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 Paris San Diego San Francisco Singapore Sydney Tokyo Academic Press is an imprint of Elsevier

Academic Press is an imprint of Elsevier 30 Corporate Drive, Suite 400, Burlington, MA 01803, USA 525 B Street, Suite 1900, San Diego, California 92101-4495, USA 84 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 any means, electronic or mechanical, including photocopy, recording, or any information storage 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 Data Gay, David. Explorations in topology / David Gay. -- 1st ed. p. cm. ISBN 0-12-370858-3 1. Topology. I. Title. QA611.G39 2006 514--dc22 2006017400 British Library Cataloguing in Publication Data A catalogue record for this book is available from the British Library ISBN 13: 978-0-12-370858-8 ISBN 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 his permission to use them. For all information on all Elsevier Academic Press publications visit our Web site at www.books.elsevier.com Printed in the United States of America 07 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

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Contents

Preface vii Acknowledgments xi Chapter Chapter Chapter Chapter

1: 2: 3: 4:

Chapter Chapter Chapter Chapter Chapter

5: 6: 7: 8: 9:

Chapter 10: Chapter 11: Chapter 12: Chapter 13: Chapter 14:

Acme Does Maps and Considers Coloring Them 1 Acme Adds Tours 27 Acme Collects Data from Maps 49 Acme Collects More Data, Proves a Theorem, and Returns to Coloring Maps 73 Acme’s Solicitor Proves a Theorem: the Four-Color Conjecture 89 Acme Adds Doughnuts to Its Repertoire 103 Acme Considers the Möbius Strip 125 Acme Creates New Worlds: Klein Bottles and Other Surfaces 149 Acme Makes Order Out of Chaos: Surface Sums and Euler Numbers 177 Acme Classifies Surfaces 205 Acme Encounters the Fourth Dimension 225 Acme Colors Maps on Surfaces: Heawood’s Estimate 253 Acme Gets All Tied Up with Knots 271 Where to Go from Here: Projects 313

Index 331

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Preface

This book will (I hope!) give you a rich experience with low-dimensional topology, get you really hooked on solving mathematical problems, and launch you into an adventure of creating ideas, solutions, and techniques—one that gives you a “research experience” with topology. Of course, I also hope you get excited about 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 will encounter in this book emerged from a long evolutionary process. I hope that this book engages you in this process while it introduces you to the topological point of view. Along the way, you will encounter many of the concrete, simply-stated problems for which much of modern topology was created to solve—problems associated with maps, networks, surfaces, and knots. The book is more than just an introduction to topological thinking. It was designed to provide you with the opportunity to ●

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

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observe new mathematical techniques springing from solutions to problems;

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see the connectivity of mathematical ideas—the solution to one problem helps 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 have frustrations, 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 vii

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solving problems, and provide you with tools to help you on your next mathematical journey. 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 questions that propel the unfolding of your own topological knowledge and understanding and 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 problemsolving process and provides you with ways to communicate and explain solutions. And then it lets you loose, as a former student of mine put it, “to write our own book.”

Format of the Chapters Each chapter (excluding chapter 14) poses a big problem and takes a stab at its solution. This takes place in the context of a story whose characters are employees of Acme Maps. The big problem and smaller related problems come up naturally as the characters carry out their work. The jobs Acme takes on expand naturally as the book unfolds. The characters fiddle around with the problems much as real problem-solvers would. They make mistakes. They find themselves in dead ends and work their way out. They create their own definitions and terminology. Sometimes they don’t solve the big problem, but they work on it and come up with partial solutions. These solutions are sometimes what a “mainstream” text would call theorems. As you follow the story, paper-and-pencil icons involve you by pointing out explicit tasks for you to carry out. These tasks are called Your Turn. A set of Investigations, Questions, Puzzles, and More replaces the traditional 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 you say? Can you explain it? In an investigation you make observations, look for patterns, make a conjecture, and prove the conjecture (or come up with a counterexample and refine the conjecture). When I teach a course that goes with this book I ask students to carry out many of these investigations. They form the heart of the course. To relate what goes on in the chapter to the rest of the world of mathematics each chapter ends with a section entitled Notes. It places the problems, concepts, and results of the chapter in their historical context. It introduces the standard terminology used by other mathematicians. Occasionally it will summarize the results of the chapter and set these results out in the form of theorems. A list of appropriate References follows the Notes section. The format of chapter 14 differs from that of chapters 1-13. This chapter is an annotated list of possible projects that extend the concepts of the book, but which may involve research beyond the materials of the book. It includes an extensive list

Preface

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of helpful resource materials and a guide for carrying out a project and communicating its results.

Overview of the Chapters ●

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

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Chapter 2. This is an introduction to networks (graphs)—to taking particularly “nice” trips on networks, and identifying those networks where such trips are possible.

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Chapter 3. This is an introduction to collecting data about maps and observing relationships among these data.

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

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

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

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

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

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

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Chapter 10. The study of surfaces becomes algebraic as symbol strings replace patterns and rules are developed for arriving at equivalent, recognizable symbols. The consequence: a classification theorem for surfaces.

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

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

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

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

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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 and take risks—by offering a solution that may turn out to be incorrect, by making a guess that might be wrong, or by using an approach to a problem that might not pan out. We will laugh at our mistakes. We will listen to others’ ideas and possibly build 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 highlights each chapter’s important aspects and points out investigations important for the development of the main ideas and it includes solutions to many of problems and hints for others. The manual contains large-size versions of diagrams and patterns that appear in the text so that the instructor could have copies made for use in class.

Acknowledgments

More than thirty years ago, I taught an experimental, low-dimensional topology course with an incredible group of talented and enthusiastic students. I wrote up notes as the course progressed. The course was so successful that I have taught it several times since, revising and supplementing my original notes each time. The notes eventually became this book. Several individuals played parts in the development and evolution of the course and book. Initial credit goes to my college teachers Albert Tucker and Ralph Fox who, when I was a mathematical tad, first fed me these topological ideas and got me hooked. Over the years, Philip Straffin and David T. Gay taught the course using 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 the The University of Arizona’s Department of Mathematics for supporting curriculum development and for valuing innovation in teaching. My colleague Olga Yparaki was particularly excited with the story format and encouraged me to continue developing 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 mathematical ideas, problems, and approaches. Most of these appear as references in the text. One author that deserves special mention is Martin Gardner. He has the uncanny ability to come up with problems that are accessible to students, that grab them, and that lead them nicely and naturally to deeper mathematical involvement. Of course, the lion’s share of thanks for this project’s development goes to the students who took the course, who struggled with the problems and projects and taught me what worked and what got them excited. It was their enthusiasm for these 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 manuscript thirty years ago and who has been an enthusiastic supporter of the project xi

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ever since. To those who reviewed the manuscript, I am grateful for the many helpful criticisms and encouraging comments during the phases of its development. Finally, let me thank my editor Tom Singer at Academic Press who believed in the project from the beginning, Michael Troy at Graphic World Publishing Services who kept me on task and let me know what was happening during production, Fritz Simon who turned my scribbles into effective illustrations and helped me make the manuscript “look like a book,” and Tulley Straub who helped render three-dimensional objects realistically. It was a pleasure and a privilege to work with all of you.

Acme Does Maps and Considers Coloring Them

CHAPTER

1

Scene 1 It is early afternoon in a small storefront company. The owner, Boss, 50-something and portly, wears a rumpled shirt with bola tie and cowboy hat. Joe, 30-something and 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 into each 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?

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2

Chapter 1 Acme Does Maps and Considers Coloring Them JOE: Hmm. Interesting. The G7 model with bold black borders oughta work for 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 a map.)

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 he was 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 same color...maybe pink? And Austria could be pink too. BOSS: Sure....Hey, wait! If you keep goin' like that, we'll be right back to where 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 different from its neighbors. So Portugal can be colored anything except the 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–

Scene 1 JOE: Stop. You're making it too complicated. Start with some country, like Switzerland, and color it red. Now you know none of its neighbors can 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 be different from red and blue. Then comes Austria; it can't be red, can't be green. OK, color it blue, just like France. And then color Italy green! 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 then Portugal 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. Then Holland can be red, and Denmark blue. Done! Four colors. Look.

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Chapter 1 Acme Does Maps and Considers Coloring Them

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 be colored....Hey, Boss, you're not listening. BOSS: (Looks at latest version on Joe's paper.) Not bad. I'm a genius! Why didn't I think of this before? Color maps with three colors. Cheap. I'll get 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 it down 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 Map To color a map properly: ● Each country must be completely colored by one color. ● Two countries sharing a common border must be colored different colors.

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?

1. Your Turn Help Joe. Can he color the map of Western Europe in three colors? There are two choices: 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 Turn Joe still needs help. Color the following map of South America properly and minimally. Boss will want to know how the coloring is minimal so provide Joe with an explanation. (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 of time before Boss will want me to color all maps properly and minimally. (He makes a list.)

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Chapter 1 Acme Does Maps and Considers Coloring Them

List of Maps ● ● ● ● ● ●

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

Hmm. The United States. That should be fun!

3. Your Turn Help Joe. Color the map of the 48 lower states properly and minimally. How many colors 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.

Scene 2 JOE: How should I color that? The four states all touch at the Four Corners so each one should be a different color. But four colors could really push the number of colors needed way up. What if a map had a ''Five Corners'' 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.)

New Rule for Coloring Maps Properly A border between two countries must consist of a positive length; it cannot consist of a single point or even two or three points or any finite number of points. With this new rule I can color the Four Corners area this way.

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

Maps ● ● ●

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

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8

Chapter 1 Acme Does Maps and Considers Coloring Them ● ● ● ● ● ●

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

Whew. The list goes on and on and I haven't even touched Asia! Then there's a new map created by a new configuration of countries in Eastern Europe and the former Soviet Union. There're maps that haven't been thought about, 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 seems to think I'd never need more than three, but I know better. What if I had a very complicated map, maybe one with thousands of countries or subdivisions? Wouldn't that need a lot of colors? (He takes out another sheet and writes.)

Big Problem Given any map of countries (one that exists or one that could exist in the future), 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 and more colors?

Scene 3

9

4. Your Turn What 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 little time. Infinitely many possibilities! This is a big problem that needs simplification. (He finds a map and pulls it out.)

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

10

Chapter 1 Acme Does Maps and Considers Coloring Them Same thing. Color all of these the same. It's as if the ''original'' map of Western Europe had been printed on a thin piece of rubber and then stretched-pulled-shrunk into the modern version. Hey, I have an idea! Why stop with ''modern''? (He draws.)

Cool. Would this sell? Western Europe as a rectangle where the countries have 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 Turn Help 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.)

Scene 4

11

Big Problem, Simple Version Any map can be transformed to a map on a rectangle where borders can be made up of straight lines. Solving the coloring problem for any map is reduced to solving the coloring problem for the transformed map.

6. Your Turn It's now easy for Joe to draw maps and practice coloring them. Here are ones he drew. Help him color them properly and minimally. For each one, write down the number of colors you used. Make up a couple of maps of your own, and do the same 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 possibilities. How about starting really simply:

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

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Chapter 1 Acme Does Maps and Considers Coloring Them Whoa! Look at that! Two colors. Amazing. I wonder if that always works. Add another line ...

Hmm. Works again. Now why does it do that? Now let's take the new map 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, reverse old 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 bunch where you only need two colors. BOSS: Ya don't say. JOE: Take a rectangle. Draw lines from one side to another, as many lines as you want. You always get a map you can color in two colors. Like this.

Scene 4 BOSS: I like the idea of two colors. But, those ''maps'' don't look like anything 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 map in two colors. The main idea works like this. Think of such a map as being made one line at a time. Here it is colored at one stage, just before adding another line that we'll call L.

You can see that the map (with line L added) is colored properly on both sides of L. The only problems are pairs of countries that have a border that's a piece of L. The solution is to reverse the colors on one side of L. You get a properly colored map. That tells you what to do when 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, and you 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 transition step to color this. Add yet another line, and use the transition step again. Keep doing this until all the lines have been added and the original map is colored in two colors.

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Chapter 1 Acme Does Maps and Considers Coloring Them

BOSS: Humph. Lot of coloring and recoloring. That's too much work. There must 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 to please. Anyway, I think we're making progress—one step at a time. (He takes a new piece of paper and writes.)

Progress on Map Color Problem Special case solved: Map on a rectangle created by drawing lines from one side to another can be colored in two colors. Moreover, I can describe exactly how to carry out the coloring. Solving the special case didn't solve the whole problem, but it's still pretty powerful. One of these maps–formed by millions of lines—can be colored properly with only two colors! Solving the special case was kind of an investigation. I didn't really know ahead of time what the outcome would be. It's worth documenting what I did. (He takes another sheet and carefully writes.)

How To Carry Out an Investigation Create 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 an argument explaining my two-color conjecture), or

Investigations, Questions, Puzzles, and More

15

2. Find a counterexample to the conjecture. (There is no counterexample to my two-color conjecture. However, Boss made the following conjecture: ''All maps can be colored in three colors.'' But I found that the map of Western Europe needed four colors. So the map of Western Europe is a counterexample to Boss' conjecture.) It had been quite a day, and it was time to go home. Before leaving, Joe wrote down some investigations and some questions, which he included in the following, for you to carry out and answer.

Investigations, Questions, Puzzles, and More 1. Investigation Joe has considered the special case of maps on rectangles formed by drawing lines from one edge to another. As a new case for investigation, he alters this prescription to allow, in addition to lines, circles drawn inside the rectangle as well as arcs of circles with 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 and the 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 and color some more until I find one. 3. Make a conjecture. 4. Justify the conjecture or find a counterexample to it. (If I find a counterexample, go find a conjecture that I can justify!)

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Chapter 1 Acme Does Maps and Considers Coloring Them

5. Try to generalize the outcomes of this investigation. That is, alter the rules for forming these maps in some simple way so that my conjecture still holds and can 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. Investigation Joe 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 three ends 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.

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Look for a pattern.

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Make a conjecture.

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Justify the conjecture.

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Tell Boss about my results.

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

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

Investigations, Questions, Puzzles, and More

17

●

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

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Joe also wonders whether or not the results of the investigation can be generalized; that is, can the rules for creating maps in this investigation be altered even further so that the number of colors needed stays the same and the arguments justifying it remain pretty much the same? (For example, in addition to Y's and circles with a single chord, what would happen if he were also to allow lines from one side of the rectangle to another?)

4. Investigation In 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 in the middle, a sort of island with lake.

Joe figures that a simple case of this type to investigate would be a map on this island 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 side of the island to the lake. Like this.

He outlines the investigation: ●

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

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

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Chapter 1 Acme Does Maps and Considers Coloring Them

●

Make a conjecture.

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Justify the conjecture or find a counterexample.

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If I find a counterexample, revise my original conjecture to account for this new evidence and try to justify the new conjecture. I want to arrive at a conjecture I can justify although I may have to revise my conjecture several times until I'm happy 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. Question Joe tells himself: ''I can draw a lot of maps that can be colored properly in two colors.'' Then he asks: ●

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

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

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

Help Joe answer these questions.

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

7. Question Joe showed Boss the method of coloring a map on a rectangle gotten by drawing lines 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 map in two colors? Could you explain (to Boss) why your method works? What Joe did has two parts: (1) a proof that, given such a map, there exists a way to color it using two colors, and (2) a method for coloring such maps that always works. You could assume part 1 and argue from there.

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

Investigations, Questions, Puzzles, and More

19

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

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

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Chapter 1 Acme Does Maps and Considers Coloring Them

Carry out this investigation.

9. Investigation You get a distorted circle if you take a pencil, and starting at a point on a piece of paper, you draw a curve without letting the pencil leave the paper and continue until you end up at the point where you started. Don't allow the curve to intersect itself. (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 simple closed curves of Investigation 6 with closed curves. What would happen? How many colors will you need? Carry out this investigation.

10. Question Suppose you have a map in a rectangle that can be colored properly in three colors. What else can you say about this map?

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

Investigations, Questions, Puzzles, and More

21

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

12. Question A 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 problem might shed some light on the ''real problem.'') What exactly would this analogous problem be? What would a three-dimensional map look like? What would the countries be? What would the rules for coloring such maps be? Do the following threedimensional configurations satisfy your definition of a three-dimensional map? How would they be colored? How many colors would they need? Is there anything you can say in general about coloring three-dimensional maps?

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Chapter 1 Acme Does Maps and Considers Coloring Them

13. Gathering Evidence Here are more maps to color properly and minimally.

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

●

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

●

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

●

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

15. Summarizing Joe knows that Boss needs to see things written down so Joe usually writes Boss a report of his activities. Help Joe out by writing a summary of what he accomplished this afternoon. Include goals, statements of problems, terminology, results, and arguments. Make sure it's well organized and convincing because Joe likes 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 making a picture of them ....He has got to paint them so you can tell them apart the minute you look at them, hain't he? Of course. Well, then, do you want him to go and paint both of them brown? Certainly you don't. He paints one of them blue, and then you can't make no mistake. It's just the same with maps. 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 Francis Guthrie (then a graduate student at the University of London) posed it to his teacher Augustus de Morgan (of de Morgan's Law fame), who in turn wrote to the Irish mathematician William Rowan Hamilton:

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Chapter 1 Acme Does Maps and Considers Coloring Them

“A student of mine asked me today to give him a reason for a fact which I did not 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 of common boundary line are differently colored—four colors may be wanted, but no more ...What do you say? And has it, if true, been noticed? My pupil says he guessed 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 rectangle, or an island—are countries, borders, and vertices. A country is the interior of a polygon or distorted polygon. The countries of a map do not overlap. A border of the map is an edge of one or more of the country polygons. Two countries may meet along one of these borders. A vertex of the map is where two or more borders meet. The map is the union of all these elements.

A major idea of this chapter is the equivalency of two maps, one a stretchedshrunk version of the other. This is an instance of ''rubber sheet'' geometry. Print your map on a thin sheet of rubber, and then distort the rubber sheet to get other equivalent maps. The distortion preserves the countries, borders, vertices, and their relationships to each other. However, the distortion does not preserve exact distances or angles. The investigation of geometric properties preserved under this notion of geometric equivalence was first proposed by Leibnitz (see Kline, p. 1163f). He called the study of such properties, geometria situs. The modern word 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 rectangle to 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 by mathematical 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

References

25

with vertices that all have even order can be colored with two colors. Neither provided 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.

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Acme Adds Tours

CHAPTER

2

Scene 1 A few days later at Acme Maps. Millie, the gum-chewing, middle-age company secretary, 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’ve got. Call you back. BOSS: What was that about?

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Chapter 2 Acme Adds Tours MILLIE: It’s a possible job, Boss, for Walking Tours, Ltd. They sell package 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 in Russia, 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 downtown Königsberg with the bridges. BOSS: Doubt it. MILLIE: And, second, could we design a walking tour of the city. He said he’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 how much he’d offer us? MILLIE: No, but he said something else. He wants the walking tour to include 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 with the bridges! JOE, MILLIE: Right, Boss! JOE: (Turns to Millie.) Well, this should be easy. Where should we start the tour?

Scene 1

MILLIE: That big church looks like a good place to start. It’s right smack in the 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 to the 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 all that scribbling?

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Chapter 2 Acme Adds Tours 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 and churches and things for the moment. Just do the bridges. That’s less confusing, calmer too. JOE: All right. Let’s start here: bridge, bridge. So then you have to cross this bridge twice, see?

I must be going out of my mind. A simple city tour. Boy, it’s just impossible! MILLIE: Impossible. Impossible? Let me see the map. Hmm. (She blows a bubble 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 what makes you so excited? MILLIE: Nobody can take a tour of Königsberg going over all the bridges unless 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 each bridge exactly once. Here, I’ll show you. Draw a plain picture of the river, the island, and bridges. Label the land masses A, B, C, and D. Like this.

Scene 1

JOE: What are the A, B, C, and D for? MILLIE: Makes things easy. You’ll see. Anyway, this is a lot less complicated than 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 important so make each hunk a dot. Then, instead of a bridge connecting two 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 each bridge.

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, tracing over each line exactly once and, without ever lifting your pencil from the paper, returning to the dot you started from is exactly the same as drawing–on the map of Königsberg–a tour crossing each bridge 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 and connecting 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.

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Chapter 2 Acme Adds Tours 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 an odd number. JOE: So? MILLIE: That’s the key. Suppose you’ve designed a tour that crosses each bridge 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 the trip 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 arrows pointing toward the dot or the number of arrows pointing away from 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 arrive on. Then there’s got to be a bridge you leave on. That means an arrow pointing away from the dot. So a “toward’’ arrow has to be paired with an “away from’’ arrow. MILLIE: What about the dot where you start and the bridge you first leave on, the first “away from’’ arrow? What “toward’’ arrow does that pair up with?

Scene 2

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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 the bridge 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 of bridges connected to each dot? JOE: Huh? Well, that depends….Oh, I see. Supposing we can take a nice tour, at any dot there are a certain number n of “toward” arrows and the same number n of “away from” arrows. The total number of 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 arrows attached 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 tour of…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 they have discovered. Boss does not appear happy.

Scene 2 Later that day. Millie and Joe are writing up what they discovered about tours. Joe writes.

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.

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Chapter 2 Acme Adds Tours

Millie writes.

A Nice Tour on a Dot-Line Diagram Ingredients 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 attached to every dot must be even. In other words, if the diagram has a dot such that the number of lines connected to it is odd, then no nice tour is possible.

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

Scene 3

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. Now we’re getting all these orders. MILLIE: We didn’t really mess up. We showed that a nice tour could never be designed in that case. Maybe customers like our honesty and our fine minds. JOE: Huh? (Boss walks in.) BOSS: That’s cool beans! We’re gettin’ all this work. After you guys failed to 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 we can’t design. BOSS: I been tellin’ ‘em to send us only situations with even numbers of bridges. Everybody knows that you mess up if there is a dot with an odd number of bridges.

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Chapter 2 Acme Adds Tours JOE: So far we’ve been lucky. What if one of those turns out to be impossible? Now that would really be embarrassing. BOSS: Waddaya mean “turn out to be impossible”? Never say can’t! Not here. Acme is a can-do eee-stab-lish-mint. We been able to do all right so far, haven’t we? Besides, you and Millie said if you could take a nice trip (start from one dot, traverse each connecting line exactly once, and return to starting dot), then there must be an even number of connecting lines attached to each dot. JOE: Yes, yes. But I think you’re concluding that if every dot has an even number 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 and cities shown on the map below. As part of setting up a schedule, Arizona Air has hired 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 all the routes in one day, and get back to Tucson at night. Clearly he doesn’t want to double up on any route; he wants to fly each route exactly once. Piper has engaged Acme, the local expert on taking efficient (i.e., nice) trips, to figure out such a flight plan. Can Acme really do the job? Help Joe and Millie out.

Scene 4

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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 a little clearer. Ya know, these dot-and-line diagrams look a lot like maps. Why don’t we use that terminology? (Millie shows Joe what she has written.)

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 the first 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 vertex and containing each edge exactly once MILLIE: Here are examples of networks with labels. (She draws.)

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

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Chapter 2 Acme Adds Tours

Nice Tours on a Network Theorem: 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.

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

MILLIE: Boss thinks that just because we’ve proved the theorem, the conjecture must also be true. I suppose he thinks that because every square is a rectangle, then every rectangle must also be a square? (Phone rings. Millie answers it. Boss walks in. Joe has become engrossed in a problem.) MILLIE: (Hangs up and turns to Boss) Boss, I just talked with a lady on the phone. She said something about inspecting doors. She had faxed us these floor plans.

Scene 4

39

BOSS: We don’t do doors! We do maps! MILLIE: Well, I told her that. But she heard about our work with tours and thought we could help her. BOSS: Huh? What do doors have to do with tours? MILLIE: The woman inspects doors. Checks hinges, locks, whether or not the doors stick or slam too fast—that sort of thing. Sometimes when she does a door inspection, she winds up passing through the same door more than once. She wants to avoid that so she wants us to design a tour of the building that goes through every door exactly 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 turn to the floor plans.)

4. Your Turn. Help Millie and Joe with the problem of designing a tour for the door inspector for each of the buildings whose floor plans are shown above. MILLIE: This Königsberg thing has really gotten outta hand. Door inspectors, your conjecture, lotta work. Let’s get to it! Lights fade. Music comes up.

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Chapter 2 Acme Adds Tours

Investigations, Questions, Puzzles, and More 1. Investigation Joe’s conjecture is that if every vertex in a connected network is even, then there is a nice trip starting at a vertex and returning to the same vertex. Part of the investigation has already taken place: Joe has taken nice trips on a lot of networks; he has a lot of examples for which he knows it can be done. Boss is convinced that it can always be done, but Joe wants to be sure. What Joe wants is a sure-fire, guaranteed method for taking a nice tour on any network of the prescribed kind. This would prove his conjecture. Right now, Acme’s method seems pretty hit or miss. Of course, it’s possible that something really nasty could happen: the presence of a counterexample to the conjecture—a connected network having every vertex even for which there is no nice tour. Of course, then there would be no sure-fire, guaranteed method. So what is 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. Investigation Considering that the door inspector may return to Acme for more help with tour design (and perhaps other door inspectors may come by), Millie would like some general rules to help her. Give her all the advice you can! (If, as Boss suggests, it’s really like the Königsberg bridge problem, it might be helpful if you could transform the problem into a nice tour on a network. If you do that, you have to describe the network pretty explicitly.) Write a report to Millie on your findings.

3. Investigation Joe, Millie, and Boss are convinced that no nice tour of the bridges of Königsberg exists. So they’re thinking of what they might come up with as an alternative. Millie suggests they design a tour with a drop-off location and a pick-up location. A van would drop off the tourists and pick them up later after they have crossed each bridge exactly once. She calls this an OK tour. Is an OK tour possible in Königsberg and what would 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 tour situations. Help her to investigate the possibility of an OK tour for any connected network. Of course, you will need to write a report to Boss about your investigation.

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

Investigations, Questions, Puzzles, and More

41

number of times he or she shakes hands. After people shake hands a bit, Boss asks how 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 explain why he got what he got.

5. Investigation Word has gotten to the postal service in Tucson about Acme’s expertise with designing tours. The service wants Acme to design a tour for the mail delivery person for the downtown area included in the map below. The delivery person delivers mail to buildings on both sides of all the streets in the map. The path of the tour should start at the corner of Franklin and Main, traverse each block exactly 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 on any connected network. (Describe the new kind of tour and the possibilities for finding one.)

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

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Chapter 2 Acme Adds Tours

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

8. Investigation Southwestern Amalgamated Circuits has the following problem. It wants to design 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 made from 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 network design” should be able to help them with this, Southwestern Amalgamated has engaged Acme to carry out the basic research for this problem. As a member of Acme’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 avoid weakening the structure of the board, no holes can be punched in it.)

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

Investigations, Questions, Puzzles, and More

43

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

10. Investigation The National Association of Museums (NAM) wants to help its member museums design tours for its visitors. One issue that has arisen is the possibility of designing a tour through a museum that begins at the museum entrance, passes through each room exactly once, and ends at the museum exit. NAM has contracted with Acme to resolve this issue. You, an Acme employee, have been given the task of working on this problem for rectangular museums with the following floor plans. (Each plan shows the entrance, exit, rooms, and doors between rooms.)

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

11. Puzzle At the open house mentioned in investigation 3, Millie set up a puzzle. On a table she lays out all the pieces of a standard domino set, 28 of them from double zero to double six—all possible pairs from 0 through 6.

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Chapter 2 Acme Adds Tours

The puzzle has two tasks: ●

Arrange the dominos in a straight line so that the dots of adjacent pieces match.

●

Arrange the dominos in a closed circle so that the dots of adjacent pieces match.

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

12. Puzzle Pinkley Smith, Acme’s legal advisor, was watching the other guests play the puzzle Millie set out in puzzle 11. He began to think of variations: “Suppose I limit myself to just those dominos with 0, 1, 2, 3, or 4 dots on them. There are 15 such dominos. 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 set and call the original a double-6 set. Then there’s a double-3 set and, hey! a double-n set! Can I solve the tasks for them all, just some, or none? Which, I wonder?’’ Help Pinkley!

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

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

Notes

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15. Connecting It is a stroke of genius that Millie noticed that the dot-line diagram looked a lot like a map and decided to adopt the terminology they had used for maps. Not that anything will come of it, but it sets the stage for making connections between the two 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 that might get you started in exploiting the connection. Does working on map coloring give you any insight on tour design, and vice versa? In addition to the overall visual similarity between networks and maps, are there other similar features that have arisen in the two problem areas? If Joe were able to prove his conjecture, would that help him solve any of the map coloring problems? Conversely, is there a map coloring problem that, were you able to solve it, would it help you with tour design? If the 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 time to 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 study of graphs is called graph theory. A graph that can be drawn on the plane so that the joining lines intersect only at vertices is called a planar graph. In this chapter the term 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, he came up with a general method for other problems of the same type. In the language of graph theory, here are the theorems he proved: Theorem 1. If a tour on a connected graph returning to one’s starting position is possible traversing every edge exactly once (such a tour is now called an Eulerian circuit), then every vertex is even. Theorem 2. If a tour on a connected graph is possible without returning to one’s starting position and traversing every edge exactly once (such a tour is now called an 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 begins the paper: “The branch of geometry that deals with magnitudes has been zealously studied throughout the past, but there is another branch that has been almost unknown up

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Chapter 2 Acme Adds Tours to now; Liebnitz spoke of it first, calling it the “geometry of position’’ (geometrica situs). This branch of geometry deals with relations dependent on position alone, and investigates the properties of position; it does not take magnitudes into consideration nor does it involve calculation with quantities. But as yet no satisfactory definition has been given of the problems that belong to this geometry of position or of the method to be used in solving them. Recently there was announced a problem that, while it certainly seemed to belong to geometry was nevertheless so designed that it did not call for the determination of a magnitude, nor could it be solved by quantitative calculation; consequently I did not hesitate to assign it to the geometry of position, especially since the solution required only the consideration 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 the Greek topos, place, + logos, word). In his paper, Euler replaced the map of the city by a simple picture similar to the one Millie and Joe created. Below is a picture of another rivers-bridges-land masses situation 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 a paper 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 of odd type….’’

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

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47

Several of the problems from this chapter have been around for a while. In 1849 Terquem proposed the domino problem (Puzzle 11) and related it to the work of Euler (see Biggs et al., p. 16). Investigation 8 first appeared in a book of puzzles by Dudeney in 1917 (see References). The problem is usually referred to as The Utilities 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 complete graph on n points and is denoted as Kn. The ideas in this chapter can be used to solve many old problems, such as how to 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.)

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Acme Collects Data from Maps

CHAPTER

3

Scene 1 Early that summer. Acme has hired Brandon, its first summer intern. Brandon is a math major at the state university. Boss and Joe are discussing what Brandon should do. JOE: Boss, you hired this kid from the university as a summer intern, but our customers are taking their vacations and things are slow. What’re we 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 could gather from ‘em? Think of somethin’.

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Chapter 3 Acme Collects Data from Maps JOE: OK, Boss. (Joe mutters to himself.) What’ll we do with a lot of data? 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 something for you to do. See that file cabinet? (Joe points.) That’s where we keep our maps. I’d like you to go through the maps and get me some 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 and have a look. (Goes to file cabinet, opens it, and pulls out map.) Every map has three ingredients at least: countries, borders between countries—we call those edges, places where three or more 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 be yours. Brandon sits down and gets to work. He takes out a piece of paper and constructs the following table.

Data on Maps Item

V

E

C

1. Your Turn Here 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 of countries (C). Enter the data in the table.

Scene 2

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 cabinet, I decided to draw some diagrams to see if I could draw some conclusions from the data.

JOE: Conclusions? Waddya mean? BRANDON: Well, I was looking at the table of data and wondering if some of those numbers were related. JOE: Related? BRANDON: Like suppose you knew that a map had 20 countries and 40 vertices. 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, like some simple ones or maybe even made-up ones. JOE: Wait a minute, Acme doesn’t deal in made-up maps. This is the REAL world! BRANDON: Hold on. Let me explain. If there’s some sort of relationship— maybe a formula—between those numbers for any map, then it should work for made-up maps, too. Look, here’s what I mean.

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Chapter 3 Acme Collects Data from Maps Draw an imaginary continent or an island. Acme’s maps are really 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 got the V, E, and C data like a real map! (Brandon writes the data for La-La Land in his table.)

Data on Maps Item

V

E

C

La-La Land

20

29

10

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 countries! Let’s call it something exotic! BRANDON: How about “20-29-10’’?

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JOE: That would really make me want to take a vacation there. BRANDON: OK. Now you understand what I mean. With those diagrams I was trying to come up with a whole bunch of maps at once. Something that would give us a lot of data—in fact, an infinite amount—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). For every pair n and m of whole numbers, there’s a map and data that goes with it. We can add all that data to my table.

Data on Maps Item La-La Land n × m rectangle

V 20 (n + 1)(m + 1)

E 29 n(m + 1) + m(n + 1)

C 10 nm

For the second diagram, you can vary the number of spokes and get another infinite family; for the third, change the sides of the large equilateral triangle and get yet another infinite family. So, two more infinite sets of data. JOE: Infinite families? Boy, you must really be a math major.

2. Your Turn Help Joe and Brandon. Figure out what V, C, and E are for the two additional infinite families of maps. Put the corresponding data you get into your Data on Maps table. Look for patterns; see if you can find a relationship among V, E, and C for all maps considered so far—by you, Brandon, and Joe.

Scene 3 An hour later. Brandon is working alone. Millie walks in.

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Chapter 3 Acme Collects Data from Maps MILLIE: Hi, ya! You must be Brandon! We haven’t met. I’m Millie. I heard you found out some weird stuff about maps. BRANDON: I’m pleased to meet you, Millie. Yes, I think I’ve found something really interesting. If V is the number of vertices, E the number of edges (borders), and C the number of countries on a continent or island map, then it looks like you’ll always have V − E + C = 1. I’ve been trying to come up with an explanation for it. MILLIE: Explanation? All ya have to do is check all the maps in the cabinet. You’ve done that, haven’t ya? BRANDON: I’ve checked a lot of them. It’s true that the formula works for all the 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 all maps, including any maps that haven’t been drawn or even thought of yet. Of course, I might not be able to do it. In fact, there might be some yet-unthought-of map for which the formula doesn’t work. But, if I can show it’s true for all maps, then the formula will express some basic and immutable principle 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 all our maps are maps on an island, OK? The map is made up of countries, borders (edges), and vertices. I’ve been looking at some really simple maps, and I think we have to be careful about what we’re going to allow in order to draw a useful conclusion. 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 formula doesn’t work?

Scene 3 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’ve got 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 to pin down when it’s true. Maybe it’s true for maps where every country 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 it wouldn’t hurt if we add a vertex to every country that doesn’t have one. The map would be pretty much the same. Then we could say that for every map on an island where every country has at least one 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 with another problem:

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

All of these are distorted versions of a polygon or a disc. MILLIE: Distorted?

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Chapter 3 Acme Collects Data from Maps BRANDON: Yeah. Imagine a disc made of rubber that you can stretch to make into any one of these countries.

. MILLIE: Then you could just prick a hole with a pin in one of those and stretch 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 country has a vertex and is a distortion of a disc. Here’s my idea. You’ve got this complicated OK map on an island. Maybe it’s got 5283 countries so you sure don’t want to count V, E, and C. Then you’ve got this one-country map on an island.

We know V − E + C = 1 for the one-country map. Is there a way we could change the 5283-country map little by little, step by step, into the one-country map? I don’t mean really do it because then we might just as well have counted V, E, and C in the 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.

Scene 3

We’d want V − E + C = V' − E' + C'. I guess at each step we’d want to change what we have into another OK map. Do you have 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, will things always work out so well? MILLIE: Why not? BRANDON: OK. Look at your map. What if we had decided to remove a different country from the island’s edge. Any problems? MILLIE: Uh, oh. I think I see one. Lookit that country:

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Chapter 3 Acme Collects Data from Maps 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 think we’d want V − E + C = 2 for two islands. MILLIE: Wait. I know what. The “bad’’ country in my map has two of its edges on the island’s shoreline.

We’ll remove a country when just one of its edges lies on the island’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 we made 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 remove one 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

Scene 3 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 sure that the vertex you remove isn’t the last vertex on the country’s boundary. MILLIE: I think we’ve got it! Start with the original. Remove a country that shares just one edge with the island’s shoreline. Then remove vertices of order two. So far no change in V − E + C. Then remove another country that borders the “new’’ island with just one edge. Keep doing this until you get an island with just 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 = 1 for the original!

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

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Chapter 3 Acme Collects Data from Maps

Map Data Theorem (almost) A map on an island is OK if every country has at least one vertex and if every country 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.

3. Your Turn Help Brandon solve the problem. Millie claims that, given an OK map on an island, you can (possibly after removing some vertices of order two) always find a country that shares exactly one edge with the “shoreline’’ of the island. Find an argument 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 for maps. Say, what’s that gizmo you’ve got? BRANDON: (Looks at Joe.) Thanks. (Turns to object on desk.) This “gizmo’’ is a calendar. My uncle gave it to me. It’s got 12 faces, one for each month.

It’s neat and I look at it a lot. I’d been thinking about maps and the formula we found and I thought, “The calendar is like a map on the surface of the earth. The month-faces are the countries. The borders are the edges of the faces, and so on.’’ That make sense 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 faces off and then stretch and flatten out the rest. JOE: C-c-cut a face off? F-f-flatten out the rest?

Scene 4

61

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

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

JOE: Gosh darn! I see what you mean. There’s the map. Except for the face you cut off, the faces of the cube are now the countries of the map. The edges of the cube are the edges. The sharp corners of the cube have become the vertices of the map. What’s your calendar look like when you flatten it out? BRANDON: That’s more complicated, but I’ve got it right here.

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

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JOE: What are you going to do with these flattened-out maps? BRANDON: Well, the formula Millie and I found has consequences for these shapes 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 an island 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 the shape has F faces, then the flattened-out part has F − 1 faces. So it 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 Shapes If a three-dimensional shape has V vertices, E edges, and F faces, then V−E+F=2

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JOE: Let me do it for your calendar gizmo. Because there are 12 months of the year, there are 12 faces. You can also see them: 1 on 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. That makes 6 more.

And that’s all: the top plus the top belt meshes nicely together with the bottom plus bottom belt….Neat. Now to count edges….Hmm, it’s complicated to keep track of everything because 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 a pentagon. Multiply 5 by 12 to give you the total number of edges; except every edge would be counted twice: once for each of the two faces that meet along the edge. So E = (12)(5)/2 = 30. Count vertices the same way. Five is the number of vertices of a pentagon. Multiply 5 by 12 to give you the total number of vertices; except that every vertex would be counted three times: once for each 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:

Data for Gizmo F = 12 E = 30 V = 20 V − E + F = 20 − 30 + 12 = 2

5. Your Turn Check out the formula for the shapes in Exercise 4. Describe how you count the elements.

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6. Your Turn A certain mystery three-dimensional shape has the following properties: all its faces 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 threedimensional shapes we’ve been looking at correspond to maps on the surface of the sphere. Take the surface of a cube, and imagine it’s made of rubber and that you can inflate it like a balloon. You’ll get a sphere. The imprint of the cube’s faces, edges, and vertices forms 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-dimensional shapes, you can remove a face and flatten it out to make a map on an island. You can do the same thing with the corresponding spherical shape: remove a “face,’’ flatten, and get a map on

Scene 5

65

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 for the 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 the sphere that isn’t the island (the sea?) into a country.

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

Formula for Maps on a Sphere If V is the number of vertices, E the number of edges, and C the number of countries of an OK map on a sphere, then V − E + C = 2.

7. Your Turn The calendar gizmo corresponds to a map on the sphere such that every country has five edges and every vertex is of order three. Something similar is true of the cube: the corresponding map on the sphere is such that every country has four edges and every vertex is of order three. Investigate maps on the sphere of the following kind: every country is a hexagon (the number of edges is six), and every vertex is of order 3 or more.

Scene 5 The next day. Brandon and Joe are working at their desks. Millie walks up to Brandon’s desk.

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Chapter 3 Acme Collects Data from Maps MILLIE: Still thinkin’ about our formula? BRANDON: Yeah. When I was hired, Boss told me about a problem you guys were working on. You were supposed to design a network on a circuit board starting with five terminals and connecting each of the five terminals once to each of the other four, without letting the 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 if you had enough smarts. JOE: (Overhears the conversation.) I don’t suppose anyone has heard of a nice tour of the bridges of Königsberg. BRANDON: Huh? OK, suppose you had a solution to the problem on the sphere. MILLIE: I feel this abstract rash coming on— BRANDON: Just concentrate. You have each of the five terminals connected to the other four without any of the connections crossing. The whole network would form a map on the sphere; the part outside all the connections would be a country. The terminals would be the vertices: V = 5. The connections would be the edges: E = 5 × 4/2 = 10 because each of 5 terminals is connected to each of the 4 other terminals, giving you 20 edges; but each edge would be counted twice, once for each endpoint. We don’t know much about the countries yet. Since V − E + C = 2, the number of countries, including the part outside all the connections, would be 2 − 5 + 10 = 7. Seven countries. Let’s pick a country and see if we can say anything about it, maybe the number of edges it can have. The boundary of the country would be formed

Investigations, Questions, Puzzles, and More

67

by a trip along edges from one terminal back to itself. You can see that no country could have exactly two edges because we aren’t connecting a terminal to itself. So every country must have at least three edges. Let Ck denote the number of countries with exactly k edges. Then C = C3 + C4 + C5 + … We can use this to count edges: 3C3 + 4C4 + 5C5 + … = 2E Since 3C = 3C3 + 3C4 + 3C5 + …, we have that 3C < 2E. Plugging the 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 be solved led us into this horrible mess! It must be that the problem can’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.

Investigations, Questions, Puzzles, and More 1. Investigation Brandon wonders if his technique for resolving the five terminals problem might be helpful in investigating the utilities problem, in which each of three houses 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 your results.

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

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3. Investigation Each of the maps on the sphere corresponding to the cube and Brandon’s calendar 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 of this to Boss. Suggest to him how you might advertise a trip to one of these locations.

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

5. Gathering Data Count 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?

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

7. Question Suppose 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 color any map on an island?

8. Question Suppose your world is a cylinder—the outside surface of a tin can with both ends removed.

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

9. Question Actually 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 an island? 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. Question One way to classify OK maps on a sphere would be by number of countries. For which whole numbers n is there a map with C = n?

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11. Investigation Consider 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 that V − 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 that V = 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 otherwise significantly different (i.e., you can’t transform the network of one into the network of the other by shrinking or stretching).

12. Investigation Acme wants to maintain a catalog of different types of maps. One type of map is one 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. Investigation Another type of map Acme is interested in for its catalog is one on a sphere having eight or fewer countries, each of which is a triangle, i.e., a three-edged country. Investigate all maps of this type.

14. Summarizing This time the job of summarizing Acme’s investigations has been given to Brandon. Help him out by writing a summary to Boss of what was accomplished these last few days. Be sure to include goals, statements of problems, new terminology introduced, results, and arguments. Brandon would like good letters of recommendation to come out of his internship with Acme, so be sure to make the report 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 be found in many of the references listed below. The formula relating vertices,

References

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edges, and countries of a map is called Euler’s formula and was originally a formula relating the edges, vertices, and faces of a polyhedron. Theorem. If V, E, F denote the number of vertices, edges, and faces (respectively) of a 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 the formula appears to be due to Adrian-Marie Legendre (1752–1833). Descartes (1596–1650) studied polyhedra and obtained an expression for the sum of the angles of all the faces of a polyhedron. An easy consequence of this expression is Euler’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 Notes section in Chapter 5). The search for a description of maps on the sphere for which Euler’s formula is true deserves a bit of discussion. As we have seen, there are some very simple maps that do not satisfy Euler’s formula—a map consisting of two countries, each a disc, is a good example: V = 0, E = 2, C = 2, V − E+ C ≠ 2. Thus, if we want an Euler’s formula theorem for maps, we need to specify those maps for which the theorem applies. Then, if we want to use the theorem, we need to make sure that the map satisfies the conditions of the theorem. So much for restrictions. Allowing vertices of order two in a map is another matter. This seems artificial because in the real world these maps have no use! However, this was useful for us as it enabled us to prove the theorem, which is a small price to pay. If it seemed we were changing the rules as we went along, we were not. Instead, we were creating the definitions appropriately in order to get the mathematics to do what we wanted. Mathematics is a human endeavor, and we were exercising our humanity; watch for this in subsequent chapters. Incidentally, “OK map’’ is not standard terminology; 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. Biggs, N.L., Lloyd, E.K., and Wilson, R.J. Graph theory 1736–1936. Oxford: Oxford University Press, 1976. 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.

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Acme Collects More Data, Proves a Theorem, and Returns to Coloring Maps

CHAPTER

4

Scene 1 A week later. A couple of days ago, Acme hired Tiffany, another math major from the other state 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 his formula, 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 countries and counting the number of edges they have. I’m trying to see if there is some kind of pattern. MILLIE: Any luck? TIFFANY: I don’t know. I’ve found something curious. Look at my data:

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Map

Number of countries with this many edges:

Western Europe Lower 48 South America La-La Land

2 2 2

3 2 6 5 3

4 5 8 4 3

5

6

7

16 1 3

13 2 1

1

8 1

9 1 1

10

11

12

13

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 what an attraction that would be for our customers! “Take your vacation on an island where every country has more than six borders!” MILLIE: Why not for the moon and let every country have 17 or more borders! The formula Brandon and I found relates V, E, and F. May be you and I could find a formula that relates the number of countries of different sizes (i.e., different number of borders) with other 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 the utility problem. The big idea was to let Ck indicate the number of countries with k edges. Then C = C2 + C3 + C4 + … The next thing was 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 can count edges from vertices: 3V3 + 4V4 + 5V5 + … = 2E.

Scene 1

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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 can put 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 the inequality 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 we started with this. We can’t find a map where every country has more than six borders. We can’t do that.

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Chapter 4 Acme Collects More Data, Proves a Theorem TIFFANY: Why not? MILLIE: Look at your last inequality. If you’ve got a map where all the countries have seven or more edges, then everything on the left side of your inequality would be zero and everything on the right side 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 edges wouldn’t work, either. Tiffany writes:

Possible “Sizes” of Countries in a Map If 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 there must be at least one country whose number of edges is five or less.

1. Your Turn Suppose a map on the sphere (all vertices of order three or more) has no countries with exactly three borders and none with exactly four borders. What can you say 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 on maps. 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 vacation 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 you and Tiffany did gives me a way to color any map on the sphere with six or few colors. Guaranteed!

Scene 2

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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 had our suspicions, but never a guarantee. Let me show you how it works. I’m going to show you how to color, in six colors, any map on the sphere that has all its vertices of order three. But, before doing that, here’s how to use that method for any map on the sphere with vertices of order three or greater: Joe wrote.

Order-Three Reduction for Map Coloring ●

Replace every vertex of order greater than three by a little country. Every vertex 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 you added.

Voilà! You’ve colored the map in six colors! MILLIE: OK. But what’s your method for coloring the altered map? And what 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, this time every vertex has order exactly three. You and Tiffany showed that 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.

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Chapter 4 Acme Collects More Data, Proves a Theorem 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 have come 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 threeedged 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 keep track, somehow, of the order in which you have eliminated edges, vertices, and countries. Then you color the last map you get—the one with six countries—properly with six or fewer colors. You can do that, right? MILLIE: Well, I can. JOE: Next, you restore the countries, vertices, and edges you’ve eliminated one country at a time, in reverse order, coloring the restored countries 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 its neighbors. A color will be available. The last country has five or fewer edges so there are at most five countries surrounding it; at most five colors are used to color those neighbors so there is a color left over (from the six colors) to color the restored country. The map with the last country eliminated restored is colored properly!

Scene 2

Repeat this process: restore a removed country (along with an edge and 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 six colors, because at each stage you color the restored country a color different from its neighbors in the coloring of the previous stage map. Here’s a map I colored using this method.

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Chapter 4 Acme Collects More Data, Proves a Theorem MILLIE: That’s a real, rootin’, tootin’ explanation if ever I seen one! Millie leaves. Joe smiles, goes to his desk, and writes.

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

2. Your Turn Try 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 a map 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 for improving on your method. JOE: Lotta thinkin’ goin’ on….Wait. Improving? You mean a faster method? 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 the fact that, when you restore a country, it’s surrounded by five or fewer other already colored countries, so that there’s a sixth color left to color the restored country. JOE: Huh? TIFFANY: Let me show you.

Scene 3

81

JOE: So, what’s your point? TIFFANY: What if the five or fewer already colored countries only use up four colors? Like this.

JOE: That would be guaranteed if there were always a country with four or fewer edges, but that’s not always the case. Check out Brandon’s gizmo: 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 use up. A restored country with four or fewer edges has four or fewer countries bordering it and uses at most four colors, so the restored country can be colored a fifth color. The problem is when the country has exactly five edges. Look, here are the pictures you drew 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.

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Chapter 4 Acme Collects More Data, Proves a Theorem And, you have to make sure the end result is an OK map. For example, you wouldn’t want to do this either.

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

JOE: Neat! Removing two edges makes three bordering countries into part 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 country and edge, you leave the rest of the map colored as it is and color the restored country with a color not used by its neighbors. A color will be available for the restored country because the countries surrounding the restored one use four or fewer colors. The resulting map with the country restored is colored properly! Repeat this process and you’ll get a sequence of maps that are each colored properly with five colors, because at each stage you color the restored country a color different from its neighbors in the coloring of the previous stage map. Here’s how I colored your map using this method.

Investigations, Questions, Puzzles, and More

83

JOE: Well, if that isn’t a gas!! Joe walks off. Tiffany writes.

Coloring Maps on a Sphere, Part Deux Theorem. Every map on a sphere can be colored in five or fewer colors. Lights fade. Music rises.

Investigations, Questions, Puzzles, and More 1. Investigation Because Tiffany and Millie can’t find vacation hideouts where all the countries have a large number of borders (six or more), maybe they could take on a new

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

2. Investigation The folks at Acme’s rival, El Grande Maps, have been looking at OK maps on the sphere 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 they haven’t been able to explain why. Investigate their claim. You’ll want to draw a few such maps and color them. Also, you’ll want either to find a counterexample or to explain why their claim is true. Of course, you will want to write up your findings in the form of a report.

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

4. Investigation In 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 think that it would be good to have a list of all OK maps in which every vertex is of order three and every country has five or fewer edges. Help them out. Of course, you will want to write a report of your findings.

5. Investigation Consider 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. (Compare this classification with those of Investigations 12 and 13 in Chapter 3.) Two maps are of the same combinatorial type if they have the same network of edges and vertices; that is, you can get one from the other by shrinking or stretching. Here are four shapes, each corresponding to maps of the same combinatorial type.

Investigations, Questions, Puzzles, and More

85

Here are three shapes, each corresponding to maps with eight countries but of a 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 results of your investigation.

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

7. Question You have an OK map on the sphere such that ●

Every vertex is of order three.

●

The map is properly colored in three colors.

What else can you say about this map?

8. Question You want to color a bunch of non-overlapping circles drawn in the plane. Circles tangent to one another must be colored different colors. What can you say about the numbers of colors necessary to color all the circles?

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Chapter 4 Acme Collects More Data, Proves a Theorem

9. Puzzle Pinkley 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 that an OK map on the sphere of the following kind can be colored in four colors: ●

Every vertex is of order three.

●

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 his claim is true: ●

The trip he mentions forms a simple closed curve on the sphere. This cuts the sphere 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 will color the map properly?

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

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Be sure organize your summary report carefully; write clearly; mention goals, problems, results, and arguments; and tie in today’s activities with those carried out 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’s green” “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 Percy John Heawood (see Biggs et al., p. 105f). Tiffany’s proof can be neatened up by using mathematical induction on the number of countries of the map (see the proofs in Stein and Beck et al.). The idea of reducing the problem of coloring maps to that of coloring maps all of whose vertices are of order three is due to Cayley and Kempe in separate papers of 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 with five or fewer borders. In addition, he seemed to know that, a map, all of whose vertices 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. Biggs, N.L., Lloyd, E.K., and Wilson, R.J. Graph theory 1736–1936. Oxford: Oxford University Press, 1976. 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.

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Acme’s Solicitor Proves a Theorem: the Four-Color Conjecture

CHAPTER

5

Scene 1 A couple of days later. Tiffany, Millie, and Brandon are in the shop, working at their desks. 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 using four colors. Pinkley says he based it on our methods, Tiffany. BRANDON: Wow! Who’s Pinkley Smith? How did he find out what we’ve been doing? MILLIE: Pinkley Smith is the company’s lawyer. I saw him the other day and told him about our stuff. BRANDON: I thought we were going to keep quiet about it. We were thinking about publishing. MILLIE: Well, we shouldn’t keep our lawyer in the dark. Anyway, he’s a good cat. 89

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Chapter 5 Acme’s Solicitor Proves a Theorem JOE: Cat? (Pinkley Smith walks in. He is wearing a seersucker jacket 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 coloring 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 also heard about the immutable fact that every map (all vertices of order three or more) has to have a country with five or fewer borders. Just yesterday Millie filled me in on the latest. You have a method for coloring a map with five colors that will work every time. Well, I asked myself, why can’t I find a method with four colors that works every time? After all, none of us—yes, my friends, I consider myself one of you—none of us has been able to find a map that requires more than four colors. So, guess what, I found a method. MILLIE: Get on with it, Pinkley. PINKLEY SMITH: Hold on to your hats, friends. My method uses the following 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 a sequence of contiguous countries colored, sequentially, α—β—α—β—α—β. Like this. If the sequence begins with country X and ends with country Y, then I call it an a-b chain from country X to

Scene 1 country Y. I can reverse the coloring of the chain and get the 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 chain and 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 X is one of the countries of the map and it’s colored α or β. Then consider 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 starts where 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 are of order three. You know that there must be a country with five or fewer edges. You choose one of these and, by erasing edges, create a map (with vertices that are all of order three) with fewer countries. You keep doing this until the altered map has four or fewer countries, which you naturally color in four or fewer colors. Then you reverse your steps, put back erased edges, and color reinstated countries.

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Chapter 5 Acme’s Solicitor Proves a Theorem BRANDON: Sounds just like our method. How do you show that a fifth color won’t be needed? PINKLEY SMITH: I was just getting to that. As you say, so far everything’s the same as your method. Let me tell you what I do differently. First, when I am erasing edges and have chosen a country with exactly four edges, I erase two edges as follows.

Thus, if the altered map is colored properly with four or fewer colors, then—when you restore the erased edges— there is a fourth color to color the reinstated country. In your method, you only erased one edge. TIFFANY: How do you know you can color the altered map in four colors? MILLIE: I think Pink’s referrin’ back to our method when we have to restore a four-edged country. He’s assumin’ he’s been able to color with four up to then. BRANDON: I think Pinkley Smith has more to do before we can believe that. PINKLEY SMITH: Yes. I said there were two differences. I’ve just described one of them. The second change in your method occurs when I reinstate a five-edged country, which is denoted by X. We assume that the rest of the map—all the countries except X—is colored properly in four colors, including the countries that border X. If only three colors are used for these bordering countries, then a fourth color is left for X itself, and so we would have a proper coloring for the map with the reinstated country.

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

Scene 1

93

I want to show you how to reduce the number of colors used (by the countries bordering X) to three. Doing this will involve altering the colors of the remainder of the map. I’ll illustrate the method referring to the diagram above and using 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 this point, countries 1 and 4 are both colored yellow. Except for X, the entire map is colored properly in four colors. Furthermore, only yellow, blue, and green are used to color countries 1 through 5. Color country X red and 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 this point, countries 1 and 3 are both colored green. Except for X, the entire map is colored properly in four colors. Furthermore, only yellow, blue, and green are used to color countries 1 through 5. Color country X red and 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 an red-yellow chain from countries 1 to 4 and an red-green chain from countries 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 blueyellow chain from countries 2 to 5. Reversing colors in the sets {U: a blue-green

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Chapter 5 Acme’s Solicitor Proves a Theorem

chain exists from country 5 to country U} and {V: a blue-yellow chain exists from 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!

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.

Investigations, Questions, Puzzles, and More 1. Gathering Evidence Use Pinkley Smith’s method to recolor the map and color country X so that the entire map is colored properly in four colors.

Investigations, Questions, Puzzles, and More

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

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

Several players may play, each taking a turn coloring any region of his or her choice with any one of four colors in such a manner that adjacent regions be colored 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?)

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4. Game This 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 different colors. The first player, the Minimizer, selects a crayon and colors a region of the map. The second player, the Maximizer, then colors another region using one of the crayons. Players continue in this way, alternately coloring a region with any one of the n colors and always following the rules for the proper coloring of a map. The Minimizer tries to avoid the need for the use of an n +1 - st color to complete 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 colored with 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 you explain it? 2. Martin Gardner makes yet another claim: There is a map such that, if you play with five colors, the Maximizer can always win. Can you find such a map and give the Maximizer’s winning strategy? 3. Consider the smallest value of n such that when the game is played on any map the Minimizer can, if both players play optimally, always win. Brams conjectures that this smallest value of n is six. What do you think?

5. Investigation When 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.

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This gave Joe an idea: “Suppose I have a map of empires. Each empire would consist 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 the others as colonies. To color a map of empires, I would color all the regions of a 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 the investigation. 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?

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6. Investigation You 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. Give a simple argument that shows four colors will always be sufficient.

7. Looking Back Go back and have a look at what you know about maps on the sphere or on an island that can be colored with three colors. In particular, have a look at Investigations 2 and 3 of Chapter 1 and Investigation 2 of Chapter 4. Can Investigation 3 of Chapter 1 be generalized to the sphere? (Compare the generalizations of two-colorable maps on an island to a sphere.) Given a map, could you tell if it is one of these? Are there easier methods for coloring such maps than those methods suggested by your earlier investigations? (Compare different methods for coloring two-colorable maps.) Suppose you have an OK map in which every 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 in general about maps on the sphere that are colorable in three colors?

Notes [Scott Parris, chief of police, is trying to sell Charlie Moon, Southern Ute detective, 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 into his head had come a theorem which might be true or might be false, and his mind darted hither and thither seeking proofs to establish its truth and counter-examples to show that it could not possibly be true. –from Kandelman’s Krim, J. L. Synge

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In the notes to Chapter 1, we mentioned that in 1852 the English graduate student Francis Guthrie first posed the problem of coloring maps and thought they could all be colored in four colors but did not have a proof of the 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 first issue of The American Journal of Mathematics that contained the following remarks: Some inkling of the nature of the difficulty of the question, unless its weak point be discovered and attacked, may be derived from the fact that a very small alteration in one part of a map may render it necessary to recolor it throughout. After a somewhat arduous search, I have succeeded, suddenly, as might be expected, in hitting upon the weak point, which proved an easy one to attack. The result is, that the experience 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 some time. Even Felix Klein, of the university at Göttingen and one of the foremost mathematicians 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 high regard for the English mathematicians of the time.) In this chapter we put Kempe’s argument into the mouth of Pinkley Smith. Kempe’s paper appears in the book by Biggs et al. (p. 96f). Not much was done with map coloring until 1890, when P. J. Heawood published a paper that began: The Descriptive-Geometry Theorem that any map whatsoever can have its divisions properly distinguished by the use of but four colors, from its generality and intangibility, seems to have aroused a good deal of interest a few years ago when the rigorous proof of it appeared to be difficult if not impossible, though no case of failure could be found. The present article does not profess to give a proof of this original Theorem; in fact its aims are so far rather destructive than constructive, for it will be shown that there is a defect in the now apparently recognized proof [of Kempe’s]....

In this paper Headwood produced a counterexample to the technique that Kempe had used in his proof. This is the map of Gathering Evidence 2. Heawood’s paper also appears in Biggs et al. (p. 105f). So Heawood found that Kempe’s “proof ” had flaws. However, in the same paper, Heawood showed that Kempe’s method could be used to prove the fivecolor theorem. This was the proof we saw in Chapter 4. For many years after Heawood’s paper appeared, no one was able to patch up Kempe’s “proof ” of the four-color theorem nor was anyone able to produce a map on the sphere that needed five colors. The four-color problem became famous but remained unsolved until 1976 when Professors Kenneth Appel and Wolfgang Haken, from the University of Illinois, announced a proof. The Times (of London) reported the following on July 23, 1976:

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Their proof, published today, runs to 100 pages of summary, 100 pages of detail and a further 700 pages of back-up work. It took each of them about 40 hour research a week [for four years] and 1,000 hours of computer time. Their proof contains 10,000 diagrams, and the computer printout stands four feet high on the floor. ‘It was terribly tedious, with no intellectual stimulation,’ Appel said. ‘There is no simple elegant answer, and we had to make an absolutely horrendous case analysis of every possibility. I hope it will encourage mathematicians to realize that there are some problems still to be solved, where there is no simple God-given answer, and which can only be solved by this kind of detailed work. Some people might think they’re best left unsolved.’

Contrast this description of Appel and Haken’s proof with the following comments by René Thom, a leading 20th-century French geometer: Bertrand Russell has said that ’mathematics is the subject in which we never know what we are talking about nor whether what we are saying is true.’ Unfortunately, the purely formal view is difficult to uphold, paradoxically for formal reasons. We know the difficulties presented by the formalization of arithmetic associated with Gödel’s Theorem....For myself, I am content with the following illustration: Let us suppose that we have been able to construct for a formal theory S an electronic machine M capable of carrying out at a terrifying speed all the elementary steps in S. We wish to verify 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 validity of such a ‘proof,’ given the impossibility of verifying all its steps?” (See Thom’s article in The American Scientist.)

In an article that appeared in the September/October 1976 issue of Scientific American, Haken and Appel were reportedly “not troubled by the unwieldiness of their proof ”: Indeed, they believe it is intrinsic to problems such as the four-color theorem. Said Appel: ’While we are excited to solve a problem as well known as the four-color theorem, we think that the real significance of our work lies in the area of mathematical proof. ’ The Illinois mathematicians believe a concise, elegant proof of the four-color theorem does not exist. They think they have found the first example of a type of problem that cannot be solved by the usual mathematical techniques. ’We believe there must be many true statements in mathematics that can only be proved by such methods,’ said Haken. ‘We hope our work on this problem will encourage other mathematicians to investigate these techniques.’

Expositions of the ideas used by Appel and Haken to prove the four-color theorem can 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, then it seems reasonable to assume that there will be five countries that each border on all of the others, thus forcing the map to be colored in five colors. So suppose there are five such countries. Put a point inside each of these—call it the capital. Join two

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capitals with an edge if the two countries share common borders. (This is an example 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 of the connections crossing. But this is impossible! (See Investigation 9 of Chapter 2 and Scene 5 of Chapter 3.) Therefore no map needs five colors. The catch here is the phrase “it seems reasonable to assume that.” The assumption that needing five colors means five countries that each border on the other four is a kind of “local” assumption; that is, the difficulty with coloring a map with four colors can be localized to a few neighboring countries. The problem with this assumption is illustrated as follows. Consider map A in the figure below. Three countries are colored.

What color should be used for the fourth blank country? We must color it either red or a fourth color. Suppose we take the second alternative and color it yellow, as in map B. Then we add another region (as in map B). Now it is impossible to color the map without using a fifth color. If we return to map A and color the blank region red, we are in trouble if a situation such as map C occurs in which two regions border the first four. Fourth and fifth colors are needed for the two blank countries in 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 to Kempe’s proof and in which he proved the five-color theorem, Heawood also considered the empire maps described in Investigation 6, in which countries can have colonies. If you allow only one colony per country, then Heawood showed that 12 colors will suffice. He also produced an empire map that requires 12 colors (see Beck et al., p. 70f. and Gardner, p. 85f.).

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Computational Complexity. Given a map on the sphere, it is easy to decide whether or not it can be colored in two colors. All you do is determine the orders of all the vertices. This can be carried out in polynomial time, that is, the time it takes is 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, countries, and adjacencies. Some problems, however, cannot be carried out in polynomial time, for example, the problem of listing all the arrangements of seating n persons at a table. Every solution takes exponential time. Consider another problem very close to the two-color problem. Given a map on the sphere, decide whether or not it can be colored in three colors. It appears that this is not an easy problem; every known algorithm for solving it takes exponential time. It turns out that the problem is also computationally equivalent to every problem in a class of problems called NP complete. Two problems are computationally equivalent if one can be transformed to the other in a time that is polynomial in the number of elements. Many NP-complete problems have been identified that are very difficult and are probably exponential. For more information on NP-complete problems, see the article by Cipra and the books by Garey and Johnson and by Neapolitan and Naimipour.

References Appel, K., and Haken W. The solution of the four-color-map problem. Scientific American, October 1977, p. 108f. 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: the millennium edition. Natick, MA: A. K. Peters, 2000. Biggs, N.L., Lloyd, E.K., and Wilson, R.J. Graph theory 1736–1936. Oxford: Oxford University Press, 1975. 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.

Acme Adds Doughnuts to Its Repertoire

CHAPTER

6

Scene 1 A week or so later. Boss and Brandon are alone in the Acme shop working. Each is working at his desk. The phone rings. BOSS: Howdy. Acme Maps and Tours. We can give you the trip of your life!...Oh, yes. Amalgamated...Yes, I remember....But we really don’t....Hole?...OK. I’ll see what we can do. Adios. (Hangs up.) Those fellers are really barkin’ at a knot. Next thing ya know they’ll be asking us to do their laundry and having us make doughnuts. (Turns to Brandon.) You remember those problems Southwestern Amalgamated Circuits sent us awhile back, the ones 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. 103

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Chapter 6 Acme Adds Doughnuts To Its Repertoire BOSS: Huh? Well, they’re kickin’ up a row about this new circuit board they’ve got. One with a hole in it. I guess the old one didn’t have a hole. They asked if we could solve their problems using the new 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, and spreads it out on his desk:

Five-terminal problem Problem: 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

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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 the connections 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 the surface of a sphere if you allow distortions. Cutting a hole through the circuit board makes the surface more like the surface of...a washer or, hey, like that box-with-a-hole puzzle we had around a while ago!

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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 doughnuts from Krusty Krums. Want one? BRANDON: Yeah, sure. (Takes one. Looks at it.) Hey, doughnuts! That’s funny; I’m working on a doughnut problem! JOE: In your dreams? BRANDON: Remember Amalgamated’s circuit board problems? (Shows Joe and Tiffany the paper.) Now they want us to solve the problem on a doughnut. JOE: Here, eat this quick before you faint. Dreams is right. I think your brain needs a little food. BRANDON: No, listen. They have a new board. It has a hole in it; with a little distortion, 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

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Chapter 6 Acme Adds Doughnuts To Its Repertoire puzzle? That’s really the surface of a doughnut, and we got V − E + F = 0. We thought that was weird. But maybe something’s going on. I mean, maybe that’s what it should be for a map on a doughnut. JOE: Ooooh. Slow down! This is getting heavy. You need to get out of the shop more. Have another doughnut!

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

Five-Terminal Problem on a Doughnut Problem: 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 a doughnut. 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 if I can solve the problem on a doughnut, then there is no contradiction. I haven’t shown anything for sure, only that my old argument isn’t an impossibility argument. Even that evidence of progress depends on the assumption V − E + C = 0. The invisible emperor is wearing invisible clothes! But I think we’ve got some ideas to

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work on. Joe, why don’t you work on seeing if you can actually draw the connections for the five terminals on the surface of a doughnut. Tiff, let’s see if we can show that V − E + C = 0 for any map on the surface of a doughnut. Joe walks back to his desk and begins to work on the problem.

1. Your Turn Joe needs help. Place five points on the surface of a doughnut. See if you can draw a connection 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 some examples to get a feel for the problem. We’ve only actually counted V, E, and C for one case. But the box-with-a-hole gives me an idea for more maps, a whole family of them.

2. Your Turn Help Tiffany figure out V, E, and C for the infinite family of maps on a doughnut suggested 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 are made of cardboard so they aren’t solid. We can cut them open like this.

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Chapter 6 Acme Adds Doughnuts To Its Repertoire Hmm. I’d really like to flatten the whole thing out. Then maybe we 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. We must be doing something wrong. We got V − E + C = 0 when it was put 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 the rectangle. Those are all the same. BRANDON: Same? TIFFANY: They come from the same vertex on the original map. When we counted V on the original, that vertex only counted for one, but when we count V on the square, we count four. One has become four!

BRANDON: So we’d have to subtract three from the vertex count on the island 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 be duplicated on the island’s border. So we have to subtract those duplications in the island count to get the correct doughnut count. BRANDON: Whew. This is getting complicated. How do we keep track of everything? That one vertex on the box turns into four. Other edges and vertices turn into two.

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TIFFANY: Chill. We can handle it. Let’s be careful with our notation. Let’s let V, E, C denote the doughnut-map data and V', E', C' the islandmap data. First, it’s easy to see that C = C'. Right? That’s easy. No duplications there. Now, suppose there are k edges along the island’s boundary. Then there are also k vertices along the island’s boundary. 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 “duplicated” four times! TIFFANY: You’re right. In the count V' − k, there’s a “vertex” that’s been counted twice. We really should write V = V' − k − 1. The final list 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 for a 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 Turn The surface of a doughnut is a new world for the Acme team. It’s only a matter of time before the team has a look at related worlds. For example, consider the surface 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 a map on a doughnut with n lakes?

Scene 2 A little later. Brandon, Tiffany, and Joe are working at their desks. Suddenly, Joe jumps up and leaps around.

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Chapter 6 Acme Adds Doughnuts To Its Repertoire JOE: Eureka! I’ve got it!

BRANDON: OK, Archimedes, what have you got? JOE: I’ve solved the five-terminals problem on the doughnut! (Hops back 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’s really a sphere at this point. I’m going to add the hole later. Put five 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 around and under the doughnut and up through the hole:

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

4. Your Turn Brandon and Tiffany have a way of looking at the surface of a doughnut that Joe does not know about, namely, a square with opposite edges identified. Now that they 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. Here is a start. (In these diagrams, a single-arrow edge gets glued to the other singlearrow edge. The method is similar for the two double-arrow edges. Edges to be glued must have arrows pointing in the same direction.)

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Scene 3 A half hour later. Brandon and Tiffany are alone in the office writing up their solution to the five-terminal problem on the surface of a doughnut. Brandon is annoyed that Tiffany has just pointed out a possible glitch in their reasoning. BRANDON: Whaddaya mean we need to prove that V − E + C = 0 on a doughnut? I thought we did that! TIFFANY: Well, we explained why we were getting V − E + C = 0 for that particular kind of map. It might be that V − E + C = 0 is only true for some maps. We should see if we can show V − E + C = 0 for every map. BRANDON: V − E + C = 2 for every map on a sphere. We know V − E + C = 0 for some maps on the doughnut. Doesn’t it follow that any other map 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 we were able to cut the surface open along edges of the map. I don’t think we can always do that and get something nice, like a square with opposite sides identified. Here’s an example of what I mean.

BRANDON: That’s pretty wild.

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5. Your Turn Cut 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. Suppose that V, E, and C are its data. Look at a circle that goes around the doughnut’s “waist.”

The circle might go through a vertex or two, and it might coincide with 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 avoid the vertices and intersect edges transversely—at single points—so that a narrow “belt” surrounding the “circle” would look like this.

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

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So, as you travel along the entire circle, the circle creates k new vertices, 2k new edges, and k new countries. With all of this added 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 as an 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 the added 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 inner tube, the vertices and edges along the shore of the island would match 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 data for 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:

Theorem about Data from Maps on a Doughnut Theorem: If V, E, and C denote the number of vertices, edges, and countries respectively of an OK map on the surface of a doughnut, then V − E + C = 0.

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Chapter 6 Acme Adds Doughnuts To Its Repertoire

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

Scene 4 Millie eats the last Krusty Krum while listening to Joe, Brandon, and Tiffany recount the 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 her doughnut.) JOE: Yeah, we did. Cool, eh? Then, while I was doin’ that, Bran and Tiff were 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 is just like the surface of a doughnut. We thought it would be easier to think about doughnuts. BOSS: (Regaining composure, trying–not quite successfully–to cover up his outburst.) You solved their problem! The next thing is the six-terminals problem and then the seven-terminals problem! Go to it, gang! Whaddaya waitin’ fer? And keep away from those doughnuts, 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 each terminal to make a country, then label the terminal countries 1, 2, 3, 4, and 5.

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

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

So. Five countries, each country touching the other four. That’s a 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 countries, you can fill the rest in with countries any way you like. (Turns to Millie.) That’s amazing! Neat idea! JOE: And you know what? We could do the same thing if we solved the six-terminal problem: We could make a map on the doughnut that requires six colors! Maybe Boss is onto something. Let’s get to work! High fives around. Music rises. Lights fade.

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Investigations, Questions, Puzzles, and More 1. Investigation Now 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 Acme team investigate this.

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

3. Question Millie 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?” Help Millie answer her question.

4. Investigation The 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 the floor 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. Investigation Brandon remembers that any OK map on the sphere having every vertex of order three or more must have a country with five or fewer edges. Because that conclusion 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 there a restriction on country “sizes” for maps on the doughnut similar to that for maps on the sphere? Help Brandon investigate this issue.

6. Investigation The “polyhedron” block-with-the-square hole corresponds to a map on the surface of a doughnut.

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D F

A

E C

G

H B

I A

D

C

F

G

H B

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

7. Question Suppose a map on the surface of a doughnut can be colored in two colors. What else can you say about the map?

8. Question Suppose a network is drawn on the surface of a doughnut on which you can take a nice trip starting and returning to a single vertex. What else can you say about the network?

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

10. Question On the sphere a map made up of a bunch of closed curves, each intersecting with itself 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?

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11. Question Take two distinct points on a simple closed curve and join the two points with another 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 a doughnut?

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

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

13. Puzzle Here are two simple closed curves on an inner tube.

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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 a simple closed curve on an inner tube so that when you cut along it the result will be a knotted band? Could you find two simple, non-intersecting closed curves on an inner tube so that when you cut along them you get two bands interlocked like the links of a chain?

14. Something to Make Cut out the patterns below and glue them together to make the threedimensional object called C´saszár’s polyhedron. The following directions will help: 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.

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Glue A to A' by the flap 1 on A.

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Glue flap 2 on D' to the base of F.

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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!)

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●

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

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

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

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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?

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

16. Summarizing Write 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. Connecting Solving 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 investigate comparing two worlds: the surface of the sphere and the surface of the doughnut, especially with respect to networks and maps. A general line of investigation might be this: If a certain problem can be solved on the sphere, can it be solved 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 sphere that makes a problem solvable on one and not the other? A specific issue might be this: although every map on the sphere can be colored properly in four or fewer 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 answer such questions gives you ways of getting acquainted with the new world; it also provides new insight on the old one. Given what we all have done, make a list of some general lines of investigation and of questions to ask comparing sphere with 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 was based 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 four of the points, you will get something like the following.

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(It may not look exactly like this. But I claim that your connections will be a distortion of this. This claim is easy to see if you start with one connection and then add the others one at a time.) The connections divide the sphere into four regions. Each region is a “triangle” with vertices that are three of the 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 side of 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 the triangle in whose interior point 5 sits. This argument seems straightforward enough, but what lies behind it is the following theorem, known as the Jordan curve theorem. Theorem. A simple closed curve in the plane divides the plane into the union of three disjoint sets: the curve, the interior of the curve, and the outside of the curve. Moreover, any path joining a point in the interior with a point in the outside must intersect the curve. Two points in the interior set can be joined by a path lying entirely in the interior set, similarly for two points in the outside set.

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(A similar theorem for the surfaces of the sphere follows easily.) This theorem seems almost self-evident, but a remarkable thing about it is that a proof of it is not trivial. In fact, many false proofs were given by outstanding mathematicians before an acceptable proof appeared in 1913. The theorem is named after the French mathematician Camille Jordan (1838–1922), who stated the theorem and claimed to have proved it in 1887. One of the reasons for the difficulty of its proof lies in the meaning of simple closed curve and the fact that the proof must cover all possible simple closed curves. For example, it must include the following simple closed curve of infinite length.

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

1 2π

1 π

For a proof of the Jordan curve theorem for simple closed polygonal curves, see Boltyanskiı˘ 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 the conclusions of the theorem.

In a paper of 1811, the Swiss mathematician Simon-Antoine-Jean L’Huilier was the first person to note that the formula V − E + C = 2 is not true for a ring-shaped polyhedron, that is,“it has but a single surface, [but] there is an opening passing right through 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. The graphs suggested by Investigation 15 are called the complete bipartite graphs Kn,m (see Biggs et al., pp. 30 and 142).

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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. Boltyanski˘ı , 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.

Acme Considers the Möbius Strip

CHAPTER

7

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 has upbeat 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 kind of fix-it shop?” Then he stomped around the shop with smoke coming 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? 125

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Chapter 7 Acme Considers the Möbius Strip 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 one side getting used up when the other is good as new. Well, the replacement is twice as wide as it should be. The airport thought we could advise them what to do. They were thinking about cutting it in half down the middle, but they thought we should so some experiments before they mess things up. SANDRA: (Brings drinks.) There’s a special today, Pescado Ranchero (fish in 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 make some while we’re waitn’ fer the victuals. BRANDON: Groan. MILLIE: (Passes out paper strips and tape to everybody.) You’re going to tape 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!

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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 out scissors to everybody.) Cut along the dotted line!

Lights fade. Music rises.

1. Your Turn Make a Möbius strip and cut it as Millie suggests. What happens? Make sure you do this before Scene 2 begins.

Scene 2 A few minutes later. Lights up. Music down. Sandra brings the food. The table is covered with the results of Möbius strip cutting. Sandra has to find a place to put the plates. SANDRA: What a mess! You sure you wanna eat? ‘Cause we’ve got customers outside waitin’ ta get in. BRANDON: Yes, we want to eat! Pardon my friends; they always have to bring their 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! Half as wide, twice as long!

BRANDON: That’s not what I got. I got two interlocking loops!

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JOE: You weren’t paying attention, Brandon. You put two twists in the strip before you taped it together. BRANDON: No, I didn’t. I twisted it all the way around, once. Isn’t that what I was supposed to do? TIFFANY: OK, OK. Calm down everybody. Actually, it’s not supposed to be twisted all the way around. Maybe Millie should have called it a half-twist. Here.

THIS

NOT THIS

Hey, that gives me an idea! What would happen if we made strips with different numbers of twists—or rather half-twists— in them and then cut them down the middle? Maybe we’d learn something. BRANDON: (Lighting up for the first time.) I have another idea! We’ve been cutting them down the middle, in half. Why don’t we try cutting them 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 a table and keep track. Twisted Strips No. of Half-Twists 1 2 1 2 1 3 3 4

No. of Divisions 2 (halves) 2 3 (thirds) 3 4 (fourths) 2 3 2

Results 1 strip twice as long 2 strips, same size, linked ?

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TIFFANY: Let’s each take a different one and try it. I’ll start with three halftwists 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, get your sister, too. (Sandra starts cutting and joins them.) SANDRA: Marcela, come on out here. (Marcela joins them. The group cuts away. Sounds of “Wow!” and “Lookit that!” break the concentration. 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’s team is also oblivious to the outraged customers as the small café overflows with the chaos of paper loops and twisted strips. Lights fade. Music comes up.

2. Your Turn Carry out the investigation suggested by the table above. Fill in the blanks. Look for 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 such would 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 going on. You cut a Möbius strip in half down the middle and you get a single strip, twice as long. (Gestures in the air with invisible scissors and paper.)

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Chapter 7 Acme Considers the Möbius Strip 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 this one-sided bit. Millie? MILLIE: OK, everybody. Make a strip. (They all go their desks and tape a strip together.) Joe, you use my nail polish. Brandon and Tiffany, you get a felt-tip pens. Ready? Start somewhere and paint the strip, lengthwise, until you meet up with where you started.

3. Your Turn Make a Möbius strip and “paint” it as Millie suggests. What happens? Do this before continuing with the story! JOE: Wow. You’re right. You paint along the strip until you get to where you started. Without turning it over, you’ve painted the other side! What a blast! MILLIE: There ain’t no other side. It’s all one side. That’s why the airport uses it 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 have one 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 always forget 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 the island itself, and then there’s the edge of the lake. Two edges.

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MILLIE: So what about the one-edged Möbius strip? BRANDON: Just like you had us paint it, Millie. Take a pen. Start at a point on the edge and run the pen along the edge until you get back to where you started. (Everybody tries this.)

4. Your Turn Join the crowd. Make a Möbius strip and follow Brandon’s instructions. What happens? Do this before returning to the story. While the others are running their pens around the Möbius strip, Brandon makes another strip with two half-twists and uses the pen on it just as he did on the one halftwist strip.

5. Your Turn Make 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. You know, maybe we ought to check the number of edges of that double-long strip we got when we cut the Möbius strip down the middle. Maybe we’ll learn something. If it’s got more than one edge, it ain’t a Möbius. (Brandon 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. TIFFANY: I’m going to add a line to our table. (She writes.) Twisted Strips

No. of Half-Twists 0

No. Divisions 2

1 2 1

2 (halves) 2 3 (thirds)

Results 2 strips, same size, unlinked; each one like the original 1 strip twice as long 2 strips, same size, linked ?

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BRANDON: Good idea, Tiff. Let’s count edges for the items in the Results column and see what we get. I’ll take the second line—the single half-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 Turn Carry out the investigation begun by the Acme team. Look at the items described in the right column of your table. For each line, there is an original strip and strips resulting from cutting the original. Figure out how many edges the original strip has and how many edges each resulting strip possesses. Add this new information to the right column of your table. Look for patterns, and make predictions.

7. Your Turn The 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 are working at their desks. Millie is painting on eye shadow at the reception desk. Joe is staring 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.

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You glue the top to the bottom to form a cylinder. Then stretch to glue 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 other twisted strips. A dressmaker’s pattern for a Möbius strip is

When you glue, you have to make the arrows match so go in the same direction. That forces you to make the half-twist.

BRANDON: What about the top and bottom of the rectangle? There aren’t any 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 going up—then you get a dressmaker’s pattern for a plain, old cylinder!

TIFFANY: Totally. Well, I thought the dressmaker’s pattern might help explain what’s going on when we cut the Möbius strip down the middle.

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BRANDON: How? TIFFANY: Instead of assembling it from the pattern and then cutting it down the middle, you could cut the pattern first and then assemble. BRANDON: But the results might be different! TIFFANY: True. But we might learn something. Let me show you what I’ve found out. Start with a pattern. Cut it.

Next, glue. But how? We’ve got to be careful to put the two pieces together as they were supposed to be originally. I came up with this scheme.

The part labeled A on one side gets glued to the part labeled A on the other side. And B gets glued to B. (I’ve numbered the two pieces just to keep track of things.) And when you glue, the arrows have to be going in the same direction. OK.You cut and get these two strips.

Now we’re ready. Use the labels and arrows to glue them together. Wait. Can’t glue yet. Got to flip piece 2 over so the arrows match. Now we can glue!

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BRANDON: It’s twice as long as the original; it’s half as wide! And, it’s a pattern for a cylinder! I told ya, that’s not what we got when we did things in the right order. TIFFANY: I thought the same thing. But what’s a pattern for a two-half-twist strip? 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 for a cylinder. All strips with an even number of half-twists have the same pattern! TIFFANY: Exactly. Cutting the pattern and then assembling doesn’t tell you everything, but it tells you something. (She writes.)

Cutting a Möbius Strip in Half If you cut a Möbius strip in half down the middle, then the result is a strip half as wide and twice as long with an even number of half-twists. BRANDON: You know, the pattern for a Möbius strip is the same as the pattern for a three-half-twist strip and a five-half-twist strip; any strip with an odd number of half-twists. So your little trick tells you more. (Brandon writes.)

Cutting a Strip with an Odd Number of Half-Twists in Half Cut a strip with an odd number of half-twists down the middle. The result is a strip half as wide and twice as long with an even number of half-twists.

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Chapter 7 Acme Considers the Möbius Strip Tiff, you’re a genius! I think we can use this idea to explain a lot of what 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 idea to tell us definitively how many edges a twisted strip will have. The edges of the strip come from the top and bottom edges of the 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.

Scene 4

(Tiffany writes.)

Number of Edges of a Twisted Strip ●

A strip with an even number of half-twists has two edges.

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A strip with an odd number of half-twists has one edge.

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BRANDON: That’s baaad! Wait’ll Joe and Millie hear about all of this! We’ll need their help to check things out when we cut twisted strips in thirds, fourths, and... Lights dim. Music comes up. The two approach Joe and Millie. Curtain falls while the four are engaged in animated conversation. Brandon and Tiffany gesture wildly while Joe and Millie alternate between eyes scrunched up and eyes open wide as they move back and forth between incomprehension and understanding.

Investigations, Questions, Puzzles, and More 1. Gathering Evidence Brandon and Tiffany have come up with ideas suggesting similarities—at least as far as cutting goes—among all strips with an even number of twists and among all strips with an odd number of twists. These ideas are new, and the Acme team would like confirming evidence. Help them out by completing the following table. Use the evidence you already have from cutting twisted strips. Cutting a Twisted Strip: How Many Pieces Do You Get? No. Half-Twists in Original 0 1 2 3 4

No. Pieces Resulting When Cut In Halves Thirds fourths 2 3 4 1

What do you observe?

2. Investigation Brandon and Tiffany have found out what happens when they cut a strip with an odd number of half-twists in half. Use their new ideas to continue the investigation. 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.

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Cut a strip with an odd number of half-twists in thirds.

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Cut a strip with an even number of half-twists in thirds.

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Cut a strip with an odd number of half-twists in fourths.

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Cut a strip with an even number of half-twists in fourths.

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Cut a strip with an odd number of half-twists in fifths.

Investigations, Questions, Puzzles, and More ●

139

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 number of half-twists.

3. Generalizing Help the Acme team generalize the results of Investigation 2. Figure out what happens 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 number of half-twists.

4. Question The Acme team suspects that every strip with an even number of half-twists is two-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 Evidence Here are more surfaces. For each one, decide how many edges it has and whether it is one- or two-sided.

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6. Investigation To test Acme’s new methods for analyzing twisted strips using patterns, consider the following dressmaker’s patterns.

As they stand, the patterns are incomplete. Each can be made complete by placing arrows 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. Investigation Brandon likes Tiffany’s simple explanation for the fact that strips with an odd number of half-twists have a single edge and strips with an even number of halftwists 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 two edges that will become the edge (or edges) of the twisted strip.

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Then, without gluing, he puts twists in the strip and looks just at these two edges. Here is what he gets in the first few cases.

Then he glues the ends together. In the cases above, here is what happened to the original edges of the rectangle.

A D

ZERO TWISTS

B C

ONE TWISTS

A C

B D

TWO TWISTS

THREE TWISTS

Brandon called these edge diagrams. Based on these edge diagrams, Brandon wrote the following.

Edges of Twisted Strips ●

The edges of a strip with zero half-twists are two unlinked, unknotted closed curves.

●

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.

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Help Brandon out with this investigation by drawing the edge diagrams for strips with 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. Investigation Brandon, Tiffany, and you (see Investigations 2 and 3) were able to gain and explain a lot of information about the results of cutting twisted strips by ignoring the exact number of twists in the original strip, while retaining only that the strip had an even or an odd number of half-twists. This simplification enables you to count the resulting number of strips, determine the size of each, and say whether each has an even or odd number of half-twists. However, a lot more seems to be going on. Sometimes a strip may be knotted, and other strips may be linked in some way. Brandon thinks the edge diagrams defined in Investigation 7 can be used to provide information of this 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 in the 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, you could say that the result of cutting a strip with three half-twists down the middle is one knotted strip. Use this idea to complete the following table and, from the table, make predictions about what would happen with more than six half-twists.

Cutting Twisted Strips Down the Middle No. Half-Twists

Knotting/Linking of Resulting Strip/Strips

0 1 2 3 4 5 6

One knotted strip

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9. Investigation Investigate the results of cutting a strip with n half-twists in thirds. Use the technique of Investigation 8 to describe how the resulting strips are knotted and linked. Make a table similar to that in Investigation 8 to display your observations.

10. Things To Make Take 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 and B to B, to get a kind of double Möbius strip. Look carefully at the result. What can you 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 whole bunch simultaneously, and then tape. What happens? Does any of this have anything to do with cutting twisted strips?

11. Something To Make: Möbius Shorts Take a T-shaped piece of paper with a long stem. Bend the top of the T to make a ring (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 and B are glued.

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How many sides does the surface have? How many edges? Guess what happens if 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 put the result back together again (see Boas).

12. Game Looking at a pattern for a Möbius strip reminded Brandon of the game inner tube tic-tac-toe which is played on a pattern for an inner tube. He thought: Why not Möbius tic-tac-toe (MT3)? The board is a three-by-three square with one pair of opposite sides “glued” (arrows in opposite directions), and the rules are otherwise the same as ordinary tic-tac-toe. As with inner tube tic-tac-toe, the gluing of the pair of opposite sides changes the look of winning configurations. On the playing board below, what would be winning moves for X?

Find a friend and play a several MT3 games. Is there a way the first player can play to guarantee that he wins? (What about the second player?)

13. Investigation Imagine 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, problems that Acme considered for the surface of the sphere and torus: the utilities problem and the five cities problem. One way of visualizing these problems is to assume 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 that what you draw on one side shows up on the other.

14. Question A Möbius strip is a surface with one side and one edge. An island is a surface with two sides and two edges. Can you find a surface that has one side and two edges?

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15. Question The original question for the Acme team was the following: Can you cut the Möbius strip down the middle to get a Möbius strip half the width? You know the answer is no. However, is there a way to cut it so that the result is a Möbius strip half its original width? There might be many ways. What’s the best? You can throw some of the original material away, but you do not want to make any cuts that you will have to glue back together. Such cutting and regluing would weaken the resulting strip.

16. Something To Make Make a photocopy of the design below. Cut it out, fold along the dotted line with the design on the outside, and glue the two halves together on the inside to form a long strip of double thickness paper with the writing on the outside. Put a halftwist in the strip and tape the ends together. Pull the printed strip through your fingers to read the message. Enjoy!

ONCE UPON A TIME THERE WAS A STORY THAT BEGAN 17. Summarizing Brandon, Tiffany, Joe, and Millie have given Tiffany the job of producing a summary of the day’s investigations to be delivered to Boss. Help her out. Include statements of problems, terminology, results, explanations, and directions for future investigations. Of course, Boss will expect the summary to be clear, organized, and understandable. He would also like to know that their time was well spent.

Notes A mathematician confided That a Möbius strip is one-sided. You’ll get quite a laugh If you cut it in half, For it stays in one piece when divided. Image reprinted from Lost in the Funhouse by John Barth.

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In 1858 two mathematicians, Augustus Ferdinand Möbius (German, 1790–1868) and Johann Benedict Listing (Czech, 1808–1882), independently discovered the Möbius strip and the fact that it is a one-sided surface. We encountered Listing in Chapter 2 (see the Notes) in reference to taking nice trips on networks. We will encounter both again as our story unfolds. Both mathematicians were students of Gauss at Göttingen. In 1816, Möbius was appointed Extraordinary Professor of Astronomy in Leipzig, where he stayed for the rest of his 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 death the sons should divide the kingdom into five regions in such a way that the boundary of each one should have frontier line in common with each of the other four. Can the terms of the will be satisfied? Because of this occurrence, many mathematicians apparently and erroneously credited Möbius with the four-color conjecture 12 years before Guthrie proposed it. For more on Möbius and the German mathematical world of his time, see the book edited by Fauvel et al. (see also Kline, p. 1165). Möbius strips have entered our culture in many ways. They appear as objects of art, as in Maurits Escher’s woodcut, Möbius Band I, showing a Möbius strip cut down the middle (See the book by Escher, picture 41). Möbius bands occur in literature, as in the short stories “No-Sided Professor,” “A. Botts and the Moebius Strip,” and “A Subway Named Moebius” collected by Clifton Fadiman in his Fantasia 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

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147

Magicians have exploited properties of twisted strips in a cloth-tearing trick called “Afghan Bands” and described in Martin Gardner’s Mathematics, Magic and Mystery (p. 70f). A number of patents have been obtained for industrial devices whose underlying 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 to wear equally on “both” sides, an abrasive belt, a self-cleaning filter belt for a drycleaning machine (the dirt can be easily washed from both “sides” as the belt goes round), a conveyer of hot material, and a nonreactive resistor. For more about Möbius strips, see chapter 9 of Martin Gardner’s Mathematical Magic 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 chemistry and molecular biology. This is partly because a DNA molecule often resembles a strip 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 ways this can happen and determine how each possibility affects chemical and biological processes of DNA, such as recombination (see the book by Sumners, as well as Chapter 13 and Investigation 8 of the present chapter). Chemists have been studying Möbius ladders, molecules similar to the DNA molecules above. In a Möbius ladder, the surface of the strip is replaced by rungs of carbon-carbon double bonds and the edge is a polyether chain of carbon and oxygen 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.” Thus an island with lake would have two boundaries, one the shore line of the island itself and the other the shore line of the lake. The limerick Brandon quotes (“A burlesque dancer...”) is by Cyril Kornbluth and is 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.

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Acme Creates New Worlds: Klein Bottles and Other Surfaces

CHAPTER

8

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 make video games. They heard that Acme has been doing things with inner tubes and “Merr-bee-us’’ strips (whatever they are) that might turn into video games. They want us to come up with new stuff….Video games? (Boss rolls his eyes then looks around to room to see if anyone is paying attention. They all ignore Boss and work more intently on whatever it was they were working on.) OK, anyone wanta tell me what’s goin’ on? JOE: I’ll try, Boss. Didja ever play airplane videos?

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BOSS: Airplane videos? JOE: You shoot down enemy planes. The planes move in straight lines across the screen. When a plane reaches one side, it doesn’t just disappear 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 screen and 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 of the screen is the same as the corresponding point on the left edge of the 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 whole left edge, at least for the airplanes.

And the same is true of the top and bottom of the screen. The left edge is the right edge, and the top is the bottom.

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151

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 “world of the airplane’’ is a little spacey. JOE: The airplane really lives in the screen. Since the left edge of the screen is really the right edge, you can pick the edges up and glue them 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 its world is a cylinder. But the top is really the bottom, so you glue them together.

BOSS: No you don’t. You’ve gone too far. You’re distorting things. How can you do that? JOE: Boss, this is not the real world. Imagine the cylinder is made of something you can stretch and shrink, like rubber. You glue the ends 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 Play More wants. Don’t bother me. Right now, my head hurts. (Walks off holding his head. Once Boss leaves, the others come out of “hiding.’’)

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BRANDON: Poor Boss. Good job, Joe. Wow! Video games. What do you suppose 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. How would 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 like that. What if we combined Möbius world with inner tube world? BRANDON: Huh? JOE: Plane goes to right edge, appears on left edge flipped over. Plane goes to top, appears on bottom flipped over. Planes wouldn’t disappear. Here’s the scheme.

BRANDON: Wait. There’s another combination; it’s kinda mixed. Plane arrives at left edge, then appears on right flipped over. Plane arrives at top, appears at bottom just as in inner tube world. JOE: Cool.

Scene 1

153

MILLIE: I sorta like planes goin’ off the edge of the world. That’s where the 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 give us 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 inner tube. 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 you execute that twist? If you line up the two ends of the cylinder this way, then the arrows match.

But if you put the two ends together, just about to be glued, the arrows don’t match.

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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 through itself. MILLIE: Ouch. JOE: If you accept that, it’s really pretty simple. You start out with Millie’s first picture of the cylinder. Grab the top end and push it 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 with his eyes closed.)

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TIFFANY: Awesome! It looks like some kind of bottle. You know, I don’t think we have a choice. We have to do something strange, just like 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 to make a cylinder. But there was another choice. If we glue left to right 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 a sphere, without any boundaries and without some trick like Joe’s, we’d get something with an inside and an outside. A twosided thing. And that would be impossible: a thing one-sided and two-sided. No good. BRANDON: (Revives. Runs around the room holding his head. The others are stunned, not so much at Brandon, but at Tiffany’s statement.) Ahhhggghhh… Lights fade. Music comes up.

1. Your Turn If 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 Turn Before 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 a Möbius strip. Right now the team is looking at other patterns in hopes of coming up with new video games and of understanding their new “bottle.’’ Here are some of the patterns. Help them assemble them.

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Scene 2 An hour later. Calm has returned to the Acme shop. Tiffany, Millie, Joe, and Brandon—especially Brandon—are at work “assembling’’ other patterns. Brandon looks 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 original idea, the one where the planes flip over going vertically and horizontally.

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 the square out and stretched it into a hemisphere. The boundary of the square becomes the hemisphere’s equator. (I’m using letters and arrows to label sides.)

Now I’ve got to do some gluing. I want to put A and A together with arrows matching. First, fold up the hemisphere a bit, like a clam shell. Then pinch it together at points labeled X, the points at the ends of the A arrows, and the beginnings of the B arrow. Those two points are really supposed to be glued together eventually anyway.

Scene 2 X

A

B X

X

X B

A

X

B

X

B

A

A X

X B

X X

157 B X A

A

X

You can see how the arrows for the B’s are going in the same direction. Glue them together.

X

A

B

B

X

B B A

X B

X

A

A

A

A

MILLIE: Know what? You’ve got yerself a Möbius strip! The two A’s make up its edge. The edge is a circle in a plane. JOE: Oh-ma-gosh! That’s right. Nice goin’. The edge wasn’t flat before. BRANDON: OK. Next, we have to glue the two B’s together. The arrows are pointing in the same direction, but we have to do something along the lines of Joe’s trick: we have to make the surface pass through itself. Here it is. Ta da!

MILLIE: It still hurts. TIFFANY: That’s really weird. Looks like a beehive. Doesn’t even resemble the weird bottle. The team stares at the diagram in disbelief. They try different angles of viewing. They shrug their shoulders and scratch their heads. They go back to their desks. Music comes up. Lights fade.

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3. Your Turn The Acme team decided it might give them some insight if they didn’t limit themselves to a square screen. They thought looking at other polygonal screens—even a di-gon!—would be useful. Help the team assemble the following patterns. (The team has decided to label every edge, with the convention that only labels that occur twice get glued.)

Scene 3 A half hour later. Members of the team are back at their desks. This time it’s Joe who looks up with the big smile. JOE: This stuff gets curiouser and curiouser. Let me show you what I found. I was working on this triangular pattern.

Scene 3

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I remembered that when we were cutting up a Möbius strip we decided to see what would happen if we cut the strip first and then glued. We found out a lot when we did that. Well, I thought, “Why not try the same thing here?’’ I cut the triangle in half down the middle. I figured I’d eventually have to glue the new edges where I 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 of doing what Brandon did: push the insides back, stretch, and turn it into a hemisphere with the triangle’s boundary becoming the equator. A A

B

Split each A side up into two halves: first X and then Y.

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Now, except for B, the part that doesn’t get glued, you have just what Brandon had. BRANDON: Yeah, and since the two ends of B are identical, you can glue them together. 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 know what that means? Joe’s triangular pattern was really a patternin-disguise for a Möbius strip. It’s also a pattern for a beehivewith-lake. Sooo…in some sense, a beehive-with-lake is the same as a Möbius strip. BRANDON: In the same sense that a strip with three half-twists is the same as a strip with one half-twist? TIFFANY: Totally. MILLIE: Tiff, I tried to do the same thing you did with Joe’s triangle pattern, but this time I started with the bottle pattern. I wanted to see if I’d get something else. I cut it down the middle and added new labels and glued A to A. Here’s what happened.

Scene 3

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B

A

C

A

C

B

A A

B

C

B

A

C

B

C

B

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. Glue them together along the shorelines of their lakes. Whaddaya get? Our “bottle!’’ We may not know much about the bottle or about the beehive, but we do know they’re related. All the weird stuff lives in the same house. But, wait, we do know more—and this’ll knock your socks off—our bottle is two Möbius strips glued together, the edge of one glued to the edge of the other. Can you believe that? (He writes.)

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Strange Facts about Patterns and Surfaces Pattern

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. two beehives-with-lake glued together along the shorelines of their lakes; a.k.a. two Möbius bands glued together along their edges)

While Brandon writes, Joe starts to clap. Millie stamps her feet. The others pick it up 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.

Investigations, Questions, Puzzles, and More 1. Puzzle Make a Möbius strip, the simple one with one half-twist. Figure out how you would cut it out so that it would flatten out into the following pattern.

2. Puzzle The following hexagon is a pattern for some surface.

Investigations, Questions, Puzzles and More C

163

D B E

E A

D

C

We’re going to try to figure out what it is by cutting it up, labeling along the newly cut edges, and then gluing the pieces back together another way. Cut along the dotted lines A and B. Label the freshly cut lines with labels A, B, and arrows as shown. You’ll get three pieces. Glue the two D’s together, then the two E’s, and finally the two C’s. What do you get? What do you conclude?

3. Investigation Here is an interesting surface.

We’re going to cut it open, label cuts, flatten it out, stretch, and bend it to see if it has a pattern that is the same as some familiar object.

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A A

untwist

A

B

BB A A A A

untwist once BB A A

B A untwist again

A B

B

A

stretch it out A A

B

B

A

glue B back together

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 it can 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 at some other place.

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4. Investigation Consider the surfaces in Gathering Evidence 5 in Chapter 7. For each one, find a pattern, just as you did for the surfaces in Investigation 3 above: cut the pattern out, label cuts, flatten it, stretch, and bend it to get something you recognize.

5. Gathering Evidence: Möbius Shorts This is the surface you made for Something to Make 11 in Chapter 7. What is it?

6. Gathering Evidence Make a double strip in the form of a cross as shown below. The idea is to glue the ends of the cross as indicated. However, the arrows have been left out. Find all the ways to put arrows on the labels so that you get really different surfaces.

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Then make them. (Did you have any trouble putting them together?) Find names for these surfaces based on what you know.

7. Pattern Recognition Here are patterns for surfaces. Cut, label, tape, bend, and stretch each one to see if it is a pattern for a familiar object. C

C

A

C

A

A

C

C

A

E

B

D

F

E

A

A

B

A

8. Investigation Here are patterns for two surfaces. Describe each one as best you can. A

B

D B

A C

C

C

B A

B B

A

A

A

F

B

B

C

C

D

A B

C

C

B

A

C

B

A

B

B

C

D

B

A

B

D

C

E F

E D

F

C

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We’ve heard a rumor that these two patterns are patterns for the “same’’ surface. Is it true?

9. Something To Make Here is a pattern for something that can actually be put together! Here are the steps. ●

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.

A

B C

GLUE

D

GLUE

A

GLUE

GLUE

When you get done, look it over and see if you can find a familiar name for the object.

B

C

D

10. Something To Make Here 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.

●

Fold away from you along the solid lines.

●

Fold toward you along the broken lines.

●

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 position and glue it into place.

●

Glue sides flaps of the cube. (Do not glue the lid down; it is more interesting to be able to look inside.)

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GLUE

GLUE

GLUE

GLUE

When you get done, look it over and see if you can find a familiar name for the object. (This model is based on ones from the books by Barr and by Jenkins and Wild.)

GLUE

GLUE

GLUE

GLUE

GLUE

GLUE

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LID

GLUE CU T

GLUE CU T

CU T

GLUE

GLUE

11. A Game Looking at a pattern for a Klein bottle reminded Brandon of the games inner tube tic-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 identified (arrows in same direction from left to right, in opposite directions from top to bottom), and the rules are the same as those in ordinary tic-tac-toe. As with the other games of tic-tac-toe, gluing of pairs of opposite sides changes the look of

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winning configurations. On each of the playing boards below for KBT3, what would be winning moves for X?

O

O O

O X

X

X

X

X

O

X

O

Find a friend and play several KBT3 games. Is there a way the first player can play to guarantee that she wins? What about the second player? (This game appears in the book by Weeks.)

12. Investigation An airplane on the “one-sided’’ bottle video screen moves off the left edge of the screen and appears on the right, flipped over. It moves off the top edge of the screen and appears on the bottom, not flipped over. In both cases it is as if the airplane 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 possibility of solving problems there that Acme considered for the surface of the sphere and torus: the utilities problem and the five-cities problem. Use the video screen with identified edges as your model, much in the way you did with the torus. See Your Turn 4 in Chapter 6.

13. Experiment

Av. 10 th

st

1 Av.

2

nd

Av.

Imagine that you are a creature living “in’’ a Möbius strip. (For ways of thinking about this, look at Investigation 12 above and Investigation 13 in Chapter 7.) You are setting up street signs in your world. Streets on “the strip’’ consist of Main Street—going right down the middle—and cross streets (First, Second, Third, …, Tenth) perpendicular to Main Street.

MAIN ST.

Starting at First and Main and heading toward Second Street, you put up the sign West First on the left side and East First on the right side. You continue to Second and do the same thing: on the left side the sign is West Second, and on the

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right it is East Second. You keep going up the street, installing signs at each intersection. Eventually you pass the corner of Tenth and Main and arrive again at the corner 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. What do you conclude?

14. Question Here is a picture of the Klein bottle as “put together’’ by Joe. Draw a closed path on this Klein bottle so that if you cut along it you would get two Möbius strips.

15. Question Here is another way to “put together’’ the Klein bottle. It involves having the surface intersect itself in another way.

Draw a closed path on this version Klein bottle so that if you cut along it you would 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.

●

Fold away from you along the solid lines.

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●

Fold toward you along the broken lines.

●

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 careful maneuvering. Have patience!)

What familiar surface does this model represent? H' J'

G' F' A'

CUT

C'

B'

D'

E

CUT

CUT E'

D B

C

CUT

A F G

J H

Notes

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17. Summarizing This time Joe has the job of producing a summary of the day’s investigations to be delivered to Boss. Help him out. Include relevant diagrams and statements of problems, definitions, results, explanations, and directions for future investigations. Of course, Boss will expect the summary to be clear, organized, and understandable. He would also like to know that the team’s time was well-spent.

Notes Three jolly sailors from Blaydon-on-Tyne They went to sea in a bottle by Klein. Since the sea was entirely inside the hull The scenery seen was exceedingly dull. – Frederick Winsor, from The Space Child’s Mother Goose A mathematician named Klein Thought the Möbius strip was divine. Said he: “If you glue The edges of two, You’ll get a weird bottle like mine.” –Anonymous The bee-hive surface, which the Acme team discovered, is usually called a cross-cap, or the projective plane. The latter needs a little explanation. As the Euclidean plane is the place where the points and lines of Euclidean geometry live, so the projective plane is the place where the points and lines of projective geometry live. In projective geometry, the usual axioms of geometry hold except for the parallel axiom, in which two lines always intersect. A model for projective geometry might be the surface of a sphere in which the lines are great circles. But a problem with this model is that a pair of great circles intersect in two antipodal points. To eliminate this difficulty, you take the sphere and identify (glue!) antipodal points. Consider this new object. A pair of antipodal points of the sphere becomes one point of the new object. A great circle of the sphere becomes a “line’’ of the new object, and the new object becomes a good model for a projective plane. To find out what this object is, start with a sphere and identify the points of the northern hemisphere with their antipodal counterparts in the southern hemisphere. (Save the equator for later.) One way to see how to do this is to slice the sphere along the equator, and then turn the northern hemisphere inside out so that it now sits snugly inside the southern hemisphere. Now, keeping the southern hemisphere rigid, rotate the northern hemisphere inside it 180°. Antipodal points are now aligned, and the northern hemisphere can be glued to the southern hemisphere.

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To complete the job, all we need to do is glue diametrically opposed points on the equator. That is just what the Acme team did when it put together the crosscap/beehive. Experiment 13 above can be captured by the following “trip.’’ Place a circle with an arrow on a point on the Möbius strip. This specifies what is meant by clockwise at that point; it defines an “orientation’’ there.

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Travel in the strip—lengthwise—accompanied by the little circle-with-arrow until you get to the point where you started. Remember you are “in’’ the strip. The little circle-with-arrow now indicates counterclockwise instead of clockwise as at the start. Your original orientation has reversed itself. As a creature living on the Möbius strip, you can have no notion of clockwise/counterclockwise or of left/right. However, if you were a creature on a torus or sphere and you took your clockwise circle-with-arrow on a trip along a closed path, every time you return to the starting point, the arrow at the end points in the same direction as the arrow at the beginning of the trip. Thus the sphere and the torus are called orientable. Because there are closed paths on the Möbius strip and Klein bottle along which the orientation changes, these surfaces are called nonorientable. In 1874 German mathematician Felix Klein (1849–1925) and Swiss mathematician Ludwig Schläfli were the first to investigate the projective plane as a surface and to show that it is nonorientable (see Pont, p. 123). Klein was a professor of mathematics 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. In 1903 the mathematician W. Boy assembled the pattern into an object, called Boy’s surface, having three-fold symmetry (see Hilbert and Cohn-Vossen, pp. 317–321 and 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.

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Acme Makes Order Out of Chaos: Surface Sums and Euler Numbers

E

CHAPTER

9

E

E

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! Professor Gomfrina Demasiadas Palabras has just finished giving the group a tour of the laboratory. Now she is explaining how Acme might be of use to her. PROFESSOR: As you can see, we are designing robots to carry out specific tasks. Of course, we are looking for efficient and effective methods of controlling the robots. We would like to develop some theoretical tools for dealing with this in general. Consequently, we have decided to focus some of our attention on some very simple robots in the hopes that they might shed some light on our problems. We are studying robots that operate in a single plane. They’re made up of rods, connector pivots that link up

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Chapter 9 Acme Makes Order Out of Chaos the rods, and fixed pivots. A fixed pivot attaches one end of a rod to the plane but otherwise allows it to move freely in the plane. A connector pivot also allows free movement of the rods it connects. We call such a system a linkage. Before I go any further, let me show you a very basic example. This may give you some idea of where all of you might come in. Are there any questions 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 the Mer-bee-us strips and other video games, such as the Klein bottle.

PROFESSOR: Yes, thank you. Actually, I have heard of your work on video games. I think the example I’m about to show you will explain why 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. This linkage has just two rods. At the end of the first rod, there is a fixed pivot. Between the first and second rods, there is a connecting pivot.

Linkage #1

FREE PIVOT

FIXED PIVOT

As you can see, the entire linkage can rotate freely about the fixed pivot, and the second rod can rotate freely about the connecting pivot. BRANDON: Can the second rod pass over the first? PROFESSOR: I was just going to mention that. These are kind of idealized rods. Infinitely thin and thick.

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BOSS: Infinitely…? TIFFANY: Shhh…

1. Your Turn Cut out pieces of a manila folder in the sizes shown below. Use a paper punch to punch 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 a feeling for the set of possible positions of the linkage. 5'' 3 4''

BRADS

PROFESSOR: We can control the movement of the linkage by moving a cursor on a computer screen. We call it a control screen. I think it behaves in a fashion similar to one of the video games I’ve heard you’ve worked on. Here’s how the control screen works. (She plops a transparency on the overhead.)

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Chapter 9 Acme Makes Order Out of Chaos

Linkage/Control Screen Coordination

Linkage

Control Screen B

B

A

ANGLE B (DEGREES)

360

270

180

90

0

0

90

180

270

360

A

ANGLE A (DEGREES)

Each rod makes an angle to the horizontal. We impose a coordinate system on the screen so that the lower left-hand corner is the origin, the bottom of the screen is the positive x-axis, and the left edge of the screen is the positive y-axis. When the cursor is at the point (s, t), the angle of the first rod to the horizontal is s and the angle of the second is t. Thus, as the cursor moves from left to right along the bottom of the screen, the angle the first rod makes to the horizontal increases smoothly to 360°. The first coordinate parameterizes the angle of the first rod to the 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 second rod remains parallel to the horizontal. Similarly, if the cursor moves from the lower left corner up the left side of the screen, the angle the second rod makes to the horizontal increases smoothly to 360°. At the same time, the first rod lies parallel to

Scene 1

181

the horizontal line. Consequently, the second coordinate parameterizes the angle of the second rod. So, for example, if the cursor is at (90, 30), then the first rod is perpendicular to the horizontal and the second makes an angle of 30° to the horizontal. (She writes on a transparency.)

30˚

(90, 30)

Linkage

Control Screen

BOSS: I get it! You move the cursor to (s, t) and the linkage moves over s 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 the second 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 the cursor is at the top of the screen, the second rod has rotated 360°. It’s back where it started. The point on the top of the screen produces the same position of the linkage as the corresponding point at the bottom. (t, 360)

360˚ t

(t, 0)

MILLIE: So if you want to do negative angles, you start at the bottom of the screen and come out the top going down. BOSS: Huh? PROFESSOR: Similarly, if the cursor hits the right of the screen, then the first rod has rotated 360° from the horizontal. It’s back where it started.

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Chapter 9 Acme Makes Order Out of Chaos The point at the right edge and the point directly to its left on the left edge both correspond to the same position of the linkage. JOE: If you look at it backwards and spun the first rod, keeping the second rod at the same angle to the horizontal, you’d see the cursor going from left to right across the screen. Each time the 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 is identified 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 the positions of the linkage: to every point on the torus, there corresponds a position of the linkage, and conversely, to every position 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 to know about the linkage. There is something more. I used the word “smoothly.’’ What I meant is, moving the cursor smoothly around the screen produces a smooth motion of the linkage. There are no lurches or bumps.

2. Your Turn Imagine that the cursor makes a path on the screen’s diagonal from lower left to upper right. What would the motion of Linkage 1 be in order to create this path of the cursor? What would the motion of Linkage 1 be in order for the cursor to make 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 linkage that’s a little bit more complicated. We’ve been able to unravel some information about a control screen, but not everything.

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183

We think you might be able to help us with this. This time there are four arms, three free pivots, and two fixed pivots.

Linkage #2

3'' 3'' 6'' 6''

10''

Let me describe to you what we know about the control screen. We thought that we might try to control the angles as we did with the first linkage. But that would give us four numbers, and we wanted something two-dimensional. Besides, we thought there might be complicated relationships between the angles. We want something simple, no? The goal is to describe a set of points corresponding to the positions (or states) of the linkage. We chose to look at the central free pivot p and describe its possible positions. Instead of telling you what we came up with, I thought I’d let you find out for yourselves. This will give 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 your exploration. Lights fade. Music comes up as Professor Gomfrina Demasiadas Palabras leaves the podium and Acme team members play with the linkages.

2. Your Turn Join the Acme team in exploring Linkage 2. Cut out pieces of a manila folder in the 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 that correspond to possible positions of the pivot p. (As with Linkage 1, assume that rods can pass over one another easily.)

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Chapter 9 Acme Makes Order Out of Chaos 7''

11''

4''

8 1 2''

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 to share? Yes. Please come up. (Tiffany comes up to the podium.) TIFFANY: First let me label the linkage: P is the central free pivot, Q and R are the other free pivots, S and T are the fixed pivots. R

P

Q S

T

I started with the rods set up like an isosceles triangle with base of 10 inches and two sides of 9 inches each. So nine inches is the farthest P can be from T and from S. The position of P has to lie inside two circles, each of radius 9 inches, one with center S and the other with center T. P 9'' 9''

Q

R

9''

9''

S S

T

10''

10''

T

Scene 2

185

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 as before, I decided to bend it as much as I could, so P would actually lie on top of the rod RT. This is the closest P could be to T. So P must lie outside a circle with a radius 3 of inches with center T. Similarly, P would have to lie outside the same size circle with center S. These are the “closest’’ circles. Now you know P is inside both of the farthest circles and outside both of the closest circles. Other than these four circles, I don’t think there are any more constraints. Here’s the picture: POSSIBLE POSITIONS FOR P 9''

3''

10''

CLOSEST CIRCLE

CLOSEST CIRCLE

FARTHEST CIRCLES

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 of point P. But there’s more to the linkage than the position of point P, no? In fact, every position of P corresponds to four different states of the linkage, depending on how the secondary elbows are bent. The left elbow could be bent up or down; the right elbow could be bent up or down. Here’s an example. Q

P

P

R

R

Q

Q R

S

T

S

T

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Chapter 9 Acme Makes Order Out of Chaos To cover these possibilities, make four copies of the hexagon and label them UU, UD, DU, and DD. The first letter describes the state of the left elbow (up or down), the second the state of the right one. So the state of the linkage in the example just shown corresponds to a point in hexagon UD with the left elbow up and the right elbow down.

UU

UD

DD

DU

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 gets identified with the edge of another hexagon—a sort of wild variation of identifying opposite edges of the single screen for our 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 is the same as the corresponding point on the top right edge of hexagon DU. These are points where the left elbow is fully extended and not bent so that “crossing’’ this edge from UU to DU (or DU to UU) smoothly switches the left elbow from up to down (or down to up, respectively).

DU

UU

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.

Scene 3

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Crossing edges TR, BR, and ML amounts to switching the left elbow smoothly from up to down (and vice versa). Thus, when two hexagons are labeled with the same second letter, the corresponding edges labeled TR, BR, and ML get identified. Call them TRU, BRU, and MLU when the second letter is U and TRD, BRD, and MLD when the second letter is D. Crossing the remaining three sides of the hexagon—TL, BL, and MR—amounts to switching the right elbow from up to down and back. Combining all this information, we obtain four hexagons having their edges labeled like this: L

L

MU

L

R

TU

R

MU

UU L

R

BU

BU

L

MU

R

MD

L

R

BU

L

MD L

R

MD R

BD

R

TU

TD DD

BD

L

R

TD

TU DU

BD

L

R

TD

TU

L

MD

TD R

MU

UD L

BU

R

BD

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 willing to 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 interesting problem! Right up our alley! We’ll get on it right away. Lights fade. Music up. Professor winces as Boss shakes her hand. The Acme teams says its good-byes and leaves the room. Professor puts away her notes.

4. Your Turn Help the Acme team. Put together the four hexagons as best you can. See if you can figure out what the assembled object should be.

Scene 3 Back at Acme headquarters, Brandon, Millie, Joe, and Tiffany are having a strategy session on how to attack the Professor’s problem. Boss is in the corner figuring out how much money this is going to bring in.

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Chapter 9 Acme Makes Order Out of Chaos JOE: I glued the hexagons together side by side. TThat Tgave me a T T single shape, with 18 edges left unglued. Then I stretched T T the whole thing into an 18-gon, like an 18-sided video T screen! T T T T T T T T R D

L U

L D

R D

R U

L TU

L

MU

R

MU

UU L

BU

R D

R U

R

BU

L U

L

MD

UD R

BD

L

BU

R D

L D

L

BD

L

R

MD

DD

L D

R U

R

BD

MU

DU R

BU

R U

L U

L

BD

L

MU

L D

L

MU L

L

BU

BD R

BU

R

BU R D

R D

B

B

L

BU

L

BD

BRANDON: That’s an answer to the Professor’s question all right. But I doubt that’s what she wants. Does anybody have a clue about what it is? When you put it together completely? TIFFANY: Looks like something we’ve seen. Sort of a doughnut, a Klein bottle. But so much bigger! Those were rectangles, and this is an 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 a definition. 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.

Definition of Surface A surface S is an object on which you can draw a map; that is, you can express the set S as a union of vertices, edges, and countries of an OK map. Now, can we connect this with something that comes from a pattern? MILLIE: Well, you can always draw a map on a pattern. Include the border of the pattern as part of the map’s edges.

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B

A

C

C

B

A

BRANDON: What about the other way? MILLIE: What other way? BRANDON: Start with a surface. Something you can draw a map on. Does it come from a pattern? JOE: You know, the Professor’s four hexagons, they’re just countries in 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 up of the edges, vertices, and countries of some map. Cut out all the countries. (Don’t forget to label all the edges with letters and arrows ahead of time.) Flatten them out and straighten out the edges. You’ll get a bunch of polygons. F

B A

F

C

C II

I E

D

A

G

E

D

III

IV

G

B

H

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! F G E

G

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E

G III

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Chapter 9 Acme Makes Order Out of Chaos

Joe writes.

Theorem Connecting Surfaces with Patterns Given 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. BRANDON: I’ve been thinking about what we were doing with the Klein bottle and the cross-cap. Remember when we found out that the Klein bottle was two Möbius strips glued together along their boundaries? Well, we also found out that a Möbius strip is a cross-cap-with-lake. So a Klein bottle is just two crosscaps-with-lake glued together along the shorelines of the lakes.

Why not do the same thing with any two surfaces? Add a lake to 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. (Brandon writes.)

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Surface Sum Let 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 resulting surface is called the sum of S and T and is denoted S#T.

TIFFANY: This is awesome. I think we might be able to detect a sum sometimes just by looking at a pattern. In fact, we did that with the Klein bottle and the cross-caps. One pattern for the Klein bottle we cut up into two cross-caps-with-lake. A

A

C

A

C

A

B

B C

B

B

That’s how we discovered the connection between Klein and cross-cap. Let’s go backward again. Take a pattern for surface S. Then it’s easy to make a pattern for S-with-lake. G

G G

F

S

H

H

F X

E

H

E

F

X E

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Chapter 9 Acme Makes Order Out of Chaos If you have a pattern for S and one for T, it’s easy to find a pattern for S#T. (She writes.)

Pattern for Surface Sum Suppose patterns for surfaces S and T are as follows: C

F

Y B

W E

Z

X

A

D

So patterns for S-with-lake and T-with-lake are C

F

Y

W G

G B

E

Z

X

A

D

Then a pattern for S#T is Y

Z

E

F

G

C B

A

D

W X

JOE: OK. Nice. So how does all this help with our problem? BRANDON: Two things. First, we can build more complicated surfaces. Like we can put two doughnuts together.

E

E

E

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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 also means we can go in the other direction and “break down’’ a surface. The Klein bottle is the sum of two crosscaps: K = C#C. This breaks the Klein bottle down into simpler surfaces. Maybe we could do that with the four-hexagon surface: break it down. (Brandon writes.)

Strategy for Analyzing a Surface To understand a surface S, see if you can make it the sum of two simpler surfaces. You can try this with the pattern: see if you can recognize the pattern as being a pattern for the sum of two surfaces. JOE: Well, let’s go! Let’s try this idea out on some of the patterns we’ve seen. Fade.

5. Your Turn Go back to the investigations, questions, etc., of Chapter 8, look at the patterns there, 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 is something 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

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Chapter 9 Acme Makes Order Out of Chaos calculate V − E + C. This is a number. For the sphere it’s always two, and for the torus it’s always zero. If V − E + C isn’t two or zero, then we know it’s not a sphere or a torus. That would be a start. MILLIE: I don’t think we know what V − E + C is for a Klein bottle, or even 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. For a 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 a bit and start small. We can build from there using the sum idea. If you started with surfaces—like the sphere and torus—for which V − E + C is the same no matter the map, maybe the surface sum would have the same property. I think we need a definition.

Euler Number Suppose that S is a surface and that, for every OK map on the surface, V − E + C is always the same. Then call that common number the Euler number of the surface. Denote the Euler number by N(S). So N(sphere) = 2 and N(torus) = 0. Now, suppose S and T are two surfaces with Euler numbers N(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 surfaces that do. S and T do. So, take a map on S#T. If we were to count V, E, and C, would there be some way we could predict what V −E + C is if we knew N(S) and N(T)?

Scene 4

195

S

T

JOE: I think we did something like this for the torus. We cut it open along its “waist,’’ flattened it out into an island-with-lake, and then 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 to the map the new vertices, edges, and countries it created. We showed that doing that didn’t change V − E + C. We could try the same thing with S#T.

S T

TIFFANY: So we add the circle. The same argument as before says V − E + C stays the same. Now cut along the circle. We get S-with-lake and

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Chapter 9 Acme Makes Order Out of Chaos T-with-lake with maps having data VS, ES, CS and VT, ET, CT, respectively.

S

T

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 the vertices and edges along the added circle have been counted twice. We need to subtract those off from the V and E counts.

S

But, look! The circle has the same number of vertices as edges. We’d be subtracting n (vertices) and adding n (edges) to V − 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 Sum If S and T are two surfaces having Euler numbers N(S) and N(T), then the surface sum has an Euler number given by the formula N(S#T) = N(S) + N(T) − 2.

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197

MILLIE: So if S is a torus and T is a torus, S#T is a two-holed doughnut and it has Euler number −2. If that don’t beat all. BRANDON: Well, we have our work cut out for us, figuring out the Euler numbers for a bunch of things: a Klein bottle (if it has one), an n-holed doughnut, and many other things we can put together with the surface sum. JOE: Before we do anything, I was thinking. That four-hexagon thing that goes with the linkage? It’s really a map on whatever surface it is. Each of the hexagons is a country. So C = 4. Each edge of a hexagon is an edge of the map, but each gets glued to 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 the same. Stack the four hexagons on top of each other so that running a pin through the four of them would correspond to the same position of the central pivot. Four vertices that lie on top of each other are really just one vertex—think about it! Those four vertices from the four hexagons correspond to the same position of the linkage. That means that the surface has only six vertices. So, V − E + C = 6 − 12 + 4 = −2. That’s new information! That number is also the Euler number of a twoholed doughnut. Now that’s the cat’s meow! Could we assemble that four-hexagon thing into a two-holed doughnut? We’ve made an amazing start. But there’s work ahead so let’s get going! High fives all around. Lights fade. Music comes up.

Investigations, Questions, Puzzles, and More 1. Gathering Evidence The Acme team would like to begin recognizing surfaces, as well as how they are built up from simpler surfaces by the surface sum, just by looking at their patterns. Given that T is a torus and C a cross-cap, help them find patterns for the following surfaces: T#T#T, C#C#C, T#T#C, and T#C.

2. Question Members of the Acme team wonder what can be said about R#Sphere where R is a surface. Help them out!

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Chapter 9 Acme Makes Order Out of Chaos

3. Question Suppose that S is a surface and that S′ is gotten by removing a disc from S. (Thus S′ is S-with-lake.) The Acme team wants to know the answers to the following questions: ●

If S has an Euler number, does S′? If the answer is yes, are the two numbers related? How?

●

If S′ has an Euler number, does S? If the answer is yes, are the two numbers related? How?

●

What about S-with-n-lakes?

4. Investigation If 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. Investigation The Acme team knows that the torus and sphere have Euler numbers and that N(sphere) = 2 and N(torus) = 0. The team would like to know the answers to the following 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 want explanations for everything.

6. Question If M is the Möbius strip, how would you describe M#M? Does M#M have an Euler number? If so, what is N(M#M)? (See Investigation 5.)

7. Question Suppose that K is a Klein bottle, C a cross-cap, and T a torus. Rumor has it that K#C is the same as T#C (see Investigation 8 from Chapter 8). If K and C have

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199

Euler numbers, then knowing that the rumor is true and knowing the formula for the Euler number of a surface sum ought to enable you to find N(C). What would N(C) be? Does your answer jibe with what other investigations have revealed?

8. Question The Acme team came up with a method for taking a pattern for surfaces A and B and coming up with a pattern for A#B. This method is flawed in some cases. Can you see when it works and when it does not?

9. Investigation Here are the first few items in an infinite sequence of surfaces. Your task is to identify as many of these as you can. Ultimately you would like to identify all of them.

10. Question The Acme team feels that the Euler number of a surface might be useful in identifying 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 know it has an Euler number, and you know N(P). Furthermore, there is a surface Q you do know completely, and it turns out that Q has an Euler number and that N(P) = N(Q). Are P and Q the same surface?

11. Investigation Although we cannot assemble the cross-cap C and the Klein bottle K without having 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].)

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12. Investigation Professor Demasiadas Palabras has given the members of the team the following linkage for them to analyze. It is very similar to the second linkage the Professor discussed in her lecture. However, the lengths of the rods have been changed. Help the team figure out a control screen. 3'' 2'' 5''

4''

13. Investigation Professor Demasiadas Palabras gave the team another linkage to investigate. Although the linkage has three fixed pivots, she feels that it can be analyzed in a fashion similar to the one with two fixed pivots. Help the team in its investigation. A

O

B

C

Notes

201

14. Investigation Here is a another linkage from Professor Demasiadas Palabras. It has one fixed pivot and is similar to the first one she presented. Find a control screen for this linkage.

15. Summarizing Write a summary of the Acme team’s activities for the day. Include statements of problems, definitions, results, explanations, relevant diagrams, and directions for future investigations. Be sure to make the summary clear, organized, and understandable. Convince Boss that the time spent was worthwhile.

Notes The greatest impetus to the study of surfaces came from within mathematics itself through the work of German mathematician Bernhard Riemann (1826–1866). In his inaugural dissertation of 1851, Riemann, a student of Gauss, saw the study of surfaces as a major ingredient in understanding certain complex functions of one variable (for a description of the Riemann surfaces of certain complex functions, 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 the surface along these curves, you obtain something that can be flattened and stretched/shrunk into a disc. A special case is the sphere, defined to have connectivity 1 (see Kline, p. 1166+). In 1861 Möbius was the first to undertake a general study of surfaces. Although he and Listing are responsible for the creation of the Möbius band, his study included only orientable (two-sided) surfaces without boundaries. In 1874 Felix Klein added possible boundaries to the surfaces and began the study of nonorientable surfaces. In the 1880s two of Klein’s students, Guido Weichold and Walter 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 characteristic of a surface and is frequently denoted by χ (S). In 1813 Simon-Antoine-Jean

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Lhuilier found a formula for the Euler characteristic of an n-holed doughnut (see Biggs, et al., p. 84+). The Euler characteristic of a cross-cap is an easy consequence of 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 the two surfaces. The notion seems to be a 20th-century one. Klein and his students used other ideas to build surfaces. They started with the surface of a sphere on which 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. (This paraphrases von Dyck using our terminology.) The study of linkages has a long history, in part associated with the real problem of turning linear motion into circular motion. The construction of a straight line by means of a linkage was a famous problem solved by Peaucellier in 1864 (for a description of this and other linkages, such as those that construct a plane, see Hilbert and Cohn-Vossen, pp. 272–275). For more on linkages similar to the ones appearing 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-dimensional world in which we live?’’ To answer these questions, it would be useful to know the answer to a related mathematical question: “What are all the possible threedimensional worlds?’’ To answer the latter, it might be good first to answer an analogous question: “What are the possible two-dimensional worlds, that is, what are the possible surfaces?’’ For more on possible three-dimensional worlds, see the book 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: Academic Press, 1982. Hilbert, D., and Cohn-Vossen, S. Geometry and the imagination. New York: Chelsea Publishing Company, 1956.

References

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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.

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Acme Classifies Surfaces

CHAPTER

10

Scene 1 Two days later. Tiffany, Millie, Joe and Brandon are having a meeting to discuss progress on their problem. BRANDON: I know we haven’t figured out what the 18-sided polygon is. We have a clue, but nothing definitive. So let’s go around the group and 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 polygon pattern into an algebraic expression. I think it might make it easier to identify what you’re looking at, like when a surface is the connected sum of simpler surfaces. BRANDON: Sounds great. Hold on to that. We’ll come back and hear more about it. Joe?

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Chapter 10 Acme Classifies Surfaces JOE: Well, I’ve been looking at patterns. I’ve developed some rules for cutting and regluing patterns that might help us figure out what the 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. (She writes.)

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. BRANDON: Wow. That’s pretty ambitious. I like it! What do the rest of you think? 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 together with the pattern for a doughnut-with-lake. A

B

B A

Should all those things be on the list? The possibilities are endless. How could you list them all? MILLIE: Keep your britches on, guys. We’ve already lumped the oddtwisted strips together, and we lumped the even-twisted strips together. We lumped things together if they had the same pattern. BRANDON: You’re right, Millie. We did do some lumping. I guess what we need to 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. We defined a surface as something made up of a map—a union of vertices, edges, and countries—with certain properties satisfied. We could say that two surfaces are the same if both have the same maps.

Scene 1

207

JOE: What does “same map’’ mean? TIFFANY: Hmm. Good question. I guess I mean that there is a one-to-one correspondence between the countries of the two maps, a oneto-one correspondence between the two sets of vertices, and another between the two sets of edges—all of these so that all the “relationships’’ match. For example, if two countries of one surface border, then the corresponding countries of the other must also border. If two edges meet at a vertex of one surface, then the corresponding edges meet at the corresponding vertex of the other. Hmm. This may be complicated to write down completely. JOE: I think I see what you mean. But I think your definition would lead to Millie’s. Take the two maps, and cut out the countries. Then reassemble each set of them like we did before to make patterns. You could reassemble them in the “same’’ way. Get the same 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. (She writes.)

Definition of “Two Maps Are the Same” Start with maps M1 and M2. Denote the sets of vertices, edges, and countries of Mi by Vi, E1, and C1, respectively for i = 1, 2. Suppose F is a 1-1 onto function F: V1 傼 E1 傼 C1 → V2 傼 E2 傼 C2 such that ● F takes vertices to vertices, edges to edges, and countries to countries. ● If v and v′ are two vertices in M joined by edge e, then F(v) and F(v′) are 1 two vertices in M2 joined by edge F(e). ● If e and e′ are two edges in M that meet in vertex v, then F(e) and F(e′) 1 are two edges of M2 that meet in vertex F(v). ● If c and c′ are two countries in M that border along edge e, then F(c) and 1 F(c′) are two countries of M2 that border along edge F(e). Then maps M1 and M2 are the same. 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 glue country c and c′ of surface S1 together along edge e, you also glue country F(c) and F(c′) of surface S2 together along edge F(e). TIFFANY: OK. Here’s where we are. (She writes.)

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Two Surfaces Are the Same Definition: 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 surfaces S1 and S2 are the same. Theorem. If surfaces S1 and S2 are the same, then they have the same pattern. I think the other direction is pretty clear. If two patterns are the same, then the two surfaces have the same map. You draw a map on one pattern and copy it onto the other. (She writes.) Theorem. If surfaces S1 and S2 have the same pattern, then they are same surface. MILLIE: What if a surface has more than one pattern? Like the Klein bottle. We took one pattern, cut it up, labeled the cuts, and glued it in another way. What would you call that? An equivalent pattern? A map on one would carry over to a map on the other. (She writes.)

Equivalent Patterns and Surfaces Patterns P1 and P2 are equivalent if you can cut up P1, label the cuts, reassemble the pieces, and glue them to get pattern P2. Theorem. Two surfaces are the same if and only if they have equivalent patterns. ALL: Way to go, Millie! Lights fade. Music comes up.

1. Your Turn Does 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 edges F(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 as borders the edges F(e1), … , F(em). In particular, c and F(c) must have the same number of edges.

Scene 2

209

Do these properties follow from Tiffany’s definition? Are there other criteria you’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. Tell us about it. TIFFANY: I propose that we replace a pattern for a surface by an algebraic symbol. It will be more efficient, because we wouldn’t have to draw the pattern, and it might be easier to identify familiar surfaces. Here’s how to create the symbol. (She writes.)

Surface Symbol Take a pattern for the surface. Pick a vertex of the polygon and travel clockwise around its perimeter, writing down the edge labels as you go, one after the other. 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 until you arrive back at the vertex where you started. The result is the surface symbol. Example: Start with vertex P on the following pattern. B

C

A

B P

Q D

A

Result: symbol ABC−1BA−1D−1.

It shouldn’t matter at which vertex you start. If you start at vertex Q on the pattern, then you’d get the symbol A−1D−1ABC−1B. This is different from the one I got before. But they’re for the same surface! 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 instead of using A, B, C, and D, I used W, X, Y, Z (respectively), I’d get WXY−1XW−1Z−1. So I’d want to write

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Chapter 10 Acme Classifies Surfaces 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 arrow is going “against the grain,’’ or pointing clockwise, the opposite of the direction of your trip. MILLIE: And you’d get another symbol if you changed the direction of all the 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 those symbols to be equivalent. In fact, you’d want the symbols you get from equivalent patterns to be equivalent themselves. For example, the Klein bottle has patterns B

C

A

A

D

C

B

D

with symbols ABAB−1 and CCDD, respectively. You’d want ABAB−1 ~ CCDD. Maybe we can make a list of rules to tell when two symbols are equivalent.

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.)

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211

Equivalent Symbols αAβAγ ~ 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 JOE: This is neat. I think we should make a list of symbols for some of our favorite surfaces. BRANDON: And what about the symbol for the sum of two surfaces? MILLIE: Well, back to the drawing board. Lights fade. Music comes up.

2. Your Turn Help 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 a surface; in that case, there would more than one symbol. Write down these extra symbols, too.

Symbols for Familiar Surfaces Surface

Pattern

Symbol

X

X

Sphere

XX−1

B A

Cylinder

A C Continued

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Symbols for Familiar Surfaces F H

H E

Möbius band

E G

J

Torus Klein bottle Cross-cap Two-holed doughnut Connected sum of three cross-caps

3. Your Turn Write down a symbol for the connected sum of three tori. Write down the symbol for the connected sum of two tori and three cross-caps. What is the general situation? That is, suppose you know symbols for surfaces S and T. Is it easy to find a symbol for S#T? If you were to be presented with a surface symbol, when could you 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 of today’s session? Well, I’ve been translating my rules for patterns into rules for symbols. Let me tell you how I got the idea for the rules. First, you all agree that in a pattern—and now, in a symbol—a given letter can appear one time or two times, but not more. If a letter happened three times, you’d have three edges to glue together and that certainly wouldn’t be a surface. TIFFANY: Good point. We need to write it down. (She writes.)

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Occurrence of Letters in Surface Symbol In 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 edge of a lake. I just had to add that other bit. JOE: Well, I noticed that, in patterns I recognized, a letter occurring twice shows up close together. Let me show you. D B

A CONNECTED SUM OF A THREE CROSSCAPS

E B

F G

G

C

C

E

CONNECTED D SUM OF TWO TORI

F

One of the things about the linkage surface pattern we’re trying to decode is that those pairs of letters are scattered all over. L

R TD

TU

L

TD

R

TD

R

TU

R

TU

L

TU

L

TD

L

MU

L

MU L

L

BU

BD R

BU

R

BU R

R

BD

L BU

L

BD

BD

I wanted to figure out some way to bring such a pair of letters closer together. Here is the first thing I came up with. (He writes.)

Rule 1 for Equivalent Surface Symbols αXβγXδ ~ αXγXβ−1δ “A string of symbols to the right of an X can be moved to the right of the other X after inverting.’’

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Chapter 10 Acme Classifies Surfaces I used Tiffany’s convention: a Greek letter stands for a string of symbols. 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 why it works. Take a pattern that goes with the symbol on the left. Then cut it, label cuts, reassemble, and glue. β

X Y

α

Y

β γ

Cut

Y

α

Y

Glue

X δ

X

δ

γ

α

γ

X

X δ

Y β

The pattern you wind up with has symbol αYγYβ−1∆. So, using Tiffany’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 closer together. 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 of equivalent symbols. The connected sum of a cross-cap with something else! All you need to do is figure out what that something 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 pair of letters that are the same and that appear as Z and Z–and not as Z 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.)

Rule 2 for Equivalent Surface Symbols αβXγXδ ~ αXγβ−1Xδ “A string of symbols to the left of an X can be moved to the left of the other X after inverting.’’ I’ll let you guys figure out why the rule works.

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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 and something 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 as A−1 and A−1. (She writes.)

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 string having 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.

JOE: Good deal, Mill! I just had a look at our linkage surface. Every letter occurs twice but none as…A…A…or…A−1…A−1.…I have a couple rules that relate to that situation. I wonder if they’ll help? (He writes.)

Rule 3 for Equivalent Surface Symbols αXβγX−1δ ~ αXγβX−1δ “In a surface symbol containing X and X−1, a string of symbols to the right of X can be moved to the left of X−1.’’

Rule 4 for Equivalent Surface Symbols αβXγX−1δ ~ αXγX−1βδ “In a surface symbol containing X and X−1, a string of symbols to the left of X can be moved to the right of X−1.’’

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Chapter 10 Acme Classifies Surfaces 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 next to 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 are two pairs of symbols: A and A−1, B and B−1. If you got both together 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.) 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. 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 thing would be to work on the “something else.’’ We could “assemble the 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

Scene 3

217

This time we want to get the “κυµ’’ out of there without messing 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.) XYX−1Y−1κυµUVU−1V−1λθ

(Left of the V−1 to right of the V.) ~

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 it with the “something else.’’ Millie, you’re a genius! Tiffany, time to chalk up another theorem. (Tiffany writes.)

Theorem: Assembling the Tori Part A. If S is a surface symbol such that every letter A that occurs twice occurs as…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 occurs as…A…A−1…or…A−1…A.…So the corresponding surface is the sum of n tori and something else. In addition, no pairs of letters X, Y occurring twice in σ also 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 the sum of m cross-caps, n tori, and something else, where the something else has symbol σ.

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’’ music comes up as the gang slowly dances about. Lights fade.

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Chapter 10 Acme Classifies Surfaces

Investigations, Questions, Puzzles, and More 1. Investigation The surface symbol for the linkage “control screen’’ the Acme team is concerned with is ABCD−1B−1EDC−1E−1A−1F−1G−1HFJ−1H−1GJ. Use Rules 3 and 4 to figure out what it is.

2. Investigation Suppose 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. Make up more examples of your own. What can you say about the surface that corresponds to such a symbol? (Of course, you will want to justify your conclusions.)

3. Investigation This is very much like Investigation 2. In this case, you are to investigate the surfaces 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 is that, here, there may be letters occurring only once.) Investigate all such surface symbols. Make up several examples. In general, what can you say about the surface that corresponds to such a symbol? (Of course, you will want to provide good explanations for your conclusions.)

4. Investigation Combine Part B of Assembling the Tori Theorem with the results of Investigations 2 and 3 to come up with a theorem that describes all possible surfaces.

Investigations, Questions, Puzzles, and More

219

5. Investigation Millie 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 torus would be the same as the sum of three cross-caps. What would this do to the theorem you came up with in Investigation 4?)

6. Question Earlier, there was a question: does every surface have an Euler number? Given what you know from Investigation 4 (and all the work leading up to it), what is the status of this question?

7. Question Joe claims that, if S is any surface, then its Euler number has the following property: N(S) ≤ 2. Is he correct?

8. Question Brandon 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 the sphere, 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. Question Do 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. Question Do 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.

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Chapter 10 Acme Classifies Surfaces

11. Question Brandon has defined a “prime’’ surface T as one in which T = U#V is impossible unless U is the sphere and V is T itself (or the other way around). What are the prime 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 factorization unique?

12. Investigation Investigate 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? Try using Rules 1 through 4 to identify this surface exactly. What do you conclude?

13. Investigation Investigate 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? Try using Rules 1 through 4 to identify this surface exactly. What do you conclude?

14. Investigation Assembling the Tori (part B) and Investigations 4 and 5 provide us with a complete list of surfaces. Identify the items in that list that can be obtained using the following two steps: 1. From a disc in the plane remove a finite number of smaller, nonintersecting discs from its interior. 2. Join the boundaries of certain pairs of these discs by tubes. Each tube may be attached in two different ways:

Investigations, Questions, Puzzles, and More

221

15. Investigation Assembling the Tori (part B) and Investigations 4 and 5 provide us with a complete list of surfaces. Which items on this list can be obtained by using soap bubble solution? (Recall that to make a soap bubble surface, you take a wire, put a knot in it, dunk it in soap solution, and poke holes where necessary in the soap film so that you get a surface with the wire as the only edge.)

16. Summarizing Write a summary of the Acme team’s activities for the day. Include statements of problems, questions, definitions, results, explanations, relevant diagrams and pictures, and directions for future investigations. Make the summary so clear, organized, and understandable that Boss will be convinced that the time spent was worthwhile.

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Chapter 10 Acme Classifies Surfaces

Notes This chapter contains two big accomplishments. The first is a definition of equivalent surfaces. This may seem like a triviality. But deciding when two surfaces are the “same’’ is an important step. We had been headed for this definition since we 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 the last word, it is just a beginning. We may want to adjust the definition later. The second accomplishment is a complete classification of surfaces. The results of this chapter plus those of Investigations 4 and 5 show that any surface must be equivalent 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 a pattern for a surface), the results of the chapter (and Investigations 4 and 5) not only tell you that the surface is one of the items on the list but also give a method for 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 of nonorientable surfaces in 1883 (Pont, p. 129f). Möbius also found it useful to think 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 is that of Lehrbuch der Topologie, by Seifert and Threlfall. Their book became the standard topology textbook soon after it appeared in 1934. In it, they think of a surface as a polygon with certain edges identified, that is, as a pattern, and use surface symbols just as they are used by the Acme team (see the English translation 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-Threlfall model are those by Blackett, Fréchet/Fan, Carlson, and Firby/Gardiner (listed below). There are several more 20th-century proofs of the classification theorem. One uses a method called “surgery’’ created by Zeeman. A treatment of this can be found in the book by Stewart. A second method is due to Conway (see article by Frances and Weeks), and a fourth can be found in the book by Boltyanskii and Efremovich.

References

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References Blackett, D.W. Elementary topology: a combinatorial and algebraic approach. London: Academic Press Limited, 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.

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Acme Encounters the Fourth Dimension

CHAPTER

11

Scene 1 The next day. Millie, Tiffany, Brandon, and Joe are recapping the previous day’s successes. BRANDON: Well, Millie, we accomplished everything you suggested. We’ve got a list of all possible surfaces. And, if we are presented with a surface, we have a method for deciding which item on the list it is. MILLIE: Thanks to Tiffany for coming up with the idea of surface symbols 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.) 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.

225

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Chapter 11 Acme Encounters the Fourth Dimension

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

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 all things we could have visualized a long time ago. And, if you give me a surface, I know it’s one of those and I can figure out which one 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 any duplications? MILLIE: Yeah. How could the sum of seven cross-caps be the same as a sphere with three lakes? TIFFANY: We were pretty excited when we discovered that the sum of a cross-cap and a torus was actually the same as the sum of three cross-caps. That duplication meant that all those “hybrid’’ surfaces—such as the sum of s cross-caps and t tori—were really not hybrid at all. The sum of s cross-caps and t tori is really the same as the sum of s + 2t cross-caps when s is positive. That revelation enabled us to whittle our list down to what we have now. The question is, can we whittle it down further? That revelation wasn’t at all obvious. Is there another obscure revelation lurking in there? BRANDON: OK, OK. I get your point. Any ideas, gang? (Silence.)

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227

JOE: I’ll stick my neck out. One thing we’ve never used is the Euler number N(S). TIFFANY: Could we use it? JOE: Well, what do we know about it? We know that two surfaces that are 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.) Every surface has an Euler number. BRANDON: And, we know how to calculate the number for everything on our list. Let’s do it. (He writes.) Euler Numbers of Surfaces Surface

Euler Number

1. 2. 3. 4. 5. 6.

2 2−r 2−s 2−s−r 2 − 2t 2 − 2t − r

Sphere Sphere with r lakes Sum of s cross-caps Sum of s cross-caps with r lakes Sum of t tori Sum of t tori with r lakes

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 two surfaces have the same Euler number... JOE: Not so fast. The Euler numbers of the torus and the Klein bottle are both zero. And, look, the number of a sphere with two lakes is 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 think they’re different? For one thing, one is orientable, the other isn’t. Can we show from the definition that an orientable surface can’t be the same as a nonorientable surface? We’d accomplish a lot if we could do that. Look at the table. It would mean that everything in the third and fourth rows of our table is different from everything else. Then there’s the torus and the sphere with two lakes. My gut feeling is that something with lakes can’t be the same as something without. Let’s take a break and think about those things. Members of Team Acme return to their desks. Lights dim.

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Chapter 11 Acme Encounters the Fourth Dimension

1. Your Turn Help 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 of surfaces 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 Turn Below are patterns for a couple of surfaces. Figure out how many lakes each one has just by looking at the patterns. P

B B

A C

A Q

C

Q

P

B

A

A B

S R

Scene 2 A half hour later. MILLIE: Gather ‘round, luvs. I think I’ve got it. (The group surrounds Millie at her desk.) BRANDON: What do you have, Millie? MILLIE: You all remember that if you were a creature living “in’’ the Möbius band, you could take a trip around the band in such a way that at the end of the trip your orientation was reversed from what it was at the beginning of the trip. Of course, the same thing can happen on other surfaces. In fact, it can happen

Scene 2

229

on any surface with a symbol that has a letter X occurring twice as ...X...X... or ...X−1...X−1..... Here is its pattern:

X

X

If you take a trip along the dotted path, your orientation reverses—just as it did on the Möbius band. The converse is also true: 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 of pieces. Glue them back together, but don’t glue the two labels marked with an X! You’ll eventually get a pattern whose symbol looks like ...X...X... or ...X−1...X−1.... So here’s what we know. (She writes.)

Nonorientable Surface A surface is nonorientable if and only if it has a pattern whose symbol is of the form ...X...X.... Of course that means that all the surfaces in Rows 3 and 4 of our table are nonorientable. BRANDON: Where are you going with this, Millie? How do you know that a surface in another row can’t have a pattern that would make it nonorientable? 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 a surface S that has a symbol where every letter that occurs twice occurs as ...X...X−1.... Certainly, no amount of stretching or shrinking of the pattern that goes with the symbol would change that fact. No amount of relabeling would change it either. The only thing that could change it would be to cut up the pattern and reglue it. What’s a typical cut and paste operation amount to? You pick a letter that occurs twice (X), cut the pattern along some path (Y) that separates the two X’s, and then paste the X’s together. Here’s what it looks like.

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Chapter 11 Acme Encounters the Fourth Dimension X

γ

α δ

β

X

β

γ

Y

Y

α

Y

γ

δ α

X δ

β

X Y

γ

X

β

δ α

X

X

The symbol for S we started with was αXβγX−1δ. Every letter Z in the symbol occurring twice occurs as ...Z...Z−1.... The symbol for the new pattern is αδXγβX−1. It’s clear that this new symbol also has the property that a letter Z occurring twice occurs as ...Z...Z−1.... That means that the surface S can’t have a symbol where, when a letter W occurs twice, it occurs as ...W...W.... The surface can’t be a nonorientable surface. It must be orientable. We’ve got a theorem! (She writes.)

Theorem Relating Orientability and Symbols If S is a nonorientable surface, then every symbol has the property that if a letter X occurs twice, it occurs as ...X...X.... In particular, if surface S has a symbol with the property that every letter occurring 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 a surface in any other row. JOE: Hey, that’s great! Tiff, whaddaya think? TIFFANY: This is just what we needed: turn our gut feelings into good thinking. Make our facts really tight. BRANDON: There’s the fact that the sphere with two lakes and the torus have the same Euler number. Shall we tackle lakes next? My gut feeling is that if two surfaces are the same, they’ve got to have the same number of lakes. Anybody have any ideas? JOE: Well I wasn’t making any headway on any of this stuff. So I got to thinking about the Euler number and how neat it was that we could use it to tell some surfaces apart. I began to wonder if there was some way to calculate the Euler number straight from the symbol, without knowing ahead of time which item on our list it was. I thought knowing that might be pretty useful. Like, suppose you knew the surface had no lakes and was nonorientable. From what Millie has done, we know how to definitively decide what is nonorientable. Then it seems pretty clear to me that it’s the sum of a bunch of cross-caps. Then, if you knew the Euler number, you’d know exactly how many in the bunch—without having to use my rules!

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BRANDON: Hey, Joe, you’re right. Might not tell us about the number of lakes, but it’ll give us something useful. Who knows? Maybe it’ll be useful. Fire away! JOE: Anyway, it still needs a little work, but here’s my idea for calculating the Euler number from the pattern. My big assumption is that counting the vertices and edges of the pattern and counting the polygon as one country would give me all the data I need to calculate the Euler number. You don’t exactly count the edges of the pattern: your count has to account for duplications, a pair of edges to be glued really counts for just one edge. You can get the adjusted count of the number of edges right from the symbol itself: 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 little trickier than counting edges. Here’s my method. It involves labeling all the vertices. (Joe writes.)

Labeling the Vertices of a Surface Symbol We 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 one letter, 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 two possibilities: B is unmatched or B is matched in the symbol. In case B is unmatched, stop. If B is matched, go to the other occurrence of B in the symbol and write ...C1B... or ...B−11C.... Now work with the “new’’ letter C. Again, there are 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 doing this until you reach an unmatched letter or you repeat a pair. (This has got to happen because there are only a finite number of letters.) If you reach an unmatched letter, go back and work with A (just as you did with B). This time you’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 the vertices 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’t yet been labeled and follow the procedure in Step 1, using a new label. Keep repeating this process until no such pairs exist. This process has got to end because there are only a finite number of letters. That does it!

232

Chapter 11 Acme Encounters the Fourth Dimension MILLIE: Gracious! Give us an example, Joe. JOE: OK. Here’s one: XYXAZ−1YWBZW−1. Start with the first pair XY... 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 an example where the sequence begins and ends with unmatched letters. Pick another pair ...YX... where the shared vertex hasn’t yet been labeled. Label it X1Y2X1AZ−11YWB1ZW−1. Proceed as before, X1Y2X1AZ−11YWB1ZW−1 → X1Y2X1AZ 1Y2WB1ZW−1 → X1Y2X1AZ−11Y2WB1ZW−12. −1

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 labeling 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 vertex hasn’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 you weren’t paying attention and just followed the rules, you’d notice that you had reached an unmatched letter, B. The rules then say to stop, go back, work on the A in ...AZ−1..., but A is also unmatched. The rules say you’re now done, which we

Scene 3

233

knew since everything was labeled!) This is another example where the sequence begins and ends with unmatched letters. BRANDON: Wow. I think it’s time for a break to practice this technique on a few of our favorites. The group disperses. Each goes back to his or her desk, on the way congratulating Joe on the new idea. Lights fade. Music comes up.

4. Your Turn Try out Joe’s technique. Label the vertices in the following symbols: LMNP EEFFGG ABA−1B−1CDC−1D−1 XYXZ−1YWZW−1 AXBYCWXDW−1Y ABA−1CDC−1E

5. Your Turn A certain surface has symbol XWYZX−1WZY−1. By using Joe’s technique, we got the following labeling of vertices: 2X1W2Y2Z1X−12Z1W2Y−12. Now use Joe’s idea for calculating the Euler number to figure out directly which surface on the list he thinks 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 Joe labels a single vertex, two things happen: You can get a loop of pairs of letters that returns back to the beginning pair. Or, you can get a string of pairs that doesn’t connect but that begins with a pair having an unmatched letter and ends with a pair having an unmatched letter. MILLIE: What’s this have to do with lakes? TIFFANY: Unmatched letters! Those form pieces of the shorelines of the lakes. A single shoreline consists of a list of unmatched letters, one after another. Consecutive letters in the list will share a single vertex. The list should form a loop: If you start with an

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Chapter 11 Acme Encounters the Fourth Dimension unmatched letter and move along the list, eventually you wind up back at the initial unmatched letter. Here’s a picture. 3 C

D

2 4 B

A 1

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 Symbol A. 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 appears in 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 the other positions for vertex 1. Eventually, you’ll arrive at another, unique unmatched symbol B with its initial vertex 1. (It might be that B = A. But that’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. The unmatched letter will appear in the symbol as ...B2Y... or ...Y2B−1.... The vertex “2’’ is the terminal vertex of B and can’t be the same as “1.’’ Follow step I with A1... replaced by B2.... Keep repeating Step A until you duplicate an unmatched letter. (This has got to happen since the number of unmatched letters 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 like this would happen: A1B2C3D4E5F6C That would mean that starting with C−16, and using Joe’s method, you’d arrive 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!

Scene 3

235

BRANDON: That’s impressive. You can count the lakes just from the symbol itself. But how do you know that the number you get will be the same if you have another, equivalent symbol? TIFFANY: Well, I thought of that. The way to really change a symbol/pattern is to do a cut-and-paste operation. The basic one goes like this. 1 D

E

2 C

2

B

1

1 D

D

E

2

2

C

A

B

1

1

2 A

C

E D

1 2

What happens is that vertices of both patterns are the same, and the edges with unmatched letters are the same. Moreover, for each unmatched letter, the initial and terminal vertices are the same. So if one pattern has ...A1B..., then the terminal vertex of A is 1 and the initial vertex of B is 1. The same will be for the other pattern, so the other pattern also has ...A1B.... The unmatched letters will line up to form lakes in exactly the same way for both patterns. So the number of lakes will be the same for both patterns. (She writes.)

Theorem Relating Number of Lakes and Symbols If a surface has two symbols S and S′, then the number of lakes counted for S equals the number of lakes counted for S′. JOE: Hey, that does it! Everything on our list is really different! MILLIE: This has been hard work. Haarrrddd work! We deserve another big theorem. (She writes.)

Theorem about Same Surfaces Two 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

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Chapter 11 Acme Encounters the Fourth Dimension Whew! Time for a little nap.

The group, dazed, does a few high fives and a little dance before leaving the Acme shop for the day. Sounds of “nite’’ and “see ya’’ are heard. Music comes up. Lights fade.

6. Your Turn Consider surfaces that have the following symbols: AXBYCWXDW−1Y AXBYCZ−1DYEZFUGXHU Four 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 Turn Tiffany showed that different symbols for the same surface should yield the same number of lakes. Her argument involved a cut-and-paste operation in which the cut went from one vertex of the pattern to another. She did not consider the possibility of the cut going from the midpoint of one edge to another vertex or even to the midpoint of another edge. Her argument might be complete if she had shown that 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 about the 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?

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BRANDON: After our hike, I started thinking again about those surfaces we can’t really put together. MILLIE: You mean the Klein bottle and the cross-cap? That’s had your knickers 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 involves putting 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 at surfaces. Think of the “problem’’ with the Klein bottle. Here’s the way we usually put it together.

The surface crosses itself along a circle. Every point on the circle represents two points on the surface: one for each of the two “branches’’ of the surface that meet there. That doesn’t happen with the sphere or the torus or a bunch of other surfaces. I got my idea for getting around the problem with the Klein bottle by thinking about a figure-eight path in the plane:

In this case, it’s the path that crosses itself. If you take a trip along the path from beginning to end, there’s that one point that really represents two instances on the trip. If you’re a train, you can’t turn corners so there’s really only one trip you can take. So how can you avoid that double point? It might even be that the train is long enough so that the double point would make the train crash into itself!

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Chapter 11 Acme Encounters the Fourth Dimension MILLIE: Well, duh. Build a bridge!

BRANDON: Exactly. But you need to go out of the plane to do that. You need three dimensions.

JOE: So your idea with the Klein bottle is to “build a bridge’’ where the 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 “build a bridge’’ in four dimensions. BRANDON: Thought you’d never ask! To show you how to build a bridge in four-space, I need to go back and show you a way to build the bridge in three-space for the figure-eight path. Very carefully. Also, I’m going to use coordinates to tell you what I mean by three-space and four-space. Three-space is the set of points of the form (x, y, z), where x, y, and z are real numbers. The values x, y, and z usually represent the distance of a point along perpendicular axes to some fixed point called the origin. Four-space is the set of objects of the form (x, y, z, w), where x, y, z, and w are real numbers. JOE: What’s “w’’? Is that time? BRANDON: Not necessarily. Depends on your interpretation. I just want you to think of four-space as something just like three-space, except that every “point’’ has four coordinates rather than three. MILLIE: Does it exist? BRANDON: Think of it as something we’re creating. Back to the threedimensional 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 are of the form (x, y, 0). Align the figure so that the part to be replaced by the bridge lies on the x-axis between −1 and +1. Make the crossing point be the origin (0, 0, 0).

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239

y

–1

0

+1

x

Here are two views of the part with a third axis, the z-axis, showing. The right-hand one is a view head on. y

z y

x

x

z

I’m going to build a bridge in the x-z plane. It might look something like this. z

–1

+1

x

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]

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Chapter 11 Acme Encounters the Fourth Dimension z

f(x)

–1

x

+1

An example of a function that satisfies these conditions is f(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 the bridge—the set {(x, 0, 0): −1 < x <1}—into three-space that satisfies these 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 bridge hooks up with the rest of the path at both ends. The function F(x, 0, 0) = (x, 0, f(x)) works. The replacement path is actually the graph of f(x).

z

y –1

+1

x

Scene 4

241

Here’s what the figure-eight looks like with the bridge replacement. z y x

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 close to 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 the set-up.

Cylinder

We’ll think of the Klein bottle as we have it now (with problem circle) 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 bottle and stretch/shrink it so that the part to be replaced is a cylinder and 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 ends of the cylinder at x = −1 and x = +1, and make the circular ends of the cylinder have radius 1. That means that the cylinder is the set C = {(x, y, z, 0): y2 + z2 = 1, −1 ≤ x ≤ 1}.

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Chapter 11 Acme Encounters the Fourth Dimension z

z x

Cylinder

–1

0

+1

x

To construct the bridge, I want a function G from C to four-space such 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-space are the two circular ends of C.

This time the range of G will be the replacement part, the bent cylinder-bridge in four-space. The third condition makes sure the bridge hooks up with the rest of the Klein bottle at both ends. 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 this replacement part, the Klein bottle is all hooked up in fourspace–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 let time 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 Klein bottle 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!

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243

MILLIE: Then, if 0 < t < 1, you’ll get two values of x on either side of 0. You’ll see two circles! OK. Here’s what you see in the whole movie. Freeze everything at t = 0. You see the Klein bottle minus the cylinder. Then let time roll, slowly. The Klein bottle minus the cylinder disappears instantly, and you see just two circles, starting at the ends of the cylinder and moving slowly to the middle. When the two circles meet in the middle, that’s t = 1 and everything disappears. End of movie! Pretty dramatic. JOE: Wow. TIFFANY: Brandon, it’s awesome what you’ve done. I think I have another way 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 bridge in the “other direction.’’ Let me explain what I mean. Put the Klein bottle 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 Klein bottle you need to fill in the lake. Arrange KBWL in three-space so that the edge of the lake is the circle {(x, y, 0): x2 + y2 = 1}. edge of lake

KBWL

The interior of the circle is the disc {(x, y, 0): x2 + y2 < 1}. The easy way out would be to plug the lake with that disc. But it intersects KBWL, and we want to avoid that. Something else we could plug the 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). This has the same problem: it intersects KBWL. Here’s my way out. Put KBWL in four-space via (x, y, z) → (x, y, z, 0). The edge of the lake is then the circle {(x, y, 0, 0): x2 + y2 = 1}. The hemisphere that creates 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’t intersect 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

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Chapter 11 Acme Encounters the Fourth Dimension the edge of the lake. The last property means that the range of K (the “hemisphere’’ in four-space) is glued to the edge of the 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 be 1 when x = y = 0—for the “center’’ of the lake. And it can be any value 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 time rolling. Immediately, KBWL will disappear, and you’ll see a circle concentric 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.

Investigations, Questions, Puzzles, and More 1. Investigation Unfinished business about Tiffany’s vertex-labeling scheme: ●

The method involves generating a sequence of pairs of adjacent letters. How do you know, when a sequence of pairs repeats, that it first does so with the initial pair? Why not earlier?

●

How do know the labeling is consistent? Why can’t two different sequences of pairs share a pair in common?

2. Investigation Unfinished business about Tiffany’s lake algorithm: Could the method result in shore lines that have vertices or edges in common?

3. Question Here 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.

Investigations, Questions, Puzzles, and More ●

Surface U has three lakes, is nonorientable and N(U) = −10

●

Surface V is orientable with N(V) = −8.

245

4. Investigation Here 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 you decide?)

●

What is the Euler number of each one? (How do you calculate N(S) for each of the surfaces?)

●

Use the information you have to figure out what each surface is in the classification scheme.

5. Investigation Joe’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 the pattern using Tiffany’s algorithm. N(S) = V − L + 1. Does this method work? Why?

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Chapter 11 Acme Encounters the Fourth Dimension

6. Question Pictures of several surfaces appear below. What is each one in the scheme?

This tube is “hidden” on the inside

Investigations, Questions, Puzzles, and More

247

These are hidden tubes on the inside.

7. Investigation On the sphere and torus, we have investigated maps of the following form. Every country 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?

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Chapter 11 Acme Encounters the Fourth Dimension

8. Investigation This 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. Investigation Here is another way to “assemble’’ the Klein bottle. A

A

B

B2

B1

B1

B2

B

A

A

B2

B1 A A

B1

B2

Surface passes through itself to glue along A. Put a twist in this "Figure-8", double-tube and glue.

Similar to the usual version, the surface passes through itself: some points in three-space that represent two points on the Klein bottle—in fact a whole circle of points. Use this figure-eight Klein bottle to create a Klein bottle in four-space without “double-points.’’

10. Investigation You know that you can think of the Klein bottle as two Möbius bands glued together along their edges. Use this to find yet another way of assembling the Klein 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 so that its edge lies in a plane.)

11. Investigation Similar 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.’’

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249

Show how you might assemble a cross-cap in four-space so that these difficulties are avoided.

12. Question A certain surface has the pattern shown below. Use several methods to figure out what it is in our classification scheme. One method might be to use the surface symbol and the rules of Chapter 10. Another method might be to use the results of this chapter together with Joe’s method for calculating N(S) (see Investigation 5). C E

D B

A

A

C

F

B F

E D

13. Summarizing Write a summary of the Acme team’s activities for the day. Include statements of problems, questions, definitions, results, explanations, relevant diagrams and pictures, and directions for future investigations. Make the summary so clear, organized, and understandable 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 no duplications. 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, and orientation—are what is sometimes called a complete set of invariants.) Our argument that number of lakes is an invariant of a surface is basically the same as that of James (see also Blackett, pp. 76–79).

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Chapter 11 Acme Encounters the Fourth Dimension

Assembling the Klein bottle in four-space leads to several questions. We’ve defined 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 the countries, flatten them out, and stretch/shrink them into discs? Whether you embrace 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 second is 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 these in the following paragraphs. In the middle of the 19th century, while mathematicians were beginning to accept the idea of the existence of non-Euclidean geometries, geometers began to explore the idea of n-dimensional geometry. One approach to this was to look at the set Rn of all n-tuples (x1, x2, ... , xn) of real numbers xi which generalizes the plane 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 with how they can be described in R2 (the Cartesian manifestation of the Euclidean plane) and R3 (the Cartesian manifestation of three-space). And they sought to prove 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 to be the set of points (x1, x2, ... , xn) such that Σxi2 = 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 they assembled the Klein bottle in four-space. For more details on the historical development 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 definition that two objects (maps, surfaces) are the same if one could be stretched/shrunk into the other without tearing. We broadened this and said that we could call two surfaces the same if a pattern for one could be stretched/shrunk into a pattern for the other. In 1863 Möbius made this definition precise (and extended the range of the definition) by saying that two figures are the “same’’ if there is a one-to-one correspondence between the two figures such that neighboring points correspond to neighboring points (see Kline p. 1164f and Pont p. 90). The modern language for this is the following. Two figures F and G are topologically equivalent if and only if there is a function f: FÆ G that is 1-1, onto, continuous, and such that the inverse function f −1 is also continuous. Such a function f is called a homeomorphism and F and G are called homeomorphic (as well as topologically equivalent). The functions that Brandon and Tiffany created are homeomorphisms. The third development occurs when, in 1895, Poincaré makes more precise the definition of a surface. First, some notions he uses in his definition: the n-disc is the set of all points (x1, x2, ... , xn) in Rn such that Σxi2 < 1; a neighborhood of a point x0 in F is the set of all points in F—a subset of Rm for some m—with a

Notes

251

distance to x0 that is less than ε, for some positive number ε. Armed with this terminology, Poincaré defines a surface (without boundary) to be a bounded subset F of Rm such that every point x in F has a neighborhood that is homeomorphic 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 homeomorphic 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 Klein bottle in four-space as a surface. We reconcile this definition with our earlier one by noting that you can think of a two-manifold as a bounded set in Rm made up of the elements of a map, except that this time an edge is homeomorphic to the unit 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 definition 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 incorporating and generalizing the properties of neighborhoods of a subset of Rm as defined above), then T would be called a topological space. By use of this idea, here is how one 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. A

B

B

A

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 of points within the rectangle itself plus neighborhoods that capture the peculiar gluing that’s supposed to occur to make the Klein bottle. Here are some examples. A

B

A

B

A

B

A

B

A

B

B

A

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Chapter 11 Acme Encounters the Fourth Dimension

If you “close up’’ this collection using the set operations of finite intersection and arbitrary union, then you get a larger collection of subsets satisfying the axioms of neighborhoods of a topological space. This makes K a topological space. It is an abstract definition of a Klein bottle. This chapter and the last were devoted to the successful classification of all surfaces. In the language of Poincaré, this is a classification of all two-manifolds (with and without boundary). An important question that we have mentioned before is “What are the possible (three-) spaces that we live in?’’ A possible space would be a three-manifold, and an associated mathematical problem would be to classify all three-manifolds. (As an introduction to this problem, see the book by Weeks.) This problem has not been solved. Solving it is the object of a lot of current research in mathematics. A lot of work on the problem has already been done (see the book by Thurston). Lots of folk have written about four-space in more informal ways than we have above. Of course, Flatland, the 1884 book by Abbott, in which a three-dimensional creature visits a two-dimensional world, is the classic introduction to thinking beyond the third dimension. A particularly good book that explores four-space from 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 the four-dimensional analogue to the three-dimensional cube. Other books that explore four-dimensional thinking, especially through the use of analogy, are those by Weeks, 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 on Surfaces: Heawood's Estimate

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12

A couple of days later. Tiffany, Brandon, and Joe are at their desks working on reports for Boss. Millie is standing at the window, painting her nails. MILLIE: Ohmagosh! Here comes Pinkley Smith. Wonder what he’s up 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 done much 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. 253

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Chapter 12 Acme Colors Maps on Surfaces: Heawood’s Estimate 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 fed us last time?

PINKLEY SMITH: Friends, friends, friends! Have mercy. Just a little slip and you 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 titillate your tummies. TIFFANY: Toes? Tummies? PINKLEY SMITH: I heard that you made a major breakthrough by classifying all surfaces. So I thought, “How could I, humble Pinkley Smith, make a contribution, however meager, to this new knowledge created by my fellow topologists?” JOE: Cut to the chase, Pink. PINKLEY SMITH: Pink? I thought, “There’s a theorem about coloring maps on a sphere: you only need four colors. How about all these other surfaces? How many colors does a map on another surface need?” So, I came to tell you the answer to this question. TIFFANY: Hmm. Good question. I think we’d really like to know the answer. PINKLEY SMITH: OK. Here it is.

Map Color Theorem Suppose that S is a surface without boundary (i.e., no lakes) and that N(S) is its Euler number. Then any map on S will need no more than 7 + 49 − 24 N(S) 2 colors in order for it to be colored properly.

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?

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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 of colors needed to color maps on S′ compared with the number of colors needed to color maps on S? PINKLEY SMITH: Take a map M on surface S. The first thing we want to do is to replace the problem of coloring the countries of M with the problem of coloring the vertices of the dual graph. JOE: Dual graph? PINKLEY SMITH: Yes. Coloring the vertices of a graph is easier. Remember that a graph is a collection of vertices and edges. Each edge joins a pair of vertices; each pair of vertices is joined by at most one edge. The rule for coloring the vertices of a graph properly is that two vertices joined by an edge must be colored different colors. Here’s how you create the dual graph to a map. Take the map and provide each country with a capital, a point in its interior. The capitals are the vertices of the dual graph. Join two capitals (vertices) with an edge exactly when the two countries share a common border. A

B VA

MAP

VB

VC

VA

DUAL GRAPH

VB

VC

C

TIFFANY: I think I understand. Coloring the vertices of the dual graph is the same as coloring the countries of the original map. The dual graph and the map need exactly the same number 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.”

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Chapter 12 Acme Colors Maps on Surfaces: Heawood’s Estimate We won’t have to worry about that with the graph. Here’s a simple example of one difference. If two countries—A and B—on the map have more than one common border, then the dual graph has the corresponding vertices vA and vB but only one edge joining them.

VA

VB

A

B

Another thing: a graph doesn’t have to be the dual graph of some map in order to color it. You’ll see in what follows that 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 some notation and a definition.

Chromatic Numbers and Critical Graphs The number of colors c(M) needed to color map M properly is called the chromatic number of M. If S is a surface, then c(S), the chromatic number of S, is the maximum of all c(M) where M is a map on S. The number of colors c(G) needed to color graph G properly is called the chromatic 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.

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Now start with a map M on surface S. Construct the dual graph G. If necessary, remove vertices and edges from G to get yet 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, but it has the same chromatic number as the dual graph. Moreover, it’s critical. Here’s a reason why a critical graph is useful:

Lemma about Critical Graphs Lemma. If graph K is critical with chromatic number c(K), then every vertex of K has order at least c(K) − 1. Proof: This is going to be a proof by contradiction. So suppose that the lemma isn’t true. Then there must be a vertex v whose order is less than c(K) − 1. From the graph K remove v and all the edges joined to it. Obtain graph K′. Because K is critical, K′ can be colored in c(K) − 1 or fewer colors. Color K′ properly. Now put the vertex v you removed back along with all the removed edges. Color K using the coloring you’ve already started by coloring K′. All you need to do is color 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 the c(K) − 1 colors with which to color v. Do so! This means that K is now colored with c(K) − 1 colors. This contradicts the fact that K needs c(K) colors for a proper coloring. We were led to this contradiction by the assumption that K has a vertex of order less than c(K) − 1. So it must be the case that every vertex of K has order greater than or equal to c(K) − 1. This proves the lemma. BRANDON: OK. Critical. That’s a nice property for a graph. But why is it good for us? PINKLEY SMITH: Glad you asked! We want to know how many colors to color all maps on a surface S. Start with a map M that requires the maximum number of colors for the surface S. Take the dual graph G, and remove edges and vertices to get a critical graph G′ with c(M) = c(S) = c(G) = c(G′). The critical graph property will enable us to connect c(G′) with the Euler number N(S) in a useful way. The first step in making this connection is: If every vertex of G′ has order ≥ c(G′) − 1, then the average order of vertices of G′ must also be ≥ c(G′) − 1.

Critical Graph and the Average Order of Vertices Suppose 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.

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Chapter 12 Acme Colors Maps on Surfaces: Heawood’s Estimate BRANDON: I still don’t…

PINKLEY SMITH: Hold on. There’s more! Let Vi denote the number of vertices of G′ of order i and let V denote the total number of vertices of G′. Then A(G′) = 2V2+3V3+4v4+… V MILLIE: Hey, there’s something familiar with that equation! The numerator counts edges—sort of. In fact, it’s 2E, where E is the number of edges of G′. So you’d get A(G′) = 2E/V ≥ c(G′) − 1 or just 2E/V ≥ c(G′) − 1.

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 of an 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 the edges of G′. What do you get? JOE: If G′ were the network of a map, you’d get a bunch of countries. I guess what you’d get in this case would be a bunch of pieces of the surface. Hey, each piece would be some kind of surface with one lake. The edges of the lake would be 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, this gives 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’d count 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 of each piece. Add them up. You’ll get 2E. Now I claim that the lake on each piece will have at least three edges from G′.

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Otherwise, if there were just one or two edges from G′ you’d have something like this in G′.

Neither can happen because G′ is critical. Consequently, 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′. BRANDON: That looks a lot like something we got when we were coloring maps on the sphere. Back then we got 3C ≤ 2E. I think we used that to prove the five-color theorem. PINKLEY SMITH: Yes, so we did! And we’ll use this new inequality to prove another coloring theorem. TIFFANY: And is there going to be some kind of Euler’s formula involving V, E, and P? PINKLEY SMITH: You fellows are sharp as tacks. That’s just what I want to do next: find a relationship between V − E + P and N(S). Here’s how you do it. To the graph G′ on S, add vertices and edges so that the surface pieces have been cut up into the countries of an OK map. At each stage let’s keep track of the number of vertices V′, edges E′, surfaces pieces P′ you’d get, as well as V′ − 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.

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Chapter 12 Acme Colors Maps on Surfaces: Heawood’s Estimate 3. Add an edge to a single vertex.

In case 1, the number of vertices and the number of edges each increases by 1 and the number of surface pieces stays the same. Result: V′ − E′ + P′ doesn’t change. In cases 2 and 3, the number of vertices stays the same, the number of edges increases by one and the number of surface 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, N(S) ≤ V − E + P 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. What we know about G′ on surface S: 3P ≤ 2E N(S) ≤ V − E + P c(S) − 1 ≤ 2E/V where G′ has V vertices, E edges, and—when cut along—P surfaces pieces. Now for a little calculating.

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.

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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

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 out why 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.

Starting with the following inequalities, c(S) − 1 ≤ 6 − 6N(S)/V c(S) ≤ V N(S) ≤ 0 we 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)!

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Chapter 12 Acme Colors Maps on Surfaces: Heawood’s Estimate

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-hand factor 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).

PINKLEY SMITH: That does it! That proves the map color theorem! (Looks at watch.) 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 this case. Go back to the inequality c(S) − 1 ≤ 6 − 6N(S)/V in the argument given by Pinkley Smith. This inequality is true no matter what N(S) is. But in 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 for this case? Does the map color theorem provide a new proof of the fourcolor problem for the sphere?

Investigations, Questions, Puzzles and More

263

2. Investigation Suppose that you can solve the n-cities problem on a p-holed doughnut. What can you say about the relationship between n and p? (First try the case in which all the faces are triangles. Then consider the case when there is no restriction on the faces.)

3. Question You are interested in coloring maps on a 17-holed doughnut. You want to have enough colors on hand to color all possible maps. But, of course, you do not want to have more colors than you need. What is your best estimate for the number of colors you should have?

4. Investigation You want to find a surface on which you can solve the 10-cities problem. Find a specific surface, and show how you would solve it there. You’d also like to know the smallest n such the 10-cities problem could be solved on an n-holed doughnut. Use the Euler number for an n-holed doughnut and counting arguments to find out what might be possible. How does what you get jibe with what the map color theorem tells you?

5. Investigation We’ve solved the five-cities problem on the Möbius strip (see Investigation 13 of Chapter 7). What about the six-cities problem?

6. Investigation If 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 is even, then the map can be colored in two colors. Is this true for maps on a p-holed doughnut (p > 0)? If this is not true, what can you say?

7. Investigation Consider this statement about a map on a surface: If every vertex of a map is of order 3 and the map can be colored in 3 colors, then every country must have an even number of edges. We know that this is true for maps on a sphere. What about maps on a p-holed doughnut (p > 0)? We know that the converse of the statement

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Chapter 12 Acme Colors Maps on Surfaces: Heawood’s Estimate

is also true for maps on the sphere (see Investigation 2 of Chapter 4). Is it true for maps on a p-holed doughnut (p > 0)? If not, what can you say?

8. Investigation Suppose you have a map on a surface with the following property: for every pair of neighboring countries at least one is a triangle. Someone said that such a map can be colored in four colors. Is this an old wives’ tale, or what?

9. Investigation a. 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. Question Given a surface, can you find a graph that you cannot draw on it without having the edges cross?

11. Question Suppose graph G with V vertices and E edges has c(G) = 2. What can you say about the relationship between V and E?

12. Investigation On a surface, a map with a sufficient quantity of small countries (small, in the sense of number of borders) can be colored in seven colors. What does “sufficient quantity” mean? Is it true?

13. Investigation Could you solve the six-cities problem on the Klein bottle? What does this say about c(Klein)?

14. Construction In a paper of 1910, the mathematician Heinrich Tietze describes the following construction of a polyhedral surface (see Biggs et al., pp. 128–129).

Investigations, Questions, Puzzles and More

265

“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 plane with any of the five triangles....The said polyhedron consists of the 10 triangles: 1. ABC, BCD, CDE, DEA, EAB 2. FAC, FBC, FCE, FDA, FEB.” What is this surface? What can you say about the map whose countries are the 10 triangles above?

15. Construction Here is a Möbius band to assemble. It has a map with six countries. (The countries are labeled.) How many colors do you need to color the map properly?

5 3 6

6 1 4

2 5

16. Construction The object of this construction is to build a paper model of a torus having a map on 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.”

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Chapter 12 Acme Colors Maps on Surfaces: Heawood’s Estimate

●

Complete the model by starting at one end of the chain and moving to the other gluing as you go.

●

The thick lines divide the torus model up into the countries of a map. Color the map properly. GLUE GLUE

GL UE GL UE GLUE A

UE

GL UE

GL

17. Question By 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, could you bore a hole through it and construct a map with eight countries, each of which is adjacent to the other seven? (The model in 14 might be a place to start.)

18. Summarizing Write a summary of the Acme team’s activities for the day. Include statements of problems, questions, definitions, results, explanations, relevant diagrams and pictures, and directions for future investigations. Make the summary so clear, organized, and understandable that Boss will be convinced that the time spent was worthwhile.

Notes

267

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 which he found a counterexample to Kempe’s earlier “proof ” of the four-color conjecture. This was also the same paper in which he proved that every map on the sphere could be colored in five colors or less (see Notes in Chapters 4 and 5). Since 1/2(7 + (49 − 24N(S)1/2) is not always an integer but c(S) is, the inequality is often written c(S) ≤

7 + 49 − 24 N(S) 2

⎤ ⎡ where [x] denotes the greatest integer in x. The number ⎢ 7 + 49 − 24 N(S) ⎥ is 2 ⎥⎦ ⎢⎣ often called Heawood’s number and is denoted h(S). In the same paper, Heawood also claimed that the inequality is actually an equality. He stated that in order to prove that c(S) = h(S), one must construct a map on S that requires h(S) colors. Here is the map he constructed for the torus.

2 1

6

5 4

2

7

1

4

3

5 3

6 7

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 pointed out a year later in a paper by Lothar Heffter (see Biggs et al., p. 118+). The equation c(S) = h(S) became known as Heawood’s conjecture. Heffter’s work was the beginning of 70 years’ effort to prove the conjecture. Heffter’s method of attack on the problem was to construct maps on surfaces in which each face is adjacent to every other one, what he called a system of neighboring regions. (In the case in which the map is a system of n-neighboring regions

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on a surface, we recognize that the dual graph is a solution to the n-cities problem on the surface.) In one part of his paper, Heffter asked for the least number g such that n neighboring regions may be constructed on a g-holed doughnut. He called this number g(n). Using Euler’s formula and some counting arguments, he showed that g(n) > {(n − 3)(n − 4)/12}. Here {x} is the least integer not less than x (see Investigation 2 above). Heffter realized that trying to draw a system of n neighboring regions on a surface was cumbersome as was drawing the solution to n-cities problem. He devised a clever, new approach; he constructed the adjacency pattern for a map. Here is how he did it. Number each country of the map from one to n. Take a country and walk around the borders of the country in a counterclockwise direction, and record, in order, the bordering countries you encounter on your trip. For example,

7

6

5

3

1

4

2

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 will arrive at the array that follows the map. 2

1

4

3

Country 1 2 3 4

Record of Bordering Countries 2 1 4 3

4 3 2 1

3 4 1 2

Notes

269

If you did this for the map on the torus below (a system of seven neighboring countries), you would get the array that follows it. 1 6

5

4

3

2 6

1

1

6 7

2 4

7 3

6

5 7

7

Country

Record of Bordering Countries

1 2 3 4 5 6 7

2 3 4 5 6 7 1

4 5 6 7 1 2 3

3 4 5 6 7 1 2

7 1 2 3 4 5 6

5 6 7 1 2 3 4

6 7 1 2 3 4 5

Heffter observed that this array exhibits an amazing pattern: you can obtain every row 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 labels for 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 create more arrays that corresponded to systems of n neighboring countries on surfaces of genus (n − 3)( n − 4)/12. He assumed that n leaves a remainder of 7 on division by 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 map requiring h(g) colors, then the h(g)-cities problem could also be solved on the surface. This confirmed that Heffter’s approach—of trying to solve the h(g)-cities problem on a g-holed doughnut—was the right one. In a program that began around the time of Dirac’s result, Ringel and Youngs and others had by 1968 constructed arrays (à la Heffter) for all cases, thus proving the Heawood conjecture. A summary of the events leading up to the final proof

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can be found in the article by Stahl. Details of the proof can be found in the book by 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 the sphere, is that it came eight years before the proof of the four-color conjecture, the coloring theorem for maps on the sphere. In coloring maps on surfaces, neither Heawood nor Heffter considered nonorientable surfaces. However, in 1910, Tietze made the appropriate Heawood conjecture for nonorientable surfaces. The formula is the one proved in this chapter. Previous work of Möebius confirmed the conjecture for the cross-cap (see Investigation 15 above). In 1934 Franklin showed that, even though h(Klein) = 7, in fact it was the case that c(Klein) = 6. In 1954, Ringel showed that the Heawood conjecture is true for all nonorientable surfaces (except the Klein bottle) by producing Heffter-style arrays for the required systems of neighboring regions. It turned out that nonorientable surfaces were easier than the orientable ones (for some details of the history 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 Up with Knots

CHAPTER

13

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 something about ladders. Möbius ladders! Doesn’t that just beat the cake? Who do 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 to the others.) Hey, guys, we’re famous! (Turns back to letter.) They think we might be able to help them. (Passes letter to the others.)

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Chapter 13 Acme Gets All Tied Up with Knots BOSS: Bosh. Ladders. I don’t want you people going on another wild goose 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 just joshing. You folks take care of it. (Returns letter to group.) Come on, 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 we can have a lot of fun with this project. If I understand things correctly, the folks at Molecule.com want us to look at a molecule that looks something like this.

The O’s stand for oxygen atoms; the black circles stand for carbon molecules. So there’s a single chain of...hmm...60 oxygen and carbon atoms hanging together by single bonds. Then there’s some carbon-carbon double bonds represented by the double lines. It’s like a Möbius band, where the edge is the chain and double bonds replace the surface. The double bonds are like the 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.

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BRANDON: What does that mean? TIFFANY: (Hands letter to Brandon.) I’m trying to sort this out. I think it’s like when we had a pattern for a Möbius band. You could really assemble the pattern in a lot of different ways. The pattern is really a pattern for a lot of things: a strip with a single half-twist or a strip with three half-twists or, in general, a strip with an odd number of half-twists. Well, it’s the same with the Möbius ladder molecule. Break some of the bonds and flatten it out, and you’d get something like this. B

A

A

B

Of course, the real molecule is what you get when you glue the ends together. But we know you can get lots of things—a single half-twist ladder, a three-half-twist ladder—anything with an odd number of half twists. Chemically, all the bonds are the same. So all of those are the same molecule. Sort of. BRANDON: They’re the same molecule. But a half-twist ladder might have different physical properties from a three-half twist ladder: one might 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 to figure out which pairs of these molecules are really different, meaning you can’t manipulate one into the other in three-space without 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 we did 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 single pattern 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 manipulated it in three-space—no cutting—and you managed to get

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Chapter 13 Acme Gets All Tied Up with Knots another ladder molecule, then the same would be true of the edge: you could take the edge of one, manipulate it, and get the edge of the other. So here’s the idea: figure out some way to classify the edges. If two edges are different, then the ladders are different. Hmm. If two edges are the same, well, who knows what you 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 is just a simple closed curve in space. I think it’s time to look at some examples. MILLIE: Good idea. You know, we can just look at the edges of oddtwisted strips. They’re really the same as the edges of the molecular ladders. Here are the edges of a one (half-)twisted strip and a three (half-)twisted strip.

JOE: Looks like the edge of the one-twist strip is just a circle. And the edge 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 simple closed curve in space a knot. The edge of an odd-twisted strip is a knot! JOE: When I tie my shoes, I make an overhand knot like this.

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Is it the same as yours, Millie? MILLIE: Looks the same. JOE: Know what? You can do twisted strips in different ways. You can twist one way, then glue. Or you could twist another way, then glue. Like this.

One half twist

Three half twists

JOE: Are the edges the same?

One half twist Three half twists

Guess what? Those are the two overhand knots we got before. MILLIE: They still look alike.

Your Turn 1 Use rope to make the two overhand knots. (Be sure to secure the ends.) See if you can manipulate one into the other. (No cutting!) What happens?

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Your Turn 2 Here 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 ropes to make the two knots. See if you can manipulate one into the other. What happens? 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 of each other. Question is: Are they the same? Can you manipulate one into the other? Or can you show you that can’t? TIFFANY: Time for some simplification. You could turn one of those knots into a real mess. Let’s see if we can keep things clean. Those pictures we had before are pretty neat. Sort of flattened out pictures of knots, or projections of knots onto a plane. Seems to me we could keep track of all the manipulations with these pictures. Let’s see what we have. If you held a knot above a plane with a light above it, you’d see a shadow on the plane—something like this.

MILLIE: Looks like a nice closed curve in the plane!

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TIFFANY: Absolutely! But a place where the projection curve crosses itself really represents two points on the original knot. Let’s call such a point a crossing. At a crossing, the piece of the closed curve nearest the light source is an overpass; and the piece farthest from the light source is an underpass. We draw the overpass in one stroke and the underpass in two. The underpass is broken to indicate that the location of the point of the original knot is hidden by the projecting light. So a crossing of the projection curve could be drawn in one of two ways: UNDERPASS

UNDERPASS

OR

CROSSING OVERPASS UNDERPASS

UNDERPASS OVERPASS

MILLIE: It’s as if the knot really lies in a single plane except at the crossings 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 the picture have? What does a picture of a granny knot look like? How many crossings does it have?

Your Turn 5 What does a picture of the edge of a five-half-twist strip look like? How many crossings does it have? BRANDON: OK. This is a good start. A picture of a knot is sort of like a pattern for a surface. Of course, unlike a pattern, we can’t “cut” the picture and 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 do with the knot in space, but very, very slowly so we still have a picture.

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BRANDON: Sort of like little baby steps? JOE: Little baby steps like these?

OR

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 to undo each one of those. TIFFANY: Yes. Good idea. We ought to include the “inverses” of each of those “moves.” Here’s a list of the moves we’re allowing. (1)

(1)–1

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(2)

(2)–1

(3)

(3)–1

(4)

(4)–1

(5)

(5)–1

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 that we know are pictures of the same knot. Here they are.

Then I tried to see if we could use just our moves to change one picture into the other. Here’s what I got.

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(4)–1

(3)

(5)

(2)

(3)–1, (1)–1

(5)

(1), (2)

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 those moves 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’s question 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 as you can, you don’t arrive at the second picture. Are they different knots? MILLIE: Seems like it. TIFFANY: Here’s an analogy. You have patterns for two surfaces. You don’t know whether or not they’re the same. You cut and reglue and cut and reglue, but somehow you can’t turn the first pattern into the 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 never find it.

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JOE: But we do have a method to find out if two surfaces are the same. 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 three things 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. Like orientability or the Euler number. More to think about. Team members go back to their desks. Lights fade. Music comes up.

Your Turn 6 Here 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, the two knots would be different. Crude, but a start. TIFFANY: A natural thing to start with. Problem is, a knot has got a lot of pictures. One picture might have one number of crossings,

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Chapter 13 Acme Gets All Tied Up with Knots another picture might have another. In fact, some of those moves 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 crossing number of the overhand knot is three. TIFFANY: Hmm. The crossing number of the circle—the un-knot, the nonknot, the not knot?—would be zero.

The not knot

JOE: I like it. The not knot! TIFFANY: Well, that one’s easy to figure out. For other knots, how do you know when you’ve got the smallest number?

Your Turn 7 The 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?

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JOE: I tried something different. We’ve been coloring maps on surfaces. I thought we could try to color the knot pictures. The idea 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 underpass and follow it until it ends at another underpass. That whole part of the picture would be a thread. You can think of a knot picture as being made up of threads.

Thread

So I figured the rule should be that at each crossing, the overpass thread and the two underpass threads should be different colors. Then I thought what would happen if you had a knot picture with all its threads colored and then transformed it using one of our moves. I started with the “pigtale” move and immediately ran into trouble.

At that new crossing, you couldn’t actually color all the threads different because the overpass thread and one of the underpass threads 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’re trying to tell knots apart.

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Chapter 13 Acme Gets All Tied Up with Knots MILLIE: Picky, picky. TIFFANY: So it looks like you’ve got two rules: (1) at each crossing either color all the threads different colors or color them all the same color, and (2) at least one crossing must use three different colors.

BRANDON: Joe, did you try coloring knots with those rules using just three colors? JOE: Yeah. It’s interesting. I found some knots you could color in three colors and some knots you couldn’t. Of course, the not knot can only be colored in one color! For example, I can color the overhand knot in three colors, but I can’t color the fivepoint-star knot in three colors.

Your Turn 8 Is Joe right? Is it true you can’t color the five-point-star knot with three colors using the rules?

Your Turn 9 Use 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 the picture or of the knot itself? MILLIE: What? BRANDON: If it’s a property of the knot, then every picture of the knot should be three-colorable. TIFFANY: So with the overhand knot, you know the picture we had is threecolorable. You’d want to show that if you alter the picture with

Scene 2

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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.

R

S

T

This move adds one crossing. Thread R becomes threads S and T 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 new picture the same color as thread R of the original picture. Color the other crossings of the new picture the same as the old picture. At the new crossing, all the threads are colored the same color. But there is some other crossing where all three colors are used. So this move is OK.

red red

red

MILLIE: Let me try one. In the second move, one thread gets pushed under (or over) another. Look at the first case. Call the two threads of the original A and B. In the new picture, A becomes A', A'', and A''', and B stays the same.

B

A

B

A'

A" A"'

Color the original in three colors. Now color the new picture this way: If B and A were originally colored the same, then color threads A', A'', and A''' the same as A was colored. Color all the other threads the same as they were in the original. At the two new crossings, all the threads are colored the same. But there is another crossing where the colors are different, because they

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Chapter 13 Acme Gets All Tied Up with Knots were different in the new one. So the new picture is threecolorable, too! If A and B were originally colored different colors, then do the following and everything will be OK.

red red

blue

blue

yellow blue

BRANDON: I want in this, too! Hmm. This third move looks a little trickier.

B

B

A F

C'

C A F D

E

D

E

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 thing to do would be to keep the coloring of A, B, D, E, and F of the new the same as the coloring of the original. Let’s try that. There’re lots of possibilities. Let’s try a couple.

Scene 3 q

287

q

p

q

r ?

p r q

q

p

p

p

p

p

r

r ?

p r q r

q r

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 can assign 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 10 Help the team with the other possible colorings of the original. That would take care of move 5. But there are other moves to consider, the “inverse” moves to the ones we’ve looked at. You want to show that if a knot picture is three-colorable, then changing 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?

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Chapter 13 Acme Gets All Tied Up with Knots JOE: I think we’ve shown that if a knot picture is three-colorable and you alter it by one of our moves, then the new picture is also three-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 overhand knot you can. That means the not knot and the overhand knot are different.

BRANDON: Yes! Our first method for telling knots apart. JOE: OK. Here are some questions. The overhand knot and the square knot are both three-colorable. Are they the same? Neither the figure-eight knot nor the not knot are three-colorable. Are they the same? TIFFANY: No and no. At least I think so. But so far, we can’t really say. All we can do so far is tell things apart; we can’t say yet when they’re the same. A problem Molecule.com has is trying to distinguish the left-hand overhead knot with the right-hand one (or show they’re the same). They’re both three-colorable. You can’t conclude they’re different, but you can’t conclude they’re the same either. BRANDON: While you were all working on showing that three-colorability was preserved when you make a “move,” I was trying to come up with a generalization of three-colorability to n-colorability for bigger values of n. I was trying to think of some rules that would work. For example, I thought maybe you could color every thread a different color. Then if a picture had 10 threads, you’d color it with 10 colors. But then if you applied move 3, that would increase the number of threads by two.

So 10-colorability wouldn’t be preserved. Also, the coloring scheme wouldn’t work if you had something like the pigtail in the picture.

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So, I looked for some kind of restriction on the coloring scheme. I started playing around and arranged the colors in a circle, like the spokes of a wheel. I noticed that if I did this with the three colors, the wheel would look like this.

p r q

The rules for three-colorable would make the color choices like this at each crossing. A,B,C

A B

A

p

C

q

p

Or

r

C

A

B

r

C

q

B

That makes coloring the overpass central: whatever the overpass thread is colored on the wheel, the two underpass threads are colored colors that are symmetrically opposite to that one. I thought, “Hey, maybe that’s the key.” For n-colorability the rules become the following: Color the threads with n colors such that, whenever an overpass is colored X and the underpasses colored Y and Z, then the colors Y and Z are mirror images of each other through the X spoke. Furthermore, you’ve got to use at least two colors on the threads.

X

Y

X

Y

Z

Z

MILLIE: Is this useful? Can you use it to distinguish knots?

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Your Turn 11 Help the team answer Millie’s question. Show that the following knot pictures are five-colorable.

JOE: Wow. Does it work? BRANDON: You mean with the moves? Well, it works fine when you use moves 1 through 4. The argument is the same as with threecolors. The hard part is with move 5. I thought of something that might simplify things. I decided to number the colors 0, 1, 2, … , n − 1 and arrange them in order around the wheel, like the figures on a clock. The advantage of this is that you can use clock arithmetic. 0 1

n–2

n–

1 2

MILLIE: Clock arithmetic? BRANDON: Yes. Addition and multiplication modulo n. Here’s how it works. Suppose an overpass and its two underpass threads are colored A, B, and C, respectively. Then the condition that B and C are mirror images through the A spoke means that the two distances A 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).

Scene 3 A

A

B

291

B

C

C

Now let’s see what that does for us with respect to move 5. B

B

A F

C'

C A F D

D

E

E

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.)

Ingredients of a Knot Picture (how to interpret a knot picture) Crossing : point of the picture that represents two points of the original knot

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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-Colorable A knot picture is n-colorable means that associated to every thread is a number from the set {0, 1, ... , n − 1} such that for every crossing, if m is the number associated with the overpass and p and q the numbers associated with the two underpass threads then 2m − p − q ⬅ 0 (mod n) m

p

q

Theorem. If knot picture K is n-colorable and knot picture K' is obtained from K by a move, then K' is also colorable.

Your Turn 12 To test the strength of n-colorability in distinguishing knots, you decide to look at the edges of odd-twisted strips. Here they are for the three-, five-, and seven-twist strips.

Scene 3

293

1. What can you say about all numbers m such that the edge of the three-twist strip is m-colorable? 2. What can you say about all numbers p such that the edge of the five-twist strip is p-colorable? 3. What can you say about all numbers q such that the edge of the seven-twist strip 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 13 You are looking the knot pictures below. You decide to use all the devices you have at your disposal to determine how many really different knots are represented by these pictures.

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Scene 4 Brandon and Tiffany are at their desks, packing up their belongings in cardboard cartons. 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 play make 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 a table? (Brandon and Joe scramble to get a table.) BRANDON: So what is going on? SANDRA: Boss said you and Tiffany were finishing your internships and going 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 you shouldn’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 to keep working on these ideas; so are Bran’ and Tiff when they go back to school. We’re going to keep in touch.

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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 calculus student who came in for help. After they had worked some problems together, the student asked, “So what kind of math do you like?” The professor replied, “Knot theory.” The student said, “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 totally awesome! (Everybody stands around eating the goodies and drinking the punch.) JOE: Here’s a question. “What’s a topologist’s excuse for not doing homework?” TIFFANY: I don’t know, what? JOE: “She locked the paper in her trunk, but a four-dimensional dog got in and ate it.” ALL: Ha! Ha! BOSS: Here’s another excuse: “She took time out to snack on a doughnut and a cup of coffee and then spent the rest of the night 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 homeomorphic 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 cross the Möbius strip?” BRANDON: This has got to be baaaad.

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Chapter 13 Acme Gets All Tied Up with Knots 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.

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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 say next.) Well, shucks. I got you a little somethin’ to remember us by. 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 call it 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 is embarrassed.)

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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 knot problems. 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-mail each other. The cool thing about math is that you’re really never done. There’s always more questions...then some answers, but more questions...then answers and more questions. It never ends. Eternal e-mailing! The above image is reprinted with permission. www.kleinbottle.com

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TIFFANY: I want to say that I’ve really learned a lot working with you all. Our different points of view shed different lights on each problem. MILLIE: It helped me to really listen and try to understand. Then I got to talk out my ideas, and you asked questions, I talked some more, you asked more questions. In the end I understood more, and ideas got clearer and clearer in my head. BRANDON: I really value the strengths each of you has. Joe, you’re really down-to-earth and concrete. You always go for working out examples. JOE: Thanks, Bran. I really appreciate what you do to make us see the big picture and keep us organized and focused on the problem. TIFFANY: And, Millie, your intuition. There you are, painting your nails, not saying a thing. Then, wham, you come up with an incredible insight. MILLIE: Then, whoosh, enter Tiff with clear definitions. Making sure the arguments click. BOSS: Myself, I’s been proud to see you guys and gals making a fuss and getting all worked up. I had the privilege of watching you spout off your ideas. I got a kick our of yer expressions 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 breathers and new stuff to think about. PINKLEY SMITH: Colleagues, colleagues! It was nothing. You are my colleagues? MILLIE: Fuss to feed the soul. Ideas to feed the brain. Food to feed the tummy. Marcella and Sandra, you fix some fine victuals! Music to Cielto Linda comes up. The group begins to dance again with allemande lefts and do-si-dos. Gradually, the party breaks up. People gather up their stuff. Lots of handshakes, high fives, and hugs. The group goes off into the night to the sounds of “see ya,” “hasta la vista,” “bye,” and “adios.”

Investigations, Questions, Puzzles, and More Investigation 1 Joe and Millie are getting familiar with the notion of n-colorability. They wonder for which numbers n ≥ 2 can they find a knot K such that K is n-colorable. Help them out.

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Investigation 2 Joe 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 of the two?)

Investigation 3 Joe and Millie have realized that, given a knot picture K, there may be several numbers 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 is the “figure-eight” knot below.

Investigation 4 Joe and Millie have discovered that if K is the trefoil knot then S(K) = {m: m is divisible by three} (see Investigation 3 for the definition of S(K)). They wonder what the sets S(K') look like for other knots K'. Here are a couple of questions they ask: Given a knot K', what can you say about the smallest number in S(K')? Is there a 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') if K' happens to be the not knot?)

Investigation 5 Millie thinks she has a method that will find her a number n such that a given knot is n-colorable. Here is how her method works. Take a knot picture, and orient the strands with arrows to show how you might take a trip along the simple closed curve that is the knot.

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Pick a crossing and label the overpass with 1; the underpasses, 0 and 2.

1 0 2

0

1 2

Follow the arrow from the labeled underpass to the next crossing. Label the strands there with A, B, C so that A − B = C − A (A is the overpass). Repeat.

9 5

0

1 2

4 3

For example, you get nine by considering

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Chapter 13 Acme Gets All Tied Up with Knots 1 5

x

then finding x so that x − 5 = 5 − 1; i.e., x = 9. When you get near the end of your trip 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 method on some of the knots in Your Turn 13. Does it work? Will the method work for any knot picture? 0

5

9 0

4 9

Investigation 6 Brandon was thinking about the kind of knot he ties and the kind of knot the people 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 without cutting the laces! But if I glue the loose ends together, I get a knot of the kind we’ve been working with.”

Brandon then looked more closely at a knot he actually ties when he wraps a package. First he does this.

Then he does this.

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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. The sum of two surfaces. Take two knots K and L.”

K

L

“Cut each one open.”

K

L

“Glue the loose ends of one knot to the loose ends of the other.”

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K#L

“That would be K#L, the sum of K and L. The package knot is the sum of these two 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 you say about K#L? Is it n-colorable or m-colorable? What else could you say? What about the other way around: if K#L is p-colorable, does that tell you anything about K or L? How does S(K#L) compare with S(K) and S(L) (see Investigation 3 for the definition of S(K))? Help Brandon out.

Investigation 7 (Continuation of Investigation 6) Brandon is still thinking about tying that knot for a package in which the final knot is really the sum of two knots. Since the sum of two knots is really like tying one knot right after the other, he wonders if one knot could “cancel” the other knot out. In other words, you first tie a knot (not the not 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), can you find another knot L so that K#L is the not knot? What do you think?

Investigation 8 Tiffany is thinking about the edges of twisted strips and knows that if the number of 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 not considered this situation; they have only dealt with one knot, one simple closed space 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 nonintersecting closed curve an n-link. She draws some examples.

2-links

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3−links

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 defines equivalence of links in this way: link L and L' are equivalent if you can get from a picture of one to a picture of the other by using a finite number of moves. She also notices 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), where A 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 independent 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.

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(This three-link is called the Borromean [Ballentine] rings. Tiffany is interested in these because every pair of closed curves in this three-link is unlinked!)

Investigation 9 Joe has been looking more closely at the assignment of the numbers 0, ... , n − 1 to 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 following assignment. 0 1

2

But Joe notices that this also makes the trefoil knot six-colorable by the following assignment. 1

3

5

Somehow Joe feels that this six-coloring of the trefoil is a special kind of sixcoloring. He calls it borrowed. He figures that a borrowed n-coloring has the following property: there is a divisor n > d > 1 of n such that only every dth number has been assigned. He calls a knot really n-colorable if it’s n-colorable and if there is an assignment of the numbers 0, 1, 2, ... , n − 1 that is not borrowed. He thought maybe an n-coloring would be borrowed if some of the n numbers were left out. Here’s an example where that does not seem to be the case: 2

0

1 4

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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 by one 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 of S(K))?

Question 10 In his investigation of R(K) Joe asked this question: How does R(K#K') compare with R(K) and R(K')?

Investigation 11 Joe thinks that the knot K that forms the edge of a 2n + 1-half-twist strip is really 2n + 1-colorable. Further, he believes that 2n + 1 is the largest number for which K is really 2n + 1-colorable. Help Joe sort this out.

Question 12 Joe has told Millie about knots that are really n-colorable. Millie claims that, if p is a prime number and a knot is p-colorable, the knot must also be really p-colorable. Is she right?

Question 13 Joe’s definition of really n-colorable also applies to links. What can you say about R(L) if L is a link but not a knot?

Investigation 14 (Continuation of Investigation 8) Tiffany has been investigating the two-links that are 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.

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“0”

“2”

“4”

“6”

She is interested in finding out for what n is each two-link n-colorable. She has come 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 2ntwisted strips. (Of course, for a “twist” she means a “half-twist.”) Help her out.

Question 15 Tiffany and Brandon have been talking about the links that form the edges of even-twisted strips. Brandon thinks that what he knows about really n-colorable and 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 say about really n-colorable and these links? Will this help with Tiffany’s project in Investigation 14?

Summary 16 Boss wants a summary of what the Acme team has done with knots and the evolution of n-colorability. You might want to include in your summary the properties of 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 medium through which light waves traveled (now considered to be a vacuum), was

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occupied by a substance called ether. The chemist Lord Kelvin proposed that atoms, the basic elements of matter, were vortices, or knots, in the ether. Different knots would correspond to different elements. Scottish physicist P. G. Tait and American mathematician C. N. Little started to construct lists of all knots as a way of compiling a table of all the elements. But by 1900, the theory of the ether had been abandoned and with it the idea of knots as atoms. However, questions posed by chemists and physicists—How can you tell when two knots are different? How can you come up with a complete list of knots?— became a mathematical challenge. Knot theory became a vigorous subfield of topology, itself a fairly new field of mathematics. A basic strategy for answering the big questions might be to do what we did with surfaces: figure out a good way to 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 not only a complete list but a list that includes each item exactly once. For surfaces, a representation is a surface symbol; two symbols are the same if one can be obtained from the other by a finite number of application of the rules, and the invariants 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 the sense that different surfaces have different invariants. For a knot, the representation is a picture, and two pictures represent the same knot if one can be obtained from the other by a finite number of moves. It appears that Tait and Little were able to come up with a complete list of knots up to nine crossings and that their techniques provide a method for extending their list to knots of more crossings and hence to a list of all possible knots (for more details on the tabulation of knots, see the book by Adams, chapter 2). However, this is not a tidy list like the one we have for all surfaces! What they were not able to do was decide definitively whether or not there were duplications. That leaves the question: How do you decide whether or not a pair of knot pictures represents the same 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 finite number of moves. All the n for which a knot K is n-colorable is encapsulated in the 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 knots K and K' have the property that S(K) = S(K').

But do the pictures in each pair actually represent the same knot? If you feel that each pair is really a pair of different knots, you are right: knot theorists have

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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 the not knot, are not n-colorable for any n! For such a knot, n-colorability will not help us decide whether or not it has an “inverse.” The upshot is that n-colorable is not the last word; knot theorists need other invariants if they are to answer the big question. What sorts of things might we be looking for? What is the nature of the invariants we have for surfaces? For surfaces, 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 biggest breakthrough in the search for knot invariants came in 1929 with an invariant that is a polynomial, called the Alexander polynomial after its creator. All the knots in the tables of knots existing at the time—of nine or fewer crossings—can be 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 that n-colorable is really an algebraic invariant. When a knot is n-colorable, we associate the number n with the knot. This invariant is a number. But thinking back, we remember that this number was obtained by associating the strands of the knot picture in a certain way with the integers module n. The invariant n is related to the algebraic structure of the integers modulo n. This is a typical approach for a topologist to finding invariants: find some way to associate an algebraic structure with the topological object so that two same objects imply same algebraic structures. (The word “same” for algebraic structures is “isomorphic.”) The philosophy behind 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 perspective on the problem and new opportunities for investigation. In 1895 Henri Poincaré defined one of the earliest algebraic invariants of a 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 to be the fundamental group of R3\K, the latter set gotten by removing the knot from

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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 article by 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 polynomial an algebraic invariant. It turns out that n-colorable can also be derived from the knot group: If K is n-colorable, then there is a group homomorphism from the knot 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 used to define two knot pictures as the same are called in the literature Reidemeister moves in honor of this mathematician. Reidemeister proved that equivalence by the moves are sufficient to show two knots are the same. The next comprehensive book on knot theory appeared in 1963 by the American mathematicians Richard Crowell and Ralph Fox. A complete set of algebraic invariants for knots was found by Waldhausen in 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 due in part to the connections between the invariants within the field itself. (For example, n-colorable and the knot group.) It is also due to connections with other fields of 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 the classification of three-manifolds. (You will recall that three-manifolds are the three-dimensional analog to two-dimensional surfaces.) The connection goes something 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 make it 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 you get 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 “specified ways” of gluing one torus inside another (see Adams, chapter 9). Recently, other connections between knot theory and other fields have brought the subject back to its original roots in the sciences. In the 1980s, in his work on operator algebras, the American mathematician Vaughan Jones came up with both a new, powerful invariant (called the Jones polynomial) and a connection between 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 knot theory by biologists and chemists, the former in part because of the knotting of DNA molecules, the latter in part because of an interest in the synthesis of knotted molecules. (For more details, see the book by Adams, chapter 7 and the article by Summers.) Some of the issues confronting chemists provided the setting for the beginning of this chapter (see the book by Flapan). Knot theory continues to be a lively, active area of mathematical research. For a 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.

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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

CHAPTER

14

Now that you have been through this book, or most of it, what should you do next? One thing is to step out on your own, find a problem or a topic related to things we have talked about here that interests you, find a group of chums with the 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 the chapter is just that: a list of project ideas. After you wrestle with the project for awhile and feel you are an expert on a topic or after you have solved some problems and have some great insights, it is great to share what you found. A second purpose of the chapter is really about this aspect of doing projects: communicating your findings. Nothing is worse than telling someone what you have done, then to have them fall asleep or roll their eyeballs. You should attempt to communicate the results of your project effectively.

Project Ideas Here is an annotated list of ideas for projects. The ideas span a wide range of project types. At one end, there are exploratory projects that are like the investigations 313

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from the earlier chapters; no references are given for these. At the other end of the spectrum are expository projects. Such a project involves addressing a known topological procedure, idea, technique, fact (or collection of related facts), or theorem in a new way, relating it to things that happened in the book and to previous experiences with mathematics and making a model (or two) that illustrates the material. There are lots of references to the expository projects. Most projects will be some combination of the two. On the one hand, solving a problem sometimes leads to finding out what others have done. On the other hand, trying to understand what others have done in an area you are interested in sometimes unexpectedly leads to solving new problems. In the list below, the annotations are not uniform. For some project ideas, there are lots of details. For others, there is just a title, a few questions, and a reference or two. This uneven presentation may not only give you ideas but also show you ideas in various stages of development—some with just the germ, some with lists of questions to ask (and answer) and suggestions for things to make, and some with 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 you need 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 directly to you. If you do not find something below that excites you, use the list to begin thinking. Perhaps one idea might suggest something else to pursue. Browse the references following this list and at the ends of the chapters, and talk with friends and 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 a chapter number in brackets. Otherwise, the reference is listed in the Resources section.

I. Map Coloring A. Four-Color Conjecture Before 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 Colonies Heawood proved a theorem about coloring maps on a sphere in which a country may 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 a particular surface, is there a relationship between the number of colors needed for

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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 Moon This is related to B, “Coloring Maps with Colonies.’’ Suppose Earth countries establish colonies on the moon. You want to color the countries on Earth as well as their colonies on the moon. Of course, you want countries and colonies to be colored 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 Maps What 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 Euler It turns out that if your world is a surface and if you count the number of valleys and mountain peaks and saddles (passes), then these numbers can be used to calculate 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 Surfaces You 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

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soap-bubble replacement: Fun-Film or Joli-Form-a-Film, or petroleum acetate–based liquid plastic that hardens. Is there some simple way to identify what each surface is (according to our classification scheme)?

C. Generalized Regular Polyhedra The surfaces of the regular polyhedra (Platonic solids) are spherelike. (If they were made of rubber, you could inflate them into spheres.) The regular polyhedra have the following property: there are numbers n and m such that every face is a regular n-gon and every vertex is of order m. What are analogous shapes when the underlying surface is something other than a sphere? You might have to relax the condition of “every face being a regular n-gon’’ to something like “the faces are congruent n-gons.’’ Can you build the corresponding “polyhedra’’? (Senechal and Fleck, p. 212+, pp. 251–253.)

D. “Nice’’ Models of a Cross-cap We have seen one embedding of a cross-cap in three-dimensions. It is not without its problems. But just as there is more than one embedding of the Möbius strip in three-dimensions, there are many embeddings of the cross-cap. Some are nice and symmetric; the Boyes surface is one example. Build some of these, explain them (explain why they might be more satisfactory than the familiar embedding), create them with computer graphics, or find them on the Web. (Hilbert and Cohn-Vossen)

E. The Geometry of Surfaces Think of a cylinder as a plane that is rolled up; certain points of the plane have been identified. (Point (x, y) is identified with point (u, v) iff y = v and x − u is an integer.) This coincides with the usual way of thinking about the cylinder geometrically: shortest paths on the cylinder are really just shortest paths in the plane. You can 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 thinking about the torus geometrically also results in shortest paths on the torus being really just shortest paths in the plane. Although this is not the usual way of thinking about 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 on the rigid geometric structure of a shape. This project is meant to “restore’’ some of that rigidity. The video Shaping Space and the book The Shape of Space by Weeks give ways of visualizing some of these “Euclidean’’ surfaces. The purpose of this project is to explore some ideas surrounding this notion. Which surfaces can be endowed with a Euclidean geometry? If Euclidean geometry doesn’t work, the two non-Euclidean geometries are candidates: hyperbolic geometry and elliptical

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(spherical) geometry. Which surfaces could be endowed with geometries of these types? The identification scheme for the torus comes about by tessellating the plane with squares. Perhaps a tessellation of the hyperbolic plane or sphere by a regular “polygon’’ would give rise to another surface. If you pursue this, you will need to learn a little bit about hyperbolic geometry and spherical geometry (and some of their key properties). It might also be helpful to 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 the three geometries and some of their subgroups. (Hilbert and Cohn-Vossen; Stillwell; Thurston; Weeks, Shape of Space)

F. Games on Surfaces The investigations of Chapters 6, 7, and 8 have described tic-tac-toe on the surface of a doughnut, on a Möbius band, and on a Klein bottle. Analyze these. Must they end in a tie? Are there winning strategies? Which player can win? (Weeks, Shape of Space and Exploring the Shape of Space)

G. Alternative Arguments You know that the connected sum of a cross-cap with the torus is the same as the connected sum of the cross-cap with the Klein bottle. You can see this by analyzing patterns. Are there other ways of seeing this? (Maybe a four-dimensional argument might work?) The same goes for showing that the connected sum of two cross-caps is 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, “Model of Ungar-Leech Map on Doughnut that Requires the Maximum Number of Colors’’; and IIIF, “‘Regular’ Maps on Cross-cap and Klein Bottle.’’) Several paper “polyhedral’’ versions of the torus, cross-cap, Möbius band, and Klein bottle have been suggested in the Investigations, etc., of Chapters 6, 7, and 8. Make nice models of these. You might think of other possibilities and make those, too, such as a “polyhedral’’ version of the Boyes surface or a “polyhedral’’ torus “hat’’ having seven-fold rotational symmetry or a “twisted’’ Klein bottle (see Investigation #9 of Chapter 11). Relate each one to some interesting properties of the surface, such as maps requiring the maximum number of colors.

J. Linkages and the Surfaces That Go with Them Analyze some linkages and the surfaces that go with them. See also the Investigations of chapter 9. (Abrams; Shimamoto; Thurston and Weeks; Toussaint; Walker)

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K. Proofs for the Classification of Surfaces Have a look at the different ways mathematicians have proved the classification theorem for surfaces, understand them, and explain how they work. See the Notes and 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 for surfaces 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 up with a proof that Heawood’s inequality (chapter 12) is actually an equality. Their proof is broken down into 12 cases. Describe their program. Give an exposition of the proof of one of the “nice’’ cases. (Ringle; Stahl)

C. Constructing Real Maps on Surfaces That Require the Maximum Number of Colors The proof of Ringle and Young (see previous project) is theoretical; few maps have actually been constructed. But their proof gives prescriptions for constructing the maps 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 require the (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 the Maximum Number of Colors The Ungar-Leech map on a doughnut is a map that not only requires seven colors for proper coloring but also has seven-fold rotational symmetry. This is an opportunity to make a really elegant model. You’ll need a plaster of Paris model of a torus.

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Alternatively, you could design a polyhedral version of the torus having seven-fold rotational symmetry. See IIH, “ Interesting Models of Familiar Surfaces.’’ (Coxeter, The Mathematics Teacher)

E. Map Coloring on the Klein Bottle The 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 on the Klein bottle requiring the maximum number of colors. (Ringle)

F. “Regular’’ Maps on Cross-cap and Klein Bottle A map is regular if whole numbers n and m exist such that every country has exactly n borders and every vertex is of order m. What are the regular maps on a cross-cap? What are the regular maps on a Klein bottle? What do these maps look like on familiar realizations of the surfaces? Are there connections between any of these maps and coloring problems on the surface? Are there connections between any 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 Small What 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? Are there connections between any of these maps and coloring problems on the surface? (Firby and Gardner, p. 83+)

IV. Three-Space A. Coloring “Maps’’ in Three-Space The 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 definition of “proper coloring’’ of a map? What about two- and three-colorable three-space maps? Are there map coloring theorems in two-dimensions that could be generalized to three dimensions?

B. Three-Dimensional Worlds What are the three-dimensional analogues to the two-dimensional surfaces we have been studying? What is a definition? What are some examples? What corresponds

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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 and are described in the book by Adams. A good project would be to understand these and 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 Space We 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 by Weeks in the Notices. Understanding it involves understanding dodecahedral space. Understanding the latter involves understanding the four-dimensional analogue to the regular dodecahedron in three dimensions. A good project would be to bring this example of a three-dimensional world to life and to explain, in layman’s terms, why astronomers are interested in it. (Coxeter, Geometry, chapter 22; Luminet et al.; Weeks, Notices)

V. Knots and Links A. Surfaces Spanning Knots Fun-Film is one way to create a surface that has a given knot as its “edge’’ (see project 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 for surfaces. 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 knot inverses (see project VJ, “Knot Inverses’’) and of unique factorization of knots into prime knots (see project VK, “Prime Factorization of Knots’’). (Adams; Fox; Kauffman, On Knots)

B. Torus Knots Some 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 Mathematical Society; Kauffman, On Knots)

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C. The Jones Polynomial of a Knot This 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 Biology In 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 of DNA knot? Find out what is going on, what they are looking for and why, and what they have found. (See notes and references in Chapters 7 and 13; Adams; Flapan; Juriˇsi´c; Kauffman, On Knots; Summers)

E. Linking Number This is an invariant of links. What is it? Why is it an invariant? Use it to distinguish even twisted strips. (Adams; Kauffman, On Knots; Livingston, American Mathematical Monthly)

F. Braids It turns out that every braid can be turned into a knot or link and vice versa. What are these things? They have a nice algebraic structure. (Adams; Gardner, Scientific American, December 1959, January 1962)

G. Ballentine (a.k.a. Borromean) Rings, Generalized In the Ballentine rings, if you remove one, the whole thing falls apart, yet the three rings “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 fundamental 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-Manifolds This is an extension of project IVB, “Three-Dimensional Worlds.’’ It turns out that every three-manifold can be constructed by starting with the complement of a knot 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 involves the lens spaces of project IVB. A good project would be to understand how this construction works and explain it. (Adams)

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J. Knot Inverses In 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 knot inverses exist for the knot sum operation? In Chapter 13 this was answered in the negative for some knots. What is the whole story? An initial discussion of this problem 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 in Fox; the other is by Conway and is given in the Mathematical Games section of Scientific American. A good project would be find out what these answers are, understand them, and explain them. (Adams, 4.3; Fox; Gardner, Scientific American, December 1972)

K. Prime Factorization of Knots The knot sum operation suggests a definition and some questions. What should the definition of a prime knot be? What are some examples of prime knots? Can every knot be factored into a finite number of prime knots? Is that factorization unique? (Compare definitions and questions for the surface sum.) Again, spanning surfaces give some answers as suggested in VA, “Surfaces Spanning Knots” (Fox; Livingston, chapter 4).

L. The Group of a Knot The group of a knot is the fundamental group of the complement of the knot in three-space. For more, see the Notes from Chapter 13. Find out about more about the definition. (Knowing a little group theory will help.) Figure out how to calculate the group of a knot given a picture of the knot. (Bertuccioni; Crowell and Fox; Neuwirth)

VI. Miscellaneous A. Hexlike Games Hex. Bridge-it. Twixt. Can any of these games end in a tie? What can you say about winning strategies for these games? What is it, topologically, that makes these games similar? In a primitive sort of way, Hex is just a map on an island with all of the vertices of order three. That opens up the possibility for lots of games! What can you say about them? (Alpern; Beck et al.; Gale; Gardner, Mathematical Puzzles and Diversions, chapter 8, Scientific American)

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B. A Variety of Additional Project Ideas with Knots and Links Self-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 the ways a rubber band can be wrapped around a cylinder (Kauffman, p. 18); analysis of knots and links by computer (do Web search.)

C. Complexity Theory Given a problem, how long will it take to solve? Some of the problems considered 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 the sphere can be colored with three colors is another. See Notes and References in Chapter 2.

D. Taking Tours in the Real World Sometimes you want to take a tour on a network and it is not possible by Euler’s theorem. You can do the next best thing if you allow “double-backs,’’ edges that you 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 nice tour possible? (Tannenbaum et al., and the book’s Web site)

E. Nonplanarity of Graphs Kuratowski’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-Space Project IIG, “Alternative Arguments’’ suggests that four-space might help to explain some of the facts we know about surfaces. Describe how you can easily carry 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.)

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G. Sprouts This 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, October 1977, 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. Boltyanskiˇι, 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.)

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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. Juriˇsi´c, 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.

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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 University Press, 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 Mathematical Monthly, 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 the shape of space?’’ “Can knots be factored uniquely into prime knots?’’ “What are some good representations of the cross-cap?’’ As in solving any problem, the response will occur in stages, such as ●

Exploring

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Gathering data

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Search for a pattern

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Working out examples

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Making a model (or two...)

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Coming up with an answer to the question, or a partial answer, or an answer to a similar or related question

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Making a convincing argument explaining the answer

When your project involves an expository component, you will likely find material written about your problem. Read and digest it, knowing that you will need 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 mathematical questions you can answer. Relate the material to things discussed in the text by analogy or by generalization or by extension or by any other method. Make a model (or two) that illustrates the material.

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Even though the material may not be new to the world at large, it will be new to you. Wrestling with the ideas, making examples, building models, and putting together an exposition of them always produces a new prospective on the original material and frequently produces new proofs and insights. I have chosen the topics above to maximize the possibilities of making connections with material in the text, possibilities for models, and possibilities to put materials together in new ways appropriate for you. After you have found an answer (or answers) you are happy with, then you must work on communicating the results of these earlier stages. This communicative aspect 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 turn would include a summary of your results: the question to be answered, your plan of attack, ideas you used/developed, and a summary of your results. This aspect of the project should convey in a visual way an overall feeling for the project (its results and its procedures), without providing too many details. This is a good place to use models, illustrations, diagrams, and charts. There is more on the display in the next section.

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An oral presentation. This is an occasion for you (and your team) to introduce to others the results of your project and to point out important features of the display (if there is one). The oral presentation should effectively communicate a few of the main ideas of the project’s topic, its flavor, its interest to other mathematicians, its applications in the real world, how it relates to the topics encountered 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 short time, 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 the problem. You want to strike a balance between the trivial and the complex: On the one 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, the listener would be bored.) If listeners want to know more than you are able to include in your presentation, let them ask questions. Your presentation will be more successful if you let them participate in it: Pick up on where they are puzzled, on what their interests are. You must remember that you have been living with your topic and are an expert on it. Your listener is not. Do not overwhelm her with jargon. What you think is trivial (relative to the project) might be obscure to the uninitiated. Practice your presentation. Time it. Practice it again. Make sure you have all your props ready and in order. (If you have to fumble around during your 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 the project; it should describe the ideas, procedures, and theorems the team used to answer the question/solve the problem; and it should include the answer(s) and written arguments justifying them. If your project involves a display

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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/oral presentation and to provide more of the details to back up what was claimed there. In the paper, you put the pieces together and show how they are connected. The paper can use as many pieces of the display or presentation as you wish, but there should be no loose ends. Make sure that your reader knows what question (or set of questions) you are answering. Describe how you (or others) explored the terrain in looking for answers. Provide some of the examples you worked out or create new examples that are useful to demonstrate your lines of attack or to point out possible blind alleys. Again, provide pictures, diagrams, and tables to help illustrate your ideas and 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, which many office supply stores sell. Alternatively, you can buy a piece of foam board for around $7. Then, with knife and a metal straight edge, score it twice to make a triptych that will stand up on it own (see pictures below).

Score along dotted lines.

Bend back along scored lines.

Stand it up: make a triptych on top of table.

Choose a title for the project. The title should be a question that your project answers. It should be immediately understandable to the viewers. Place the title in a prominent place on the foam board. Outline main aspects of the project. Here is a sample outline: ●

Detailed statement of problem

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Significance of problem

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Procedures used (steps taken) to solve problem

Project Evaluation ●

Results obtained

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Arguments used

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Illustrative examples

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Descriptions of models accompanying the display

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Conclusions

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You could think of this as an outline for the project’s paper as well. Projects will differ. 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 attach to the display board. Each “window’’ would correspond to an item in your outline and be labeled by the outline heading. Each window should be as much of a visual display of information as possible, employing diagrams, photos, and graphs and using 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, you may 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 make it stand out. You want your display to draw in viewers and get them interested and intrigued. You are selling your project. You want the viewer to become curious about it and to ask questions.

Project Evaluation Here are criteria (the four C’s) for evaluating every aspect of your project, whether it 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 the presentation carefully organized? Can an outsider see how the various pieces of the project are connected? Completeness. Have all reasonable lines of inquiry been investigated? If some lines have been used and not others, have the choices been explained (e.g., “there was not enough time to pursue such and such a line of reasoning,’’ “such and such approach appeared to be too difficult’’): Have connections been made between the project and other areas of study, especially areas covered in the course? Is the project problem 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 to the project problem similar to the solution of a problem seen before? What are related problems that need further study? Why is the project problem important? Does its solution have applications to some part of mathematics? Is its solution

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Chapter 14 Where to go from Here: Projects

useful to another discipline or to the real world? Did the problem remain unsolved for a long time? Does the project incorporate effective uses of models, charts, or diagrams where appropriate? Does the project incorporate effective use of technology where appropriate? Correctness. Are all parts of the project mathematically correct? Are definitions correctly stated? Are the arguments correct? Have all aspects of the project been communicated 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? Are there unusual ways of putting ideas together? Are the means of communicating project’s topic or its solution new? Does the project include the effective use of a clever 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

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The experience of having a real context (more or less self created) out of which to expound on mathematical ideas, put together an effective visual display, and make an oral presentation

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The experience of communicating your insight to others

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Making models and appreciating their value, for explanation, for illustration, and for understanding

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Experience working with a team: learning from and communicating with your peers

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Sense of pride in a job well done

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Passion: Really getting into something you’re interested in

(Much of the material in this chapter on carrying out a project has been adapted from the project chapter of the author’s Geometry by Discovery, published in 1998 by John Wiley and Sons.)

Index

A Adjacency pattern, 268 Airplane videos cylinder (inner tube) and, 150-151, 153, 155 edges in, 150 flat screen and, 150 Alexander polynomial, 310, 311 Appel, Kenneth four-color problem and, 99-100

B Ballentine rings, 306, 321 Bee-hive surface, 173. See also Cross-cap; Projective plane Beehive-with-lake Möbius strip as, 160-162 Border defined, 24. See also Edge(s) Border rule between two countries, 7 Borromean rings, 306 Boundary, 147. See also Edge(s) Box-with-square hole correspondence with map on doughnut, 116-117 Boy’s surface, 175 Braids, 321

Bridge building in three-space for two-dimensional figure-eight, 238-241 in three-dimensional space, 238 Bridges of Konisberg problem Euler’s analysis of, 46 Euler’s theorems for, 45 converses of, 46

C Cauchy, Augustin-Louis proof of Euler’s formula for network, 71 Checkerboard domino puzzle, 23 Chromatic numbers and critical graphs, 256 Circles coloring non-overlapping, 85-86 in maps on rectangles, 15-20 Circuit board with hole, 104-109 comparison with box with hole, 104-106 Circuit board design investigation of, 121 Circuit board networks with five terminals problem resolution of, 66-67 investigations of, 42-43, 47

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332

Index

Colonies coloring maps with, 314-315 Coloring maps G7 map problem, 1-4 list of maps for 6 rules and terminology for, 6 Complexity theory, 102, 323 Countries coloring problem for, 8 defined, 24 with edges, 74 on real maps, 9 sizes of, 76 on transformed maps, 8-9 Critical graphs and average order of vertices, 257-258 network of map on sphere, 258-261 chromatic numbers for, 256 Lemma about, 257 Cross-cap(s), 173, 174 arguments, 317 assembly of theorem for, 215 in three-space, 248-249 Euler number for, 198 in four-space, 249 “nice” models of, 316 possibilities for, 247 “regular” maps on, 319 Crossing(s) coloring rules for, 284 defined, 291 in five-half-twist strip, 277-278 inverses of, 278-279 ingredients of, 292 in knots, 277, 278, 281-282 of not knot, 282 overpass and underpass, 277 on square knot, 277 three-colorable, 284-285 Csázsár’s polyhedron construction of, 119-120 Cube, data for, 62 Cylinder airplane video game and, 151-152, 153, 154-155 as bridge for Klein bottle, 241-242

D de Morgan, Augustus, 24 Dirac, G. A., 269 Displays of projects, 327 putting together, 328-329

Domino problem, 47 Domino puzzles, 43-44 Door inspector problem, 38-40, 47 investigation on doughnut, 116 Dot-line diagram. See also Network; Graph even number of lines to dots, 34, 36 of Konisberg tour problem, 31-33 of New York City, 34 nice tour on, 34, 35, 36 odd number of lines to dots, 34 of Paris, 34 Doughnut map data from, 109, 111-113 theorem about, 113 Euler formula for, 114 Doughnut(s). See also Inner tube; Torus colors for, 318 connecting, 192 five terminal problem on, 106-107 17-holed colors needed for, 263 infinite family of maps on, 107-108 as inner tube, 113 map of, 109, 111-113 n-holed colors for, 318 “regular” maps on, 319 seven-terminals investigation, 116 six-terminals investigation, 116 three-holed, 193 two-holed, 193 colors for, 318 Ungar-Leech map on, 318-319 Doughnut-with-lake, exercise, 109

E Earth coloring maps on, 315 Edge(s), 37, 50 counting from vertices, 74-75 theorem for, 76 of countries, 73-74 of twisted strips, 136-137, 140-141 Empire maps, 97-98, 101 Equivalence geometric, 24 of surface symbols, 210-211 exercise for, 212 prime surfaces and, 220 questions, 219 rules for

Index Equivalence (Continued) investigations of, 218-219 topological, 250 Euler, Leonhard bridges of Königsberg problem theorems of, 45 geometry of position and, 45-46 Euler formula(s). See also Euler number for network, 71 for polyhedra, 70-71 Eulerian circuit, 45 Eulerian path, 45 Euler number, 201-201 from symbol, 230-231 problem of pairs not same, 228 for S, S’, and S-with-lake, 198 of surfaces, 199, 219, 227 exercise, 236 for sphere, 194, 198 for torus, 194, 198 of surface sum formula for, 196 proof of, 193-196 Exponential time vs. polynomial time, 102

F Figure-eight in plane, 237 Five-cities problem. See also Five-terminal problem on Möbius strip, 144 Five-color maps on doughnut, 114-115 on sphere, 81-83, 99, 878 Five-color theorem, 80-83, 87, 99 Five-terminal problem on doughnut, 106-107 solution of, 110 on sphere, 104-105 Euler’s formula and, 102 Jordan curve theorem and, 102-103 simple closed curve and, 122-123 on square, 110-111, 113 Four-color conjecture, 314 Bram’s map coloring game, 96 investigations of map of empires, 96-98 pennies in plane, 98 method for altered map in, 92-93 applications of, 94-95 β-α sequence in, 91 α-β sequence in, 90-91

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Four-color conjecture (Continued) steps to color map, 93-94 Tetrachrome map coloring game, 95 Four-color maps history of, 99 Four-color problem, 24 difficulty in resolution of, 100-101 solution of, 99-100 Four-color theorem real proof of, 315 Four-space, 323 defined, 238 Klein bottle in, 237-238, 241-243, 248, 250 Klein bottle-with-lake in, 243-244 publications on, 252

G Game(s) hexlike, 322 map coloring, 95, 96 on surface, 317 Geometery n-dimensional, 250 Geometric equivalence of maps, 24 Geometry of position, Euler, 45-46 Gizmo (calendar) problem, 60-61 calendar gizmo exercise, 65 data for, 63 formula for, 63 Graph(s), 45. See also Network complete bipartite, 121 critical, 256, 257 defined, 255 dual, 255-256, 256 nonplanarity of, 323 planar, 45 Guthrie, Francis, and four-color maps, 99

H Haken, Wolfgang four-color problem and, 99-100 Heawood, P. J. empire maps and, 97-98, 101 five-colors for maps on sphere proof of, 87 and four-color maps, 95, 99 Heawood’s conjecture, 267 for nonorientable surfaces, 270 proofs of, 267-268, 269-270 Heawood’s inequality in map color theorem, 260-262, 267

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Index

Hexagon, as pattern, 162-163 Hexlike games, 322

I Inequality(ies) in map color theorem, 260-262, 267 Infinite families of maps, 53, 67, 315 Inner tube, 113. See also Doughnut; Torus airplane video game and, 150-151, 153, 155 with lake investigation of surfaces for, 163-164 simple closed curves puzzle, 118-119 tic-tac-toe game, 118 Inner tube tic-tac-toe (IT4), 118 Invariants for knots, algebraic, 310 for surfaces, 249, 309 Island. See Maps on island

J Jones polynomial, 321

K Kelvin, Lord, knots and, 309 Kempe, Sir Alfred and four-color maps, 99 Klein, Felix, 175 nonorientable surfaces, 201 Klein bottle bridge for, 238-241 definitions of, 251 Euler number for, 198-199 exercises, 155 in four-space, 237-238 assembling, 241-243, 248, 250 map coloring on, 319 pattern for, 154-155 possibilities for, 248 “regular” maps on, 319 six-cities problem on, 264 some things to make, 167-168, 171-172 symbols for, 211 Klein bottle tic-tac-toe (KBT3), 169-170 Klein-bottle-with lake (KBWL) in four-space, 243-244 in three-space, 243 Knot pictures determination of different knots in, 293-294 ingredients of, 291-292 Knot(s) and biology, 321

classification of three-manifolds classification and, 311, 321 as closed curve in plane, 276 colorability of investigations of, 299-300 method for questions, 300-302 crossings in, 277, 278, 281-282 defined, 274 five-colorable, 290-291 modular-clock arithmetic in, 290 generalization of three-colorability to n colorability, 288-289 group of, 322 invariants for, 310 inverses of, 322 investigations of, 299-309 Jones polynomial of, 321 Kelvin Lord, and, 309 and links, 320-322, 323 lists of, 309 with loose ends, 302 manipulation of one into other, 274, 275, 276-277, 281, 283 n-colorability of, 291, 292, 304 exercise in distinguishing, 292-293 n-colorability rules for, 289 n-link in, 304-305 not knot, 282 overhand, 274, 275, 276, 284-285 prime factorization of, 322 questions, 300, 307, 308 sum of two n-colorability of, 304 surfaces spanning, 320 three-colorable, 284-287 rules for, 284 trefoil, 306-307 for wrapping package, 302-303 Knot theory and statistical mechanics, 311 and topology, 311 Königsberg with bridges tour, 27-33

L Lakes colors needed for surfaces with, 255-256 identifying from surface symbol, 233-234 number as invariant of surface, 249 patterns for numbers of, 228 theorem relating number of, for symbols, 235, 236

Index Legendre, Adrian-Marie proof of Euler’s formula for polyhedra, 71 L’Huilier, Simon-Antoine, 123, 202 Lines. See Edge(s) Linkage control of, 179-180 coordinate system for, 180 edge identification and, 182 first case, two rods, 178 for investigation, 200-201 negative angles in, 181-182 postive angles in, 180-181 of rods and pivots, 177-178 second case, 183-184 exploration of, 184-187 and surfaces, 317 surfaces and patterns, 188-190 theorem connecting, 190 as torus, 182 Linking number, 321 Links equivalence of, 305 n-link picture, unlinked, 304-305 unlinked, 305-306 rules for, 305 two-link n-colorable for even half-twist strips, 307-308 Listing, Johann Benedict Möbius strip and, 146 Little, C. N. lists of knots, 309

M Map coloring games, 95, 96 G7 map problem, 1-4 list of maps for, 6 projects, 314-315 rules and terminology for, 4-5 rules for, 7 Map coloring problem de Morgan, Augustus, 24 origin of, 23 Map color theorem, 254 four-color problem for sphere and, 262 inequalities in, 260-262, 267 notation and definition for, 256-257 proof of, 256-262 Map data formula from, 53, 54 relationships in, 51 vertices (V), edges (E), countries (C), 50 altered, 57

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Map data (Continued) one country map on an island, 50 simplified, 57-58 two-island case, 58 vertices of order two, 59 Map of empires investigations of, 96-98 Maps formulas for, 53, 54 imaginary maps, 52-53 infinite sets (families) of, 53 ingredients of, 24, 50 same, properties of, 208 on sphere, rectangle or island, 24 two are the same, defined, 207, 208 Maps on island colors needed for, 69 construction of dual graph to, 256-258 data for, 109 distortions of polygon or disc, 55-56 as doughnut, 108 on doughnut, 108 formula for, 55 with lake colors needed for, 69 data formula for, 68 OK map on, 56 defined, 60 theorem for, 55, 60 side and edges of, 144 from three-dimensional shapes, 61, 64-65 Maps on rectangles investigations of circles in, 15-16, 18-20 circles in with one chord, 16-17 curves in, 20 rectangular lake in, 17-18 regions in, 18-20 steps in, 14-15 Y-shaped figures in, 16 line drawing from side to side, 11-14, 24-25 questions number of colors for, 18, 20 Maps on sphere, 62, 64-65. See also Sphere average order of vertices on, exercise for, 258-261 calendar gizmo and, 68 circuit board network with five-terminals problem, 66, 67 colors needed for, 69 countries as hexagons, 65 cube and, 61, 68 five or fewer colors for, 80-83

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Index

Maps on sphere (Continued) proof of, 80-83, 87 theorem for, 83 formula for, 65 four-color problem and, 262 with quadrilateral countries, 70 six or fewer colors for, 78-79, 80 elimination of edges, vertices, countries method for, 77-78, 79 restoration of eliminations in reverse order, 78-79 theorem for, 80 three-color, 98 exponential vs. polynomial time for, 102 of simple closed curves, 128 with triangular countries, 70 two-color, 102 vertex order three classification by combinatorial type, 84-85 country with five or fewer edges, 84 country with five or six edges, 84 questions, 85-86 vertex order two, 58-59 Möbius, Augustus Ferdinand, 146 orientable surfaces and, 222 Möbius band. See also Möbius strip six countries on, coloring, 265 Möbius ladders, 271-272 odd half-twists of, 283, 284-285 in three-space, 273-274 Möbius shorts construction of, 143-144 gathering evidence, 165-166 Möbius strip airplane video game and, 153-154, 155, 156, 157 applications of theory, 147 as beehive (cross-cap) with lake, 157, 160, 161 cultural appearance of, 146 cut in half, 127, 135, 145 with odd number of half-twists, 135 in thirds, 128-129 discovery of, 146 edge diagrams of, 141 edges and sidedness of gathering evidence, 139 Euler number for, 198 exercises, 130-132 experiment as living in, 170-171 facts about, 131 half-twists in cut in half, 128-129 cut in nths, 139

Möbius strip (Continued) cut in thirds, 128-129, 143 even number of, 136-137, 139 investigation of cuts in, 138-139 number of edges of, 136-137 odd number of, 136-137, 138-139 one-edged, 130, 131 one-sided, 126, 130, 131 pattern for, 133-134 dressmaker’s, 132-133, 140 side and edge of, 144 tic-tac-toe, 144 tour around reversal of orientation on, 228 twist in, 126 cut down middle, investigation of, 142 edges of, 136-137, 140-141 investigation of, 137-138, 141 patterns for, 135 pieces from cutting, 138 two-edged, 130, 131 two from Klein bottle, 171 two half-twists results, 131 zero half-twist, 130 Möbius tic-tac-toe (MT3), 144 Moon coloring maps on, 315

N Network. See also Graph connected, 37 example, 37 disconnected example, 37 on doughnut surface, 117 nice tour of, 40 OK tour of, 40 path in, 37 on sphere, 117 Nice tour defined, 37 on dot-line diagram, 34, 35, 36 of Königsberg bridges needed for, 44 on a network theorem and conjecture, 38, 40 Nonorientability. See Surface

O Odd-twisted strips edges of, 274, 275

Index OK map on island defined, 60 theorem for, 55, 60 OK maps on sphere classification of, 69-70 four-color puzzles, 86 investigation on surface of doughnut, 116 vertices order three questions, 85 OK tour, of connected networks, 40 Order-three reduction for map coloring, 77 Order-two reduction for map coloring, 77 Orientabile surface theorem for detection of in symbol, 230 Overhand knot, 274, 275, 276, 284-285 three-colorable, 285

P Path, 37 Pattern(s) adjacency, 268 for an object, 167 beehive-with-lake, 160 connecting surfaces with, 188-190 equivalent, and surfaces, 208 facts about, strange, 162 investigation of, 166-167 for Klein bottle, 154-155 recognition of, 166 for surface, 167-168 for surface-with-lake, 191 triangle, 160-162 Poincaré, Henri definition of surface, 250-251 group of knots, 310-311 Polygonal screens pattern exercise for, 158 Polyhedral surface construction of, 264-265 Polyhedron(a) Csázsár’s, construction of, 119-120 Euler’s formula for, 70-71 regular, generalized, 316 ring-shaped formula for, 123 Polynomial time, exponential time vs., 102 Projective geometry surface of sphere and, 173-174 Projective plane, 173, 175

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Projects, 313-326 communicating results, 326-328 displays of, 327, 328-329 evaluation criteria for, 329-330 n-link, nonintersecting closed curves, 304-305

R Rectangles. See Maps on rectangles Reidemeister, Kurt, knots and, 311 Riemann, Bernhard, surfaces and connectivity, 201

S Schäfli, Ludwig, and projective plane, 175 Seifert, H., 222 Six-cities problem on Klein bottle, 264 on Möbius strip, 263 Six-countries map. See also Six-terminal problem on Möbius band, coloring, 265 Six-terminal problem, 116 South America real map of, 5 transformed map of, 10, 11 Space shape of, 320 topological, 251-252 Sphere. See also Maps on sphere connected network on, 117 Euler number for, 194, 198 Möbius strip problem for, 144 orientable surface of, 175 Sprouts, 324 Square boundary as equator of hemisphere, 156 Surface(s) algebra of, 205-209 analyzing, 193 Boy’s, 175 classification of, 249 proofs of, 318 classification theorem for, 222 proofs of, 222-223 connecting with patterns, 188-190 correspondence between, 207 defined, 188, 206, 250-251 duplication of, 226 equivalent patterns and, 208 Euler numbers for, 194, 198, 227 exercise, 236

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Index

Surface(s) (Continued) facts about, strange, 162 familiar, models of, 317 in four-space developments in, 250-251 fun-film, 315-316 games on, 317 geometry of, 316-317 identification of, 246-247, 249 with Euler number, 199 invariants for, 249, 309 n-colorable, 309 investigations of, 220, 244, 245 of Klein bottle problem solving on, 170 lakes on colors needed for, 255-256 investigations of, 244, 245 list of, 217, 218-219 complete, 226 as maps, 206-207 with maximum colors, 318 two are the same, defined, 207, 208 mystery, questions about, 244-245 nonorientable classification of, 222 defined, 229 Klein bottle, 175 Möbius strip, 175 studies of, 201, 202 orientable classification of, 222 sphere, 175 torus, 175 pits, peaks, passes, and Euler, 315 prime, 220 same application of theorem, 236 theorem about, 235 two as, 250 for solution of 10-cities problem, 263 sum of, algebraic properties of, 219 two-color theorems on, 318 what it might be, 206 Surface(s)’,(prime) equivalence of, 222 Surface sum, 190-191 gathering evidence, 197 pattern for, 192 Surface symbols construction of, 209 equivalent rule 1 for, 213-214 rule 2 for, 214-215

Surface symbols (Continued) rule 3 for, 215, 218 rule 4 for, 215 exercises for, 218 identifying lakes from, 233-234 lakes and exercise, 236 letters in occurrence of, 213 theorem for pairs of, 216 vertices of, labeling of, 231-233 Surface-with-lake number of colors needed for, 255-256 Symbol(s) equivalent, 210-211 exercise for, 212 for familiar surfaces, 211-212 surface, 209 theorem relating number of lakes and, 235, 236 theorem relating orientability of, 230

T Tait, P. G. lists of knots, 309 Thom, René, 100 Threlfall, W. A., 222 Three-color map(s) South America, 5 Western Europe and, 5 Three-dimensional map coloring countries of, 21-22 Three-dimensional shapes construction of map on island, 61, 64-65 formulas for, 62 arguments for, 60-62 maps on, 60-62 Three-dimensional worlds, 319-320 Three-space defined, 238 Klein bottle-with-lake in, 243 map coloring in, 319 Möbius ladders in, 273-274 Tietze, Heinrich Heawood’s conjecture for nonorientable surfaces and, 270 polyhedral surface and, 264-265 Torus Euler number for, 194, 198 linkage and, 182 map on surface of construction of map with eight countries, 266

Index Torus (Continued) on Möbius strip, 144 with seven-color map on construction of, 265-266 surface as sum of, 216 theorem for assembling, 217 investigation with, 217 Torus knots, 320 Tour(s) of building for door inspector, 38-39 investigation of, 40 question, 42 Central Park problem, 38 dots-line diagram of, 31-33 essential ingredients of, 33, 34 flights between Arizona towns and cities, 36-37 investigations of, 40, 42-43, 44 circuit board networks, 42-43, 47 connected networks, 40 hand shaking odd number of times, 40-41 mail delivery for downtown Tucson, 41 rectangular museums, 43 streets (edges) one-way, 41 Königsberg with bridges, 27-33 on network, 40 nice, 34 in real world, 323 terminology for, 37 Transformed map(s) countries on, 8-9 South America, 10, 11 Western Europe, 8-9 Trefoil knot coloring of, 306-307 Triangle as hemisphere, 160 making from Möbius strip, 162 as Möbius strip, 159-160

Two-color maps, 11-13, 14 on doughnut, 117 Two-color theorems, 318 for map coloring, 24

U Ungar-Leech map, on doughnut, 318-319 Utilities problem, 42, 47 five-terminals problem and, 66-67 on Möbius strip, 144 on surface of doughnut, 116

V Vertex (vertices), 37 coloring, 255 defined, 24, 45 labeling of to count lakes, 233-235 exercise, 236 investigation of, 244 for surface symbol, 231-233 odd and even, 37 order of, 24-25, 37 on critical graph, 257-258 order three edges of countries and, 76 in four-color conjecture, 91 reduction of, 77 order two reduction of, 58-59

W Weichold, Guido nonorientable surfaces and, 222 Western Europe ancient map of, 9 transformed map of, 8-9

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