space,except that every “point’’ has four coordinates rather thanthree.

MILLIE: Does it exist?

BRANDON: Think of it as something we’re creating. Back to the three-dimensional bridge for our two-dimensional “track.’’ Think of the figure-eight track—before the bridge—as lying in the x-y plane of three-space. So all the points of the figure-eight areof the form (x, y, 0). Align the figure so that the part to bereplaced by the bridge lies on the x-axis between −1 and +1.Make the crossing point be the origin (0, 0, 0).


Here are two views of the part with a third axis, the z-axis, showing.The right-hand one is a view head on.

I’m going to build a bridge in the x-z plane. It might look something like this.

JOE: That’s just the graph of a function.

BRANDON: Correct. Let’s call the function f(x). It has these properties:

● It has domain [−1, +1].

● f(−1) = f(+1) = 0

● f(x) > 0 on (−1, +1)

● f(x) is continuous on [−1, +1]

–1 +1

z

x

yy

x x

z

z

0–1 +1

y

x

Scene 4 239

An example of a function that satisfies these conditions isf(x) = (x − 1)2(x + 1)2. To use this idea for the Klein bottle, what I really want is a function F from the part to be replaced by thebridge—the set {(x, 0, 0): −1 < x <1}—into three-space that satisfiesthese properties:

● F is continuous

● F is 1-to-1

● F(−1, 0, 0) = (−1, 0, 0) and F(1, 0, 0) = (1, 0, 0)

● The only places that the range of F intersects the x-y plane is at (−1, 0, 0) and (1, 0, 0).

I want the range of F to be a path in three-space playing the role of the bridge. The third condition makes sure the bridgehooks up with the rest of the path at both ends. The functionF(x, 0, 0) = (x, 0, f(x)) works. The replacement path is actuallythe graph of f(x).

y

–1 +1

z

x

f(x)

–1 +1

z

x


Here’s what the figure-eight looks like with the bridge replacement.

MILLIE: Seems like a lot of work just to build that little bridge.

BRANDON: I wanted to be really careful so that when we get to the “strange’’bridge, it won’t be so strange.

JOE: We’re ready. Lay it on us, Bran.

BRANDON: OK. Give me some time to set things up. First the bridge.In four-space I want to build a bridge “over’’ the problem circle.The part that intersects the problem circle is a cylinder, or closeto one. So the bridge for the Klein bottle will be a bent cylinder,not a path as it was with the figure-eight. Here’s a picture of theset-up.

We’ll think of the Klein bottle as we have it now (with problemcircle) as points of the form (x, y, z, 0)—what we might call the x-y-z three-space sitting in four-space. Line up the Klein bottleand stretch/shrink it so that the part to be replaced is a cylinderand has its axis coinciding with the x-axis in x-y-z three-space.Just to have everything be concrete and explicit, put the two endsof the cylinder at x = −1 and x = +1, and make the circular endsof the cylinder have radius 1. That means that the cylinder is theset C = {(x, y, z, 0): y2 + z2 = 1, −1 ≤ x ≤ 1}.

Cylinder

y

x

z

Scene 4 241

To construct the bridge, I want a function G from C to four-spacesuch that

● G is continuous.

● G is 1-to-1.

● G(−1, y, z, 0) = (−1, y, z, 0) and G(1, y, z, 0) = (1, y, z, 0) .

● The only points where the range of G intersects x-y-z three-spaceare the two circular ends of C.

This time the range of G will be the replacement part, the bentcylinder-bridge in four-space. The third condition makes surethe bridge hooks up with the rest of the Klein bottle at bothends. The function G(x, y, z, 0) = (x, y, z, f(x)) works, where f(x)is the function we used for the figure-eight bridge. With thisreplacement part, the Klein bottle is all hooked up in four-space–without any problem points. Isn’t that neat?

MILLIE: Aren’t you going to draw a picture?

BRANDON: That’s a problem with objects in four-space. Hard to draw pictures.

JOE: I think we ought to make the fourth coordinate be time. Then lettime roll. See what we get. Sort of like a movie.

MILLIE: Hmm. The little function f(x) measures time. If we use f(x) =(x − 1)2(x + 1)2 with −1 ≤ x ≤ 1, then f(x) ranges between 0 and 1.So time would go from 0 to 1. Not a very long movie.

JOE: We can slow it down. Let’s see. When t = 0, you see all of the Kleinbottle minus the cylinder. When t = 1, that’s when f(x) reaches a peak at x = 0, half-way through the cylinder. You’ll see {(0, y, z): y2 + z2 = 1}—a circle, a cross-section of the cylinder!

0–1 +1

z

z

x

xCylinder


MILLIE: Then, if 0 < t < 1, you’ll get two values of x on either side of 0. You’llsee two circles! OK. Here’s what you see in the whole movie. Freezeeverything at t = 0. You see the Klein bottle minus the cylinder. Thenlet time roll, slowly. The Klein bottle minus the cylinder disappearsinstantly, and you see just two circles, starting at the ends of thecylinder and moving slowly to the middle. When the two circlesmeet in the middle, that’s t = 1 and everything disappears. End ofmovie! Pretty dramatic.

JOE: Wow.

TIFFANY: Brandon, it’s awesome what you’ve done. I think I have anotherway of assembling the Klein bottle in four-space. Let me try it out.I start with the same problem circle, but I want to build a bridgein the “other direction.’’ Let me explain what I mean. Put the Kleinbottle together, sort of, with a lake.

Call this Klein-bottle-with-lake, or KBWL for short.

MILLIE: Isn’t that the name of a radio station?

TIFFANY: So right now the circle isn’t a problem. But to complete the Kleinbottle you need to fill in the lake. Arrange KBWL in three-space sothat the edge of the lake is the circle {(x, y, 0): x2 + y2 = 1}.

The interior of the circle is the disc {(x, y, 0): x2 + y2 < 1}. The easyway out would be to plug the lake with that disc. But it intersectsKBWL, and we want to avoid that. Something else we could plugthe lake with is the hemisphere {(x, y, z): x2 + y2 + z2 = 1, 0 < z}.This is the range of the function H(x, y, 0) = (x, y, 1 − x2 − y2). Thishas the same problem: it intersects KBWL. Here’s my way out. PutKBWL in four-space via (x, y, z) → (x, y, z, 0). The edge of the lakeis then the circle {(x, y, 0, 0): x2 + y2 = 1}. The hemisphere thatcreates problems would be {(x, y, z, 0): x2 + y2 + z2 = 1, 0 < z}.However, {(x, y, 0, w): x2 + y2 + w2 = 1, 0 < w} is another “hemisphere,’’ just like it. And it wouldn’t create problems; it doesn’tintersect KBWL! We could attach it to KBWL in four-space using the function K(x, y, 0, 0) = (x, y, 0, 1 − x2 − y2) for all points(x, y, 0, 0) such that x2 + y2 ≤1. It’s continuous, 1–to–1, and has the property that K(x, y, 0, 0) = (x, y, 0, 0) for every point (x, y, 0, 0) on

KBWL

edge of lake

Scene 4 243

the edge of the lake. The last property means that the range of K (the “hemisphere’’ in four-space) is glued to the edge ofthe lake.

MILLIE: My turn to do the movie! The expression 1 − x2 − y2 measures time.It can be 0 when x2 + y2 = 1—for points on the lake’s edge. It can be1 when x = y = 0—for the “center’’ of the lake. And it can be anyvalue in between—for a circle concentric with the edge of the lake.So we’re going to have another movie with time ranging from 0 to 1.Freeze everything with t = 0. You’ll see KBWL. Then start timerolling. Immediately, KBWL will disappear, and you’ll see a circleconcentric with the lake-edge getting smaller and smaller until, at t = 1, you’ll see a point, the center of the lake. After that, nothing.All traces of the Klein bottle vanish—forever!

Looks of amazement all around, followed by a lot of high fives. Lights fade.Music comes up.


1. InvestigationUnfinished business about Tiffany’s vertex-labeling scheme:

● The method involves generating a sequence of pairs of adjacent letters. Howdo you know, when a sequence of pairs repeats, that it first does so with theinitial pair? Why not earlier?

● How do know the labeling is consistent? Why can’t two different sequences ofpairs share a pair in common?

2. InvestigationUnfinished business about Tiffany’s lake algorithm: Could the method result inshore lines that have vertices or edges in common?

3. QuestionHere are clues about certain mystery surfaces. What are the possibilities for each one?

● Surface S has N(S) = −7.

● Surface T is orientable, has two lakes, and N(T) = −11.


● Surface U has three lakes, is nonorientable and N(U) = −10

● Surface V is orientable with N(V) = −8.

4. InvestigationHere are some surfaces you have seen before (see Investigation 5 of Chapter 7).You may have already calculated the number of lakes each one has.

● Which are orientable, and which are nonorientable? (How do youdecide?)

● What is the Euler number of each one? (How do you calculate N(S) for eachof the surfaces?)

● Use the information you have to figure out what each surface is in the classification scheme.

5. InvestigationJoe’s method for calculating N(S) from a pattern for S says: count the number of distinct letters L in the symbol and count the number of vertices V in thepattern using Tiffany’s algorithm. N(S) = V − L + 1. Does this method work? Why?


6. QuestionPictures of several surfaces appear below. What is each one in the scheme?

This tube is“hidden” on the

inside


7. InvestigationOn the sphere and torus, we have investigated maps of the following form. Everycountry has the same number (n) of edges, and at every vertex the same number(m) of edges meet. What are the possibilities for a cross-cap? Can you draw them?

These are hidden tubes on the inside.


8. InvestigationThis is the same as Investigation 7 but with the cross-cap replaced by a Klein bottle.What are the possibilities for a Klein bottle? Can you draw them?

9. InvestigationHere is another way to “assemble’’ the Klein bottle.

Similar to the usual version, the surface passes through itself: some points inthree-space that represent two points on the Klein bottle—in fact a whole circleof points. Use this figure-eight Klein bottle to create a Klein bottle in four-spacewithout “double-points.’’

10. InvestigationYou know that you can think of the Klein bottle as two Möbius bands gluedtogether along their edges. Use this to find yet another way of assembling theKlein bottle in four-space without having the surface pass through itself. (Idea:The model in Investigation 9 of Chapter 8 shows how to build a Möbius band sothat its edge lies in a plane.)

11. InvestigationSimilar to the Klein bottle, we have “assembled’’ the cross-cap in three-space by having the surface pass through itself, and thus producing “double-points.’’

A

B

A

B

A

A

Surface passes through itself to glue along A.Put a twist in this "Figure-8", double-tube and glue.

A

AB2B1

B1B2

B2B1

B1B2


Show how you might assemble a cross-cap in four-space so that these difficultiesare avoided.

12. QuestionA certain surface has the pattern shown below. Use several methods to figure outwhat it is in our classification scheme. One method might be to use the surfacesymbol and the rules of Chapter 10. Another method might be to use the results ofthis chapter together with Joe’s method for calculating N(S) (see Investigation 5).

13. SummarizingWrite a summary of the Acme team’s activities for the day. Include statements ofproblems, questions, definitions, results, explanations, relevant diagrams and pictures,and directions for future investigations. Make the summary so clear, organized, andunderstandable that Boss will be convinced that the time spent was worthwhile.

Notes

In the previous chapter, we found a complete list of all possible surfaces.In the first half of this chapter, we were able to show that this list contains noduplications. Along the way, we arrived at a slightly different way to identify a surface using the Euler number, orientation, and number of lakes. The key theorem in this is: two surfaces are the same if and only if they have the same Euler number, the same number of lakes, and either both are orientable or both are nonorientable. (The three items—Euler number, number of lakes, andorientation—are what is sometimes called a complete set of invariants.) Our argu-ment that number of lakes is an invariant of a surface is basically the same as thatof James (see also Blackett, pp. 76–79).

A

B

DC

E

A

C

B

ED

F

F

Notes 249

Assembling the Klein bottle in four-space leads to several questions. We’vedefined a surface as something made up of a map—a union of countries, edges,and vertices. But how can you draw a map in four-space? How can you cut out thecountries, flatten them out, and stretch/shrink them into discs? Whether youembrace them or not within the framework that has been set out in this book,several developments in the 19th century might explain what “surface in four-space’’means. The first development is the study of n-dimensional geometry; the secondis a definition of “same’’ for topological objects; and the third is a broader (yet natural) definition of the notion of surface. We’ll comment on each of thesein the following paragraphs.

In the middle of the 19th century, while mathematicians were beginning toaccept the idea of the existence of non-Euclidean geometries, geometers began toexplore the idea of n-dimensional geometry. One approach to this was to look atthe set Rn of all n-tuples (x1, x2, ... , xn) of real numbers xi which generalizes theplane R2 of all ordered pairs (x1, x2) and the space R3 of all ordered three-tuples(x1, x2, x3). (The notation is a 20th-century one.) Mathematicians defined notions(such as length, lines, planes, angle, generalized “volume’’) in Rn by analogy withhow they can be described in R2 (the Cartesian manifestation of the Euclideanplane) and R3 (the Cartesian manifestation of three-space). And they sought toprove theorems about these notions. In particular, they called each element of Rn

a point, and defined the distance between two points (x1, x2, ... , xn) and (y1, y2, ... , yn)to be (Σ(xi − yi) 2)1/2. Thus, by analogy, they were led to define the unit n-sphere tobe the set of points (x1, x2, ... , xn) such that Σxi

2 = 1. These definitions have persisted up to the present in defining and describing Euclidean n-space Rn.Euclidean four-space R4 is what Brandon and Tiffany were using when theyassembled the Klein bottle in four-space. For more details on the historical develo-pment of n-, dimensional geometry see the introduction of the book by Manning(see also the article by Cajori).

The second development is due to Möbius. We started out with the definitionthat two objects (maps, surfaces) are the same if one could be stretched/shrunkinto the other without tearing. We broadened this and said that we could call twosurfaces the same if a pattern for one could be stretched/shrunk into a pattern forthe other. In 1863 Möbius made this definition precise (and extended the range ofthe definition) by saying that two figures are the “same’’ if there is a one-to-onecorrespondence between the two figures such that neighboring points correspondto neighboring points (see Kline p. 1164f and Pont p. 90). The modern languagefor this is the following. Two figures F and G are topologically equivalent if and onlyif there is a function f: FÆ G that is 1-1, onto, continuous, and such that the inversefunction f−1 is also continuous. Such a function f is called a homeomorphism and Fand G are called homeomorphic (as well as topologically equivalent). The functionsthat Brandon and Tiffany created are homeomorphisms.

The third development occurs when, in 1895, Poincaré makes more precise thedefinition of a surface. First, some notions he uses in his definition: the n-discis the set of all points (x1, x2, ... , xn) in Rn such that Σxi

2 < 1; a neighborhood ofa point x0 in F is the set of all points in F—a subset of Rm for some m—with a


distance to x0 that is less than ε, for some positive number ε. Armed with thisterminology, Poincaré defines a surface (without boundary) to be a boundedsubset F of Rm such that every point x in F has a neighborhood that is homeomor-phic to a two-disc. Moreover, he generalized this: a bounded set (in Rm for some m)is an n-manifold (without boundary) if every point has a neighborhood homeo-morphic to an n-disc. In this language, a surface (without boundary,i.e., no lakes) is a two-manifold. It is in this sense that we can think of the Kleinbottle in four-space as a surface. We reconcile this definition with our earlier oneby noting that you can think of a two-manifold as a bounded set in Rm made upof the elements of a map, except that this time an edge is homeomorphic to theunit interval [0, 1] and a country is homeomorphic to the set {(x, y): x2 + y2 ≤ 1}.Incidentally, Poincaré also defined a manifold with boundary in a similar manner(see Klein, p. 1170f).

Additional historical steps occurred that created an even more abstract defini-tion of a surface. In the early part of the 20th century, Hausdorff (Klein, p. 1160)took a set T together with a certain collection of its subsets called neighborhoods.If these neighborhoods satisfied a certain set of axioms (basically incorporatingand generalizing the properties of neighborhoods of a subset of Rm as definedabove), then T would be called a topological space. By use of this idea, here is howone might define an “abstract’’ Klein bottle. The set K is the rectangle shown below.K includes the interior of the rectangle and some but not all of its boundary, as noted.

Now we need to describe the neighborhoods of K. (We will not do them all—just enough to give you an idea.) Start with all the “usual’’ neighborhoods ofpoints within the rectangle itself plus neighborhoods that capture the peculiargluing that’s supposed to occur to make the Klein bottle. Here are some examples.

A

B

A

B

A

A

B B

A

A

B B

A

B

A

B

Notes 251

If you “close up’’ this collection using the set operations of finite intersectionand arbitrary union, then you get a larger collection of subsets satisfying theaxioms of neighborhoods of a topological space. This makes K a topologicalspace. It is an abstract definition of a Klein bottle.

This chapter and the last were devoted to the successful classification of allsurfaces. In the language of Poincaré, this is a classification of all two-manifolds(with and without boundary). An important question that we have mentionedbefore is “What are the possible (three-) spaces that we live in?’’ A possible spacewould be a three-manifold, and an associated mathematical problem would be to classify all three-manifolds. (As an introduction to this problem, see the bookby Weeks.) This problem has not been solved. Solving it is the object of a lot ofcurrent research in mathematics. A lot of work on the problem has already beendone (see the book by Thurston).

Lots of folk have written about four-space in more informal ways than we haveabove. Of course, Flatland, the 1884 book by Abbott, in which a three-dimensionalcreature visits a two-dimensional world, is the classic introduction to thinkingbeyond the third dimension. A particularly good book that explores four-spacefrom many perspectives is the one by Banchoff. In particular, he discusses ways of visualizing four-dimensional shapes by using computer graphics. He puts his techniques to work in the video The Hypercube, an animated “look’’ at thefour-dimensional analogue to the three-dimensional cube. Other books that explorefour-dimensional thinking, especially through the use of analogy, are those byWeeks, Kinsey and Moore, and Rucker.

References

Abbott, E.A. Flatland. New York: Dover, 1952.Banchoff, T.F. Beyond the third dimension. New York: W. H. Freeman, 1990.Banchoff, T.F. The hypercube: projections and slicing. Video. Providence, RI. Thomas Banchoff

Productions, 1978.Blackett, D.W. Elementary topology. San Diego: Academic Press, 1982.Cajori, F. Origins of fourth dimensional concepts. American Mathematical Monthly, 1926,

397–406.James, R.C. Combinatorial topology of surfaces. Mathematics Magazine, September/October 1955,

pp. 1–39.Kinsey, L.C., and Moore, T.E. Symmetry, shape, and space. Emeryville, CA: Key College Publishing,

2002.Kline, M. Mathematical thought from ancient to modern times. New York: Oxford, 1972.

pp. 1164–1166.Manning, H.P. Geometry of four dimensions. New York: Macmillan, 1914.Pont, J.C. La Topologie Algébrique des Origines á Poincaré. Paris: Presses Universitaires de France,

1974.Rucker, R. Geometry, relativity, and the fourth dimension. New York: Dover Publications, 1977.Rucker, R. The fourth dimension: a guided tour of the higher universe. Boston: Houghton Mifflin

Company, 1984.Thurston, W.P. Three-dimensional geometry and topology, Vol. 1. Princeton: Princeton University Press,

1997.Weeks, J.R. The shape of space, 2nd ed. New York: Marcel Dekker, 2002.


Acme Colors Maps onSurfaces: Heawood'sEstimate

253

A couple of days later. Tiffany, Brandon, and Joe are at their desks working onreports for Boss. Millie is standing at the window, painting her nails.

MILLIE: Ohmagosh! Here comes Pinkley Smith. Wonder what he’sup to? (She puts her nail polish back in her desk in a hurry.)

JOE: He must’ve heard what we’ve been doin’

MILLIE: Well, he’s here! (Pinkley Smith walks in.)

PINKLEY SMITH: Hello, fellow map folk!

BRANDON: Map folk?

JOE, MILLIE,TIFFANY: Hello, Pinkley Smith!

TIFFANY: To what do we owe this visit, Pinkley? We haven’t donemuch with map coloring lately.

PINKLEY SMITH: So I’ve heard. That’s just why I’ve come, to bring a little“color” back into your lives.

C H A P T E R

12

ALL: Groan...

MILLIE: Out with it, Pinkley. What have you got up your sleeve?Come to make up for that lawyerly sleight-of-hand you fedus last time?

PINKLEY SMITH: Friends, friends, friends! Have mercy. Just a little slip andyou take it out on me forever?

BRANDON: Little slip. You really pulled the wool...

PINKLEY SMITH: Peace, my colleagues—and you are my colleagues! I think I have something for you that will tickle your toes and titil-late your tummies.

TIFFANY: Toes? Tummies?

PINKLEY SMITH: I heard that you made a major breakthrough by classifyingall surfaces. So I thought, “How could I, humble PinkleySmith, make a contribution, however meager, to this newknowledge created by my fellow topologists?”

JOE: Cut to the chase, Pink.

PINKLEY SMITH: Pink? I thought, “There’s a theorem about coloring maps ona sphere: you only need four colors. How about all theseother surfaces? How many colors does a map on anothersurface need?” So, I came to tell you the answer to this question.

TIFFANY: Hmm. Good question. I think we’d really like to know theanswer.

PINKLEY SMITH: OK. Here it is.

BRANDON: Wow! That is interesting. Are you going to show us why?

PINKLEY SMITH: Of course, fellow mappies. I thought you’d never ask.

MILLIE: Mappies?

254 Chapter 12 Acme Colors Maps on Surfaces: Heawood’s Estimate

Map Color TheoremSuppose that S is a surface without boundary (i.e., no lakes) and that N(S) isits Euler number. Then any map on S will need no more than

colors in order for it to be colored properly.

7 49 24

2

+ − N S( )

JOE: What about surfaces that have boundaries?

PINKLEY SMITH: I’m not going to tell you everything! I’ll leave that to you.

1. Your Turn

Suppose that S′ is surface S with a lake. What can you say about the number ofcolors needed to color maps on S′ compared with the number of colors neededto color maps on S?

PINKLEY SMITH: Take a map M on surface S. The first thing we want to do isto replace the problem of coloring the countries of M withthe problem of coloring the vertices of the dual graph.

JOE: Dual graph?

PINKLEY SMITH: Yes. Coloring the vertices of a graph is easier. Remember thata graph is a collection of vertices and edges. Each edge joins apair of vertices; each pair of vertices is joined by at most oneedge. The rule for coloring the vertices of a graph properly isthat two vertices joined by an edge must be colored differentcolors. Here’s how you create the dual graph to a map. Takethe map and provide each country with a capital, a point inits interior. The capitals are the vertices of the dual graph.Join two capitals (vertices) with an edge exactly when the twocountries share a common border.

TIFFANY: I think I understand. Coloring the vertices of the dualgraph is the same as coloring the countries of the originalmap. The dual graph and the map need exactly the samenumber of colors for them both to be colored properly.If that’s the case, then why do you need this new object,the dual graph?

PINKLEY SMITH: Good question. It turns out that the graph is simpler. Also,you were always worried about the map being “OK.”

VA VB

VC

VA VB

VC

C

BA

DUALGRAPH

MAP

Chapter 12 Acme Colors Maps on Surfaces: Heawood’s Estimate 255

We won’t have to worry about that with the graph. Here’s asimple example of one difference. If two countries—A andB—on the map have more than one common border, thenthe dual graph has the corresponding vertices vA and vB butonly one edge joining them.

Another thing: a graph doesn’t have to be the dual graph ofsome map in order to color it. You’ll see in what followsthat there are other advantages to using the dual graph.

2. Your Turn

Construct the dual graph to the following map on an island.

PINKLEY SMITH: To prove the map color theorem, I'm going to need somenotation and a definition.

Chromatic Numbers and Critical GraphsThe number of colors c(M) needed to color map M properly is called thechromatic number of M.If S is a surface, then c(S), the chromatic number of S, is the maximum of allc(M) where M is a map on S.The number of colors c(G) needed to color graph G properly is called thechromatic number of G.If removing edges and/or vertices from a graph G always results in a graph G'such that c(G') < c(G), then G is called a critical graph.

VA VB

BA


Now start with a map M on surface S. Construct the dualgraph G. If necessary, remove vertices and edges from G to getyet another graph G′ on S that has the following properties:

● c(G′) = c(G) = c(M).

● G′ is a critical graph.

So G′ is a graph that is possibly simpler than the dual graph, butit has the same chromatic number as the dual graph. Moreover,it’s critical. Here’s a reason why a critical graph is useful:

BRANDON: OK. Critical. That’s a nice property for a graph. But why isit good for us?

PINKLEY SMITH: Glad you asked! We want to know how many colors to colorall maps on a surface S. Start with a map M that requires themaximum number of colors for the surface S. Take the dualgraph G, and remove edges and vertices to get a critical graphG′ with c(M) = c(S) = c(G) = c(G′). The critical graph property will enable us to connect c(G′) with the Eulernumber N(S) in a useful way. The first step in making thisconnection is: If every vertex of G′ has order ≥ c(G′) − 1, thenthe average order of vertices of G′ must also be ≥ c(G′) − 1.

Critical Graph and the Average Order of VerticesSuppose that K is a graph. Let A(K) denote the average order of the vertices of K.If K is critical, then A(K) ≥ c(K) − 1.

Lemma about Critical GraphsLemma. If graph K is critical with chromatic number c(K), then every vertexof K has order at least c(K) − 1.Proof: This is going to be a proof by contradiction. So suppose that the lemmaisn’t true. Then there must be a vertex v whose order is less than c(K) − 1. Fromthe graph K remove v and all the edges joined to it. Obtain graph K′. Because Kis critical, K′ can be colored in c(K) − 1 or fewer colors. Color K′ properly. Nowput the vertex v you removed back along with all the removed edges. Color Kusing the coloring you’ve already started by coloring K′. All you need to do iscolor the vertex v. Since the order of v is less than c(K) − 1, the vertices joined to v have been colored with fewer than c(K) − 1 colors. This leaves one of thec(K) − 1 colors with which to color v. Do so! This means that K is now coloredwith c(K) − 1 colors. This contradicts the fact that K needs c(K) colors for aproper coloring. We were led to this contradiction by the assumption that K hasa vertex of order less than c(K) − 1. So it must be the case that every vertex ofK has order greater than or equal to c(K) − 1. This proves the lemma.


BRANDON: I still don’t…

PINKLEY SMITH: Hold on. There’s more!

MILLIE: Hey, there’s something familiar with that equation! Thenumerator counts edges—sort of. In fact, it’s 2E, where E isthe number of edges of G′. So you’d get

3. Your Turn

If K is the network of a map on the sphere, can you say anything about A(K)?

TIFFANY: Now I can see a little bit where you might go with this.The E and V are the E and V of Euler’s formula. But what I don’t see first off is what you’re going to use for C because G′is just a graph. Then, since G′ isn’t necessarily the network ofan OK map, how can you write V − E + “C ” = N(S)?

PINKLEY SMITH: One step at a time. What do we use for C? The idea is this:the graph G′ sits on surface S. Cut the surface along all theedges of G′. What do you get?

JOE: If G′ were the network of a map, you’d get a bunch of coun-tries. I guess what you’d get in this case would be a bunchof pieces of the surface. Hey, each piece would be somekind of surface with one lake. The edges of the lake wouldbe made up of edges from G′.

PINKLEY SMITH: Yes! Suppose there are P of these pieces. In fact, every edge of G′ will be an edge of one of those lakes. Moreover, thisgives us a way to compare P with E, the number of edges of G′.Remember how, in an OK map with data V, E, and C we’dcount the edges of each country, add them up, and get 2E?

MILLIE: Sure do, Pinkley.

PINKLEY SMITH: Same thing with the pieces. Count the edges of the lakes ofeach piece. Add them up. You’ll get 2E. Now I claim that thelake on each piece will have at least three edges from G′.

A(G′) = 2E/V ≥ c(G′) − 1

or just

2E/V ≥ c(G′) − 1.

Let Vi denote the number of vertices of G′ of order i and let V denote thetotal number of vertices of G′. Then

A(G′) = 2V2+3V3+4v4+…V


Otherwise, if there were just one or two edges from G′ you’dhave something like this in G′.

Neither can happen because G′ is critical. Consequently,

BRANDON: That looks a lot like something we got when we were color-ing maps on the sphere. Back then we got 3C ≤ 2E. I thinkwe used that to prove the five-color theorem.

PINKLEY SMITH: Yes, so we did! And we’ll use this new inequality to proveanother coloring theorem.

TIFFANY: And is there going to be some kind of Euler’s formulainvolving V, E, and P?

PINKLEY SMITH: You fellows are sharp as tacks. That’s just what I want to donext: find a relationship between V − E + P and N(S). Here’show you do it. To the graph G′ on S, add vertices and edgesso that the surface pieces have been cut up into the countriesof an OK map. At each stage let’s keep track of the numberof vertices V′, edges E′, surfaces pieces P′ you’d get, as well asV′ − E′ + P′. At each stage one of the following happens.

1. Add a vertex of order 2 and an edge.

2. Add an edge joining two vertices.

Cut surface S along the edges of the graph G′ to obtain P surfaces. Then

3P ≤ 2E,

where E is the number of edges of G′.


3. Add an edge to a single vertex.

In case 1, the number of vertices and the number of edgeseach increases by 1 and the number of surface pieces staysthe same. Result: V′ − E′ + P′ doesn’t change.In cases 2 and 3, the number of vertices stays the same, thenumber of edges increases by one and the number ofsurface pieces may increase by one, but not necessarily.Result: V′ − E′ + P′ ≤ what it was for the previous stage.So V′ − E′ + P′ can only decrease. In the end, you’ll have data V′, E′, and P′ with V′ − E′ + P′ ≤ V − E + P and V′ − E′ + P′ = N(S). Consequently,

JOE: Lotta stuff. Could you do a little review of where we are so far?

PINKLEY SMITH: Of course! Here’s a list of the pieces we have.

Now for a little calculating.

What we know about G′ on surface S:3P ≤ 2EN(S) ≤ V − E + Pc(S) − 1 ≤ 2E/Vwhere G′ has V vertices, E edges, and—when cut along—P surfaces pieces.

N(S) ≤ V − E + P


The second inequality gives us

3N(S) ≤ 3V − 3E + 3P.

The first inequality then gives us

3N(S) ≤ 3V − 3E + 3P ≤ 3V − 3E + 2E.

From that we get

3N(S) ≤ 3V − E.

BRANDON: Wow. We’ve got an expression just involving c(S), N(S),and V!

PINKLEY SMITH: Time to get rid of V! First off, c(S) ≤ V. I’ll let you figure outwhy that’s the case.

4. Your Turn

Why is it that c(S) ≤ V?

PINKLEY SMITH: Second off, I’m going to assume that N(S) ≤ 0. More calculations.


That eliminates P! Another way of writing this is

E ≤ 3V − 3N(S)

or

2E ≤ 6V − 6N(S)

(we're heading towards using the third inequality...).From that we get

2E/V ≤ 6 − 6N(S)/V

The third inequality then says

c(S) − 1 ≤ 6 − 6N(S)/V

Starting with the following inequalities,

c(S) − 1 ≤ 6 − 6N(S)/Vc(S) ≤ V

N(S) ≤ 0we get

c(S) − 1 ≤ 6 − 6N(S)/V ≤ 6 − 6 N(S)/c(S)

or

c(S) − 1 ≤ 6 − 6N(S)/c(S),

an inequality involving only c(S) and N(S)!

PINKLEY SMITH: That does it! That proves the map color theorem! (Looks atwatch.) My, my, time flies when you’re having fun. Must be off!Theorems to prove! Uh, I mean cases to solve. Ta, ta, mappies!

BRANDON: Ta, ta, mappies? But what happens when N(S) > 0? Pinkley,come back!

MILLIE: He’s gone.

Lights fade. Music comes up. The Acme team is stunned.

Investigations, Questions, Puzzles and More

1. Investigation

Help Brandon figure out what happens when N(S) > 0.

1. In case N(S) = 1, what could S be? Here’s an idea for an argument for thiscase. Go back to the inequality c(S) − 1 ≤ 6 − 6N(S)/V in the argumentgiven by Pinkley Smith. This inequality is true no matter what N(S) is. Butin case N(S) = 1, the inequality is c(S) − 1 ≤ 6 − 6/V. Good luck!

2. In case N(S) = 2, S is the sphere. What does the map color theorem say forthis case? Does the map color theorem provide a new proof of the four-color problem for the sphere?


Another way to write the inequality is

c(S) − 7 + 6N(S)/c(S) ≤ 0.

If you multiply both sides of the inequality by c(S) you’ll get

c(S)2 − 7c(S) + 6 N(S) ≤ 0.

The left hand side is a quadratic! Let’s factor it:

[c(S) − 1/2(7 + (49 − 24N(S))1/2)] [c(S) − 1/2(7 − (49 − 24N(S))1/2)] ≤ 0.

Since N(S) ≤ 0, the root of the quadratic in the right hand factor is negative,making the whole right-hand factor positive. That means that the left-handfactor must be less than or equal to zero:

c(S) − 1/2(7 + (49 − 24N(S))1/2) ≤ 0

or

c(S) ≤ 1/2(7 + (49 − 24N(S))1/2).

2. InvestigationSuppose that you can solve the n-cities problem on a p-holed doughnut. What canyou say about the relationship between n and p? (First try the case in which all thefaces are triangles. Then consider the case when there is no restriction on the faces.)

3. QuestionYou are interested in coloring maps on a 17-holed doughnut. You want to haveenough colors on hand to color all possible maps. But, of course, you do not wantto have more colors than you need. What is your best estimate for the number ofcolors you should have?

4. InvestigationYou want to find a surface on which you can solve the 10-cities problem. Find aspecific surface, and show how you would solve it there. You’d also like to knowthe smallest n such the 10-cities problem could be solved on an n-holed dough-nut. Use the Euler number for an n-holed doughnut and counting arguments tofind out what might be possible. How does what you get jibe with what the mapcolor theorem tells you?

5. InvestigationWe’ve solved the five-cities problem on the Möbius strip (see Investigation 13 ofChapter 7). What about the six-cities problem?

6. InvestigationIf a map can be colored in two colors, we know that every vertex must be even.We also know that the converse is true for maps on the sphere: if every vertex iseven, then the map can be colored in two colors. Is this true for maps on a p-holeddoughnut (p > 0)? If this is not true, what can you say?

7. InvestigationConsider this statement about a map on a surface: If every vertex of a map is oforder 3 and the map can be colored in 3 colors, then every country must have aneven number of edges. We know that this is true for maps on a sphere. What aboutmaps on a p-holed doughnut (p > 0)? We know that the converse of the statement



is also true for maps on the sphere (see Investigation 2 of Chapter 4). Is it true formaps on a p-holed doughnut (p > 0)? If not, what can you say?

8. InvestigationSuppose you have a map on a surface with the following property: for every pairof neighboring countries at least one is a triangle. Someone said that such a mapcan be colored in four colors. Is this an old wives’ tale, or what?

9. Investigationa. Suppose that G is the graph whose vertices and edges are the vertices and edges

of an n-gon. What can you say about c(G)?b. Suppose that G is the graph of the utilities problem with n houses and

m utilities. What can you say about c(G)?

10. QuestionGiven a surface, can you find a graph that you cannot draw on it without havingthe edges cross?

11. QuestionSuppose graph G with V vertices and E edges has c(G) = 2. What can you sayabout the relationship between V and E?

12. InvestigationOn a surface, a map with a sufficient quantity of small countries (small, in thesense of number of borders) can be colored in seven colors. What does “sufficientquantity” mean? Is it true?

13. InvestigationCould you solve the six-cities problem on the Klein bottle? What does this sayabout c(Klein)?

14. ConstructionIn a paper of 1910, the mathematician Heinrich Tietze describes the followingconstruction of a polyhedral surface (see Biggs et al., pp. 128–129).


“Let A, B, C, D, E be the points with rectangular coordinates (0, 0, 0), (1, 0, 0),(0, 1, 0), (2/3, 2/3, 2/3), (0, 0, 1), and let F be any point which does not lie in a planewith any of the five triangles....The said polyhedron consists of the 10 triangles:

1. ABC, BCD, CDE, DEA, EAB2. FAC, FBC, FCE, FDA, FEB.”

What is this surface? What can you say about the map whose countries are the10 triangles above?

15. ConstructionHere is a Möbius band to assemble. It has a map with six countries. (The countriesare labeled.) How many colors do you need to color the map properly?

16. ConstructionThe object of this construction is to build a paper model of a torus having a mapon it that requires seven colors.

How to build this model:

● Make seven copies of the shape below on heavy paper.

● Cut out all pieces precisely.

● Score along all fold lines, but not the thick diagonal lines.

● Fold away from you along each fold line and crease firmly.

● Glue together a chain of the seven pieces by using the flaps marked “A.”

5

3

6 4 5

1 2

65

3

6 4 5

1 2

6


● Complete the model by starting at one end of the chain and moving to theother gluing as you go.

● The thick lines divide the torus model up into the countries of a map. Colorthe map properly.

17. QuestionBy boring a hole through a solid sphere, you get an object with a surface that is a torus.On that surface you can construct a map with seven countries, each of which is adjacent to the other six. Starting with that map on the surface of a solid torus, couldyou bore a hole through it and construct a map with eight countries, each of which isadjacent to the other seven? (The model in 14 might be a place to start.)

18. SummarizingWrite a summary of the Acme team’s activities for the day. Include statements of prob-lems, questions, definitions, results, explanations, relevant diagrams and pictures, anddirections for future investigations. Make the summary so clear, organized, andunderstandable that Boss will be convinced that the time spent was worthwhile.

GLU

E

GLU

EG

LUE

GLU

E

GLU

E

GLUE

GLU

E

A

Notes

The inequality in the map color theorem is often called Heawood’s inequality.Heawood proved this for orientable surfaces in 1890 in the same paper in whichhe found a counterexample to Kempe’s earlier “proof” of the four-color conjec-ture. This was also the same paper in which he proved that every map on thesphere could be colored in five colors or less (see Notes in Chapters 4 and 5). Since1/2(7 + (49 − 24N(S)1/2) is not always an integer but c(S) is, the inequality is oftenwritten

where [x] denotes the greatest integer in x. The number isoften called Heawood’s number and is denoted h(S).

In the same paper, Heawood also claimed that the inequality is actually anequality. He stated that in order to prove that c(S) = h(S), one must construct amap on S that requires h(S) colors. Here is the map he constructed for the torus.

The torus is the only case for which he constructed a map. For the other cases,he simply stated “there are generally contacts enough and to spare.”

The fact that Heawood really showed c(S) = h(S) only for S = torus was pointedout a year later in a paper by Lothar Heffter (see Biggs et al., p. 118+). The equa-tion c(S) = h(S) became known as Heawood’s conjecture. Heffter’s work was thebeginning of 70 years’ effort to prove the conjecture.

Heffter’s method of attack on the problem was to construct maps on surfacesin which each face is adjacent to every other one, what he called a system of neigh-boring regions. (In the case in which the map is a system of n-neighboring regions

2

6

6

1

1

5

53

3

4

4

27

7

7 49 24

2

+ −⎡

⎣⎢⎢

⎤

⎦⎥⎥

N S( )

c(S)N S≤ + −7 49 24

2

( )

Notes 267


on a surface, we recognize that the dual graph is a solution to the n-cities problemon the surface.)

In one part of his paper, Heffter asked for the least number g such that n neigh-boring regions may be constructed on a g-holed doughnut. He called this number g(n). Using Euler’s formula and some counting arguments, he showedthat g(n) > {(n − 3)(n − 4)/12}. Here {x} is the least integer not less than x (seeInvestigation 2 above).

Heffter realized that trying to draw a system of n neighboring regions on a surfacewas cumbersome as was drawing the solution to n-cities problem. He devised aclever, new approach; he constructed the adjacency pattern for a map. Here is howhe did it. Number each country of the map from one to n. Take a country and walkaround the borders of the country in a counterclockwise direction, and record, inorder, the bordering countries you encounter on your trip. For example,

Then, the record of countries bordering country 5 is 1, 6, 7, 3, 2, 4.Do this for all countries of the map. You will obtain an array of adjacencies. For

the map on the sphere below (a system of four neighboring countries), you willarrive at the array that follows the map.

Country Record of Bordering Countries1 2 4 32 1 3 43 4 2 14 3 1 2

2

1

3

4

5

7 6

2 4

13

Notes 269

If you did this for the map on the torus below (a system of seven neighboringcountries), you would get the array that follows it.

Heffter observed that this array exhibits an amazing pattern: you can obtain everyrow from the one just above it by adding 1 to each entry. (Arithmetic is modulo 7.)This is certainly not something you would have noticed by looking at the original map. Another surprising fact is that the numbers 1, 2, 3, … are just labelsfor the countries. They’re arbitrary! But when we use them, the array manifests a rigid arithmetical pattern!

Heffter exploited his observation and was able, in some special cases, to createmore arrays that corresponded to systems of n neighboring countries on surfacesof genus (n − 3)( n − 4)/12. He assumed that n leaves a remainder of 7 on divisionby 12 so that this fraction is a whole number.

The first major breakthrough to follow Heffter’s pioneering work was made by G. A. Dirac in 1952 when he showed that if a g-holed doughnut had a maprequiring h(g) colors, then the h(g)-cities problem could also be solved on thesurface. This confirmed that Heffter’s approach—of trying to solve the h(g)-citiesproblem on a g-holed doughnut—was the right one.

In a program that began around the time of Dirac’s result, Ringel and Youngsand others had by 1968 constructed arrays (à la Heffter) for all cases, thus provingthe Heawood conjecture. A summary of the events leading up to the final proof

Country Record of Bordering Countries1 2 4 3 7 5 62 3 5 4 1 6 73 4 6 5 2 7 14 5 7 6 3 1 25 6 1 7 4 2 36 7 2 1 5 3 47 1 3 2 6 4 5

15

3

162

4

6

6 2

4

6

75

377

1

7

can be found in the article by Stahl. Details of the proof can be found in the bookby Ringel (see also the book by Gross et al.) A remarkable fact about this proof,which tells you all you need to know about all orientable surfaces other than thesphere, is that it came eight years before the proof of the four-color conjecture, thecoloring theorem for maps on the sphere.

In coloring maps on surfaces, neither Heawood nor Heffter considered nonori-entable surfaces. However, in 1910, Tietze made the appropriate Heawood conjecturefor nonorientable surfaces. The formula is the one proved in this chapter. Previouswork of Möebius confirmed the conjecture for the cross-cap (see Investigation 15above). In 1934 Franklin showed that, even though h(Klein) = 7, in fact it was thecase that c(Klein) = 6. In 1954, Ringel showed that the Heawood conjecture is truefor all nonorientable surfaces (except the Klein bottle) by producing Heffter-stylearrays for the required systems of neighboring regions. It turned out that nonori-entable surfaces were easier than the orientable ones (for some details of thehistory of this problem, see Biggs et al., chapter 7, and Stahl).

References

Biggs, N.L., Lloyd, E.K., and Wilson, B.J. Graph theory: 1736–1936. Clarendon Press, Oxford, 1976.Boltyanskiı, V.G., and Efremovich, V.A. Intuitive combinatorial topology. Springer-Verlag, New York, 2001.Gross, J.L., and Tucker, T. Topological graph theory. New York: Wiley, 1987.Ringel, G. Map color theorem. Springer-Verlag, New York, 1974.Stahl, S. The other map coloring theorem. Mathematics Magazine, May 1985, pp. 131–145.


Acme Gets All Tied Upwith Knots

271

Scene 1

The gang is busy at work at its desks. Boss is reading the mail, concentrating on one item. He turns it over and upside down, holds it up to the light, squints his eyes, screws up his face like a prune, shrugs his shoulders, and turns to Brandon ...

BOSS: This stuff is just plumb loco. These critters want us to do somethingabout ladders. Möbius ladders! Doesn’t that just beat the cake? Whodo they think we are? (Tosses the letter in the wastebasket.)

BRANDON: Hold on, Boss! Let me see that. (Retrieves letter from trash.)Hmm. You’re right. Möbius ladders. Interesting. These guys,Molecules.com, heard what we did with twisted strips. (Turns tothe others.) Hey, guys, we’re famous! (Turns back to letter.) They think we might be able to help them. (Passes letter to the others.)

C H A P T E R

13

BOSS: Bosh. Ladders. I don’t want you people going on another wildgoose chase.

MILLIE: Waddaya mean, Boss, “wild goose chase”?

BOSS: You people have enough to do. You think this is some kind of“think tank”? We’re here to make dinero.

JOE: (Looking at letter.) Says here they want to use our expertise.They’d pay us for stuff we’d do on their problem.

BOSS: Huh? (Takes back letter, looks again.) Well, you know I was justjoshing. You folks take care of it. (Returns letter to group.) Comeon, get on with it! Figure this ladder stuff out. Don’t have all day.(Stomps out.)

TIFFANY: Hmm. “...just joshing...” (Starts to read letter.)

BRANDON: (Looks over Tiffany’s shoulder.) This does look cool. I think wecan have a lot of fun with this project. If I understand thingscorrectly, the folks at Molecule.com want us to look at a mole-cule that looks something like this.

The O’s stand for oxygen atoms; the black circles stand forcarbon molecules. So there’s a single chain of...hmm...60 oxygenand carbon atoms hanging together by single bonds. Thenthere’s some carbon-carbon double bonds represented by thedouble lines. It’s like a Möbius band, where the edge is the chainand double bonds replace the surface. The double bonds are likethe rungs of a ladder. So, a Möbius ladder!

JOE: What do they want us to do?

TIFFANY: You can put the ladder together in many ways.

272 Chapter 13 Acme Gets All Tied Up with Knots

BRANDON: What does that mean?

TIFFANY: (Hands letter to Brandon.) I’m trying to sort this out. I think it’slike when we had a pattern for a Möbius band. You could reallyassemble the pattern in a lot of different ways. The pattern isreally a pattern for a lot of things: a strip with a single half-twistor a strip with three half-twists or, in general, a strip with an oddnumber of half-twists. Well, it’s the same with the Möbius laddermolecule. Break some of the bonds and flatten it out, and you’dget something like this.

Of course, the real molecule is what you get when you glue theends together. But we know you can get lots of things—a singlehalf-twist ladder, a three-half-twist ladder—anything with anodd number of half twists. Chemically, all the bonds are thesame. So all of those are the same molecule. Sort of.

BRANDON: They’re the same molecule. But a half-twist ladder might havedifferent physical properties from a three-half twist ladder: onemight be like olive oil, the other might be like Jello.

JOE: I still don’t understand what they want us to do.

BRANDON: Hold on, I’m getting there. The folk at Molecule.com want us tofigure out which pairs of these molecules are really different,meaning you can’t manipulate one into the other in three-spacewithout cutting one and regluing it to make the other.

MILLIE: Are you allowed to stretch or shrink?

BRANDON: They probably wouldn’t like it. But maybe we can...

JOE: So this is a little different from what we did with surfaces. All wedid there was look at patterns and differentiate things that way.Now we’ve got to look at all the things you can get from a singlepattern and figure out some way of classifying them!

BRANDON: I think that’s what we need to do.

TIFFANY: I have an idea for how to get started. It will make things simpler,I think. If you took one of these ladder molecules and manipu-lated it in three-space—no cutting—and you managed to get

B

A

A

B

Scene 1 273

another ladder molecule, then the same would be true of theedge: you could take the edge of one, manipulate it, and get theedge of the other. So here’s the idea: figure out some way to clas-sify the edges. If two edges are different, then the ladders aredifferent. Hmm. If two edges are the same, well, who knows whatyou could conclude. Might be like a torus and the Klein bottle:they both have same Euler number, but they’re different surfaces.

BRANDON: Let’s go with the edges idea. An edge of one of those ladders isjust a simple closed curve in space. I think it’s time to look atsome examples.

MILLIE: Good idea. You know, we can just look at the edges of odd-twisted strips. They’re really the same as the edges of the molec-ular ladders. Here are the edges of a one (half-)twisted strip anda three (half-)twisted strip.

JOE: Looks like the edge of the one-twist strip is just a circle. And theedge of the three-twist strip is some kind of knot?

MILLIE: Looks like an overhand knot with the ends glued.

TIFFANY: That’s what we can call these things: knots. So we’ll call a simpleclosed curve in space a knot. The edge of an odd-twisted strip isa knot!

JOE: When I tie my shoes, I make an overhand knot like this.


Is it the same as yours, Millie?

MILLIE: Looks the same.

JOE: Know what? You can do twisted strips in different ways. You cantwist one way, then glue. Or you could twist another way, thenglue. Like this.

JOE: Are the edges the same?

Guess what? Those are the two overhand knots we got before.

MILLIE: They still look alike.

Your Turn 1Use rope to make the two overhand knots. (Be sure to secure the ends.) See if youcan manipulate one into the other. (No cutting!) What happens?

One half twist

Three half twists

Three half twists

One half twist

Scene 1 275

Your Turn 2Here are four simple knots. Are there pairs of these that are the same?

Your Turn 3

Make two five-half twist strips: one strip made with twists in one direction,the other with twists in the other direction. Look at the edges you get. Use ropesto make the two knots. See if you can manipulate one into the other. Whathappens?

BRANDON: (Looks at the two overhand knots.) Hmm. Now that’s a real problem.Those do look alike. I guess you’d say they’re mirror images ofeach other. Question is: Are they the same? Can you manipulateone into the other? Or can you show you that can’t?

TIFFANY: Time for some simplification. You could turn one of those knotsinto a real mess. Let’s see if we can keep things clean. Those pictureswe had before are pretty neat. Sort of flattened out pictures ofknots, or projections of knots onto a plane. Seems to me we couldkeep track of all the manipulations with these pictures. Let’s seewhat we have. If you held a knot above a plane with a light aboveit, you’d see a shadow on the plane—something like this.

MILLIE: Looks like a nice closed curve in the plane!


TIFFANY: Absolutely! But a place where the projection curve crosses itselfreally represents two points on the original knot. Let’s call sucha point a crossing. At a crossing, the piece of the closed curvenearest the light source is an overpass; and the piece farthestfrom the light source is an underpass. We draw the overpass inone stroke and the underpass in two. The underpass is broken toindicate that the location of the point of the original knot ishidden by the projecting light. So a crossing of the projectioncurve could be drawn in one of two ways:

MILLIE: It’s as if the knot really lies in a single plane except at the cross-ings where the curve passes for a bit under the plane.Underpasses. Hmm. Clever.

Your Turn 4

What does a picture of a square knot look like? How many crossing does thepicture have? What does a picture of a granny knot look like? How many crossingsdoes it have?

Your Turn 5

What does a picture of the edge of a five-half-twist strip look like? How manycrossings does it have?

BRANDON: OK. This is a good start. A picture of a knot is sort of like a patternfor a surface. Of course, unlike a pattern, we can’t “cut” the pictureand reglue it. But what sorts of things can we do with it?

TIFFANY: What we can do with the picture ought to reflect what we can dowith the knot in space, but very, very slowly so we still have a picture.

CROSSING

UNDERPASS OVERPASS

UNDERPASSUNDERPASS

UNDERPASSOVERPASS

OR

Scene 1 277

BRANDON: Sort of like little baby steps?

JOE: Little baby steps like these?

MILLIE: Hey. Not bad. (Bad knot!) How about this?

TIFFANY: Here’s another one.

BRANDON: How about sliding a hunk of the picture over a crossing. Like this.

MILLIE: What if you didn’t like what you did? You ought to be able toundo each one of those.

TIFFANY: Yes. Good idea. We ought to include the “inverses” of each ofthose “moves.” Here’s a list of the moves we’re allowing.

(1)–1

(1)

OR


JOE: Hey. Look what I did. While you guys were writing all that down,I took a pair of knot pictures that look pretty different but thatwe know are pictures of the same knot. Here they are.

Then I tried to see if we could use just our moves to change onepicture into the other. Here’s what I got.

(5)–1

(5)

(4)–1

(4)

(3)–1

(3)

(2)–1

(2)

Scene 1 279

MILLIE: Fifty-two skidoo! Will that always work?

JOE: Will what always work?

MILLIE: If two pictures are of the same knot, then could you use thosemoves to change one picture into the other?

BRANDON: Good question. Looks like it might be. Let’s think about it.

TIFFANY: Here’s another thought. Suppose that these moves really were enough. In other words, suppose the answer to Millie’squestion is “yes.” Now suppose you had pictures of two knots.You don’t know off hand whether they’re the “same” or not.So you use the moves on one of the pictures, and try as much asyou can, you don’t arrive at the second picture. Are they differ-ent knots?

MILLIE: Seems like it.

TIFFANY: Here’s an analogy. You have patterns for two surfaces. You don’tknow whether or not they’re the same. You cut and reglue andcut and reglue, but somehow you can’t turn the first pattern intothe second one. Are the two surfaces different?

JOE: If they’re the same, there is a way.

TIFFANY: But you just haven’t found it. Maybe you’ll be unlucky and neverfind it.

(3)

(5) (2)

(1), (2)

(5)

(4)–1

(3)–1, (1)–1


JOE: But we do have a method to find out if two surfaces are thesame. The same method tells when they’re not the same. Oh,I see. We need some kind of method for knot pictures.

MILLIE: For surfaces we can count lakes, figure out orientability, find the Euler number. We can do that for both patterns. If the threethings are the same, then the surfaces are the same. If not,then not!

TIFFANY: Bingo. We’ve got to find some easy things for knots. Likeorientability or the Euler number. More to think about.

Team members go back to their desks. Lights fade. Music comes up.

Your Turn 6Here are two pictures of the same knot.

Try using the moves to turn one picture into the other.

Scene 2

The team members are working hard at their desks. Brandon turns around.

BRANDON: Whew. This is hard work. Do you guys have anything?

MILLIE: Well, we talked about crossings. Why don’t we just count those?Then if one knot has four crossings and another one three, thetwo knots would be different. Crude, but a start.

TIFFANY: A natural thing to start with. Problem is, a knot has got a lot ofpictures. One picture might have one number of crossings,

Scene 2 281

another picture might have another. In fact, some of thosemoves we talked about actually adds a crossing or two:

MILLIE: Hmm. Here’s an idea. Take the smallest number you get.

TIFFANY: You mean, if you look at all the possible pictures for a knot,count the crossings for each and take the smallest number?

MILLIE: Yep. Call it the crossing number of the knot. I bet the crossingnumber of the overhand knot is three.

TIFFANY: Hmm. The crossing number of the circle—the un-knot, thenonknot, the not knot?—would be zero.

JOE: I like it. The not knot!

TIFFANY: Well, that one’s easy to figure out. For other knots, how do youknow when you’ve got the smallest number?

Your Turn 7The crossing number of the not knot is zero. What can you say about knots with a crossing number that is one? What about knots whose crossing number is two?

The not knot


JOE: I tried something different. We’ve been coloring maps onsurfaces. I thought we could try to color the knot pictures. Theidea would be to color the threads of a knot picture.

MILLIE: Threads?

JOE: You know, you start at a crossing and pick one side of the under-pass and follow it until it ends at another underpass. That wholepart of the picture would be a thread. You can think of a knotpicture as being made up of threads.

So I figured the rule should be that at each crossing, the overpassthread and the two underpass threads should be different colors.Then I thought what would happen if you had a knot picturewith all its threads colored and then transformed it using one ofour moves. I started with the “pigtale” move and immediatelyran into trouble.

At that new crossing, you couldn’t actually color all the threadsdifferent because the overpass thread and one of the underpassthreads are actually the same! So then I changed the rule. Either all the threads are colored different or they’re all colored the same.

MILLIE: So color all the threads in the whole picture the same color.That’s easy.

JOE: But not very interesting. You could do that with any knot. We’retrying to tell knots apart.

Thread

Scene 2 283

MILLIE: Picky, picky.

TIFFANY: So it looks like you’ve got two rules: (1) at each crossing eithercolor all the threads different colors or color them all the same color, and (2) at least one crossing must use three differentcolors.

BRANDON: Joe, did you try coloring knots with those rules using just threecolors?

JOE: Yeah. It’s interesting. I found some knots you could color inthree colors and some knots you couldn’t. Of course, the notknot can only be colored in one color! For example, I can colorthe overhand knot in three colors, but I can’t color the five-point-star knot in three colors.

Your Turn 8Is Joe right? Is it true you can’t color the five-point-star knot with three colorsusing the rules?

Your Turn 9Use the rules and try to color each of the following knot pictures with three colors.With which ones are you successful? Which are not successful?

BRANDON: Hmm. Three-colorable. Is three-colorable a property of thepicture or of the knot itself?

MILLIE: What?

BRANDON: If it’s a property of the knot, then every picture of the knotshould be three-colorable.

TIFFANY: So with the overhand knot, you know the picture we had is three-colorable. You’d want to show that if you alter the picture with


one of our moves, the new picture would also be three-colorable.That would tell you the overhand knot is three-colorable.

JOE: Let’s try it. Here’s the first move.

This move adds one crossing. Thread R becomes threads S andT of the new picture. Color the original picture in three colors;we’ve assumed we can do that. Color threads S and T of the newpicture the same color as thread R of the original picture. Colorthe other crossings of the new picture the same as the oldpicture. At the new crossing, all the threads are colored the samecolor. But there is some other crossing where all three colors areused. So this move is OK.

MILLIE: Let me try one. In the second move, one thread gets pushedunder (or over) another. Look at the first case. Call the twothreads of the original A and B. In the new picture, A becomesA', A'', and A''', and B stays the same.

Color the original in three colors. Now color the new picture thisway: If B and A were originally colored the same, then colorthreads A', A'', and A''' the same as A was colored. Color all theother threads the same as they were in the original. At the twonew crossings, all the threads are colored the same. But there isanother crossing where the colors are different, because they

BB

A A'

A"A"'

red

red red

RS T

Scene 2 285

were different in the new one. So the new picture is three-colorable, too! If A and B were originally colored differentcolors, then do the following and everything will be OK.

BRANDON: I want in this, too! Hmm. This third move looks a little trickier.

For a start, let’s label the threads of the original by A, B, C, D, E,and F. So the threads of the new picture would be A, B, C', D, E,and F. So color the original in colors p, q, and r. The easy thingto do would be to keep the coloring of A, B, D, E, and F of thenew the same as the coloring of the original. Let’s try that.There’re lots of possibilities. Let’s try a couple.

A

B

C

E

F

D

F

A

B

C'

ED

red

red

blue blue

blue

yellow


Hmm. In each case, the only thread of the new picture that doesn’t have a color is C'. But it looks like, in both cases, you canassign a color so that the new picture is colored in three colors.(In the first case, color C' with color r. In the second, color C'with color p.) Let’s work on the other cases.

Members of the team go back to their desks. Lights fade. Music comes up.

Your Turn 10Help the team with the other possible colorings of the original. That would take careof move 5. But there are other moves to consider, the “inverse” moves to the oneswe’ve looked at. You want to show that if a knot picture is three-colorable, thenchanging it by one of these moves would leave the picture still three-colorable.

Scene 3

Members of the team are huddled excitedly around Brandon.

BRANDON: Well, what do we have?

p

p

p

q q

q

q

r

r

pq

?

p

p

r

p p

r

q

r

r

rq

?

Scene 3 287

JOE: I think we’ve shown that if a knot picture is three-colorable andyou alter it by one of our moves, then the new picture is alsothree-colorable.

TIFFANY: The same is true if you alter it by any number of moves.

MILLIE: You can’t color the not knot with three colors, but the overhandknot you can. That means the not knot and the overhand knotare different.

BRANDON: Yes! Our first method for telling knots apart.

JOE: OK. Here are some questions. The overhand knot and the squareknot are both three-colorable. Are they the same? Neither thefigure-eight knot nor the not knot are three-colorable. Are theythe same?

TIFFANY: No and no. At least I think so. But so far, we can’t really say. Allwe can do so far is tell things apart; we can’t say yet when they’rethe same. A problem Molecule.com has is trying to distinguishthe left-hand overhead knot with the right-hand one (or showthey’re the same). They’re both three-colorable. You can’tconclude they’re different, but you can’t conclude they’re thesame either.

BRANDON: While you were all working on showing that three-colorabilitywas preserved when you make a “move,” I was trying to come upwith a generalization of three-colorability to n-colorability forbigger values of n. I was trying to think of some rules that wouldwork. For example, I thought maybe you could color everythread a different color. Then if a picture had 10 threads, you’dcolor it with 10 colors. But then if you applied move 3, thatwould increase the number of threads by two.

So 10-colorability wouldn’t be preserved. Also, the coloringscheme wouldn’t work if you had something like the pigtail inthe picture.


So, I looked for some kind of restriction on the coloring scheme.I started playing around and arranged the colors in a circle, likethe spokes of a wheel. I noticed that if I did this with the threecolors, the wheel would look like this.

The rules for three-colorable would make the color choices likethis at each crossing.

That makes coloring the overpass central: whatever the overpassthread is colored on the wheel, the two underpass threads arecolored colors that are symmetrically opposite to that one.I thought, “Hey, maybe that’s the key.” For n-colorability therules become the following:

Color the threads with n colors such that, whenever an overpass iscolored X and the underpasses colored Y and Z, then the colors Yand Z are mirror images of each other through the X spoke.Furthermore, you’ve got to use at least two colors on the threads.

MILLIE: Is this useful? Can you use it to distinguish knots?

X

Z

Y X

ZY

A

C

B

Or

A

C B

p

qr

A

C

B

A,B,C

p

qr

p

qr

Scene 3 289

Your Turn 11Help the team answer Millie’s question. Show that the following knot pictures arefive-colorable.

JOE: Wow. Does it work?

BRANDON: You mean with the moves? Well, it works fine when you usemoves 1 through 4. The argument is the same as with three-colors. The hard part is with move 5. I thought of somethingthat might simplify things. I decided to number the colors 0, 1,2, … , n − 1 and arrange them in order around the wheel, like thefigures on a clock. The advantage of this is that you can use clockarithmetic.

MILLIE: Clock arithmetic?

BRANDON: Yes. Addition and multiplication modulo n. Here’s how it works.Suppose an overpass and its two underpass threads are coloredA, B, and C, respectively. Then the condition that B and C aremirror images through the A spoke means that the two distancesA to B and A to C are the same. In modular-clock arithmetic,that amounts to B − A � A − C (mod n). Or, what’s the same:2A − B − C � 0 (mod n).

n–2

n–10

1

2


Now let’s see what that does for us with respect to move 5.

Let C' = 2A − B. You want to show that also C' � 2D − E (mod n).From the original, you know that

1. 2D − A − F � 0 (mod n).

2. 2F − C − E � 0 (mod n).

3. 2D − B − C � 0 (mod n).

Consequently, C' = 2A − B � (4D − 2F) + (2D − C) (mod n) (from Equations 1 and 3)

� 2D − 2F + C (mod n)

� 2D − (C + E) + C (mod n) (from Equation 2)

� 2D − E (mod n) (Yes! What we wanted to show!)

So that does it! n-colorability is a knot invariant! (Brandon writes.)

A

B

C

E

F

D

F

A

B

C'

ED

A

C

BA

C

B

Scene 3 291

Ingredients of a Knot Picture(how to interpret a knot picture)

Crossing : point of the picture that represents two points of the original knot

Your Turn 12To test the strength of n-colorability in distinguishing knots, you decide to look at theedges of odd-twisted strips. Here they are for the three-, five-, and seven-twist strips.


Ingredients of a Crossing:

overpass: the part of the knot on top

underpass: the part of the knot underneath

Thread: maximal connected subset of the picture

n-ColorableA knot picture is n-colorable means that associated to every thread is a numberfrom the set {0, 1, ... , n − 1} such that for every crossing, if m is the numberassociated with the overpass and p and q the numbers associated with the twounderpass threads then

2m − p − q � 0 (mod n)

Theorem. If knot picture K is n-colorable and knot picture K' is obtained fromK by a move, then K' is also colorable.

m

q

p

1. What can you say about all numbers m such that the edge of the three-twiststrip is m-colorable?

2. What can you say about all numbers p such that the edge of the five-twiststrip is p-colorable?

3. What can you say about all numbers q such that the edge of the seven-twiststrip is q-colorable?

4. Can you use n-colorability to distinguish the three knots?5. What about the knots that form the edges of all odd-twisted strips? Can you

use n-colorability to show that no two of them are the same?

Your Turn 13You are looking the knot pictures below. You decide to use all the devices you haveat your disposal to determine how many really different knots are represented bythese pictures.

Scene 3 293


Scene 4

Brandon and Tiffany are at their desks, packing up their belongings in cardboardcartons. Joe is at his desk trying to work. Millie is at her desk filing her nails.Pinkley Smith and Boss enter the room, carrying balloons and party hats, followed by Sandra and Marcella from Café Ecantado, who are carrying punch and goodies.The newcomers sing some song appropriate for the occasion.

PINKLEY SMITH: Up from those desks, you dullards. All work and no playmake Jack a dull boy!

BRANDON: What’s going on?

PINKLEY SMITH: Don’t dawdle, boys and girls, it’s party time!

TIFFANY: Party? I didn’t know you guys had parties.

BOSS: You think I fancy standing here holding this stuff ? Where’s atable? (Brandon and Joe scramble to get a table.)

BRANDON: So what is going on?

SANDRA: Boss said you and Tiffany were finishing your internships andgoing back to school.

MARCELLA: Yes. Your last day! We need to give you a good sendoff !

BRANDON: You’re right! Our last day. This is very nice, Boss, but youshouldn’t have.

BOSS: Maybe now we’ll get back to the simple life. Just sell maps.Something I can understand.

JOE: Awe, Boss. You wouldn’t want that. Millie and I are going tokeep working on these ideas; so are Bran’ and Tiff when theygo back to school. We’re going to keep in touch.

PINKLEY SMITH: Speaking of going back to school, I have a story to tell. Gather‘round all. A math professor at the University had a calculusstudent who came in for help. After they had worked someproblems together, the student asked, “So what kind of mathdo you like?” The professor replied, “Knot theory.” The studentsaid, “Yeah, me either.”

ALL: Groan...

JOE: Hey, look at this: pretzel knots! Cool!

BRANDON: And doughnuts! Two-holed doughnuts! Dude!

TIFFANY: Hmm. Little fried wonton Möbius bands. Like totallyawesome! (Everybody stands around eating the goodies anddrinking the punch.)

JOE: Here’s a question. “What’s a topologist’s excuse for not doinghomework?”

TIFFANY: I don’t know, what?

JOE: “She locked the paper in her trunk, but a four-dimensionaldog got in and ate it.”

ALL: Ha! Ha!

BOSS: Here’s another excuse: “She took time out to snack on adoughnut and a cup of coffee and then spent the rest of thenight trying to figure which one to dunk.”

ALL: Ho! Ho! That’s a good one, Boss!

TIFFANY: OK. Here’s another question, “What’s green and homeo-morphic to the open unit interval?”

MILLIE: Hmm. I think you’re going to tell us anyway.

TIFFANY: Yep. The real lime.

ALL: Groan...

BRANDON: OK, gang, what’s nonorientable and lives in the sea?

JOE: Spill it, Bran.

BRANDON: Möbius Dick.

ALL: Groan...

MILLIE: Here’s another question for you: “Why did the chicken crossthe Möbius strip?”

BRANDON: This has got to be baaaad.

Scene 4 295


MILLIE: To get to the other...er, um...

ALL: Ohhhh....

BRANDON: (He starts to sing.)

(He grabs Millie and starts to dance.)

(Tiffany joins Brandon and Millie in singing and dancing.)

A mathematician named Klein

Thought the Möbius Band was divine

Said he, “If you glue

The edges of two,

You get a weird bottle like mine.”

Ay, ay, etc.

(Joe joins the others.)

The Möbius strip is a pain

When you cut it again and again.

But if you should wedge

A large disk in the edge

Then you’ll get the projective plane.

Ay, ay, etc.

(Pinkley Smith joins the group)

A knife meets a cross-cap that’s near

And cuts out a disk. Have no fear!

Just shake it about,

Take the tangle out,

And a Möbius band will appear!

Ay, ay, etc.

(Finally, Boss joins in. They all circle round.)

A burlesque dancer, a pip,

Named Virginia, could peel in a zip

She read science fiction

And died of constriction

Attempting a Möbius strip.

Ay, ay, etc.

BOSS: (Collapsing in a chair.) Whew! Some fandango! Them legs ain’t as young as they yusta! (Sits up.) Well. Time fer a few words.

MILLIE: Words?

BOSS: Brandon and Tiffany. (Pauses. Boss doesn’t know what to saynext.) Well, shucks. I got you a little somethin’ to remember usby. Brandon, for you.

BRANDON: It’s a belt! It’s really nice! Thanks!

BOSS: That’s not any belt. Put a twist in it before you buckle it up. I callit a Mer-bee-us belt. (Boss demonstrates; Brandon is speechless.)Tiffany, for you.

TIFFANY: Omagosh! A glass Klein bottle full of M&M’s! Thanks, you’re the sweetest, Boss! (She gives him a kiss on the cheek. Boss isembarrassed.)

Scene 4 297


JOE: Jeepers, it’s been great having you here. We’re going to miss you.But, you know, Mill and I are going to keep working on knotproblems. We’ll let you know what we come up with.

BRANDON: I think Tiff and I will be working on them, too. We’ll all e-maileach other. The cool thing about math is that you’re really neverdone. There’s always more questions...then some answers, butmore questions...then answers and more questions. It neverends. Eternal e-mailing!

The above image is reprinted with permission. www.kleinbottle.com

TIFFANY: I want to say that I’ve really learned a lot working withyou all. Our different points of view shed different lightson each problem.

MILLIE: It helped me to really listen and try to understand. Then Igot to talk out my ideas, and you asked questions, I talkedsome more, you asked more questions. In the end I under-stood more, and ideas got clearer and clearer in my head.

BRANDON: I really value the strengths each of you has. Joe, you’rereally down-to-earth and concrete. You always go forworking out examples.

JOE: Thanks, Bran. I really appreciate what you do to make ussee the big picture and keep us organized and focused onthe problem.

TIFFANY: And, Millie, your intuition. There you are, painting yournails, not saying a thing. Then, wham, you come up withan incredible insight.

MILLIE: Then, whoosh, enter Tiff with clear definitions. Makingsure the arguments click.

BOSS: Myself, I’s been proud to see you guys and gals making afuss and getting all worked up. I had the privilege of watch-ing you spout off your ideas. I got a kick our of yer expres-sions when someone came up with somthin’ surprisin’.

BRANDON: If it weren’t for you, Boss, we wouldn't be here. Thank you!

TIFFANY: Thank you, too, Pinkley Smith, for giving us breathersand new stuff to think about.

PINKLEY SMITH: Colleagues, colleagues! It was nothing. You are mycolleagues?

MILLIE: Fuss to feed the soul. Ideas to feed the brain. Food to feedthe tummy. Marcella and Sandra, you fix some fine victuals!

Music to Cielto Linda comes up. The group begins to danceagain with allemande lefts and do-si-dos. Gradually, theparty breaks up. People gather up their stuff. Lots of hand-shakes, high fives, and hugs. The group goes off into the nightto the sounds of “see ya,” “hasta la vista,” “bye,” and “adios.”


Investigation 1Joe and Millie are getting familiar with the notion of n-colorability. They wonderfor which numbers n ≥ 2 can they find a knot K such that K is n-colorable. Helpthem out.



Investigation 2Joe and Millie have noticed for some specific cases that if a knot is n-colorable,then it is also mn-colorable for any whole number m. They wonder if this is true for any knot. Help them resolve this problem. (They also wonder what they can say about “the other direction.” That is, if a knot is pq-colorable, is it also p-colorable and/or q-colorable? Could it be neither or at least one ofthe two?)

Investigation 3Joe and Millie have realized that, given a knot picture K, there may be severalnumbers m such that K is n-colorable. They put all these numbers together in a set S(K); that is, S(K) = {m: K is m-colorable}. Help them find S(K) where K isthe “figure-eight” knot below.

Investigation 4Joe and Millie have discovered that if K is the trefoil knot then S(K) = {m: m isdivisible by three} (see Investigation 3 for the definition of S(K)). They wonderwhat the sets S(K') look like for other knots K'. Here are a couple of questions theyask: Given a knot K', what can you say about the smallest number in S(K')? Is therea number c such that S(K') = {n: n is divisible by c}? As K' ranges over all knots,what kinds of sets S(K') can you expect to encounter? (By the way what is S(K') ifK' happens to be the not knot?)

Investigation 5Millie thinks she has a method that will find her a number n such that a given knotis n-colorable. Here is how her method works. Take a knot picture, and orient thestrands with arrows to show how you might take a trip along the simple closedcurve that is the knot.

Pick a crossing and label the overpass with 1; the underpasses, 0 and 2.

Follow the arrow from the labeled underpass to the next crossing. Label thestrands there with A, B, C so that A − B = C − A (A is the overpass). Repeat.

For example, you get nine by considering

1

2

3

4

5

9

0

1

2

0

1

2

0


then finding x so that x − 5 = 5 − 1; i.e., x = 9. When you get near the end of yourtrip with two remaining crossings, dotted in the picture above and also below,you’ll want to find n such that 0 − 9 � 9 − 5 (mod n) and 4 − 0 � 0 − 9 (mod n)or 13 � 0 (mod n). So n = 13 and the knot is 13-colorable. Try out this methodon some of the knots in Your Turn 13. Does it work? Will the method work forany knot picture?

Investigation 6Brandon was thinking about the kind of knot he ties and the kind of knot thepeople at Acme have been looking at. He noticed, “A knot I tie has loose ends.”

“In fact, if there were no loose ends, I wouldn’t be able to untie my shoes with-out cutting the laces! But if I glue the loose ends together, I get a knot of the kindwe’ve been working with.”

Brandon then looked more closely at a knot he actually ties when he wraps apackage. First he does this.

Then he does this.

0

0 4

59

9

51

x


Finally, to make it a knot Acme studies, he would glue the ends together.

“Hmm,” he thought. “That reminds me of something we did with surfaces. Thesum of two surfaces. Take two knots K and L.”

“Cut each one open.”

“Glue the loose ends of one knot to the loose ends of the other.”

K L

K L



“That would be K#L, the sum of K and L. The package knot is the sum of thesetwo knots.”

Brandon wonders how n-colorability would interact with the sum of two knots.For example, if you know that K is n-colorable and L is m-colorable, what can yousay about K#L? Is it n-colorable or m-colorable? What else could you say? Whatabout the other way around: if K#L is p-colorable, does that tell you anythingabout K or L? How does S(K#L) compare with S(K) and S(L) (see Investigation 3for the definition of S(K))? Help Brandon out.

Investigation 7(Continuation of Investigation 6) Brandon is still thinking about tying that knotfor a package in which the final knot is really the sum of two knots. Since the sumof two knots is really like tying one knot right after the other, he wonders if oneknot could “cancel” the other knot out. In other words, you first tie a knot (not thenot knot); then you tie another right after that; finally, you pull on the loose ends.Could both knots pull free? Put another way, given a knot K (not the not knot), canyou find another knot L so that K#L is the not knot? What do you think?

Investigation 8Tiffany is thinking about the edges of twisted strips and knows that if the numberof half twists is even, then the twisted strip has two edges. Generally the two edges(knots) seem to be linked in some way. She realizes that the folk at Acme have notconsidered this situation; they have only dealt with one knot, one simple closedspace curve. She decides to look at several, nonintersecting closed curves at once.She calls such a system a link. In particular, she calls a system of n nonintersectingclosed curve an n-link. She draws some examples.

2-links

K#L

She observes that all the terminology that applies to knots also applies to links:link picture, thread, crossing, overpass and underpass of a crossing. She definesequivalence of links in this way: link L and L' are equivalent if you can get from apicture of one to a picture of the other by using a finite number of moves. She alsonotices that the idea of link L being n-colorable would make sense:

● One of the numbers (colors) 0, ... , n − 1 is assigned to each thread.

● At least two numbers are assigned.

● At each crossing the numbers assigned satisfy 2A − B − C � 0 (mod n), whereA is the number assigned to the overpass.

She realizes that if L and L' are equivalent links, then L is n-colorable implies L'is n-colorable, too. She calls an n-link picture unlinked if it can be separated into(at least) two parts. For example, this is an unlinked three-link.

And this is an unlinked four-link.

Here are some things that Tiffany observed for which she would like independ-ent verification:

● Any unlinked n-link is n-colorable for any n.

● The two-link below cannot be three-colored. (If true, what would this tell you?)

Help her out with this.Finally, she wants to show that the three-link below cannot be unlinked.

3−links



(This three-link is called the Borromean [Ballentine] rings. Tiffany is interestedin these because every pair of closed curves in this three-link is unlinked!)

Investigation 9Joe has been looking more closely at the assignment of the numbers 0, ... , n − 1to strands of a knot (or link) picture that enables one to call the knot (or link) n-colorable. For example, the trefoil knot is three-colorable by the followingassignment.

But Joe notices that this also makes the trefoil knot six-colorable by the followingassignment.

Somehow Joe feels that this six-coloring of the trefoil is a special kind of six-coloring. He calls it borrowed. He figures that a borrowed n-coloring has thefollowing property: there is a divisor n > d > 1 of n such that only every dthnumber has been assigned. He calls a knot really n-colorable if it’s n-colorable andif there is an assignment of the numbers 0, 1, 2, ... , n − 1 that is not borrowed. Hethought maybe an n-coloring would be borrowed if some of the n numbers wereleft out. Here’s an example where that does not seem to be the case:

0

4

2

1

1

35

0

1

2

So the figure-eight knot appears to be really five-colorable.Help Joe figure out some of the properties of this new notion: really

n-colorable. Here are some questions to start with:

● If knot picture K is really n-colorable and knot picture K' is gotten from K byone of the moves, is it also true that K' is really n-colorable?

● What can you say about the set R(K) = {m: K is really m-colorable}? How does this set differ from S(K) (see Investigation 3 for the definition ofS(K))?

Question 10In his investigation of R(K) Joe asked this question: How does R(K#K') comparewith R(K) and R(K')?

Investigation 11Joe thinks that the knot K that forms the edge of a 2n + 1-half-twist strip is really2n + 1-colorable. Further, he believes that 2n + 1 is the largest number for whichK is really 2n + 1-colorable. Help Joe sort this out.

Question 12Joe has told Millie about knots that are really n-colorable. Millie claims that, ifp is a prime number and a knot is p-colorable, the knot must also be really p-colorable. Is she right?

Question 13Joe’s definition of really n-colorable also applies to links. What can you say aboutR(L) if L is a link but not a knot?

Investigation 14(Continuation of Investigation 8) Tiffany has been investigating the two-links thatare the edges of 2n-half-twist strips. Here are her pictures of the edges of the 0-half-twist, 2-half-twist, 4-half-twist, and 6-half-twist strips.



She is interested in finding out for what n is each two-link n-colorable. She hascome up with the following conjectures.

1. “2” is n-colorable if and only if n is even.2. “4” is really four-colorable but is not three-colorable.3. “6” is three-colorable.4. “6” is really six-colorable but is not five-colorable.

Help Tiffany. For each conjecture, find either a proof or a counterexample.Tiffany would like to use her results to distinguish the two-links “0,”“2,”“4,” and

“6”. Help her with this project.Finally, Tiffany would like to generalize her results to the edge links of all 2n-

twisted strips. (Of course, for a “twist” she means a “half-twist.”) Help her out.

Question 15Tiffany and Brandon have been talking about the links that form the edges ofeven-twisted strips. Brandon thinks that what he knows about really n-colorableand the knots forming the edges of odd-twisted strips also applies to these links.(Joe has e-mailed Brandon his thoughts; see Investigation 11.) What can he sayabout really n-colorable and these links? Will this help with Tiffany’s project inInvestigation 14?

Summary 16Boss wants a summary of what the Acme team has done with knots and the evolu-tion of n-colorability. You might want to include in your summary the propertiesof n-colorability that you and your friends have derived in these investigations.

Notes

In the 1870s the prevailing theory of physical matter was that space, the mediumthrough which light waves traveled (now considered to be a vacuum), was

“0” “2” “4” “6”

occupied by a substance called ether. The chemist Lord Kelvin proposed thatatoms, the basic elements of matter, were vortices, or knots, in the ether. Differentknots would correspond to different elements. Scottish physicist P. G. Tait andAmerican mathematician C. N. Little started to construct lists of all knots as a wayof compiling a table of all the elements. But by 1900, the theory of the ether hadbeen abandoned and with it the idea of knots as atoms.

However, questions posed by chemists and physicists—How can you tell whentwo knots are different? How can you come up with a complete list of knots?—became a mathematical challenge. Knot theory became a vigorous subfield oftopology, itself a fairly new field of mathematics. A basic strategy for answeringthe big questions might be to do what we did with surfaces: figure out a good wayto represent them, decide when two representations are the same, and find“invariants” that are easy to calculate and use and that enable you to obtain notonly a complete list but a list that includes each item exactly once. For surfaces, arepresentation is a surface symbol; two symbols are the same if one can beobtained from the other by a finite number of application of the rules, and theinvariants are orientability, number of crosscaps/tori, and number of lakes.(Alternatively, we could use orientability, Euler number, and number of lakes.)These invariants form what is called a complete set of invariants for surfaces in thesense that different surfaces have different invariants.

For a knot, the representation is a picture, and two pictures represent the sameknot if one can be obtained from the other by a finite number of moves. It appearsthat Tait and Little were able to come up with a complete list of knots up to ninecrossings and that their techniques provide a method for extending their list toknots of more crossings and hence to a list of all possible knots (for more detailson the tabulation of knots, see the book by Adams, chapter 2). However, this is nota tidy list like the one we have for all surfaces! What they were not able to do wasdecide definitively whether or not there were duplications. That leaves the ques-tion: How do you decide whether or not a pair of knot pictures represents thesame knot? In this chapter the Acme team came up with one type of invariant,n-colorable. That n-colorable is an invariant means that, if a knot picture is n-colorable, then so is any knot picture obtained from the original by a finitenumber of moves. All the n for which a knot K is n-colorable is encapsulated inthe set S(K) = {n: K is n-colorable}, which is studied in the investigations. Is S(K)is a complete set of invariants? It is easy to show that the following pairs of knotsK and K' have the property that S(K) = S(K').

But do the pictures in each pair actually represent the same knot? If you feelthat each pair is really a pair of different knots, you are right: knot theorists have

Notes 309

shown that they are different. Here is a more startling pair K, K' for which S(K) =S(K') but K and K' represent different knots.

This is not so obvious (see Rolfson, p. 157). It tells us that some knots, not thenot knot, are not n-colorable for any n! For such a knot, n-colorability will nothelp us decide whether or not it has an “inverse.”

The upshot is that n-colorable is not the last word; knot theorists need otherinvariants if they are to answer the big question. What sorts of things might we belooking for? What is the nature of the invariants we have for surfaces? Forsurfaces, a complete set of invariants is a triple (a, b, c) of whole numbers: a = ± 1(+1 for orientable, −1 for nonorientable), b ≥ 0 (number of crosscaps/tori), c ≥ 0(number of lakes). So these invariants are numbers. It turns out that the biggestbreakthrough in the search for knot invariants came in 1929 with an invariantthat is a polynomial, called the Alexander polynomial after its creator. All theknots in the tables of knots existing at the time—of nine or fewer crossings—canbe distinguished using this invariant.

In the latter part of the 19th century and the early part of the 20th century,topologists began to develop what are called algebraic invariants. It turns out thatn-colorable is really an algebraic invariant. When a knot is n-colorable, we associ-ate the number n with the knot. This invariant is a number. But thinking back, weremember that this number was obtained by associating the strands of the knotpicture in a certain way with the integers module n. The invariant n is related tothe algebraic structure of the integers modulo n. This is a typical approach for atopologist to finding invariants: find some way to associate an algebraic structurewith the topological object so that two same objects imply same algebraic struc-tures. (The word “same” for algebraic structures is “isomorphic.”) The philosophybehind this is that an algebraic structure may be simpler than a topological one.Whether or not it is simpler, the algebraic structure provides a different perspec-tive on the problem and new opportunities for investigation.

In 1895 Henri Poincaré defined one of the earliest algebraic invariants ofa topological space called the fundamental group. (A group is a set with a single,associative operation with identity and inverses.) The group of knot K is defined tobe the fundamental group of R3\K, the latter set gotten by removing the knot from


all of space. Unlike the group of integers modulo n associated with n-colorability,the knot group is infinite. However, knot groups have nice descriptions (see articleby Neuwirth), and from them certain more accessible invariants can be derived.Knot theorists were able to derive the Alexander polynomial from the knot group.(Alexander did not do it in this way, however.) That makes the Alexander polyno-mial an algebraic invariant. It turns out that n-colorable can also be derived fromthe knot group: If K is n-colorable, then there is a group homomorphism from theknot group onto Zn, the integers modulo n.

More on n-colorable can be found in the articles by Crowell and by Anderson.Knotentheorie, the first major treatise on knot theory, appeared in 1932 written by

the German mathematician Kurt Reidemeister. The moves that the Acme team usedto define two knot pictures as the same are called in the literature Reidemeister movesin honor of this mathematician. Reidemeister proved that equivalence by the movesare sufficient to show two knots are the same. The next comprehensive book on knottheory appeared in 1963 by the American mathematicians Richard Crowell andRalph Fox. A complete set of algebraic invariants for knots was found by Waldhausenin 1968; however, the set is unwieldy and the invariants are not easy to compute.

Knot theory has evolved into a rich subject over the past 100 years. This is duein part to the connections between the invariants within the field itself. (For exam-ple, n-colorable and the knot group.) It is also due to connections with other fieldsof enquiry. One of these connections is within the broader field of topology.It turns out that the classification problem in knot theory is closely related to theclassification of three-manifolds. (You will recall that three-manifolds are thethree-dimensional analog to two-dimensional surfaces.) The connection goessomething like this. The three-sphere, the set S3 = {(x, y, z, w): x2 + y2 + z2 + w2 = 1},is a three-manifold sitting in R4. Start with a knot K, and thicken it slightly to makeit into a solid torus T. Imagine the set T embedded in S3. Now remove T from S3

and glue a new torus back in its place in one of many “specified ways.” What youget is a new three-manifold. It turns out that every three-manifold is one of these.None of this is obvious, and details need to be filled in, especially the “specifiedways” of gluing one torus inside another (see Adams, chapter 9).

Recently, other connections between knot theory and other fields have broughtthe subject back to its original roots in the sciences. In the 1980s, in his work onoperator algebras, the American mathematician Vaughan Jones came up withboth a new, powerful invariant (called the Jones polynomial) and a connectionbetween knot theory and statistical mechanics (for details, see the book by Adams,sections 6.1 and 7.4). The 1980s also began a period of intense interest in knottheory by biologists and chemists, the former in part because of the knotting ofDNA molecules, the latter in part because of an interest in the synthesis of knottedmolecules. (For more details, see the book by Adams, chapter 7 and the article bySummers.) Some of the issues confronting chemists provided the setting for thebeginning of this chapter (see the book by Flapan).

Knot theory continues to be a lively, active area of mathematical research. Fora brief survey of the history of knot theory, see Livingston (chapter 1); for a good,accessible survey of the various approaches to knot theory see the book by Adams.

Notes 311


For information about the existence of an “inverse” for a knot (cf. Investigation 7),see the article by Martin Gardner.

References

Adams, C. Why Knot? An introduction to the mathematical theory of knots. Emeryville, CA,Key-Curriculum Press, 2002.

Adams, C. The knot book: an elementary introduction to the mathematical theory of knots. New York,Freeman, 1994.

Anderson, P. The color invariants for knots and links. American Mathematical Monthly, May 1995,pp. 442–448.

Crowell, R. Knots and wheels. In Enrichment mathematics for high school, 28th Yearbook. Reston, VA:National Council of Teachers of Mathematics, 1963: pp. 339–354.

Crowell, R., and Fox, R. Introduction to knot theory. Boston: Ginn and Company, 1963.Flapan, E. When topology meets chemistry. New York: Cambridge University Press, 2000.Gardner, M. Mathematical games, Scientific American, December 1972, pp. 104–105.Livingston, C. Knot theory. Washington, DC: Mathematical Association of America, 1993.Neuwirth, L. The theory of knots. Scientific American. June 1979, pp. 110–124.Reidemeister, K. Knotentheorie. In Ergebnisse der Matematik un ihrer Grenzgebiete. Berlin: Springer, 1932.

(English translation: L. Boron, C. Christenson, and B. Smith. Moscow, ID: BCS Associates, 1983.)Rolfsen, D. Knots and links. Berkeley, CA: Publish or Perish, 1976.Summers, D. W. Untangling DNA. Mathematical Intelligencer, vol. 12, no. 3, Summer 1990, pp. 71–80.Waldhausen, F. On irreducible 3-manifolds. Annals of Mathematics, vol. 87, no. 1, 1968, pp. 56–88.

Where To Go from Here: Projects

313

Now that you have been through this book, or most of it, what should you donext? One thing is to step out on your own, find a problem or a topic related tothings we have talked about here that interests you, find a group of chums withthe same interests (if you want company), and “go to town’’ and have some fun.One purpose of this chapter is to give you some suggestions. So the bulk of thechapter is just that: a list of project ideas. After you wrestle with the project forawhile and feel you are an expert on a topic or after you have solved some problemsand have some great insights, it is great to share what you found. A secondpurpose of the chapter is really about this aspect of doing projects: communicatingyour findings. Nothing is worse than telling someone what you have done, then tohave them fall asleep or roll their eyeballs. You should attempt to communicate theresults of your project effectively.

Project Ideas

Here is an annotated list of ideas for projects. The ideas span a wide range of proj-ect types. At one end, there are exploratory projects that are like the investigations

C H A P T E R

14

from the earlier chapters; no references are given for these. At the other end of thespectrum are expository projects. Such a project involves addressing a knowntopological procedure, idea, technique, fact (or collection of related facts), ortheorem in a new way, relating it to things that happened in the book and to previ-ous experiences with mathematics and making a model (or two) that illustratesthe material. There are lots of references to the expository projects. Most projectswill be some combination of the two. On the one hand, solving a problem some-times leads to finding out what others have done. On the other hand, trying tounderstand what others have done in an area you are interested in sometimesunexpectedly leads to solving new problems.

In the list below, the annotations are not uniform. For some project ideas, thereare lots of details. For others, there is just a title, a few questions, and a referenceor two. This uneven presentation may not only give you ideas but also show youideas in various stages of development—some with just the germ, some with listsof questions to ask (and answer) and suggestions for things to make, and somewith a detailed outline.

Sometimes the references are sketchy or not complete; they are suggestions.I hope that what I list here will get you started and that you will find what youneed from there.

Lots of good project ideas are not on this list! Items in the earlier chapters,especially in the Investigations, etc., and Notes sections, may suggest ideas directlyto you. If you do not find something below that excites you, use the list to beginthinking. Perhaps one idea might suggest something else to pursue. Browse thereferences following this list and at the ends of the chapters, and talk with friendsand teachers. Happy hunting!

Suggested references are given in parentheses following each project idea.If the reference has already appeared in an earlier chapter, then it is followed by achapter number in brackets. Otherwise, the reference is listed in the Resourcessection.

I. Map Coloring

A. Four-Color ConjectureBefore it was proven, the four-color conjecture had been studied for a long time.Lots of statements are equivalent to it. What are they? (Saaty)

B. Coloring Maps with ColoniesHeawood proved a theorem about coloring maps on a sphere in which a countrymay have a colony. Investigate this theorem. Could this be generalized to maps on a torus with colonies? Could it be generalized to other surfaces? Given aparticular surface, is there a relationship between the number of colors needed for

314 Chapter 14 Where to go from Here: Projects

a all maps on the surface to the number of colors needed for maps with colonies?I do not know the answer to these questions! (Barnette p. 158+; Beck et al., p. 70+;Hutchinson; Hudson)

C. Coloring Maps on the Earth and MoonThis is related to B, “Coloring Maps with Colonies.’’ Suppose Earth countriesestablish colonies on the moon. You want to color the countries on Earth as wellas their colonies on the moon. Of course, you want countries and colonies to becolored the same color.

What if you have the Earth, Moon, and a few planets? (Hutchinson)

D. What’s Involved in the Real Proof of the Four-Color Theorem?A summary or exposition of the ideas and techniques involved in the 1976 proof of the four-color theorem and in more recent simplifications of the proof.(See references to Chapter 5; Barnette; Saaty and Kainen; Appel and Haken;Wilson).

E. Coloring Infinite MapsWhat if you have an infinite map in the plane, how many colors would you need? What about infinite maps that are two-colorable or three-colorable?(Barnette, p. 160+)

II. Surfaces

A. Pits, Peaks, Passes, and EulerIt turns out that if your world is a surface and if you count the number of valleysand mountain peaks and saddles (passes), then these numbers can be used tocalculate the Euler characteristic of the surface. Find out how and why! (Blackett,p. 138+; Boltyanskii and Efremovitch, p. 55+; Fauvel et al., p. 133–139;Pits, Peaks, and Passes video with Marston Marris)

B. Fun-Film SurfacesYou can make lots of surfaces by dunking a wire form (a simple closed curve in three dimensions) in soap bubble solution and poking out holes in suitable places. You can make permanent versions of these by using a

II. Surfaces 315

soap-bubble replacement: Fun-Film or Joli-Form-a-Film, or petroleumacetate–based liquid plastic that hardens. Is there some simple way to identifywhat each surface is (according to our classification scheme)?

C. Generalized Regular PolyhedraThe surfaces of the regular polyhedra (Platonic solids) are spherelike. (If they weremade of rubber, you could inflate them into spheres.) The regular polyhedra havethe following property: there are numbers n and m such that every face is a regularn-gon and every vertex is of order m. What are analogous shapes when the underly-ing surface is something other than a sphere? You might have to relax the conditionof “every face being a regular n-gon’’ to something like “the faces are congruentn-gons.’’ Can you build the corresponding “polyhedra’’? (Senechal and Fleck,p. 212+, pp. 251–253.)

D. “Nice’’ Models of a Cross-capWe have seen one embedding of a cross-cap in three-dimensions. It is not with-out its problems. But just as there is more than one embedding of the Möbiusstrip in three-dimensions, there are many embeddings of the cross-cap. Some arenice and symmetric; the Boyes surface is one example. Build some of these,explain them (explain why they might be more satisfactory than the familiarembedding), create them with computer graphics, or find them on the Web.(Hilbert and Cohn-Vossen)

E. The Geometry of SurfacesThink of a cylinder as a plane that is rolled up; certain points of the plane havebeen identified. (Point (x, y) is identified with point (u, v) iff y = v and x − u is aninteger.) This coincides with the usual way of thinking about the cylinder geometri-cally: shortest paths on the cylinder are really just shortest paths in the plane. Youcan also think of a torus as a plane with certain points identified. Point (x, y) is identified with point (u, v) iff x − u and y − v are both integers. This way of thinkingabout the torus geometrically also results in shortest paths on the torus beingreally just shortest paths in the plane. Although this is not the usual way of thinkingabout a torus, it is a way of endowing the torus with a Euclidean geometry.In most of our discussions, we have ignored properties of shapes that depend onthe rigid geometric structure of a shape. This project is meant to “restore’’ someof that rigidity. The video Shaping Space and the book The Shape of Space byWeeks give ways of visualizing some of these “Euclidean’’ surfaces. The purpose ofthis project is to explore some ideas surrounding this notion. Which surfaces canbe endowed with a Euclidean geometry? If Euclidean geometry doesn’t work, thetwo non-Euclidean geometries are candidates: hyperbolic geometry and elliptical


(spherical) geometry. Which surfaces could be endowed with geometries of thesetypes? The identification scheme for the torus comes about by tessellating theplane with squares. Perhaps a tessellation of the hyperbolic plane or sphere by aregular “polygon’’ would give rise to another surface.

If you pursue this, you will need to learn a little bit about hyperbolic geometryand spherical geometry (and some of their key properties). It might also be helpfulto bring in the notion of Gaussian curvature and its relation to these geometries.It might also be helpful to bring in knowledge of the groups of isometries of thethree geometries and some of their subgroups. (Hilbert and Cohn-Vossen;Stillwell; Thurston; Weeks, Shape of Space)

F. Games on SurfacesThe investigations of Chapters 6, 7, and 8 have described tic-tac-toe on the surfaceof a doughnut, on a Möbius band, and on a Klein bottle. Analyze these. Must theyend in a tie? Are there winning strategies? Which player can win? (Weeks, Shape of Space and Exploring the Shape of Space)

G. Alternative ArgumentsYou know that the connected sum of a cross-cap with the torus is the same as theconnected sum of the cross-cap with the Klein bottle. You can see this by analyzingpatterns. Are there other ways of seeing this? (Maybe a four-dimensional argumentmight work?) The same goes for showing that the connected sum of two cross-capsis the same as a Klein bottle. (Bethards et al.)

H. Interesting Models of Familiar Surfaces(This might be combined with IID, “‘Nice’ Models of a Cross-cap’’; IIID, “Modelof Ungar-Leech Map on Doughnut that Requires the Maximum Number ofColors’’; and IIIF, “‘Regular’ Maps on Cross-cap and Klein Bottle.’’) Several paper“polyhedral’’ versions of the torus, cross-cap, Möbius band, and Klein bottle havebeen suggested in the Investigations, etc., of Chapters 6, 7, and 8. Make nicemodels of these. You might think of other possibilities and make those, too, suchas a “polyhedral’’ version of the Boyes surface or a “polyhedral’’ torus “hat’’ havingseven-fold rotational symmetry or a “twisted’’ Klein bottle (see Investigation #9 ofChapter 11). Relate each one to some interesting properties of the surface, such asmaps requiring the maximum number of colors.

J. Linkages and the Surfaces That Go with ThemAnalyze some linkages and the surfaces that go with them. See also the Investigationsof chapter 9. (Abrams; Shimamoto; Thurston and Weeks; Toussaint; Walker)

II. Surfaces 317

K. Proofs for the Classification of SurfacesHave a look at the different ways mathematicians have proved the classificationtheorem for surfaces, understand them, and explain how they work. See the Notesand References in Chapter 10.

III. Surfaces and Maps

A. Two-Color Theorems on Surfaces Other Than the Sphere?What can you say about a map on the torus that can be colored in two colors?What about two-colorable maps on other surfaces? Are there theorems forsurfaces other than the sphere that read “If a map on surface X has property Y,then it can be colored in Z colors’’? I do not know the answer to all these questions.You might want to do a literature search.

B. How Many Colors for a Doughnut? A Two-Holed Doughnut? An n-Holed Doughnut?In 1968, Ringle and Young completed several years of investigation and came upwith a proof that Heawood’s inequality (chapter 12) is actually an equality. Theirproof is broken down into 12 cases. Describe their program. Give an exposition ofthe proof of one of the “nice’’ cases. (Ringle; Stahl)

C. Constructing Real Maps on Surfaces That Require the MaximumNumber of ColorsThe proof of Ringle and Young (see previous project) is theoretical; few maps haveactually been constructed. But their proof gives prescriptions for constructing themaps if you choose to construct them. The idea for this project is to take one (or more)of these prescriptions and draw maps on n-holed doughnuts—n small—that requirethe (theoretical) maximal number of colors. Ditto for connected sum of n cross-caps(with lake, since you want to be able to make them) for n small. (Ringle; Stahl)

D. Model of Ungar-Leech Map on Doughnut That Requires theMaximum Number of ColorsThe Ungar-Leech map on a doughnut is a map that not only requires seven colors forproper coloring but also has seven-fold rotational symmetry. This is an opportunityto make a really elegant model. You’ll need a plaster of Paris model of a torus.


Alternatively, you could design a polyhedral version of the torus having seven-foldrotational symmetry. See IIH,“ Interesting Models of Familiar Surfaces.’’ (Coxeter,The Mathematics Teacher)

E. Map Coloring on the Klein BottleThe Klein bottle does not seem to fit into any general scheme for map coloring.So it needs a special argument all its own. Provide it. Make a model of a map onthe Klein bottle requiring the maximum number of colors. (Ringle)

F. “Regular’’ Maps on Cross-cap and Klein BottleA map is regular if whole numbers n and m exist such that every country hasexactly n borders and every vertex is of order m. What are the regular maps on across-cap? What are the regular maps on a Klein bottle? What do these maps looklike on familiar realizations of the surfaces? Are there connections between any ofthese maps and coloring problems on the surface? Are there connections betweenany of these maps and regular maps on the sphere or on the torus? (Biggs et al.,p. 129+; Firby and Gardner, p. 83+)

G. Regular Maps on n-Holed Doughnut, n SmallWhat are the regular maps on one-holed, two-holed, three-holed doughnuts (for definition of regular, see IIIF, ‘Regular’ Maps on Cross-cap and Klein Bottle’’)?What do these maps look like on familiar representations of the surfaces? Arethere connections between any of these maps and coloring problems on thesurface? (Firby and Gardner, p. 83+)

IV. Three-Space

A. Coloring “Maps’’ in Three-SpaceThe analogue to an island in the plane might be a solid sphere in three-space.What would be a definition of “map’’ in a solid sphere? What would be a definitionof “proper coloring’’ of a map? What about two- and three-colorable three-spacemaps? Are there map coloring theorems in two-dimensions that could be generalizedto three dimensions?

B. Three-Dimensional WorldsWhat are the three-dimensional analogues to the two-dimensional surfaces we havebeen studying? What is a definition? What are some examples? What corresponds

IV. Three-Space 319

to a pattern? Is there a nice list of such objects? Does orientability come into play?Is there an analogue to Euler’s formula? (Banchoff; Blackett; Ellis; Meyerhoff;Thurston and Weeks; Weeks, Shape of Space;)

A particular collection of three-dimensional worlds are called lens spaces andare described in the book by Adams. A good project would be to understand theseand to give good descriptions of what they are. They also relate to knots. See VH,“Knots and the Classification of Three-Manifolds’’ below. (Adams)

C. The Shape of SpaceWe live in a three-dimensional world. Which one is it? Astronomers are interested.Investigate (and report on) what the possibilities are, what they arre looking for, and what they have found. See also the previous project and project IIE,“The Geometry of Surfaces’’ (Kaku; Jackson; Luminet et al.; Meyerhoff; Osserman;Thurston; Thurston and Weeks; Weeks, American Mathematical Monthly, Notices,Shape of Space; Shape of Space video)

A particular candidate for the space of the universe is discussed in the paper byWeeks in the Notices. Understanding it involves understanding dodecahedralspace. Understanding the latter involves understanding the four-dimensionalanalogue to the regular dodecahedron in three dimensions. A good project wouldbe to bring this example of a three-dimensional world to life and to explain, inlayman’s terms, why astronomers are interested in it. (Coxeter, Geometry, chapter 22;Luminet et al.; Weeks, Notices)

V. Knots and Links

A. Surfaces Spanning KnotsFun-Film is one way to create a surface that has a given knot as its “edge’’ (see proj-ect IIB, “Fun-Film Surfaces’’). The Seifert surface is another. Find out what it is.Build a few models. Figure out how the surface fits in the classification scheme forsurfaces. Find out how it can be used to prove things about knots. In particular,the genus of a knot is related to these surfaces; the genus is a knot invariant.Furthermore, the genus of knots can be used to deal with the issues of knotinverses (see project VJ,“Knot Inverses’’) and of unique factorization of knots intoprime knots (see project VK, “Prime Factorization of Knots’’). (Adams; Fox;Kauffman, On Knots)

B. Torus KnotsSome knots lie on the surface of a torus. Some of these are called torus knots.What are they? Can you tell them apart? (Adams; Cipra, American MathematicalSociety; Kauffman, On Knots)


C. The Jones Polynomial of a KnotThis strong, easy device for distinguishing knots was created in the past 20 years.Find out what it is, how to calculate it, and why it does what it is supposed to.(Adams; Kauffman, American Mathematical Monthly; Lickorish et al.)

D. Knots and BiologyIn the past 20 years, biologists have been looking at the mathematics of knots.The interest is mostly at the molecular level. For example, how could a strand ofDNA knot? Find out what is going on, what they are looking for and why, andwhat they have found. (See notes and references in Chapters 7 and 13; Adams;Flapan; Jurisic; Kauffman, On Knots; Summers)

E. Linking NumberThis is an invariant of links. What is it? Why is it an invariant? Use it to distinguisheven twisted strips. (Adams; Kauffman, On Knots; Livingston, American MathematicalMonthly)

F. BraidsIt turns out that every braid can be turned into a knot or link and vice versa. Whatare these things? They have a nice algebraic structure. (Adams; Gardner, ScientificAmerican, December 1959, January 1962)

G. Ballentine (a.k.a. Borromean) Rings, GeneralizedIn the Ballentine rings, if you remove one, the whole thing falls apart, yet the threerings “hang’’ together. This can be generalized to n rings in interesting ways.It might be nice to connect the solution of this with the construction of the funda-mental group of a knot/link mentioned in project VL, “The Group of a Knot.”(Livingston, p. 105+; Penny, chapter 2.)

H. Knots and the Classification of Three-ManifoldsThis is an extension of project IVB,“Three-Dimensional Worlds.’’ It turns out thatevery three-manifold can be constructed by starting with the complement of aknot in the three-sphere and then gluing the knot back in the hole—in a special way.(Compare with the description in the Notes to Chapter 13.) The construction involvesthe lens spaces of project IVB. A good project would be to understand how thisconstruction works and explain it. (Adams)

V. Knots and Links 321

J. Knot InversesIn Chapter 13 the following question came up: given a knot (not the not knot),can you tie another knot in it and get the not knot? In other words: do knotinverses exist for the knot sum operation? In Chapter 13 this was answered in thenegative for some knots. What is the whole story? An initial discussion of thisproblem can be found in Adams (1.2). An answer related to the genus of a knot(see VA, “Surfaces Spanning Knots’’) can be found in Adams. There are two additional, accessible answers that I know of. One has to do with infinite knots inFox; the other is by Conway and is given in the Mathematical Games section ofScientific American. A good project would be find out what these answers are,understand them, and explain them. (Adams, 4.3; Fox; Gardner, ScientificAmerican, December 1972)

K. Prime Factorization of KnotsThe knot sum operation suggests a definition and some questions. What shouldthe definition of a prime knot be? What are some examples of prime knots? Canevery knot be factored into a finite number of prime knots? Is that factorizationunique? (Compare definitions and questions for the surface sum.) Again, spanningsurfaces give some answers as suggested in VA, “Surfaces Spanning Knots”(Fox; Livingston, chapter 4).

L. The Group of a KnotThe group of a knot is the fundamental group of the complement of the knot inthree-space. For more, see the Notes from Chapter 13. Find out about more aboutthe definition. (Knowing a little group theory will help.) Figure out how to calcu-late the group of a knot given a picture of the knot. (Bertuccioni; Crowell and Fox; Neuwirth)

VI. Miscellaneous

A. Hexlike GamesHex. Bridge-it. Twixt. Can any of these games end in a tie? What can you say aboutwinning strategies for these games? What is it, topologically, that makes thesegames similar? In a primitive sort of way, Hex is just a map on an island with allof the vertices of order three. That opens up the possibility for lots of games! Whatcan you say about them? (Alpern; Beck et al.; Gale; Gardner, Mathematical Puzzlesand Diversions, chapter 8, Scientific American)


B. A Variety of Additional Project Ideas with Knots and LinksSelf-weaving belts. (See braids, above.) classification of all twisted strips(Kauffman, p. 179+); Conway polynomial of a knot (Kauffman, On Knots, p. 41,ex 3-13); Conway’s tangles (Adams; Kauffman, On Knots); classification of theways a rubber band can be wrapped around a cylinder (Kauffman, p. 18); analysisof knots and links by computer (do Web search.)

C. Complexity TheoryGiven a problem, how long will it take to solve? Some of the problems consid-ered in this course are thought to be ones that take the longest time. The traveling salesman problem (this problem involves touring a network so that every vertex is visited exactly once; the museum tour problem is a problem like this) is one of these. Deciding whether or not a map on thesphere can be colored with three colors is another. See Notes and Referencesin Chapter 2.

D. Taking Tours in the Real WorldSometimes you want to take a tour on a network and it is not possible by Euler’stheorem. You can do the next best thing if you allow “double-backs,’’ edges thatyou can traverse twice. Of course, you do not want to double back too much.Given a network, what is the minimum number of double-backs that makes a nicetour possible? (Tannenbaum et al., and the book’s Web site)

E. Nonplanarity of GraphsKuratowski’s Theorem states that a nonplanar graph must “contain’’ either a solution to the utilities problem or a solution to the five-cities problem. What does this mean and what is the proof? Generalizations to other surfaces? (Biggs et al.)

F. Four-SpaceProject IIG, “Alternative Arguments’’ suggests that four-space might help toexplain some of the facts we know about surfaces. Describe how you can easilycarry out in four-space some of the “impossible’’ maneuvers we have made with surfaces in three-space. See the Investigations in Chapter 11. (Banchoff;Bethards et al.)

VI. Miscellaneous 323


G. SproutsThis is another topological two-person game. Does it end in a finite number of moves?(How many?) Is there a winning strategy for one of the players? (Copper; Gardner,Scientific American, July 1961)

Resources

The issues of the journal Structural Topology contain many articles related to the ideas above, and more.The Web is also a source of lots of neat stuff. More resources appear at the ends of the chapters.

Abrams, A. Finding topology in a factory: configuration spaces. American Mathematical Monthly,February 2002, pp. 140–150.

Adams, C. The knot book: an elementary introduction to the mathematical theory of knots. New York:Freeman, 1994.

Alpern, S., and Beck, A. Hex games and twist maps on the annulus, American Mathematical Monthly,November 1991, pp. 803–811.

Appel, K., and Haken, W. The solution of the four-color-map problem. Scientific American, October1977, p. 108f.

Banchoff, T.F. Beyond the third dimension. New York: W. H. Freeman, 1990.Barnette, D. Map coloring, polyhedra and the four color problem. Washington, DC: Mathematical

Association of America, 1984.Beck, A., Bleicher, M.N., and Crowe, D.W. Excursions into mathematics. New York: Worth Publishers,

1970.Bertuccioni, I. A topological puzzle. American Mathematical Monthly, December 2003, pp. 937–939.Bethards, S., and Dolle, E. Exploring surfaces. Web site: http://math.arizona.edu/~ura/013/

bethard.steven/Biggs, N.L., Lloyd, E.K., and Wilson, R.J. Graph theory: 1736–1936. Oxford: Oxford University Press,

1976.Blackett, D.W. Elementary topology: a combinatorial and algebraic approach. San Diego: Academic

Press, 1982.Boltyanski ι, V.G. and Efremovich, V.A. Intuitive combinatorial topology. New York: Springer-Verlag,

2001.Carter, H.G. Messages from a distant sky. Princeton Alumni Weekly, October 25, 2000, pp. 24–27.Cipra, B. Advances in map coloring: complexity and simplicity. SIAM News, December 1996, p. 20f.Cipra, B. From knot to unknot: What’s Happening in the Mathematical Sciences. American

Mathematical Society, vol. 2, 1994, pp. 9–13.Cipra, B. Map-coloring theorists look at new worlds: What’s Happening in the Mathematical Science.

American Mathematical Society, vol. 1, 1993, pp. 43–46.Copper, M. Graph theory and the game of sprouts, American Mathematical Monthly, May 1993,

pp. 478–482.Cornish, N.J., and Weeks, J.R. Measuring the shape of the universe. Notices of the AMS, December

1998, pp. 1463–1471.Courant and Robbins. What is mathematics? Oxford: Oxford University Press, 1978.Coxeter, H.S.M. Introduction to geometry. New York: John Wiley and Sons, 1969.Coxeter, H.S.M. The four-color map problem, The Mathematics Teacher, vol. 52, 1959, pp. 283–289.Crowell, R., and Fox, R. Introduction to knot theory. Boston, Ginn and Company, 1963.Cundy, H.M., and Rollett, A.P. Mathematical models, Oxford: Oxford University Press, 1961.Ellis, G.F.R., The shape of the universe. Nature, October 9, 2003, pp. 566–567.Fadiman, C., editor. Fantasia mathematica. New York: Simon and Schuster, 1958.Fauvel, J., Flood R., and Wilson, R., editors. Möbius and his band. Oxford: Oxford University Press, 1993.Firby, P.A., and Gardiner, C.F. Surface topology. West Sussex, England: Ellis Horwood Limited, 1982.Flapan, E. When topology meets chemistry: a topological look at molecular chirality. Washington, DC:

Mathematical Association of America, 2000.Fox, R. A quick trip through knot theory. In Topology of three-manifolds. Englewood Cliffs, NJ: Prentice

Hall, 1962. (editor. M. K. Fort, Jr.)

Fritsch, R., and Fritsch, G. The four-color theorem: history, topological foundations, and idea of proof.Springer-Verlag: New York, 1998.

Gale, D. The game of Hex and the Brouwer fixed-point theorem. American Mathematical Monthly,vol. 86, 1979, pp. 818–827.

Gardner, M. Mathematical games. Scientific American, December 1959, pp. 166–177.Gardner, M. Mathematical games. Scientific American, July 1961, pp. 150–152.Gardner, M. Mathematical games. Scientific American, January 1962, pp. 136–143.Gardner, M. Mathematical games. Scientific American, July 1967, pp. 112–116.Gardner, M. Mathematical games. Scientific American, December 1972, pp. 104–105.Gardner, M. Mathematical magic show. New York: Knopf, 1977.Gardner, M. Mathematical puzzles and diversions. New York: Simon and Schuster, 1959.Gardner, M. No sided professor. In Fantasia mathematica. New York: Simon and Schuster, 1958.

(editor, C. Fadiman.)Hilbert, D., and Cohn-Vossen, S. Geometry and the imagination. New York: Chelsea, 1956.Hudson, H. Four colors do not suffice. American Mathematical Monthly, May 2003, pp. 417–423.Hutchinson, J.P. Coloring ordinary maps, maps of empires, and maps of the moon. Mathematics

Magazine, October 1993, pp. 211–225.Jackson, A. A geometer’s view of space-time (review of Osserman’s Poetry of the Universe), Notices

AMS, June 1995, pp. 675–677.Jacobs, K. Invitation to mathematics. Princeton, NJ: Princeton University Press, 1992.Jurisic, A. The Mercedes knot problem. American Mathematical Monthly, November 1996,

pp. 756–770.Kaku, M. Hyperspace. New York: Anchor Books, 1994.Kauffman, L.H. New invariants in the theory of knots. American Mathematical Monthly, March 1988,

pp. 195–242.Kauffman, L.H. On knots. Princeton, NJ: Princeton University Press, 1987.Lickorish, W.B.R., and Millett, K.C. The new polynomial invariants of knots and links. Mathematics

Magazine, February 1988, pp. 3–23.Livingston, C. Enhanced linking numbers, American Mathematical Monthly, May 2003, pp. 361–385.Livingston, C. Knot theory. Washington, DC: Mathematical Association of America, 1993.Luminet, J.-P., Weeks, J.R., Raizuelo, A., Lehoucq, R., and Uzan, J.P. Dodecahedral space topology as an

explanation for weak wide-angle temperature correlations in the cosmic microwave background.Nature, October 9, 2003, pp. 593–595.

Meyerhoff, R. Geometric invariants for three-manifolds. The Mathematical Intellegencer, 1992,pp. 37–53.

Neuwirth, L. The theory of knots. Scientific American. June 1979, pp. 110–124.Osserman, R. Poetry of the universe. New York: Anchor Books, 1995.Penny, D.E. Perspectives in mathematics. Menlo Park, CA: Benjamin, 1972.Pits, Peaks, and Passes. Video with Marsten Morse. Washington, DC: Mathematical Association of

America, 1993.Ringel, G. Map color theorem. Springer-Verlag, New York, 1974.Rolfson, D. Knots and links. Providence, RI: AMS Chelsea Publishing, 2003.Saaty, T.L. Thirteen colorful variations on Guthrie’s four-color conjecture American Mathematical

Monthly, January 1972, pp. 2–43.Schwarz, G.E. The dark side of the Möbius strip. American Mathematical Monthly, December 1990,

pp. 890–897.Senechal, M., and Fleck, G. Shaping space. Boston: Birkhauser, 1988.Shape of Space. Video created by Jeffrey Weeks. Emeryville, CA: Key Curriculum Press, 1998.Shimamoto, D. and Vanderwaart, C. Spaces of polygons in the plane and Morse theory. American

Mathematical Monthly, April 2005, pp. 289–311.Stahl, S. The other map coloring theorem. Mathematics Magazine, May 1985, pp. 131–145.Steen, L. ed. For all practical purposes. New York: W. H. Freeman and Co., 1988.Steinhaus, H. Mathematical snapshots. New York: Oxford University Press, 1950.Stillwell, J. Geometry of surfaces. New York: Springer-Verlag, 1992.Sullivan, M.C. Knot factoring. American Mathematical Monthly, April 2000, pp. 297–315.Summers, D.W. Untangling DNA. Mathematical Intelligencer, vol. 12, no. 3, 1990, pp. 71–80.Sumners, D.L., et al., editors. New scientific applications of geometry and topology. Proceedings

of Symposia in Applied Mathematics, vol. 45. Providence, RI: American Mathematical Society,1992.

Resources 325

Tannenbaum, P., and Arnold, R. Excursions in Modern Mathematics. Upper Saddle River, NJ:Prentice-Hall, 1998.

Thomas, R. An update on the four-color theorem. Notices of the American Mathematical Society,August 1998, pp. 848–859.

Thurston, W. Three-dimensional geometry and topology, Vol. 1. Princeton, NJ: Princeton UniversityPress, 1997.

Thurston, W.P., and Weeks, J.R. The mathematics of three-dimensional manifolds. Scientific American,July 1984, pp. 108–120.

Toussaint, G. Simple proofs of a geometric property of four-bar linkages. American MathematicalMonthly, June/July 2003, pp. 482–494.

Vesyolov, A.P. Flexible in the face of adversity. Quantum, September/October 1990, pp. 13–18.Walker, K. Configuration spaces of linkages. Senior thesis. Princeton, NJ: Princeton University, 1985.Weeks, J.R. Exploring the shape of space. Emeryville, CA: Key Curriculum Press, 2001.Weeks, J.R., The Poincare dodecahedral space and the mystery of the missing fluctuations. Notices

of the AMS, vol. 51, pp. 610–619.Weeks, J.R. The shape of space, 2nd edition. New York: Marcel Decker, 2002.Weeks, J.R. The twin paradox in a closed universe. American Mathematical Monthly, August/September

2001, pp. 585–590.Why is there life? Discover, November 2000, pp. 64–69.Wilson, R. Four colours suffice: How the map problem was solved. Princeton, NJ: Princeton University

Press, 2002.

Communicating Project Results

Think of your project as the response to a question. For example, “What is theshape of space?’’ “Can knots be factored uniquely into prime knots?’’ “What aresome good representations of the cross-cap?’’ As in solving any problem, theresponse will occur in stages, such as

● Exploring

● Gathering data

● Search for a pattern

● Working out examples

● Making a model (or two...)

● Coming up with an answer to the question, or a partial answer, or an answerto a similar or related question

● Making a convincing argument explaining the answer

When your project involves an expository component, you will likely findmaterial written about your problem. Read and digest it, knowing that you willneed to explain it to someone effectively. Put together the material in a new way.Turn the mathematical explanations into ones you understand and can explain.Mine the material for mathematics if it is not evident. Ask simple mathematicalquestions you can answer. Relate the material to things discussed in the text byanalogy or by generalization or by extension or by any other method. Make amodel (or two) that illustrates the material.


Even though the material may not be new to the world at large, it will be new toyou. Wrestling with the ideas, making examples, building models, and puttingtogether an exposition of them always produces a new prospective on the originalmaterial and frequently produces new proofs and insights. I have chosen thetopics above to maximize the possibilities of making connections with material inthe text, possibilities for models, and possibilities to put materials together in newways appropriate for you.

After you have found an answer (or answers) you are happy with, then you mustwork on communicating the results of these earlier stages. This communicativeaspect of the project could have one or more of the following parts:

● A display. This could take the form of a display (science fair style) that in turnwould include a summary of your results: the question to be answered, yourplan of attack, ideas you used/developed, and a summary of your results. Thisaspect of the project should convey in a visual way an overall feeling for theproject (its results and its procedures), without providing too many details.This is a good place to use models, illustrations, diagrams, and charts. Thereis more on the display in the next section.

● An oral presentation. This is an occasion for you (and your team) to intro-duce to others the results of your project and to point out important featuresof the display (if there is one). The oral presentation should effectively commu-nicate a few of the main ideas of the project’s topic, its flavor, its interest to othermathematicians, its applications in the real world, how it relates to the topicsencountered in the rest of the book, and perhaps an esoteric detail or two.

Here is where you can use a cute trick or gadget. You want to convey, in a shorttime, the flavor of the project—an aspect of its solution that’s easy to communicate,a connection with other problems the listener will know, the significance of theproblem. You want to strike a balance between the trivial and the complex: On theone hand, you do not want to spend time belaboring the obvious; on the other,you do not want to subject your listener to an involved proof. (In both cases, thelistener would be bored.) If listeners want to know more than you are able toinclude in your presentation, let them ask questions. Your presentation will bemore successful if you let them participate in it: Pick up on where they arepuzzled, on what their interests are. You must remember that you have been livingwith your topic and are an expert on it. Your listener is not. Do not overwhelm herwith jargon. What you think is trivial (relative to the project) might be obscure tothe uninitiated. Practice your presentation. Time it. Practice it again. Make sureyou have all your props ready and in order. (If you have to fumble around duringyour presentation for a piece of paper or a model, you lose precious time and your audience.)

● A paper. Like any other paper, it should be an exposition of the results of theproject; it should describe the ideas, procedures, and theorems the team usedto answer the question/solve the problem; and it should include the answer(s)and written arguments justifying them. If your project involves a display

Communicating Project Results 327

and/or oral presentation, the paper is the opportunity for you (and your team)to expand on the broad ideas and impressions indicated in the display/oralpresentation and to provide more of the details to back up what was claimedthere. In the paper, you put the pieces together and show how they areconnected. The paper can use as many pieces of the display or presentation asyou wish, but there should be no loose ends.

Make sure that your reader knows what question (or set of questions) you areanswering. Describe how you (or others) explored the terrain in looking foranswers. Provide some of the examples you worked out or create new examplesthat are useful to demonstrate your lines of attack or to point out possible blindalleys. Again, provide pictures, diagrams, and tables to help illustrate your ideasand summarize your results.

You want the arguments to be valid, and you want what you write to make sense.If you quote source materials, make sure you understand what those sources say.Document your sources.

Nitty-Gritty on Putting Together an Effective Display

Get a backdrop for your display. The easiest thing to do here is to buy a coated,corrugated cardboard triptych, about 36 inches high by 48 inches wide, whichmany office supply stores sell. Alternatively, you can buy a piece of foam board foraround $7. Then, with knife and a metal straight edge, score it twice to make atriptych that will stand up on it own (see pictures below).

Choose a title for the project. The title should be a question that your projectanswers. It should be immediately understandable to the viewers. Place the title ina prominent place on the foam board.

Outline main aspects of the project. Here is a sample outline:

● Detailed statement of problem

● Significance of problem

● Procedures used (steps taken) to solve problem

Score alongdotted lines.

Bend backalong scoredlines.

Stand it up:make a triptychon top of table.


● Results obtained

● Arguments used

● Illustrative examples

● Descriptions of models accompanying the display

● Conclusions

You could think of this as an outline for the project’s paper as well. Projects willdiffer. Depending on the project, you may want to delete or add to the list above.

Decide on windows. Plan on having 6 to 10 “windows of information’’ to attachto the display board. Each “window’’ would correspond to an item in your outlineand be labeled by the outline heading. Each window should be as much of a visualdisplay of information as possible, employing diagrams, photos, and graphs andusing only brief verbal descriptions.

Make it spiffy. To make your display attractive, you may want to use large,computer-generated or stick-on letters for the title. To accompany the title, youmay want to have a single, large, catchy, and representative picture or diagram.You may want to place each “window’’ on a background of colored paper to makeit stand out. You want your display to draw in viewers and get them interested andintrigued. You are selling your project. You want the viewer to become curiousabout it and to ask questions.

Project Evaluation

Here are criteria (the four C’s) for evaluating every aspect of your project, whetherit is an oral presentation, display, or paper.

Clarity. Is the project topic (or problem) clear? Are project goals—how the topic is to be dealt with, what kind of problem resolution is sought—clear? Is thepresentation carefully organized? Can an outsider see how the various pieces ofthe project are connected?

Completeness. Have all reasonable lines of inquiry been investigated? If somelines have been used and not others, have the choices been explained (e.g., “therewas not enough time to pursue such and such a line of reasoning,’’ “such and suchapproach appeared to be too difficult’’): Have connections been made between theproject and other areas of study, especially areas covered in the course? Is the projectproblem analogous to a problem discussed in the course? Is it a generalization?Does it carry a problem (and its solution) one or two steps beyond what the book did? Does the project shed light on a problem seen before? Is the solution tothe project problem similar to the solution of a problem seen before? What arerelated problems that need further study? Why is the project problem important?Does its solution have applications to some part of mathematics? Is its solution

Project Evaluation 329

useful to another discipline or to the real world? Did the problem remainunsolved for a long time? Does the project incorporate effective uses of models,charts, or diagrams where appropriate? Does the project incorporate effective useof technology where appropriate?

Correctness. Are all parts of the project mathematically correct? Are definitionscorrectly stated? Are the arguments correct? Have all aspects of the project beencommunicated effectively in correct English? Are outside sources acknowledged,and is there an appropriate bibliography?

Creativity. Does the team use an original approach to solving a problem? Arethere unusual ways of putting ideas together? Are the means of communicatingproject’s topic or its solution new? Does the project include the effective use of aclever model? Has the project used technology (computers, videos, overheads, etc.)creatively?

The bottom line

In the end, what should you get out of the project?

● The experience of doing research: choosing a topological problem/topic,working independently to solve it/understand a body of material related to it,and making the related ideas hang together

● The experience of having a real context (more or less self created) out of whichto expound on mathematical ideas, put together an effective visual display, andmake an oral presentation

● The experience of communicating your insight to others

● Making models and appreciating their value, for explanation, for illustration,and for understanding

● Experience working with a team: learning from and communicating withyour peers

● Sense of pride in a job well done

● Passion: Really getting into something you’re interested in

(Much of the material in this chapter on carrying out a project has been adaptedfrom the project chapter of the author’s Geometry by Discovery, published in 1998by John Wiley and Sons.)


Index

331

A

Adjacency pattern, 268Airplane videos

cylinder (inner tube) and, 150-151, 153, 155edges in, 150flat screen and, 150

Alexander polynomial, 310, 311Appel, Kenneth

four-color problem and, 99-100

B

Ballentine rings, 306, 321Bee-hive surface, 173. See also Cross-cap;

Projective planeBeehive-with-lake

Möbius strip as, 160-162Border

defined, 24. See also Edge(s)Border rule

between two countries, 7Borromean rings, 306Boundary, 147. See also Edge(s)Box-with-square hole

correspondence with map on doughnut,116-117

Boy’s surface, 175Braids, 321

Bridgebuilding in three-space

for two-dimensional figure-eight, 238-241in three-dimensional space, 238

Bridges of Konisberg problemEuler’s analysis of, 46Euler’s theorems for, 45

converses of, 46

C

Cauchy, Augustin-Louisproof of Euler’s formula for network, 71

Checkerboard domino puzzle, 23Chromatic numbers

and critical graphs, 256Circles

coloring non-overlapping, 85-86in maps on rectangles, 15-20

Circuit boardwith hole, 104-109

comparison with box with hole,104-106

Circuit board designinvestigation of, 121

Circuit board networkswith five terminals problem

resolution of, 66-67investigations of, 42-43, 47

Coloniescoloring maps with, 314-315

Coloring mapsG7 map problem, 1-4list of maps for 6rules and terminology for, 6

Complexity theory, 102, 323Countries

coloring problem for, 8defined, 24with edges, 74on real maps, 9sizes of, 76on transformed maps, 8-9

Critical graphsand average order of vertices, 257-258

network of map on sphere, 258-261chromatic numbers for, 256Lemma about, 257

Cross-cap(s), 173, 174arguments, 317assembly of

theorem for, 215in three-space, 248-249

Euler number for, 198in four-space, 249“nice” models of, 316possibilities for, 247“regular” maps on, 319

Crossing(s)coloring rules for, 284defined, 291in five-half-twist strip, 277-278

inverses of, 278-279ingredients of, 292in knots, 277, 278, 281-282of not knot, 282overpass and underpass, 277on square knot, 277three-colorable, 284-285

Csázsár’s polyhedronconstruction of, 119-120

Cube, data for, 62Cylinder

airplane video game and, 151-152, 153, 154-155as bridge for Klein bottle, 241-242

D

de Morgan, Augustus, 24Dirac, G. A., 269Displays

of projects, 327putting together, 328-329

Domino problem, 47Domino puzzles, 43-44Door inspector problem, 38-40, 47

investigation on doughnut, 116Dot-line diagram. See also Network;

Grapheven number of lines to dots, 34, 36of Konisberg tour problem, 31-33of New York City, 34nice tour on, 34, 35, 36odd number of lines to dots, 34of Paris, 34

Doughnut mapdata from, 109, 111-113

theorem about, 113Euler formula for, 114

Doughnut(s). See also Inner tube;Torus

colors for, 318connecting, 192five terminal problem on, 106-10717-holed

colors needed for, 263infinite family of maps on, 107-108as inner tube, 113map of, 109, 111-113n-holed

colors for, 318“regular” maps on, 319

seven-terminals investigation, 116six-terminals investigation, 116three-holed, 193two-holed, 193

colors for, 318Ungar-Leech map on, 318-319

Doughnut-with-lake, exercise, 109

E

Earthcoloring maps on, 315

Edge(s), 37, 50counting from vertices, 74-75

theorem for, 76of countries, 73-74of twisted strips, 136-137, 140-141

Empire maps, 97-98, 101Equivalence

geometric, 24of surface symbols, 210-211

exercise for, 212prime surfaces and, 220questions, 219rules for

332 Index

Equivalence (Continued)investigations of, 218-219

topological, 250Euler, Leonhard

bridges of Königsberg problemtheorems of, 45

geometry of position and, 45-46Euler formula(s). See also Euler number

for network, 71for polyhedra, 70-71

Eulerian circuit, 45Eulerian path, 45Euler number, 201-201

from symbol, 230-231problem of pairs not same, 228for S, S’, and S-with-lake, 198of surfaces, 199, 219, 227

exercise, 236for sphere, 194, 198for torus, 194, 198

of surface sumformula for, 196proof of, 193-196

Exponential time vs. polynomial time, 102

F

Figure-eightin plane, 237

Five-cities problem. See also Five-terminalproblem

on Möbius strip, 144Five-color maps

on doughnut, 114-115on sphere, 81-83, 99, 878

Five-color theorem, 80-83, 87, 99Five-terminal problem

on doughnut, 106-107solution of, 110

on sphere, 104-105Euler’s formula and, 102Jordan curve theorem and, 102-103simple closed curve and, 122-123

on square, 110-111, 113Four-color conjecture, 314

Bram’s map coloring game, 96investigations of

map of empires, 96-98pennies in plane, 98

method foraltered map in, 92-93applications of, 94-95β-α sequence in, 91α-β sequence in, 90-91

Four-color conjecture (Continued)steps to color map, 93-94Tetrachrome map coloring game, 95

Four-color mapshistory of, 99

Four-color problem, 24difficulty in resolution of, 100-101solution of, 99-100

Four-color theoremreal proof of, 315

Four-space, 323defined, 238Klein bottle in, 237-238, 241-243, 248, 250Klein bottle-with-lake in, 243-244publications on, 252

G

Game(s)hexlike, 322map coloring, 95, 96on surface, 317

Geometeryn-dimensional, 250

Geometric equivalence of maps, 24Geometry of position, Euler, 45-46Gizmo (calendar) problem, 60-61

calendar gizmo exercise, 65data for, 63formula for, 63

Graph(s), 45. See also Networkcomplete bipartite, 121critical, 256, 257defined, 255dual, 255-256, 256nonplanarity of, 323planar, 45

Guthrie, Francis, and four-color maps, 99

H

Haken, Wolfgangfour-color problem and, 99-100

Heawood, P. J.empire maps and, 97-98, 101five-colors for maps on sphere

proof of, 87and four-color maps, 95, 99

Heawood’s conjecture, 267for nonorientable surfaces, 270proofs of, 267-268, 269-270

Heawood’s inequalityin map color theorem, 260-262, 267

Index 333

Hexagon, as pattern, 162-163Hexlike games, 322

I

Inequality(ies)in map color theorem, 260-262, 267

Infinite families of maps, 53, 67, 315Inner tube, 113. See also Doughnut; Torus

airplane video game and, 150-151, 153, 155with lake

investigation of surfaces for, 163-164simple closed curves puzzle, 118-119tic-tac-toe game, 118

Inner tube tic-tac-toe (IT4), 118Invariants

for knots, algebraic, 310for surfaces, 249, 309

Island. See Maps on island

J

Jones polynomial, 321

K

Kelvin, Lord, knots and, 309Kempe, Sir Alfred

and four-color maps, 99Klein, Felix, 175

nonorientable surfaces, 201Klein bottle

bridge for, 238-241definitions of, 251Euler number for, 198-199exercises, 155in four-space, 237-238

assembling, 241-243, 248, 250map coloring on, 319pattern for, 154-155possibilities for, 248“regular” maps on, 319six-cities problem on, 264some things to make, 167-168, 171-172symbols for, 211

Klein bottle tic-tac-toe (KBT3), 169-170Klein-bottle-with lake (KBWL)

in four-space, 243-244in three-space, 243

Knot picturesdetermination of different knots in, 293-294ingredients of, 291-292

Knot(s)and biology, 321

classification ofthree-manifolds classification and,

311, 321as closed curve in plane, 276colorability of

investigations of, 299-300method for questions, 300-302

crossings in, 277, 278, 281-282defined, 274five-colorable, 290-291

modular-clock arithmetic in, 290generalization of three-colorability to n

colorability, 288-289group of, 322invariants for, 310inverses of, 322investigations of, 299-309Jones polynomial of, 321Kelvin Lord, and, 309and links, 320-322, 323lists of, 309with loose ends, 302manipulation of one into other, 274, 275,

276-277, 281, 283n-colorability of, 291, 292, 304

exercise in distinguishing, 292-293n-colorability rules for, 289n-link in, 304-305not knot, 282overhand, 274, 275, 276, 284-285prime factorization of, 322questions, 300, 307, 308sum of two

n-colorability of, 304surfaces spanning, 320three-colorable, 284-287

rules for, 284trefoil, 306-307for wrapping package, 302-303

Knot theoryand statistical mechanics, 311and topology, 311

Königsberg with bridges tour, 27-33

L

Lakescolors needed for surfaces with, 255-256identifying from surface symbol,

233-234number as invariant of surface, 249patterns for numbers of, 228theorem relating number of, for symbols,

235, 236

334 Index

Legendre, Adrian-Marieproof of Euler’s formula for polyhedra, 71

L’Huilier, Simon-Antoine, 123, 202Lines. See Edge(s)Linkage

control of, 179-180coordinate system for, 180

edge identification and, 182first case, two rods, 178for investigation, 200-201negative angles in, 181-182postive angles in, 180-181of rods and pivots, 177-178second case, 183-184

exploration of, 184-187and surfaces, 317surfaces and patterns, 188-190

theorem connecting, 190as torus, 182

Linking number, 321Links

equivalence of, 305n-link picture, unlinked, 304-305

unlinked, 305-306rules for, 305two-link n-colorable

for even half-twist strips, 307-308Listing, Johann Benedict

Möbius strip and, 146Little, C. N.

lists of knots, 309

M

Map coloringgames, 95, 96G7 map problem, 1-4list of maps for, 6projects, 314-315rules and terminology for, 4-5rules for, 7

Map coloring problemde Morgan, Augustus, 24origin of, 23

Map color theorem, 254four-color problem for sphere and, 262inequalities in, 260-262, 267notation and definition for, 256-257proof of, 256-262

Map dataformula from, 53, 54relationships in, 51vertices (V), edges (E), countries (C), 50

altered, 57

Map data (Continued)one country map on an island, 50simplified, 57-58two-island case, 58vertices of order two, 59

Map of empiresinvestigations of, 96-98

Mapsformulas for, 53, 54imaginary maps, 52-53infinite sets (families) of, 53ingredients of, 24, 50same, properties of, 208on sphere, rectangle or island, 24two are the same, defined, 207, 208

Maps on islandcolors needed for, 69construction of dual graph to, 256-258data for, 109distortions of polygon or disc, 55-56as doughnut, 108on doughnut, 108formula for, 55with lake

colors needed for, 69data formula for, 68

OK map on, 56defined, 60theorem for, 55, 60

side and edges of, 144from three-dimensional shapes, 61, 64-65

Maps on rectanglesinvestigations of

circles in, 15-16, 18-20circles in with one chord, 16-17curves in, 20rectangular lake in, 17-18regions in, 18-20steps in, 14-15Y-shaped figures in, 16

line drawing from side to side, 11-14,24-25

questionsnumber of colors for, 18, 20

Maps on sphere, 62, 64-65. See also Sphereaverage order of vertices on, exercise for,

258-261calendar gizmo and, 68circuit board network with five-terminals

problem, 66, 67colors needed for, 69countries as hexagons, 65cube and, 61, 68five or fewer colors for, 80-83

Index 335

Maps on sphere (Continued)proof of, 80-83, 87theorem for, 83

formula for, 65four-color problem and, 262with quadrilateral countries, 70six or fewer colors for, 78-79, 80

elimination of edges, vertices, countriesmethod for, 77-78, 79

restoration of eliminations in reverseorder, 78-79

theorem for, 80three-color, 98

exponential vs. polynomial time for, 102of simple closed curves, 128

with triangular countries, 70two-color, 102vertex order three

classification by combinatorial type, 84-85country with five or fewer edges, 84country with five or six edges, 84questions, 85-86

vertex order two, 58-59Möbius, Augustus Ferdinand, 146

orientable surfaces and, 222Möbius band. See also Möbius strip

six countries on, coloring, 265Möbius ladders, 271-272

odd half-twists of, 283, 284-285in three-space, 273-274

Möbius shortsconstruction of, 143-144gathering evidence, 165-166

Möbius stripairplane video game and, 153-154,

155, 156, 157applications of theory, 147as beehive (cross-cap) with lake, 157, 160, 161cultural appearance of, 146cut

in half, 127, 135, 145with odd number of half-twists, 135in thirds, 128-129

discovery of, 146edge diagrams of, 141edges and sidedness of

gathering evidence, 139Euler number for, 198exercises, 130-132experiment as living in, 170-171facts about, 131half-twists in

cut in half, 128-129cut in nths, 139

Möbius strip (Continued)cut in thirds, 128-129, 143even number of, 136-137, 139investigation of cuts in, 138-139number of edges of, 136-137odd number of, 136-137, 138-139

one-edged, 130, 131one-sided, 126, 130, 131pattern for, 133-134

dressmaker’s, 132-133, 140side and edge of, 144tic-tac-toe, 144tour around

reversal of orientation on, 228twist in, 126

cut down middle, investigation of, 142edges of, 136-137, 140-141investigation of, 137-138, 141patterns for, 135pieces from cutting, 138

two-edged, 130, 131two from Klein bottle, 171two half-twists

results, 131zero half-twist, 130

Möbius tic-tac-toe (MT3), 144Moon

coloring maps on, 315

N

Network. See also Graphconnected, 37

example, 37disconnected

example, 37on doughnut surface, 117nice tour of, 40OK tour of, 40path in, 37on sphere, 117

Nice tourdefined, 37on dot-line diagram, 34, 35, 36of Königsberg

bridges needed for, 44on a network

theorem and conjecture, 38, 40Nonorientability. See Surface

O

Odd-twisted stripsedges of, 274, 275

336 Index

OK map on islanddefined, 60theorem for, 55, 60

OK maps on sphereclassification of, 69-70four-color

puzzles, 86investigation on surface of

doughnut, 116vertices order three

questions, 85OK tour, of connected networks, 40Order-three reduction

for map coloring, 77Order-two reduction

for map coloring, 77Orientabile surface

theorem for detection of in symbol, 230Overhand knot, 274, 275, 276, 284-285

three-colorable, 285

P

Path, 37Pattern(s)

adjacency, 268for an object, 167beehive-with-lake, 160connecting surfaces with, 188-190equivalent, and surfaces, 208facts about, strange, 162investigation of, 166-167for Klein bottle, 154-155recognition of, 166for surface, 167-168for surface-with-lake, 191triangle, 160-162

Poincaré, Henridefinition of surface, 250-251group of knots, 310-311

Polygonal screenspattern exercise for, 158

Polyhedral surfaceconstruction of, 264-265

Polyhedron(a)Csázsár’s, construction of, 119-120Euler’s formula for, 70-71regular, generalized, 316ring-shaped

formula for, 123Polynomial time, exponential time vs., 102Projective geometry

surface of sphere and, 173-174Projective plane, 173, 175

Projects, 313-326communicating results, 326-328displays of, 327, 328-329evaluation criteria for, 329-330n-link, nonintersecting closed curves,

304-305

R

Rectangles. See Maps on rectanglesReidemeister, Kurt, knots and, 311Riemann, Bernhard, surfaces and

connectivity, 201

S

Schäfli, Ludwig, and projective plane, 175Seifert, H., 222Six-cities problem

on Klein bottle, 264on Möbius strip, 263

Six-countries map. See alsoSix-terminal problem

on Möbius band, coloring, 265Six-terminal problem, 116South America

real map of, 5transformed map of, 10, 11

Spaceshape of, 320topological, 251-252

Sphere. See also Maps on sphereconnected network on, 117Euler number for, 194, 198Möbius strip problem for, 144orientable surface of, 175

Sprouts, 324Square

boundary as equator of hemisphere, 156Surface(s)

algebra of, 205-209analyzing, 193Boy’s, 175classification of, 249

proofs of, 318classification theorem for, 222

proofs of, 222-223connecting with patterns, 188-190correspondence between, 207defined, 188, 206, 250-251duplication of, 226equivalent patterns and, 208Euler numbers for, 194, 198, 227

exercise, 236

Index 337

Surface(s) (Continued)facts about, strange, 162familiar, models of, 317in four-space

developments in, 250-251fun-film, 315-316games on, 317geometry of, 316-317identification of, 246-247, 249

with Euler number, 199invariants for, 249, 309

n-colorable, 309investigations of, 220, 244, 245of Klein bottle

problem solving on, 170lakes on

colors needed for, 255-256investigations of, 244, 245

list of, 217, 218-219complete, 226

as maps, 206-207with maximum colors, 318two are the same, defined, 207, 208

mystery, questions about, 244-245nonorientable

classification of, 222defined, 229Klein bottle, 175Möbius strip, 175studies of, 201, 202

orientableclassification of, 222sphere, 175torus, 175

pits, peaks, passes, and Euler, 315prime, 220same

application of theorem, 236theorem about, 235two as, 250

for solution of 10-cities problem, 263

sum of, algebraic properties of, 219two-color theorems on, 318what it might be, 206

Surface(s)’,(prime) equivalence of, 222Surface sum, 190-191

gathering evidence, 197pattern for, 192

Surface symbolsconstruction of, 209equivalent

rule 1 for, 213-214rule 2 for, 214-215

Surface symbols (Continued)rule 3 for, 215, 218rule 4 for, 215

exercises for, 218identifying lakes from, 233-234lakes and

exercise, 236letters in

occurrence of, 213theorem for pairs of, 216

vertices of, labeling of, 231-233Surface-with-lake

number of colors needed for, 255-256Symbol(s)

equivalent, 210-211exercise for, 212

for familiar surfaces, 211-212surface, 209theorem relating number of lakes and,

235, 236theorem relating orientability of, 230

T

Tait, P. G.lists of knots, 309

Thom, René, 100Threlfall, W. A., 222Three-color map(s)

South America, 5Western Europe and, 5

Three-dimensional mapcoloring countries of, 21-22

Three-dimensional shapesconstruction of map on island, 61, 64-65formulas for, 62

arguments for, 60-62maps on, 60-62

Three-dimensional worlds, 319-320Three-space

defined, 238Klein bottle-with-lake in, 243map coloring in, 319Möbius ladders in, 273-274

Tietze, HeinrichHeawood’s conjecture for nonorientable

surfaces and, 270polyhedral surface and, 264-265

TorusEuler number for, 194, 198linkage and, 182map on surface of

construction of map with eight countries, 266

338 Index

Torus (Continued)on Möbius strip, 144with seven-color map on

construction of, 265-266surface as sum of, 216theorem for assembling, 217

investigation with, 217Torus knots, 320Tour(s)

of building for door inspector, 38-39investigation of, 40question, 42

Central Park problem, 38dots-line diagram of, 31-33essential ingredients of, 33, 34flights between Arizona towns and cities,

36-37investigations of, 40, 42-43, 44

circuit board networks, 42-43, 47connected networks, 40hand shaking odd number of

times, 40-41mail delivery for downtown Tucson, 41rectangular museums, 43streets (edges) one-way, 41

Königsberg with bridges, 27-33on network, 40nice, 34in real world, 323terminology for, 37

Transformed map(s)countries on, 8-9South America, 10, 11Western Europe, 8-9

Trefoil knotcoloring of, 306-307

Triangleas hemisphere, 160making from Möbius strip, 162as Möbius strip, 159-160

Two-color maps, 11-13, 14on doughnut, 117

Two-color theorems, 318for map coloring, 24

U

Ungar-Leech map, on doughnut, 318-319Utilities problem, 42, 47

five-terminals problem and, 66-67on Möbius strip, 144on surface of doughnut, 116

V

Vertex (vertices), 37coloring, 255defined, 24, 45labeling of

to count lakes, 233-235exercise, 236investigation of, 244for surface symbol, 231-233

odd and even, 37order of, 24-25, 37

on critical graph, 257-258order three

edges of countries and, 76in four-color conjecture, 91reduction of, 77

order tworeduction of, 58-59

W

Weichold, Guidononorientable surfaces and, 222

Western Europeancient map of, 9transformed map of, 8-9

Index 339

Explorations in Topology: Map Coloring, Surfaces and Knots

Documents

Transcript of Explorations in Topology: Map Coloring, Surfaces and Knots