Knots, Molecules, and the Universe: an Introduction to Topology

KNOTS, MOLECULES, AND THE UNIVERSE: AN INTRODUCTION TO TOPOLOGY

AMERICAN MATHEMATICAL SOCIETY

https://doi.org/10.1090//mbk/096

KNOTS, MOLECULES, AND THE UNIVERSE: AN INTRODUCTION TO TOPOLOGY

ERICA FLAPAN with

Maia Averett David Clark Lew Ludwig Lance Bryant Vesta Coufal Cornelia Van Cott Shea Burns Elizabeth Denne Leonard Van Wyk Jason Callahan Berit Givens Robin Wilson Jorge Calvo McKenzie Lamb Helen Wong Marion Moore Campisi Emille Davie Lawrence Andrea Young

AMERICAN MATHEMATICAL SOCIETY 2010 Mathematics Subject Classiﬁcation. Primary 57M25, 57M15, 92C40, 92E10, 92D20, 94C15.

For additional information and updates on this book, visit www.ams.org/bookpages/mbk-96

Library of Congress Cataloging-in-Publication Data Flapan, Erica, 1956– Knots, molecules, and the universe : an introduction to topology / Erica Flapan ; with Maia Averett [and seventeen others]. pages cm Includes index. ISBN 978-1-4704-2535-7 (alk. paper) 1. Topology—Textbooks. 2. Algebraic topology—Textbooks. 3. Knot theory—Textbooks. 4. Geometry—Textbooks. 5. Molecular biology—Textbooks. I. Averett, Maia. II. Title.

QA611.F45 2015 514—dc23 2015031576

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Preface xi

Acknowledgments xv

Part 1. Universes

Chapter 1. An Introduction to the Shape of the Universe 3 1. To Inﬁnity and Beyond 3 2. A. Square and His Universe 5 3. Straightest Paths in Flatland 8 4. Exploring the Shape of a Cave 11 5. Creating Universes by Gluing 13 6. Games on a Gluing Diagram for a Cylinder and a Torus 15 7. Extended Diagrams 17 8. Introducing the 3-Dimensional Torus and Sphere 19 9. Distinguishing a 2-Sphere from a 2-Torus 22 10. Distinguishing a 3-Sphere from a 3-Torus 24 11. Exercises 26

Chapter 2. Visualizing Four Dimensions 29 1. Teaching A. Square About the Third Dimension 29 2. Projections and Perceptions 31 3. Movies and Movement 33 4. Inductively Deﬁning Cubes of Dimensions 0, 1, 2, and 3 36 5. A. Square Learns About a 3-Dimensional Cube 37 6. A 4-Dimensional Cube 40 7. Tetrahedra in Various Dimensions 42 8. Exercises 44

Chapter 3. Geometry and Topology of Diﬀerent Universes 49 1. Intrinsic and Extrinsic Properties of a Space 49 2. Intrinsic and Extrinsic Geometry 52 3. Straightest Paths and Geodesics 53 4. The Deﬁnition of a Triangle 59 5. The Sum of the Angles of a Triangle 60 6. Triangles on a Flat and Curved Torus 61 7. Triangles on a Flat and a Curved Cylinder 62 8. Extrinsic and Intrinsic Topology 64 9. Using Loops to Understand Intrinsic Topology 67 10. Local and Global Properties 69

v vi CONTENTS

11. Manifolds 71 12. Assumptions about Universes 72 13. A 2-Dimensional Sphere with a Point Removed 74 14. A 3-Dimensional Sphere with a Point Removed 77 15. Curves on a Torus 78 16. How to Draw a (p, q)-Curve on a Flat Torus 80 17. Lines in the Extended Diagram of a Flat Torus 83 18. Exercises 84

Chapter 4. Orientability 87 1. The M¨obius Strip 87 2. Orientation Reversing Paths 89 3. The Klein Bottle 90 4. The 3-Dimensional Klein Bottle 92 5. Tic-Tac-Toe on a Torus and Klein Bottle 93 6.TheProjectivePlane 96 7. The Projective Plane with a Disk Removed 98 8. Projective 3-Dimensional Space 100 9. 1-Sided and 2-Sided Surfaces 101 10. Non-orientability and 1-Sidedness 103 11. Exercises 106

Chapter 5. Flat Manifolds 109 1. Flat Surfaces 109 2. Flat Gluing Diagrams 110 3. The Point of a Cone 114 4. Using Extended Diagrams to Understand Cone Points 116 5. Anti-cone Points 119 6. We Show that the 3-Torus Is Flat 120 7. A Method to Determine if a Glued Up Cube Is Flat 122 8. Extended Diagrams of Glued Up Cubes 123 9. Other Types of Gluings of a Cube 127 10. Exercises 128

Chapter 6. Connected Sums of Spaces 135 1. Einstein-Rosen Bridges 135 2. Connected Sums of Surfaces 136 3. Arithmetic Properties of the Connected Sum 137 4. Gluing Diagrams for nT 2 and nP 2 138 5. The Classiﬁcation of Surfaces 140 6. Dividing a Surface into Vertices, Edges, and Faces 142 7. The Euler Characteristic of a Surface 144 8. The Euler Characteristic of Connected Sums 146 9. The Genus of a Surface 147 10. The Genus of nT 2 and nP 2 150 11. Connected Sums of 3-Manifolds 151 12. Exercises 153

Chapter 7. Products of Spaces 157 1. Products of Sets 157 CONTENTS vii

2. Products of Spaces 159 3. A. Square and A. Pentagon Are Products 160 4. Some Products where One Factor Is a Circle 162 5. Examples of Spaces Incorrectly Expressed as Products 164 6. The Topological Uniqueness of Products 166 7. The Dimension of Product Spaces 167 8. Visualizing S2 × S1 and nT 2 × S1 168 9. Geometric Products 170 10. Geometric Products of Flat Spaces 171 11. A Flatland-Friendly Geometric S1 × I 173 12. Flat 3-Dimensional Spaces as Geometric Products 174 13. A Geometric nT 2 × S1 177 14. A Geometric S2 × S1 179 15. Isotropic and Non-isotropic Spaces 180 16. Exercises 181

Chapter 8. Geometries of Surfaces 185 1. Euclid’s Axioms 185 2. Flat Surfaces and Euclidean Geometry 191 3. Some Alternative Axioms 192 4. Spherical Trigonometry 193 5. The Area of a Disk in a Sphere 196 6. Maps of the Earth 199 7. Hyperbolic Geometry 200 8. A. Square Learns to Draw a Hyperbolic Plane 202 9. Homogeneous Geometries for all nT 2 with n ≥ 2 205 10. A Homogeneous Geometry for P 2 207 11. Uniqueness of Homogeneous Geometries for Surfaces 208 12. Exercises 209

Part 2. Knots

Chapter 9. Introduction to Knot Theory 215 1. 1-Dimensional Universes 215 2. When Are Two Knots Equivalent? 217 3. The Mirror Image of a Knot or Link 219 4. The Connected Sum of Two Knots 220 5. A Brief History of Knot Theory 221 6. Reidemeister Moves 223 7. Coloring Knots with Three Colors 226 8. Tricolorability and Knot Equivalence 229 9. Oriented Knots and Invertibility 232 10. Connected Sums of Non-invertible Knots 233 11. Exercises 235

Chapter 10. Invariants of Knots and Links 239 1. What’s an Invariant? 239 2. Crossing Number, Tricolorability, and Number of Components 241 3. Positive and Negative Crossings 242 viii CONTENTS

4. Writhe 244 5. Linking Number 245 6. Nugatory Crossings and Alternating Knots and Links 247 7. Tait’s Conjectures about Alternating Knots 248 8. What Proportion of Knots Are Alternating? 250 9. Seifert Surfaces 252 10. The Genus of a Knot 254 11. Using Euler Characteristic to Compute Genus 257 12. A Potpourri of Knot Invariants 258 13. Exercises 263

Chapter 11. Knot Polynomials 269 1. An Introduction to Polynomial Invariants 269 2. The Rules for the Bracket Polynomial 270 3. The Bracket Polynomial and Reidemeister II Moves 272 4. The Bracket Polynomial with Only One Variable 274 5. The Bracket Polynomial and Reidemeister I Moves 275 6. The X-Polynomial 277 7. The Jones Polynomial 279 8. The State Model for Computing the Bracket Polynomial 282 9. Exercises 285

Part 3. Molecules

Chapter 12. Mirror Image Symmetry from Diﬀerent Viewpoints 291 1. Mirror Image Symmetry 291 2. Geometric Symmetry 293 3. Geometric Chirality and Achirality 294 4. Chemical Chirality and Achirality 295 5. Chemical Achirality and Geometric Chirality of Figure 8 298 6. Euclidean Rubber Gloves 301 7. Geometrically Chiral and Achiral Knots 302 8. A Topological Rubber Glove 304 9. Topological Chirality 306 10. Exercises 308

Chapter 13. Techniques to Prove Topological Chirality 311 1. Topological Chirality 311 2. Molecular Knots 312 3. Technique 1: Knot Polynomials 313 4. Technique 2: 2-Fold Branched Covers 315 5. Technique 3: Chiral Subgraphs 318 6. Some Graph Theory 320 7. Technique 4: A Combinatorial Approach 322 8. Exercises 325

Chapter 14. The Topology and Geometry of DNA 329 1. Synthetic versus Biological Molecules 329 2. The Biology of DNA 330 3. The Problem of Packing DNA into a Cell 332 CONTENTS ix

4. Supercoiling 333 5. Visualizing DNA as a Circular Ribbon 335 6. The Linking Number of the Backbones 337 7. The Average Writhe of the Axis 339 8. The Twist of a Backbone around the Axis 340 9. The Relationship between Linking, Twisting, and Writhing 343 10. Representing the Supercoiling with a Single Number 345 11. Replication 346 12. Site Specific Recombination 348 13. Knotted and Linked Products of Recombination 349 14. An Introduction to Tangles 350 15. Rational Tangles 352 16. Operations on Tangles 354 17. The Tangle Model of Site Specific Recombination 355 18. Applying the Model to the Enzyme Tn3 Resolvase 357 19. Exercises 358 Chapter 15. The Topology of Proteins 361 1. An Introduction to Proteins 361 2. Trapping Knots in Proteins 362 3. Which Knots Occur in Proteins? 364 4. Examples of Knotted and Linked Proteins 366 5. Non-planar Graphs in Metalloproteins 369 6. The Protein Structure Nitrogenase 373 7. Möbius Ladders in Metalloproteins 376 8. Möbius Ladders in Small Proteins 378 9. Goodbye for Now 380 10. Exercises 381 Index 383

Preface

There are a quite a few excellent topology textbooks written for undergraduate mathematics majors in their junior or senior year of college. Such books usually begin with an introduction to point set topology, followed by geometric topics such as homotopy, covering spaces, knot theory, the classification of surfaces, or some area of applied topology. While not all undergraduate topology texts follow this precise structure, almost all presuppose experience with proofs and exposure to a rigorous definition of continuity. For math departments that do not have a large number of majors or have an inordinate burden of service courses to teach, offering a topology course which presupposes such a background is almost impossible. As a result, many mathematics majors graduate without having the opportunity to take a topology class. In addition to textbooks written for math majors, there are a number of noteworthy intuitive introductions to topology intended for a wider audience. Such books are great for curious students and for the general public, but can be difficult to teach out of because they rarely include definitions, statements of theorems, or a sufficient number of elementary exercises (without solutions in the back). What seems to be absent from the available topology books is one that is easy to teach from, includes a wide range of topics, has no prerequisites, and makes no assumptions about the mathematical sophistication or motivation of the reader. Such a book could be the basis for a course that would attract students to the math major who might not be excited by the analytical and algebraic arguments that they typically see in their math courses during the first two years of college. This book is intended to fill this gap. It is an elementary introduction to geometric topology and its applications to chemistry, molecular biology, and cosmology. It does not assume any particular mathematical or scientific background, sophistication, or even motivation to study abstract mathematics. It is meant to be fun and engaging while at the same time drawing students in to learn about fundamental topological and geometric ideas. Though the book can be read and enjoyed by non-mathematicians, college students, or even eager high school students, it is intended to be used as an undergraduate textbook. With this in mind, all of the concepts are introduced with explicit definitions and the theorems are clearly stated so that it is easy for an instructor to develop lectures and for students to refer to when doing the homework. Some theorems are proved formally, others are proved intuitively, and for those results that aren’t proved, students are told in which advanced course they might expect to see a proof. In addition, each chapter concludes with numerous exercises which are at a level appropriate for students in their first two years of college, and whose solutions are only available to instructors.

xi xii PREFACE

In contrast with most topology textbooks, the style of the book is informal and lively. Though all of the definitions and theorems are explicitly stated, they are given in an intuitive rather than a rigorous form. For example, rather than introducing the formal definition of an isotopy, two knots are defined to be equivalent if one can be deformed to the other. This style of presentation allows students to develop intuition about topology and geometry without getting bogged down in technical details. In addition, the topics in the book were chosen for their visual appeal, which is highlighted by the abundance of illustrations throughout. In order to make the text more fun and engaging, early on it introduces a cast of characters who reappear in multiple chapters. This includes the 1-dimensional character A. Dash, the 2-dimensional characters A. Square and B. Triangle, and the 3-dimensional character A. 3D-Girl, each of whom is learning about geometry and topology in order to understand his or her own universe. The characters are illustrated with amusing pictures that entice readers to think about the experience from the character’s point of view. The book is divided into three parts corresponding to the three areas referred to in the title. Part 1, consisting of Chapters 1–8, concerns two and three dimensional universes, though there is a brief foray into the fourth dimension in Chapter 2. The goal of Part 1 is to develop techniques that enable creatures in a given space to visualize possible shapes for their universe, and to use topological and geometric properties to distinguish one such space from another. The second part of the text, consisting of Chapters 9–11, provides an introduction to knots and links. Building on the ideas developed in Part 1, the emphasis in Part 2 is on deformations of knots and links, and the use of invariants to distinguish inequivalent knots and links. Tricolorability is introduced in Chapter 9 as an invariant which is relatively simple to understand and apply but does not distinguish many knots. Then Chapter 10 surveys a collection of invariants which are not hard to understand but in some cases are hard to compute, and in the exercises we ask the reader to figure out what invariants should be used to distinguish a given pair of knots or links. Finally, Chapter 11 introduces the Kauffman bracket and Jones polynomial, and provides a step-by-step explanation of how to compute them, as well as some theorems that illustrate their power. The third part of the text, consisting of Chapters 12–15, presents applications of topology and geometry to chemistry and molecular biology. In particular, Chapter 12 compares the concepts of mirror image symmetry from geometric, topological, and chemical viewpoints. It also introduces the idea of using embeddings of graphs in 3-dimensional space as models of non-rigid molecules. Then Chapter 13 presents techniques that can be used to show that some non-rigid molecules are topologically distinct from their mirror images. Chapter 14 explores the topology and geometry of DNA, developing material about rational tangles as a way to model DNA recombination. Finally, Chapter 15 describes topologically complex protein structures, giving examples of knotted and linked proteins, as well as proteins containing non- planar graphs. While Chapter 13 depends on Chapter 12, Chapters 14 and 15 are independent of one another and of Chapters 12 and 13. Thus an instructor can choose which of the applied topics they would like to cover. The three parts of the book make it usable as a textbook for several different types of courses. For example, Part 1 of the book could be the basis for a seminar for first year college students on a topic such as “Dimensions and the Shape of PREFACE xiii the Universe.” Since the book is written with students in mind, the instructor could give daily reading assignments and then lead the class in a discussion of the material followed by breaking the class into small groups to work through some of the exercises. Such a seminar would also lend itself well to open ended writing projects on diverse topics: Describe a sport that could take place in a 2-dimensional universe. Pick a well-known game (for example chess, checkers, Othello, Hex, or Go), and explain how the strategy for the game would change if it were played on a flat torus or a flat Klein bottle. Design a 4-dimensional version of such a game, and explain how the strategy compares to the usual 2- or 3-dimensional version. Imagine the form of a 4-dimensional person, and explain what we would see if the person passed through our space in different ways. If we had a spaceship that could travel in any direction as fast and far as we would like, what evidence should the spaceship seek to help us figure out the shape of our universe? These are a few of the topics. The book could serve as the text for a topology course for math majors with or without prerequisites. The instructor could either cover all three parts of the text, or pick and choose topics from each of the three parts and supplement with point set topics from a more traditional topology text. Another option for a course for math majors would be for the instructor to give lectures to introduce each chapter, and then have students take turns giving lectures on each section within a chapter. The sections were intentionally made short to enable students to read and understand a single section at a time. The text could also be used for an interdisciplinary course linking topology with molecular biology and chemistry. In this case, the course would focus on Part 2 and Part 3 (comprising Chapters 9–15). Part 2 would give students enough background in knot theory to understand the applications to molecular symmetry and the topology of DNA and proteins developed in Part 3. Note that it is not necessary for students to read Part 1 in order to understand Part 2, nor is it necessary for students to have a background in biology or chemistry in order to understand Part 3. A geometry course for future teachers could also be based on this text. Such courses often consist of an axiomatic approach to Euclidean and non-Euclidean geometry. However, approaching geometry this way can be tedious and may actually kill any natural interest the student has in geometry. By contrast, a course for future teachers based on this text would focus on developing geometric thinking and visual intuition together with problem solving and mathematical reasoning. Such a course would cover Part 1, culminating in Chapter 8 which presents Euclid’s five Axioms and alternatives to the Fifth Axiom that are valid on a sphere and a hyperbolic plane. The instructor for such a course would want to present the material slowly in order to emphasize visualization and logical deduction throughout. The nearly 450 illustrations in the text are designed to help students develop their visual intuition, and the many exercises at the end of each chapter would give students necessary practice writing mathematical arguments. Furthermore, the exercises are designed to encourage students to explain their reasoning in complete sentences and integrate illustrations into their explanations whenever possible. This course could culminate with a project such as designing a museum exhibit featuring the student’s choice of the ten most important images from the text with an introduction to the exhibit and short explanatory descriptions associated with each image. xiv PREFACE

The students could also be asked to include some 3-dimensional models constructed themselves. Finally, because the text is readable by students and is self-contained, the book would be well suited for an independent study or senior year project. Also, because of its lack of prerequisites and its division into three parts and the subdivision of each chapter into short sections, it would work well as the textbook for a four-week winter term or a summer course. We hope that at least in some small way this book contributes to one or more of the following large goals: it motivates some undergraduates to major in math; it motivates some math majors to study topology; it motivates some math students to read their textbooks; it convinces some future K–12 teachers that geometry is fun; or it convinces some future scientists, administrators, or government oﬃcials that even the purest areas of mathematics can have important applications. At the very least, we hope that you and your students enjoy reading it. Acknowledgments

As is apparent from the eighteen co-authors listed on the cover, this book has an unusual history. It began as lecture notes I developed for an intensive course I taught at the Summer Mathematics Program (SMP) at Carleton College for eleven summers between 2000 and 2014. The SMP was a very successful four-week, NSF- funded program for women students in their first and second year of college, with the goal of motivating, encouraging, and mentoring these students to get PhD’s in mathematics. The students in the program had some prior exposure to proofs, but there was no specific prerequisite material. The aim of my course was to introduce them to the beauty and applications of topology and geometry, with the hope that the visual types of thinking they would experience might motivate those who were not as excited by algebraic and analytical arguments to continue in mathematics. My course was originally inspired by three books. I used ideas from Jeffrey Weeks’ wonderful book TheShapeofSpaceto introduce the geometry and topology of 2- and 3-manifolds; I used ideas from Colin C. Adams’ excellent book The Knot Book to expose students to knot theory; and I used ideas from my own book When Topology Meets Chemistry to explore applications of topology to molecular structures. However, over the fourteen years that I taught the course, my lectures took on a life of their own, veering away from these books. In particular, my approach to 2- and 3-manifolds became more formal than that of Weeks, though some of the arguments and organization in Chapters 1 through 8 are similar to his. For example, the proof in Chapter 4 that a surface in an orientable 3-manifold is 1- sided if and only if it is non-orientable is essentially the same as his, simply because I couldn’t imagine a better proof. My approach to knot theory became focused on invariants in contrast with Adams’ comprehensive introduction to knot theory, and my approach to applications of topology became less rigorous and more light- hearted than that of When Topology Meets Chemistry, and included the topology of proteins which was not in my earlier book. While my lecture notes worked well for the intensive course that I taught at SMP, I did not originally foresee them becoming a book that could be used more widely. However, my ideas changed when I led the Undergraduate Faculty Program (UFP) at the Park City Mathematics Institute (PCMI) in 2011. As the leader of the UFP, I was charged with guiding a group of faculty to collaboratively produce materials that could be used at a wide range of institutions. I helped select the 18 UFP participants largely based on their interest in developing an undergraduate topology course with few or no prerequisites. These faculty members, who ranged from post-docs to full professors, came from a great variety of institutions. Some of their institutions were small while others were large, some were private and others public, and some had significant numbers of students headed for graduate school

xv xvi ACKNOWLEDGMENTS in mathematics, while others had few math majors, almost none of whom would go on to graduate school. During the UFP, I divided my lectures notes into twelve parts that would become twelve chapters of the manuscript. I then divided the eighteen participants into six groups of three and assigned each group two parts of my lecture notes to make into two chapters, one in the first half of the book and the other in the second half. Starting with their parts of the notes, the groups added topics, examples, exercises, figures, and more complete explanations. In addition, all eighteen UFP participants helped to write solutions to the exercises that were in the 2011 draft of the manuscript. After the program was over, some of the participants continued working on the manuscript in various capacities including editing chapters and adding more exercises and solutions. Ultimately, I revised, edited, rewrote, and added to the final manuscript so that it would all be in the same style and level, and would flow well. Below, I list the eighteen UFP participants together with their individual contributions to the manuscript during and after the program. Maia Averett co-wrote Chapters 2 and 9, and edited Chapter 8. Lance Bryant co-wrote Chapters 3 and 10, and edited solutions for exercises in Chapter 3. Shea Burns co-wrote Chapters 3 and 10. Jason Callahan co-wrote Chapters 3 and 10, edited Chapters 6, 7, and 11, co-created the table of contents and index, clarified throughout the book which proofs would be included in the manuscript and which would be seen in more advanced courses, standardized numbering of results throughout, and contributed the book’s title. Jorge Calvo co-wrote Chapters 2 and 9, edited Chapter 5, created some of the most difficult figures, and formatted and standardized word usage in Chapters 1–7. Marion Moore Campisi co-wrote Chapters 4 and 11, and added and edited figures throughout. David Clark co-wrote Chapters 2, 9, 10, and 14, revised and added exercises to Chapters 9, 10, and 11, edited solutions for exercises in Chapters 4 and 14, added and edited figures throughout, ensured all figures complied with AMS graphics guidelines, helped develop ideas for the graphic on the cover, and compiled this list of contributions. Vesta Coufal co-wrote Chapters 5 and 13, edited Chapters 1 and 10, added exercises to Chapters 6 and 7, checked lengths of sections, created common preambles for the chapters, co-created the table of contents and index, created a driver file, and worked out the technical details of formatting the final manuscript. Elizabeth Denne co-wrote Chapters 6, 7, and 12, edited Chapters 6, 7, 12, and 14, edited solutions for exercises in Chapters 6, 7, and 12, added exercises to Chapter 9, added references and URLs, and proofread the entire manuscript twice. Berit Givens co-wrote Chapters 5 and 13, edited Chapters 1 and 3, and edited solutions for exercises in Chapters 6, 7, and 13. McKenzie Lamb co-wrote Chapters 4 and 11, and edited solutions for exercises in Chapter 8. ACKNOWLEDGMENTS xvii

Emille Davie Lawrence co-wrote Chapters 1 and 8, edited the preface, introduction, and Chapters 5–15, added exercises to Chapter 3, and designed the graphic for the cover. Lew Ludwig co-wrote Chapters 6, 7, and 12, edited Chapter 9, formatted the manuscript as a book, and compiled a list of URLs used in the book. Cornelia Van Cott co-wrote Chapters 1 and 8, edited Chapters 10 and 11, and edited solutions for exercises in Chapters 9, 10, and 11. Leonard Van Wyk co-wrote Chapters 6, 7, and 12. Robin Wilson co-wrote Chapters 4, 11, and 14, and created figures for the solutions. Helen Wong co-wrote Chapters 5, 8, and 13, edited Chapters 1, 3, 7, 8, 14, and 15. Helen has also been my unofficial sounding board and helper for so many aspects of this book that I cannot list them all. Andrea Young co-wrote Chapters 1 and 8, edited Chapter 4, and edited solutions for exercises in Chapters 1 and 8. In addition to the work of the UFP participants, several undergraduate and graduate students contributed to the development of the manuscript and solutions. I list them below together with their affiliation at the time when they worked on the project. Bryan Brown (undergraduate at Pomona College) proofread a draft of the entire manuscript and the solutions. Dwayne Chambers (PhD student at Claremont Graduate University) provided technical support during PCMI, TEXed solutions and created figures for the solutions during PCMI. Gabriella Heller (undergraduate at Pomona College) wrote the first version of Chapter 15 and created the protein drawing using the program Pymol. Indra Elizabeth Kumar (undergraduate at Scripps College) created and edited figures throughout the manuscript. In particular, the pictures of A. 3D-Girl were designed and created by her. Becky Patrais (PhD student at University of Minnesota) proofread and edited solutions to exercises. Benjamin Russell (undergraduate at Carleton College) wrote and edited some solutions to exercises in Chapters 8 and 15. I also want to take this opportunity to thank Jeff Weeks and Colin Adams. Much of the manuscript was inspired by their outstanding books. In addition, I am grateful to Jeff Weeks for encouraging me to pursue this project. When I was afraid this manuscript would be too similar to his book TheShapeofSpace,he insisted that what is important is bringing this material to more students, not who was first to do so. I am also indebted to Colin Adams for his suggestions along the way, and for helpful comments on the figures throughout the manuscript. I hope this book is anywhere near as clear and inspiring as the many wonderful books that Colin has written. Finally, I want to thank Grace Hibbard for generously sharing her mother Helen Wong with me for the last year of this project. Last, but certainly not least, I want to thank Francis Bonahon and Laure Flapan, whose love and support sustained me throughout this project and always. Erica Flapan

Part 1

Universes

https://doi.org/10.1090//mbk/096/01

CHAPTER 1

An Introduction to the Shape of the Universe

Topics: • 2-dimensional universes • Straightest paths in different spaces • Gluing diagrams • The flat torus and the curved torus • Games on a flat cylinder and a flat torus • The extended diagram for a flat torus • The 3-torus • The 3-sphere

1. To Inﬁnity and Beyond

People normally assume that our universe goes off infinitely in all directions, but the true shape and size of our universe is not known. It is hard to believe that we really don’t know something as basic as the shape of our universe, and it’s hard to imagine how the universe could not go on forever in all directions. If the universe were finite, would that mean that the universe has an end that you can’t go beyond? Let’s imagine the following scenario, featuring an astronaut whose name is A. 3D-Girl. Like her name, she is extraordinary and super-duper in every way. She’s smart, funny, charming, adventurous, brave, and anything else you’d want in an interstellar explorer. She’s been sent off deep into space to investigate a potential colonization site that was identified by astronomers through super-duper technologically advanced telescopes. The astronomers were pretty excited when they saw this planet, because it seemed to have many of the same characteristics as Earth. They even named it “Earth 2”. A. 3D-Girl’s spaceship, aptly named A. Spaceship, has a super-duper hyper-warp drive that allows it to zoom off in any direction, going as fast as she wants and as far as she wants. After months of planning and training for the mission, A. 3D-Girl blasts off in her spaceship, traveling in a straight line past Mars, Jupiter, Saturn, Uranus, Neptune, and Pluto, beyond Alpha Centauri, in fact beyond all known stars and planets, until Earth 2 finally comes into view. As the spaceship gets close to Earth 2, A. 3D-Girl notices more similarities with our earth. Earth 2 is a blue and green planet with a single moon. It’s also the third planet from its sun. After landing, she cautiously descends from her spaceship and looks around. To her great relief, the place feels comfortable and

3 4 1. AN INTRODUCTION TO THE SHAPE OF THE UNIVERSE

Figure 1. A. 3D-Girl waving to the crowd as her spaceship blasts oﬀ.

familiar. After looking around a bit more, she does a double take. This place is too familiar. There’s even a doughnut shop that looks exactly like the one that A. 3D-Girl stopped at before she boarded her spaceship. She checks and rechecks her flight log. Nope, she didn’t make a mistake in navigation, and didn’t make any U-turns. In fact, she didn’t make any turns at all in getting here from Earth. She set A. Spaceship to full throttle and went along the straightest, shortest path from Earth to the exact coordinates where the astronomers had told her to go. As crowds of people gather around her, A. 3D-Girl boldly remarks, “Huh? What happened?” Although it is hard to imagine ever developing a spaceship as technologically advanced as A. 3D-Girl’s, this story is not completely unbelievable. Many elementary schools teach children that people in the middle ages thought the earth was flat, and it was not until 1492 when Christopher Columbus sailed the ocean blue that people found out that the earth is round. In fact, this is a myth that was put forth in 17th-century works of fiction including Washington Irving’s fictionalized biography The Life and Voyages of Christopher Columbus. But what is noteworthy about this myth is that to this day people accept it as completely reasonable. We can imagine ourselves in the middle ages, looking out in all directions and conclud- ing that either the Earth must have an edge way off in the distance or it must be a plane that goes on forever. The story about A. 3D-Girl’s voyage isn’t that different from the myth that Christopher Columbus discovered that the earth is round. When we look out into space in all directions, we can’t imagine how space could wrap around and come back to where it started. So we assume that either the universe must end way off in the distance or it must go on forever. But just because we don’t know how to visualize a universe that closes back on itself like the Earth, doesn’t mean it isn’t possible. Though the actual shape of the universe is not known, in this and the next seven chapters, we’ll develop our intuition about some of the possibilities for the shape of our universe. We’ll learn how to visualize universes of different dimensions, and how we might distinguish one such universe from another. While you may end 2. A. SQUARE AND HIS UNIVERSE 5 up with more questions than you started with about our own universe, we hope you’ll find the voyage intriguing.

2. A. Square and His Universe

Sometimes, when you’re faced with a hard math problem, it’s a good idea to start by solving an easier version of the same problem. So instead of thinking about the shape of our universe, we begin one dimension down, by considering how 2-dimensional creatures living in a 2-dimensional universe might think about possible shapes for their universe. You might have heard of the short novel titled Flatland: A Romance of Many Dimensions which was written in 1884 by the English school teacher Edwin A. Ab- bott. The book was meant to be a commentary on Victorian society, but it can give us some insight into the world of 2 dimensions. Whether or not you’ve read Flatland does not matter here (though if you would like to read it, you can down- load a copy from the internet for free). We will simply introduce you to the main character, whose name is A. Square. His portrait is given in Figure 2.

Figure 2. A portrait of A. Square.

A. Square lives in a 2-dimensional universe. He and his space have no thickness, like a shadow or an image projected on a screen. In the book Flatland,A.Square’s universe is a plane. However, there are other possibilities for his 2-dimensional universe. For example, his universe could be a sphere, a torus (that is the surface of a doughnut), a 2-holed torus, a 3-holed torus, etc. (see Figure 3). Continuing in this way, we ﬁnd that there are inﬁnitely many 2-dimensional universes.

Figure 3. A. Square on the surfaces of a sphere, a torus, and a 2-holed torus.

While you might think of these surfaces as 3-dimensional objects in our space, keep in mind that we are only interested in the surfaces of the doughnuts (or the doughnut hole when he’s on a sphere) and not the cake inside. This means that a 2-dimensional being like A. Square could live in one of these universes, but a 3-dimensional person could not. Furthermore, if A. Square lived in one of these universes, he could only move back and forth and up and down, staying in the icing of the doughnut with no way to get to (or even known about) the cake underneath. 6 1. AN INTRODUCTION TO THE SHAPE OF THE UNIVERSE

We say that A. Square’s universe is 2-dimensional because he can only move in two perpendicular directions. Back and forth is one direction, and up and down is another direction. Of course he can combine these two directions to go in any diagonal direction he wants. In general, the number of dimensions of a space is the number of perpendicular directions in which creatures in the space can move. A line and a circle each have only 1 dimension because a creature living inside one of these spaces can only move backward and forward. This is true no matter how the line or circle is situated in our space. In particular, even though we need 3 dimensions to tie a knot, if we consider a knotted circle as a space by itself, it would only be 1-dimensional. Figure 4 illustrates A. Dot who lives in a circle and her cousin A. Dash who lives in a knotted circle. Each of the cousins is only able to move backwards and forwards in her space. They seem happy, but imagine what boring lives they must have. They talk on the phone on a daily basis to report how many laps around their space they run. However, they have no idea that their spaces are diﬀerent because the diﬀerence can only been seen with our 3-dimensional perspective.

Figure 4. A. Dot lives in a circle and A. Dash lives in a knotted circle.

Notice that even though the surfaces in Figure 3 have no boundaries, they all have finite area. Figure 5 illustrates some 2-dimensional universes with infinite area. Since there is only a finite amount of space on a page, we use dotted lines to mean that the spaces go on forever. The illustration on the left is an infinite cylinder. The illustration next to it has two planes joined with a tube and a handle on the lower plane. The illustration on the bottom has a tube and two such handles. Keep in mind that just like the surfaces in Figure 3, these universes consist only of surfaces. In particular, A. Square and his friends can’t get to the 3-dimensional space inside of the tubes or handles. Observe that we could create a 2-dimensional universe with any number of planes joined together with any number of tubes, and each plane could have any number of handles stuck to it. Pushing this idea even further, we could have infinitely many planes joined with infinitely many tubes, and each plane could have infinitely many handles. This gives us infinitely many 2-dimensional universes all with infinite area. Of course, A. Square can’t picture any of these spaces. He thinks there is only one possible universe, and it’s a plane that goes on forever in all directions. We denote such an infinite plane by R2, since we can think of it as the coordinate space with an x-axis and a y-axis. Similarly, we use R3 to denote the coordinate space 2. A. SQUARE AND HIS UNIVERSE 7

Figure 5. Some 2-dimensional universes with infinite area. with x, y,andz axes. Knowing that there are infinitely many finite and infinite 2- dimensional universes gives us the idea that perhaps our own 3-dimensional universe could be something besides R3. But we will have to wait until later in the chapter to see some possibilities. Maybe we should consider the disk illustrated in Figure 6 as a possible 2- dimensional universe. The problem is that a disk has a boundary. If A. Square reached the boundary, he would hit a wall and could go no further. He could not fall off the edge of the universe, because the universe is all there is. He can’t step outside his universe. He also couldn’t go around the edge of the disk and come back on the bottom because, since the disk is only 2-dimensional, it doesn’t have a top and a bottom. Imagine if there was a similar wall in our 3-dimensional universe. Suppose we sent out a spaceship and it smashed into a wall at the edge of our universe. This seems neither pleasant nor likely.

Figure 6. If A. Square lived in a disk, when he reached the edge he could go no further.

As we think about possible universes for A. Square and for ourselves, we do not want to consider the possibility that universes might have walls that you could smash into. This just seems too implausible to be true. Thus, from now on whenever we use the word universe, we will mean a space that has no boundary. Note that this does not mean that we require every universe to be infinite. Keep in mind that 8 1. AN INTRODUCTION TO THE SHAPE OF THE UNIVERSE there are universes like those in Figure 3 which have finite area but no boundary. By contrast, when we use the word space, we won’t make any assumptions about whether or not the space has a boundary. For example, we say that a disk like the one in Figure 6 is a 2-dimensional space but not a 2-dimensional universe. Now that we know that there are so many possibilities for 2-dimensional universes, we want to help A. Square learn about his own universe. Ideally, he would like to create a list of all possible 2-dimensional universes. Even though such a list would be infinite, if it was set up in a systematic way, it would give him an idea what the possibilities were. Even if his list isn’t complete, he would like to determine some property of his universe that would enable him to distinguish it from other spaces on the list or at least to reduce the number of possibilities. For example, if he could determine that his universe had finite area, he could eliminate all of the spaces with planes and tubes and handles. You can think about properties of a space the way we think about characteristics of people. For example, suppose you had a list of all of the students in your math class, and you wanted to find out the name of the girl who sits next to you. Knowing that she’s a girl might help you to reduce the number of possibilities, but you still wouldn’t know which one was her. If the list included major and year of graduation, and you were able to find out this information about her, you could reduce the possibilities even further. We’d like to do the same thing with properties of a universe. For example, suppose that A. Square somehow determines that his universe is one of the infinite surfaces illustrated in Figure 5. The number of planes, tubes, and handles that the surface has is a property which would enable us to distinguish them. However, if A. Square were living in one of these universes, he would not be able to see the planes, tubes, and handles like we can. So we need to think of properties that A. Square could use to distinguish among these universes. If we can succeed at that, perhaps it will give us some insight about properties we could use to learn about our own universe.

3. Straightest Paths in Flatland

One of the tools that A. Square can use in investigating his universe is the notion of a “straight line” between two points. You’ve probably heard people say “the shortest path between two points is a straight line”. This is true if the two points are in a plane. However, if A. Square is on a sphere, what does “straight line” even mean? There are no straight lines or even straight line segments on a sphere. A path which appears to A. Square to be the shortest path between two points will appear to us as a curve since every path on a sphere is a curve. One way to think about what “straight” means to A. Square is to imagine that he has a 2-dimensional flashlight that he is shining in front of him. We could picture his flashlight like the lamp that’s attached to a miner’s helmet. A 2-dimensional ray of light from his flashlight stays within his universe, illuminating the shortest path to get to a nearby point. So if he follows the beam of his flashlight, his path will appear to him to be a straight line. We will call the path of a light ray within his universe a straightest path, even if it doesn’t actually follow a straight line in our 3-dimensional universe. 3. STRAIGHTEST PATHS IN FLATLAND 9

Another approach that A. Square could take to finding the straightest path between two points would be to attach a piece of string to one of the points in his space and then go to the other point and pull the string tight. Then if he walks along the string, he knows he will be going as straight as he can. If A. Square lives in a plane, then the straightest path between two points is a line segment. But if he lives in a sphere, then the straightest path between two points is an arc on what’s known as a great circle. The great circles on a sphere are circles whose radius is the same as the radius of the sphere. An equator is an example of a great circle. Great circles can be thought of as the intersection of a sphere with a plane that divides the sphere into two equal halves. Note that if the plane does not divide the sphere exactly in half, then the circle of intersection is not a great circle. Paths which follow great circles curve as little as possible while staying within the surface of the sphere. Airplanes and ships follow great circles around the earth because these are the shortest paths between two points on earth. If you have looked at flight paths in airline magazines, you may have noticed that the flight paths are drawn as curves rather than as straight lines. These flight paths appear to be longer than necessary, but that’s because a flat map of our spherical earth will necessarily be distorted. Let’s consider how A. Square might use the idea of straightest paths to distinguish between a plane and a sphere. Suppose his universe is a plane, and he leaves home to take a trip following a straightest path (making use of his headlamp to go straight). Much to the disappointment of his family, if he continues following this straightest path, he will never return home. However, if A. Square lived on a sphere and left on a similar voyage following a straightest path, after some time he would happily return to his awaiting loved ones (see Figure 7). At this point he would know that his universe could not be a plane. It’s important in this experiment that A. Square use a taut string or a light ray to follow a straightest path. Otherwise, he could just walk in a circle on the plane and return to his starting point.

Figure 7. Straightest paths on a plane and a sphere.

In order to see if a similar method would enable A. Square to differentiate between a sphere and a torus, consider the path illustrated in Figure 8. A string could be pulled taut along this path so that A. Square would know that it’s indeed a straightest path. If A. Square travels along this path, he will eventually return to his starting point as he did on the sphere. Thus finding a straightest path that returns to its starting point will enable A. Square to distinguish between a torus and a plane, but will not give him enough information to distinguish between a torus and a sphere. In search of some other property that a torus has and a sphere doesn’t have, A. Square tries to find a different type of straightest path on the torus. The problem 10 1. AN INTRODUCTION TO THE SHAPE OF THE UNIVERSE

Figure 8. A straightest path on a torus. is that ﬁnding such paths on the torus is not as straightforward as it seems. For example, take a look at the grey circle on the left of Figure 9. At ﬁrst glance, we might think that it is a straightest path. It’s nice and round and rests just at the top of the surface. However, if A. Square marks points p and q on the grey circle and pulls a string tight between these points, the string will not lie on the grey path. The black string between points p and q on the left in Figure 9 is just a bit shorter than the grey path going between the points. So the grey path on the left of Figure 9 isn’t a straightest path on the torus after all. While it’s hard to identify a straightest path on the top of a torus, an inner circle going around the hole of the torus is a straightest path because the circle cannot be made any shorter. Such a shortest path is illustrated on the right of Figure 9

Figure 9. The grey circle on the left is not a straightest path, but thegreycircleontherightis.

When looking at the illustration on the left of Figure 9, it’s important to keep in mind that to us the black string is curved, though not as curved as the grey string. A truly straight segment would cut through the inside of the torus, but A. Square’s string could not leave the surface of the torus to follow such a path. Now let’s imagine that A. Square lives on a torus and has two balls of string, one black and the other one grey. He pins down the end of the black string, then follows the straightest path illustrated on the left in Figure 10 pulling the string taut as he goes. After returning to his starting point, he ties the two ends of the black string together. Then he repeats this process with the ball of grey string going around the inner circle of the torus, again as illustrated on the left in Figure 10. He notices that the black and grey circles are both straightest paths, and they intersect at precisely one point (indicated by a dark grey dot). On the other hand, if A. Square lives on a sphere, any two straightest paths will intersect in two points as illustrated by the black and grey circles on the right 4. EXPLORING THE SHAPE OF A CAVE 11

Figure 10. Two straightest paths can meet in exactly one point on a torus, but not on a sphere. in Figure 10. Thus this experiment with colored string will allow him to distinguish between a torus and a sphere. Could A. Square use additional balls of colored string to distinguish between tori with different numbers of holes? He would have to begin by finding some straightest paths on a torus with two holes. You may want to experiment with this yourself. Get a bagel, a giant pretzel, some pins, and a few skeins of colored yarn, and use them to try to distinguish the bagel from the pretzel. (How many holes does a pretzel have?) Afterwards, when you eat the bagel and pretzel make sure you first remove the yarn and the pins.

4. Exploring the Shape of a Cave

Having watched A. Square use straightest paths to compare different universes, we would like to use similar ideas to compare 3-dimensional universes. But before considering such a hard problem, let’s check on A. 3D-Girl, who has been suffering from nightmares and insomnia ever since her confusing voyage to Earth 2. During a restless night of sleep, A. 3D-Girl thinks she’s woken up in a deep, dim, dank cave. Feeling her way along the wall, she discovers that the cave has no exit and no entrance to any other passageway. While exploring the entire cave, she starts thinking about a book she read a long time ago about a square trying to figure out the shape of his universe, and suddenly a ball of grey string appears in her hand. She places one end of the string under a big rock so it can’t move, and trails the string behind her as she walks. Eventually she returns to the big rock where she started without ever retracing her steps. From this exploration she concludes that the cave is like the inside of a doughnut. The cave has some nooks and crannies as all caves do, but the overall shape is as illustrated in Figure 11. Depressed at the thought of a life walking around in circles trailing grey string, she decides to go back to sleep hoping to wake up in her own comfortable bed. Unfortunately, the next thing she knows she’s in a different cave, again with no exit and again with a ball of grey string in her hand. After exploring for a while in the dark, she discovers that this cave has a single four-way intersection. She winds the string around both circular passageways making it into a figure-eight. She pictures this cave as illustrated in Figure 12. She becomes agitated as she dreams that she is waking up in more and more complicated caves. What if the cave had two three-way intersections? The cave could look like two circular loops connected by a passageway (as illustrated on the left of Figure 13), or it could be one circular loop with a passageway cutting across 12 1. AN INTRODUCTION TO THE SHAPE OF THE UNIVERSE

Figure 11. A cave which looks like the inside of a doughnut.

Figure 12. A cave with one four-way intersection.

Figure 13. These caves each have two three-way intersections. the middle (as illustrated on the right of Figure 13). How would she know which one it was? Could she use a ball of string to help ﬁgure it out? After some thought, A. 3D-Girl decides that she could distinguish these two caves by seeing if there is a circular path from one three-way intersection back to itself that doesn’t pass near the other three-way intersection, as there would be in the cave on the left in Figure 14. There would be no such path if she were in the cave on the right. But what if she were in a cave that had many diﬀerent types of intersections, and included some branches which seemed to go on forever and others which looped back on themselves? The thought of listing all the possibilities and coming up with a plan to distinguish them makes her head spin so much that she passes out. Luckily, the next time she wakes up she is actually back in her own bed. But her night adventures leave her wondering what could help her answer these sorts of questions. Perhaps if she checks back with A. Square, she can get some ideas from him. 5. CREATING UNIVERSES BY GLUING 13

Figure 14. On the left there’s a path from a three-way intersection back to itself that doesn’t go near the other three-way intersection.

5. Creating Universes by Gluing

A. Square was able to use the notion of straightest paths to distinguish between a plane (denoted by R2), a sphere (denoted by S2), and a torus (denoted by T 2). Note that we use the exponent 2 whenever we want to indicate that a space is 2- dimensional. This is in contrast with ordinary 3-dimensional space which is denoted by R3. The terms “ball” and “sphere”, and “disk” and “circle” are often used interchangeably in normal life. But when talking about spaces of different dimensions, we want to use these terms more precisely. A sphere S2 is the 2-dimensional surface that is the boundary of a 3-dimensional ball B3.AcircleS1 is the 1-dimensional boundary of a 2-dimensional disk D2. We will now introduce the idea of gluings and instant transport to give A. Square a way to visualize some of the possibilities for his universe. Later, we will use similar techniques to enable us to visualize some analogous 3-dimensional universes. We begin by trying to help A. Square visualize a sphere. Drawing the usual picture of a sphere (like the one on the right side of Figure 10) won’t help A. Square at all, since he doesn’t have the 3-dimensional perspective necessary to understand that the dotted arcs are behind the rest of the picture. Instead we want to break up the sphere into pieces that A. Square can understand. As 3-dimensional creatures we know that a sphere can be obtained by gluing two hemispheres together along their boundaries. To draw a picture of two hemispheres being glued together, we put arrows on the boundaries of each hemisphere to indicate how they should be matched up (as illustrated in Figure 15). In particular, the head of the arrow on one hemisphere is glued to the head of the arrow on the other hemisphere. A diagram which uses arrows to indicate how the boundaries of shapes are glued together is called a gluing diagram. We can think of each hemisphere as being a disk that has been curved, like a disk made of play dough that you mold around half of a ball or even around your fist. Of course, A. Square cannot visualize disks being curved around a 3- dimensional object, because that would require him to have 3-dimensional vision. However, he can draw two flat disks with arrows on their boundaries. In order to help him understand that the boundaries of the two disks are glued together, we explain to him that when he arrives at the boundary of one disk, he will be instantly transported to the boundary of the other disk (see Figure 16). We show him the gluing diagram in Figure 16 to help him understand. A. Square reads a lot 14 1. AN INTRODUCTION TO THE SHAPE OF THE UNIVERSE

Figure 15. A sphere consists of two hemispheres that are glued together along their boundaries. of science ﬁction so he accepts the idea that instant transport can take him from one disk to the other.

Figure 16. When A. Square arrives at the boundary of one disk, he is instantly transported to the boundary of the other.

Another way to think about gluing diagrams is as pages of a road atlas. When you drive oﬀ the right side of the map on one page of an atlas, there is a number or a letter to indicate that you drive onto the left side of a map on another page. If there were a road that went all the way around the earth, you could follow the pages of the atlas going from the map on one page to a map on another page and another and another, until eventually you would come back to the page where you started. Now that A. Square can visualize a sphere as two disks glued together, we would like to use a gluing diagram to help him visualize a torus. Let’s consider a 2-dimensional universe consisting of a square where opposite sides of the square are glued together. We put arrows on the sides of the square to indicate how they are attached. The head of the single arrow is glued to the head of the other single arrow, and the head of the double arrow is glued to the head of the other double arrow, as illustrated on the left side of Figure 17.

Figure 17. Gluing up opposite sides of a square. 6. GAMES ON A GLUING DIAGRAM FOR A CYLINDER AND A TORUS 15

As 3-dimensional people, we can physically glue up opposite sides of the square (if the square is made of a sufficiently flexible material). After gluing together one pair of sides we get a cylinder. Then if we glue together the double arrows on the ends of the cylinder, we get a torus as illustrated on the right side of Figure 17. You might want to check for yourself that it doesn’t matter which pair of sides we glue together first—we’ll still end up with a torus after both pairs of sides are glued. A. Square can’t visualize gluing up the sides of the square. So we have to explain to A. Square that if he goes through the top edge of the square, he is instantly transported to the bottom edge, and if he goes through one side, he is instantly transported to the other side (see Figure 18). It’s just like a video game where a character who goes off the screen on one side reappears on the opposite side. When A. Square is not reading science fiction, he likes to play video games. So this explanation also makes sense to him.

Figure 18. When A. Square exits from the right edge, he returns on the left edge.

If we want to physically glue up opposite sides of a square, the square needs to be made out of a flexible material. In particular, if the square is made of paper, we can only glue together one pair of opposite sides at a time (you should try this to see for yourself what happens). In fact, there is no way to make a nice round torus out of paper. For this reason, it is more convenient for us to think of the torus the way A. Square does, as an abstractly glued up square. We use the word “abstractly” here to mean that we’re thinking about it in terms of instant transport rather than as a square with sides that are physically glued together. When we think of a torus this way, we call it a flat torus; whereas when we imagine a torus as the surface of a doughnut or a bagel, we call it a curved torus. A flat torus is a “better” universe than a curved torus because a flat torus is the same everywhere. For example, if A. Square watches his friend B. Triangle move around on a flat torus, she will always have the same area even if she goes out one edge and comes in another (as illustrated on the left side of Figure 19). This is not the case if she moves around on a curved torus. For example, on the right side of Figure 19 we see that B. Triangle might get a stomach ache if she eats a big dinner at a restaurant near the outer circle of a torus and then walks towards the inner circle. It wouldn’t be very pleasant to live in a place where you would get a stomachacheeverytimeyouwalkedinacertainneighborhood.

6. Games on a Gluing Diagram for a Cylinder and a Torus

One way to check your understanding of abstract gluings is to play games on a gluing diagram. For example, you can play Connect Four on a gluing diagram for a cylinder (see Figure 20). The normal game of Connect Four is like tic-tac-toe but 16 1. AN INTRODUCTION TO THE SHAPE OF THE UNIVERSE

Figure 19. B. Triangle changes her shape when she moves around on a curved torus. with gravity. One player has black tokens and the other player has grey tokens. The players take turns dropping their tokens into a vertical plastic grid which has seven columns and six rows. As you drop in a token, it falls down due to gravity. The game ends when one player has four in a row either horizontally, vertically, or diagonally.

Figure 20. The game of Connect Four on a cylinder.

While normal Connect Four is played in a vertical rectangular grid, we can also play Connect Four on such a grid where the pair of vertical sides have been glued together as indicated by the arrows in Figure 20. This gives us a cylindrical grid. There still is gravity which causes the tokens to drop down to the bottom of the cylindrical grid as illustrated in Figure 20. As with normal Connect Four, you win if you are the first player to get four in a row horizontally, vertically, or diagonally. For example, consider the game in progress in Figure 20. It’s Grey’s turn, and Grey can win in one move. Can you see how? You should find a partner and play Connect Four a few times on the gluing diagram for a cylinder. In 1988, Victor Allis “solved” Connect Four in the sense that he determined a strategy wherein the first player can always win. How might your strategy for Connect Four on a cylinder be different from your strategy in a normal Connect Four game? We can also play games on a flat torus. In Figure 21, we see a game of tic-tac- toe on a torus where X has won. This game would be much harder to draw (and to think about) on a curved torus. You should stop reading now to play a game of torus tic-tac-toe with a friend. 7. EXTENDED DIAGRAMS 17

Figure 21. X has won this game of torus tic-tac-toe.

There are many other games that you can play on a flat torus. Could you play Connect Four on a flat torus? You should check out the website http://www.geometrygames.org/TorusGames/ where you can play tic-tac-toe, mazes, crossword puzzles, word search puzzles, jigsaw puzzles, chess, pool, and gomoku all on a flat torus.

7. Extended Diagrams

One way to plan your strategy when playing games on a gluing diagram is to draw an extended diagram to help you see where you go when you go off one side of the board. In order to construct an extended diagram for the flat torus, let’s first imagine that A. Square lives all by himself in a flat torus. To keep track of his experience, let’s imagine that curtains are hung at the edges of the gluing diagram. This gives him the impression that his neighborhood is a square, but he can walk through the curtains. He puts on his hat to go out for a walk. When he walks through the curtains on the right side of the gluing diagram he is instantly transported to the left side. However, he thinks he has entered a second square neighborhood where there is another guy with a hat just like his, who is also taking a walk at the same speed that he is, and that guy is going through the curtain on the right just as A. Square enters through the curtain on the left (see Figure 22). In fact, the square that he sees taking a walk ahead of him is really just A. Square himself.

Figure 22. A. Square sees a copy of himself up ahead.

A. Square continues his walk in the same direction, believing he is leaving the second square and entering a third square, and so on. Continuing in this fashion, he believes that there are infinitely many squares in front of him, one after another. Similarly, he believes that there are infinitely many squares on top of him stacked one above the other, and infinitely many squares going off on a diagonal. Altogether, 18 1. AN INTRODUCTION TO THE SHAPE OF THE UNIVERSE

Figure 23. A portion of an extended diagram of a flat torus with A. Square in it. he imagines there is an entire plane full of squares glued together left to right and top to bottom, containing other creatures who look just like him, wearing hats like his, and moving at exactly the same rate as he is. The space he imagines he lives in is called the extended diagram of the flat torus. We see a portion of the extended diagram of the flat torus in Figure 23. Keep in mind that all of the square gluing diagrams in the extended diagram are really just a single flat torus, and all of the guys with hats on are really just A. Square himself.

Figure 24. The three black X’s in an extended diagram represent three in a row in ﬂat torus tic-tac-toe. 8. INTRODUCING THE 3-DIMENSIONAL TORUS AND SPHERE 19

It’s a bit like when you are in a dressing room with opposing mirrors and it seems like there are infinitely many copies of you, though one is blocking another so you can’t really see them all. You know that really you are the only you, and the copies of you in the mirror are just your images. If we draw an extended diagram of a flat torus tic-tac-toe board, then it is easier to see all the possible ways you can get three in a row. For example, the row of X’s that we saw in Figure 21 looks like an ordinary three in a row in the extended diagram (see the three black X’s in Figure 24). In fact, there are infinitely many copies of these three X’s in the extended diagram and we could choose any three adjacent ones to represent the 3 in a row on the tic-tac-toe board from Figure 21. You should try to see if an extended diagram can help you find a winning move for grey in the game of Connect Four illustrated in Figure 20. Note that the extended diagram for a cylinder will look like a strip of paper that goes on indefinitely to the left and the right but not on the top and the bottom.

8. Introducing the 3-Dimensional Torus and Sphere

Now that we have some experience helping A. Square understand some possibilities for his 2-dimensional universe, let’s consider some analogous possibilities for our 3-dimensional universe. It is natural for us to assume that our universe goes off infinitely in all directions, just as it was natural for A. Square to assume that his universe was an infinite plane. We saw that while it is easy for us to think of many different possibilities for A. Square’s universe, it is hard for him to visualize these possibilities. We used the idea of gluing up disks and squares in order to help him understand what it would be like if his universe were a sphere or a torus. This gives us the idea that our own 3-dimensional universe might be an analogously glued up 3-dimensional shape. We begin by exploring a 3-dimensional version of the flat torus. Suppose that you live in a large cubical room with pairs of opposite walls glued together and the floor and the ceiling glued together. We imagine this space “abstractly” just as A. Square imagines a flat torus. We don’t assume that the room is flexible and exists in a higher dimension where opposite walls curve around so that they can be glued together. Rather, we imagine that a person is instantly transported from one wall to the same point on the opposite wall. This 3-dimensional universe is called a3-torus or a 3-dimensional torus and is denoted by T 3. In order to help us visualize this, let’s imagine that A. 3D-Girl lives in a 3-torus. If she floats through the ceiling, she comes back through the floor (see Figure 25). This is similar to what we saw in Figure 18 when A. Square travelled through the right edge of the square and reappeared on the left edge. Even though we visualize the 3-torus as a glued up cube, there is no boundary to this space. We never have to worry about walking “off the edge” or bumping into a wall. Figure 26 illustrates a gluing diagram for a 3-torus where pairs of identical arrows are used to indicate that opposite walls are glued together. Notice that the head of an arrow is glued to the head of the similar arrow with the colored part glued to the colored part and the white part glued to the white part. We use three different types of arrows because there are three pairs of opposite walls that are glued together. 20 1. AN INTRODUCTION TO THE SHAPE OF THE UNIVERSE

Figure 25. If we glue up opposite faces of a cube, we get a 3-torus.

Figure 26. The gluing diagram of a 3-torus.

A very, very large version of the 3-torus could actually be the shape of our universe. If this were the case and we had a spaceship with a super-duper hyper- warp drive that allows it to zoom off as fast as we want and as far as we want, then we could send it off in one direction and it would eventually come back to Earth from the opposite direction. In reality, though, such a spaceship would probably take billions of years to return to Earth. In fact, if our universe were a 3-torus and we had a really powerful telescope, we could see a copy of our own Earth (though the image we would see could be billions of years old since it would take the light billions of years to traverse the entire universe to get back to us). Another possibility for the shape of our universe is a 3-dimensional sphere (also known as a 3-sphere), which is denoted by S3. Itisveryimportanttoremember that S3 is NOT a solid 3-dimensional ball (which we denote by B3). Rather S3 is the 3-dimensional analog of a 2-dimensional sphere S2. The ball B3 cannot be a universe because it has a boundary, and we are assuming that no universe can have a boundary. There are several ways to visualize S3, each of which is analogous to an approach that A. Square could use to visualize S2. The first way to think about S3 is to formally define it in terms of coordinates using an equation. With this approach, we would define S2 = {(x, y, z)|x2 + y2 + z2 =1} as a subset of R3. Keep in mind that S2 is 2-dimensional even though it’s a subset of R3. This definition of S2 makes it easy for 3-dimensional creatures to picture S2 as the outside of a round ball with radius 1. By analogy, we could define S3 in 8. INTRODUCING THE 3-DIMENSIONAL TORUS AND SPHERE 21 terms of coordinates as S3 = {(x, y, z, w)|x2 + y2 + z2 + w2 =1} as a subset of R4. However, this doesn’t give us much intuition about how to visualize S3 since it’s hard enough to picture R4. A. Square has the same problem if he tries to picture S2 = {(x, y, z)|x2 + y2 + z2 =1} as a subset of R3. For a more visual approach, recall that we helped A. Square visualize a sphere by telling him that a sphere is made of two disks which have been glued together along their boundaries (see Figure 16). In order to construct S3 in an analogous way, we need to first find the 3-dimensional analogue of a disk. One way of defining a disk is as the set of points in R2 whose distance from the origin is 1. We would express this symbolically as D2 = {(x, y)|x2 + y2 ≤ 1} as a subset of R2. Thus the 3-dimensional analogue of a disk is the set of points in R3 whose distance from the origin is 1. We express this symbolically as B3 = {(x, y, z)|x2 + y2 + z2 ≤ 1} as a subset of R3. Hence the analogue of a disk is just a solid ball.Nowthe3- dimensional analogue of constructing S2 by gluing two disks together along their boundaries is constructing S3 by gluing two solid balls together along their boundaries. If A. 3D-Girl lives in a 3-sphere, then she can travel through the boundary of one of the solid balls into the other solid ball as illustrated in Figure 27.

Figure 27. S3 consists of two balls that are glued together along their boundaries.

No matter how we imagine contorting or smashing the balls in Figure 27, it is impossible for us to visualize the boundaries of the balls actually being glued together. This is because we would need four dimensions in order to do the gluing. This is the same problem that A. Square would have if he tried to visualize gluing together the boundaries of the two disks in Figure 16. The best way for us to imagine the gluing is to think about it point by point, so that each point on the boundary of one ball is glued to the corresponding point on the boundary of the other ball. The gluing arrows on the spheres can help us remember how one ball is glued to the other. Now we know that in addition to R3, our universe could be a 3-torus T 3 or a 3-sphere S3. As soon as we discover new possibilities for our universe, we want 22 1. AN INTRODUCTION TO THE SHAPE OF THE UNIVERSE to know how we could determine whether our universe has one of these forms. If we could somehow map out every cubic inch of our universe, we would know that our universe was ﬁnite. So we could conclude that it wasn’t R3. But how could we distinguish between T 3 and S3? As usual, it’s a good idea to start by answering the analogous question for 2-dimensional universes before we attempt to answer it for our own universe.

9. Distinguishing a 2-Sphere from a 2-Torus

In Figure 10, we saw that A. Square could ﬁgure out whether he was living in S2 or T 2 by considering diﬀerent straightest paths. In particular, he observed that two straightest paths can meet in exactly one point on T 2, but not on S2. Another way to distinguish between S2 and T 2 is to consider what would happen if a wall were built along a straightest path. For example, on the left side of Figure 28 we see that A. Square is in the northern hemisphere of S2 and B. Triangle is in the southern hemisphere. The Great Circular Wall which was built along the equator divides S2 into two pieces preventing A. Square from visiting B. Triangle. On the other hand, as we see on the right side of Figure 28, a wall along a straightest path on T 2 might make it inconvenient for A. Square to visit B. Triangle, but if he’s willing to walk around the long way he will eventually get there. Of course, walls can be built on a torus which separate it into two regions as we see in Figure 29. However, such a wall does not follow a straightest path.

Figure 28. A wall along a straightest path in S2 divides the space into two regions, but it doesn’t in T 2.

Figure 29. This wall is not along a straightest path. 9. DISTINGUISHING A 2-SPHERE FROM A 2-TORUS 23

In order to explain to A. Square how he can use walls along straightest paths to distinguish between a torus and a sphere, we need to use gluing diagrams rather than the illustrations in Figure 28, since he can’t understand 3-dimensional illustrations. Also, it’s easier to draw straightest paths in the gluing diagram than on the curved torus, since straightest paths in the gluing diagram just consist of line segments. To show him a wall along a straightest path in T 2, we draw Figure 30. It’s then easy for him to check that the wall doesn’t divide T 2 into two separate regions.

Figure 30. This wall doesn’t separate T 2 into two regions.

Next we want to show A. Square that any wall along a straightest path in S2 will separate it into two regions. The problem is it’s not so easy to draw a straightest path in the gluing diagram for S2, because the gluing diagram for S2 flattens the hemispheres into disks which distorts distances. This means that if we draw a line segment in one of the disks, it won’t necessarily correspond to a great circle on S2. Instead, let’s start out with a great circle which goes between A. Square and B. Triangle (illustrated as a thick black circle in the left image of Figure 31). This great circle will be our wall. To make it easier to draw this circle in the gluing diagram, we divide the sphere into two hemispheres so that half of the black circle is in each hemisphere, as illustrated in the second image of Figure 31. Then we flatten out the hemispheres into disks so that the black semi-circles are now diameters of the disks, as illustrated in the right image of Figure 31. We only show A. Square the picture of two disks on the right, since the other pictures in the figure wouldn’t make sense to him anyway. We then explain to A. Square that even if he goes from one disk to the other (as indicated by the dotted lines and arrows) he will still be on the opposite side of the wall from B. Triangle.

Figure 31. This wall separates S2 into two regions. 24 1. AN INTRODUCTION TO THE SHAPE OF THE UNIVERSE

Figure 32. The grey circle will become a diameter in each disk if we cut along this great circle.

You may be concerned that we have only shown A. Square that this particular wall separates a sphere into two regions. But given any great circle, we can cut the sphere into hemispheres so that half of the great circle is in each hemisphere, and hence each half of a great circle becomes a diameter of one of the disks. For example, if we want to build a wall along the great circle which is grey in Figure 32, we can cut the sphere into hemispheres along the black circle, giving us the same picture of disks that we had in Figure 31. So we have in fact shown that any wall along a straightest path on the sphere separates it into two regions.

10. Distinguishing a 3-Sphere from a 3-Torus

Now let’s see if we can use walls along straightest paths to distinguish between T 3 and S3. Since we are inhabitants of our space, we have to use gluing diagrams, just as we did for A. Square. To build the wall, we lay a row of bricks along a straightest path, expanding the wall by laying more bricks up, down, left, and right, just as you would if you were building a brick wall to keep deer out of your vegetable garden. Recall that for the walls we built in T 2 and S2, one side of the wall met the other side so the wall was circular. Since we now have one more dimension, not only do the sides of the wall meet, but the top and bottom of the wall also meet. Figure 33 illustrates such a wall in T 3. We didn’t put the arrows on opposite faces of the cube, but you should remember that opposite sides of the cube are glued together. This means that the top of the wall is indeed glued to the bottom of the wall and the left side of the wall is glued to the right side of the wall. If A. 3D-Girl walks with her back to the wall, she can go visit her friend on the other side of the wall. Thus this wall does not separate T 3 into two regions, just as the wall in Figure 30 didn’t separate T 2 into two regions. We have to be a little more careful with the gluing diagram for S3,sincethetwo balls have been “flattened” to fit in three dimensions just as the two hemispheres of S2 had to be flattened to fit in two dimensions. Recall that we saw in Figures 31 and 32 that given any wall along a straightest path in S2, we can divide S2 into two disks so that the wall cuts each disk exactly in half. By analogy we now see that given any wall along a straightest path in S3, we can divide S3 into two balls so that the wall splits each ball exactly in half. In Figure 34 we illustrate a wall along a straightest path in S3 which splits each ball exactly in half. As we can see in the figure, even if A. 3D-Girl walks away from the wall into the next ball, she 10. DISTINGUISHING A 3-SPHERE FROM A 3-TORUS 25

Figure 33. A. 3D-Girl is going to visit her friend on the other side of this wall in T 3.

Figure 34. No matter where A. 3D-Girl goes, she can’t visit her friend on the other side of the wall. will remain on the same side of the wall. In fact, no matter where she goes in S3, her friend will always be on the other side of the wall.

Thinking about 3-dimensional universes like T 3 and S3 reminds A. 3D-Girl about her nightmare with the caves, and she starts to worry that she’ll have the same nightmare when she goes to sleep later. She has a momentary panic about losing her ball of string, and forces herself to take a few deep breaths. Once she’s a bit more calm, she starts thinking about whether she could build walls along straightest paths to distinguish different caves. Recall that in Section 4, A. 3D-Girl was faced with the problem of distinguishing between the two caves in Figure 35. To solve this problem, she had figured out that for the cave on the left she could lay out a loop of string from one three-way intersection back to itself that didn’t go near the other three-way intersection. This wouldn’t be possible for the cave on the right of Figure 35. Now she realizes that she could also distinguish between the two caves in Fig- ure 35 by building walls along straightest paths. In particular, for the cave on the left in Figure 36, she could build a wall that separates it into two doughnut-shaped regions which she couldn’t do for the cave on the right. A. 3D-Girl reasons that even if she didn’t want to go to the trouble of building a complete wall, she could use rocks to mark where the wall should go. Knowing two different techniques to distinguish the caves reassures A. 3D-Girl enough that 26 1. AN INTRODUCTION TO THE SHAPE OF THE UNIVERSE

Figure 35. On the left, there’s a path from a three-way intersection back to itself that doesn’t go near the other three-way intersection.

Figure 36. The wall on the left separates the cave into two doughnut-shaped regions. she can go to sleep, during which has the most wonderful, peaceful, blissful sleep she’s had since her voyage to Earth 2.

11. Exercises

1. Suppose A. Square lives alone in a small flat torus and each of his sides is a different color. Imagine that he can see in all directions at once. Draw an extended diagram representing how A. Square might imagine his universe. Explain your drawing. 2. Suppose A. Cube lives alone in a small 3-torus and each of her sides is a different color. Imagine that she can see in all directions at once and has depth perception. Draw an extended diagram of A. Cube’s universe. Explain your drawing. 3. Suppose A. Square lives alone in a small sphere. What direction would he look if he wanted to see the top of his head? 4. How could A. Square detect the difference between a plane, an infinite cylinder, and an infinite cone? Is an infinite cone or an infinite cylinder a reasonable 2- dimensional universe for A. Square to live in? Why or why not? 5. How do we create an infinite cylinder and an infinite cone by gluing up a planar surface? What could be the 3-dimensional analogue of an infinite cylinder? What about an infinite 3-dimensional cone? 11. EXERCISES 27

6. Suppose we tell A. Square that his universe is either a ﬂat torus, a sphere, or a plane. What information would help him to determine which it really is? 7. Suppose we were told that our universe was either a 3-torus or R3.What information would we look for to determine which it really was? 8. Draw a circular path on a torus which intersects both the thick black and the thick grey circle in Figure 10 in just one point. Draw a second circular path also with this property, but which can’t be deformed on the torus to the ﬁrst one you drew. Your circular paths do not have to be straightest paths. 9. Draw a circular path on the torus in Figure 10 which intersects the thick black curve in two points and intersects the thick grey curve in three points. 10. In the Connect Four game on a cylinder illustrated in Figure 37, can either player win in one move if that player goes next?

Figure 37. Illustration for Exercise 10.

11. In tic-tac-toe, we say two moves are equivalent if the strategy for the rest of the game after one of the moves is analogous to the strategy for the rest of the game after the other move. For example, in regular tic-tac-toe, a first move in one corner is equivalent to a first move in any of the other three corners; and a first move in the middle of one side is equivalent to a first move in the middle of any side. Thus in regular tic-tac-toe, there are three inequivalent moves for the first player: the center, a corner, or the middle of a side. How many inequivalent first moves are there for tic-tac-toe on a cylinder (see Figure 38).

Figure 38. Illustration for Exercise 11.

12. In torus tic-tac-toe how many inequivalent moves are there for the ﬁrst player? 28 1. AN INTRODUCTION TO THE SHAPE OF THE UNIVERSE

13. In torus tic-tac-toe, how many inequivalent moves are there for the second player? 14. Is either player guaranteed to win torus tic-tac-toe if both players use optimal strategies? Justify your conclusion. 15. Is it possible to have a tie in torus tic-tac-toe if both players wanted to? Justify your conclusion. 16. If you play torus chess, with the pieces set in the usual starting positions, explain how the ﬁrst player can win in just one move. Now, design a better starting position that does not lead to either checkmate or a stalemate in just one move. 17. The wall built along a straightest path in Figure 30 does not divide T 2 into two regions. Suppose you built two walls in T 2 both along straightest paths. How many regions would the walls divide T 2 into? (Hint: Your answer may depend on whether the walls intersect or not.) 18. It is possible to build three intersecting walls along straightest paths in T 3 that still do not separate T 3 into diﬀerent regions. Explain or draw a picture showing how this can be done. 19. Explain how A. 3D-Girl can use walls to distinguish between a doughnut-shaped cave (see Figure 11) and a cave with one four-way intersection (see Figure 12). 20. In Figure 32, we started with a wall along the thick black circle on the sphere, then cut along a perpendicular equator to get the gluing diagram on the right. What would the thick black circle look like in a gluing diagram if we cut along an equator that was not perpendicular to it? Draw a sketch and explain. https://doi.org/10.1090//mbk/096/02

CHAPTER 2

Visualizing Four Dimensions

Topics: • How A. Square can visualize 3 dimensions • Projections and perception • Movies and movement • Inductively deﬁning n-dimensional cubes • Inductively deﬁning n-dimensional tetrahedra

1. Teaching A. Square About the Third Dimension

To help us think about 4-dimensional space, let’s begin by imagining how Flat- landers might think about 3-dimensional space. A. Square is probably convinced that 3-dimensional space can’t possibly exist. To prove to him that it does exist we could play some 3-dimensional tricks on him. Suppose A. Square has a flower in his garden. We could remove the flower from his 2-dimensional space just by pulling it up into the third dimension, as we’ve done in Figure 1. To A. Square, the flower would seem to instantly disappear into thin air, leaving him mystified. Similarly, if a 4-dimensional creature pulled my desk out of 3-dimensional space, it would seem to me that my desk had disappeared into thin air.

Figure 1. If we lifted A. Square’s ﬂower into the third dimension, it would seem to disappear.

If I were evil, I could lift A. Square’s heart right out of his body. But if I did that, a 4-dimensional creature might remove my heart from my body. So let’s not do that. On the other hand, if A. Square has a heart condition, we could perform “closed heart surgery” to repair one of his heart valves without making any cuts in his skin just by reaching into his chest from the third dimension. We can even help A. Square perform non-invasive surgery on himself. In Fig- ure 2, we see an unhappy A. Square suﬀering from appendicitis and in desperate need of an appendectomy. Unfortunately, the nearest Flatland hospital is far out of town. He’s got a hand, of course, but he wouldn’t want to cut himself with

29 30 2. VISUALIZING FOUR DIMENSIONS a scalpel. Certainly, A. Square can’t reach into himself to remove his appendix without cutting through his skin.

Figure 2. A. Square needs to have his appendix removed.

But we can help him by pulling his hand into the third dimension (something he certainly couldn’t do by himself), as in Figure 3. With our help, he can then grab and remove his own appendix, just like I could grab and remove his heart. If another Flatlander were observing this appendectomy happening, what would she see?

Figure 3. We could pull A. Square’s arm into the third dimension and help him remove his own appendix.

We could play tricks on A. Square that don’t involve surgery. Suppose we lifted A. Square up into the third dimension, turned him over, and put him back in Flatland (see Figure 4). Could he tell what happened? What would the other Flatlanders think?

Figure 4. A. Square has been turned over.

What would happen if a 4-dimensional creature turned me over? Would I know what happened? Would my friends notice any changes in me? If a 4-dimensional creature turned a clock over, what would it look like? 2. PROJECTIONS AND PERCEPTIONS 31

Of course playing tricks is not the best way to teach A. Square about the third dimension. Instead, let’s help him imagine what some 3-dimensional objects might look like. Before we discuss objects in various dimensions, we want to clarify the differences between the words “universe”, “space”, and “object”. You may remember from Chapter 1, that while a space may or may not have a boundary, a universe has no boundary. For example, in Chapter 1 we saw that A. Square’s universe could be a sphere, a torus, an infinite plane, or any other finite or infinite surface that doesn’t have a boundary. On the other hand, a finite cylinder with two circular boundaries is a 2-dimensional space but not a 2-dimensional universe. We use the word object to refer to something which is contained in a space. My pencil and paper and even my friends are objects in my space. In fact, I’m an object in my own space. Figure 5 illustrates some objects that A. Square might find in his space.

Figure 5. A. Square, a ﬂower, C. Circle, and H. Hexagon are all objects in Flatland.

The dimension of an object can be smaller than the dimension of the space (for example, a shadow which has no 3-dimensional thickness is an object in our space). Similarly, a sphere with no 3-dimensional thickness, like a holograph, could be an object in our space. Even though such a sphere curves around in 3 dimensions, the sphere itself is still 2 dimensional because creatures in it can only move in two perpendicular directions. We saw in Chapter 1 that a 2-dimensional sphere could be the universe for Flatland. This means that Flatland could be an object inside of our 3-dimensional universe. Analogously perhaps our universe is a 3-dimensional object with no 4-dimensional thickness living inside some 4-dimensional universe. In fact, perhaps there is a 1-dimensional universe living inside a 2-dimensional universe which lives inside our 3-dimensional universe which lives inside a 4-dimensional universe, and so on with the dimensions getting bigger and bigger forever. We could even imagine that all of these diﬀerent dimensional universes live inside a single inﬁnite-dimensional universe. But even if this were true, we would have no way of knowing.

2. Projections and Perceptions

When A. Square encounters another Flatlander, what does he actually see? If every object appeared to him as a line segment, how could A. Square tell the diﬀerence between B. Triangle (who he’s infatuated with) and any other shape out for a walk? Perhaps his brain uses shadows to help him determine distances. Or perhaps he even has a second eye which gives him a sense of depth and perspective, just as our two eyes give us depth perception (see Figure 6). 32 2. VISUALIZING FOUR DIMENSIONS

Figure 6. How does A. Square perceive B. Triangle?

In fact, our visual abilities as 3-dimensional creatures are not so different from those of Flatlanders. When viewing a 3-dimensional object, our eyes create a 2- dimensional projection of the object, very much like a photograph. You probably don’t think about it this way when you look around, but that’s because your brain has lots of tricks to turn a 2-dimensional projection into 3-dimensional perception. Without conscious thought, your brain synthesizes a great deal of information from light, shadows, motion, and relative size. Also, your two eyes together give a sense of depth and distance to an otherwise flat image. This is the difference between what we see and what we perceive. If our brains did not have the ability to perceive depth and distance, we would find all of the 2-dimensional images that we see somewhat confusing. Walking on a snowy, icy path is difficult no matter what. But if you want a real challenge, the next time you encounter an icy path, try walking on it with one eye shut. With only one eye you lose a lot of your depth perception, making it harder to anticipate the little bumps and valleys in the ice, which in turn makes it harder to navigate. Those of us who live in warm climates are sadly deprived of the opportunity to try walking with one eye closed on an icy path. So let’s consider another example. If we had no depth perception, one coil of a spring might look to us like one of the projections in Figure 7. If we only had one of these projections and we had never seen a spring before, it might be difficult to go from this flat image to knowing the actual form of a spring.

Figure 7. Without any depth perception a coil of a spring might look like one of these images.

In the image on the right, the spring seems to be intersecting itself. But our depth perception allows us to understand that one part of the spring is actually behind the other. When we are making a sketch of an object in 3 dimensions, we often use a gap to indicate that one part of the object passes behind another part. For example, we could redraw the image on the right of Figure 7 as in Figure 8. Of course, there is no actual gap in the spring, but the drawing with the gap is easier for our brains to interpret than either of the drawings in Figure 7. 3. MOVIES AND MOVEMENT 33

Figure 8. A gap in the drawing helps our brains realize that one part of the spring passes behind the other.

3. Movies and Movement

A. Square has just had his ﬁrst date with B. Triangle and is on his way home humming softly to himself, while thinking peaceful, happy thoughts about what a great day he had. Meanwhile, as we can see in Figure 9, A. Ball is lurking just above him in the third dimension preparing to pay him a surprise visit.

Figure 9. A. Ball preparing to visit A. Square.

A. Ball pushes herself through Flatland from above so that A. Square sees only slices of her. Figure 10 shows what A. Square might perceive is happening at each second from t = 0 when she ﬁrst touches Flatland through t =4whenonlyacross section of the bow on her head is in Flatland. Notice that we’ve drawn t =0atthe bottom of the picture and t = 4 at the top of the picture. Drawing the ﬁgure from bottom to top may seem backwards. However, we draw it this way so that if we put all of the slices of A. Ball together, we will see a right-side-up 3-dimensional image of her. Note that A. Ball is solid, and the disks that A. Square sees are actually cross sections of her body. We draw them as ovals in order to make the picture look more 3 dimensional. As you can imagine, A. Square is more than a bit disturbed to notice a black dot appear, grow into a larger and larger circle, then shrink into a smaller and smaller circle, turn into two rounded triangles glued together at a dot, and then disappear. This all seems quite spooky and mysterious to him. In Figure 11, we’ve put together a sequence of frames (like the frames of an animated movie) to illustrate what A. Square sees as A. Ball passes through Flatland. As with Figure 10, the images start at the bottom and go upward so that if the images are stacked one on top of the other we get a right-side-up picture of A. Ball. Since we are now viewing the scene from above, the ovals that we drew in Figure 10 now appear as circles. Note that if we were really drawing all of the images that 34 2. VISUALIZING FOUR DIMENSIONS

Figure 10. A. Ball passing through Flatland.

A. Square would see, there would be many more frames. In this way, Flatlanders can view a 3-dimensional object passing through Flatland as an animated movie of gradually changing 2-dimensional shapes. These examples show us that Flatlanders can get a sense of 3-dimensional objects by watching 2-dimensional slices of them pass through Flatland. In an analogous way, we can get a sense of 4-dimensional objects by watching 3-dimensional slices of them pass through our space. For example, in Chapter 1 we learned that by analogy with the deﬁnition of a 2-dimensional sphere in R3, the 3-dimensional sphere is deﬁned by

S3 = {(x, y, z, w)|x2 + y2 + z2 + w2 =1} as a subset of R4. In order to help A. Square visualize S2, we told him that S2 can be obtained by gluing two disks together along their boundaries. However, in order to physically glue the disks together, we would ﬁrst have to stretch them into two hemispheres in R3, which would be impossible for A. Square to picture. Similarly, we can think 3. MOVIES AND MOVEMENT 35

Figure 11. Frames of a movie of A. Ball passing through Flatland. of S3 as a gluing of two 3-dimensional balls along their boundaries. However, if we wanted to physically glue the balls together, we would need to stretch them into two “hemiballs” in R4. Since this is impossible for us to picture, let’s instead think about what we would see if S3 passed through our space. It’s always a good idea to begin by considering the same situation one dimension down. What would A. Square see if a 2-sphere passed through Flatland? It would be similar to how A. Square perceived A. Ball passing through Flatland, except that a 2-dimensional sphere would be hollow (like a holograph) and wouldn’t have a bow on its head. In particular, A. Square would see a point appear out of thin air, and then he would see it grow into a small hollow circle. The circle would grow until it had radius 1, which is the radius of S2 = {(x, y, z)|x2 + y2 + z2 =1}.

Then the circle would shrink, finally becoming a single point again before disap- pearing altogether. In a similar way, if S3 passed through our 3-dimensional universe, we would see a point appear out of thin air, and then grow into a small hollow sphere, like a round soap bubble. The hollow sphere would continue to grow until it had radius 1, and then it would shrink back down until it finally became a single point again, after which it would disappear altogether. While we can’t put the slices of S3 together physically, watching the slices change over time enables us to visualize S3 in a 4-dimensional space made up of three physical dimensions together with time as the fourth dimension. Similarly, A. Square can visualize A. Ball in a 3-dimensional space made up of two physical dimensions together with time as the third dimension. In fact, no matter how many spatial dimensions you live in, you can always add time as the next dimension. At first this seems like a satisfying way of visualizing objects in higher dimensions. However, our relationship with time is very different from our relationship with spatial dimensions. We can move forwards and backwards in space at various speeds. But we can’t go back in time, stay still in time, or move faster in 36 2. VISUALIZING FOUR DIMENSIONS time. We can only move forward in time at a fixed “speed”. Since mathematicians look for patterns and symmetry, they prefer all dimensions to be interchangeable. There is no particular dimension which is the first dimension, and another which is the second dimension, and so on. Rather, dimensions 1, 2, and 3 are all interchangeable. Therefore, from a mathematician’s point of view, dimension 4 should be interchangeable with the others as well, and hence should not be time. Nonetheless, time is useful because it enables us to use movies and slicing as one way to help us visualize higher-dimensional objects. Also, time is a necessary coordinate. If I want to meet a friend for dinner, then I have to specify not only three spatial coordinates (e.g., on the second floor of the building at the corner of Broadway and 95th street), but also a time coordinate (e.g., at 7:00 PM on June 29th). A mix-up in any one of these four coordinates will prevent us from meeting.

4. Inductively Deﬁning Cubes of Dimensions 0, 1, 2,and3

In this section we will explain how to define cubes of different dimensions inductively. This means that we will define a 0-dimensional cube, and explain a method to construct a 1-dimensional cube from a 0-dimensional cube. Then we will use the same method to construct a 2-dimensional cube from a 1-dimensional cube. Continuing on, we again use this method to construct a 3-dimensional cube from a 2-dimensional cube. And so on. We begin by defining a 0-dimensional cube to be a single point (we don’t have much choice here; any 0-dimensional object is a single point). We construct a 1-dimensional cube by dragging the 0-dimensional cube along a straight path of unit length in a perpendicular direction. Since all directions are perpendicular to a point, we can choose any direction for our path. In Figure 12, we have dragged our 0-dimensional cube in the horizontal direction, so that our 1-dimensional cube is a horizontal line segment of unit length.

Figure 12. A 1-dimensional cube.

Now we use the same method to construct a 2-dimensional cube, starting with a 1-dimensional cube. This means we drag our 1-dimensional cube along a straight path of unit length in a direction which is perpendicular to the 1-dimensional cube. We see from Figure 13 that a 2-dimensional cube is a square. Note that even though we have not colored the interior of the square, every point inside the square is part of our 2-dimensional cube.

Figure 13. A 2-dimensional cube.

Continuing with this method, we construct a 3-dimensional cube by dragging a 2-dimensional cube along a straight path of unit length in a direction which is 5. A. SQUARE LEARNS ABOUT A 3-DIMENSIONAL CUBE 37

Figure 14. A 3-dimensional cube. perpendicular to the 2-dimensional cube. In particular, by dragging Figure 13 in a direction which is perpendicular to the page of the paper we obtain a 3-dimensional cube, which is what we usually call a cube (see Figure 14). Again, remember that this is a solid cube and not just a wire frame or a cardboard box in the shape of a cube.

5. A. Square Learns About a 3-Dimensional Cube

Before we construct a 4-dimensional cube, we want to check in on A. Square who was so disturbed by his visit from A. Ball that he has gone to bed. However, A. Ball is not ready to leave him alone yet. She has the idea that perhaps she can teach him about the third dimension without scaring him if she talks to him in a soothing voice while he is dozing off. This works well because A. Square assumes her voice is part of a dream, and he is quite used to having bizarre dreams. In order to help A. Square visualize a 3-dimensional cube, A. Ball is planning to describe Figure 14 to him. However, she suddenly realizes that A. Square would see the figure as a hexagon with a bunch of strange dotted line segments inside. So instead, A. Ball takes the approach of describing what A. Square would see if cubes of dimensions one, two, and then three passed through Flatland. First, she tells him that if a line segment (a 1-dimensional cube) pierced through Flatland end-on he would see a 1-dimensional cross section, which would appear, stay for a while, and then disappear suddenly. Because he can picture cross sections of a line segment in his own space, he can understand when she explains that this is actually a continuous progression of distinct points which are cross sections of the line segment. Similarly, if a square (a 2-dimensional cube) passed through Flatland edge first, A. Square would see the cross sections of the square as a single segment which appeared out of nowhere, stayed for a while, and then disappeared. Again because he can visualize cross sections of a square, A. Square can understand that this is actually a continuous progression of distinct segments which are “stacked together” to form a square. At this point, A. Ball tells A. Square that if a 3-dimensional cube passed through Flatland, he would see the cross sections of the cube as a single square which appeared out of nowhere, stayed for a while, and eventually disappeared. A. Square thinks that this sounds reasonable after having thought about the cross sections of the line segment and square. But it’s still hard for him to visualize how a continuous progression of squares could be put together to create a 3-dimensional cube. A. Ball considers showing him the pictures of cross sections in Figure 15. But then she realizes that these pictures would be as incomprehensible to him as Figure 14 would have been. 38 2. VISUALIZING FOUR DIMENSIONS

Figure 15. Two cross sections of a 3-dimensional cube.

One reason that the pictures appearing in Figures 14 and 15 are incomprehensible to A. Square is that they contain segments which pass behind other segments. We may have trouble seeing what goes behind what in the picture, but A. Square doesn’t have the perspective to be able to do this at all. We can avoid drawing such segments if we are willing to use size to represent distance from the viewer. For example, in the illustration of a room in Figure 16, the grey square in the back of the room is drawn smaller than the black square (which we are looking through) in order to indicate that the back wall is farther away from the viewer than the front wall. In art, this type of drawing is called one-point perspective and is often described in “learn-to-draw” books.

Figure 16. A drawing of a 3-dimensional room with one-point perspective.

Figure 17 is a drawing of a 3-dimensional cube using one-point perspective. At first glance, this figure just looks like an arrangement of line segments in the plane. However, if you think about the picture in three dimensions as you did the room in Figure 16, you can force your mind to see it as a cube. To see the six faces of the cube, observe that the center square is the back face of the cube and the four trapezoids around it are the four sides of the cube. The large outer square is the front face of the cube that we are looking through. Notice that the front face is actually superimposed on the other five faces. A. Square cannot see Figure 17 as we see it. However, he can understand what a square is, and he can understand what it means for one square to be smaller than another. He can even understand what A. Ball means if she describes the drawing as a smaller square inside of a larger square with the corners connected. She then explains to A. Square that the smaller square is not actually inside the larger square, but rather it is behind the larger square in the third dimension. She also tells him that both squares are actually the same size, and that one just looks smaller than the other because it’s farther away in the third dimension. Even though he doesn’t understand the notions of “behind in the third dimension” or 5. A. SQUARE LEARNS ABOUT A 3-DIMENSIONAL CUBE 39

Figure 17. A one-point perspective drawing of a cube.

“farther away in the third dimension”, he is willing to accept this picture. A. Ball decides to call pictures of this type “Flatland-friendly” pictures, because they don’t require 3-dimensional perspective in order to understand them. Returning to her earlier description of what A. Square would see if a 3-dimensional cube passed through Flatland, A. Ball now draws the Flatland-friendly pictures of slices of a cube in Figure 18. When we look at this ﬁgure, we can imagine a vertical plane moving from the front to the back of a cube, creating identical grey slices at each stage. By contrast, when A. Square looks at this ﬁgure he sees a grey square in the left image that is bigger than the grey square in the middle, which in turn is bigger than the grey square on the right. Nonetheless, A. Square seems willing to accept the idea that this drawing represents slices of a 3-dimensional cube.

Figure 18. A Flatland-friendly picture of slices of a cube as it passes through Flatland.

At this point, A. Ball feels like she’s on a roll (which, after all, is one of her favorite feelings). So she decides to give A. Square even more insight into a 3- dimensional cube by drawing projections of the cube rotating around a vertical axis in 3-dimensional space (see Figure 19). In order to avoid confusing A. Square, she does not use gaps to indicated when one segment passes behind another. However, this makes the drawings somewhat more confusing for a 3-dimensional person. If we set aside our 3-dimensional intuition for the moment, we can see each image of Figure 19 through A. Square’s eyes as a collection of quadrilaterals. If we track the grey shape from one image to the next, we will observe that its edges stretch and shrink, and the angles between them increase and decrease. In particular, the grey square in the center of the first image becomes a trapezoid starting in the third image, since its left edge shrinks while its right edge stretches. Similarly, the white trapezoid that is on the left side of the image at time t =0 will slowly morph into a square. Most interesting of all, the white trapezoid on the right side of the figure at t = 0 first becomes thinner, then when t =2,itsshort side switches from the left to the right, and finally it gets bigger and more square. 40 2. VISUALIZING FOUR DIMENSIONS

Figure 19. A cube rotating around a vertical line.

A. Square is a bit uncomfortable with these images of a rotating cube. However he says he accepts them so that A. Ball will shut up and let him dream peacefully about B. Triangle.

6. A 4-Dimensional Cube

Now that A. Square has learned about the 3-dimensional cube, it’s time for us to take the leap and learn about the 4-dimensional cube. Continuing the inductive construction that we began in Section 4, we build a 4-dimensional cube by dragging a 3-dimensional cube along a straight path of unit length in a direction which is perpendicular to the cube itself. Now, it’s pretty tough for us to understand how something could be perpendicular to all three of our dimensions, just as it was pretty tough for A. Square to understand how something could be perpendicular to his two dimensions. But we can take our inspiration from A. Square’s experience and draw a “3-dimension-friendly” picture of a 4-dimensional cube by drawing two 3-dimensional cubes, one inside the other, and connecting analogous vertices in the outside and inside cubes with line segments (see Figure 20). Remember that the inner cube is not smaller than the outer cube, it’s just “farther away in the fourth dimension”. Notice that this visualization is analogous to the one-point perspective of a 3-dimensional cube that we illustrated in Figure 17.

Figure 20. A “3-dimension-friendly” drawing of a 4-dimensional cube. 6. A 4-DIMENSIONAL CUBE 41

Let’s examine Figure 20 more carefully using A. Square’s visualizations of a 3- dimensional cube as a guide. A 4-dimensional cube has eight 3-dimensional “solids” that are the analogues of the six 2-dimensional faces of a 3-dimensional cube. These solids are actually 3-dimensional cubes, though most of them look distorted in the figure. This is analogous to the distortion of the faces of the 3-dimensional cube that we saw in Figure 17. One of these solids is the small cube in the center of the figure, another one is the large cube containing the figure, and the other six solids are the prisms that connect each face of the small cube to the corresponding face of the large cube. By analogy with our discussion of slicing a 3-dimensional cube, we can understand the 4-dimensional cube as a continuous progression of 3-dimensional cubical cross sections. For example, the grey cubes in the middle of the images in Figure 21 are 3-dimensional cubical cross sections of the 4-dimensional cube. Note that each cross section consists not only of the six faces of a cube, but of the cubical volume enclosed by those faces. Just as A. Square had to accept that the cross sections of a cube in Figure 18 were not nested, though they appeared to him as if they were, we have to accept that the cubical cross sections of the 4-dimensional cube in Figure 21 are not nested though they appear to us as if they are.

Figure 21. Two 3-dimensional cross sections of a 4-dimensional cube.

You can find animations of a rotating 4-dimensional cube on the internet. How- ever, when you look at such animations you should remember the distortions that A. Square saw in Figure 19. We were able to understand the projections in Fig- ure 19 because of our experience with solid cubes. In particular, we knew that all of the faces of the cube were really squares, that all of the edges have the same length, and that every angle is a right angle. The same is true for animations of a rotating 4-dimensional cube. The various edges appear to stretch and shrink, while the angles between pairs of adjacent edges seem to change as the edges pivot about the vertices. As a consequence, the faces and solids appear to rotate and flex. Some of the solids even look like they are turning inside out during the rotation. However, we have to keep in mind that the faces are always flat squares and the solids are always cubical boxes, no matter how they look. Don’t worry if you are uncomfortable with Figure 21 or the animations of a rotating 4-dimensional cube that you can find on the internet. A. Square knows exactly how you feel. 42 2. VISUALIZING FOUR DIMENSIONS

7. Tetrahedra in Various Dimensions

Cubes are not the only geometric ﬁgure that can be constructed inductively in diﬀerent dimensions. We will use a similar procedure to construct a 4-dimensional tetrahedron. Again, we start with a single point which we call a vertex. This is a 0-dimensional tetrahedron (remember a point is the only 0-dimensional shape). To get a 1-dimensional tetrahedron, we add a new vertex one unit away from the previous one and connect the two vertices with a line segment. As we see in Figure 22, a 1-dimensional tetrahedron is the same as a 1-dimensional cube.

Figure 22. A 1-dimensional tetrahedron is the same as a 1- dimensional cube.

Next, we add a third vertex which is one unit from each of the previous two vertices. Notice that we cannot put the new vertex on the same line as the previous two. To get a 2-dimensional tetrahedron, we add line segments that connect the new vertex with the vertices of the 1-dimensional tetrahedron. This creates an equilateral triangle (including its 2-dimensional interior) as seen in Figure 23.

Figure 23. A 2-dimensional tetrahedron.

Continuing in a similar fashion, we build a 3-dimensional tetrahedron by adding a new vertex that is one unit from each of the previous three vertices. In order to do this, we are forced to place our new vertex outside the plane of the paper. We connect this new vertex to the vertices of the 2-dimensional tetrahedron by means of line segments, thus creating a solid triangular pyramid like the one shown in Figure 24(a). This is what we usually think of as a tetrahedron.

Figure 24. Two projections of a 3-dimensional tetrahedron.

Flatlanders prefer the image in Figure 24(b), which shows the tetrahedron from above and does not have one segment behind another. They think of this as an equilateral triangle with a vertex in the center and line segments connecting it 7. TETRAHEDRA IN VARIOUS DIMENSIONS 43 to the other three vertices. We explain to the Flatlanders that even though the segments on the inside of the triangle look shorter than the sides of the triangle, the central vertex is actually in a diﬀerent plane in the third dimension so that all of the segments have exactly the same length. To give the Flatlanders more intuition, we explain that if a 3-dimensional tetrahedron passed through Flatland bottom ﬁrst, they would see a 2-dimensional tetrahedron (an equilateral triangle) which appeared out of nowhere and gradually got smaller until it eventually shrank down to a point. In Figure 25, we illustrate two of the cross sections that Flatlanders would see. Again they prefer the images in Fig- ure 25(b) over those in Figure 25(a) since they cannot understand the 3-dimensional perspective in Figure 25(a). However, we still have to remind them that the two grey triangles in Figure 25(b) are not nested, even though they appear that way.

Figure 25. Cross sections of a tetrahedron, viewed from two different perspectives.

Finally, to construct a 4-dimensional tetrahedron, we add a new vertex which is one unit away from the previous four vertices. This is impossible for us to accomplish in three dimensions, but poses no problems in 4-dimensional space. If we momentarily suspend our disbelief and connect this new vertex to all of the vertices in the 3-dimensional tetrahedron using line segments, then we will create a 4-dimensional tetrahedron. From our 3-dimensional perspective, we can construct a 4-dimensional tetrahedron by placing the new vertex in the center of the 3-dimensional tetrahedron, as in Figure 26. However, we should realize that the central vertex is not actually inside the tetrahedron; rather, it is poking out of our 3-dimensional space into the fourth dimension. Furthermore, all of the edges in the 4-dimensional tetrahedron are the same length even though the ones containing the central vertex look shorter. If a 4-dimensional tetrahedron passed through our 3-dimensional space bottom ﬁrst, we would see a 3-dimensional tetrahedron which appeared out of nowhere and gradually got smaller until it eventually shrank down to a point. One of these cross sections is illustrated in Figure 27. Remember that the tetrahedra are not nested, even though the picture makes it look like the grey tetrahedron is inside of the white tetrahedron. 44 2. VISUALIZING FOUR DIMENSIONS

Figure 26. A 3-dimensional drawing of a 4-dimensional tetrahedron.

Figure 27. A 3-dimensional cross section of a 4-dimensional tetrahedron.

8. Exercises

1. Another way to explain a 3-dimensional cube to a Flatlander uses the gluing technique we discussed in the last chapter. We cut open the cube along some of the edges (like collapsing a box), and then label the cut-open edges with numbers so that we know how to reassemble the box. Thus, edge 1 is glued to the other edge 1, edge 2 is glued to the other edge 2, and so on (as illustrated on the left side of Figure 28).

Figure 28. Illustration for Exercise 1. 8. EXERCISES 45

Now consider the unfolded 4-dimensional cube illustrated on the right side of Figure 28. Fill in the missing numbers to show how the faces are glued together to produce a 4-dimensional cube.

2. What does A. Square perceive if a 3-dimensional person slowly puts his/her hand into Flatland? What might we perceive if a 4-dimensional person slowly stuck his/her hand in our space?

3. A 3-dimensional ball is deﬁned by

B3 = {(x, y, z)|x2 + y2 + z2 ≤ 1}

as a subset of R3. By analogy we deﬁne a 4-dimensional ball by

B4 = {(x, y, z, w)|x2 + y2 + z2 + w2 ≤ 1}

as a subset of R4. Describe the slices that we would see if B4 passed through our 3-dimensional space.

4. What would A. Square see if a line segment passed through Flatland at an angle other than 90◦?

5. What would A. Square see if a square passed through Flatland corner ﬁrst?

6. Suppose a cube passes through Flatland in various diﬀerent ways. What are all the possible 2-dimensional shapes that a Flatlander might see?

7. Consider the frame sequences in Figure 29, and imagine that A. Square is experiencing these movies frame-by-frame (from bottom to top). In each case, describe the shape of the object that is passing through Flatland.

Figure 29. Illustration for Exercise 7. 46 2. VISUALIZING FOUR DIMENSIONS

8. What if a solid torus (in other words, a doughnut) passes through Flatland, as in Figure 30(a)? What shapes would A. Square see, and how would they change? What sequence of frames would he perceive if the solid torus was oriented as in Figure 30(b)?

Figure 30. Illustration for Exercise 8.

9. Suppose that A. Square is watching a circle move in a straight line away from him as in Figure 31(a). Would A. Square perceive the same thing if a slanted cylinder passed through Flatland as in Figure 31(b)? Could he tell the diﬀerence between these two scenarios? If so, how?

Figure 31. Illustration for Exercise 9.

10. Design a 2-dimensional room with a lock on the door so that the Flatlanders inside can lock themselves in so that nobody will rob them while they are sleeping. Note that gravity and magnets don’t exist in Flatland, so you cannot use them in designing your lock. 11. Design a digestive system for A. Square that does not cause him to fall apart. For the sake of health, A. Square’s digestion must have a separate entrance for food and exit for waste, both of which might be open at the same time in case of illness. Also you cannot assume that food is simply absorbed through a membrane. 8. EXERCISES 47

12. If we lift A. Square above Flatland and hold him perpendicular to it, what will he see? 13. If Flatland and everything in it had the exact same amount of 3-dimensional thickness (including A. Square’s eyes), could A. Square detect it? If everything in our universe had the same small 4-dimensional thickness, could we detect it? 14. What would we see if another 3-dimensional universe intersected ours perpen- dicularly? How would we see the 3-dimensional creatures of that space? 15. How many vertices, edges, faces, and solids does a 4-dimensional cube have and why? What about a 4-dimensional tetrahedron? Find inductive formulae for the number of vertices, edges, faces, and solids of an n-dimensional cube and an n-dimensional tetrahedron in terms of these numbers for an (n−1)-dimensional cube and (n − 1)-dimensional tetrahedron. Justify your conclusions. 16. An n-dimensional pyramid consists of an (n − 1)-dimensional cube with all the vertices attached to a single additional vertex in the nth dimension (see Figure 32). Determine how many vertices, edges, faces, and 3-dimensional solids a 4-dimensional pyramid and 5-dimensional pyramid have. Justify your conclusions.

Figure 32. Illustration for Exercise 16.

https://doi.org/10.1090//mbk/096/03

CHAPTER 3

Geometry and Topology of Diﬀerent Universes

Topics: • Intrinsic and extrinsic geometry • Geodesics • Triangles and geometric properties • Intrinsic and extrinsic topology • Loops and topological properties • Local and global properties • Homogeneity • Manifolds • Assumptions about universes • 2- and 3-dimensional spheres with a point removed • Curves on a torus

1. Intrinsic and Extrinsic Properties of a Space

Early one Saturday morning your phone rings. Half-asleep, you grab the phone and mumble “hello”. You don’t catch the person’s name, but she claims to be your fourth cousin twice removed, or something like that. “Oh, yes, nice to hear from you” you say, though you actually have no idea what it means to be a fourth cousin twice removed. She says she just found out that she’s related to you and she’s eager to get to know you. She wants to know everything about your life, starting with where you live. You tell her you live in New York City. But she asks where New York City is. This seems strange since everyone in the US (and in a large part of the world) knows where New York City is. Things get even stranger when you tell her it’s in the Northeastern part of the US and she doesn’t seem to know what you mean. You ﬁnd yourself describing the geography of the US, while wondering how such an ignorant person could possibly be any kind of a cousin of yours no matter how many times removed. Eventually, after you have explained that the earth is a planet in the universe she asks “what’s the shape of the universe?” At this point, you are wide awake and decide this is a prank call. You slam the phone down in exasperation. The problem is that her last question keeps going around and around in your head. What is the shape of the universe? In Chapter 1, we learned that in addition to R3, our universe could be T 3 or S3. But it’s unlikely that there are only three possibilities for the shape of our universe. Especially since in Chapter 1 we saw that there are inﬁnitely many 2-dimensional

49 50 3. GEOMETRY AND TOPOLOGY OF DIFFERENT UNIVERSES universes. However, coming up with other possibilities for our 3-dimensional universe seems much harder than coming up with more 2-dimensional universes. Of course, for A. Square it’s much easier to think of possibilities for 1-dimensional universes than 2-dimensional universes. In general, it’s easier for an external observer in a higher dimension to understand a space than for inhabitants to understand their own space. To help us keep track of the distinction between properties of a space that only an outsider can observe and properties that inhabitants can figure out themselves we make the following definition. Definition. Apropertyofaspaceissaidtobeextrinsic if it can be detected by an observer outside of the space but it can NOT be detected from within the space. Otherwise, the property is said to be intrinsic. Recall from Chapter 1 that properties of spaces are like characteristics that we use to tell people apart. For example, suppose that I am talking on the phone to A. Person who is in A. Faraway Place and has never heard of Santa Claus or the Easter Bunny. If I said that Santa always wears a red suit and the Easter Bunny is naked except for its fur, A. Person would agree that they must be different beings. Thus how they are dressed or not dressed is a characteristic that distinguishes them. Properties of spaces play the same role. They are useful because if you know that one space has a property and another space doesn’t, then you can conclude that they must be different. Extrinsic properties have to do with how a space is situated in a larger or higher- dimensional space. For example, suppose that A. Square is meeting B. Triangle for lunch. Whether they live on a flat plane or a plane that is bent like the one illustrated on the right in Figure 1, A. Square has to walk the same distance to meet her. In fact, there is no way that he can tell the difference between the two planes illustrated in Figure 1. On the other hand, to us he seems much closer to her if the plane is bent than if it’s flat. But he cannot jump through 3-dimensional space to go from one side of the bend to the other. So the fact that the two of them are closer in 3-dimensional space is irrelevant (and meaningless) to him. Thus whether or not a plane is bent is an extrinsic property.

Figure 1. A. Square has to walk the same distance to get to B. Triangle whether his plane is ﬂat or bent.

Now suppose that A. Square somehow knows that Flatland is either a sphere or a curved torus. From our 3-dimensional perspective we have no trouble telling these spaces apart since we can see that the curved torus has a hole in it while the sphere does not. Because we see the hole easily from the outside, we might be tempted to say that the existence of such a hole is an extrinsic property. However, for a property to be extrinsic it must be undetectable by the inhabitants of the 1. INTRINSIC AND EXTRINSIC PROPERTIES OF A SPACE 51 space. You might recall from Chapter 1 that A. Square could in fact distinguish a sphere and a torus either by using straightest paths or by using walls. Thus the existence of the hole in the torus is actually an intrinsic property, though A. Square certainly wouldn’t use the word “hole” in his explanation of the difference between a torus and a sphere. This example illustrates that even though a property may be easier for creatures outside of the space to detect, if the creatures in the space can detect it at all (even with great difficulty and without real understanding or insight into what’s going on), then the property is intrinsic. Definition. We say that two spaces are intrinsically the same if a being in one space would have no way of knowing if he or she were suddenly transported from one space to the other. We say that say two spaces are extrinsically the same if external beings would be unable to distinguish the two spaces. In other words, two spaces are intrinsically the same if all of their intrinsic properties are the same. Since external beings can observe both intrinsic and extrinsic properties of a space, two spaces are extrinsically the same if all of their intrinsic and extrinsic properties are exactly the same. This means that if two spaces are extrinsically the same, they are automatically intrinsically the same. For example, the flat plane and the bent plane in Figure 1 are intrinsically the same but extrinsically different. On the other hand, a sphere and a curved torus are intrinsically different (and extrinsically different) because of the intrinsic properties involving straightest paths and walls that we used to distinguish them in Chapter 1. The language can be a bit confusing here, and it’s worth taking some time to carefully think through the definitions to make sure you understand them. In particular, keep in mind the following important observations. • A property of a space cannot be both intrinsic and extrinsic, since it is only extrinsic if it is not intrinsic. • Two spaces (like a sphere and a curved torus) can be both intrinsically and extrinsically different. • Two spaces that are identical (like two spheres of the same size) are both intrinsically and extrinsically the same. • If two spaces are intrinsically different, then one has an intrinsic property that the other does not have. Hence they are also extrinsically different. • If two spaces are extrinsically different (like a flat plane and a bent plane), they may or may not be intrinsically different. The goal of this chapter is to develop ways to determine if two spaces are the same or different, both from the outside and from the inside. To do this we will need to introduce a lot of definitions which will build on one another. In order to absorb all of this material, you should read the chapter slowly, and at the end of each section you should stop and check that you understand and remember earlier definitions before you continue reading. For example, right now you should check that you understand the definitions in this section before starting the next section. 52 3. GEOMETRY AND TOPOLOGY OF DIFFERENT UNIVERSES

2. Intrinsic and Extrinsic Geometry

One way to distinguish spaces is to use some kind of measuring device like a ruler to measure lengths or a protractor to measure angles. More formally, we have the following definition. Definition. A property of a space that can be numerically measured is said to be a geometric property. Recall from the last section that whether a plane is flat or bent like in Figure 1 is an extrinsic property. We can now be more specific and say that whether or notaplaneisbentisanextrinsic geometric property because we can measure the distance between A. Square and B. Triangle in 3-dimensional space and see that they are closer together when they are on the bent plane than when they are on the flat plane. Definition. We say that two spaces have the same intrinsic geometry if there is no measurement that would enable a being in one space to know if he or she were suddenly transported to the other space. We say that two spaces have the same extrinsic geometry if there is no measurement which would enable external beings to distinguish the two spaces. This definition means that two spaces have the same intrinsic geometry if all of their intrinsic geometric properties are the same, and two spaces have the same extrinsic geometry if all of their intrinsic and extrinsic geometric properties are the same. For example, let’s consider the two spheres illustrated in Figure 2. These spheres have different intrinsic geometry because if A. Square were transported from one to the other, he could measure the total area of his space to know that it had changed. Since total area is an intrinsic geometric property, they automatically also have different extrinsic geometry. However, this is not the only way to distinguish their extrinsic geometry. For example, the two spheres enclose different volumes in R3, which is an extrinsic geometric property that distinguishes them.

Figure 2. These spheres have diﬀerent intrinsic and extrinsic geometry.

Suppose that A. Square and B. Triangle live on a rigid steel sheet. If we moved the sheet without bending it, would the distance between them be any diﬀerent? What if we rotated it or turned it over and then remeasured it? These types of movements of the steel sheet are examples of rigid motions. Any motion of an 3. STRAIGHTEST PATHS AND GEODESICS 53 object in space that doesn’t involve stretching, shrinking, or bending is said to be a rigid motion. If one object can be taken to another by a rigid motion, then the two objects are identical. In particular, this means both their extrinsic geometry and their intrinsic geometry are the same. However, the following theorem (whose proof you might see in a more advanced geometry class) tells us even more.

Theorem 3.1. Two objects in R3 have the same extrinsic geometry if and only if there is a rigid motion of R3 taking one object to the other.

It’s not surprising that if there is a rigid motion taking one object to another, then the two objects are identical and hence have the same extrinsic geometry. What’s more surprising is that if two objects have the same extrinsic geometry, then there is actually a rigid motion taking one to the other. In other words, saying that two objects have the same extrinsic geometry is the same as saying they are identical in every respect. So if you can find any aspect of them that is different, then you can conclude that they have different extrinsic geometry.

3. Straightest Paths and Geodesics

Almost every geometric property has something to do with straightest paths. We need straightest paths to be able to measure length, area, volume, and even angles. In Chapter 1, we defined straightest paths in a practical way, saying that light travels along a straightest path and a taut string follows a straightest path. But now that we are going to be using straightest paths to help us distinguish the geometry of different spaces, we need to be a little clearer about exactly what we mean by a straightest path. Thus we introduce a new word for a straightest path and a more formal definition of what it means. Definition. A path in a space is said to be a geodesic, if for any small enough region which intersects the path, and points p and q in the small region on the same segment of the path, then the shortest path from p to q is on that segment. As a first example, suppose we have a cylinder made out of paper and a path on the cylinder which is a geodesic. Let’s see what we can learn about the path. According to the definition, we need to focus on a “small enough region” of the cylinder which intersects the path. So let’s choose our region to be a really, really, really small disk which intersects the path in a really, really, really small arc, and let p and q be points on the really, really, really small arc. Because the disk is so very small and the path is a geodesic, we know that the shortest path from p to q on the cylinder goes along the arc in the disk (see Figure 3). Now let’s cut the disk out of the cylinder and lay it flat on a table. Because paper is not stretchy, the lengths of paths in the disk are the same whether the disk is part of the cylinder or lying flat on the table. This means that the geodesic arc from p to q in the disk will also be the shortest path from p to q after the disk has been cut out and is lying flat on the table. But we know that for the flat disk the shortest path from p to q is a line segment. Thus the geodesic arc in the disk becomes a line segment when the disk is lying flat on the table. Since this is true for every really, really, really small disk that intersects the geodesic, it follows that if we cut open the entire cylinder and lay it flat on the table, then the entire geodesic will become one or more line segments on the flat 54 3. GEOMETRY AND TOPOLOGY OF DIFFERENT UNIVERSES

Figure 3. p and q are points on the really really really small arc in the really really really small disk.

Figure 4. When we cut open the cylinder along the thick black line, this geodesic becomes three line segments. page. In Figure 4, we cut open the cylinder along the thick black line and see that our geodesic becomes three line segments. We can make a similar argument going in the reverse direction. That is, starting with a line segment on a piece of paper, if we roll the paper into a cylinder, then the line segment will become a geodesic on the cylinder. This is exactly what we will do in our next example, except we will roll the paper into a cone rather than a cylinder. In this case, we will start with a piece of paper in the shape of a wedge whose top angle is 120◦. Thenwedrawahorizontallinesegmentfromoneedgeofthe paper to the corresponding point on the other edge (see the left side of Figure 5). Now when we roll the wedge into a cone, the ends of the grey segment will touch. In Figure 5, we glue the edges of the paper in the front of the picture to make it easier to see how the ends of the grey segment come together at the junction. Now we draw a really, really, really small disk on the cone which intersects the grey segment as illustrated in Figure 6, and we let p and q be points on the segment in the disk. As we saw with the cylinder, we can cut out the disk and lay it ﬂat, and it won’t change the length of any path. Since the grey arc is a line segment on the ﬂat disk, we know that the grey arc is the shortest path from p to q on the cone. Next let’s consider a small disk centered around the point where the endpoints of the grey arcs meet, and let p and r be points on the grey segment as illustrated in Figure 7. In this case, we can see that the shortest path from p to r would go above the above the grey arcs rather than along them. However, this does not 3. STRAIGHTEST PATHS AND GEODESICS 55

Figure 5. When we attach the sides of the cone, the endpoints of the grey segment will come together at the junction.

Figure 6. The shortest path from p to q is along the grey arc in the disk. contradict our definition of a geodesic since the points p and r are on different arcs in the disk. Since we only have to consider pairs of points which are on the same arc in the disk, we can use the same argument as we did for Figure 6 for any really, really, really small disk intersecting the grey segment. Hence it follows that the grey path is in fact a geodesic on the cone. Thus for both the cone and the cylinder, a line segment in the gluing diagram becomes a geodesic once we glue up the appropriate edges. Returning to the cone, we see that the geodesic in Figure 7 doesn’t end when the endpoints of the grey segment come together. To see where it goes, let’s see what happens when A. Square takes a walk on the gluing diagram following the grey segment as it goes off to the right. Since the right and left edges of the wedge are glued together, we know A. Square will re-enter the wedge on the left. But we have to be careful about what direction he’s facing when he re-enters. 56 3. GEOMETRY AND TOPOLOGY OF DIFFERENT UNIVERSES

Figure 7. The points p and r are on diﬀerent arcs in the disk.

In Figure 8, we can see that as he is exiting on the right, the right edge of the diagram cuts across him from just in front of the feather in his hat to his mouth. So when he re-enters the diagram on the left, the left edge of the diagram has to cut across him in the exact same way, from just in front of the feather in his hat to his mouth. Also, after he exits on the right, there is a 30◦ angle between his eye and the gluing edge. Thus after he re-enters on the left, there must also be 30◦ angle between his eye and the gluing edge. Since 180◦ − 30◦ − 30◦ = 120◦,we can now conclude that when A. Square re-enters in the left, he has been rotated by 120◦, as illustrated in Figure 8.

Figure 8. When A. Square goes oﬀ the right side of the diagram, he re-enters on the left rotated by 120◦.

A similar thing happens if A. Square takes a walk on the gluing diagram following the grey segment as it goes oﬀ to the left. Thus in the gluing diagram, we can extend the grey segment on both sides as illustrated on the left in Figure 9. In particular, suppose A. Square starts on the grey segment marked 1, walking in the direction of the arrow. This is the reverse of the way he walked in Figure 8. Now, when he crosses the gluing edge, he continues on the grey segment marked 2, again walking in the direction of the arrow. Then after again crossing the gluing edge, this time as he did in Figure 8, he continues on the grey segment marked 3 3. STRAIGHTEST PATHS AND GEODESICS 57

Figure 9. The grey segment can be extended on both sides.

Figure 10. The points p and q are on the same arc in the disk, but r is on a diﬀerent arc.

again in the direction of the arrow. His path looks more normal when we look at the grey segments on the cone. Now that we’ve extended the grey segment on both ends, let’s return to the really really really small disk cutting across the gluing line on the cone that we saw in Figure 7. As we see in Figure 10, this disk intersects the grey path in two arcs, and the points p and q areononearcwhilethepointr is on the other arc. In order to see what the disk in Figure 10 looks like in the gluing diagram, we need to put the two half-disks together. To do this, we ﬁrst split the wedge down the middle, as illustrated in the ﬁrst image in Figure 11. Then we move the left half-wedge to the right side and rotate both half-wedges so that the thin black edges are lined up as in the second image. Finally, we glue the half-wedges together along the thin black edges as in the third image. Thus we see that the grey segments form an X and the disk from Figure 10 becomes a disk in the gluing diagram. 58 3. GEOMETRY AND TOPOLOGY OF DIFFERENT UNIVERSES

Figure 11. We cut apart and reattach the gluing diagram to see the disk from Figure 10 in the gluing diagram.

Now because the length of the grey arc from p to q on the cylinder is the same as it is in the gluing diagram, the shortest path between p and q on the cylinder must go along the grey arc. Thus the grey path on the cone is indeed a geodesic. Before we end this section, let’s revisit a sphere using our new word “geodesic” in place of “straightest path”. In Chapter 1 we saw that on a sphere the great circles are “straightest paths”. In particular, for any two points on a sphere, there is a great circle containing them, and the shortest route from one point to the other will be along a great circle. Thus the geodesics on a sphere are the great circles. Observe that if two points are not opposite one another on the sphere, then there is only one great circle containing them and one arc on this great circle will be shorter than the other. For example, in Figure 12 we see that there is only one shortest path from point a to point b. By contrast, for points x and y,whichareexactly opposite one another on the sphere, there are inﬁnitely many geodesics between them, and the lengths of all of these geodesics are the same.

Figure 12. There is only one geodesic from a to b, but inﬁnitely many geodesics from x to y. 4. THE DEFINITION OF A TRIANGLE 59

4. The Deﬁnition of a Triangle

In addition to measuring distances and size, measuring the angles of a triangle is a good way to study the intrinsic geometry of a space. In a plane, a triangle is a ﬁnite region bounded by three line segments. Using the idea of a geodesic, we can now talk about triangles in any space.

Definition. We deﬁne a triangle in a space as a ﬁnite region bounded by three geodesic segments such that the region can be deformed to a disk.

For example, in Figure 13, the grey region bounded by three grey segments is ﬁnite, but the white region bounded by the same three segments is inﬁnite. Thus the white region can’t be a triangle.

Figure 13. The white region is not a triangle because it is inﬁnite.

Also notice that we can deform the grey region to a disk as illustrated in Figure 14. Hence we can conclude that the grey region is in fact a triangle and the white region is not. This is important, since the sum of the angles of the grey region is 180◦, but the sum of the angles of the white region is much more.

Figure 14. The grey region is a triangle because we can deform it to a disk.

Next observe that in Figure 15, the three arcs on the torus on the left do not form a triangle, since they don’t bound a region on either side. On the other hand, for the torus on the right, the grey region is a triangle, but the white region is not a triangle since it cannot be deformed to a disk. 60 3. GEOMETRY AND TOPOLOGY OF DIFFERENT UNIVERSES

Figure 15. On the left, the arcs don’t form a triangle. On the right, the grey region is a triangle but the white region isn’t.

5. The Sum of the Angles of a Triangle

We can measure the angle between a pair of intersecting geodesics by projecting the geodesics onto a plane which is tangent to the surface at the point of intersection. For example, Figure 16 illustrates two triangles whose sides are segments of great circles on a sphere. For the large triangle, the sum of the angles is 270◦, since all three angles are 90◦. In fact, on a sphere the sum of the angles of any triangle is greater than 180◦, and the bigger the triangle relative to the area of the sphere, the greater the sum of its angles. As we can see in Figure 16, a really small triangle looks almost ﬂat, and hence its angle sum is almost equal to 180◦. In fact, the complement of each of these triangles is also a triangle, with an even bigger angle sum.

Figure 16. On a sphere, the bigger a triangle is, the greater the sum of its angles will be.

However, a sphere is not the only surface where the sum of the angles of a triangle is not equal to 180◦. Figure 17 illustrates a saddle surface where any triangle has angle sum which is less than 180◦, and the bigger the triangle is, the smaller the angle sum will be. Notice that on the sphere in Figure 16, each of the groups of three geodesic segments bound two triangles—the one that’s shaded grey and the one on the outside of the grey region. By contrast in Figure 17, the three geodesic segments only bound one triangle, since only one of the two regions is finite. Since A. Square can measure angles with his protractor, he can tell that the sums of angles of triangles in a plane, a sphere, and a saddle surface are all different. From this he can conclude that these three surfaces have different intrinsic geometry. By using Theorem 3.1, we also know that these three surfaces have different extrinsic geometry. 6. TRIANGLES ON A FLAT AND CURVED TORUS 61

Figure 17. A triangle on a saddle surface whose angle sum is less than 180◦.

6. Triangles on a Flat and Curved Torus

Let’s use triangles to compare the intrinsic geometry of a flat torus and a curved torus. On a flat torus, any triangle that does not cross the edges of the gluing diagram is an ordinary planar triangle, and hence the sum of its angles is 180◦.In order to check that triangles that do cross the edges of the gluing diagram also have angle sum 180◦, we use the idea of an extended diagram which we introduced when we were studying tic-tac-toe on a flat torus in Chapter 1. Figure 18 illustrates an extended diagram of A. Square in a flat torus. The grid lines and the arrows are merely to indicate how the space is constructed; A. Square would not perceive them at all. In fact, depending on how big the flat torus is, he may or may not realize that the beings he sees are all just copies of himself.

Figure 18. An extended diagram of A. Square in a ﬂat torus.

A triangle in the flat torus behaves just as A. Square does. Since the extended diagram is a plane, any triangle in the gluing diagram corresponds to infinitely many identical triangles in the plane (see Figure 19). In particular, the angles of the triangle in the flat torus are the same as they are for each of the triangles in the extended diagram. Since planar triangles have an angle sum of 180◦, every triangle in the flat torus must also have an angle sum of 180◦, regardless of whether or not it crosses the edges of the gluing diagram. Now let’s consider the two triangles in the curved torus illustrated in Figure 20. First observe that the outer region of the torus seems to curve like a sphere, while 62 3. GEOMETRY AND TOPOLOGY OF DIFFERENT UNIVERSES

Figure 19. A triangle in the flat torus corresponds to infinitely many triangles in the extended diagram. the region near the hole seems to curve like a saddle. As a result, triangles in the outer region have angle sums that are greater than 180◦, while triangles in the inner region have angle sums that are less then 180◦. If you look carefully enough, you might even find some triangles whose angle sums are precisely 180◦.So,there is no general statement that we can make about angle sums of triangles on the curved torus. The fact that not all triangles on a curved torus have angle sums equal to 180◦ implies that the flat torus and the curved torus have different intrinsic geometry.

Figure 20. The triangle on the outside has an angle sum greater than 180◦, while the one near the hole has an angle sum less than 180◦.

Note we can’t compare the extrinsic geometry of a flat torus and a curved torus because a flat torus is not contained in R3. It’s an abstractly glued up surface that exists in and of itself, rather than as a subset of a larger space. Hence it has no extrinsic geometry. So we can’t say that its extrinsic geometry is the same or different from that of the curved torus. Similarly, the 3-torus T 3 is an abstractly glued up space, which is not contained in a larger space, and hence has no extrinsic geometry.

7. Triangles on a Flat and a Curved Cylinder

Next let’s compare the intrinsic geometry of a flat infinite cylinder with that of a curved infinite cylinder. We illustrate these two surfaces in Figure 21, where we use dotted boundaries to indicate that the surfaces go on indefinitely. Just like on 7. TRIANGLES ON A FLAT AND A CURVED CYLINDER 63 the flat torus, the geodesics on the flat infinite cylinder are ordinary line segments (or infinite lines) and every triangle has angle sum 180◦, even those that cut across the gluing line. You should check this by drawing an extended diagram of the flat infinite cylinder. Note that just as we saw with the flat torus, the flat cylinder has no extrinsic geometry since it’s an abstractly glued up surface.

Figure 21. Triangles on a ﬂat and a curved inﬁnite cylinder.

Next we want to consider triangles on a curved infinite cylinder. To think about the geodesics on a curved infinite cylinder, you should make a curved (finite) cylinder by rolling up a piece of paper and taping a pair of opposite sides together. Draw two points on your cylinder and then draw what seems like a geodesic segment P between the points. Now cut out a little region around your geodesic and lay it flat on a table. Since paper is not stretchy, flattening out your region doesn’t lengthen or shrink the path P that you drew. Thus if P was a geodesic on your curved cylinder, then P will be a line segment when the region is flattened out. Now that we know what a geodesic is on the curved cylinder, let’s consider triangles on the curved cylinder. Observe that if we draw a triangle on the curved cylinder, we can cut it out and lay it flat on a table as we did with the region around our geodesic. Since each of the edges of the triangle becomes a line segment when it’s flattened out, the cut-out triangle becomes a triangle in the plane when it’s flattened out. Thus we know that its angles add up to 180◦. Just as geodesics don’t lengthen or shrink when we cut them out and lay them flat, angles don’t get bigger or smaller when we cut out our triangle and lay it flat. Thus the angles of any triangle on a curved cylinder must also add up to 180◦. In fact, since distance and angles are the same in the flat and curved cylinder, any intrinsic measurement that we make in a curved cylinder will be the same as the corresponding measurement within a flat cylinder. Thus the intrinsic geometry of the flat cylinder is the same as that of the curved cylinder. This may seem strange since the flat and curved torus have different intrinsic geometries. The reason is that a curved cylinder can be made out of a piece of paper while a curved torus cannot. If you try to make a curved torus from a piece of paper, you can start by making a cylinder without any problem. But it’s impossible to make the cylinder into a torus without some stretching or shrinking. We can see this in the third picture of Figure 22 where the grey arc that will become a circle going around the outside of the torus has to become longer than the black arc that will become the circle going around the hole of the torus. Thus a curved torus can’t be made from a piece of paper. You should try it yourself with a piece of paper and some tape to see that indeed it can’t be done. 64 3. GEOMETRY AND TOPOLOGY OF DIFFERENT UNIVERSES

Figure 22. We can’t make a curved torus out of a piece of paper because paper can’t stretch or shrink.

As we have seen in these examples when we stretch or shrink an object, we change its intrinsic geometry. But, as we’ll see in the next section, geometry isn’t everything.

8. Extrinsic and Intrinsic Topology

Since A. Square has been spending a lot of time with B. Triangle, he’s been going out to eat a lot. His friends see that he has put on a little extra weight in the rear and tease him by making comments about noticing a change in his extrinsic geometry (see Figure 23). He can’t deny that his intrinsic geometry has changed a little because his clothes have been feeling tight lately. But he insists that his shape is essentially thesameasitalwayswas.

Figure 23. A. Square has changed his geometry.

In order to explain what he means by “essentially the same” he reminds his friends of what happened to their old classmate B. Rectangle. She was sitting on her porch using super glue to repair a broken vase, when a huge dog ran at her growling aggressively. In the shock of seeing it coming right at her, she squeezed the tube of glue in her hand as she was bending over the vase. The dog’s owner grabbed its leash just in time. But B. Rectangle’s head had touched the hand that held the glue, attaching her head and hand together forever. A. Square even brought out the article with “Before” and “After” photos that he had cut out and saved from the local newspaper (see Figure 24). A. Square concluded his story by saying, “Now that’s what I would call an essential change in shape”. The memory of this sad story led to an awkward silence lasting several minutes. Nobody ever teased A. Square about his shape again. Comparing the types of bodily changes that happened to A. Square and B. Rec- tangle motivates us to introduce a word to refer to the “essential” shape of an object or space. Definition. Two objects living in the same space have the same extrinsic topology if one can be deformed to the other. Deforming an object includes any kind of stretching, shrinking, twisting, bending, sliding, rotating, etc., but it does not include tearing, cutting, or gluing two parts together, like what happened to B. Rectangle. Thus when A. Square put on 8. EXTRINSIC AND INTRINSIC TOPOLOGY 65

Figure 24. Headline News: Local rectangle gets bent out of shape as a result of an out-of-control dog. some weight he still had the same topology, but B. Rectangle’s accident changed her topology because it attached her hand to her head. We illustrate a doughnut and a coﬀee cup in Figure 25. Though it’s hard to see at ﬁrst, one can be deformed to the other. You can watch a video of this deformation at http://www.youtube.com/watch?v=dwrhCSORERA

Figure 25. A doughnut has the same extrinsic topology as a coffee cup.

Thus a doughnut and a coffee cup have the same extrinsic topology. This is why some people say “topologists are people who can’t tell the difference between a doughnut and a coffee cup”. Let’s consider the knotted and unknotted tori illustrated in Figure 26. Try as we might, we cannot deform the knotted torus on the left into the unknotted torus on the right. This is because the two tori have different extrinsic topology.

Figure 26. These tori have diﬀerent extrinsic topology.

On the other hand, Flatlanders living in the surface of one of these tori would have no way of knowing whether their torus was knotted or not. This is because the knotting has to do with how the torus is situated in space, rather than the surface itself. Hence knotting is an extrinsic property of a torus. In fact, a Flatlander would not even understand the concept of a knot. Imagine trying to explain what a knot is to A. Square. This motivates our next deﬁnition. 66 3. GEOMETRY AND TOPOLOGY OF DIFFERENT UNIVERSES

Definition. Two objects living in the same space have the same intrinsic topology if one can be obtained from the other by some combination of deforming the object and cutting it apart then deforming and regluing it so that the same points are glued together.

To clarify this definition, imagine that you have a bed frame made of wooden boards that fit neatly together into slots. In order to move to a new apartment, you have to take apart your bed so that it will fit out the door. You label each pair of pieces that fit together, so that when you try to reassemble your bed in the new apartment it will be easy to know what pieces go where. We do the same thing when we cut apart an object, deform it in space, and then glue it back together. In particular, we need to make sure that the pieces go together exactly the way they were before. We saw that a knotted and unknotted torus have different extrinsic topology. However, in Figure 27 we cut open a knotted torus, undo the knot, and then reglue the ends exactly as they were before to get an unknotted torus. This illustrates that a knotted and unknotted tori have the same intrinsic topology.

Figure 27. A knotted torus can be cut apart, deformed, and then reglued to get an unknotted torus.

In general, whether we’re talking about topology or geometry, an outsider has more perspective on a space than an inhabitant of the space has. Thus if two objects are situated in the same space in such a way that they are extrinsically the same, then they are necessarily intrinsically the same. This is true regardless of whether we’re talking about topology or geometry. We summarize the relationships between extrinsic and intrinsic topology and geometry as follows:

Relationship between intrinsic and extrinsic sameness: • If two objects have the same extrinsic geometry, then they are identical. Hence they must have the same intrinsic geometry, intrinsic topology, and extrinsic topology.

• If two objects have the same extrinsic topology, then they have the same intrinsic topology.

You might learn the proofs of the above assertions in a more advanced geometry or topology class. 9. USING LOOPS TO UNDERSTAND INTRINSIC TOPOLOGY 67

9. Using Loops to Understand Intrinsic Topology

If we could understand the extrinsic topology or geometry of our universe, we would be able to understand its intrinsic topology or geometry. But unfortunately, as inhabitants of the universe, we can only detect intrinsic properties. There is no way for us to know if our universe is contained in a larger or higher-dimensional space, let alone to understand how it might be situated in such a space (or for that matter to know if some 4-dimensional beings are currently discussing how to teach us something about our space). Thus we have no choice but to think about our space intrinsically. For example, we can think about what it might be like to live in a 3-torus. Astronomers might even be able to design an experiment to try to to determine if our universe is indeed a 3-torus. But even if it is a 3-torus, we would have no way of knowing whether or not it is glued up in some 4-dimensional space. Thus if we want to have any hope of understanding the geometry or topology of our own universe, we need to develop intrinsic methods to gather information. In the last section, we said that two objects in the same space have the same intrinsic topology if one can be obtained from the other by some combination of deforming the object and cutting it apart, then deforming and regluing it so that the same points are glued together. This definition is fine for 3-dimensional beings to determine if two surfaces are intrinsically the same or different. But it doesn’t help the inhabitants of a surface to determine if their own surface is intrinsically different from some other surface, since there is no way for them to know if their surface can be cut apart, deformed in three dimensions, and reglued to get another surface. It similarly doesn’t help us learn about the intrinsic topology of our own space. The following definition can help us. Definition. Any property of a space that can be detected without making any measurements or making use of geodesics is said to be a topological property. Suchapropertyisintrinsic if it can be detected by inhabitants of the space and extrinsic if it cannot be detected by inhabitants of the space. Topological properties are useful because if one space has a particular topological property and another space does not, then the two spaces cannot be topologically the same. In this section, we will see that the types of loops in a space is a topological property that enables us to compare the intrinsic topology of many different spaces. We saw in Chapter 1 that A. Square and B. Triangle can distinguish a sphere from a torus by using geodesics. Now we’ll see that they can do so even more easily with loops that aren’t necessarily geodesics. In particular, suppose that A. Square

Figure 28. A. Square and B. Triangle can pull back any loop. 68 3. GEOMETRY AND TOPOLOGY OF DIFFERENT UNIVERSES

Figure 29. On a torus a loop can get stuck around the hole. holds one end of a string while B. Triangle carries the ball of string with her as she takes a walk around their space. When she returns to where she started, the two of them simultaneously pull on the string. If their universe is a sphere, then no matter where the string went, together they can pull it back as illustrated in Figure 28 But if their universe is a torus and B. Triangle has walked around the hole, as in Figure 29, the string will get stuck, so they can’t pull it back. Definition. If a loop can be pulled in, we say the loop is shrinkable.Other- wise, we say the loop is unshrinkable. As we see in Figure 30, there are loops on a curved torus, a knotted torus, and a flat torus that are unshrinkable. The existence of such loops is an intrinsic topological property because it does not involve measurements, does not require the loops to be geodesics, and can be detected by creatures within the space. Since every loop in a sphere is shrinkable, a sphere is topologically different from a torus (whether knotted, curved, or flat).

Figure 30. The curved torus, knotted torus, and ﬂat torus all have unshrinkable loops.

On a torus we can divide all of the loops into two distinct types: those that are unshrinkable and those that bound a region. We can see in Figure 31 that A. Square and B. Triangle can trap a 2-dimensional rabbit in a region bounded by a loop. But an unshrinkable loop on the torus can’t trap anything. Definition. If a loop in a surface bounds a region, then we say the loop is separating. Otherwise we say the loop is non-separating. In contrast with the torus, on a 2-holed torus there are three diﬀerent types of loops. In Figure 32, we see that a loop of type 1 is unshrinkable and is non- separating, a loop of type 2 is separating and shrinkable, and a loop of type 3 is separating and unshrinkable. We can see in Figure 32 that the third type of loop separates the surface into two pieces each containing one of the holes. The existence 10. LOCAL AND GLOBAL PROPERTIES 69

Figure 31. On a torus some loops bound a region and others do not. of separating unshrinkable loops is a topological property that a 2-holed torus has but a 1-holed torus does not. This shows that a 1-holed torus and a 2-holed torus have diﬀerent intrinsic topology.

Figure 32. A 2-holed torus has three types of loops.

10. Local and Global Properties

In this section, we will consider geometry as well as topology and explain how to divide properties into those that are local and those that are global. Let’s begin by considering a point on a sphere. In a small region around this point, we can draw various triangles and add up the sums of their angles. No matter how small the region is, and no matter what triangles we choose, this sum will always be greater than 180◦. Thus we say that the property of having a triangle with angle sums greater than 180◦ is a local property of the sphere. By contrast, the total surface area of a sphere cannot be measured in a small region around a point. Hence, we say that total surface area is a global property of the sphere. More generally, we have the following definition. Definition. A property of a space is said to be local if it can be detected in arbitrarily small regions around points. Otherwise, the property is said to be global. We would like to be able to combine the concepts of local and global with the concepts of geometry and topology and intrinsic and extrinsic to help us distinguish spaces. For example, consider two flat tori whose gluing diagrams are of different sizes. Their total area is different, and total area is a measurement that cannot be detected in a small region around a point. Furthermore, this difference is intrinsic because we aren’t considering the flat torus inside of a larger space. Thus we can conclude that the two tori have different intrinsic global geometry. On the other 70 3. GEOMETRY AND TOPOLOGY OF DIFFERENT UNIVERSES hand, any small region around a point on either a big or small flat torus will have the same geometry as a similar small region of R2. Thus both the big and small flat torus have the same local intrinsic geometry as R2. In the last section, we showed that a torus is topologically different from a sphere because the torus has unshrinkable loops but every loop on a sphere is shrinkable. Let’s think about whether this loop property is local or global. If the property is local, then we should be able to detect it in any region around a point. However, no matter where we are on the torus, if we pick a small enough region, it will not include the hole. Thus any loop in such a small region is shrinkable (see Figure 33). This leads us to conclude that having unshrinkable loops is a global intrinsic topological property.

Figure 33. Every loop in the dotted region is shrinkable.

On the other hand, if we look at a very small region around a point in either a sphere or a torus we can deform the region to a disk. It follows that the sphere and the torus have the same local intrinsic topology, but different global intrinsic topology. In order to compare the local intrinsic geometry of a sphere with that of a curved torus, we consider the sums of angles of triangles. However, we have to be careful not to refer to any specific triangle because a specific triangle cannot be detected in an arbitrarily small region. For example, we saw that on a sphere there is a triangle with three right angles. But such a triangle does not occur in a small region of the sphere. Thus the existence of this particular type of triangle is a global property of the sphere. In order to consider the local geometry of a sphere and a torus, we want to consider sums of angles of triangles, no matter how small. On a sphere the sum of the angles of any triangle is always greater than 180◦, no matter where we are on the sphere and no matter how small the triangle is. On the curved torus, the story is quite different. As we saw in Figure 20, in some regions the sum of the angles of a small triangle is greater than 180◦, but in other regions the sum of the angles of a small triangle is less than 180◦. This means that the local intrinsic geometry of a curved torus is different in different regions. It follows that the sphere and the curved torus have different local intrinsic geometry. The local intrinsic geometry of the curved torus is difficult to study precisely because it changes as we move from one point to another. We would like to focus on spaces where the geometry doesn’t depend on where you are. In particular, we make the following definition. 11. MANIFOLDS 71

Definition. A space is said to be homogeneous if its local intrinsic geometry is the same at all points. The word “homogeneous” may remind you of homogenized milk. In fact, dairies homogenize their milk so that the cream will be evenly mixed and milk from different cows will be evenly mixed. This means that every sip you take of the milk will taste the same, from one carton to the next and from the top of the carton to the bottom of the carton. Thus the words “homogeneous” for spaces and “homogenized” for milk are actually closely related, though their pronunciation is somewhat different. The curved torus is not homogeneous, because its geometry is not the same at every point. On the other hand, a sphere, a flat torus, a flat infinite cylinder, and Rn are all homogeneous spaces. Even the 3-torus that we constructed in Chapter 1 by gluing opposite sides of a cube is homogeneous. Homogeneous spaces are easier to study than non-homogeneous spaces because we only have to look at the local intrinsic geometry in regions around one point in order to understand the local intrinsic geometry everywhere. In particular, if a space is homogeneous, the sum of the angles of a triangle of a given size is the same no matter where the triangle is.

11. Manifolds

Though cosmologists don’t know the global topology of our universe, it is reasonable to assume that in any small region, the local topology is the same as in a small region of R3. This is certainly true near the Earth and in the areas of space that have been explored. Definition. A space is said to be an n-manifold if its local intrinsic topology is the same as Rn at every point. This means that a 2-manifold is a space where any point has a region around it that can be deformed to a disk with that point at the center. In Figure 34 we illustrate some 2-manifolds. On the curved torus, around any point there is a curved disk that can be deformed to an ordinary disk. In the ﬂat torus the disks may be split by the gluing wall, but they are still disks. The illustration on the right is a plane with the origin removed. This is also a 2-manifold, but we will have to choose smaller and smaller disks as we consider points closer and closer to the missing origin. Note that because the origin is not in the space, you can’t choose it as one of the points whose local intrinsic topology you’re considering.

Figure 34. A curved torus, a ﬂat torus, and a plane with the origin removed are all 2-manifolds. 72 3. GEOMETRY AND TOPOLOGY OF DIFFERENT UNIVERSES

Figure 35. Neither a disk nor two intersecting planes is a manifold.

A region around a point in R3 can be deformed to a ball with that point at the center. So a space is a 3-manifold if every point in the space has a region around it that can be deformed to a ball with the point at the center. For example, a 3-torus is a 3-manifold, though some balls may be split by the gluing wall, just as some disks on a ﬂat torus were. Figure 35 illustrates two spaces that are not manifolds. The disk on the left is not a 2-manifold because a region around p on the boundary will look like a half of a disk. While this region could be deformed to a round disk, the point p will still be on the boundary of the region rather than in the center. So the local intrinsic topology around this point is not the same as on a plane. The space on the right in the ﬁgure consists of two intersecting planes. This space is not a 2-manifold because a region around the point q will look like two intersecting disks rather than a single disk. A 2-manifold is often referred to more simply as a surface. Unfortunately, there is no such convenient word to refer to a 3-manifold. So we have to call a 3-manifold a 3-manifold.

12. Assumptions about Universes

While A. Square cannot know for certain the shape of his universe, there are some assumptions that seem reasonable for him to make. For instance, it would be very strange if his universe had two or more disconnected pieces, preventing him from ever visiting some parts of his universe without the magical powers of teleportation. Thus, it seems reasonable for A. Square to assume that his universe consists of a single connected space such that given enough time, suﬃciently super- duper technologically advanced transportation, and the appropriate visa, he could travel to any part of it. Furthermore, it seems unlikely that his universe has an edge which forms an invisible wall that nobody can pass through. So it’s safe to assume that his universe isn’t a disk or any other 2-dimensional shape with a boundary. It also seems unlikely that Flatland contains two intersecting planes, because at the line of intersection a vertical shape could cut through the middle of a horizontal shape. For example, in Figure 36, A. Square has a terrible stomach ache because C. Oval is passing through his stomach. Perhaps all of our undiagnosed aches and pains are caused by other 3-dimensional creatures passing through our insides in this way. But this seems unlikely. So it’s reasonable to assume that Flatland is a 2-manifold. 12. ASSUMPTIONS ABOUT UNIVERSES 73

Figure 36. A. Square has a stomach ache because C. Oval is passing through him.

Figure 37. When B. Triangle takes a walk on a curved torus her angles change.

If Flatland were a curved torus, other problems could occur. In particular, as we’ve seen, if B. Triangle took a walk, her area and angles could dramatically change causing serious internal pain, as illustrated in Figure 37 . It would be pretty strange to live in a universe where every time you walked in a certain area you got a stomach ache. In order to avoid this kind of situation, it also seems reasonable to assume that Flatland is homogeneous. The above assumptions about Flatland make sense for any universe. This gives us the following list of assumptions that we will make about all universes, including our own.

List of Assumptions about Universes: (1) Every universe is connected. (2) Any universe where n-dimensional beings live is an n-manifold. (3) All universes are homogeneous.

Recall that the creatures in a universe can only study properties of their space that are intrinsic, since they cannot leave their universe to get an extrinsic perspective. For this reason, we focus on trying to discover the intrinsic topology and geometry of our universe, rather than assuming our space is contained in a higher- dimensional space and trying to imagine what its extrinsic topology and geometry could be. By our second assumption about universes every n-dimensional universe has the local intrinsic topology of Rn. Thus, what remains to discover is its global intrinsic topology. While it would be nice to also know about the global intrinsic geometry of our universe, often such information is hard to ﬁnd because it requires measuring 74 3. GEOMETRY AND TOPOLOGY OF DIFFERENT UNIVERSES something very large like the total volume of the universe. By contrast, since we’re assuming our universe is homogeneous, understanding the local intrinsic geometry at one point will tell us the local intrinsic geometry at every point. Thus our eﬀorts might be better spent exploring the local intrinsic geometry of our universe rather than trying to determine its global intrinsic geometry. In summary, we are most interested in learning about the global intrinsic topology and the local intrinsic geometry of our universe.

13. A 2-Dimensional Sphere with a Point Removed

Now that we have developed our vocabulary, we would like to explore the topology of some specific spaces in more detail. As our first example we will show that a sphere with a point removed has the same extrinsic topology as a plane when considered as objects in R3.Webeginby removing a point from a sphere in Figure 38. We’ve stretched open the hole a little bit to make it easier to see, and we use a dotted boundary to indicate that there is no actual boundary after removing the point. Next, we stretch out the surface, and flatten it so it lies in a plane (as illustrated in the center and right side of Figure 38). In this way, we get a disk whose boundary is not included. We call this an open disk. We can further stretch or shrink the open disk so that it has radius 1.

Figure 38. A sphere with a point removed can be stretched out to get a ﬂat disk with no boundary.

1 3 7 The next step is to subdivide the open disk into rings with radius 2 , 4 , 8 ,and so on forever (see the left side of Figure 39). This makes the open disk look like a target with an infinite number of rings that get thinner and thinner as we approach 1 but never reach the boundary. In particular, the bulls-eye has radius 2 , the first 1 1 ring has width 4 , the next ring has width 8 ,etc. We will now stretch the open disk so that it covers the whole plane. Leaving 1 the bulls-eye fixed, we stretch the first ring so that its width is 2 , which is the same as the radius of the bulls-eye. Then we stretch the second ring so it has the same width and so on, stretching one ring after another so that each ring has the same width as the previous one. We do this forever to get a target with an infinite number of concentric rings all of the same width as illustrated on the right side of Figure 39. Since there are an infinite number of rings with the same width, the target will now fill up the entire plane. To summarize, we have removed a point from a sphere, deformed it into an open disk, and then stretched the open disk to cover the entire plane. It follows 13. A 2-DIMENSIONAL SPHERE WITH A POINT REMOVED 75 …

… … …

Figure 39. We subdivide an open disk into concentric rings, then stretch the rings so that they ﬁll up the plane. that a sphere with a point removed has the same extrinsic topology as a plane. This implies that they have the same intrinsic topology as well. In order to see this another way, we introduce the method of stereographic projection. We start with a sphere tangent to a plane R2 so that it is only touching the plane at the South Pole. Then we extend a line from the North Pole through each point on the sphere to a point on the plane as illustrated in Figure 40. This assigns to each point on the sphere other than the North Pole, a point in the plane. We can imagine that we have a very bright light at the North Pole and each point on the sphere casts a shadow on the plane. All of the circles on the sphere that are are parallel to the equator cast circular shadows in the plane. For example, the equator in Figure 40 casts the circular shadow shown. The closer such a circle is to the North Pole, the larger its circular shadow will be.

Figure 40. Every point in S2 except the North Pole can be paired with a point in R2.

We can’t extend this projection to the whole sphere because there’s no point on the plane for the North Pole to go. To solve this problem, we send the North Pole to a point outside of the plane that we call the “point at ∞”. You get to this point if you go inﬁnitely far in any direction in R2. In this way, we have assigned every point on the sphere to a point in the plane together with ∞. We write this 76 3. GEOMETRY AND TOPOLOGY OF DIFFERENT UNIVERSES

Figure 41. Gluing together these two regions gives us a sphere with a point removed. as S2 = R2 ∪{∞}. Thinking about S2 this way allows us to imagine that R2 is contained in S2,becauseifweremovethepointattheNorthPolefromS2,weare left with R2. We would like to explain to A. Square that a sphere with a point removed has the same intrinsic topology as a plane. He can understand Figure 39 together with the argument about how we stretch an open disk into a plane, but he can’t understand Figure 38. So we’ll have to use the gluing diagram for a sphere that we introduced in Chapter 1. Figure 41 illustrates the two disks that are glued together to get a sphere, but now we have removed a single point from the center of one of the disks. A. Square remembers the gluing diagram for a sphere, so he has no trouble understanding Figure 41. We can deform the light grey region in R3 as follows. We pull the hole up to get an inverted funnel, stretch open the hole, and then we push it down into the plane to get an ring with no outer boundary, as illustrated in Figure 42.

Figure 42. We deform the disk with a point removed to get a ring with no outer boundary.

Finally, we glue the inner boundary of the ring to the outer boundary of the darker disk as illustrated in Figure 43. Since the darker disk ﬁlls the hole of the ring, we end up with an open disk. A. Square can understand the gluing in Figure 43, but the deformation in Figure 42 took place in 3-dimensional space, and hence A. Square can’t follow it. So instead we tell him that we turn the ring inside out in 3-dimensional space so that the inside and outside are switched. In order to understand this he tries to think about something analogous that is one dimension down. He has the idea that if a line in Flatland has an arrow on it, he can turn the line around so that the arrow will switch directions (see Figure 44). But the idea of switching the direction of an arrow would seem impossible to a 1-dimensional creature like A. Dash. Keeping this in mind, A. Square is willing to accept the idea that the inside and outside of a ring can be switched in 3-dimensional space. 14. A 3-DIMENSIONAL SPHERE WITH A POINT REMOVED 77

Figure 43. We glue the disk into the hole in the ring to get an open disk.

Figure 44. An arrow can be turned around in Flatland, but not in Lineland.

This convinces A. Square that a sphere with a point removed has the same intrinsic topology as an open disk. We then use Figure 39 to show him that an open disk has the same intrinsic topology as a plane. Thus he agrees that a sphere with a point removed has the same intrinsic topology as a plane. Note that it would be pointless to try to convince A. Square that in R3 the sphere with a point removed has the same extrinsic topology as a plane, since there’s no way that he can know anything about the extrinsic topology of a 2-dimensional space.

14. A 3-Dimensional Sphere with a Point Removed

Using ideas similar to those in the last section, we now see as follows that S3 with a point removed has the same intrinsic topology as R3. To do this, we need to imagine 3-dimensional analogues of Figures 39, 41, 42, and 43. First recall, that S3 is obtained by gluing two balls together along their boundaries, just as S2 is obtained by gluing together two disks along their boundaries. The topology of these glued up balls is the same as S3, but the geometry is different since the balls had to be deformed to fit into our 3-dimensional space. But we are currently only interested in the topology, so the geometric differences are not important to us here. We remove a point from the center of one of the gluing balls and stretch open the hole so it’s easier to see, as we did for the disks in Figure 41. You can visualize this as an avocado with nothing where the pit would normally be. We now visualize the gluing space for S3 with a point removed as one solid ball and one hollow ball that looks like an avocado which has an outer boundary (where the skin is) but no inner boundary (where the pit would begin). 78 3. GEOMETRY AND TOPOLOGY OF DIFFERENT UNIVERSES

Next we accept the fact that the hollow ball can be turned inside out in 4- dimensional space so that it has no outer boundary but has an inner boundary. This deformation is as impossible for us to picture just as it was impossible for A. Square to picture how we turned the ring inside out in Figure 42. But it seems reasonable that such a deformation can take place in 4-dimensional space since we saw how to turn a ring inside out in 3-dimensional space. Finally, we glue the solid ball to the inner boundary of our hollow ball, as we did with the disk and the ring in Figure 43. This gives us an open solid ball, analogous to the open disk in Figure 43. In particular, we now have a ball whose boundary is not included. 1 Next we subdivide our solid ball into a ball of radius 2 , and concentric hollow balls of smaller and smaller widths like the concentric rings of smaller and smaller widths on the left side of Figure 39. Finally, we stretch these hollow balls so that 1 they all have width 2 , as we did with the concentric rings in Figure 39. Since there 1 are infinitely many concentric hollow balls all of width 2 surrounding the inner ball 1 R3 of radius 2 , together they cover all of just as the bulls-eye and rings on the right side of Figure 39 covered all of R2. This shows us that the space we get by removing a point from the gluing diagram of S3 has the same intrinsic topology as R3. Recall that in the last section we observed (with the help of stereographic projection) that in addition to thinking of R2 as S2 with a point removed, we can think of S2 as R2 together with a point at infinity. We wrote this as S2 = R2 ∪{∞}. We can now think of both R3 as S3 with a point removed, and S3 as R3 together with a point at infinity. We write this as S3 = R3 ∪{∞}. You get to the point at ∞ in S3 by going infinitely far in any direction. This way of visualizing S3 is convenient because it enables us to imagine that R3 is contained in S3,justas visualizing S2 in the analogous way enabled us to imagine that R2 is contained in S2.

15. Curves on a Torus

In this section, we want to use loops to explore the topology of a torus in more detail. Recall that in Figure 31 we saw that a torus has two types of loops, those which are separating and those which are non-separating. It may seem like this is all there is to know about loops on a torus. However, it turns out that on a torus, there are actually inﬁnitely many non-separating loops whose extrinsic topology are all distinct. In order to analyze loops on a torus, we begin by picking two special loops that we’ll refer to. A circle that runs once around the curved torus the long way (as in Figure 45) is called a longitude, while one that runs once around the torus the short way is called a meridian. We are thinking of a longitude and meridian as topological objects in the torus. So any loop which can be deformed to a longitude is also a longitude, and any loop which can be deformed to a meridian is also a meridian. In order to remember which circle on the curved torus is a meridian and which is a longitude, you can use the following trick. If you want to make a bagel sandwich, youﬁrsthavetocutthebagelopenthelong way, along an inner and outer longitude so that you can put meat, cheese, lox, or tofu together with whatever condiments you like between the two halves. Finally, after closing up your sandwich, you cut 15. CURVES ON A TORUS 79

Figure 45. A meridian and a longitude on a curved and flat torus. it in half by slicing along two meridians which makes you merry because you can now eat your sandwich. In the flat torus pictured on the right of Figure 45, the sides of the square represent loops, but both have the same length. We designate the loop on the left and right sides of the square as the meridian and the loop on the top and bottom sides as the longitude. However, it doesn’t actually matter which is which, since we can rotate the square by 90◦ to interchange the pairs of sides. This is an important difference between the curved torus, which is an object in R3, and the flat torus which has no extrinsic topology. We are now interested in looking at loops on the torus which don’t intersect themselves. We’ll call such loops on the torus curves. We begin by considering the grey curves illustrated in Figure 46. These curves wrap once around the torus meridionally and once around the torus longitudinally. However, by following the arrows we can see that one curve goes around the back of the torus from the outside longitude to the inside longitude, while the other curve goes around the back of the torus from the inside longitude to the outside longitude. Because one of these curves cannot be deformed to the other on the torus, the two curves have different extrinsic topology.

Figure 46. Two topologically distinct loops on a torus.

We orient the longitude and meridian, and then we refer to the curve whose orientation agrees with the orientations of both the meridian and the longitude as a(1, 1)-curve and refer to the curve whose orientation agrees with the longitude but not the meridian as a (−1, 1)-curve. Of course, which is which depends on how we choose our arrows. More generally, for a given pair of positive or negative numbers p and q,we would like to deﬁne a (p, q)-curve on the torus as a curve which goes p times around the torus meridionally and q times around the torus longitudinally. For example, consider the grey curve illustrated in Figure 47. If you trace it out with your ﬁnger, you can see that it winds around the torus three times meridionally while 80 3. GEOMETRY AND TOPOLOGY OF DIFFERENT UNIVERSES

Figure 47. A(3, 2)-curve on a torus. winding two times around the torus longitudinally. Thus this curve is referred to as a (3, 2)-curve. You should notice that the grey curve intersects the meridian in two points and intersects the longitude in three points (marked with dots in the picture). Keep in mind that the dashed arcs are on the under side of the torus and hence are not intersecting the longitude. Furthermore, there is no way to deform the grey curve to intersect the meridian or longitude in fewer points. In general, if a curve on the torus intersects the meridian in p points and the longitude in q points and can’t be deformed to intersect either in fewer points, then the curve must be a (p, q)-curve.

16. How to Draw a (p, q)-Curve on a Flat Torus You may find that it’s not so easy to draw a specific (p, q)-curve on the curved torus. For example, you might pause here and try to draw a (3, 4)-curve on a curved torus. By contrast, we will now learn an easy method for drawing (p, q)-curves on aflattorus. We begin by explaining how to draw a (3, 2)-curve on a flat torus. We will use the observation that a (3, 2)-curve intersects a longitude three times and a meridian two times. Thus in our flat torus the (3, 2)-curve will have three points of intersection along the top and bottom sides of the square (representing the longitude) and two points of intersection along the left and right sides of the square (representing the meridian). In order to draw the (3, 2)-curve on the flat torus, we first label the points on the edges of the square where we want the (3, 2)-curve to intersect the meridian and longitude as illustrated in Figure 48. Since a (3, 2)-curve never intersects itself, it must be a collection of non- intersecting paths that start and end at the labeled points on the top, bottom,

Figure 48. A ﬂat torus with points labeled to indicate where the (3, 2)-curve intersects a meridian and a longitude. 16. HOW TO DRAW A (p, q)-CURVEONAFLATTORUS 81 and sides of the gluing square. Hence the paths making up our (3, 2)-curve must adhere to the following rules:

Rule 1: Every labeled point must be an endpoint of a path. Rule 2: No pair of paths intersect.

Since every labeled point is an endpoint of a path and together the paths make a loop, it doesn’t matter which labeled point we begin with. So let’s say we choose a2 on the bottom edge as a starting point for our first path. The first path must go from a2 on the bottom edge to some labeled point either on the top or on one of the sides. First, let’s see what happens when the path from a2 on the bottom edge goes to a3 on the top edge. Since a3 on the top edge is glued to a3 on the bottom edge, the second path must start at a3 on the bottom edge. Any path from a3 to a labeled point to the left of a3 would intersect the first path, as we see in the left image of Figure 49. Thus, the ending point of the second path must be either b1 or b2 on the right edge, as illustrated on the right in Figure 49.

Figure 49. Problems that could arise if we choose the wrong ending point for the second path.

However, if b1 is the ending point of the second path, then any path starting at b2 on the right side would have nowhere to go without intersecting the path from a3 to b1. A similar problem would occur if b2 were the ending point of the second path. Thus, once we choose a3 as the ending point of the first path, we are eventually forced to isolate one of our labeled points so that it cannot be an endpoint of any path (as illustrated on the right of Figure 49) This contradicts Rule 1. So a3 cannot be the ending point of the first path. An analogous argument on the left edge of the square shows that a1 cannot be the ending point of the path from a2. Furthermore, if we chose a2 on the top edge as the ending point of the path from a2 on the bottom, we would again violate one of the rules. Would it violate Rule 1 or Rule 2? Thus, starting at a2 on the bottom edge, the first path must end at one of the labeled points on the side of the square, since every other possibility will result in a violation of Rule 1 or Rule 2. Moreover, if the path from a2 ended at b2 on either side, then we would isolate one of the points on the bottom edge. (Which one in each case?) Therefore, we have no choice but to make b1 be the ending point of the 82 3. GEOMETRY AND TOPOLOGY OF DIFFERENT UNIVERSES

first path, though it doesn’t matter if it’s the b1 on the right edge or the b1 on the left edge. We choose b1 on the right edge as the ending point of the first path (see the first image in Figure 50). Then b1 on the left edge is the starting point of the second path. In order to avoid isolating any labeled points or intersecting the path from a2, we must then choose a1 on the top edge as the ending point of the second path (see the second image in Figure 50). This means that the third path will start at a1 on the bottom edge. Again, keeping with our strategy of not isolating any point or intersecting any existing paths, a3 on the top edge must be the ending point of the third path (see the third image in Figure 50). What would have happened if b2 on the left edge were chosen as the ending point of the third path? Continuing in this way, the fourth path has starting point a3 on the bottom edge and ending point b2 on the right edge (see the fourth image in Figure 50). The fifth and final path has starting point b2 on the left edge and ending point a2 on the top edge, and we have come full circle (see the fifth image in Figure 50). In this way, we obtain the set of paths making up a (3, 2)-curve.

Figure 50. How we construct the paths that make up a (3, 2)-curve.

Note that if we had started with b1 on the left edge as the ending point of the first path and continued from there, we would end up with the mirror image of the last drawing in Figure 50. Also if we had made our labeled points equidistant on each edge, we could make the paths in the square parallel line segments as in Figure 51. What would the slope of these parallel line segments be? Try to use the above method to draw a (3, 4)-curve on a flat torus. What happens if you try to draw a (2, 2)-curve on the flat torus? What about a (4, 6)- curve? Explore these questions by placing the appropriate number of points along the edges of a square and follow a procedure similar to the one above. Can you deduce anything about the relationship between the integers p and q in a (p, q)-curve on the torus? 17. LINES IN THE EXTENDED DIAGRAM OF A FLAT TORUS 83

Figure 51. A(3, 2)-curve drawn on the ﬂat torus with lines of equal slope. What is the common slope of these lines?

17. Lines in the Extended Diagram of a Flat Torus

Another way to explore curves on a flat torus is to start with a line in an extended diagram and then see what curve it represents on the flat torus. For 2 example, consider the line with slope 1 illustrated on the left in Figure 52. When we draw this line in the flat torus it becomes just two segments, because the intersection of the grey line in Figure 52 with the square in the top and bottom rows of the extended diagram represent the same segment on the flat torus. The two segments in the flat torus together form a (2, 1)-curve because the loop intersects a longitude of the flat torus in two points and a meridian of the flat torus in one point. Notice that the points in the lower left corner and the top right corner of the flat torus count as both one intersection point with a meridian and one intersection point with a longitude.

Figure 52. 2 A line with slope 1 in the extended diagram represents a(2, 1)-curve on the ﬂat torus.

p In general, we can consider any line in the extended diagram with slope q where p and q are integers with no common factors. Even though our line is infinitely long, in the flat torus it has only a finite number of segments, and together these segments form a (p, q)-curve. But what happens if we consider a line in the extended diagram whose slope p cannot be written as q where p and q are integers. In this case, no pair of points on 84 3. GEOMETRY AND TOPOLOGY OF DIFFERENT UNIVERSES the line will represent the same point in the flat torus. So the line in the extended diagram will give us infinitely many different segments in the flat torus. These segments form an infinitely long path rather than a closed loop in the torus. Thus, even though a torus has finite area, it contains lines that never end just like lines in the plane never end. What would happen if A. Square went for a walk on such a line? Would he ever return to his starting point?

18. Exercises

1. Consider the space R2 −{(0, 0)} illustrated in Figure 53. Is there a geodesic between the points p and q? If so, ﬁnd it. If not, why not?

Figure 53. Illustration for Exercise 1.

2. Draw the extended diagram for a flat infinite cylinder and explain how we know that the sum of the angles of every triangle in the flat infinite cylinder is 180◦.

3. Make a conical party hat and draw two points on it. Does the distance between the points change when the hat is cut open into a flat surface? What does a geodesic segment on the hat look like when the hat is cut open? Is a circle parallel to the circular base of the hat a geodesic? 4. Using one or more pieces of paper and some tape, can you make a surface containing a triangle whose angles add up to more than 180◦? 5. Can you make a surface out of paper and tape with a triangle whose angles add up to less than 180◦? 6. Can you make a surface out of paper and tape that contains some triangles whose angles add up to 180◦, some which add up to less than 180◦,andsome which add up to more than 180◦? Explain your constructions. 7. Prove that the ratio of the perimeter of a hemisphere to its area is different from the ratio of the perimeter of a disk to its area. Conclude from this that the intrinsic geometry of a hemisphere is different from that of a disk. 8. In Figure 40, the shadows cast by latitudinal circles are circles in the plane. What kinds of shadows are cast by longitudinal circles? 9. If the sphere in Figure 40 were the Earth, compare the sizes of the shadows cast on the plane by Antarctica, Hawaii, Greenland, and the polar ice caps at the North Pole. 18. EXERCISES 85

10. Is the property of being homogeneous global or local? Is the property of being an n-manifold global or local? 11. Consider the circular strips in Figure 54, which you should imagine were created using three identical rectangles of paper. Compare their extrinsic geometry and topology. Then make a table comparing their intrinsic properties (geometric and topological, local and global).

Figure 54. Illustration for Exercise 11.

12. Compare the local and global intrinsic geometry of a disk and a conical party hat. Then compare their local and global intrinsic topology. How could Flat- landers distinguish between these two spaces? 13. Give examples (different from the ones given in the chapter) of pairs of spaces with the following properties. (a) The same local intrinsic topology, but different local intrinsic geometry. (b) The same local intrinsic geometry, but different global intrinsic topology. (c) The same global intrinsic topology, but different global extrinsic topology. (d) The same local intrinsic geometry, but different global intrinsic geometry. (e) The same global intrinsic geometry, but different extrinsic geometry. 14. Is the surface of a cube homogeneous? Why or why not? 15. Compare the local and global intrinsic geometry and topology of a 3-torus made from a room whose length is longer than its width and one made from a room whose length is equal to its width. 16. Do two spheres of different sizes have different global intrinsic geometry? Do two spheres of different sizes have different local intrinsic geometry? Explain your conclusions. 17. How would you define a 1-manifold? Give two examples of 1-manifolds and explain how you know they are different. 18. Do the two objects in Figure 55 have the same extrinsic topology in R3?Do they have the same extrinsic geometry in R3?

Figure 55. Illustration for Exercise 18.

19. Show that every capital letter in our alphabet can be deformed to one of the capital letters: A, B, or C. Thus topologically, there are only three capital letters.

https://doi.org/10.1090//mbk/096/04

CHAPTER 4

Orientability

Topics: • M¨obius strips • Orientation reversing paths • The Klein bottle • Tic-Tac-Toe on a Klein bottle • The 3-dimensional Klein bottle • Theprojectiveplane • Cross-caps • Projective 3-dimensional space • Sidedness and orientability of surfaces in 3-manifolds

1. The M¨obius Strip

The person you see in the mirror each morning looks quite similar to you, and yet when you raise your right hand the person in the mirror raises his or her left hand. Is it possible to move into a position so that when you raise your left hand your mirror image does the same? What if the mirror is on the ceiling? What if you are standing on your head? If you look in the mirror while wearing a t-shirt that says “Math is Cool”, you will see yourself with a t-shirt that says “ Cool is Math ”. If you leave your t-shirt on and you take a walk, no matter where you go, your t-shirt will never change from saying “Math is Cool” to saying “ Cool is Math ”. We want to understand what it would be like to live in a universe where you could go on a trip and come back as your mirror image. We begin by considering what happens to A. Square when he takes a long walk around Flatland. He puts on his walking hat and leaves town. He never crosses the road or changes direction. After walking for longer than ever before, he sees that he is approaching his home- town from the opposite direction. However, when he gets close to the doughnut shop with his favorite cinnamon twists, the sign in front of the shop looks diﬀerent than usual (see Figure 1). Figure 2 shows a picture of the road that A. Square walked along. A. Square does not feel like he has changed at all, but he thinks that everything else has become its mirror image. On the other hand, his friends think that he has become mirror reversed. It’s not just that he is upside down. He could turn around, but then he would be facing the opposite direction. The only way for him to ﬁx himself

87 88 4. ORIENTABILITY

Figure 1. The sign in front of A. Square’s favorite doughnut shop.

Figure 2. Walking along this path reverses A. Square. is to go for a second walk all the way around Flatland. But he is certainly too tired for that. The road that A. Square has walked along is what is called a Möbius strip (also known as a Möbius band). You can make a Möbius strip out of a belt by adding a half-twist and then buckling the inside to the outside. We illustrate a gluing diagram for a Möbius strip in Figure 3. Notice that the bottom corner of the gluing diagram is attached to the top corner of the other side. As a result, the boundary of the Möbius strip is a single circle. You should check this for yourself by tracing your finger along the boundary in Figure 2.

Figure 3. Gluing diagram of a M¨obius strip.

Once we glue the two edges of the strip together, the “seam” that we glued along disappears completely just as it does on the flat torus. For the Möbius strip this means that there is no specific location where the mirror-flip happens all of a sudden. A. Square had to make a complete circle around the Möbiusstrip for this phenomenon to occur. If he travelled part of the way or even all of the way and then retraced his steps, he would not be reversed. 2. ORIENTATION REVERSING PATHS 89

2. Orientation Reversing Paths

In order to talk about paths like the one A. Square traveled along and the kinds of spaces that contain them, we introduce the following definition. Definition. We say a circular path in an n-dimensional space is orientation reversing if an n-dimensional being can walk along it to become mirror reversed. We illustrate the importance of the dimension in the definition as follows. The path that A. Square followed in Figure 2 was an orientation reversing path in Flatland. In Figure 4, A. 3D-Girl is taking a walk along a Möbius strip in R3.Her left hand is swollen because it was stung by a bee. Her right hand is OK, but she’s keeping it in her pocket to protect it from any bees she might encounter. Notice that she comes back upside down and underneath her starting point, but she is not reversed. In particular, her left hand is still swollen and her right hand is still in her pocket. Thus the orientation reversing path that A. Square walked along in the Möbius strip is not orientation reversing for 3-dimensional people like A. 3D-Girl.

Figure 4. A. 3D-Girl walks along a M¨obius strip but is not mirror reversed.

We have a special name for spaces like a M¨obius strip that contain orientation reversing paths. Definition. An n-dimensional space is said to be non-orientable if it contains a path which is orientation reversing for n-dimensional creatures. If it contains no such path, then it is said to be orientable. 90 4. ORIENTABILITY

When we look at a gluing diagram for a surface, it is generally hard to tell what the surface would look like if we glued it up. Nonetheless, it is possible to check if the surface is non-orientable just by looking for an orientation reversing path. Any path that does not cross the boundary of a gluing diagram cannot be orientation reversing since such a path would be contained in a region of R2.This means that we just have to check whether a path that goes out one edge and in another edge is orientation reversing. To find such a path, we look for the gluing diagram of a Möbius strip within our gluing diagram. Any Möbius strip will contain an orientation reversing path. For example, in Figure 5 we see that the grey strip is a a gluing diagram for a Möbius strip, and hence we know the surface containing it is non-orientable.

Figure 5. A non-orientable gluing diagram containing a M¨obius strip.

In what follows we see that there is no good way to define left and right in a non-orientatible surface. In Figure 6, A. Jester takes the same walk around the Möbius strip that A. Square did. At his starting position, he is shown with his left arm and left leg painted grey and his right arm and right leg painted black. However, when he comes back to his starting position after one trip around the Möbius strip, his left arm and left leg are painted black and his right arm and right leg are painted grey. Since left and right can morph into one another in this way, it is meaningless to define left and right in a surface containing a Möbius strip.

Figure 6. When A. Jester returns home, his left and right are reversed.

3. The Klein Bottle

Though a M¨obius strip is a non-orientable space, it’s not a 2-dimensional universe because it has a boundary. In order to get rid of the boundary of a M¨obius strip, we can glue together the top and bottom sides of the gluing diagram in Fig- ure 3. The surface that we obtain is called a Klein bottle, named after Felix Klein 3. THE KLEIN BOTTLE 91 who introduced this surface in 1882. This surface, denoted by K2, is illustrated in Figure 7.

Figure 7. A Klein bottle is a non-orientable 2-manifold.

Notice that the gluing diagram for the Klein bottle in Figure 7 is a square, whereas the gluing diagram for a Möbius strip in Figure 5 is a long thin rectangle. If we had used a rectangle as a gluing diagram for the Klein bottle, its intrinsic topology and local intrinsic geometry would be the same as in Figure 7. However, its global intrinsic geometry would be different. Whether defined on a square or a rectangle, since the Klein bottle contains a Möbius strip, it contains an orientation reversing path and hence is non-orientable. Also, since the Klein bottle has no boundary, it’s a 2-manifold. Thus, Flatland could theoretically be a Klein bottle. While there is nothing wrong with defining the Klein bottle intrinsically as a glued up square, you may have the feeling that you don’t really understand it because you haven’t seen a picture of it glued up in 3-dimensional space. But before attempting to glue up the Klein bottle, let’s look at Figure 8 to remember how we glued up a torus.

Figure 8. This is how we glued up a torus.

We start to glue up the Klein bottle in Figure 9 by gluing the two arrows that point in the same direction as we did for the torus in Figure 8. Next, we would like to glue together the circles on the ends of the cylinder in the central drawing of Figure 9. However, when we try to do this we run into a problem because the arrows don’t match up as they did in Figure 8. ?

Figure 9. Our attempt to glue up a Klein bottle.

In order to align the arrows, the surface would have to pass through itself so that one circle could approach the other from the back. (You might try this for yourself with a slinky.) But if the surface intersects itself, then it would not be a 2-manifold and hence not a legitimate universe. We can solve this problem if 92 4. ORIENTABILITY the Klein bottle is in R4 rather than R3. In this case, we can bring one end of the cylinder up into the fourth dimension so that it can pass into the inside of the cylinder without intersecting the cylinder (just as a 4-dimensional creature could remove my heart without piercing my skin). We cannot draw an accurate picture of this in 3-dimensional space, let alone on this 2-dimensional page. The best we can do is to draw Figure 10 where the Klein bottle appears to intersect itself.

Figure 10. Gluing up a Klein bottle in 4-dimensional space.

When looking at Figure 10, you should keep in mind that there really is no “circle of intersection”. Analogously, imagine holding two crisscrossing pencils up to a light so that their shadows on the wall look like they intersect when in reality the two pencils don’t intersect at all. When we draw the Klein bottle in Figure 10, we are discarding one dimension’s worth of information, just like when we look at the shadow of two crisscrossing pencils. Notice that the local intrinsic geometry of the glued up Klein bottle in Figure 10 is non-homogeneous just like that of the curved torus. On the other hand, the gluing diagram of the Klein bottle in Figure 7 is nice and flat and homogeneous just like the flat torus. So although trying to imagine a curved Klein bottle sitting in 4-dimensional space is an interesting mental exercise, we prefer to use the gluing diagram of the flat Klein bottle as in Figure 7. So from now on whenever we refer to the Klein bottle, you should think of the gluing diagram rather than of the picture illustrated in Figure 10.

4. The 3-Dimensional Klein Bottle

Now that we understand the gluing diagram for the 2-dimensional Klein bottle, we would like to create a gluing diagram for a 3-dimensional Klein bottle K3.We start with a rectangular room where the left wall is glued to the right wall and the front wall is glued to the back wall, just as we did for the 3-torus. But in this case, we glue the ceiling to the floor with a left-right flip, rather than gluing it straight as we did for the 3-torus. In Figure 11, we indicate how the different pairs of walls are 5. TIC-TAC-TOE ON A TORUS AND KLEIN BOTTLE 93

Figure 11. The gluing diagram for a 3-dimensional Klein bottle. glued with different types of arrows. Note that the arrows on the top and bottom are mirror images of each other. Let’s try to imagine what it would be like to live in a 3-dimensional Klein bottle. First recall that in a 3-torus T 3 when A. 3D-Girl floated through the ceiling, she come back through the floor. Now in K3 when she floats through the ceiling and comes back through the floor, she is mirror reversed. For example, in Figure 12, when A. 3D-Girl comes back through the floor she is still facing forward, but her left and right sides have switched. Also, notice that she floats through the left part of the top face and comes back through the right part of the bottom face.

Figure 12. When A. 3D-Girl goes through the ceiling, her left and right sides are reversed.

It’s pretty strange to think about living in a non-orientable space. There would be no way to distinguish left and right since your left and right hands could switch as they did for A. 3D-Girl in Figure 12. Imagine that you take a clock with you when you walk along an orientation reversing path. What will the clock look like when you return? In what direction would the hands be moving?

5. Tic-Tac-Toe on a Torus and Klein Bottle

We can play tic-tac-toe on a flat Klein bottle, as we did on a flat torus. But now it’s a little trickier. It’s useful to draw an extended diagram to see all the different ways to have three in a row. To check that you’ve drawn the extended diagram correctly, check that the same squares are adjacent in the extended diagram and are 94 4. ORIENTABILITY

Figure 13. A Klein bottle tic-tac-toe board together with an extended diagram. adjacent in the gluing diagram. For example, in both the gluing diagram and the extended diagram of Figure 13, if you go up from square 3 you enter square 7, and if you go down from square 9 you enter square 1. Although the extended diagram continues infinitely in all directions, we only draw nine boards because the pattern repeats. Before we analyze Klein bottle tic-tac-toe, let’s remember what we learned about tic-tac-toe and torus tic-tac-toe in Chapter 1. In particular, we learned that two moves are said to be equivalent if the strategy for the rest of the game after one of the moves is analogous to the strategy for the rest of the game after the other move. For example, in regular tic-tac-toe, a first move in one corner is equivalent to a first move in any of the other three corners; and a first move in the middle of one side is equivalent to a first move in the middle of any side. Thus in regular tic- tac-toe, there are exactly three inequivalent moves for the first player: the center, a corner, and the middle of a side. What’s important in any kind of tic-tac-toe is which squares are adjacent to which. If you did the exercises on torus tic-tac-toe in Chapter 1, then you probably came up with an argument explaining why all first moves of torus tic-tac-toe are equivalent. As one step in your argument, you might have observed that if we rotate the numbers on the board for torus tic-tac-toe by 90◦ without changing the gluing arrows, then we don’t change which squares are adjacent to which. For example, we can see from Figure 14 that after rotating the numbers in the gluing diagram for torus tic-tac-toe by 90◦, square 1 is still adjacent to squares 2, 4, 3, and 7, square 4 is still adjacent to squares 1, 7, 5, and 6, and so on. Since this rotation takes 1 to 7, 7 to 9, 9 to 3, and 3 to 1, whatever strategy we have beginning with square 1, can be rotated to get an analogous strategy beginning with square 7, 9, or 3. Thus we can conclude that a first move in one corner square is equivalent to a first move in any other corner square. What happens if we rotate the numbers on a Klein bottle tic-tac-toe board by 90◦ without changing the gluing arrows? We see in Figure 15 that on the original 5. TIC-TAC-TOE ON A TORUS AND KLEIN BOTTLE 95

Figure 14. A rotation of torus tic-tac-toe by 90◦ does not change which squares are adjacent. board square 1 is adjacent to squares 2, 4, 3, and 9, but after the rotation square 1 is adjacent to squares 2, 4, 7, and 9. Thus a strategy starting with square 1 cannot be rotated to get a strategy starting at square 7. But this doesn’t mean that the four corners of Klein bottle tic-tac-toe are necessarily NOT equivalent. It just means that if we want to show that the corners are equivalent, we need to ﬁnd another way to go from a strategy starting at one corner to a strategy starting at a diﬀerent corner.

Figure 15. A rotation of Klein bottle tic-tac-toe by 90◦ does change what squares are adjacent.

Another operation we can do with the numbers on torus tic-tac-toe is to scroll down one row (again without changing the gluing arrows). By scrolling down one row, we mean moving the numbers in the bottom row to the top while pushing the other rows of numbers down. We can see in Figure 16 that again which squares are adjacent to which hasn’t changed. So whatever strategy we have for making the first move in one row can be scrolled down to get a strategy for making the first move in a different row. Thus a first move in any row is equivalent to a first move in another row, assuming the move is in the same column.

Figure 16. Scrolling the torus board down one row does not change what squares are adjacent.

In Figure 17, we scroll the Klein bottle board down one row. Notice that the bottom row is ﬂipped over when it moves to the top. You can see in Figure 17 that scrolling down does not change which squares are adjacent. Thus, as we saw on 96 4. ORIENTABILITY

Figure 17. Scrolling the Klein bottle board down one row does not change what squares are adjacent. the torus tic-tac-toe board, a first move in any row on the Klein bottle tic-tac-toe board is equivalent to a first move in another row, assuming the move is in the same column. However, you should notice that the columns on the left and the right together form one long column with six entries. In the Exercises, you will further analyze which first moves are equivalent in Klein bottle tic-tac-toe.

6. The Projective Plane

The Klein bottle is not the only non-orientable 2-dimensional surface. There are actually inﬁnitely many non-orientable surfaces that we will learn about in Chapter 6. Here we focus on a particularly important non-orientable surface known as the projective plane, denoted by P 2. The gluing diagram for the projective plane is a disk with every pair of diametrically opposite points on the boundary glued together. For example, in Figure 18 each pair of points with the same letter are glued together.

Figure 18. In a projective plane every pair of opposite points on the boundary are glued together.

In order to understand what happens along the gluing circle, let’s watch as A. Square takes his pet Starry for a walk in a projective plane (see Figure 19). The middle image of the ﬁgure shows A. Square after he has crossed the gluing circle but Starry has not. At this moment, they look at each other across the inside of the disk and each thinks that the other is mirror reversed. However, if they look at each other back across the gluing circle, they will seem normal. Once Starry has followed A. Square across the gluing circle in the third image, when he and A. Square see each other across the inside of the disk, they will again seem normal. B. Triangle has been watching A. Square and Starry as they take their walk. From her perspective, in the third image they are both mirror reversed. We can see from our 3-dimensional vantage point that B. Triangle is in fact correct, A. Square 6. THE PROJECTIVE PLANE 97

Figure 19. A. Square takes his pet Starry for a walk in a projective plane. and Starry have taken an orientation reversing path and are now both mirror reversed. This tells us that the projective plane is non-orientable. What happens if A. Square and Starry go for a second walk across the gluing circle? Of course, we’d like to be able to visualize the projective plane after we’ve glued up all of the pairs of opposite points on the boundary of the disk. But it turns out that it’s even more difficult to visualize the glued up projective plane than it was to visualize the glued up Klein bottle. We start with the gluing diagram on the left in Figure 20. The disk only has two arrows because the entire left arc is glued to the entire right arc with a twist as indicated by the direction of the arrows. In order to see how the boundary is glued up, we pretend the inside of the disk isn’t there. First we bring the top and bottom dots together to get a figure eight. We want to glue the circles together so that the arrows match. To do this, we have to first twist one of the circles at the vertex so that the arrows are lined up (as we’ve done to the right circle in the third image of the figure). At this point it’s easy to glue the circles together.

Figure 20. We can glue up the boundary of P 2 if we pretend the inside of the disk isn’t there.

Now we want to see what happens to the inside of the disk during this process. On the left side of Figure 21 we have deformed the disk into a hemisphere to give us a 3-dimensional view of what’s going on. The first two pictures in the figure follow the first two pictures of Figure 20. But in the third picture we see that the inside of the disk gets in the way of twisting the right circle so that the arrows are lined up. Thus we won’t be able to glue the two circles together as we did in Figure 20. Instead, in the fourth image of Figure 21 we pinch each circle into a line segment so that we can glue the two line segments together. Finally, we obtain the fifth image. Keep in mind that the fifth image in Figure 21 is not actually the projective plane since it contains segments where there should be circles. However, you may recall that Figure 10 was also not actually a Klein bottle since it intersected itself along a circle. 98 4. ORIENTABILITY

Figure 21. We pinch the two gluing circles into line segments so that we can glue them together.

7. The Projective Plane with a Disk Removed

We saw in the last section that we can either think of a projective plane in terms of its gluing diagram or as the object we would get by gluing together the two circles in the third image of Figure 21. We now introduce another way to think about the projective plane. In Figure 22, we’ve cut out a piece of a disk from each side of the projective plane, like taking bites out of opposite sides of an apple. We can stretch each of these pieces into a half-disk and then glue the two half-disks together to get the disk on the right of Figure 22. Note that we had to ﬂip one of the half-disks over in order to line up the black and white arrows. The double headed arrow and the triangular arrow on the boundary of the disk tell us how to glue it back to the apple core.

Figure 22. We cut two half-disks out of a projective plane and then glue them together.

In Figure 23 we see that the apple core can be deformed to a gluing diagram for aMöbius strip by straightening the edges so that it looks like a rectangle with the top and bottom glued together with a twist. In this way, we see that a projective plane can be thought of as a Möbius strip with a disk glued along its boundary. You could try to do this process in reverse to create a projective plane. Start with a Möbius strip together with a large floppy disk, made of cloth or rubber. Use tape or staples to start attaching the boundary of your disk to the boundary of your Möbius strip. Continue attaching the boundaries for as long as you can until you get stuck and can’t continue. This experiment will give you a better sense of what goes wrong when you try to glue up a projective plane in R3. 7. THE PROJECTIVE PLANE WITH A DISK REMOVED 99

Figure 23. A projective plane can be cut up into a M¨obius strip and a disk.

In Figure 22, we chose a particular disk to cut out of a projective plane so that we could easily recognize that what remained was a Möbius strip. In fact, it doesn’t matter what disk we remove from a projective plane, what remains will still be a Möbius strip. It just might be harder to recognize. In Figure 24, we cut a disk out of the center of the gluing diagram of the projective plane. We know that the washer that remains is supposed to be a gluing diagram for a Möbius strip. But we can’t glue up this Möbius strip in R3.

Figure 24. A projective plane with a disk cut out in the center of the gluing diagram.

In order to better understand the Möbius strip in Figure 24, we redraw the projective plane as we drew the hemisphere in Figure 21. Then the disk that we’ve removed is the southern polar cap of this hemisphere. In Figure 25, we try to glue up the Möbius strip as we did the projective plane in Figure 21. The picture on the right of Figure 25 is not really a Möbius strip because we have squeezed the two gluing circles into line segments just as we did in Figure 21.

Figure 25. A cross-cap represents a projective plane with a disk removed. 100 4. ORIENTABILITY

We call this picture a cross-cap. We can think of a cross-cap as representing either a Möbius strip or a projective plane with a disk removed. The one advantage of drawing a cross-cap over the usual drawing of a Möbius strip is that the boundary of a cross-cap is a flat circle rather than a twisted circle. This will be useful in Chapter 6 when we want to attach a projective plane with a disk removed to other surfaces.

8. Projective 3-Dimensional Space

Now that we understand the projective plane, we would like to create an analogous 3-dimensional space. First, recall that the 3-dimensional analog of a disk is a solid ball (which we denoted by B3 in Chapter 1). To obtain the projective 3-dimensional space P 3, we glue together every pair of opposite points along the boundary of a ball, just as we glued together opposite points along the boundary ofadisktoobtainP 2. Figure 26 illustrates some of the pairs of opposite points that are glued together along the boundary of the ball.

Figure 26. To obtain P 3, we glue opposite points on the boundary of a ball.

In Figure 27, we illustrate a Möbius strip that is contained in a horizontal cross section of P 3. We saw in Figure 2 that if A. Square walks along the center line of a Möbius strip, he will come back as his mirror image. But when A. 3D-Girl walked along a Möbius strip in Figure 4 she did not come back as her mirror image. Let’s consider what happens to A. 3D-Girl when she walks along the Möbiusstrip in Figure 27. In the third picture of the figure, she comes back to the spot on the strip where she started, but she is now upside down and underneath the strip (as indicated by the dotted lines). Because of the gluing, her left hand has gone from the right side of the picture to the left side of the picture. However, if you compare the third picture with the first picture, you can see that though she is under the strip, her left hand is still a left hand. Thus just as in Figure 4, she has not become her mirror image. In the Exercises you will show that no matter what path A. 3D-Girl takes in P 3, she will not be reversed. Thus, in fact, P 3 is orientable. This seems surprising since P 2 is non-orientable and P 3 was constructed analogously. 9. 1-SIDED AND 2-SIDED SURFACES 101

Figure 27. A. 3D-Girl walks along a M¨obius strip inside P 3.

9. 1-Sided and 2-Sided Surfaces

We saw in Figures 4 and 27 that when A. 3D-Girl walked along a Möbiusstrip, she ended up standing on the opposite “side” of the strip from where she started. Since she can get from one “side” of the Möbius strip to the other by walking along it in this way, we would like to say that the Möbius strip has only one side.On the other hand, as we see in Figure 28, no matter what path A. 3D-Girl follows on the surface of a disk she remains on the same side. In particular, she cannot climb over the edge of the disk. For this reason, we would like to say that a disk in R3 has two sides.

Figure 28. No matter where A. 3D-Girl walks on a disk she stays on the same side. 102 4. ORIENTABILITY

Another way to think about the distinction between a Möbius strip and a disk is to observe that we could seamlessly paint a disk with two colors, blue on top and red on the bottom. But we cannot seamlessly paint a Möbius strip with two colors. Before we define 1-sided and 2-sided, let’s recall from Chapter 3 that a surface has the same local topology as R2 and a 3-manifold has the same local topology as R3. Suppose that a surface is in a 3-manifold and p is a point in the surface. Then there is a small 3-dimensional region around p which looks like a ball whose intersection with the surface looks like a disk. We can see this ball and disk for a point in a Möbius strip in R3 in Figure 29.

Figure 29. A ball around the point p which intersects the M¨obius stripinadisk.

So when we say that a 3-dimensional person walks along a M¨obius strip and comes back on the other side, we mean that the person starts on one side of this disk and comes back on the other side of the disk. In particular, we now have the following deﬁnition.

Definition. A surface contained in a 3-manifold is said to be 1-sided if a 3-dimensional creature standing on a small disk in the surface can walk along a path in the surface and come back on the other side of that disk. If there is no such path, then the surface is said to be 2-sided.

Using this definition, we can see that the Möbiusstrip in Figure 4 is 1-sided, and the disk in Figure 28 is 2-sided. In fact, any Möbius strip in R3 will be 1-sided, and any disk, sphere, or curved torus in R3 will be 2-sided (we’ll see why in the next section). It is important to note that sidedness is an extrinsic property of a surface in a 3-manifold, since it has to do with the experience of a 3-dimensional creature and hence cannot be detected by a 2-dimensional creature in the surface. On the other hand, orientability of a surface is an intrinsic property because it can be detected from within the surface. People often think that saying a surface is 1-sided is equivalent to saying it’s non-orientable, but is this always true? We’ll learn the answer to this question in the next section. 10. NON-ORIENTABILITY AND 1-SIDEDNESS 103

10. Non-orientability and 1-Sidedness

Figure 30 illustrates a 3-dimensional Klein bottle K3 obtained by gluing the top and bottom of a cube with a left-right ﬂip and gluing the other pairs of opposite sides straight.

Figure 30. The gluing diagram for a 3-dimensional Klein bottle.

Because of the gluing arrows on the cube, the grey square cutting the cube in half in Figure 31 is the gluing diagram for a torus, which we know is orientable. To figure out if this torus is 1-sided or 2-sided, we illustrate A. 3D-Girl walking upward along the right side of the torus. Before passing through the ceiling, we see that her left hand is still swollen from her bee sting. Because the ceiling is glued to the floor with a left-right flip, as she passes through the ceiling, she emerges from the floor on the left side of the torus. Thus, this torus is a 1-sided orientable surface in K3. Observe that both before and after she passes through the ceiling, the hand on the tail side of the black and white arrow is the one that’s swollen. Has A. 3D-Girl been reversed?

Figure 31. A. 3D-Girl walks along a 1-sided torus in K3.

In Figure 32, we illustrate three perpendicular surfaces in the K3 whose gluing diagram is in Figure 30. For each surface, we have labeled it according to whether it’s 1-sided or 2-sided and whether it’s orientable or non-orientable. You should check that the surfaces each have the orientability and sideness properties listed. This example shows that a single 3-manifold can contain surfaces that are orientable and 1-sided, orientable and 2-sided, and non-orientable and 2-sided. InspiteoftheabovesurfacesinK3, the following theorem shows that for surfaces in orientable 3-manifolds, 1-sidedness and non-orientability are in fact equivalent. 104 4. ORIENTABILITY

Figure 32. Orientability is independent of sidedness in K3.

Theorem 4.1. Let S be a surface in an orientable 3-manifold M.ThenS is 1-sided if and only if S is non-orientable. The statement of our theorem contains the phrase “if and only if”. This means that to prove it we need to show both “if S is 1-sided, then S is non-orientable” and “if S is non-orientable, then S is 1-sided”. We think of the ﬁrst statement as the forwards direction of “if and only if” because we are assuming the beginning of the sentence and proving the end of the sentence. In the proof, we will use the symbol (=⇒) to indicate that we are about to prove the forwards direction. We think of the statement “if S is non-orientable, then S is 1-sided” as the backwards direction of “if and only if” because we are assuming the end of the sentence and proving the beginning of the sentence. We indicate that this is what we are about to prove by using the symbol (⇐=).

Proof. (=⇒) Suppose that S is a 1-sided surface in an orientable 3-manifold M.WeprovethatS is non-orientable with the help of A. 3D-Girl. Since S is 1-sided, A. 3D-Girl can find path on S to walk along which will take her to the other side of the surface. Before beginning her journey, she dips her left foot in black paint so that it leaves footprints as she walks. In fact, there’s so much black paint on her foot that her footprints are significantly bigger than her foot. Since M is orientable, no matter what path she walks on, she will not be reversed. So her left foot will remain a left foot throughout her journey. We don’t know what the surface is or what the 3-manifold looks like, but we do know that A. 3D-Girl’s journey has taken her from one side of the surface to the other side of the surface, and hence at the start and finish of her trip her left footprints must look like those in Figure 33. In the meantime, A. Square who lives in the surface, thinks her footprint is a 2-dimensional creature whose name is A. Footprint. He observes that A. Footprint has gone for a walk and has become reversed. From this he concludes that the surface he lives in must be non-orientable. Hence indeed S is non-orientable. (⇐=) Now suppose that S is a non-orientable surface in M.WeprovethatS is 1-sided in M, again with the help of A. 3D-Girl. Since S is non-orientable, it contains an orientation reversing path that reverses any 2-dimensional creature who walks along it. A. 3D-Girl follows this path, leaving footprints as she goes. Since the footprints are 2-dimensional creatures, they will become reversed as the result of following an orientation reversing path. In 10. NON-ORIENTABILITY AND 1-SIDEDNESS 105

Figure 33. A. 3D-Girl comes back on the other side of the surface but her left foot remains a left foot. particular, her left footprint will have become a right footprint by the end of her trek. If A. 3D-Girl stayed on the same side of the surface throughout her journey and her left footprint became a right footprint, her left foot would have to have become a right foot as illustrated in Figure 34. But this could only happen if A. 3D-Girl had become reversed, which is impossible since the 3-manifold she lives in is orientable. Thus, in fact, her journey must have taken her to the other side of the surface so that her footprint could become reversed without her becoming reversed. It now follows that the surface is indeed 1-sided.

Figure 34. A. 3D-Girl stays on the same side of the surface, but her left footprint becomes a right footprint.

Notice that at the end of a proof we use the symbol to tell the reader that the proof is done. You will see this at the end of the proofs in this book as well as in many other math books. 106 4. ORIENTABILITY

11. Exercises

1. A tube connecting two spaces is sometimes called an Einstein-Rosen bridge (see Figure 35).

Figure 35. Illustration for Exercise 1.

(a) What evidence could A. Square look for that might indicate that Flatland was two planes with an Einstein-Rosen bridge connecting them? (b) Suppose two parallel planes in R3 are connected by an Einstein-Rosen bridge such that all Squares in the two planes are oriented in a parallel way from the perspective of a 3-dimensional person looking down on the two planes. Do the Squares in one plane appear to the Squares in the other plane to be reversed? (c) Is the surface consisting of two parallel planes connected by an Einstein- Rosen bridge an orientable surface? (d) If two planes are connected by two Einstein-Rosen bridges, is the surface necessarily orientable? (e) By analogy with our description of a 2-dimensional Einstein-Rosen bridge, describe a 3-dimensional Einstein-Rosen bridge. Suppose there is a 3- dimensional Einstein-Rosen bridge attaching one R3 to another copy of R3. Is the space orientable? 2. We saw in the chapter that left and right can morph into one another in a non-orientable surface. Is this also true for clockwise and counterclockwise? 3. Suppose A. Square lives on a Möbius strip. (a) How does he see himself in the distance? Does he appear to be reversed? Draw an extended diagram. (b) Suppose A. Square stands still and watches B. Triangle walk around and around the strip. Just as B. Triangle is passing him after going once around the strip, A. Square looks ahead at all the copies of himself and her. Which, if any, of them will appear to be reversed? Draw an extended diagram of a Möbius strip containing A. Square and B. Triangle at this moment in time. 4. Suppose A. Square lives alone in a small Klein bottle and each side of A. Square is colored a different color. Draw an extended diagram of A. Square’s space. 5. Suppose you live in a 3-dimensional Klein bottle with the front and back wall glued with a left-right flip and the other pairs of opposite walls glued straight. 11. EXERCISES 107

(a) As you look around, you will observe many images of yourself. Which of these images are reversed? (b) Suppose that A. 3D-Girl walks once through the front wall and returns (from the back wall) to stand next to you. Which images of her appear to be reversed? 6. Does a 3-dimensional Klein bottle necessarily contain a 2-dimensional Klein bottle? If so, where? How many Klein bottles does it contain? Does it contain a pair of perpendicular Klein bottles?

7. List the numbered squares in every diagonal in the Klein bottle tic-tac-toe board illustrated in Figure 13. Do some diagonals contain more numbers than others? 8. Recall that two moves in Klein bottle tic-tac-toe are said to be equivalent if the strategy for the rest of the game is precisely analogous. (a) Explain why two board positions that are related by a vertical or horizontal reflection of the board are equivalent. (b) Start with the Klein bottle tic-tac-toe board illustrated in Figure 13. Is the board we obtain by scrolling up one row equivalent to the original board? Is the board we get by scrolling to the right one column equivalent to the original board? Explain. (c) List all first moves that are equivalent to placing an X in the upper left corner of the board. (d) List all first moves that are equivalent to placing an X in the center of the board. (e) How many non-equivalent first moves are there in Klein bottle tic-tac-toe? 9. For each of the non-equivalent first moves that you found in Exercise 8, list the optimal moves for each player. Determine whether the first player will necessarily win Klein bottle tic-tac-toe if both players play optimally. 10. Draw a picture illustrating a P 2 contained in P 3.IsyourP 2 1-sided or 2-sided? Explain. 11. Explain why A. 3D-Girl cannot be reversed by any path she takes in P 3. 12. How would you define a projective 1-dimensional space P 1 that is analogous to P 2?IsP 1 orientable or non-orientable? 13. For each of the surfaces listed below, state one or more intrinsic topological and/or geometric properties that enables you to distinguish that surface from ALL of the other surfaces in the list. State whether each property you refer to is topological or geometric. • aplane • aflattorus • a flat Klein bottle • a sphere • the surface bounding a cube 108 4. ORIENTABILITY

Figure 36. Illustration for Exercise 14.

14. Discuss the sidedness and orientability of the grey surfaces in the 3-manifolds in Figure 36. 15. For each of the gluing diagrams in Figure 36, draw two intersecting surfaces perpendicular to the grey surface and determine their sidedness and orientability. 16. Determine whether the surface in Figure 37 is orientable. Is it 1-sided or 2-sided in R3?

Figure 37. Illustration for Exercise 16. https://doi.org/10.1090//mbk/096/05

CHAPTER 5

Flat Manifolds

Topics: • Flat surfaces • Polygonal gluing diagrams • Cone points and anti-cone points • Extended diagrams of cones • Flat manifolds with cubical gluing diagrams • Turn manifolds • Extended diagrams of ﬂat 3-manifolds

1. Flat Surfaces

By now you’re probably comfortable with the flat torus and flat Klein bottle obtained by abstractly gluing opposite sides of a square. One reason that we like these gluing diagrams is that their local geometry is the same as that of a plane. By contrast, the local geometries of the curved torus in R3 and the curved Klein bottle in R4 are not nearly so nice. For this reason, when we refer to a torus or Klein bottle, you should always think of the gluing diagram illustration of it, rather than imagining it as a curved surface in 3 or 4 dimensions. In general, we’ll say a surface is flat if its local intrinsic geometry is the same as that of a plane. This makes sense since a plane corresponds to our intuitive notion of the word “flat”. In Chapter 3, we saw that the sum of the angles of any triangle on a sphere is greater than 180◦, and the sum of the angles of a triangle on a curved torus depends on where the triangle is located on the torus. Since the sum of the angles of a triangle is a local intrinsic geometric property, this tells us that neither a sphere nor a curved torus has the same local intrinsic geometry as R2, and hence neither is flat. This again corresponds to our intuition since we wouldn’t want to call a sphere or a curved torus “flat”. Although our intuition about flatness only applies to surfaces, we can extend the concept to spaces of any dimension. Definition. An n-dimensional manifold is said to be flat if its local intrinsic geometry is the same as that of Rn. In particular, this means that the geometry of any flat manifold is homogeneous and the sum of the angles of any triangle in a flat manifold is 180◦.Sotoshow that a surface is not flat, all we need to do is find one triangle in the surface whose angle sum is not 180◦. The following theorem tells us that we can also prove that asurfaceis flat just by looking at angle sums of triangles.

109 110 5. FLAT MANIFOLDS

Theorem 5.1. A surface is flat if and only if the sum of the angles of any triangle in the surface is equal to 180◦. This theorem is quite useful, but it’s not easy to prove. In particular, why does simply knowing about the sums of angles of triangles tell us that all of the local intrinsic geometric properties of a surface are the same as they are in R2?You might see the proof of this result if you take a course in differential geometry. We saw in Chapter 3 that the sum of the angles of every triangle in the flat torusis180◦. It now follows from this theorem that the flat torus is indeed flat, which is what our intuition told us anyway. To see how we can apply Theorem 5.1 to a new surface, let’s consider the hexagonal gluing diagram in Figure 1. Any triangle which is in the interior of the hexagon is planar and hence the sum of its angles is 180◦. If a triangle goes out one edge and in another but doesn’t contain a corner, we can cut out the two pieces of the triangle and tape them together to get a planar triangle as illustrated in Figure 1. Thus the sum of the angles would again be 180◦. Can you think of how you would you check that the sum of the angles of a triangle containing a corner is yet again 180◦? After checking that this is the case for all three types of triangles, we can apply Theorem 5.1 to conclude that the surface in Figure 1 is indeed flat.

Figure 1. The sum of the angles of this triangle in the glued surface is equal to 180◦.

In addition to triangles on a flat torus, in Chapter 3 we saw that the sum of the angles of any triangle on an infinite cylinder is 180◦. Thus we can also use Theorem 5.1 to conclude that the infinite cylinder is also flat. Since the glued up cylinder appears to be curved, our intuition about what flat means has led us somewhat astray.

2. Flat Gluing Diagrams

We would like to find an easier method to determine whether a surface is flat than by checking the angle sum of every triangle. In Chapter 3, we saw that every triangle in the flat torus corresponds to infinitely many identical triangles in the extended diagram (see Figure 2). Since the extended diagram is a plane, we concluded that every triangle in the flat torus has an angle of sum 180◦. But how did we know that the extended diagram of a surface is a plane? To answer this question, observe that when we exit through one side of the square gluing diagram, we re-enter from the opposite side. This means that we get the extended diagram by gluing together one square after another as we go up, down, back, and forth forever. These squares fit neatly together to form a plane 2. FLAT GLUING DIAGRAMS 111

Figure 2. A triangle in the ﬂat torus and in the extended diagram. because the edges of the squares are glued together in pairs and four squares come together at each corner. We illustrate the way the four corners of the square ﬁt together in Figure 3.

Figure 3. Copies of the square gluing diagram for a torus ﬁt neatly together to form a plane.

The theorem below shows that we can use this idea more generally to check if surfaces are flat without having to draw an extended diagram or consider the angle sum of every triangle in the surface. Theorem 5.2. Let S be a surface with a polygonal gluing diagram. Then S is flat if and only if the edges of the polygon are glued together in pairs and the sum of the angles around each glued up vertex is 360◦. Recall from Chapter 4 that if we want to prove a theorem containing the phrase “if and only if”, we have to prove two statements. In this case, the statements that we have to prove are the following. (1) If S is flat, then the edges of the polygon are glued together in pairs and the sum of the angles around each glued up vertex is 360◦. (2) If the edges of a polygon are glued together in pairs and the sum of the angles around each glued up vertex is 360◦,thenS is flat. We think of the first statement as the forwards direction of the proof because we are assuming the beginning of the “if and only if” statement and proving the 112 5. FLAT MANIFOLDS end. We use the symbol (=⇒) to indicate that we are about to prove the forwards direction. We think of the second statement as the backwards direction of the proof because we are assuming the end of “if and only if” and proving the beginning. In this case, we indicate what we are about to prove by using the symbol (⇐=).

Proof. (=⇒) Suppose that the surface S defined by the polygonal gluing diagram is flat. We prove as follows that the edges must be glued together in pairs and that the sum of the angles around each glued up vertex is 360◦. Recall that if a surface is flat, then its local intrinsic geometry at every point is the same as that of R2. In particular, every point must have exactly 360◦ around it. To find the angles around a vertex v in the glued up surface, we have to add together the angles around all of the vertices that are glued to v. This means that in order to have 360◦ around each vertex in the glued up surface, the sum of the angles around any collection of vertices that are glued together must be 360◦. Also, each point on an edge which is not a vertex has precisely 180◦ around it in the polygon before the gluing. Thus we must glue the edges together in pairs in order to have 2 × 180◦ = 360◦ around each such point. (⇐=) Suppose that in the gluing diagram for S the edges are glued together in pairs and the sum of the angles around each glued up vertex is 360◦.Weproveas follows that S is flat. Observe that since the polygonal gluing diagram is drawn on a plane, any point in the inside of the polygon already has the same local intrinsic geometry as R2.So let’s consider a point p which is not a vertex but is on an edge of the polygon. Since edges are glued together in pairs, there are two edges in the glued up surface that contain p. To evaluate the local intrinsic geometry around p, we can tape together a little semicircle around p on each of these edges. Because the two semicircles are flat and we are gluing them together along straight edges, we get a flat disk around p in the glued up surface. Thus the local intrinsic geometry around p is the same as it is around a point in R. Finally, to evaluate the local intrinsic geometry around a vertex v in the glued up polygon, we cut out a wedge around each vertex glued to v andalignthewedges one after the other, as we would if we had served too many plates of pizza and we were putting the untouched slices back together into the pizza box. Since the sum of the angles around the vertices glued to p is 360◦, the circular wedges around these vertices will fit nicely together so that the region around v has the same local intrinsic geometry as a point in R. We illustrate this in Figure 4 for a hexagonal gluing diagram where a group of three vertices are glued together.

Figure 4. When these wedges are glued together we get 360◦ around the black vertex. 2. FLAT GLUING DIAGRAMS 113

Since every point in the glued up surface has the same local intrinsic geometry as a point in R2, we can conclude that the surface is indeed ﬂat.

As an example, let’s use Theorem 5.2 to show that the surface illustrated in Figure 5 is ﬂat. Since the edges are glued together in pairs, we only need to check that the sum of the angles around any collection of vertices that are glued together is 360◦. Before we compute these angle sums, let’s determine which vertices are glued together.

Figure 5. We use Theorem 5.2 to show that this surface is ﬂat.

Since the triple arrows are glued together, vertices 2 and 6 are glued together and vertices 3 and 5 are glued together. Now since the solid black arrows are glued together, vertices 2 and 4 are glued together and vertices 1 and 5 are glued together. Finally, since the white triangular arrows are glued together, vertices 4 and 6 are glued together and vertices 1 and 3 are glued together. Altogether, this means that vertices 2, 4, and 6 are all glued together, and vertices 1, 3, and 5 are all glued together. We indicate this in Figure 5, by coloring the even numbered vertices grey and the odd numbered vertices black. Now since all of the angles of a regular hexagon are 120◦, the sum of the angles around every glued up vertex is 360◦. So even though we don’t know what our surface would look like if we tried to physically glue it up, we can conclude from Theorem 5.2 that the surface is indeed flat. Note that we will refer to this method of determining which vertices are glued together as “matching up arrows”. For the special case when our gluing diagram is a square, we can restate The- orem 5.2 more simply. We call the following result a corollary because it follows from Theorem 5.2. Corollary 2. A surface whose gluing diagram is a square is flat if and only if its edges are glued in pairs and all four vertices are glued together. As an application of Corollary 2, we consider the gluing diagram of the Klein bottle in Figure 6. Observe that opposite edges are glued together. Hence edges are indeed glued together in pairs. Also, by matching up arrows as we did for the hexagon in Figure 5, we can check that all four vertices are glued together. Thus by Corollary 2 we can conclude that the Klein bottle is indeed flat, without having to construct the extended diagram. Next, let’s consider the surface we get from the gluing diagram in Figure 7. The edges are again glued together in pairs, though the edges that are glued are adjacent rather than opposite one another. By matching the arrows, we see that the vertices are not all glued together. Rather the pair of white vertices are glued 114 5. FLAT MANIFOLDS

Figure 6. We can apply Corollary 2 to show that this gluing diagram for the Klein bottle is ﬂat. together, and the remaining two vertices are not glued to any other vertex. Thus by Corollary 2, the surface in Figure 7 is not ﬂat. Actually, the surface is topologically the same as a surface that we’ve seen before. Can you identify it?

Figure 7. We can apply Corollary 2 to show that the surface with this gluing diagram is not ﬂat.

3. The Point of a Cone

We now consider what happens if we have a polygonal gluing diagram where edges are glued in pairs, but the sum of the angles around some vertex is not 360◦. In this case, the surface would be flat everywhere except at these glued up vertices. We introduce the following definition in order to be able to describe such “bad” points. Definition. If the sum of the angles around a point is less than 360◦, then the point is said to be a cone point. If the sum of the angles around a point is more than 360◦, then the point is said to be an anti-cone point. A cone point looks like the point of a cone. In Figure 8, we illustrate a cone point with 90◦ around it. The surface is flat except at the cone point. In Figure 8, we removed 270◦ from the plane in order to get a cone angle of 90◦. In general, we can create a cone point with angle θ by removing an angle of 360◦ − θ from a region of the plane and then gluing together the edges of the wedge that remains. Don’t panic if the symbol θ looks foreign to you—it should! It’s a lower-case Greek letter, which is often used for angles in trigonometry. The letter θ is spelled “theta”, which is pronounced to rhyme with the cheese feta and the word “laytah” as in “catch you laytah”. A. Square has heard that there is a cone point in Flatland. But it’s in a region of Flatland that is supposed to be a bad neighborhood, so none of his friends or family have dared to go there. He has no idea what a cone point is, but decides to risk going to a bad neighborhood to see one for himself. On the left side of Figure 9, 3. THE POINT OF A CONE 115

Figure 8. A cone whose cone point has 90◦ around it.

Figure 9. A. Square tries unsuccessfully to explore a cone point. we can see A. Square as he is starting his trek up the back of the cone towards the cone point. In the left-hand picture in Figure 9, A. Square’s sides are illustrated with dashed lines because he’s on the back side of the cone. In the middle picture, he thinks he sees the back of another square who’s also walking towards the cone point. As he moves closer to the cone point, the other square seems to be deliberately bumping into him. A. Square goes to the left to move out of the other guy’s way, but the guy moves to the left too so that their top corners are still touching. With this guy blocking him, it’s impossible for A. Square to get any closer to the cone point. A. Square thinks that this other square is a real jerk because even when A. Square tries to take a completely diﬀerent path to the cone point, that other square gets in his way. No wonder people say this is a bad neighborhood if the squares who live here are all jerks like this guy. Eventually, A. Square gives up and goes home, murmuring under his breath about never wanting to come to this neighborhood again. A cone is not the only surface with a cone point. For example, consider the gluing diagram for the surface in Figure 10. Although the edges of the gluing diagram are glued in pairs, we saw in Figure 7 there is less than 360◦ around each of the glued up vertices. As a result, this surface is ﬂat except at these three 116 5. FLAT MANIFOLDS

Figure 10. This gluing diagram has three cone points. cone points. If you glue the surface up, it will resemble the surface of an apple turnover. Topologically this surface is the same as a sphere, but geometrically it’s quite diﬀerent.

4. Using Extended Diagrams to Understand Cone Points

An extended diagram can help us visualize what it would be like to live in a universe with a cone point. As a first example, let’s consider an infinite cone with a cone angle of 90◦. We can think of the gluing diagram for the cone as an infinite wedge. To obtain the extended diagram, we glue together copies of the wedge along their edges, as we did to obtain the extended diagram of the flat torus. However, since the cone angle is 90◦, once we’ve glued together four of these wedges, we can just glue the remaining edge of the fourth wedge to the remaining edge of the first edge to get a plane, as illustrated in Figure 11.

Figure 11. The extended diagram for a cone with a cone angle of 90◦.

Since the extended diagram of a cone with cone angle 90◦ is a plane, we might be tempted to think that the cone is ﬂat. In particular, we know that every triangle in the extended diagram has an angle sum of 180◦. However, in the Exercises in Chapter 3 we saw that a triangle containing the cone point has an angle sum of more than 180◦. To understand this apparent contradiction, below we consider triangles containing the cone point in the cone and in the extended diagram. In the gluing diagram on the left side of Figure 12, we see a path of three segments with right angles between them. The endpoints of the path come together as a result of the gluing, creating a triangle in the cone whose angle sum is more than 180◦. However, because the extended diagram consists of four wedges which are each identical to the gluing diagram, the three segments making up the triangle in the cone are repeated in each of these wedges. This creates a cross shape rather than a triangle in the extended diagram, as illustrated on the right in Figure 12. 4. USING EXTENDED DIAGRAMS TO UNDERSTAND CONE POINTS 117

Figure 12. A triangle in the gluing diagram containing the cone point is not a triangle in the extended diagram.

Thus, the fact that the extended diagram is a plane does not imply that the angle sum of every triangle in the cone is 180◦. Now let’s consider the triangle in the extended diagram illustrated on the left in Figure 13. Since the extended diagram is a plane, this triangle has an angle sum of 180◦. But all four wedges in the extended diagram are superimposed in the gluing diagram. Hence the sides of the triangle in the extended diagram intersect in the gluing diagram as illustrated on the right in Figure 13. (You should trace the image on the left, then cut out the four wedges, and put them one on top of the other to check that you agree with the image on the right.) Thus the triangle in the extended diagram does not represent a triangle in the gluing diagram.

Figure 13. A triangle in the extended diagram containing the cone point is not a triangle in the gluing diagram.

These examples should convince you that triangles that contain the cone point do not correspond to triangles in the extended diagram, and triangles that contain the cone point in the extended diagram do not correspond to triangles in the cone. In particular, the cone with a cone angle of 90◦ is not ﬂat, even though its extended diagram is a plane. We can use the extended diagram of a cone to determine how many images A. Square sees of himself. For example, Figure 14 illustrates A. Square in the extended diagram of the cone with a cone angle of 90◦. In this case, there are four copies of A. Square in the extended diagram, each rotated 90◦ from the previous one. A. Square can see three images of himself and believes that he is one of four squares in Flatland. He thinks that the other three squares are so shy that they always run away whenever he tries to approach them. 118 5. FLAT MANIFOLDS

Figure 14. In a cone with a cone angle of 90◦, A. Square sees three other squares.

Next we consider what A. Square sees in a cone with a cone angle of 73◦.In this case, it is not as easy to create an extended diagram because 73 does not divide evenly into 360. As a result, the extended diagram spirals around and around the cone point before the ﬁnal wedge is attached to the original wedge. But we don’t need to draw the whole extended diagram in order to determine what A. Square sees. Since light travels in a straight line, his vision cannot wrap around the cone point, just like your vision cannot wrap around a building to see who’s approaching on a perpendicular street. In fact, as we can see in Figure 15, A. Square can only see four images of himself, two on each side of the cone point. The squares adjacent to him on either side are rotated by 73◦ relative to him, and the next squares after those are rotated by 73◦ relative to the previous ones.

Figure 15. In a cone with a cone angle of 73◦, A. Square sees four other squares.

We have numbered the wedges to make it easier to talk about them. In the diagram it looks like the “real” A. Square is standing in wedge 3. But in fact, it’s not that there is one real A. Square and the others are fake. We have just picked the image of A. Square in wedge 3 to use to trace his line of sight. Observe that since each wedge has an angle of 73◦, wedge 1 and wedge 5 overlap in a region with angle (5 × 73◦) − 360◦ =5◦. This gives A. Square the impression that the squares in regions 1 and 5 are closer together than the squares in any other two adjacent regions. But keep in mind that A. Square does not see the numbers or the lines separating one wedge from the next. He just sees the other squares who are always 5. ANTI-CONE POINTS 119 running away from him; and wherever he goes, the whole picture moves with him. In particular, the overlap region will always be exactly opposite him. For our next example, let’s imagine that A. Square lives in a cone with a cone angle of 300◦. This means each wedge has an angle of 300◦. Hence the next images of him are in wedges that are 300◦ away in either direction. However, as we can see on the left side of Figure 16, when A. Square is looking to the left of the cone point, he can’t see the square which is at an angle of 300◦ clockwisearoundthecone point from him. On the other hand, as we can see on right side of the ﬁgure, when A. Square is looking to the right of the cone point, he can’t see the square which is at an angle of 300◦ counterclockwise around the cone point from him. Thus in this space, A. Square doesn’t see any copies of himself.

Figure 16. In a cone with a cone angle of 300◦,A.Squaredoesn’t see any other squares.

5. Anti-cone Points

In contrast to a cone point which has an angle with less than 360◦ around it, an anti-cone point has an angle with more than 360◦ around it. One way we can construct an anti-cone point is by cutting a slit in a region of the plane and inserting a wedge. If we add a wedge with angle θ, we will obtain an anti-cone point with an angle of 360◦ + θ around it. For example, in Figure 17 we inserted a wedge with an angle of 45◦ giving us an anti-cone point with angle 360◦ +45◦ = 405◦ around it.

Figure 17. We add a wedge of 45◦ to create an anti-cone point with an angle of 405◦ around it.

In Figure 18, A. Square and B. Triangle go on an expedition in search of an anti-cone point. They never do ﬁnd an anti-cone point, but as they walk side- by-side each following a straight line, they seem to be getting farther apart. This 120 5. FLAT MANIFOLDS

Figure 18. A. Square and B. Triangle search in vain for an anti- cone point. makes A. Square worry about whether B. Triangle really likes him as much as he likes her. The hexagon in Figure 19 is the gluing diagram for a torus. (Can you see why?) However, it has a cone point of 120◦ at vertex p and an anti-cone point of 5 × 120◦ = 600◦, made up of the vertices labeled q (you should convince yourself that indeed all of the vertices labeled q are in fact glued together). This hexagonal gluing diagram yields a torus whose local geometry is diﬀerent from both the curved torus and the ﬂat torus.

Figure 19. A gluing diagram for a torus with a cone point and an anti-cone point.

6. We Show that the 3-Torus Is Flat

We can construct flat 3-manifolds using a method similar to the one we used to construct flat surfaces. However, instead of starting with a polygonal gluing diagram in the plane, we start with a 3-dimensional polyhedron in R3 in which faces are glued together in pairs. In order for the glued up space to be flat, every point must have a region around it whose local intrinsic geometry is the same as R3. This means that each point must have a region around it that looks like a ball. We’ll see how this works with the gluing diagram for the 3-torus T 3, illustrated in Figure 20. Since the top and bottom faces of the cube in Figure 20 are glued together, the top front edge is glued to the bottom front edge. Since the back and front faces are glued together, the bottom front edge is glued to the bottom back edge. Finally, since the top and bottom faces are glued together, the bottom back edge is glued to the top back edge. In this way, we can conclude that the four dotted edges are glued together. You should use a similar argument to convince yourself that the 6. WE SHOW THAT THE 3-TORUS IS FLAT 121

Figure 20. A gluing diagram for T 3. four thick grey edges are glued together, and the four thin black edges are glued together. You should also check that all eight vertices are glued together. To see if this glued up cube has the same local intrinsic geometry as R3,we consider four types of points in the gluing diagram: interior points of the cube, points on a face of the cube, points on an edge of the cube, and points around a vertex of the cube. In Figure 21, we illustrate a region around a point of each type. We see as follows that each of these glued up regions is a ball.

Figure 21. There is a ball around each type of point in T 3.

• A point of Type 1: This type of point is in the interior of the cube, so a region around it is a ball. • A point of Type 2: Because faces are glued in pairs, two points at the same spot on opposite faces of the cube get glued together. Around each of these points, there’s half a ball inside the cube. These two halves are glued together to produce a whole ball around the glued up point. • A point of Type 3: A region around an edge point of the cube looks like a quarter of a solid ball. As we saw in Figure 20, edges are glued in groups of four. Thus four quarters of a ball are glued together to produce a complete ball around the glued up point. • A point of Type 4: A region around each vertex of the cube looks like an eighth of a solid ball. Since all eight vertices are glued together, these eight eighths of a ball are glued together to produce a complete ball around the glued up point. Since every point in T 3 has a region around it that looks like a ball, T 3 has the same local intrinsic geometry as R3.ThusT 3 is indeed a ﬂat 3-manifold. 122 5. FLAT MANIFOLDS

7. A Method to Determine if a Glued Up Cube Is Flat

We would like to have a result like Theorem 5.2 to check whether a 3-dimensional gluing diagram is flat. It’s a little more complicated in 3 dimensions than it was for surfaces, so we focus on gluings of a cube. Since we want our glued up space to be a manifold, it can’t have a boundary. Thus we must have pairs of faces glued together. But there are additional requirements needed to make sure the manifold is flat. In particular, as we saw in Figure 21, the region around an edge of a cube looks like a quarter of a solid ball. So we need precisely four such edges to be glued together so that the local intrinsic geometry of the glued up manifold will be the same as that of R3. Also, a region around a vertex of a cube looks like an eighth of a solid ball. So we need precisely eight such regions to be glued together to obtain the local intrinsic geometry of R3. If all of these requirements are met, then the local intrinsic geometry of our glued up manifold should be exactly the same as that of R3. These arguments should give you an idea of why the following theorem is true. You should look back at Corollary 2 to see that Theorem 5.3 is the 3-dimensional analogue. Theorem 5.3. A 3-manifold whose gluing diagram is a cube is flat if and only if its faces are glued in pairs, its edges are glued together in groups of four, and all eight vertices are glued together. As an example of how to use Theorem 5.3, let’s consider the 3-dimensional Klein bottle K3, illustrated in Figure 22. From the arrows we see that the top and bottom faces of the cube are glued together with a back-front flip, and the other two pairs of opposite faces are glued together straight.

Figure 22. A gluing diagram for K3.

In order to use Theorem 5.3, we need to check that the edges of the cube are glued together in groups of four and all eight vertices are glued together. As we consider each edge, we “color” it so that the ones that are glued together are colored in the same way. Similarly, as we consider each vertex, we color those which are glued together with the same color. As we see in Figure 23, all eight vertices are colored black and the edges are colored in groups of four. This shows us that we can apply Theorem 5.3 to conclude that the 3-dimensional Klein bottle is flat. Next we consider the gluing of the cube in Figure 24. In this case, the top and the bottom faces of the cube are glued together with a front-back flip and the front and back faces are glued together with a top-bottom flip. Such a 3-manifold is not a 3-dimensional Klein bottle because two pairs of opposite faces are glued with a flip. We have colored the edges in the figure to indicate which ones are glued 8. EXTENDED DIAGRAMS OF GLUED UP CUBES 123

Figure 23. Edges are glued together in groups of four, and all eight vertices are glued together.

Figure 24. This 3-manifold is not ﬂat because its edges are not glued together in groups of four. together. Observe that there are only two thick black edges and only two dotted edges. Since Theorem 5.3 is an “if and only if” statement, we can conclude that the 3-manifold obtained from this gluing diagram is not ﬂat.

8. Extended Diagrams of Glued Up Cubes

Before we construct the extended diagram of a glued up cube, let’s recall the extended diagrams of a flat torus and flat Klein bottle. In particular, we saw in Chapter 1 that the extended diagram of T 2 is a plane made up of infinitely many copies of the gluing diagram. If A. Square lives in a flat T 2, he will see copies of himself going off forever upwards, downwards, leftwards, and rightwards. Figure 25 illustrates a 3 × 3 grid of the extended diagram with A. Square in it. In Chapter 4 we saw that the extended diagram of the flat Klein bottle is also a plane made up of infinitely many copies of the gluing diagram. If we glue the top of the square to the bottom of the square with a flip and glue the left and right sides straight, then the extended diagram for K2 with A. Square in it is illustrated in Figure 26. We are now ready to consider the extended diagram of a flat 3-manifold obtained by gluing up a cube. We know by Theorem 5.3 that such a 3-manifold is flat if and only if the faces of the cube are glued in pairs, the edges are glued in groups of four, and all eight vertices are glued together. This corresponds exactly to how infinitely many cubes could be glued together to create R3. That is, if we imagine R3 as made out of infinitely many cubes, each face would be shared by two cubes, each edge would be shared by four cubes, and each vertex would be shared by eight 124 5. FLAT MANIFOLDS

Figure 25. A3× 3 grid of the extended diagram for T 2 with A. Square in it.

Figure 26. A3× 3 grid of the extended diagram for K2 with A. Square in it. cubes. Thus if a glued up cube is flat, then its extended diagram is R3,madeup of infinitely many copies of the gluing diagram. We can get a sense of what it’s like to live in such a 3-manifold by looking at copies of A. Cube in an extended diagram. To make it easier to see what’s going on, we color the top face of A. Cube white, her bottom face orange, her front face black, her back face purple, her right face grey, and her left face yellow. Because this book is in black and white, we denote the faces that are orange, yellow, and purple, by the letters O, Y, and P. If a white face has no letter on it, then it’s white. We illustrate the colored faces of A. Cube in Figure 27. Now let’s suppose that A. Cube lives in T 3. If she looks towards the front wall of the gluing diagram, her line of vision goes through the front wall and into the back wall, hitting her in the back. So imagines she is seeing the back of another cube. Similarly, if she looks towards the back, the top, the bottom, the right, and the left, she will see her opposite side. In fact, A. Cube sees identical images of herself going off forever upwards, downwards, backwards, forwards, leftwards, and rightwards. In Figure 28, we illustrate a 3 × 3 × 3 block of the extended diagram of T 3 with A. Cube in it. Since you can’t see all sides of a cube at once, it’s helpful 8. EXTENDED DIAGRAMS OF GLUED UP CUBES 125

Figure 27. The colored faces of A. Cube.

P P P

Figure 28. The extended diagram of T 3 with A. Cube in it. to remember which colors are opposite which. In particular, we can’t see her back side, but we remember that purple is opposite black. So her back side must be purple. Also, we can’t see her right side, but we remember that grey is opposite yellow. So we know that her right side is grey. You can refer back to Figure 27 if you forget how her faces are colored. All of the cubes in Figure 28 are identical, because each wall of the gluing diagram for T 3 is glued to the opposite wall straight. By contrast, Figure 29 illustrates an extended diagram of K3 with A. Cube in it. We know that K3 is obtained from a cube by gluing two pairs of opposite faces straight and one pair with a flip. We can determine which pair is glued with a flip by analyzing the colored cubes in the extended diagram. Let’s start with the cube in the front lower left corner of the figure with an arrow pointing at it. We know that the back of the cube is purple since the front is black, the right is grey because the left is yellow, and the bottom is orange because the top is white. 126 5. FLAT MANIFOLDS

Figure 29. The extended diagram of K3 with A. Cube in it.

Now let’s consider the cube just to the right of the cube with the arrow pointing at it. Since the top of this cube is white, we know that there couldn’t have been a top-bottom flip of the right and left walls of the gluing diagram. Also, since the front of this cube is black, we know there couldn’t have been a back-front flip of the right and left walls of the gluing diagram. From this we can conclude that the two side walls are glued straight. Next we consider the cube just above the one with the arrow pointing at it. Since the left face of this cube is grey, but the left face of the original cube was yellow, there must have been a left-right flip of the top and bottom faces in the gluing diagram. On the other hand, since the upper cube and lower cube both have black in front, there could not have been a front-back flip of the top and bottom walls of the gluing diagram. Finally, when we look out the back of the cube in the lower left, we see an identical cube just behind it. Thus it follows that the front and back walls must be glued straight. This tells us everything about how the walls of the gluing diagram were glued. Using this type of reasoning, if you’re given an extended diagram of a glued up cube, you should be able to determine what the gluing is, and if you’re given a gluing diagram for a cube, you should be able to draw an extended diagram. In the Exercises, you will have a chance to practice going back and forth between a gluing diagram and an extended diagram for a glued up cube. 9. OTHER TYPES OF GLUINGS OF A CUBE 127

9. Other Types of Gluings of a Cube

So far, we have only considered gluing the faces of a cube straight or with a flip. Now we will see what happens when we glue a pair of faces of a cube with a 1 rotation. In particular, we define a 4 -turn manifold as the manifold obtained by gluing one pair of opposite faces with a 90◦ rotation, while gluing the other pairs of opposite faces straight. In Figure 30, the top and bottom faces are glued with a 90◦ rotation. This means that when A. 3D-Girl goes out the top of the cube, she comes in the bottom of the cube rotated by 90◦. By checking that the edges are glued in groups of four and all eight vertices are glued together, we see that the resulting manifold is indeed flat. Is this manifold orientable?

Figure 30. Since the top and bottom of this cube are glued with a90◦ rotation, when A. 3D-Girl goes out the top she comes in the bottom rotated by 90◦.

1 3 In a similar way, we can create 2 -turn and 4 -turn manifolds by gluing opposite faces straight except for one pair which is glued with a 180◦ or a 270◦ rotation. Observe that we can’t create turn manifolds in two dimensions since if you start with a polygon, the edges can only be glued straight or with a flip. In particular, gluing one edge to another with a 90◦ rotation wouldn’t make sense, and a 180◦ rotation of an edge is the same as a flip. All of the glued up cubes we’ve considered so far have pairs of opposite faces glued. We can also have a gluing diagram where some number of adjacent faces of the cube are glued together. For example in Figure 31, the top face is glued to the front face, the back face is glued to the bottom face, and the right and left faces are glued together. In the figure, we have colored the edges and vertices to indicate which ones are glued together. You can imagine gluing adjacent faces together by folding along the edge between them. Thus the gluing in Figure 31 is obtained by folding along the dashed line and the dotted line before gluing up the left and right faces. Since there are only two thick black edges and only one dashed edge and one dotted edge, this gluing diagram is definitely not flat. Notice that the endpoints of the thick black edge are glued together, as are the endpoints of the dashed edge, as well as those of the dotted edge. This means that the thick black edge becomes a circle of cone points with 180◦ around each point on it. The dashed edge and 128 5. FLAT MANIFOLDS

Figure 31. The top and front faces are glued, the back and bottom faces are glued, and the right and left faces are glued. the dotted edge each form another circle of cone points with only 90◦ around each point. A cube is not the only 3-dimensional gluing diagram. We can create 3-manifolds by gluing up pairs of faces on any polyhedron. In Exercises 18 and 19 we consider 3-manifolds obtained by gluing together opposite faces of a dodecahedron. As you glue up pairs of faces of a polyhedron, remember that you can check if the glued up 3-manifold is ﬂat by determining whether every point on each edge and vertex has a 3-dimensional ball around it. In particular, you might want to look at Figure 21 to help you.

10. Exercises

1. Draw a gluing diagram for a 2-manifold that is homogeneous at all but ﬁve points. Draw a 2-manifold that is homogeneous at all but n points. 2. For each of the conditions listed below, draw an octagonal gluing diagram with edges glued in pairs and indicate whether your glued up surface is orientable or non-orientable. (a) The surface has least one cone point. (b) The surface has at least one anti-cone point. (c) The surface has both a cone point and an anti-cone point. 3. Explain how you know that the surface whose gluing diagram is illustrated in Figure 19 is a torus, and how you know it has one cone point and one anti-cone point. 4. What surface is obtained by gluing up the hexagon in Figure 32? Does this surface have any cone points or anti-cone points?

Figure 32. Illustration for Exercise 4. 10. EXERCISES 129

5. Suppose A. Hexagon lives in the surface illustrated in Figure 32 and has a different color on each side. Draw an extended diagram of A. Hexagon in her universe. 6. Compare the global topology and geometry of a flat torus and the hexagonal gluing diagram in Figure 32. 7. What surface do you get by gluing each side of a square to itself by folding the side in half. Does the surface have any cone points or anti-cone points? If so, how many? 8. Imagine a large square with a smaller square cut out of it. Glue together opposite sides of each square. What surface do you get? 9. Suppose A. Square lives alone in the space illustrated in Figure 33. (a) Draw an extended picture of his space. (b) Is his space flat?

Figure 33. Illustration for Exercise 9.

10. Suppose that A. Square lives alone in his space, and imagine that he is translu- cent so he can see through himself. For each of the spaces listed below, determine how many squares A. Square can see in addition to himself. In each case, draw a picture to illustrate your conclusion. (a) An infinitely long cylinder. (b) An infinite cone with a cone angle of 45◦. (c) An infinite cone with a cone angle of 50◦. (d) An infinite cone with a cone angle of 200◦. 11. Draw a polyhedral gluing diagram which is homogeneous everywhere except along one circle where every point on that circle has less than 360◦ around it and along another circle where every point is has more than 360◦ around it. Note this means that the space has a circle of cone points and a circle of anti-cone points.

1 12. Suppose that you live in a 4 -turn manifold and you talk on the phone to some- 3 one who lives in a 4 -turn manifold. Could you determine that your manifolds 1 were diﬀerent? What if you talk to someone in a 2 -turn manifold? 130 5. FLAT MANIFOLDS

13. Draw three gluing diagrams of a cube whose extended diagrams are different, in which each have two pairs of opposite faces glued with flips and the remaining pair of opposite faces glued straight. Explain how you know that the extended diagram of each of your spaces is distinct from the others. 14. How can you tell the difference between a glued up cube in which one pair of opposite faces is glued with a flip and the rest are glued straight, from a glued up cube in which one pair of opposite faces is glued with a 180◦ rotation and the rest are glued straight? 15. Suppose that we glue opposite faces of a cube in each of the ways described below. Is it possible to get a flat manifold? If so explain how. (a) One pair of sides is glued with a flip, and the other two pairs are each glued with a 180◦ rotation. (b) One pair of faces is glued with a flip, and the other two pairs are each glued with a 90◦ rotation. (c) Each pair of faces is glued with a flip. 16. Start with a cube and glue each of the vertical sides to itself by folding the side in half along a floor to ceiling line. Then glue the bottom of the cube to the top 1 ofthecubewitha 4 -turn. How many circles does this manifold have where it is not homogeneous? What is the angle around each of these non-homogeneous circles? 17. Consider the 3-manifold whose gluing diagram is illustrated in Figure 24. If we take a cross section of the gluing diagram parallel to the left and right faces, we obtain a square gluing diagram. Draw a picture of the gluing diagram with arrows to indicate the gluings along the edges. Determine whether the corresponding surface is flat, and identify the surface. 18. A dodecahedron is a polyhedron with 12 faces that are pentagons (see Fig- ure 34). If the pentagons are transparent, you can look through one and see 1 × ◦ ◦ that the opposite pentagon is rotated by 10 360 =36.Inordertoglue opposite faces, the vertices of the pentagons must line up. So we must rotate afacebyanoddnumbertimes36◦ before gluing it to the opposite face.

Figure 34. Illustration for Exercise 18. 10. EXERCISES 131

Consider the space obtained from a dodecahedron by gluing each pair of 1 ◦ opposite faces with a 10 -turn. In other words we rotate the faces by 36 before we glue them to their opposite faces. How many vertices are glued together? How many edges are glued together? Is the 3-manifold flat? 19. Suppose you live in a dodecahedron with opposite pairs of faces glued with 1 a 10 -turn and you are suddenly transported to a dodecahedron with opposite 3 pairs of faces glued with a 10 -turn. Can you tell the difference between your new space and your original one? 20. Figures 35, 36, and 37 are extended diagrams for spaces created by gluing opposite faces of a cube. A. Cube is alone in each space. Her coloring is illustrated in Figure 27. Explain how the faces of the cube were glued together to create the extended diagram for each figure.

Figure 35. Illustration for Exercise 20. 132 5. FLAT MANIFOLDS

Figure 36. Illustration for Exercise 20.

Figure 37. Illustration for Exercise 20. 10. EXERCISES 133

Figure 38. Illustration for Exercise 21.

21. Trace Figure 38 onto a piece of paper and then color it to illustrate the view 1 inside of a 4 -turn manifold containing A. Cube. 22. Create a gluing of a cube where pairs of opposite faces are glued, but not all eight vertices are glued together.

https://doi.org/10.1090//mbk/096/06

CHAPTER 6

Connected Sums of Spaces

Topics: • Connected sums of surfaces • Classiﬁcation of surfaces versions 1 and 2 • Euler characteristic • Genus • Connected sums of 3-manifolds

1. Einstein-Rosen Bridges

A first step in trying to understand the shape of our universe is to make a list of all the possibilities. If we had such a list, we might be able to compare evidence about the shape of our universe to topological and geometric properties of spaces on the list in order to make a short list of serious contenders. However, so far, the list of 3-dimensional universes that we have learned about is not very long. In particular, the only 3-manifolds we have discussed are R3, S3, T 3, K3, together with other manifolds obtained by gluing up a cube. But these are not the only possibilities. Just as we can put together bricks to make complicated buildings, we can put together familiar spaces to make new more complicated spaces. As usual, before we describe how to do this in 3 dimensions, we consider the analogous problem in two dimensions. Perhaps the simplest way to combine two surfaces is to put them next to each other, and then find a way to connect them to form a single surface. In one of the Exercises in Chapter 4, we joined two planes together with a cylindrical tube known as an Einstein-Rosen bridge (see Figure 1 below). This tube provides a way for A. Square and his friends to get from one of the planes to the other, just as an ordinary bridge enables us to get from one island to another. We can describe the construction of an Einstein-Rosen bridge between two planes as removing the interior of a disk from each plane and then adding a tube that goes from the boundary of the hole in one plane to the boundary of the hole in the other plane. Notice that after we attach the tube, the circles at the top and the bottom along which it’s glued are different from normal circles in the plane because they bound infinite regions on both sides. In this chapter, we will not pay any attention to the geometry of our spaces. This means we are free to shorten an Einstein-Rosen bridge between two planes until it has essentially no length at all. The result is two planes which are joined

135 136 6. CONNECTED SUMS OF SPACES

Figure 1. An Einstein-Rosen bridge joining two planes. together along a single circle. We can visualize this by removing the tube in Figure 1 and gluing the circle at the top of the tube to the circle at the bottom of the tube. Figure 2 illustrates a gluing diagram for this space, where the grey regions have been removed from the two planes and the arrows indicate how to connect the boundary circles together. When A. Square goes through one of these circles, he is instantly transported to the other circle, as illustrated in the ﬁgure.

Figure 2. These two planes with holes are joined together along the circle.

The same idea enables us to connect any two surfaces along a circle. In particular, we remove the interior of a disk in each surface, and then we attach a tube that goes from the boundary of the hole in one surface to the boundary of the hole in the other surface as illustrated on the left side of Figure 3. Because we are concerned with the topology of the new surface and not the geometry, we can then shorten the tube between the surfaces until it has no length at all. In this way, the surfaces are glued together along a circle rather than joined by a tube, as illustrated on the right side of Figure 3.

Figure 3. We can shrink the tube joining two surfaces until it’s just a circle.

2. Connected Sums of Surfaces

We formalize the ideas from the last section with the following deﬁnition. Definition. Let A and B be surfaces. The connected sum of A and B, written as A#B, is the surface obtained by removing the interior of a disk from each of A and B and then gluing the remaining pieces together along the boundaries of the holes. 3. ARITHMETIC PROPERTIES OF THE CONNECTED SUM 137

Figure 4 illustrates the connected sum of two tori. In the ﬁrst step, we put the two tori side by side with the connected sum symbol between them. When we remove a disk from each one, it looks like we have two hollow doughnuts with a bite taken out of each. Finally, we stick the two partially eaten donuts together along their boundaries to get the double doughnut. The resulting surface is a 2-holed torus denoted by T 2#T 2, or more succinctly by 2T 2.

Figure 4. The connected sum of two tori.

To take the connected sum of three tori, we can first add two tori to get the 2- holed torus 2T 2, and then add another torus T 2 to get a 3-holed torus. Symbolically we can express this as: T 2#T 2#T 2 =(T 2#T 2)#T 2 =2T 2#T 2 =3T 2. In a similar way we can write the connected sum of two copies of any surface A as A#A =2A, and we can write the connected sum of n copies of A as: ··· nA = A #A# #A . n terms You might wonder what would happen if the two tori in Figure 4 were glued together along a different circle. Since we are focusing on topology here, the disks that we remove from A and B can slide around on each of the surfaces so that A#B does not depend on where the gluing takes place. We say that the connected sum operation is well defined to mean that it does not depend on anything other than A and B.

3. Arithmetic Properties of the Connected Sum The word “sum” gives us the idea that the connected sum is, at least in some ways, similar to normal addition of numbers. In the Exercises, you will show that the connected sum operation obeys the commutative property (expressed as a + b = b + a) and the associative property (expressed as a +(b + c)=(a + b)+c)). We’ll explore some of the other properties of the connected sum below, and we’ll see that connected sums do share many (but not all) of the properties of sums of numbers. In ordinary addition, the number 0 is said to be an identity element because for any number a we have 0 + a = a and a +0 = a. We see as follows that the sphere S2 serves as an identity element under the connected sum operation. Figure 5 illustrates the connected sum of a torus T 2 and a sphere S2. Since we are only considering topology here, we can deform the resulting surface to our heart’s delight. In particular, we can shrink the blister on the torus to a small lump, which eventually becomes absorbed into the surface of the torus itself. Thus, we see that T 2#S2 = T 2.Infact,S2#T 2 = T 2 is also true, and the torus in Figure 5 can be replaced with any surface X.ThusweseethatS2 is indeed an identity element for the connected sum. 138 6. CONNECTED SUMS OF SPACES

Figure 5. The connected sum of a torus and a sphere is a torus.

In ordinary addition, any number a has an additive inverse −a such that a +(−a) = 0. Since S2 plays the role of 0 for the connected sum, we might wonder whether every surface has an inverse under the connected sum operation. In other words, if we are given a surface A, can we ﬁnd a surface B such that A#B = S2? If such an inverse surface exists, we might want to call it −A rather than B. We can ask the related question of whether we are allowed to “cancel” under the connected sum operation. That is, if we know that A#B = A#C,can we conclude that B = C? For ordinary addition, we know that if 2 + x =2+3, then we can cancel the 2’s to get x = 3. Before we answer these questions, let’s consider how the connected sum works with gluing diagrams.

4. Gluing Diagrams for nT 2 and nP 2

We start with the gluing diagram of a torus. Figure 6 shows some pictures of how we might remove a disk from the gluing diagram. Note that in the diagram on the right, the disk that we have removed is shaped like an eye rather than a disk. But that’s OK because an eye shape can be deformed to a disk, and we are only interested in topology right now rather than geometry.

Figure 6. Diﬀerent types of disks on the gluing diagram for a torus.

Since we saw above that the connected sum is well deﬁned, it doesn’t matter where we remove the disk from our surface. Thus we can choose to remove whichever disk is most convenient. You might guess that the best disk to remove is the one right in the center of the picture. But, in fact, we will use the picture on the right because after removing the disk, we can cut open the gluing diagram so that the circle around the disk we’ve removed becomes an edge in a pentagon (see Figure 7). This is allowed because all ﬁve vertices in the pentagon are glued together (you should check this), so the two endpoints of the new edge represent the same point. In particular, the new edge still represents a circle in the pentagonal gluing diagram. The connected sum of the two tori is then formed by attaching two copies of this pentagonal gluing diagram together along the two new edges (corresponding 4. GLUING DIAGRAMS FOR nT 2 AND nP 2 139

Figure 7. Changing the gluing diagram from a square with a hole in it to a pentagon.

∪ ∪

Figure 8. A gluing diagram for the connected sum of two tori. to the boundaries of the two disks that were removed). In Figure 8 we see that the gluing diagram for 2T 2 is an octagon. Notice that in Figure 8, we use the union symbol ∪ to mean that we want the two pieces glued together. This symbol is diﬀerent from the connected sum symbol #, which we put between two surfaces before we have removed the disks from each surface. For example, look at the way the symbol # is used on the left side of Figure 5. Next we use a similar method to obtain the gluing diagram for the connected sum of two projective planes. We start with two copies of the gluing diagram for P 2. Then we remove the inside of a small disk at the corner of each gluing diagram as illustrated on the left side of Figure 9. This enables us to open up the diagrams into triangles. Finally, these triangles are glued together along their new edges to form the quadrilateral gluing diagram on the right of Figure 9.

∪

Figure 9. A gluing diagram for the connected sum of two projective planes.

We can repeat this process to get a hexagonal gluing diagram for 3P 2 = P 2#P 2#P 2. In fact, for any number n, we can similarly obtain a gluing diagram with 2n sides for nP 2 and one with 4n sides for nT 2. We can also ﬁnd a gluing diagram for the connected sum of some number of tori together with some number of projective planes. 140 6. CONNECTED SUMS OF SPACES

5. The Classification of Surfaces The following theorem gives us the surprising result that all finite surfaces can be obtained as connected sums of the three surfaces S2, T 2,andP 2.You might see the proof of this theorem in a more advanced topology class. The word “classification” in the title of the theorem is often used in mathematics to mean that the result describes all possibilities.

Theorem 6.1 (The Classiﬁcation of Surfaces, version 1). Every surface with ﬁnite area is either S2 or the connected sum nT 2#mP 2 for a pair of non-negative integers m and n, at least one of which is positive.

The Classification of Surfaces together with the method we learned in the last section enables us to create a gluing diagram for any finite surface. However, the gluing diagrams that we create in this way are not the only ones for a given surface. For example, in Figure 9 we illustrated a gluing diagram for 2P 2 = P 2#P 2.We obtain a different gluing diagram for 2P 2 as follows. On the left side of Figure 10 we’ve redrawn the gluing diagram from Figure 9 as a square. Then in the second step of Figure 10, we’ve cut the gluing diagram along a diagonal to obtain two triangles. We then rotate the top triangle and flip over the bottom triangle so that the double arrows line up in the third step. Finally, we glue the double arrows together and then reshape the diagram into a new square. We can now recognize this gluing diagram as that of a Klein bottle. This wasn’t apparent from Figure 9. The technical term for this kind of manipulation of a gluing diagram is cut and paste. In particular, we say that Figure 10 uses cut and paste to show that P 2#P 2 = K2.

Figure 10. We use cut and paste to show that P 2#P 2 = K2.

In the Exercises, you will use cut and paste to show that 3P 2 = T 2#P 2. From this equation together with the equation P 2#P 2 = K2, we obtain the following lemma. Lemma 1. K2#P 2 =(P 2#P 2)#P 2 =3P 2 = T 2#P 2. This result can be used to answer our earlier question about whether cancelation works for the connected sum operation. In particular, observe that since K2 is non- 5. THE CLASSIFICATION OF SURFACES 141 orientable and T 2 is orientable, we know that K2 = T 2. Thus we cannot cancel the P 2 from both sides of the equation K2#P 2 = T 2#P 2. It is important to keep this in mind as you work with the connected sum operation. Earlier we had also asked whether the connected sum operation had an inverse. Starting again with the result that K2#P 2 = T 2#P 2,weshowbelowthatP 2 cannot have an inverse. Suppose that P 2 has an inverse Q. Then we could add Q to both sides of the equation K2#P 2 = T 2#P 2 and use the associative property (which you will prove in the Exercises) to get K2#(P 2#Q)=(K2#P 2)#Q =(T 2#P 2)#Q = T 2#(P 2#Q).

Since Q is the inverse of P 2, we obtain the equation K2#S2 = T 2#S2. From this we would again get the false equation K2 = T 2.ThusP 2 cannot have an inverse. Finally, notice that the equation 3P 2 = T 2#P 2 (which you will prove in the Exercises) allows us to avoid mixing tori and projective planes in connected sums. For example, the connected sum 2T 2#P 2 canbewrittenasfollows: 2T 2#P 2 = T 2#(T 2#P 2)=T 2#3P 2 =(T 2#P 2)#2P 2 =3P 2#2P 2 =5P 2. In the Exercises, you are asked to generalize this computation to show that we can rewrite any mixed sum of tori and projective planes as a sum of just projective planes. This result together with the Classification of Surfaces, version 1, gives us the following. Theorem 6.2 (The Classification of Surfaces, version 2). Every surface with finite area is either S2, nT 2,ornP 2 for some positive integer n. It now follows that every finite orientable surface is either S2 or nT 2,and every finite non-orientable surface is nP 2.SinceS2#T 2 = T 2 and S2#P 2 = P 2, we can get any finite surface by starting with S2 and adding on some number of tori or some number of projective planes. We can visualize these two possibilities as follows. Starting with S2#T 2 in Figure 11, we absorb the base of T 2 into S2 until the intersection of T 2 and S2 splits into two circles. Then we separate the circles so that the torus becomes a tube that goes from the sphere back to itself. Thus instead of visualizing S2#T 2 as a sphere with a torus attached, we can visualize it as a sphere with a handle attached. By attaching n handles to S2,weobtainnT 2.

Figure 11. S2#T 2 is topologically the same as S2 with a handle. 142 6. CONNECTED SUMS OF SPACES

Figure 12. We can represent a projective plane with a disk removed as a cross-cap.

We would like to ﬁnd an operation analogous to attaching handles that will enable us to visualize adding projective planes to S2. In Chapter 4, we learned that we can represent a projective plane with a disk removed as a cross-cap (see Figure 12). Keep in mind that a cross-cap is not really a projective plane with a disk removed because in the fourth image of Figure 12 we squeezed the two gluing circles into line segments before gluing them together. However, drawing a projective plane with a disk removed in this way allows us to represent S2#P 2 as a sphere with a cross-cap attached. Thus we can visualize S2#nP 2 as a sphere with n cross-caps attached. We can now visualize the Classiﬁcation of Surfaces, version 2, as saying that every surface can be obtained from a sphere by attaching some number of handles or some number of cross-caps. For example, in Figure 13 we illustrate 3T 2 as a sphere with three handles and 2P 2 as a sphere with two cross-caps. Note that the number of handles or cross-caps could be zero in the case where the surface is just a sphere.

Figure 13. We represent 3T 2 by the image on the left and 2P 2 by the image on the right.

6. Dividing a Surface into Vertices, Edges, and Faces

Suppose that A. Square knows that his space is ﬁnite and he knows the Clas- siﬁcation of Surfaces, version 2. He would like to determine whether he is living 6. DIVIDING A SURFACE INTO VERTICES, EDGES, AND FACES 143

Figure 14. An atlas of a cube with one page for each face. in S2, nT 2,ornP 2. So he decides to systematically map out every region of his universe and collect the maps together into an atlas. Like an atlas of the earth, each page represents some region. The gluings would only occur between one page and another. For example, suppose that his space is the surface of a cube. Each face of the cube could correspond to one page of his atlas. Altogether, the atlas would have six pages corresponding to the six faces of a cube. In Figure 14, the center of each page has two words to indicate which face of the cube it represents. These labels describing the position of the page in three dimensions (top, bottom, front, back, etc.) would be meaningless to A. Square, but is useful for us to visualize how the pages form a cube. If you’ve ever used a road atlas to navigate, you might recall that a country is divided into square regions and there is a detailed road map of each region on a separate page. If you want to go off the top of one page, there is a number indicating the page you should turn to in order to continue going in that direction. A similar thing holds for A. Square’s atlas. If he wants to know where he would be if he went off the top of a particular page, he would look for the letter on another page that matched the letter on the top of his page. For example, if he went off the top of the page labeled “right face”, he would see the letter j which would tell him that to continue going in that direction he should enter the right side of the page labeled “back face”. The pages also contain arrows on the edges to indicate how the pair of pages with the same letter are glued together. As A. Square is making an atlas of his space, he decides to keep track of the number of pages, the number of edges, and the number of vertices in his atlas. In order to avoid having to keep track of the topology of the regions represented by

Figure 15. A region in the shape of a washer must be split into two pages so that each page can be deformed to a disk. 144 6. CONNECTED SUMS OF SPACES

Figure 16. A tetrahedron and a dodecahedron. different pages, he makes the rule that the region on each page must be topologically the same as a disk. This means for example that if a region was shaped like a washer (which mathematicians call an annulus), he would have to split it into two pages, as illustrated in Figure 15. We will refer to the pages of A. Square’s atlas as faces, keeping in mind the rule that every face can be deformed to a disk. For the atlas of the cube in Figure 14, there are 8 vertices, 12 edges, and 6 faces. In Figure 16 we illustrate a tetrahedron and a dodecahedron. You should check that the tetrahedron has 4 vertices, 6 edges, and 4 faces, and the dodecahedron has 20 vertices, 30 edges, and 12 faces. Since A. Square is trying to determine the topology of his universe, what’s important to him is how the faces are glued together, not the specific size or shape of each face. In fact, the surfaces of a cube, a tetrahedron, and a dodecahedron are all topologically the same as a sphere. Yet the information he is collecting on the number of faces, edges, and vertices of each are different. In the next section we’ll learn about a wonderful formula that A. Square can put these numbers into to obtain a single number that will help him determine the topology of his universe.

7. The Euler Characteristic of a Surface

In the middle of the eighteenth century the mathematician Leonhard Euler (whose name is pronounced “Oiler”) introduced the following formula for surfaces with finite area. Definition. Let S be a surface which is divided into a finite collection of v vertices, e edges, and f faces, where a face is a region that can be deformed to a disk. We define the Euler characteristic of S to be the number χ(S)=v − e + f. Note that the symbol χ is a Greek letter, which is pronounced “kai” to rhyme with “sky”. The Euler characteristic is represented by χ because both characertistic and χ start with a hard k sound. Let’s compute the Euler characteristics of a cube, tetrahedron, and dodecahedron by using the numbers of faces, edges, and vertices that we determined in the last section. χ(cube) = 8 − 12+6=2. χ(tetrahedron) = 4 − 6+4=2. χ(dodecahedron) = 20 − 30+12=2. It’s not a coincidence that all three of these values are equal to 2. It turns out that the Euler characteristic is a topological property of a surface. Since the 7. THE EULER CHARACTERISTIC OF A SURFACE 145 cube, the tetrahedron, and the dodecahedron are all topologically equivalent, any topological property that one of these three surfaces has, the other two surfaces must also have. More generally, we have the following powerful theorem, whose proof you might see in a more advanced topology class. Theorem 6.3 (Euler Characteristic Theorem). The Euler characteristic is a topological property of a surface with finite area. In particular, the value of the Euler characteristic is the same no matter how the surface is divided into faces, edges, and vertices. If it really doesn’t matter how we divide a surface into faces, edges, and vertices, we might as well pick an easy way of dividing it up. In fact, the easiest way is to draw a gluing diagram and consider the whole region of the gluing diagram as a single face. Then we just have to count how many distinct edges and vertices there are and plug these numbers into the Euler characteristic formula. For instance, let’s compute Euler characteristics using the gluing diagrams for the surfaces S2, T 2, P 2,andK2, as illustrated in Figure 17. We aren’t used to seeing S2 represented by a gluing diagram. To see that the first gluing diagram represents a sphere, we just fold over the top to attach the arrows. This gives us the surface of an apple turnover, which is topologically the same as S2.

Figure 17. We use these gluing diagrams to compute Euler characteristics.

Using the gluing diagram in Figure 17, we see that the sphere has 1 face, 1 edge (since the two arrows are glued together), and 2 vertices (one at each end of the arrows), so we have χ(S2)=2− 1+1=2. Note that this agrees with our computations for the cube, the tetrahedron, and the dodecahedron, which are each topologically the same as the sphere. Next, we see that the torus consists of 1 square face, 2 edges (one horizontal and the other vertical), and just 1 vertex (since all four corners of the face are glued together), so we have χ(T 2)=1− 2+1=0. In the case of the projective plane, we have 1 face, 1 edge, and 1 vertex, so χ(P 2)=1− 1+1=1. Finally, for the Klein bottle there is 1 square face, 2 edges, and 1 vertex (since all four corners of the face are glued together), so χ(K2)=1− 2+1=0. Since P 2 and K2 have diﬀerent Euler characteristics, by the Euler Characteris- tic Theorem (Theorem 6.3) they must be topologically distinct. On the other hand, the Euler characteristics of T 2 and K2 are the same even though we know that T 2 and K2 are diﬀerent surfaces, since T 2 is orientable and K2 is not. 146 6. CONNECTED SUMS OF SPACES

8. The Euler Characteristic of Connected Sums

The following theorem gives us an easy way to compute the Euler characteristic of a connected sum.

Theorem 6.4. Let M1 and M2 be ﬁnite area surfaces. Then

χ(M1#M2)=χ(M1)+χ(M2) − 2.

Proof. Since M1 and M2 are ﬁnite area surfaces, they can each be divided into some number of faces, edges, and vertices. Suppose that M1 has f1 faces, e1 edges, and v1 vertices, and M2 has f2 faces, e2 edges, and v2 vertices. For each of M1 and M2, we add a circle attached to a vertex such that the circle is contained entirely in a single face. Since the face containing the new circle is topologically the same as a disk, the new circle will bound a region that is also topologically the same as a disk. In Figure 18, the new circle bounds the grey region.

Figure 18. In both M1 and M2, we add a circle attached to a vertex such that the circle is contained in a single face.

In both M1 and M2, this new circle counts as an additional edge and bounds an additional face. Thus M1 will now have f1 + 1 faces, e1 + 1 edges, and v1 vertices, and M2 will have f2 + 1 faces, e2 + 1 edges, and v2 vertices. Notice that the Euler characteristics of M1 and M2 do not change. In order to create the connected sum M1#M2,weremoveadiskfromeachone and then glue the surfaces together along the circular boundaries. Since it doesn’t matter what disk we remove, we might as well remove the new faces bounded by the circles we’ve added to M1 and M2. Now M1#M2 has f1 + f2 faces, since we have removed the new faces from each of M1 and M2. When we count the edges of M1#M2,wehavetoaccountforthe original edges of M1, the original edges of M2, and the new circle in each. However, the new circles are glued together in M1#M2. Thus we should only count this new edge once. This means that the total number of edges in M1#M2 is e1 + e2 +1. Similarly the vertex on the boundary of the face that we’ve removed from M1 is glued to the vertex on the boundary of the face that we’ve removed from M2. Hence, the number of vertices of M1#M2 is v1 + v2 − 1. We can now compute the Euler characteristic of M1#M2 as follows:

χ(M1#M2)=(v1 + v2 − 1) − (e1 + e2 +1)+(f1 + f2)

=(v1 − e1 + f1)+(v2 − e2 + f2) − 2=χ(M1)+χ(M2) − 2, as desired. 9. THE GENUS OF A SURFACE 147

Recall from the previous section that χ(T 2)=0andχ(P 2)=1.Thusthe theorem tells us that if we add T 2 to a surface M,thenwehave χ(M#T 2)=χ(M)+0− 2=χ(M) − 2, and if we add P 2 to M,wehave χ(M#P 2)=χ(M)+1− 2=χ(M) − 1. Using this result together with induction, you will prove the following corollary in the Exercises.

Corollary. The Euler characteristic of the orientable surface nT 2 is given by the formula χ(nT 2)=2− 2n, and the Euler characteristic of the non-orientable surface nP 2 is given by the formula χ(nP 2)=2− n.

Observe that the corollary together with the Euler Characteristic Theorem (Theorem 6.3) gives us the following theorem. Theorem 6.5. Two ﬁnite area surfaces are topologically the same if and only if they have the same Euler characteristic and either both are orientable or both are non-orientable. This result is very convenient for A. Square. If he can explore all of his universe and create an atlas, he can count up the faces, edges, and vertices in his atlas and compute the Euler characteristic. Then he can compare the Euler characteristic that he found with the formulas in the corollary. If he gets an odd number, he knows right away that his universe is nP 2, and he can determine the value of n from his computation of the Euler characteristic. For example, suppose he determines that his universe has an Euler characteristic of −47. Then he can use the second formula from the corollary to get 2 − n = −47, and from this he can conclude that his universe is 49P 2. But if the number that he gets is even, then he has to do further explorations to determine whether or not his universe has an orientation reversing path. Once he ﬁnds this out, he can use the correct formula from the corollary to determine exactly what surface he lives in. For example, if he determines that his universe is orientable and has Euler characteristic −10, then he uses the formula 2 − 2n = −10 from which he concludes that his universe is 6T 2.

9. The Genus of a Surface

After discovering Theorem 6.5, A. Square is so excited that he rushes to find B. Triangle to tell her about it. However, to his surprise, B. Triangle has been busy developing her own approach to distinguishing finite area surfaces. Instead of dividing surfaces into faces, edges, and vertices as A. Square did, B. Triangle has been considering the topological properties of different types of circles that can occur on a surface. In particular, she’s interested in non-separating circles,which are those you could cut along without dividing the surface into two regions (see Figure 19). In Chapter 3, we saw that there are infinitely many non-separating 148 6. CONNECTED SUMS OF SPACES circles on a torus. On the other hand, if we cut along any circle on a sphere, we will separate it into two pieces. Note that we are using the word “circle” here to mean a loop that is topologically the same as a circle. In particular, we are not allowing these circles to intersect themselves.

Figure 19. This circle does not separate the torus into two regions.

B. Triangle has discovered that the more complicated a surface is, the more non-separating circles it contains. She makes the following deﬁnition. Definition. The genus of a surface is the maximum number of disjoint circular cuts that can be made without disconnecting it. The condition that the circular cuts be disjoint is important. For example, we saw that there are inﬁnitely many non-separating loops on a torus, but any two of them intersect. In fact, as we can see in Figure 20, even the two simplest non-separating circles on a torus (a meridian and a longitude) will intersect.

Figure 20. After cutting a torus open along a meridian, any additional circular cut will disconnect it.

If we cut a torus along a meridian, it will become a cylinder. Now any circular cut that we make on the cylinder will disconnect it. In fact, no matter what non- separating circle we use to cut open a torus, we’ll get a cylinder; and once we get a cylinder, we can’t make another circular cut without disconnecting it. According to B. Triangle’s deﬁnition this means that the genus of a torus is 1. Next let’s consider the 2-holed torus 2T 2. Figure 21 shows two disjoint non- separating circles in 2T 2, each of which wraps meridionally around one of the tori. The existence of these two circles shows that the genus of 2T 2 is at least 2. When we cut along the circles in Figure 21 it doesn’t disconnect the surface. But we’d like to know if we can make another cut in Figure 21 without disconnecting the surface. To ﬁgure this out, let’s deform the cut open 2-holed torus into a sphere with four holes. We’ve illustrated this deformation in Figure 22. We can then check that cutting the sphere with four holes along any circle will disconnect it. So it seems reasonable that the genus of 2T 2 is actually equal to 2. But we don’t know 9. THE GENUS OF A SURFACE 149

Figure 21. A pair of disjoint non-separating circles in 2T 2. whether we could have made three circular cuts in 2T 2 without disconnecting it if we had made our ﬁrst two cuts diﬀerently. We’ll come back to this later, so let’s not worry about it for the moment.

Figure 22. This cut open 2-holed torus can be deformed to a sphere with four holes.

Recall that K2 =2P 2, and as we saw in Section 1, we can visualize the Klein bottle as a sphere with two cross-caps attached. Recall that a cross-cap is made by squeezing two gluing circles into line segments and gluing them together, as in Figure 12. On the left side of Figure 23 we illustrate K2 = S2#P 2#P 2, where we use cross-caps in place of projective planes with holes in them. Now we cut each cross- cap along the line segment representing the gluing circle. On the right side of Figure 23, we see the result of cutting open K2 = S2#P 2#P 2 along these two circles. Since this does not separate the Klein bottle, we know that the genus of the Klein bottle is at least 2.

Figure 23. Cutting along the two gluing circles in K2 = S2#P 2#P 2 does not separate the surface.

In fact, we can see from Figure 24 that once we cut our Klein bottle open along these two circles, we can deform it to get a cylinder, which can be further deformed to get a sphere with two holes. Now if we cut along any circle in the sphere with two holes, we’ll disconnect it. So it seems likely that the genus of the Klein bottle equals 2. But again we don’t know if we could have made more circular cuts without disconnecting it if we had chosen our ﬁrst two cuts diﬀerently. 150 6. CONNECTED SUMS OF SPACES

Figure 24. We can deform this cut open Klein bottle to a sphere with two holes.

Figure 25. A non-separating circle in the Klein bottle.

10. The Genus of nT 2 and nP 2

We can generalize the strategies illustrated in Figures 21 and 23 to cut along n circles in any nT 2 or nP 2.FornT 2 we can cut along a meridional circle on each of the n tori without disconnecting it. For nP 2, we can cut along each of the n gluing circles represented by line segments running down the middle of each of the n cross-caps. In both cases, the collection of circles we cut along are disjoint and don’t disconnect the surface. These constructions show that the genus of nT 2 and of nP 2 is at least n. Thus we have the inequalities genus nT 2 ≥ n and genus nP 2 ≥ n.

Furthermore, we know that after cutting along these circles, each of our surfaces becomes a sphere with some number of holes. So any additional circles will necessarily separate it into pieces. However, we don’t know whether there is a better choice of circles which would allow us to cut along more than n circles without disconnecting the surface. For example, we might ﬁnd more circles if we allowed the circles to wrap multiple times around several handles or cross-caps, like the (p, q)-curves on a torus that we saw in Chapter 3. After all, if we had looked at the projection of the Klein bottle shown in Figure 25, we may have imagined that we could not do better than a single circle wrapping meridionally around the main cylinder forming the bottle; nevertheless, Figure 23 shows that two circles are possible. The following theorem, whose proof you might see in a more advanced topology class, tells us that in fact there is no way to make more than n circular cuts on nT 2 or nP 2 without disconnecting it. Theorem 6.6. The genus of all ﬁnite area surfaces is given by the formulas genus S2 =0, genus nT 2 = n, and genus nP 2 = n. 11. CONNECTED SUMS OF 3-MANIFOLDS 151

When B. Triangle explains the idea of genus to A. Square, the two of them get to work to see if there is a relationship between the Euler characteristic and the genus of a surface. They know that a sphere has Euler characteristic 2 and genus 0, and nT 2 has Euler characteristic 2 − 2n and genus n. On the other hand, for nP 2, the Euler characteristic is 2 − n and the genus is n. They summarize this relationship in the following theorem. Theorem 6.7. The Euler characteristic of an orientable surface S of genus n is given by the formula χ(S)=2− 2n. The Euler characteristic of a non-orientable surface of genus n is given by the formula χ(S)=2− n.

Notice that the ﬁrst formula works even if the surface is a sphere which has genus 0.

11. Connected Sums of 3-Manifolds

We would like to define a 3-dimensional connected sum operation which is analogous to the connected sum of surfaces. Such a definition would enable us to combine 3-manifolds into more complicated ones, just as we used the connected sum to combine surfaces into more complicated surfaces. Recall that we defined the connected sum of two surfaces as the surface we get by removing the interior of a disk from each and then gluing what remains together along the boundaries of the holes. Since the analogue of a disk is a ball, if we are working with 3-manifolds, we have to remove the interior of a ball rather than a disk from each. In particular, we make the following definition. Definition. Let A and B be 3-manifolds. The connected sum of A and B, written A#B,isthe3-manifold obtained by removing the interior of a ball from each of A and B and then gluing the remaining parts together along the boundaries of the holes. As with surfaces, the connected sum of two 3-manifolds is a topological notion, so it does not depend on where the balls are removed. Just as we did with surfaces, we can use connected sums to construct an infinite number of 3-manifolds. For instance, we are already familiar with the 3-torus T 3, the projective 3-dimensional space P 3, and the 3-dimensional Klein bottle K3. Combining T 3 with each of these manifolds, we obtain T 3#T 3, T 3#P 3,andT 3#K3. But visualizing connected sums of 3-manifolds is not as easy as it was for surfaces. In fact, we have the same problem in visualizing connected sums of 3-manifolds that Flatlanders have in visualizing connected sums of surfaces. A. Square has been reading this chapter along with you. So he understands connected sums of surfaces, but his friends don’t. Figure 26 illustrates a good way for A. Square to explain to his friends what it would be like to live in the connected sum of two tori. A. Square tells his friends that you start with two gluing diagrams for a torus. Then you remove a grey disk from each, and glue the surfaces together along their boundaries. This means that if you live in one of the tori and you 152 6. CONNECTED SUMS OF SPACES

Figure 26. When A. Square enters one grey circle, he will be instantly transported to the other grey circle. enter the grey circle you will be instantly transported to the other torus. Once his friends understand T 2#T 2, he can use similar drawings with other gluing diagrams to explain the connected sum of any pair of surfaces. We can use a similar picture and explanation to help us visualize the connected sum of 3-manifolds. In Figure 27, we obtain 2T 3 by removing a grey ball from each T 3 and then gluing the two spaces together along their spherical boundaries. This means that if A 3D-Girl lives in one T 3 and enters the grey sphere she will be instantly transported to the other T 3. Recall from earlier in the chapter that S2 is an identity element for the connected sum of surfaces. In other words, for any surface X, we have the connected sum X#S2 = X. We see as follows that S3 plays the same role for connected sums of 3-manifolds. Suppose that X is a 3-manifold, and we want to evaluate X#S3. ThismeanswehavetoremoveaballfromeachofX and S3 and then glue what remains together. However, we learned in Chapter 1 that just as S2 is topologically the same as two disks glued together along their boundaries, S3 is topologically the same as two balls glued together along their boundaries. Thus when we remove a ball from S3, we are left with another ball, which we will call B.NowtogetX#S3, we remove a ball from X and then glue in B.SinceB is just replacing the ball we removed from X, we end up with X#S3 = X. While we have been comparing the connected sum operation to ordinary addition, we can also think of it terms of multiplication. In this case, S2 plays the role of the number 1, which is the identity in multiplication. Since any surface with

Figure 27. When A. 3D-Girl puts her arm through one grey sphere, it comes out of the other grey sphere. 12. EXERCISES 153

finite area is either S2, nT 2,ornP 2,andT 2 and P 2 cannotbebrokendownany further, we can think of T 2 and P 2 as prime numbers. However, unlike the integers which contain infinitely many primes, T 2 and P 2 are the only prime surfaces. On the other hand, like the integers, there are infinitely many prime 3-manifolds. While there is no purely topological classification of 3-manifolds like the Classification of Surfaces, there is a geometric classification of 3-manifolds. This result, known as the Geometrization Conjecture, was proved by the Russian mathematician Grigori Perelman, earning him the two most prestigious prizes in mathematics, the Fields Medal (in 2006) and the Clay Millennium Prize (in 2010), the latter is worth one million dollars. Surprisingly, Perelman declined both awards, stating that he was “not interested in money or fame”.

12. Exercises

1. For each n>1, construct nT 2 as a glued up polygon. Then determine how many cone points and anti-cone points your glued up surface has. 2. For each n>1, construct nP 2 as a glued up polygon. Then determine how many cone points and anti-cone points your glued up surface has. 3. In this exercise we consider whether the connected sum operation obeys the same rules as ordinary addition of numbers. (a) Is the connected sum operation commutative? In other words, is it always true that X#Y = Y #X for any surfaces X and Y ? (b) Is the connected sum operation associative? In other words, is it always true that X#(Y #Z)=(X#Y )#Z for any surfaces X, Y ,andZ? (c) Explain why no surface other than S2 has an inverse with respect to the connected sum operation? 4. (Hard.) Use cut and paste operations to show that the two surfaces illustrated in Figure 28 have the same topology. Explain how this proves that T 2#P 2 = 3P 2.

Figure 28. Illustration for Exercise 4. 154 6. CONNECTED SUMS OF SPACES

5. We saw in the chapter that the connected sum of n tori and m projective planes can be written as a connected sum of some number of projective planes, as long as m is at least 1. Write a formula for the number of projective planes that must be used. Use induction together with the equation T 2#P 2 =3P 2,to prove that your formula is correct. (Hint: Try to induct on only one variable.) Why does this imply that every surface other than S2 is either nT 2 or nP 2? 6. Explain how T 2 can be obtained by gluing pairs of sides of any polygon with an even number of sides. 7. Suppose that Flatland is a non-trivial connected sum. How could A. Square detect it topologically? Suppose our universe is a non-trivial connected sum. How could we detect it topologically? 8. Using a strip of paper, construct a M¨obius strip. Cut the strip along the center circle; this is the dotted line in the gluing diagram on the left in Figure 29. Now cut along the other curve; this is the dotted line in the gluing diagram on the right Figure 29. What do you notice in each case?

Figure 29. Illustration for Exercise 8.

9. Use the theorem about the Euler characteristic of connected sums together with induction to prove that the Euler characteristic of the orientable surface nT 2, is given by the formula χ(nT 2)=2− 2n, and the Euler characteristic of the non-orientable surface nP 2 is given by the formula χ(nP 2)=2− n. 10. Can you find two finite surfaces which have the same Euler characteristic but are topologically distinct? How could you tell the surfaces apart? 11. Use the Euler characteristic and the Classification of Surfaces, version 2, to determine what surface is represented by the hexagonal gluing diagram in Fig- ure 30.

Figure 30. Illustration for Exercise 11. 12. EXERCISES 155

12. We have seen that K2#P 2, T 2#P 2,and3P 2 are diﬀerent representations of the same surface. We did this using cut and paste operations. Find the number of vertices, edges, and faces, and the Euler characteristic for the gluing diagrams associated with each of these three representations. 13. Suppose that X is a surface and M is obtained by removing n disks from X. What is the relationship between the Euler characteristics of M and X? 14. B. Triangle is trying to identify a surface by making disjoint circular cuts. Her third cut disconnects the surface. What surface might she have? 15. Explain why cutting a sphere with four holes along any circle will disconnect it. 16. Show that for any n, cutting a sphere with n holes along a circle will disconnect it. 17. Find the Euler characteristic of a genus 5 non-orientable surface. Characterize this surface as one of the surfaces listed in the Classiﬁcation of Surfaces, version 2. 18. Let M be a surface with genus m, and let N be a surface with genus n. Find a formula for the genus of the connected sum M#N in terms of m and n.

https://doi.org/10.1090//mbk/096/07

CHAPTER 7

Products of Spaces

Topics: • Products of sets • Products of spaces • Checklist to show a space is a product • Topological uniqueness of products • Geometric products • Geometric products of ﬂat spaces • Visualizing geometric products • Isotropic and non-isotropic spaces • A 3-manifold which is homogeneous but not isotropic

1. Products of Sets

While the connected sum operation enabled us to combine two surfaces or two 3-manifolds to create new ones, we cannot use it to combine two manifolds of different dimensions. In this chapter we will see how to use the product operation to combine spaces and then we’ll construct some new 3-manifolds from 2-manifolds and 1-manifolds. We begin by defining products of sets, then move on to study the topology of products of spaces, and finally consider the geometry of products later in the chapter. Before we give a formal definition of the product of sets, let’s consider the illustration of R2 in Figure 1. The black point in the figure has coordinates (1, 2) because a vertical line through the point hits the X-axis at 1 and a horizontal line through the point hits the Y -axis at 2. The idea of a product of two sets is motivated by the idea that every point in R2 can be expressed by a pair of coordinates in this way. Definition. Let X and Y be sets. The product X × Y is defined to be the set of ordered pairs (x, y) where x is in X and y is in Y . We refer to the sets X and Y as the factors of the product X × Y . According to this definition, we see that R × R is the set of ordered pairs (x, y) where both x and y are real numbers. This corresponds exactly to the definition of R2. Thus we can write R × R = R2. Observe that since the product X × Y is defined as the set of ordered pairs, if the pairs (x, y)and(a, b) differ in either of their coordinates, then they must be

157 158 7. PRODUCTS OF SPACES

Figure 1. We use vertical and horizontal lines through a point in R2 to find its coordinates. different elements of the set X × Y . For example, the point (1, 2) in R2 can’t be written in any other way as an ordered pair. As our next example, let’s consider the sets A = {1, 2, 3, 4} and B = {1, 2, 3}. The product A × B is obtained by listing all ordered pairs of points whose first coordinate is in the set A and whose second coordinate is in the set B.Wecan write this product as the set

{(1, 1), (2, 1), (3, 1), (4, 1), (1, 2), (2, 2), (3, 2), (4, 2), (1, 3), (2, 3), (3, 3), (4, 3)}.

Since A and B are each a finite set of numbers, it is easy to check that the above set of ordered pairs satisfies the definition of A × B. However, it is easier to think about the points in A × B if we organize the set into the rectangular formation illustrated in Figure 2. We can then describe the product both as three rows of four points and as four columns of three points.

Figure 2. The product of A = {1, 2, 3, 4} and B = {1, 2, 3} written in a rectangular formation.

While the product of A = {1, 2, 3, 4} and B = {1, 2, 3} is just a product of sets, the product R×R can also be thought of as a product of spaces. That is, if we think of R as a 1-dimensional space rather than as an inﬁnite set of numbers, then the product R × R = R2 is a 2-dimensional space whose topological form comes from the topological forms of the factors. However, the above deﬁnition of the product of sets does not tell us how to visualize a product of spaces. This is what we will explore in the next section. 2. PRODUCTS OF SPACES 159

2. Products of Spaces

We already know how to visualize the product R × R,sinceweknowthat this product equals R2 and we can visualize R2 as a plane. We will now use our understanding of R2 as a product of two lines as the basis for a more general method of visualizing products of spaces. We begin by analyzing the plane in more detail below. The ﬁrst step is to choose two lines in the plane to be our axes. We pick a horizontal line which we call the X-axis, and a vertical line which we call the Y - axis. Note that we will use capital letters for the axes to avoid confusing the axes with points, whose coordinates will be written with lower case letters. Observe that the X-axis and the Y -axis intersect in precisely one point. Also, for each point x on the X-axis, there is a vertical line disjoint from the Y -axiswhichintersectsthe X-axis in the point x.Wedenotethislineby{x}×R. Similarly, for each point y on the Y -axis, there is a horizontal line disjoint from the X-axis which intersects the Y -axis in the point y. WedenotethislinebyR ×{y}.Furthermore,each horizontal line intersects each vertical line in precisely one point, and this point has coordinates (x, y). We illustrate this in Figure 3.

Figure 3. The lines {x}×R and R ×{y} meet at the point (x, y).

For example, {3}×R is a vertical line which intersects the X-axis at the point 3, and R ×{8} is a horizontal line which intersects the Y -axis in the point 8. Also, the vertical line {3}×R intersects the horizontal line R ×{8} at the point (3, 8). Now let p be a point in the plane. Then there is exactly one vertical line {x}×R which goes through p and exactly one horizontal line R ×{y} which goes through p. This gives us a way to write every point p in the plane as a coordinate pair (x, y) in precisely in one way, which we know is required by the deﬁnition of R × R as a product of sets. Generalizing the above analysis of the plane as the product of two lines, we now have the following Checklist, which will be referenced frequently below.

Checklist to show that a space S is the product X × Y

(1) Choose copies of X and Y in S intersecting in precisely one point, which will be the X-axis and Y -axis, respectively. 160 7. PRODUCTS OF SPACES

(2) Check that for every point x on the X-axis, S contains a copy of Y disjoint from the Y -axis whose intersection with the X-axis is the point x.We denote this copy of Y by {x}×Y . (3) Check that for every point y on the Y -axis, S contains a copy of X disjoint from the X-axis whose intersection with the Y -axisisthepointy.We denote this copy of X by X ×{y}. (4) Check that each {x}×Y intersects each X ×{y} in precisely one point. (5) Check that each point p in S is contained in precisely one of the {x}×Y and precisely one of the X ×{y}.Wedenotep bythecoordinatepair (x, y).

Let’s verify that if we follow these steps, then S will indeed be the product of X and Y .Letx be in X and y be in Y . By Step (2) we know that S contains {x}×Y , and by Step (3) we know that S contains X ×{y}.ByStep(4){x}×Y and X ×{y} intersect in precisely one point p. And finally by Step (5) the point p is (x, y). Thus every coordinate pair (x, y)isapointinS. Furthermore, by Step (5) we know that every point in S can be written as a coordinate pair (x, y)in precisely in one way. Thus it follows from the definition of the product of sets in Section 1 that S = X × Y . Before doing some examples, let’s recall that in Section 1 we wrote the product of the sets A = {1, 2, 3, 4} and B = {1, 2, 3} in the form of a rectangle, and we then described the set A × B as both three rows of four points and four columns of three points. We can use similar language to describe Steps 2–5 of the Checklist. In particular, we can rephrase Step (2) as saying that S is an “X of Y ’s” where each copy of Y meets the X-axis in one point, and we can rephrase Step (3) as saying that S is a “Y of X’s” where each copy of X meets the Y -axis in one point. Then we rephrase Step (4) as saying that every copy of X meets every copy of Y in precisely one point. Finally, we rephrase Step (5) as saying that every point in S is on precisely one of the copies of X and one of the copies of Y . Using this terminology and some pictures, we can now more easily check that a plane is the product of two lines. In particular, after picking our axes, we observe in Figure 4(a) that there is a vertical line through each point on the X-axis meeting the X-axis in a single point. Note that we cannot really draw a line through every point on the X-axis. So we just draw a finite collection of lines and leave the rest to your imagination. Then we observe in Figure 4(b) that there is a horizontal line through each point on the Y -axis meeting the Y -axis in a single point. Finally, we check that each horizontal line meets each vertical line in precisely one point and each point of the plane is on precisely one horizontal and one vertical line. In the next two sections, we use a similar method to show that several 2- dimensional spaces can be expressed as products of 1-dimensional spaces.

3. A. Square and A. Pentagon Are Products

We begin by using the Checklist to show that a square is the product of two line segments. We denote a line segment by the letter I to stand for interval,since we can think of a line segment as an interval in R. From now on, we will use the 3. A. SQUARE AND A. PENTAGON ARE PRODUCTS 161

Figure 4. R2 = R×R is both a union of vertical lines and a union of horizontal lines. words interval, segment,andline segment interchangeably. Thus we show that a square is I × I. First we choose the axes to be the line segments on the bottom and left side of the square. Then in Figure 5 we do Steps (2) and (3) by illustrating the square is both a union of horizontal segments which each intersect the vertical axis in precisely one point, and a union of vertical segments which each intersect the horizontal axis in precisely one point. Furthermore, since each vertical segment meets each horizontal segment in one point, we have done Step (4).

Figure 5. I × I written as a union of horizontal intervals and a union of vertical intervals.

Finally, we illustrate Step (5) in Figure 6 by observing that every point in the square is contained in precisely one horizontal segment and one vertical segment. This shows that our representation of I × I as a square is correct.

Figure 6. A square is the product I × I.

After reading the beginning of this section, A. Square starts bragging to his friends that squares and rectangles are superior to other shapes because they’re the only shapes that are products. His rectangular and square friends are feeling 162 7. PRODUCTS OF SPACES

Figure 7. A pentagon is also the product I × I. pretty good, since they always secretly believed they were superior. And now they knew why. Several polygonal shapes have been arguing with A. Square, demanding that he explain what’s so great about being a product. A. Square responds that products are more beautiful, and ends with, “If you won’t admit that, then you’re just being defensive.” Finally A. Pentagon pushes her way to the front of the crowd and says, “Well, you’re just ignorant, because all polygons are products.” At which point she draws Figure 7, which even A. Square has to admit is beautiful. In fact, A. Pentagon is right. As we see in the image on the left side of Figure 7, a pentagon is a black interval of dark grey intervals. Note that we call the dark grey arcs “intervals” because they are topologically the same as intervals even though they aren’t straight. We’ll discuss the geometry of products later in the chapter. Also, we see in the center image of Figure 7 that the pentagon is also a black interval of light grey intervals. Furthermore, as we see on the right side of the ﬁgure, each dark grey interval intersects each light grey interval in exactly one point, and every point in the pentagon is contained in precisely one dark grey interval and one light grey interval. This includes all of the points on the edges of the pentagon. For example, a point on the top left edge of the pentagon is an endpoint of one of the dark grey intervals and is contained in one of the light grey intervals. In the Exercises you will show that a triangle and a hexagon are also products of two intervals.

4. Some Products where One Factor Is a Circle

For our next example, we show that a cylinder can be expressed as the product of a circle S1 and a line segment I. In Figure 8, we illustrate a circle of intervals going along the horizontal S1-axis and an interval of circles going along the vertical I-axis. We can see that each of the intervals intersects the S1-axisinonepointand each of the circles intersects the I-axis in one point In Figure 9, we do Step (4) by checking that each circle meets each segment in a single point. Finally, we do Step (5) by observing that every point in the cylinder is contained in precisely one horizontal circle and one vertical segment. Thus we can conclude that the cylinder is in fact the product S1 × I. We now show that an annulus (which, as we saw in Chapter 6, is the mathematical word for a washer) can also be expressed as the product S1 × I. In Figure 10, we illustrate the S1-axisasabigcircleandtheI-axisasahorizontallinesegment. Here we have a circle of intervals which are radial spokes going along the S1-axis 4. SOME PRODUCTS WHERE ONE FACTOR IS A CIRCLE 163

Figure 8. The cylinder is circle of intervals and an interval of circles.

Figure 9. A cylinder is the product S1 × I. and an interval of concentric circles going along the I-axis. Each of the intervals intersects the S1-axis in one point and each of the circles intersects the I-axis in one point. Also each circle meets each segment in a single point, and every point in the annulus is contained in an interval. Thus an annulus is also the product S1 × I.

Figure 10. An annulus is also the product S1 × I.

As our last example, we see as follows that a torus can be expressed as the product S1 × S1. In Figure 11, we illustrate the product for both the curved torus and the flat torus. On the curved torus, one S1-axis is a longitudinal circle and the other S1-axis is a meridional circle. We can see that each meridional circle intersects each longitudinal circle in precisely one point. So the curved torus is indeed a longitude of meridional circles and a meridian of longitudinal circles. Furthermore, every point of T 2 is on precisely one longitudinal circle and one meridional circle. For the flat torus in Figure 11, the two vertical edges of the gluing diagram represent a single S1-axis, and the two horizontal edges of the gluing diagram represent the other S1-axis. Because of the gluings, the horizontal and vertical segments in the square are actually circles. Thus again we have a circle of circles in two different ways, and every point is on precisely one vertical segment and one horizontal segment. 164 7. PRODUCTS OF SPACES

Figure 11. The curved and ﬂat torus T 2 are each the product S1 × S1.

5. Examples of Spaces Incorrectly Expressed as Products

In order to better understand the Checklist, in this section we will explore some 2-dimensional spaces which are incorrectly expressed as products. In each case we identify which steps of the Checklist are satisﬁed and which ones are violated. We begin by considering the shape in Figure 12 which looks like a rectangle with a semicircle on each end.

Figure 12. Can this shape be expressed as S1 × I?

Figure 13. This attempt at expressing the shape in Figure 12 as S1 × I violates Steps (2), (4), and (5) of the Checklist.

Let’s try to express this shape as the product S1 × I with a vertical S1-axis and a horizontal I-axis as illustrated on the left in Figure 13. Since the axes meet in just one point, Step (1) is satisﬁed. Now observe that the space is a circle of horizontal intervals. However, with the exception of the top and bottom intervals, all of these horizontal intervals meet the S1-axis in two points rather than just one, violating Step (2). On the other hand, the space is an interval of circles which each meet the I-axis in precisely one point. Thus Step (3) is satisﬁed. But again with the exception of the top and bottom intervals, all of the horizontal intervals meet each of the circles in two points. Thus Step 4 is violated. Finally, all of the points in the interior of the shape in Figure 12 are on more than one of the circles. For example, we see two circles containing the point p. Thus Step (5) is also violated. From this we can conclude that Figure 13 is not a representation of S1 × I. Note that we actually only need to know that one of the steps of the Checklist is violated to reach this conclusion. 5. EXAMPLES OF SPACES INCORRECTLY EXPRESSED AS PRODUCTS 165

Figure 14. This attempt at expressing a M¨obius strip as S1 × I violates Steps (3) and (4) of the Checklist.

Next, let’s consider whether a Möbius strip could be represented as the product S1 × I as illustrated in Figure 14. Here, the S1-axis is the center circle in black in the left image of the figure, and the I-axis is the vertical interval in black in the right image of the figure. Since the two axes intersect in a single point, Step (1) is satisfied. Also on the left side of the figure we see that the Möbius strip is a circle of intervals where each interval meets the S1-axis in one point. Hence Step (2) is satisfied. On the right side we see that it is also an interval of circles, but every circle except the center circle intersects the I-axis in two points rather than just one. For example, we see that the grey circle wraps twice around the Möbius strip, intersecting the I-axis in the points p and q. Thus Step (3) is violated. Similarly, apart from the center circle, all of the circles intersect all of the intervals in two points. Thus Step (4) is violated. On the other hand, every point on the Möbius strip is on precisely one interval and one circle. Thus Step (5) is satisfied. We conclude that Figure 14 is not a representation of the product S1 × I. Our final example is based on the Earth, where locations are often described by their longitude and latitude. A longitude is a great circle going north-south and a latitude is a horizontal circle going east-west. We will now consider whether S2 is a product of two circles where one axis is the equator (which is a latitude) and the other axis is a longitude. We illustrate our axes in Figure 15. In addition to the equator, we’ve drawn four other latitudes, and in addition to our longitude axis we’ve drawn two other longitudes. However, you should imagine longitudes and latitudes through every point.

Figure 15. S2 is not the product of an equator and a longitude.

Observe that since the axes meet in two points, Step (1) is violated. In addition, for every point on the equator, there is a longitude, but these longitudes are not disjoint from the longitude axis, and each longitude meets the equator in two points rather than one. So Step (2) is (doubly) violated. Also, for every point on the 166 7. PRODUCTS OF SPACES longitude axis there is a latitude, but again each latitude meets the longitude axis in two points. So Step (3) is also violated. In fact, all of the longitudes meet all of the latitudes in two points. Thus Step (4) is also violated. Finally observe that the North and South Poles are on every longitude and on no circle of latitude. Thus Step (5) is (doubly) violated. Hence this attempt at writing S2 as S1 × S1 violates every Step of the Checklist. In the Exercises you will have the chance to consider some other spaces which are incorrectly expressed as products, and determine which steps of the Checklist are and are not violated.

6. The Topological Uniqueness of Products

In Section 3, we saw that both a cylinder and an annulus can be expressed as the product S1 × I. It may seem odd that two different spaces can be expressed by the same product. If we applied this to products of numbers, it would mean that you could get more than one result by multiplying the same two numbers. This seems ridiculous. We know that 2 times 3 equals 6, no matter whether you do the multiplication on the North Pole, in a Halloween costume, or standing on your head. So what’s the catch? How can we get two different spaces by taking the product of the same two factors? The answer is that your intuition is right. It’s impossible to get different results. Rather, we had mistakenly assumed that a cylinder and an annulus were different; when they are actually topologically the same. In Figure 16, we can see a deformation taking a cylinder to an annulus by pulling open the top circle while keeping the bottom circle fixed. So a cylinder and an annulus have the same intrinsic topology, and there is only one space which describes S1 × I after all. Phew. These spaces are geometrically different, but we’ll get to that later in the chapter.

Figure 16. A cylinder has the same intrinsic topology as an annulus.

In fact, this result is true for all products as we see in the following theorem, whose proof you could see if you take a rigorous course on topology. Theorem 7.1. Any pair of spaces which can be expressed as the product of the same two factors has the same intrinsic topology. Theorem 7.1 is useful for showing that a given space is not a particular product. For example, in the last section we showed that S2 is not the product of an equator and a longitude. However, this proof required ﬁnding at least one step of the Checklist which is violated, and when we were done we could only conclude that 7. THE DIMENSION OF PRODUCT SPACES 167

S2 was not a product of these particular circles. We didn’t know that S2 couldn’t be expressed as S1 × S1 in some other way. Now since we know that a sphere is topologically distinct from a torus, we know immediately from Theorem 7.1 that there is no way that the sphere can be expressed as S1 × S1.

7. The Dimension of Product Spaces

So far, we have only been considering products of two 1-dimensional spaces. You may have noticed that all of spaces that we obtained in this way are 2- dimensional. Also, we know that 1 + 1 = 2 (even if you’re on the North Pole, in a Halloween costume, or standing on your head). It turns out that the same rule of addition of dimensions holds for products of any dimension. We state this in the next theorem, whose proof you might see if you take an advanced course on point set topology. Theorem 7.2. If X is a space of dimension n and Y is a space of dimension m, then the product X × Y is a space of dimension n + m. If it seems strange to you that when we take the product of spaces we add the dimensions rather than multiplying them, think about what we do to exponents when we multiply numbers raised to powers. For example, 102 × 103 is 105 and not 106. This is consistent with the product R × R = R2. According to Theorem 7.2, the product of a 1-dimensional space X and a 0- dimensional space should be a space of dimension 1 + 0 = 1. You might have thought that no space could have fewer than one dimension. In fact, the only 0- dimensional space is a single point y. So we can visualize the product X ×{y},as the copy of X that we described in Step 2 of the Checklist (illustrated in Figure 3 for R2). So it’s not surprising that X ×{y} has the same dimension as X.Thisis true regardless of the dimension of X. Since such a product is just a copy of X,we saythatitistrivial, meaning it’s not interesting. This is similar to the fact that we don’t factor a number n into n × 1. Every number could be written this way, but it wouldn’t be interesting. So we don’t bother doing it. Thus if we want to write a 2-dimensional space as a non-trivial product, we know that each factor must be a 1-dimensional space. However, the only 1- 1 dimensional spaces are S , I, R,andR+ (which represents a ray that starts at 0 and includes all the the non-negative real numbers). Note that we don’t include the ray R− (of non-positive numbers) in our list, because it is topologically the same as R+. Hence the only 2-dimensional spaces which are non-trivial products are obtained as a product of two of these 1-dimensional spaces. So far, we have seen the products R × R, I × I, S1 × I,andS1 × S1. In the Exercises, you will 1 1 construct the products I × R+, S × R, S × R+,andR+ × R+. If we want to construct a 3-dimensional space as a non-trivial product, then one factor must be 2-dimensional and the other factor must be 1-dimensional. In order for our 3-dimensional space to be a manifold, it can’t have any boundary. This means that neither factor can have any boundary. If in addition we want the 3-manifold to have finite volume, then both of the factors have to be finite. In particular, this means the 1-dimensional factor must be S1 and the 2-dimensional space must be a surface with finite area. Recall from Chapter 6 that by the Clas- sification of Surfaces, version 2, every surface with finite area is either S2, nT 2, or nP 2 for some positive integer n. Thus the only 3-manifolds with finite volume 168 7. PRODUCTS OF SPACES that can be obtained as non-trivial products have one of the three forms S2 × S1, nT 2 × S1,ornP 2 × S1. We explain how to visualize S2 × S1 and nT 2 × S1 in the next section. You will explore the product nP 2 × S1 in the Exercises.

8. Visualizing S2 × S1 and nT 2 × S1 As we often do, we begin by tackling an easier problem. In this case, the easier problem is to visualize S2 ×I. Imagine an intact hard-boiled egg where the yolk has completely disappeared without a trace, or possibly never existed. Now suppose that you pierce the egg white all over by toothpicks that go from the outside where the shell would be to the inside where the yolk would begin. In Figure 17, we illustrate some of the horizontal toothpicks, but you should imagine that there is a toothpick piercing through every point from the outer layer towards the center where the yolk would be. This yolkless hard-boiled egg is S2 × I, where the outer layer is the S2-axis, and one of the toothpicks is the I-axis. You should check that this satisﬁes the ﬁve steps of the Checklist.

Figure 17. S2 × I can be visualized as the egg white of a hard- boiled egg that has been pierced everywhere by toothpicks.

Now that we can visualize S2 × I,wecanuseittocreateS2 × S1. Recall that if we glue together the endpoints of an interval, it becomes a circle. So in order to turn S2 × I (which is an S2 of intervals and an interval of S2’s) into S2 × S1 (which is an S2 of circles and a circle of S2’s), we want to glue together the endpoints of each toothpick so that it becomes a circle. Thus we abstractly glue the outer layer oftheeggwhitetotheinnerlayer.Weusetheword“abstractly”heretoemphasize that this gluing cannot be done in R3. In Figure 18, we see A. 3D-Girl who is on vacation in S2 × S1. She trips over a stone feels like she’s losing her balance, so she grabs hold of a nearby stick to steady herself. Unfortunately, at just that moment, someone starts pulling her down by the leg. When she returns from her vacation, she tells all of her friends never to go to S2 × S1 because the people there are so aggressive and rude. Having visualized S2 × S1, we are now ready to imagine nT 2 × S1 for all n ≥ 1. We begin again with an easier visualization, which in this case is to picture a chocolate bunny. Such Easter treats are usually made of a thick layer of chocolate which is hollow inside. The mathematical name for shapes of this form is thickened surfaces. In fact, our yolkless hard-boiled egg is an example of a thickened surface, 8. VISUALIZING S2 × S1 AND nT 2 × S1 169

Figure 18. Someone is pulling A. 3D-Girl’s leg. where the surface is a sphere. Now imagine a thick layer of chocolate in the shape of an n-holed torus, which is empty inside just like a chocolate bunny. Note that most chocolate bunnies have an opening at the bottom where you can stick your finger to make the chocolate bunny into a finger puppet (though if you’re not careful the chocolate bunny could break into pieces or melt while you’re doing your puppet show). Our thickened n-holed torus has no such opening. A thickened n-holedtorusisinfactnT 2 × I.ThenT 2-axisistheouterlayer of chocolate and the I-axis is a segment (or toothpick) perpendicular to the nT 2- axis that could be used to measure the thickness of the chocolate. For example, in Figure 19 we illustrate half of the thickened torus T 2 × I. This half looks like a piece of macaroni. Hence you can imagine T 2 × I either as two pieces of macaroni glued together on their ends or as a bagel made of chocolate which is hollow inside (like a chocolate bunny, but in the shape of a bagel). This description of nT 2 × I is both an nT 2 of intervals (or toothpicks) and an interval of nT 2 (chocolate layers). Again you should check that this description satisfies the steps of our Checklist.

Figure 19. Half of a thickened torus.

In order to get nT 2 × S1, we again want to glue together the endpoints of each interval so that it becomes a circle just as we did to get S2 ×S1 from S2 ×I.Wedo this by abstractly gluing the inside copy of nT 2 to the outside copy of nT 2.Thus at each point of the surface nT 2 there is now a circle where there had been a line segment. It’s as if we glued the inside surface of a chocolate bunny to the outside surface of the chocolate bunny, again assuming the bunny has no ﬁnger hole in the bottom. The problem with such a glued up bunny is that, because it has no boundary, there would be no way to take the ﬁrst bite. Wouldn’t that be frustrating if someone 170 7. PRODUCTS OF SPACES

Figure 20. A. Square cannot give B. Triangle a chocolate sphere, because a sphere can’t be contained in Flatland. gave you a (chocolate bunny) × S1 as a present and you couldn’t even take a bite? But for that matter, nobody could give you such a present because one 3-manifold can’t ﬁt inside of another. A. Square would have the same problem if he wanted to give B. Triangle a chocolate sphere as a present (see Figure 20). Even though a sphere is 2 dimensional, there would be no way that a sphere could ﬁt inside of Flatland unless Flatland itself were the sphere. But then A. Square and B. Triangle would be part of the sphere, and if it were made of chocolate, then a 3-dimensional person might eat Flatland with A. Square and B. Triangle in it. But we have to stop this line of reasoning because it’s becoming entirely too gruesome for a math book.

9. Geometric Products

So far, we’ve learned what a product is and how to correctly and incorrectly construct products. We’ve also learned how to construct and visualize various 2-dimensional and 3-dimensional product spaces. We now want to consider the geometry of products as well as their topology. Once we understand the definition of a geometric product, we’ll go back through some of the examples of 2-dimensional and 3-dimensional products to see if we can construct them in a geometric way. In order to be able to say anything about the geometry of a product, we need to construct the product more carefully than we’ve been doing so far. In particular, we need to add restrictions on the size of the copies of the two factors as well as on the angles between them. Definition. AproductX × Y is said to be geometric if all of the following conditions hold: (1) All subsets of the form X ×{y} have the same size. (2) All subsets of the form {x}×Y have the same size. (3) Every subset of the form X ×{y} is perpendicular to every subset of the form {x}×Y . As a first example, let’s consider the product S1 × I. We saw earlier, that S1 × I could be viewed either as a cylinder or as an annulus . In the cylinder, all of the circles S1 ×{y} are the same size, and all of the intervals {x}×I are the same length (see the left side of Figure 21). Thus conditions (1) and (2) of the definition of a geometric product are satisfied. Moreover, each interval {x}×I is perpendicular to each circle S1 ×{y}, as required by condition (3). Hence, the cylinder is indeed a geometric representation of the product S1 × I. Let’s see what happens when we view an annulus as the product S1 ×I.Inthis case, the intervals {x}×I areallthesamesizeandintersectthecirclesS1 ×{y} 10. GEOMETRIC PRODUCTS OF FLAT SPACES 171

Figure 21. A cylinder is a geometric S1 × I, but an annulus is not. at right angles. Thus, conditions (2) and (3) are satisﬁed. However, as can be seen on the right side of Figure 21, the concentric circles S1 ×{y} are not all the same size, failing condition (1) of the Deﬁnition. Thus topologically an annulus is S1 × I but the product is not geometric.

10. Geometric Products of Flat Spaces

We’d like to see how we can use the geometry of the factors in a geometric product to understand the geometry of the product. Since we understand flat geometry best, let’s look at geometric products of flat spaces first. In Chapter 5, we defined an n-dimensional manifold to be flat if its local intrinsic geometry is the same as that of Rn. In other words, any measurement you can make in an arbitrarily small region of a flat n-manifold would have the same value if you made the same measurement in an arbitrarily small region of Rn. This includes lengths, areas, angles, volumes, etc.. Since we want to be able to consider spaces which have boundaries, we need to extend the definition of flat to n-dimensional spaces that are not necessarily manifolds. Definition. An n-dimensional space is flat if it has the same local intrinsic geometry as a subset of Rn. It follows from this definition that any 2-dimensional space that you can draw on a piece of paper is flat. For example, a square, an octagon, and a rectangle are all flat. However, be careful. Just because these shapes are all flat, doesn’t mean they all have the same local intrinsic geometry. They each have the same local intrinsic geometry as a subset of the plane, but these subsets are not the same. Suppose we want to compare the local intrinsic geometry of a square, an octagon, and a rectangle. If we consider a point in the interior of any of these shapes, a small region around it will look exactly like a comparable small region of R2.So the local intrinsic geometry around interior points of all three shapes is the same. But now let p be one of the corners of the square. Then any small region around p is a 90◦ wedge that looks like a quarter of a pizza (which you could eat as a single slice if either the pizza is very small or you are very hungry). However, there is no point on an octagon which has a region around it that looks like a 90◦ wedge. Rather, a region around one of the corners of an octagon looks like a 135◦ wedge. On the other hand, on a rectangle a small region around a corner is a 90◦ wedge, just like on a square. Thus the local intrinsic geometry of a square and a rectangle are the same, and are both different from that of an octagon. 172 7. PRODUCTS OF SPACES

We saw in Section 6 that there are only four 1-dimensional spaces, and they 1 are R, R+, I,andS . The question is which of these are flat? Certainly R itself is flat, and R+ and I are also flat since they are both subsets of R. But what about S1?SinceS1 is a 1-manifold, it is flat if its local intrinsic geometry is the same as R. This doesn’t mean that S1 has to be contained in R (which it isn’t). Rather, it meansthateverypointofS1 has to have a small region around it whose intrinsic geometry is the same as that of a comparable small region of R. At first this may seem unlikely. Since a round circle is curved, how could it also be flat? One way to see that S1 is indeed flat is to represent it with a gluing diagram which is an interval whose endpoints are glued together. Any point other than an endpoint certainly has the same local intrinsic geometry as R. In Figure 22, we consider a small region around the endpoints of the interval. Since the endpoints p and q are glued together, a region around this glued up point consists of the two grey segments illustrated in the figure. However, the grey segments are glued together at the points p and q. Thus they become a single grey segment, which is identical to the grey segment around the point r on R. Thuseverypointonthe glued up segment has the same local intrinsic geometry as R. Hence this gluing diagram for S1 is flat.

Figure 22. The local intrinsic geometry at the endpoints of the interval is the same as at the point r.

Right now you may be saying “OK sure, a gluing diagram for S1 is flat just like a gluing diagram for a torus is flat, but what about a round circle?” In fact, unlike a torus, we can take the gluing diagram for S1 andglueitupintoaround circle without shrinking it or stretching it in any way. Keep in mind that the gluing diagram for S1 is 1-dimensional—it’s not like gluing up the ends of a piece of string cheese. It may help to recall from Chapter 3 that we can also glue up the diagram for a cylinder without shrinking or stretching it in any way. In fact, we can see this for ourselves by gluing together one pair of opposite sides of a piece of paper. Thus the local intrinsic geometry of a flat gluing diagram for a cylinder is exactly the same as that of a curved cylinder. Now, we can think of our round circle as one of the boundary circles of a piece of paper that we’ve glued up into a cylinder. Thus the local intrinsic geometry of the round circle is exactly the same as that of the gluing diagram. Hence, whether we think of the round circle as a glued up a line segment or as one of the boundary circles of a cylinder, we reach the conclusion that the round circle is flat. This means that in fact every 1-dimensional space is flat. The following theorem (whose proof you might see in a differential geometry course) tells us that taking geometric products of flat spaces is as nice as it could be. Theorem 7.3. A geometric product of flat spaces is flat. For example, in addition to the cylinder in Figure 21, the four spaces illustrated in Figure 23 are geometric products because the horizontal and vertical factors all have the same length and meet at right angles. Hence by Theorem 7.3 we can immediately conclude that each of these spaces is flat. 11. A FLATLAND-FRIENDLY GEOMETRIC S1 × I 173

Figure 23. By Theorem 7.3, these geometric products of ﬂat spaces are ﬂat.

You may recall that we know from Theorem 5.1 in Chapter 5 that a surface is flat if and only if the sum of the angles of any triangle in the surface is equal to 180◦. However, using Theorem 7.3 is easier than checking the sums of angles of every triangle, and applies to spaces of any dimension with boundary as well as to surfaces (remember that we use the word “surface” interchangeably with 2- manifold). In fact, since every 1-dimensional space is flat, the geometric product of any two 1-dimensional spaces is flat. Thus any geometric product which has factors 1 chosen from I, S , R or R+ is flat. How many flat 2-dimensional spaces does this give us?

11. A Flatland-Friendly Geometric S1 × I

A. Square is a little frustrated because he heard us say that S1 × I is flat, but when he pictures it as an annulus, it isn’t flat. Sure he can picture a flat S1 × I using a gluing diagram like the second picture in Figure 23. But he wants to know if there’s a way to interpret his normal picture of an annulus as a geometric product. In fact, there is a way for A. Square to visualize the geometric product S1 × I without a gluing diagram that is reminiscent of how we explained a 3-dimensional cube to him back in Chapter 2. Recall that we drew the “Flatland-friendly” picture of a cube in Figure 24 and told him that the inner square is the same size as the outer square, but farther away in the third dimension. While the third dimension doesn’t really mean anything to him, he was able to suspend his disbelief long enough that he could accept this explanation. As 3-dimensional people we can see Figure 24 as a tunnel with a square cross section, where the smaller square is just deeper inside the tunnel.

Figure 24. A Flatland-friendly drawing of a 3-dimensional cube. 174 7. PRODUCTS OF SPACES

We now show A. Square the annulus in Figure 25 and explain that he can see this as a geometric product of S1 and I by imagining that all of the concentric circles are really the same size, the inner ones are just farther away in the third dimension. In fact, this picture of the annulus is not much diﬀerent from that of the cube. We can see it as a tunnel with a circular cross section, where the smaller circles are deeper inside the tunnel. Furthermore, since parallel circles on a cylinder are geodesics, we tell A. Square that the concentric circles on the annulus are actually geodesics. This means that if two points are on the same circle, then the shortest path between them follows the circle. For example in Figure 25 the grey arc is the shortest path between its endpoints.

Figure 25. The grey arc is a geodesic in this Flatland-friendly illustration of a geometric S1 × I.

It would be hard for A. Square to check that the sums of angles of any triangle in this geometric S1 × I are all 180◦, since the sides of a triangle have to be geodesics. For example, if he considers the grey shape in Figure 26, only one of the three arcs in the boundary looks like a geodesic. But if we draw the same shape on the cylinder, we see that all three arcs are geodesics and hence the shape is indeed a triangle. This will seem quite strange to A. Square. However, he remembers that in Figure 24 the trapezoidal shapes were actually squares, so he accepts that this bulging shape is actually a triangle.

Figure 26. A triangle in the geometric S1 × I.

12. Flat 3-Dimensional Spaces as Geometric Products

In this section, we will learn to visualize flat 3-dimensional spaces which are geometric products of a flat 2-dimensional space and a flat 1-dimensional space. Any 1-dimensional space will do, since all of them are flat. For the flat 2-dimensional 12. FLAT 3-DIMENSIONAL SPACES AS GEOMETRIC PRODUCTS 175 space, we can take any of the geometric products of two 1-dimensional spaces or one of the spaces T 2 or K2 which we already know are flat. As a first example, let’s consider a geometric product of a square X = I × I andanintervalY = I. On the left side of Figure 27, we draw a copy of the square as the horizontal X-axis on the bottom and copies of the interval vertically at every point on the square. On the right side of Figure 27, the vertical interval in the left corner is the Y -axis, and you should imagine that at every point on it there is a horizontal square. As we can see from the figure, the product (I ×I) ×I is a square of intervals and an interval of squares, satisfying all of the steps of our Checklist. Since the product is geometric, by Theorem 7.3 the cube is flat. In fact, we already know a cube is flat because it has the same local intrinsic geometry as a subset of R3.

Figure 27. The geometric product of a square and an interval is a cube.

Observe that we will get the same product if we put the square factor on any one of the faces of the cube, as long as the interval factor is perpendicular to the square. Also, notice that we read the product (I × I) × I as “the product of a square and an interval”. Whereas, we read the product I × (I × I) as “the product of an interval and a square”. In either case, the geometric product is a cube. For our next example, let’s take the flat torus T 2 as our 2-dimensional space and then use any one of the 1-dimensional spaces as the other factor. For example, in Figure 28, we illustrate the geometric product of a flat torus T 2 and an interval I. In both pictures, the torus factor is on the bottom and the interval factor is vertical. At every point in the torus we can visualize a perpendicular interval, and at every point on one of these intervals we can visualize a perpendicular torus. Thus each of the horizontal cross sections of this space is a flat torus that is identical to the one on the bottom.

Figure 28. The geometric product of a ﬂat torus and an interval. 176 7. PRODUCTS OF SPACES

In Chapter 5 we visualized ﬂat 3-manifolds using a cubical gluing diagram with 2-dimensional arrows on the faces. We can do the same thing for any geometric product of a 2-dimensional space whose factors are a square gluing diagram and a 1- dimensional space. For example, we can turn the illustration of T 2 × I in Figure 28 into a cubical gluing diagram with 2-dimensional arrows as follows. Since each horizontal cross section of T 2 × I is a ﬂat torus with the same gluing arrows as the square at the bottom, the front and back faces of the cube must be glued together straight, and the left and right faces of the cube must be glued together straight. Also, since there is an interval perpendicular to each horizontal cross section, none of the endpoints of the intervals should be glued together or to an endpoint of another interval. This means that the top of the cube cannot be glued to the bottom of the cube. Thus the gluing arrows on the cube must be as illustrated in Figure 29.

Figure 29. A cubical gluing diagram for T 2 × I.

Now we can use the gluing diagram for T 2 × I to create gluing diagrams for geometric products of T 2 and other 1-dimensional spaces. For example, the gluing diagram for T 2 × R is like that of T 2 × I, except instead of being a cube with a top and a bottom, the gluing diagram would have the form of an inﬁnite column with a square cross section. This is analogous to the gluing diagram for S1 × R that we saw in Figure 23 which was an inﬁnite strip with an interval as the cross section. 2 Similarly, the gluing diagram for T ×R+ would be a column which went on forever upward but had a square face on the bottom. In Figure 30, we use dotted lines to indicate that the gluing diagram goes on forever in that direction.

1 1 Figure 30. Gluing diagrams for S × R and S × R+ with dotted lines indicating the gluing diagram goes on forever.

The one remaining geometric product which has the ﬂat torus as one factor and a 1-dimensional space as the other factor is T 2 ×S1. In this case, we start with the gluing diagram for T 2 × I, and then we want each interval to become a circle. 13. A GEOMETRIC nT 2 × S1 177

Figure 31. The product T 2 × S1 is T 3.

This means that we want to glue together the endpoints of each of the intervals in Figure 28. As a result, the top face of the cube in Figure 29 must be glued to the bottom face. We can see from Figure 31 that the resulting 3-manifold is the 3-torus T 3. Note that just as we could write the product of a square and an interval as either (I × I) × I or I × (I × I), we can write the product of a torus and a circle as either (S1 ×S1)×S1 or S1 ×(S1 ×S1). Thus we don’t actually need any parenthesis in these products. We can simply say that a cube is I ×I ×I and T 3 is S1 ×S1 ×S1. In the Exercises, you will explore geometric products where one factor is a ﬂat K2 and the other factor is a 1-dimensional space.

13. A Geometric nT 2 × S1

Observe that when we take a geometric product of two homogeneous spaces, the local intrinsic geometry of the product will also be homogeneous. By contrast, in Section 8, we learned to visualize nT 2 ×S1 for n>1 as a thickened n-holed torus made of chocolate where the inside layer is glued to the outside layer. While such a visualization is appealing (as chocolate always is), the geometry of this nT 2 × S1 is hard to analyze. The one thing we can say is that it’s not homogeneous, since the curved n-holed torus itself is non-homogeneous. However, this is not the only way to visualize nT 2 × S1. In Chapter 6, we learned that for n>1 we can obtain nT 2 abstractly as a glued up 4n-gon where all of the vertices are glued together into a single anti-cone point. While this glued up surface still isn’t homogeneous, it’s only “bad” at the anti-cone point and otherwise it’s flat. So let’s see if we can understand the geometry of a geometric product of this nT 2 and an interval. We illustrate the geometric product 2T 2 × I in Figure 32. You should imagine that there is an identical octagonal gluing diagram at every horizontal level of the product as there was an identical square gluing diagram at every horizontal level of Figure 28. Let v denote the anti-cone point of the glued up 4n-gon. We saw in Section 7 that the product of any space X with a single point is just a copy of X.Thusthe product of the anti-cone point v with an interval is an interval. It follows that the product nT 2 × I contains the interval {v}×I of anti-cone points and otherwise it’s flat. What we mean by an interval of anti-cone points is that for any point x on I, the cross section of nT 2 × I of the form nT 2 ×{x} contains the anti-cone point (v, x) which has 180◦(4n − 2) degrees around it (you may want to see if you can prove that this is indeed the sum of the angles of a 4n-gon). So we can think of {v}×I as an “anti-cone interval” inside of the geometric product nT 2 × I.It’sthe only “bad interval” in nT 2 × I and otherwise nT 2 × I is flat. 178 7. PRODUCTS OF SPACES

Figure 32. The product 2T 2 × I.

As we saw in the last section, to go from the product of a surface and an interval to the product of a surface and a circle, we simply glue together the top and bottom face of our 3-dimensional gluing diagram. Thus we can obtain the geometric product nT 2 × S1 by gluing together the top and bottom 4n-gon in nT 2 ×I (illustrated for the octagon in Figure 33). This will give us nT 2 ×S1 which is ﬂat everywhere except along an “anti-cone circle”, which we obtain by gluing together the endpoints of the “anti-cone interval” in nT 2 × I.

Figure 33. The product 2T 2 × S1 is ﬂat except for an anti-cone circle along the glued up vertices.

We saw in Chapter 6 that we can obtain every finite surface as a glued up polygon which is flat except for at some number of cone points or anti-cone points. Thus we can take the geometric product of the gluing diagram for any finite surface and S1 to get a 3-manifold which has flat geometry everywhere except at a finite number of cone circles or anti-cone circles. On one hand, such 3-manifolds are nice because they are flat almost everywhere. But on the other hand, such 3-manifolds are not nice because they’re not homogeneous. If we started with a homogeneous surface and took a geometric product, we would end up with a homogeneous 3-manifold. In the next chapter we’ll learn how to create homogeneous geometries for nT 2 and nP 2. 14. A GEOMETRIC S2 × S1 179

14. A Geometric S2 × S1

We would now like to create S2 × I and S2 × S1 as geometric products. We begin by recalling from Section 8 that we visualized S2 ×I as a yolkless hard-boiled egg (illustrated in Figure 17). Since we want all of the layers of our S2 × I to have homogeneous geometry, let’s imagine that the hard-boiled egg is perfectly round so that each of the layers from the outside to the inside is geometrically a sphere. This is nice, but it’s still not a geometric product, since the concentric spheres get smaller as we go towards where the yolk would be. Before we try to visualize S2 × I as a geometric product, let’s remember that when we taught A. Square to picture S1 × I as a geometric product, we started by reminding him of the Flatland-friendly picture of a cube. So we begin our discussion of a geometric S2 ×I by recalling our drawing of a 4-dimensional cube from Chapter 2 (see Figure 34). In particular, since we have no 4-dimensional perspective, we had to accept that the inner cube is the same size as the outer cube but farther away in the fourth dimension. In fact, the 3-dimensional cross sections which are parallel to the outer cube are all cubes of exactly the same size.

Figure 34. A 3-dimensional drawing of a 4-dimensional cube.

Now we want to think of the geometric product S2 × I in the same way. In particular, we imagine that, in spite of the way it looks, our yolkless hard-boiled egg is spherical and all of the concentric spheres from the outside to the inside are exactly the same size. When we think of S2 × I in this way, we have to accept that if two points are on the same spherical layer, then the geodesic between them will be part of a great circle contained in that sphere, rather than an arc which cuts across many spherical layers. Thus the edges of a triangle whose vertices are all on the same spherical layer must all be arcs of great circles contained in that sphere (see Figure 35). This means that the sum of the angles of such a triangle will necessarily be greater than 180◦. In particular, this shows that the geometry of S2 × I is not ﬂat. But this is not a surprise, since we knew that S2 was not ﬂat. By contrast with our spherical cross sections, we now consider a horizontal cross section of the geometric S2 × I. In the drawing on the left of Figure 36, this cross section looks like an annulus. However, since all of the concentric spheres in S2 × I are actually the same size, all of the circles in this cross section should also be the same size. Thus we can either visualize this annulus the way that A. Square imagines a geometric S1 × I, or we can visualize it as a cylinder rather than as an 180 7. PRODUCTS OF SPACES

Figure 35. The sum of the angles of this triangle in a spherical layer of S2 × I is greater than 180◦.

Figure 36. A horizontal cross section of a geometric S2 × I is a geometric S1 × I. annulus. Either way, this means that the sum of of the angles of every triangle in this cross section must be 180◦. Thus at any point in S2 ×I, the local geometry is ﬂat if you look in the direction of a cross section of the form S1 × I (as in Figure 36). Whereas, the local geometry is spherical if you look in the direction of a spherical cross section (as in Figure 35). In other words, the local geometry at a point depends on the direction in which you look. We’ll discuss this strange phenomenon in the next section. But in the meantime, let’s create the geometric product S2 × S1 by abstractly gluing the inside and outside spheres of our geometric S2 ×I. This means that each line segment (which could be a toothpick measuring the thickness of the egg white) becomes a circle of S2 × S1. As we saw earlier, the geometry of S2 is homogeneous and the geometry of S1 is homogeneous, and their product is geometric. Thus we end up with a nice homogeneous, geometric representation of the 3-manifold S2 × S1, even if there is something a little strange going on when you look in diﬀerent directions.

15. Isotropic and Non-isotropic Spaces

In order to better understand how the geometry of a space can change depending on the direction in which you look, let’s suppose that we are talking about the amount of illumination in a region rather than the local geometry of a region. Suppose that you are in the dessert at noon with the sun overhead, there is not a cloud in the sky, and there are no houses or trees to create shadows. No matter where you walk, the amount of light is the same. In this sense you could say 16. EXERCISES 181 that the illumination is “homogeneous”. However, if you look directly up at the sun, the light will be more intense than if you look straight ahead; and if you look straight ahead there will be more light than if you look down at your shoes. In this situation, you could say that the amount of light that hits your eye depends on the direction in which you look. While this seems normal for illumination, it might seem strange to live in a universe where the geometry changed when you looked in different directions. It order to talk about such strange spaces more easily, we make the following definition. Definition. A space is said to be isotropic if the local geometry at every point is the same in all directions. The word isotropic comes from the Greek words iso (meaning equal) and tropos (meaning direction). Next time you are out with a friend on a cloudless day, you could remark that the amount of light is “homogeneous but not isotropic”. Your friend could either be impressed with your vocabulary or think you’re crazy. But either way, it would give you a chance to talk about the math you’re learning. We saw in the last section that the geometric product S2 × I is not isotropic because if you look in the direction of a spherical layer the geometry is different than if you look towards the inside of the egg white. Also, the local geometry at a point on the inside sphere or outside sphere of S2 × I will be different from that of a point in the interior of S2 × I because such points are in the boundary of the space. Thus S2 × I is neither isotropic nor homogeneous. On the other hand, we saw in the last section that the geometric S2 × S1 is homogeneous, and since the geometric S2 × I was not isotropic, it now follows that the geometric S2×S1 is not isotropic. This means that at any point in S2×S1 if you look in different directions, the geometry is different, but this strange directional geometry will be exactly the same at every point in the manifold. The idea of a manifold which is homogeneous but not isotropic is indeed strange. This is quite different from T 3, K3, and the other flat 3-manifolds that we studied in Chapter 5. In particular, every flat 3-manifold is isotropic as well as homogeneous, since its local geometry is the same as R3. Even the 3-dimensional sphere S3 turns out to be isotropic as well as homogeneous.

16. Exercises

1. Show that an inﬁnite cylinder can be expressed as the product of S1 and R.

2. Recall that we use the symbol R+ to represent the ray consisting of all real numbers greater than or equal to 0. Describe the following 2-dimensional products: 1 I × R+, S × R+, R+ × R+. 3. How many topologically distinct 2-dimensional spaces are there which are products of two 1-dimensional spaces? How many topologically distinct 2-manifolds are there which are products of two 1-dimensional manifolds? 4. List all 2-manifolds with ﬁnite area which can be written as a non-trivial product. Explain how you know you have them all. 182 7. PRODUCTS OF SPACES

5. Explain what would go wrong if you tried to express a Möbiusstrip as S1 × I where the S1-axis is a circle other than the center circle of the Möbius strip and the I-axis is the vertical interval illustrated in black on the right in Figure 14. 6. Explain why neither S2 nor a Möbius strip can be a product of two 1-dimensional spaces. 7. Could a pentagon be expressed as I × I with all of the segments straight as illustrated in Figure 37. If not, name a step of the Checklist which is violated.

Figure 37. Illustration for Exercise 7.

8. Illustrate a triangle and a hexagon as the product of two intervals. Then explain why neither of these products is geometric. 9. Are the descriptions of nT 2 × I and nT 2 × S1 in Section 8 geometric? Explain why or why not. 10. In this problem we want to determine what goes wrong if we try to express a plane as a product of a circle and a line as illustrated in Figure 38. In particular, list all of the steps from the Checklist which are violated and explain how.

Figure 38. Illustration for Exercise 10.

Note that the axes in Figure 38 are in bold black. Besides the S1-axis, which has radius 1, we’ve drawn parallel circles with radii 2 and 3. Similarly, we’ve drawn a few lines parallel to the R-axis. Even though the ﬁgure includes only a few circles and lines, you should imagine that there are circles of every radius and a vertical line through every point. 11. Suppose S = X × Y and a and b are distinct points of X. Prove that {a}×Y and {b}×Y are disjoint. 16. EXERCISES 183

12. Create a sphere as a glued up square. Then describe the geometry of a geometric product of your glued up square and S1. In particular, discuss where it’s flat and where it’s not flat. 13. Draw a picture of a geometric nP 2 × S1 which is flat everywhere except along some number of circles. How many circles of cone points or anti-cone points does the space have? 14. Describe the geometric products (S1 × I) × S1 and (S1 × S1) × I, and illustrate each with a cubical gluing diagram. Are the two products the same or different?

15. Draw a picture of what you would get if you physically glued up the faces of the gluing diagram in Figure 29. 2 2 2 2 1 16. Draw cubical gluing diagrams for K × I, K × R, K × R+,andK × S .Is one of the spaces a familiar flat manifold? 2 2 17. Describe the following 3-dimensional products: S × R+, S × R, R+ × R+ × R+. 18. Imagine that the picture on the left in Figure 36 is supposed to be a geometric S2 × S1 rather than a geometric S2 × I. Describe the cross section illustrated in the picture. Is the cross section flat? 19. Can you find a manifold other than S2 × S1 which is homogeneous but not isotropic? 20. Can you find a manifold which is isotropic but not homogeneous? 21. Let A be a surface which is not flat, and suppose that A × S1 is a geometric product. Prove that A × S1 is not flat. 22. Suppose that we live in a universe that we knew was one of S2 × S1, P 2 × S1, or T 3. How could we tell the difference between these three spaces? 23. Is the product operation distributive over the connected sum operation? In other words, for any spaces X, Y ,andZ,isitalwaystruethatX × (Y #Z)= (X × Y )#(X × Z)? Give a proof or counterexample.

https://doi.org/10.1090//mbk/096/08

CHAPTER 8

Geometries of Surfaces

Topics: • Euclid’s axioms • Spaces that don’t satisfy one of the axioms • Alternative axioms • Spherical geometry • Areas of disks in diﬀerent surfaces • Hyperbolic geometry • Helping A. Square visualize a hyperbolic plane • Homogeneous geometries for all ﬁnite surfaces

1. Euclid’s Axioms

Recall that a surface is flat if it has exactly the same local intrinsic geometric properties as a plane. This means that all flat surfaces are homogeneous and isotropic because the plane is homogeneous and isotropic. Also, all triangles in a flat surface have angle sum equal to 180◦ because all triangles in the plane have angle sum equal to 180◦. With all of our attention on the local geometry of flat surfaces, we might assume that the only global geometric property which distinguishes a plane from other flat surfaces is that it has infinite area. But we saw in Chapter 3 that an infinite cylinder is flat, yet it also has infinite area. In order to distinguish the global geometry of an infinite cylinder from that of a plane, we could observe that some of the geodesics on a cylinder are finite whereas no geodesics on the plane are finite. Perhaps what distinguishes the global geometry of a plane from all other flat surfaces is that its area is infinite and all of its geodesics are infinite. Let’s consider the geometry of the surface we get by removing a single point from the plane. We can’t analyze the local geometry at the missing point, because the missing point isn’t part of the space. If we look at the local intrinsic geometry at any other point, it’s still exactly the same as it is at any point in the plane. In particular, no matter how close a point is to the hole, there is a region around it which is small enough that it doesn’t contain the hole. This means that the local intrinsic geometry of a region around any point in our surface is identical to that of a region in the plane. Hence the plane with a point missing is also flat. Also, all of its geodesics are infinite, just like the geodesics in the plane. So what distinguishes the global geometry of a plane with a hole from that of a plane without a hole?

185 186 8. GEOMETRIES OF SURFACES

We could keep coming up with global geometric properties that enable us to distinguish one flat surface after another from a plane, but why do all this work when it’s been done for us more than two thousand years ago? Euclid, who was a philosopher around 300 BC, wrote a list of five axioms that completely describe the local and global geometry of a plane. All other statements about the plane, such as the fact that the sum of the angles of a triangle is 180◦ or the fact that a plane has infinite area, can be proven from just these five axioms. We now consider Euclid’s axioms one by one to understand what they mean and to get a sense of what sorts of spaces don’t satisfy them. Axiom 1. For any two distinct points, there is a geodesic that passes through both points. Even though Axiom 1 seems quite innocuous, it is easy to find spaces which do not satisfy it. In fact, A. Square found such a space. Enticed by an ad that promised a “once-in-a-lifetime trip to a far away and exotic place”, A. Square took a vacation to the plane with the point at the origin missing. When he asked at the hotel desk if they could tell him the shortest path to the beach, the person replied, “No. There is no shortest path to the beach”. At first he thought the person was making a joke. But as he found out in Figure 1, every path from the hotel to the beach has to go around the origin, and there is no single path which is shorter than all others. Thus the plane with the origin removed is an example of a space where Axiom 1 fails.

Figure 1. There is no geodesic that A. Square can take to get to the beach.

Axiom 2. Any geodesic can be extended to a geodesic line. Note that in this chapter we use the term geodesic line for a geodesic which doesn’t terminate in either direction. On a plane the geodesic lines are straight lines, whereas on a sphere the geodesic lines are great circles. We’ll use the term geodesic segment for a geodesic that terminates in both directions, and geodesic ray for a geodesic which terminates in one direction but not in the other. On a plane the geodesic segments are line segments and the geodesic rays are rays that go off arbitrarily far in one direction but not the other. On a sphere the geodesic segments 1. EUCLID’S AXIOMS 187 are arcs of great circles, and there are no geodesic rays. The term geodesic will be used for any geodesic line, geodesic segment, or geodesic ray. Certainly in a plane, any geodesic segment or geodesic ray can be extended to get a geodesic line. On the other hand, as A. Square discovers on his next vacation, not all 2-dimensional spaces satisfy Axiom 2. A. Square had not intended to take a vacation over Thanksgiving because the holiday is too short and it’s considered the heaviest travel period in the year. However, he saw an ad that caught his eye: Tired of pointless vacations? Want a vacation that includes all of the positives and none of the negatives? Love chocolate? If you answered yes to all three of these questions, then Chocotopia is your ideal vacation destination. This ad seemed like it was targeted directly at him. He was willing to fight the crowds to spend four full days at Chocotopia. Chocotopia was located on a half- plane, which seemed fine. At least it didn’t have any points missing! He was so excited that he booked a package deal which included all of the chocolate you could eat. Almost immediately after he arrived, he began helping himself to small samples of chocolate bars, chocolate mousse, brownies, chocolate truffles, chocolate sundaes, and chocolate chip cookies. But he saved his favorite—chocolate cupcakes—for last. Unfortunately, the chocolate cupcake that he ate on top of all the samples caused a bit of indigestion. Based on past experience, he decided that taking a long walk would be the perfect remedy. Unfortunately, as we see in Figure 2, after walking only a short way, he was unable to continue walking in the same direction. Even veering off slightly, was no better. Resigned, he went back to the hotel to take a nap and rest up for his next chocolate opportunity.

Figure 2. A. Square can’t continue walking in this direction.

Axiom 3. Given any point a and any positive number r, there is a circle of radius r with center a. In order to understand this axiom, we need to be clear about what we mean by a circle. Euclid defined a circle as follows. Definition. A circle with center a and radius r is defined to be a single curve which bounds a region containing the point a such that the distance from a to every point on the curve is equal to r. We will see as follows that an infinite cylinder violates Axiom 3. In particular, we show that it does not have circles of every possible radius. 188 8. GEOMETRIES OF SURFACES

Suppose that c is the circumference of the infinite cylinder, let a be a point c in the cylinder, and choose r>2 . On the left side of Figure 3, we illustrate a gluing diagram for the infinite cylinder in which we have attempted to draw a circle centered at a with radius r. However, since the circumference of the infinite c cylinder is the width of the gluing diagram and r>2 , the diameter of the circle we are trying to draw is larger than the width of the gluing diagram. Thus our “circle” will cross over the gluing line and intersect itself as we see on the left side of the figure.

Figure 3. Thesetofpointswhosedistancefroma is r consists of two curves which together bound the grey region.

Now observe in Figure 3 that the shortest paths from a to x and from a to y are along the dotted segments, though the segment going from a to x crosses over the gluing line so it looks like it’s in two pieces. The lengths of these segments are shorter than r, and hence the distance from a to the points x and y is less than r. Similarly, we see that of all the points on the vertical eye shape except for the top and bottom points are less than r away from a. Hence the points which are precisely r units from a are those lying on the two grey curves which together bound the grey region illustrated on the right in Figure 3. Since neither of these curves by itself bounds a region, neither curve satisfies Euclid’s definition of a circle. Thus the infinite cylinder does not have a circle with radius r and center a, and hence violates Axiom 3.

Axiom 4. All right angles are equal. In order to understand this axiom, we need to be clear about what we mean by a right angle and what we mean by two angles being equal. We define the second one first. In particular, we say that two angles in a surface are equal if there is a rigid motion of the surface taking one to the other. This means that even if the two angles are made of steel rods which are completely rigid, you could still move one to the other. It may seem like it’s not necessary to define a right angle, since we all know that right angles are those that have 90◦. This definition of a right angle works just fine for the plane, but we want to consider right angles on other surfaces where it might not be so easy to measure 90◦. So instead we define a right angle as the angle formed when two geodesics intersect in a T-shape so that the two resulting 1. EUCLID’S AXIOMS 189 angles are equal. This definition makes sense in any surface since we know what it means for two angles to be equal. In Figure 4, we see two T-shapes. In each case, the angles on either side of the T are equal, and hence they are right angles.

Figure 4. Two right angles in the plane.

Thus Axiom 4 is saying that we can move any right angle to any other right angle by a rigid motion. You may recall from Chapter 3 that two objects in a space have the same extrinsic geometry if and only if there is a rigid motion of the space taking one object to the other. If we think of a right angle is an object in our space, then we can restate Axiom 4 as saying that any pair of right angles have the same extrinsic geometry. After his stressful vacations on the plane with a point missing and the half- plane, A. Square decides that he doesn’t want to waste his next vacation going to some other part of a plane. The following year he sees an ad that says, “Build your own windsurf board, and learn how to sail on the beautiful blue seas of a curved torus”. The idea of building a windsurf board and sailing it in the ocean sounds great, and a curved torus is a complete surface without missing parts or boundaries, so what could possibly go wrong? A. Square’s first day on the torus is great. Listening to the surf crashing in the background, he eagerly builds his board and attaches the mast. He finds it a bit odd that the materials for the mast and board seem quite pliable. When he asks about this, the instructor responds by saying “the materials need some flexibility so they can communicate with different regions of the ocean”. This doesn’t make any sense to A. Square, since a windsurf board can’t talk like a square or a triangle can. But his instructor has a thick toroidal accent, and A. Square is too embarrassed to ask him to repeat. The instructor does manage to convey to A. Square that he should keep the mast at a right angle to the board at all times, which A. Square promises he will do. After his instructor checks that his windsurf board is sturdy and ocean worthy, A. Square rushes to the beach to give it a whirl. For a while, he is whipping through the waves having a great time, but just as the sea seems to get calmer, he finds that he is having a harder and harder time staying on the board. Suddenly he realizes that the problem is that the board has changed from being curved towards his body to being curved away (see Figure 5). No wonder he keeps sliding away from the mast. He can’t figure out what he did wrong, since the mast and board are still at right angles. Scared that he might fall off and get eaten by sharks, A. Square quickly sails back to shore. He is about to call the instructor over to show him what happened to the board, when he notices that in fact the board is exactly the way it was when he built it. He had so much fun sailing, that he shrugs it off as an illusion created 190 8. GEOMETRIES OF SURFACES

Figure 5. A. Square’s windsurf board has changed from being curved towards his body to being curved away. by too much sun. A. Square didn’t realize it at the time, but the “diﬀerent regions of the ocean” that his instructor referred to were what caused his boat to bend, and what causes the curved torus to violate Axiom 4.

Axiom 5. Given any point a and geodesic line L not containing a,thereis exactly one geodesic line containing a and that does not intersect L.

Figure 6. In a plane, there is always a line through a parallel to L.

Axiom 5 is illustrated for the plane in Figure 6. The above statement of Axiom 5 is not in the original form stated by Euclid. Rather it is a simpler version of the axiom given by John Playfair in the fifth century AD. To see that Axiom 5 doesn’t hold for a sphere, we need look no further than A. Square’s next vacation. He enjoyed windsurfing so much on the torus that this year he’s eager to take another windsurfing vacation. After hearing that the windsurfing on a sphere is out of this world, he buys a ticket and packs a bag. When A. Square rents his windsurf board on the sphere, he has to sign a statement that he won’t cross into the roped off swimming area. Anyone caught sailing over the rope has to pay a hefty fine. He paddles with his board to a few feet outside of the swimming area and plans to sail parallel to the rope. Once he begins sailing, he notices that there’s a good tailwind which is pushing the board straight along a geodesic. Mistakenly, he figures that he doesn’t have to worry about crossing into the swimming area because the tailwind will keep him going along the black circle in Figure 7. He’s having a great time riding the waves, happy that he’s having no trouble staying on his board. 2. FLAT SURFACES AND EUCLIDEAN GEOMETRY 191

Figure 7. A. Square assumes the tailwind wind is pushing him along the black circle.

Figure 8. The tailwind actually pushes him along a geodesic which crosses into the swimming area.

But he suddenly notices that the geodesic he’s following is taking him straight into the rope rather than parallel to it, as illustrated in Figure 8. In a panic, he quickly capsizes his windsurf board just as it hits the rope. Fortunately, the lifeguard was so busy ﬂirting with a cute pentagon that she didn’t notice A. Square’s close call.

2. Flat Surfaces and Euclidean Geometry

If a surface satisfies Axioms 1–5, then we say that it has Euclidean geometry or simply that it’s Euclidean. For example, as you probably learned when you took geometry, the plane satisfies Axioms 1–5 and hence is Euclidean. This is no surprise since Euclid’s axioms were intended to describe the geometry of the plane. On the other hand, a space violating one or more of the Axioms is not Euclidean. So none of A. Square’s unfortunate vacation destinations is Euclidean, which was why they each had such strange geometric properties. Since some of these non- Euclidean surfaces are flat and others aren’t, we might think that being flat and 192 8. GEOMETRIES OF SURFACES being Euclidean are completely independent characteristics of surfaces. However, we see as follows that this is not the case. In Theorem 5.1 from Chapter 5, we saw that a surface is flat if and only if the sum of the angles of any triangle in the surface is equal to 180◦.Nowwehave the following theorem which relates the sum of the angles in a triangle to Euclid’s Axioms. Theorem 8.1. Let S be a surface which satisfies Axioms 1–4. Then S satisfies Axiom 5 if and only if the sum of the angles of any triangle in S is equal to 180◦. This result together with Theorem 5.1 shows that every Euclidean surface is flat and every flat surface which satisfies Axioms 1–4 is Euclidean. Keep in mind that there are examples like the plane with a point removed, the half-plane, and an infinite cylinder which are flat but violate at least one of Axioms 1–4. In the Exercises, you will explore some other examples of flat surfaces which violate one or more of the axioms. In addition to the property of Euclidean surfaces that the sum of the angles of a triangle is 180◦, you may recognize some other properties of Euclidean surfaces which have names like the Pythagorean Theorem or the Side-Angle-Side Theorem, and you may even have proved them in a geometry class. We won’t present the proofs of these results or of Theorem 8.1, but you can read about them and other properties of Euclidean surfaces in a textbook on Euclidean geometry.

3. Some Alternative Axioms

In many languages the word for the article “a” and the word for the number “one” are the same. If we replace the word “a” by the word “one” in Axiom 1, then Axiom 1 would become the following. Alternative Axiom 1: For any two distinct points, there is one geodesic that passes through both points. Both Axiom 1 and Alternative Axiom 1 are true for the plane. But not all spaces which satisfy Axiom 1 satisfy this stronger alternative version of the axiom, as A. Square discovered when he and his twin brother B. Square went to the flat torus for their winter vacation. A. Square and B. Square are mirror images of one another, which has left some people wondering if they were identical at birth and then somehow one took an orientation reversing path without the other. The Square brothers spent most of their vacation on the flat torus building elaborate snow forts and having snowball fights. In Figure 9, A. Square realized that if he positioned himself exactly halfway around the flat torus from his brother, he could throw two snowballs at once to hit the front and back of B. Square’s head at exactly the same time. He thought this was a great discovery, but B. Square didn’t. In fact many books on Euclidean geometry use Alternative Axiom 1 instead of Axiom 1. Euclid himself assumes in some of his proofs that there is a unique line through any pair of points. Thus according to many historians it is likely that Euclid intended to include uniqueness in the statement of Axiom 1. Axiom 5 is the axiom which has gotten the most attention from mathematicians and historians because it is regarded as the least self-evident of the axioms. For 4. SPHERICAL TRIGONOMETRY 193

Figure 9. A. Square throws snowballs to hit B. Square from the back and front simultaneously. hundreds of years, mathematicians tried unsuccessfully to derive Axiom 5 from the other four axioms. We now know that this is impossible. In fact, there are two ways that Axiom 5 can be violated. To understand these two ways, let’s see what happens when we negate Axiom 5. In particular, consider the statement: It is not the case that, given any point a and geodesic line L not containing a, there is exactly one geodesic line containing a and that does not intersect L. We could say this more simply as follows. There is a point a and geodesic line L not containing a such that it is not the case that there is exactly one geodesic line containing a and that does not intersect L. This statement implies that either there is no geodesic line containing a that is disjoint from L or there is more than one such line. In the ﬁrst case, every line through a intersects L. In the second case, there are multiple lines through a that do not intersect L. We refer to these two possibilities as “First Alternative Axiom 5” and “Second Alternative Axiom 5”. For the rest of this chapter, we focus on spaces in which one of these two alternatives to Axiom 5 is satisﬁed.

4. Spherical Trigonometry

As A. Square saw in Figure 8, a sphere does not satisfy Euclid’s fifth axiom. In fact, every pair of great circles on a sphere intersects in two points. In Figure 10, we illustrate a great circle L and a point a disjoint from L.Aswecanseeinthe figure, any great circle containing a will intersect L. Thus a sphere satisfies the following alternative axiom. First Alternative to Axiom 5. Given any point a and any geodesic line L not containing a, every geodesic line through a intersects L. Because a sphere satisfies this alternative to Axiom 5, the geometry of a sphere is quite different from that of a plane. For example, recall from Chapter 3 that the sum of the angles of any triangle in a sphere is more than 180◦; and the bigger 194 8. GEOMETRIES OF SURFACES

Figure 10. There is no great circle containing a which doesn’t intersect L. the triangle is, the greater the angle sum. Consider a spherical triangle whose angles are θ1, θ2,andθ3, and whose complementary angles are ϕ1, ϕ2,andϕ3,as illustrated in Figure 11. Recall from Chapter 5 that θ is a Greek letter pronounced “theta”, which rhymes with the cheese feta. The Greek letter ϕ is “phi”, which is pronounced “fee” and sometimes “ﬁe”, but never “fo” or even “fum”.

Figure 11. Two complementary triangles in a sphere.

Notice that the three geodesic segments in Figure 11 actually bound two triangles: one on the inside with angles θ1, θ2,andθ3, and one on the outside with angles ϕ1, ϕ2,andϕ3. We say that this pair of triangles are “complementary”. But be careful never to confuse the complementary triangles in Figure 11 with the complimentary triangles in Figure 12.

Figure 12. Two complimentary triangles.

What happens to the angle sum ϕ1 + ϕ2 + ϕ3 of the outer triangle if the angles ◦ θ1, θ2,andθ3 of the inner triangle get smaller? Since there are 360 around every 4. SPHERICAL TRIGONOMETRY 195

◦ point, we see that the angles in the complementary triangle are ϕ1 = 360 − θ1, ◦ ◦ ϕ2 = 360 −θ2,andϕ3 = 360 −θ3. We know that θ1 +θ2 +θ3 will always be bigger than 180◦, but if we make the triangle smaller and smaller, this sum will get closer ◦ ◦ and closer to 180 . Ifwedothis,thesumϕ1 +ϕ2 +ϕ3 =3×360 −(θ1 +θ2 +θ2)= ◦ 1080−(θ1 +θ2 +θ2)getsverycloseto900 . So the angle sum of a spherical triangle is always between 180◦ and 900◦. A. Square loves having snowball ﬁghts with his brother B. Square. Even though they don’t realize it, they are unconsciously using trigonometry all the time. When A. Square aims a snowball at B. Square, A. Square’s brain must ﬁrst determine the distance between them. A. Square looks at B. Square, and his eye registers the angle θ formed by the light rays coming from the top and bottom of B. Square. In Figure 13, this angle θ is 45◦.

Figure 13. A. Square sees B. Square in a ﬂat universe.

The light rays from the top and bottom of B. Square together with the edge of B. Square’s face form a triangle. A. Square’s brain uses the angle θ together with trigonometry to estimate how far to throw a snowball. Your brain does this all the time without you being aware of it. This happens even if you hated trigonometry, swore you’d never use it, and couldn’t tell a sohcahtoa from a cohsahtoa if you met one walking down the street. If A. and B. Square are in a flat surface as they do in Figure 13, then the sum of the angles of any triangle is 180◦. Hence the angle between the bottom light ray and B. Square’s face must be 45◦, since the triangle is a right triangle and θ =45◦. Using Euclidean geometry, we can now deduce that the triangle in Figure 13 is isosceles. This means that the length of the top light ray is exactly equal to the height of B. Square. Since A. Square is the mirror image of B. Square, A. Square knows B. Square’s height like the back of his hand (or better yet, like the front of his face), and hence also the distance between them. A. Square’s brain cranks out computations nearly instantaneously and A. Square launches his snowball before B. Square even notices that A. Square has a snowball in his hand. Our brains make similar computations when we throw a ball, a snowball, or even a frisbee. However, the same calculations wouldn’t apply if A. Square and B. Square were on a sphere; which is exactly where they go for their next vacation. After listening to A. Square talk endlessly about his great windsurfing vacation on the sphere (without mentioning his close call with the swimming area), B. Square wants to see the sphere for himself. Though B. Square pretends to be disappointed that the only time they can go to the sphere is over winter break, he’s actually eager to go 196 8. GEOMETRIES OF SURFACES somewhere snowy so that he can get back at A. Square with his own well-placed snow balls. As soon as they check in at their hotel on the sphere, they run outside to have a snowball fight. We see in Figure 14 that the light rays from the top and bottom of B. Square together with B. Square’s face form a triangle as they did before. Also, again the angle between the top light ray and B. Square’s face is a right angle and θ =45◦. But, in this case, the sides of the triangle are arcs of great circles rather than line segments, and the triangle has two right angles in addition to the 45◦ angle. Furthermore, the geodesic segment between A. Square and the top of B. Square’s head is almost half of a great circle, whereas B. Square’s height is only a quarter of a great circle. This means that the distance between A. Square and B. Square is almost twice the height of B. Square.

Figure 14. A. Square sees B. Square in a sphere.

In A. Square’s eagerness to throw the ﬁrst snowball, his brain uses Euclidean trigonometry to aim. As a result, A. Square thinks that the distance between them is the same as B. Square’s height, and his snowball lands only halfway to its intended target. B. Square teases him about this for the rest of their vacation on the sphere.

5. The Area of a Disk in a Sphere

After their snowball fight, the Square brothers see an ice-skating rink in the distance. Thinking that maybe ice-skating will be less frustrating than a snowball fight, A. Square rushes up to the entrance of the rink. They don’t see where to go to rent ice-skates, and after some arguing about which way it should be, they decide to go left. The rental counter was actually to the right of the entrance, and hence they have to go all the way around what seems to be a fairly small ice-skating rink to get there. With skates and tickets in hand, they are about to enter the rink and start skating when A. Square sees a sign that says “8 skaters in the rink”. Always an ace at math, A. Square is eager to figure out the area of the skating rink in order to know if it will be crowded or not. Since A. Square’s edges are straight and he is a bit self-centered, he declares that his height and his width are each 1 unit. B. Square is perfectly happy to use A. Square’s edges as the unit of measurement because he has exactly the same dimensions as his brother. This means that the area of each of the brothers is 1 square unit, which sounds perfect to the Square brothers. 5. THE AREA OF A DISK IN A SPHERE 197

Remembering the geometry he learned in high school, A. Square tells his brother that since the rink is circular with radius r, its circumference is given by C =2πr and its area is given by A = πr2. He then solves the circumference equation for r to get C r = . 2π From their walk around the rink, A. Square guesses that the circumference of the rink is 10 units. He then plugs C = 10 into the equation for the radius, and computes the radius of the rink to be 10 5 r = = . 2π π Next, he plugs this value for r into the equation A = πr2, and finds that the area of the rink is 5 2 25 A = πr2 = π = . π π Using his calculator he finds that the area of the skating rink is roughly equal to 8 square units (you should check A. Square’s computation with your own calculator). If there are 8 skaters in the rink, each of which has area equal to 1 square unit like the Square brothers, then there would be no room to even enter the rink. If the 8 skaters are somewhat thinner than the Square brothers, then they might be able to enter the rink, but skating with them would still be like sardines skating together in a can. Proud of his geometric prowess, A. Square declares that they should go do something else a bit roomier, like watch the snow fall. Just as they are about to give up on skating, B. Square thumps his forehead and then thumps A. Square’s forehead for good measure. A. Square has made a mistake in his calculations! (Can you see where?) A. Square mistakenly used Euclidean rather than spherical geometry to find the area of the rink. A circle of circumference 2πr encloses an area of πr2 in a plane, but the same circumference encloses more area in a sphere. For example, suppose that the skating rink takes up exactly one hemisphere. Looking at Figure 15, we see that the area of the top hemisphere is larger than the area of the planar disk with the same circumference.

Figure 15. The area of a hemisphere is larger than the area of a ﬂat disk with the same circumference.

We can ﬁnd the exact area of the hemisphere as follows. Suppose that r is the radius of the sphere. Then the length of the equator is C =2πr. Hence the radius of the rink is again C r = . 2π 198 8. GEOMETRIES OF SURFACES

Since the surface area of a sphere is 4πr2, the area of one hemisphere must be A =2πr2. Using A. Square’s observation that the circumference of the rink is roughly 10 units, we plug in C = 10 to again get that the value of the radius is 10 5 r = = . 2π π But now we plug this value into the formula for the area of a hemisphere to ﬁnd that the area of the rink is 5 2 50 A =2πr2 =2π = , π π which we ﬁnd with a calculator is approximately 16 square units. So the ice-skating rink could be as much as twice as spacious as A. Square had previously thought. Having twice your body size to move in a skating rink is better than skating in a can of sardines, but it’s still not great. Since the other skaters all appear to be quite skinny, A. Square agrees to enter the skating rink (see Figure 16).

Figure 16. Because the skating rink is on a sphere, it is more spacious than A. Square had thought.

In his calculations for both the snowball ﬁght and the ice skating rink, A. Square mistakenly assumed that the geometry of a sphere would be the same as that of a plane, forgetting that triangles and circles in a sphere have more area because they bulge outwards. In fact, the same is true for any geometric shape in a sphere. As a result, angle sums of polygons will be bigger in a sphere than in a plane. Also, just as for spherical triangles, the bigger the spherical n-gon is compared to the size of the sphere, the bigger its angle sum will be. For instance, if all of the vertices of an n-gon lie on a great circle, then the n-gon will have angle sum n × 180◦. Figure 17 6. MAPS OF THE EARTH 199

Figure 17. This hemisphere has six vertices and hence is a hexagon. shows a hexagon whose angle sum is 6×180◦ = 1080◦. Furthermore, such an n-gon will be a hemisphere, and hence will have area 2πr2 (where r is the radius of the sphere).

6. Maps of the Earth

After returning to their flat universe, A. Square and B. Square decide to make a scrapbook with drawings of the skating rink and the other things they saw during their visit to the sphere. But they keep arguing about how to draw the skating rink. The problem is that since spherical and flat geometry are so different, there is actually no way to draw spherical distances and angles accurately on a piece of paper. Mapmakers encounter the same problem as the Square brothers when they make flat maps of the Earth like the one shown in Figure 18. To explain this map to a Flatlander, we would need to first explain that if we cross the left edge of the map, we reappear on the right edge. This is not hard for Flatlanders to understand because they are used to drawings of the flat torus. But we also tell them that the entire top edge is shrunk to a single point which is called the North Pole, and the entire bottom edge is shrunk to another point which we call the South Pole. Each vertical line on the flat map (known as a longitude) corresponds to half a great circle through the North and South Poles. The horizontal lines (known as latitudes) correspond to parallel circles on the Earth whose circumferences decrease as you move away from the equator towards either of the poles. Apart from the line corresponding to the equator, none of the latitudes are great circles, and hence do not represent geodesics. Looking at Figure 18, let’s compare the distance between points A and B to the distance between points C and D. Although these two distances look like they are equal in length on the map, in reality the distance between C and D is shorter than the distance between A and B. This is because the shortest path between points A and B is an arc of a great circle on the sphere which follows the curved grey arc illustrated in Figure 18. On the other hand, since points C and D are on the equator, the line segment between them on our map corresponds exactly to the arc of a great circle on the sphere. If you have ever looked at flight paths on airline maps, you have probably noticed that they are represented by curved arcs rather than straight lines. This is because airlines take the shortest route, which means flight paths follow great circles on the Earth. 200 8. GEOMETRIES OF SURFACES

Figure 18. A planar map of the Earth cannot be accurate.

The map we have drawn in Figure 18 is known as a Mercator map.Thead- vantage of this map is that it preserves angles and the shapes of small regions. For instance, longitudes and latitudes form right angles on the Mercator map, as they do on a globe. Also, if you compare a small patch of the Mercator map with the corresponding small region on a globe, then you’ll notice that shapes within that small region are nearly the same. But this is not true for large regions. Can you find a large area in Figure 18 whose shape is different from what it is on a globe? The Mercator map has been used as the primary map for steering ships since the 1500s because it preserves angles and is relatively accurate on a small scale. Even today, Google Maps uses a variant of the Mercator map to make its maps. Maps other than the Mercator map might have less distortion on a large scale, but some distort angles while others divide up the Earth in strange ways. There is always a trade-off. A flat map of the Earth will never be completely free of distortion, because as we have seen spherical geometry is quite different from Euclidean geometry.

7. Hyperbolic Geometry

In this section, we consider an axiom system consisting of Euclid’s ﬁrst four axioms but the ﬁfth axiom is now replaced by: Second Alternative to Axiom 5. For any given line L and any point a not on L, there are multiple lines through a that do not intersect L. 7. HYPERBOLIC GEOMETRY 201

The hyperbolic plane illustrated in Figure 19 satisfies such an axiom system. Note that for the rest of this chapter we will use the term “flat plane” to refer to the usual plane in order to distinguish it from the hyperbolic plane. Topologically the hyperbolic plane is the same as the flat plane, but geometrically it is like a saddle at every point. The best word to describe the hyperbolic plane is “floppy”. In Figure 19, we illustrate two geodesics through the point a neither of which intersects the geodesic L. In fact, there are infinitely many geodesics through a that don’t intersect L.

Figure 19. There are inﬁnitely many geodesics through the point a which do not intersect the geodesic line L.

Furthermore, all triangles on a hyperbolic plane have angle sums which are less that 180◦, and the bigger the triangle is, the smaller its angle sum will be. This makes hyperbolic triangles bulge inwards and look thin like the white triangle on the right side of Figure 19. Recall that when A. and B. Square were having a snowball fight in the sphere, A. Square’s Euclidean brain thought that B. Square was closer than he really was. In Figure 20, we see that in the hyperbolic plane the light rays going from the top and bottom of B. Square to A. Square’s eye together with B. Square’s face make a triangle. However, the triangle is thinner than it would be in a flat plane. As a result, A. Square’s Euclidean brain would deduce that B. Square is further away than he really was, and he would throw a snowball flying past B. Square’s head. If you want to make your own hyperbolic plane so that you can get a feel for how floppy it is, there are many fun ways to do so. You can learn to crochet a hyperbolic plane and to make a hyperbolic blanket by following the directions on these websites: • http://www.math.cornell.edu/∼dwh/papers/crochet/crochet.html

• http://geometrygames.org/HyperbolicBlanket/index.html Just as a spherical circle encloses more area than a circle of the same circumference in the ﬂat plane, a hyperbolic circle encloses less area than a circle of the 202 8. GEOMETRIES OF SURFACES

Figure 20. A. Square thinks B. Square is farther away than he really is. same circumference in the ﬂat plane. The circular ice-skating rink in Figure 21 has a circumference of roughly 10 times the height of A. Square. Even though there are only 8 other skaters in the rink, all of whom are much skinnier than A. Square and B. Square, it is too crowded for them to skate. In this case, A. Square and B. Square really would be better oﬀ watching snow fall than trying to ice skate in this crowded rink.

Figure 21. This hyperbolic skating rink encloses less area than a rink with the same radius in the ﬂat plane.

8. A. Square Learns to Draw a Hyperbolic Plane

Although we can visualize the hyperbolic plane as a floppy plane in space, it is much more difficult for A. Square to visualize a floppy plane. If you try to visualize a “floppy” 3-dimensional space, you’ll see the problem. 8. A. SQUARE LEARNS TO DRAW A HYPERBOLIC PLANE 203

To help A. Square visualize the hyperbolic plane, let’s ﬁrst recall how we explained the geometric product S1 × I to him in Chapter 7. We showed him the picture of an annulus in Figure 22, and told him to imagine that all of the concentric circles are actually geodesics of the same size. In addition, we told him that the shortest path between two points on one of these circles goes along the circle, as illustrated by the grey arc in the ﬁgure. This seemed counterintuitive to him, since he expected the shortest path to be along a straight segment. However, he was willing to ignore his intuition in order to visualize S1 × I as a geometric product. We used a similar technique ourselves to visualize the 3-dimensional space S2 × I as a geometric product.

Figure 22. In the geometric product S1 × I, the shortest path between these points is along the grey arc.

Since there’s not enough space in a flat plane to draw a floppy plane, we’ll have to use a similar type of distorted drawing to enable A. Square to visualize a hyperbolic plane. We begin by drawing a circle in the plane and telling A. Square that we’re going to draw the hyperbolic plane inside of this circle. Since a hyperbolic plane has no boundary, the hyperbolic plane won’t include the circle bounding the region. At first A. Square objects to the idea that a region bounded by a circle could have infinite area. But then he remembers that in Chapter 6 when we defined the connected sum of two planes, the circle where the planes were glued together bounded an infinite area on both sides (see Figure 23).

Figure 23. A gluing diagram for the connected sum of two planes.

Next, we tell him that for the hyperbolic plane there is not only an infinite amount of area inside of the circle, but some of the geodesics there are pretty strange as well. He accepts the idea of strange geodesics, since he accepted the strange geodesics for the annulus in Figure 22. We tell him that now there are two different types of geodesics: diameters of the disk and pieces of circles meeting the boundary of the disk at right angles. Since we’re not including the boundary of the region, these geodesics have no endpoints. Figure 24 shows some geodesics in the hyperbolic plane. Figure 24 includes two geodesics through a point a together with a disjoint geodesic L. Once he has accepted the strange geodesics, A. Square can see that the Second Alternative to Axiom 5 is satisfied by the hyperbolic plane. 204 8. GEOMETRIES OF SURFACES

Figure 24. Geodesics are diameters of the disk and circular arcs meeting the boundary circle at right angles.

Figure 25. The grey segments between points a and b and points c and d are longer than the circular arcs between them.

In order to help A. Square remember that straight line segments which are not part of a diameter are not geodesics, we show him Figure 25, and tell him that the grey line segments between points a and b and points c and d are actually longer than the black circular arcs between these pairs of points. This is reminiscent of the geometric S1 × I in Figure 22, where the shortest path was along an arc of a circle rather than along a straight segment. However, because we’ve had to squish the infinite hyperbolic plane into the space of a disk, the closer to the boundary circle we get, the more distances are distorted. For example, we can see in Figure 25 that the hyperbolic distance between a and b is actually greater than the hyperbolic distance between c and d, even though their Euclidean distances are identical. Observe that the shaded region in the lower right of Figure 25 is bounded by three circular arcs which all meet the boundary circle in right angles. Such arcs are geodesics, and hence the shaded region is a hyperbolic triangle. Because two of the sides of the triangle curve inward, the angles are much smaller than they would be in a Euclidean triangle. In fact, if all three of the vertices of a triangle are very close to the boundary circle, the angle sum of the triangle will be very close to 0◦. Thesameistrueforn-gons in the hyperbolic plane. In particular, an n-gon can have an angle sum as close to 0◦ as we like. Figure 26 shows an octagon in the hyperbolic plane whose angle sum is 8 × 45◦ = 360◦. This will be important to us in the next section when we try to find homogeneous geometries for all finite surfaces. 9. HOMOGENEOUS GEOMETRIES FOR ALL nT 2 WITH n ≥ 2 205

Figure 26. A hyperbolic octagon whose angles are 45◦.

9. Homogeneous Geometries for all nT 2 with n ≥ 2

It turns out that if a finite surface has homogeneous geometry, then its local intrinsic geometry must be the same as that of either a flat plane, a sphere, or a hyperbolic plane. We would like to figure out how to construct each finite surface so that it has one of these three homogeneous geometries. Recall from the Classification of Surfaces, version 2 (in Chapter 6), that every finite surface is topologically the same as either S2, nT 2,ornP 2 for some n>0. Since we know that K2 =2P 2 and T 2 can both be constructed to have flat geometry and S2 has spherical geometry, we are left with the task of finding homogeneous geometries for the surfaces nT 2 for all n>1andnP 2 for all n =2. Recall that we saw the following theorem in Chapter 5. Theorem 8.2. Let S be a surface with a polygonal gluing diagram. Then S is flat if and only if the edges of the polygon are glued together in pairs and the sum of the angles around each glued up vertex is 360◦. This theorem is useful for showing that a surface obtained from a polygonal gluing diagram is flat. But if the sum of the angles around a glued up vertex is not 360◦, the theorem only tells us that the glued up surface isn’t flat. It doesn’t tell us if there is another way to construct the surface to make it homogeneous. For example, we saw in Chapter 6 that 2T 2 is obtained from the octagonal gluing diagram illustrated in Figure 27. However, each vertex of an octagon has 135◦ around it, and all of the vertices in the diagram are glued together (you should remind yourself of how we know this). This means that the glued up octagon has a single anti-cone point with 135◦ × 8 = 1080◦ around it. Thus we know by Theorem 8.2 that this gluing diagram is not flat. However, there could be another gluing diagram for 2T 2 which is homogeneous, though not necessarily flat. In order to allow us to create homogeneous gluing diagrams which are not flat, we use the following generalization of Theorem 8.2 to polygonal gluing diagrams in other homogeneous surfaces. The proof of this theorem requires techniques that you might learn in a differential geometry class. Theorem 8.3. Let S be a surface with a polygonal gluing diagram in a homogeneous surface. Then S has the same geometry as the homogeneous surface if and only if its edges are glued together in pairs and the sum of the angles around every glued up vertex is 360◦. 206 8. GEOMETRIES OF SURFACES

Figure 27. All of the vertices in this gluing diagram for 2T 2 are glued together.

We’ll see how to use Theorem 8.3 by applying it to an octagonal gluing diagram for T 2 with the same gluing arrows as the one illustrated in Figure 27 but in a hyperbolic plane rather than a flat plane. Recall from the last section that polygons in the hyperbolic plane have smaller angles than corresponding polygons in a flat plane, and the closer the vertices are to the boundary of the disk, the smaller the angles will be. Thus if we draw the octagonal gluing diagram for 2T 2 on a hyperbolic plane rather than on a flat plane, its angles will be smaller. Moreover, as we saw in Figure 26, we can choose the vertices of the octagon on the hyperbolic plane in such a way that each of the angles is equal to precisely 45◦. While A. Square has to accept the drawing of the octagonal gluing diagram on the left side of Figure 28, we can also draw the gluing diagram on the floppy hyperbolic plane illustrated on the right. Now since all eight vertices of the hyperbolic octagon are glued together and all eight angles are 45◦, the glued up vertex will have a total of 45◦ × 8 = 360◦ around it. Thus we can apply Theorem 8.3 to conclude that the 2T 2 obtained from this gluing diagram has homogeneous hyperbolic geometry. Recall from Chapter 6 that for any n ≥ 2, the n-holed torus nT 2 can be obtained from a gluing diagram which is a 4n-gon. However just as we saw with

Figure 28. In the hyperbolic plane we can ﬁnd an octagonal gluing diagram for 2T 2 whose angles are 45◦. 10. A HOMOGENEOUS GEOMETRY FOR P 2 207 the octagon, all of the vertices in the 4n-gon are glued together. Thus we can create an nT 2 which has homogeneous hyperbolic geometry by drawing the 4n-gon on a hyperbolic plane and by moving the vertices apart until each vertex angle is 360◦ ◦ exactly 4n . In this way, the glued up vertex will have exactly 360 around it, and by Theorem 8.3 the nT 2 obtained from this gluing diagram will have homogeneous hyperbolic geometry.

10. A Homogeneous Geometry for P 2

Next we would like to construct a projective plane P 2 which has homogeneous geometry. The usual gluing diagram for P 2 is on the left in Figure 29. We can’t apply Theorem 8.3 to this diagram, since the diagram is not a polygon. However, by splitting each of the edges in half and then straightening the four edges out, we obtain the square gluing diagram on the right. At this point, we still can’t apply Theorem 8.3, since the vertices of the gluing diagram are glued together in pairs and hence there is only 90◦ × 2 = 180◦ around each glued up vertex.

Figure 29. This give us a polygonal gluing diagram for the projective plane.

Thus we need to increase the angle around each vertex from 90◦ to 180◦ so that when we glue the vertices together in pairs, there will be 360◦ around each glued up vertex. Since any polygon on a sphere has larger angles than the corresponding polygon in a plane, in this case we want to draw our a gluing diagram on a sphere as illustrated in Figure 30.

Figure 30. A gluing diagram for P 2 on a sphere.

Now the bigger the polygon is on the sphere, the bigger its angles will be. Since we want the angle around each vertex to be 180◦,wewanttomovetheverticesof the square apart until they all lie on a single great circle. This stretches the square out to cover an entire hemisphere as illustrated in Figure 31. The hemisphere may not look like a “square”, but it has four vertices and four geodesic edges, and hence it is in fact a “spherical square” (not to be confused with a “squarical sphere”). 208 8. GEOMETRIES OF SURFACES

Figure 31. A square gluing diagram for P 2 in which the angle around each vertex is 180◦.

The angle at each vertex is now 180◦. So after gluing the vertices together in pairs, there will be exactly 360◦ around each vertex, just as we wanted. Thus we can apply Theorem 8.3 to conclude that the P 2 obtained from this gluing diagram has spherical homogeneous geometry. Recall that 2P 2 = K2 has a square gluing diagram with ﬂat geometry. In the Exercises, you will construct an nP 2 with homogeneous geometry for each n>2. Thus in fact every ﬁnite surface can be constructed so that it has homogeneous geometry.

11. Uniqueness of Homogeneous Geometries for Surfaces

We have seen that it is possible to construct every finite surface so that it has one of the three homogeneous geometries. Also, we know that different constructions of the same surface can have different geometry. For example, the flat torus has flat geometry while the curved torus has non-homogeneous geometry. Perhaps a surface could be constructed in two different ways, so that one has one homogeneous geometry and the other has a different homogeneous geometry. Maybe we could even find a surface which can have each of the three homogeneous geometries depending on how it was constructed. As an example, let’s consider two different gluing diagrams for a torus. In the Exercises in Chapter 5, we showed that the hexagonal gluing diagram on the right in Figure 32 is a torus. Perhaps the two tori in Figure 32 have two different homogeneous geometries. The square gluing diagram has all four vertices glued together, and the hexagonal gluing diagram has two groups of three vertices glued together. Since 4 × 90◦ = 360◦ and 3 × 120◦ = 360◦, it follows from Theorem 8.2 that in fact both of these diagrams have flat geometry. Drawing either of these gluing diagrams on a hyperbolic plane will cause the glued up vertices to have less

Figure 32. These are two different gluing diagrams for a torus. 12. EXERCISES 209 than 360◦, and hence the surface will have one or two cone points. Drawing either of these gluing diagrams on a sphere will cause the glued up vertices to have more than 360◦, and hence the surface will have one or two anti-cone points. Perhaps with more work we could find a completely different gluing diagram for a torus that could be drawn on a hyperbolic plane or a sphere with no cone points or anti-cone points. In fact, the following theorem shows that this is impossible. Theorem 8.4 (The Uniformization of Surfaces). Every finite surface can be constructed with exactly one homogeneous geometry. This theorem follows from a beautiful result known as the Gauss-Bonnet For- mula, which you might learn about if you take differential geometry.

12. Exercises

1. Prove that a rectangular flat torus whose sides have different lengths does not satisfy Euclid’s Axiom 3. 2. Prove that a square flat torus does not satisfy Axiom 3. 3. Prove that a flat Klein bottle does not satisfy Axiom 3. 4. Prove that a sphere does not satisfy Axiom 3. 5. Consider a sphere with radius r.Leta be a point on the sphere. Is there a circle on the sphere with center at a and radius πr? Are there circles with radius less than πr? 6. Prove that the sphere and the infinite cylinder don’t satisfy Alternative Axiom 1. 7. Prove that the surface in Figure 33 violates Axiom 5.

Figure 33. Illustration for Exercise 7.

8. Prove that an infinite cone violates Axiom 5. 9. Prove that a flat torus satisfies Axioms 1, 2, 4, and 5. 10. Given any two points in a flat torus, show that there are infinitely many geodesic segments between them. 11. Do all of the geodesic lines in a flat torus have finite length? 12. Explain why the flat plane with the origin removed does not satisfy Axiom 5. 210 8. GEOMETRIES OF SURFACES

13. Describe a surface together with a pair of points p and q, such that there are four geodesic segments between p and q with the same length. 14. Show that the flat Klein bottle satisfies Axioms 1, 2, and 4. 15. Explain why the flat Klein bottle does not satisfy Axiom 5. 16. Show that a sphere satisfies Axioms 1, 2, and 4. 17. For each of Euclid’s five axioms, determine whether it is a statement about the local geometry of a surface or the global geometry of a surface. 18. In what ways is the map of Earth in Figure 34 accurate or inaccurate?

Figure 34. Illustration for Exercise 18.

19. Prove that the sum of the angles of a flat n-gon is (n − 2) × 180◦. 20. Suppose A. Square and B. Square are in a small sphere but have had a bit of brotherly disagreement. So A. Square is camped out at the North Pole and B. Square is at the South Pole. If we rotate the sphere to compare the North Pole and the South Pole, to us A. Square and B. Square look identical, except that B. Square is the mirror image of A. Square. What does B. Square look like to A. Square? 21. Suppose B. Square couldn’t stand being in the sphere with A. Square anymore and has somehow transported himself home to the flat torus, leaving A. Square in the sphere by himself. Would A. Square be able to see himself? And if so, how would he look? In particular, in what ways would the view be the same or different from the view of B. Square when he was at the opposite pole? 22. Suppose A. Square and his brother B. Square are again both in a small sphere. A. Square watches as his brother walks away from him. Does the image of B. Square get bigger or smaller as he walks away? At what point is the image the smallest and at what point is it the largest? 23. We say that two triangles are congruent if they have the same intrinsic geometry. This means that each of the sides and each of the angles in one have to equal each of the corresponding sides and angles of the other. In Euclidean geometry, there are four congruence theorems: • Side-Side-Side (SSS) says that if two triangles have all three side lengths equal, then the two triangles are congruent. 12. EXERCISES 211

• Angle-Angle-Side (AAS) says that two triangles are congruent if two pairs of angles are equal and the lengths of a pair of corresponding sides not in between the two angles are also equal. • Side-Angle-Side (SAS) says that if two pairs of sides have equal length and the corresponding angles between the two sides have equal length, then the triangles are congruent. • Angle-Side-Angle (ASA) says that two triangles are congruent if two pairs of angles are equal and the corresponding sides between the two angles have equal length. Which, if any, of the SSS, AAS, SAS, and ASA theorems hold for triangles in a sphere? 24. Explain how to create an nP 2 with hyperbolic geometry for any n>2. Explain why this method won’t work for n =1orn =2.

Part 2

Knots

https://doi.org/10.1090//mbk/096/09

CHAPTER 9

Introduction to Knot Theory

Topics: • Knots and links • Projections • Crossing number • Prime and composite knots • Knot tables • A brief history of knot theory • Reidemeister moves • Tricolorability • Invertibility

1. 1-Dimensional Universes

Our 1-dimensional friend A. Dash is feeling sad. Just yesterday, A. Square told her about all the different 2-dimensional universes he had learned about: the plane, the sphere, the torus, the Klein bottle, and infinitely many more. A. Dash began pondering the different 1-dimensional universes, and could only think of two: an infinite line and a circle. How boring! “Don’t worry, I’m sure there must be others”, exclaims A. Square, and he begins sketching the squiggly shapes in Figure 1. But after a few minutes, A. Square realizes that all of the shapes he has drawn could actually be deformed to ordinary lines or circles. Poor A. Dash really did have only two possible universes. A. Square began to feel very lucky that he did not have to live in a boring 1-dimensional universe.

Figure 1. Some 1-dimensional universes in the plane.

215 216 9. INTRODUCTION TO KNOT THEORY

While the intrinsic topology of 1-dimensional universes might be a bit boring, the extrinsic topology of 1-dimensional spaces in R3 is anything but boring. We tried to explain to A. Dash and A. Square that there are inﬁnitely many diﬀerent knotted universes that she could live in, like the ones illustrated in Figure 2. She was very excited to hear this, and wanted to visualize a knotted universe. But unfortunately, neither A. Dash nor A. Square can understand what a knot is, since they can’t imagine how one arc can wrap around another arc. This is because, by its very nature, knottedness is an extrinsic property of circles in 3-dimensional space.

…

Figure 2. A. Dash in some knotted universes.

Even though it might sound like it’s not a theory, knot theory is an important area of topology that has some surprising applications, including to the study of DNA and proteins, as we’ll see in Chapters 14 and 15. The broad goal of knot theory is to develop methods to determine whether a given knot can be deformed to another knot, or even to an ordinary circle. This is not as easy as it might seem. Think about how hard it can be to untangle ﬁshing line or thread when it’s tangled up, and such strings have loose ends so you should be able to unthread the tangles. When we study knots, we require that the ends of our string be glued together so that if there is a knot, it will be “trapped” in the loop. Otherwise, every knotted string could be deformed to an unknotted arc, as we see in Figure 3.

Figure 3. A knotted arc is topologically the same as an unknotted arc.

Mathematically speaking, we deﬁne a knot as a circular loop in R3,andwe deﬁne a link as one or more disjoint loops in R3. Note that this means that a knot can also be considered to be a link with only one component. Each loop of a link is 2. WHEN ARE TWO KNOTS EQUIVALENT? 217 called a component . For example, in Figure 4, A. Dash is in one component of a link, and it would be impossible for her to get to the other component, or even to know that it existed. In fact, since you can’t get from one to the other, each component of a link is its own universe, regardless of how they may be linked together in 3-dimensional space. Imagine if our universe were somehow linked together with another universe in a higher dimension but we had no way of knowing that the other universe even existed.

Figure 4. A. Dash in one component of a link.

2. When Are Two Knots Equivalent?

We say that two knots or links are equivalent or even the same if one can be deformed to the other in R3. Figure 5 shows two knots that are equivalent. Can you visualize a deformation that transforms the knot on the left into the knot on the right? At this point it would be a good idea to beg, borrow, or steal an extension cord (the shorter the better). Such a cord will come in handy to do deformations in this and the next two chapters. Tie your extension cord into the knot on the left in Figure 5, then stick the plug end of it into the socket end to make a circle. Now see if you can deform your cord into the knot on the right in Figure 5.

Figure 5. These two knots are equivalent.

Rephrasing this notion of sameness in the language of Chapter 3, we say that two knots or links are equivalent if they have the same extrinsic topology in R3.In contrast to earlier chapters, where we focused on the intrinsic properties of a space, the study of knots and links is focused on extrinsic topology. In particular, every knot has the same intrinsic topology as an unknotted circle—they are all the same to A. Dash—and every link with n components has the same intrinsic topology as n unknotted and unlinked circles. It is because they exist in 3-dimensional space that knots and links are interesting. Unfortunately, since the pages of this book are only 2 dimensional, we need to represent knots and links in a 2-dimensional way. A projection is a 2-dimensional drawing of a knot or link where we indicate undercrossings (i.e., places where part of the knot or link passes under another part in R3) with gaps in the arcs. All the pictures of knots and links you’ve seen so far in this chapter have been projections. You can think of a projection as the shadow cast by shining a bright light on the 218 9. INTRODUCTION TO KNOT THEORY knot or link from above, with gaps to indicate crossings. As we deform a knot or link in R3, its shadow will change, giving us many diﬀerent projections of the same knot or link. For example, Figure 5 shows two projections of the same knot. Any knot that can be deformed to the projections in Figure 5 is called the trefoil knot. Notice that both of the projections in Figure 5 have three crossings. Do all projections of the trefoil have three crossings? Well, no; it’s pretty easy to deform the trefoil so it will have additional crossings, as illustrated in Figure 6.

Figure 6. Projections of the trefoil with diﬀerent numbers of crossings.

However, it’s often more diﬃcult to remove crossings from a projection than to add crossings. Is there a projection of the trefoil with fewer than three crossings? We use c(K) to denote the fewest number of crossings of any projection of a knot or link K,andrefertoc(K)asthecrossing number of K. We might guess that the crossing number of the trefoil is three. But looks can be deceiving. In Figure 7, we see a projection of a knot with six crossings. But the crossing number of this knot is less than six. Can you see how you might deform it to remove some crossings? Again, your extension cord may come in handy here.

Figure 7. The crossing number of this knot is less than six.

The simplest knot is an ordinary unknotted circle, which is called the unknot. It is the only knot that has crossing number zero. Similarly, any link with crossing number zero is called an unlink. We say that the unknot and unlink are trivial, while all other knots and links are said to be non-trivial. Figure 8 shows a projection of a trivial knot. See if you can deform it so that it has no crossings.

Figure 8. A complicated projection of the unknot.

One goal of knot theory is to be able to tell if a knot is trivial or not. You might think that a computer program could do this for us: just feed the computer a knot projection, and after a few seconds, out pops an answer! Unfortunately, no 3. THE MIRROR IMAGE OF A KNOT OR LINK 219 such computer program exists. It is an unsolved problem of knot theory to ﬁnd and implement an algorithm that distinguishes trivial knots from non-trivial knots in a reasonable amount of time.

3. The Mirror Image of a Knot or Link

Even if you don’t know whether a knot is trivial, you can still do some interesting things with it. For example, imagine holding a knot up to a mirror. Now you have two knots: the one in your hand, and the one in the mirror. It’s natural to wonder whether you can deform the knot in your hand so that it looks like its mirror image. In Figure 9, A. Square is standing in front of a projection of the trefoil looking in a mirror.

Figure 9. A. Square standing in front of a knot projection looking in a mirror.

Let’s say our original knot is called K.Thenwecallitsmirror image K∗. Starting with a projection of K, we can draw a projection of K∗ without using a mirror. We just switch all of the crossings in K from over to under, and under to over. To see why this works, imagine the mirror is in the plane of the paper. Every place the knot goes above the page, it’s mirror image goes below the page, and every place the knot goes below the page, it’s mirror image goes above the page. Figure 10 illustrates a link and its mirror image.

Figure 10. To get the mirror image, we switch all of the crossings.

So, is a knot or link really diﬀerent from its mirror image? It’s not hard to show that the links in Figure 10, known as Hopf links, are actually the same. Just turn over one component of K without moving the other, and you’ll get K∗. However, not all knots and links are equivalent to their mirror images; in fact, most are not. The trefoil knot, and its mirror image (in Figure 9) are not equivalent, though we won’t be able to show this until Chapter 10. 220 9. INTRODUCTION TO KNOT THEORY

4. The Connected Sum of Two Knots

In Chapter 6, we used connected sums to build complicated manifolds from simpler ones. We will now use a similar idea to take familiar knots and build more complicated ones. To take the connected sum of knots K1 and K2,wecuteach knot into a knotted arc and glue the resulting endpoints together as illustrated in Figure 11. This will yield a new knot, denoted by K1#K2.

Figure 11. The connected sum of two knots.

The knot on the left side of Figure 11 is the trefoil knot that we saw previously, and the knot on the right is the figure eight knot. These knots come up often. So it’s good to know their names. In particular, if you ran into one of them at a party, it would be embarrassing not to know its name. The connected sum of knots can be thought of as a 1-dimensional version of the connected sum of two surfaces or two 3-manifolds that we studied in Chapter 6. Just as we can create infinitely many different surfaces by taking connected sums of tori and projective planes, we can use the connected sum operation to create infinitely many different knots. For example, for any number n, we can create a knot which is the connected sum of n trefoil or figure eight knots. In Figure 12, we illustrate a knot which is the connected sum of three figure eight knots.

Figure 12. This knot is the connected sum of three ﬁgure eight knots.

In Chapter 6, we compared the properties of the connected sum operation for surfaces with ordinary addition for numbers. In particular, recall that a sphere is an identity element for connected sums of surfaces because the connected sum of 5. A BRIEF HISTORY OF KNOT THEORY 221 any surface with a sphere will be topologically the same as the original surface. Similarly, the unknot is an identity element for connected sums of knots. We can also compare the connected sum operation to multiplication of integers. In particular, a positive integer is said to be prime if it cannot be expressed as a product of two integers neither of which is equal to the number 1. Every positive integer which is not 1 and which is not prime is said to be composite. For example, the number 3 is prime, but the number 6 is composite. If we applied this concept to connected sums of finite surfaces, we could say that a surface is prime if it cannot be expressed as the connected sum of two surfaces, unless one of the two surfaces is topologically equivalent to a sphere. By the Classification of Surfaces we know that every finite surface has the form S2, nT 2,ornP 2 for some n. Hence the only prime surfaces are T 2 and P 2. Now we apply this concept to connected sums of knots. In particular, we say a knot is prime if it can’t be expressed as a connected sum of two knots, unless one of them is the trivial knot (the unknot). Every knot which is not the trivial knot and which is not prime is said to be composite. For example, the trefoil knot and the figure eight knot are both prime, whereas the knot in Figure 12 is composite. However, while the study of finite surfaces which are prime is not very interesting because there are only two prime surfaces, there are infinitely many prime knots. You can think of prime numbers, together with the operation of multiplication, as forming the building blocks of all positive integers. This is why prime numbers play a big role in number theory. Prime knots play a similar role in knot theory. Since every non-trivial knot is either prime or a connected sum of prime knots, we study prime knots as a way to understand all knots. In fact, just as every number can be expressed uniquely as product of prime numbers, every knot can be expressed uniquely as the connected sum of prime knots. Figure 13 is a table listing every prime knot with fewer than nine crossings. Each knot in the table is denoted with two numbers: the first number is the crossing number of the knot, and the subscript is just a way to keep track of different knots with the same crossing number. To save space, mirror images are not listed in the table. For example, the trefoil is denoted by 31. There is no 32, because the only other knot with crossing number 3 is the mirror image of 31, which is denoted as ∗ 31. Similarly, the figure eight knot is the only knot whose crossing number is 4. On the other hand, there are two different knots with crossing number 5: one is denoted by 51 and the other is denoted by 52. As you might guess, there are significantly more 9 crossing knots than 8 crossing knots, and still more 10 crossing knots than 9 crossing knots. Knot theorists have compiled tables of all prime knots with up to 16 crossings – a total of 1, 701, 935 distinct knots. Not surprisingly we don’t include the table here. In fact, such a table is only available online.

5. A Brief History of Knot Theory

Today’s knot tables trace their origins back to 19th-century Scotland. While the United States was embroiled in civil war, Scottish physicists were trying to develop a model for the atom. We know today that atoms are made of a dense, positively charged nucleus surrounded by a cloud of electrons. But in the 19th century, physicists had very diﬀerent theories about atoms. In particular, their 222 9. INTRODUCTION TO KNOT THEORY

Figure 13. A table of prime knots with up to 8 crossings. ideas grew out of a theory, introduced by Descartes in 1644, that all of space and matter were filled with an invisible substance called ether. This theory took off in the 19th century when physicists believed space needed to be filled with some kind of medium in order to explain how light could behave like a wave and yet travel outside of the atmosphere. They referred to this medium as luminiferous ether to indicate its role in light propagation. At the time, scientists knew that elements, such as carbon, oxygen, etc., existed and also that matter was composed of atoms. However, the actual structure of atoms was still a mystery. One prominent theory came from the work of physicists Peter Guthrie Tait and Sir William Thomson (a.k.a. Lord Kelvin, of the Kelvin temperature scale). Inspired by the behavior of smoke rings that Tait had observed in his laboratory, they theorized that atoms were swirling tubes, or “vortices”, in the 6. REIDEMEISTER MOVES 223 ether. But these tiny tubular whirlpools were not simple geometric circles. Rather, they were knotted or linked, and the type of knot or link in an ether whirlpool determined the element. Thomson thought the spectroscopic properties of each element were reflected in the crossings of its projection. For example, he hypothesized that a sodium atom, which emits two prominent spectral lines when energized, should correspond to a vortex with crossing number 2. Perhaps a trefoil in the ether was really a helium atom, and a figure eight knot could be a lithium atom (see Figure 14).

Figure 14. Atoms were believed to be knotted or linked vortices in the ether.

In the mid-19th century, physicists were beginning to create tables of elements that much later would be replaced by the modern periodic table. Tait, Thomson, and their colleague James Maxwell (famous for his equations of electromagnetism) realized that if this knotted vortex theory was correct, a table of distinct knots was equivalent to a table of the elements. Creating such a table would be an important scientific contribution, and Tait thought an analysis of crossing number would be agoodplacetostart. With the help of a few other early knot theorists, Tait worked for many years to create an accurate knot table based on crossing number. While their work was imperfect and non-rigorous, they noticed many important patterns. In particular, Tait made several conjectures about the properties of knots and their crossings, two of which we’ll explore in the next chapter. Unfortunately, all of this scientific interest in knots would soon come to an abrupt end. The ether vortex model for the atom was falling out of favor among physicists, and by 1887 the famous Michelson-Morley experiment shook the scientific world by showing conclusively that luminiferous ether did not exist! Sadly, since the primary motivation for studying knots was to create a periodic table based on the vortex model, knot theory was abandoned for many years. It wasn’t until the 1920s that interest in knots was re-ignited by topologists, and Tait’s conjectures— still unproven—were rediscovered.

6. Reidemeister Moves

A. Square has been listening quietly on the sidelines while we’ve been discussing knot tables and the history of knot theory. He’s interested in the table of knots in Figure 13 because it’s ﬁlled with 2-dimensional pictures that he can appreciate. The problem is that when A. Square looks at a projection, he sees it as a collection of disjoint arcs in Flatland. For example, A. Square sees the projection of a trefoil in Figure 15 as three disconnected arcs that are free to move away from each other. In order to help him understand that the arcs are actually joined together, without losing the important 3-dimensional information of which strand goes under 224 9. INTRODUCTION TO KNOT THEORY

Figure 15. A. Square sees a projection of a knot as a collection of disconnected arcs in Flatland. and which strand goes over, we add a grey arc where the gaps are. We’ll call the projection with the grey arcs a Flatland-friendly projection of a knot rather than just a projection. Now we explain to A. Square that if he can deform one Flatland- friendly projection of a knot to another, then the knots are equivalent. For example, he can deform the Flatland-friendly projection on the left in Figure 16 to the one on the right. Keep in mind that this is a planar deformation, meaning that the deformation takes place in the plane. He is not sliding a grey arc under one of the arcs of the projection. Rather, he sees the black and grey together as forming a little cross that moves along as he deforms and rotates the Flatland-friendly projection in the plane.

Figure 16. A. Square can deform the Flatland-friendly projection on the left to the one on the right.

A. Square is disappointed to learn that you can’t get from any knot to any equivalent knot by just using those planar deformations that he can do. However, A. Square is in luck. The German mathematician Kurt Reidemeister showed in the 1920s that two knots are equivalent to one another if and only if you can go from a projection of one knot to a projection of the other by using combinations of three small “moves” in the plane together with planar deformations. The reason that we say these moves are small is because, when we do one of the moves, we don’t change anything else about the projection. By applying these moves one after the other to different parts of the projection, we can make big changes in the projection. But each individual move is just a small change in a very specific spot. We will illustrate the three moves on knot projections and explain what we’re doing in terms of 3-dimensional deformations. But we tell A. Square to fill in the gaps in the projections with grey arcs and to think about any moves as allowable changes in order to make a knot into a Flatland-friendly projection in the plane, even if he can’t see with his 2-dimensional vision why these moves are valid.

Reidemeister I move: A kink can be added or removed as in Figure 17. The double headed arrows in the ﬁgure indicate that you can do the move going from 6. REIDEMEISTER MOVES 225 either of the kinks to no kink, or from no kink to either kink. With this move, a crossing is removed or added to the projection. We denote this move by (R1).

Figure 17. (R1) A kink can be added or removed.

Reidemeister II move: An arc can be slid over or under another arc as in Figure 18. With this move, two crossings are removed or added to the projection. We denote this move by (R2).

Figure 18. (R2) An arc can be slid over or under another arc.

Reidemeister III move: An arc can be slid under or over a crossing as illustrated by the thick black arc in Figure 19. With this move, the number of crossings remain the same, but the arc is positioned diﬀerently in the projection. We denote this move by (R3).

Figure 19. (R3) An arc can be slid under or over a crossing.

It’s easy to check that any two projections related by a Reidemeister move (or a sequence of finitely many such moves) must represent equivalent knots. However, it’s not easy to see why repeatedly using just these three moves together with planar deformations would give you all possible deformations of a knot. The proof of the following important theorem is quite technical, so it is unlikely that you would see it in a course. However, if you are interested, you can find several different proofs of it on the web. Theorem 9.1 (Reidemeister’s Theorem). Any deformation of a knot or link in 3-dimensional space can be achieved by planar deformations together with a finite number of the three Reidemester moves. Reidemeister’s Theorem tells us that if we know that two projections represent equivalent knots, then we can go from one to the other with a finite sequence of Reidemeister moves, denoted by (R1), (R2), and (R3), along with planar deformations denoted by (P). For example, we observed that the two projections in Figure 5 represent equivalent knots. It follows that there must be a sequence of Reidemeister moves and planar deformations taking one to the other. Figure 20 shows one such sequence, but there are many other possibilities. 226 9. INTRODUCTION TO KNOT THEORY

Figure 20. A sequence of Reidemeister moves and planar deformations.

While Figure 20 illustrates how you could deform a projection using Reidemeis- ter moves, the Reidemeister moves should not actually be used for this purpose. In particular, deforming a projection with Reidemeister moves is much slower, harder, more confusing, and more tedious than deforming the knot or link in 3-dimensional space with the help of an extension cord. So when you’re doing the Exercises, remember that you should NEVER use Reidemeister moves to prove that two knots are equivalent unless you are speciﬁcally told to do so. If we don’t use Reidmeister moves to deform one knot to another, than why do we even care about Reidemester’s Theorem? The point of the theorem is that it gives us a way to prove that certain properties of knots and links are preserved under any deformation. In particular, if we can show that a property does not change under any of the three Reidemeister moves or planar deformation, then we’ll know that the property doesn’t change under any of the inﬁnitely many possible deformations of a knot or link in 3-dimensional space. Hence if one knot has such a property and another knot doesn’t, then we can conclude that the knots are not equivalent. We’ll see how this works in the next section.

7. Coloring Knots with Three Colors

You’re so excited about all the cool math you’ve been learning lately that you decide to take your knot table with you to Thanksgiving dinner at your grand- parents’ house. While you’re busy stuffing yourself with your second helping of stuffing, your eight-year-old cousin, Felicity, finds your knot table and starts coloring on it with her crayons. She traces the arcs of the knot projections with her three favorite colors (black, grey, and striped). After noticing that at each crossing there are three arcs which come together, she makes the following list of rules for herself. Felicity’s Very Special Coloring Rules:

(1) More than one color must be used for every knot. (2) At each crossing either the three arcs that come together must all be the same color or each arc must be a distinct color. 7. COLORING KNOTS WITH THREE COLORS 227

She skips the unknot because it’s boring and starts coloring the other knot projections. She’s pretty happy with her coloring of the projection of the trefoil knot (illustrated in Figure 21) because she succeeded in following her rules.

Figure 21. Felicity’s coloring of a projection of the trefoil.

The next knot in the table is the ﬁgure eight knot. She starts with the arc on the right, and colors it black (as in Figure 22). Then she looks at the long arc on the left and decides to color it grey.

Figure 22. Felicity colors the right arc black.

Since black and grey come together at the bottom left crossing, by Felicity’s Rule (2), the third arc at this crossing must be striped (see Figure 23).

Figure 23. Because black and grey come together at this crossing, by Rule (2) the third arc must be striped.

Now there’s only one arc left to color—the short arc in the middle. At the top crossing, Felicity has already colored one arc grey and one black, so by Rule (2) she has to color the middle arc striped. But this means that the middle crossing has two arcs that are striped and one that is grey, which violates Rule (2) (see Figure 24). Felicity gets frustrated and wads up the paper. You go over to console her and ask why she’s upset. She smooths out the crumpled paper and explains her game and what didn’t work. You try to be helpful and observe that she hasn’t tried all the options yet. You suggest that she color the left arc black instead of grey. But then you see that this implies by Rule (2) that she must also color the bottom arc black, as illustrated in Figure 25. Again she is left with one arc to color. Felicity doesn’t want a monochromatic picture because that would violate Rule (1). But because of Rule (2) she is forced 228 9. INTRODUCTION TO KNOT THEORY

Figure 24. The crossing in the center violates Rule (2).

Figure 25. Since two arcs at the bottom left crossing are black, thethirdarcmustbeaswell. to color the middle arc black. Before she has a complete meltdown, you teach her to play torus tic-toe-toe and let her go ﬁrst. After winning torus tic-tac-toe, Felicity quickly forgets about coloring knots. The coloring game that Felicity was playing turns out to be useful for distinguishing knots. Let’s take a moment to formalize Felicity’s rules.

Definition. A projection of a knot or link is said to be tricolorable if there is a way to color the arcs with one of three colors such that: (1) More than one color is used. (2) At each crossing, either only one color is used or all three colors are used as illustrated in Figure 26. Aknotorlinkissaidtobetricolorable if it has a tricolorable projection.

Figure 26. Each crossing has either only one or all three colors.

Notice that we have to color a projection of a knot rather than a knot itself because a knot itself is a curve in space and doesn’t have crossings. It is projections of the knot that require one arc to pass in front of another and hence have crossings. However, this deﬁnition is telling us that if a single projection of a knot is tricolorable, then we will say that the knot itself is tricolorable even if some other projection of the knot is not tricolorable. 8. TRICOLORABILITY AND KNOT EQUIVALENCE 229

8. Tricolorability and Knot Equivalence

Our experience with Felicity shows us that the trefoil knot is tricolorable. But we haven’t shown that the figure eight knot is not tricolorable. We’ve only shown that the particular projection of the figure eight in Figure 22 is not tricolorable. Perhaps we can deform the figure eight to a position that has a tricolorable projection. After all, there are infinitely many different ways to deform a knot. This is where Reidemeister’s Theorem is useful, because it tells us that any two projections of a given knot can be connected by deformations in the plane together with a finite sequence of Reidemeister moves. Deforming the arcs in the plane certainly doesn’t change whether a projection is tricolorable. If, in addition, we could show that each of the three Reidemeister moves doesn’t change whether a projection is tricolorable, then we would know that no deformation, no matter how complicated, could cause a tricolorable projection to change into one which was not tricolorable or vice versa. This would tell us that for a given knot, either all projections are tricolorable or none of them are. Theorem 9.2. The tricolorability of a knot or link projection does not change when a single Reidemeister move is applied. Proof. Let’s suppose that we start with a knot or link projection which is tricolored and we do a single Reidemeister move. We want to show that the new projection is also tricolorable. Since a Reidemeister move only changes a projection in a small disk (which we will refer to as “the disk”), we want to show that it’s possible to color the new projection within this disk so that the arcs exiting the disk are the same color as they were before and at least two colors are still used in the overall projection. If we can do this, we’ll have a tricoloring of our new projection, and hence we’ll be able to conclude that doing this one Reidemeister move did not change the tricolorability of the projection. We’ll prove this for Reidemeister moves (R1) and (R2) and give some hints about how to prove it for (R3). You can fill in the details for (R3) in the Exercises. Reidemeister I move: First suppose that we want to apply an (R1) move to get rid of a kink (this means going from left to right in Figure 27). Observe in Figure 27 that just two arcs come together at the crossing of a kink. By Felicity’s Rule (2), the projection within the disk must be a single color, say black, and hence the arcs leaving the disk are both colored black. Now coloring the single arc that’s left black will be consistent with the coloring of the arcs leaving the disk. Also, we have not changed the total number of colors used in the overall projection. Thus, this gives us a tricoloring of the overall projection after the (R1) move. Next suppose we want to apply an (R1) move to add a kink (this means going from right to left in Figure 27). Then the original projection in the disk is a single

Figure 27. Tricoloring within a small region before and after an (R1) move. 230 9. INTRODUCTION TO KNOT THEORY arc, and hence must be colored with a single color, say black. When we add the kink we color its two arcs black so that again the coloring in the region is consistent with the coloring of the arcs leaving the disk. Again we have not changed the total number of colors used in the overall projection. Thus, we still have a tricoloring of the overall projection after the (R1) move. So doing a single (R1) move will not change a tricolorable projection into a not tricolorable projection. Reidemeister II move: First suppose that the arcs in the disk where we’re going to do an (R2) move are originally all the same color, say black. Then, as we saw in the proof for (R1) moves, whether we are going from the projection on one of the sides of Figure 28 to the projection in the middle or vice versa, by coloring the new arcs in the disk black the coloring will be consistent with the coloring of the arcs leaving the disk and we won’t have changed the number of colors used in the overall projection. Thus we will still have a tricoloring of the new projection.

Figure 28. If the arcs in the disk are all black, then we can color them all black after doing an (R2) move.

Next suppose that the arcs in the disk are not all a single color, and we are doing an (R2) move to remove two crossings. Since one arc goes over both crossings, it must have only one color (say black as illustrated on the far left or far right of Figure 29). Say the top of the other arc is grey. Then the middle arc is forced to be striped, and hence the remaining bottom arc is forced to be grey. Thus prior to the (R2) move, the arcs in the disk must be colored as illustrated on the far left or far right in Figure 29. After doing the (R2) move to remove the two crossings, we color one of the new arcs black and the other new arc grey so that they match up with the colors of the arcs leaving the disk. Also, since the coloring in the disk uses more than one color, the overall coloring satisﬁes Felicity’s Rule (1) of tricolorability. Thus we have a tricoloring of our new projection.

Figure 29. The arcs in the disk are not all the same color, and we do an (R2) move.

Finally, suppose that the arcs in the disk are not all a single color, and we are doing an (R2) move to add two crossings. Then originally there are no crossings in the disk, and we can assume one arc is black and the other is grey as illustrated in the middle two drawings in Figure 29. Now after we do the (R2) move, we can 8. TRICOLORABILITY AND KNOT EQUIVALENCE 231 color the new arc that is created striped. This will give us a tricoloring within the disk which matches up with the coloring of the arcs exiting the disk. Also since the coloring in the disk uses all three colors, the overall coloring satisﬁes Felicity’s Rule (1) of tricolorability. This again gives us a tricoloring of the new projection. Reidemeister III move: This proof is one of the Exercises. It’s similar to the proof for Reidemeister II moves. However, for (R2) moves we saw that there are two cases depending on whether or not all of the arcs in the disk are the same color. Since there are more crossings involved in an (R3) move, there are more cases that you have to consider. For example, in Figure 30 we illustrate three diﬀerent types of colorings inside of the disk where an (R3) move will take place.

Figure 30. Three diﬀerent colorings inside of the disk where an (R3) move will take place.

Try to list all possible colorings of the arcs in the disk. Then see if one of your diagrams can be turned around or the colors can be interchanged to get another one of your diagrams. If so, then these two cases are analogous and, hence, don’t need separate proofs. Once you eliminate the analogous cases, there should be ﬁve cases remaining. In each case, you need to show that it’s possible to color the new projection within the disk so that the arcs exiting the disk are the same color as they were before the (R3) move and at least two colors are still used in the projection.

In the arguments above for moves (R1), (R2), and (R3) we showed that if a projection of a knot or link is tricolorable and we do a single Reidemeister move, then the resulting projection will still be tricolorable. But if we start with a projection which is NOT tricolorable and do a single Reidemeister move, could we end up with a projection which IS tricolorable? In fact, this follows from what we have already shown. In particular, if we ended up with a tricolorable projection, we would just start with that projection and do the Reidemeister move in reverse to show that our original projection was also tricolorable. Thus once you have completed the argument for (R3) moves, we will have shown that doing a single Reidemeister move will neither change a tricolorable projection to one which is not tricolorable nor will it change a not tricolorable projection to one which is tricolorable. Now we can use Theorem 9.2 together with Reidemeister’s Theorem to prove the following corollary. Corollary. One projection of a knot or link is tricolorable if and only if every projection of the link is tricolorable. Since our corollary contains the phrase “if and only if” we need to prove both “if some projection is tricolorable, then every projection is tricolorable” and “if every 232 9. INTRODUCTION TO KNOT THEORY projection is tricolorable then a particular projection is tricolorable”. Recall that we use the symbol (=⇒) to indicate that we are about to prove the ﬁrst statement, and (⇐=) to indicate that we’re about to prove the second statement.

Proof. (=⇒) Suppose that some projection of a knot or link is tricolorable. By Reidemeister’s Theorem, any other projection of the link can be obtained by a finite sequence of Reidemeister moves together with planar deformations. As we remarked earlier, a planar deformation will not have any effect on tricolorability. Also by Theorem 9.2, after each subsequent Reidemeister move, the new projection is still tricolorable. Hence, after a finite sequence of Reidemeister moves and planar deformations, the projection is still tricolorable. Thus every projection of the knot or link is tricolorable. (⇐=) If every projection of a knot or link is tricolorable, then certainly any particular projection is tricolorable.

We could restate this corollary as saying that if a pair of knots or links is equivalent and a projection of one of the knots or links is tricolorable, then any projection of the other one must also be tricolorable. Also, it follows from the corollary that if one projection of a knot or link is not tricolorable, then no projection is tricolorable. Thus we can now conclude that the figure eight knot is not tricolorable, since Felicity showed us that one projection of it is not tricolorable. Since the trefoil is tricolorable and the figure eight is not, we’ve thus shown that the trefoil and the figure eight knot are not equivalent. Observe that the unknot is not tricolorable because it has a projection with no crossings which would have to be colored with a single color, violating Rule (1) of tricolorability. On the other hand a trivial link with two components is tricolorable, since even though the link has a projection with no crossings, each component can be colored with a different color. Since the figure eight and the unknot are both not tricolorable, we don’t yet know how to prove that the figure eight knot is not equivalent to the unknot. Similarly, if two knots or links are both tricolorable, then we cannot distinguish them from each other in this way.

9. Oriented Knots and Invertibility

A. Dash is fond of taking walks around her universe, stopping occasionally for a picnic. Sometimes, though, after some wine and cheese, she forgets which direction she is going. She’d prefer not to retrace her steps, so she decides to mark the direction she’s walking with an arrow every time she stops. A knot with a chosen direction like this is said to be oriented. The direction, or orientation, is shown by one or more arrows on the knot, drawn consistently like signs on a one-way street (see Figure 31). In order to orient a link, we have to put an orientation on each component of the link. If K is an oriented knot or link, then −K denotes K with the opposite orientation on every component. Figure 32 shows that when the trefoil knot is turned over, we get the same projection that we started with, but with the arrow reversed. Since the trefoil knot is denoted by 31,wesaythat31 is equivalent to −31. However, not every oriented knot is equivalent to itself with its orientation reversed. 10. CONNECTED SUMS OF NON-INVERTIBLE KNOTS 233

Figure 31. A. Dash gives her knotted universe an orientation.

Figure 32. When we turn over the trefoil knot its projection is the same but the direction of the arrow is reversed.

Definition. If an oriented knot K can be deformed so that it has an identical projection but with its orientation reversed, then we say the knot is invertible.If no such deformation exists, then we say the knot is non-invertible. We saw in Figure 32 that the trefoil knot is invertible. In fact, most knots in the table in Figure 13 are invertible. It was not until 1963 that the ﬁrst non-invertible knot was even discovered. The knot with the fewest crossings that is non-invertible is 817 (illustrated in Figure 33), though this was not the ﬁrst knot which was shown to be non-invertible.

Figure 33. 817 is the non-invertible knot with the fewest number of crossings.

10. Connected Sums of Non-invertible Knots When working with non-invertible knots, the deﬁnition of the connected sum operation is slightly more complicated than what we gave in Section 4. In particular, if K1 and K2 are non-invertible knots, then we have to orient the knots before we can take their connected sum so that we know which end of one knotted arc to attach to which end of the other knotted arc. This is not a problem for the connected sum of invertible knots, since an invertible knot can be reversed by a deformation.

Definition. Let K1 and K2 be oriented knots. The connected sum K1#K2 is formed by cutting each knot and gluing the endpoints of K1 to the endpoints of K2 in such a way that the arrow going out of K1 is attached to the arrow going into K2. 234 9. INTRODUCTION TO KNOT THEORY

The following examples involving the non-invertible knot 817 show why it’s important to specify an orientation when forming the connected sum of non-invertible knots. We begin by giving 817 an orientation, and then use the above deﬁnition to create the knot 817#817 in Figure 34. The original direction of the orientation doesn’t matter, but once we assign an orientation, we have to stick with it. In particular, both copies of 817 have to have the same orientation before we take their connected sum. If we want to use the opposite orientation on one of the knots, then we write −817 rather than 817. Observe in Figure 34 that since we have glued the arrow going out of one 817 to the arrow going into the other 817, the arrows on the connected sum match up so that 817#817 has a single orientation.

Figure 34. 817#817.

If we had glued the arrow going out of one 817 to the arrow going out of the other 817, we would end up with conﬂicting orientations on the connected sum. You wouldn’t want to be driving down a narrow one-way street and encounter two one-way signs pointing head to head, just as you see another car speeding towards you. In Figure 35 we illustrate the connected sum 817# − 817. Notice that since the knots have opposite orientations, we need to turn one knot over in order to line up the endpoints so that we can glue them correctly. Since 817 and −817 are not equivalent, the connected sums 817#817 and 817# − 817 are not equivalent.

Figure 35. 817# − 817.

One way to illustrate connected sums of oriented knots without having to keep track of all of the crossings is as beads with pictures on them. If a knot is invertible, then its bead has a picture that is the same regardless of which hole is up and which hole is down. If the knot is non-invertible, then its bead looks different if we turn it upside down. Figure 36 illustrates a bead representing an invertible knot and another bead representing a non-invertible knot. We can make a necklace out of two non-invertible beads in two different ways, as illustrated in Figure 37. If your beads represent the knot 817, the necklace on the left represents the connected sum 817#817 because the two beads are both oriented stem to flower. The necklace on the right represents the connected sum 817# − 817 because the left bead is oriented stem to flower and the right bead is oriented flower to stem. This example shows that orientation really does matter for connected sums of non-invertible knots. 11. EXERCISES 235

Figure 36. An invertible bead and a non-invertible bead.

Figure 37. We can make a necklace out of two non-invertible beads in two diﬀerent ways.

Let’s say Felicity asked you and A. Square to each show her the connected sum of two unoriented 817 knots. You might come up with the knot in Figure 34, without any arrows, while A. Square might draw the knot in Figure 35 (again, without any arrows). These knots are non-equivalent, but both you and A. Square would have done everything correctly. This shows us that the connected sum operation without orientations is ambiguous for non-invertible knots. Thus, while it still makes sense to talk about the connected sums of unoriented knots in general, we have to be careful to always use orientations if one of our knots might be non-invertible.

11. Exercises

1. Use Reidemeister moves to show that the links in Figure 38 are equivalent. In other words, show a sequence of moves I, II, and III taking one projection to the other.

Figure 38. Illustration for Exercise 1.

2. Use Reidemeister moves to show that the knot in Figure 39 is really the unknot. 3. Prove that every knot has a projection with over 1000 crossings. 236 9. INTRODUCTION TO KNOT THEORY

Figure 39. Illustration for Exercise 2.

4. Show that the third Reidemeister move preserves tricolorability. This proof is similar to the proof in the text for Reidemeister II moves. However, there are more cases because you have to consider all the ways that the arcs exiting a small disk around the (R3) move might be colored. By turning the diagram around or interchanging colors, you can see that some of the cases are analogous and hence don’t need separate proofs. Once you eliminate the analogous cases, there will only be ﬁve cases remaining. In each case, you need to show that it’s possible to color the new projection within the disk so that the arcs exiting the disk are the same color as they were before the (R3) move and at least two colors are still used in the projection. 5. Consider all of the seven crossing knots in the tables and determine which ones are tricolorable.

6. Draw a picture of 31#41, and show that it is tricolorable. If you used some other knot in place of 41 in this connected sum, would the connected sum still be tricolorable?

7. Let K1 be any tricolorable knot, and let K2 be any knot (which may or may not be tricolorable). Prove that K1#K2 is tricolorable. 8. There are many diﬀerent ways to color a tricolorable knot. For example all of the black arcs could be colored grey, while all of the grey arcs are changed to striped, and all of the striped arcs are colored black. (a) Show that the projection of the trefoil knot on the left in Figure 40 has exactly six tricolorings. (b) Find two diﬀerent tricolorings of the trefoil knot where one can be obtained from the other by simply permuting the three colors. Such tricolorings are called equivalent. How many non-equivalent tricolorings does the trefoil knot have? (c) How many tricolorings does the square knot (shown on the right in Fig- ure 40 right) have? (d) How many non-equivalent tricolorings does the square knot have? 9. Show that the two projections of the link in Figure 41 are equivalent. 10. Prove that the link in Figure 41 is not tricolorable and therefore is not the trivial link of two components. 11. EXERCISES 237

Figure 40. Illustration for Exercise 8.

Figure 41. Illustration for Exercises 9 and 10.

11. Draw two crossings in the plane, and consider all possible ways to connect the diﬀerent endpoints. Then use all of the ways of connecting the endpoints as cases in a proof that there can be no non-trivial knots with only two crossings. How does this imply that the crossing number of a trefoil knot is 3? 12. Give an upper bound for the number of links with crossing number n (don’t try to ﬁnd the least upper bound).

13. Prove that for all unoriented knots K1 and K2, we have the inequality

c(K1#K2) ≤ c(K1)+c(K2). ∗ 14. Here we will show that 41 is equivalent to 41. To do this, draw a deformation of the ﬁgure eight knot (shown in Figure 42) to its mirror image. Don’t use Reidemeister moves! Instead, try to make your deformation short and easy to follow. You may want to color the arcs of your knot with diﬀerent colors to make it clear which part of the knot is being deformed at each step.

Figure 42. Illustration for Exercise 14.

15. Prove that the two links illustrated in Figure 41 are equivalent. 16. Determine how many ways there are to orient a link with n components.

17. Recall that the knot 817 (shown in Figure 43) is non-invertible. Are the knots 817#817 or (−817)#817 invertible? You may want to draw beads to represent the knots in this problem.

18. Is it true that −(31#817)=−31# − 817? Justify your answer. 238 9. INTRODUCTION TO KNOT THEORY

Figure 43. Illustration for Exercise 17.

19. Is 31#817 equivalent to (−31)#817?Is31#817 invertible? You may want to draw beads to represent the knots in this problem.

20. If K1 is an invertible prime knot and K2 is a non-invertible prime knot, are K1#K2 and (−K1)#K2 equivalent? Is K1#K2 necessarily invertible? You may want to draw invertible and non-invertible beads to represent the knots in this problem. 21. Prove that any knot can be unknotted by changing some number of crossings. Hint: Think about how a water slide always goes down, and use that as a model for your projection. https://doi.org/10.1090//mbk/096/10

CHAPTER 10

Invariants of Knots and Links

Topics: • Linking number • Writhe • Tait conjectures • Unknotting number • Bridge number • Stick number • Lattice number • Seifert surfaces and genus

1. What’s an Invariant?

After a long day of studying, you settle into a comfy chair and flip on the television. Surfing through the channels, you stop at a new reality show. One of the participants is named Buttercup. “Funny”, you say to yourself. “The only person I’ve ever known named Buttercup is my cousin.” You haven’t seen your cousin since she was a kid, so who knows what she looks like now. You pause and look more closely at the TV screen trying to figure out if it could really be her. Unfortunately, as much as you would like to be related to a reality TV star, having the first name of your cousin is no proof that this woman is your cousin. You continue watching the show, making a mental note of everything you see or hear about Buttercup. She has brown eyes and black hair just like your cousin. She mentions she was born in New York, and she has a fear of heights. You search your brain trying to remember if the same things were also true about your cousin. “Fiddlesticks”, you mutter and flip off the TV. It has been too long. You won’t be able to figure this out without asking your mother. Even so, the time spent watching the reality show has (for perhaps the first time ever!) not been a waste. It reminded you of your other favorite pastime—comparing knots. The primary goal of knot theory is to develop methods to determine whether two knots are equivalent or not. This goal is similar to what you were just trying to do with the two Buttercups. Consider the two knot projections in Figure 1. Do these projections represent the same knot or different knots? The Buttercup problem might take some time to solve (and a little help from your mother), but it’s no match in difficulty to this knot puzzle. This particular puzzle took over 50 years to solve.

239 240 10. INVARIANTS OF KNOTS AND LINKS

Figure 1. Perko showed that these two projections represent the same knot.

If we can find a deformation from one knot in Figure 1 to the other, then they are equivalent. However, looking for a deformation won’t help us show that the knots are different. Suppose you sit at a table, trying to deform one knotted extension cord into the other, for years, without ever succeeding. This might provide strong evidence that the knots are not equivalent, however you can’t be 100% sure. Maybe just one more day of playing with the cord would enable you to show that the knots are equivalent. In fact, the pair of knots in Figure 1, were listed separately in the original knot tables. But in 1974, the two projections were shown to be equivalent by Kenneth Perko, a lawyer who had studied knot theory as an undergraduate. Since then, these two projections have been known as the Perko pair. While the Perko pair turned out to be the same knot, we’d like a method that would enable us to prove that knots that seem different really are different. For example, so far we haven’t even been able to prove that the figure eight knot cannot be deformed to the unknot. Recall from Chapter 3 that we said that a topological property of a space is a characteristic of the space that is unchanged when the space is deformed. Such properties enable us to distinguish spaces. For example, we know that the torus and the Klein bottle are topologically distinct because the torus has the property of being orientable whereas the Klein bottle does not. Looking for distinguishing characteristics was also what you were trying to do with the two Buttercups. If the TV Buttercup was born in New York and your cousin Buttercup was born in California, then you would know they must be different people. We would like to find topological properties that enable us to distinguish different knots in a similar way. Definition. An invariant of a knot or link is a property that does not change when the knot or link is deformed in space. This means that in order to show that something is an invariant, we have to check that it won’t change when we deform the knot or link. If a quantity is defined over all possible projections of a knot or link, then it’s an invariant by definition. Before we introduce some specific knot invariants, let’s think about how we use characteristics to distinguish people that we don’t know well. We often recognize people based on physical traits like eye color, hair color, facial hair, glasses, or height. We can imagine that each of these characteristics sorts people into different buckets. Of course, these buckets are figurative and not literal, but that’s immaterial—the idea is the same. For instance, if we consider eye color, we might have different buckets filled with blue-eyed people, green-eyed people, brown-eyed people, people with bright purple contact lenses that they never remove, and so on. 2. CROSSING NUMBER, TRICOLORABILITY, AND NUMBER OF COMPONENTS 241

The buckets might be labeled according to the eye color for easy reference. If you have two acquaintances who belong to different buckets (maybe one has green eyes and the other has brown eyes), then you can tell them apart immediately using this “bucket system”. However, if two people belong to the same bucket because they both have bright purple contact lenses that they never remove, then you will need to move on to a different bucket system—maybe hair color—to distinguish between them. Invariants work the same way for knots and links. We can think of an invariant as a bucket system in which knots and links are sorted into different buckets according to a particular invariant. If we have two equivalent knots or links, then they will always be placed in the same bucket no matter how many bucket systems we use because the value of any invariant will be the same. On the other hand, if we find an invariant that places a pair of knots or links in different buckets (because they have different values of the invariant), then we can be certain that the knots or links are not equivalent to one another. But be careful. Just because a pair of knots or links are in the same bucket or even several of the same buckets doesn’t mean they are necessarily equivalent. This is just like the fact that two different people can be placed in the same two buckets because they both have brown eyes and black hair.

2. Crossing Number, Tricolorability, and Number of Components

Our first example of an invariant of knots and links is the crossing number. Recall from Chapter 9 that the crossing number c(K) of a knot or link K is the fewest number of crossings of any projection of K. This means no matter how we deform K, a projection of it will have at least c(K) crossings. To compute the crossing number, we start with any projection K and count the number of crossings. We then try to deform K in such a way that the number of crossings is reduced. If we find a projection with fewer crossings, then we repeat the process with this new projection. If there is no projection of K with fewer than n crossings, then c(K)=n. Since the crossing number is defined in over all possible projections of the knot or link, by definition the crossing number is an invariant. The problem with this method of computing the crossing number is that we might not know whether in fact there is a projection of K with fewer than n crossings. For example, Figure 2 illustrates a knot with 15 crossings. Perhaps if we deformed this knot, we would find a projection with fewer than 15 crossings. Maybe if we worked at it long enough, we could even show that this knot could be deformed to the unknot. But just because we can’t figure out how to reduce the number of crossings doesn’t mean there isn’t some way to do it. For this reason the crossing number is difficult to compute, and hence it’s not very usable for distinguishing knots. Tricolorability which was introduced in Chapter 9, is another example of a knot and link invariant. Recall that in Chapter 9 we showed that tricolorability is preserved by all three Reidemeister moves as well as by planar deformation. Since Reidemeister’s Theorem tells us that we can get from one projection of a knot or link to any other projection via a finite sequence of Reidemeister moves and planar deformations, we know that tricolorability is unchanged by any deformation. It follows that tricolorability is indeed an invariant. 242 10. INVARIANTS OF KNOTS AND LINKS

Figure 2. Can we ﬁnd a projection of this knot with fewer than 15 crossings?

Also, in contrast with crossing number, tricolorability is not hard to determine. There are only so many ways to color a projection following the rules of tricolorability. So if we have a little more patience than Felicity had, we can try each possibility and reach a definite conclusion about whether or not a projection is tricolorable. However, because tricolorability only divides knots and links into those that are tricolorable and those that aren’t, it too has limited usefulness. For example, since the trefoil knot is tricolorable and the figure eight knot is not, we know that they are topologically distinct. But since both the figure eight knot and the unknot are not tricolorable, tricolorability won’t help us determine if they are topologically the same or different. In particular, just because two knots are both tricolorable or both are not tricolorable doesn’t mean they are the same. Similarly, just because two people are both called Buttercup or both have brown eyes doesn’t mean they’re the same person. Another example of an invariant is the number of components of a link. Each component of a link is a disjoint loop and a deformation of a link cannot add or remove a component. Thus, every projection of a link will have the same number of components. Also, like tricolorability, the number of components of a link is easy to compute. But also like tricolorability, the number of components doesn’t help us distinguish many knots or links. In fact, all knots are links with one component. So the number of components won’t help us distinguish knots at all. In addition, the number of components of a link is so easy to see that you wouldn’t even be wondering if two links with different numbers of components are equivalent. It’s like if we had a bucket system for distinguishing people, and there was one bucket for mannequins and another for humans. This wouldn’t help us much since the difference between a person and a mannequin is pretty obvious. So far we have three invariants—crossing number, tricolorability, and the number of components—none of which is particularly helpful. Each one is either hard to compute and hence not very usable, or it doesn’t distinguish many knots or links and hence is not very useful. In fact, there is no known invariant that is computable and distinguishes every pair of inequivalent knots or links. We may have to try several invariants before we are able to distinguish a particular pair of knots or links. Thus, we want to have lots of invariants at our disposal—the more, the better!

3. Positive and Negative Crossings

You are deep in thought trying to imagine an invariant that would be both easy to compute and useful, when your phone rings. It’s our 1-dimensional friend 3. POSITIVE AND NEGATIVE CROSSINGS 243

A. Dash who loves talking to higher-dimensional beings because they seem so complicated and mysterious. “Remember how I like to use arrows to keep myself from backtracking on my afternoon walks?” she asks. “I have a feeling arrows are useful for other things, too. Maybe you can use them to come up with a knot invariant?” A. Dash’s enthusiasm is infectious, but she doesn’t have any concrete plans for what to do. Rather, she asserts that with all of your complicated 3-dimensional thoughts you should be able to ﬁgure that out yourself. It’s understandable that A. Dash is not sure how to proceed. After all, she can’t even visualize a knot or link. How could she possibly distinguish a pair of knots or links? But we might have more luck. First let’s notice in Figure 3 that if there are no arrows on a projection, then every crossing looks like every other crossing if you just rotate it properly. However, we will see below that if we orient our knot or link, all the crossings in a projection can be divided into two diﬀerent types which we can use our hands to distinguish.

Figure 3. If we rotate the crossing on the left by 90◦,weobtain the crossing on the right.

A crossing is said to be positive or right-handed if when you put your right thumb on the over strand so that it points in the direction of the arrow, then the rest of the ﬁngers on your right hand point under the strand in the direction of the arrow on the undercrossing (see Figure 4).

Figure 4. A positive (right-handed) crossing.

A crossing is said to be negative or left-handed if when you put your left thumb on the over strand so that it points in the direction of the arrow, then the rest of the fingers on your left hand point under the strand in the direction of the arrow on the undercrossing (see Figure 5). In contrast with Figure 3, in Figure 6 we see that no matter how we rotate a positive crossing, we won’t get a negative crossing. A similar thing is true if we rotate a negative crossing. So even though you may have to rotate a picture (or your head) to figure out if a crossing is positive or negative, you don’t have to worry that you might get a different result depending on how you rotate it. Figure 7 illustrates an oriented projection of the trefoil knot. We can now go through the crossings one by one, using our hands to determine which ones are 244 10. INVARIANTS OF KNOTS AND LINKS

Figure 5. A negative (left-handed) crossing.

Figure 6. A positive crossing is still positive, no matter how it’s rotated. positive and which ones are negative. In fact, this particular projection has three negative crossings and no positive crossings (you should check this). What if you oriented the knot the other way? Would the crossings still be negative or would they become positive?

Figure 7. This oriented projection has three negative crossings.

When we assign a positive or negative sign to a crossing of an oriented knot or link, we call it a signed crossing. Signed crossings are going to play an important role in helping us distinguish knots and links, but it will take us a couple of sections to get there.

4. Writhe

Now instead of simply counting the total number of crossings of a projection of a knot or link, we can consider an oriented projection and count the number of positive crossings and the number of negative crossings separately. We see as follows that we can combine these two numbers into a single number, which tells us something about the way the knot is twisting in space. Definition. The writhe w(L) of an oriented link projection L is given by the formula total number of total number of w(L)= − . positive crossings negative crossings 5. LINKING NUMBER 245

Since the knot projection in Figure 7 has no positive crossings and three negative crossings, its writhe is −3. If we reverse the orientation on the knot (changing the direction of all the arrows) as in Figure 8, the negative crossings remain negative. Thus the writhe does not change.

Figure 8. The orientation of this projection is opposite that of Figure 7, but the writhe is the same.

At this point, you might be tempted to rush oﬀ and phone A. Dash to tell her we’ve found the invariant she was hoping for. But we need to slow down. We’ve made progress on her challenge, but we haven’t ﬁgured it out yet. The writhe is not actually an invariant, even though it is independent of how a knot is oriented. Consider the two projections of the same knot in Figure 9. One has writhe −3and the other has writhe 0. In fact, by adding twists to a projection, we can create a projection of an equivalent knot with any writhe we want. Thus the writhe depends on the particular projection and hence is not an invariant. Despite this, we will see in Section 6 that the writhe can help us distinguish knots or links which are in a special family.

Figure 9. These two projections of the same knot have diﬀerent writhes.

5. Linking Number

While the writhe is not an invariant, it turns out that a closely related value is an invariant of two component links. Now, instead of counting all positive and negative crossings, we will only count the crossings between the two components of the link.

Definition. Consider an oriented projection of a link with components K1 and K2.Thelinking number of K1 and K2 is deﬁned by the formula ⎡⎛ ⎞ ⎛ ⎞⎤ total number of total number of 1 lk(K ,K )= ⎣⎝ positive crossings ⎠ − ⎝ negative crossings ⎠⎦ . 1 2 2 between K1 and K2 between K1 and K2 246 10. INVARIANTS OF KNOTS AND LINKS

Figure 10 illustrates two oriented links together with their linking numbers. Note that we don’t consider all of the crossings of the projection on the right, only the four crossings between the diﬀerent components. You might be surprised that the link on the right has linking number 0, since it’s not the trivial link. However, the linking number is measuring the number of times that one component wraps around the other component, not whether or not the link is trivial. In particular, for the link on the right, the long component goes up and back through the small component, so it’s not really wrapping around at all.

Figure 10. Some links with their linking numbers.

In order to prove that the linking number is an invariant of oriented links, we use Reidemeister’s Theorem as we did in Chapter 9 to prove that tricolorability is an invariant. In particular, we have to show that the linking number will not change as a result of a planar deformation or doing any of the three Reidemeister moves. Proving this is one of the Exercises. While the linking number is an invariant of an oriented link, changing the orientation of just one of the components might change the linking number. For example, reversing the orientation of one of the components of the link on the left in Figure 10 gives us the link in Figure 11, which has linking number 1 rather than −1.

Figure 11. This link has linking number 1.

Note that reversing the orientation on one of components of a link will change the signs of all of the crossings between the two components, but will not otherwise aﬀect the linking number. This means that the absolute value of the linking number will be the same regardless of the orientation on the link. Thus we can deﬁne the following invariant of unoriented links.

Definition. Consider an unoriented link with components K1 and K2.The unsigned linking number of K1 and K2,is|lk(K1,K2)| for any orientation on K1 and K2. In some sense, the unsigned linking number tells us how much the two components are intertwined with each other; the larger the number, the more intertwined. 6. NUGATORY CROSSINGS AND ALTERNATING KNOTS AND LINKS 247

And since the linking number is unchanged by Reidemeister moves and the unsigned linking number is independent of orientation, the unsigned linking number is an invariant of unoriented two component links. For example, since the links in Figure 10 have diﬀerent unsigned linking numbers, we know they must be distinct. The unsigned linking number is an invariant for two component links that is more powerful than tricolorability. Now you can call A. Dash and impress her with how complicated and mysterious you are to have ﬁgured out this new invariant.

6. Nugatory Crossings and Alternating Knots and Links

It has been a few weeks since that evening when you thought you saw your cousin Buttercup on TV. By a strange coincidence your Mom tells you that your cousin Buttercup will be in town for a few days. So you invite her over to see for yourself if she was the woman that you saw on the TV show. As soon as she walks in, it’s obvious that she was not that woman. But you laugh about it together and then tell her what you’ve been learning about knot theory. You explain the idea of invariants, giving her the example of the crossing number of a knot. Buttercup sees the 15 crossing projection from Figure 2 lying on your desk and asks, “So, what’s the crossing number of this knot?” Truth be told, all you know about this knot is that its crossing number is no more than 15. In order to avoid having to admit this to her, you quickly try to flip through this chapter to see if you can learn more about the crossing number of this knot. In fact, you’re in luck. While the crossing number is difficult to compute for most knots and links, there is a special family of knots and links for which it’s easy to compute. And the knot from Figure 2 is a member of this family. Before we describe the family, we need a few more definitions. Definition. A crossing in a knot or link projection is said to be nugatory if it can be removed by twisting part of the knot (see Figure 12).

Figure 12. We can twist the parts of the knots in the dotted circles to remove these crossings.

According to an English dictionary, the word “nugatory” means “worthless, unimportant, or invalid”. For example, you could use it in a sentence such as “All of your excuses for not wanting to study math with me are nugatory”. The knot theory deﬁnition of nugatory is not all that diﬀerent in the sense that a nugatory crossing contributes nothing to a link. For example, each projection in Figure 12 has a nugatory crossing (indicated by an arrow) that can easily be eliminated by twisting the part of the knot in the dotted circle. None of the other crossings in these projections are nugatory. 248 10. INVARIANTS OF KNOTS AND LINKS

Definition. A projection of a link is said to be reduced if it contains no nugatory crossings. Just because a projection is reduced doesn’t mean you can’t remove more crossings by deforming it in some other way. For example, in Figure 13 we illustrate a reduced projection of the unknot. See if you can deform it to a planar circle.

Figure 13. An unknot with no nugatory crossings.

Definition. A projection of a knot or link is said to be alternating if consecutive crossings on each component alternate between going over and going under. Aknotorlinkissaidtobealternating if it has an alternating projection. For example, in Figure 14 we illustrate an alternating and a non-alternating knot.

Figure 14. An alternating and a non-alternating knot.

7. Tait’s Conjectures about Alternating Knots

It turns out that link projections which are both alternating and reduced are really great to work with. In 1870, P. G. Tait, (who constructed the first knot tables) made several important assertions, including two about reduced alternating projections. Since Tait gave intuitive arguments rather than proofs of these assertions, they came to be known as Tait’s Conjectures. His first conjecture about alternating knots was finally proven in 1986, more than 100 years after he first asserted it. While it is still known as “Tait’s Conjecture”, since it is now proven, we refer to it as a theorem. You might see the proof of this theorem if you take a more advanced knot theory course. 7. TAIT’S CONJECTURES ABOUT ALTERNATING KNOTS 249

Theorem 10.1 (Tait Conjecture #1). A reduced alternating projection of a knot or link has the fewest number of crossings of any projection of that knot or link. This theorem should certainly help you impress your cousin Buttercup with your vast knowledge of knot theory. In particular, since the knot projection in Figure 2 is reduced and alternating, we know by Theorem 10.1 that the number of crossings is as small as possible. Thus you can confidently tell Buttercup that the knot on your desk indeed has crossing number 15. Similarly, since the projection of the figure eight knot on the left side of Figure 14 is reduced and alternating, we know immediately that it has crossing number 4. This enables us to distinguish the figure eight knot from the unknot, which of course has crossing number 0. Note that the crossing number is still hard to compute for knots and links that don’t have reduced alternating projections, and remains a mysterious invariant in other ways. In fact, the following remains an unproven conjecture.

Conjecture: For any knots K1 and K2, c(K1#K2)=c(K1)+c(K2). For example, the crossing number of a trefoil knot is 3 and the crossing number of a ﬁgure eight knot is 4. In Figure 15 we illustrate a projection of the connected sum of a trefoil and a ﬁgure eight knot which has seven crossings. Is it possible to draw it in some other way with only six or even fewer crossings? Can we we apply Theorem 10.1 in this example?

Figure 15. Is there a projection of this connected sum with fewer than seven crossings?

Among P. G. Tait’s other ideas was a conjecture about the writhe. We saw in Section 4 that the writhe is not an invariant of knot or links. In particular, Figure 9 illustrated how we can use Reidemeister I moves to make a given oriented knot or link have whatever writhe we want. However, we see in the following theorem that if we restrict ourselves to reduced alternating oriented projections, then writhe actually is an invariant. You might see the proof of this theorem in a more advanced course on knot theory. Theorem 10.2 (Tait Conjecture #2). All reduced alternating projections of a given oriented knot or link have the same writhe. Since the writhe of a knot projection does not depend on its orientation, it follows from this theorem that if a pair of knots have reduced alternating projections with diﬀerent writhes, then the knots are not equivalent.1 For example, both of the projections in Figure 16 have six crossings and are reduced and alternating. You can check that the knot on the left has writhe equal to −2, while the knot on the right has writhe equal to 0. Thus we know that the knots are not equivalent.

1The situation is slightly more complicated for links, since then the writhe depends on the orientation of each component. 250 10. INVARIANTS OF KNOTS AND LINKS

Figure 16. These knots are not equivalent since they have reduced alternating projections with diﬀerent writhes.

Theorem 10.2 can also help us distinguish links and their mirror images. In the Exercises for Chapter 9, you showed that the ﬁgure eight knot and its mirror image are equivalent. However, many knots are not equivalent to their mirror images. For example, we can use Theorem 10.2 to show that the trefoil knot and its mirror image are distinct. Exercises 22, 23, and 24 will help you further explore the relationship between alternating knots and links and their mirror images.

8. What Proportion of Knots Are Alternating?

Remember that we originally started discussing alternating knots because your cousin Buttercup asked you how you knew that a particular projection of a knot with 15 crossings couldn’t be deformed to have fewer crossings. It’s a good thing the projection was reduced and alternating. That made things easy since we had Theorem 10.1 in our back pocket. But what are the chances that if Buttercup chooses another knot projection at random it would also be alternating? Let’s consider the knot table from Chapter 9. Working our way from the unknot up through the eight-crossing knots, we see that every knot on the table is alternating except for 819,820,and821. From this information we might conclude that almost all knots are alternating. As nice as this conclusion seems, it is not actually true. Table 1 shows what happens as we look at prime knots with more and more crossings (question marks indicate quantities that are not currently known). Look at the column labeled “Proportion alternating”, and observe that this number falls markedly after the ﬁrst nine lines. Though the actual proportion of alternating knots is not known for crossing numbers greater than 16, it was proven in 2009 that this proportion gets closer and closer to zero as crossing numbers get bigger and bigger. So you were a bit lucky with Buttercup’s question. If Buttercup randomly chose another knot with a high crossing number, the likelihood of it being an alternating knot would be low. 8. WHAT PROPORTION OF KNOTS ARE ALTERNATING? 251

Table 1. Thenumberofprimeknotsofagivencrossingnumber and the proportion of them that are alternating.

Crossing Known to be Known to be Total Proportion number alternating non-alternating number alternating 0 1 0 1 1 3 1 0 1 1 4 1 0 1 1 5 2 0 2 1 6 3 0 3 1 7 7 0 7 1 8 18 3 21 .86 9 41 8 49 .84 10 123 42 165 .75 11 367 185 552 .66 12 1288 888 2176 .59 13 4878 5110 9988 .49 14 19536 27436 46972 .42 15 85263 168030 253293 .34 16 379799 1008906 1388705 .27 17 1769979 ??? ??? ??? 18 8400285 ??? ??? ??? 19 40619385 ??? ??? ??? 20 199631939 ??? ??? ??? 21 990623857 ??? ??? ??? 22 4976016485 ??? ??? ??? 252 10. INVARIANTS OF KNOTS AND LINKS

9. Seifert Surfaces

Our next invariant will connect knots and surfaces—two worlds which we have studied separately until now. First, we need to specify what we will mean by the word “surface” in this context. Previously, we used the word “surface” interchangeably with “2-manifold”. In particular, a surface was not allowed to have a boundary, because we were studying surfaces as possible 2-dimensional universes. Now that we will be using surfaces to analyze knots, all of our surfaces will have boundaries.

Definition. We deﬁne a Seifert surface for a knot to be an orientable surface whose boundary is the knot.

For example, a disk is a Seifert surface for the unknot. (In fact, the unknot is the only knot which has a disk as a Seifert surface.) Figure 17 shows another Seifert surface for the unknot. This Seifert surface was obtained by attaching a handle to a disk. Similarly, we can attach two, three, or any number of handles to a disk to get distinct Seifert surfaces for the unknot. This creates an inﬁnite collection of Seifert surfaces for the unknot.

Figure 17. A Seifert surface for the unknot.

While it was easy to list infinitely many Seifert surfaces for the unknot, finding Seifert surfaces for non-trivial knots requires a little more work. For example, suppose we want to find a Seifert surface for the trefoil knot. The first surface we think of is the one illustrated in Figure 18. This is a surface whose boundary is the knot, but it won’t be a Seifert surface unless it’s orientable. Unfortunately, as we can see in the figure, when A. Square goes for a walk around the surface, he comes back as his mirror image. A. Square remembers this phenomenon—it happened when he walked around the Möbius strip in Chapter 4. It means that the surface is non-orientable, and hence it’s not a Seifert surface. Let’s try again to find a Seifert surface for a trefoil knot. Recall from Theo- rem 4.1 in Chapter 4 that since R3 is an orientable 3-manifold, a surface in R3 is 2-sided if and only if it’s orientable. To check whether a surface in R3 is 2-sided, we can shade it with two colors using the rule that whenever we come to a crossing, we have to switch colors. We call such a coloring a checkerboard shading because it reminds us of a checkerboard where squares alternate between white and black. Notice that we can’t color the projection in Figure 19 with a checkerboard shading because we’ll be forced to have two adjacent regions that are the same color. 9. SEIFERT SURFACES 253

Figure 18. This surface is non-orientable, and hence it’s not a Seifert surface.

Figure 19. We can’t color this surface with a checkerboard shading.

In our quest to find an orientable surface with the trefoil knot as its boundary, let’s start with a different projection of the knot. On the left side of Figure 20, we illustrate a checkerboard shading of a different projection of the trefoil knot. Notice that at each crossing we switch colors, and hence this surface has two sides. You can also check that the surface is orientable by sending A. Square for walks around it, noticing that no matter what path A. Square follows, when he returns to his starting point he is not mirror reversed. Thus, the surface on the left in Figure 20 is a Seifert surface for the trefoil knot. Once we have one Seifert surface for a knot, we can create infinitely many Seifert surfaces by adding handles to our surface, just as we did for the unknot. For example, on the right in Figure 20, we illustrate a new Seifert surface that we get by attaching a handle to the surface on the left. Observe that adding handles in this way does not change the orientability of the surface. Remember how A. Dash was convinced that orienting her knot could lead to a useful invariant? We are in the same situation here with surfaces. The Seifert surfaces we’ve constructed ought to give us some helpful information about knots. 254 10. INVARIANTS OF KNOTS AND LINKS

Figure 20. Two Seifert surfaces for a trefoil knot.

Just like A. Dash, we may not see how we should proceed right away, but what we learned about surfaces in Chapter 8 will come in handy now.

10. The Genus of a Knot

In Chapter 6, we defined the genus of a finite surface without boundary to be the maximum number of disjoint circular cuts that can be made in a surface without disconnecting it. Then Theorem 6.6 told us that the genus of all finite area surfaces is given by: genus S2 =0, genus nT 2 = n, genus nP 2 = n.

Since for orientable surfaces the genus tells us exactly what the surface is, the genus seems like it might help us analyze Seifert surfaces for knots. However, in order to apply the definition of genus to a Seifert surface, we first need to get rid of the boundary. Definition. Let S be a Seifert surface for a knot K. We define the genus of S to be the genus of the surface that we get by abstractly gluing a disk along K. Keep in mind that this is an abstract gluing, since a non-trivial knot cannot bound a disk in R3. Instead we can think of the gluing as instant transport between the boundary of the Seifert surface and the disk. With this definition of the genus of a Seifert surface, we are ready to define our invariant. Definition. The genus of a knot is the smallest genus of any Seifert surface for the knot. Notice that by its definition, the genus of a knot is a non-negative integer. Also, since the genus is defined over all possible Seifert surfaces for any projection of the knot, then by definition it’s an invariant. However, because each knot has infinitely many Seifert surfaces, finding the one with the smallest genus seems like it could be a daunting task. However, once you have one Seifert surface for a knot, you at least know that the genus of the knot can’t be any bigger than the genus of that surface. Below, we tackle two examples which are not so daunting. For our first example, let’s determine the genus of (you guessed it!) the unknot. As we remarked above, the unknot has a disk as a Seifert surface. In order to find the genus of this Seifert surface, we have to glue another disk along its boundary. Recall from Chapter 1 that gluing two disks together along their boundary results in a sphere. Now, since genus S2 = 0 and 0 is the smallest possible value for the 10. THE GENUS OF A KNOT 255

Figure 21. A deformation of our surface to two light grey disks with three twisted bands joining them. genus of any surface, we can conclude that the unknot has genus equal to 0. In fact, the unknot is the only knot which has genus equal to 0, since the unknot is the only knot which bounds a disk in R3. Though you might have to wait until you take an advanced topology course before you see a proof of this seemingly simple fact. Next, let’s see if we can determine the genus of a trefoil knot. We begin by finding the genus of the Seifert surface illustrated on the left in Figure 21. This Seifert surface looks complicated, but we know that once we glue a disk to the boundary we will get a finite orientable surface without boundary. Hence, according to the Classification of Surfaces that we saw in Chapter 6, it is a connected sum of some number of tori. Before we glue a disk onto the boundary, we will do some manipulations of the Seifert surface to make it easier to recognize. We will describe our steps as we go along and illustrate each with a picture. As you read this, make sure that you understand how to get from each picture to the next one. Our first step is to deform the surface on the left in Figure 21 so that it looks like two disks joined together by three twisted bands as illustrated in the center of the figure. Next we flip the right disk over, turning it 180◦ along the dotted circle in the direction of the arrow. This gives us the image on the right. Notice that this step removes the twist in the bottom band, but adds a twist to each of the other bands and causes the top and middle bands to switch places on the disk on the right side. Since we are trying to determine the genus of the surface rather than its position in R3, we only care about its intrinsic topology. Thus we are free to cut the surface apart, as long as we carefully glue the exact same points back together. In Figure 22 we cut open the two doubly twisted bands from Figure 21, undo the twists, and then glue them back together. This makes the surface on the right a lot simpler than the one on the left. Notice that when we reattach the bands in the drawing on the right, we have to be careful to glue together exactly the same points that we cut apart. This means that the top two bands still have to criss-cross even after we untwist them individually. Next starting with the image on the right in Figure 22, we fatten the bottom band to get the image at the top left in Figure 23. Then we stretch out the bottom and curve it up in the back to create a bowl shape with a dark inside. Next we smooth out the picture so that it looks like a basket with two handles. Finally, we fatten the lower handle, and slide the top handle so that it goes all the way from the left side of the basket to the right side of the basket, lifting it so that we only see its dark side. 256 10. INVARIANTS OF KNOTS AND LINKS

Figure 22. We cut the two doubly twisted bands, untwist them, and then reattach them.

Figure 23. We deform the surface to get a basket with two handles.

Next in Figure 24, we make the lower handle shorter and shorter while making the top handle wider and wider, to get the picture on the right.

Figure 24. We make the lower handle shorter and the top handle wider.

Next, on the left side of Figure 25, we have pulled up the light grey underside around the hole. Then in the center of Figure 25, we enlarge the light grey while decreasing the size of the dark grey, until we obtain a torus with a long rectangular strip removed from the top. Finally, we shrink the rectangular hole until it becomes a small round hole. Viola! We have a torus with a disk removed. If we glued the disk back in, we would obtain a torus, which we know has genus 1. Thus, the genus of the Seifert surface that we started with in Figure 21 is 1. It follows that the genus of the trefoil knot is at most one. While you might think that we still need to consider the remaining inﬁnite number of other Seifert surfaces for the trefoil, we can actually stop here. Since the unknot is the only 11. USING EULER CHARACTERISTIC TO COMPUTE GENUS 257

Figure 25. We enlarge the light grey while shrinking the dark grey to get a torus with a disk removed. knot with genus equal to 0, we know there is no Seifert surface for the trefoil whose genus is less than 1. So the trefoil knot must have genus exactly equal to 1.

11. Using Euler Characteristic to Compute Genus

If all of the pictures in Figures 21 through 25 overwhelmed you, then you’ll probably prefer the following technique to compute the genus of a Seifert surface. We begin by recalling the definition of the Euler characteristic of a surface from Chapter 6. Definition. Let S be a surface without boundary which is made up of v vertices, e edges, and f faces, where a face is defined as a shrinkable region. We define the Euler characteristic of S to be the number χ(S)=v − e + f. In Chapter 6, A. Square and B. Triangle determined that an orientable surface with genus n has Euler characteristic χ =2− 2n. We will now go backwards and use the Euler characteristic of a Seifert surface to determine its genus. To see how this works, let’s start with the Seifert surface on the left in Figure 21. We divide it into vertices, edges, and faces as shown on the left in Figure 26. Counting each, we find that there are 12 vertices, 18 edges, and 5 faces. However, to determine the genus, we need to glue a disk to the boundary of the Seifert surface. We divide the boundary of this disk into the same number of vertices and edges that are on the boundary of the Seifert surface. Then we abstractly glue the new disk onto the boundary of the Seifert surface so that vertices are glued to vertices and edges are glued to edges. This does not change the total number of vertices or edges. but it adds one to the total number of faces. This gives us a total of 6 faces. Since the vertices and edges have already been counted when we counted those of the Seifert surface, we don’t need to add them into our totals. Thus, the Euler characteristic of the Seifert surface together with the disk

Figure 26. We divide the Seifert surface into vertices, edges, and faces, and then add a disk along the boundary. 258 10. INVARIANTS OF KNOTS AND LINKS is χ = v − e + f =12− 18 + 6 = 0. Since an orientable surface with genus n has Euler characteristic χ =2− 2n,weset2− 2n = 0 and solve to ﬁnd that the genus of the Seifert surface is n =1. This method is a lot easier than following all of the pictures to see that the Seifert surface is a torus with a disk removed. On the other hand, manipulating a Seifert surface to get a surface that we recognize gives us more intuition about the surface than adding and subtracting numbers. So both methods are useful.

12. A Potpourri of Knot Invariants

So far, we have learned about several knot invariants in some detail. We are like spectators at a fireworks show where we watch with attention as each of the fireworks ascends, bursts, and spreads across the sky. Now we close this chapter with a quick discussion of some additional knot invariants. This is not meant to be a complete list of invariants or even a detailed explanation of the invariants described. Rather, the goal of this section is to simply to give you a sense of the variety of invariants out there, which we hope will motivate you to learn more knot theory. It’s like the finale of a firework show, where fireworks burst one upon another in contrasting heights, shapes, and colors. You might even find yourself spontaneously saying a few oooh’s and aaah’s. Enjoy!

12.1. Unknotting Number. We kick off our “fireworks finale” with an invariant which reminds us of the winter holidays. Think beyond your warm fuzzy memories and recall the yearly fight you have with the strings of holiday lights stuffed in a box in a closet. Each year you struggle to untangle the mess and ask yourself, “How many times do I have to pull the whole string of lights through a tiny loop to untangle this stupid thing?” You wish you could just magically pass one piece of wire through another to instantly untangle the whole thing. This age-old holiday untangling problem can be recast in the setting of knot theory. Suppose you start with a non-trivial knot (similar to your tangled up string of lights if one end was plugged into the other). You want to magically convert this knot into an unknot. In one of the Exercises from Chapter 9, you showed that any knot projection can be turned into a projection of the unknot by changing some number of crossings from undercrossings to overcrossings and vice versa as illustrated in Figure 27. Keep in mind that changing crossings is not a deformation. So being able to change a knot to the unknot in this way does not mean your original knot was equivalent to the unknot.

Figure 27. This is how we change a crossing.

Definition. The unknotting number of a knot K,denotedu(K), is deﬁned as the smallest number of crossings that must be changed to obtain the unknot among all possible projections of K. 12. A POTPOURRI OF KNOT INVARIANTS 259

Since the unknotting number is deﬁned over all possible projections of a knot, it’s an invariant. We compute it for the trefoil knot as follows. First observe that because we know that the trefoil knot is truly knotted, every one of its projections will require at least one crossing change to produce a projection of the unknot. Recall that we denote the trefoil knot by 31. Hence, we write u(31) ≥ 1. We can prove that the unknotting number of the trefoil is exactly 1 by ﬁnding a projection in which one crossing change produces a projection of an unknot. In Figure 28, we change the crossing inside of the small grey circle to go from the left picture (which is the trefoil knot) to the right picture (which is the unknot). In fact, changing any crossing in this projection of the trefoil will give us the unknot.

Figure 28. Switching this crossing will change the trefoil into an unknot.

It has been shown that any knot with unknotting number 1 is necessarily prime. However, there are prime knots whose unknotting numbers are more than 1. For example, in Figure 29 we make a crossing change on the knot 71, and then deform it to get the knot 51. Then we make a crossing change on 51, and then deform it to get the trefoil knot 31. As we saw above, we can then make a crossing change on the trefoil to get the unknot. Thus, we can state with certainty that u(71) ≤ 3 and u(51) ≤ 2. However, perhaps some smaller number of crossing changes will also do the trick. Perhaps if we start with a different projection of 71 or 51 and do a single crossing change, we could get the unknot. In fact, the knot 51 really does have unknotting number 2, and the knot 71 really does have unknotting number 3. Butthesefactsarenoteasytoprove. According to the definition, in order to determine the unknotting number of a knot, we need to find a projection which requires the fewest possible crossing changes to get the unknot. Surprisingly, such a projection is not necessarily one with the minimal number of crossings. For example, the illustration of the knot 108 on the left in Figure 30 has ten crossings, which is the minimal crossing number of the knot. For this projection, we would need to change three crossings to get the unknot. However, if we deform the knot to the 14-crossing projection on the right, then we only need to change two crossings to get the unknot. Examples like this tell us that just seeing what happens when you change different crossings in a minimal crossing projection might not be sufficient to determine the unknotting number of a knot. Since it’s not possible to consider infinitely many different projections of a knot to check which one requires the fewest number of crossing changes, the unknotting number is generally difficult to compute. 260 10. INVARIANTS OF KNOTS AND LINKS

Figure 29. We see that u(71) ≤ 3andu(51) ≤ 2.

Figure 30. The projection on the left requires more crossing changes to get the unknot than the one on the right.

12.2. Bridge Number. Your friends love going to amusement parks, but you do not. Each roller coaster seems to be hurtling you either to an early death or to losing your lunch. To hide your fright, you take up the habit of counting the number of peaks on each ride. Other riders wonder why you’re shouting out numbers on the ride, but you don’t care. Any diversion from these near-death experiences is welcome! Moreover, you are getting good practice for our next knot invariant, the so-called bridge number of a knot. The bridge number is deﬁned as follows. Starting with a knot projection, we pick a direction which we will call up. As we trace along the projection, we will sometimes be moving upwards and sometimes downwards, just like when you’re riding on a roller coaster. We pay no attention to when we’re going forwards and backwards or side to side. Points at which we change from going upwards to going downwards—similar to the peaks of the roller coaster—are called bridges.These look like local maxima of a function. Note that valleys where we change direction from going downwards to going upwards, are not bridges.

Definition. The bridge number of a knot is deﬁned as the smallest number of bridges among all possible projections of the knot and all possible choices for the up direction of a projection. 12. A POTPOURRI OF KNOT INVARIANTS 261

Figure 31. This designation of up makes the ﬁgure eight knot have three bridges.

Since the bridge number is defined over all possible projections, by definition it’s an invariant. However, like the crossing number and the unknotting number, the bridge number is difficult to compute.For instance, suppose that we consider the projection of the figure eight knot shown in Figure 31 using the indicated direction as up. We highlight the bridges with thick grey arcs. Since this projection and direction has three bridges, we know that the bridge number of the figure eight knot is at most 3. On the other hand, in Figure 32 we rotate our projection in order to designate a different direction as up. In this new projection, we find that there are only two bridges. It can be shown that the only knot which has a projection with just one bridge is the unknot. Since we know that the figure eight knot is not equivalent to the unknot (because the projection in Figure 32 is reduced and alternating), this means that the figure eight knot has no projection with fewer than two bridges. In particular, the projection in Figure 32 has the minimum number of bridges. Thus the bridge number of the figure eight knot is indeed 2.

Figure 32. This projection of the ﬁgure eight knot has two bridges.

12.3. Stick Number. Until now, we have drawn our knots as curves in space. This certainly makes knot theory visually appealing. But nothing is preventing us from considering polygonal constructions of knots. In other words, we can construct knots from straight line segments (called sticks) that are glued at their endpoints to form a single knotted polygon in 3-dimensional space. Every knot can be constructed in this way by using lots of very short sticks. Definition. We define the stick number of a knot as the smallest number of sticks needed to construct that particular knot. Since the stick number is again defined over all possible stick constructions of a knot, the stick number is another invariant. Note that there is no requirement that all of the sticks be the same length, and you should not assume that the sticks have any particular thickness. The unknot has stick number 3 since it is impossible to make a circle with fewer than three sticks. In Figure 33, we illustrate a stick construction of the trefoil knot with six sticks and a stick construction of the figure eight knot with seven sticks. 262 10. INVARIANTS OF KNOTS AND LINKS

Figure 33. Stick constructions of the trefoil and the ﬁgure eight knot.

In order to prove that the trefoil has stick number 6 and the figure eight has stick number 7, we would have to show that each of the following statements holds: (1) Every 3-stick, 4-stick, and 5-stick configuration represents an unknot. (2) The illustration on the left side of Figure 33 is a 6-stick configuration of the trefoil knot. (3) The illustration on the right side of Figure 33 is a 7-stick configuration of the figure eight knot, but there is no 6-stick configuration of the figure eight knot. The difficulty involved in proving each of these steps is exacerbated by the fact that we need to consider polygonal configurations in 3-dimensional space and cannot rely entirely on 2-dimensional projections. For instance, even though Figure 34 looks like a 5-stick projection of a trefoil knot and an 8-stick projection of the knot 818, this is an optical illusion. In fact, it’s impossible to make these constructions with real sticks in 3-dimensional space. In both of these examples, the choice of over- and under-crossings in the projections will force one or more of the sticks to bend. You should try making these configurations yourself using matchsticks, toothpicks, or popsicle sticks to see what happens.

Figure 34. Some conﬁgurations that are impossible to construct with real sticks.

12.4. Lattice Number. The lattice number is similar to the stick number, but now in addition to using sticks to build our knots, the sticks are not allowed to go any which way. Rather, we require the sticks to be of equal length and parallel to the x, y,orz-axes in R3. Another way of thinking about this, is to imagine that all of R3 is ﬁlled with a grid like you would see on graph paper but in three dimensions instead of two dimensions. We can only create knots out of edges that go along the 3-dimensional grid, as illustrated in Figure 35. We call this a lattice conﬁguration of a knot. 13. EXERCISES 263

Figure 35. A lattice conﬁguration of a trefoil knot with 24 vertices.

Definition. The lattice number of a knot is the smallest number of vertices in any lattice configuration of that knot. Again we know this is an invariant because it’s defined over all possible lattice configurations. Perhaps it is surprising to realize just how large lattice numbers tend to be. In particular, Figure 35 shows a lattice configuration of the trefoil knot which has a minimum number of vertices. If you count the number of vertices in this configuration, you will find that there are in fact 24.

Looking back over this chapter, we’ve done all kinds of things to knots and links to help us distinguish them. We colored them, we built them out of sticks, we put arrows on them, we counted their peaks, and we related them to surfaces. Each of these ideas led us to an invariant. Nevertheless, a common theme in this journey has been that the most useful invariants are often hard to compute. We will spend the next chapter developing a single invariant which is both useful and not too diﬃcult to compute.

13. Exercises

1. Prove that the writhe is preserved under Reidemeister moves II and III, but not Reidemeister move I. 2. Prove that the writhe of a knot projection is independent of the choice of orientation.

3. Draw three diﬀerent projections of the knot 52: one with writhe −5, one with writhe −4, and one with writhe −6. 4. Prove that the linking number of an oriented link projection will not change as a result of a planar deformation or doing any of the three Reidemeister moves. Then use Reidemeister’s Theorem to conclude that the linking number is an invariant. Hint: You may want to look back in Chapter 9 to see how we proved that tricolorability would not change as a result of the ﬁrst two Reidemeister moves. 264 10. INVARIANTS OF KNOTS AND LINKS

5. (Hard.) Prove that the linking number is always an integer. In other words, prove that ⎛ ⎞ ⎛ ⎞ total number of total number of ⎝ positive crossings ⎠ − ⎝ negative crossings ⎠ between K1 and K2 between K1 and K2

is always an even number. 6. For any n, prove that there are inﬁnitely many non-equivalent oriented two- component links with linking number n, each of whose components is unknotted. 7. Let L denote an oriented link with components J and K.Let−J and −K denote J and K with their orientations reversed, and let L∗ denote the mirror image of L with oriented components J ∗ and K∗. Suppose that lk(J, K)=n. Find lk(−J, K), lk(J, −K), lk(−J, −K), and lk(J ∗,K∗). Justify your conclusions. 8. Suppose that an oriented link with components J and K can be deformed to its mirror image while preserving the orientation on each component. Prove that lk(J, K)=0. 9. Find an unoriented link with components J and K which can be deformed to its mirror image, but regardless of how we orient the link, lk(J, K) =0. 10. Find the unsigned linking number of each of the links in Figure 36. Can you use the linking number to determine if some of the links are non-equivalent?

Figure 36. Illustration for Exercise 10.

11. The three links in Figure 37 are all diﬀerent. Can you prove it? Try to use two diﬀerent invariants in your analysis.

Figure 37. Illustration for Exercise 11.

12. The unsigned linking number is an invariant of two component links. In Fig- ure 38 you’ll ﬁnd two links, each with three components. How might the unsigned linking number help you show that these links are non-equivalent? Hint: Every three component link contains several two component links. 13. EXERCISES 265

Figure 38. Illustration for Exercise 12.

13. The left link in Figure 38 is special because if you delete any component, the link falls apart. In particular, the linking number between any pair of components is 0. This link is considered to be very beautiful and has even appeared in various works of art and architecture. More generally, we have the following deﬁnition.

Definition. A link with n components, L1,...,Ln,issaidtobeBrunnian if the link is non-trivial, but removing any one component gives the trivial link.

For every n>2, ﬁnd a Brunnian link with n components. 14. How many non-equivalent two component links are there having a projection with precisely two crossings? Justify your conclusion.

15. The projection of 63 in the knot table is alternating. Find another projection of 63 that is non-alternating. Can you ﬁnd a non-alternating projection of 63 which has no nugatory crossings? 16. The projection in Figure 39 is not reduced. Find another projection of this knot that is reduced.

Figure 39. Illustration for Exercise 16.

17. Find the crossing number of the knot in Figure 40. Carefully explain how you arrived at your answer and include pictures.

Figure 40. Illustration for Exercise 17.

18. Find the crossing number of the knot in Figure 41. Carefully explain how you arrived at your answer and include pictures. 266 10. INVARIANTS OF KNOTS AND LINKS

Figure 41. Illustration for Exercise 18.

19. Let K1 and K2 be knots. Prove that if K1 and K2 are alternating, then their crossing numbers satisfy the equation c(K1#K2)=c(K1)+c(K2). Hint: First show that since K1 and K2 are alternating, the connected sum K1#K2 can be deformed to an alternating projection. 20. Find the writhes of all the prime 7-crossing knot projections given in the knot table. Using this information, which knots can you conclude must be distinct from one another?

21. Compute the writhes of the projections of 820 and 821 given in the table. Why can’t we use the writhe to conclude that 820 and 821 must be diﬀerent knots? 22. Let L denote an oriented knot or link projection, and let L∗ denote the mirror image of L.Howisw(L∗) related to w(L)? 23. Let L be a reduced alternating oriented knot or link projection, and let L∗ be its mirror image. (a) If L and L∗ are equivalent, what does Tait’s Second Conjecture say about their writhes? (b) Combining your results from part (a) and Exercise 22, what must be true of w(L)?

(c) Use your observation to prove that the knot 51 cannot be deformed into its mirror image. List all prime knots with up to six crossings that cannot be equivalent to their mirror images. 24. Let L be an alternating link with odd crossing number. Show that L cannot be deformed into its mirror image. 25. Show that any knot projection can be changed into an alternating projection (of a possibly different knot) by switching some number of crossings. If the projection you started with had n crossings, what is the largest number of crossings you would have to switch to make it alternating? Explain. 26. In Section 9 we saw that once you have one Seifert surface for a knot, you can create infinitely many Seifert surfaces for the knot by adding handles. Is there a way to add a handle to an orientable surface to make it non-orinetable? 27. Determine the unknotting number of the figure eight knot. 28. Determine the bridge number of the trefoil knot. 29. Show that there are infinitely many distinct knots with bridge number 2. 30. Show that if a knot projection has only one bridge, then it must be a projection of the unknot. (Hint: Draw the one bridge, and consider how the two strands that come down from it interact with one another.) 31. Show that every 4-stick knot is the unknot. 13. EXERCISES 267

32. Create three non-trivial links with diﬀerent linking numbers such that in each case one component has three sticks and the other component has four sticks. 33. Show that it is impossible to construct the 5-stick trefoil knot shown in Fig- ure 34(a). 34. For each prime 7-crossing knot projection in the knot table, use shading to ﬁnd a surface whose boundary is the knot. Which of your surfaces are orientable and which are non-orientable? Which surfaces are Seifert surfaces?

35. Calculate the genus of the knot 52 using the projection given in the knot table.

36. Calculate the genus of the knot 72 using the projection given in the knot table. 37. Complete the outline below to prove that every knot has a lattice configuration. (a) Suppose that we start with a knot projection with n crossings. Show that, after an appropriate planar deformation, we can find a collection of horizontal lines H0,H1,...,Hn and a collection of vertical lines V0,V1,...,Vn such that: (i) H0 lies below the knot projection and does not intersect the projection; (ii) Hn lies above the knot projection and does not intersect the projection; (iii) Hi+1 lies above Hi, and there is exactly one crossing of the projection lying between them; (iv) V0 lies to the left of the knot projection and does not intersect the projection; (v) Vn lies to the right of the knot projection and does not intersect the projection; and (vi) Vj+1 lies the the right of Vj, and there is exactly one crossing of the projection lying between them. (b) Each pair of consecutive horizontal lines Hi and Hi+1 together with each pair of consecutive vertical lines Vj and Vj+1 defines a rectangle containing at most one crossing of the projection. Show that the arcs of the projection inside this rectangle can be deformed so that they are made up of only vertical and horizontal line segments. Note that the entire deformation should be accomplished without moving the endpoints of the arcs (where they meet Hi, Hi+1, Vj,orVj+1) and without introducing any new crossings. (c) Show that the projection can be deformed so that every horizontal line segment lies along the line y = k for a different integer value of k.Also, show that the projection can be deformed so that every vertical line segment lies along the line x = m for a different integer value of m. Show why this means that after these deformations every line segment in the projection consists of some union of unit-length edges parallel to either the x-ory-axis. (d) Complete the lattice configuration of the knot by adding unit-length edges parallel to the z-axis.

https://doi.org/10.1090//mbk/096/11

CHAPTER 11

Knot Polynomials

Topics: • The bracket polynomial • The X-polynomial • The Jones polynomial • The span of a Jones polynomial • Reading the Jones polynomial from a knot table • The state model of the Jones polynomial

1. An Introduction to Polynomial Invariants

Let’s say you cross paths with a shady character under a bridge and he holds you at knifepoint, demanding that you compute the crossing number of a non- alternating knot. (More likely, this is a nightmare after a late night of knot studying.) Under the circumstances, you might be better off begging him to take your money instead. Even if, instead of a crossing number, the shady character de- manded that you compute another invariant from Chapter 10, you’d likely fare no better. Most of the knot invariants we’ve learned about so far can be pretty tough to calculate. That is, except for tricolorability, which doesn’t distinguish many knots. In this chapter, we will finally meet a useful invariant that you can compute for any knot simply by following a list of rules. Your success is guaranteed (though your speed may not be). We hope you never find yourself computing knot invariants for shady characters under bridges, but if you do, we advise that you offer to compute the invariant we’ll discuss here. This new invariant is not a numerical value, like most of the invariants we saw in Chapter 10. It is a special type of polynomial. By arranging the terms in order from lowest exponent to highest exponent and listing the coefficients, a polynomial can be represented by a finite list of numbers. For example, the polynomial 1−4x+3x3 could be expressed as the list (1, −4, 0, 3). Note that the third entry is 0 because there is no x2 term in the polynomial. Since a list of numbers can encode more information than a single number, a polynomial invariant seems like it might be more useful than an invariant consisting of a single number. In fact, this new invariant is more useful in distinguishing knots than any of our previous invariants. One last note before we get started. The invariant that we will introduce in this chapter is not technically a polynomial because it can have negative as well as

269 270 11. KNOT POLYNOMIALS positive integer exponents. The technical name for such an expression is a Laurent polynomial. But knot theorists simply call it a polynomial, so we’ll do the same.

2. The Rules for the Bracket Polynomial

Our goal in this chapter is to define a polynomial invariant of knots and links. However, as the first step we define the bracket polynomial of a link projection, which is not actually an invariant. Then in Section 6, we will use the bracket polynomial to create the X-polynomial, which will be an invariant. Defining the bracket polynomial requires introducing three rules which we will explain as we go along.

2.1. Rule 1. The bracket polynomial L will be defined for any link projection L, but we’ll start out slowly and just define it on a projection of a circle with no crossings at all. It seems natural that the simplest possible projection should be assigned a very simple polynomial. Indeed, we begin by defining: Rule 1 : =1.

2.2. Rule 2. For a link projection L which has one or more crossings, we want to remove the crossings one at a time until we end up with a link with no crossings so that we can apply Rule 1. We start by choosing one crossing and cutting the overcrossing into two arcs as shown in Figure 1. This gives us four endpoints of arcs where the crossing had been.

Figure 1. Cutting a crossing.

We then want to reconnect the dangling arcs in the center of the picture without reintroducing a crossing. We do this by attaching each endpoint of the former overcrossing (the thick arc in the ﬁgure) to an endpoint of the former undercrossing (the thin arc in the ﬁgure). We can do this in two ways, depending on whether we go in the clockwise or counterclockwise direction from the endpoint of an overcrossing (illustrated in Figure 2). The dotted lines are just there to help you see the clockwise or counterclockwise direction. Let Lclock be the link projection produced from a projection of L by cutting an overcrossing and reconnecting the ends clockwise, and let Lcounter be the link projection produced from L by cutting the same overcrossing and reconnecting the ends counterclockwise. Notice that the projections Lclock and Lcounter each contain one fewer crossing than L. 2. THE RULES FOR THE BRACKET POLYNOMIAL 271

Figure 2. Re-attaching the ends after cutting the overcrossing.

We now introduce variables A and B and an equation that uses these variables to relate the bracket polynomial of L to the bracket polynomials of the simpler link projections Lclock and Lcounter:

L = A Lclock + B Lcounter .

This equation is saying that the bracket polynomial of L is the sum of A times the bracket polynomial of Lclock and B times the bracket polynomial of Lcounter.It may seem strange that the variables are A and B rather than x and y. The capital letters A and B have always been used for knot and link polynomials, so we stick with them despite their strangeness. The above equation is Rule 2, but we prefer to express it symbolically as Rule 2 : = A + B .

The little dotted circles inside of the bracket symbols mean that the projections of the associated links (L, Lclock,andLcounter) are all identical outside of these circles. Figure 3 shows how to apply Rule 2 to one of the crossings in a trefoil knot. Keep in mind that when we connect the ends of the former overcrossing clockwise, we multiply by A, and when we connect the ends counterclockwise, we multiply by B. This is true no matter how the crossing is turned in the projection.

= A + B = A + B

Figure 3. Resolving one crossing of the trefoil knot.

Applying Rule 2 to get rid of a crossing is called resolving the crossing. By repeatedly resolving crossings, we will break our projection into a collection of link projections with fewer and fewer crossings, which will eventually become a collection of circles with no crossings. 272 11. KNOT POLYNOMIALS

2.3. Rule 3. Our third (and ﬁnal) rule says that we can remove circles with no crossings, if we multiply by yet another variable, which we will call C. Rule 3 : L = C L .

For example, suppose that we start with a link projection L consisting of two circles in a plane with no crossings at all. By applying Rule 3 followed by Rule 1, we see below that the bracket polynomial of L is just the variable C. = C = C(1) = C.

3. The Bracket Polynomial and Reidemeister II Moves

At this point, we have rules that can be applied repeatedly to a knot or link projection to get rid of crossings and extra circles one after the other until we eventually get a polynomial without any bracket symbols in it. But a big question remains—is this polynomial a knot or link invariant? If it’s not an invariant, why would we want to spend the time applying the rules over and over to get some complicated expression with variables A, B,andC which wouldn’t help us distinguish one knot from another? To be an invariant, the bracket polynomial needs to be unchanged under the three Reidemeister moves. In this section, we will see what happens when we apply Reidemister II moves. We will consider Reidemeister I moves later, and you will deal with Reidemeister III moves in the Exercises. Below we illustrate what happens when we resolve the crossings that occur in a Reidemeister II move. Make sure you can follow the steps as we describe them line by line. = A + B = A A + B + B A + B = A A + BC + B A + B = A2 + ABC + AB + B2 = A2 + ABC + B2 + AB . 3. THE BRACKET POLYNOMIAL AND REIDEMEISTER II MOVES 273

• Line 1: The left picture illustrates what the projection looks like near the place where we will do a Reidemeister II move. We apply Rule 2 to get rid of the top crossing. This means we multiply by A when we connect clockwise and we multiply by B when we connect counterclockwise. • Line 2: We apply Rule 2 to get rid of the bottom crossing. This means that each expression on Line 1 is broken into two expressions, giving us a total of four terms. • Line 3: We deform the drawings so they look nicer and we use Rule 3 to eliminate the little circle in the second term by multiplying by the variable C. • Line 4: We multiply the variables and write the products of variables in alphabetical order. • Line 5: We collect together the three terms that are being multiplied by the symbol and put them inside of parentheses.

In order for the bracket polynomial to be invariant under Reidemeister II moves, the polynomial needs to be the same before and after we do a Reidemeister II move. Thus we need to know that = .

By the above computation, we know that = A2 + ABC + B2 + AB . Wesetthisequalto to get the equation A2 + ABC + B2 + AB = Since cannot turn into , we can split this into two separate equations: A2 + ABC + B2 =0 and AB = .

Thisgivesustheequations: A2 + ABC + B2 =0 and AB =1. The second equation implies that B = A−1. Hence we can substitute A−1 for B in the ﬁrst equation to get the equation A2 + AA−1C + A−2 =0. 274 11. KNOT POLYNOMIALS

Now we can solve this equation for the variable C to get C = −A2 − A−2. These computations show that if the bracket polynomial is going to be invariant under Reidemeister II moves (which we want it to be), then the equations B = A−1 and C = −A2 − A−2 must both be true. To check that, with the substitutions B = A−1 and C = −A2 − A−2, the bracket polynomial will be invariant under Reidemeister II moves, we plug these values for B and C into our computation of the bracket polynomial for the crossings in a Reidemeister II move, and then simplify to get the following: = A + A−1 = A A + A−1 + A−1 A + A−1 = A A + A−1(−A2 − A−2) + A−1 A + A−1 = A2 + AA−1(−A2 − A−2) + AA−1 +(A−1)2 = A2 + AA−1(−A2 − A−2)+(A−1)2 + AA−1 =(A2 − A2 − A−2 + A−2) + = .

This shows that our variable substitutions are exactly what we need in order to guarantee that the bracket polynomial will be invariant under Reidemeister II moves.

4. The Bracket Polynomial with Only One Variable We now plug the expressions B = A−1 and C = −A2 − A−2 into Rules 2 and 3 of the bracket polynomial to get the deﬁnition of the bracket polynomial with only one variable. 5. THE BRACKET POLYNOMIAL AND REIDEMEISTER I MOVES 275

Definition. The bracket polynomial L of a link projection L is a Laurent polynomial with variable A, deﬁned by three rules: (1) =1. (2) = A + A−1 . (3) L =(−A2 − A−2) L .

We know from Section 3 that the bracket polynomial is invariant under Reide- meister II moves, and you will prove in the Exercises that it’s also invariant under Reidemeister III moves. Unfortunately, as we will soon see, this polynomial is not invariant under Reidemeister I moves. Don’t despair, however. This problem will eventually be ﬁxed. But in the meantime, let’s practice using the rules to compute bracket polynomials of a couple of link projections. We previously calculated the bracket polynomial of a projection L consisting of two planar circles and found that L = C. Now that we only have the variable A in our polynomial, we recalculate this polynomial using Rules 1 and 3 as follows: =(−A2 − A−2) =(−A2 − A−2)(1) = −A2 − A−2.

Next, we compute the bracket polynomial of a projection of a Hopf link. Make sureyouknowhowtojustifyeachstepinthecomputationbelow. = A + A−1 = A A + A−1 + A−1 A + A−1 = A2(−A2 − A−2)+1+1+A−2(−A2 − A−2) = −A4 − 1+2− 1 − A−4 = −A4 − A−4.

5. The Bracket Polynomial and Reidemeister I Moves

In order to see what goes wrong with the bracket polynomial when we do Reidemeister I moves, we compute the bracket polynomial of a projection of an unknot with a single crossing. Keep in mind that the bracket polynomial is deﬁned on a projection of a link, not on the link itself. So we can’t just assume that every projection of the unknot has bracket polynomial equal to 1. 276 11. KNOT POLYNOMIALS

Using the rules we have = A + A−1 = A(−A2 − A−2)+A−1 = −A3. The projection differs from a circle without crossings by a single Reidemeister I move. Thus we see that doing a Reidemeister I move to a circle without any crossings changes the bracket polynomial from 1 to −A3. This means that the bracket polynomial is not invariant under Reidemeister I moves, and hence equivalent knots do not necessarily have the same bracket polynomials. Thus we need to be very careful to only talk about the bracket polynomial of a knot or link projection rather than of a knot or link itself. Recall from Chapter 10 that we defined an oriented crossing to be positive if when you point the thumb of your right hand in the direction of the overcrossing, you fingers curl in the direction of the under crossing. We illustrate a negative and positive crossing in Figure 4.

Figure 4. A negative and positive crossing.

Using the designation of positive and negative crossings, we can designate a twist as positive or negative. In particular, we refer to the twist in Figure 5 as a “positive” twist because regardless of how we orient it, the crossing is positive. You should use your right hand to check for yourself that both crossings illustrated in Figure 5 are indeed positive.

Figure 5. A positive twist.

By contrast, the two twists illustrated in Figure 6 are negative.

Figure 6. A negative twist. 6. THE X-POLYNOMIAL 277

We saw in the beginning of this section that adding a positive twist to a circle with no crossings changes the bracket polynomial from 1 to −A3.Wenowseeas follows that starting with any knot or link projection L and adding a positive twist to get a new link projection L multiplies the bracket polynomial by a factor of − 3 A : L = = A + A−1

= A L (−A2 − A−2)+A−1 L = L (−A3 − A−1 + A−1)= L (−A3). You will show in the Exercises that starting with a knot or link projection L and adding a negative twist to get a new link projection L multiplies the bracket polynomial by a factor of −A−3.

6. The X-Polynomial

We saw in the last section that adding a twist to a link projection multiplies the bracket polynomial by −A3 or −A−3. Sadly, this means that the bracket polynomial is not invariant under Reidemeister I moves. We will solve this problem by making use of the writhe that we introduced in Chapter 10. Recall the following definition. Definition. The writhe w(L) of an oriented link projection L is given by the formula total number of total number of w(L)= − . positive crossings negative crossings In Chapter 10, we saw that the writhe isn’t an invariant. In fact, we could make a knot have any writhe that we want by adding some number of positive or negative twists. What’s amazing is that we will now use the writhe (which isn’t an invariant) and the bracket polynomial (which isn’t an invariant) to define a new polynomial which will be a link invariant. Sometimes two wrongs actually do make a right! But don’t tell your parents. Definition. Let L be an oriented link projection with writhe w(L). We define the X-polynomial of L to be X(L)=(−A3)−w(L) L . We did not need a knot or link projection to be oriented in order to compute the bracket polynomial. However, since the writhe is only defined for oriented projections, X(L) is only defined if L is an oriented projection. Recall that for a knot projection, both orientations give the same writhe, so which orientation we assign won’t matter for the X-polynomial. But this is not true for a link projection. As an example, we compute the X-polynomial for the oriented Hopf link projection L illustrated in Figure 7.

Figure 7. An oriented Hopf link projection. 278 11. KNOT POLYNOMIALS

Recall that in Section 4 we computed the bracket polynomial of this (unoriented) link projection to be L = −A4 − A−4. Also you should check Figure 7 to see that w(L)=−2. Thus we can compute the X-polynomial of this oriented projection to be X(L)=(−A3)−(−2)(−A4 − A−4) = A6(−A4 − A−4) = −A10 − A2. We are now ready to prove the marvelous, magniﬁcent, miraculous result that the X-polynomial is indeed an invariant of oriented links. This is so awesome that wewanttostateitasatheoremsowecansitandadmireitbeforeweproveit.

Theorem 11.1. The X-polynomial is an invariant of oriented links. Proof. Let L be an oriented link projection. If we can show that X(L)is preserved by planar deformations and the three Reidemeister moves, then it will follow from Reidemeister’s Theorem that X(L) is indeed an invariant. Since the writhe and the bracket polynomial are both deﬁned in terms of crossings, planar deformations will have no eﬀect on them. Thus the X-polynomial will also be invariant under planar deformations. So we can move on to the Reidemeister moves. We saw that the bracket polynomial L is preserved by Reidemeister II moves, and you will verify in the Exercises that L is also preserved by Reidemeister III moves. In the Exercises for Chapter 10, you proved that the writhe is also preserved by Reidemeister II and III moves. Since X(L) is the product of two quantities that are each invariant under Reidemeister II and III moves, it follows that X(L) is also invariant under Reidemeister II and III moves. At this point, all that is left to do is to prove that X(L) is invariant under Reidemeister I moves. Let L denote L with a positive twist added. This means that L has exactly one more positive crossing than L. Hence w(L)=w(L)+1. Also,wesawin Section 5 that L = L (−A3). So we have X(L)=(−A3)−w(L ) L =(−A3)−w(L)−1 L (−A3) =(−A3)−w(L)(−A3)−1(−A3) L =(−A3)−w(L) L = X(L). Thus, the X-polynomial is invariant under adding or removing a positive twist. As you will check in the Exercises, by an analogous argument, the X-polynomial is invariant under adding or removing a negative twist. Therefore, the X-polynomial is invariant under planar deformation and all three Reidemeister moves, and hence it is an invariant of oriented links.

Time for a celebration. At long last, we have defined the X-polynomial and proven that it is indeed an invariant. You are now armed with a knot invariant which you can compute by yourself or with a friend, for any oriented knot or link (even in dark alleys or under bridges if necessary). Since the X polynomial is an invariant, we can now talk about the X-polynomial of an oriented link instead of always having to say the X-polynomial of an oriented link projection. Nonetheless, 7. THE JONES POLYNOMIAL 279 we have to fix a projection in order to compute the bracket polynomial as a step in the computation of the X-polynomial. However, the particular projection we choose will not affect the value of the X-polynomial. In order to see what might happen to the X-polynomial when the orientation of a single component of a link is reversed, let’s consider the oriented Hopf link projection L in Figure 7, and let L denote the oriented Hopf link projection in Figure 8. These two oriented projections differ only in the orientation of one of their components. Since L and L are the same if we ignore the orientations, L = L . However, w(L)=−2andw(L)=2.Thus X(L)=(−A3)−2 L = A−6 L and X(L)=(−A3)2 L =(−A3)2 L = A6 L . Thus the X-polynomial can be affected by changing the orientation of one component. In the Exercises, you will explore what happens when the orientations on all components of a link are reversed.

Figure 8. A Hopf link whose orientation is diﬀerent from that of Figure 7.

7. The Jones Polynomial

The X-polynomial was originally discovered by Vaughan Jones in a different form known as the Jones polynomial, which is denoted by V (L) for an oriented knot or link L. In spite of being called a “polynomial”, it’s not actually a polynomial or even a Laurent polynomial like the X-polynomial, because its exponents can be fractions as well as integers. The Jones polynomial and the X-polynomial represent the same information about a link though the rules for computing them are somewhat different. In fact, you can get from one to the other by an easy substitution. To change the X-polynomial X(L) (whose variable is the letter A)into the Jones polynomial V (L) (whose variable is the letter q), we use the substitution A = q−1/4. We can also go from the Jones polynomial to the X-polynomial by using the substitution q = A−4. For example, since the we know the X-polynomial of the oriented Hopf link in Figure 7 is X(L)=−A10 − A2, we find that the Jones polynomial of this oriented Hopf link is V (L)=−q−5/2 − q−1/2. 280 11. KNOT POLYNOMIALS

In your previous math classes, the variable of a function was probably written inside of the parentheses after the name of the function. For example, you’ve probably worked with the function f(x)=x2. By contrast, even though knot and link polynomials have variables like A and q, we write the knot or link itself inside of the parentheses next to the name of the function rather than the variable. In other words, if L is an oriented knot or link, we write X(L)fortheX-polynomial of L and V (L) for the Jones polynomial of L, rather than writing X(A)andV (q). The reason for this distinction is that a function like f(x)=x2 takes in a real number x and spits out a new real number x2, whereas a knot and link polynomial takes in an oriented knot or link and spits out a polynomial rather than a single real number. Of course once you have found the polynomial of a particular knot, you could plug in numbers for the variable and get numbers out. But we’re not interested in the specific numbers you would get out—we’re interested in comparing the polynomials of different oriented knots or links in order to know that one can’t be deformed to the other. The Jones polynomial is a powerful tool in knot theory because it’s not hard to compute and it distinguishes a lot of knots. For example, we know that the knots 51 and 52 are distinct because they have different polynomials. The same is true for the non-alternating knots 819 and 820, illustrated in Figure 9. Note that, since the Jones polynomial of a knot is independent of the orientation of the knot, we are free to omit the orientation, as we’ve done in Figure 9.

Figure 9. We know that the knots 819 and 820 are distinct because they have diﬀerent polynomials.

You may wonder whether the X-polynomial (or, equivalently, the Jones polynomial) distinguishes every pair of inequivalent links. Unfortunately, this is not the case. For example, the two knots illustrated in Figure 10 have the same Jones polynomial but are distinct. If you look carefully, you’ll see that they look quite similar but are not identical. Of course this doesn’t prove that they aren’t equivalent. In fact, knot theorists used the genus to distinguish them by showing that the one on the left has genus 3, while the one on the right has genus 2. We might at least hope that the Jones polynomial could distinguish all non- trivial knots from the unknot. This may be possible. As of 2015, no non-trivial knot with Jones polynomial equal to 1 has been found. But who knows? Maybe there is some complicated knot with Jones polynomial equal to 1 somewhere out there just waiting to be discovered. In spite of the fact that Jones polynomials can in theory have fractional powers, Jones polynomials of knots actually only have integer exponents. By contrast, links with at least two components have Jones polynomials whose exponents are multiples of 1/2. 7. THE JONES POLYNOMIAL 281

Figure 10. These knots have the same Jones polynomial but are distinct.

Many knot tables include Jones polynomials compactly by displaying the lowest exponent in the polynomial together with a list of numbers representing the coefficients of all of the terms. For example, next to the trefoil knot in Table 1 is the code {−4}−1101.The{−4} indicates that −4 is the lowest exponent in the Jones polynomial of the trefoil, and the list of numbers that follow are the coefficients of the variables of each subsequent exponent. So −1 is the coefficient of q−4, 1 is the coefficient of q−3, 0 is the coefficient of q−2, 1 is the coefficient of q−1, and that’s it. This means that the Jones polynomial of the trefoil knot is

−4 −3 −1 V (31)=−q + q + q .

Note that since the coeﬃcient of q−2 is 0, we omit the q−2 term from the polynomial. Now, by substituting q = A−4 into the Jones polynomial, we ﬁnd that the X-polynomial of the trefoil knot is

16 12 4 X(31)=−A + A + A .

Since Jones polynomials have been calculated (with the help of computers) for knots with up to 11 crossings, it is now easy to find the X-polynomials of these knots by simply doing a substitution of q = A−4,aswedidforthetrefoilknotabove. If you look at Table 1, you will see that the lists of coefficients in the table get longer as the crossing number increases—until we reach the non-alternating knots 819,820,and821 at the end of our table. This gives us the impression that the more crossings a knot has, the more terms are in its Jones polynomial. To help keep track of the number of terms in the Jones polynomial (including terms with coefficient 0), we define the span of a Jones polynomial V (L), written as span(V (L)), to be the difference between the highest power and the lowest power of the variable q in V (L). For example, for the trefoil knot, since the Jones polynomial −4 −3 −1 is V (31)=−q + q + q ,wehavespan(V (31)) = −1 − (−4) = 3. It turns out that for any alternating knot K, the span of its Jones polynomial is equal to its crossing number. We could express this symbolically by saying that for alternating knots, we have span(V (K)) = c(K). In Chapter 10, we learned about two of Tait’s Conjectures which took more than 100 years to be proved. The first of these conjectures said that a reduced alternating projection of a knot or link has the smallest number of crossings of any projection. This important result was proved using the span of the Jones polynomial. 282 11. KNOT POLYNOMIALS

Table 1. Jones polynomials for knots with up to 8 crossings.

Knot Jones coeﬃcients Knot Jones coeﬃcients

31 {-4} -1 1 0 1 85 {0} 1-13-33-43-21 41 {-2} 1-11-11 86 {-7} 1-23-44-43-11 51 {-7} -11-1101 87 {-2} -12-24-44-32-1 52 {-6} -1 1 -1 2 -1 1 88 {-3} -12-35-44-32-1 61 {-4} 1-11-22-11 89 {-4} 1-23-45-43-21 62 {-5} 1-22-22-11 810 {-2} -12-35-45-42-1 63 {-3} -1 2 -2 3 -2 2 -1 811 {-7} 1-23-55-44-21 71 {-10} -1 1 -1 1 -1 1 0 1 812 {-4} 1-24-55-54-21 72 {-8} -1 1 -1 2 -2 2 -1 1 813 {-3} -13-45-55-32-1 73 {2} 1-12-23-21-1 814 {-7} 1-34-56-54-21 74 {1} 1-23-23-21-1 815 {-10} 1-34-66-55-21 75 {-9} -1 2 -3 3 -3 3 -1 1 816 {-6} -13-56-66-43-1 76 {-6} -1 2 -3 4 -3 3 -2 1 817 {-4} 1-35-67-65-31 77 {-3} -1 3 -3 4 -4 3 -2 1 818 {-4} 1-46-79-76-41 81 {-6} 1-11-22-22-11 819 {3} 10100-1 82 {-8} 1-22-33-22-11 820 {-5} -1 1 -1 2 -1 2 -1 83 {-4} 1-12-33-32-11 821 {-7} 1-22-33-22 84 {-5} 1-23-33-32-11

8. The State Model for Computing the Bracket Polynomial

Hopefully you are feeling confident with your newfound understanding of the X-polynomial and the Jones polynomial. But before you get too comfortable, ask yourself the following question: Is there anything better than having a step-bystep procedure to compute the X-polynomial? Tough question, right? But there is something better: knowing not just one but two step-by-step procedures to compute the X-polynomial. Yes, there is another way to compute it, and we would be remiss if we did not end this chapter with a description of this alternate approach. Recall that we previously determined the bracket polynomial of a knot or link projection by resolving crossings one at a time and each time applying Rule 2. We did this repeatedly until we obtained a collection of circles with no crossings, whose bracket polynomial we could find using Rules 1 and 3. Our new method for computing the bracket polynomial of a knot or link projection will work much faster. Instead of eliminating the crossings one at a time, we simultaneously eliminate every crossing in the projection. Then we find all possible ways to reconnect the four dangling ends of each crossing. We call this process splitting the crossings. For example, in Figure 11, we illustrate a Hopf link projection at the top, and all the different ways that we can split the crossings under it. Since the Hopf link projection has two crossings and each crossing can be split in two different ways, we end up with four diagrams, each of which has zero crossings. 8. THE STATE MODEL FOR COMPUTING THE BRACKET POLYNOMIAL 283

Figure 11. A Hopf link projection and its four states.

Figure 12. Rotating the upper arc counterclockwise sweeps out the A regions. The other regions are B regions.

Each of the diagrams with zero crossings will be called a state of our original projection. In Figure 11, we obtain four states. In general, a link projection with n crossings has 2n states. Next we go on a labeling spree. We want to label each split crossing in each state with either an A or a B. To do this we use Figure 12. In particular, starting with the arc going over a crossing, we rotate it counterclockwise until it lines up with the arc going under the crossing. This will sweep out two regions, which are both labeled with the letter A. We then label the other two regions with the letter B. If we split a crossing so that the two A-regions are joined together, we call the split an A-split. If we split a crossing so that the two B-regions are joined together, we call the split a B-split. We illustrate these two types of splits in Figure 13.

Figure 13. A crossing can be split in these two ways.

After labeling each split in this fashion, we then label each state by indicating which splits came from A-splits and which splits came from B-splits. In Figure 14, we have labeled all of the states for the Hopf link projection. Next, for each state S,leta(S) denote the number of A-splits in S,letb(S) denote the number of B-splits in S, and let |S| denote the total number of unknotted circles in S. In Figure 15, we determine these values for each of the states of our Hopf link projection. We are now ready to put all of these quantities together to compute the bracket polynomial. 284 11. KNOT POLYNOMIALS

Figure 14. The A-splits and B-splits in the states of this Hopf link projection.

Figure 15. The values of a(S), b(S), and |S| for the states of the Hopf link projection.

Definition. Let L be a knot or link projection. For each state S in L,we deﬁne a term of the form Aa(S)A−b(S)(−A2 − A−2)|S|−1.

The sum of all of these terms is said to be the state model of the bracket polynomial of L.

To see how this works, in Figure 16 we compute the term that each state of the Hopf link projection contributes to the state model. Adding up the terms in Figure 16 we get 1+(−1 − A4)+1+(−A4 − 1) = −A4 − A−4.

Observe that this polynomial agrees with the bracket polynomial that we computed for this projection in Section 4. In fact, this is true in general. For any knot or link projection, we get the same polynomial whether we use the state model or the rules for the bracket polynomial described in Section 4. You might see a proof of this if you take a more advanced knot theory class. In any case, now you have 9. EXERCISES 285

Figure 16. The terms that each state of the Hopf link projection contributes to the state model. two methods to compute the bracket polynomial of a knot or link projection, and either way you can use the result to get the X-polynomial for the knot or link, and once you have that by a simple substitution you can ﬁnd the Jones polynomial.

9. Exercises

1. Show that starting with a knot or link projection L and adding a negative twist to get a new link projection L multiplies the bracket polynomial by a factor of −A−3. 2. (a) Find the bracket polynomial of each of the unknot projections in Figure 17.

Figure 17. Illustration for Exercise 2(a).

(b) Find the bracket polynomial of each of the unknot projections in Figure 18.

Figure 18. Illustration for Exercise 2(b).

3. Find the bracket polynomial of the Hopf link projection in Figure 19. 4. Using the deﬁnition of the bracket polynomial and your answers to Exercise 2, ﬁnd the bracket polynomial of the link projection given in Figure 20. 5. Use Rules 1 and 3 together with induction to show that the bracket polynomial of a planar link projection L (that is, a projection with no crossings) which has n components is L =(−A2 − A−2)n−1. 286 11. KNOT POLYNOMIALS

Figure 19. Illustration for Exercise 3.

Figure 20. Illustration for Exercise 4.

Figure 21. Illustration for Exercise 6.

6. The goal of this problem is to compute the Jones polynomial of the trefoil knot illustrated in Figure 21 by following the outline below. (a) First, compute the bracket polynomial for this projection. (i) Start by applying Rule 2 to any crossing. This gives us two new projections. (ii) Next, use planar deformations to deform the two new projections to look like projections we’ve seen in the previous exercises. Since you already know the bracket polynomials for those projections, you can plug them in. (iii) Once you have plugged in the bracket polynomials for both projections, simplify the bracket polynomial and write it so that the exponents are in descending order. (b) Now, ﬁnd the writhe of the projection in Figure 21 and use it to compute X(31). −1/4 (c) Finally, use the substitution A = q to compute V (31), the Jones polynomial of the trefoil. The exponents of your polynomial should now be in ascending order so that you can compare your answer to the Jones polynomial for 31 given in Table 1. 7. Prove that the bracket polynomial is invariant under a Reidemeister III move.

8. Prove that the X-polynomial is invariant under adding or removing a negative twist. 9. If L is an oriented knot or link, then −L denotes the knot or link with the orientation of every component reversed and L∗ denotes the mirror image of L. How is X(L) related to X(−L)? How is X(L∗) related to X(L)? Justify your conclusions. 9. EXERCISES 287

10. Use Exercise 9 to show that the trefoil knot cannot be deformed to its mirror image. 11. Compute the Jones polynomial of the ﬁgure eight knot, then check your answer using Table 1. Hint: Start by applying bracket polynomial Rule 2 to the crossing indicated in Figure 22, and remember that you can plug in bracket polynomials that you’ve already computed.

Figure 22. Illustration for Exercise 11.

12. Use the Jones polynomials given in Table 1 to ﬁnd a knot with eight crossings that cannot be deformed to its mirror image. (Remember that illustrations of these knots are given in the knot table in Chapter 9.) Explain your reasoning. 13. Find an oriented link that cannot be deformed to its mirror image, yet has linking number zero. Explain how you know that your link cannot be deformed to its mirror image.

14. Let L+, L−,andL0 be three oriented link projections that are identical except at a single crossing where the three links diﬀer as indicated in Figure 23.

Figure 23. Illustration for Exercise 14.

The goal of this problem is to follow the steps below to prove the equation −1 − 1 1 q V (L+) − qV (L−)+(q 2 − q 2 )V (L0)=0.

(a) First write equations for the bracket polynomials L+ and L− in terms of the variable A and the bracket polynomial of the link with the crossing removed. (b) Use the equations from step (a) to get an equation relating L+ , L− , and L0 . (c) Use the equation from step (b) to get an equation relating X(L−), X(L+), and X(L0). (d) Then use the substitution q = A−4 in the equation from step (c) to get the required equation.

Part 3

Molecules

https://doi.org/10.1090//mbk/096/12

CHAPTER 12

Mirror Image Symmetry from Diﬀerent Viewpoints

Topics: • Mirror image symmetry • Geometric chirality • Chemical chirality • Understanding diagrams of molecules • Euclidean rubber gloves • Topological rubber gloves • Topological chirality

1. Mirror Image Symmetry

In Chapter 3, we investigated spaces from both a topological and a geometric viewpoint, and we explored the consequences of each. In Chapters 9, 10, and 11 we considered the topology of knots and links, and we saw that knot theory is historically linked to chemistry. In Chapters 12–15, we apply techniques we’ve learned in our study of topology, geometry, and knot theory to the study of non- rigid molecules. Don’t worry, you don’t need to know any chemistry or biology to understand these chapters. If you are left handed, you have probably had the frustrating experience of trying to take notes in a classroom that has only right-handed desks (see Figure 1). Sitting at a right-handed desk makes writing with your left hand more challenging. You could stand on your head on the seat of a right-handed desk so that your left-hand would be next to the table, but if you’ve done this, you know that taking notes while standing on your head presents other challenges. If you were designing your own classroom, it might make sense to select desks which have mirror image symmetry, so that both right- and left-handed students could use them. An object that does not have mirror symmetry has two mirror forms, which we can think of as the left-handed form and the right-handed form. For example, the mirror image of a right-handed desk is a left-handed desk. Generalizing from our experience with left-handed students having to use right-handed desks, we see that when two objects without mirror image symmetry come together, the way they interact depends on their handedness. For example a right foot and a left foot interact diﬀerently with a left shoe. On the other hand, a right foot and a left

291 292 12. MIRROR IMAGE SYMMETRY FROM DIFFERENT VIEWPOINTS

Figure 1. A right-handed and left-handed desk. foot interact the same way with a sock, because a sock has mirror image symmetry. That is, unless it’s a toe sock whose bottom and top are different. Our observations about asymmetric objects can also be applied to the molecules in our bodies. In particular, most of the molecules in living things don’t have mirror image symmetry. For example, DNA spirals in a left-handed helix, and DNA, RNA, and proteins are all made of left-handed building blocks. In fact, scientists who study the origins of life believe that life can’t exist without such molecular asymmetry. As a result of this asymmetry, we react differently to the right-handed and left-handed forms of an asymmetric molecule, just as a right foot reacts differently to a right shoe and a left shoe. For example, one form of the molecule carvone smells like spearmint and the mirror form smells like caraway. The difference in smells is the result of our olfactory system (which is asymmetric) reacting differently to the mirror forms of carvone. Similarly, most naturally occurring amino acids taste bitter, while their mirror images taste sweet. Pharmaceutical companies pay particular attention to mirror image symmetry, because both the effectiveness of a drug and the severity of its side effects can depend on which of the two forms is given. The pain reliever ibuprofen has two forms only one of which is effective, while the mirror form has no significant effect on your body. Ibuprofen, like almost all medications, is synthesized in a 50:50 mix of the two mirror forms. Since it is expensive to separate the two forms, ibuprofen is sold as a mixture. The fact that half of the molecules making up each pill are the useless mirror forms is not important. On the other hand, the mirror image of the pain reliever naproxen is actually toxic to your liver. So pharmaceutical companies are obliged to go through the expense of separating the two forms so that they only sell the harmless form. Sometimes when the two forms of a medication are separated, it turns out that they each have a different use. For example, the medication Darvon is a painkiller, while its mirror form Novrad is used as a cough medicine. From these examples we see that knowing whether a molecule has mirror image symmetry is important. But what exactly is meant by “mirror image symmetry” is not quite as a simple as it sounds. For example, consider a stretchy right-handed glove. If we turn it inside out, we will get a left-handed glove. Suppose that the inside and outside of the glove are identical. Would we say that the glove has mirror image symmetry? In the next few sections, we introduce three different approaches to the study of mirror image symmetry: geometric, chemical, and topological. 2. GEOMETRIC SYMMETRY 293

2. Geometric Symmetry

We begin with geometric symmetry because this corresponds most closely to our intuition. Recall from Chapter 3 that two objects in R3 have the same extrinsic geometry if there is a rigid motion taking one to the other. In the same chapter, we deﬁned a rigid motion to be any motion of an object in space that doesn’t involve stretching, shrinking, or bending. Such motions can also be thought of as a combination of rotations and translations of the object in space. We are now interested in whether an object and its mirror image are geometrically the same. That is, we want to know whether we can go from an object to its mirror image by a motion that doesn’t involve stretching, shrinking, or bending. For example, could such a motion change the right-handed desk in Figure 1 to a left-handed desk? If it could, left-handed students could just move the chair in the right way, and they wouldn’t have to stand on their heads. Consider the knot projection illustrated on the left side of Figure 2. The mirror image (illustrated on the right) is obtained by switching all of the undercrossings and overcrossings. At ﬁrst glance you might not say that this projection has mirror symmetry. Yet, if we rotate the knot on the left by 90◦ we get the knot on the right. So geometrically this knot has mirror image symmetry.

Figure 2. A knot and its mirror image.

When we ask whether or not an object is geometrically the same as its mirror image, it may seem like it could depend on where we place the mirror. Actually, at this very moment, A. Square is wondering the same thing about mirror images in Flatland. In Figure 3, we see A. Square experimenting with mirrors in diﬀerent positions to determine whether all of his mirror images are the same. This experiment also provides A. Square with the opportunity to admire himself in the mirror from all diﬀerent angles, which is something he always enjoys.

Figure 3. A. Square and his image in diﬀerent mirrors.

Note that the mirrors in the ﬁgure are illustrated as dotted line segments and A. Square’s mirror images are illustrated in grey. From this ﬁgure, we can see 294 12. MIRROR IMAGE SYMMETRY FROM DIFFERENT VIEWPOINTS that A. Square can get from any one of his mirror images to any other by a rigid motion of the mirror image in Flatland. Thus all of A. Square’s mirror images are geometrically the same, regardless of the position of the mirror. This is also true in 3-dimensional space. No matter where you are relative to a mirror, your mirror image will be the same. But you might want to verify this for yourself.

3. Geometric Chirality and Achirality

In order to determine whether a molecule is the same or different than its mirror image, we need to physically represent the molecule in space. We do this with a graph where each vertex represents an atom or a group of atoms, and each edge represents a molecular bond. Chemists have developed a nice way to draw pictures of these graphs which includes information about what is in front and what is behind the plane of the paper. For example, consider the amino acid L-alanine represented by the graph in Figure 4. The line segments in the figure indicate edges which lie in the plane of the paper. The dark triangular segment indicates an edge which comes out of the plane of the paper towards you with the widest part of the triangle closest to you. The dashed triangular segment indicates an edge which goes back behind the plane of the paper with the widest part of the triangle farthest from you. Thus the C, CH3, and H are in the plane of the paper, while the NH2 is in front of the paper and the CO2H is behind the plane of the paper. In 3-dimensional space, the vertices labelled CH3,H,NH2,andCO2H lie at the corners of a regular tetrahedron. Don’t worry if you don’t know what each of these letters stands for. Just realize that different letters represent different atoms.

Figure 4. This graph represents the 3-dimensional structure of the amino acid L-alanine.

The molecule L-alanine together with its mirror image are illustrated in Figure 5. If we imagine that the molecule is rigid, no matter how we rotate or translate it, we will never obtain its mirror image. Thus geometrically this molecule is different from its mirror image. Now let’s consider the molecule illustrated in Figure 6. This molecule is different from L-alanine because it has two H’s (representing hydrogen) rather than just one. Is this molecule geometrically different from its mirror image? In order to be able to refer to molecules and other objects which are geometrically different from their mirror image we introduce the following definition. Definition. An object is said to be geometrically chiral if there is no rigid motion taking it to its mirror image. Otherwise, it’s said to be geometrically achiral. 4. CHEMICAL CHIRALITY AND ACHIRALITY 295

Figure 5. The molecule L-alanine together with its mirror image.

Figure 6. Is this molecule geometrically diﬀerent from its mirror image?

For example, the molecule in Figure 5 is geometrically chiral, while the molecule in Figure 6 is geometrically achiral. At first this definition may seem confusing because an object is chiral if it is NOT the same as its mirror image and it is achiral if it IS the same as its mirror image. Normally, we add the prefix “a” to a word to indicate that something is not the case. For example, a person is said to be apolitical if he/she is NOT political. So our definitions of chiral and achiral seem backwards. However, like many definitions, the definition of the word chiral makes more sense if you understand its etymology. The word chiral (which rhymes with spiral) comes from the ancient Greek word χειρ which means hand. There is no rigid motion taking a left hand to a right hand. Thus a hand is a prototypical example of a chiral object. So an object which is chiral has the trait of being “like a hand”, while an object which is achiral has the trait of being “not like a hand”. In this sense, the prefix “a” added to the word chiral has the same meaning as it does when it is added to the word political. If you ever get confused between chiral and achiral, just remember that the word without the “a” means hand.

4. Chemical Chirality and Achirality

As we mentioned in Section 1, because of the asymmetry of the molecules in our bodies, we may react differently to the right-handed and left-handed forms of a medication. Chemists want to know whether a given molecule is the same as its 296 12. MIRROR IMAGE SYMMETRY FROM DIFFERENT VIEWPOINTS mirror image in order to predict how it will react with other molecules. This can be determined experimentally because molecules which are different from their mirror image rotate polarized light at room temperature, while molecules which are the same as their mirror image do not. Here by “the same” we mean chemically the same. We formalize this with the following definition.

Definition. A molecule is said to be chemically chiral if it cannot transform itself to its mirror image at room temperature. Otherwise, it is said to be chemically achiral.

By transform itself, we mean the chemical structure of the molecule enables it to change into its mirror image. Such a transformation is not a chemical reaction, but rather a conversion of the molecule back and forth between the two mirror forms of the structure. Saying that a molecule is chemically achiral says something about the behavior of the molecule, not necessarily its geometry. Before we look at a molecular example, let’s consider the behavior of a right- handed glove. A right-handed glove is geometrically chiral because it cannot be rotated or translated to get a left-handed glove, which is its mirror image. This means that as a rigid object, it is distinct from a left-handed glove. But suppose we are interested in how the glove interacts with a hand rather than its geometry. We might observe that our right-handed glove can be turned inside out and then it becomes a perfectly wearable left-handed glove (assuming there aren’t a lot of threads and what-not on the inside). This means that the glove is achiral in a practical sense even though it’s geometrically chiral. Similarly, our definition of being chemically achiral is a practical definition about molecular behavior rather than a geometric definition. If a molecule is geometrically achiral, then as a rigid object its structure is identical to that of its mirror image. It follows that it must be chemically achiral as well, since no chemical transformations are necessary to change it into its mirror image. Equivalently, we could argue that if a molecule is chemically chiral, then no chemical transformations allow it to transform itself into its mirror image, and hence it is geometrically chiral as well. We illustrate the relationship between the collection of all chemically achiral molecules and the collection of all geometrically achiral molecules in Figure 7. This picture tells us that the set of all geometrically achiral molecules is contained in the set of all chemically achiral molecules. The question is, are there molecules which are chemically achiral, but not geometrically achiral? Or are the ovals in Figure 7 actually equal? We have seen that a glove can be geometrically chiral yet practically achiral. But can there be a molecule that has this property? We will answer this question in the next section. Chemists normally use the word “chiral” for both geometrically chiral molecules and chemically chiral molecules. Lord Kelvin, who created the first knot tables, coined the term chiral to refer to an object that is geometrically different from its mirror image. In particular, in 1884, Kelvin wrote the following:

I call any geometrical ﬁgure, or group of points chiral,andsay it has chirality, if its image in a plane mirror, ideally realized, cannot be brought to coincide with itself. 4. CHEMICAL CHIRALITY AND ACHIRALITY 297

Figure 7. The set of all geometrically achiral molecules is contained in the set of all chemically achiral molecules.

Modern organic chemistry textbooks often paraphrase Lord Kelvin’s deﬁnition as follows:

Definition. A chiral molecule is one that is not superimposable on its mirror image. This definition means that a molecule is chiral if as a rigid object it is not identical to its mirror image. This is precisely the same as our definition of being geometrically chiral. The problem with applying this definition to molecules is that it assumes that all molecules are rigid. But some molecules are large enough that they can be floppy. For example, DNA is floppy and can even be knotted (as we’ll see in Chapter 14). There are other synthetic and biological polymers that are not as long as DNA, but can still be somewhat flexible. Furthermore, even some small molecules have certain bonds around which a piece of the molecule can rotate. For example, let’s consider the molecule in Figure 8.

Figure 8. In the next section, we’ll show that this molecule is geometrically chiral but chemically achiral.

In order to understand the 3-dimensional structure of Figure 8, recall that a dark triangular segment indicates an edge which comes out of the plane of the paper towards you with the widest part of the triangle closest to you. A dashed triangular segment indicates an edge which goes behind the plane of the paper with the widest part of the triangle farthest away from you. These rules help us understand the positions of the four hexagons in the figure. In particular, we see that the left-most and right-most hexagons are behind the plane of the paper, because of the way the 298 12. MIRROR IMAGE SYMMETRY FROM DIFFERENT VIEWPOINTS dashed segments connect them to the central part of the molecule. The hexagon which is the second from the left in the figure is in the plane of the paper, and the hexagon which is second from the right is perpendicular to the plane of the paper with half of it in front and half of it behind the page. You should carefully check that you agree with all of the notes on the figure saying what’s behind and what’s in front of the page. The molecule in Figure 8 is rigid except for the two pieces on the sides which rotate simultaneously as indicated by the arrows. We call these rotating pieces propellers. In contrast with molecules we’ve seen earlier in the chapter, the molecule in Figure 8 has corners with no letters on them. Such corners should be seen as vertices of the graph that represent carbon atoms. The letter C is often omitted from molecular graphs because carbons are so common. Note that the NO2 and the O2N represent the same group of atoms. The order in which the letters are written is meant to indicate that the N is attached to the adjacent hexagon.

5. Chemical Achirality and Geometric Chirality of Figure 8

The twins, Leftie and Rightie, have been begging to appear in this book. They think it’s not fair that A. 3D-Girl is practically the only 3-dimensional person who is in the book. They have even accused us of not wanting them because they’re not ambidextrous. So we’ve agreed to include them in this section to help us explain why the molecule in Figure 8 is chemically achiral but geometrically chiral. We begin by drawing the molecule in Figure 9 with the pieces on the left and right sides replaced by Leftie and Rightie, respectively. Because of the striped triangles in Figure 8, attaching the left-most and right-most hexagons to the central part of the molecule, Leftie and Rightie are standing behind the plane of the paper. Thus Leftie has her left arm forward to hold on to her oxygen and Rightie has her right arm forward to hold on to her oxygen. Since the molecule is rigid everywhere except for the rotating propellers, Leftie must always have her left arm forward, and Rightie must always have her right arm forward.

Figure 9. Leftie and Rightie have replaced the left and right pieces of the molecule.

Now let’s imagine a vertical mirror to the right of the molecule in Figure 10. We’ve seen that it doesn’t matter where the mirror is, so we’ve chosen this one because it makes the mirror image easy to draw. The molecule in the mirror (illustrated in Figure 10) looks the same as the original except that the positions 5. CHEMICAL ACHIRALITY AND GEOMETRIC CHIRALITY OF FIGURE 8 299

Figure 10. The mirror image of the molecule from Figure 9. of the vertical and horizontal hexagons are switched. Also, in the mirror image, Leftie is wearing a dress with a backwards R and Rightie is wearing a dress with a backwards L. This may seem confusing, but since Leftie only has a left arm and Rightie only has a right arm, even when they wear each other’s clothes turned inside out, there is no ambiguity about which twin is which. We show as follows that the molecule in Figure 9 can transform itself to its mirror image in Figure 10. First we rotate the molecule in Figure 9 by 90◦ around a horizontal axis as illustrated in Figure 11. Since the molecule is in solution, it can rotate freely.

Figure 11. We rotate the molecule by 90◦ around a horizontal axis.

Figure 12. This is the result of rotating the molecule by 90◦. 300 12. MIRROR IMAGE SYMMETRY FROM DIFFERENT VIEWPOINTS

As we see in Figure 12, this causes the vertical hexagon to become horizontal and the horizontal hexagon to become vertical. We’re happy about this because this is the way the hexagons are in the mirror image of the molecule. However, the oxygens are now in back and in front of the page, and Leftie and Rightie are now horizontal. To ﬁx the oxygens and the twins, we simply let the propellers spin until they are as they should be. In this way, we end up with the mirror image that was illustrated in Figure 10. Since rotating the molecule by 90◦ and letting the propellers spin are both chemically possible, this shows that the molecule is chemically achiral. Next we want to show that the molecule is geometrically chiral. This means we have to show that no rigid motion can take the molecule to its mirror image. We begin by freezing the molecule in its original position so that the propellers can’t spin. In Figure 13, we see that in the original molecule, Leftie is parallel to the adjacent hexagon. Hence, as a rigid object, she will always have her left arm forward and be parallel to the adjacent hexagon. In fact, whether or not she’s rigid, she doesn’t have a right arm so she can’t put her right arm forward.

Figure 13. Leftie is parallel to the adjacent hexagon in the original form.

In the mirror image (illustrated in Figure 14), Leftie is perpendicular to the adjacent hexagon, whereas Rightie is parallel to the adjacent hexagon. Since no rigid motion can take a pair of parallel hexagons to a pair of perpendicular hexagons, and Leftie cannot become Rightie, as a rigid object the original molecule is diﬀerent from its mirror image. Thus the molecule is geometrically chiral.

Figure 14. Leftie is perpendicular to the adjacent hexagon in the mirror image. 6. EUCLIDEAN RUBBER GLOVES 301

Figure 15. Not every chemically achiral molecule is geometrically achiral.

This molecule illustrates that the concepts of chemical achirality and geometric achirality are not equivalent. In particular, the question that was in Figure 7 can now be replaced by the molecule in Figure 8. The new diagram is illustrated in Figure 15.

6. Euclidean Rubber Gloves

There is a name for molecules like the one illustrated in Figure 8, which we deﬁne below.

Definition. AmoleculeissaidtobeaEuclidean rubber glove if it is chemically achiral, but it is chemically impossible for it to attain a position which is geometrically achiral. In the last section, we demonstrated that the molecule in Figure 8 is a Euclidean rubber glove. In order to see why these types of molecules are called “rubber gloves” remember that a right-handed rubber glove becomes a left-handed rubber glove when it is turned inside out. However, despite it being made of rubber or even latex, a rubber glove does not have the flexibility to get into a position which is geometrically achiral. It’s the physical characteristics of a rubber glove which prevent it from attaining a position which would be rigidly the same as its mirror image. The word “Euclidean” in this definition refers to physical, chemical, or geometric constraints that prevent the molecule from getting into a geometrically achiral position. For example, if we ignored all such constraints for the molecule in Figure 8, we could flatten it into the page of the paper as we’ve done in the image on the bottom of Figure 16. Notice that the drawing at the bottom of Figure 16 has no 302 12. MIRROR IMAGE SYMMETRY FROM DIFFERENT VIEWPOINTS

Figure 16. Ignoring chemical constraints, we could flatten this molecule into a plane. black triangles or dashed triangles because no part of it is in front of or behind the plane of the paper. This includes the horizontal hexagon in the molecule which has now become vertical in the flattened picture. If we imagine that the mirror is in the plane of the paper, we can see that the flattened out molecule would be its own mirror image. In fact, any object that is contained in a plane is its own mirror image in 3-dimensional space. Thus a flattened out drawing of the molecule in Figure 8 would be geometrically achiral. However, flattening out the molecule is not chemically possible.

7. Geometrically Chiral and Achiral Knots

It is very tempting to think that every object which can change into its mirror image can be deformed into a geometrically achiral position if it’s sufficiently flexible. In particular, we might wonder whether every molecule which is a Euclidean rubber glove can be flattened out so that it’s contained in a plane like the molecule in Figure 16. To explore whether this is true, let’s first consider some knots. We saw in Chapter 9 that the figure eight knot can be deformed to its mirror image. However, there is no rigid motion taking the usual projection of a figure eight knot to its mirror image (see Figure 17). For example, if you turn over the projection of the figure eight knot on the left, you will obtain the same projection again rather than its mirror image. In particular, this projection of the figure eight knot is not geometrically achiral. With some effort you can deform the figure eight knot to the projection illustrated in Figure 18, which we saw previously can be rotated by 90◦ to obtain its mirror image. Thus Figure 18 is a geometrically achiral projection of the figure eight knot. What makes this projection difficult to find is that it has twice as many crossings as the usual projection of the figure eight knot. 7. GEOMETRICALLY CHIRAL AND ACHIRAL KNOTS 303

Figure 17. Turning over the ﬁgure eight knot gives us the same projection rather than the mirror image.

Figure 18. A geometrically achiral projection of the ﬁgure eight knot together with its mirror image.

Figure 19. 817 can be deformed to its mirror image but cannot be deformed to a geometrically achiral position.

It turns out that the eight crossing knot 817 (illustrated in Figure 19) is the ﬁrst example in the knot tables of a knot which can be deformed to its mirror image but cannot be deformed to a geometrically achiral position. However, the proof of this is not easy. In fact, if an inﬁnitely long string contains 817, then it can be deformed to a projection which is geometrically achiral. We illustrate such a projection in Figure 20, where we see that the knotted line can be rotated by 180◦ to obtain its mirror image. However, as a knotted circle, 817 cannot be deformed to a projection which can be transformed to its mirror image by a rigid motion. For example, imitating Figure 20, we can deform 817 to the projection in Figure 21. However, if we rotate this projection by 180◦, then even though the knotted arc goes to its mirror image, the semi-circular arc will go to the top when it should remain at the bottom. This

Figure 20. This knotted line can be rotated by 180◦ to obtain its mirror image. 304 12. MIRROR IMAGE SYMMETRY FROM DIFFERENT VIEWPOINTS

Figure 21. If we rotate this projection by 180◦, the arc at the bottom goes to the top. knot demonstrates that not every object that can change into its mirror image necessarily can be deformed to a position which is geometrically achiral. Let’s compare the properties of the knot 817 to those of the molecule illustrated in Figure 8. Since the molecule is a Euclidean rubber glove, it is chemically achiral but geometrically chiral. However, we saw that if we were to ignore all chemical constraints, the molecule could be ﬂattened out into a geometrically achiral position. By contrast, the knot 817 can be deformed to its mirror image, but cannot be deformed to a geometrically achiral position.

8. A Topological Rubber Glove

We might wonder if there could be a molecule that behaves like the knot 817. That is, is there a molecule which is chemically achiral but can’t be deformed to a geometrically achiral position? This question was asked in 1983, but the first example of such a molecule wasn’t synthesized until 1997. The molecule (illustrated in Figure 22) is a link with a pair of adjacent hexagons at the top that can rotate around the bonds connecting them to the rest of the molecule (as indicated by the arrows). Note that the H3CandtheCH3 represent the same group of atoms. The order in which the letters are written is meant to indicate that the C is attached to the adjacent hexagon in both molecules. We see as follows that this molecule is chemically achiral. Since the two components of the link are not connected to one another, the lower ring can turn over without affecting the top ring. When the lower ring does turn over, the crossings between the two rings switch and the H3C goes from sticking out on the left to sticking out on the right. In this way, the turned over lower ring looks exactly like the lower ring in the mirror image. Since the pair of adjacent hexagons on the top ring rotates, the pair can change from being slanted up to being slanted down. This means the pair of hexagons can go from their position in the original molecule to their position in the mirror image. Thus the molecule can chemically transform itself into its mirror image, and hence it is indeed chemically achiral. Next we want to convince ourselves that, even if the molecule in Figure 22 were completely flexible, it could not be deformed to a position which is geometrically achiral. Suppose it could be deformed into a position which is geometrically achiral. Once it’s in such a position, it’s frozen. This means the slanted hexagons can no longer rotate. Since the components of the link are not identical, the rigid motion taking the deformed molecule to its mirror image must take each component to itself. But the bottom component doesn’t have mirror image symmetry because the H3C is only on one side, and the top component doesn’t have mirror image 8. A TOPOLOGICAL RUBBER GLOVE 305

Figure 22. A molecular link and its mirror image. symmetry because the pair of slanted hexagons are not symmetric. So the only way for the molecule to be geometrically achiral would be if both components were actually lying in the plane of the mirror. But because the components are linked, they cannot simultaneously lie in a plane. Hence the molecule cannot be deformed to a position which is geometrically achiral. By analogy with our deﬁnition of a Euclidean rubber glove, we now make the following deﬁnition.

Definition. A molecule is said to be a topological rubber glove if it is chemically achiral, but even if it were completely ﬂexible it could not be deformed to a position that is geometrically achiral.

Recall that a molecule is a Euclidean rubber glove if it is chemically achiral, but it is chemically impossible to attain a position which is geometrically achiral. In other words, it is chemically impossible for a Euclidean rubber glove to get into a geometrically achiral position. A topological rubber glove has the stronger property that even if we ignore all chemical constraints and imagine that the molecule is a completely ﬂexible graph, then the molecule still couldn’t get into a position that is geometrically achiral. The molecule illustrated in Figure 22 is an example of a topological rubber glove. Figure 23 provides an illustration of the relationship between chemical achirality and geometric achirality. rubber glove. Note that the molecules from both Figure 8 and 22 are in the larger oval but not the smaller oval because they are chemically achiral but not geometrically achiral. The diﬀerence between these two molecules is that chemical constraints prevent the molecule in Figure 8 from attaining a geometrically achiral position, while even ignoring chemical constraints the molecule in Figure 22 cannot be deformed to a geometrically achiral position. 306 12. MIRROR IMAGE SYMMETRY FROM DIFFERENT VIEWPOINTS

Figure 23. The relationship between chemically achiral and geometrically achiral molecules.

9. Topological Chirality

When we defined geometric chirality, we were taking a geometric viewpoint by treating all molecules as if they were completely rigid. As we have seen, some molecules are rigid, while others have pieces that can rotate around certain bonds, and large molecules (like DNA) can even be pretty flexible. So a geometric viewpoint does not always correspond to chemical reality. In fact, a mathematical char- acterization of chirality for all molecules is impossible because the level of rigidity of a molecule depends on its chemistry not just its geometry. In this section, we will take a topological viewpoint by treating all molecules as if they were completely flexible. The chemical reality is wedged between the geometric viewpoint and the topological viewpoint. Definition. A molecule is said to be topologically chiral if it cannot be deformed to its mirror image assuming complete flexibility. Otherwise, it is said to be topologically achiral. Any chemical motion of a molecule (whether rigid or not) can be thought of as a deformation of the molecule. Thus if a molecule can chemically transform itself into its mirror image, then it can be deformed into its mirror image. This means that chemical achirality implies topological achirality. It follows that if a molecule is topologically chiral, it is also chemically chiral. Furthermore, a molecule whose chirality comes from its rigidity can be forced to change into its mirror image by heating it enough. However, if a molecule is topologically chiral, then bonds would have to break for it to change into its mirror image. Thus it will not change into its mirror image even when heated. So topological chirality is a more enduring type of chirality than geometric chirality. For this reason, topological chirality is an interesting concept for chemists. On the other hand, let’s consider the L-alanine molecule in Figure 24. This molecule is small and hence quite rigid. Thus it cannot chemically transform itself 9. TOPOLOGICAL CHIRALITY 307

Figure 24. This molecule is chemically chiral but topologically achiral. into its mirror image. However, if we pretend that the molecule is completely ﬂexible, we could grab the CO2H and push it to the left, while pulling the H to the right. In this way, we could deform the molecule to its mirror image. Hence the L-alanine molecule is chemically chiral but topologically achiral. The diagram in Figure 25 illustrates the relationship between topologically achiral molecules, chemically achiral molecules, and geometrically achiral molecules. In particular, we have seen that if a molecule is geometrically achiral, then it is chemically achiral. But if it’s geometrically chiral, it may or may not be chemically chiral, depending on the rigidity of the molecule. On the other hand if a molecule is topologically chiral, then it is chemically chiral, independent of how rigid the

Figure 25. The relationship between topologically achiral, chemically achiral, and geometrically achiral molecules. 308 12. MIRROR IMAGE SYMMETRY FROM DIFFERENT VIEWPOINTS molecule is. But if a molecule is topologically achiral, it may or may not be chemically achiral, depending on the rigidity of the molecule. So geometry is useful to prove chemical achirality, and topology is useful to prove chemical chirality. The molecular Möbius ladder illustrated in Figure 26 is an example of a topologically chiral molecule. This molecule is called a Möbius ladder because it looks like a Möbius strip except it is made from a ladder with three rungs rather than from a strip of paper. But how can we show that a molecule like the molecular Möbius ladder is topologically chiral? In order to do this, we would need to demonstrate that no deformation could take the molecule to its mirror image. This reminds us of the task of showing two knots are not equivalent. In Chapters 9, 10, and 11 we addressed this problem by developing invariants. However, there aren’t as many invariants for molecules as there are for knots and links. In the next chapter, we will learn several different methods for showing that a molecule is topologically chiral.

Figure 26. A molecular M¨obius ladder is an example of a topologically chiral molecule.

10. Exercises

1. Explain how we know that geometric chirality does not imply chemical chirality, and chemical chirality does not imply topological chirality. 2. Prove that if the molecule illustrated in Figure 27 were completely ﬂexible, it could be deformed into a plane. Is the molecule geometrically chiral? Is it topologically chiral? Given that this molecule is rigid, is it chemically chiral?

Figure 27. Illustration for Exercise 2. 10. EXERCISES 309

3. (Hard.) Prove that if we don’t distinguish between the rungs and sides of the molecular M¨obius ladder illustrated in Figure 26, then the graph could be deformed to its mirror image. 4. Draw a step-by-step deformation taking a M¨obius ladder with four rungs (illustrated in Figure 28) to a position where the circle representing the sides of the ladder lies in a plane.

Figure 28. Illustration for Exercise 4. 5. Consider a closed ladder like the molecular Möbius ladder in Figure 26, but with three half-twists instead of just one. Prove that there is no deformation that takes the closed ladder to its mirror image taking sides to sides and rungs to rungs. 6. We can define an object (not just a molecule) to be a Euclidean rubber glove if it can be deformed to its mirror image, but for physical, chemical, or geometric reasons it cannot be deformed to a geometrically achiral position. Explain why a real rubber glove is a Euclidean rubber glove. Give another example of a familiar object which is a Euclidean rubber glove, and explain why. 7. We can define an object to be a topological rubber glove if it can be deformed to its mirror image, but even if it’s completely flexible, it cannot be deformed to a geometrically achiral position. Is a trefoil knot a topological rubber glove? What about a figure eight knot? 8. Prove that if a molecule is a topological rubber glove, then it is a Euclidean rubber glove, but not the converse. 9. Is the molecule illustrated in Figure 8 a topological rubber glove? Explain your reasoning. 10. (Hard.) Prove that the Kuratowski cyclophane molecule (illustrated in Fig- ure 29) can be deformed to a geometrically achiral position. Note that the dotted edge goes behind the solid edges.

Figure 29. Illustration for Exercise 10.

https://doi.org/10.1090//mbk/096/13

CHAPTER 13

Techniques to Prove Topological Chirality

Topics: • A molecular knot • Using knot polynomials to prove topological chirality • Using 2-fold branched covers to prove topological chirality • Using chiral subgraphs to prove topological chirality • Graph automorphisms and their properties • A combinatorial method to prove topological chirality

1. Topological Chirality

In Chapter 12, we learned about three different types of chirality: geometric, chemical, and topological. In order to determine if a given molecule is geometrically chiral, we treat it as a rigid object and look at it from all different angles to see if there is a rigid motion from it to its mirror image. Chemical chirality is determined experimentally. In order to prove that a molecule is topologically chiral, we have to prove that it is impossible to deform to its mirror image assuming complete flexibility. This is most difficult type of chirality to prove, because there are infinitely many possible deformations of a given molecular graph. However, by treating molecules as topological objects, we can use techniques which build on material we’ve learned about knot theory to show particular molecules are topologically chiral. In this chapter, we will present four different techniques that have been used to prove specific molecules are topologically chiral. None of the techniques presented here (or any technique currently known) will work for all possible molecules. The best we can do is to develop a collection of techniques that we can choose among when confronted with a molecular graph. The more techniques we know, the better the odds are that one of them will work for the given molecule. This is analogous to what we learned about knot invariants in Chapters 9, 10, and 11. We would like to be able to prove that a particular knot is different from another knot. If we can show that a particular invariant of the two knots is different, then we know the knots are different. But there is no single invariant that we can use to distinguish every possible pair of knots. So the more invariants we know, the more likely it is that one of them will work to distinguish a given pair of knots.

311 312 13. TECHNIQUES TO PROVE TOPOLOGICAL CHIRALITY

2. Molecular Knots

Knots have been found to occur naturally in both DNA and proteins (which we will learn about in Chapters 14 and 15). However, it took over 50 years of attempts by organic chemists before the first molecular knot was successfully synthesized in 1989. Not surprisingly, the first molecular knot to be synthesized was a trefoil knot (illustrated in Figure 1). If we want to be precise, the molecular trefoil knot is not actually a knot according to the definition we gave in Chapter 9 because it has clumps of hexagons (known by chemists as aromatic rings) at various places along what should just be a circle. We call it a molecular knot because the overall structure of the molecule is that of a knot. That is, if you look at the picture from far away without your glasses or contact lenses, you’ll see the knot rather than all the hexagons and the many copies of the letter O.

Figure 1. A molecular trefoil knot.

In fact, this molecular graph contains many trefoil knots. Each hexagon has two different paths going from one end to the other. Depending on which of these paths we choose, we will trace a out a different trefoil. For example, in Figure 2, we use thick grey lines to illustrate two of the trefoils contained in the graph. Can you see where the two knots differ? You should also see if you can find other trefoils in the graph.

Figure 2. Two trefoil knots contained in this molecular graph. 3. TECHNIQUE 1: KNOT POLYNOMIALS 313

Notice that all of the non-trivial knots in the molecular graph are identical trefoil knots. In particular, the graph contains many left-handed trefoil knots (illustrated on the left in Figure 3), but no right-handed trefoil knots (illustrated on the right in Figure 3). In addition to all of the trefoil knots, the graph contains some small planar circles made of one, two, or three hexagons. But planar circles are identical to their mirror images, so they aren’t important to us here.

Figure 3. A left-handed trefoil and a right-handed trefoil.

3. Technique 1: Knot Polynomials

In order to prove that the molecular trefoil knot is topologically chiral we need to review what we learned about knot polynomials in Chapter 11. In particular, recall that the X-polynomial is defined for an oriented knot or link projection L by the formula X(L)=(−A3)−w(L) L ,wherew(L) denotes the writhe of L,and L is the bracket polynomial of L. The writhe of an oriented link is the difference between the number of positive crossings and the number of negative crossings. Also, recall that the bracket polynomial is defined by the rules below: Rule 1. =1. Rule 2. = A + A−1 . Rule 3. L =(−A2 − A−2) L .

In one of the Exercises in Chapter 11 you were asked to show that the trefoil knot cannot be deformed to its mirror image. If you did this exercise, you might have proved the following more general theorem. Theorem 13.1. Let L beaknotwithmirrorimageL∗. Then the X-polynomial X(L∗) is obtained from X(L) by reversing the signs of all of the exponents. For a projection of a knot, the value of the writhe is the same regardless of the orientation on the knot, which is why we don’t need to orient our knot in the statement of Theorem 13.1. Proof. We begin by comparing the writhe of a knot projection L to that of its mirror image L∗. When we take the mirror image of the projection, the positive and negative crossings switch, causing the writhe to change sign. Thus we have 314 13. TECHNIQUES TO PROVE TOPOLOGICAL CHIRALITY

∗ w(L∗)=−w(L). It follows that (−A3)−w(L ) =(−A3)w(L) =(−A−3)−w(L).Thus the X-polynomial of the mirror image is given by ∗ X(L∗)=(−A3)−w(L ) L∗ =(−A−3)−w(L) L∗ . Next let’s consider how the bracket polynomial of L∗ compares with that of L. Suppose that Rule 2 of the bracket polynomial is used to remove a particular crossing. If we apply Rule 2 to the same crossing in the mirror image, we get the same result except that A and A−1 are switched. Furthermore, replacing A by A−1 has no eﬀect on the formula in Rule 3, since −(A−1)2 − (A−1)−2 = −A2 − A−2. Also, switching A and A−1 has no eﬀect on the outcome of Rule 1. It follows that L∗ can be obtained from L by switching A and A−1. ∗ The formula (−A3)−w(L ) =(−A−3)−w(L) that we derived above can be re- ∗ stated by saying that we obtain (−A3)−w(L ) from (−A3)−w(L) by switching A and A−1. Putting all of this together, we conclude that X(L∗)isthesameastheX(L) except that the signs of the exponents have been reversed. This is exactly what the theorem says.

Returning now to the task of showing that the molecular trefoil knot is topologically chiral, let T be a left-handed trefoil knot. From Chapter 11, we know that the X-polynomial of T is X(T )=−A16 + A12 + A4. Theorem 13.1 tells us that the X-polynomial of the mirror image is X(T ∗)=−A−16 + A−12 + A−4.Thiscould be also shown by direct computation, but it’s easier to use the theorem. Since X(T ) = X(T ∗), we conclude that the left-handed trefoil knot cannot be deformed to the right-handed trefoil knot. Hence the trefoil knot is topologically chiral. Fi- nally, since the molecular trefoil knot in Figure 1 contains left-handed trefoils but no right-handed trefoils, and a trefoil knot is topologically chiral, our molecular trefoil knot must also be topologically chiral. We say that a knot polynomial is symmetric with respect to the signs of its exponents if switching the sign of every exponent does not change the polynomial. For example, the polynomial −A16 + A12 + A4 is not symmetric with respect to the signs of its exponents, whereas the polynomial −A2 − A−2 is symmetric with respect to the signs of its exponents. If we compute the X-polynomial of a knot and observe that it is not symmetric with respect to the signs of its exponents, then we immediately know that the knot is topologically chiral. The same statement is also true for the Jones polynomial, since the X-polynomial and the Jones polynomial are related by the change of variable q = A−4. This means that we can determine that a knot is topologically chiral just by looking at a knot table containing Jones polynomials. For instance, a knot table shows that the Jones polynomial for the trefoil knot is q + q3 − q4, which has only positive exponents, and hence does not have exponents which are symmetric with respect to sign. Hence we immediately know from the knot table that the trefoil is topologically chiral. These arguments should convince you of the truth of the following theorem. Theorem 13.2. If the Jones polynomial (or X-polynomial) of a knot is not symmetric with respect to the signs of the exponents, then the knot is topologically chiral. Though this is a useful technique to detect topological chirality, it does have some limitations. First of all, the technique can only be used on a molecule which contains a knot or a link. Secondly, even if the molecule does contain a knot or 4. TECHNIQUE 2: 2-FOLD BRANCHED COVERS 315 link, if its Jones polynomial happens to be symmetric with respect to the signs of its exponents, we can’t conclude anything about whether or not the knot is topologically achiral. Finally, in the case of the molecular trefoil knot, all of the knots in the molecular graph were left-handed trefoils. Hence, proving that a trefoil knot is topologically chiral is suﬃcient to know that the molecule is topologically chiral. But if a molecule contains both a knot and its mirror image, it may or may not be topologically chiral. For example, Figure 4 illustrates a molecular knot which contains both a left-handed trefoil and a right-handed trefoil. This molecule is chemically achiral, and hence topologically achiral.

Figure 4. A chemically achiral molecular knot.

Now imagine a vertical mirror dividing the knot in half. The mirror would simply interchange the two sides of the molecule, taking the molecule to itself. Thus the molecule is actually geometrically achiral.

4. Technique 2: 2-Fold Branched Covers

In 1983, the chemist David Walba synthesized a molecular M¨obius ladder by forcing the ends of a 3-rung molecular ladder to join with a half-twist (see Figure 5). Remember from Chapter 12 that corners in molecular drawings should be treated as vertices of the graph representing carbon atoms. The three vertical pairs of edges that we see in the illustration represent double bonds joining a pair of carbon atoms.

Figure 5. The molecular M¨obius ladder. 316 13. TECHNIQUES TO PROVE TOPOLOGICAL CHIRALITY

Walba wanted to assert that some of his molecules did indeed have the form of aMöbius ladder, but the molecules were too small to see in a microscope. Walba showed experimentally that some of his molecules were chemically chiral, and he wanted to argue that this was consistent with their having the form of Figure 5. He could prove that a Möbius ladder was geometrically chiral, but the molecules that he had synthesized were long enough that they had some flexibility. So knowing that the molecules were geometrically chiral wasn’t sufficient to conclude that they were chemically chiral. Walba suspected that the molecular Möbius ladder was topologically chiral, but he couldn’t prove it. He knew that a Möbius strip with the twist going one way could not be deformed to a Möbius strip with the twist going the other way. However, there are more ways to deform a Möbius ladder than a Möbius strip. In 1986, the topologist Jon Simon solved Walba’s problem by finding a link in the 2-fold branched cover of the Möbius ladder whose topological chirality he could use to prove the topological chirality of the molecular Möbius ladder. This was the first time deep topological techniques were used to solve a chemically motivated question, and as such marked the beginning of the interdisciplinary field of chemical topology. We won’t define 2-fold branched covers because the definition is too technical. You could see the definition if you took a geometric topology course. Here we give a sketch of the main idea’s of Simon’s proof, without going into the technicalities. The first step of the proof is to simplify the graph of the Möbius ladder by omitting all of the atoms and the double bonds, and distinguishing the rungs from the sides by coloring the rungs black and the sides grey (see Figure 6).

Figure 6. A colored M¨obius ladder.

In fact (as you may have shown in the Exercises in Chapter 12), if we don’t distinguish between rungs and sides, then there is actually a deformation from the graph of the Möbiusladder to its mirror image. However, such a deformation interchanges a pair of rungs of the ladder (which represent carbon-carbon double bonds) with a pair of chains of carbons and oxygens contained in the sides of the ladder. Interchanging carbon-carbon bonds with chains of carbons and oxygens makes no sense from a chemical point of view. Hence we will only allow deformations of the graph in Figure 6 which take grey edges to grey edges and black edges to black edges. We draw the different rungs with different types of segments, not because the rungs are different from one another, just to make it easier to keep track of what’s going on later in our argument. Since we’re trying to prove that the Möbius ladder graph is topologically chiral, we can assume the graph is completely flexible. With this flexibility, we can deform the circle making up the sides of the ladder so that it lies in a plane, as illustrated 4. TECHNIQUE 2: 2-FOLD BRANCHED COVERS 317

Figure 7. We deform the Möbius ladder so that the grey circle lies in a plane. on the right in Figure 7. You should check for yourself that the graph can indeed be deformed to look like this. We have numbered the rungs in Figure 7 to help us keep track of their vertical order. In particular, observe that in Figure 7, rung 1 is at the bottom, rung 2 is in the middle, and rung 3 is at the top. In Figure 8, we illustrate two copies of the Möbius ladder, one above the other. To get the 2-fold branched cover of the Möbius ladder, we need to glue the top and bottom copies together along the grey circle. In order to be able to fit the two copies together, in the bottom graph we have pushed all of the rungs down below the level of the plane containing the grey circle. Nonetheless, in the bottom graph, rung 3 is still in the top position, rung 2 is still in the middle position, and rung 1 is still in the lowest position.

Figure 8. Both of these M¨obius ladders have rung 3 on top, rung 2 in the middle, and rung 1 on the bottom.

Now we glue the top and bottom copies of the M¨obius ladder together along the grey circle in such a way that the endpoints of identical rungs are lined up as illustrated on the left side of Figure 9. This is the 2-fold branched cover of the M¨obius ladder. 318 13. TECHNIQUES TO PROVE TOPOLOGICAL CHIRALITY

Figure 9. If we erase the grey circle in the 2-fold branched cover, then we get a link.

Observe in the left image in Figure 9 that the 2-fold branched cover consists of one grey circle and three black circles. Each black circle is the result of gluing together a pair of identical rungs along their endpoints. By erasing the grey circle, we obtain the link with three components illustrated on the right. This link has three components but they are not the Borromean rings. Rather it is a link with the property that each component is linked with each of the other two components. Jon Simon proved that this link is topologically chiral by analyzing the linking numbers of each pair of components. His proof required considering all possible ways of orienting each of the components and computing all of the linking numbers in each case. An outline of the proof is given in the Exercises. By proving that this link is topologically chiral, Simon was able to conclude that the molecular M¨obius ladder is topologically chiral as well. The topological chirality of the M¨obius ladder is important not only because it’s an interesting molecule, but because it can be used to prove that more complicated molecules are topologically chiral. We see an example of this in the next section.

5. Technique 3: Chiral Subgraphs

This method was developed by the chemist Kurt Mislow to prove that the molecule triple layered naphthalenophane (illustrated in Figure 10) was topologically chiral. A naphthalene molecule consists of a pair of adjacent hexagons and is the primary ingredient in mothballs. The name “triple layered naphthalenophane” is easy to remember because the molecule does indeed have three layers, each consisting of a naphthalene. The idea of this technique is to search for a subgraph contained in our molecule which is known to be topologically chiral and use it to prove that the entire graph is topologically chiral. For this method to work, the topologically chiral subgraph must have the property that any deformation of the molecule to its mirror image would necessarily take this subgraph to a mirror image of itself. In the case of triple layered naphthalenophane, we will show that there is M¨obius ladder hiding in the molecule which has the required property. We deﬁne the length of a path or a circle in a graph as the number of vertices contained in it. Remember that each corner in a molecular graph represents a carbon atom, and hence counts as a vertex. A key observation that we will use in 5. TECHNIQUE 3: CHIRAL SUBGRAPHS 319

Figure 10. Triple layered naphthalenophane. our proof that triple layered naphthalenophane is topologically chiral is that the graph contains a special circle C that is longer than any other circle in the graph. This special circle is illustrated in grey in Figure 11. The length of the grey circle is 35 (you should check this yourself). In the Exercises, you are asked to show that this circle is indeed longer than any other circle in the graph. Observe that in Figure 11 the three black edges in the center of the graph are the only edges which have both vertices on the circle C.SobothC and these three edges play a special role in our graph.

Figure 11. C is the longest circle in triple layered naphthalenophane.

We will prove that triple layered naphthalenophane is topologically chiral by contradiction. This means that we begin by assuming that the graph of triple layered naphthalenophane can be deformed to its mirror image, assuming complete flexibility. Observe that no matter how we deform a graph, we do not change the number of vertices in a circle or a path. So our deformation of the graph to its mirror image must take the circle C to a circle of the same length in the mirror image, and hence must take the edges with both vertices on C to edges with both vertices on the corresponding circle in the mirror image. This means that our deformation taking triple layered naphthalenophane to its mirror image must also take the colored graph on the left in Figure 12 to its mirror image. However, the graph in Figure 12 can be deformed to a Möbius ladder by smooth- ing out the circle C and spreading out the three grey edges around C. In the illustration on the right in Figure 12 we erase all of the corners in order to make the Möbius ladder look like the ones we saw in the previous section. Notice that the rungs of our Möbius ladder are black and the sides are grey as they were in our proof 320 13. TECHNIQUES TO PROVE TOPOLOGICAL CHIRALITY

Figure 12. C together with these three edges can be deformed to aMöbius ladder. that the molecular Möbius ladder is topologically chiral. Since we know that there is no deformation of this colored Möbius ladder to its mirror image, the graph of triple layered naphthalenophane also could not be deformed to its mirror image. In particular, this means that triple layered naphthalenophane is topologically chiral.

6. Some Graph Theory

The next technique that we introduce uses the graphical structure of a molecule to prove that no matter how the structure is situated in 3-dimensional space, it will be topologically chiral. In order to explain this technique, ﬁrst we need to learn some terminology from graph theory. An abstract graph is a collection of vertices, together with a collection of edges joining some of the vertices. In particular, an abstract graph is independent of how the graph is situated in space, just like an abstract gluing of a surface is independent of how the surface is situated in space. If we are interested in talking about how a graph is sitting in space, we refer to its position in space as an embedding of the graph. In the language of Chapter 3, if we are talking about an abstract graph, then we are considering the graph intrinsically. Whereas if we are talking about an embedding of a graph in space, then we are considering the graph extrinsically. In Chapters 1–8, we saw that when we study 3-manifolds we always think of them intrinsically, because as 3-dimensional beings we are incapable of knowing anything about how our 3-dimensional space could be sitting inside of a higher- dimensional space. Similarly, A. Square always studies 2-manifolds intrinsically, and A. Dash always studies 1-manifolds intrinsically. We can consider 2-dimensional spaces extrinsically and A. Square can consider 1-dimensional spaces extrinsically. But poor A. Dash can only consider 0-manifolds extrinsically, which is pretty boring since every 0-manifold is just a point. In Chapters 9–11, we saw that knots have to be studied extrinsically, because intrinsically every knot is just a circle. Molecular graphs are noteworthy because they can either be studied extrinsically as objects in space as well as intrinsically just in terms of what is connected to what. So far, in this chapter and the previous one, we have only been considering molecular graph extrinsically. The fourth technique that we will present for proving that a molecule is topologically chiral treats molecular graphs intrinsically. 6. SOME GRAPH THEORY 321

The valence of a vertex in a graph is the number of edges that contain that vertex. The distance between two vertices in a graph is the fewest number of edges contained in a path from one of the vertices to the other. Two vertices are adjacent if there is an edge in the graph joining them. Before we introduce more terminology, let’s practice using these definitions on the Möbius ladder graph illustrated in Figure 13. If we want to talk about the abstract graph of the Möbius ladder, then we ignore the twist in the front of the graph. On the other hand, if we are considering the embedding of the graph, then this twist is a crucial piece of information. To determine the valence of a vertex, we only need to look at an abstract graph. For the Möbius ladder graph, the valence of each vertex is 3. We can further observe that vertex 1 in the Möbius ladder is adjacent to vertices 2, 4, and 6; whereas, vertices 1, 3, and 5 are each a distance 2 apart.

Figure 13. An embedding of the M¨obius ladder graph.

If you’re confident about the terminology we’ve introduced so far, then it’s time to introduce a little more. Definition. An automorphism of an abstract graph is a permutation of the vertices which takes each pair of adjacent vertices to a pair of adjacent vertices. An automorphism of a molecular graph is also required to take atoms of a given type to atoms of the same type. When we look for automorphisms of a graph, the embedding of the graph is not important. What is important is which vertices are connected to which. As an example, we again consider the Möbius ladder graph in Figure 13. The permutation which interchanges vertices 2 and 6 and interchanges vertices 3 and 5 is an automorphism. The fact that this automorphism can’t be achieved just by turning the graph upside down in space is irrelevant, since we are only considering the abstract graph. Observe that the permutation which just interchanges vertices 2 and 5 and doesn’t move any other vertices is not an automorphism of Figure 13 because vertex 2 is adjacent to vertex 3, while vertex 5 is not adjacent to vertex 3. The following properties of automorphisms may seem obvious. But proving them from the above definitions is worth doing in order to check your understanding. You’ll have the opportunity to prove them in the Exercises. Properties of Automorphisms:

(1) Any automorphism of an abstract graph takes vertices of a given valence to vertices of the same valence. (2) Any automorphism of an abstract graph takes a pair of vertices which are a certain distance apart to a pair of vertices which are the same distance apart. 322 13. TECHNIQUES TO PROVE TOPOLOGICAL CHIRALITY

All right, now we’re ready to learn one final definition. Definition. The order of an automorphism is the smallest number n such that doing the automorphism n times takes each vertex back to its original position. For example, the automorphism of the Möbius ladder in Figure 13 which interchanges vertices 2 and 6 and vertices 3 and 5 has order 2, because if we do the automorphism twice, then every vertex returns to its original position; and if we only did the automorphsim once, every vertex would not stay in its original position. The identity automorphism (also known as the trivial automorphism)which doesn’t move any vertex, is the only automorphism that has order 1. In Figure 14 we illustrate two very important graphs: the complete graph on five vertices K5, and the complete bipartite graph on two sets of three vertices K3,3. Note that all of the vertices in the figure are represented by dots. In particular, the places where edges appear to intersect in Figure 14 are neither vertices nor actual intersections. We draw the graphs this way because we are considering them as abstract graphs, and hence we don’t want to specify a particular embedding of the graphs in space.

Figure 14. The graphs K5 and K3,3.

The graphs K5 and K3,3 are special because they are the “smallest graphs” which cannot be embedded in a plane without any edges intersecting. We are interested in embeddings of graphs in a plane because such an embedding is its own mirror image, where the mirror is the plane itself. This means that any embedding of a graph in a plane is topologically achiral. In fact, such an embedding is geometrically achiral.

7. Technique 4: A Combinatorial Approach

The following theorem, whose proof uses advanced techniques in topology, allows us to prove that many abstract graphs have no topologically achiral embedding in 3-dimensional space.

Theorem 13.3. If a graph has no order 2 automorphisms and contains either 3 of the graphs K5 or K3,3, then any embedding of the graph in R is topologically chiral. 7. TECHNIQUE 4: A COMBINATORIAL APPROACH 323

Figure 15. The molecule ferrocenophane.

We will now use Theorem 13.3 to show that the molecule ferrocenophane (illustrated in Figure 15) is topologically chiral. To help you understand this figure, you should note that the Fe represents an iron atom, the O represents an oxygen atom, and the rest of the atoms are carbons. While it is certainly doesn’t look like it at first glance, ferrocenophane contains the complete graph K5. To see this, we first observe in Figure 14 that as an abstract graph, K5 is just a set of five vertices together with edges joining every pair of vertices. We illustrate a K5 which is contained in ferrocenophane in Figure 16. In the K5 on the right in Figure 16, only the dots with numbers on them represent vertices. Though the corners in the graph of ferrocenophane represent carbon atoms, in this K5 the corners have no significance other than to help us see that K5 is contained in the molecule.

Figure 16. Ferrocenophane contains the graph K5.

In order to use Theorem 13.3, we must also show that ferrocenophane has no order 2 automorphisms. We will actually show that the only automorphism that ferrocenophane has is the identity automorphism. We begin by letting f be an automorphism of the abstract graph of ferrocenophane. Recall that an automorphism of a molecular graph is required to take atoms of a given type to atoms of the same type. This means that f must take the single oxygen atom to itself as illustrated in Figure 17. If an automorphism takes a vertex to itself, we will say it ﬁxes the vertex. 324 13. TECHNIQUES TO PROVE TOPOLOGICAL CHIRALITY

Figure 17. f must ﬁx the single oxygen atom.

Figure 18. f must ﬁx the carbon atom adjacent to the oxygen.

Now it follows from the Properties of Automorphisms that f must take a pair of adjacent vertices to a pair of adjacent vertices. In particular, f must ﬁx the only carbon which is adjacent to the oxygen atom, as shown in Figure 18. Next observe that there are two vertices in addition to the oxygen atom which are also adjacent to this carbon. However, one vertex has valence 2 and the other vertex has valence 4. Recall that we know from the Properties of Automorphisms

Figure 19. f must fix each of these vertices. 8. EXERCISES 325 that an automorphism must preserve the valence of a vertex. Hence f must take each of these vertices to itself, as illustrated in Figure 19. We can continue arguing in this way, going through the vertices one after the other, to conclude that in fact every vertex of ferrocenophane must be fixed by the automorphism f. But this means that f is actually the identity automorphism. Thus we have shown that the identity is the only automorphism of ferrocenophane. In particular, ferrocenophane cannot have an automorphism of order 2. Since we have now shown that ferrocenophane satisfies both of the hypotheses of Theorem 13.3, we can conclude that not only is ferrocenophane topologically chiral, but any molecule which has the same abstract graph as ferrocenophane must be topologically chiral as well. In fact, Theorem 13.3 has nothing to do with a particular embedding of a graph in R3. Once we check that the conditions of Theorem 13.3 are satisfied for a graph, as we did for ferrocenophane, then even if we change the embedding of the graph in space it will still be topologically chiral. This seems surprising since topological chirality has to do with comparing an embedding of a graph with the mirror image embedding of the same graph. This observation motivates us to make the following definition. Definition. A graph is said to be intrinsically chiral if every embedding of it in R3 is topologically chiral. The idea of this definition is that a graph is intrinsically chiral if its topological chirality is an intrinsic property of the graph, rather than depending on the extrinsic topology of the graph in R3. Using this terminology, we now restate Theorem 13.3 as follows. Theorem 13.4. If a graph has no order 2 automorphisms and contains either of the graphs K5 or K3,3, then it is intrinsically chiral. In the Exercises, you will use Theorem 13.4 to show that some other molecules are intrinsically chiral.

8. Exercises

1. In this exercise you will follow the steps below to complete the proof that the molecular M¨obius ladder is topologically chiral by proving that the link in Figure 17 is topologically chiral.

Figure 20. Illustration for Exercise 1. 326 13. TECHNIQUES TO PROVE TOPOLOGICAL CHIRALITY

(a) Assign an orientation to each component of the link and compute the linking numbers lk(J, K), lk(J, L), and lk(K, L). (b) Compute the linking numbers of the mirror image of the oriented link. (c) Show that changing the orientation of a single component in the mirror image will change the number of negative linking numbers by an even number. (d) Use the above steps to prove that the link in Figure 20 is topologically chiral. 2. Prove that the circle C in Figure 11 is longer than any other circle in the graph. Hint: Any longest circle must travel through all three pairs of adjacent hexagons. 3. Illustrate a deformation from the graph on the left of Figure 12 to the colored M¨obius ladder. 4. Consider a graph like triple layered naphthalenophane, but with a row of four hexagons on each of the three layers instead of just two. Prove that this graph is topologically chiral.

5. Prove that the graph K3,3 is not intrinsically chiral by ﬁnding a topologically 3 achiral embedding of K3,3 in R . 6. Figure 21 illustrates the Simmons-Paquette molecule. All of the atoms in this molecule are carbons except for the three oxygens labeled with O’s. Prove that this molecule is intrinsically chiral.

Figure 21. Illustration for Exercise 6.

7. The molecular sphere in Figure 22 is called a fullerene. The ﬁgure illustrates a fullerene together with a cap which has 3-fold symmetry. That is, rotating the molecule around a vertical axis by 120◦ will not change the way it looks. In particular, the way that the cap is attached in the back of the molecule is the same as on the two sides. Prove that this molecule contains either K5 or K3,3, and contains a “special circle” such that any automorphism of the molecule takes that circle to itself. 8. Use Exercise 7 to prove that the molecule illustrated in Figure 22 is intrinsically chiral. 9. Prove that the protein illustrated in Figure 23 is topologically chiral. You may use the fact that any deformation of the protein to its mirror image must send the N to the N in the mirror image and the C to the C in the mirror image. 8. EXERCISES 327

Figure 22. Illustration for Exercises 7 and 8.

Figure 23. Illustration for Exercise 9.

10. Prove that any automorphism of a graph takes vertices of a given valence to vertices of the same valence. 11. Prove that any automorphism of a graph takes a pair of vertices which are a given distance apart to a pair of vertices which are the same distance apart. 12. Prove that a 3-rung molecular M¨obiusladder has an automorphism of order 6. 13. Let C denote a circle with six vertices. Prove that the order of any automorphism of C divides 6.

14. Find topogically achiral embeddings of the abstract graphs K5 and K6 illustrated in Figure 24. Note there are no intersections between the edges. We draw them with intersecting segments because we don’t want to specify a particular embedding of the graphs in space.

15. Figure 25 illustrates a M¨obius ladder with four rungs both as an embedded graph in the form of a M¨obius strip (on the left) and as an abstract graph (on the right). In the picture of the abstract graph, note that the edges do not intersect and there is no vertex in the center of the picture. Draw an embedding of this graph which is topologically achiral. 328 13. TECHNIQUES TO PROVE TOPOLOGICAL CHIRALITY

Figure 24. Illustration for Exercise 14.

Figure 25. Illustration for Exercise 15.

16. Show that the molecule illustrated in Figure 26 contains the abstract graph of aM¨obius ladder with four rungs. 17. Prove that the graph illustrated in Figure 26 is topologically achiral.

Figure 26. Illustration for Exercises 16 and 17. https://doi.org/10.1090//mbk/096/14

CHAPTER 14

The Topology and Geometry of DNA

Topics: • The biological structure of DNA • Supercoiling • The linking of the two backbones • The average writhe of the axis • The twist of one backbone around the axis • The equation Lk = Tw+ Wr • How to measure supercoiling • Replication • Site speciﬁc recombination • Rational tangles • Tangle operations • The tangle model of site speciﬁc recombination • Solving the mystery of Tn3 resolvase

1. Synthetic versus Biological Molecules

In Chapters 12 and 13 we learned that organic chemists have created molecular versions of knots, links, Möbiusladders, and other interesting 3-dimensional graphs. Synthesizing such topologically complex molecules is no mean feat. In fact, despite many years of trying to synthesize a variety of different molecular knots, the only knots which have been synthesized as of 2015 (without using DNA or other biologically based polymers) are the five illustrated in Figure 1.

Figure 1. These are the only knots that have been synthesized by chemists.

In fact, most molecules that are created in a laboratory are small and relatively rigid, so that their geometry is important, but their topology is not. If you have ever looked at the leaﬂets that come with medications, you might have noticed that the chemical diagrams on them are pretty simple. Perhaps at some point, there

329 330 14. THE TOPOLOGY AND GEOMETRY OF DNA might be a revolutionary new medicine synthesized in the form of a knot. But in view of the difficulty involved, it’s unlikely to happen any time in the near future. In contrast with molecules designed and created from scratch by chemists, biological molecules like DNA1 and proteins are sufficiently long and floppy that they can become tangled up all by themselves, just like your headphones get tangled when you stuff them into your pocket. In fact, while organic chemists are trying to synthesize molecules which are more knotted, molecular biologists are trying to better understand how DNA keeps from becoming too knotted. This means that topology and knot theory are not only important tools for understanding the chirality of synthetic molecules (as we saw in Chapter 13), they are also important tools for understanding the structure and behavior of biological molecules that occur naturally.

2. The Biology of DNA

Imagine if you had the most important job on the planet. No, not being the President of the United States. That’s way too minor. The job is making sure that life continues. The job description involves designing and operating a system of reproduction that passes on complex genetic information while introducing genetic diversity and allows cells to duplicate and grow during reproduction and throughout the life cycle. While this job seems much too diﬃcult and important for you or me to take on, DNA is happy to do it, and its success at the job is entirely due to its knowledge of topology and geometry. There is no doubt that the job is stressful, but DNA even has its own system to relieve its stress without having to join a gym. The rest of this chapter will be devoted to learning a few of the secrets of its success, though we won’t include enough information that DNA might worry that we’re thinking of trying to take over its job. Rather, we’ll learn just enough for us to appreciate how incredibly smart and competent DNA is. In this section and the next we’ll learn a little bit about the biological aspects of the job of DNA. Then, starting in Section 4, we’ll learn about the topology and geometry that DNA uses to do its job. The most important information that DNA has is coded in a sequence of letters, known as the genetic code. The more complicated the organism, the longer the sequence needs to be to keep track of all the information. For humans, this sequence contains roughly 3 billion letters. The letters making up the sequence are chosen from among A, T, C, and G. These letters stand for four bases known as adenine, thymine, cytosine, and guanine, which are attached to a long string of alternating sugar and phosphate molecules making up a strand of DNA. However, DNA is so clever and good at its job that right from the beginning it instituted a system of having a backup copy of everything. In particular, there are actually two strands of DNA wound together, which each keep track of all of the important information. But the sequence of bases on the two strands aren’t identical. Such clever DNA would think that’s much too simple a scheme. Rather, each base A on one strand bonds with the base T on the other strand, and each base C on one strand bonds with the base G on the other. So if the sequence ATGGA occurs on one strand, then it must be bonded with the sequence TACCT

1In case you were wondering, DNA is short for DeoxyriboNucleic Acid. 2. THE BIOLOGY OF DNA 331 on the other strand. Since the sequence of A’s, T’s, C’s, and G’s that occurs on one strand determines the sequence of A’s, T’s, C’s, and G’s that occurs on the other strand, we only need to list one of these two sequences in order to specify the entire genetic code. But if somehow there are bases on one strand that don’t bond with the corresponding bases on the other string, then the DNA knows right away that some funny business has happened on one of the strands, and it can try to figure out who made a mistake and fix it. Figure 2 illustrates the sugars, phosphates, and bases on two strands of a small segment of DNA. Each sugar is denoted by the letter S and each phosphate is denoted by the letter P . The bases that are bonded together on a DNA molecule are called base pairs. For example the segment of DNA illustrated in Figure 2 has five base pairs. Looking at the figure, we can think of the base pairs as rungs on a molecular ladder.

Figure 2. The sequence ATGGA on one strand binds with the sequence TACCT on the other strand.

Occasionally, one can ﬁnd a cell in which there is a DNA molecule that only has one strand instead of two. In order to distinguish this single stranded form of DNA from the usual DNA molecules, we refer to the two stranded form as duplex DNA. Figure 3 illustrates how the two strands of duplex DNA are twisted together in the form of a right-handed double helix.

Figure 3. The two strands of DNA are twisted together in a right-handed double helix.

In Figure 3, the bonds between the bases are illustrated as vertical bars, so that it looks more like a ladder than in the close up picture in Figure 2. If you count the number of vertical bars in one full turn of the helix, you will ﬁnd that there are about 10.5. In general, duplex DNA is said to be in a relaxed state if the double helix makes one complete right-hand turn for every 10.4–10.5 base pairs. This means that on average the strands of the duplex DNA will make two complete turns for every 21 = 2 × 10.5 base pairs. If there are more than 10.5 base pairs per turn or fewer than 10.4 base pairs per turn, then the DNA molecule is said to be stressed. 332 14. THE TOPOLOGY AND GEOMETRY OF DNA

We call the two strands the backbones of the DNA molecule. This terminology may be confusing because you would naturally expect that DNA should have only one backbone, just like a person or an animal or even a fish. Imagine you go out to dinner with some friends, and you order a whole fish to share. When it arrives steaming, smelling delicious, and looking at you with one of its eyes, you confidently offer to cut the flesh off the bones and serve it to your friends in nice neat portions. However, when you cut the fish open, you discover that there are actually two backbones that are twisted together in a double helix with the flesh stuck between them. Getting the flesh out without causing bits of fish to start flying around seems virtually impossible. So you say “just kidding” and pass the platter to someone else to cut up. There are two possible forms for a DNA molecule: closed up in a loop or open like a wiggly arc in space. DNA of the first type is said to be circular, while DNA of the second type is said to be linear. Note that circular DNA is not necessarily planar. In fact, it can be quite wiggly like some of the complicated unknots that we saw in Chapter 9. For circular DNA, the twisting of the double helix is trapped in the molecule unless one or both strands of the molecule are cut. While the DNA of organisms such as bacteria and viruses is normally circular, this is not the case for human DNA. You might think this means that duplex DNA easily becomes untwisted. However, even when DNA isn’t circular, the ends are usually fixed to a membrane so that the double helix cannot untwist. Thus, to make it easier to study the topology of DNA, we imagine that all DNA is circular even though technically this is not always the case.

3. The Problem of Packing DNA into a Cell

If the above job description for DNA did not sound challenging enough, there are further difficulties which you only realize when you look at the details of the job more closely. The problem is that the length of each DNA molecule ranges from 1,000 to 500,000 times the diameter of the cell which contains it. In order to help us visualize the problem of packing this much DNA into a cell, let’s imagine scaling the cell up so that it’s the size of a basketball. Then human duplex DNA would be a 120 mile long thread made from two super thin threads (the backbones) twisted together. If this much thread were randomly stuffed into a basketball, it would likely result in a tangled mess that would be impossible to undo. Think about how tangled up your headphones get when you stuff them into your pocket, and then imagine how hard it would be to untangle them if they were 120 miles long and were inside of a basketball. One of the most important aspects of the job of DNA is to enable each cell to split into two identical cells. This is necessary for any organism to grow and create offspring. When cells divide, the DNA molecules within the cells split into two identical DNA molecules through a process known as replication.Thisinvolves the two strands of the double helix separating and each creating a new strand of DNA which binds to it. Since the sequence of bases in one backbone determines the sequence of bases in the other backbone, one strand is all that’s needed to create a duplex DNA molecule which is a copy of the original. It’s impressive that DNA has the capability to duplicate itself in this way, since no other molecules are known 4. SUPERCOILING 333 to have this capability. But in order for DNA to be able to replicate, it can’t be tangled up randomly in the cell like 120 miles of thread stuffed into a basketball. Another important aspect of the job of DNA is creating genetic diversity. This means it’s not enough for the DNA to just copy the genetic code of one of the parents to create the genetic code of an offspring. Rather each time an offspring is created, the genetic codes of the parents are combined in a new way. Have you ever wondered how you can be so different from your siblings when you have the same parents? How is it possible for you to have light brown hair when your mother has black hair and your father has strawberry blond hair? Even if you imagine the genes for hair color are like paint, if you mix black and yellow paint, you are likely to get an ugly shade of green rather than a normal light brown hair color. Why can’t a mother with black hair and a father with orange hair produce a child with black and orange striped hair? Cats seem to be able to do this, so why can’t we? A key component to answering all of these questions is that during reproduction, segments of DNA are rearranged to create new genetic combinations in a process known as recombination. One way this works is that distant sections of DNA with identical codes are brought together and the endpoints are cut and interchanged. But how can two identical sections of DNA end up near one another if the DNA is randomly packed into a cell like a 120 mile long thread stuffed into a basketball? Furthermore, this cutting and reattaching has to be done without causing the DNA to become knotted or tangled so badly that it makes it impossible for it to do the other aspects of its job. You may recall from our discussion of the unknotting number in Chapter 10 that we wanted to change a crossing in order to create a new knot whose minimal crossing number was less than that of the original. But there was no good way to figure out what projection to use and which crossing to change. In the case of recombination, somehow the DNA is able to know how to cut and reattach to make the tangling better rather than worse. This means DNA has to be able to think globally but act locally. But how is it able to do this when it’s packed so tightly into a cell?

4. Supercoiling

As we saw in the last section, DNA needs a good way to organize itself so that it can be packed into a cell without uncontrolled tangling occurring. To understand how DNA solves this problem, let’s consider a similar problem that yarn manufacturing companies face. If yarn were sold as a tangled mess stuﬀed inside of a basketball, nobody would buy it. Instead, yarn is wound up into forms known as cakes, skeins,andhanks. Though must people call all three forms a skein. In fact, a cake is what you would get if you wound yarn around a stick, a skein is an oblong shape where the yarn is pulled out from the center, and a hank is illustrated in Figure 4. To make a hank, you wind the yarn around two pegs or two arms to create a loose coil. Then you twist the two sides of the coil tightly together. Finally, you fold the twisted up yarn in half and again twist the two sides together. This gives the yarn a nice compact form which holds its shape and doesn’t tangle. In order to ﬁt neatly into a cell, DNA needs to have a compact shape which holds together and doesn’t tangle, just like yarn. Duplex DNA is naturally wound up in a double helix, as illustrated for circular DNA on the left side of Figure 5. If 334 14. THE TOPOLOGY AND GEOMETRY OF DNA

Figure 4. These are the steps to winding yarn into a hank. two parts of the double helix come together and twist, as illustrated on the right side of Figure 4, the DNA will become more compact, which will allow it to ﬁt more easily into a cell without tangling. We refer to twists like those on the right side of Figure 5 as supercoils.

Figure 5. When duplex DNA twists up like this it becomes compact and doesn’t tangle, just like a hank of yarn.

In order to see for yourself how supercoiling occurs, you can do the following experiment with two pieces of rope and a friend. Grab the end of one rope in each hand and tell your friend to hold the other two ends together in one hand. Then begin twisting up your ends so that what you get looks like a double helix as illustrated on the left of Figure 6. Now keep twisting the two ropes together tighter and tighter. Eventually, it will be harder to keep twisting because you are creating stress on the ropes. If you keep twisting anyway, two pieces of the double helix rope will come together and start to supercoil, as illustrated on the right side of Figure 6. This supercoiling relieves some of the stress on the twisted ropes. The more you twist up the two ropes, the more supercoiled it will become. You and your friend can forcibly remove the supercoils by moving away from each other to pull the ropes tight. But as soon as you come back together so that the ropes are no longer taut, the supercoils return. For this reason, we say that the supercoiling of the two ropes is stable. Next, imagine making a loop out of a pair of ropes which are loosely twisted up in a double helix but do not supercoil even when the ropes are not taut. You could then turn part of the double helix over to add a supercoil as illustrated on the right of Figure 7. But this supercoil will go away if you lift the ropes up. Hence, in this case, we say that the supercoiling of the two ropes is unstable. Supercoiling is certainly one of DNA’s most brilliant ideas. Maybe it should even win a Nobel Prize for this idea. Supercoiling keeps the strands neat and 5. VISUALIZING DNA AS A CIRCULAR RIBBON 335

Figure 6. If you keep twisting ropes together, they will eventually become supercoiled.

Figure 7. We can create unstable supercoiling by turning over part of the twisted rope. organized, and packed eﬃciently in a cell. It allows distant parts of the DNA to be close together to facilitate recombination. It also facilitates the unwinding of the two strands to allow replication to take place. In the next six sections, we’ll learn about the geometry and topology of duplex DNA as a result of supercoiling.

5. Visualizing DNA as a Circular Ribbon

In order to best describe the geometry and topology of DNA, we visualize a duplex DNA molecule as a twisted ribbon whose edges are the two backbones. We illustrate a short piece of such a ribbon in Figure 8. If you imagine duplex DNA this way, then the center of the ribbon is what is known as the axis of the DNA. Remember from Section 2 that we deﬁne duplex DNA to be in a relaxed state if the double helix makes one complete right-hand turn for every 10.4–10.5 base pairs. When an unknotted DNA molecule is in such a state, its axis will be roughly planar, just like the rope we saw on the left side of Figure 7. However, if a knotted DNA molecule is in a relaxed state, its axis still cannot be planar since a knot cannot lie in a plane. For circular DNA, you may wonder if the two ends of the ribbon can be joined so that we get a M¨obius strip or some other non-orientable surface. To answer this question, let’s remember that the backbones of a DNA molecule consist of alternating sugar and phosphate molecules, with one base attached to each sugar. When sugars and phosphates bond together, each sugar has a special site on one end denoted by a 3 and a special site on the other end denoted by a 5.To visualize this, imagine that each sugar molecule is a white extension cord, with a 3 designating the socket on one end and a 5 designating the plug on the other end, as illustrated in Figure 9. 336 14. THE TOPOLOGY AND GEOMETRY OF DNA

Figure 8. We visualize DNA as a ribbon that twists around its central axis.

Figure 9. We think of a sugar molecule as an extension cord with a3 designating the socket and a 5 designating the plug.

Next imagine that each phosphate molecule is a dark grey extension cord with a3 designating its socket and a 5 designating its plug. The 5 site on a phosphate molecule bonds to the 3 site on a sugar molecule, and the 3 site on a phosphate molecule bonds to the 5 site on a sugar molecule. Thus the backbone of a DNA molecule can be pictured as a long chain of alternating white and grey extension cords, where the plug of one cord ﬁts neatly into the socket of the next cord, and so on as illustrated in Figure 10.

Figure 10. We can think of a backbone of a DNA molecule as a chain of alternating white and grey extension cords.

Observe that no matter how many extension cords we put together, one end of the chain will always be a socket and the other end of the chain will always be a plug. For DNA, this means that one end of each backbone will be a 3,andthe other end will be a 5. Furthermore, for a duplex DNA molecule, one backbone ends in 3 while the adjacent backbone ends in a 5 as illustrated in Figure 11. Now let’s return to our visualization of DNA as a twisted ribbon, but this time we label the ends of each edge of the ribbon with a 3 and a 5 as in Figure 12. As you can see in the ﬁgure, the 3 and the 5 on the black edge, and the 3 and the 5 on the grey edge are on opposite ends of the ribbon. 6. THE LINKING NUMBER OF THE BACKBONES 337

Figure 11. At each end of duplex DNA, one backbone ends in 3 while the other backbone ends 5.

Figure 12. The 3 and the 5 on the black edge, and the 3 and the 5 on the grey edge are on opposite ends of the ribbon.

In order to close the ribbon to make a loop, the 3 site at the end of one of the backbones has to bond with a 5 site at the other end of the ribbon. As we can see in Figure 12, the only way this can happen is if the ends of the grey strand join together and the ends of the black strand join together. So each backbone of a closed up duplex DNA molecule forms its own loop. This means that the DNA ribbon cannot be a M¨obius strip.

6. The Linking Number of the Backbones

We saw above that if we think of a duplex DNA molecule as a circular ribbon twisting around its axis, then each backbone forms its own loop. Furthermore, because of the twisting of the strands in the double helix, the two backbones will be the components of a non-trivial link. Hence we can think of a circular DNA molecule either in terms of the knot (or unknot) in the axis of the ribbon or in terms of the link between the two backbones. In Figure 13, the knotted axis of the DNA molecule is illustrated with a thick grey line and the linked backbones are illustrated with thin black lines. We arbitrarily give the axis of the DNA an orientation, and then orient each of the backbones so that they go in the same direction as the axis (see Figure 14). Now that the backbones are oriented, we can keep track of their linking with the linking number that was introduced in Chapter 10. In particular, we make the following deﬁnition. 338 14. THE TOPOLOGY AND GEOMETRY OF DNA

Figure 13. We can think of this DNA molecule in terms of either the knotted axis or the link of the two backbones.

Figure 14. We put an orientation on the axis, and then orient each of the backbones in the same direction as the axis.

Definition. Consider an oriented projection of the backbones K1 and K2 of a DNA molecule. The linking number Lk is deﬁned by the formula ⎡⎛ ⎞ ⎛ ⎞⎤ total number of total number of 1 Lk = ⎣⎝ positive crossings ⎠ − ⎝ negative crossings ⎠⎦ . 2 between K1 and K2 between K1 and K2

The most important thing that we learned about the linking number in Chapter 10 is that it’s a topological invariant, which means that deforming the link will not change its linking number. For example, suppose that the axis of our DNA is an unknot like the one on the left in Figure 5. Putting some supercoils in the axis, as illustrated on the right, does not change the linking number between the two backbones. Changing the orientation of the axis also does not change the linking number, since it would change the orientation of both backbones. The linking number Lk keeps track of important topological information about a DNA molecule, which we will return to later after we introduce two other quantities which give us some important geometric information about the molecule. 7. THE AVERAGE WRITHE OF THE AXIS 339

7. The Average Writhe of the Axis

Recall the following definition from Chapter 10. Definition. The writhe w(L) of an oriented knot projection K is given by the formula total number of total number of w(K)= − . positive crossings negative crossings This number does not depend on the orientation of the knot. However, as we saw in Chapter 10, the writhe is only an invariant for knot projections which are alternating and reduced. In particular, the writhe may change when we deform a knot, and it may even change when we consider a projection from a different angle. Hence the writhe is not a topological property of a knot. But we will see how to use the writhe to get a geometric property of DNA. Recall from Chapter 3 that an extrinsic property of an object in space is geometric if it doesn’t change when the object undergoes a rigid motion. We see as follows that the writhe is not a geometric property because it can change when a knot is rotated. Consider the unknot obtained by wrapping a rigid steel band around two perpendicular soda cans as illustrated in Figure 15. The projection that we see on the left side has no crossings and thus has writhe equal to 0. However, if we remove the cans and rotate the the steel band by 45◦, we’ll get the projection illustrated on the right side of Figure 15. This second projection has a single negative crossing regardless of how it’s oriented (you should check this), and thus has a writhe of −1. In order to convince yourself that the second projection can be obtained by rotating the first projection by 45◦, you could try letting your index fingers play the role of the cans and a rubber band wrapped around your fingers play the role of the steel band.

Figure 15. These two projections of a rigid loop have diﬀerent writhe.

Thus sadly the writhe is neither a topological nor a geometric property of an arbitrary knot, and hence might seem pretty useless. However, the following definition turns the writhe of a rigid loop into a useful geometric property. Definition. The average writhe of a rigid loop, denoted by Wr, is the average of the writhes over all possible projections of the loop. We want to emphasize that the average writhe is only defined for rigid loops. That is, we’re not considering the writhe over all projections of any deformation of a loop. Rather, we imagine that the loop is made of steel. We can turn it in different ways to get different projections, but we can’t deform it at all. The notion of taking the average of infinitely many different projections might seem a bit strange. However, if you have taken calculus, you might have seen this kind of average as an application of integration. But don’t worry if you haven’t 340 14. THE TOPOLOGY AND GEOMETRY OF DNA taken calculus. Since Wr is an average over all possible projections, the quantity Wr is unchanged when a loop is rigidly rotated. This means that Wr is indeed a geometric property of rigid loops. Unfortunately, for most loops, Wr can only be computed with the aid of a computer. However, to get a feel for it, we can compute the value of Wr for some very simple examples by hand. To begin with let’s imagine a planar circle with no crossings. To compute its average writhe we are supposed to consider the writhes of all possible projections. But if we consider a plane perpendicular to the plane containing the circle cutting the circle in half, one semi-circle will be on top of the plane and the other semi-circle will be below the plane. Thus the projection of the two semi-circles will be on top of each other so that the projection looks like a line segment. In Figure 16, both semi-circles project to the grey line segment in the plane. This type of projection is not allowed because it doesn’t have crossings that we can count. You can check that any other projection of our planar circle will have zero crossings. Thus we can conclude that the average writhe is Wr =0.

Figure 16. Both semi-circles project to the grey line segment.

Next let’s determine the average writhe for the steel band illustrated in Fig- ure 15 (again a rubber band around your index ﬁngers may be helpful). Observe that for some projections, the loop will have no crossings at all, while for others it will have a single negative crossing. In fact, these are the only possibilities for legitimate projections, that is where one arc does not lie directly on top of another. Thus, for every projection of the steel band, the writhe is either 0 or −1. This tells us that the average writhe for the loop is somewhere between −1 and 0. Though we can’t assign it a precise value, we can see from this example that in general, the average writhe of a loop is not an integer, even though for any given projection the writhe is an integer. This makes sense since the average of a collection of integers is not always an integer. For example, the average of the numbers 1 and 2 is 1.5.

8. The Twist of a Backbone around the Axis

We have now seen that for any closed loop of duplex DNA, we can obtain information about the geometry of the molecule by computing the average writhe Wr of its central axis (though we may need the help of a computer to do so). Also, we can obtain topological information about the molecule by computing the linking number of the two backbones oriented in the direction of the axis. But we need to introduce one more quantity before we can completely describe the geometry and topology of circular duplex DNA. Definition. The twist of a circular duplex DNA molecule, denoted by Tw, is the total amount that one of the backbones of the DNA winds about the central axis. 8. THE TWIST OF A BACKBONE AROUND THE AXIS 341

At ﬁrst glance, the amount that one backbone winds around the axis seems like it should equal the linking number between the two backbones. To see why this is not always the case, let’s consider the ribbon illustrated in Figure 17. Observe that for this ribbon, neither backbone twists around the central axis, and so Tw =0. However, you should check that after orienting the two backbones in the same direction, they are indeed linked with Lk = −1.

Figure 17. For this ribbon, Lk = −1andTw =0.

The following visualization may help you understand the twist better. We start by drilling a hole in the center of a penny, and threading the oriented axis of the DNA through the hole from the tail side of the penny to the head side of the penny. You can picture the penny as a bead on a necklace where the chain is the DNA axis. Now imagine that the top of Lincoln’s head is touching one of the backbones of the DNA, as illustrated in Figure 18. We slide the penny up along the axis in the direction of the arrow, always keeping the top of Lincoln’s head against the backbone. In Figure 18 we see that as the penny moves up the axis, Lincoln will have to rotate counterclockwise so that the top of his head remains on the backbone.

Figure 18. The twist measures the total rotation of Lincoln’s head as the penny moves along the axis.

The twist Tw will measure the total rotation of Lincoln’s head as the penny goes all the way around the axis until it returns to the place on the axis where it started. Each full (360◦) counterclockwise rotation contributes +1 to the twist, and each full clockwise rotation contributes −1 to the twist. Note that the penny may return to where it started on the axis without Lincoln’s head having the same 342 14. THE TOPOLOGY AND GEOMETRY OF DNA orientation it had at the beginning. If this is the case, the twist will not be an integer. We compute the twist for the simple ribbons illustrated in Figure 19. In the two illustrations on the left, the axis of the ribbon is in a plane and the backbones don’t wind around the axis at all. So in these cases, Tw = 0. In the third illustration from the left, the axis is again in a plane, but each backbone winds negatively around the axis. Thus in this case, Tw = −1. Finally, in the example on the right, the axis is not in a plane, but even so the backbones don’t wind around it. So again we have Tw =0.

Figure 19. Some ribbons and their twists.

What is interesting to note about these examples is that the two ribbons illustrated on the left of Figure 19 are topologically equivalent, and the two ribbons illustrated on the right are topologically equivalent. However, it is easier to find a deformation between the left two ribbons than between the right two ribbons. You can do the following experiment with a belt to convince yourself that in fact the right two ribbons are topologically equivalent. When your belt is not around your pants (or your waist), put a negative twist in it and then buckle it so that it looks like the picture that is third from the left in Figure 19. Putting this twist in the belt creates stress. When you let go of the belt, this stress will cause the belt to deform itself so that it looks like the picture on the far right side of Figure 19. In spite of the simple examples illustrated in Figure 19, in general, the twist is not an integer, and (like the average writhe) it is not easy to calculate by hand. For example, consider the ribbon illustrated in Figure 20. The axis is represented by the dotted black loop and one of the backbones is represented by the solid black loop. The line segments between the axis and black backbone are there to help you visualize how the black backbone is twisting around the axis. If you look at the figure carefully, you will see that the twist of the ribbon is somewhere between 0 and 1, though we can’t determine the precise number just by looking at it. Since the two pictures on the right in Figure 19 are topologically equivalent but have different values of Tw, we know that the twist of a ribbon is not a topological

Figure 20. The twist of this ribbon is between 0 and 1. 9. THE RELATIONSHIP BETWEEN LINKING, TWISTING, AND WRITHING 343 property. It turns out that, like the average writhe, the twist is actually a geometric property of a ribbon.

9. The Relationship between Linking, Twisting, and Writhing

Looking back at what we’ve done in the last three sections, we have deﬁned three quantities that each tell us something about the geometry or the topology of circular duplex DNA. The linking number Lk of the two backbones is a topological property, whereas both the average writhe of the axis Wr and the twist Tw between a backbone and the axis are geometric properties. As a result, any deformation of the molecule will leave Lk ﬁxed, but may change Tw and Wr. Surprisingly, it turns out that these three quantities are related by the equation: Lk = Tw+ Wr.

We won’t prove this equation is true. After taking a course on differential geometry, you could read a proof that this equation holds. Rather, we will focus on computing some examples. In particular, in Figure 21 we find the values of Lk and Wr for the ribbons whose Tw we found in Figure 19. In the first and second illustrations on the left, the axis is a planar circle, and so the average writhe is Wr = 0. Also, since the backbones don’t twist around each other or around the axis, we have Lk = Tw = 0. In the third picture, the axis is again a planar circle, but in this case the backbones twist around each other and around the axis. Thus we have Wr =0andLk = Tw = −1. Finally, we observe that the ribbon on the right is nearly flat. It comes out of the page ever so slightly so that one part of the ribbon can pass over the other. This means that in almost every direction the writhe is −1. We might imagine that the writhe is zero when we project in a direction perpendicular to the plane of the page. However, in that direction, pieces of the projection lie on top of one another as they did for the planar circle in Figure 16. Hence this projection is not allowed. In fact, there are almost no projections where the writhe is zero. So averaging over all projections, we see that Wr = −1. Also, we already know from Figure 17 that Lk = −1andTw = 0. Thus for each of the ribbons in Figure 21, the given values satisfy the equation Lk = Tw+ Wr.

Figure 21. Lk, Tw,andWr for the ribbons in Figure 19.

For our next example, let’s consider a circular unknotted duplex DNA molecule with 315 base pairs which is in a relaxed state. Recall, that because it’s unknotted and in a relaxed state, the backbones make one complete right-hand turn for every 10.4–10.5 base pairs and the axis is roughly planar. Thus we can divide the total number of base pairs by 10.5 to determine that the linking number between the 344 14. THE TOPOLOGY AND GEOMETRY OF DNA two backbones is Lk = 315/10.5 = 30. Also, since the axis is roughly planar, we can assume that Wr = 0. Hence it follows from the equation Lk = Tw+ Wr that Tw = 30. Now suppose that we want to twist up the axis of our molecule so that Wr = −3, as illustrated on the right side of Figure 22, and we want our conﬁguration to be stable. If a supercoil is added to the axis of our duplex DNA molecule without increasing the linking between the two backbones, then the supercoil will be unstable, like the supercoil that we added to our circular ropes in Figure 7.

Figure 22. We want to twist up the axis so that Wr = −3.

In order to create stable supercoiling as we did in Figure 6, we have to increase the linking between the two backbones to put enough stress on the DNA that it will be forced to supercoil. To do this, we cut one backbone, twist it around the other backbone, and then re-attach it as in Figure 23.

Figure 23. To increase Lk we cut one backbone, twist it around the other backbone, and then re-attach the ends.

Let’s say that we perform the operation illustrated in Figure 23 to add +1 to Lk. Then one or both of Tw and Wr must increase in order to add +1 to the right side of the equation Lk = Tw+ Wr. Remember that even though Lk is always an integer, the values of Tw and Wr are not necessarily integers. It turns out that if Lk is increased by +1, the most stable configuration of DNA occurs when the value of Wr increases by 0.75 and the value of Tw increases by 0.25. Since 3 × .25 = .75, this means that for each full turn added between the two backbones, Wr increases by 3 times as much as Tw increases. Similarly, for each full turn removed from the two backbones, Wr decreasesby3timesasmuchasTw decreases. Now let’s return to our example of a circular unknotted duplex DNA molecule with 315 base pairs, which we want to have a stable configuration with Wr = −3. Since we’ve decreased Wr by 3, we must also decrease Tw by 1. However, by the equation Tw + Wr = Lk, this means that the linking number between the two backbones must be decreased by (−1) + (−3) = −4. Hence, the linking number between the backbones in the new configuration will be Lk =30− 4 = 26. 10. REPRESENTING THE SUPERCOILING WITH A SINGLE NUMBER 345

10. Representing the Supercoiling with a Single Number

We would like to use the linking number of the two backbones as a way to measure how supercoiled a DNA molecule is. However, just knowing what the linking number is doesn’t give us enough information. Analogously, if all you know about a dog is its weight, you couldn’t reach any conclusions about its shape. For example, suppose you know that a full grown dog weighs 20 pounds. This would be an enormous amount for a Yorkshire Terrier but very little for a German Shepherd. If you knew the size of the dog in addition to its weight, you would be able to have a sense of whether it was underweight or overweight. Similarly, the linking number only gives us meaningful information about supercoiling if we put it in the context of how long the molecule is, which we measure by its number of base pairs. Thus we want to compare the linking number of an arbitrary duplex DNA molecule to the linking number of a relaxed DNA molecule with the same number of base pairs, which we denote by Lkrelax. Since relaxed DNA has the form of a right-handed double helix and the two backbones have parallel orientations, all of the twists between the two backbones are positive. This means that Lkrelax > 0. Now the linking difference, Lk −Lkrelax is the number of full turns between the two backbones that need to be added or removed in order to change an arbitrary DNA molecule with linking number Lk into a relaxed form. If we divide this difference by Lkrelax, we get the number of turns that need to be added or removed per full turn between the two backbones to change it into a relaxed form. Thus, the amount of supercoiling of an arbitrary DNA molecule can be represented by the value Lk − Lk relax . Lkrelax We say a molecule is negatively or positively supercoiled according to whether this number is negative or positive. Since this value is per full turn, we are able to compare the supercoiling of DNA molecules with different numbers of base pairs. The bigger the absolute value − Lk Lkrelax Lkrelax is, the more tightly supercoiled the molecule. As an example, let’s return to the unknotted DNA molecule that we considered at the end of the last section which had 315 base pairs and Lk = 26. Recall that in a relaxed state, the linking number of this molecule was 30. Thus, the supercoiling is represented by 26 − 30 ≈−0.133. 30 Now let’s compare the supercoiling of this molecule to that of an unknotted DNA molecule that has 630 base pairs and has Lk = 56. To find the supercoiling of our second DNA molecule, we first compute the linking number of a relaxed DNA molecule which has 630 base pairs as Lkrelax = 630/10.5 = 60. Thus we find that the supercoiling of our second molecule is given by 56 − 60 ≈−0.067. 60 We know both molecules are negatively supercoiled since −0.133 and −0.067 are both negative numbers. However, since |−0.133| > |−0.067|, we can conclude 346 14. THE TOPOLOGY AND GEOMETRY OF DNA that the molecule with 315 base pairs is more supercoiled than the molecule with 630 base pairs. To explain this, let’s observe that the difference in the linking number of the given molecule and a relaxed molecule of the same length is Lk − Lkrelax = −4 in both cases. However, the molecule with only 315 base pairs has to absorb this linking difference over a shorter length of DNA than the molecule with 630 base pairs. Thus the shorter molecule will encounter more torsional stress and, as a result, will have more supercoiling. We finish this section by introducing some terms which capture what we’ve learned. A relaxed DNA molecule has Lk − Lkrelax = 0; and if the molecule is unknotted, then we also have Wr = 0. If a DNA molecule has Lk < Lkrelax, then we say that the DNA is underwound because the winding between the two backbones is less than it would be if the molecule were in a relaxed state. This means that the molecule has more than 10.5 base pairs per full turn between the two backbones. To relieve the stress that this creates, the value of Wr decreases. If the molecule is unknotted, this means that Wr becomes negative, in which case we say the molecule is negatively supercoiled. On the other hand, if Lk > Lkrelax, then we say that the DNA molecule is overwound because now there are fewer than 10.5 base pairs per full turn between the two backbones. To relieve the stress that this creates, the value of Wr increases. If the molecule is unknotted, this means that Wr becomes positive, and we say the molecule is positively supercoiled. In fact, in the cells of almost all organisms the DNA is negatively supercoiled, and it is this supercoiling that allows it to fit compactly and efficiently into cells. Nonetheless, in general, the values of Lk and Tw are positive, as they are for relaxed DNA molecules.

11. Replication

Recall that in order to duplicate itself for reproduction and cell growth, a duplex DNA molecule has to split into two identical DNA molecules through a process known as replication. The ﬁrst step in this process is that one after another the bonds between the pairs of bases on the two backbones break apart. If we think of these bonds as the teeth of a zipper, breaking these bonds is like unzipping the zipper. Since the linking number between the backbones doesn’t change while they are being pulled apart, the twists between them are all being pushed into the part of the DNA that is still zipped up. This means there are fewer and fewer base pairs for the same number of twists that there were before replication began (see the middle picture in Figure 24). Observe that the fewer base pairs that a DNA molecule has, the smaller Lkrelax will be. We remarked at the end of the last section that DNA in a cell is negatively supercoiled as a result of having more than 10.5 base pairs between each full turn of the backbone. This means that the twisting between the two backbones of DNA in a cell is more spread out than it would be for DNA of the same length in a relaxed state. It follows that before replication, we have Lk < Lkrelax. Once replication begins and the backbones start to unzip, the twists between them are forced into the zipped up part, which has fewer and fewer base pairs, as illustrated in Figure 24. Let Lkrelax represent the linking number of a relaxed DNA 11. REPLICATION 347

Figure 24. Unzipping the backbones pushes the twisting into a shorter segment, which leads to positive supercoiling. molecule which has the same number of base pairs as the zipped up part. Then Lkrelax will get smaller and smaller as the DNA unzips. Since Lk is unchanged throughout the unzipping process, Lkrelax eventually becomes smaller than Lk.As we saw in the last section, when Lk > Lkrelax, DNA is positively supercoiled in order to relieve the stress created by having fewer than 10.5 base pairs per full twist of the backbones. Thus the shorter and shorter the zipped up part becomes, the more it becomes positively supercoiled, as illustrated on the right of Figure 24. Eventually, when there are too many positive supercoils to fit in the remaining zipped up part, the unzipping process is forced to stop. This would put an end to replication and the future of life as we know it, if it weren’t for the work of a special enzyme that comes to the rescue in the nick of time. But first we pause to bring you a word from our sponsors—the topoisomerase enzymes. In general, enzymes are large protein molecules that facilitate important molecular changes in living organisms. There are many types of enzymes, all of whose names end with the suffix ase.Thetopoisomerases are a family of enzymes which maintain the intrinsic structure of DNA molecules while changing their extrinsic topology, which is why they have the prefix topo. There are two types of topoisomerases: Type I which cuts one backbone of the DNA to remove twists between the two backbones and then reconnects the backbone to itself; and Type II which cuts both backbones to change the topology of the axis and then reconnects each backbone to itself. Topoisomerase I comes to the rescue, solving the problem of excessive supercoiling during replication by cutting one backbone, allowing it to untwist around the other backbone, and then regluing the two ends as illustrated in Figure 25. In this way, positive supercoiling is reduced, allowing the unzipping process and thus replication to continue. By regulating supercoiling during replication, Type I topoisomerases enable cells to grow and reproduce. However, not all cell growth and reproduction is a good thing. The growth of cancerous cells in our bodies is one of the scariest things 348 14. THE TOPOLOGY AND GEOMETRY OF DNA

Figure 25. Topoisomerase I cuts one backbone, untwists it around the other backbone, then reglues the ends. that can happen to us. Topoisomerase I inhibitors have been used as cancer ﬁght- ing drugs because they prevent cancer cells from replicating. Unfortunately, such medications can also prevent certain healthy cells from replicating. In particular, topoisomerase I inhibitors interfere with the growth of hair cells, which is why chemotherapy often causes hair loss.

12. Site Speciﬁc Recombination

While replication plays a crucial role in reproduction, it is not unique in its importance. DNA is also involved in the equally important process of recombination. In fact, recombination is what enables genetic diversity and allows species to evolve over time. There are several types of recombination which you might learn about in a course on genetics. In this section, we discuss a particular type of recombination known as site specific recombination in which two small pieces of duplex DNA are brought together, cut, and then reattached in a different way. We focus on this type of recombination because it can change the extrinsic topology of DNA, and can even create knotted or linked molecules. In our study of site specific recombination, we will focus exclusively on topology and ignore the geometry of the molecules. Note that in contrast with the earlier part of this chapter, we are now thinking of a duplex DNA molecule as a single string, ignoring the fact that it actually consists of two backbones which twist together to form a double helix. If you find it hard to forget about the the two strands, keep in mind that you are used to thinking about a rope as a single string, even though if you look closely you can see that it too consists of multiple strands wound together. We begin with a little vocabulary to help us talk about the steps of site specific recombination. Prior to the start of recombination, a duplex DNA molecule is called the substrate. The first step of recombination is that two identical short pieces of genetic code known as sites come together, most often as a result of supercoiling together with random motion inside of the cell. When they’re close enough together, a recombinase enzyme binds to both of the sites at once. At this point we can visualize the enzyme as a small ball with the two sites inside it. The DNA molecule together with this ball is called the synaptic complex.Nextthe recombinase cuts open the two sites, moves the endpoints of the cut open sites 13. KNOTTED AND LINKED PRODUCTS OF RECOMBINATION 349

Figure 26. The steps of site specific recombination. around within the ball, and then reattaches them, possibly in a different way. The resulting DNA molecule is said to be the product of the recombination. We illustrate these steps in Figure 26, where we’ve drawn the substrate as an unknot, and the two sites as small black rectangles. In the first picture, the sites are far apart. But as a result of supercoiling and random motion, the sites are brought together so that the enzyme can bind to both of them at once to obtain the synaptic complex in the third picture. In the fourth picture the recombinase has cut the ends of the two sites. Then in the final picture the ends are interchanged and glued to the new strands to obtain the product. Observe that in regluing the ends of the sites, a new crossing has been introduced. In this particular example, the product of recombination is knotted though the substrate was not. Thus the recombinase has indeed changed the topology of the molecule.

13. Knotted and Linked Products of Recombination

What makes site specific recombination tricky to study is that molecular biologists can observe the substrate and the products, but they can’t observe what is happening during the actual process of recombination. In particular, they don’t know what the synaptic complex looks like, and they can’t tell whether there are twists that are caught inside or outside of the enzyme ball. For example, in Fig- ure 26 we can see that there are three twists that are outside of the enzyme ball that end up as crossings in the knotted product. Imagine that you’re a scientist and all you can see is the first and last picture of Figure 26. It would be hard to figure out exactly how the enzyme managed to change the unknotted substrate into the particular knotted product. Sometimes scientists get lucky and can find clues that help them piece together what the process might have been. For example, suppose a recombinase acts on an unknotted substrate to obtain a knotted product that scientists can identify, then the recombinase acts on that knotted product to obtain a link that scientists can again identify. Now scientists have two pieces of information rather than just one. If the recombinase acts a third time, the scientist can get even more information. It’s like a detective story—the more times the bad guy commits his crime, the easier it is for the detective to figure out the pattern. However, in the case of knotted and linked products, scientists want to know the order in which the knots and links are produced so that they can guess how one could lead to another. But they can’t watch the process or stop it and restart it. 350 14. THE TOPOLOGY AND GEOMETRY OF DNA

All their information has to be obtained from the collection of knotted and linked products they have at the end of the process. The trick to figuring out the order in which the knots and links were produced is by determining the quantities of each. The largest number of knotted or linked products are the result of recombination on the original unknotted substrate. The second largest number of knotted or linked products are the result of recombination on the first products, and so on. Knowing what knots and links are produced and in what order gives the scientists a major clue as to what the enzyme did to create them. But the mystery is still not solved. In order to understand the difficulty of this problem, let’s consider an experiment conducted by the molecular biologists Wasserman, Dungan, and Cozzerelli in 1985. They observed that the enzyme Tn3 resolvase acted on an unknotted circular substrate, and the product of one recombination was the Hopf link. The second recombination produced the figure eight knot, and the third resulted in the Whitehead link. There were also some 62 knots that resulted from the fourth recombination. They could not detect any products beyond these. These knotted and linked products are illustrated in the order they were produced in Figure 27.

Figure 27. Consecutive products of the enzyme Tn3 resolvase.

You can see from Figure 27 that knowing the knotted and linked products of recombination and the order in which they were produced does not reveal the process. You might want to pause here to see if you can ﬁgure out how the enzyme did it. (Hint: It was not in the library with the candlestick.) Before we can present the solution to this mystery, we need to take a detour in the next three sections to learn about tangles. As you will see, the study of tangles is closely related to the study of knots and links. So the next few sections should bring back fond memories of Chapters 9, 10, and 11.

14. An Introduction to Tangles

While most people use the word “tangle” to refer to a string in a messy jumbled heap, mathematicians use the term in a more precise way. In particular, we deﬁne a tangle to be a ball containing two strings with endpoints along the equator at the compass points NE, NW, SW, and SE. In Figure 28, we illustrate a tangle where the endpoints of one string are at the NE and SE points, and the endpoints of the other string are at the NW and SW points. We say that two tangles are equivalent if we can deform the strings of one to the strings of the other while keeping the four endpoints of the strings ﬁxed at the compass points. The simplest tangles are those with two vertical strings or two horizontal strings with no crossings (see Figure 29). The tangle with two horizontal strings is known as the 0-tangle because horizontal lines have slope 0, and the tangle with two vertical 14. AN INTRODUCTION TO TANGLES 351

Figure 28. An example of a tangle.

Figure 29. The 0-tangle and the ∞-tangle. strings is known as the ∞-tangle because vertical lines have slope ∞.Notethat the strings in these tangles do not have to be straight lines. We can create new inequivalent tangles from the 0-tangle and the ∞-tangle by moving the endpoints of the strings around within the boundary of the ball. For example, let’s start with the 0-tangle and twist the NE and SE endpoints around each other within the boundary of the ball to get the new tangle illustrated on the right in Figure 30. Note that the tangles we create in this way will not in general be equivalent to the 0- or ∞-tangle, because a pair of tangles is only equivalent if we can get from one to the other by a deformation keeping the endpoints of the strings fixed. We can get infinitely many different tangles by twisting the endpoints of the strings of the 0- and ∞-tangles in the boundary of the ball in different ways. The exact tangle that we get will depend on how we twist. To keep track of our twisting, we designate a twist as positive if the overcrossing has positive slope, and we designate a twist as negative if the overcrossing has negative slope. If we twist the NE and SE endpoints n times around each other so that we get n crossings and the slope of each overcrossing is positive we label our tangle (+n)orsimply(n). If we twist the NE and SE endpoints n times around each other and the slope of each overcrossing is negative, we label our tangle (−n). We refer to tangles obtained in this way as integer tangles. We illustrate some examples in Figure 31.

Figure 30. If we twist the NE and SE endpoints around each other, we get a new tangle. 352 14. THE TOPOLOGY AND GEOMETRY OF DNA

Figure 31. Some integer tangles.

15. Rational Tangles

We can form more complicated tangles by starting with a 0-tangle or an ∞- tangle and alternating between twisting the NE and SE endpoints of the strings around each other some number of times and twisting the SW and SE endpoints around each other some number of times. When we twist the NE and SE endpoints around each other we say that we are adding horizontal twists, and when we twist the SW and SE endpoints around each other we say that we are adding vertical twists. If we start with the 0-tangle, we must begin by twisting the NE and SE endpoints; and if we start with the ∞-tangle we must begin by twisting the SW and SE endpoints. Notice that we never twist the NE and NW endpoints or the NW and SW endpoints around each other. For example, the tangle on the far right in Figure 32 was obtained by starting with the 0-tangle and twisting the NE and SE endpoints horizontally around each other three times in the positive direction, followed by twisting the SW and SE endpoints vertically around each other twice in the negative direction, and then twisting the NE and SE endpoints horizontally around each other once in the positive direction. In order to keep track of the twisting and the order in which we did it, we denote this tangle by the tangle vector (3, −2, 1). If you haven’t studied vectors before, don’t worry. A vector is just a list of numbers in a particular order.

Figure 32. The steps to creating a (3, −2, 1)-tangle from a 0-tangle.

In order to make it clear from a tangle vector which twists are horizontal and which twists are vertical, we make a rule that the last entry in any tangle vector must correspond to some number of horizontal twists, even if that number is 0. For example, in the third step of Figure 32, there is a 0 in the last entry of the tangle vector because the last thing we did was add vertical twists. Because of this rule, whenever we start with the 0-tangle, the tangle vector we get will have an odd number of entries. You can see in Figure 32 that all of the vectors have an odd number of entries. 15. RATIONAL TANGLES 353

If we start with the ∞-tangle, we begin by twisting the SW and SE endpoints vertically around each other. Since the last entry of the tangle vector still has to be some number of horizontal twists, in this case the vector will always have an even number of entries. Thus looking at whether a tangle vector has an even number or an odd number of entries tells us which of the two tangles we started with. For example, in Figure 33 we construct a tangle with the same number of twists as we did in Figure 32, but this time we start with the ∞-tangle instead of the 0-tangle. Now we begin by twisting the SW and SE endpoints vertically around each other three times in the positive direction, then we twist the NE and SE endpoints horizontally around each other twice in the negative direction, and ﬁnally we twist the SW and SE endpoints horizontally around each other once in the positive direction. We illustrate our construction of this tangle in Figure 33. We can see that the vectors representing the steps in this construction are diﬀerent from those in Figure 32. In particular, now they all have an even number of entries.

Figure 33. The steps to creating a (3, −2, 1, 0)-tangle from the ∞-tangle.

Any tangle that we can obtain in this manner starting with the 0 or ∞-tangle is called a rational tangle and is represented by a vector of the form (a1,a2,...,an), where each entry corresponds to some number of twists of the NE and SE or SW and SE endpoints and the last entry represents some number of horizontal twists. Note that an is the only term in the tangle vector that is allowed to be zero. We call tangles constructed in this way “rational” because to each such tangle β we can associate a rational number α by computing the so-called continued fraction of the vector (a1,a2,...,an) using the following rule. β 1 = a − α n 1 a − − n 1 1 ···− 1 a2 − a1 To get a feel for what is going on, we compute the continued fraction corresponding to the (3, −2, 1)-tangle in Figure 32 as follows. 10 1 =1− 7 1 −2 − 3 John H. Conway showed that two rational tangles are equivalent if and only if their continued fractions are equal. This means that to determine if two rational tangles are the same or diﬀerent, we only need to compute these continued fractions 354 14. THE TOPOLOGY AND GEOMETRY OF DNA to see if they are equal. This is a very powerful result that makes it easy to distinguish rational tangles. Rational tangles are incredibly nice. They can be coded up by vectors, and distinguished easily by computing their continued fractions. But not all tangles are rational. For example, in Figure 34 we illustrate a non-rational tangle. You can tell that this tangle isn’t rational because no matter how you move the endpoints around on the boundary of the tangle ball, you can’t undo the tangle.

Figure 34. A non-rational tangle.

16. Operations on Tangles

Just as we have the connected sum operation that enables us to add two knots together to get a more complicated knot, we have an addition operation for tangles that we can use to combine two tangles to get a more complicated tangle. In particular, we add tangles A and B together by connecting the NE and SE endpoints of A to the NW and SW endpoints of B and extending the remaining endpoints to the boundary of a larger ball, as illustrated in Figure 35. The result is called the sum of A and B and is denoted by A + B. Observe that even though tangles A and B in Figure 35 are rational, the tangle A + B is not rational.

Figure 35. In this example, the sum of rational tangles is not rational.

We can create a simpler example by adding together two ∞-tangles. In this case, as we see in Figure 36, the sum has an extra component in the center, and hence is not rational. We can also start with a tangle A, and then form a knot or link by adding an arc to the top and the bottom of the tangle as illustrated in Figure 37. The result is called the numerator closure of A and is denoted by N(A). As we see in Figure 37, we can also combine these two operations to get N(A + B). With this background in the theory of rational tangles and tangle operations, we are now ready to use tangles to analyze how site speciﬁc recombination produces particular knots and links. 17. THE TANGLE MODEL OF SITE SPECIFIC RECOMBINATION 355

Figure 36. The sum of two ∞-tangles has an extra component, and hence is not rational.

Figure 37. The numerator closure of the tangles A and A + B.

17. The Tangle Model of Site Speciﬁc Recombination

The idea of a mathematical model is that you state a number of reasonable assumptions about a real situation, and then you use mathematics to derive conclusions. We would like to use tangles to model site specific recombination. The first assumption of our model is that a recombinase enzyme has the form of a ball, and that this ball together with the two sites inside of it (which we refer to as the enzyme ball) looks like a tangle. If you look back at the drawing of the synaptic complex in Figure 26, you can see that we drew the enzyme ball with the sites inside of it as a 0-tangle. In reality, we don’t know that the enzyme together with the sites looks the way we drew it. Maybe there are additional pieces of the substrate that go in and out of the enzyme and are tangled together with the two sites, or maybe the two sites themselves are twisted together. If we knew exactly what was going on inside the enzyme, we wouldn’t need a model. But thinking of the enzyme as a ball and the pieces of the substrate inside of the enzyme as the two strings of a tangle seems like a reasonable assumption. In our model, we will refer to the tangle inside of the enzyme ball prior to recombination as the site tangle and denote it by T . Next we model the twisted part of the substrate outside of the enzyme as another tangle that we refer to as the substrate tangle and denote it by S. Finally, we model the substrate as the numerator closure N(S + T ). Note that this means all of the crossings of the projection must be in one of the two tangles. We illustrate a substrate expressed as the numerator closure N(S + T ) on the left side of Figure 38. The next assumption of our model is that the action of the enzyme takes place entirely inside of the enzyme ball and only affects the pieces of DNA which are in the site tangle T . In particular, the part of the molecule outside of the enzyme ball remains fixed while the enzyme is acting on T . This means that the enzyme 356 14. THE TOPOLOGY AND GEOMETRY OF DNA

Figure 38. According to our model, the enzyme changes the substrate N(S + T ) into the product N(S + R). causes the tangle T to be replaced by a new tangle R that we will refer to as the recombination tangle. Hence the enzyme changes the numerator closure N(S + T ) representing the substrate into the numerator closure N(S + R)representingthe product as illustrated in Figure 38. For a given recombinase, if we know what knot or link the substrate and the product are, then we can set N(S + T ) equal to the knot or link of the substrate and N(S + R) equal to the knot or link of the product. This gives us two tangle equations with the tangles S, R,andT as the unknowns. Recall from when you took algebra that if you have two equations and three unknowns you won’t be able to solve to get specific values for the unknowns. The same is true for tangle equations. In particular, with two tangle equations and three unknown tangles we cannot solve the equations for the tangles S, R,andT . However, if there is a second or even a third knotted or linked product, then we can obtain more equations. But first we have to add one more assumption to our model, which is that if the enzyme acts multiple times, it always binds at the same place and always acts the same way. This means that each recombination event adds the same tangle R next to the previous one. Hence the first product of recombination is N(S + R), the second product of recombination is N(S + R + R), the third product is N(S + R + R + R) and so on. Using this assumption, Figure 39 illustrates the products we would get from Figure 38 after one and two recombination events.

Figure 39. The ﬁrst product is N(S +R) and the second product is N(S + R + R). 18. APPLYING THE MODEL TO THE ENZYME TN3 RESOLVASE 357

18. Applying the Model to the Enzyme Tn3 Resolvase

We now apply the tangle model to analyze the products of the enzyme Tn3 resolvase. In fact, the tangle model was originally developed by Claus Ernst and De Witt Sumners precisely to explain how Tn3 resolvase acted on an unknotted substrate to produce first the Hopf link, then the figure eight knot, then the Whitehead link, and finally the 62 knot, as illustrated in Figure 27. Based on this consecutive list of knotted and linked products, our model gives us the tangle equations: N(S + T ) = unknot, N(S + R) = Hopf link, N(S + R + R) = figure eight knot, N(S + R + R + R) = Whitehead link,

N(S + R + R + R + R)=62 knot. Ernst and Sumners then used advanced techniques from topology to solve these tangle equations for the tangles R, S,andT . Their solution, illustrated in Figure 40, shows the steps that the enzyme took to get from one knot or link to the next, and thus solves the mystery of Figure 27.

Figure 40. How the products of Tn3 resolvase occurred.

The following theorem expresses the same result in terms of tangle vectors. Theorem 14.1. Suppose that S, T ,andR are rational tangles that satisfy the following equations: (1) N(S + T )= unknot, (2) N(S + R)= Hopf link, (3) N(S + R + R)= ﬁgure eight knot, (4) N(S + R + R + R)= Whitehead link.

Then S =(3, 0), R =(−1),andN(S + R + R + R + R) is the 62 knot. 358 14. THE TOPOLOGY AND GEOMETRY OF DNA

YoucancheckinFigure40thatS is indeed a vertical tangle with three positive crossings and R is a single negative crossing. The tangle model together with this theorem shows that knowing the knots and links in the substrate together with the ﬁrst three products of Tn3 resolvase is enough to not only determine what the tangles S and R are, but to correctly predict that the fourth product of recombination will be the 62 knot.

19. Exercises

1. Explain why a closed duplex DNA molecule always contains an even number of half twists.

2. Find Lk, Wr,andTw for the ribbon in Figure 41, then, check that the equation Lk = Tw+ Wr holds.

Figure 41. Illustration for Exercise 2.

3. Draw a picture of a ribbon that has Lk =0andWr =2.

4. Suppose that a duplex DNA molecule is in a stable conﬁguration. If you add 3tothevalueofWr, how would you need to change Lk so that the new conﬁguration is also stable?

5. Consider two unknotted DNA molecules with stable conﬁgurations. The ﬁrst one has 630 base pairs and Wr = 6, and the second one has 210 base pairs and Wr = −3.

(a) For each molecule ﬁnd the value of Lk − Lk relax . Lkrelax (b) Determine which molecule is more tightly supercoiled.

6. Consider the ribbon illustrated in Figure 42 which is like the one illustrated on the right of Figure 21 but with two twists instead of one. The ribbon is almost entirely in the plane of the page, but comes out of the plane ever so slightly in two places so that one piece can cross over the other. The enzyme Topoisomerase II works by cutting both backbones of duplex DNA and reattaching them so as to make a crossing change in the central axis as illustrated in Figure 42. What eﬀect will this enzyme action have on the values of Lk, Wr, and Tw? 19. EXERCISES 359

Figure 42. Illustration for Exercise 6.

Figure 43. Illustration for Exercise 7.

7. Recall that in site speciﬁc recombination two identical short segments of DNA are brought together by the recombinase. These segments are strings of letters which must be lined up precisely in the enzyme ball. Rather than writing out the letters in each segment, we can represent them as arrows as illustrated in the middle pictures of Figure 43. Prior to recombination, the two arrows could either represent the same or opposite orientations on the circular substrate. If the orientation of the two arrows represents the same orientation on the circular substrate, then we say that the sites are direct repeats . If the arrows on the sites represent opposite orientations on the substrate, then we say that the sites are inverted repeats. These are illustrated in Figure 43. Explain how we know that if a recombinase acts once on a circular substrate with inverted repeats, then the product is a knot.

8. If a recombinase acts on a circular substrate an even number of times with direct repeats then the product must be a knot, and if it acts an odd number of times the product must be a link. Explain why this must be the case. 360 14. THE TOPOLOGY AND GEOMETRY OF DNA

9. Draw pictures of the rational tangles (−2, 3, 4) and (−2, 3, 4, 0). 10. Find the continued fractions for the tangles (−2, 3, 4) and (−2, 3, 4, 0). 11. Give examples of tangles A and B other than (0) and (∞) such that N(A)is the unknot and N(B)istheunlink. 12. Give examples of tangles A and B other than (0) and (∞) such that N(A + B) is the unknot. Note A and B do not have to be rational. 13. Figures 40 and 44 illustrate that the products of recombination with Tn3 resolvase can be written as numerator closures of rational tangles in two diﬀerent ways. Consider the continued fractions of the two diﬀerent tangles whose closure is the 62 knot, and determine whether the tangles are equivalent.

Figure 44. Illustration for Exercise 14.

14. Illustrate deformations taking the knots and links in Figure 44 to those in Figure 40. https://doi.org/10.1090//mbk/096/15

CHAPTER 15

The Topology of Proteins

Topics: • The structure of proteins • Which knots occur in proteins • Knots and links in proteins • Non-planar graphs in metalloproteins • Non-planar graphs in the protein nitrogenase • M¨obius ladders with three and four rungs in metalloproteins • M¨obius ladders in small proteins • Recommendations for further reading

1. An Introduction to Proteins

In Chapter 14, we learned that DNA is a long floppy molecule whose topology is just as important as its geometry. But DNA isn’t the only floppy biological molecule whose topology is important. Proteins are made up of long chains of amino acids, just as DNA is made up of long chains of sugars and phosphates. The chain of amino acids, known as the backbone of a protein, is not as long as the backbone of a DNA molecule, but it is long enough that it lacks a rigid structure and behaves like a floppy coil before folding up into a rigid functional shape. Scientists want to understand how protein folding occurs because many illnesses including Alzheimer’s disease, Parkinson’s disease, type 2 diabetes, and even many allergies are the result of proteins folding incorrectly. It is believed that the topology of a protein structure can affect its ability to fold properly. To shed light on the role topology might play, scientists are interested in studying proteins containing knots, links, and non-planar graphs. Such structures are not very common among all proteins, but they are by no means unheard of. In fact, knotted proteins have been found in a wide range of organisms from yeast to humans. Furthermore, when a protein loses or gains a knot or link through the process of evolution, it seems to cause a change in the behavior or function of the protein. For example, proteins containing knots and links have been observed to have increased stability relative to unknotted and unlinked proteins. Yet the precise biological function of knotting and linking in proteins is still unknown. In this chapter, we will learn about protein structures which cannot be deformed into a plane. In order to help us focus on the topology of such structures, we will treat them as if they are completely flexible, even though in reality proteins are only partially flexible. Nonetheless, if we can show that a completely flexible model

361 362 15. THE TOPOLOGY OF PROTEINS cannot be deformed to lie in a plane, then the actual protein, with somewhat less flexibility than the model, also cannot be deformed to lie in a plane. Thus, for the purpose of identifying non-planarity within protein structures, we can assume complete flexibility, even though this is not physically accurate. In order to study the topology of a particular protein, we first need to draw a picture of it. On left side of Figure 1, we see an illustration of a protein that you might find in a biochemistry paper or textbook. Unfortunately, this type of illustration, while relatively accurate, is so complicated that it’s a challenge to see which pieces of the protein pass behind and which pieces pass in front. Such information is crucial if we are looking for knots or links in the drawing. Thus we will deform our proteins to create much simpler drawings that omit many of the twists and turns, but make it easy to identify all of the crossings. We will refer to these simplified drawings as the underlying topological structure of the protein. For the protein in Figure 1, we can remove all of the crossings so that we obtain a squiggly segment in the plane as its underlying topological structure. The two ends of this segment are known as the termini of the protein. In order to distinguish one endpoint from the other, one is labeled as the C-terminus and the other is labeled as the N-terminus.

Figure 1. We simplify the biochemical diagram on the left to get the underlying topological structure on the right.

2. Trapping Knots in Proteins

Recall that in Chapter 9 we deﬁned a knot to be a circular path in R3.The requirement that the path be circular is so that the knot will be trapped within the circle, as the knot on the left side of Figure 2 is. If we don’t require this, a knot like the one on the right side of Figure 2 can be easily undone. You probably have ample experience with pulling your earbuds out of your pocket only to ﬁnd them completely tangled up. In spite of how annoying this can be, you know that with enough patience you can remove all the knots by systematically unthreading one of the ends back through any loops. While the backbones of protein structures are normally open segments like shoelaces rather than circular paths, if a knot is far away from the endpoints of the segment, then from a biochemical viewpoint, it is trapped within the structure. In 2. TRAPPING KNOTS IN PROTEINS 363

Figure 2. The knot on the left is trapped, whereas the knot on the right can be removed. particular, the energy required for the termini to unthread the knot is so large that this is unlikely to occur. So even though most proteins are open segments like the one on the right side of Figure 2, we want to create a model of them which prevents any knots from wriggling off the ends. Many approaches have been developed to creating such a model. On the left side of Figure 3 we start with a protein in the form of an open segment. Then we replace the termini by arrows indicating a straight ray that goes on indefinitely. These infinite rays prevent the knot in the middle figure from slipping off the structure. The problem with this approach is that if one or both of the termini are near the tangling of the knot, then different knots can be trapped in the structure depending on the direction in which the rays are extended. For example, in the right image of Figure 3, if we extend the bottom endpoint by a ray that goes under the arc, we will get a trefoil knot, but if we extend at a slightly different angle so that it goes over the arc, we will get the unknot.

Figure 3. The knot on the left is trapped in the structure, while the knot on the right may or may not be.

Another approach to modeling a knotted linear protein structure is to place it in the center of a large ball. The termini are then extended by straight segments going to the spherical boundary of the ball and are finally joined by an arc in the boundary to create a closed loop. In Figure 4, the extensions to the boundary are drawn as grey dashed lines, and the arc in the boundary is drawn as a grey curve. However, the same problem can occur that we saw on the right side of Figure 3 when we extended the termini indefinitely. In particular, different knots may be obtained depending on how the arcs are extended to the boundary of the ball. One way to resolve the problems with both approaches is to list all the knots that can be obtained by extending the ends in every possible direction, and then the 364 15. THE TOPOLOGY OF PROTEINS

Figure 4. We put the knotted arc in a ball, extend the endpoints to the boundary, and then join them by an arc. knot that occurs the most frequently is considered to be the one in the structure. To do this more formally, a probability is assigned to each knot and the knot with the highest probability is the one considered to be in the structure. When the termini of a protein are quite close to each other, adding a small arc between them gives the knot with the highest probability. For example, in Figure 5, we connect the termini together with a short arc to obtain the knot on the right. In this chapter, when we are looking at knots in open linear proteins, we require the termini to be close to each other so there is no ambiguity about which knot is formed.

Figure 5. Because the termini are close together, we can add a short arc joining them to obtain a knot.

3. Which Knots Occur in Proteins?

Whatever method is taken to trapping knots in proteins, amongst the tens of thousands of proteins identified as of 2015, only a few hundred have been recognized to contain knots. Furthermore, among the hundreds of knotted proteins, all contain one of the four knots 31,41,52,and61, illustrated in their circular forms in Figure 6. Note that Figure 6 includes both mirror forms of the trefoil knot, since both have been identified in proteins. Since the figure eight knot 41 can be deformed to its mirror image, only one form of it is listed in Figure 6. By contrast, the 52 and 61 knots are distinct from their mirror images, yet surprisingly these knots have only been found in proteins in the forms illustrated in Figure 6. Perhaps even more surprising is the fact that the relatively simple 51 knot (illustrated in Figure 7) has yet to be observed in any protein structure. 3. WHICH KNOTS OCCUR IN PROTEINS? 365

Figure 6. As of 2015, these are the only knots that have been found in proteins.

Figure 7. This ﬁve crossing knot has never been identiﬁed in a protein structure.

Various theories have been proposed as to why certain knots have been found and other knots have not. One such theory is based on how proteins might become knotted in the ﬁrst place. According to this theory, the protein’s underlying topological structure looks like a twisted bobby pin with a loop at the top. If one of the ends of the protein threads through this loop and the other end comes close to the threaded end, then adding a short arc between the two termini traps the knot, as illustrated for the 52 knot in Figure 8.

Figure 8. The 52 knot could be the result of threading one terminus through the loop of a twisted bobby pin.

You may recall that among the various knot invariants that we introduced in Chapter 10, we deﬁned the unknotting number to be the smallest number of crossings that must be changed to obtain the unknot among all possible projections of a given knot. If a knot is obtained by threading one terminus through the loop of a twisted bobby pin, then changing one of the crossings of the arc joining the termini will result in an unknot. This means that any non-trivial knot that occurs in a protein in this way, has unknotting number 1. In Figure 9 we illustrate how to unknot the knot from Figure 8 by changing a single crossing of the threading arc. 366 15. THE TOPOLOGY OF PROTEINS

Figure 9. Changing the crossing created by the threading gives us the unknot.

It turns out that the knot 51 has unknotting number 2 (though this is not easy to prove). It follows that 51 can’t be formed in this way. So if the only way a knot could form is through threading one end of a protein through the loop of a twisted bobby pin, then this would explain why the knot 51 has never been found. However, it’s not known whether there are in fact other ways that proteins can become knotted. The other big mystery that remains is why the trefoil knot occurs in proteins in both mirror forms, but the 52 and 61 knots only occur in one of their mirror forms.

4. Examples of Knotted and Linked Proteins

Roughly half of all proteins are known as metalloproteins because they contain metal atoms. Because of the extra structure provided by the metal atoms, the graphs of these molecules are more complex than other proteins and hence more likely to contain knots and links. The trefoil knot 31 was the first knot to be discovered in a protein structure, and it was found in the metalloprotein illustrated in Figure 10. In contrast with the protein knots we saw in the last section, which were entirely contained in the backbones of the proteins, the knot in Figure 10 includes two disulfide bonds between different pieces of the backbone (illustrated as thick grey segments). In general, disulfide bonds are relatively strong bonds that join different parts of the protein backbone with each other or with metal atoms which are attached to the protein. Disulfide bonds are important biologically because their formation guides how the protein folds and, once the folding is completed, these bonds hold the protein in its folded form. The trefoil knot is by far the most common knot identified in proteins and has been found in some forms of carbonic anhydrase, a metalloprotein present in virtually all living organisms from humans to animals to plants. In particular, trefoil knots appear in one type of human carbonic anhydrase found in red blood cells. If we only consider knots in the backbones of proteins rather than allowing disulfide bonds to be included as part of the knot, then there are only about forty proteins which contain knots and almost all of these knots are trefoils. However, there are a few such knotted proteins which contain the 41 knot, and only single proteins have been found to contain the 52 and 61 knots. Interestingly, the protein with the 52 knot, known as ubiquitin hydrolase,is contained in human brains. So if someone tells you that your brain is tied in knots, 4. EXAMPLES OF KNOTTED AND LINKED PROTEINS 367

Figure 10. This knotted protein includes the thick grey segments representing disulﬁde bonds. they’re right! This protein is quite important in preventing neurodegenerative disorders. In particular, a mutation in the gene encoding it seems to be correlated with Parkinson’s and Alzheimer’s diseases. Some researchers believe that the knot in this structure is what helps prevent it from degrading. This theory is based on the fact that protein degradation can occur when proteins are threaded through a narrow hole. If there’s a knot in the protein, it could make the protein too fat to ﬁt through such a hole. If you sew, you’ve probably encountered a similar situation when trying to thread a needle with a piece of thread containing a knot (see Figure 11). However, the existence of such knots in proteins is good because it prevents disease, while such knots in thread are just annoying.

Figure 11. An unknotted thread can pass through the hole of a needle, but a knotted thread cannot.

Topological links in proteins are believed to provide stability to help organisms survive in harsh environments. For example, protein links have been found in organisms that live in the extreme heat found in volcanos. The Hopf link is by far the most common of the protein links. It occurs in several distinct ways, a couple of which are illustrated in Figure 12. The image on the left side of Figure 12 shows two protein backbones (one in thin grey and the other in thin black) that form a link if you include the disulﬁde bonds illustrated as thick grey segments. The image on the right shows a linked protein structure in which a heme porphyrin ring makes up one ring (illustrated in black) and the protein backbone together with a central iron atom make up another 368 15. THE TOPOLOGY OF PROTEINS

Figure 12. Some Hopf links in protein structures.

(illustrated in grey). Note the Fe represents an iron atom and the N represents a nitrogen atom. Other types of links can occur as a result of multiple heme porphyrin rings binding to a protein. In Figure 13, each heme porphyrin ring is illustrated as a black squares and the protein backbone is contained in the dark grey together with dotted grey curve. Note that the black squares represent the same structure as the black ring with the nitrogens in the link on the right of Figure 12. The dotted lines are not part of the link. We illustrate the underlying topological structure of the link on the right. This link is called a key ring link becauseitlookslikeabiggrey key chain with four black keys hanging from it.

Figure 13. This protein structure is a key ring link with four keys.

Perhaps the most interesting example of a protein link is the structure that protects the virus bacteriophage HK97, illustrated in Figure 14. While you might think that viruses are always bad, bacteriophages are good viruses because they ﬁght bacterial infections. So we should be happy that this virus has a link that looks like chainmail protecting it against the evil bacteria. The chainmail around the bacteriophage is actually a 72 component link known as a capsid whose stability comes from the linking. It is made up of 12 pentagonal rings and 60 hexagonal rings, linked into a spherical shape. In Figure 14, we highlight a pentagon in grey to make it easier to see it among all the hexagons. 5. NON-PLANAR GRAPHS IN METALLOPROTEINS 369

Figure 14. The front hemisphere of a protein link made up of 12 pentagonal rings and 60 hexagonal rings.

Whether a protein link contains 2, 5, or 72 components, linking seems to greatly increase the stability of the resulting structure.

5. Non-planar Graphs in Metalloproteins

As we saw above, protein structures can be non-planar because they contain knots or links. However, these are not the only types of non-planarity that can occur in proteins. In particular, some proteins are non-planar because they are attached to a metal cluster, like the iron-sulfur cluster illustrated in Figure 15. So if you see an iron-sulfur box with four legs walking down the street looking like a four-legged tarantula, you’ll recognize it as a metal cluster whose nitrogen atoms might be attached to a protein.

Figure 15. An iron-sulfur cluster resembles a four-legged bug.

Notice that in spite of its cube-like form, if the iron-sulfur cluster in Figure 15 isn’t attached to anything, it could be deformed into a plane. It’s the combination of the metal cluster together with the protein backbone that makes a structure non-planar. For example, let’s consider the protein structure in Figure 16 which contains the iron-sulfur cluster from Figure 15. We’ve drawn the backbone of the protein as a straight line to make it easier to see, not because it’s actually straight. If we imagine that the protein in Figure 16 is completely ﬂexible, maybe we could deform it into the plane. But what if we try unsuccessfully to deform it into a plane? We might suspect that it can’t be done, but we’d like to be sure. With 370 15. THE TOPOLOGY OF PROTEINS

Figure 16. Is this protein structure non-planar? a bit of work we can check that Figure 16 doesn’t contain any non-trivial knots or links. So if it can’t be deformed to lie in a plane, there must be some other reason. Before showing that the protein in Figure 16 is indeed non-planar, we need to learn a little bit more about non-planar graphs. First let’s recall from Chapter 13 that when we talk about an abstract graph, we are ignoring how the graph is situated in space. If we want to consider the speciﬁc way that a graph is sitting in R3 we refer to it as an embedding of the graph in space. Thus a knot is a particular embedding of a circle in space. Using the language we learned in Chapter 3, we would say that all knots have the same intrinsic topology, but whether or not a knot is non-trivial depends on its extrinsic topology. For graphs, the situation is a little more complicated. A graph in 3-dimensional space can be non-planar because of either its extrinsic or its intrinsic topology. In Figure 17 we illustrate three graphs whose non-planarity is entirely due to their extrinsic topology. This means that none of these graphs can be deformed to lie in a plane, but each has a planar embedding. In other words, each could be taken apart and reassembled in the plane. You should see if you can draw each graph in the plane without any crossings. Note that while it’s easy to see that the ﬁrst graph contains a knot and the second graph contains a link, showing that the graph on the right can’t be deformed to lie in a plane is not easy.

Figure 17. These graphs are non-planar because of their extrinsic topology. 5. NON-PLANAR GRAPHS IN METALLOPROTEINS 371

In the Exercises in Chapter 10, we deﬁned a link to be Brunnian if the link is non-trivial but removing any single component leaves a trivial link. Graphs, such as the one on the right in Figure 17, are said to be Brunnian because they are non- planar, but if we remove any single edge, the graph that remains can be deformed to lie in a plane. You should test this out by removing each edge in turn. In contrast with the graphs in Figure 17 whose extrinsic topology makes them non-planar, there are some graphs whose non-planarity comes from their intrinsic topology. This means that if you took them apart and reassembled them, you still wouldn’t be able to put them in a plane. For example, you may recall from Chapter 13 that the graphs K5 and K3,3 are special because they are the “smallest graphs” which cannot be embedded in a plane without any edges intersecting. For K5 this means that no matter how you put ﬁve vertices in a plane and start drawing edges between pairs of vertices, you would not be able to draw non-intersecting edges between every pair of vertices. You might want to try this to see how many edges you can draw before two of them are forced to cross. Because the non-planarity of K5 and K3,3 is independent of how they are embedded in space, we say that these graphs are intrinsically non-planar.

Figure 18. The graphs K5 and K3,3 cannot be embedded in a plane.

Note that in Figure 18, the vertices of the graphs are represented by dots. This is to distinguish them from the places where edges appear to intersect, which are neither vertices nor actual intersections. We draw the graphs as if they had intersecting edges because we are thinking of them as abstract graphs, and hence we don’t want to specify a particular embedding of the graphs in 3-dimensional space. Now we return to the task of showing that the protein structure in Figure 16 is non-planar. If we can prove that it contains one of the intrinsically non-planar graphs K5 or K3,3, then the protein itself would be intrinsically non-planar. This is because if there were a planar embedding of Figure 16, then it would contain a planar embedding of K5 or K3,3, which we know is impossible. In Figure 19, we show that the protein structure in Figure 16 contains the graph K3,3. To do this, we label two sets of three vertices such that every vertex in one set is connected by a disjoint path to every vertex in the other set. In the center image, we have deleted all of the edges that won’t be part of our K3,3.Oneofthe two sets of three vertices (shown as black dots in the center image) consists of three of the four iron atoms that occur on the corners of the cube-like structure. The other set of three vertices (shown as dark grey dots) consists of two sulfur atoms on the corners of the cube-like structure together with one atom on the backbone. We delete the light grey vertices from the center image and then observe that the dotted edges, grey edges, and black edges form disjoint paths connecting every black dot 372 15. THE TOPOLOGY OF PROTEINS

Figure 19. This protein contains a K3,3 graph and hence is non-planar. to every dark grey dot. On the right in Figure 19, we illustrate K3,3 as an abstract graph with the same color coding of vertices and edges. It now follows that this protein structure is indeed intrinsically non-planar. So it’s not surprising that we weren’t able to deform it to lie a plane. As we saw in the above example, if a subgraph of a protein structure is non- planar, then the protein structure itself is non-planar. However, this is not the only way of showing that a structure is non-planar. If a smaller graph can be obtained from a larger graph by collapsing edges and/or omitting edges or vertices, we say the smaller graph is a minor of the larger graph. For example, in Figure 20, the graph on the right is obtained from the graph on the left by collapsing the dotted edges and deleting the grey vertices and the thick grey edge. Thus the graph on the right is a minor of the graph on the left.

Figure 20. The graph on the right is obtained by collapsing the dotted edges and deleting the grey vertices and grey edge.

We can use the following theorem, whose proof you might see in a graph theory course, to show that certain protein structures are intrinsically non-planar. Theorem 15.1 (Kuratowski’s Theorem). A graph is intrinsically non-planar if and only if it contains K5 or K3,3 as a minor. We saw above that the protein structure in Figure 19 is intrinsically non-planar because it contains K3,3. In Figure 21, we see that this protein also contains the complete graph K5 as a minor. In particular, starting with the graph on the left, we collapse the dashed edges and delete the grey vertices and grey edges to obtain the 6. THE PROTEIN STRUCTURE NITROGENASE 373

Figure 21. We collapse the dashed edges and delete the grey vertices and edges to obtain K5. graph in the middle which has vertices numbered 1 through 5 and disjoint paths between every pair of vertices. We can compare the graph in the middle to the one on the right to see that indeed we have K5. Thus by Kuratowski’s Theorem, this gives us another way of showing that this protein structure is intrinsically non-planar.

6. The Protein Structure Nitrogenase

In this section, we analyze the protein structure nitrogenase. This structure is important because it’s responsible for turning nitrogen in the atmosphere into nitrogen that can be consumed by bacteria, which are then consumed by other organisms. However, nitrogenase is so complicated that biochemists studied it for decades before it was ﬁnally completely mapped out in 2011.

Figure 22. A schematic drawing of the proteins in nitrogenase and the metal clusters in the MoFe protein. 374 15. THE TOPOLOGY OF PROTEINS

Figure 23. An M-cluster of the MoFe protein in nitrogenase.

Nitrogenase actually consists of two proteins, known as the Fe protein and then MoFe protein, which work together. We focus here on the MoFe protein, which itself contains four metal clusters: two of which are known as M-clusters and two of which are known as P-clusters. To help you keep track of what is contained in what, you should refer to the schematic drawing in Figure 22. Note that the clusters that we’re interested in are the ones represented by grey disks in the ﬁgure. We will show below that both the M-clusters and the P-clusters of the MoFe protein are intrinsically non-planar.

Figure 24. An M-cluster contains K3,3 and hence is non-planar. 6. THE PROTEIN STRUCTURE NITROGENASE 375

Figure 25. A P-cluster attached to the backbones of nitrogenase.

We start with Figure 23, which is an M-cluster. Recall from Chapters 12 and 13 that thick wedges in a molecular diagram indicate bonds that are coming out of the page. Note that the dashed lines in the ﬁgure indicate that the carbon atom in the center is bound to the iron atoms around it. In Figure 24, we show that the M-cluster is intrinsically non-planar by speci- fying two sets of three vertices in a K3,3 graph together with edges between them. In particular, in the second image, we label the top left sulfur atom, the central carbon, and the top right sulfur atom by the letters a, b and c, and we label the top left iron, bottom left iron, and central right iron by the numbers 1, 2, and 3. We use dotted, grey, and black lines to highlight disjoint paths from the vertices a, b,andc to the vertices 1, 2, and 3. In the third image we have deleted the edges that are not part of the K3,3. The last image shows the same color scheme for the

Figure 26. The P-cluster together with part of the backbone contains K3,3. 376 15. THE TOPOLOGY OF PROTEINS usual illustration of K3,3. This shows that the M-cluster contains a K3,3 graph, and hence by Kuratowski’s Theorem it is indeed intrinsically non-planar. Now we turn our attention to a P-cluster of the MoFe protein. While the M- cluster is intrinsically non-planar by itself, the P-cluster is only non-planar if we consider it together with the backbone of the protein. Figure 25 illustrates one of these clusters and its attachment to the two protein backbones, illustrated with dotted curves. In order to prove that this structure is non-planar, we show that it contains a K3,3 graph. Starting with the structure from Figure 25, we remove all of the edges that are not part of the K3,3 to get the middle illustration of Figure 26. Notice that the arc joining vertex 2 to vertex a uses the grey part of the backbone. The ﬁnal illustration of K3,3 is color coded to look like the middle illustration. In the Exercises you will show that the P-cluster not only contains an additional K3,3, but also contains K5 as a minor.

7. M¨obius Ladders in Metalloproteins

In Chapters 12 and 13 we learned that a Möbius ladder is a graph that can be embedded in space in the form of a Möbius strip. Now we will see that such graphs can actually be found in some proteins. In Figure 27 we label the vertices of a Möbius ladder with three rungs to show that, as abstract graph, it is identical to K3,3. While the figure on the left looks quite different from the one on the right, you should check that exactly the same vertices are connected by edges in both graphs. In particular, in both graphs, every vertex with a letter is connected to every vertex with a number, and no two letters or two numbers are connected to each other.

Figure 27. As an abstract graph, a M¨obiusladder with three rungs is the same as K3,3.

We saw in Figure 24 that the M-cluster of the MoFe protein in nitrogenase contains the abstract graph K3,3. Figure 28 illustrates a step-by-step deformation of the K3,3 in the M-cluster to the Möbius ladder in Figure 27 that has the form of a Möbius strip. In the Exercises you are asked to explain the steps of the deformation showninFigure28. Since a Möbius ladder with three rungs is an embedding of the non-planar graph K3,3, any protein structure containing a Möbius ladder with at least three rungs is necessarily non-planar. However, just as knowing which particular knot is contained in a protein structure gives us more information about the structure than knowing just that it contains some unspecified knot, knowing that a protein structure contains a K3,3 intheformofanembeddedMöbius ladder gives us more 7. MOBIUS¨ LADDERS IN METALLOPROTEINS 377

Figure 28. A deformation of the K3,3 in the M-cluster of nitro- genasetoaMöbius ladder with three rungs. information than just knowing that it contains some unspecified embedding of the abstract graph K3,3. Even more surprising than the fact that nitrogenase contains a Möbius ladder with three rungs in the form of a Möbiusstrip is the fact that protein structures can contain Möbius ladders with four rungs in the form of Möbius strips. In particular, Figure 29 depicts a protein structure containing an iron-sulfur cluster which we will show contains such a Möbius ladder. We begin by illustrating the underlying

Figure 29. The underlying topological form of a protein structure with an iron-sulfur cluster. 378 15. THE TOPOLOGY OF PROTEINS

Figure 30. A deformation taking the iron-sulfur cluster from Fig- ure29toaMöbius ladder with four rungs. topological form of the protein structure on the right with numbers to help us see the Möbius ladder. In Figure 30 we omit the four thin grey edges from the cube-like form in the center of the molecule. We can then shorten the edge connecting vertices 1 and 8 and the edge connecting vertices 4 and 5 to obtain the black infinity symbol with four dotted curves connecting the two lobes in the second illustration in Figure 30. We can then push the two lobes together in the back and rotate the graph to get aMöbius ladder with four rungs in the form of a Möbius strip illustrated on the right in Figure 30.

8. M¨obius Ladders in Small Proteins

Möbius ladders can even be found in small proteins that do not contain metal clusters. This includes some proteins in the cyclotide family. These small proteins were originally discovered in the Democratic Republic of Congo where they are a chemical component of a medicinal plant that was traditionally used to stimulate labor and facilitate childbirth. Because cyclotides are much smaller than normal proteins, some biochemists do not even consider them to be proteins. Nonetheless, cyclotides are important because they have tremendous potential in the development of medications to fight HIV, cancer, and antibiotic-resistant bacterial infections. Because of their stability, they have also been proposed as a scaffolding to carry medications to a specified site in the body. For example, a cyclotide might be used to deliver a chemotherapy drug directly to the site of a tumor and only release it locally so it doesn’t get into the bloodstream. This might enable chemotherapy drugs to have fewer side effects.

Figure 31. A cyclotide with three disulﬁde bonds. 8. MOBIUS¨ LADDERS IN SMALL PROTEINS 379

Figure 32. A deformation taking nerve growth factor to a M¨obius ladder with three rungs.

In the Exercises, you will show that the cyclotide illustrated in Figure 31 can be deformed to a Möbius ladder with three rungs that has the form of a Möbius strip. Notice that the backbone of this protein is actually circular, in contrast with the backbones of most proteins. This cyclotide is often referred to as a “cysteine knot motif” though it does not actually contain a knot. In order to make sense of this terminology, we could interpret “knot motif” to simply mean that the cyclotide is topologically non-planar. Recall from Section 1 that when the termini of a protein molecule are close to each other, adding a small arc between them gives the knot with the highest probability. We use a similar method in Figure 32 to join the termini of the protein known as nerve growth factor, after which we can deform it to a Möbius ladder. In the Exercises you are asked to explain the steps of this deformation. Nerve growth factor is a small protein that is important for growth and func- tioning of nerve cells. It was discovered in the 1950s by the scientists Rita Levi- Montalcini and Stanley Cohen, who were later awarded a Nobel Prize for their discovery. Like the cyclotide in Figure 31, nerve growth factor is said to have a

Figure 33. This graph cannot be deformed into a M¨obius ladder. 380 15. THE TOPOLOGY OF PROTEINS

“cysteine knot motif”, which we can again interpret as referring to the non-planarity of the structure. The middle steps of the deformations illustrated in Figures 28, 30, and 32 each include a loop with three or four edges joining different parts of the loop. It might be tempting to assume that any such embedded graph can be deformed to a Möbius ladder. But this is not always the case. In particular, you will show in the Exercises that the graph illustrated in Figure 33 cannot be deformed into a Möbius ladder.

9. Goodbye for Now

We hope that reading this book has given you the background necessary to explain to your friends and family what topology is and why it’s important. In particular, if scientists want to understand the universe, they need to begin with an understanding of the ideas of geometry and topology that were introduced in Chapters 1–8. Knots are so ubiquitous in everyday life (think about your shoelaces or computer cords, not to mention the 52 knotsinyourbrain)thatitseemslike everyone should want to learn the theory of them. But even beyond untangling our shoelaces and cords, as we saw in Chapters 12–15, knots, links, and non-planar graphs can be found in surprising places, such as molecular structures ranging from small synthetic molecules to DNA and proteins. Perhaps the most important thing you have gotten out of this book is the realization that what seems like abstract esoteric mathematics at ﬁrst glance may turn out to be fun to study and useful in the real world. As we say “goodbye”, we hope this book has whet your appetite to learn more mathematics, with the knowledge that no matter how theoretical an area of mathematics seems, there is probably an application of it just waiting to be discovered. If you are interested in reading other books about topology and geometry and knot theory and their applications, then you might enjoy the following: • Weeks, J.R. (2002). TheShapeofSpace. Boca Raton, FL: CRC Press, Taylor & Francis Group. • Armstrong M.A. (1983). Basic Topology. New York: Springer Science+ Business Media, Inc. • Messer, R., & Staﬃn, P. (2006) Topology Now! Washington DC: Mathe- matical Association of America. • Goodman, S.E. (2005). Beginning Topology. Providence, RI: American Mathematical Society. • Adams, C. (2004). The Knot Book. Providence, RI: American Mathe- matical Society. • Cromwell, C. (2004). Knots and Links. Cambridge: Cambridge Univer- sity Press. • Livingston, C. (1993). Knot Theory. Washington DC: Mathematical Association of America. • Murasugi, K. (1996). Knot theory and Its Applications.NewYork: Springer Science+Business Media, Inc. • Flapan, E. (2000). When Topology Meets Chemisty: A topological look at molecular chirality. Cambridge: Cambridge University Press and the Mathematical Association of America. 10. EXERCISES 381

10. Exercises

1. The structure in Figure 34 represents a protein backbone. Show that this may or may not represent a knotted structure depending on how the endpoints of the structure are extended.

Figure 34. Illustration for Exercise 1.

2. Illustrate how a trefoil knot could be the result of threading one end of a protein backbone through a twisted hairpin.

3. Show that the 51 knot could be obtained from a twisted hairpin by doing one threading followed by another.

4. Draw a single threading of the unknot that creates a 61 knot. Draw two thread- ings that together go from the unknot to a 62 knot. 5. Recall that an abstract graph of a Möbius ladder with three rungs is topologically equivalent to K3,3. Is the abstract graph of a Möbiusladder with four rungs topologically equivalent to K4,4? 6. Draw embeddings of Möbius ladders with three and four rungs which cannot be deformed to the usual embedding that looks like a Möbius strip. 7. (a) Show that the iron-sulfur cluster in Figure 15 can be deformed to lie in a plane. (b) Show that without the backbones, the P-cluster in Figure 25 can be deformed to lie in a plane.

8. The P-cluster in Figure 25 contains multiple K3,3 graphs. Find one that is distinct from the one presented in Figure 26.

9. Show that the P-cluster of nitrogenase contains K5 as a minor.

10. Explain the steps of the deformation illustrated in Figure 28.

11. Explain the steps of the deformation illustrated in Figure 32.

12. The structure in Figure 31 represents the “cyclotide knot motif”. Show that this structure can be deformed to a M¨obius ladder with three rungs in the form of a M¨obius strip. 382 15. THE TOPOLOGY OF PROTEINS

13. In Figure 31, we say that the disulfide bonds have connectivity (1, 4), (2, 5), (3, 6). This notation indicates that there are disulfide bonds between vertices 1 and 6, between vertices 2 and 4, and between vertices 3 and 5. Draw a protein structure with disulfide connectivity (1, 3), (2, 5), (4, 6). Is your drawing intrinsically non-planar, or does whether or not it can be deformed into a plane depend on the embedding of the structure in space?

14. Using the same labeling system as in the above problem, list the connectivity of a protein that contains a M¨obius ladder with four rungs.

15. Consider the graph of the protein in Figure 33 with the N and C termini connected. Does this protein contain a M¨obius ladder? If not, is it planar? (a) In the exercises in Chapter 10, we deﬁned a link to be Brunnian if the link is non-trivial, but removing any single component gives a trivial link. Is the chain mail protein link shown in Figure 14 Brunnian? Why or why not?

(b) How many rings have non-zero linking number with a given pentagonal ring?

16. Illustrate a planar embedding of each of the graphs in Figure 17.

17. Show that if you remove any single edge from the graph on the right side of Figure 17, then graph can be deformed to lie in a plane.

18. Show that if you remove or collapse a single edge of the graphs K5 and K3,3, then you obtain a graph that can be embedded in the plane. Index

2-fold branched cover, 316 crossing negative (left-handed), 243, 313 Abbott, Edwin A., 5 nugatory, 247 achiral overcrossing, 270, 293, 351 chemically, 296, 315 positive (right-handed), 243, 313 geometrically, 294 resolving, 271 topologically, 306, 322 split (A-split or B-split), 283 Allis, Victor, 16 splitting, 282 angle, 39, 52, 170 undercrossing, 217, 243, 270, 293 right, 171, 188 crossing number, 218, 241, 269, 281 sum, 59, 109, 185 cube, 19 anti-cone point, 114, 205, 209 0-dimensional, 36 area 1-dimensional, 36 finite, 84, 140 2-dimensional, 36 infinite, 6 3-dimensional, 36, 173 associative, 137, 153 4-dimensional, 40, 179 atom, 221, 294, 315 n-dimensional, 47 automorphism, 321 solid, 41 identity, 322 cut and paste, 140 order, 322 cycloid, 378 cylinder, 15 B3,13,see also ball, 3-dimensional Connect Four, 15 ball, 21 curved, 62 3-dimensional, 13, 128 finite, 31 boundary, 7 flat, 62 bridge number, 260 infinite, 6, 62, 110 slanted, 46 chiral chemically, 296, 311, 316 D2,13,see also disk geometrically, 294, 311 dimension, 6, 13 intrinsically, 325 disk, 7, 21, 59, 96 topologically, 306, 311 open, 74 circle, 6 distance, 31, 50 great, 9, 58 distributive, 183 circumference, 197 DNA, 216, 292, 297, 312, 330 commutative, 137, 153 axis, 335 cone point, 114, 209 backbone, 332 cone, infinite, 26 base pairs, 331 connected sum duplex, 331 of knots, 220, 233, 249 overwound, 346 of surfaces, 203, 255 relaxed, 331, 335 continued fraction, 353 site, 335, 348 Conway, John H., 353 stressed, 331 cross-cap, 100, 142 substrate, 348

383 384 INDEX

twist, 340 alternating, 248, 281 underwound, 346 composite, 221 equivalent, 217 Einstein-Rosen bridge, 106, 135 figure eight, 220, 227, 240, 249, 261, 302, enzyme, 347 350, 357 ball, 355 invariant, 240, 269, 278 recombinase, 348 left or right handed, 313 Euler characteristic, 144, 257 mirror image, 219 Euler, Leonhard, 144 non-alternating, 248, 269, 280 extended diagram, 17, 61, 93 non-trivial, 218, 365 oriented, 232, 243, 286 Flatland,5 prime, 221 genus, 148 projection, 217 of a knot, 254 square, 236 of a Seifert surface, 254 trefoil, 218, 242, 253, 271, 281, 286, 312, geodesic, 53, 174 363 Geometrization Conjecture, 153 trivial, 218 geometry lattice Euclidean, 191 configuration, 262 extrinsic, 52, 293 number, 263 intrinsic, 52, 91, 109 line, 6 local, 70, 180 geodesic, 186, 192 spherical, 197 straight, 8 gluing, 13 link, 216 abstractly glued, 15, 62 alternating, 248 gluing diagram, 13, 61, 88, 109, 138, 145 Brunnian, 265 hexagonal, 110, 154, 208 component, 217 octagonal, 128, 206 Hopf, 219, 275, 350, 357, 367 polygonal, 111, 205 invariant, 272 polyhedral, 122 oriented, 244, 277 graph, 294, 298, 311 reduced, 248 complete, 322 trivial, 218, 265 complete bipartite, 322 Whitehead, 350, 357 embedding, 320, 370, 376 linking number, 287 non-planar, 361, 376 longitude, 78, 148, 199 handedness (left and right handed), 291, loop, 67 301 non-separating, 147 handle, 6, 141 unshrinkable, 68 adding, 252 Lord Kelvin, 222, 296 hemisphere, 13, 84, 97 Möbius homogeneous, 71, 92, 109, 181, 185, 205 ladder, 308, 315, 329, 376 I, see also interval, 160 strip, 88, 335, 376 identity element, 137, 220 manifold, 71 induction, 147, 154 1-manifold, 85 inverse, 138, 153 2-manifold, 71, 91, 128 isotropic, 181 3-manifold, 72, 120, 151 n-manifold, 71 Jones, Vaughan, 279 flat, 109, 171 turn, 127 2 K , see also Klein bottle, 91 mathematical model, 355 3 K , see also Klein bottle, 3-dimensional, 92 Maxwell, James, 223 Klein bottle, 90, 113, 145 meridian, 78, 148 3-dimensional, 92, 122, 151 mirror reversed, 87 flat, 93, 109 Mislow, Kurt, 318 tic-tac-toe, 93 molecule, 291, 311, 329, 361 Klein, Felix, 90 knot, 216 non-orientable, 89, 141 INDEX 385 orientable, 89, 127, 141 S1,13,see also circle S2,13,see also sphere, 2-sphere (p, q)-curve, 79 S3,20,see also sphere, 3-sphere 1 P , see also projective space, saddle surface, 60 1-dimensional, 107 Seifert surface, 252 2 P , see also projective plane, sidedness 2-dimensional, 96 1-sided, 102 3 P , see also projective space, 2-sided, 102, 252 3-dimensional, 100 Simon, Jon, 316 parallel, 190 space, 8 path sphere, 5 orientation reversing, 89, 147 2-sphere, 22 shortest, 8, 58, 186 3-sphere, 21 straightest, 8 state model, 283 Perko pair, 240 stereographic projection, 75 Perko, Kenneth, 240 stick number, 261 plane, 5 supercoil, 334 flat, 4, 50 negatively supercoiled, 345 hyperbolic, 201 positively supercoiled, 345 point stable, 334 at infinity, 75 unstable, 334 polynomial symmetry X-, 270, 277, 313 3-fold, 326 bracket, 270, 313 mirror image, 291 Jones, 279, 314 synaptic complex, 348 product, 157 geometric, 170, 203 T 2,13,see also torus, 2-torus projective plane, 96, 139, 207 T 3,19,see also torus, 3-torus projective space Tait’s Conjectures, 248 1-dimensional, 107 Tait, Peter Guthrie, 222, 248 3-dimensional, 100, 151 tangle, 350 property 0-tangle, 350 extrinsic, 50, 102, 216 ∞-tangle, 351 geometric, 52, 109, 339, 343 equation, 356 global, 69, 70 equivalent, 350 local, 69 integer, 351 topological, 67, 144, 240, 339 numerator closure, 354 protein, 216, 361 rational, 353 backbone, 361 recombination, 356 Pythagorean Theorem, 192 site, 355 R2,13,see also plane substrate, 355 R3,13 sum, 354 R4,21,34 vector, 352 Rn,71 tetrahedron, 42 recombination, 333 topology product, 349 extrinsic, 64, 217, 347 Reidemeister moves, 224, 272 intrinsic, 66, 91, 217 Reidemeister’s Theorem, 225, 241 torus, 5 Reidemeister, Kurt, 224 2-holed, 5, 68, 137, 148 repeats 2-torus, 22 direct, 359 3-holed, 5, 137 inverted, 359 3-torus, 19, 62, 120 replication, 332, 346 curved, 15, 50, 102 right angle, 41 extended diagram, 18 rigid motion, 52, 293, 300, 339 flat, 15, 61, 92, 107, 109 rubber glove knotted, 65 Euclidean, 301 unknotted, 65 topological, 305 triangle, 42 386 INDEX

complementary, 194 congruent, 210 equilateral, 42 hyperbolic, 201 isosceles right, 195 planar, 61, 110 spherical, 194 tricolorable, 228, 242 twist, 351 horizontal, 352 negative, 351 positive, 276, 351 vertical, 352

Uniformization of Surfaces, 209 universe, 3, 7 1-dimensional, 31 2-dimensional, 5, 31 3-dimensional, 8, 31 4-dimensional, 31 inﬁnite-dimensional, 31 unknot, 218 unknotting number, 258, 365 unlink, 218 vertex, 315 distance, 321 valence, 321

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