Symmetry: A Mathematical Exploration (Texts for Quantitative Critical Thinking) 3030516687, 9783030516680

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Texts for Quantitative Critical Thinking

Kristopher Tapp

Symmetry A Mathematical Exploration Second Edition

Texts for Quantitative Critical Thinking

Series Editors Stephen Abbott, Middlebury College, Middlebury, VT, USA Kristopher Tapp, St. Joseph's University, Philadelphia, PA, USA

Texts for Quantitative Critical Thinking (TQCT) is a series of undergraduate textbooks, each of which develops quantitative skills and critical thinking by exploring tools drawn from mathematics or statistics in the context of real-world questions. Topics are high sophistication, low prerequisite, offering students of all disciplines the opportunity to build skills in the understanding, evaluation, and communication of quantitative information. These are books about mathematical and statistical thinking, not computation and procedure. Each book explores an application or idea in depth, offering students in non-STEM fields a focused, modern context to develop quantitative literacy, while students with a technical background will gain a broader perspective to bridge beyond procedural proficiency. Books in Texts for Quantitative Critical Thinking develop writing and communication skills, which are essential to such cross-disciplinary discourse and are highly-transferable for students in any field. These titles are ideal for use in General Education courses, as well as offering accessible quantitative enrichment for an independent reader.

More information about this series at http://www.springer.com/series/15949

Kristopher Tapp

Symmetry A Mathematical Exploration

Second Edition

Kristopher Tapp Department of Mathematics Saint Joseph's University Philadelphia, PA, USA

ISSN 2523-8647 ISSN 2523-8655 (electronic) Texts for Quantitative Critical Thinking ISBN 978-3-030-51668-0 ISBN 978-3-030-51669-7 (eBook) https://doi.org/10.1007/978-3-030-51669-7 © Springer Nature Switzerland AG 2012, 2021 This work is subject to copyright. All rights are reserved by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors, and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, expressed or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. This Springer imprint is published by the registered company Springer Nature Switzerland AG The registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland

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Preface

Vitruvian Man by Da Vinci

Symmetry is a ubiquitous concept in mathematics and science. Certain shapes and images seem more symmetric than others, yet it is not immediately obvious how to best measure and understand an object’s symmetry. In fact, the quest to understand symmetry more precisely has been a driving force in science and mathematics. It is that quest which will form the central theme of this book. You will learn the ways in which mathematicians study the topic of symmetry. If you are curious about the mathematical patterns underlying the symmetry you observe in the physical world, then this book is for you. Why is the honeycomb in Figure 1 hexagonal? Why are bubbles spherical? Why did the HIV virus evolve the icosahedral shape seen in Figure 2? What is the shape of the universe, and how might this shape be related to the shape of a virus? How can you understand the symmetry of molecules or crystal formations? Parts of these answers are found in other disciplines— biology, chemistry, physics—but the common thread is mathematics. Mathematics provides the tools to understand and classify the possible types of symmetry that objects may possess, which is a crucial prerequisite for addressing questions like those above. The mathematical topics in this book are drawn from diverse fields, including graph theory, abstract algebra, linear algebra, and topology, all of which offer essential tools for undertaking a rigorous study of symmetry. Although some of these topics are advanced, understanding this book requires no background beyond the level of middle-school algebra. The presentation here is intended to be precise and rigorous, yet accessible to a general audience. The only real prerequisite is that you discard any preconceived notions of what math is and isn’t, and begin this mathematical journey with an open mind and a willingness to begin actively doing what mathematicians do: discovering

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Preface

Figure 1: Hexagonal honeycomb structure

Figure 2: An icosahedral HIV virus

patterns, inventing precise language for discussing the mathematical principles underlying those patterns, forming conjectures, and eventually proving beautiful theorems.

Intended Audience This book is primarily intended as a textbook for a math course for math or non-math majors, including humanities majors, with the goal of encouraging effective analytical thinking and exposing students to elegant mathematical ideas. Students find it comforting that there are no prerequisites; previous struggles with math won’t come back to haunt them because the content of this book is very different from the math with which they struggled. The book includes some of the topics that are commonly found in buffet-style sampler textbooks, such as Platonic solids, Euler’s formula, irrational numbers, countable sets, permutations, and a proof of the Pythagorean Theorem. But in this book, all of these topics serve a single compelling goal: to understand the mathematical patterns underlying the symmetry that we observe in the physical world. I hope that students from all majors will enjoy the many beautiful mathematical topics herein, and will come to better appreciate the powerful cumulative nature of mathematics as these topics are woven together into a single story about symmetry.

vii Preface

Improvements in the Second Edition This book is about symmetry, but it’s also about mathematics. The new edition more explicitly embraces the goal of teaching mathematical thinking. A mathematical exploration of any topic (in this case, symmetry) involves precise language, abstraction, generalization, conjecture, proof, disproof, and much more. The new edition more explicitly highlights and discusses these meta-aspects of mathematics as they arise in the study of symmetry. In particular, a new “Elements of Mathematics” section at the end of most chapters highlights a specific nuts-and-bolts aspect of mathematics that arose in the chapter. These elements include: the contrapositive, the converse, sets versus lists, counterexamples, equivalence relations, well-defined definitions, indirect proofs, and inductive proofs. The other changes are too numerous to list. The new edition represents a major overhaul, not a minor tweak. There are a substantial number of new worked-out examples within chapters and exercises at the end of chapters. I’ve woven into the new edition all of the tricks that I’ve stumbled upon over many years of teaching symmetry to non-math majors—tricks for rendering abstract mathematical ideas more understandable and engaging. You’ll see! To make the book appropriate not just for humanities majors, but also for beginning math majors, I’ve added optional sections and exercises that engage advanced topics and proofs. I believe that math majors would be better prepared for abstract algebra if they first explored some explicit symmetry groups by studying this book. Symmetry is all about motion—yet in the first edition I was limited to still images. I’m thrilled that Springer’s new multimedia app lets me include animated images and movies into the second edition! For some readers, watching an animation is more helpful than reading a description of the motion. These readers should download the app and keep a smartphone or tablet on hand while reading the book.

Advice to Instructors Forget about the chalkboard. It’s not the right tool for teaching symmetry. Instead, I strongly recommend forming a course around the PowerPoint presentations and class activities available here: https://sites.google.com/sju.edu/Symmetry This instructor resource page includes animated PowerPoint presentations, handouts, suggestions for manipulatives, class activities, and other teaching suggestions.

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Acknowledgements I am delighted to thank Rachel Hall for many valuable suggestions, resources, and images, and Paul Klingsberg and Alan Gibbons for useful feedback. I am also grateful to the following artists who generously permitted me to include their images in my book: Manuel Arala Chaves, Robert Fathauer, Brian Sanderson, Paul Söderholm, and Ken Tapp. Philadelphia, PA, USA

Kristopher Tapp

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Contents 1

2

3

4

5

Introduction to Symmetry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1

1A Counting Symmetries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1B Objects . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1C Rigid Motions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1D Bounded and Unbounded Objects (Revisited) . . . . . . . . . . . . . . . . . . . 1E Elements of Mathematics: The Contrapositive . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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The Algebra of Symmetry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

29

2A Composition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2B Cayley Tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2C Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2D Symmetry Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2E One Reflection Is Enough . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2F An Improved Classification of Rigid Motions . . . . . . . . . . . . . . . . . . . . . 2G Elements of Mathematics: The Counterexample . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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The Classification Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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3A Rigid Equivalence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3B Bounded Objects . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3C Border Patterns . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3D Wallpaper Patterns (optional) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3E Elements of Mathematics: Equivalence (optional) . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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Isomorphic Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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4A The Definition of an Isomorphism . . . . . . . . . . . . . . . . . . . . . . . . . . . 4B Isomorphism Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4C A Better Notation for Cyclic Groups . . . . . . . . . . . . . . . . . . . . . . . . . . 4D Elements of Mathematics: The Converse . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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Subgroups and Product Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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5A Subgroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5B Generated Subgroups (optional) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5C Product Groups (optional) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5D Elements of Mathematics: Sets versus Lists . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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31 34 37 39 41 45 47

57 60 63 67 70

84 87 89 90

99 101 106 107

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6

7

8

9

10

Contents

Permutation Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

111

6A Counting Permutations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6B Cycle Notation and Composition . . . . . . . . . . . . . . . . . . . . . . . . . . . 6C Even and Odd Permutations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6D Symmetries and Permutations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6E Elements of Mathematics: Well-Defined . . . . . . . . . . . . . . . . . . . . . . 6F Elements of Mathematics: The Inductive Proof (optional) . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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111

Symmetries of 3D Objects . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

135

7A Basics of 3D . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7B Essentially Two-Dimensional Objects . . . . . . . . . . . . . . . . . . . . . . . . 7C A Trick for Counting Symmetries . . . . . . . . . . . . . . . . . . . . . . . . . . . 7D The Tetrahedron . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7E The Cube . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7F The Dodecahedron . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7G The Classification Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7H Chirality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7I The Full Story (optional) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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135

The Five Platonic Solids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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8A The Classification of Platonic Solids . . . . . . . . . . . . . . . . . . . . . . . . . . 8B Counting Their Parts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8C Duality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8D Euler’s Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8E The Euler Characteristic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8F The Platonic Solids Through the Ages . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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171

Symmetry and Optimization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

191

9A Minimal Surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9B The Circle Wins . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9C Elements of Mathematics: Proof by Contradiction . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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191

What Is a Number? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

201

10A Natural Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10B Integers and Rational Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10C Irrational Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10D Real Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10E Which Real Numbers Are Rational? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10F Real Numbers and Symmetry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10G Elements of Mathematics: Construction of Rational and Real Numbers . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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114 117 122 124 127 129

140 143 145 147 149 150 153 156 163

176 177 180 184 185 187

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203 203 205 207 209 211 213

xi Contents

11

Excursions in Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

217

11A How Many Prime Numbers Are There? . . . . . . . . . . . . . . . . . . . . . . . 11B The Meaning of “Same Size” . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11C Are the Rational Numbers Countable? . . . . . . . . . . . . . . . . . . . . . . . 11D Cantor’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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Rigid Motions as Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

231

12A Measuring Distance in Euclidean Space . . . . . . . . . . . . . . . . . . . . 12B Naming the Points on the Unit Circle . . . . . . . . . . . . . . . . . . . . . . 12C The Dot Product and Perpendicularity . . . . . . . . . . . . . . . . . . . . . 12D Using the Dot Product to Find a Friend . . . . . . . . . . . . . . . . . . . . 12E Rigid Motions Are Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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Rigid Motions as Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

243

13A Matrix Computations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13B Representing Rigid Motions as Matrices . . . . . . . . . . . . . . . . . . . . . 13C Orthogonal Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13D Concluding Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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Image Credits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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1

1

Introduction to Symmetry

Our journey starts with the question: what does “symmetry” mean? Look at the four objects in Figure 3, and rank them from least to most symmetric.

Figure 3: Which object is most symmetric?

How do we interpret this question in a manner that is precise enough to lead us to a justifiable ranking of the four objects? For this, instead of common English syntax in which “symmetry” is a quality or property of an object, we must adopt a mathematical syntax in which the “symmetries” of an object can be counted and listed. The question then becomes: which of the above four objects has the most symmetries? Let’s begin!

1A Counting Symmetries Let’s start by focusing on a single object, namely, the square in Figure 4. In common English usage, why does the word “symmetry” seem appropriate for a square? Some people would answer that the square looks the same from multiple positions; for example, Figure 4: Why does the word “symmetry” seem appropriate for a square?

Electronic supplementary material The online version of this chapter (https://doi.org/10.1007/ 978-3-030-51669-7_1) contains supplementary material, which is available to authorized users. The videos can be accessed individually by clicking the DOI link in the accompanying figure caption or by scanning this link with the SN More Media App

© Springer Nature Switzerland AG 2021 K. Tapp, Symmetry, Texts for Quantitative Critical Thinking, https://doi.org/10.1007/978-3-030-51669-7_1

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1

Chapter 1 • Introduction to Symmetry

a square painted on the ground looks the same whether approached from the north, south, east, or west. Other people might answer that the square has repeated parts; for example, all of its corners look the same. But these intuitive indicators of symmetry—looking the same from multiple positions and having repeated parts—are explained by a more fundamental observation that cuts to the heart of the matter: there are motions that leave the square unchanged.
One such motion is the 90 counterclockwise rotation around its center point, illustrated in the first filmstrip (Figure 5). Another such motion is a vertical reflection, which means a flip over its vertical center line (Figure 6). Yet another such motion is the reflection over a diagonal line (Figure 7). Each of these motions is a “symmetry of the square,” according to our first attempt to precisely define the word “symmetry.”

Rough-Draft Definition A symmetry of an object is a motion that leaves it unchanged.

Figure 5: Rotation symmetry of a square (▶ https://doi.org/10.1007/000-256)

Figure 6: Vertical reflection symmetry of a square (▶ https://doi.org/10.1007/000-252)

Figure 7: Diagonal reflection symmetry of a square (▶ https://doi.org/10.1007/000-253)

3  1A

Counting Symmetries

What does it mean for a motion to leave an object “unchanged”? It roughly means that, if a viewer were to close her eyes during the motion, she would not detect a difference when she opened her eyes. The object would look exactly the same. In each filmstrip, the square appears the same in the first frame as in the last. This rough-draft definition will require fine tuning, but it’s already pretty good. It captures the essential idea and empowers us to answer precise quantitative questions like: how many symmetries does the square have? The answer is 8. It has four rotation



symmetries (by angles 0 , 90 , 180 , and 270 about its center point). Notice that the


180 rotation is really just two repetitions of the 90 rotation, while the 270 rotation is
three repetitions. We will later explain the convention that 0 always counts as a rotation symmetry. The square’s remaining four symmetries are reflections, namely, the reflections over the four lines shown in Figure 8.

Figure 8: The four reflection lines of a square

The reflections over the first and third lines were illustrated in the previous filmstrip images, while the other two can be visualized similarly.

▶ Exercise

Count and describe the symmetries of the star in Figure 9.

Figure 9: How many symmetries does this 5-pointed star have?

Solution It has 5 rotation symmetries because it is a 5-pointed star. The smallest nonzero
rotation angle is 360/5 ¼ 72 . The five rotation angles are the multiples of this smallest angle:




0 , 72 , 144 , 216 , 288 . It also has 5 reflection symmetries over the reflection lines in Figure 10. We conclude that this star has 10 total symmetries.

These 10 symmetries provide the underlying reason that the star exhibits some of the previously mentioned intuitive indicators of symmetry. The star must have repeated parts (its five arms look the same) because the star looks unchanged when these five arms rotate or reflect onto each other.

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Chapter 1 • Introduction to Symmetry

Figure 10: The five reflection lines of a star Figure 11: A pentagon has 5 rotation symmetries and 5 reflection symmetries

Think about why the pentagon in Figure 11 also has 10 symmetries. In fact, it has the same 5 rotation angles as the star and has a similar arrangement of its 5 reflection lines.

1B Objects For a rigorous mathematical investigation of symmetry, we must be more precise about the type of “objects” whose symmetries we are studying. When we counted and listed the square’s 8 symmetries, the filmstrip images encouraged us to think of the square as cut out of cardboard. But any actual cardboard square has imperfections and therefore looks slightly different after being reflected or rotated. Only a mathematical idealization of a square has 8 symmetries, not an actual cardboard square. This is a math book, so the “objects” whose symmetries we study will be not the craftproject kind, but the mathematical kind, as given in the following definition.

Definition The plane is the set of all ordered pairs (x, y) of real numbers. A two-dimensional object is a nonempty subset of the plane.

“The plane” is also called “the xy-plane”. We visualize it as an infinite glass wall that extends indefinitely up, down, right, and left. The numbers (x, y) just identify all the possible locations on this glass wall where one could put a drop of paint. We visualize an “object” as painted onto the plane, but more technically, it’s the subset of locations where there is paint.1 The adjective “nonempty” means that there is at least one drop of paint.

1 A “set” roughly means a collection, and a “subset” of a set is composed of some (but possibly not all) of the set’s members. For example, the set of points with distance 1 from a given point (say, from the point (0,0), known as “the origin”) is a subset of the plane (an object), namely, a circle. Here we are really only defining a single-color object (visualized as made with only one color of paint). Think about how to precisely define a multi-color object.

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5  1B

Objects

For brevity, we will henceforth write “object” instead of “two-dimensional object,” at least until we begin studying three-dimensional objects in Chapter 7. Perhaps the most natural objects are the regular polygons:

Definition For n  2, a regular n-sided polygon is the part of the plane enclosed by n equal-length straight sides assembled so that all n of its angles are equal.

Thus, a regular 3-sided polygon means an equilateral triangle, a regular 4-sided polygon means a square, a regular 5-sided polygon means a regular pentagon, a regular 6-sided polygon means a regular hexagon, and so on (Figure 12). We will interpret a “regular 2-sided polygon” to mean a line segment, as pictured. This convention is consistent with our definition, since two equal-length line segments assembled to meet at equal angles will meet at zero-degree angles, and hence lie on top of each other. Table 1 indicates the following pattern: A regular n-sided polygon has n rotation symmetries and n reflection symmetries, for a total of 2n symmetries. The smallest angle of a rotation symmetry (other than zero) is 360/n. You previously verified the cases n ¼ 4 and n ¼ 5 of this pattern. The general claim is not much harder to verify, but notice that counting reflection symmetries requires you to understand the arrangement of reflection lines, which works differently depending on whether n is even or odd. Think about it.

n=2

n=3

n=4

n=6

n=5

n=7

Figure 12: Regular polygons

Table 1: Counting the symmetries of regular polygons

n=2 smallest angle # rotations # reflections # symmetries




180 2 2 4

120 3 3 6




90 4 4 8

n=6

n=5

n=4

n=3





72 5 5 10




60 6 6 12

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Chapter 1 • Introduction to Symmetry

Figure 13: Oriented pentagons

n=2

n=3

n=4

n=5

n=6

n=7

Figure 14: Oriented polygons

Each regular polygon in Figure 12 has both rotation and reflection symmetries. Can you think of a way to orient each of these polygons, which means to decorate it in such a way that its rotation symmetries are preserved but it no longer has any reflection symmetries? For example, Figure 13 shows a few artistic ways to orient the pentagon. There are many other possibilities; whichever you choose, the result is called an oriented pentagon. Each oriented pentagon pictured in Figure 13 has 5 rotation symmetries but no reflection symmetries. Do you see why? A reflection would reverse the issue of whether it appears to spin clockwise or counterclockwise, and would therefore not be a symmetry. Since the first method of orienting a pentagon in Figure 13 is the simplest, we’ll use that method to orient the remaining polygons shown in Figure 12. The result is shown in Figure 14. The regular and oriented polygons together form a large collection of objects to which we will later compare other objects. You may be taking for granted a property that they all have in common: they are all bounded objects. If so, you are about to stop taking it for granted, because now, for the sake of some other kinds of symmetry, we need to look at some objects that are not bounded. Our precise definition of “object” is broad enough to include unbounded objects. Intuitively, an object is called unbounded if it extends indefinitely to infinity in at least one direction. We will study two important types of unbounded objects—a wallpaper pattern and a border pattern. First, a wallpaper pattern intuitively means an object that extends indefinitely in all directions (left, right, up, and down) according to some repeating scheme. Figure 15 shows two examples. Imagine that each pattern continues indefinitely in all directions to fill the whole infinite plane. Second, a border pattern intuitively means an object that extends indefinitely only along a line (usually the x-axis) according to some repeating scheme. Figure 16 shows

7  1B

Objects

two examples. Imagine that each pattern continues indefinitely to the left and right (but not up or down). Unbounded objects sometimes have rotation symmetries. For example, the left
wallpaper pattern of Figure 15 can be rotated 60 about the center of any hexagon, or
(a corner where three hexagons meet). The top 120 about any point that looks like
. The border pattern of Figure 16 can be rotated 180 about any point that looks like
bottom border pattern of Figure 16 doesn’t have any rotation symmetries (except by 0 ). Unbounded objects also sometimes have reflection symmetries. For example, the top border pattern of Figure 16 can be reflected over its horizontal center line. Both border patterns of Figure 16 can be reflected over certain vertical lines. Can you find reflection lines for either wallpaper pattern of Figure 15? The PowerPoint presentation for Chapter 1 (and the movies linked to these pages) includes animations that help you visualize the symmetries of border patterns and wallpaper patterns. Rotations and reflections don’t tell the whole story here. Each of the four unbounded objects we have considered so far has symmetries of a type that we haven’t yet encountered—namely, translation symmetries. A translation means a shifting (sliding) of the plane. For example, visualize the “Seahorses and Eels” wallpaper pattern in

Figure 15: Two wallpaper patterns

Figure 16: Two border patterns

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Chapter 1 • Introduction to Symmetry

Figure 15 painted onto the glass plane, with both the glass plane and the painting extending indefinitely in all directions. This requires infinitely many seahorses and infinitely many eels. Now imagine translating (shifting) the whole glass plane, together with the image painted on it. If the direction and distance of this shift are carefully chosen, then the image will appear exactly the same after the repositioning. One translation that works is depicted by the length (about an inch) and direction (up) of the red arrow in Figure 17. Several copies of the red arrow are included to demonstrate that each composition element (the eye of an eel, the center of the purple tail spiral, etc.) moves exactly onto an identical element. A viewer who closed his eyes during this repositioning would not notice any difference after opening his eyes. That’s why this particular translation is a symmetry of this wallpaper pattern. What other translations are symmetries of it? Border patterns also have translation symmetries. The direction (right) and length (about 1 inch) of the red arrow in Figure 18 depicts a translation that happens to be a symmetry of the illustrated border pattern. In fact, this border pattern has infinitely many translation symmetries, because the length of the arrow could be doubled, tripled, quadrupled, etc. Or it could point left instead of right.

Figure 17: A translation symmetry of this wallpaper pattern (▶ https://doi.org/10.1007/000-254)

Figure 18: A translation symmetry of this border pattern (▶ https://doi.org/10.1007/000-255)

9  1C

Rigid Motions

Unbounded objects are mathematical abstractions that don’t really exist in the world, so why should we study them? One answer is that abstract mathematics is intrinsically beautiful. A more practical answer is that it helps us understand real-world objects. Consider the wallpaper that you might purchase at a hardware store to cover your dining room walls. The manufacturer designed it as a repeating pattern so that it is capable of covering any size wall. Your intuitive sense of the pattern’s symmetry derives from the repetition of images accounted for by translation symmetries (of its infinite extension).

1C Rigid Motions We decided that a “symmetry” of an object roughly means a motion that leaves it unchanged. We will now fine-tune this definition to make it precise enough to form the foundation of our mathematical study of symmetry. The problem is that the word motion is too vague. The vagueness might already have caused you confusion, even back when we decided (correctly) that a square has 8 sym
metries. Why did we count the 0 rotation? Why didn’t we separately count all multiples








of 90 (namely, 0 , 90 , 180 , 270 , 360 , 450 , 540 , 630 , . . .) and thereby conclude that the square has infinitely many rotation symmetries? Why didn’t we count two different vertical reflections (depending on whether you flip the right or left side up towards you)? Why didn’t we count the activity of cutting the square into four quadrants and reassembling them in a different order? The answers to these questions will follow once we clarify what types of “motions” we should allow, namely, the “rigid” ones.

Definition A rigid motion of the plane is a repositioning that preserves distances.

Much later in the book we will make this definition even more precise using matrices. For now, it’s a workable definition when combined with the following interpretations: 1. Rigid motions are done to the whole plane, so it makes sense to discuss them even when there is no object painted on the plane. Picture the plane as a glass wall and a rigid motion as something done to the wall, like a translation, rotation, or reflection. 2. A rigid motion “preserves distances.” This means that any pair of drops of paint on the plane must be the same distance apart before and after the repositioning. Can you visualize why rotations, translations, and reflections all preserve distances? If two drops of paint are 10 inches apart before the plane is rotated (or reflected or translated), they will still be 10 inches apart after. 3. Hence, a rigid motion may not break, bend, stretch, compress, or otherwise distort distances on the glass wall. It’s not a rigid motion if it requires glass cutters or a blow torch, or if it could be done only to a wall made out of stretchy bendy material. The term “rigid” is intended to suggest something that can be done only to a rigid material like glass.

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4. Our previous filmstrip images depicted a motion as a movie, but that was an imprecise metaphor. A rigid motion is a “repositioning” of the plane, which means just the ending position, not the movie of how it got there. Only the last frame of the movie is relevant. Think of a rigid motion as an instantaneous repositioning. In other words, a rigid motion is completely determined by its effect on each point of the plane (each speck of dust on the glass wall). Two motions that do the same thing to every point are
considered the same motion. For example, rotating 90 is the same as rotating


90 + 360 ¼ 450 around the same point. When enumerating the symmetries of an

object, we would not list both a 90 and a 450 rotation, or any other such redundancies. What matters is the effect of a motion, not the middle frames of the movie. Since an object is technically a subset of the plane (pictured as painted onto the glass wall), it goes along for the ride when a rigid motion of the plane is performed. This entire book is based on the following improvement of our previous rough-draft definition.

Definition A symmetry of an object is a rigid motion of the plane that leaves the object unchanged.

What does it mean to leave the object unchanged? This just means that the rigid motion moves it to the same subset of the plane. Here’s the best way to visualize it: the plane is a glass wall and the object is painted onto the glass, but the x-axis and y-axis are fixed and never-moving (they are not painted onto the glass, but rather are drawn on a blackboard lying behind the glass). So in Figure 19, the translation to the right is not a symmetry of the circle because it noticeably shifts the center point of the circle away from the y-axis. When we previously decided that a square has 8 symmetries, we visualized these symmetries as motions performed with a cardboard square. This gave the right answer, but it’s technically more correct to visualize the whole plane (glass wall) rotating and reflecting along with the square. It’s even more technically correct to ignore the middle frames of the movie filmstrips and thus visualize each symmetry as an instantaneous repositioning. In summary, a rigid motion is always a motion of the whole plane (the whole glass wall). One may then ask whether it is a symmetry of any object in the plane. For example,
the 72 rotation around the red point in Figure 20 is a rigid motion of the plane that is a symmetry of the purple pentagon, but is not a symmetry of the other three shapes. Notice

y

y x

Figure 19: This translation to the right is not a symmetry of the circle (▶ https://doi.org/10.1007/000-251)

x

11

y

TE

Rigid Motions

y

R OT A

 1C

ROTATE x

x

Figure 20: This rotation is a symmetry of the pentagon, but not of the other shapes (▶ https://doi.org/10.1007/000-257)

ER

y

TCEL F

y

REFLECT x

x

Figure 21: This reflection is a symmetry of the pentagon, but not of the other shapes (▶ https://doi.org/10.1007/000-258)

that the red point itself is a fixed point (it stays in the same position), while all other points of the plane rotate around it. Similarly, the reflection over the red line in Figure 21 is a symmetry of the purple pentagon but not of the other shapes. Notice that each point of the red line itself is a fixed point (it doesn’t move). Each of the other points moves to its “mirror image” across the red line. The simplest rigid motion of the plane is the “do nothing” motion. It might seem boring, but it has a special name: it is called the identity. It can be considered a rotation (by zero degrees) or a translation (by zero distance). Every point is a fixed point of the identity (because the identity doesn’t move anything). Thus, the identity is a symmetry of every object; in fact, it’s the only rigid motion that can make this claim. An object is called asymmetric if it has exactly one symmetry, namely, the identity. This is the fewest symmetries that an object could possibly have. The gnome in Figure 22 (by Paul Söderholm) is asymmetric. A haphazard doodle will almost certainly be asymmetric. Let us summarize this in a definition.

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Chapter 1 • Introduction to Symmetry

Figure 22: I’m asymmetric. I have no symmetries other than the identity

Definition The identity is the rigid motion that fixes (leaves unmoved) every point of the plane. An object is called asymmetric if the identity is its only symmetry.

To describe to me your favorite rotation or translation or reflection, what information must you give me? A rotation is specified by its center point and its (counterclockwise) angle. That is all the information that is needed. Similarly, a translation is specified by the length and direction of a single arrow. A reflection is specified by its reflection line. We are using the words “identity” and “translation” the way mathematicians use them, which doesn’t match very well with common English, but serves our purpose of giving definite expression to ideas about symmetry. We will need a few more such terms. For example, another word that mathematicians are funny about, for the same reason, is “order.”

Definition The order of a rotation is the minimum number of times it must be repeatedly performed to yield the identity.





For example the order of a 90 rotation equals 360/90 ¼ 4, while the order of a 60 rotation equals 360/60 ¼ 6. Rotations and translations are called “proper” rigid motions. Reflections are called “improper” rigid motions. If you think of the glass plane as having two faces (like the exterior and interior faces of a glass window of your house), the intuitive difference is that improper motions exchange the two faces. But since our technical definition of “the plane” doesn’t give it any thickness, this intuition doesn’t lend itself to a precise definition. Instead, we will characterize proper and improper rigid motions according to their effect on the palm of a right hand.

Definition A rigid motion is called proper if, after it is applied, the image of a right palm still looks like a right palm, while a rigid motion is called improper if it turns a right palm into a left palm.

13  1C

Rigid Motions

y

y

x

x

Figure 23: The reflection over the red line turns a right palm into a left palm (▶ https://doi.org/10.1007/000-259)

Rotations and translations are clearly proper: imagine what would happen if shifting or rotating a right hand turned it into a left hand—it would be dangerous to wave hello to a friend. A reflection is improper, as shown in Figure 23. A reflection followed by a translation is also improper. We’re now ready for the book’s first theorem! A theorem is a mathematical fact that has been proven to be true. In this book, definitions are in blue-framed boxes (as you’ve already noticed) and theorems are in blue shaded boxes. Here is our first:

Classification of Plane Rigid Motions (Version 1) Every proper rigid motion of the plane is a translation, a rotation, or a rotation followed by a translation. Every improper rigid motion of the plane is a reflection or a reflection followed by a translation. In a much later chapter of the book, you’ll learn a more precise definition of “rigid motion” involving matrices, which can be used to prove the above theorem. For now, we unfortunately don’t yet have the tools to prove this theorem, so you need to just accept that it’s true. The theorem says that there are no rigid motions other than the types we have already considered and combinations thereof. Moreover, only certain “combinations thereof” are needed. Consider a rotation followed by a translation followed by a different rotation followed by a reflection. This four-step repositioning is a rigid motion, so the theorem tells us that it can be re-described in a simpler way. It is improper (think about why), so it must be the same as either a reflection or a reflection followed by a translation. The theorem says that you never need a many-step sequence of rigid motions because there is always a oneor two-step sequence that has the same effect on every speck of dust of the plane. As previously mentioned, rigid motions are done to the plane, so it makes sense to discuss them even when there is no object in the plane. If there is an object, then each of its symmetries is a rigid motion of the plane (that leaves it unchanged), and is therefore of one of the types enumerated in the classification theorem.

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Chapter 1 • Introduction to Symmetry

We end this section with a precise definition of a previously encountered word. The oriented polygons shown in Figure 14 on page 6 had only rotation symmetries, not reflection symmetries. This previous usage of the word “oriented” is consistent with its general meaning, as given in the following definition.

Definition An object is called oriented if it has no improper symmetries.

When an oriented object has been reflected, you can always detect that it has been reflected. The knotted blue object in Figure 24 is oriented; hold it up to a mirror, and notice how its over/under crossing pattern differs from that of its mirror image. To understand the difference, trace the blue path counterclockwise starting from one of the three outermost points: do you encounter an over- or under-crossing first?

Figure 24: An oriented object doesn’t match its mirror image (▶ https://doi.org/10.1007/000-25a)

An improper rigid motion would make the knotted blue object in Figure 24 look like its mirror image, and therefore, it could not be a symmetry of this object. How do we know that an improper rigid motion would make it look like its mirror image? The classification theorem tells us: any improper rigid motion is a reflection (which makes it look like its mirror image) possibly followed by a translation (which would just shift this mirror image).

1D Bounded and Unbounded Objects (Revisited) You might now feel that you understand the terms bounded, unbounded, border pattern, and wallpaper pattern because we previously provided examples, pictures, and vague verbiage about them. But that’s not enough! In this section we will formulate precise definitions for these terms. Why? The value of precise language in other areas of science is commonly appreciated. In biology, for example, you can’t catalog mammals until you specify their defining characteristics (warm blooded, hair or fur, mammary glands, etc.). How else could you decide whether a whale is a mammal? Precise definitions are even more important in mathematics. The ideal is to say exactly what each new term means,

15  1D

Bounded and Unbounded Objects (Revisited)

using only words whose meanings are already precisely understood. If we hope to eventually classify the symmetry types of border patterns and wallpaper patterns, then we will need an exact characterization of their defining properties. Let’s start with the term “unbounded.” We previously called an object unbounded if it “extends indefinitely to infinity in at least one direction.” That’s the right intuition, but it’s not the right basis of a precise definition, since the word “infinity” is vague. It’s much better to sidestep this complication by focusing instead on the more concrete issue of whether the object can be framed (say, by a square frame).

Definition An object is called bounded if there exists a square in the plane that fully contains it. Otherwise it is called unbounded.

The meaning would remain unaltered if the word “square” were replaced by “circle” or “pentagon” or many other possibilities. If an image can be framed by one of these frame shapes, then it can be framed by all of them. We previously characterized both wallpaper patterns and border patterns as “extending indefinitely according to some repeating scheme.” But there is no manageable way to precisely define the terms “scheme” and “pattern.” Try to explain the scheme/ pattern of exactly how the seahorses and eels fit together like puzzle pieces in Figure 15. You’ll quickly realize how difficult it would be to concisely characterize all possible schemes/patterns. To sidestep this difficulty, we will define border patterns and wallpaper patterns, not according to the puzzle pieces out of which they are built, but rather according to the translational symmetries they possess. This strategy works because translation symmetries force repeating patterns. For example, imagine that an object has a translation symmetry one inch to the right. This forces it to be a horizontally repeating pattern— each part looks the same as the part one inch to its right or left. If it additionally has a translation symmetry one inch up, this forces the pattern to repeat vertically also. Every one-inch-by-one-inch square looks the same as the square above, below, right, or left of it. This leads us to our definition of border patterns and wallpaper patterns.

Definition An object that has at least one translation symmetry besides the identity (and that has a smallest one) is called

▬ a border pattern if all of its translations are parallel to a single line, and is called ▬ a wallpaper pattern otherwise. A few comments are in order. The phrase “besides the identity” is necessary because the identity technically qualifies as a translation (shifting zero distance), and it is a symmetry of every object. Among its translation symmetries other than the identity, our definition requires a border pattern or wallpaper pattern to have “a smallest one.” This means that a horizontal

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Chapter 1 • Introduction to Symmetry

line (like, say, the x-axis) does not qualify as a border pattern, since its symmetries include translations to the right by arbitrarily small amounts: a millionth of an inch, a trillionth of an inch, etc. No one choice is the smallest. Similarly, the whole plane is an object (visualized as filling the whole plane with paint) that does not qualify as a wallpaper pattern. It would have been redundant to include the adjective “unbounded” in our definition of border/wallpaper patterns, because, as the following theorem shows, they are automatically unbounded.

Theorem If an object has at least one translation symmetry besides the identity, then it has infinitely many translation symmetries, and the object is unbounded.

Recall that an “object” means a nonempty subset of the plane, visualized as painted onto the plane. So as you read the following proof, interpret the expression “point of the object” visually as meaning “position on the plane where there is a drop of paint.” Proof Repeatedly applying the one translation symmetry any number of times (once, twice, thrice, etc.) yields a translation symmetry. So there are infinitely many of them. Consider a single point of the object. The one translation moves it to another point of the object because it’s a symmetry. Repeated applications of the one translation move it to equally spaced points of the object extending indefinitely along a line. No square in the plane could contain all infinitely many of these points. The object is therefore unbounded (there does not exist a square that contains it). □

A proof means a logical argument that establishes the truth of a theorem. If this proof convinced you that the theorem is true, then it did its job. Notice that the precise definitions of terms like “object” and “unbounded” were necessary to precisely state and prove this theorem. Without precise language, one can’t even accurately describe an observed pattern, much less prove it. Here is one consequence of this theorem: If an object is bounded, then it doesn’t have any translation symmetries besides the identity. This is because having translation symmetries would make it unbounded. You might not be surprised—the bounded objects we’ve encountered so far have had only rotation and reflection symmetries. It can be proven that the same is true of all bounded objects.

The Center Point Theorem: Every bounded object has a center point such that

(1) every proper symmetry is a rotation around this center point, and (2) every improper symmetry is a reflection over a line through this center point.

17  1E

Elements of Mathematics: The Contrapositive

Figure 25: The star’s center point (black) is where all of its reflection lines (red) cross

A center point of a bounded object is defined as a point satisfying properties (1) and (2) of the theorem. The center point of the object in Figure 25 is the black dot. It is the star’s rotation center, and all of the star’s reflection lines (shown in red) pass through it. If you cut the star out of cardboard and wish to balance it on your fingertip, this is the correct place to position your finger.

1E Elements of Mathematics: The Contrapositive This book is about symmetry, but also about mathematics. Unfortunately, some people have internalized the notion that math is just about performing rote algebraic skills. One goal of this book is to exhibit a truer portrait of what it really means to do mathematics. A mathematical exploration of a topic (in this case symmetry) involves abstraction, creativity, logic, proofs, quantitative reasoning, and much more. We will address aspects of mathematical thinking in the final section of this (and other) chapters. We have already repeatedly emphasized the chapter’s main mathematical theme: the crucial value of precise language. In fact, the primary goal of this chapter was to precisely define some of the vocabulary with which we will study symmetry throughout the book: the plane, object, regular polygon, rigid motion, symmetry, bounded/unbounded, proper/ improper, border pattern, wallpaper pattern, the identity, asymmetric, oriented, rotation, reflection, translation, fixed point, center point. Learn these terms now! We have also encountered a few theorems (true mathematical statements) and proofs (logical arguments that establish the truth of theorems). We will see many more theorems and proofs throughout the book. Writing proofs is a creative endeavor. There is no algorithm for it, any more than there is an algorithm for writing a novel or investigating a crime. But there are some generally useful techniques that are worth identifying. In the remainder of this section, we’ll zoom in on a particular mathematical construction: the contrapositive. Recall the following two statements from the previous section:

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Chapter 1 • Introduction to Symmetry

STATEMENT 1: If an object has at least one translation symmetry besides the identity, then it is unbounded. STATEMENT 2: If an object is bounded, then it doesn’t have any translation symmetries besides the identity. These two statements are logically equivalent—they say exactly the same thing, just phrased in different ways. If you prove either statement, then you are allowed to conclude that the other is true because it says the same thing. We chose to prove statement 1 (which was organized in such a way that the hypothesis led fairly naturally to the conclusion) and we learned for free that statement 2 was true (which was what we were really after). Formally, each of these statements is the contrapositive of the other. A common trick for proving a statement is to instead prove its contrapositive, which is sometimes easier. What is the general meaning of “contrapositive”? Mathematicians often find themselves studying a set and considering two properties—let’s call them property A and property B—that some of the members of this set might have. In this setting, the following two statements (about an arbitrary member from the set) are logically equivalent: STATEMENT 1: If it has property A, then it has property B. STATEMENT 2: If it doesn’t have property B, then it doesn’t have property A. Each statement is called the contrapositive of the other. The previous example was about the set of all objects. Property A was “having at least one translation symmetry besides the identity.” Property B was “being unbounded.”

▶ Exercise

My neighbor owns lots of pets. Each pet is either a cat or a dog. Determine the contrapositive of “All of my neighbor’s cats are females.” Solution This statement is about the set of my neighbor’s pets. Property A is “being a cat,” and property B is “being female.” The contrapositive says that if an animal lacks property B (it’s not female) then it lacks property A (it’s not a cat). More concisely: “All of my neighbor’s male pets are dogs.”

Do you see why the contrapositive is logically equivalent to the original statement? If one is true then the other must be true. They say the same thing.

19  Exercises

Exercises (1) Write a precise definition of each of the following terms: the plane, object, rigid motion, symmetry, bounded/unbounded, proper/improper, border pattern, wallpaper pattern, the identity, asymmetric, oriented. (2) What information is needed to describe a specific . . . (a) rotation? (b) reflection? (c) translation?


(3) What is the smallest angle (other than 0 ) of a rotation symmetry of a regular 10-sided polygon? (4) Which are proper and which are improper? (a) (b) (c) (d)

A proper rigid motion followed by a proper rigid motion A proper rigid motion followed by an improper rigid motion An improper rigid motion followed by an improper rigid motion The identity

(5) Suppose that 20 rigid motions are performed one after the other. Suppose 13 of them are proper and the other 7 are improper. Will the net result be proper or improper? Explain your answer. (6) Draw two different objects that have exactly the same collection of symmetries. (7) Make sketches of several bounded objects that have interesting collections of symmetries. Try to sketch a bounded object whose collection of symmetries is significantly different from that of a regular polygon or an oriented polygon. (8) Decide whether each statement is TRUE or FALSE and explain your answer. (a) (b) (c) (d) (e) (f) (g) (h) (i) (j) (k)

Every asymmetric object is oriented. Every oriented object is asymmetric. Every border pattern has infinitely many translation symmetries. A circle is asymmetric. Every wallpaper pattern is unbounded. Every border pattern is unbounded. Every unbounded object is either a border pattern or a wallpaper pattern. Every object has at least one symmetry. If an object is not oriented, then it has at least two symmetries. Every proper rigid motion of the plane has a fixed point. Every improper rigid motion of the plane has a fixed point.

(9) For each capital letter in the English language (in the font shown here),

ABCDEFGHIJKLMNOPQRSTUVWXYZ

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Chapter 1 • Introduction to Symmetry

count the number of proper symmetries, count the number of improper symmetries, and determine whether it is oriented. Organize your work by grouping letters with the same answers. (10) Each capital letter in the English language (in the font from the previous exercise) can be used to create a border pattern like those shown here.

…A A A A A A A A A…. …B B B B B B B B B B … …C C C C C C C C C C … For each of the 26 letters, decide whether the resulting border pattern (a) (b) (c) (d)

has a reflection symmetry across a horizontal line, has reflection symmetries across any vertical lines, has rotation symmetries other than by zero degrees, is oriented.

Organize your work by grouping letters with the same answers. (11) Consider the border pattern shown here to be extended indefinitely to the right and left.

Explain why it is not oriented, even though it does not have any reflection symmetries. HINT: According to the classification theorem on page 13, reflections are not the only types of improper rigid motions. (12) For each of the following objects, count the number of proper symmetries, count the number of improper symmetries, and determine whether the object is oriented.

(13) For each of the following objects, count the number of proper symmetries, count the number of improper symmetries, and determine whether the object is oriented.

21  Exercises

(14) For each of the following objects, count the number of proper symmetries, count the number of improper symmetries, and determine whether the object is oriented.

(15) The triskelion, which is the symbol of the Isle of Man, is shown in Figure 26. How many rotation symmetries does it have? How many reflection symmetries? Is it oriented? List the angles of its rotation symmetries. (16) You have infinitely many identical copies of the triskelion (from the previous exercise) from which you wish to build a wallpaper pattern. Describe (a) an arrangement that gives a wallpaper pattern with no rotation symmetries
(other than 0 ), and
(b) an arrangement that gives a wallpaper pattern with 120 rotation symmetries. (17) Draw a border pattern whose symmetries include horizontal but not vertical reflections. (18) Draw a wallpaper pattern whose symmetries include vertical but not horizontal reflections. (19) I have a bucket full of balls. Every ball is either red or green. Every ball has a number between 1 and 10 painted on it. Consider the statement “Every green ball has an odd number.” The contrapositive of this statement is . . . (choose one). . . Figure 26: The triskelion

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Chapter 1 • Introduction to Symmetry

(a) Every odd-numbered ball is green. (b) Every red ball has an even number. (c) Every even-numbered ball is red. Draw an example bucket for which the statement is true. Verify that its contrapositive is also true. Draw an example bucket for which the statement is false. Verify that its contrapositive is also false. (20) In a card game, I am dealt a hand of seven cards from a standard deck. Consider the statement “Every black card in my hand is a face card.” The contrapositive of this statement is . . . (choose one). . . (a) Every non-face card in my hand is red. (b) Every face card in my hand is black. (c) Every non-face card in my hand is black. Draw an example hand for which the statement is true. Verify that its contrapositive is also true. Draw an example hand for which the statement is false. Verify that its contrapositive is also false.
(21) What is the contrapositive of this statement? “If an object has a 10 rotation
symmetry, then it has a 20 rotation symmetry.” (22) What is the contrapositive of this statement? “If a rigid motion of the plane which is not the identity has a fixed point, then it is not a translation.” (23) What is the contrapositive of this statement? “If a rigid motion of the plane has exactly one fixed point, then it is a rotation.” (24) What is the contrapositive of this statement? “If an object has infinitely many symmetries, then it’s not a regular polygon.” (25) Do you think that a circle can be oriented? In other words, do you think there is an oriented object that has the same proper symmetries as a circle? Guessing is fine— you do not need to prove your answer. (26) Count the proper and improper symmetries of each star pictured in Figure 27. Which are oriented? (27) Perform a web image search for “M.C. Escher symmetry.” Many of Escher’s paintings are wallpaper patterns (if you imagine them infinitely extended). How

Figure 27: Three styles of stars for Exercise 26 (▶ https://doi.org/10.1007/000-25b)

23  Exercises

Figure 28: Images for Exercises 28 and 29

many different angles of rotational symmetry can you find among his wallpaper pattern paintings? (28) Consider infinitely many concentric squares filling the plane, each successive one a centimeter larger than the previous one. The first few are shown in Figure 28 (left). Is the resulting object . . . (a) (b) (c) (d)

bounded? a border pattern? a wallpaper pattern? oriented?

Explain your answers. How many symmetries does it have? (29) Consider the wallpaper pattern obtained by filling the whole plane with infinitely many copies of the letter O arranged in a regular grid extending indefinitely up, down, left, and right. Then choose a single row, and replace each O in this row with a red Q, as shown in Figure 28 (right). Is the resulting object . . . (a) a border pattern? (b) a wallpaper pattern? Explain your answers. (30) For each of the seven border patterns shown here (considered to extend indefinitely to the left and right), decide (a) whether it has any reflection symmetries over horizontal lines, (b) whether it has any reflection symmetries over vertical lines, and
(c) whether it has any 180 rotation symmetries. For which pattern is the following true: it has vertical reflection symmetries and
180 rotation symmetries, but none of its rotation centers lies on any of its vertical reflection lines?

1

24

1

Chapter 1 • Introduction to Symmetry

(31) For each of the four illustrated border patterns (considered to extend indefinitely to the left and right), decide (a) whether it has any reflection symmetries over horizontal lines, (b) whether it has any reflection symmetries over vertical lines, and
(c) whether it has any 180 rotation symmetries. For which pattern is the following true: it has vertical reflection symmetries
and 180 rotation symmetries, but none of its rotation centers lies on any of its vertical reflection lines?

(32) Describe the symmetries of a single point (a single drop of paint). (33) The bounded object pictured in Figure 25 has a unique center point. Draw

25  Exercises

Figure 29: Two wallpaper patterns for Exercise 34

▬ a bounded object of which every point of the plane qualifies as a center point, and ▬ a bounded object of which every point on the x-axis qualifies as a center point. (34) For each wallpaper pattern illustrated in Figure 29, (a) (b) (c) (d)

identify some reflection lines, if there are any;
identify some 120 rotation points, if there are any;
identify some 60 rotation points, if there are any; and decide whether the wallpaper pattern is oriented.

(35) What is wrong with the following?

Theorem If an object has at least one rotation symmetry besides the identity, then it has infinitely many rotation symmetries.

Proof Repeatedly applying the one rotation symmetry any number of times (once, twice, thrice, etc.) yields a rotation symmetry. So there are infinitely many. □

(36) There is a bucket full of balls. Every ball is either red or green. Every ball has a number between 1 and 10 painted on it. Suppose I know that the following statement is true: “Every red ball has an even number.”

Which of the following statements can I therefore conclude is guaranteed to be true?

1

26

1

Chapter 1 • Introduction to Symmetry

(a) (b) (c) (d) (e) (f)

Every odd-numbered ball is green. Every red ball has an even number. Every even-numbered ball is red. Every odd-numbered ball is red. All of the above statements are guaranteed to be true. None of the above statements is guaranteed to be true.

(37) There is a classroom full of students. Every student is either male or female. Every student either wears glasses or doesn’t wear glasses. Suppose I know that the following statement is true: “Every female student wears glasses.”

Which of the following statements can I therefore conclude is guaranteed to be true? (a) (b) (c) (d) (e) (f)

Every student who wears glasses is female. None of the male students wears glasses. Every student who does not wear glasses is male. Every student who wears glasses is male. All of the above statements are guaranteed to be true. None of the above statements is guaranteed to be true.

(38) Name a bounded object that has . . .


(a) a rotation symmetry by 36 .
(b) a rotation symmetry by 91 .

(c) a rotation symmetry by 45 and a rotation symmetry by 60 . (39) On page 4, we defined an object to mean a nonempty subset of the plane. How was the “nonempty” assumption used in the proof of the theorem on page 16? Would this theorem still be true if “nonempty” were removed from the definition of an object? (40) A kindergarten classroom has a collection of pet mice. Each mouse is either white or black (not both). Each mouse is either male or female (not both). Suppose I know that the following statement is true: “Every female mouse is white.”

Which of the following statements can I therefore conclude is guaranteed to be true? (a) (b) (c) (d)

Every white mouse is female. Every male mouse is black. Every black mouse is male. Every female mouse is black.

27  Exercises

(41) What is the contrapositive of the following statement about words in the English language? “If a word’s first letter is Q then its second letter is U.”





(42) What is the order of a 91 rotation? What is the order of a 95 rotation? Explain your answers.

1

2

29

The Algebra of Symmetry

You learned as a child how to count, and thus became familiar with the numbers, but that wasn’t nearly enough. To really understand numbers, you also needed to learn how to perform algebraic operations with them, namely, addition and multiplication. These are ways of combining two numbers to get a number back as the answer. These algebraic operations greatly expanded your ability to understand and work with numbers. It’s the same with symmetries. In the previous chapter, you listed all of the symmetries of an object (like a triangle or a square). But just listing them is not nearly enough. To really understand and work with them, you will need to learn a crucial algebraic operation called “composition.” It is a way of combining two symmetries to get a symmetry back as the answer. The composition of two rigid motions just means one followed by the other. If they are both symmetries of the object you are studying, then so is their composition. This chapter is about the surprisingly profound consequences of that simple observation.

2A Composition

Definition If A and B are rigid motions, then A
B denotes the rigid motion obtained by first performing B and then performing A. It is called “the composition of A with B ” or “A composed with B.”

Notice that the motion written on the right of the
symbol, namely B, is performed first. The order is the opposite of what you might have expected. It’s important to keep this straight, because the order sometimes matters. In the illustrations shown in Figure 30, T is  the translation encoded by the vertical arrow, and R45 is the 45 counterclockwise rotation around the origin. The composition R45
T (left image) does not equal the composition T
R45 (right image). The gnome helps you see the difference. From his starting position

Electronic supplementary material The online version of this chapter (https://doi.org/10.1007/ 978-3-030-51669-7_2) contains supplementary material, which is available to authorized users. The videos can be accessed individually by clicking the DOI link in the accompanying figure caption or by scanning this link with the SN More Media App.

© Springer Nature Switzerland AG 2021 K. Tapp, Symmetry, Texts for Quantitative Critical Thinking, https://doi.org/10.1007/978-3-030-51669-7_2

29

30

Chapter 2 • The Algebra of Symmetry

2

y

y R45

T T R45 x

x

Figure 30: Order sometimes matters! Here R45
T (left image) does not equal T
R45 (right image) (▶ https://doi.org/10.1007/000-25f)

standing on the x-axis, the two orders put him in different ending positions, so they must be different rigid motions: R45
T 6¼ T
R45. In these images, the rigid motions are applied to the whole plane, and the gnome goes along for the ride because he’s painted onto the plane. But the x-axis and y-axis remain fixed; they are not painted onto the plane. So even if R45 is the second motion performed (as in the left image), its rotation center (the origin) is unaffected by the first motion. Similarly, even if T is the second motion performed (as in the right image), its direction (up) is unaffected by the first motion. The information given in order to prescribe a rigid motion (the center of a rotation, the direction of a translation, the line of a reflection) is always given with respect to fixed, never-changing x- and y-axes, and is therefore unaffected by prior rigid motions. There are also pairs of rigid motions for which the order does not matter. For example, when you compose two rotations about the same point, the order doesn’t matter (think about why). Nor does the order matter when you compose a horizontal translation with a reflection over a horizontal line; as the image in Figure 31 shows, both options move the gnome from his bottom-left starting position to the same top-right ending position. There is a special name for this type of composition: it is called a “glide reflection.”

Definition A glide reflection means the composition of a translation (other than the identity) with a reflection over a line that is parallel to the direction of the translation.

It’s important here that the reflection line be parallel to the direction of the translation, as it is in Figure 31 where both are horizontal. If you were to compose a horizontal translation with a reflection over a diagonal line, then the order would matter. There

31  2B

Cayley Tables

y

Reflection line

x Figure 31: A glide reflection (▶ https://doi.org/10.1007/000-25d)

would be two different ending positions for the two choices of order, and the composition (in either case) would not be called a glide reflection.

2B Cayley Tables If A and B are both symmetries of an object (say, a triangle or rectangle or wallpaper pattern), then A
B is also a symmetry of that object. Thus, composition is an algebraic operation on the collection of symmetries of that object. It combines two symmetries of the object and gives an answer that is a symmetry of the object. We can therefore make “composition tables” analogous to addition and multiplication tables. Let’s start with a square. Its 8 symmetries are fI, R90 , R180 , R270 , H, V, D, D0 g: Here I means the identity, H means the horizontal reflection, V means the vertical reflection, D means the first of the two diagonal reflections, and D0 (or “D prime”) means the other diagonal reflection. Here and throughout the book, R means a counterclockwise rotation by the subscripted angle. Figure 32 shows the effects of these 8 symmetries on a square that’s decorated with a gnome. Table 2 exhibits the composition of every pair of these symmetries. This table is called a Cayley table. You find a composition A
B in a Cayley table by locating A (the motion performed second) along the left edge and locating B (the motion performed first) along the top edge. For example, the yellow cell in the table says that H
R90 ¼ D. That is, first performing R90 and then performing H results in the same final position as D, as illustrated in Figure 33. Notice that the H-reflection line does not rotate along with R90 because the notion of “horizontal” is relative to fixed x- and y-axes. We strongly recommend that you pause now to confirm all of the cells of the Cayley table in Table 2. Use a cardboard square, or print a gnome-square from the resource

2

Chapter 2 • The Algebra of Symmetry

32

2 I

H

R90

V

R180

D

R270

D⬘

Rotations

Reflections

Figure 32: The 8 symmetries of a square

Table 2: The Cayley table

∗ 90 180 270

90

180

270

⬘

180

270

⬘

90

90 180

270

180

270

270

90

⬘ ⬘

⬘

90

⬘ ⬘

⬘ ⬘

180

⬘

180 180 270

90

90

270

270

90 180

33  2B

Cayley Tables

D R90

H

Figure 33: H
R90 ¼ D (▶ https://doi.org/10.1007/000-25e)

website of this book. This site also has a gnome-triangle and a blank Cayley table for the 6 symmetries of a triangle; we recommend that you take time now to complete it. The Cayley table provides a wealth of information about the symmetries of the square. What patterns do you see? In particular, think about: 1. 2. 3. 4.

What is the significance of the grey shading? Does order matter? How does the Cayley table remind you of Sudoku? Why were the identity row and column so simple to complete?

▶ Exercise

Solve for X in the following equation: X
R270 ¼ H.

Solution Scan down the R270 column of the Cayley table until you find H. The row in which you found it is D, so the answer is X ¼ D.

▶ Exercise

What symmetry of the square must be performed before R90 to get D0?

Solution This is the same as solving the equation R90
X ¼ D0 for X. The solution is to scan across the R90 row of the Cayley table until you find D0. The column in which you found it is H, so the answer is X ¼ H.

Each of the previous two questions had a unique answer. Why? Because when you scan a particular row or column of the Cayley table looking for a particular symmetry, you always find it, and you only find it once. In other words, the Cayley table is like a Sudoku game. We will learn soon that there’s nothing special about the square. Start with any object, and its collection of symmetries work the same way; they form the kind of system in which equations can be uniquely solved. There is a unique way to solve for X in any equation of the form: X
A ¼ B or of the form A
X ¼ B. We can solve algebra equations involving symmetries! That’s why this chapter is titled “The Algebra of Symmetry.” How will we prove that the symmetries of any object form this kind of system? With abstraction! First, we will think of other systems in which equations can be uniquely solved. One such system is addition of integers: an equation like X + 4 ¼ 10 has a unique integer solution. Another is multiplication of nonzero real numbers: an equation like X  5.7 ¼ 67.97 has a unique real-number solution. Second, we will identify the

2

34

2

Chapter 2 • The Algebra of Symmetry

underlying properties these systems possess that make them work like this. In mathematics, the word for a system possessing these properties is “group.” The technical definition of a group abstracts the features that the above-mentioned systems have in common, so it turns out to roughly mean a system in which equations can be uniquely solved. Finally, we will verify that the collection of symmetries of any object qualifies as a group.

2C Groups To begin the abstraction, let’s first review the familiar algebraic properties of multiplication and addition. ▬ The order in which a pair of numbers is added (or multiplied) does not affect the result. This is called the commutative property. In symbols: A þ B ¼ B þ A and A  B ¼ B  A:

▬ The order in which a pair of additions (or a pair of multiplications) is performed does not affect the result. This is called the associative property. In symbols: ðA þ BÞ þ C ¼ A þ ðB þ CÞ and ðA  BÞ  C ¼ A  ðB  CÞ:

▬ Adding 0 to a number has no effect. Multiplying a number by 1 has no effect. In symbols:

A þ 0 ¼ 0 þ A ¼ A and 1  A ¼ A  1 ¼ A:

We call 0 and 1 the identities; 0 is the additive identity and 1 is the multiplicative identity.

▬ For any number A, the sum of A and A equals 0, the additive identity. Similarly, for any nonzero number A, the product of A and its reciprocal 1/A equals 1, the multiplicative identity. We call A the additive inverse of A, and 1/A the multiplicative inverse of A. Composition of symmetries shares all of these features except for the commutative property. Do you see the similarities? Imagine three friends, Adam, Michelle, and Chris. 1. Adam likes to add integers. Adam has a special integer, 0, that has no effect when he adds it to other numbers. That’s his identity. He can find any integer’s “inverse” (the thing that adds to it to give his identity). For example, the inverse of 35 is 35. 2. Michelle likes to multiply nonzero real numbers. Michelle has a special number, 1, that has no effect when she multiplies it by other numbers. That’s her identity. She can find any number’s “inverse” (the thing that multiplies with it to give her identity). For example, the inverse of 35 is 1/35. 3. Chris likes to compose symmetries of the square. Chris has a special symmetry, I, that has no effect when he composes it with other symmetries. That’s his identity. He can find any symmetry’s “inverse” (the thing that composes with it to give his identity). For example, the inverse of R90 is R270, while the inverse of H is H.

35  2C

Groups

What do these three stories have in common, aside from corny alliteration? Each of these three friends is working with a system that mathematicians refer to as a “group.” In mathematics, the word “group” has a very specific technical meaning designed to capture the commonalities between the systems studied by Adam, Michelle, and Chris in the above stories. All three systems involve a set (which roughly means a collection) together with an algebraic operation (which combines any pair of members of the set and gives back a member of the set as the answer). In all three systems, the algebraic operation satisfies the three properties set forth in the definition of “group.”

Definition A group is a set (denoted G) with an algebraic operation on that set (generically denoted ) that satisfies the following properties:

(1) There is a member of G, called its identity, that has no effect on other members: A  identity ¼ A, and identity  A ¼ A for each member, A, of G. ~ which means a member of (2) Each member, A, of G has an inverse (denoted A), G that combines with it to give the identity: A  A~ ¼ identity, and A~  A ¼ identity:

(3) The associative property holds: (A  B)  C ¼ A  (B  C) for every triple A, B, C of members of G. If the commutative property also holds, which means A  B ¼ B  A for all pairs A, B of members of G, then we call G a commutative group; otherwise we call G a noncommutative group.

Think of “” as a generic symbol that Adam replaces with “+”, Michelle replaces with “” and Chris replaces with “
”. Example Adam likes the infinite set of all integers, f. . . , 3, 2, 1, 0, 1, 2, 3, . . .g with the algebraic operation of addition. This is a commutative group, called the additive group of integers. The identity is 0, and the inverse of any integer A is A; that is, A~ ¼ A. We will henceforth denote this group as “ℤ” after the German word for integer, Zahl. Non-Example The set of all real numbers with the algebraic operation of multiplication is not a group. It does have an identity, namely 1, but it includes one poorly behaved member, namely 0, that doesn’t have an inverse: no number times zero equals 1.

This minor problem is solved by removing zero from the set, yielding the next example.

2

36

2

Chapter 2 • The Algebra of Symmetry

Example Michelle likes the set of all nonzero real numbers with the algebraic operation of multiplication. This is a commutative group. The identity is 1, and the inverse of any member A is its reciprocal 1/A; that is, A~ ¼ 1=A. Non-Example Is the set of integers a group with the algebraic operation of division? This question doesn’t make sense because division is not an algebraic operation in this context. More precisely, division is not an algebraic operation on the set of integers. Most integer divisions, like 5/7, do not result in integers, while others like 4/0 are undefined. A valid algebraic operation on a set must be a method of combining each pair of members of the set to always give an answer that’s a member of the set. Non-Example The set of integers with the algebraic operation of multiplication is not a group. Although it has an identity, namely 1, most members do not have inverses. For example, the integer 3 does not have an inverse because no integer times 3 equals 1.

A “group” is defined to essentially mean a system with the underlying algebraic properties needed to allow equations to be uniquely solved. Using the precisely stated properties given in the definition above, we can put this statement in the form of a theorem.

The Unique Solution Theorem Suppose G is a group and A, B are members of G. Then the equation X  A ¼ B has a unique solution, and so does the equation A  X ¼ B.

“Unique solution” means there is one and only one member of the group that can be substituted for X to make the equation true. Proof ~ This works because The solution to X  A ¼ B is X ¼ B  A.

B  A~  A ¼ B  A~  A ¼ B  identity ¼ B: Similarly, the solution to A  X ¼ B is X ¼ A~  B. The claim that these solutions are unique is left to the exercises. □

Notice that the above proof uses all of the defining properties of a group, including the associative property. One can make a Cayley table for any finite group. It works just like Table 2 on page 32, so that A  B is the member of the group given at the intersection of the row of A and the column of B. The fact that equations have unique solutions means that every Cayley table will have the “each appears once” feature of a Sudoku game, which gives us the name for the following theorem.

37  2D

Symmetry Groups

The Sudoku Theorem Every member of a finite group appears exactly once in each row and each column of its Cayley table.

Proof Finding a particular member, B, of the group in the column of another particular member, A, is the same as solving the equation X  A ¼ B. According to the previous theorem, there is one and only one solution. Thus, B appears exactly once in the column of A. By similar reasoning, B also appears exactly once in the row of A. □

2D Symmetry Groups Chris likes the 8 symmetries of a square, which form a noncommutative group. It is noncommutative because, for example, H
R90 6¼ R90
H, as seen in Table 2 on page 32. There is nothing special about a square; the collection of symmetries of any object forms a group!

Theorem The collection of symmetries of any object is a group with the algebraic operation of composition.

This group is called the symmetry group of that object. Idea of Proof The symmetry group’s identity can be identified as the rigid motion that we previously called “the identity,” because composing with it leaves every other rigid motion unchanged. Every rigid motion of the plane has an inverse, which means a rigid motion that undoes it (composes with it to give the identity). Although we are not yet equipped to give a formal proof of this fact, it should seem believable. No matter what rigid motion I apply to the plane, you can always move it back to its starting position. If I rotate 27 degrees clockwise, then you can rotate 27 degrees counterclockwise. If I flip and then translate, then you can translate back and then flip back. If the original motion is video recorded, then its inverse can be visualized as running the video backwards. If a rigid motion is a symmetry of an object, then so is its inverse. It remains to verify the associative property. If A, B, and C are rigid motions, it is difficult to visualize why (A
B)
C is the same rigid motion as A
(B
C). The trick is to study their effects on any single point of the plane (any single drop of paint on the infinite glass wall). Both (A
B)
C and A
(B
C) affect a single point by first performing C, then B, then A. They have the same effect on each single point, so they must be the same rigid motion. □

2

Chapter 2 • The Algebra of Symmetry

38

2

The symmetry groups of the regular polygons and oriented polygons are so important, we have special names and symbols for them:

Definition Suppose that n  2. The symmetry group of a regular n-sided polygon is denoted as Dn and is called the nth dihedral group. The symmetry group of an oriented n-sided polygon is denoted as Cn and is called the nth cyclic group.

We will generally use bold typeface for names of groups and non-bold for members of groups. Recall from Chapter 1 that Dn has n rotations and n reflections, while Cn has n rotations and no reflections (the identity counts as a rotation). This pattern can be extended to n ¼ 1 by defining D1 and C1 to mean the symmetry groups of the smiley face and gnome respectively, as shown in Figure 34. The Cayley table for D4, the symmetry group of a square, is Table 2 on page 32. Notice that C4 ¼ {I, R90, R180, R270} is the symmetry group of an oriented square, which is the same as the rotation-only symmetries of a regular square, so the Cayley table for C4 is just the upper left quadrant of the Cayley table for D4, reprinted as Table 3. More generally, the Cayley table for Cn is the upper left quadrant of the Cayley table for Dn, provided it is arranged (like D4’s) with rotations first followed by reflections.

Its symmetry group is Dn (n rotations, n reflections)

n=1

n=2

n=3

n=4

n=5

n=6

n=7

Its symmetry group is Cn (n rotations, NO reflections)

Figure 34: Objects whose symmetry groups are Dn and Cn Table 3: The Cayley table for C4

∗

90

180

270

90

180

270

270

90

90

180

180

180

270

90

39  2E

One Reflection Is Enough

2E One Reflection Is Enough The main statement presented in this chapter so far is that the symmetries of any object form the type of system in which equations can be solved, namely, a group. This new group viewpoint is indispensable for studying symmetry. Building upon this idea, the remainder of this chapter exemplifies mathematics at its best: ideas building upon ideas and culminating in powerful new discoveries. We begin by formalizing a pattern that you might have conjectured in Chapter 1. None of the objects we studied had mismatched counts, like, say, 5 proper and 7 improper symmetries. That’s because, as the following theorem shows, mismatched counts are impossible.

The Zero-or-Equal Theorem The number of improper symmetries of an object either is zero or is equal to the number of proper symmetries.

So if an object has 5 proper symmetries, then it must have either zero or 5 improper symmetries. If an object has infinitely many proper symmetries, we interpret the theorem as saying that it must have either zero or infinitely many improper symmetries. Let’s warm up to the proof with a partially true technical puzzler. I used Microsoft Word to compile the gnome-square image first presented in Figure 32, a portion of which is reproduced in Figure 35. Constructing the horizontally reflected image was easy because the software has menu options for performing vertical and horizontal reflections. But how did I create the diagonally reflected image? Imagine you work for Microsoft tech support: YOU: Technical support, how may I help you? ME: I’m writing a Symmetry book, and I need to reflect my gnome-square over the diagonal, but your software only has choices for horizontal and vertical reflection. What should I do?

H

D

Figure 35: How do I reflect diagonally?

2

40

Chapter 2 • The Algebra of Symmetry

2 =H I

H

R90

= D⬘

R180 =V

I

H

R90

D⬘

R180

V

R270

D

*

I

R90

R180 R270

H

V

D

I

I

R90

R180 R270

H

V

D

D⬘

R90

R90

I

D⬘

D

H

V D

R180 R270

R180 R180 R270

R270 =D

D⬘

I

R90

V

H

D⬘

R270

R270

I

R90

R180

D

D⬘

V

H

H

H

D

V

D⬘

I

R180

R90

R270

V

V

D⬘

H

D

R180

I

R270

R90

D

D

V

D⬘

H

R270

R90

I

R180

D⬘

D⬘

H

D

V

R90

R270 R180

I

Figure 36: One reflection is enough because of the Sudoku Theorem!

How would you reply? Since the software allows rotations by arbitrary angles, the key is to compose the horizontal reflection with rotations: YOU: Just compose the horizontal reflection with the square’s four rotation symmetries. This yields the square’s four reflection symmetries. Figure 36 shows these compositions. The software punch line is: one reflection is enough. Microsoft didn’t even need to provide two reflection options because (together with rotations) one is enough to get every reflection. The math punch line is: composing with H gives a one-to-one matching between the rotation symmetries and the reflection symmetries. The matching is depicted on the scroll in Figure 36. The left column of the scroll lists all the rotation symmetries. The right column shows the result of composing each of these rotations with H. How could you have known in advance that the right column would turn out to list all of the reflection symmetries with each listed exactly once? Because of the Sudoku Theorem! The right column is a copy of the improper half of the H column of the Cayley table, as illustrated by the Cayley table within Figure 36. We have stumbled upon the key idea needed to prove the theorem.

2

41  2F

An Improved Classification of Rigid Motions

Figure 37: A single improper symmetry, F, induces a one-to-one matching between all of the object’s proper symmetries (left) and all of its improper symmetries (right)

P1

P1*F

P2 P3

P2*F P3*F

. . .

. . .

P4

P4*F

Proof Of The Zero-Or-Equal Theorem Suppose that the object has at least one improper symmetry. Choose one and call it F. Our job is to prove that the object has equal numbers of proper and improper symmetries. We will prove this under the added hypothesis that it has finitely many proper symmetries (if it has infinitely many, we will save for Chapter 11 the proof that it therefore also has infinitely many improper symmetries). On the left column of a scroll of ancient parchment, list all of the proper symmetries. In Figure 37, these are denoted P1 , P2 , etc. On the right column, list the result of composing each of these proper symmetries with F. The right column is actually a list of all the improper symmetries. The scroll is therefore a one-to-one matching of the proper and improper symmetries, so there are equal numbers of each. Why are the symmetries in the right column all improper? Because an improper symmetry composed with a proper symmetry is always improper. Why does every improper symmetry appear somewhere in the right column, with no repetitions? Because every symmetry appears exactly once in F’s column of the Cayley table. □

2F An Improved Classification of Rigid Motions The fact that any object’s symmetries form a group has powerful consequences. In the previous section, we used this fact (and the Sudoku property of Cayley tables) to prove the Zero-or-Equal Theorem. In this section, we will use it to strengthen our classification of rigid motions. When we constructed the Cayley table for D4 (Table 2 on page 32), the gnome image helped us detect rigid motions. The eight symmetries of the square all affected the gnome differently, so his ending position revealed which symmetry had been applied. The gnome was a good choice because he’s asymmetric. In fact, any asymmetric object is capable of differentiating, not just one symmetry from each of the other seven symmetries of the square, but any rigid motion from any other rigid motion.

42

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Chapter 2 • The Algebra of Symmetry

The Rigid-Motion Detector Theorem If an object is asymmetric, then it distinguishes rigid motions; that is, any two different rigid motions have different effects on it.

To understand this theorem, let’s first think about why it’s not true for a symmetric object, like, say, an equilateral triangle. Suppose you close your eyes and then reopen them to discover that the triangle has moved 3 inches to the right. From this, you cannot tell what motion I performed while your eyes were closed. I might have translated 3 inches to  the right, or I might have rotated 120 and then translated 3 inches to the right. You have no way of knowing. The rotation part is undetectable because it’s a symmetry of the triangle. The following proof shows that, if the triangle in this story is replaced by an asymmetric object like a gnome, then there couldn’t be any undetectable part. For a visualization of the difference asymmetry makes, see the movie linked to Figure 38. Proof We will prove the contrapositive: If there are two different rigid motions that have the same effect on the object, then the object is not asymmetric. Suppose that two different rigid motions, called A and B, have exactly the same effect on the object. This implies that A~
B is a symmetry of the object that’s different from the identity, so the object is not asymmetric. □

For variety, we’ll detect rigid motions with a baby photo instead of a gnome in our next story. Since the baby photo is asymmetric, the previous theorem implies that there is only one rigid motion that moves it from its bottom-right position to its top-left position in  Figure 39. It seems to have been rotated counterclockwise by 27 (you could measure this angle with a protractor) and then translated up and left. In fact, you could achieve that  same repositioning of the photo by only doing one thing: rotating by 27 about the correctly chosen point. How do you find this point? Figure 40 shows how. First, choose a distinguished point on the photo, like the nose, and draw a red line connecting the two noses. Next, draw a green perpendicular bisector. The point we seek lies on this green line, and it is chosen so that the illustrated blue hinge   measures exactly 27 . Rotating the plane by 27 about this special point will achieve the

End position

Start position

Figure 38: Video explanation of the Rigid Motion Detector Theorem (▶ https://doi.org/10.1007/000-25c)

43  2F

An Improved Classification of Rigid Motions

illustrated repositioning of the baby photo, since it moves the nose to the proper place and it tilts the photo the proper amount! If you don’t own a protractor, there are alternative methods you could use to solve this problem. For example, you could select several distinguished points (nose, earlobe, eyelash), and locate the intersection of all of the corresponding green lines. But with our protractor method, we can easily explain why the method works. As you slide down the green line, there is clearly a unique point where the blue-hinge angle will change from too large to too small. Animations to help you visualize this discussion are available in the PowerPoint presentation for Chapter 2 (and in the video linked to Figure 40). We now have the key idea for proving the following improved version of the theorem on page 13. Figure 39: How did it move?

27°

More than 27°

Exactly 27° Less than 27°

Figure 40: Finding the rotation point (▶ https://doi.org/10.1007/000-25g)

2

44

Classification of Plane Rigid Motions (Version 2) Every proper rigid motion of the plane is a translation or a rotation. Every improper rigid motion of the plane is a reflection or a glide reflection.

Proof Our starting point is version 1 of the classification theorem in Chapter 1. So, for the claim about proper rigid motions, all we must prove is this: A rotation (other than the identity) followed by a translation can be re-described as a single rotation. We’ll use the baby photo (or any other asymmetric image) as our rigid-motion detector. A rotation followed by a translation has the same effect on the baby photo as a single rotation whose center is found using the previously explained protractor method. Since it has the same effect on the baby photo, it must be the same rigid motion! For the claim about improper rigid motions, all we must prove is this: A reflection followed by a translation (other than the identity) can be re-described as a glide reflection. For example, in Figure 41, the translation direction is not parallel to the reflection line. We must identify a new reflection and a new translation such that the direction of the new translation is parallel to the new reflection line. These new motions must be chosen so that the resulting composition (which is a glide reflection) has the same effect on the gnome. We leave it to the reader to explain how to make these choices. HINT: The new reflection line should be vertical and centered between the gnome’s starting and ending positions. □

Reflection line

2

Chapter 2 • The Algebra of Symmetry

e

lat

ns

Tra

Flip

Figure 41: Translation
reflection ¼ glide reflection

45  2G

Elements of Mathematics: The Counterexample

2G Elements of Mathematics: The Counterexample We have already repeatedly emphasized the chapter’s main mathematical theme: the crucial value of abstraction. The Latin root of the word abstraction means to draw out, as in “the abstraction of metal from ore.” Our definition of group was set up to abstract (pull out) the essential properties of familiar systems (addition of integers, multiplication of nonzero real numbers)—namely, those properties which ensure that equations can be uniquely solved. This definition then guided our way towards proving that symmetries work like that too. The resulting “symmetry group” viewpoint allowed us to prove several important new theorems. With these theorems under our belt now, it’s a good time to ponder this question: before you spend time trying to prove something, how do you know it has a good chance of being true? What if it’s false? In mathematics, it is equally important to know how to disprove a statement, which means demonstrating that it is false. In practice, disproving is often easier than proving. Mathematical statements often have the form “All members of a certain set have a certain property.” Disproving such a statement just requires you to find a single counterexample, which means a single member of the set that does not have that particular property.

▶ Exercise

Disprove the statement: “All integers are odd.”

Solution The number 42 is a counterexample. Comment Of course there are lots more non-odd (even) integers, but one counterexample is enough to disprove a statement like this.

There is nothing deep here. You use this type of logic frequently in day-to-day life. If your grandpa claims that all senators are male, you only need to name one counterexample (one female senator) to show him he’s wrong. We’re just formalizing commonsense reasoning. The formalization will be helpful as the problems get harder.

▶ Exercise

Disprove the statement: “If an object is bounded, then it has only finitely many symmetries.” Solution A circle is a counterexample. It is bounded, yet it has infinitely many symmetries.

For a statement of the form “If a member of a set has property A, then it has property B,” a counterexample means a member of the set that has property A but not property B. This was illustrated in the previous example about the set of all objects, where property A was “being bounded” and property B was “having finitely many symmetries.”

2

46

2

Chapter 2 • The Algebra of Symmetry

▶ Exercise

Disprove the statement: “The symmetry group of the illustrated border pattern is commutative.”

Solution A counterexample to the commutative property means a pair of symmetries for which the order matters. Let F denote the reflection over the illustrated dashed blue line. Let T denote the translation determined by the illustrated red arrow (about an inch to the right).

We claim that F
T 6¼ T
F. The border pattern itself can’t detect the difference because F
T and T
F are both symmetries of it, but the gnome detects the difference in the following illustrations. From the same starting position, T
F repositions him to the right while F
T repositions him to the left. A single drop of paint would have detected the difference between the two orders just as well as the gnome.

2

47  Exercises

F1

I

R120

F2

R240

F3

*

I

R120

R240

I R120 R240 F1 F2 F3

F1

F2

F3

F2

Figure 42: Symmetries of the triangle and Cayley table for its symmetry group

Exercises (1) What is the size (the number of members) of C27? What about D27? (2) If an object has 19 proper symmetries, what are the possibilities for its number of improper symmetries? (3) If an object has 19 improper symmetries, what are the possibilities for its number of proper symmetries? (4) Write a precise definition of each of the following terms: composition, glide reflection, group, commutative, symmetry group, cyclic group, dihedral group, ℤ. (5) Fill in the Cayley table for the dihedral group D3, which is the symmetry group of the triangle. See Figure 42. It is helpful to use the gnome-triangle provided on the book’s resource website. (6) Is D3 a commutative group? (7) Use the Cayley table for D3 to solve the following: (a) (b) (c) (d) (e) (f) (g)

F2
F3 ¼ _______. _______
F3 ¼ R120. R240
_______ ¼ F1. Which symmetry must be performed after F2 to yield R120? Which symmetry must be performed before F1 to yield R240? What is the inverse of F2? What is the inverse of R240?

(8) Construct a Cayley table for D1 = {I, V}. Is D1 a commutative group? (9) Construct a Cayley table for D2 = {I, R180, H, V}. Is D2 a commutative group? (10) Construct a Cayley table for Cn for each n ¼ 1, 2, 3, 4, 5. Describe any patterns and similarities you see. Explain why all cyclic groups are commutative. (11) Use the Cayley table for D4 (Table 2 on page 32) to solve the following: (a) (b) (c) (d) (e)

H
D0 ¼ _______. _______
V ¼ R270. R180
_______ ¼ H. Which symmetry must be performed after H to yield R90? Which symmetry must be performed before D to yield R270?

48

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Chapter 2 • The Algebra of Symmetry

(f) What is the inverse of H? (g) What is the inverse of R270? (12) Describe a general method for using a Cayley table to determine the inverse of a member of a finite group. (13) Construct a Cayley table for D5. HINT: Construct a cardboard pentagon with labeled vertices. Use your pentagon to fill in some of the Cayley table, and then save time by using the Sudoku property to fill in the rest. (14) Explain why the inverse of any improper rigid motion is improper. (15) Consider the following definition:

Definition A glide reflection in the symmetry group of an object is called decomposable if the two rigid motions out of which it’s built (the reflection and the translation) separately are symmetries of the object. Otherwise it is called indecomposable.

For example, the border pattern illustrated in Figure 43 has an indecomposable glide-reflection symmetry, namely, the reflection over the horizontal blue center line followed by the translation encoded by the red arrow.

Figure 43: Border pattern with indecomposable glide-reflection symmetry (▶ https://doi.org/10.1007/000-25h)

For each border pattern illustrated in Figure 44, decide whether it has decomposable and/or indecomposable glide reflections in its symmetry group. (16) Does the G border pattern (an infinite strip of Gs) have a commutative symmetry group?

What about the C border pattern? What about Z? What about Y? Explain your answers. (17) Does the border pattern illustrated here have a commutative symmetry group? Explain your answer.

(18) Do the 4 proper symmetries of a square form a group with the algebraic operation of composition? What about the 4 improper symmetries? Explain your answers.

49  Exercises

Figure 44: Border patterns for Exercise 15

(19) Is the set of even integers a group with the algebraic operation of addition? What about the set of odd integers? Explain your answers. (20) Is the set of positive integers {1, 2, 3, 4, . . .} a group with the algebraic operation of addition? What about the set of nonnegative integers {0, 1, 2, 3, 4, . . .}? Explain your answers. (21) What does the scroll in the proof of the Zero-or-Equal Theorem look like if the object is a square and F ¼ D? What if the object is a triangle and F ¼ F1? (22) Define the algebraic operation “¤” on the set of integers so that A¤B equals one more than the product of A and B. (a) Does this algebraic operation satisfy the associative property? (b) Is there an identity? (23) What is wrong with this statement? “The integers are a group with the algebraic operation of subtraction.” (24) What type of symmetry results from the composition of two reflections if (a) the two reflection lines are parallel? (b) the two reflection lines intersect? Explain your answers. HINT: In each case, use the classification theorem on page 44 to limit the possibilities.

2

50

2

Chapter 2 • The Algebra of Symmetry

(25) Consider the statement “If a bounded object has more than one reflection symmetry, then it has more than one rotation symmetry.” (a) Explain why this statement is an immediate consequence of the Zero-or-Equal Theorem. (b) Without using the Zero-or-Equal Theorem, prove this statement by instead using the Center Point Theorem and the solution to the previous exercise. (26) Prove that a group has only one identity; that is, there is only one member that has no effect on other members. (27) Prove that a member of a group has a unique inverse; that is, there is only one member of the group that combines with it to give the identity. (28) Prove the uniqueness claim of the Unique Solution Theorem on page 36. (29) Disprove the following statement: “If a real number x satisfies the inequality |x  7| > 10 then it also satisfies the inequality x  7 > 10.” (30) Disprove the following statement: “If an integer is a multiple of 3 then it is a multiple of 5.” (31) Disprove the following statement: “If a border pattern has glide-reflection symmetries, then it has reflection symmetries.” (32) Disprove the following statement: “The symmetry group of every wallpaper pattern is commutative.” (33) Disprove the following statement: “The symmetry group of every wallpaper pattern is noncommutative.” (34) Consider the statement “If an object is unbounded, then it has infinitely many symmetries.” (a) Disprove this statement. (b) Write the contrapositive of this statement. (c) Is there a difference between a counterexample to the statement and a counterexample to the contrapositive of the statement? (35) Consider the statement “If an object has any reflection symmetries, then it has a noncommutative symmetry group.” (a) Disprove this statement. (b) Write the contrapositive of this statement. (c) Is there a difference between a counterexample to the statement and a counterexample to the contrapositive of the statement? 



(36) If a bounded object has a 95 rotation symmetry, prove that it also has a 5 rotation  symmetry. If a bounded object has a 91 rotation symmetry, prove that it also has a  1 rotation symmetry.

51  Exercises

(37) Suppose you have a bounded object with finitely many symmetries. Let α denote the smallest angle (other than zero) of a rotation symmetry of this object. 

(a) Give an example where α ¼ 15 .  (b) Explain why α could not equal 95 . (c) Characterize the possibilities for α. 

(38) If 10 is the smallest angle (other than zero) of a rotation symmetry of a bounded  object, prove that the object does not have a 37 rotation symmetry. List all of the object’s rotation angles and explain why your list is complete. (39) Suppose you have a bounded object with finitely many symmetries. Let α denote the smallest angle (other than zero) of a rotation symmetry of this object. Prove that α divides evenly1 into 360 and that the multiples of α are the object’s only rotation symmetry angles. HINT: Refer to the previous three exercises. (40) Prove that if the total number of symmetries of an object is odd, then the object must be oriented. HINT: Prove the contrapositive. (41) Let F denote a reflection over a vertical line, and let T denote a translation L inches to the right. (a) Prove that T
F is a reflection over a vertical line that is L/2 inches to the right of F’s reflection line. (b) Prove that F
T is a reflection over a vertical line that is L/2 inches to the left of F’s reflection line. HINT: Use the classification theorem on page 44 to limit the possibilities, and identify fixed points. (42) You marked some of the vertical reflection lines of the border pattern here as blue lines, but you missed some. How could the previous exercise help you know that you missed some? Draw the rest.



(43) Let R denote a 180 rotation and let T denote a translation L inches to the right. (a) Prove that T
R is a rotation whose center is L/2 inches to the right of R’s center. (b) Prove that R
T is a rotation whose center is L/2 inches to the left of R’s center. HINT: Use the classification theorem on page 44 to limit the possibilities, and identify the fixed point.

1

This means that 360/α is an integer (which makes sense even if α is not an integer).

2

52

Chapter 2 • The Algebra of Symmetry



2

(44) You marked some of the 180 rotation centers of the border pattern here as yellow dots, but you missed some. How could the previous exercise help you know that you missed some? Mark the rest.

(45) In the following diagram, let F denote the reflection over the vertical blue line, and  let R denote the 180 rotation around the yellow point (which is L inches to the right of the blue line).

L inches

(a) Prove that R
F is the glide reflection obtained by reflecting over the horizontal red line and translating 2L inches to the right. (b) Prove that F
R is the glide reflection obtained by reflecting over the horizontal red line and translating 2L inches to the left. 

(46) In this border pattern, you marked the 180 rotation centers as yellow dots and the vertical reflections as blue lines. How could the previous exercise help you understand the glide-reflection symmetries?

(47) There is a bucket full of balls. Every ball is either red or green. Every ball has a number between 1 and 10 painted on it. Suppose my friend Charlie makes the following assertion: “Every red ball has an even number.”

Which of the following would I need to find in the bucket in order to prove that Charlie is incorrect? (a) (b) (c) (d)

A green ball. A red ball. A green ball with an odd number. A green ball with an even number.

53  Exercises

(e) A red ball with an odd number. (f) A red ball with an even number. (48) There is a classroom full of students. Every student is either male or female. Every student either wears glasses or doesn’t wear glasses. Suppose my friend Charlie makes the following assertion: “Every female student wears glasses.”

Which of the following would I need to find in the room in order to prove that Charlie is incorrect? (a) (b) (c) (d) (e) (f)

A female student. A male student. A female student who wears glasses. A female student who does not wear glasses. A male student who wears glasses. A male student who does not wear glasses.

(49) Draw arrows matching each member of this cyclic group with its inverse: C5 ¼ fI, R72 , R144 , R216 , R288 g (50) Draw arrows matching each member of this cyclic group with its inverse: C6 ¼ fI, R60 , R120 , R180 , R240 , R300 g (51) Fill in the blank: If n is _______, then the cyclic group Cn has a member (other than the identity) that is its own inverse. (52) Is the set {1, 1} a group with the algebraic operation of multiplication? Explain your answer. (53) Since a regular (equilateral) triangle is not asymmetric, the conclusion of the RigidMotion Detector Theorem is not guaranteed to be true for it. How many different rigid motions are there that have the same effect of moving the illustrated triangle from its start to its end position? Prove your answer.

y

End

Start

x

2

3

55

The Classification Theorems

Leonardo Da Vinci’s self-portrait

Historically, the concept of symmetry evolved slowly from a vague idea to a precise notion as scientists and mathematicians sought to study the symmetry of their world using ever more precise language and methods. The classification theorems in this chapter represent some of the pinnacles of this journey.

Electronic supplementary material The online version of this chapter (https://doi.org/10.1007/ 978-3-030-51669-7_3) contains supplementary material, which is available to authorized users. The videos can be accessed individually by clicking the DOI link in the accompanying figure caption or by scanning this link with the SN More Media App.

© Springer Nature Switzerland AG 2021 K. Tapp, Symmetry, Texts for Quantitative Critical Thinking, https://doi.org/10.1007/978-3-030-51669-7_3

55

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Chapter 3 • The Classification Theorems

3A Rigid Equivalence

3

The goal of this chapter is to classify the ways in which 1. bounded objects, 2. border patterns, and 3. wallpaper patterns can be symmetric. What does this mean? In each of these three categories, our goal is to provide a list of model objects and a theorem that guarantees every other object in the category is “equivalent” to one of those model objects. But what should it mean for two objects to be “equivalent”? It should roughly mean that the two objects are symmetric in the same way, but how do we make this notion precise? We could require that the two objects have exactly the same collection of symmetries, but this meaning would be very restrictive. In the left image of Figure 45, the blue and orange squares do have exactly the same collection of symmetries. But in the
right image, they do not; for example, a 90 rotation of the plane about the center of the orange square would noticeably reposition the blue square. Yet even in the right illustration, you probably want to say, “Yes, the two squares are equivalent.” Who cares that they have different tilts and different centers? When you completed a square’s Cayley table in the previous chapter, the location and tilt of the desk on which you worked didn’t matter. One precise notion of equivalence that doesn’t care about location or tilt is called “rigid equivalence.”

Definition Two objects are called rigidly equivalent if there exists a rigid motion of the plane which, when applied to the first object, repositions it so that afterwards the two objects have exactly the same collection of symmetries.

In the right illustration in Figure 45, the orange and blue squares are rigidly equivalent. Imagine the blue square is drawn on a whiteboard while the orange square is drawn on a glass sheet (the plane) held in front of the whiteboard. There is a rigid motion of the

y

y

x

x

Figure 45: Left: The two squares have the same symmetries. Right: They do not (▶ https://doi.org/10.1007/000-25j)

3

57  3B

Bounded Objects

y

Figure 46: Are they rigidly equivalent?

x plane (a repositioning of the glass sheet) that slides and tilts the plane until the two squares become concentric and aligned and therefore have the same collection of symmetries.

▶ Exercise

Are the orange square and blue triangle in Figure 46 rigidly equivalent?

Solution No. Applying a rigid motion of the plane to reposition one on top of the other wouldn’t change the fact that the square has 8 symmetries while the triangle has 6.

In general, if two objects are rigidly equivalent, then they must have the same number of proper symmetries and the same number of improper symmetries (because these counts are unaffected by repositioning one on top of the other).

▶ Exercise

Are these two symbols rigidly equivalent?

Ψ ‘

Solution Yes. In addition to the identity, one symbol has a vertical reflection symmetry while
the other has a horizontal one. Rotating Ψ by 90 will align the reflection lines of the two symbols, after which they will have the same collection of symmetries.

We technically should have drawn the x-axis and y-axis around the two symbols in the previous example, because objects are subsets of the plane. But we felt it was OK to leave the axes out of the picture; after all, the axis location wouldn’t affect the conclusion. The whole point is that rigid equivalence is insensitive to location and tilt.

3B Bounded Objects The description of all possible symmetry types of bounded objects is attributed to Leonardo Da Vinci (1452–1519).

58

Chapter 3 • The Classification Theorems

Da Vinci’s Theorem Any bounded object with finitely many symmetries must be rigidly equivalent to one of these model objects:

3

The “. . .” symbols mean that the double list of model objects does not stop at 7, but goes on indefinitely. It’s the familiar list of regular polygons and oriented polygons (plus a smiley face and a gnome). Notice that Da Vinci’s Theorem tells you nothing about a bounded object with infinitely many symmetries. You might guess that such an object must be a circle, or at least must be rigidly equivalent to a circle, but there are some nuances here that we will not be equipped to discuss until Chapter 10. To prepare for the proof of Da Vinci’s Theorem, we recommend that you review the “one reflection is enough” idea from the proof of the Zero-or-Equal Theorem. We also recommend that you solve Exercise 39 from Chapter 2. Proof of Da Vinci’s Theorem Imagine you draw in the plane a bounded object with finitely many symmetries. Count its proper and improper symmetries. According to the Zero-or-Equal Theorem, one of the model objects has these same counts. Our goal is to prove that your object is rigidly equivalent to that model object. Your object and that model object each have a center point (according to the Center Point Theorem). There is a translation of the plane that repositions your object on top of that model object so that afterwards the two objects have the same center point. So now the two objects have the same center point and the same number of rotation symmetries, which we’ll call n. We first wish to show that they have the same list of proper symmetries; that is, the same list of rotation angles. For this, let α denote the smallest rotation angle (other than zero) of your object. Since Rα is a symmetry, so is the result of composing Rα with itself any number of times, so { I, Rα, R2α, R3α, . . .} are all symmetries. We claim α divides evenly1 into 360. Why couldn’t α be a number like 91 that doesn’t divide evenly into 360 ? Because then the list would look like { I, R91, R182, R273, R364, . . .}, but since R364 ¼ R4, we would have been wrong about 91 being the smallest angle. There’s nothing special about

1

This means that 360/α is an integer (which makes sense even if α is not an integer).

59  3B

Bounded Objects

91 here; by the same logic, no number that fails to divide evenly into 360 could ever qualify as a smallest rotation angle. Thus, α divides evenly into 360. We next claim that the multiples of α are your object’s only rotation angles. To understand why, let’s consider the example α ¼ 10, in which case {I, R10, R20, R30, . . ., R350} are symmetries. If there were any other rotation symmetries, then we would have been wrong about 10 being the smallest. For example, if R37 were also a symmetry, then so would be (inverse of R30)  R37 ¼ R7. This logic works in general (not just when α ¼ 10) to conclude that the multiplesof α are the only rotation angles. In summary, α divides evenly into 360, and your object’s only rotation angles are the multiples of α, so in particular α ¼ 360/n. These conclusions are equally true of the model object, so your object and the model object have the same list of rotation angles. So now your object has been repositioned on top of the model object such that the two objects have the same list of proper symmetries. If they are oriented, then we are done—they are rigidly equivalent. If they have reflection symmetries, then choose one reflection line of each object. There is a rotation (about the objects’ common center point) that, when applied to your object, matches up these chosen reflection lines. We claim that all the other reflection lines are now automatically matched up. Why? The answer is the title of a section from the previous chapter: one reflection is enough! The remaining reflections are the same because they are all just compositions of the common reflection with the common list of rotations. □

Figure 47 shows the key idea of the proof. Select the model object that has the same number of rotation and reflection symmetries as your object. Move your object on top of the model object such that the (red) rotation centers match and the chosen (blue) reflection lines match. After this repositioning, your object is guaranteed to have the same list of symmetries as the model object. Its rotation angles and other reflection lines will automatically match. Figure 47: Da Vinci’s proof idea: Your object and the model object will have the same symmetries after their (red) center points and (blue) reflection lines are matched

model object your object

▶ Exercise

Identify the symmetry type of the object in Figure 48. In other words, identify the model object that it matches. Figure 48: To identify this object’s symmetry type, just count its proper and improper symmetries

3

60

3

Chapter 3 • The Classification Theorems

Solution It has 2 proper (rotation) symmetries and 0 improper (reflection) symmetries. The proof of Da Vinci’s Theorem shows that it is guaranteed to be rigidly equivalent to the model object with these same counts, namely, the oriented 2-sided polygon: .

Figure 49: Four border patterns

3C Border Patterns The border patterns you encounter in art and architecture, like those in Figure 49, exhibit a seemingly infinite variety of artistic motifs. However, from a mathematical point of view, we will prove there are only 7 different types of border patterns. The seven model border patterns are illustrated in Figure 50. We will prove that any other border pattern is guaranteed to be equivalent to one of them. But what should “equivalent” mean here? Rigid equivalence does not correctly capture the idea that the two patterns are symmetric in the same way. Why not? Imagine making a copy of one of the seven model patterns and enlarging the copy by a factor of, say, 37% (like you could do using a copy machine with an enlarge/shrink button). The enlarged copy is not rigidly equivalent to the original, even though we certainly want to consider them symmetric in the same way. No matter how the copy is repositioned on top of the original, its smallest translation distance will be 37% longer than that of the original. So the original and the enlarged copy would not have the same list of translation symmetries no matter how they were repositioned. To solve this problem, we will consider two border patterns to match if one is rigidly equivalent to a rescaling (which means an enlarging or shrinking) of the other. This allows us to state a classification theorem.

61  3C

Border Patterns

The Classification of Border Patterns Any border pattern is rigidly equivalent to a rescaling of one of the seven model border patterns in Figure 50.

Figure 50: The seven model border patterns

A “rescaling” (enlarging or shrinking) is needed to ensure that the border pattern has the same smallest translation distance as the model pattern to which it’s being matched. If you have a border pattern, how do you know which of the seven model patterns yours matches? All you must do is fill out the identification card in Figure 51. Figure 51: Border-pattern identification card

H – Any horizontal reflection symmetry? V – Any vertical reflection symmetries? R – Any 180° rotation symmetries? G – Any glide reflection symmetries?

There is guaranteed to be exactly one model pattern that has the same yes/no responses to all four questions, and your pattern (after rescaling) is guaranteed to be rigidly equivalent to it. The term “horizontal” in Figure 51 should be interpreted to mean the direction of your border pattern’s translations, while the term “vertical” means the direction perpendicular to your border pattern’s translations.

3

62

Chapter 3 • The Classification Theorems

▶ Exercise

Identify the symmetry type of this border pattern:

3

Solution We must answer the four identification-card questions:

H – No, the pattern would look different after being reflected over the horizontal center line (brown in Figure 52). V – Yes, it has infinitely many vertical reflection lines; several are shown as blue lines in Figure 52. R – Yes, it has infinitely many 180
rotation centers; several are shown as pink dots in Figure 52. G – Yes, reflecting over the brown horizontal center line is not a symmetry, but it becomes a symmetry when composed with the translation depicted by the red arrow in Figure 52.

Figure 52: The border pattern’s rotation centers and reflection lines

So our pattern’s identification card is NYYY. Exactly one of the seven model border patterns has the same identification, namely, model #6. Try to visualize how our pattern (after rescaling) can be repositioned on top of model pattern #6 so that all symmetries match up.

Here are the steps involved in proving the Classification of Border Patterns Theorem: 1. Prove that a border pattern can’t have any types of symmetries other than translations and those types mentioned on the identification card. 2. Among all 16 ways in which one could fill out the identification card with yes/no responses, show that only 7 of these ways yield the identification card of an actual border pattern; the remaining 9 yield impossible cards. 3. Prove that two border patterns with the same identification-card responses must be rigidly equivalent (after rescaling one so that they have the same smallest translation distance). The complete proofs of these three steps are addressed (with ample hints) in Exercises 33–39 at the end of this chapter.

63  3D

Wallpaper Patterns (optional)

Figure 53: Two wallpaper patterns

3D Wallpaper Patterns (optional) Wallpaper patterns in art and architecture, like those in Figure 53, exhibit endless artistic variety, yet we will soon learn that there are only 17 symmetry types of wallpaper patterns. Wallpaper patterns are classified according to their answers to the questions on the identification card at the top of Figure 54. For example, one question asks whether there are any indecomposable glide-reflection symmetries (as defined on page 48). Another question asks the maximum order of a rotation symmetry of the pattern (as defined on page 12). It is quite surprising and profound that the only possible orders for the rotation symmetries of wallpaper patterns are 1, 2, 3, 4, and 6. This fact is the starting point for classifying wallpaper patterns. We will not prove this fact or the classification theorem because the proofs are difficult and beyond the scope of this book. In fact, a complete proof did not appear until 1891. When you use the flow chart in Figure 54, you do not necessarily need to answer all of the questions on the identification card, but just the questions that come up on the path you follow. In Figure 54, we have numbered the model patterns 1–17 in blue. Even though this order is arbitrary, it is how we will index and reference the patterns in this book. We also include (in red) one of several widely recognized systems for code-naming the 17 wallpaper patterns. We will not use the red code names in this book; we only include them for compatibility with other texts.

The Classification of Wallpaper Patterns Any wallpaper pattern is rigidly equivalent to a transformation of one of the seventeen model wallpaper patterns in the flow chart in Figure 54.

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Chapter 3 • The Classification Theorems

What is the maximum Order of a rotation symmetry? orange: Does it have any Reflection symmetries? purple: Does it have any indecomposable Glide-reflection symmetries? green: Does it have any rotations centered ON reflection lines? ON OFF Does it have any rotations centered OFF reflection lines? 1 (p1)

2 (pg)

4 (cm)

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15(p31m) 10 (p4)

14 (p3m1)

13 (p4)

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11 (p4m)

Figure 54: Flow chart for the 17 model wallpaper patterns (by Brian Sanderson)

Suppose you have a wallpaper pattern. The flow chart tells you which one of the 17 model patterns it matches. The theorem asserts that your pattern is rigidly equivalent to a transformation of that model pattern. What does “transformation” mean? It depends on the model number: For model numbers 10–17 (the ones for which the maximum order of a rotation symmetry is 3, 4, or 6), a transformation just means a rescaling (enlarging/shrinking). So the matching of these wallpaper patterns works just like the matching of border patterns from the previous section. For model numbers 2, 3, 4, 6, 7, 8, and 9 (the non-oriented ones for which the maximum order of a rotation symmetry is 1 or 2), a transformation means a squeeze, which is a rescaling with separate (possibly different) horizontal and vertical scaling

65  3D

Wallpaper Patterns (optional)

factors. In the illustration shown in Figure 55, model pattern #3 is squeezed by vertical factor 0.5 and by horizontal factor 1.5. For model numbers 1 and 5 (the oriented ones for which the maximum order of a rotation symmetry is 1 or 2), a transformation means a squeeze possibly followed by a shear. A shear alters the angle between the x-axis and the y-axis. The illustration in Figure 56 shows a shear applied to model pattern #1.

Figure 55: A squeeze transformation

Figure 56: A shear transformation

Notice that the squeeze and the shear illustrated in Figures 55 and 56 do not change anything essential about the way in which the pattern is symmetric. That’s why we must incorporate them into our notion of “matching” to achieve a classification theorem that is correct and meaningful.

▶ Exercise

Identify the symmetry type of the wallpaper pattern in Figure 57.

Solution Figure 58 shows the pattern marked up with purple reflection lines and with blue and yellow rotation points. All rotation symmetries (other than the identity) have order 3; the difference is that the blue dots lie on reflection lines, while the yellow ones do not. In the flow chart (Order¼3, Reflections¼yes, OFF¼yes), we arrive at pattern #15. Visualize how this pattern (after rescaling) can be repositioned on top of pattern #15 so that all symmetries match up.

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Chapter 3 • The Classification Theorems

Figure 57: Identify this pattern’s symmetry type; that is, decide which of the 17 model patterns it matches

Figure 58: The blue rotation centers lie ON the (purple) reflection lines, while the yellow rotation centers lie OFF them

The symmetry groups of the 17 model patterns are often called “wallpaper groups.” Chemists call them “plane crystallographic groups” because they represent the possible configurations into which two-dimensional crystal structures can form. M.C. Escher incorporated many of these patterns into his paintings. They also occur throughout nature, for example in honeycombs. We recommend the iOrnament app for creating your own wallpaper patterns.

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67  3E

Elements of Mathematics: Equivalence (optional)

3E Elements of Mathematics: Equivalence (optional) In this chapter, we did something remarkable! We described all possible ways in which (1) a bounded object, (2) a border pattern, and (3) a wallpaper pattern can be symmetric. For each of these three categories, we now have a list of models and a theorem guaranteeing that every object in the category is “equivalent” to one of the models. What it means for two objects to be “equivalent” is roughly intended to capture the idea that they are symmetric in the same way (they have the same symmetry type). The precise meaning of “equivalence” depends on the category. For bounded objects, it means they are rigidly equivalent. For border patterns, it means that one is rigidly equivalent to a rescaling of the other. For wallpaper patterns, it’s complicated. We’ll use the “~” symbol to denote any of these various types of equivalence; for example, in Figure 59, the ~ symbol means that the two objects are rigidly equivalent. Mathematicians have a useful framework for formalizing the concept of equivalence in general. To encourage a richer understanding of equivalence in this and future chapters, we will now do a deep dive into this framework. Let’s start with a story. The host of a multigenerational game of lawn darts announces, “Everyone please organize yourselves into teams according to the last digit of your age.” In Figure 60, each yellow dot is a player, and the numbers tell their ages. The orange circles are the teams they form based on these instructions. Mathematically we model this story by considering the set of all 14 players. The host’s instructions are what’s called a relation on this set, which we’ll denote with the “~” symbol. In general, a relation is defined as follows:

Figure 59: Generally, “~” denotes equivalence; in this case it denotes rigid equivalence

Figure 60: The players divide into teams according to the last digit of their age

35 47

7 17

75 34 14 54

8 18 11

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Chapter 3 • The Classification Theorems

Definition A relation on a set is a criterion for determining, given any pair A, B of members of the set, whether it’s true or false that A~B.

3 In the previous example, A~B is true if the ages of players A and B have the same last digit, otherwise it is false. But not every relation you make up would be the kind that works to create teams. A relation that works to create teams is called an equivalence relation. The relation “having ages with the same last digit” worked—it is an equivalence relation. To understand what’s needed, let’s think about some relations that do not work. Example The host announces, “Everybody please organize yourselves into teams such that your teammates are the people younger than you.” That is, “A~B” means that A < B. This relation is not an equivalence relation. Do you see why these instructions will cause mayhem? The 59year-old will insist on being teammates with the 8-year-old, but the 8-year-old will refuse to be teammates with the 59-year-old. Furthermore, the 8-year-old is not allowed to be her own teammate, since she isn’t younger than herself. They will argue all day and never get to play lawn darts.

We must insist that “~” have the symmetric property: if A~B, then B~A. That is, if I’m on your team, you’re on mine. We must also insist on the reflexive property: A~A. That is, each person is on his/her own team. The “less than” relation from the previous example lacked both of these properties, which made it impossible to form teams. Example The host announces, “Everybody please organize yourselves into teams such that your teammates are the people with ages less than 10 years apart from your age.” That is, “A~B” means that jA  B j < 10. This relation is not an equivalence relation. Do you see why these instructions will cause mayhem? The 7-year-old will insist on being teammates with the 14year-old, and the 14-year-old will insist on being teammates with the 18-year-old, but this incorrectly puts the 7-year-old on the same team as the 18-year-old.

We must insist that “~” have the transitive property: if A~B and B~C, then A~C. That is, if Alfred is on Bernie’s team and Bernie is on Charlie’s team, then Alfred must be on Charlie’s team. The “within 10 years” relation from the previous example lacked this property, which made it impossible to form teams. These considerations lead to this definition:

69  3E

Elements of Mathematics: Equivalence (optional)

Definition A relation “~” on a set is an equivalence relation if it has the following properties (true for all members A, B, C of the set):

Reflexive: A~A. Symmetric: If A~B, then B~A. Transitive: If A~B and B~C, then A~C. These are exactly the right properties to insure “~” is the kind of criterion that partitions the set into teams. These teams will henceforth be called equivalence classes. Each equivalence class (team) is a subset whose members are all equivalent to each other. The “same last digit” idea works not just for the set of player ages but also for the set of all natural numbers. Example On the infinite set of all natural numbers, {1, 2, 3, 4, . . .}, define “~” to mean “same last digit.” It is straightforward to verify the reflexive, symmetric, and transitive properties, so this is an equivalence relation. There are exactly 10 equivalence classes; for example, one of these equivalence classes is {3, 13, 23, 33, . . .} (the team of numbers with last digit equal to 3). Since there are infinitely many natural numbers, it is interesting that there are only finitely many equivalence classes in this example.

In general, an equivalence class can be represented by any single one of its members. For example, the equivalence class {3, 13, 23, 33, . . .} from the previous example can be described as “all natural numbers that are equivalent to 33.” There’s nothing special about 33; it could equally well be described as “all natural numbers that are equivalent to 143.” Any single member works to represent the whole equivalence class. Equivalence relations provide a more precise framework for understanding the classification theorems in this chapter. For example, Da Vinci’s Theorem is related to this example:

▶ Exercise

On the set of all bounded objects, prove that rigid equivalence is an equivalence

relation. Solution We’ll use the letters A, B, C to denote bounded objects, and we’ll write “A~B” to mean that A is rigidly equivalent to B; that is, there is a rigid motion that repositions A so it has the same symmetries as B. We must verify three properties:

Reflexive: The identity repositions A so that it has the same symmetries as itself; thus, A~A. Symmetric: Suppose A~B, which means there is a rigid motion that repositions A so that it has the same symmetries as B. Then the inverse of that rigid motion repositions B so that it has the same symmetries as A; thus, B~A. Transitive: Suppose A~B (there is a rigid motion that repositions A so that it has the same symmetries as B) and B~C (there is another rigid motion that repositions B so

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Chapter 3 • The Classification Theorems

that it has the same symmetries as C). The composition of these two rigid motions will reposition A so that it has the same symmetries as C; thus, A~C.

3

Da Vinci’s Theorem describes the equivalence classes: there is one equivalence class for each model bounded object. Imagine all of the bounded objects in the conceivable world are sorted into piles such that every object is rigidly equivalent to all of the other object within its pile. These piles are the equivalence classes. Da Vinci’s Theorem says there is one pile for each model bounded object. Example When A and B are border patterns, define A~B to mean that there exists a rescaling of A that is rigidly equivalent to B. You will prove in the exercises that this is an equivalence relation on the set of all border patterns. Since this is an infinite set, it is interesting that there are only finitely many equivalence classes; in fact, there are exactly 7 equivalence classes, one for each model border pattern.

There is even an equivalence relation on the set of wallpaper patterns with respect to which the classification theorem can be rephrased as: any wallpaper pattern is equivalent to one of the seventeen model wallpaper patterns. Think about what it is.

Exercises (1) Write a precise definition of each of the following terms: rigidly equivalent, indecomposable, order, equivalence relation. (2) Identify the symmetry type of each bounded object pictured; in other words, identify the model bounded object that it matches.

(3) Identify the symmetry type of each bounded object pictured; in other words, identify the model bounded object that it matches.

71  Exercises

(4) Group these 26 capital letters (in this font) into rigid-equivalence classes:

ABCDEFGHIJKLMNOPQRSTUVWXYZ (5) For each of the seven model border patterns, answer yes/no to each of the four questions on the identification card. (6) Identify the symmetry type of each illustrated border pattern; that is, identify the model pattern it matches.

(7) Identify the symmetry type of each illustrated border pattern; that is, identify the model pattern it matches.

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Chapter 3 • The Classification Theorems

3

(8) For each capital English letter (in the font shown), decide which of the 7 model border patterns is formed by repeating it indefinitely along a line.

ABCDEFGHIJKLMNOPQRSTUVWXYZ (9) Identify the symmetry type of each border pattern in Figure 49. (10) Which of the 7 model border patterns have commutative symmetry groups? Explain your answer. (11) Which of the 7 model border patterns are oriented? Explain. (12) For each of the seven model border patterns, decide: (a) Does it have any 180
rotations centered on reflection lines? (b) Does it have any 180
rotations centered off reflection lines? (c) Does it have any indecomposable glide-reflection symmetries? (13) Which of the 17 model wallpaper patterns are oriented? Explain. (14) Black and white versions of the paintings Seahorses and Eels and Three Fishes by Robert Fathauer are illustrated in Figure 61. Consider each painting to be a wallpaper pattern (indefinitely extended). Identify the symmetry type of each; that is, identify the number of the model it matches. (15) Must any two asymmetric objects be rigidly equivalent? Explain. (16) Identify the symmetry type of each wallpaper pattern in Figure 53. (17) Identify the symmetry type of each wallpaper pattern in Figure 62. (18) Identify the symmetry type of each wallpaper pattern in Figure 63. (19) Identify the symmetry type of each wallpaper pattern in Figure 64. (20) Many paintings by M.C. Escher are wallpaper patterns (if you imagine them infinitely extended). Perform a web image search for the following paintings, and identify the symmetry type of each:

73  Exercises

Figure 61: Images for Exercise 14

Figure 62: Images for Exercise 17

Figure 63: Images for Exercise 18

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Chapter 3 • The Classification Theorems

3

Figure 64: Images for Exercise 19 Figure 65: Image for Exercise 21

(a) (b) (c) (d) (e) (f) (g) (h)

Horses (No. 8) Sea Horse (No. 11) Fish (No. 20) Clowns (No. 21) Shells & Starfish (No. 42) Angel-Devil (No. 45) Fish/Duck/Lizard (No. 69) Butterfly (No. 70)

(21) What is the maximum order of a rotation symmetry of the wallpaper pattern in Figure 65? Is it oriented? Identify its symmetry type.

75  Exercises

Figure 66: Image for Exercise 22

(22) On the wallpaper pattern in Figure 66: (a) Find several 180
rotation centers, and mark them with “♦”. (b) Find several reflection lines, and indicate them with solid lines. (c) Find several indecomposable glide-reflection lines, and indicate them with dashed lines. (23) Is it possible for a bounded object to be rigidly equivalent to a border pattern? Is it possible for a border pattern to be rigidly equivalent to a wallpaper pattern? Explain your answers. (24) On the set of integers {. . ., 3, 2, 1, 0, 1, 2, 3, . . .}, is “same number of digits” an equivalence relation? If so, describe some of the equivalence classes. (25) On the set of all fractions that have been reduced to lowest terms, is “same denominator” an equivalence relation? If so, describe some of the equivalence classes. (26) Two objects are called “intersecting” if they share at least one point of the plane in common. On the set of bounded objects, is “intersecting” an equivalence relation? Explain your answer. (27) Consider the set of all circles in the plane. Define “A~B” to mean that the centers of circles A and B are less than an inch apart. Is this an equivalence relation? Explain your answer. (28) Consider the set of all words in the English dictionary. Define “A~B” to mean that words A and B share at least one letter in common. For example, HELLO~RACE because both words contain the letter E. Is this an equivalence relation? Explain your answer. (29) Consider the set of all words in the English dictionary. Define “A~B” to mean that words A and B have the same number of vowels. For example, HI~RUN because both words contain one vowel. Is this an equivalence relation? Explain your answer.

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Chapter 3 • The Classification Theorems

(30) When A and B are border patterns, define A~B to mean that a rescaling of A is rigidly equivalent to B. Verify that this is an equivalence relation on the set of border patterns. (31) Consider rigid equivalence as an equivalence relation on the set of border patterns. Are there infinitely many or finitely many equivalence classes? Explain your answer. (32) Consider “rigid equivalence after rescaling” as an equivalence relation on the set of wallpaper patterns. Are there infinitely many or finitely many equivalence classes? Explain your answer. Exercises 33–35 traverse the first step of the proof of the classification theorem for border patterns (as outlined on page 62). (33) Prove that every rotation symmetry of a border pattern is by zero or 180 degrees. HINT: Suppose that some rotation (called R) by a different angle were a symmetry. If T is a translation symmetry, explain why the symmetry R  T  (inverse of R) equals a translation symmetry that is not parallel to T. What can you conclude? (34) Prove that every glide-reflection symmetry of a border pattern is horizontal. HINT: Compose it with itself. (35) Prove that every reflection symmetry of a border pattern is either horizontal or vertical. HINT: modify the hint from Exercise 33. Exercises 36–37 proceed through the second step of the proof of the classification theorem for border patterns (as outlined on page 62). (36) Without using the classification theorem, prove that each of the following rules is true of all border patterns: I. If H, then G (that is, if it has a horizontal reflection symmetry, then it must have a glide-reflection symmetry). II. If H and V, then R. III. If H and R, then V. IV. If V and R, then either H or G. HINT: For I, what is H composed with a translation? For II, what is H * V? For IV, consider V * R in two cases, depending on whether the rotation center lies on or off the vertical reflection line. Look to the model border patterns for examples of each possibility. (37) Table 4 shows all 16 ways to fill out the border-pattern identification card. For each possibility, either identify the model pattern with that identification card, or identify a rule (like the rules from the previous exercise) by which you can conclude that there is no border pattern with that identification card. Exercises 38–39 complete the third step of the proof of the classification theorem for border patterns (as outlined on page 62). (38) Without using the classification theorem, prove the following statements about an arbitrary border pattern (whose smallest nonzero translation distance is denoted L ).

77  Exercises

Table 4: Combinations of answers to border-pattern identification questions H Y Y Y Y Y Y Y Y N N N N N N N N

V Y Y Y Y N N N N Y Y Y Y N N N N

R Y Y N N Y Y N N Y Y N N Y Y N N

G Y N Y N Y N Y N Y N Y N Y N Y N

#Illustration of border pattern or reason why there is none#

(a) The set of all of its translation distances equals the set of multiples of L, namely, {. . ., 3L, 2L, L, 0, L, 2L, 3L, . . .}. HINT: Copy an idea from the proof of Da Vinci’s Theorem. (b) If it has indecomposable glide-reflection symmetries, then the smallest translation distance among them is L/2. Every other glide-reflection symmetry can be obtained by composing the smallest one (or its inverse) with itself some odd number of times. Every translation symmetry can be obtained by composing this smallest glide-reflection symmetry (or its inverse) with itself some even number of times. (c) If it has vertical reflection symmetries, then it has infinitely many whose reflection lines are equally spaced L/2 apart. HINT: Choose one vertical reflection and call it F. Prove that each of the other vertical reflections can be obtained by composing F with a translation symmetry, then use Exercise 41 from Chapter 2. (d) If it has 180
rotation symmetries, then it has infinitely many whose centers are equally spaced L/2 apart. HINT: Choose one rotation symmetry and call it R. Prove that each of the other rotation symmetries can be obtained by composing R with a translation symmetry, then use Exercise 43 from Chapter 2. (e) If it has 180
rotation symmetries and vertical reflection symmetries, then either all rotation centers lie on vertical reflection lines, or all rotation centers lie halfway between two vertical reflection lines. HINT: Use Exercise 45 from Chapter 2.

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Chapter 3 • The Classification Theorems

Figure 67: Image for Exercise 45

y

3

x (39) Suppose you have two border patterns with the same identification card, and they have the same smallest translation distance. Prove they are rigidly equivalent. (40) On the set of 3-digit integers, is “same first digit AND same second digit” an equivalence relation? Explain your answer. (41) On the set of 3-digit integers, is “same first digit OR same second digit” an equivalence relation? Explain your answer. (42) Use the iOrnament app to do the following: (a) Create an artistic representative of each of the 17 wallpaper-pattern types. (b) Create artistic border patterns by cropping strips out of wallpaper patterns. Can representatives of all 7 border-pattern types be created in this way? (43) Use the GeCla app (https://www.atractor.pt/soft-_en.html) to design artistic representatives of each of the 17 wallpaper-pattern types. What other functionality does this app provide for studying wallpaper patterns? (44) Disprove the following statement: “If two objects have four symmetries each, then these objects are rigidly equivalent.” (45) The orange and blue square in Figure 67 are rigidly equivalent. How many different rigid motions of the plane are there which, when applied to the orange square, repositions it so that afterwards the two squares have the same collection of symmetries? Prove your answer. (46) Which of the following are equivalence relations on the set of humans alive on the earth right now?

▬ “born in the same month” ▬ “have the same biological mother” ▬ “have the same biological grandmother” ▬ “have more Facebook friends” ▬ “are cousins” ▬ “have the same first or last name”

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Isomorphic Groups

Leonardo Da Vinci’s helicopter design

Are we done yet? This book’s goal was to classify the ways that objects can be symmetric. That’s exactly what the classification theorems achieved in the previous chapter. Mission accomplished! What’s left? The primary goal of Chapters 4, 5, and 6 is to better understand symmetry groups. What’s the most elegant way to understand a group’s algebraic operation? What patterns are found in its Cayley table? The story begins in this chapter with a notion of equivalence for groups. Two groups are called “isomorphic” basically if it’s like they are really a single group expressed in two different notation systems. In the case of finite groups, this essentially means that they have the same patterns in their Cayley tables. If you understand one, then you understand the other. When you study a new object, the goal will be to prove that its symmetry group is isomorphic to a group with which you are already familiar.

Electronic supplementary material The online version of this chapter (https://doi.org/10.1007/ 978-3-030-51669-7_4) contains supplementary material, which is available to authorized users. The videos can be accessed individually by clicking the DOI link in the accompanying figure caption or by scanning this link with the SN More Media App.

© Springer Nature Switzerland AG 2021 K. Tapp, Symmetry, Texts for Quantitative Critical Thinking, https://doi.org/10.1007/978-3-030-51669-7_4

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Chapter 4 • Isomorphic Groups

Next, Chapters 5 and 6 will introduce some new types of groups. The more groups with which you are familiar, the better your odds that one of them will match the symmetry group of the next object you study. This strategy is illuminating for two-dimensional objects, but more importantly, it will later be absolutely essential for understanding symmetry groups of three-dimensional objects, beginning in Chapter 7.

4 4A The Definition of an Isomorphism In Chapter 2, you learned that D4 means the symmetry group of a square. Did you respond by asking, “Which square? Where in the plane is it centered? Is it upright or tilted?” You probably didn’t ask these questions because you intuitively sensed that their answers don’t matter. You already know one sense in which these things don’t matter: all squares are rigidly equivalent to each other. But what does that say about their symmetry groups? The red and green squares in Figure 68 are rigidly equivalent. How are their symmetry groups related? Imagine that Rene is studying the symmetry group of the red square (on the left). It has the following eight symmetries: ⬘ . Her Cayley table looks exactly like the Cayley table in Chapter 2 (typeset in red). Meanwhile, Gretchen is studying the symmetry group of the green square (on the right). Her square has the following eight symmetries: . Since her reflection lines are not quite horizontal, vertical, or diagonal, she chose different symbols for naming them, as shown in Figure 69.

y

Figure 68: How are their symmetry groups related?

x

D

V

D⬘

F4

F3

F2 F1

H

Figure 69: Matching up the reflection lines (▶ https://doi.org/10.1007/000-25k)

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The Definition of an Isomorphism

Even her rotations differ from Rene’s rotations. For example, Rene’s is a rotation is a rotation around the center of around the center of the red square, while Gretchen’s the green square—these are different rigid motions of the plane. Nevertheless, Gretchen’s Cayley table is essentially the same as Rene’s in the following precise sense. Starting with Rene’s red Cayley table, if I ask my word processor to convert red symbols into green symbols using the symbol-replacement dictionary shown below (with the yet-to-be-explained title “An isomorphism”), then I will end up with a correct Cayley table for Gretchen’s green group, as shown in Figure 70.

* I R90 R180 R270 H V D D⬘

* I R90 R180 R270 F1 F3 F4 F2

I

R90

R180

R270

H

I R90 R180 R270 H V D D⬘

R90 R180 R270 I D D⬘ V H

R180 R270 I R90 V H D⬘ D

R270 I R90 R180 D⬘ D H V

H D⬘ V D I R180 R270 R90

I

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R270

F1

I R90 R180 R270 F1 F3 F4 F2

R90 R180 R270 I F4 F2 F3 F1

R180 R270 I R90 F3 F1 F2 F4

R270 I R90 R180 F2 F4 F1 F3

F1 F2 F3 F4 I R180 R270 R90

V

D

V D D H H D⬘ D⬘ V R180 R90 I R270 R90 I R270 R180

F3

F4

F3 F4 F4 F1 F1 F2 F2 F3 R180 R90 I R270 R90 I R270 R180

D⬘ D⬘ V D H R270 R90 R180 I

F2 F2 F3 F4 F1 R270 R90 R180 I

Figure 70: The dictionary converts a correct red Cayley table into a correct green Cayley table

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Chapter 4 • Isomorphic Groups

An isomorphism 90

↕

↕ 90

4

180

↕ 180

’

270

↕ 270

↕

↕

1

↕ 2

↕ 3

4

Do you see why the resultant green table is correct (each of its 64 cells is filled in correctly)? If you start checking cells, you’ll quickly see why things are working out. Our dictionary was very carefully constructed. The reflection lines listed counterclockwise around the red square were matched with the reflection lines listed counterclockwise around the green square (see Figure 69). Said differently, this dictionary is induced by the rigid motion that slides and tilts the red square on top of the green square, matching up the two sets of reflection lines. In summary, each cell in the green table is filled in correctly, which really means this: Our dictionary converts every true red equation into a true green equation! For example, , which is a true red the yellow highlighted cell of the red table says , which is a equation. This equation is symbol-by-symbol converted into true green equation, represented by the yellow highlighted cell of the green table. The red and green groups are isomorphic. The dictionary is an isomorphism. Here is the definition:

Definition An isomorphism between two groups means a one-to-one matching (dictionary) between their members that converts each true equation in one group into a true equation in the other group. We say two groups are isomorphic if there exists an isomorphism between them.

An isomorphism between finite groups converts a correct Cayley table for one group into a correct Cayley table for the other. An isomorphism is a very special kind of matching. The next exercise indicates that a haphazard matching is unlikely to qualify as an isomorphism:

▶ Exercise

Prove that the following dictionary is not an isomorphism between the previously discussed red and green groups.

90

↕

↕ 90

180

↕ 180

′

270

↕ 270

↕ 1

↕ 2

↕ 3

↕ 4

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The Definition of an Isomorphism

Solution We need only exhibit a single true red equation that gets converted into a false green equation. The reflections seem haphazardly matched, so we’ll consider equations involving reflections. Trial and error leads to:

↕

*

= ↕

TRUE

180

↕

FALSE. This demonstrates that the matching is not an isomorphism. The matching that did produce an isomorphism was not haphazard, but was based on the rigid equivalence between the red and green squares. “Rigid equivalence” exactly means that there is a repositioning of one square on top of the other that matches up their symmetries; this matching provides an isomorphism. It always works like that:

Theorem If two objects are rigidly equivalent, then their symmetry groups are isomorphic.

Proof Let’s call the objects X and Y. There is a rigid motion (which we’ll call M ) that repositions X so that afterwards the two objects have the same symmetries. In Figure 71, it is visually apparent that M matches every symmetry of X with a symmetry of Y. For example, M moves each reflection line of X to a reflection line of Y, and M moves the rotation center of X to the rotation center of Y. Here is a more precise way to describe the manner in which M matches each symmetry of ~ denotes the inverse of M. Each symmetry, A, of X gets X with a symmetry of Y. Recall that M ~ Do you see why M  A  M ~ is a symmetry of Y? First M ~ is matched with M  A  M. performed (which moves Y to X), then A is performed (which moves X to X because it’s a symmetry of X), and then M is performed (which moves X back to Y ). The net result is a symmetry of Y. Think on your own about why this matching is a one-to-one correspondence; it matches each symmetry of X with exactly one symmetry of Y and vice versa. Figure 71: The motion M aligns the two objects so as to match their symmetries

Y

X

M

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Chapter 4 • Isomorphic Groups

To verify that this matching is an isomorphism, we must check that it converts true equations into true equations. Suppose A  B ¼ C is a true equation in the symmetry group

~  MBM ~ ¼ M  C  M, ~ of X. Our matching converts this equation into M  A  M which is a true equation because its left side simplifies to its right side as follows:

4 □

4B Isomorphism Examples The previous theorem related “rigid equivalence” and “isomorphism,” but at face value these are very different concepts. We ask whether two objects are rigidly equivalent. We ask whether two groups are isomorphic, and they can be any groups, not necessarily symmetry groups. When two groups are isomorphic, we think of them as essentially the same; they have the same algebraic structure. Among finite groups, this means they have the same patterns in their Cayley tables. An observer who doesn’t know or care what the names of the group members represent would study the two Cayley tables and discover exactly the same patterns. From this observer’s perspective, the two groups would look like a single group represented in two different notational systems.

▶ Exercise

Are the groups with these Cayley tables isomorphic?

G1 = { A B C } * A B C

A

B

C

A B C

B C A

C A B

G2 = {Ω Ψ Ѳ} * Ω Ψ Ѳ

Ω Ψ

Ѳ

Ω Ψ

Ѳ

Ψ Ѳ

Ω Ψ

Ѳ Ω

Solution Yes, because the following dictionary converts the red Cayley table into the green Cayley table:

↕

↕ ↕ Ɵ

In fact, here is a description that characterizes both groups simultaneously: in addition to the identity, there are two more members. Each is the inverse of the other. Each combines with itself to give the other. These groups are not only isomorphic to each other, but isomorphic to the familiar group C3 (the symmetry group of an oriented triangle).

4

85
4B

Isomorphism Examples

Figure 72: C2 is isomorphic to D1

C2 = { I , R180}

D1 = { I , V }

*

I

R180

*

I

V

I

I

R180

I

I

V

I

V

V

I

R180 R180

Even simpler than the previous exercise is this important fact: C2 is isomorphic to D1. Their Cayley tables are illustrated in Figure 72. The isomorphim matches with and with . It doesn’t matter that C2 has a rotation while D1 has a reflection. matches They are isomorphic because they have the same pattern in their Cayley tables. An observer who didn’t know the meanings of the symbols would study the Cayley tables and characterize both groups this same way: in addition to the identity, it has one other member that combines with itself to give the identity. Is the isomorphism between C2 and D1 a one-off, or are there other isomorphic pairs among the cyclic and dihedral groups? The following facts help answer this question: 1. Isomorphic groups always have the same size. 2. A commutative group could never be isomorphic to a noncommutative group. For example, C5 is not isomorphic to D3 because they have different sizes: C5 has 5 members while D3 has 6 members. Furthermore, C6 is not isomorphic to D3 because, even though they have the same size, the former is commutative while the latter is not. Doing more examples like this helps expose the general fact:

Theorem Among all of the cyclic and dihedral groups, the only isomorphic pair is {C2 , D1}.

Proof The illustration enumerates the cyclic and dihedral groups (in black) and their sizes (in red). The commutative ones are lassoed in purple; they include all of the cyclic groups and the first two dihedral groups, as discussed in Chapter 2. The noncommutative ones are lassoed in green; they include the rest of the dihedral groups.

86

Chapter 4 • Isomorphic Groups

commutative C1 (1)

C2 (2)

D1 (2)

D2 (4)

S YE

C3 (3) NO D3 (6)

C4 (4) D4 (8)

C5 (5)

C6 (6)

D5 (10) D6 (12) non-commutative

C7 (7) . . . D7 (14) . . .

4 Among the noncommutative ones, there aren’t any pairs with the same size. Among the commutative ones, there are exactly two pairs with the same size. The first is {C2, D1}, which is an isomorphic pair. The second is {C4, D2}, which is not an isomorphic pair, as you will prove in the exercises at the end of this chapter.□

Consider the collection of all conceivable objects that are bounded and have finitely many symmetries. Imagine sorting them into piles according to whether they are rigidly equivalent. Da Vinci’s Theorem says there would be one pile for each bounded model object. Imagine instead sorting according to whether their symmetry groups are isomorand piles would phic. This would lead to exactly the same piles, except the merge into a single pile. Most people want to keep these two piles separate, and we agree that sorting according to rigid-equivalence classes (as we did in Chapter 3) is more natural than sorting according to whether symmetry groups are isomorphic. Sorting is not the main purpose for isomorphisms in this book. In this book, the main purpose for isomorphisms is to better understand symmetry groups. To understand a new object, we will often try to prove that its symmetry group is isomorphic to a group with which we are already familiar. This will be particularly useful for the three-dimensional objects that we will study beginning in Chapter 7. For now, we will illustrate the strategy with some two-dimensional examples.

▶ Exercise

Prove that the symmetry group of this border pattern is isomorphic to ℤ (the additive group of integers).

…. G G G G G

G G G G G G

G G G ….

Solution This border pattern has only translation symmetries. Let’s enumerate them like , where denotes the translation by n letters to this: the right (if n is positive) or to the left (if n is negative). Here is the natural dictionary matching symmetries of the border pattern (in green) with integers (in purple):

…

−4

↕ … −4

−3

↕ −3

−2

↕ −2

−1

↕ −1

0

↕ 0

1

↕ 1

2

↕ 2

3

↕ 3

4 … ↕ 4 …

The pattern is that matches with . Even though we only showed a portion of this matching, it’s clear that it continues left and right in such a way that it matches all of the infinitely many symmetries with all of the infinitely many integers in a one-to-one manner.

87
4C

A Better Notation for Cyclic Groups

Why is this matching an isomorphism? Because composing translation symmetries is , which our dictionary really just integer addition in disguise. For example, . Notice that the converted equation has “+” converts into the integer equation instead of “” because that’s the algebraic operation in ℤ. There is nothing special about 5 and 8; this dictionary converts every true equation about composing symmetries into a true equation about adding integers.

In the previous exercise, did the isomorphism help you to better understand the symmetry group? Perhaps the symmetry group was simple enough that you understood it well enough already, so let’s consider a more difficult example.

▶ Exercise

Prove that the symmetry group of this border pattern is isomorphic to ℤ.

Solution Let’s refer to the length of the illustrated red arrow as “one letter.” This border pattern’s symmetries include a glide reflection to the right or left by 1 letter, a translation by 2 letters, a glide reflection by 3 letters, a translation by 4 letters, etc. Let’s enumerate all of these symmetries like this: , where denotes the symmetry that shifts n letters to the right (if n is positive) or to the left (if n is negative). Notice that is a translation if n is even, or is a glide reflection if n is odd. This notation system exposes the key observation: composing these symmetries is just integer addition in disguise. For example, here is a true equation: . The clever notation system doesn’t make a big deal about the distinction that and are glide reflections while is a translation; all that we need to keep track of is the shift amounts, and composition just adds these shift amounts. So just like in the previous exercise, the dictionary that matches with is an isomorphism.

In the previous exercise, constructing the isomorphism entailed inventing a notation system for naming the symmetries that simplified our understanding of the algebraic operation (composition). Do you see the value? For example, the isomorphism exposes something that you might not have previously noticed: that the border pattern’s symmetry group is commutative (because integer addition is commutative).

4C A Better Notation for Cyclic Groups In this section, we introduce a convenient and simple new notation system for members of cyclic groups. For example, C5 ¼ {I, R72, R144, R216, R288} contains the 5 rotations of the oriented star in Figure 73. Keeping track of all of these angles is cumbersome. Instead, it is common convention to name the members of C5 like this: C5 ¼ {0, 1, 2, 3, 4}. Think of “3” as representing

4

Chapter 4 • Isomorphic Groups

88

Figure 73: An oriented star

4 ∗

I

R72

R144

R216

R288

I R72

I R72

R72 R144

R144 R216

R216 R288

R288 I

R144 R216 R288

R144 R216 R288

R216 R288 I

R288 I R72

I R72 R144

R72 R144 R216

+ 0 1 2 3 4

0 0 1 2 3 4

1 1 2 3 4 0

2 2 3 4 0 1

3 3 4 0 1 2

4 4 0 1 2 3

Figure 74: The convention of counting turns instead of angles in the Cayley table for C5

“R216” which is a rotation of the star by 3 counterclockwise “turns.” We’re using the word “turn” here to mean the smallest possible rotation angle, 72 , which moves each starpoint to its counterclockwise neighbor. Thus, we’re counting turns instead of angles. The Cayley table for C5 looks particularly simple when converted into this concise new notation system, as shown in Figure 74. It is also common to use “+” to denote the algebraic operation in a cyclic group. The yellow highlighted cells in Figure 74 represent the equation that we previously wrote as ”. “R216  R288 ¼ R144” and which we will henceforth write as “ This notational change encourages you to think of the algebraic operation in C5 as encoding a kind of “addition with wrap-around.” As you count 4 past 3 on your fingers, remember that when you reach 5, you must wrap back around to zero, so your count goes: 4, 0, 1, 2. In our future study of Cn, we will always use this abbreviated notation system, that is, Cn ¼ {0, 1, 2, . . ., n  1}, and we will always write “+” instead of “”. For example, we will write “5 + 7 = 2 (in C 10)” to indicate that, when you rotate a 10-sided oriented polygon by 5 turns and then by 7 more turns, you have done 2 turns more than going all of the way around, so the result is the same as rotating just two turns (here the word “turn” means a rotation by 360/10 ¼ 36 ). Alternatively, just add on your fingers, wrapping back to zero when you reach 10, so your count of 7 past 5 goes “6, 7, 8, 9, 0, 1, 2.” Alternatively, just add 5 + 7 ¼ 12, and then subtract 10 to get 2. It’s all the same thing. . The algebraic operaHere is another illuminating example: tion in C12 works like familiar “clock arithmetic”: 7 hours after 9 o’clock, it will be 4 o’clock. This new notation system for Cn is not exactly an isomorphism between two groups because there was only ever one group in the story. Nevertheless, it fits well into a chapter

89
4D

Elements of Mathematics: The Converse

about isomorphisms. When a pair of groups is an isomorphic pair, it is often best to think of them as a single group described in two different notation systems.

4D Elements of Mathematics: The Converse In this section, we’ll zoom in on a particular mathematical construction called the converse. Consider the following two statements: STATEMENT 1: If two objects are rigidly equivalent, then their symmetry groups are isomorphic. (TRUE) STATEMENT 2: If two objects have isomorphic symmetry groups, then they are rigidly equivalent. (FALSE) The first statement is a true theorem that we proved in the chapter. The second , }, as we discussed in this statement is false; a counterexample is the pair { chapter. Each of these statements is the converse of the other. What is the general meaning of “converse”? Mathematicians often find themselves studying a set, and considering two properties—let’s call them property A and property B—that some of the members of this set might have. In this setting, each of the following statements (about an arbitrary member of the set) is the converse of the other: STATEMENT 1: If it has property A, then it has property B. STATEMENT 2: If it has property B, then it has property A. The previous example was about the set of all pairs of objects. Property A was “being rigidly equivalent.” Property B was “having isomorphic symmetry groups.” Don’t confuse “converse” with “contrapositive” (from Chapter 1). A statement and its converse generally do not say the same thing. If a statement is true, then its converse might be true or might be false. It depends on the statement. In the previous example, the statement was true but its converse was false. After you prove a new theorem, it is often valuable to also think about whether its converse is true or false.

▶

Exercise Determine the converse of the following statement about an arbitrary real number x: “If x2 > 4, then x <  2 or x > 2.”

Solution “If x <  2 or x > 2, then x2 > 4.”

In the previous exercise, both the statement and its converse are true. In other words, property A (x2 > 4) and property B (x <  2 or x > 2) say exactly the same thing. Mathematicians describe this situation by saying: “Property A is true if and only if property B is true.”

4

90

Chapter 4 • Isomorphic Groups

For example, the fact that the statement and its converse were both true in the previous exercise can be expressed this way: “x2 > 4 if and only if x <  2 or x > 2.”

4

The phrase “if and only if” allows us to simultaneously make two claims (that a statement is true and that its converse is true) in a single concise sentence.

Exercises (1) Write a precise definition of the following terms: isomorphism, isomorphic, converse, if and only if. (2) Fill in each blank:

5 + 7 ¼ _____ (in C9), 6 + _____ ¼ 2 (in C10), 80 + 35 ¼ _____ (in C100), _____ + 5 ¼ 8 (in C10). (3) Fill in each blank:

10 + 8 ¼ _____ (in C13), 7 + _____ ¼ 2 (in C17), 2 + 3 ¼ _____ (in C100), _____ + 8 ¼ 5 (in C30). (4) Using the shorthand notation, construct a complete Cayley table for each of these cyclic groups: C2, C4, and C6. (5) Using the shorthand notation, construct a complete Cayley table for each of these cyclic groups: C3, C5, and C7. (6) Circle all the letters that are rigidly equivalent to the letter C.

D X R N A Z V H (7) Circle all the letters whose symmetry groups are isomorphic to the symmetry group of the letter C.

D X R N A Z V H (8) Among the following objects, find all pairs that are rigidly equivalent. Find all pairs that have isomorphic symmetry groups.

91
Exercises

(9) Sort the 26 capital English letters (in this font) into piles according to whether. . . (a) . . .they are rigidly equivalent, (b) . . .their symmetry groups are isomorphic.

ABCDEFGHIJKLMNOPQRSTUVWXYZ (10) Explain why these two border patterns do not have isomorphic symmetry groups.

…A A A A A A A A A A…. …G G G G G G G G G G… (11) Explain why the symmetry group of wallpaper pattern #1 is not isomorphic to the symmetry group of wallpaper pattern #17. (12) Verify that the following one-to-one matching is not an isomorphism between the red and green groups discussed in this chapter.

I R90 R180 R270 H D⬘ V D ↕ ↕ ↕ ↕ ↕ ↕ ↕ ↕ I F1 R180 R270 R90 F2 F3 F4 (13) Prove that an isomorphism between two groups always matches the identity of one group with the identity of the other group. (14) Prove that a commutative group could never be isomorphic to a noncommutative group. (15) Verify that the following one-to-one matching between the members of is not an isomorphism. and

↕

180

↕

90

↕

↕

180

270

(16) Prove that D2 is not isomorphic to C4. HINT: for the matching in the previous is converted into a false equation. For an arbitrary exercise, matching, consider replacing from D2 with whatever matches with . from C4. (17) If one object is rigidly equivalent to a rescaling of another object, prove the objects have isomorphic symmetry groups. (18) Prove that ℤ (the additive group of all integers) is isomorphic to the additive group of all even integers. (19) Determine the converse and the contrapositive of the following statement about an arbitrary integer n: “If n is a multiple of 4, then n is even.”

Give a counterexample to the statement’s converse.

4

92

Chapter 4 • Isomorphic Groups

(20) I have a bucket full of balls. Every ball is either red or green. Every ball has a number between 1 and 10 painted on it. Determine the converse of the following statement: “Every green ball has an odd number.”

4

Draw an example bucket for which. . . (a) (b) (c) (d)

The statement is true but the converse is false. The statement is false but the converse is true. The statement and its converse are both true. The statement and its converse are both false.

(21) Consider the statement “Every green ball has an odd number” from the previous exercise. (a) Write the converse of the contrapositive of the statement. (b) Write the contrapositive of the converse of the statement. Why are your answers to (a) and (b) the same? (22) In a card game, I am dealt a hand of seven cards from a standard deck. Determine the converse of the following statement: “Every black card in my hand is a face card.”

Draw an example hand for which. . . (a) (b) (c) (d)

The statement is true but the converse is false. The statement is false but the converse is true. The statement and its converse are both true. The statement and its converse are both false.

(23) Consider the statement “Every black card in my hand is a face card” from the previous exercise. (a) Write the converse of the contrapositive of the statement. (b) Write the contrapositive of the converse of the statement. Why are your answers to (a) and (b) the same? (24) Disprove the converse of the following statement: “If an object has a 10 rotation symmetry, then it has a 20 rotation symmetry.” (25) Disprove the converse of the following statement: “If a rigid motion of the plane has exactly one fixed point, then it is a rotation.” (26) Disprove the converse of the following statement: “If two commutative groups are an isomorphic pair, then they have the same size.” (27) TRUE or FALSE (and explain why): An object is unbounded if and only if it is a border pattern or a wallpaper pattern.

93
Exercises

(28) TRUE or FALSE (and explain why): A real number is greater than or equal to zero if and only if it equals its absolute value. (29) TRUE or FALSE (and explain why): A border pattern has reflection symmetries if and only if it has glide-reflection symmetries. (30) TRUE or FALSE (and explain why):

▬ An object has at least one translation symmetry if and only if it has infinitely many translation symmetries.

▬ An object has at least one translation symmetry besides the identity if and only if it has infinitely many translation symmetries. (31) You have two bounded objects, and each object has finitely many symmetries. TRUE or FALSE (and explain why): They are rigidly equivalent if and only if they have the same number of proper symmetries and the same number of improper symmetries.

4

5

95

Subgroups and Product Groups

In this chapter, we learn how to find small groups inside of large groups, and then how to build large groups out of small groups. The point is to better understand symmetry groups. If we can recognize an object’s symmetry group as having been built out of smaller groups, then this realization might help us to more clearly understand its underlying algebraic structure.

5A Subgroups First, we’ll learn how to find small groups inside of large groups. For example, inside D4 ¼ ⬘ , let’s separately consider the rotations and the reflections ⬘ . The tables in Figure 75 show the results of composing any pair of the rotations (left) and any pair of the reflections (right). , which is a self-contained The left table is the Cayley table for group called C4. The right table isn’t a Cayley table at all, because ⬘ is not a group. Why not? If the green symbols were a self-contained group, then combining any pair of them would yield a green symbol, making the whole table green. That’s how the algebraic operation in a group works; it combines any pair of members of the group and always gives an answer that’s a member of the group. The rotations form a very special type of collection called a subgroup, defined below; the reflections do not.

Definition A collection of a group’s members is called a subgroup if it is a self-contained group on its own, which means that the collection passes all three of these tests:

(1) Identity: The collection includes the identity. (2) Combinations: When any pair of members of the collection is combined (with the group’s algebraic operation), the answer lies in the collection. (3) Inverses: The inverse of every member of the collection lies in the collection.

Electronic supplementary material The online version of this chapter (https://doi.org/10.1007/ 978-3-030-51669-7_5) contains supplementary material, which is available to authorized users. The videos can be accessed individually by clicking the DOI link in the accompanying figure caption or by scanning this link with the SN More Media App.

© Springer Nature Switzerland AG 2021 K. Tapp, Symmetry, Texts for Quantitative Critical Thinking, https://doi.org/10.1007/978-3-030-51669-7_5

95

96

Chapter 5 • Subgroups and Product Groups

*

I

R90 R180 R270

*

H

I

I

R90 R180 R270

H

I

I

V

R180

R90

D

R270 R90

R90 R180 R270

R90

R180 R180 R270 I

5

I

V

D

D⬘

R180 R90 R270 I

R270 R90 I

D⬘

R180 I

Figure 75: Composition of pairs of rotations (left) and pairs of reflections (right) from D4

Look back at the definition of group and verify that passing these tests insures that the collection is a group on its own. We’ve seen that the rotations (that is, the proper symmetries) form a subgroup of D4. Much more generally:

Theorem The proper symmetries of any object form a subgroup of its symmetry group (called its proper symmetry group).

Proof The collection of proper symmetries passes the three tests:

1. The identity is proper. 2. The composition of every pair of proper symmetries is proper. 3. The inverse of every proper symmetry is proper.

□

For example, the symmetry group of a regular n-sided polygon is Dn, while its proper symmetry group is Cn (which we already knew is a self-contained group). As another example, the symmetry group of the following border pattern includes translations and vertical reflections, while the proper symmetry group includes only the translations. …W W W W W W W W W W W W W W W W W …

The following exercise is about a collection that includes some of the proper and some of the improper symmetries.

▶ Exercise subgroup?

In D4 ¼

, do the four symmetries in red form a

5

97  5A

Subgroups

Solution Yes, because they pass the three tests:

1. The identity is red. 2. The composition of any two red symmetries is red, as shown in Table 5. 3. Every red symmetry is its own inverse, so its inverse is red. The red subgroup in the previous exercise is familiar: it is D2, the symmetry group of a regular 2-sided polygon. Think of the 2-sided polygon as the horizontal stripe in Figure 76. Visually, D2 comprises the symmetries of the square that are also symmetries of the striped square. Table 5: red  red ¼ red, which confirms the combinations test

*

I

H

V

R180

I

I

H

V

R180

H

H

I

R180

V

V

V

R180

I

H

V

H

I

Figure 76: The collection of red symmetries is a subgroup (a self-contained group) because it is the symmetry group of the striped square

This observation provides a quicker solution to the previous exercise: “Yes, it is a subgroup (a self-contained group) because it is the symmetry group of the striped square.” We know from Chapter 2 that the symmetries of any object form a selfcontained group.

▶ Exercise

In C9 ¼

, do the red numbers form a subgroup?

Solution Yes, C9 is the symmetry group of the 9-pointed star in Figure 77, while the red numbers form the symmetry group of the red-dotted version of the star.

In the previous exercise, C9 was described using the shorthand notation, whereby the numbers represented the turns of the 9-pointed star. We cleverly avoided verifying the three subgroup tests because the collection of symmetries of any object (in this case the red-dotted star) forms a self-contained group.

98

Chapter 5 • Subgroups and Product Groups

Figure 77: To visually confirm that the red numbers form a subgroup, note that they form the symmetry group of the red dotted star

5

▶ Exercise

In C9 ¼

, do the red numbers form a subgroup?

Solution No, because they fail tests 2 and 3:

2. Table 6 shows the combination of all pairs of red numbers. Not all answers are red. 3. Figure 78 has arrows matching each member of C9 with its inverse. Not all red members have red inverses. Table 6: The second test fails: red plus red is not always red

Figure 78: The third test fails: some red numbers have black inverses

+

0

2

4

6

8

0

0

2

4

6

8

2

2

4

6

8

1

4

4

6

8

1

3

6

6

8

1

3

5

8

8

1

3

5

7

C9 =

We did much more than necessary to solve the previous exercise. It would have sufficed to provide a single example of the second test’s failure, like the fact that (in C9), or a single example of the third test’s failure, like the fact that the inverse of 2 is 7.

5

99  5B

Generated Subgroups (optional)

▶ Exercise

In ℤ (the additive group of integers), do the even numbers form a subgroup?

Solution Yes, because the even numbers pass the three tests:

1. The identity (zero) is even. 2. The sum of any two even numbers is even. 3. The inverse (which means negative) of any even number is even. The second test generically requires the “combination” of any two members of the collection to lie in the collection. In the previous exercise, “combination” meant “sum” because the group’s algebraic operation was addition.

▶ Exercise

In ℤ (the additive group of integers), do the odd numbers form a subgroup?

Solution No, because the odd numbers fail tests 1 and 2:

1. The identity (zero) is not odd. 2. It is not true that the sum of any two odd numbers is odd. For example, which is even.

,

5B Generated Subgroups (optional) In this optional section, we describe a method for constructing a subgroup of a group. The idea is to choose one member of the group and build the smallest subgroup that includes that member. For example, what’s the smallest subgroup of D4 that includes R90? Since it’s a subgroup, it must include I. Since it includes R90, it must also include R90  R90 ¼ R180 and R90  R90  R90 ¼ R270. In summary, it must include {I, R90, R180, R270}. Since this list is already a subgroup, the list is done. It’s called “the subgroup of D4 generated by R90” and it is denoted as hR90i.

Definition If G is a finite group, and A is a member of G, then hAi ¼ f I, A, A  A, A  A  A, . . .g is called the subgroup of G generated by A.

100

5

Chapter 5 • Subgroups and Product Groups

In Exercise 27, you will prove that hAi really is a subgroup of G. You will also prove that the list I, A, A  A, A  A  A, . . . begins repeating as soon as one of its members equals I, and has no repetitions before that. So here is how you list the members of hAi: you write the identity, then write A (which is called “the generator”), then A  A, and so on until the next thing on the list would be I, at which point you’re done. In any particular example, the generic symbol “” should be replaced with the symbol for the algebraic operation of the group in question (like “+” in an additive group, or “” in a symmetry group).

▶ Exercise

In C10 ¼ { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 }, list the members of each of the following subgroups: h2i, h3i, h4i, h5i, h0i.

Solution The identity is 0 and the algebraic operation is “addition with wrap-around.” So our list is formed by starting with 0 and repeatedly adding the generator to the previous thing on the list, stopping when the next thing on the list would be 0, like this:

h2i ¼ {0, 2, 4, 6, 8}, h3i ¼ {0, 3, 6, 9, 2, 5, 8, 1, 4, 7}, h4i ¼ {0, 4, 8, 2, 6}, h5i ¼ {0, 5}, h0i ¼ {0}. The order in which the members of a set are listed doesn’t matter; for example, in the previous exercise, h2i and h4i are the same subgroup. Figure 79 is a nice way to visualize the subgroups of C10 from the previous exercise. Imagine 10 friends (numbered 0 through 9) standing in a circle. They pass a ball amongst themselves. Friend number 0 starts with the ball and passes it to her right. The subgroup h2i is the collection of friends who touch the ball if everyone passes 2 to their right. The subgroup h3i is the collection of friends who touch the ball if everyone passes 3 to their right, and so on.

0 1

0 1

9

8

2

2

8

in C10

in C10 7

3

6

4 5

9

3

7

4

6 5

Figure 79: Circle diagrams help to visualize generated subgroups of cyclic groups

5

101  5C

Product Groups (optional)

Try drawing your own such circle diagram to illustrate h4i and h5i in C10. Also try other cyclic groups, like C9 and C11. These diagrams reveal how cyclic groups got their names: it is often useful to imagine their members arranged in a circle (rather than listed in a line). We previously defined the “order” of a rotation (on page 12). For example, R90 has order 4 because it must be performed 4 times in a row to equal the identity. Here is a more general definition of the word “order”:

Definition The order of a member of a finite group is the size of the subgroup it generates.

This definition is consistent with our previous definition of the order of a rotation symmetry. For example, the order of R90 equals 4 (the same answer as before) because hR90i ¼ {I, R90, R180, R270} has 4 members. But now it makes sense to ask about the order of other things. For example, the order of any reflection equals 2 because if F is a reflection, then hFi ¼ {I, F} has 2 members. From the previous exercise we see that the members of C10 have the following orders: member of C10 order

0 1

1 10

2 5

3 10

4 5

5 2

6 5

7 10

8 5

9 10

5C Product Groups (optional) In this optional1 section, we consider an important construction that uses a pair of groups, called G1 and G2, to build a single new group, which is denoted G1  G2 and is called the product of G1 and G2. The members of this new group are all of the possible ways of pairing together a member of the first group with a member of the second group and (wrapped in parentheses and separated by a comma). For example, if G1 ¼ , then is one member of G1  G2. Here are all of the members: G2 ¼

×

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

Thus, G1  G2 catalogs all ways to pair a (red) member of G1 with a (green) member of G2. The coloring is optional but helpful for understanding the idea here. If the red letters are men dancers and the green numbers are women dancers, then G1  G2 catalogs all ways in which a mixed-gender dance couple could be formed.

1

This section is optional for readers who intend to skip the final section of Chapter 7.

Chapter 5 • Subgroups and Product Groups

102

In this example, G1 has 3 members, G2 has 4 members, so G1  G2 has 3  4 ¼ 12 members. This isn’t surprising, since the above list has its members arranged in a 3-by-4 grid. The general rule is expressed in the following theorem.

Theorem The size of the product of two finite groups equals the product of their sizes.

5 We will denote the algebraic operation in any product group as “”. How do you think it works? You simply do the G1 part and the G2 part separately. Let’s practice!

▶ Exercise

In the product group ℤ  D4, find (5, H )  (7, V ).

, we are asked to fill in the Solution Let’s use color to keep organized. In the group blanks: . The left (red) blank is found by combining the two left (red) things with the algebraic operation of the left (red) group (addition). The right (green) blank is found by combining the two right (green) things with the algebraic operation of the . right (green) group (composition). So the answer is

For more practice, let’s construct the Cayley table for the product of the groups . and C2 ¼ C3 ¼ The product of these two groups has the following six members:

×

= {(0,0), (0,1), (1,0), (1,1), (2,0), (2,1)}.

The Cayley table is shown in Table 7. . Just think of this The yellow cell in the table represents in C3) packaged together with a right equation as a left (red) equation ( in C2). (green) equation ( An ambidextrous person might construct the Cayley table shown in Table 7 by holding a red pen in the left hand and a green pen in the right hand. The red parts are unrelated to the green parts. Working with a product group always feels like doing two unrelated things simultaneously (the red part and the green part). So what are product groups good for? They are perfect for modelling situations in which two unrelated things happen simultaneously. Here is a key example:

▶ Exercise

Prove that the symmetry group of the illustrated border pattern is isomorphic to

ℤ  C2.

… D D D D D D D D D D D D D …

103  5C

Product Groups (optional)

Table 7: Cayley table showing the results of the algebraic operation in the product group C3  C2

• (0,0) (0,1) (1,0) (1,1) (2,0) (2,1)

(0,0) (0,0) (0,1) (1,0) (1,1) (2,0) (2,1)

(0,1) (0,1) (0,0) (1,1) (1,0) (2,1) (2,0)

(1,0) (1,0) (1,1) (2,0) (2,1) (0,0) (0,1)

(1,1) (1,1) (1,0) (2,1) (2,0) (0,1) (0,0)

(2,0) (2,0) (2,1) (0,0) (0,1) (1,0) (1,1)

(2,1) (2,1) (2,0) (0,1) (0,0) (1,1) (1,0)

Solution The proper symmetries of this border pattern are the translations, which we’ll denote {. . ., T2, T1, T0, T1, T2, . . .}. As before, Tn means the translation by n letters to the right (if n is positive) or to the left (if n is negative). The improper symmetries are: f. . . , H  T 2 , H  T 1 , H  T 0 , H  T 1 , H  T 2 , . . .g, where H denotes the reflection over the horizontal center line. In other words, the list of improper symmetries is obtained by composing H with each of the proper symmetries.2 They are all glide reflections, except for H  T0 ¼ H which is just a reflection. will match the proper symmetry Tn with and will The isomorphism with match the improper symmetry H  Tn with . In other words, the component counts reflections: “ ” means “no reflection” and “ ” means “one reflection.” Why is this matching an isomorphism? Here is an example of how it converts a true equation in the symmetry group into a true equation in the product group:

In the top equation, the composition of two glide reflections is a translation because the reflection parts cancel each other; one reflection plus one reflection equals zero reflections. Here is another example, in which the composition of a glide reflection with a translation is a glide reflection:

2

Here we’re using the “one reflection is enough” idea from the proof of the Zero-or-Equal Theorem on page 40.

5

104

Chapter 5 • Subgroups and Product Groups

C2 = {0, 1}

Table 8: The Cayley table for C2 is just like adding reflections; doing two reflections is the same as none because they cancel each other out

*

0

1

0

0

1

5 A few more examples like this will convince you that the matching converts every true equation in the symmetry group into a true equation in the product group.

The isomorphism in the previous exercise made it clear that composing symmetries of the border pattern involved doing two things simultaneously: adding integers (to find the total translation amount) and performing a C2 calculation (to decide whether it ends up reflected). Think of this C2 calculation as adding the total number of reflections. Doing two reflections is the same as none because they cancel each other, and that’s exactly how the Cayley table for C2 works (in Table 8). In the previous exercise, the border pattern’s proper symmetry group is isomorphic to ℤ, while its symmetry group is isomorphic ℤ  C2. Thus, its symmetry group is isomorphic to the product of its proper symmetry group and C2. Let’s shorthand this observation as

FULL ≅ PROPER × where “FULL” means the (full) symmetry group, “PROPER” means the proper symmetry group and “ffi” means “is isomorphic to.” This magic happened because the border pattern had an improper symmetry (namely H ) with some very special properties. To identify these special properties, think about why we previously decided that ðH  T 8 Þ  ðH  T 7 Þ ¼ T 15 : We explained it visually, but could have justified it algebraically like this: ðH  T 8 Þ  ðH  T 7 Þ ¼ ðH  H Þ  ðT 8  T 7 Þ ¼ I  T 15 ¼ T15 : The special algebraic properties being used here are (1) H is its own inverse, so H  H ¼ I, and (2) the order doesn’t matter when H is composed with translations. Another word mathematicians use for this second property is that H commutes with all translations.

5

105  5C

Product Groups (optional)

Definition A pair of members of a group (called A and B) are said to commute if the order in which they are combined doesn’t matter; that is, A  B ¼ B  A.

Our next theorem is an important generalization of the previous exercise.

The Full-Vs-Proper Theorem If an object has an improper symmetry that is its own inverse and that commutes with all of the object’s proper symmetries, then FULL ffi PROPER  C2.

In words, the theorem’s conclusion is that the object’s symmetry group is isomorphic to the product of its proper symmetry group and C2. The proof involves copying the logic of the previous exercise, and is left to the reader in the exercises.

▶

Exercise For the illustrated FULL ffi PROPER  C2?

border

pattern,

is

it

true

or

false

that

… V V V V V V V V V V V V … Solution False. The (full) symmetry group is noncommutative, while PROPER  C2 is commutative for the following reasons: (1) the proper symmetry group is commutative because it contains only translations, and (2) the product of two commutative groups is always commutative.

The border pattern in the previous exercise does not have an improper symmetry with the special properties required by the Full-vs-Proper Theorem. Vertical reflections do not commute with translations, as discussed on page 46 at the end of Chapter 2. In truth, there aren’t many substantially different two-dimensional objects to which the Full-vs-Proper Theorem applies, so the theorem’s main purpose will be for the study of three-dimensional objects (beginning in Chapter 7).

▶ Exercise

Find the order of (1, 1) in C4 3 C3.

Solution Using color to keep organized in , we find in the usual way: by , and then continuing to combine with the generator first listing the identity, which is until the next combination result would be the identity again. This procedure reveals equals that

The answer is the size of this subgroup, which is 12.

106

Chapter 5 • Subgroups and Product Groups

In the previous list, the red pattern is independent from the green pattern. This is another illustration of the mantra that product groups model situations where two unrelated things happen simultaneously. The list must be continued until both colors would next equal zero.

5

5D Elements of Mathematics: Sets versus Lists This chapter provided a few nice examples of an activity common in mathematics called generalization. For example: 1. We generalized our definition of order, so it now makes sense to discuss the order of any member in any finite group (not just the order of a rotation in a symmetry group). 2. We generalized the fact that FULL ffi PROPER  C2 was true of the border pattern …D D D D…, so we now know this equation is true of any object possessing an improper symmetry with certain special properties. Imbued as you are with this mathematical point of view, no doubt you have guessed that to generalize a definition or theorem means to make it more general; that is, to find a broader setting in which it applies. For example, Pythagoras undoubtedly began by observing that his famous theorem was true of some particular right triangles, and was then wise enough to generalize the observation to all right triangles. Generalization is so ubiquitous in mathematics, it is woven into even the starting pages of most math books. For example, Chapter 1 of this book includes several definitions and theorems about two-dimensional objects. These definitions and theorems were historically first discovered only for specific kinds of objects (like bounded ones), so even our starting point was a generalization. Moreover, Chapter 1 was worded with careful forethought so that most of its definitions and theorems will further generalize to the three-dimensional objects that we will begin studying in Chapter 7. For the remainder of this section, we will shift gears from the concept of “generalization” to a nuts-and-bolts notational issue that arose in this chapter in relation to sets and lists. As previously mentioned, a “set” roughly means a collection, like the set of all real numbers, the set of all points in the plane, or the set of all three-letter words. A finite set is often described by listing its members within curly brackets separated by commas. The order in which the members are listed is irrelevant. For example, {11, 12, 13, 14} and {12, 14, 13, 11} are different ways of describing the same set. The size (number of members) of this set is 4. It could be described in words as “the set of integers that are greater than 10 and less than 15.” As another example, h2i and h4i are the same subset of C10, even though we previously listed the members in different orders. This set could be described as “the even members of C10.” Describing the set in words helps to emphasize the point that a set is an unordered collection.

107  Exercises

In contrast, mathematicians use round parentheses for lists in which the order does matter, like the coordinates (x, y) of a point of the plane. Order matters because, for example, (2, 7) and (7, 2) are different points of the plane. Round parentheses are used to name a member of a product group because order and are different members of the product group matters. For example, × . The rule is that every member of G1  G2 must have a member of G1 on the left and a member of G2 on the right. In summary, curly brackets are used with sets, which are unordered collections. Round parentheses are used with lists, which are ordered.

Exercises (1) Write a precise definition of each of the following terms: subgroup, product group, commute, generated subgroup, order. (2) For each of the decorated squares illustrated here, list the members of D4 ¼ {I, R90, R180, R270, H, V, D, D0} that are also symmetries of the decorated square. COMMENT: Notice that each of your lists is a subgroup.

(a)

(b)

(c)

(d)

(3) For each subgroup of D4 ¼ {I, R90, R180, R270, H, V, D, D0} below, draw a decorated square whose symmetry group equals that subgroup. (a) (b) (c) (d)

{I, R90, R180, R270} {I, D} {I, V} {I}

(4) Why is there no decorated square whose symmetry group equals {I, H, V}? (5) For each group listed below, decide whether the red members form a subgroup: (a) (b) (c) (d)

C7 ¼ C10 ¼ C12 ¼ D4 ¼

(e) D4 ¼

⬘ ⬘

5

108

Chapter 5 • Subgroups and Product Groups

(f) D4 ¼

5

⬘

(6) In ℤ (the additive group of integers), do the positive integers form a subgroup? Explain your answer. (7) In the multiplicative group of nonzero real numbers, do the positive real numbers form a subgroup? Explain your answer. (8) In the multiplicative group of positive real numbers, do the positive integers form a subgroup? Explain your answer. (9) In the product group ℤ  C12, find (100,8)  (300,9). (10) In the product group D4  C15, find (R90, 10)  (R270, 11). (11) List all members of the product group C2  C3 and construct a Cayley table for this product group. COMMENT: The chapter did this for C3  C2, which is not quite the same. (12) List all members of the product group C2  C2 and construct a Cayley table for this product group. (13) List all members of the product group C3  C4. (14) For each product group, decide whether the red members form a subgroup: (a) C3 3 C2 ¼ (b) C3 3 C2 ¼ (c) C3 3 C2 ¼ (15) Suppose that G1 and G2 are groups. Prove that G1  G2 has a subgroup that is isomorphic to G1 and also has a subgroup that is isomorphic to G2. HINT: See parts (a) and (b) of the previous exercise. (16) In C4 ¼ { 0, 1, 2, 3 }, list the members of each of the following subgroups: h0i, h1i, h2i, h3i. Draw circle diagrams to illustrate these subgroups. Determine the order of each member of C4. (17) In C5 ¼ { 0, 1, 2, 3, 4 }, list the members of each of the following subgroups: h0i, h1i, h2i, h3i, h4i. Draw circle diagrams to illustrate these subgroups. Determine the order of each member of C5. (18) In C6 ¼ { 0, 1, 2, 3, 4, 5 }, list the members of each of the following subgroups: h0i, h1i, h2i, h3i, h4i, h5i. Draw circle diagrams to illustrate these subgroups. Determine the order of each member of C6. (19) In C7 ¼ { 0, 1, 2, 3, 4, 5, 6 }, list the members of each of the following subgroups: h0i, h1i, h2i, h3i, h4i, h5i, h6i. Draw circle diagrams to illustrate these subgroups. Determine the order of each member of C7. (20) In C8 ¼ { 0, 1, 2, 3, 4, 5, 6, 7 }, list the members of each of the following subgroups: h0i, h1i, h2i, h3i, h4i, h5i, h6i, h7i. Draw circle diagrams to illustrate these subgroups. Determine the order of each member of C8. (21) In C9 ¼ { 0, 1, 2, 3, 4, 5, 6, 7, 8 }, list the members of each of the following subgroups: h0i, h1i, h2i, h3i, h4i, h5i, h6i, h7i, h8i. Draw circle diagrams to illustrate these subgroups. Determine the order of each member of C9.

109  Exercises

(22) Based on the previous exercises, make a general conjecture about the possible orders of members of Cn. (23) Prove that C12 is not isomorphic to C2  C6, even though they have the same size and are both commutative. (24) Identify a wallpaper pattern whose symmetry group is isomorphic to ℤ  ℤ. (25) Determine the following orders: (a) (b) (c) (d) (e) (f)

The order of (1, 2) in C2  C7. The order of (1, 2) in C2  C8. The order of (1, 5) in C7  C10. The order of (3, 1) in C9  C4. The order of (3, 1) in C9  C6. The order of (3, 3) in C10  C6.

(26) Based on the previous exercise, make a conjecture about how the order of (A, B) in G1  G2 relates to the order of A in G1 and the order of B in G2. (27) If G is a finite group and A is a member of G, prove that hAi really is a subgroup. Also prove that the list I, A, A  A, A  A  A, . . . has no repetitions until the identity is repeated, at which point the entire list repeats. (28) In a finite group, prove that a member and its inverse generate the same subgroup. (29) Disprove this statement: “The proper symmetry group of every border pattern is commutative.” (30) Prove the Full-vs-Proper Theorem. (31) TRUE or FALSE (and explain why): A member of a finite group has order 2 if and only if it equals its own inverse. (32) DISPROVE: “A member of a symmetry group has order 2 if and only if it is a reflection.” (33) THEOREM: D2 is isomorphic to C2  C2. Prove this theorem in two different ways: (a) Explain why it’s a consequence of the Full-vs-Proper Theorem. (b) Construct an explicit isomorphism that matches each symmetry of the cross illustrated in Figure 80 with a member of . HINT: Think of the red (respectively green) number as coding whether the symmetry exchanged the ends of the red (respectively green) rectangle, with 0 ¼ no and 1 ¼ yes.

Figure 80: A cross that illustrates the product group isomorphism in Exercise 33 (▶ https://doi.org/10.1007/000-25m)

5

110

Chapter 5 • Subgroups and Product Groups

(34) Write the converse of the Full-vs-Proper Theorem. (35) Decorate the illustrated hexagon so that its symmetry group is isomorphic to D3.

5 (36) Given any object and any subgroup of its symmetry group, prove that the number of improper symmetries in the subgroup either is zero or is equal to the number of proper symmetries. (37) TRUE or FALSE: (A, B, C) ¼ (B, A, C). (38) TRUE or FALSE: {A, B, C} ¼ {B, A, C}. (39) For the D-border pattern … D D D D D D D D D D D …, how many different subgroups of its symmetry group can you think of that are isomorphic to ℤ? HINT: (40) How many different subgroups of ℤ 3 C2 can you think of that are isomorphic to ℤ? (41) Suppose that a particular object has an improper symmetry, F, that is its own inverse and that commutes with all of the object’s proper symmetries (as in the Full-vsProper Theorem). Prove that F also commutes with all of the object’s improper symmetries.

6

111

Permutation Groups

In Figure 81, you don’t need the gnome to know that R240 was performed; just the ending positions of the three letters are enough to reveal the symmetry. This ending position is “CAB” (reading clockwise around the triangle from the top). Each other symmetry of the triangle could similarly be represented as a 3-letter word according to the ending positions in which the symmetry leaves the letters. For example, do you see why R120 is represented as “BCA”? Representing symmetries as words is the key technique we’ll use in the next chapter to understand three-dimensional objects; for example, each symmetry of the tetrahedron in Figure 82 will be represented by a 4-letter word. Here is the fundamental question: given two words (each representing a symmetry), how do we determine the word represented by the composition of the two symmetries? In this chapter, we will introduce an algebraic operation on the set of 3-letter words (perhaps representing symmetries of a triangle) or 4-letter words (perhaps representing symmetries of a tetrahedron) or n-letter words (perhaps representing symmetries of some object with n parts). This algebraic operation mimics composition of symmetries. It turns the set of words into a group that we’ll call a permutation group.

6A Counting Permutations A permutation means a rearrangement. For example, there are 6 permutations of the letters A, B, and C; namely, Electronic supplementary material The online version of this chapter (https://doi.org/10.1007/ 978-3-030-51669-7_6) contains supplementary material, which is available to authorized users. The videos can be accessed individually by clicking the DOI link in the accompanying figure caption or by scanning this link with the SN More Media App.

© Springer Nature Switzerland AG 2021 K. Tapp, Symmetry, Texts for Quantitative Critical Thinking, https://doi.org/10.1007/978-3-030-51669-7_6

111

112

Chapter 6 • Permutation Groups

Figure 81: R240 is represented by the word “CAB”

6

A

Figure 82: In the next chapter, we'll similarly represent symmetries of a tetrahedron by 4-letter words

B

D C ABC, ACB, BAC, BCA, CAB, CBA:

Think of A, B, and C as letters on a magnet board, and think of a permutation as a word (not necessarily a real English word) that you can spell using all three of the letters.

Definition The collection of all permutations of n ordered things is denoted Pn and is called the nth permutation group.

We will soon see that Pn is a group, as its name suggests. In this book, the “n ordered things” will always be the first n capital English letters. So here are the first few permutation groups: P1 ¼ fAg, P2 ¼ fAB, BAg, P3 ¼ fABC, ACB, BAC, BCA, CAB, CBAg:

= {ABCD, ABDC, ACBD, ACDB, ADBC, ADCB, BACD, BADC, BCAD, BCDA, BDAC, BDCA, CABD, CADB, CBAD, CBDA, CDAB, CDBA, DABC, DACB, DBAC, DBCA, DCAB, DCBA}. Count them. The size of P1 is 1. The size of P2 is 2. The size of P3 is 6. The size of P4 is 24. Do you see the pattern?

113  6A

Counting Permutations

Theorem For each n  1, the size of Pn equals n!.

The symbol “n!” means the product of all of the integers between 1 and n. It is pronounced “n factorial.” Here are the first few factorials: 1! 2! 3! 4! 5! 6!

= = = = = =

1, 1 × 2 = 2, 1 × 2 × 3 = 6, 1 × 2 × 3 × 4 = 24, 1 × 2 × 3 × 4 × 5 = 120, 1 × 2 × 3 × 4 × 5 × 6 = 720.

Informal Proof We already verified the theorem when n is 1, 2, and 3. We also verified it when n is 4, but let’s think about how we could instead have used the n ¼ 3 answer to derive the n ¼ 4 answer. Color is used in the following reprinting of P3 and P4 to call attention to their relationship:

= {ABC, ACB, BAC, BCA, CAB, CBA}. = {ABCD, ABDC, ACBD, ACDB, ADBC, ADCB, BACD, BADC, BCAD, BCDA, BDAC, BDCA, CABD, CADB, CBAD, CBDA, CDAB, CDBA, DABC, DACB, DBAC, DBCA, DCAB, DCBA}. The members of P4 are listed in 4 rows (one for each starting letter). The four rows all have the same length. The bottom D-row makes it clear that this common length equals the size of P3. Thus, ðSize of P4 Þ ¼ 4  ðSize of P3 Þ ¼ 4  6 ¼ 24: We can similarly use the n ¼ 4 answer to derive the n ¼ 5 answer; in fact, the same logic gives ðSize of P5 Þ ¼ 5  ðSize of P4 Þ ¼ 5  24 ¼ 120: Do you see how we can verify that the theorem is true for each successive integer based upon its truth for the previous integer? Since the theorem is true for n ¼ 5, it must be true for n ¼ 6. Since the theorem is true for n ¼ 6, it must be true for n ¼ 7, and so on. At the end of this chapter, we will introduce the method of inductive proof, which just formalizes what you probably already feel in your heart: this reasoning suffices to prove that the theorem is true for all integers n  1. □

6

114

Chapter 6 • Permutation Groups

6B Cycle Notation and Composition

6

Listing and counting permutations is not nearly enough. Our next goal is to define an algebraic operation that turns Pn into a group. This operation will be called composition and denoted “” exactly as in symmetry groups because it’s designed to mimic composition of symmetries. When A and B are permutations, we want A  B to mean the same thing it means when A and B are symmetries; namely, the result of first performing B and then performing A. What does it mean to “perform” a permutation? A permutation is a word, but it can also be regarded as a performable action; namely, the exchanging/moving/cycling of letters necessary to build that word on the magnet board from the alphabetical starting position. Cycle notation concisely describes this action. Here is an example in P6.

▶ Exercise

Express BADEFC in cycle notation.

Solution The arrow picture in Figure 83 shows the action that converts letters on a magnet board from ABCDEF to BADEFC. The cycle notation is (12)(3654), which concisely describes this action.

More explicitly, the cycle notation (12)(3654) means:

▬ Swap the first and second letters. ▬ Simultaneously cycle the remaining letters in the 3 ! 6 ! 5 ! 4 ! 3 order; that is, the letter that started 3rd ends up 6th, the letter that started 6th ends up 5th, the letter that started 5th ends up 4th, and the letter that started 4th ends up 3rd. Figure 84 is a better way to illustrate this cycle notation. It requires too much typesetting space, but it’s what you should think when you see “(12)(3654)”. In particular, “)” always indicates a return to the first number in the cycle.

A

B

C

D

E

1

2

3

4

5

6

B

A

D

E

F

C

2

3

4

5

6

1

F

Figure 83: The exchanging/moving/cycling of letters necessary to build BADEFC (▶ https://doi.org/10.1007/000-25r) Figure 84: (12)(3654) is shorthand that represents these circular diagrams

1 2

4

3 5

6

6

115  6B

Cycle Notation and Composition

The length of a cycle means the number of letters involved. For example, (12)(3654) has a cycle of length 2 and a cycle of length 4. Here are more practice problems with cycle notation.

▶ Exercise

Express the following members of P6 in cycle notation: FABCED, CEAFBD, and CABEFD.

Solution

A B 1

2

F

A

1

2

C

D

E

F

3

4

5

6

A B 1

B

C

E

D

C

3

4

5

6

1

C

D

E

F

C

D

E

F

2

3

4

5

6

A B 1

2

3

4

5

6

E

A

F

B

D

C

A

B

E

F

D

2

3

4

5

6

1

2

3

4

5

6

A number that doesn’t occur in the cycle notation indicates a letter that remains in its original position. For example, “5” doesn’t occur in FABCED ¼ (12346) because the E remains in its original 5th position. If a permutation has multiple cycles, like CEAFBD ¼ (13)(25)(46), which has three cycles, then these cycles are always disjoint (they have no numbers in common). These disjoint cycles decompose the action into its independent simultaneously performed parts. With a bit of practice, you’ll be able to express a word in cycle notation without needing to first draw an arrow picture, and you’ll come to appreciate how cycle notation concisely encodes the key features of the action. In summary, a member of a permutation group can be regarded either as a word or an action; namely, the action that moves a magnet board from alphabetical to that word. This action is described by an arrow picture, or much more concisely by cycle notation. Since permutations can be viewed as performable actions, the following definition makes sense: when A and B are permutations, A  B means the result of first performing B and then performing A. It’s just like composing symmetries because that’s what it’s designed to mimic. For practice, we’ll compose two of the members of P6 whose cycle notations were found in the previous exercise.

▶ Exercise

Find FABCEDCABEFD.

Solution Starting from the alphabetical word ABCDEF, we first perform the action for CABEFD ¼ (123)(465) and next the action for FABCED ¼ (12346). See Figure 85. So the solution is FABCED * CABEFD ¼ DCABFE.

But wait—we did too much work! The above approach nicely highlights the analogy between composing permutations and composing symmetries, but its first step (the grey-

Chapter 6 • Permutation Groups

116

A

B

C

D

E

F

1

2

3

4

5

6

First perform CABEFD = (123) (465)

C

A

B

E

F

D

1

2

3

4

5

6

D

C

A

B

F

E

1

2

3

4

5

6

Next perform FABCED = (12346)

6

Figure 85: Composition of FABCED ¼ (12346) with CABEFD ¼ (123)(465) in P6 (▶ https://doi.org/10.1007/000-25p)

boxed portion) is completely unnecessary. If you start alphabetical and perform the action for CABEFD, of course you get CABEFD. For practical purposes, it’s best to erase the grey-boxed portion of the above solution, so that the technique simplifies to this: To find the composition of two words, copy the word on the right and do to it the action for the word on the left.

In particular, you really only need the action for the word on the left, and you can understand this action via either the arrow picture or the cycle notation. Let’s practice.

▶ Exercise

Find CEAFBDFEDCBA.

Solution In a previous exercise, we found the action for the word on the left: CEAFBD ¼ (13)(25)(46). We must perform that action to the word on the right, like this:

F

E

D

C

B

A

1

2

3

4

5

6

D

B

F

1

2

3

Perform (13) (25) (46)

A 4

E 5

C 6

So the answer is CEAFBD * FEDCBA ¼ DBFAEC.

The answer might seem like a dreamlike intermixing of math class with spelling class. To compose permutations, you must alternate between regarding them as words and as actions. It’s what’s needed to make a group:

Theorem Pn is a group.

We will not include a complete proof of this theorem, but we hope it is believable. The identity is the alphabetical word, which represents the “do nothing” action. Every permutation has an inverse because, no matter how I scramble the letters on a magnet

117  6C

Even and Odd Permutations

Cayley table for

∗

ABC BCA CAB BAC ACB

ABC = BCA = (132) CAB = (123) BAC = (12) ACB = (23) CBA = (13)

CBA

CAB

Figure 86: Permutations of the letters A, B, and C and Cayley table for the permutation group P3 (▶ https://doi.org/10.1007/000-25q)

board, it is always possible for you to unscramble them back to alphabetical. One can verify the associative property for composing permutations using roughly the same trick we previously used to verify it for composing symmetries. For practice, take time now to fill out the Cayley table for P3 shown in Figure 86. We recommend using magnet letters. For each cell of the table, start with the word on the top and do to it the action for the word on the left. To help with this, the Cayley table shown in Figure 86 conveniently includes the cycle notation for each word on the left. Notice the convention that the cycle notation for the identity (alphabetical) word is .

6C Even and Odd Permutations Other than the identity, the simplest permutations are the swaps:

Definition A swap is a permutation whose cycle notation is a single cycle of length 2.

For example, ABEDCF = (35) is a swap in P6. Thinking of permutations as words, a swap is a word with two letters out of place. Thinking of permutations as actions, a swap is the act of exchanging two letters. Both are valid viewpoints, and they say the same thing.

▶ Exercise

Express EADCFB as a composition of swaps in P6.

Solution With our magnet board starting alphabetical, we must perform a sequence of swaps (letter exchanges) to arrive at the ending position EADCFB. We’ll work from left to right, moving the E to the first position, then the A to the second, and so on, as shown in Figure 87. Here the swaps are color coded, starting with positions 1 & 5, then 2 & 5, then 3 & 4, and finally 5 & 6. Thus:

EADCFB = (56) ∗ (34) ∗ (25) ∗ (15).

6

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Chapter 6 • Permutation Groups

ABCDE F

®

E BCD A F E BC D AF

®

E AC D BF E ACD BF

®

E ADC BF E ADC BF

® E ADC FB .

Figure 87: A sequence of swaps (letter exchanges) arriving at the permutation EADCFB (▶ https://doi.org/10.1007/000-25n)

6

To avoid potential confusion, we mention that the previous equation has nothing to do with cycle notation; it is a composition of four permutations, not the cycle notation of any single permutation. The “*” symbols are not removable because the cycles are not disjoint. There is clearly nothing special about the word EADCFB. The same left-to-right strategy would work to express any permutation as a composition of swaps. This is expressed in terms of the group viewpoint in the following theorem.

Theorem Every permutation in Pn equals a composition of swaps.

In fact, there are many ways to express a permutation as a composition of swaps. In the previous EADCFB example, how many swaps are needed if we instead work from right to left? What if we only use adjacent swaps (exchanges of adjacent letters)? What if we correctly position the vowels first and then the consonants? What about a wasteful strategy like swapping our way from alphabetical to some randomly chosen 6-letter word and then to EADCFB? Try several different strategies, and count the number of swaps needed for each strategy. Some strategies require 4 swaps (including the left-to-right strategy discussed above). Others might require 6 or 8 or 10 swaps. A grossly inefficient strategy might require 76 swaps. Here is the important point: no matter what strategy you employ, you will require an even number of swaps. This is because EADCFB is an even permutation.

Definition A permutation that can be obtained by composing an even number of swaps is called an even permutation. A permutation that can be obtained by composing an odd number of swaps is called an odd permutation.

The importance stems from the following theorem.

The Even-or-Odd Theorem A permutation cannot be both even and odd.

119  6C

Even and Odd Permutations

CDAEB

ABCDE AB BC CD DE

AC BD CE

AD BE

AE

Figure 88: Video proof of the Even-or-Odd Theorem (▶ https://doi.org/10.1007/000-25s)

In other words, if you obtain your favorite word in Pn from the alphabetical starting position using an even number of swaps, then this word is even—anyone else would also require an even number of swaps to obtain it. For practice, verify using several different swap strategies that CADEFB requires an odd numbers of swaps, and is therefore an odd permutation. We recommend using magnet letters (or just cut-out paper letters). Since this theorem is of fundamental importance to higher mathematics, we will include a proof, but the proof is optional in the sense that nothing later in the book will depend on understanding it (Figure 88). Proof (optional) The key idea is the following alternative way to test whether a word is even or odd: first list all pairs of letters, then circle the pairs that appear in reverse alphabetical order within the word, then check whether you circled an even or odd number of pairs. For example, CDAEB is an odd word in P5 because there is an odd number (five) of circled reverse-alphabetical pairs here:

CDAEB:

AB AC AD AE BC BD BE CD CE DE

Note that the list of 10 pairs would be the same for any 5-letter word, but which pairs get circled depends on the word. Performing a single swap will always change whether the answer is even or odd. For example, the swap turns the previous odd word into an even word:

CEADB:

AB AC AD AE BC BD BE CD CE DE

A swap changes the answer because it changes the circled status of an odd number of pairs. Don’t worry about the two separate types of changes: from circled to non-circled and vice versa. All that matters is that the total number of changes (including both types) is odd. To understand why it’s odd, these changes are marked with stars in the previous example. Since the letters D and E were swapped, the pair DE changed. Furthermore, all other changes

6

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Chapter 6 • Permutation Groups

come in twos. Specifically, each letter between D and E (in this example, A is the only such letter) is responsible for two changes because its pairing with D and its pairing with E are both changed. Thus, the total number of changes is odd. The alphabetical word is even because 0 pairs would be circled, and 0 is an even number. So if one begins alphabetical and then performs a sequence of swaps, the following two questions are guaranteed to have the same answer:

1. Was the number of swaps even or odd? 2. Does the ending word have an even or odd number of circled pairs?

6

The answer to the first question is our definition of even/odd. The answer to the second only depends on the ending word, not on the sequence of swaps, and thus the same must be true of the answer to the first question! In summary, whether the number of swaps is even or odd depends only on the ending word, not on the choice of swap sequence that was used to arrive at the ending word. □

The following follow-up result is also important.

Theorem If n  2, then exactly half of the permutations in Pn are even. Furthermore, the even permutations form a subgroup of Pn. Proof To show that the even permutations form a subgroup, we must verify three things. First, the identity is an even permutation because we can obtain it using 0 swaps, and 0 is an even number. Second, the composition of two even permutations is an even permutation. For example, if the first permutation uses 8 swaps and the second uses 12 swaps, then the composition uses 8 + 12 ¼ 20 swaps (the sum of two even numbers is even). Finally, the inverse of an even permutation is even. For example, the inverse of (56) ∗ (34) ∗ (25) ∗ (15) is the same swaps done in the reverse order, namely,

(15) ∗ (25) ∗ (34) ∗ (56). If your friend performs a sequence of swaps, you can return to the starting position by undoing your friend’s swaps in reverse order. This verifies that the even permutations form a subgroup of Pn. We leave it to the reader in the exercises to prove that exactly half of them are even (borrowing the key idea used to prove the Zero-or-Equal Theorem). □

Definition The subgroup of all even permutations in Pn is denoted An and is called the nth alternating group.

According to the previous theorem, the size of An equals half the size of Pn. Here are the sizes of some permutation and alternating groups:

121  6C

Even and Odd Permutations

Size of P2 ¼ 2

Size of A2 ¼ 1

Size of P3 ¼ 6 Size of P4 ¼ 24

Size of A3 ¼ 3 Size of A4 ¼ 12

Size of P5 ¼ 120

Size of A5 ¼ 60

The groups P4, A4, and A5 will play starring roles in the next chapter, so remember their sizes, and keep an eye out for them. We end this section with an alternative method for deciding whether a permutation is even or odd. Instead of counting swaps, you can decide using the cycle notation for the permutation.

Theorem Any cycle of length m can be obtained by composing m  1 swaps.

For example, the word BCDEFA in P6 is expressed in cycle notation as (165432). This is a cycle of length 6. Using a magnet board, check that BCDEFA ¼ (56) ∗ (45) ∗ (34) ∗ (23) ∗ (12), so it is obtained from 5 swaps. These 5 swaps “bubble” the first letter A to the end of the word (Figure 89).

▶ Exercise

Is BADEFC even or odd?

Solution Use the cycle notation for BADEFC like this:

{

{

BADEFC = + The number of swaps required for each cycle is one less than the length of the cycle. So it takes 1 swap for the first cycle (which literally is 1 swap) plus 3 swaps for the second cycle for a total of 4 swaps. Since 4 is an even number, the permutation is even.

ABCDEF 1

2

3

4

5

6

B CDE FA 1

2

3

4

5

6

Figure 89: How many swaps are needed for a cycle of length 6? (▶ https://doi.org/10.1007/000-25t)

6

122

Chapter 6 • Permutation Groups

6D Symmetries and Permutations Let’s return to the idea with which we introduced this chapter: coding symmetries as words. In Figure 90, the triangle’s vertices (corners) are labeled A, B, C. Each member of the triangle’s symmetry group, the dihedral group D3, is coded as a 3-letter word according to the ending positions of these vertices after the symmetry is applied (read clockwise around the triangle from the top). This matching of symmetries with 3-letter words is an isomorphism. This is expressed in terms of the group viewpoint in the following theorem.

6 Theorem The dihedral group D3 is isomorphic to the permutation group P3.

Idea of Proof The matching in Figure 90 is clearly a one-to-one matching between the six symmetries of the triangle and the six 3-letter words. We therefore need only verify that this matching converts each true equation into a true equation. In other words, the way that the composition of two symmetries permutes the vertices must equal the composition of the ways that those symmetries individually permute the vertices. Our definition of composition of

I

R120

R240

= ABC I

= BCA (132)

= CAB (123)

Figure 90: An isomorphism between D3 and P3

F1

F2

F3

= BAC (12)

= ACB (23)

= CBA (13)

6

123  6D

Symmetries and Permutations

permutations in this chapter was custom designed to mimic composition of symmetries in exactly this sense. □

In summary, since the symmetries of a triangle permute its 3 vertices, the triangle’s symmetry group is isomorphic to P3. How eloquent! But to highlight the care needed to draw other conclusions similar to this in the future, let’s consider some things that can go wrong when we play this game.

▶ Exercise

What is wrong with the following reasoning? “Since the symmetries of a square permute its 4 vertices, the symmetry group of a square is isomorphic to P4.”

Solution The conclusion is incorrect because D4 (the symmetry group of a square) has 8 members while P4 has 24 members. Different-sized groups are never isomorphic. But what is wrong with the reasoning? As we did with a triangle, we could try matching each symmetry of a square with the 4-letter word that codes the ending positions of the square’s 4 vertices after the symmetry is applied (say, read clockwise around the square from the top left, as in Figure 91). But this strategy does not give a one-to-one matching between D4 and P4. Among the 24 four-letter words, only 8 can be achieved by symmetries; the remaining 16 words are left unmatched. For example, there is no symmetry that leaves the vertices in the ABDC order. Figure 91: Where will A, B, C, and D appear after one of the square’s symmetries is applied?

A

B

D

C

In problems like the previous exercise, the characterizing property of an isomorphism (matching true equations with true equations) is always satisfied, because we customdefined “composition” of permutations to align with “composition” of symmetries. So the only real issue is whether the correspondence between symmetries and words is a one-to-one matching. In the previous exercise, there were more words than symmetries. In the following exercise, there are more symmetries than words.

▶ Exercise

What is wrong with the following reasoning? “Since the symmetries of a square permute its 2 diagonals, the symmetry group of a square is isomorphic to P2.”

Solution The conclusion is incorrect because D4 has 8 members while P2 ¼ {AB, BA} only has 2 members. But what is wrong with the reasoning? In Figure 92, the diagonals are labeled A (amber) and B (blue). We could try matching each symmetry of the square with the 2-letter word that codes the ending positions of the two colors after the symmetry is applied (say, starting at the

124

Chapter 6 • Permutation Groups

Figure 92: Which of the square’s symmetries leave the diagonals in their original order and which do not?

6

A

B

top left). But this strategy does not give a one-to-one matching between D4 and P2. In fact, among the 8 members of D4 ¼ ⬘ , all four of the red symmetries leave the diagonals in their original AB order, while all four black members move them into the BA order.

The next chapter, Chapter 7, will include several fundamental statements of this form: “Since the symmetries of [some object] permute its [some n parts], the object’s symmetry group is isomorphic to Pn.”

The previous exercises help clarify the condition under which such a claim is valid: the symmetries must match with the permutations in a one-to-one manner. Every permutation (word) must correspond to one and only one symmetry. The examples in Chapter 7 will involve three-dimensional objects, but we hope that our previous two-dimensional examples prepared you by spotlighting this essential issue.

6E Elements of Mathematics: Well-Defined Life is full of ambiguous instructions, like when a pirate will tell you that the treasure is buried 500 paces north without specifying whether he means your small paces or his own large swaggers. Mathematicians are more careful than pirates. All definitions in mathematics must be free of ambiguity. When a definition requires arbitrary choices to be made, we must prove that the final answer doesn’t depend on these choices, so that everyone is guaranteed to get the same final answer. A definition is called well-defined if it is unambiguous, or more specifically, if what is being defined is independent of the way in which any arbitrary choices are made. For example, the definition of even and odd permutations can be rephrased as follows: “To decide whether a word is even or odd, count swaps going from alphabetical to that word, and check whether your count is even or odd.” This definition is welldefined (that is, independent of the swap strategy used) according to the Even-or-Odd Theorem. Different swap strategies yield different counts, but all strategies agree about whether the count is even or odd. We next consider some related well-definedness issues that can arise when we match symmetries with permutations, as we did in the previous section. The first is an issue that arises when it is colors that are permuted. Care is needed when claiming that the symmetries of an object permute a color arrangement, as the next exercise demonstrates.

125  6E

Elements of Mathematics: Well-Defined

▶ Exercise

For which of the squares illustrated in Figure 93 is it true that the symmetries of the square permute the two colors in a well-defined manner?

1

2

3

4

Figure 93: Various two-color arrangements in a square

Solution All except the 4th square.

To explain this solution, let’s call the colors A (amber) and B (blue). In the 4th square, R90 is one example of a symmetry that doesn’t permute the colors in a well-defined way. Whether you think R90 moves A to A or moves A to B depends on which amber circle you look at. As seen in Figure 94, there is an amber circle that rotates onto an amber circle, but there’s another amber circle that rotates onto a blue circle. The process of associating R90 with a color permutation is not well-defined here. Which amber circle you look at is an arbitrary decision, and how you make this decision affects the final answer. But only the 4th square is a dud. Squares 1, 2, and 3 do have well-defined matchings of symmetries with color permutations. In fact, these three squares have the same matching, which works just like Figure 92; namely, the red symmetries in D4 ¼ ⬘ match with AB (the identity permutation that preserves the colors) while the black ones match with BA (which swaps the colors). In summary, when it is colors that are permuted, one must check that each symmetry really does permute the colors in a well-defined way. A different issue arises when one asks whether symmetries permute parts in even or odd ways. That is the question in the next exercise.

▶ Exercise ways?

A to A A to B

Figure 94: The 90 degree rotation does NOT permute the two colors in a welldefined manner

Which symmetries of a triangle permute its vertices in even ways? Which in odd

6

126

Chapter 6 • Permutation Groups

F2

F2 F3

A

C

I

R120

6

R240 F1

B

B

=I BCA = (132) ABC

I

F1 C

ABC

R120

CAB

I (123)

BCA

(132)

= (12)

R240 F1

CBA

(13)

ACB

(23)

BAC

(12)

CAB = BAC

A

F3

F1

(123)

F2

ACB

= (23)

F2

F3

CBA

= (13)

F3

original isomorphism

alternative isomorphism

Figure 95: Two isomorphisms from D3 to P3 (▶ https://doi.org/10.1007/000-25v)

Solution In Figure 90 on page 122, the rotations match with even permutations and the reflections with odd permutations.

The previous solution is complete and correct, but it hides an arbitrary choice, so we must confirm that this choice didn’t affect the final answer. What choice? In Figure 90, we arbitrarily chose to read words clockwise around the triangle starting from the top. What if instead we had read words counterclockwise starting from the top? This strategy would have yielded a different matching between symmetries and words, as shown in Figure 95. The alternative isomorphism gives the same answer: rotations match with even permutations and reflections with odd permutations. The matching changed, but here’s what stayed the same: the non-identity rotations matched with cycles of length 3 (which are even), while the reflections matched with cycles of length 2 (which are odd). The vertex order is irrelevant because, even without nailing down the explicit matching, it is visually clear that each rotation cycles three vertices, while each reflection swaps two vertices. The core issue here is that the vertices of a triangle don’t have any best order, yet they must be given some order to study their permutations. We will soon engage more questions of this form: “Which symmetries of [some object] permute its [some parts] in even ways?”

Such questions and the process for answering them are well-defined (independent of how the parts are ordered). The part-ordering doesn’t affect the essential structure of the cycle notation, and hence doesn’t affect the even-or-odd question.

127  6F

Elements of Mathematics: The Inductive Proof (optional)

6F Elements of Mathematics: The Inductive Proof (optional) Look back at the theorem on page 113 and its proof. The theorem claimed that a certain fact was true for all integers n  1 (namely, that the size of Pn equals n!). We proved this with an informal version of an inductive proof. Here are the steps of a formal inductive proof (also called a proof by induction): To prove by induction that a certain fact is true for all integers n  1: (1) BASE CASE: Verify that the fact is true for n ¼ 1. (2) INDUCTIVE STEP: Consider an arbitrary integer n  1, and show that the fact’s truth for n implies the fact’s truth for n + 1. That is, assume the fact is true for that particular choice of n, and use this assumption to help you conclude that the fact is also true for n + 1. We previously did the inductive step a bit too informally. We explained why the fact’s truth for n ¼ 3 implied its truth for n ¼ 4, and why the fact’s truth for n ¼ 4 implies its truth for n ¼ 5. But to really convince the reader that this pattern continues forever, we should have used an arbitrary variable n instead of just a few specific choices for n in our proof. So our proof should have looked like the “re-proof” below.

Theorem For each n  1, the size of Pn equals n!. Re-Proof We will prove the theorem by induction. Base Case The theorem is true for n ¼ 1 because P1 ¼ {A} has exactly 1! ¼ 1 member. Inductive Step Consider an arbitrary integer n  1, and assume that the theorem is true for n; that is, assume we know for this particular choice of n that the size of Pn equals n! (this assumption is called the inductive hypothesis). Using this, we wish to prove that the theorem is also true for n + 1; that is, we wish to prove that the size of Pn + 1 equals (n + 1)!. For this, first notice that the members of Pn + 1 can be naturally arranged into a grid with n + 1 equal-length rows (one for each starting letter). The length of each row equals the size of Pn, because the final row is formed by sticking the final letter in front of each member of Pn. Thus,

Notice that the equality of the two red expressions above is exactly the inductive hypothesis. □

6

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Chapter 6 • Permutation Groups

Figure 96: Induction works like dominoes

6 Wow, this is a whole new type of proof! It’s a whole new method for proving that a certain fact is true for all of the infinitely many integers n  1. Visualize these integers as the infinite row of dominoes in Figure 96. Visualize the ones for which the fact is true as the ones that get knocked over. The base case is like a promise that the first domino will get knocked over. The inductive step is like a promise that no matter which domino you look at, it is properly aimed at the next one in line, so that if it falls then so will the next one. Do you see why we can conclude from this that all of the dominos will fall? At the point in the proof where you assume the inductive hypothesis, you might worry to yourself, “Aren’t I assuming exactly what I’m trying to prove?” But the answer is no. You’re trying to prove that a certain fact is true for all infinitely many positive integers. The inductive hypothesis is just the assumption that the fact is true for one specific (arbitrarily chosen) integer n. You then must use this hypothesis to prove that the fact is also true for the next integer. The method of induction is powerful. It can be used to prove non-obvious theorems. Sometimes it’s the only method by which an important theorem can be proven.

▶ Exercise

At the beginning of the day, a gumball machine contains an odd number of red gumballs and an odd number of yellow gumballs. The machine dispenses two gumballs at a time. It is used throughout the day until it becomes empty. Prove that at some point in the day it dispensed a mixed gumball pair (which means a red and a yellow). Solution Let n denote the number of gumball pairs in the machine at the beginning of the day. For example, if it begins with 11 red and 9 yellow (so 20 total gumballs), then there are n ¼ 10 pairs. This is the number of times it will be used in the day. We will prove by induction that no matter how large n is (no matter how full the machine starts), the theorem is true—it will at some point dispense a mixed pair. Base Case If n ¼ 1, then the machine contains a single mixed pair, so that’s what it will dispense the one and only time it is used in the day.

129  Exercises

Inductive Step Let n  1 be an arbitrary integer, and assume that the theorem is true for n; that is, assume we know for this particular choice of n that any machine beginning with n pairs is guaranteed to at some point in the day dispense a mixed pair. This is the inductive hypothesis. Now consider a machine that begins with n + 1 pairs. We wish to prove that it must at some point in the day dispense a mixed pair. The first time it is used, if it dispenses a mixed pair, then the proof is done. On the other hand, if it dispenses two of the same color, then there will be n pairs left in the machine (and still an odd number of each color), so the inductive hypothesis implies that one of the remaining n uses of the machine in the day will yield a mixed pair.

Exercises (1) Write a precise definition of each the following terms: permutation group, composition (of permutations), swap, even, odd, alternating group, well-defined. (2) If you have not yet done so, fill in the Cayley table for P3 in Figure 86 on page 117. (3) What is the size of P7? What is the size of A7? (4) Express each of these members of P6 in cycle notation: (a) (b) (c) (d) (e)

CBAFDE, BCDAFE, FABCDE, EDAFCB, AFCBED.

(5) Determine whether each permutation in the previous exercise is even or odd. (6) Express each of these members of P6 in word notation (that is, find the word that corresponds to the given cycle notation): (a) (b) (c) (d)

(136)(245), (13524), (13)(26)(45), (163)(25)

(7) Determine whether each permutation in the previous exercise is even or odd. (8) Find each composition in P6: (a) (b) (c) (d)

CBAFDEBCDAFE, BCDAFECBAFDE, FABCDEEDAFCB, EDAFCBFABCDE.

6

Chapter 6 • Permutation Groups

130

(9) Find each composition in P5: (a) (b) (c) (d)

EDBCAEABCD, BCDEAABDEC, EBADCEDACB, EDACBAEBCD.

(10) What is the order of a cycle of length m? Explain your answer. (11) List all 24 members of P4. Express each permutation in word notation and in cycle notation. Then do the following with your list:

6

(a) Draw a star next to each even permutation (each member of A4). (b) Label the vertices of a square as in Figure 97. For each of the 8 symmetries of the square, write the symmetry’s name next to the word on your list that represents the way it permutes the vertices (reading clockwise around the square starting at the top left). (c) Verify that the symmetries that permute the vertices in even ways are: {I, R180, H, V}. (12) In the previous exercise, instead of reading words clockwise around the square, what if we read in the “page of a book” order, so that the identity position is as in Figure 98? Verify that the solution to part (b) changes, but the solution to part (c) stays the same. (13) Which of the symmetries of a regular n-sided polygon permute its vertices in even ways? Which in odd ways? Answer this question for (a) (b) (c) (d) (e)

n ¼ 2, n ¼ 3, n ¼ 4, n ¼ 5, n ¼ 6:

Explain your answers. Conjecture a pattern. HINT: The form of the answer repeats every four steps. (14) Explain why every permutation in Pn can be expressed as a composition of adjacent swaps. Figure 97: What permutation of ABCD comes from each of the square’s symmetries?

Figure 98: What permutation of ABCD comes from each symmetry when reading in the “page of a book” order?

A

B

D

C

A

B

C

D

131  Exercises

(15) Express the identity permutation in P7 in several different ways as a composition of swaps. Check that you always have an even number of swaps. (16) Prove that exactly half of the permutations in Pn are even (if n  2). HINT: Use the main idea in the proof of the Zero-or-Equal Theorem from Chapter 2. (17) Given any subgroup of Pn, prove that either all or half of its members are even. (18) On a final exam, students are asked to match 5 paintings with 5 artists. What do you think is the fairest method for grading such a matching problem? For example, try regarding each student’s answer as a permutation in P5 and scoring based on how “close” it is to the identity permutation (the correct answer). How should “close” be measured? (19) Is P2 isomorphic to C2? Is A3 isomorphic to C3? Explain. (20) Find the inverse of each of the following members of P6: (a) (b) (c) (d) (e)

CBAFDE, BCDAFE, FABCDE, EDAFCB, AFCBED.

Describe a general method (using cycle notation) to find the inverse of a member of a permutation group. (21) Find the order of each of the following members of P6: (a) (b) (c) (d) (e)

(22) (23) (24) (25) (26)

CBAFDE, BCDAFE, FABCDE, EDAFCB, AFCBED.

Describe a general method (using cycle notation) to find the order of a member of a permutation group. Is the reverse-alphabetical word in Pn (BA in P2, CBA in P3, DCBA in P4, etc.) even or odd? Explain your answer. Prove that Pn is noncommutative for n  3. Prove or disprove: A permutation is a swap if and only if it has order 2. List all members of the generated subgroup h(3, CAB)i in the product group C4 3 P3. What is wrong with the following reasoning about the illustration? “Since the symmetries of a square permute its 4 colored lines, the symmetry group of a square is isomorphic to P4.”

6

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Chapter 6 • Permutation Groups

More precisely, answer the following questions: (a) Are there any permutations that correspond to zero symmetries? (b) Are there any permutations that correspond to multiple symmetries? (27) What is wrong with the following reasoning about the illustration? “Since the symmetries of a hexagon permute its 3 colored diagonals, the symmetry group of a hexagon is isomorphic to P3.”

6 More precisely, answer the following questions: (a) Are there any permutations that correspond to zero symmetries? (b) Are there any permutations that correspond to multiple symmetries? (28) For which of the illustrated hexagons is it true that the symmetries of the hexagon permute the colors (purple, green, and orange) in a well-defined manner?

(29) Here is a definition I just made up and named after myself: “To decide whether a permutation (word) is Kris or non-Kris, count swaps going from alphabetical to that word, and check whether or not your count is a multiple of three.” Is this definition well-defined? Explain your answer. (30) Does there exist a permutation or alternating group that is isomorphic to the symmetry group of a square? HINT: Consider sizes. (31) Prove by induction: Any set with size n  1 has exactly 2n subsets (including the empty subset). (32) Prove by induction: In a line of at least 2 adults, if the first is a woman and the last is a man, then somewhere in line there is a man standing immediately behind a woman. (33) A bead necklace is made from equal numbers of red and yellow beads. Going around the necklace clockwise, you keep track of the number of red and yellow beads you have passed. You call the trip around the necklace “red happy” if at all times during the trip the number of red beads you have passed is at least as big as the number of yellow beads you have passed. Prove that it is possible to find a starting bead that will result in a red happy trip around the necklace. HINT: Use an inductive proof with n denoting the number of pairs of beads.

133  Exercises

(34) Prove by induction that the following is true for all n  1: 1þ2þ⋯þn¼

nðn þ 1Þ : 2

(35) Prove by induction that the following is true for all n  1: 1 þ 2 þ 22 þ 23 ⋯ þ 2n ¼ 2nþ1  1: (36) Prove by induction that the following is true for all n  1: 1 þ 3 þ 5 þ ⋯ þ ð2n  1Þ ¼ n2 : (37) Prove by induction that the following is true for all n  1: 2n1  2n  1: (38) Prove by induction that the following is true for all n  1: 1  1! þ 2  2! þ ⋯ þ n  n! ¼ ðn þ 1Þ!  1: (39) A finite number of lines are drawn in the plane. They divide the plane into regions. Prove that it is possible to color these regions with just two colors (so each region will be either black or white) such that adjoining regions (those sharing a line segment as a common border) have different colors. HINT: Use induction with n denoting the number of lines. When a new line is added to an already successfully colored map, what happens when you reverse the colors on one side of the new line? (40) Find the flaw in the following incorrect proof of a false theorem: Theorem: In any gumball machine, all of the gumballs have the same color. Proof: We prove this by induction, with n denoting the number of gumballs in the machine. When n ¼ 1, the theorem is certainly true. Let n  1 be an arbitrary integer, and assume that the theorem is true for n; that is, assume we know for this particular choice of n that in any machine with n gumballs, the gumballs must all have the same color. Now consider a machine containing n + 1 gumballs. Randomly choose one gumball and remove it. By the inductive hypothesis, the remaining gumballs all have the same color. Let’s call this color green. Now put the one gumball back in and remove a different one. By the inductive hypothesis, the remaining ones all have the same color, and since we already said that some of them are green, this common color must be green. Thus, all n + 1 gumballs are green. (41) (The Handshaking Lemma) During a party event one evening, many handshakes occurred between pairs of guests. Prove that the number of guests who shook hands with an odd number of other guests was even. HINT: Use induction, with n denoting the total number of handshakes that occur (not the number of guests).

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Now at last we begin our study of three-dimensional (3D) objects, whose definition is analogous to our previous definition of two-dimensional (2D) objects.

Definition Space is the set of all ordered triples (x, y, z) of real numbers. A three-dimensional object is a nonempty subset of space.

For example, the 3D objects pictured in Figure 99 are built from red and yellow faces, silver edges, and gold vertices. But to match the technical definition, one should think of each object as located at some particular (x, y, z) coordinates, and notice that we’re only modelling the regions they fill, not their colors or the materials out of which they are built. The crowning achievement of this chapter will be a 3D version of Da Vinci’s Theorem; that is, a classification of the ways in which bounded 3D objects can be symmetric. No matter what object I print on my 3D printer or sculpt from clay, you soon will be able to determine how it fits into the classification scheme. Figure 99: Some 3D objects

7A Basics of 3D Luckily, many of our previous definitions and theorems generalize effortlessly from the 2D to the 3D setting. In this section, we take a quick overview of the 3D versions of the stuff that works about the same way it worked in 2D, beginning with several familiar definitions.

Electronic supplementary material The online version of this chapter (https://doi.org/10.1007/ 978-3-030-51669-7_7) contains supplementary material, which is available to authorized users. The videos can be accessed individually by clicking the DOI link in the accompanying figure caption or by scanning this link with the SN More Media App.

© Springer Nature Switzerland AG 2021 K. Tapp, Symmetry, Texts for Quantitative Critical Thinking, https://doi.org/10.1007/978-3-030-51669-7_7

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Definitions ▬ A rigid motion of space is a repositioning that preserves distances. ▬ A symmetry of a 3D object means a rigid motion of space that leaves the object unchanged. ▬ The identity is the rigid motion that fixes (leaves unmoved) every point of space. ▬ A 3D object is called asymmetric if the identity is its only symmetry. ▬ A 3D object is called bounded if there exists a cube in space that fully contains it; otherwise it is called unbounded. ▬ If A and B are rigid motions of space, their composition (denoted A
B) is the rigid motion of space obtained by first performing B and then performing A.

7 Intuitively, a rigid motion of space is a moving/repositioning of all of space that does not compress, expand, or otherwise distort distances. Imagine that space is completely filled with transparent ice, with an object entombed somewhere within. A rigid motion moves the entire infinite expanse of ice, and it is called a symmetry of the object if it leaves the object unchanged. Just as before, a rigid motion is best regarded as an instantaneous repositioning; only the ending position of each speck of ice matters, not the middle frames of a movie showing how it got there. Here are some familiar-sounding examples: Translation A translation moves every point of space the direction and distance specified by a single arrow. For example, imagine that the ball pattern in Figure 100 is continued infinitely in all directions, so that the balls fill up all of space (yielding a 3D analog of a wallpaper pattern). Each of the three arrows represents a translation symmetry of this infinite ball pattern. For example, the red arrow translates each ball 5 ball-positions in the direction it points. Since each ball moves exactly on top of another ball, an observer would not notice that the translation had occurred—that’s what makes it a symmetry. Translations are not very important in this chapter because we will only study symmetries of bounded objects. Just as in the 2D setting, bounded objects never have translation symmetries (other than the identity). Rotation In space, we rotate about an axis (which means a line), not about a point. Imagine children rotating about a maypole. The axes of some rotation symmetries of a cube are illustrated in Figure 101. Try to do each of these rotations to a cardboard cube in order to convince yourself that each one really is a symmetry. It’s technically correct to imagine the cube entombed in an infinite expanse of ice that rotates along with the cube. But for practical purposes, it’s fine to think of each rotation as something that’s done just to the cube.

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Figure 100: Each arrow represents a translation symmetry of the ball pattern, assuming the pattern is indefinitely extended in all directions

90°

180° 120°

order 4

order 3

order 2

Figure 101: Three axes of rotation symmetries of the cube (▶ https://doi.org/10.1007/000-25z)

The order of an axis means the order of the smallest non-identity rotation symmetry about that axis (the number of times it must be composed with itself to get the identity). For example, the axes pictured in Figure 101 have orders 4, 3 and 2 respectively, as labeled. Reflection In space, we reflect across a plane (not a line). Think of the plane as a mirror; the reflection moves each point of space (each speck of ice) to the position of its mirror image on the opposite side of the mirror. In Figure 102, the reflection across the green plane exchanges the ice above and below this plane; it is not a symmetry of the human figure because, after this reflection occurred, he would appear upside down. Nor is the reflection across the blue plane a symmetry of this figure, because such a reflection would leave him facing backwards. Only the reflection across the orange plane is a symmetry of the human figure.

It is easy to visualize a translation or a rotation being physically done to a solid object (and to the surrounding ice, if you wish). However, reflections are different. A reflection cannot be physically done to a solid object. There is nothing you can physically do to the

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Figure 102: Reflection across the orange plane is a symmetry of the human figure, while reflection across either of the other two planes is not

7 human figure that will exchange his right and left halves. The reflected image of his right hand looks like his left hand, but there is nothing you can physically do to a solid right hand to turn it into a solid left hand. Why should we care about reflections if we can’t do them? Even though reflections cannot be physically performed, they still help explain why 3D objects appear to be symmetric. The man has bilateral symmetry, which means that his only symmetry (other than the identity) is a single reflection. That’s why his left half looks the same as (or at least like the mirror image of) his right half.

Definition A rigid motion of space is called proper if, after it is applied, a solid right hand is still a right hand (or improper if it turns a solid right hand into a left hand).

Proper rigid motions are the ones that can be physically done to a solid object, like rotations and translations. Improper rigid motions are the ones that cannot be physically done to a solid object, like reflections. The following result (and its proof) works just like the corresponding statement in the 2D setting.

Theorem The collection of symmetries of any 3D object forms a group (with the algebraic operation of composition). The proper symmetries form a subgroup.

As with 2D objects, these are respectively called the (full) symmetry group and the proper symmetry group of the object.

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The 3D version of the classification theorem on page 13 says that any rigid motion of space can be obtained by composing rotations, reflections, and translations (moreover, no more than one of each kind is needed). In other words, there are no rigid motions of space other than the types we considered above and certain combinations thereof. But most of the time, we will care only about rigid motions that can be symmetries of a bounded object. Just as in the 2D setting, this restriction removes translations from the list.

The 3D Center Point Theorem Any bounded 3D object has a “center point” that is fixed (unmoved) by each of its symmetries. Moreover, each proper symmetry of the object is a rotation around an axis through this center point.

This matches the 2D Center Point Theorem on page 16, except that the 3D version doesn’t say much about improper symmetries. That’s because in 3D, the improper symmetries of a bounded object are not necessarily all reflections. We’ll postpone examples until a later section of this chapter, in order to keep the focus of this section on the stuff that works the same in 2D and 3D. The main goal of this chapter is to obtain a 3D version of Da Vinci’s Theorem. That is, we want a list of model 3D objects and a guarantee that any bounded 3D object is “equivalent” to one of the models. Our precise formulation of equivalence works just like before:

Definition Two 3D objects are called (fully) rigidly equivalent if there exists a rigid motion of space which, when applied to the first object, repositions it so that afterwards the two objects have exactly the same collection of symmetries.

For example, the green frame and dog toy illustrated in Figure 103 are rigidly equivalent: they can be superimposed and aligned just right so that afterwards they have the same proper and improper symmetries. Since the symmetries of 3D objects are harder to grasp, we’ll sometimes have to settle (at least initially) for understanding only an object’s proper symmetries. This also means focusing only on its proper symmetry group and the model object to which it is properly rigidly equivalent.

Definition Two 3D objects are called properly rigidly equivalent if there exists a rigid motion of space which, when applied to the first object, repositions it so that afterwards the two objects have exactly the same collection of proper symmetries.

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Figure 103: After repositioning these rigidly equivalent objects, they have the same symmetries

7

Just as in the 2D setting, we have a close connection between these ideas of rigid equivalence and the existence of an isomorphism.

Theorem If two 3D objects are rigidly equivalent, then their symmetry groups are isomorphic. If two 3D objects are properly rigidly equivalent, then their proper symmetry groups are isomorphic.

7B Essentially Two-Dimensional Objects This section is about the simplest types of 3D objects. We will call a 3D object essentially two-dimensional if the study of its proper symmetry group reduces to just studying something two-dimensional. Before turning this into a precise definition, let’s consider the two examples illustrated in Figure 104. The difference in their edges is important: if you turned them both upside down, the thick square would look unchanged but the beveled square would not. In approaching these examples, recall from the 3D Center Point Theorem that a bounded object’s proper symmetries are all rotations. In other words, the object’s proper symmetry group is just its rotation symmetry group. As always, we will ignore the wood grain and other color variations and imperfections of these sample objects.

Figure 104: The beveled square (left) and thick square (right) are quintessential examples of essentially two-dimensional objects

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Essentially Two-Dimensional Objects

Example The proper symmetry group of the beveled square in Figure 104 (left) is isomorphic to C4. Why? It has only a single rotation axis (the red arrow in Figure 105). This axis has order 4,     which means the angles of rotation symmetries about this axis are: 0 , 90 , 180 , and 270 . Composing these rotations works just like composing the rotation symmetries of a  two-dimensional square: you just add the angles with 360 wrapping back to zero. In summary, the proper symmetry group of the beveled square is isomorphic to the proper symmetry group of a two-dimensional square, which is C4.

Figure 105: The only rotation axis of the beveled square

Next, we’ll study the thick square on the right in Figure 104, which is also related to the two-dimensional square, but in a more complicated way. Example The proper symmetry group of the thick square is isomorphic to D4. Why? Just as in the previous exercise, there is a rotation axis of order 4 (the red arrow in Figure 106). Let’s call this the main axis. Composing rotations about this main axis works just like composing the rotation symmetries of a two-dimensional square. So far this seems like C4, but not so fast! In addition to the main axis, there are 4 rotation axes of order 2 (illustrated in 4 different colors: green, purple, yellow, and blue). Let’s call them flip axes because rotating 180 about one of them has the effect of flipping the square upside down. This flipping motion is a proper symmetry (because it’s a rotation), which might seem confusing because it reminds you of an improper symmetry (a reflection) of the two-dimensional square. That’s exactly the point here! The four flip axes of the thick square match with the four reflection lines of the two-dimensional square.

Figure 106: The main axis of the thick square, plus its four flip axes (▶ https://doi.org/10.1007/000-25x)

In summary, the proper symmetry group of the thick square is isomorphic to the full symmetry group of the 2-dimensional square (which is D4). The isomorphism matches symmetries with symmetries as follows:

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Thick square main-axis rotations flip-axis rotations

7

2D square $ $

rotations reflections

There is nothing special about the square. For any n  2, we define the beveled regular n-sided polygon and the thick regular n-sided polygon in the analogous way. By reasoning just like what we did in the previous examples, their proper symmetry groups are isomorphic respectively to Cn (beveled) and Dn (thick). This definition requires a clarification when n ¼ 2. We consider the term “regular 2-sided polygon” to mean a rectangle, so here we’re talking about a thick or beveled rectangle, and in the former case we must assume the thickness doesn’t equal the rectangle’s length or width.1 Notice that a thick rectangle has exactly three rotation axes, all of which have order 2 (shown as purple, green, and yellow in Figure 107). Which one is the main axis? It doesn’t matter, but it is convenient to (randomly) choose one as the main axis and refer to the other two as flip axes. It is interesting to note that the word “dihedral” comes from the Greek “dihedron” which means a solid with two faces, like a thick polygon. We originally defined the dihedral group Dn to mean the full symmetry group of a regular n-sided polygon, but the name actually derives from our new 3D viewpoint: regarding Dn as the proper symmetry group of a thick regular n-sided polygon. Beveled and thick polygons are examples of essentially two-dimensional objects, which we can now define precisely.

Figure 107: This thick rectangle has three rotation axes, all with order 2

1

Similarly with a thick square, we must assume that the thickness doesn’t equal the square’s side length, because that would make a cube.

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Definition A 3D object is called essentially two-dimensional if (not counting axes of order 2) its number of rotation axes is either 0 or 1.

In other words, if we ignore flip axes, then it has either 0 axes or 1 axis. In fact, beveled and thick polygons are just about the only examples of essentially two-dimensional objects, according to the following theorem.

Theorem Given any essentially two-dimensional 3D object with finitely many symmetries, one of the following is true:

(1) It has no rotation axes, so its only symmetries are the identity and possibly also one improper symmetry. (2) It has exactly one rotation axis, in which case it is properly rigidly equivalent to a beveled regular polygon. (3) It has one main rotation axis plus additional flip axes, in which case it is properly rigidly equivalent to a thick regular polygon.

Notice that in all three cases, the object’s proper symmetry group is isomorphic to a cyclic or dihedral group. We must also acknowledge what the theorem does not do. First, it doesn’t help us understand essentially two-dimensional objects with infinitely many symmetries, like those in Figure 108. Second and more importantly, the theorem ignores improper symmetries. The theorem only guarantees that certain objects must be properly (not necessarily fully) rigidly equivalent to beveled polygons or thick polygons. The full version is addressed in the final exercise at the end of this chapter.

Figure 108: The hockey puck and cone are essentially two-dimensional objects with infinitely many symmetries

7C A Trick for Counting Symmetries Our next major goal is to understand the full symmetry groups (or at least the proper symmetry groups) of the three fundamental objects pictured in Figure 109: the tetrahedron, the cube, and the dodecahedron. Take time now to build these three shapes. Cardboard and tape work fine. Alternatively, you could build them with Magformers or Zometools, or the cut-out templates provided on this book’s resource website.

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Figure 109: The three fundamental 3D objects

Tetrahedron

7

Cube

Dodecahedron

Notice that none of these objects are essentially two-dimensional. That’s what makes them complicated. Each has multiple rotation axes of order at least 3. The good news is that they are the only complicated objects we’ll need to study. They are the model objects needed for the classification we aim at in this chapter. The classification theorem will basically say. . .(spoiler alert). . .that every other bounded 3D object either is essentially two-dimensional or matches one of these three fundamental objects. That’s why they are so important. That’s why we will devote the next three sections to carefully studying these three objects one at a time. But first, we will discuss a clever trick for discovering and remembering how many symmetries each one has. For this, we’ll denote by “F” the number of faces, and by “S” the number of sides per face. For example, a tetrahedron has F ¼ 4 faces that are identical equilateral triangles (S ¼ 3). A cube has F ¼ 6 faces that are identical squares (S ¼ 4). A dodecahedron has F ¼ 12 faces that are identical pentagons (S ¼ 5). This table is the punch line:

tetrahedron cube dodecahedron

F 4 6 12

S 3 4 5

# proper symmetries 12 24 60

# total symmetries 24 48 120

Notice the pattern: each object’s number of proper symmetries equals F  S, while its total number of symmetries is double that. To understand why, first recall why we drew a gnome on the square back in Chapter 2: the gnome’s ending position encoded the symmetry that was performed. The same idea works here: to avoid the messy work of counting up all the rotation axes and rotation angles about those axes, we instead just count up the ending positions into which the object can be rotated. For example, let’s count the ending positions into which the cube can be rotated. Take a pen and scribble some decorations on the cube’s faces. For example, you could draw a different gnome on each face, or just draw dots to make it look like a die.2 Choosing an ending position is like picking the cube up from the table and choosing how to set it back down into its original footprint. You really have only two choices to make: there are

2

Any decoration that makes it asymmetric will allow it to detect rigid motions because the proof of the RigidMotion Detector Theorem on page 57 is equally valid for 3D objects.

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The Tetrahedron

6 possibilities for which face goes down, and then 4 possibilities for how this bottom face is rotated. The number of ways to make these two choices in succession equals 6  4. The tetrahedron and the dodecahedron work similarly, so our new rule (# proper symmetries ¼F  S) is true for all three objects, the tetrahedron, cube, and dodecahedron. We’re not claiming the formula is true (or even makes sense) for all possible objects, but that’s OK because right now we only care about these three objects. But why is each object’s total number of symmetries equal to twice its number of proper symmetries? Because the Zero-or-Equal Theorem from page 39 is equally valid in the 3D setting. We restate it here for both 2D and 3D objects.

The Zero-or-Equal Theorem The number of improper symmetries of any (2D or 3D) object either is zero or is equal to the number of proper symmetries.

You should look back at the original proof to check that it is valid for 3D objects as well. With that settled, we need only verify that each of our three fundamental objects (the tetrahedron, cube, and dodecahedron) has at least one improper symmetry. For this, take your cardboard model and identify a reflection plane. If you sliced the model in half along this plane, the halves would be mirror images of each other. Counting symmetries is a good start, but it is not enough. We will now look in more depth at the symmetry group (or at least the proper symmetry group) of each of these three objects. Before continuing, take time now to complete the tetrahedron study, cube study, and dodecahedron study worksheets available on the book’s resource website; these worksheets will lead you to discover for yourself the major theorems from next three sections of the book.

7D The Tetrahedron Illustrated in Figure 110 are three types of symmetries of a tetrahedron. Pictured first is an axis of order 3, which connects a vertex to the midpoint of its  opposite face. There are 4 such axes (one for each vertex), with rotations by 120 and  240 about each, yielding a total of 8 non-identity rotations about such axes. ction

180

Refle

120 or 240

Figure 110: Rotation and reflection symmetries of a tetrahedron (▶ https://doi.org/10.1007/000-25y)

plane

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Pictured second is an axis of order 2, which connects the midpoints of a pair of  opposite edges. There are 3 such axes (one for each pair of opposite edges), with a 180 rotation around each, yielding a total of 3 non-identity rotations about such axes. We have identified 8 + 3 + 1 = 12 total rotation symmetries (counting the identity). Since 12 is the expected total number, we know that these are all of the tetrahedron’s proper symmetries. Pictured third is a reflection plane. There are 6 such planes, one for each edge. These are all of the tetrahedron’s reflection planes; however, they are not all of the tetrahedron’s improper symmetries, since we know from the previous section that it has 12 improper symmetries. Unlike the result that prevailed in the 2D setting, the 3D Center Point Theorem does not guarantee that every improper symmetry is a reflection! We wish to fully understand the tetrahedron’s symmetry group. To build a Cayley table, we must first choose a name for each symmetry. Here’s an illuminating naming system. Let’s label the 4 vertices A, B, C, and D, as in Figure 111. Think of this figure as the identity position, which we call ABCD. Name each of the other symmetries with the 4-letter word that you see after that symmetry is performed. Always read words in the order indicated by the illustration (top, bottom-right, bottom-front, bottom-left). Thus, we name symmetries according to way they permute the tetrahedron’s vertices. This system will inevitably lead you to stumble upon the following theorem.

Theorem A tetrahedron’s symmetry group is isomorphic to P4, and its proper symmetry group is isomorphic to A4.

Proof The previously described naming strategy is a matching between the 24 symmetries of the tetrahedron and the 24 members of P4 (the four-letter words). This matching is a one-to-one correspondence because no two different symmetries could permute the vertices in the same way. That is, if you know how the vertices were permuted, then you know what symmetry was performed (equivalently, if you paint the four vertices with four different colors, you have an asymmetric object). Since the matching is a one-to-one correspondence, it’s an isomorphism. As in Chapter 6, we don’t need to look any further to be sure that it matches true equations with true equations, because the definition of “composition” of permutations was custom designed to mimic

A

Figure 111: The four vertices of the tetrahedron are permuted by its symmetries

B

D C

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The Cube

A

Figure 112: Rotations permute the vertices in even ways

A 180

120

B

D C

B

D C

“composition” of symmetries in exactly this sense. Thus, the tetrahedron’s symmetry group is isomorphic to P4. It remains to verify that the proper (rotation) symmetries match with the even permuta tions. In Figure 112, the 120 rotation on the left matches with CBDA, which is even because  it’s a cycle of length 3: (143). The 180 rotation on the right matches with BADC, which is even because it’s two swaps: (12)(34). We did too much work here. Without explicitly finding the word and cycle notation, you can see at a glance that the rotation on the left permutes three vertices (the ones around the green face), while the rotation on the right swaps two pairs of vertices (the blue pair and the purple pair). Each remaining rotation symmetry is about an axis of order 3 (like the left image in Figure 112) or order 2 (like the right image) and therefore matches with an even permutation for the same reason. Thus, the tetrahedron’s proper symmetry group is isomorphic to A4. □

This theorem is a beautiful 3D analog of the following 2D fact from Chapter 6: a triangle’s symmetry group is isomorphic to P3 and its proper symmetry group is isomorphic to A3. The 2D and 3D versions are both proven by matching symmetries with how they permute the vertices.

7E The Cube Pictured in Figure 113 are the three types of rotation axes of a cube. How do we know there are no other kinds? Because these kinds account for all 24 proper symmetries, as follows: Left image: There are 3 axes of order 4 (one for each pair of opposite faces), with    rotations by 90 , 180 , and 270 about each, yielding a total of such rotations. Middle image: There are 4 axes of order 3 (one for each pair of furthest-apart   vertices), with rotations by 120 and 240 about each, yielding a total of such rotations. Right image: There are 6 axes of order 2 (one for each pair of furthest-apart edges),  with rotations by 180 about each, yielding a total of such rotations. (counting the We have identified identity). Since 24 is the expected total number, we know that these are all of the cube’s proper symmetries. Our next goal is to identify the proper symmetry group of a cube (we’ll worry about the full symmetry group in a later section of this chapter). The cube has 24 proper

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order 4

order 3

order 2

Figure 113: Axes of all three types of rotation symmetries of a cube (▶ https://doi.org/10.1007/000-25w)

7

symmetries. The only permutation group or alternating group with this size is P4. So if we go out on a limb and guess that the cube’s proper symmetry group might match with a permutation or alternating group, the only possible guess is P4. This guess turns out to be right, as the following theorem states.

Theorem The cube’s proper symmetry group is isomorphic to P4.

Proof A cube has 6 faces, 8 vertices, and 12 edges. To mimic the idea of the previous proof, we must find something of which the cube has 4. The clever solution is shown in Figure 114: it has 4 diagonals (colored green, blue, red, and yellow). Our isomorphism will match each proper symmetry of the cube with the way it permutes the 4 colored diagonals. We need only verify that this matching is a one-to-one correspondence, which means that

(1) every permutation of these colored diagonals can be achieved by a proper symmetry, and (2) no two different proper symmetries ever permute the colored diagonals in the same way.

Figure 114: The cube’s four diagonals are permuted by its proper symmetries

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The Dodecahedron

Actually, since the number of proper symmetries equals the number of permutations (24 of each), it suffices to verify (1) or (2), and then the other is true for free. We leave this verification to the reader, with the recommendation that you use a physical model like the cut-out template provided on this book’s resource website. □

Here is an alternative phrasing of the key idea from the previous proof. The largest possible stick (line segment) that’s able to fit inside of a cube can actually fit in exactly 4 different ways, and these 4 ways are permuted by the proper symmetries of the cube.

7F The Dodecahedron Finally, let’s turn our attention to the dodecahedron (reprinted in Figure 115), which we already know has 60 proper symmetries. Like those of the tetrahedron and cube in the previous sections, all of the dodecahedron’s proper symmetries can be accounted for by rotations about the “obvious” axes, which include 6 axes of order 5 (one for each pair of opposite faces), 10 axes of order 3 (one for each pair of furthest-apart vertices), and 15 axes of order 2 (one for each pair of furthest-apart edges). Our next goal is to identify the proper symmetry group of a dodecahedron (we’ll worry about the full symmetry group in a later section of this chapter). It has 60 proper symmetries. The only permutation group or alternating group with this size is A5. So if we go out on a limb and guess that the dodecahedron’s proper symmetry group might match with a permutation or alternating group, the only possible guess is A5. This guess is also right, as the following theorem states.

Theorem The dodecahedron’s proper symmetry group is isomorphic to A5. Idea of Proof We must find something of which the dodecahedron has 5. Here is the clever answer: Figure 116 shows a coloring of its edges using 5 colors (blue, green, pink, purple, and yellow). Our isomorphism will match each proper symmetry with the way it permutes these 5 colors. For this to work, we must verify the following:

Figure 115: Dodecahedron

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Figure 116: The 5 colors are permuted in even ways by the dodecahedron's proper symmetries (▶ https://doi.org/10.1007/000-260)

(1) Every proper symmetry permutes the colors in a well-defined manner. For example, if a certain rotation moves one pink edge to where a green edge used to be, then that rotation must also move each of the other pink edges to where a green edge used to be. (2) Every proper symmetry permutes the colors in an even way. (3) Every even permutation of the colors can be achieved by a proper symmetry. (4) No two different proper symmetries ever permute the colors in the same way. Verifying these assertions is time-consuming, and is left to the reader. It requires a physical model, and also requires our previous description of all of the rotation axes. □

Here are explicit instructions for correctly coloring your dodecahedron model so that you can verify the claims in the previous proof. You will need a total of 6 pink edges. Color some first edge pink, and the remaining 5 pink edges should be the edges that are either parallel or perpendicular to the first. Repeat with the other colors. Since the 6 edges of a single color are chosen to be mutually parallel or perpendicular, each color determines a “frame.” Each such color frame instructs you how to position a cube that you might wish to inscribe inside of the dodecahedron (so that every edge of the cube is parallel to an edge of the dodecahedron with that color). Figure 117 shows a cube positioned by the pink frame. In fact, a largest possible cube that’s able to fit inside of a dodecahedron can actually fit in 5 different ways (determined by the 5 color frames), and these 5 ways are permuted by the proper symmetries of the dodecahedron. This draws a nice analogy between the cube’s story and the dodecahedron’s story. Fitting largest possible sticks into a cube is analogous to fitting largest possible cubes into a dodecahedron.

7G The Classification Theorem In the last several sections, each new object was more complicated than the previous. The tetrahedron had 12 proper symmetries, the cube had 24 and the dodecahedron had 60. Can we continue building more and more complicated 3D objects with more and more symmetries? We did so with 2D objects when we considered regular polygons with

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Figure 117: Each color determines one of the five positionings of a largestpossible cube inside the dodecahedron; pink is shown here

2 sides, 3 sides, 4 sides, 5 sides, . . .; this list goes on indefinitely, with each object having more symmetries than the previous one. But quite surprisingly, we cannot build solid objects that are any more complicated than the objects that we have already considered! The 3D analog of Da Vinci’s Theorem is much more restrictive than you might have expected! We present this theorem now (at least the version for someone who only cares about proper symmetries).

The Classification Theorem Every bounded 3D object with finitely many symmetries either is essentially two-dimensional or is properly rigidly equivalent to a tetrahedron, cube, or dodecahedron.

It is remarkable that there are so few possibilities for the ways in which 3D bounded objects can be symmetric! To appreciate what this means, let’s think about people who might wish that there were more possibilities. For example, imagine a medieval blacksmith who is commissioned by a knight to build a spike ball weapon, like the one shown in Figure 118, with 12 sharp spikes. The knight insists that the 12 spikes be spread out with no bald spots to best smite his enemies. That is, he wants the weapon to look the same to an enemy facing one spike as it does to an enemy facing any other spike. He is really requesting here that the spike arrangement be symmetric: ignoring the chain, there should be a proper symmetry moving any spike to any other spike. Since a dodecahedron has 12 faces, the blacksmith is in luck—he can arrange the 12 spikes at the midpoints of the faces of a dodecahedron. The next year, the knight demands a new weapon built with 20 spikes. The blacksmith is again in luck, because the dodecahedron has 20 vertices, so he can arrange the spikes at the vertices of a dodecahedron. The blacksmith could also manage 30 spikes, arranged at the midpoints of the 30 edges of the dodecahedron, or even 60 spikes (think about how). But if the knight requests a weapon with 100 spikes, then the blacksmith is out of luck. It is impossible to build a perfectly symmetric spike ball with more than 60 spikes because such a weapon would have more proper symmetries than a dodecahedron, contradicting the classification theorem. The only way it could be done is with an essentially two-dimensional arrangement of spikes around the equator of the ball, but this

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Figure 118: There is a limit to how many spikes can be symmetrically distributed around the ball

7

Figure 119: The arrangement of a golf ball’s dimples is only approximately symmetric

arrangement would have large bald spots, and would not be what the knight really wanted. This is not an engineering restriction or an issue of manufacturing imperfections—it is a purely mathematical limitation. There is a limit to the number of points that can be symmetrically arranged around a sphere! Making the sphere larger or the points smaller does not help. It just can’t be done. There are many modern versions of this tale. For example, a golf ball manufacturer is unable to distribute more than 60 dimples symmetrically around a golf ball, and must therefore settle for an approximately symmetric arrangement of the 250 – 450 dimples found on most balls today, like the one in Figure 119. Unfortunately, this means that the aerodynamics could in principle change depending on how the ball is set on the tee. On the other hand, a lack of dimple symmetry is not all bad. Certain intentionally asymmetric dimple patterns, like those used on Polara brand balls, can cause a ball to fly straighter. The USGA banned the use in tournament play of asymmetric Polara balls, which led Polara to file a lawsuit against the USGA. The USGA was in an interesting legal and mathematical situation here; it was trying to regulate that balls must be as symmetric as possible, even while the classification theorem says that no ball could ever be perfectly symmetric. Architects and nature are also limited by the classification theorem. For example, the arrangement of triangles around a geodesic dome structure, like the Epcot Center in Figure 120, is approximately symmetric at best. Similarly, the arrangement of more than 60 seeds around a dandelion could not be perfectly symmetric.

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Figure 120: The Epcot Center and dandelion are only approximately symmetric

In biology, proteins have been discovered representing all possible types of proper symmetry groups allowed by the classification theorem: cyclic, dihedral, A4, P4, and A5. It may not be obvious how to account for the symmetry of a particular protein, but the serious lack of options is certainly part of the answer.

7H Chirality So far in this chapter, we’ve mostly settled for understanding proper symmetries only. We don’t yet know the full symmetry group of a cube or dodecahedron. Moreover, the classification theorem in the previous section guarantees only that things are properly (not fully) rigidly equivalent to certain model objects. The final two sections of this chapter are devoted to the full story. Since our goal now is to also understand the improper symmetries of objects, the simplest objects are the ones that don’t have any:

Definition A 3D object is called chiral if all of its symmetries are proper.

Recall that a 2D object is called oriented if all of its symmetries are proper. It’s unfortunate that a different word is used to name this same property in the 3D setting, but that’s what’s done. One way to make a tetrahedron, cube, or dodecahedron become chiral is to draw identical oriented polygons on all of its faces, as illustrated. The result is called a chiral tetrahedron, chiral cube, or chiral dodecahedron. It has the same proper symmetries as it used to, but none of the improper symmetries. An improper symmetry would make it look like its mirror image,3 but you can see in Figure 121 that the mirror image looks

3 Since some 3D objects have improper symmetries that aren’t reflections, how can we assert that “an improper symmetry would make it look like its mirror image”? The answer comes from the “one reflection is enough” idea from Chapter 2: every improper rigid motion is the composition of a single fixed reflection (which makes it look like its mirror image) with a proper rigid motion (which doesn’t change it).

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Figure 121: Chiral tetrahedron, chiral cube, and chiral dodecahedron (reflected in a mirror)

7 different than the original: its face decorations are all clockwise instead of counterclockwise! The previous classification theorem unfortunately only guaranteed that every bounded 3D object (that has finitely many symmetries and is not essentially two-dimensional) must be properly rigidly equivalent to a tetrahedron, cube, or dodecahedron. To what must it be fully rigidly equivalent? The full story is found in this table: If it’s properly rigidly equivalent to a . . .tetrahedron. . . . . .cube. . . . . .dodecahedron. . .

then it’s rigidly equivalent to a . . .tetrahedron, chiral tetrahedron, or volleyball. . . .cube or chiral cube. . . .dodecahedron or chiral dodecahedron.

Probably the only surprise here is the volleyball. Who expected that? By volleyball, we mean the arrangement of seam lines along which a standard volleyball is stitched together, shown in Figure 122 (left). Figure 122 (right) shows a 3D-printed sculpture of a volleyball aligned snugly inside a tetrahedron. By studying the sculpture, one can see that the two objects share the same proper (rotation) symmetries but have different improper symmetries. The reflection planes of the tetrahedron do not yield reflection symmetries of the volleyball (nor vice versa). The remainder of this section explores some applications of chirality to chemistry. These applications are best understood by regarding “chiral” as meaning that the object differs from its mirror image. Hold an object up to a mirror, and imagine that its reflected image is a real object that you could pluck out of the mirror. The object is chiral if the plucked-mirror-version is different—it can’t be aligned to look like the original version, no matter how you rotate or translate it. This conceptualization of chirality is famously found in Lewis Carroll’s novel Through the Looking Glass and What Alice Found There. The term “looking glass” is an old English term for a mirror. In the opening chapter (see Figure 123), Alice imagines that the reflected image of her room in the looking glass is part of a real other world. She asks her cat, “How would you like to live in looking-glass house, Kitty? I wonder if they’d give you milk in there? Perhaps looking-glass milk isn’t good to drink.” This comment is far more interesting than it might at first appear.

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Figure 122: The volleyball and tetrahedron are only properly rigidly equivalent

Figure 123: Alice goes through the looking glass

All chiral objects act and look differently in looking-glass world. For example, the writing on books is backwards. Also, looking-glass screws are “counter screws,” so looking-glass carpenters learn the rule “righty loosey, lefty tighty,” which doesn’t even rhyme. This makes it difficult to graduate from looking-glass carpenter school. More serious differences lurk at the microscopic level. Many molecules are chiral, which means that their reflections in looking-glass world will look and act differently. In Figure 124, the reflected image of the left-handed amino acid is the right-handed amino acid. In fact, the word “chiral” derives from the Greek word for “hand.” Naturally occurring organic chiral molecules are almost always found only in the left- or righthand version, not both. The right-hand version of a molecule may interact differently with mechanisms in living cells that evolved to interact with the left-hand version. For example, the molecule carvone is responsible for the smell of caraway seeds, but the mirror-image version of carvone smells like spearmint! Thalidomide is a deadly example of the difference between the left- and right-hand versions of molecules. It was prescribed in the late 1950s to control morning sickness, but it turned out to cause birth defects. Before being recalled in 1961, thalidomide was

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Figure 124: A chiral molecule doesn’t match its mirror image

COOH

H

C

R NH2

COOH

R

C

H

NH2

7 responsible for birth defects in tens of thousands of infants worldwide. This tragedy may have been caused by the manner in which thalidomide was produced. The synthetic manufacturing of chiral molecules yields equal amounts of the left- and right-hand versions of the molecule. It is thought that one version of thalidomide controls morning sickness, while the other causes birth defects, and that the two were not properly separated in the manufacturing process. Evidently, it is not safe for Kitty to drink looking-glass milk.

7I The Full Story (optional) The Zero-or-Equal Theorem always helps you know many improper symmetries an object has, but counting isn’t enough. There is at least one situation in which we can precisely understand how the proper and improper symmetries fit together into a single Cayley table. Recall the following theorem from Chapter 5. We restate it here for both 2D and 3D objects.

Full-Versus-Proper Theorem If a (2D or 3D) object has an improper symmetry that is its own inverse and that commutes with all of the object’s proper symmetries, then FULL ffi PROPER  C2.

Look back and check that our previous 2D proof idea also works in 3D with no modification necessary.

▶Exercise

Identify the full symmetry group of a thick hexagon.

Solution We previously learned that its proper symmetry group is isomorphic to D6. We claim now that its full symmetry group is isomorphic to D6  C2. To justify this claim using the Full-vs-Proper Theorem, we must find an improper symmetry that is its own inverse and that commutes with all 12 proper symmetries. The

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Figure 125: The reflection over the “slice it like a bagel” plane is special – it commutes with all rotation symmetries

B

a A

C

F f E e

D d

answer is the reflection across the plane that slices it in half the way one would slice a bagel. With the 12 vertices labeled as in Figure 125, this reflection exchanges lower- with upper-case letters (a $ A, b $ B, etc.). This reflection is its own inverse because every reflection is its own inverse. To puzzle out why it commutes with the proper symmetries, it helps to notice that the 12 proper symmetries can be coded as upper-case 6-letter words according to how they permute the 6 upper-case vertices (ignoring the lower-case letters because each one always stays paired with its matching upper-case letter). Do you see why the swapping of upper with lower case is independent from (and unaffected by) the permuting of the upper-case vertices?

▶Exercise

Identify the full symmetry group of a beveled square.

Solution We previously learned that its proper symmetry group is isomorphic to C4. We claim now that its full symmetry group is isomorphic to D4. The Full-vs-Proper Theorem doesn’t apply here because the beveled square does not have an improper symmetry that commutes with all of its proper symmetries. So instead we reason as follows. The reflection plane in Figure 126 slices its top face along a reflection line (of this 2D top face square). Similarly, all four reflection planes of the beveled square correspond to reflection lines of the 2D square. Do you see why this means a Cayley table for the beveled square will match a Cayley table for the 2D square? To see this explicitly, code the symmetries of the beveled square as 4-letter words, so for example, the reflection across the illustrated plane is BADC, and notice that this same word naturally codes the reflection of the 2D square across the corresponding line.

The logic of the previous two exercises generalizes to prove:

▬ The full symmetry group of a thick n-sided regular polygon is isomorphic to Dn  C2. ▬ The full symmetry group of a beveled n-sided regular polygon is isomorphic to Dn.

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Figure 126: This reflection plane of the beveled square should remind you of a reflection line of a 2D square

A B D C

7

However, to obtain the most important application of the Full-vs-Proper Theorem, we first must learn about a new type of improper rigid motion of space. In addition to reflecting across a plane, one can also “invert across a point.”

Definitions Inversion across a point is the improper rigid motion of space that relocates every speck of ice to its “mirror image” across the point. A bounded object is called centrally symmetric if inversion across its center point is a symmetry of the object.

To interpret this, it’s best to think of space as filled with ice, and think of the point as a mirrored disco ball embedded within. A speck of ice 3 inches below the ball ends up 3 inches above the ball. A speck of ice 5 inches to the right of the ball ends up 5 inches to the left of the ball. To decide where a speck of ice will end up, make a straight line from the speck to the ball and continue this line the same distance on the other side. In summary, inversion can be visualized as the mirroring of space across a disco ball. But wait—we’re not allowed to make up a new kind of rigid motion, because we previously asserted that all rigid motions are compositions of the old kinds: translations, rotations, and reflections. Indeed, Figure 127 shows that inversion across the green point is the same as a reflection composed with a rotation, specifically this composition: first  rotate 180 degrees about an axis (shown purple) through the green point, and then reflect across the plane through the green point that is perpendicular to this purple axis. This composition moves each speck of ice (like the yellow one) to its “mirror image” across this green point, as desired. We recommend always visualizing inversion via the disco ball image, not as a composition. But the composition picture is occasionally useful. For example, it confirms that inversion is improper! Try to visualize why inversion across the center point of a centrally symmetric object (1) is its own inverse, and (2) commutes with all of the object’s proper symmetries. These two features lead to the following conclusion.

Corollary For any centrally symmetric object, it is true that FULL ffi PROPER  C2.

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180

end

reflection plane reflect second

rotate first

start

Figure 127: Inversion ¼ reflection
rotation

Figure 128: A centrally symmetric object is balanced across its center point, like this explosion

A corollary means a theorem whose proof is a quick application of another more significant or more general theorem (in this case it’s a corollary of the Full-vs-Proper Theorem). “Centrally symmetric” roughly means that the object is balanced across its center point, like the explosion in Figure 128. An orange speck on one side of the explosion’s center is balanced by an orange speck on the other side, and similarly for the white specks and green specks. That’s basically how explosions work. The following are the key examples:

▬ Centrally symmetric: cube, dodecahedron, volleyball. ▬ Not centrally symmetric: tetrahedron.

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Use physical models to visualize why the cube, dodecahedron, and volleyball are centrally symmetric (balanced across their center points). On the other hand, Figure 129 shows why the tetrahedron is not centrally symmetric: inversion would change the red tetrahedron by moving it to the position of the yellow tetrahedron (and vice versa). But if superimposed and painted a single color, these two tetrahedrons would together form a centrally symmetric object. The corollary completes our understanding of the full symmetry groups of all of our fundamental objects, as summarized in Table 9. Notice that the sizes of these groups are indicated in parentheses. The main results in this chapter are combined and summarized in Figure 130. This flow chart achieves the goal of the chapter: classifying the ways that 3D bounded objects (with finitely many symmetries) can be symmetric! The question marks in the flow chart indicate that we do not yet understand all of the possible model objects to which essentially two-dimensional objects can be fully rigidly equivalent. This gap is partially filled in by the final exercise at the end of the chapter. Figure 129: Inversion exchanges the red and yellow tetrahedrons

Table 9: Symmetry groups of fundamental 3D objects (numbers in parentheses are symmetry group size)

Proper sym. group. tetrahedron chiral tetrahedron

(24) (12)

volleyball cube chiral cube dodecahedron chiral dodecahedron

Full sym. group.

(24) (60)

Beveled -gon

( )

Thick -gon

(2 )

(12) ×

(24)

×

(48)

(24) ×

(120)

(60) (2 ) ×

(4 )

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Prop. rig. equiv to...

xes No a One axis ltip

le

Ess ent ially 2D

Mu

ax es

Rig. equiv to...

Gnome

Beveled Polygon

? ?

Thick Polygon Chiral Tetra. Tetrahedron

Any bounded 3D object with finitely many symmetries

2D ly ial nt sse tE No

Vollelyball Tetrahedron Chiral Cube Cube Cube

Chiral Dodec. Dodecahedron Dodecahedron Figure 130: Classification flow chart

▶Exercise

Identify the symmetry type of the object in Figure 131.

Solution The object has three obvious axes of order 4 (up-down, left-right, and front-back). Now use the flow chart (Figure 130 ). Since it has multiple high-order axes, it is not essentially two-dimensional, so it must be properly rigidly equivalent to a tetrahedron, cube, or dodecahedron. Of these choices, only the cube has axes of order 4, so it matches the cube. The object

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Figure 131: Identify its symmetry type

7

has some obvious reflection planes, so it’s not chiral. According to the flow chart, it therefore must be fully rigidly equivalent to the cube. Can you visualize how to position the object inside a cube so that it has the same proper and improper symmetries as the cube?

▶Exercise

A “striped cube” is obtained by striping each face of a cube in such a way that all  120 rotation symmetries of the cube remain symmetries of the striped cube. Two models are shown in Figure 132. Identify the symmetry type of a striped cube.

Figure 132: Identify the symmetry type of the striped cube

Solution The striped cube is not essentially two-dimensional because it has multiple axes of order 3 (the same as a cube). According to the flow chart (Figure 130 on page 161), it must be properly rigidly equivalent to a tetrahedron, cube, or dodecahedron. It doesn’t match a dodecahedron because its symmetry group is a subgroup of the cube’s, which doesn’t contain any axes of order 5. It doesn’t match the cube because it has fewer proper symmetries than the  cube (for example, none of the cube’s 90 rotation symmetries preserve the striping). By process of elimination, it must be properly rigidly equivalent to a tetrahedron. According to the flow chart, it therefore must be fully rigidly equivalent to a tetrahedron, chiral tetrahedron, or volleyball. Among these choices, only the volleyball shares the striped cube’s property of being centrally symmetric. Therefore, the striped cube is rigidly equivalent to a volleyball! Figure 133 might help you visualize how to superimpose the striped cube and the volleyball in such a way that their symmetries align.

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163  Exercises

Figure 133: This volleyball is rigidly equivalent to the striped cube; focus on the orange stripes to see how they match

Exercises (1) Write a precise definition of each of the following terms: space, 3D object, bounded 3D object, order of an axis, proper rigid motion of space, properly rigidly equivalent, essentially two-dimensional, chiral, inversion across a point, centrally symmetric. (2) TRUE or FALSE: A typical golf ball has enough proper symmetries to move any dimple to any other dimple. (3) TRUE or FALSE: If a 3D object is not chiral, then it must have equal numbers of proper and improper symmetries. (4) Which are proper rigid motions of space? (a) (b) (c) (d) (e)

A reflection across a plane followed by a reflection across another plane. A translation followed by a rotation around an axis. Inversion across a point. A rotation around an axis followed by inversion across a point. The identity.

(5) Fill in the following table summarizing the number of rotation axes of the three fundamental objects. Number of axes with order. . . 2 3 Tetrahedron Cube Dodecahedron

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Chapter 7 • Symmetries of 3D Objects

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Figure 134: Thick oriented polygon

7

(6) Is a thick hexagon centrally symmetric? What about a thick pentagon? What about a thick n-sided regular polygon? (7) Is a beveled regular polygon centrally symmetric? (8) The 3D object pictured in Figure 134 is called a thick oriented 3-sided polygon (because it’s a thickening of the 2D recycling logo, which is itself rigidly equivalent to an oriented 3-sided polygon). (a) Identify its proper and full symmetry groups. (b) Is it centrally symmetric? Is it chiral? (c) What’s wrong with the following reasoning? “It’s chiral because its mirror image spins counterclockwise.” (d) Generalize: Identify the proper and full symmetry groups of a thick oriented nsided polygon. (e) Identify the proper and full symmetry groups of a beveled oriented n-sided polygon. (9) How many different 3D objects can you think of with. . . (a) (b) (c) (d)

60 total symmetries? 24 total symmetries? 14 total symmetries? 19 total symmetries?

Here “different” means “not rigidly equivalent to each other.” (10) If a 3D object has an odd number of symmetries, what’s the strongest conclusion you can make about its symmetry group? (11) Is looking-glass milk safe for looking-glass Kitty to drink? (12) For each object pictured in the table below, identify its proper and full symmetry groups. Is it essentially 2-dimensional? chiral? centrally symmetric? For all of the images, ignore imperfections and ignore color and shadow effects.

165  Exercises

(a) bar stool

(b) thick triangle

(c) octahedron

(d) water park double-tube

(e)

(f) hexagon wedding cake

(g) water park triple-tube

(h) fortune teller

(i) stool

(j) leggo block

(k) soccer ball

(l) pentagon house

(13) Group all of the objects in the previous exercise according to proper rigid equivalence. (14) Disprove the converse of the following fact about bounded 3D objects: “If it’s centrally symmetric then FULL ffi PROPER  C2.” (15) The product of three groups is defined analogously to the way we defined the product of two groups in Chapter 5. Prove that the full symmetry group of a thick rectangle (Figure 135) is isomorphic to C2  C2  C2. Refer back exercise 33 of Chapter 5 on page 109 in which you proved that the full symmetry group of a 2D rectangle is isomorphic to C2  C2, then prove the new assertion in two different ways: (a) As a corollary of the old exercise plus the Full-vs-Proper Theorem. (b) By constructing an explicit isomorphism, like the one used in part (b) of the old exercise.

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Figure 135: A thick rectangle has three rotation axes, each of order two

7 Figure 136: Sonobe origami

(16) Use online resources to learn how to make sonobe origami, and construct the three pieces shown in Figure 136. Identify the full symmetry group of each piece, assuming it’s constructed with a single paper color. How does the answer change if you paint the surface with a thick coat of paint to hide the lines on the faces along which the pieces of origami paper are inserted into each other? (17) Identify the full symmetry group of the left object in Figure 137. Notice that each of the four human figures has her right hand placed on top of the shoulder of her right neighbor and her left hand placed below the shoulder of her left neighbor. How would the answer change with a more symmetric hand placement? (18) The blue frame and baseball in Figure 137 (right) are rigidly equivalent to each other. Identify the proper symmetry group of the baseball. Is it chiral? Is it centrally symmetric? Conclude that the baseball is properly (but not fully) rigidly equivalent to a thick rectangle.

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Figure 137: Images for Exercises 17 and 18

Figure 138: Do the rotation symmetries permute the colors in a well-defined way?

(19) Consider the statement: “If a 3D object is chiral, then it is not centrally symmetric.” (a) Write this statement’s contrapositive. (b) Explain why this statement is true. (c) Disprove the converse of this statement. (20) We learned that the symmetry group of the tetrahedron is isomorphic to P4, thought of as the group of all permutations of its 4 vertices. Could we alternatively think of this as the group of all permutations of its 4 faces? In other words, can we replace vertices with faces in the proof? (21) Explain what is wrong with the following reasoning: “The 6 faces of the cube get permuted by each symmetry of the cube; therefore, the full symmetry group of the cube is isomorphic to P6.” (22) Referring to Figure 114, explain what is wrong with the following reasoning: “The 4 colored diagonals of the cube get permuted by each symmetry of the cube; therefore, the full symmetry group of the cube is isomorphic to P4.” HINT: Which other symmetry of the cube has the same effect on the colored diagonals as the identity? (23) Explain what is wrong with the following reasoning about Figure 138: “The four colors get permuted by each proper symmetry of the cube; therefore, the proper symmetry group of the cube is isomorphic to P4.” How could you re-color the vertices to make this reasoning correct? (24) Identify the symmetry type of several of Vladimir Bulatov’s metal sculptures, which can be viewed at: http://bulatov.org.

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Figure 139: Construction of a stellated dodecahedron

7 Figure 140: Construction of a two-colored stellated octahedron (▶ https://doi.org/10.1007/000-261)

Figure 141: An octahedron, an icosahedron, and a soccer ball

(25) A stellated dodecahedron results from attaching a pyramid (built from 5 triangles) to each pentagon face of a dodecahedron, as illustrated in Figure 139. Other objects with polygon faces can be stellated in an analogous manner, and the stellated version is always rigidly equivalent to the original. The illustration in Figure 140 shows a stellated octahedron constructed using pyramids of two different colors (red and purple). Use this illustration to find a visual answer to the following challenge: Inside the proper symmetry group of the octahedron, describe a subgroup that is isomorphic to the proper symmetry group of the tetrahedron. (26) How can you decorate the faces of a thick regular polygon to make it become chiral? What about a beveled regular polygon? (27) Use the counting trick discussed in Section 7C of this chapter, which said that # proper symmetries ¼F  S, to count the proper and improper symmetries of an octahedron and an icosahedron. Modify the trick to count the symmetries of a soccer ball. These objects are illustrated in Figure 141.

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(28) The iOrnament app allows users to create 3D images by mapping some of the 17 wallpaper patterns to the surface of a sphere, resulting in a painted sphere. Use this app to create artistic representatives of each allowable type of painted sphere. Determine the proper symmetry group of each. (29) If a bounded 3D object with finitely many symmetries has an axis of order 6, then what is the strongest conclusion you can draw about its symmetry type? What is the largest number of symmetries that it could have? (30) When a bounded 3D object is properly rigidly equivalent to a thick square or a beveled square, the question marks in the flow chart (Figure 130 on page 161) indicate that we have not yet classified all of the different model objects to which it could be fully rigidly equivalent. List as many different possibilities as you can. HINT: Start by considering Exercises 8 and 26 from this chapter. (31) This table lists all of the types of rotation symmetries of a cube. Number

Even?

Preserve stripes?

Identity   90 or 270   120 or 240  180 about an order-2 axis  180 about an order-4 axis

For each rotation type, fill in the row of the table by deciding (1) the number of such rotations, (2) whether such rotations permute the cube’s four diagonals in an even manner, and (3) whether such rotations are symmetries of the striped cube in Figure 132. What do you learn? (32) Construct a 3D object with exactly two symmetries: the identity and inversion across the center point. (33) Matt Devos created beautiful 3D object templates available at http://www.sfu.ca/ ~mdevos/notes/geom-sym/nets/. Construct each object and identify its symmetry type. (34) This exercise refers to the dodecahedron template found on the book’s resource website. (a) How are the templates for the top and bottom halves related to each other? How could you construct one from the other? (b) How does inversion across the center point permute the five letters? (c) Identify the flaw in the following incorrect reasoning: The symmetries of the dodecahedron permute the five letters, so its full symmetry group is isomorphic to P5. (35) The seven bracelets illustrated in Figure 142 are obtained by wrapping the seven border patterns around cylindrical bands. Each bracelet is essentially two-dimensional. (a) Which of these bracelets are chiral? (b) Which of these bracelets have rotation axes of order 2?

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Figure 142: Seven bracelets made by wrapping the seven border patterns around cylindrical bands

7

1

2

3

4

5

6

7

(c) The proper symmetry group of each bracelet is isomorphic to either a cyclic group or a dihedral group. Which are cyclic and which are dihedral? (d) Identify an improper symmetry of the 6th bracelet that commutes with all of its proper symmetries. (e) The first bracelet has exactly 3 proper symmetries and 3 improper symmetries. Describe all of them. Notice that none of the improper symmetries are reflections. To what familiar group is this bracelet’s symmetry group isomorphic? (f) Find a pair of bracelets whose full symmetry groups are isomorphic. COMMENT: Each bracelet was obtained from a border pattern by wrapping some number (usually 6) of iterations of the border pattern’s basic design element around a band. By using different numbers of iterations, you obtain not just seven bracelets, but seven families of bracelets. These families of bracelets represent all possible ways in which essentially two-dimensional bounded objects can be symmetric. More precisely, every essentially two-dimensional bounded object with a finite symmetry group is rigidly equivalent to one of the bracelets in one of the families. This result removes the question marks from the flow chart (Figure 130 on page 161).

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The regular polygons played starring roles in our study of 2D objects. They were defined on page 5 as built from equal-length straight lines assembled so that all angles are equal. In other words, a regular polygon is built from identical copies of regular parts assembled in the most regular possible way. Some examples of regular polygons are shown in Figure 143. The 3D analog of a regular polygon is defined as follows.

Definition A regular polyhedron (also called a Platonic solid) means a bounded 3D object whose faces are all identical regular polygons assembled so that each vertex has the same number of faces meeting at it.

We have already encountered three Platonic solids, those shown in Figure 144: the tetrahedron, cube, and dodecahedron. Study these three familiar objects to check that they fit the bill. We begin this chapter by answering the first key question: how many other Platonic solids are there?

Regular polygons Figure 143: Regular polygons

8A The Classification of Platonic Solids Since there is an infinite list of regular polygons (each successive one looking more like a circle), you might expect that there is also an infinite list of Platonic solids (each successive one looking more like a sphere). Or perhaps you expect the opposite—that Electronic supplementary material The online version of this chapter (https://doi.org/10.1007/ 978-3-030-51669-7_8) contains supplementary material, which is available to authorized users. The videos can be accessed individually by clicking the DOI link in the accompanying figure caption or by scanning this link with the SN More Media App.

© Springer Nature Switzerland AG 2021 K. Tapp, Symmetry, Texts for Quantitative Critical Thinking, https://doi.org/10.1007/978-3-030-51669-7_8

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Figure 144: Are these the only Platonic solids?

8

the tetrahedron, cube, and dodecahedron are the only Platonic solids, since these were the only objects referenced in the classification theorem of the previous chapter. Or perhaps you suspect that there are exactly five Platonic solids because the chapter title ruined the surprise. Whatever your guess, we will now attempt to systematically classify the Platonic solids. You will need cardboard and tape.1 Build a pile of identical equilateral triangles, a pile of identical squares, and a pile of identical pentagons. Start by choosing a pile, which we’ll call “S” (for Sides). For example, S ¼ 3 means you are building your Platonic solid entirely out of triangles, S ¼ 4 means squares, and S ¼ 5 means pentagons. Next, decide how many copies of this shape will meet at each vertex. We’ll call this choice “C” (for Copies). Assemble C identical copies of the S-sided polygon together to form your first vertex, and tape up the sides. Then to each newly created vertex that doesn’t yet have C polygons meeting, add more polygons until it does. Continue this process and see whether you end up with a valid Platonic solid. As you work, you won’t have any flexibility. The rule requires that C copies of the S-sided polygon must meet at each vertex. This restriction might seem mild (it doesn’t even say anything about the angles at which things meet), but in fact it’s very strong. You’ll see that it leaves you no creative freedom and no choice about what you end up building. For example, the choice S ¼ 3 and C ¼ 3 yields a tetrahedron, as seen in Figure 145. The tetrahedron is not the only Platonic solid that you can build with triangles. The choice S ¼ 3 and C ¼ 4 yields a new Platonic solid called an octahedron, as seen in Figure 146. The choice S ¼ 3 and C ¼ 5 yields another new Platonic solid called an icosahedron, as seen in Figure 147. What else can be built with triangles? The choice S ¼ 3 and C ¼ 6 does not work, because 6 triangles meeting at a vertex leaves no extra room (Figure 148). There are no gaps between the 6 triangles, so your construction will never bend upwards to become

1

Alternatively, Magformers work much better than cardboard and tape because their edges stick to each other magnetically.

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Figure 145: The choice S ¼ 3 and C ¼ 3 (triangles meeting 3 at a vertex) yields a tetrahedron

Figure 146: The choice S ¼ 3 and C ¼ 4 (triangles meeting 4 at a vertex) yields an octahedron (▶ https://doi.org/10.1007/000-265)

Figure 147: The choice S ¼ 3 and C ¼ 5 (triangles meeting 5 at a vertex) yields an icosahedron (▶ https://doi.org/10.1007/000-263)

three-dimensional; rather, no matter how many triangles you add, you’ll end up with a flat triangular tiling of a portion of the plane. Neither can you build a Platonic solid with S ¼ 3 and C > 6. Such a choice would be overcrowded; this means that C is too large—there are more copies of the polygon than

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Figure 148: Six triangles just fit around one vertex, so this choice doesn’t yield a Platonic solid

8

Figure 149: The choice S ¼ 4 and C ¼ 3 (squares meeting 3 at a vertex) yields a cube (▶ https://doi.org/10.1007/000-264)

can meet together at a point in the plane. Overcrowded choices of S and C never yield valid Platonic solids, so we won’t waste time considering any more overcrowded choices. What can be built out of squares? The cube is one familiar possibility, illustrated in Figure 149. The choice S ¼ 4 and C ¼ 4 does not work, because 4 squares meeting at a vertex leaves no extra room, and would therefore yield a square-tiling of a portion of the plane. Choices with S ¼ 4 and C > 4 do not work due to overcrowding. What can be built from pentagons (S ¼ 5)? The only possibility is C ¼ 3, since C > 3 would be overcrowded. Thus, only the dodecahedron can be built from pentagons. The assembly of a dodecahedron is illustrated in Figure 150. Are there any possibilities with S > 5? Exactly 3 hexagons fit together in the plane with no gaps (Figure 151), so the choice S ¼ 6 and C ¼ 3 would lead to a hexagonal tiling of the plane, not a Platonic solid. Any attempt to build a Platonic solid with S > 6 would fail because of overcrowding. The Platonic solids our systematic exploration has ended up with are illustrated in Figure 152. We have arrived at an important theorem, usually attributed to Plato.

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Figure 150: The choice S ¼ 5 and C ¼ 3 (pentagons meeting 3 at a vertex) yields a dodecahedron (▶ https://doi.org/10.1007/000-262)

Figure 151: Three hexagons just fit around one vertex, so this choice doesn’t yield a Platonic solid

Tetrahedron

Cube

Octahedron

Dodecahedron

Icosahedron

Figure 152: The five regular polyhedrons, also called the five Platonic solids

Plato’s Theorem There are exactly five Platonic solids: the tetrahedron, cube, octahedron, dodecahedron, and icosahedron.

The above tape-and-cardboard discussion provides very strong evidence that this theorem is true, but we must acknowledge that more work would be required to achieve a completely airtight proof of this theorem. In particular, it is still necessary to prove more rigorously that overcrowded choices for S and C never yield valid Platonic solids. Figure 153 helps clarify what you need to worry about here. Notice that each vertex along the inner white rim is overcrowded. This object is not a Platonic solid, because the

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Figure 153: Each vertex on the inner white rim of this object is overcrowded. That’s something that could never happen in a Platonic solid

8 number of triangles meeting varies from vertex to vertex (and because the triangles are not quite all equilateral). But still the image might lead you to decide that maybe a proof is needed to rule out the possibility of anyone ever building an overcrowded Platonic solid. Also, it is necessary to prove that the tetrahedron, cube, octahedron, dodecahedron, and icosahedron are all perfect (not just approximate) Platonic solids. For example, what if it turned out that a dodecahedron could only be assembled using slightly irregular pentagons, whose angles vary a degree or two from being all the same? Our cardboard and tape were too sloppy to detect such a minor issue. We will not discuss these remaining details. Mathematicians have taken care of these issues, so we can move forward with assurance that the theorem is true—there are exactly five Platonic solids!

8B Counting Their Parts If you have not already done so, take time now to build the remaining Platonic solids out of cardboard (or Magformers or Zometools). You need to build them in order to count their parts. Use your physical models to verify that the numbers of vertices (V ), edges (E), and faces (F) for each Platonic solid are as shown in Table 10. As before, “S” means the number of sides that each face has, and “C” means the number of faces meeting at each vertex. Examine your Platonic solid models, and notice that “C” also equals the number of edges emanating from each vertex. That’s another way to say the same thing. In case we miscounted the edges, let’s check our work by deriving E from the other numbers in the above table. There are actually two useful methods for deriving E: 1. What’s wrong with this logic? “Since the cube has 6 faces, each with 4 edges, the cube must have 6  4 ¼ 24 total edges.” Why did this computation mistakenly double the

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Table 10: Platonic solids with numbers of parts

Tetrahedron

4

6

4

3

3

Cube

8

12

6

4

3

Octahedron

6

12

8

3

4

Dodecahedron

20

30

12

5

3

Icosahedron

correct answer that the cube has a total of 12 edges? Each edge was mistakenly double-counted because each edge belongs to two faces. The correct formula is this: the number of faces times the number of edges per face equals twice the number of edges. In symbols: 2. What’s wrong with this logic? “Since the cube has 8 vertices, each with 3 edges emanating from it, the cube must have 8  3 ¼ 24 total edges.” As before, each edge was mistakenly double-counted, this time because each edge emanates from two vertices. The correct formula is: the number of vertices times the number of edges emanating from each vertex equals twice the number of edges. In symbols:

Check that both blue-boxed formulas above are true in each row of the Platonic solid table (Table 10).

8C Duality What other patterns do you notice in the Platonic solid table (Table 10)? The color coding is a hint. The cube and octahedron rows are colored purple. The dodecahedron and icosahedron rows are colored green. How are the two purple rows related to each other? How are the two green rows related to each other? The pattern is this: same-colored rows have identical values for E, swapped values for {V, F}, and swapped values for {S, C}. These numerical patterns are not a coincidence. Rather, they derive from a fundamental geometric relationship between the color-coded pairs of Platonic solids. This relationship is called “duality.”

Definition Duality is a procedure for starting with one Platonic solid and constructing another Platonic solid. The new solid is built from the old solid via these steps:

1. The vertices of the new solid are the centers of the faces of the old solid. 2. Two vertices of the new solid are connected via a new edge exactly when the faces of the old solid (on which these vertices are centered) share an old edge.

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Figure 154: The dual of a cube is an octahedron

8

If the old solid is a cube, let’s imagine what the new solid will look like. To help you visualize the process, enter a roughly cube-shaped room. Paint a yellow dot on the center of its roof, and one on the center of its floor and of each of its four walls. These dots are the vertices of the new solid. Use kite string for the edges of the new solid. Run a piece of kite string from the ceiling dot to each wall dot (because the ceiling shares an edge with each wall), and similarly from the floor dot to each wall dot. Run a piece of kite string between the dots on each pair of walls that share an edge (front-right, front-left, backright, and back-left). Do not run a piece of kite string between dots on pairs of walls that do not share an edge (ceiling-floor, right-left, front-back). As the image in Figure 154 shows, the new solid is an octahedron! Figure 154 helps us account for all of the relationships that we previously discovered between the cube and octahedron rows of the table. For example, the number of vertices of the new solid (the octahedron) equals the number of faces of the old solid (the cube) because the new solid’s vertices are the centers of the old solid’s faces. Further, the new and old solid have the same number of edges, because the duality procedure required us to construct one (kite string) edge of the new solid for each (ceiling-wall, floor-wall, or wall-wall boundary) edge of the old solid. Finally, the number of faces of the new solid equals the number of vertices of the old solid. If 8 spiders built their webs at the 8 vertices of the old solid (the corners of the room), then each spider would have a different face of the new solid at which to spend its day staring. The duality procedure undoes itself! If the old solid is the octahedron, then the new solid is the cube. To see this, imagine yourself in an octahedron-shaped room, painting dots and running kite string. Can you visualize the cube taking shape? It is illustrated in Figure 155. We say that the cube and octahedron form a dual pair or that each is the dual of the other. If you start with a dodecahedron, duality yields an icosahedron, and vice versa, as illustrated in Figure 156; these two Platonic solids form another dual pair. Don’t settle for these still images; they are just frames of beautiful animated rotatable graphics created by ATRACTOR that are available at: https://www.atractor.pt/mat/ Polied/index-_en.html. If you start with a tetrahedron, then the duality procedure yields a tetrahedron, as illustrated in Figure 157. The tetrahedron is therefore called self-dual.

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Figure 155: The cube and octahedron form a dual pair

Figure 156: The dodecahedron and icosahedron form a dual pair

Figure 157: The tetrahedron is self-dual

In each of the previous duality illustrations, it is easy to see that the old solid and the new solid have the same proper and improper symmetries. Every symmetry of the old solid is also a symmetry of the new solid inscribed inside it, and vice versa. In particular, the octahedron has the same (proper and full) symmetry group as the cube, while the

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icosahedron has the same (proper and full) symmetry group as the dodecahedron. With no extra work required, we have identified the proper and full symmetry groups of our two new Platonic solids!

8D Euler’s Formula

8

Here is a reprinting of our Platonic solid table, with two new columns added at the end to celebrate our new understanding of the proper and full symmetry groups of dual pairs, and with a mysterious new white column added in the middle (Table 11, extending Table 10). As the white column shows, each of the five Platonic solids has the mysterious property that its vertex count plus its face count minus its edge count equals 2. In . Why always 2? Is this a coincidence, or an indication of symbols: some underlying geometric principle waiting to be discovered? Hoping for the latter, let’s hunt for the most general setting in which this formula is valid. First, this formula appears in the study of connected planar graphs. What are those? A graph means a finite collection of vertices (dots in the plane) together with a finite collection of edges (straight or curved paths beginning and ending at the vertices). Think of the vertices as towns and the edges as roads between the towns. A graph is called connected if it is possible to travel between any pair of vertices (towns) along the edges (roads). A graph is called planar if the edges only meet each other at vertices (the roads may meet each other at towns but don’t otherwise cross each other). In Figure 158, the left graph (a) is a connected and planar graph. The right image (b) shows how to change it into a non-planar graph (by adding a new edge that crosses another edge at a non-vertex location) and how to change it into a non-connected graph (by adding a new cluster of edges and vertices with no bridge to the original cluster). Take out a pen and paper, and draw your own connected planar graph. Make it as complicated as you like. When you finish, count the number of vertices (V ), edges (E), and faces (F) that your graph has. . A face For example, our count for Figure 159 came out , , means one of the grassy pastures into which the network of towns and roads divides the landscape. If a farmer placed one cow on each face, then the cows could not share each Table 11: Platonic solids with symmetry groups and Euler expression values

+ – Tetrahedron Cube Octahedron Dodecahedron

4 8 6 20

Icosahedron

12

6 12 30

Prop

4 6 8 12

2 2 2 2

3 4 3 5

3 3 4 3

20

2

3

5

Full

× ×

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Not connected

Not planar

This graph is connected and planar

This graph is NOT

Figure 158: (a) This graph is connected and planar. (b) This graph is neither connected nor planar 1

Figure 159: Counting the vertices, edges, and faces

3

2

1 2

3

2

1

5

4

4

5

7

5

3

10

8

4

8

6 12 14

9 6

11

7

7

6

13

other’s grass, because the type of cow we’re talking about here is afraid to cross roads. One of the faces is always unbounded. By convention, this “surrounding ocean of grass” (number 5 in our graph) always counts as one and only one face (so one lucky cow will have an unlimited supply of grass to munch). If you respect this convention and count This formula carefully, your numbers will satisfy Euler’s magic formula , and it is true for your graph also, no matter how is true for our graph ( ) complicated. This is guaranteed by the following formal statement, which really has the status of a theorem but is conventionally called “Euler’s formula” (the name “Euler’s Theorem” having been given to a different result of Euler’s in a different area of mathematics).

Euler’s Formula for the Plane Every connected planar graph satisfies Euler’s formula: V + F  E ¼ 2.

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Figure 160: Video proof of Euler’s formula (▶ https://doi.org/10.1007/000-266)

8

Proof Imagine drawing any connected planar graph and then placing a piece of tracing paper over it. The idea is to copy your graph onto the tracing paper one edge at a time in such a way that V + F  E starts equal to 2 and remains equal to 2 at each step. The PowerPoint presentation for Chapter 8 (and the video linked from Figure 160) includes a useful animation of this stepby-step process. Start by tracing any one vertex. At this point, V ¼ 1, F ¼ 1 and E ¼ 0, so V + F  E ¼ 2 for this starting graph on your tracing paper. Next, trace any edge that starts at that vertex, and also trace the vertex at which this edge ends. Continue tracing one step at a time. At each step, you must trace a new edge that begins at a previously traced vertex, and also trace the vertex at which this edge ends (unless the ending vertex was previously traced). Since your graph was connected, you will be able to completely trace it in this “one edge at a time” fashion. The expression V + F  E started equal to 2, and we claim that it will continue to equal 2 after each new edge is traced. Why? Each time you trace a new edge, there are two possibilities:

1. If the new edge terminates at a previously untraced vertex, then you trace it, so at this step you just added one new edge and one new vertex and no new faces. 2. If the new edge terminates at a previously traced vertex, then this new edge will divide one large face into two smaller faces, so at this step you just added one new edge and one new face and no new vertices. So either you increased E and V by one each, or you increased E and F by one each. Neither of these changes affects the expression V + F  E. If this expression equaled 2 before you made the change, then it equals 2 afterwards as well. As you trace the graph, the expression V + F  E starts at 2 and never changes, so after the last edge is traced, it still equals 2. □

This theorem could also be proven by induction, as we will discuss in Exercise 14 at the end of this chapter. Are there other settings in which Euler’s formula is true? Let’s try balloons. Blow up a balloon and draw a graph on it with a permanent marker. As before, your graph must be

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connected, and its edges may meet each other only at its vertices. Now count V, F, and E. If you counted carefully, then your numbers will satisfy Euler’s formula!

Euler’s Formula for the Sphere For every connected graph embedded on a sphere, V + F  E ¼ 2.

“Embedded on a sphere” means that the graph is drawn on the surface of a sphere with its edges meeting each other only at its vertices. You might visualize it as a network of towns and roads on the surface of a planet. We don’t need a new proof here; just check that the previous proof works equally well on a sphere. The formula for a sphere is more natural than for the plane. With the plane, we needed to clarify the special convention that the unbounded “surrounding ocean of grass” counts as exactly one face. With the sphere, no special clarification is needed because all faces are bounded and are on equal footing. For this reason, it’s perhaps best to think of a planar graph as really living on a sphere. If the planar graph is drawn on a piece of paper, then just place the paper on the ground and now it’s on a sphere (the surface of the earth). How is Euler’s formula for Platonic solids related to Euler’s formula for the plane and the sphere? Try this. Instead of cardboard triangles, squares, and pentagons, use rubber ones to build your Platonic solids. For example, build an icosahedron out of 20 rubber triangles, carefully glued together so that the seams are airtight. Blowing air into your rubber icosahedron will balloon it into a spherical shape, and its seam-lines will form a connected graph embedded on the sphere, whose vertices, edges, and faces correspond to those of the original flat-faced icosahedron. Thus, Euler’s formula is valid for the five Platonic solids because it is valid for the balloon graphs that they determine on the sphere. Alternatively, if you hold the hollow frame of a Platonic solid over a piece of paper, shine a light from above, and trace the resulting edge-shadows on the paper, your tracing will be a connected graph. Choose a good angle from which to shine the light so that your resulting graph is planar. After you straighten their edges and pretty them up a bit, your tracings might look like the ones shown in Figure 161. Can you tell which graph corresponds to which Platonic solid? The vertices, edges, and faces of the edge-shadow graph correspond to those of the Platonic solid. Thus, Euler’s formula is valid for the five Platonic solids because it is valid for the corresponding edge-shadow graphs.

Edge-shadow graphs of the 5 Platonic solids Figure 161: Edge-shadow graphs of the 5 Platonic solids (▶ https://doi.org/10.1007/000-267)

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8E The Euler Characteristic

8

We have seen that Euler’s formula works in at least three settings: the plane, the sphere, and the Platonic solids. We have described natural connections between these three settings. But there are other settings in which Euler’s formula does not work. Try this. Instead of drawing your connected graph on a plane or a balloon, draw it on an inner tube (the kind used to float down a lazy river), or on a double inner tube (for couples who like to float together), or on a triple inner tube; see Figure 162. Use a permanent marker, but draw your graph when the water park attendant isn’t looking. What is V + F  E? It depends! Consider Figure 163 (left) as a graph on an inner tube, by imagining dots drawn for vertices everywhere lines cross. For this graph, we carefully counted that F ¼ 576, V ¼ 576, and E ¼ 1152. Therefore, the expression V + F  E equals zero (not 2). In fact, the expression V + F  E equals zero for any connected graph embedded on an inner tube, provided the graph satisfies a very mild restriction that was automatically satisfied for graphs on spheres. In vague terms, this restriction requires the graph to spread over the whole surface. It disallows Figure 163 (right), which uses only a small part of the rubber surface, and therefore yields the same V, E, and F counts as when the same graph was drawn on a piece of paper (giving V + F  E ¼ 2 as before). In more precise terms, the restriction requires all of the faces of the graph to be shaped like deformed polygons (bent and stretched triangles, squares, etc.). In Figure 163 (left), all faces were rectangular (deformed squares). But in Figure 163 (right), the large “surrounding ocean” face is funny; it is shaped like an inner tube with a hole cut out, not like a deformed polygon. That’s why the graph is disallowed. . It gets better. Any such graph on the double inner tube will satisfy . These observations Any such graph on the triple inner tube will satisfy are the beginning of an entire field of mathematics called topology. Different surfaces can Figure 162: V + F  E might not equal 2 on one of these inner tubes!

Figure 163: V + F  E equals zero for the left graph, whereas the right graph doesn't follow the rules

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be distinguished by their value of V + F  E. This value is called the Euler characteristic of the surface. Thus, the Euler characteristic of the sphere equals , of the inner tube equals , of the double inner tube equals , and of the triple inner tube equals . Do , and you see the pattern? The Euler characteristic of a quadruple inner tube equals so on. The Euler characteristic is unaffected by bending and stretching. You could stretch your spherical balloon into an egg-shape, and this wouldn’t change its Euler characteristic because it wouldn’t change the V, E, or F counts for a graph drawn on the balloon. Thus, the Euler characteristic of a surface measures some essential quality of its shape that is unaffected by bending and stretching. Surfaces with different Euler characteristics cannot be bent or stretched into each other. If you meet a friend while floating down a lazy river, you can’t make room for her by stretching your single inner tube into a double, because the single and double tubes have different Euler characteristics.

8F The Platonic Solids Through the Ages Human fascination with the Platonic solids extends back thousands of years. These solids probably first appeared in the artwork of the Neolithic people of Scotland (~1400 BC). They were studied extensively by Pythagoras (570–495 BC) and the ancient Greeks. They are named after Plato (428–328 BC), who associated them with earth (cube), air (octahedron), water (icosahedron), fire (tetrahedron), and the universe (dodecahedron). Euclid’s Elements (~300 BC) includes mathematical constructions for all five Platonic solids, and a proof that there are no others. There are only five of these perfect shapes, which elevated their importance in scientific and theological writings. These five shapes found their way into all manner of theories. They were claimed to represent the fundamental pieces out of which all matter is formed. In 1659, Kepler explained the motions of the known planets using the illustrated model of the solar system based on the five Platonic solids inscribed inside each other (Figure 164). Today, scientists do not view the Platonic solids as directly related to the motions of the planets or the fundamental building blocks of matter. Nevertheless, these five solids maintain an important status within math and science. As we saw in Chapter 7, every bounded 3D object (that has finitely many symmetries and is not essentially two-dimensional) has the same proper symmetry group as one of the Platonic solids. Thus, the Platonic solids are models for the possible ways in which the 3D objects around us, like molecules in chemistry and cell structures in biology, can be symmetric. Why did the HIV virus evolve the icosahedral shape shown in Figure 165? A biologist’s answer to this question is informed by the mathematical fact that there are so few possibilities. Does modern science still regard the Platonic solids as representing earth, air, water, fire, and the universe? Not really, but here is an intriguing update. Scientists and mathematicians are currently using data collected by the Hubble telescope to attempt to discover the shape of the universe. One of the top contenders is an abstract 3-dimensional shape that is called “Poincaré dodecahedral space” because its geometry is intimately related to the symmetry group of a dodecahedron. Perhaps the dodecahedron does represent the universe!

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Figure 164: Kepler’s model of the solar system

8

Figure 165: An icosahedral virus

187  Exercises

Exercises (1) Write a precise definition of each of the following terms: Platonic solid, overcrowded, duality, Euler characteristic. (2) Count the vertices, edges, and faces of the planar graph in Figure 166, and verify Euler’s formula for this graph. (3) If a connected planar graph has 12 vertices and 14 faces, how many edges does it have? (4) If a connected planar graph has 17 edges and 9 faces, how many vertices does it have? (5) If a connected planar graph has 13 vertices and 27 edges, how many faces does it have? (6) Is it possible for a connected planar graph to have 20 vertices, 20 edges, and 20 faces? (7) Draw a connected planar graph with 7 vertices and 4 faces. (8) Same-colored rows of the Platonic solid table have exchanged values of S and C. Explain how duality accounts for this relationship. (9) Duality can be applied to certain objects that are not Platonic solids. What solid results if you apply duality to a square-based pyramid? (10) You are asked to draw a connected planar graph with exactly 10 edges. Use Euler’s formula to decide the largest and smallest possible number of faces that your graph could have. Then draw two graphs, one with the largest and one with the smallest possible number of faces. (11) We claimed that we carefully counted F ¼ 576, V ¼ 576, and E ¼ 1152 for the inner tube graph in Figure 163 on page 184. That was a lie. Really, we just guessed the number of faces. We then knew that the graph had as many vertices as faces, and twice as many edges as faces. How did we know this? HINT: Are the blue-boxed formulas on page 177 valid here? COMMENT: With these relationships, V + F  E ¼ F + F  2F ¼ 0, so the Euler characteristic comes out right even if our face count guess was wrong. (12) A soccer ball (Figure 167) is made from 12 pentagon faces and 20 hexagon faces. Use this information to figure out how many vertices and edges it has, and then use this to verify Euler’s formula. HINT: Use the two blue-boxed formulas on page 183. One of these two formulas must be modified to apply to an object with two types of faces. Figure 166: Planar graph for Exercise 2

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188

Figure 167: Soccer ball

Figure 168: Planar graph for Exercise 13

8 (13) The planar graph illustrated in Figure 168 is not connected; rather, it is built from two connected components, with no bridge between them. Compute V, E, and F for this graph, and verify that V + F  E ¼ 3. Prove that any planar graph with exactly two connected components will satisfy V + F  E ¼ 3. Conjecture a formula for V + F  E in any planar graph with exactly k connected components. HINT: What is the simplest such graph you can draw? From the simplest one, you can construct any other such graph “one edge at a time.” (14) Instead of the “tracing paper” proof given in the book, Euler’s formula for the plane could be proven by induction. But it is actually easier to instead prove by induction the generalization of Euler’s formula that you discovered in the previous exercise: for any planar graph with k connected components, V + F  E ¼ 1 + k. Prove this formula by induction on the number of edges. In other words, if you know the formula is true for all planar graphs with E edges, prove that it’s also true for all planar graphs with E + 1 edges. (15) For each Platonic solid, the number of proper symmetries equals F  S. The reason for this was discussed on page 144. Use similar reasoning to explain why the number of proper symmetries is also equal to V  C. (16) In each row of the Platonic solid table, the values {V, E, F, S, C} satisfy these four relations: 1. 2. 3. 4.

F  S ¼ 2  E. V  C ¼ 2  E. V + F  E ¼ 2. S  3 and C  3.

Prove that there are only five ways to choose positive integers {V, E, F, S, C} so that these four relationships are all satisfied; namely, the five rows of the Platonic solid table.

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Does this observation provide an alternative proof that there are only five Platonic solids? Discuss. (17) The edge-shadow graphs illustrated on page 183 are reprinted here:

Match these five graphs with the five Platonic solids. (18) TRUE or FALSE: Every Platonic solid is rigidly equivalent to its dual. (19) Based on how duality works, explain why a self-dual Platonic solid must have equal numbers of faces and vertices. (20) Make physical models of a single, a double, and a triple inner tube (for example by sticking bagels together and covering the resulting surface with masking tape so it can be drawn on). Compute the Euler characteristic of each surface by drawing a graph on it and counting its parts. (21) Obtain the following three toys produced by Hoberman: Switch Pitch, Flip Out, and Super Flip Out (or find online videos showing how these toys work). Explain how these toys are related to duality. (22) Blow up five balloons and use a marker to draw graphs on them corresponding to the five Platonic solids. (23) Prove from scratch that the proper symmetry group of an octahedron is isomorphic to P4 by coloring its faces with four colors. (24) Color the faces of an octahedron with two colors such that the proper symmetry group of the colored object is isomorphic to A4.

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Symmetry and Optimization

Symmetry is beautiful, but that’s not why there’s so much of it in the world. Viruses evolved their icosahedral shape, not to be pretty, but because this shape optimizes performance. Bees evolved the behavior of building hexagonal honeycombs for functional, not aesthetic, reasons. The solution to an optimization problem is often highly symmetric. In other words, symmetric shapes are often the best shapes, and this principle helps account for their prevalence in nature.

9A Minimal Surfaces When you blow a bubble, it might be cigar-shaped at the instant it leaves the wand, but then it immediately snaps into a spherical shape. Why? What optimization problem is nature almost instantaneously solving here? Soap film is like elastic stretched taut. It wants to get smaller; that is, it wants to decrease its surface area. But it’s forced to enclose your breath of air because surface tension prevents it from popping. Given that it must enclose a fixed volume of air, it finds the least-surface-area way to do so. The sphere is not just the best solution that the bubble can find; mathematicians have proven that it’s the best among all conceivable shapes.

The Bubble Theorem The sphere is the least-surface-area way to enclose a given volume.

Electronic supplementary material The online version of this chapter (https://doi.org/10.1007/ 978-3-030-51669-7_9) contains supplementary material, which is available to authorized users. The videos can be accessed individually by clicking the DOI link in the accompanying figure caption or by scanning this link with the SN More Media App.

© Springer Nature Switzerland AG 2021 K. Tapp, Symmetry, Texts for Quantitative Critical Thinking, https://doi.org/10.1007/978-3-030-51669-7_9

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Figure 169: Least-area surfaces stretching over the saddle and tetrahedron frames (▶ https://doi.org/10.1007/000-269)

9

The sphere is also the most symmetric bounded 3D object, so at least in this case, the most symmetric shape is the optimum shape. As you read this chapter, keep in mind two principles: 1. The most symmetric shape is often the best. 2. The solution to an optimization problem often has the same symmetries as the problem itself. When it doesn’t, that’s interesting too. Soap solution can answer other optimization problems as well. If you dip a plastic frame into soap solution, nature finds the least-area surface stretching over that frame. Two examples of this are shown in Figure 169. At the instant that the tetrahedron frame was removed from the bucket of soap solution, the soap film probably coated each of its 4 faces separately, but then it almost instantaneously snapped into the improved configuration pictured on the right in Figure 169. The surface could reduce its surface area further by letting go of the plastic frame, but chemical bonds prevent this. Given that it’s compelled to cling to the plastic frame, it finds the least-area surface that does so. Notice that the soap-film surface inside each frame pictured in Figure 169 has the same symmetries as the frame it clings to. This illustrates our second principle: the solution (the soap shape) has the same symmetries as the problem (the frame). A least-area surface stretching over a cube frame is illustrated in Figure 170, viewed from two different angles. You might have expected the small square film in the center to collapse to a point, but a small square turns out to be better than a point. A good calculus student could verify this, but not nearly as quickly as the soap film figured it out. The solution to an optimization problem often, but not always, has the same symmetries as the problem itself. Unlike the soap films in the saddle and tetrahedron frames, the soap film in the cube frame does not have the same symmetries as the frame itself. This is possible because there are actually three soap-film configurations that tie as least-surfacearea winners, corresponding to whether the small square film in the center is parallel to

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The Circle Wins

Figure 170: A least-area surface stretching over a cube frame, viewed from two angles (▶ https://doi.org/10.1007/000-268)

the front, right, or top face of the cube frame. These three configurations are permuted by the symmetries of the cube (or you can actually shake the frame to make it switch between these three configurations). Situations like this, where the solution does not have the same symmetries as the problem, typically arise because there are multiple tied solutions, which is interesting too. In fact, we saw how interesting it can be when we mentioned the following two optimization problems in Chapter 7:

▬ [Page 149] What is the largest possible stick (line segment) that’s able to fit inside a cube?

▬ [Page 150] What is the largest possible cube that’s able to fit inside a dodecahedron? Both problems had multiple tied solutions that were permuted by the proper symmetries. That was exactly what allowed us to determine the proper symmetry groups of the cube and dodecahedron! The study of least-area surfaces is currently an extremely active area of mathematics research. If you perform a web image search for minimal surface, you will find a gallery of beautiful soap-film-like images, including bounded surfaces stretching over frames and also unbounded surfaces extending indefinitely in all directions.

9B The Circle Wins There is a natural 2D analog of the Bubble Theorem. Suppose that a farmer wants to enclose exactly 25 acres of grass land for his cows to graze on. How can he do so with the smallest possible length of fence? If you expect the most symmetric shape to be the best shape, then you will correctly guess the following theorem.

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Figure 171: We wish to prove that Farmer Don’s winning fence is a circle, but for all we know now, it could have a crazy shape like this

The Circle Theorem The circle is the least-perimeter way to enclose a given area in the plane.

Thus the circle, which is the most symmetric bounded 2D object, is also the solution to the farmer’s optimization problem. He should build a circular fence. The proof of this theorem is all about symmetry.

9

Sketch of Proof All of the farmers in the land competed in a contest to design the leastperimeter fence enclosing a given area (say 25 acres, although this number does not matter). Farmer Don won! His fence not only beat the other farmers’ fences, it also beat all possible other fences. In other words, Farmer Don found a least-perimeter possible way to enclose the given area. We wish to prove that Farmer Don’s winning fence is a circle, but for all we know now, it could have a shape as crazy as Figure 171. Consider the horizontal line that exactly divides the area of Farmer Don’s fence in half. We claim that this line must also divide the perimeter of his fence in half. Why? Because if, say, the top had more perimeter than the bottom, then his fence would not have really been a winner—replacing the long top with the mirror reflection of the short bottom would produce a fence that beats Farmer Don’s original fence, contradicting our assumption that his original fence is a winner.1 In summary, because his fence is a winner, we know that the horizontal line that divides its area in half must also divide its perimeter in half. There are equal amounts of grass on each side and equal lengths of fence on each side. Now consider the new fence obtained by replacing the top half of Farmer Don’s fence with the mirror reflection of the bottom half (over this horizontal halving-line), which for our supposed shape looks like the new shape in Figure 172. This new fence has the same area and same perimeter as the original, so it is a tied winner! It ties with the original, but it is guaranteed to have at least two symmetries, whereas the original might have had only one (the identity). Next, consider the vertical line that divides the area of this new fence in half. As before, it must also divide the perimeter in half, so if we replace the right side with the mirror reflection of the left side, as pictured in Figure 173, then we obtain yet another tied winner. This new tied winner is guaranteed to have at least four symmetries, namely, {I, H, V, R180}. Notice that R180 is a symmetry because it is the composition of H and V, which are both

1

This is an example of indirect reasoning, which will be further explained in the next section.

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Figure 172: Replacing the top half with the mirror reflection of the bottom half yields a tied winner!

Figure 173: Replacing the right half with the mirror reflection of the left yields another tied winner!

Figure 174: If the blue line met the fence at a non-right angle, then replacing one side with the mirror image of the other would produce something impossible: a winner with an innie-point

symmetries. The white dot at which the horizontal and vertical lines cross is the center point of this new winner. Observe that every line through this white center point divides the perimeter and area of the new winner in half, simply because R180 exchanges the two sides of such a line. But this implies that every such line must meet the fence at right angles. Why is a non-right angle, like the angle at which the blue line in Figure 174 meets the fence, impossible? Because the tied winner obtained by replacing one side of the blue line with the mirror reflection of the other side has an “innie-point” (two fence segments meeting at an angle pointing into the pasture side). But this is impossible; winners never have innie-points.2 The fence near any innie-point can be rounded off (as illustrated in purple) to yield a fence that’s better on all counts: the purple modification encloses more area and has a smaller perimeter than the original.

2

This is a second example of indirect reasoning, which will be further explained in the next section.

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Figure 175: Only a circle could meet all radial lines at right angles

9

In summary, starting with Farmer Don’s winning fence, we built a tied winner with four symmetries. This tied winner is guaranteed to meet every line through its center point at a right angle. Now let’s brainstorm. Think of a fence shape that meets every radial line (meaning every line through its center point) at a right angle. How many different such shapes can you come up with? In fact, the only solution is a circle! To convince yourself of this, imagine tracing along the fence edge with a compass that’s anchored at the center point, expanding and contracting the compass width as necessary to stay on the fence path (as illustrated in Figure 175). Notice that expanding the compass creates obtuse angles with the radial lines, while contracting creates acute angles. The only way to avoid acute and obtuse angles is to never expand or contract your compass, which only happens when you are tracing a perfect circle. Thus, the tied winner with at least four symmetries must be a perfect circle. This winner looks like four copies of the bottom-left quadrant of Farmer Don’s original winner, so the original winner’s bottom-left quadrant must be a quarter circle. There’s nothing special about the bottom left. A slight modification to the above proof establishes that the bottom-right, top-left, and top-right quadrants are also quarter circles. Thus, Farmer Don’s original winning fence must have been a circle. □

That was a long proof! Hopefully it convinced you that the theorem is true. It is. Nevertheless, we must confess that more work is needed to make this proof precise and rigorous enough to satisfy mathematicians today. For example, the compass-tracing part is really a calculus problem that involves verifying that the polar coordinate function, r(θ), has zero derivative (if you don’t know calculus, then please ignore that sentence). A subtler issue is this: our proof showed only that a winning fence must be a circle. This conclusion is vacuous unless we can verify that a winning fence exists; in other words, that there exists a fence enclosing the given area that has smaller perimeter than any other fence (except for possible ties). This is true, and probably seems obvious, but it does require a separate proof. After all, there does not exist a largest-perimeter fence enclosing a given area.

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Elements of Mathematics: Proof by Contradiction

It’s not a bad thing that our proof was insufficiently rigorous. Beautiful new ideas are often discovered when we strive to put visual proofs and vague heuristic arguments on more solid and rigorous footing. In this case, proving that a winner exists led to important ideas in the field of analysis. In any case, the details have been filled in, and we now know beyond doubt that the theorem is true: the circle wins!

9C Elements of Mathematics: Proof by Contradiction A proof by contradiction (also called an indirect proof) is an extremely useful construction that works like this: In order to prove an assertion is true, you assume that the assertion is false, and demonstrate that this assumption leads to a contradiction (which means an incompatibility with a known fact or with a previous hypothesis).

There is nothing deep here. Anyone who has watched detective movies has encountered this type of logic. For example, imagine you’re at the point in the movie where you know that the man either committed suicide or was murdered. The detective might reason: “Let’s assume he committed suicide. . .but that assumption contradicts the known scientific fact that no human can shoot himself 27 times repeatedly in the head. . .so the assumption must be wrong. . .he didn’t commit suicide. . .he was murdered!” Indirect reasoning is so natural, you may not have noticed that it was used in this chapter. Look back at the two points marked with footnotes in the previous section: 1. To prove that the line divides the perimeter in half, we assumed it doesn’t, and argued that this assumption led to a contradiction of our hypothesis that the fence is a winner. 2. To prove that every radial line meets the fence at a right angle, we assumed there was one that didn’t, and argued that this assumption contradicted the provable fact that winning fences never have innie-points. Each is an example of an indirect proof used to prove an assertion that is a step of a larger proof. An indirect proof is closely related to a proof by contrapositive (page 17). In fact, some indirect proofs can be rephrased as contrapositive proofs. For example, point (1) above could be rephrased as establishing “if the fence is a winner then the line divides its perimeter in half” by proving the contrapositive of this statement. Nevertheless, phrasing it as an indirect proof was more natural here. The remaining chapters of this book include several famous examples of indirect proofs.

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Exercises

9

(1) Find the perimeter of the circle, square, and equilateral triangle enclosing area 25. Order these three shapes by increasing perimeter. (2) Farmer Ann wants to fence off 25 acres of grass land along a straight river. She does not need to fence along the river’s edge because her cows are afraid of water. Guess the least-perimeter fence shape. (3) When two bubbles collide, they form a double bubble configuration that encloses and separates two (possibly different) volumes of air, as in Figure 176. Do actual soap bubbles find the least-surface-area way to do so, or do mathematicians know of a better solution that physical soap bubbles are unable to attain? Use online resources as needed to learn the answer to the famous double bubble problem. (4) Four cities lie at the corners of a square. We wish to connect them by a network of roads so that it is possible to drive between any pair of cities. Which of the three road configurations pictured in Figure 177 uses the smallest total length of roadway? Invent a road configuration that beats all three of these configurations. (5) “The circle is the most-perimeter way to enclose a given area in the plane.” This sentence is completely false, but my friend Paul tried to prove it anyway. To prove it, he copied our proof of the Circle Theorem, replacing “least” with “most,” and making other such modifications as needed. For example, he easily changed the purple path near the innie-point to make it enclose less area and have larger perimeter than the original. What goes wrong with his attempt?

Figure 176: A double bubble configuration that encloses and separates two different volumes of air – did the soap solution find the least-surface-area way to do this?

Figure 177: Road configurations for Exercise 4

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(6) The Circle Theorem is more commonly called the Isoperimetric Theorem, from the Greek word for “same perimeter.” This name is more appropriate for the following alternative formulation of the theorem: Among all curves with the same given perimeter, the circle encloses the most area. In other words, if a farmer has a fixed length of fence to work with, and he wishes to enclose the most possible grass, he should build a circular fence. Explain why the two formulations of the theorem are equivalent. HINT: Explain why a counterexample to one version could be rescaled to become a counterexample to the other.

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What Is a Number?

By now you understand the importance of precise language. The history of mathematics is, among other things, a story about the invention of ever more precise language and techniques to explore the abstract ideas required to model the physical world. New ideas force us to look back and more precisely re-define our old vocabulary. This backtracking might sound tedious, but it has led to unexpected discoveries and truths and has sparked some of the greatest breakthroughs in mathematics. The remaining four chapters of this book are all about backtracking (plus some beautiful excursions). They are organized as follows: Chapter 10: We previously defined the plane as the set of all ordered pairs of real numbers, and space as the set of all ordered triples of real numbers. But what is a real number? This chapter precisely answers this question and relates the answer back to our study of symmetry. Chapter 11 (optional): This chapter explores some peripheral topics related to the number systems studied in the previous chapter, including the Prime Number Theorem and Cantor’s Theorem. These topics are optional (because they’re not really about symmetry), but they are some of the most beautiful results in the history of math. If you have time for an excursion, this is the one to take! Chapters 12 and 13: We previously defined a rigid motion as a repositioning of the plane (or space) that preserves distances. These chapters put this definition on more solid footing by precisely defining repositioning and preserves distances (first in terms of functions and then in terms of matrices) and then relate these definitions back to our study of symmetry. We now ease into our study of real numbers by first considering the simpler number systems that historically paved the way to the invention/discovery of real numbers.

Electronic supplementary material The online version of this chapter (https://doi.org/10.1007/ 978-3-030-51669-7_10) contains supplementary material, which is available to authorized users. The videos can be accessed individually by clicking the DOI link in the accompanying figure caption or by scanning this link with the SN More Media App.

© Springer Nature Switzerland AG 2021 K. Tapp, Symmetry, Texts for Quantitative Critical Thinking, https://doi.org/10.1007/978-3-030-51669-7_10

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10A Natural Numbers In school, you first studied the natural numbers:

ℕ = {1, 2, 3, 4, 5, 6, … }

“the natural numbers”

You learned that the most important natural numbers are the prime numbers: 2, 3, 5, 7, 11, 13, 17, 19, 23, and so on.

Definition A prime number is a natural number greater than 1 that cannot be expressed as a product of two smaller natural numbers.

You probably learned that prime numbers (or, simply, “primes”) are the building blocks of all natural numbers. Here is a theorem that expresses more precisely how this building-up happens.

10 Theorem Every natural number greater than 1 either is prime or can be expressed in a unique way as a product of primes.

For example, the prime factorization of 300 is 2
2
3
5
5. We figured this out by breaking 300 down step by step until the pieces could not be further broken down. . .

…like this: 300 = 3 × 100 = 3 × 4 × 25 = 3 × 2 × 2 × 5 × 5. …or this: 300 = 10 × 30 = 2 × 5 × 3 × 10 = 2 × 5 × 3 × 2 × 5. The two answers above become the same if the primes are re-listed in increasing order. That’s what “in a unique way” means. Each natural number has only one prime factorization! By the way, once you know the prime factorization of a number, it is easy to find the prime factorization of its square. For example, the square of 300 is 3002 ¼ 300
300 ¼ 90000, which has the following prime factorization:

90000 = 300 × 300 = 2 × 2 × 3 × 5 × 5 × 2 × 2 × 3 × 5 × 5 (re-order) = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5 × 5 × 5. Notice that 3002 has twice as many of each prime as 300 has in its prime factorization. More broadly, the square of any natural number greater than 1 has an even number of occurrences of each prime in its prime factorization. This observation will be important later in this chapter. You’ll see.

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Irrational Numbers

10B Integers and Rational Numbers After becoming acquainted with the natural numbers, you learned about the integers (which include zero and negatives):

ℤ = {… , – 3, – 2, – 1, 0, 1, 2, 3, … }

“the integers”

22 Next, you studied rational numbers, which means fractions like 35, 17 8 , and 3 :

Of course, some fractions are the same as others; for example, 23 ¼ 46. If you were finding all solutions to an equation, like 3x ¼ 2, you would not separately list 23 and 46, because these are the same number. Here is the general rule: a c ¼ whenever ad ¼ bc: b d Regard this rule as an addendum to our definition of . Thus,  means the set of all quotients “ab” of integers with b 6¼ 0, with the understanding that certain quotients are really the same as others as specified by the above rule.

10C Irrational Numbers If you studied mathematics twenty-five centuries ago in ancient Greece, you would have learned that all numbers are rational. What else could a number be? All physical matter was thought to be finitely dividable (that is, every object was thought to be made up of finitely many undividable building blocks), so rational numbers could describe the exact size of any portion of any object. Furthermore, the Greeks regarded rational numbers as divine gifts from their gods. When clues first appeared that other types of numbers might be necessary, it was deemed blasphemous to pursue such thoughts. These early Greek mathematicians developed number theory and geometry in parallel. They required numbers to represent not just portions of physical objects, but also lengths in geometric constructions. For example, they knew that the diagonal of the 1-by-1 square in Figure 178 has a length that squares to 2 (when multiplied times pffiffiffi itself, the answer is 2). Today, we call this length “the square root of two,” denoted 2. They searched in vain for a fraction that squares to two, but their search led instead to a proof that no such fraction could ever be found.

Theorem There is no rational number that squares to 2.

pffiffiffi Today, we say “ 2 is a number that is not rational—an irrational number.” This sentence would have puzzled the early Greek mathematicians because, to them, “number” meant “rational number” (and also because they spoke Greek).

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Figure 178: Clues begin to appear that other types of numbers besides the rationals might be necessary

es

to

2

ar

th

is

Th

u sq

ng

1

le

1

Proof Suppose your uncle Pete claims to have found two positive integers, p and q, such that the  2 fraction pq squares to exactly 2. That is, pq ¼ 2. How do we know that Pete is mistaken? Let’s explore the consequences of his claim, and demonstrate that it leads to a contradiction. Rewrite Pete’s claim with some simple algebra like this:

10

Now ask the question: how many 2s are in the prime factorizations of the left and the right sides of the final equation? The left side (p2) has an even number of 2s because it has twice as many as p has. The right side (2
q2) has an odd number of 2s because it has one more than twice as many as q has. But the prime factorization is unique. If the left and right sides really equaled each other, there would not be a difference between their prime factorizations. There is only one possible conclusion: Pete was mistaken. There is no rational number that squares to 2. □ This is a beautiful and famous example of an indirect proof. We wished to prove the assertion that there is no rational number that squares to 2. To do so, we assumed that the assertion is false (that is, we assumed that there is a rational number that squares to 2) and demonstrated that this assumption led to a contradiction of the uniqueness of prime factorizations. Since assuming the assertion was false led to a contradiction, the only possibility is that the assertion must be true. When the early Greek mathematicians discovered this theorem, it was an abrupt challenge to their mathematical and religious belief systems, so they closely guarded this uncomfortable truth. Only a select few mathematicians were privy to the secret, and those who were caught sharing it with the uninitiated were executed!

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Real Numbers

10D Real Numbers Inevitably the secret got out, and it led to the invention of a number system pffiffiincluding ffi more than just fractions. What do rational numbers like 23 and lengths like 2 have in common, so that both can be incorporated into a more general concept of number? Today, we put them on equal footing by writing them as decimal expressions: pffiffiffi 2 ¼ 0:66666666⋯and 2 ¼ 1:41421356237⋯: 3 In the decimal expression for 23, the symbol “⋯” indicates a continuation of the pffiffiffi pattern (an unending string of 6s). In the decimal expression for 2, the symbol “⋯” just indicates some unending string of digits, which might not follow any pattern you recognize. We are led to a precise definition of a more general type of number:

Definition A real number is a decimal expression; that is, an integer followed by a decimal point followed by infinitely many digits. The set of all real numbers is denoted ℝ.

A real number like 45 ¼ 0:8 should have appended to it an unending string of zeros ¼ 0:800000⋯) so that it has infinitely many digits after its decimal point, as required by the definition. The purpose of this requirement is to put all real numbers on equal footing, so that pairs of them can be more easily added or multiplied. Speaking of which, you might be embarrassed to learn that you pffiffiffi don’t really know how to add or multiply real numbers. For example, what is 23 þ 2? Familiar rules of addition require you topbegin at the rightmost digit and work left, but the above decimal ffiffiffi expressions for 23 and 2 go on indefinitely to the right. There’s no rightmost digit at which to begin. Precisely defining real-number addition, multiplication, subtraction, and division involves technical intricacies that are beyond the scope of this book, but here is the rough idea. To approximate the answer, truncate the numbers involved (say two digits after their decimal points, or 3 or 100 or 1000), so you have a finite problem that can be solved with grade-school methods. The more digits used, the better your approximation. The exact answer is defined as the limit of the approximate answers as the number of digits used goes to infinity. For now, we’ll settle for illustrating the idea by discussing only one important subtraction problem: (45

Can you guess the answer? The difference between the red number and the green number is clearly. . .

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…less than 12.75 – 12.74 = 0.01 …and less than 12.750 – 12.749 = 0.001, …and less than 12.7500 – 12.7499 = 0.0001, and so on. Since this difference is less than arbitrarily small numbers, the only reasonable guess is that the difference equals zero. In fact, zero is the correct answer because it is the limit of the approximations. But if the difference between two numbers equals zero, then those numbers must equal each other:

The red and green decimal expressions are different ways of writing the same real number, just as 23 and 46 are different ways of writing the same rational number. Here is the general rule for when two decimal expressions represent the same real number:

Real Number Redundancy Rule

10

A digit (other than 9) followed by an unending string of 9s can be replaced by the next larger digit followed by an unending string of 0s. There are no other redundancies among real numbers.

The strings of 0s and 9s are even allowed to straddle the decimal point. So if a car dealer quotes you $19,999.9999⋯, then you’ll be making out your check for exactly 20 thousand. The Real Number Redundancy Rule is in a blue-outlined box because it is a definition, not a theorem. More precisely, the rule is an addendum to our definition of a real number. Thus, ℝ means the set of all decimal expressions, with the understanding that certain decimal expressions are really the same as others, as specified by the Real Number Redundancy Rule. The above reasoning didn’t prove that 12.749999⋯¼ 12.7500000⋯, but rather it indicated that we’d better choose to define them to be equal if we wish for the set of real numbers to end up having natural desirable properties (like the property that two real numbers are equal if and only if their difference is zero). Our “decimal expression” definition of a real number fits nicely with our visual intuition that the real numbers should represent all possible (positive and negative) lengths; that is, a real number should be a point on an idealized (infinitely thin) line, which we call the “real number line,” pictured here:

-4

-3

-2

-1

0

1

2

3

4

The digits of a real number tell us how to locate it on this number line. For example, to locate √2 = 1.41421356237…, the first digit “ ” tells us to look between 1 and 2. If

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Which Real Numbers Are Rational?

1

-4

-3

-2

-1

0

1 2 magnify

3

4

4

1.0

1.4

magnify

1

2.0

1.5 magnify 4

1.41

1.42 and so on...

Figure 179: The digits of a real number tell us how to locate it on the real number line (▶ https://doi.org/10.1007/000-26a)

we divide this interval into 10 equal bins (numbered 0 through 9), the next digit “ ” tells us which bin to look in. If we subdivide that bin into 10 equal sub-bins, the next digit “ ” tells us which sub-bin to look in, and so on. A picture of the first few steps of this process is shown in Figure 179. Each successive digit provides a 10-fold increase in the accuracy with which p weffiffiffi know the number’s location on the real number line. In fact, if you were asked to find 2 without a calculator, you would pffiffiffi probably identify it digit by digit in a manner like what is shown in Figure 179. First, 2 lies 1 and 2 because 1 squares to less than 2 while pffiffibetween ffi 2 squares to more than 2. Next, 2 lies between 1.4 and 1.5 because 1.4 squares to less than 2 while 1.5 squares to more than 2, and so on.

10E Which Real Numbers Are Rational? Every rational number is a real number, because long division can be used to convert any fraction into a decimal expression. The following exercise works out an example of such a conversion.

▶ Exercise

Convert 37 into a decimal expression.

Solution

(6-digit string repeats), as demonstrated by

the illustrated long division problem.

The long division problem is shown in Figure 180. The digits started repeating as soon as the orange numbers (the remainders) started repeating. Since there were only 7 possibilities for this orange remainder (0 through 6), you knew in advance that it would repeat after not more than 7 steps. Similar reasoning proves this: long division converts any fraction into a decimal expression that is eventually repeating. This means that, possibly after some initial digits, the tail of the decimal expression is formed from an indefinite repetition of a

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Figure 180: Long division converts any fraction into a decimal expression that is eventually repeating (because the remainders, shown in orange, must eventually repeat)

10

.42857142 ... 7 3.00000000 28 20 14 60 56 40 35 50 49 10 7 30 28 20

⋯ is eventually single finite string of digits. For example, ⋯. repeating, and so is 45 ¼ 0:8 because it gets rewritten as We can also convert in the other direction: from an eventually repeating decimal to a fraction.

▶ Exercise

Convert

⋯ into a fraction.

Solution We know from the previous exercise that the answer is 37, but what if we had never solved the previous exercise? Then we could use the following subtraction trick: Since N has a 6-digit repeating string, we will multiply it by 1000000 (which has 6 zeros), and then subtract N from the answer, which removes the repeating string and allows us to obtain a fraction. Here are the details:

3 We learn that N ¼ 428571 999999 (which reduces to N ¼ 7 ).

There is nothing special about the particular number in the previous exercise. After doing a few more similar examples, you should be able to convince yourself that the subtraction trick converts any eventually repeating decimal expression into a fraction. The diagram in Figure 181 summarizes the punch lines of the previous two exercises. What you just learned is more than just a computational skill. The theoretical significance is astounding: an equivalence concisely expressed in the following theorem.

209  10F

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Real Numbers and Symmetry

Figure 181: Any eventually repeating decimal expression can be converted into a fraction and vice versa

Long division

FRACTION

Eventually repeating decimal expression

Subtraction trick

Theorem A real number is rational if and only if its decimal expression is eventually repeating.

Because of this theorem, it is easy to find irrational numbers (real numbers that are not rational). Just design a decimal expression whose digits follow a pattern that’s more complicated than “eventually repeating.” For example, the following number is irrational:

… (pattern continues). Try to create your own examples of irrational numbers. It’s easy. Just make up a pattern that’s more complicated than “eventually repeating.” Or if you’re not feeling creative, just stick the above pattern on the end of any initial string of digits you like. You will quickly see that there are lots of irrational numbers, and they are everywhere. You will prove a precise version of this sentence in the exercises at the end of this chapter. Although it is easy to make up irrational numbers, it is often difficult to prove that particular numbers are irrational. For example, the famous numbers π and e are both irrational, pbut ffiffi proving this is far beyond the scope of this book. It is still unknown 2 whether π is rational or irrational, although most everyone expects it to be irrational.

10F Real Numbers and Symmetry Let’s bring all of this back to our study of symmetry. Recall the following definition from page 4.

Definition The plane is the set of all ordered pairs (x, y) of real numbers. A two-dimensional object is a nonempty subset of the plane.

Now that we know exactly what a real number is, this old definition has a more solid footing. But why should we care? Here is a very practical reason. We can finally create examples of objects that justify the fine print in some of our previous definitions and theorems.

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▶ Exercise

Consider the set of all rational points on the x-axis; that is, the set of all points in the plane of the form (x, 0) where x is a rational number. Which translations of the plane are symmetries of this object? Solution A translation right or left by any rational length is a symmetry (because the sum of any two rational numbers is rational, so the object translates onto itself). A translation right or left by any irrational length is not a symmetry (because a rational plus an irrational is irrational, so the object doesn’t translate onto itself).

This object helps justify the fine print in our definition of “border pattern” on page 15. Why did we insist in this definition that a border pattern must have a smallest translation symmetry (besides the identity)? To insure that crazy objects like the one in the previous exercise don’t qualify as border patterns. It doesn’t qualify because the nonzero distances that it can be translated to the right or left include arbitrarily small rational numbers: 12, 1 1 1 3, 4, 5, and so on. There is no smallest among them. It’s good that it doesn’t qualify as a border pattern because it doesn’t match any of the seven model border patterns. Neither does it match the whole x-axis, which is not a border pattern either. Here is another exotic example.

10 ▶ Exercise

Take the circle in the plane with radius 1 centered at the origin, and consider the set of all points on this circle that make a rational angle with the point (1, 0), like the silver point in Figure 182. Which rotations of the plane about the origin are symmetries of this object? Solution All rotations by rational angles are symmetries, whereas all rotations by irrational angles are not. Why? Because a rotation affects a point of the circle by adding the rotation angle to the angle made by the point. So the conclusion comes from the previously mentioned observation that rational+rational¼rational, whereas rational+irrational¼irrational.

Figure 182: An exotic example that justifies the fine print in Da Vinci’s Theorem

y

a ion rat le g an

l

x (1,0)

211  10G

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Elements of Mathematics: Construction of Rational and Real Numbers

This object helps explain why Da Vinci’s Theorem on page 58 only claimed to classify the symmetry types of bounded objects with finitely many symmetries. The exotic object from the previous exercise is bounded and has infinitely many symmetries. It does not match any of Da Vinci’s model objects (or a circle). How should we regard exotic examples like the objects from the previous two exercises? It’s tempting to ignore them. We originally intended our mathematical definition of “object” to model real-world objects that can be created with paint or clay. The exotic objects in the present section were built with equations rather than paint, but please don’t just discount them as meaningless artifacts of an imperfect mathematical model of the world. Examples that are exotic, surprising, or counterintuitive play a crucial role in mathematics. They help delineate the true from the false. For example, they showed why the fine print couldn’t be removed from our previous classification theorems. Investigating exotic examples is an important way to improve our intuition about mathematics. It leads to rich and interesting truths. Mathematics that’s rich and interesting always comes back around to being relevant to studying the real world, although not always in the way we originally expected.

10G Elements of Mathematics: Construction of Rational and Real Numbers One of the primary achievements of this chapter was to more precisely define rational numbers and real numbers. Here is a summary of how we did it.

Informal Definitions  means the set of all quotients “ab” of integers with b 6¼ 0, with the understanding that certain quotients are really the same as others, as specified by the rule: ab ¼ dc whenever ad ¼ bc. ℝ means the set of all decimal expressions, with the understanding that certain decimal expressions are really the same as others, as specified by the Real Number Redundancy Rule (on page 206).

These definitions are fine for most practical purposes, but are too imprecise to satisfy modern mathematicians. The expression “with the understanding that” is just too vague. The purpose of this section is to rephrase these bits more precisely with the help of equivalence relations. Take time now to review the concepts of “equivalence relation” from page 69 and “well-defined” from page 124. Let’s begin with the rational numbers. First, define  to mean the set of all members (a, b) of ℤ
ℤ with b 6¼ 0. As a minor cosmetic change, we’ll henceforth write “ab” as shorthand for “(a, b)” to better match the familiar notation for fractions that you learned in elementary school. Next, consider the following equivalence relation on :

Chapter 10 • What Is a Number?

212

a c  whenever ad ¼ bc: b d Take time now to verify that this is an equivalence relation. Since it’s an equivalence relation, it partitions  into disjoint equivalence classes. Now define  to mean the set of these equivalenceclasses.  7 100 , 200 , . . . is a single member of ; it is the set of all quotients For example, 12, 24 , 14 that reduce to one half. A representative of this member of  means a single one of the 7 quotients it includes, like 12 or 14 . In summary, a single member of  is technically defined as an infinite set, but usually one describes it just by choosing a single representative. The next job is to precisely define addition, subtraction, multiplication, and division in . For example, here is a multiplication problem:

10

Choosing representatives, this multiplication problem is more familiarly expressed as . Notice that we’re multiplying a pair of members of  by multiplying representatives of them via the familiar grade-school rules. That is exactly how we define multiplication in . Addition, subtraction, and division are defined analogously. Are these definitions well-defined? That’s the key question. The answer is yes, but this must be proven. The definition of multiplication asked us to arbitrarily choose representatives. We must verify that the final answer didn’t depend on how these arbitrary choices were made. Choosing the green representatives in the above example would have The green and red choices yielded the same answer because yielded and

represent the same member of . In the exercises, you will verify that the

choice of representatives never affects the final answer, and therefore that multiplication is well-defined. So are addition, division, and subtraction. The definition of ℝ is formalized in a way that’s so similar to , we will only outline it briefly here. First consider the set of all decimal expressions. Next, verify that the Real Number Redundancy Rule is an equivalence relation on this set. Finally, define ℝ as the set of equivalence classes. As previously mentioned, precisely defining addition, subtraction, multiplication, and division in ℝ requires a technical limit process that is beyond the scope of this book, but once this is done, it can be proven that the resulting definitions are well-defined. For example, multiplying your favorite real number times 12.749999⋯ gives the same answer as multiplying it times 12.7500000⋯. Once all of this is done, one can prove that the familiar algebraic rules (like the commutative and distributive properties) are true in  and ℝ, and also that the operations are consistently defined (for example, when you add two fractions, converting them to decimals first would give a consistent answer). Since you undoubtedly computed with fractions and decimals long before reading this book, is it really worth all the trouble it takes to go back and precisely define  and

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 Exercises

ℝ? It depends on your perspective. If you still regard mathematics as a collection of computation skills, then your answer is probably no. But if you’ve come to regard mathematics as conceptual, rigorous, and precise, then you might feel angry that your former teachers made you memorize computation rules for number systems that they didn’t define.

Exercises (1) Convert the following fractions into eventually repeating decimal expressions: 14 73 2 309 101 (a) 57 (b) 23 21 (c) 13 (d) 7 (e) 3 (f) 8 (g) 5 (2) Convert the following eventually repeating decimal expressions into fractions:

(3) (4) (5)

(6) (7) (8)

(9)

(10)

(a) 10.792929292… (b) 5.2003003003003… (c) 0.34444… pffiffiffi Prove that 3 is an irrational number; in other words, there is no rational number pffiffiffi that squares to 3. HINT: Copy the proof that 2 is an irrational number. pffiffi Prove that 5þ7 2 is an irrational number. HINT: Use an indirect proof. If you could pffiffiffi write it as a fraction, how could you use this to write 2 as a fraction? pffiffiffi pffiffiffi Prove that 2 þ 3 is an irrational number. HINT: Use an indirect proof. Assume it pffiffiffi pffiffiffi is rational, which means 2 þ 3 ¼ pq, then square both sides of this equation and find a contradiction. Prove this fact that was mentioned and used in Section 10F: the sum of a rational number and an irrational number is always irrational. As in the exercise on page 210, consider the set of all rational points on the x-axis. Prove that this object has infinitely many improper symmetries. Consider the set of all rational points on the x-axis that can be expressed as a fraction whose denominator is a power of 2. Describe all of the translation symmetries of this object. Is this object a border pattern? Is this object rigidly equivalent to the object described in the exercise on page 210? and Find a rational number between (to answer this, you do not need to know any more of the digits of M and N than are shown). Can you find infinitely many different rational numbers between M and and ? N? What about between and Find an irrational number between (to answer this, you do not need to know any more of the digits of M and N than are shown). Can you find infinitely many different irrational numbers between and ? M and N? What about between

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(11) Prove that there are infinitely many different rational numbers and infinitely many different irrational numbers between any pair of distinct real numbers. HINT: Generalize the previous two exercises. (12) In your solution to the previous exercise, why doesn’t your proof work when the two real numbers are equal, like M ¼ 12.7499999⋯ and N ¼ 12.750000⋯? (13) What is the smallest rational number larger than 0? If you don’t think there is one, then explain why. What about the smallest irrational number larger than 0? (14) What can you say about the decimal expression for a fraction whose denominator has 2 digits? What is the longest the repeating string could possibly be? (15) Convert N ¼ 12.74999999999⋯ into a fraction using the subtraction trick. What do you learn? (16) Fill in the blanks with the correct numbers: When members of  are precisely defined as equivalence classes, each member (equivalence class) is an infinite set. When members of ℝ are precisely defined as equivalence classes, each member (equivalence class) is a set of size _____ or _____. (17) Prove the claim (from the last section of this chapter) that multiplication in  is well-defined. (18) Precisely define the symbol “” on  and prove that your definition is well-defined. (19) Precisely define the symbol “” on ℝ and prove that your definition is well-defined. (20) To intelligently understand political news, you often need a frame of reference for comprehending big numbers. Filling in the following blanks (using internet resources as needed) may help you develop such a frame of reference:

A billion equals _______ million. The United States population equals about ______ million. If the United States government spends a billion dollars, this amount averages to about $______ per citizen. A trillion equals _____ billion. If the government spends a trillion dollars, this amount averages to about $______ per citizen. The United States national debt is about $______ trillion, which averages to about $________ per citizen. The most recent national deficit equals about $______ billion, which averages to about $_______ per citizen. The difference between the meanings of the words “debt” and “deficit” is:_______________________________________________________. (21) Here is a silly definition: A real number is called sevenless if it has no sevens in its decimal expression. Is this definition well-defined? (22) Here is a silly definition: A rational number is called balanced if its numerator and denominator have the same number of digits. Is this definition well-defined? (23) Suppose that X and Y are real numbers. For X, the third digit after its decimal point equals 5. For Y, the third digit after its decimal point equals 7. You do not know anything else about X or Y. In other words, X and Y look like this: X ¼  .   5     ⋯ and Y ¼  .   7     ⋯ . From this information, can you conclude that X and Y are different real numbers? What if X and Y instead look like this:

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 Exercises

X ¼  .   9     ⋯ and Y ¼  .   0     ⋯ ? What if X and Y instead look like this: X ¼  .   3     ⋯ and Y ¼  .   4     ⋯ ? State a general rule for guaranteeing that two numbers are different based on the fact that they differ at a single decimal position. (24) If X and Y are real numbers that differ at one or more decimal positions, and Y does not end in an infinite string of 9s or an infinite string of 0s, explain why X and Y must be different real numbers.

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Excursions in Numbers

The previous chapter was about numbers. This chapter follows up with some excursive topics related to numbers. Specifically, we explore the size of the set of prime numbers and the size of the set of real numbers. This chapter is optional because it is not directly related to symmetry, but is included in order to showcase some of the most beautiful proofs in the history of mathematics.

11A How Many Prime Numbers Are There? The prime numbers played a very minimal role in the previouspchapter. In case you ffiffiffi missed their relevance, we used prime factorization to prove that 2 is irrational, which was a step towards understanding the real number line. This section is for readers who desire to learn a few more fundamental and beautiful facts about prime numbers before returning to symmetry. The key question is: how many prime numbers are there? We take it as self-evident that there are infinitely many natural numbers. That’s what the “…” meant when we wrote . But are there infinitely many primes? This is less obvious. After all, there are only finitely many Lego block shapes, yet there are infinitely many different Lego constructions that could be built from unlimited supplies of them. It is similarly conceivable that there might be only finitely many prime numbers out of which all of the infinitely many natural numbers can be built. This matter was settled by Euclid around 300 BC.

Euclid’s Theorem There are infinitely many prime numbers.

Proof Imagine that your friend Andy insists that there are only finitely many prime numbers, and to prove it, he lists all of them together on a piece of paper: {p1, p2, p3, . . ., pn}. How do we know that Andy is wrong? No matter how lengthy his list, we’ll describe a strategy for identifying a prime number that is not on his list. Thus, no finite list of prime numbers could ever be complete.

© Springer Nature Switzerland AG 2021 K. Tapp, Symmetry, Texts for Quantitative Critical Thinking, https://doi.org/10.1007/978-3-030-51669-7_11

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Here’s how we’ll identify a prime number that’s missing from Andy’s list (or from any finite list of prime numbers). First compute the result of multiplying all of Andy’s primes together and adding 1. Call this number “L” because it’s so large: L ¼ p1  p2  p3  . . .  pn þ 1: Notice that L is not on Andy’s list because it’s much larger than anything on his list. If L happens to be prime, then it is exactly what we seek: a prime number that’s not on Andy’s list. If L happens not to be prime, then any single prime number that appears in L’s prime factorization is exactly what we seek: a prime number that is not on Andy’s list. This is because none of Andy’s primes divide evenly into L. In fact, L was custom-built so that each of Andy’s primes leaves a remainder 1 when divided into L. Thus, no finite list of prime numbers could ever be complete, which means there must be infinitely many prime numbers. □

11

This is a classic example of an indirect proof. We wished to prove the assertion that there are infinitely many primes. To do so, we assumed that the assertion was false (that is, we assumed there are only finitely many primes) and demonstrated that this assumption leads to a contradiction. Here is a quick example to help you understand the logic of the above proof. Suppose Andy’s list is {2, 3, 5, 7, 11, 13}. We consider the large number L ¼ 2  3  5  7  11  13 + 1 ¼ 30031. It turns out that L is not prime; its prime factorization is 30031 ¼ 59  509. The numbers 59 and 509 are primes missing from Andy’s list. So now we know that there are infinitely many primes, but we still might wonder how frequently occurring the prime numbers are among the natural numbers. Are primes in abundance, or are they a rare breed? Do most United States citizens have a prime social security number, or very few? To warm up to questions like these, let’s make a list of all natural numbers between 1 and 10, and highlight the primes, like this:

1, 2, 3, 4, 5, 6, 7, 8, 9, 10. Notice that 40% of the numbers on this list are prime, which we’ll write as 0.4. Next, we’ll go all the way to 100, like this:

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100.

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How Many Prime Numbers Are There?

Notice that 25% of the numbers on this list are prime, which we’ll write as 0.25. As you travel further and further down the list of natural numbers, prime numbers occur in these proportions: N 10 100 1,000 10,000 100,000 1,000,000

Fraction of numbers up to N that are prime 0.4 0.25 0.168 0.1229 0.09592 0.078498

The question is: what is the pattern in this table, and how does the pattern continue for larger and larger choices of N? Depending on how accurately you intend to answer this question, it is either difficult or extremely difficult or worth a million dollars. The first (merely difficult) answer is this: as N increases, the fraction approaches zero. Conjecturing this based on the table is easy, although proving it is a bit beyond the scope of this book. The second (extremely difficult) issue is specifying approximately how quickly the fraction approaches zero. This is achieved by the following famous theorem, which was conjectured in the early 1800s and not successfully proven until 1896.

The Prime Number Theorem The fraction of numbers up 1 Fraction  twice the number of digits in N

to

N

that

are

prime

is

approximately:

This theorem is difficult to prove, but it is easy to use, like this:

▶ Exercise

Approximately what fraction of numbers between 1 and 1,000,000 are prime?

1  0:0714, so about Solution Since there are 7 digits here, the answer is approximately 14 7% of the numbers less than a million are prime. Notice that this approximation is close to the exact value from the previous table.

Our crude version of the Prime Number Theorem is not the most accurate version. For readers familiar with logarithms, we mention the more accurate and much more famous version, which says this: 1 The fraction of numbers up to N that are prime is approximately . Our ln crude version follows from this more accurate version, because

twice the number digits of N

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Our crude version sacrifices some accuracy (partly because is not very close to ), but it has the advantage of being understandable to readers who are not familiar with logarithms. Even the logarithm version of the Prime Number Theorem can be fine-tuned and improved on. In fact, the quest to more precisely understand exactly how the prime numbers are distributed among the natural numbers is intertwined with some of the most difficult theorems in mathematics, and also some of the most infamous unsolved problems. These include the Riemann Hypothesis, which today stands as the most famous unsolved math problem, with a one-million-dollar prize promised to the person who first solves it.

11B The Meaning of “Same Size” We previously introduced the following important sets of numbers:

“the natural numbers” “the integers” “the rational numbers” (fractions) “the real numbers”

11

Which of these sets is the largest? You might respond that ℝ is the largest because it contains the others as subsets. Or you might respond that they all have the same size, namely, infinity. Until a little more than a century ago, mathematicians were content with the decision that every infinite set has the same size as every other infinite set. They were not right or wrong—this is simply what they meant by the phrase “same size.”

Old-Fashioned Definition of “Same Size”

A pair of sets is said to have the same size if either (1) they are both finite and have the same number of members, or (2) they are both infinite.

This definition probably seems reasonable, but you are about to learn a beautiful truth about infinity to which this definition blinds you. Mathematicians who used this definition did not understand their blind spot any more than the ancient Greek mathematicians understood the truths to which they were blinded when they defined “number” to mean “rational number.” In the history of mathematical thought, this “infinity” blind spot was just as significant as the “number” blind spot, and its removal unleashed a rich world of fundamentally new ideas. What else could the phrase “same size” possibly mean? To answer this question, let’s think more carefully about how we compare the sizes of sets. When my niece was a toddler, I gave her 10 candles and 10 candle holders, and I asked her whether there were as many candles as candle holders. An adult would have separately counted the candles and holders and compared the answers, but my niece did not yet know how to count to 10.

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The Meaning of “Same Size”

So instead, she simply placed one candle into each holder. Since the candles and holders matched up perfectly, she knew there were equal numbers of each. If you are given two infinite sets and asked whether they have the same size, then your situation is very analogous to my niece’s. You do not have the ability to separately count each set, because you do not know how to “count to infinity.” Your most reasonable solution is the one my niece used—you should try to find a one-to-one correspondence (a matching) between the members of the two sets. This idea is not child’s play—it is so important, it will become our new meaning of “same size.”

Modern Definition of “Same Size”

A pair of sets is said to have the same size if their members can be matched with a oneto-one correspondence.

It’s time to forget the old-fashioned definition, and from now on, use only the modern definition. To decide whether two sets have the same size, your only job is to determine whether their members can be matched with a one-to-one correspondence. For example, to decide whether you have the same number of fingers as the stranger who you just met at the aquarium, you may not count and compare; rather, like the child and the diver in Figure 183, you must attempt a finger-to-finger matching. It is often very natural to compare sizes by matching rather than counting. For example, in a truckload of new pairs of shoes, you know that the number of right shoes equals the number of left shoes without knowing how many of either are in the truck. Still, it is difficult to change old habits into new habits, so let’s practice. Example and have the same How do we verify that the sets size? If you answered “they both have 5 members,” then you haven’t yet let go of the

Figure 183: We have the same number of fingers

We have the same number of fingers.

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Chapter 11 • Excursions in Numbers

old-fashioned definition. From now on, the only way to confirm that two sets have the same size is to exhibit a one-to-one correspondence between their members, like this:

↕

↕

↕

↕

↕

Example (the set of all natural numbers) and How do we decide if (the set of all even natural numbers) have the same size? Here are some wrong answers:

“They have the same size because they are both infinite.” “ is larger than E because has all of E ’s members plus more.”

11

These answers are wrong because they do not refer to the modern definition of “same size.” If you are tempted by these wrong responses, then you need to let go of your previous associations with the phrase “same size” and let the modern definition above become your only meaning for this phrase. To answer this question, your only job is to decide whether the members of and can be put into a one-to-one correspondence. After some trial and error, you will find that they can, like this: A one-to-one correspondence between the natural numbers and the even numbers

1 ↕ 2

2 ↕ 4

3 ↕ 6

4 ↕ 8

5 ↕ 10

6 ↕ 12

7 ↕ 14

8 ↕ 16

9 ↕ 18

… …

In case the pattern is not clear, we could describe it with a formula:

Do you see why this pattern is a one-to-one correspondence? For any even number you ask me about, I can find the natural number that matches with it. For example, if you ask me about 100, I report back that 50 matches with it. I will only ever have one choice for the number I report back, because this pattern never allows multiple natural numbers to match with the same even number. Thus, it’s a one-to-one correspondence! We learn that and have the same size. How strange that an infinite set can have the same size as a subset of itself!

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The Meaning of “Same Size”

Example Does the set of natural numbers

have the same size as the set of integers To answer this question, your only job is to decide whether the members of and can be put into a one-to-one correspondence. This seems difficult at first because the members of extend indefinitely in both the right (positive) and the left (negative) direction. But the following clever matching overcomes this difficulty: A one-to-one correspondence between the natural numbers and the integers

1 ↕ 0

2 ↕ 1

3 ↕ –1

4 ↕ 2

5 ↕ –2

6 ↕ 3

7 ↕ –3

8 ↕ 4

9 ↕ –4

… …

Do you see the pattern? The even natural numbers get matched with the positive integers, while the odd natural numbers get matched with the negative integers. In case the pattern is not clear, we could clarify it using a formula:

. Thus, and have the same size! Finding a one-to-one correspondence can require cleverness and persistence. In the previous example, you might have first tried the matching , but then realized that this matching misses all of the negative members of . But just because one attempted matching fails, you can not conclude that the sets have different sizes; a cleverer attempt might still succeed.

Definition An infinite set is called countable if it has the same size as ℕ (the set of natural numbers).

In the previous two examples, we concluded that and are both countable. In general, to prove that an infinite set is countable, you must match its members with the natural numbers. That is, you must decide which 1st member of your set matches with , which 2nd member matches with , which 3rd member matches with , and so on. For and , our matchings looked like this:

ℤ

1st

2nd

3rd

4th

5th

6th

7th

…

2 0

4 1

6 –1

8 2

10 –2

12 3

14 –3

… …

In summary, to prove that an infinite set is countable, we must find an infinite listing of its members, {1st, 2nd, 3rd, . . . }, which is organized so as to eventually include each member.

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11C Are the Rational Numbers Countable? Does  (the set of all rational numbers) have the same size as ℕ ? In other words, is  a countable set? If you believe that  is countable, then to prove it you must find an infinite listing {1st rational, 2nd rational, 3rd rational, . . .} organized so that your list eventually includes each rational number. Here is a first attempted pattern:

This attempt fails because it only includes positive fractions whose numerators equal 1. Let’s improve this attempt by squeezing in more numerators:

removed because

11

This attempt is better—it eventually includes all positive fractions whose numerators are smaller than their denominators. Next, let’s squeeze in their reciprocals:

The pattern here is: 1st fraction from the previous list, then its reciprocal, then the 2nd, then its reciprocal, and so on. This new pattern is better still; it eventually includes all positive fractions. All that remains is to insert 0 at the front, and intersperse the negatives:

The pattern here is: zero, then the 1st fraction from the previous list, then its negative, then the 2nd, then its negative, and so on. Behold the power of trial and error! Here is the theorem we just proved.

Theorem The set of rational numbers, , is countable.

The previous proof is perfectly valid, but the following alternative proof is also important to understand.

11

225  11D

Cantor’s Theorem

Table 12: The positive rational numbers arranged on an infinite grid

1 2 3 4 5 :

1

2

3

4

5

…

1/1 1/2 1/3 1/4 1/5 :

2/1 2/2 2/3 2/4 2/5 :

3/1 3/2 3/3 3/4 3/5 :

4/1 4/2 4/3 4/4 4/5 :

5/1 5/2 5/3 5/4 5/5 :

… … … … …

Figure 184: This path through the infinite grid turns it into an infinite list

Alternative Proof Here is an alternative method for listing all of the positive rational numbers. This means organizing them all into an infinite list, but we first settle for organizing them all into an infinite grid which (like a computer spreadsheet grid) has a top edge and left edge, but extends indefinitely down and right. The most natural arrangement is like the one in Table 12, with the column determining the numerator and the row determining the denominator. Now we organize the cells of this infinite grid into an infinite list by meandering through the grid in a way like the one shown in Figure 184. If we record the fractions we visit along this meandering purple path, and remove the redundant ones as we go, we will have successfully listed all of the positive rational numbers. Just as before, we can insert zero at the front and intersperse the negatives. □

11D Cantor’s Theorem Our next goal is to decide whether ℝ (the set of all real numbers) is countable. To appreciate the question, try to construct an infinite listing {1st real number, 2nd real number, 3rd real number, . . .}. You might start with a listing pffiffiffi of the rational numbers and then insert some famous irrational numbers like π and 2 at the front of your list. But what about the less famous irrationals, like the ones you made up yourself in the last chapter? The more you add to your list, the more you discover are missing. Are there too many real numbers to squeeze into a single infinite list? The answer to this difficult question was discovered by Georg Cantor around 1872.

226

Chapter 11 • Excursions in Numbers

Cantor’s Theorem The set of real numbers, ℝ, is not countable (so we call it uncountable).

I know lots of ways to construct an infinite listing of real numbers that fails to include them all. But this doesn’t prove Cantor’s Theorem, since someone cleverer might someday succeed in including them all. To prove his theorem, Cantor had to show that no listing, no matter how cleverly constructed, could ever succeed in including all real numbers. In other words, he had to prove that every attempted listing is doomed in advance. Here is how he did it. Proof We will prove that any listing of real numbers is incomplete. No matter how scrupulously the list was organized, some real numbers were definitely left off. More precisely, we will describe a concrete procedure for identifying a real number that is missing from any given listing of real numbers. Imagine a listing of real numbers. Maybe it was created by your Aunt Clair, who tried her best to include all of the real numbers on her list. Maybe it begins as shown in Figure 185. Here is a concrete procedure for identifying a real number that is missing from the list. We’ll call this missing number M. It will lie between 0 and 1, so it will have the form

11

M ¼ 0:d 1 d 2 d 3 d 4 d 5 d 6 d 7 d 8 ⋯ where each dn is a digit (0 through 9). How should we choose these digits to insure that M is not on the list? The answer is ingenious, and is hinted at by the red digits in Aunt Clair’s list. Here it is: Choose M’s first digit, d1, to be anything other than the first digit (after the decimal point) of the first number on the list. This insures that M is different from the 1st number on the list, since it has a different first digit. Choose M’s second digit, d2, to be anything other than the second digit of the second number on the list. This insures that M is different from the 2nd number on the list, since it has a different second digit. Do you see the idea? Chose M’s nth digit, dn, to be anything other than the nth digit of the nth number on the list, which insures that M is different from the nth number on the list, since it has a different nth digit. In the Aunt Clair example, the red diagonal includes the numbers {1 3, 4, 0, 7, 7,…}, so we must choose

st

1 nd 2 rd 3 th 4 th 5 th 6

↔ 3.1415926635… ↔ 0.3333333333… ↔ 1.41421356237… ↔ 256655643.00000000000… ↔ 509.73737373737… ↔ 5.04749726737…

Figure 185: Aunt Clair’s list of the real numbers

( ) (1/3) (√2) (Aunt Clair’s SSN) (Her favorite number) ( √2 )

227  11D

11

Cantor’s Theorem

= 0.(not 1)(not 3)(not 4)(not 0)(not 7)(not 7) …. This leaves us a lot of freedom. The number M ¼ 0.258163⋯ works fine, as would many other choices. With each digit, there are 10 choices (0 through 9), and only one choice is disallowed, which still leaves us 9 options. To be on the safe side, we’ll also avoid 0s and 9s, which still leaves at least 7 options for each digit. See the last exercise at the end of Chapter 10 (on page 215) to understand the reason for avoiding 0s and 9s. In summary, we can use this diagonal procedure to build a real number, M, that is missing from any given listing of real numbers. Therefore, no listing of real numbers could possibly be complete. Thus, the real numbers could never all be arranged into a single list; they are uncountable. □

Cantor’s Theorem says that, in a very precise sense, the infinite sets ℕ and ℝ do not have the same size. Thus, the modern definition of “same size” leads to this truth: not all infinite sets have the same size—some are genuinely larger than others! This is a surprising and remarkable phenomenon. In popular writing, it is described with phrases like “different sizes of infinity.” A portrait of Georg Cantor is shown in Figure 186. During Cantor’s life, his work was criticized by theologians who considered it a challenge to the notion of God as the one and only infinite, and also by mathematicians who were uncomfortable with his counterintuitive conclusions. But in the end, you can’t argue with a solid proof. Cantor’s conclusions were eventually accepted, causing a paradigm shift in the way mathematicians thought about fundamental concepts like numbers and sets. The famous

Figure 186: Georg Cantor

228

Chapter 11 • Excursions in Numbers

mathematician David Hilbert predicted the long-lasting importance of Cantor’s work when he wrote: “No one shall expel us from the Paradise which Cantor has created.”

Exercises

11

(1) About what fraction of natural numbers between 1 and one trillion are prime? About how many natural numbers between 1 and one trillion are prime? (2) About what fraction of United States citizens have a prime social security number? About how many United States citizens have a prime social security number? (3) Your brother-in-law has a list of four 4-digit numbers, but his handwriting is so poor, you can only make out one digit of each: 3   , 7  ,   9,    0. Without knowing any more information, can you find a 4-digit number that’s not on your brother-in-law’s list? How is this problem related to Cantor’s proof? (4) Decide whether these sets have the same size: {all odd natural numbers} and {all natural numbers greater than 9}. (5) Is it possible to list all of the positive rational numbers in increasing order? (6) If a spreadsheet grid extended indefinitely left, right, up, and down, would its cells be countable? HINT: Can you find a path that meanders through all of its cells? (7) Prove that the set of irrational numbers is uncountable. HINT: If it were countable, then interspersing a listing of all irrational numbers with a listing of all rational numbers would produce a listing of all real numbers. (8) If you replace the word “real” with “rational” throughout the proof of Cantor’s Theorem, you get a faulty proof that the rational numbers are uncountable. Why is it faulty? Which step is incorrect? (9) Is the set of all real numbers between 0 and 1 countable or uncountable? Why? (10) Is the set of all decimal expressions that contain no 5s countable or uncountable? Why? (11) Is ℕ  ℕ (the set of all pairs of natural numbers) countable or uncountable? Why? (12) You have infinitely many piles, one for each natural number. The first pile has one marble, the second has 2 marbles, the third has 3 marbles, and so on. Is the total number of marbles a countable set? (13) You have infinitely many piles—one for each natural number. Each pile has infinitely many marbles—one for each natural number. Is the total number of marbles a countable set? (14) If you remove one member from an infinite set, will the new set always have the same size as the original set? HINT: First consider the case when the original set is countable.

229

11

 Exercises

(15) Which of the following objects have countable symmetry groups? (a) a circle (b) the border pattern ⋯H H H H H⋯ (c) a sphere Explain your answers. (16) Precisely state and prove a version of the Zero-or-Equal Theorem (on page 39) that applies even when the object has infinitely many symmetries.

12

231

Rigid Motions as Functions

The final two chapters of this book are devoted to more precisely defining rigid motions, first as functions (in this chapter) and then as matrices (in the next chapter). This new viewpoint is the right starting point, not only for finally going back and proving some of the important theorems in this book, but also for exploring whole new vistas in higher mathematics.

12A Measuring Distance in Euclidean Space Our previous definitions of the plane and space are special cases of the following generalization.

Definition n-dimensional Euclidean space, denoted ℝn, means the set of all ordered n-tuples of real numbers. 2 In particular,ℝ pffiffiffiis the  plane—the set of all ordered 2-tuples (pairs) of real numbers, like (1,
7) and 2, π . Similarly, ℝ3 is space—the set of all ordered 3-tuples (triples) of real numbers, like (1,
7, 1.35). Don’t be intimidated by higher-dimensional Euclidean spaces. Forexample, ℝ4 just pffiffiffi  means the set of all ordered 4-tuples (x, y, z, w) of real numbers, like 2, π,
18, 19 . Although higher-dimensional Euclidean spaces might be difficult to visualize, they are still of practical importance. For example, a food manufacturer who records his sales of 7 different products each week is really recording a point of ℝ7. The most important fact about the plane, ℝ2, was discovered by the pffiffiPythagoreans— ffi the same group of ancient Greek mathematicians who discovered that 2 is irrational. It is a relationship between the lengths of the sides of a right triangle.

Electronic supplementary material The online version of this chapter (https://doi.org/10.1007/ 978-3-030-51669-7_12) contains supplementary material, which is available to authorized users. The videos can be accessed individually by clicking the DOI link in the accompanying figure caption or by scanning this link with the SN More Media App.

© Springer Nature Switzerland AG 2021 K. Tapp, Symmetry, Texts for Quantitative Critical Thinking, https://doi.org/10.1007/978-3-030-51669-7_12

231

232

Chapter 12 • Rigid Motions as Functions

The Pythagorean Theorem The lengths, a, b, and c, of the sides of a right triangle (listed in increasing order) satisfy the relationship: a2 + b2 ¼ c2.

For example, if a ¼ 3 and b ¼ 4, then c2 ¼ 32 + 42 ¼ 25, so c ¼

pffiffiffiffiffi 25 ¼ 5.

Proof For the proof, you will need some supplies. Cut out four copies of the same right triangle. We’ll call its side lengths a, b, and c (in increasing order). Next, cut out a square whose side length is a + b. These supplies are pictured in Figure 187. The key observation is that there are two very different patterns in which to arrange the four triangles onto the square. These two patterns are shown in Figure 188. For the first pattern, the uncovered orange area equals a2 + b2 (because it is made of two squares with side lengths a and b respectively). For the second pattern, the uncovered orange area equals c2 (because it is made of a single square with side length c). Since rearranging the pattern could not change the uncovered area, we conclude that a2 + b2 must equal c2. If we had started with a fatter or narrower right triangle, think about why the two arrangements still work out, with corners meeting perfectly as in Figure 188, and with right angles at the corners of the orange c-by-c square. □

12

An animated illustration of this proof is found in the PowerPoint presentation for Chapter 12 (and in the movie linked to these pages).

b

c a

c

b

a+b

a a+b

Figure 187: Four copies of a right triangle, and a square to fit them onto (▶ https://doi.org/10.1007/000-26b) Figure 188: Two different patterns in which to arrange the four right triangles onto the square

a

a

c b

b a

b c

b a a

a c

b

c

b a

c c b

b a

12

233  12A

Measuring Distance in Euclidean Space

Why is the Pythagorean Theorem important? Because it allows us to measure distances between points in the plane.

Distance Formula for the Plane The distance between (x1, y1) and (x2, y2) in ℝ2 equals

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðx2
x1 Þ2 þ ðy2
y1 Þ2 .

Proof This distance formula is nothing more than the Pythagorean Theorem in disguise. The distance between (x1, y1) and (x2, y2) equals the length, c, of the hypotenuse (the longest side) of the right triangle in Figure 189 (left). Notice that a ¼ x2
x1 and b ¼ y2
y1. The Pythagorean pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi □ Theorem says c ¼ a2 þ b2 , which is exactly the distance formula.

The distance formula in space is analogous.

Distance Formula for Space The distance between (x1, y1, z1) qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 ðx2
x1 Þ þ ðy2
y1 Þ þ ðz2
z1 Þ2 .

and

(x2, y2, z2)

in

ℝ3

equals

Figure 189: Left: The distance formula for the plane is the Pythagorean Theorem in disguise. Right: The distance formula for space is a double application of the Pythagorean Theorem.

234

Chapter 12 • Rigid Motions as Functions

Proof The proof involves a clever double application of the Pythagorean Theorem. In Figure 189 (right), applying the Pythagorean Theorem to the red triangle gives that the distance between the two orange points in ℝ2 equals Distance ¼ c ¼

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a2 þ b2 :

Now simplify this expression for c after substituting the following: b ¼ z2
z1 , qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a ¼ ðx2
x1 Þ2 þ ðy2
y1 Þ2 : This expression for a comes from applying the Pythagorean Theorem to the blue triangle. □

Our distance formulas generalize naturally to ℝn in the most obvious manner. For example, the distance between the points (x1, y1, z1, w1) and (x2, y2, z2, w2) in ℝ4 equals qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðx2
x1 Þ2 þ ðy2
y1 Þ2 þ ðz2
z1 Þ2 þ ðw2
w1 Þ2 ,

12

and so on for ℝ5, ℝ6, or ℝn for any n. We do not need to prove that the n-dimensional version of this formula agrees with any previous meaning of distance in ℝn, because when n > 3, there is no previous meaning. We can therefore take this formula as our definition of the word distance in ℝn. For example, a food manufacturer who records his sales of 7 different products each week might use the distance formula in ℝ7 to compare this week’s sales to last week’s sales.

12B Naming the Points on the Unit Circle We will now take a closer look at the circle—more specifically, the unit circle in ℝ2, which means the circle with radius 1 centered at the origin, (0, 0). It is pictured in Figure 190. How many points on the unit circle can you identify? The easy ones to identify are its intersections with the x and y axes: (1, 0), (
1, 0), (0, 1),and (0,
1). y

Figure 190: The unit circle

(0,1)

x (−1, 0)

(1,0)

(0, −1)

12

235  12B

Naming the Points on the Unit Circle

Figure 191: t ¼ angle position

y (0,1)

x

t (−1, 0)

(1,0)

(0, −1)

Do you know the x and y coordinates of any other points on the unit circle? You could probably use the distance formula to find more, but to really understand the unit circle, it is not enough to just haphazardly name some points. What we really need to do is parameterize all of the unit circle’s points in a systematic way. Here is the idea. Beginning at the point (1, 0), you walk counterclockwise around the unit circle. In your left hand, you hold one end of a purple rope. The other end of the rope is tethered to (0, 0); see Figure 191. Consider the angle that the positive x-axis makes with the purple rope. As you walk one full trip around the circle, this angle steadily increases   from 0 to 360 . We’ll call this angle your angle-position (because it determines where you are) and we’ll denote this angle-position with the letter “t”. The x and y coordinates of your position (when your angle-position equals some   value of t between 0 and 360 ) have special names; see the following definition.

Definition If you are at angle-position t on the unit circle, cos(t) means your x-coordinate (pronounced “the cosine of t”), and sin(t) means your y-coordinate (pronounced “the sine of t”).

Most calculators can compute the cosine and sine of any angle-position with dedicated buttons labeled “cos” and “sin”. Here is a table showing the cosine and sine of all    multiples of 15 between 0 and 180 (the values are rounded to 2 decimal places): Cosine and Sine Table t cos(t) sin(t)



0 1 0



15 .97 .26



30 .87 .50



45 .71 .71



60 .50 .87



75 .26 .97



90 0 1



105
.26 .96



120
.50 .87



135
.71 .71



150
.87 .50



165
.97 .26



180
1 0

What this table really tells you is the x and y coordinates of the thirteen points on the top half of the unit circle whose angle-positions are labeled in Figure 192. Could you have predicted which numbers in the table would be positive and which would be negative?

236

Chapter 12 • Rigid Motions as Functions

Figure 192: Anglepositions around the top half of the unit circle

y 105 90 75 60 120 135 45 150 30 15 165 180 0

x

12C The Dot Product and Perpendicularity We will next learn a simple method for testing perpendicularity. To get started, we must introduce the terms norm and dot product.

Definition If p ¼ (a, b, c) and q ¼ (x, y, z) are points of ℝ3,

(1) the norm of p (denoted jpj) means the distance from p to the origin (0, 0, 0), pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi which is computed as j p j¼ a2 þ b2 þ c2 , and (2) the dot product of p and q (denoted p • q) is defined as

12

p • q ¼ ax þ by þ cz:

Norms and dot products of points in ℝn are defined analogously, as these examples demonstrate. Example in ℝ2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffi If p ¼ (3, 4) and q ¼ (2, 7) in ℝ2, then the norm of p is j p j¼ 32 þ 42 ¼ 25 ¼ 5, and the dot product of p and q is p • q ¼ 3  2 + 4  7 ¼ 6 + 28 ¼ 34. Example in ℝ3 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffi If p ¼ (2,
3, 5) and q ¼ (5, 7,
1), then the norm of p is j p j¼ 22 þ ð
3Þ2 þ 52 ¼ 38  6:16, and the two points’ dot product is p • q ¼ 2  5 +
3  7 + 5 
1 ¼ 10
21
5 ¼
16. Example in ℝ4 If p ¼ (2,
3, 4, 1) and q ¼ (5, 10,
1, 0), then the norm of p is j p j¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffi 22 þ ð
3Þ2 þ 42 þ 12 ¼ 30  5:48, and the two points’ dot product is p • q ¼ 2  5 þ
3  10 þ 4 
1 þ 1  0 ¼
24:

12

237  12C

The Dot Product and Perpendicularity

p

Figure 193: Two points, their arrows, and the angle between their arrows

q

α y x Norms and dot products are easy to calculate, but what are their geometric meanings? To answer this, draw a purple arrow from the origin to p and draw an orange arrow from the origin to q, as in Figure 193. In this picture, jpj is simply the length of the purple arrow. Their dot product, p • q, has a geometric meaning that depends on the angle (which we’ll call α) between the purple and orange arrows.

The Meaning of the Dot Product If p and q are points in ℝn, and α is the angle between the arrows pointing to them from the origin, then p • q ¼ |p| |q| cos (α).

That is, the dot product can be found by multiplying these three numbers together: the norm of p, the norm of q, and the cosine of the angle between them. We won’t prove this formula, but we will tell you why you should care about it. Our main purpose for this dot-product formula is the following quick and easy method for testing whether or not the purple and orange arrows are perpendicular.

Perpendicularity Test



If p • q > 0, then α is acute (less than 90 ).  If p • q ¼ 0, then α is right (equal to 90 ).  If p • q < 0, then α is obtuse (greater than 90 ).

In particular, the orange and purple arrows are perpendicular if and only if the dot product equals zero. This perpendicularity test works because the cosine of an acute angle is positive, the cosine of a right angle is zero, and the cosine of an obtuse angle is negative. Think about why. Example If p ¼ (2,
3, 5) and q ¼ (5, 7,
1) in ℝ3, then their dot product is p • q ¼
16, so the angle is obtuse.

238

Chapter 12 • Rigid Motions as Functions

Example If p ¼ (3, 4) and q ¼ (2, 7) in ℝ2, then p • q ¼ 34, so the angle is acute. Example If p ¼ (4, 2) and q ¼ (
3, 6) in ℝ2, then their dot product is p • q ¼ 4 
3 + 2  6 ¼
12 + 12 ¼ 0, so the angle is right, which means that the arrows are perpendicular.

See how easy it is to determine whether two arrows are perpendicular? But what if you wish to know, not just whether the angle is acute or obtuse, but exactly what the angle equals? For this, just solve the “Meaning of the Dot Product” formula for cos(α), which yields cos ðαÞ ¼

p•q : j pkq j

This tells you what the cosine of α equals. Then use the “cos
1” button on your calculator, which undoes the cosine function, to find what α equals. In the plane ℝ2 and space ℝ3, this answer is exactly the angle that you would measure with a protractor. In higher-dimensional Euclidean spaces, protractors don’t make sense, so we simply take this answer as our definition of the word angle.

12 12D Using the Dot Product to Find a Friend Here is an activity that might give you a better feeling for distances, dot products, and angles. This activity will also help you understand the need to compute these quantities in high-dimensional Euclidean spaces. But let’s start with ℝ2. Write down your own personal point (x, y) in ℝ2 that encodes x ¼ how much you love cats, and y ¼ how much you love sushi, each on a scale from
5 (hate) to +5 (love). For example, my personal point is K ¼ (
3, 5) because I’m allergic to cats and I love raw salmon. I named my point “K” because my first name is Kris. Next, ask a friend to write down his or her personal point. Draw your point and your friend’s point in ℝ2, and draw arrows to them from the origin. Next, do three calculations: (1) Calculate the dot product of your point and your friend’s point. To what extent do you think that this dot product measures the compatibility of your interests? If the dot product is largely positive, does this mean that your interests are closely aligned with your friend’s? If the dot product is largely negative, does this mean that the two of you have opposite interests? What does it mean if the dot product equals zero? (2) Calculate the angle between the arrows. To what extent do you think that this angle measures the compatibility of your interests? Does an acute angle mean that your interests are closely aligned? Does an obtuse angle indicate opposite interests? What does a right angle indicate?

239  12E

12

Rigid Motions Are Functions

(3) Calculate the distance between your point and your friend’s point. To what extent do you think that this distance measures the compatibility of your interests? If the distance is small, does this mean that your interests are closely aligned? If the distance is large, does this indicate that you have opposite interests? Which quantity does the best job of measuring the compatibility of your interests: (1) the dot product, (2) the angle, or (3) the distance? There is no right answer—each strategy has advantages and disadvantages. But notice that all three quantities can be computed using dot products. Here’s why.

Norms, Distances, and Angles Can All Be Computed Using Dot Products pffiffiffiffiffiffiffiffiffi (1) jpj ¼ p • p (the norm of a point is the square root of its dot product with itself). (2) The distance between the points p ¼ (a, b, c) and q ¼ (x, y, z) equals |p
q| ¼ |(a
x, b
y, c
z)| ¼

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ða
x, b
y, c
zÞ • ða
x, b
y, c
zÞ:

p•q (3) cos ðαÞ ¼ jpkqj .

In summary, knowing how to compute dot products allows you to also compute norms, angles, and distances. All three strategies for measuring the compatibility of your interests are really just based on dot-product calculations. The previous activity might remind you of those dating websites that help single people find other single people with compatible interests. How does such a website match you up with your perfect future spouse? First, the site asks you about your interest level in more than just cats and sushi. Let’s suppose the site asks you 50 questions, so that your personal point lies in ℝ50. As discussed above, there are several reasonable strategies for measuring how “close” your personal point is to the personal point of someone else, and the dot product is at the heart of all such strategies.

12E Rigid Motions Are Functions We’ve been studying rigid motions since Chapter 1. We have so far described rigid motions using words to explain how the motion affects the plane or space. At long last, we are finally equipped to describe rigid motions in a much more precise way—with functions. The key idea is to regard a rigid motion of ℝn as a function from ℝn to ℝn. That is, a rigid motion is a rule or formula that allows one to determine the output point in ℝn associated to each input point in ℝn. Visually, the output point represents where the motion “moves” the input point to.

240

Chapter 12 • Rigid Motions as Functions

y 4 3 R 90 (0,0)

= (0,0)

2

R 90 (1,0)

= (0,1)

1

R 90 (3,1)

= (−1,3)

R 90 (0,2)

= (−2,0)

R 90 (0,−2) = (2,0)

x -2

-1

0

1

2

3

4

-1

-2

Figure 194: How several points are moved by a certain rigid motion 

12

For example, one of the first rigid motions of ℝ2 that we studied was R90—the 90 counterclockwise rotation of the plane about the origin. For practice, let’s regard R90 as a function from ℝ2 to ℝ2. It inputs a point of the plane, called p, and it outputs a point of the plane, called R90( p). The illustration in Figure 194 shows how several colored points are moved by this rotation. Do you see that pattern? You might guess that the pattern is this: R90 ðx, yÞ ¼ ð
y, xÞ: In the next chapter, we will prove that this formula is correct. Assuming this for now, this example helps us get used to thinking of a rigid motion as a function. Do you see how R90 is a definite rule that associates an output point to each input point? The rule is described by the formula R90(x, y) ¼ (
y, x). This formula empowers you to quickly determine R90’s effect on any input point you like. For example, R90(37, 55) ¼ (
55, 37). This is a whole new way to think about R90. Translations are even easier than rotations to describe as functions. For example, let T3 denote the translation of the plane a distance 3 to the right. Considered as a function, T3 simply increases the x-coordinate of each point by 3. A general formula is: T 3 ðx, yÞ ¼ ðx þ 3, yÞ: For example, T3(3, 5) ¼ (6, 5) and similarly T3(10,
3) ¼ (13,
3). When regarded as a function, the defining property of a rigid motion, F, is this: it preserves distances. This means that the distance between any pair of points is the same before and after the rigid motion is applied. That’s what makes it rigid! In other words, the distance from p to q must equal the distance from F( p) to F(q), for any pair of points

12

241  Exercises

y

Figure 195: R90 preserves distances

4

F(q)

3 2

q

1

F(p) -2

-1

p 0

1

x 2

3

4

-1

p and q. For example, in Figure 195, the two dashed lines have the same length because R90 preserves the distance between the pink and green points. At last, we can precisely define the vocabulary words upon which our entire study of symmetry has been built.

Definitions A rigid motion of ℝn means a function, F, from ℝn to ℝn that preserves distances. An object in ℝn means a nonempty subset of ℝn. A symmetry of an object in ℝn means a rigid motion, F, of ℝn that moves the object onto itself; that is, F( p) is a point of the object if and only if p is a point of the object.

This new definition of rigid motion agrees with our previous definitions (on page 9 for 2D and on page 136 for 3D), but it is much more concise and precise. The new definition of symmetry agrees with the old definition, but it more precisely clarifies what it means for a rigid motion to leave an object unchanged: it moves the object onto itself. Another advantage of the new definition is its generality—it applies to all dimensions (including the plane, space, and beyond).

Exercises (1) In the proof of the Pythagorean Theorem, explain why the orange square with side length c is really a square. (2) If p ¼ (2, 7) and q ¼ (3,
5) in ℝ2, find the following quantities: jp j , j q j , p • q, the distance from p to q, and the angle between the arrows from the origin to p and q.

242

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Chapter 12 • Rigid Motions as Functions

(3) If p ¼ (1, 4, 3) and q ¼ (
1, 2, 7) in ℝ3, find the following quantities: jp j , j q j , p • q, the distance from p to q, and the angle between the arrows from the origin to p and q. (4) In ℝ5, do the arrows from the origin to (1,
2, 3,
4, 5) and (1, 0,
1, 3, 2) form an acute, obtuse, or right angle? (5) Let p ¼ (3, 7). Show that q1 ¼ (
7, 3) is perpendicular to p and has the same norm as p. Show that q2 ¼ (7,
3) is also perpendicular to p and has the same norm as p.  Draw these points. Which one of q1, q2 is obtained by rotating p by 90 clockwise?  Which is obtained by rotating p by 90 counterclockwise? Guess a formula for the  90 clockwise rotation of an arbitrary point p ¼ (x, y). (6) A formula for R90 is: R90(x, y) ¼ (
y, x). Guess an analogous formula for R180(x, y) and for R270(x, y). (7) Guess a formula for H(x, y) and V(x, y), where H means the horizontal flip over the x-axis and V means the vertical flip over the y-axis in ℝ2.  (8) If R is a rotation by 90 about the z-axis in ℝ3, guess a formula for R(x, y, z). (9) If F is the reflection across the xy-plane in ℝ3, guess a formula for F(x, y, z). Do the same for reflections across the xz-plane and the yz-plane. (10) If α is any angle, find a formula in terms of α for Rα(1, 0) and Rα(0, 1). (11) In the Cosine and Sine Table in the chapter, the values are rounded to two decimals, but these values can be determined exactly. Determine the exact values of cos     (45 ) ¼ sin (45 )  .71 and of sin(60 )  . 87. HINT: cos(60 ) equals exactly 12. (12) Fill in the following table with rounded values for the cosine and sine of angles   between 180 and 360 . HINT: Use the same numbers that are in the table in the chapter; namely, plus and minus 0, .26, .50, .71, .87, .97, and 1. 

t 180 cos (t) sin (t)



195



210



225



240



255



270



285



300



315



330



345



360

(13) If a cube with side length 2 is centered at the origin (0, 0, 0), then the locations of its 8 vertices are: (1, 1, 1), (1, 1,
1), (1,
1, 1), (1,
1,
1), (
1, 1, 1), (
1, 1,
1), (
1,
1, 1), (
1,
1,
1). Draw a picture of this cube. Prove that it is a Platonic solid by verifying that all 6 of its faces are identical regular 4-sided polygons. In particular, you must verify that the edges meet at right angles.

243

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Rigid Motions as Matrices

In the early chapters of this book, we described rigid motions with phrases like “the 90-degree rotation around a point.” Then we learned that rigid motions are really functions, so we described them with formulas like F(x, y) ¼ (y, x). In this chapter, we’ll learn that rigid motions are a very special kind of functions—the kind that are so simple, they can be described by matrices. For many purposes, the very best way to describe a rigid motion is with a matrix. In fact, matrices are the starting point of most advanced books about symmetry, and this chapter is intended as a bridge towards such books.

13A Matrix Computations A matrix simply means a grid of real numbers.

Definition An n-by-n matrix means n2 real numbers arranged into a square grid.

For example, here are a few 2-by-2 matrices:
A¼

2 3 1

7

 ,B ¼

!

pffiffiffi 2

0

51

3=5

,C ¼




20 3 π

0

 ;

and here are a few 3-by-3 matrices:

We’ll think of each row and each column of an n-by-n matrix as a point in ℝn. For , , and example, the rows of the matrix E above are , and . The entries of a matrix are while the columns of F are the numbers out of which it is built. For example, the (2, 3)-entry of D is , while the (3, 2)-entry of D is . Notice the convention of indexing the entry’s row first and then its column.

© Springer Nature Switzerland AG 2021 K. Tapp, Symmetry, Texts for Quantitative Critical Thinking, https://doi.org/10.1007/978-3-030-51669-7_13

243

244

Chapter 13 • Rigid Motions as Matrices

There are three important types of matrix computations, which we will now describe. First, we will describe how to multiply a pair of matrices. The answer is another matrix.

How to Multiply a Pair of Matrices If A and B are n-by-n matrices, then A  B is the n-by-n matrix whose (i, j)-entry equals the dot product of the ith row of A with the jth column of B.

2-by-2 Example

Each of the four entries of the answer equals the dot product of the same-colored row of the first matrix with same-highlighted column of the second matrix. 3-by-3 Example

13 As before, each of the nine entries of the answer equals the dot product of the samecolored row of the first matrix with same-highlighted column of the second matrix. Second, we will describe how to compute the determinant of a matrix.

The Determinant of a 2-by-2 Matrix The determinant of a 2-by-2 matrix A, denoted det(A), is the real number defined as
det

a c

b d

 ¼ ad  bc:

This is simple enough. For example,
det

2

5

7

3

 ¼ 2  3  5  7 ¼ 6  35 ¼ 29:

The determinant of a matrix is a number, which might be positive or negative or zero.

245  13A

13

Matrix Computations

The Determinant of a 3-by-3 Matrix The determinant of a 3-by-3 matrix A, denoted det(A), is the real number defined as

You don’t need to memorize this formula, because there is a simple way to visualize it. Just append copies of the first two columns of A to the right edge of A and draw red and green circles like those in the illustration here.

a d g

b e h

c a f d i g

b e h

Notice that the three green circles correspond to the three added terms in the determinant formula, while the three red circles correspond to the subtracted terms. There is an analogous (but messier) formula for the determinant of a 4-by-4 matrix, and one for a 5-by-5 matrix, and so on. In this book, you will only need to compute determinants of 2-by-2 and 3-by-3 matrices, but be aware that the word determinant does make sense for larger matrices. Third, we will describe how to multiply a matrix times a point. The answer is a point.

How to Multiply a Matrix Times a Point If M is an n-by-n matrix and p is a point in ℝn, then M  p means the point in ℝn whose ith coordinate equals the dot product of the ith row of M with p.

Example If

, then

, and

p

Example If

and

, then

246

Chapter 13 • Rigid Motions as Matrices

In summary, you just learned three algebraic skills: (1) multiplying a matrix times a matrix, (2) computing the determinant of a matrix, and (3) multiplying a matrix times a point. All three computations involve only adding and multiplying. In the next section, we will discuss the geometric meanings of these computations, and the relationship between matrices and rigid motions.

13B Representing Rigid Motions as Matrices Matrices are great, but they really are useful only for studying the symmetries of a bounded object. Here’s why. The Center Point Theorem (page 16 for 2D and page 139 for 3D) generalizes to objects in ℝn in the way shown in the following theorem.

General Center Point Theorem Any bounded object in ℝn has a “center point” that is fixed by each of its symmetries.

13

In other words, each of its symmetries leaves its center point unmoved. Any bounded object can be translated so that its center point becomes the origin (0, 0, . . ., 0) of ℝn. After this repositioning, all of its symmetries will be rigid motions that fix the origin (they will leave the origin unmoved). This is exactly the type of rigid motion that matrices can help us study. In this chapter, we will only consider bounded objects centered at the origin. Each symmetry of such an object fixes the origin. You previously learned to regard a rigid motion as a function and represent it using a formula. For example, R90 has the formula R90(x, y) ¼ (y, x). Now, we will re-describe a rigid motion as a matrix. Here’s how.

The Matrix That Represents a Rigid Motion Let F be a rigid motion of ℝn that fixes the origin. The matrix that represents F is the n-by-n matrix whose columns are F(1, 0, . . ., 0), F(0, 1, 0, . . ., 0), . . ., F(0, . . ., 0, 1), in this order.

So the matrix representing a rigid motion, F, of ℝ2 is the 2-by-2 matrix whose two columns are F(1, 0) and F(0, 1). Similarly, the matrix representing a rigid motion, F, of ℝ3 is the 3-by-3 matrix whose three columns are F(1, 0, 0), F(0, 1, 0), and F(0, 0, 1). Example The rigid motion R90 is represented by the 2-by-2 matrix whose first column equals . Thus, and whose second column equals

247  13B

13

Representing Rigid Motions as Matrices

In what way exactly does this matrix represent R90? At first glance, this matrix seems to tell us only which outputs R90 associates to the inputs (1, 0) and (0, 1). Magically, this matrix also tells us the outputs that R90 associates to every possible input point of ℝ2. Remember how to multiply a matrix times a point? Watch:


0 1

1 0

  ðx, yÞ ¼ ðy, xÞ:

See what happened here? When the matrix that represents R90 was multiplied times a point of ℝ2, the answer was exactly the output that R90 associates with that input point. For example,

This is exactly what always happens!

Theorem If F is a rigid motion of ℝn that fixes the origin, and M is the matrix that represents F, then for any point p of ℝn, we have M  p ¼ F ðpÞ:

In words: “To learn the output point that F associates to the input point p, you multiply M by p.” This is a powerful theorem. For example, when n ¼ 3, the columns of M are defined to record where F moves the three points (1, 0, 0), (0, 1, 0), and (0, 0, 1). It is surprising that this is enough information to determine where F moves all points of ℝ3, and that this determination is achieved via something as simple as matrix-point multiplication. Let’s revisit D4 ¼ the symmetry group of the square. Here are the matrices representing all 8 of its members (assuming the square is centered at the origin of ℝ2):
I¼

1 0


 0 0 , R90 ¼ 1 1


1 1 , R180 ¼ 0 0


 0 0 1 , R270 ¼ , 1 1 0

Look back at the pictures of these 8 symmetries (Figure 3 on page 32) and verify that these 8 matrices are all correct. For example, the matrix for D is correct because its , as shown in Figure 196. and columns are

248

Chapter 13 • Rigid Motions as Matrices

Figure 196: How two points are moved by the rigid motion D

D

y

(0,1) (1,0)

(1,0)

x

What happens when you multiply two of these matrices? For example, what is the matrix that represents R90 times the matrix that represents D? Try it. You’ll discover the answer is the matrix that represents H, agreeing with the fact that R90  D ¼ H. It seems that matrix multiplication achieves composition of symmetries! Indeed, this is the general rule.

Matrix Multiplication Achieves Composition If F1 and F2 are rigid motions of ℝn that fix the origin, and M1 and M2 are the matrices that represent them, then M1  M2 is the matrix that represents their composition F1  F2.

13

We now have a very effective way to convert visual questions about symmetries into algebraic questions about matrices. Do you see how it works? If your cousin Henry had never heard of D4 , you could describe this group to him simply by showing him the 8 matrices and nothing else. He could use matrix multiplication to build the Cayley table, without ever cutting out a cardboard square or visualizing a rotation or flip. The rotations in D4 were all multiples of 90 degrees, which made it easy to find the matrices representing them. To do the same for other cyclic and dihedral groups, the rotations are more complicated, so you will need this theorem giving the matrices for other rotation angles.

Theorem The matrix that represents a counterclockwise rotation around the origin of ℝ2 by
 cos ðt Þ  sin ðtÞ t degrees is . sin ðt Þ cos ðtÞ

This matrix’s first column is correct because cosine and sine are exactly defined so that the rotation sends (1, 0) to (cos(t), sin(t)). Think about why the second column is also correct.

13

249  13C

Orthogonal Matrices

13C Orthogonal Matrices Not just any matrix represents a rigid motion. The matrices that do represent rigid motions have a special property called being orthogonal.

Definition A matrix is called orthogonal if the norm of each column equals 1 and the dot product of each pair of different columns equals 0.

Suppose that F is a rigid motion of ℝ3 that fixes the origin, and M is the matrix that represents F. Let’s think about why we might expect M to be orthogonal. Imagine watching F move the orange, green, red, and purple points in the illustration shown in Figure 197. Since F is a rigid motion, the distances between pairs of these points are the same before and after the motion. The orange point stays put because it is the origin. Imagine the colored arrows moving along with F to aim at the new locations of the green, red, and purple points. These new locations are the three columns of M. After the motion, the arrows will still have length 1, which is to say that the columns of M have norm 1. This is simply because the distance between the orange point and each of the other points is unchanged by the motion. After the motion, these arrows will still be mutually perpendicular, which is to say that the dot product of each pair of different columns of M equals 0. Why? If the angle between two arrows became acute, the points they aim at would have grown closer together, so the motion would not have been rigid. If the angle became obtuse, the points would have grown further apart. Can you picture this? This visual discussion is not a proof, but it should help you believe the following theorem.

Theorem If M is a matrix that represents a rigid motion that fixes the origin, then M is an orthogonal matrix. Conversely, every orthogonal matrix represents a rigid motion that fixes the origin.

Figure 197: Four points for the rigid motion F to act upon

Z (0,0,1)

(0,1,0) (1,0,0)

X

y

250

Chapter 13 • Rigid Motions as Matrices

It turns out that orthogonal matrices have very limited possibilities for their determinants. Calculate the determinant of each of the 8 matrices for D4. You will discover that the determinant of each rotation is 1, while the determinant of each reflection is 1. The general rule extends this pattern and does not add any other possible values of the determinant.

Theorem The determinant of any orthogonal matrix equals either 1 (if it represents a proper rigid motion) or 1 (if improper).

This theorem should perhaps be blue-framed instead of blue-shaded, because it is in part a definition. It is a more precise definition of the terms proper and improper for rigid motions of the plane and space. For rigid motions of higher-dimensional Euclidean spaces, it is our only definition of these terms. What about rigid motions that do not fix the origin? Can we use determinants to distinguish whether they are proper or improper? This turns out to be easy because of the following theorem.

Theorem Every rigid motion of ℝn equals a rigid motion that fixes the origin followed by a translation.

13

So a rigid motion is called proper or improper depending on whether its originfixing part has determinant equal to 1 or 1. This is a good definition. Defining “proper” using determinants is much more precise and unambiguous than our previous verbiage about right hands.

13D Concluding Remarks In this chapter, we intended only to briefly describe how matrices are related to symmetry. We did not include any proofs, because the proofs belong in a linear-algebra book. We also did not carry out the important work of using matrices to prove the previously unproven theorems scattered throughout this book. Thus, this book ends with a beginning—a more rigorous starting point from which you can revisit the topic of symmetry with more precise definitions and more complete proofs. We hope we have piqued your interest in someday reading the books and taking the classes in which this idea is fully developed. Here is a brief glimpse of some of what’s left to learn about symmetries and matrices. Matrices provide a precise definition of rigid motion from which one can quickly prove the classifications of plane and space rigid motions found in this book. Some of the most difficult theorems in this book can then be rephrased and proved using matrices. For

251

13

 Exercises

example, the classification of symmetry groups of bounded 3D objects boils down to understanding the possible finite subgroups of the group of all orthogonal 3-by-3 matrices. Things get really interesting when you move into higher-dimensional Euclidean spaces. Consider this challenge: Classify the possible symmetry groups of bounded objects in ℝn You already know the answer for n ¼ 2 and n ¼ 3 (or at least you know the possible finite symmetry groups). What about general n? This question turned out to be very difficult and very important. It motivated some of the most significant mathematics of the past century. First, this question is intertwined with the problem of classifying all possible finite groups, which led to one of the most celebrated achievements of modern mathematics, aptly called “The Enormous Theorem.” Second, this question is related to the classification of compact Lie groups, which is a beautiful piece of mathematics upon which much of modern physics depends.

Exercises
(1) Consider: A ¼ (a) (b) (c) (d) (e) (f)


2 3 3 ,B¼ 1 7 5


 0 10 3 ,C¼ . 1=2 4 0

Compute the determinant of each matrix. Compute A  B and B  A. Are they equal? Verify that det(A  C) ¼ det (A)  det (C). Compute A  (2, 3) and A  (2, 7). Verify that A is not orthogonal. The function from ℝ2 to ℝ2 determined by A (sending P to A  P) is not a rigid motion because A is not orthogonal. Show this directly by finding a pair of points whose distance is not preserved. 0

2 5

9

1

0

0

1

B C B (2) Consider: A ¼ @ 0 4 6 A, B ¼ @ 5 2 1 2 1 0 1 (a) (b) (c) (d)

3

1

C 10 A. 0

Compute the determinants of A and B. Compute A  B and B  A. Are they equal? Compute A  (1, 2, 3) and A  (0, 4, 2). The function from ℝ3 to ℝ3 determined by A (sending P to A  P) is not a rigid motion because A is not orthogonal. Show this directly by finding a pair of points whose distance is not preserved.

252

Chapter 13 • Rigid Motions as Matrices

(3) Write the matrix that represents each of the six symmetries in D3. (4) Determine the matrix that represents each of the following rigid motions of ℝ3: (a) A 90-degree rotation around the z-axis or the x-axis. (b) A reflection across the yz-plane or the xz-plane. (5) This problem is about inversion across the origin, as defined on page 158. (a) Find the 3-by-3 matrix, M1, that represents F1 ¼ the reflection across the xyplane. (b) Find the 3-by-3 matrix, M2, that represents F2 ¼ the 180-degree rotation around the z-axis. (c) Verify that F1 and F2 commute by checking that M1  M2 ¼ M2  M1. Can you picture this using Figure 127? (d) The composition F ¼ F1  F2 is called inversion. Find the matrix M ¼ M1  M2 that represents F. Verify that F has the formula F(x, y, z) ¼ (x, y, z). (e) Verify that F is improper by showing det(M ) ¼  1. (f) Verify that F commutes with every rigid motion of ℝ3 that fixes the origin. Do this by checking that A  M ¼ M  A for every 3-by-3 matrix A. (g) Explain why F is a symmetry of the cube and the dodecahedron but not of the tetrahedron. (6) Describe in words the rigid motion of ℝ3 represented by 0 B @

13

cos ðtÞ 0 0 sin ðtÞ

1 0

 sin ðtÞ

1

C 0 A: cos ðtÞ

253

Image Credits

Page v: “The Vitruvian Man” by Leonardo da Vinci, accessed at https://commons. wikimedia.org/wiki/File:0_The_Vitruvian_Man_-_by_Leonardo_da_Vinci.jpg/ Public domain Page vi, Figure 1: Honeycomb photo by Ken Tapp, reproduced with permission Page vi, Figure 2: Image of virus by jscreationzs, accessed at freedigitalphotos.net, reproduced with permission Page 7, Figure 15: Seahorses and Eels by Robert Fathauer, reproduced and modified with permission, accessed at http://members.cox.net/fathauerart/. Page 21, Figure 22: Gnome image by Paul Söderholm, accessed at www.gnurf.net, reproduced with permission, license acquired from https://www.clipartof.com/portfo lio/pasoderholm/illustration/cute-little-gnome-man-in-a-red-hat-carrying-a-lantern21319.html Page 21, Figure 26: Triskelion from the flag of the Isle of Man, accessed at https:// commons.wikimedia.org/wiki/File:Three_Legs_of_Man_-_Triskelion.jpg/ Public domain Page 24: Seven model border patterns by AndrewKepert, accessed at https://commons. wikimedia.org/wiki/File:Frieze2b.png/ Made available under CC-BY-SA 3.0 license, reproduced and modified with permission. Page 55: Self-portrait by Leonardo Da Vinci, Biblioteca Reale, accessed at https:// commons.wikimedia.org/wiki/File:Leonardo_da_Vinci_-_presumed_self-portrait_-_ lossless.png/ Public domain Page 60, Figure 49 (top left): “Fragment from a Frieze with Meander Pattern and Diamond-Shaped Rosettes,” Metropolitan Museum of Art, accessed at https://com mons.wikimedia.org/wiki/File:Fragment_from_a_Frieze_with_Meander_Pattern_ and_Diamond-Shaped_Rosettes_MET_sf09-217-1abs1.jpg/ Made available under CC0 1.0 Universal Public Domain Dedication

© Springer Nature Switzerland AG 2021 K. Tapp, Symmetry, Texts for Quantitative Critical Thinking, https://doi.org/10.1007/978-3-030-51669-7

253

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Image Credits

Page 60, Figure 49 (top right, bottom left, bottom right): “Frieze (USA), 1905-15,” Smithsonian Design Museum, accessed at https://commons.wikimedia.org/wiki/File: Frieze_(USA),_1905%E2%80%9315_(CH_18500493).jpg/ Public domain Page 63, Figure 53 (left): “Qubbah Ba’adiyim in Marrakesh” by amerune, accessed at https://www.flickr.com/photos/amerune/551496150/in/photolist-QJyAj-QJmXJPvxT5-PvxU1-Pw5WM-TY2aa-PgQMY/ Made available under CC-BY 2.0 license Page 63, Figure 53 (right): Wallpaper pattern, accessed at https://en.wikipedia.org/wiki/ File:Wallpaper_group-cmm-2.jpg/ Public domain Page 64, Figure 54: Flow chart by Brian Sanderson, accessed at http://www.warwick.ac. uk/~maaac// Reproduced and modified with permission Page 66, Figure 57: Example of wallpaper group type p31m, from The Grammar of Ornament (1856) by Owen Jones, accessed at https://commons.wikimedia.org/wiki/ File:Wallpaper_group-p31m-3.jpg/ Public domain Page 73, Figure 61 (right): Three Fishes by Robert Fathauer, accessed at http://mem bers.cox.net/fathauerart// Reproduced and modified with permission Page 74, Figure 64 (left): Example of wallpaper group type p4g, from The Grammar of Ornament (1856) by Owen Jones, accessed at https://commons.wikimedia.org/wiki/ File:Wallpaper_group-p4g-2.jpg/Public domain Page 74, Figure 64 (right): Example of wallpaper group type p4m, from The Grammar of Ornament (1856) by Owen Jones, accessed at https://commons.wikimedia.org/ wiki/File:Wallpaper_group-p4m-5.jpg/ Public domain Page 74, Figure 65: Fish themed wallpaper pattern by Rachel Hall, reproduced with permission Page 75, Figure 66: Fish themed wallpaper pattern by Rachel Hall, reproduced with permission Page 79: Helicopter illustration by Leonardo da Vinci, accessed at https://commons. wikimedia.org/wiki/File:Leonardo_da_Vinci_helicopter.jpg/ Public domain Page 111: Photograph of magnet letters by Ken Tapp, reproduced with permission Page 128, Figure 96: Domino image by Pokipsy76~commonswiki, accessed at https:// commons.wikimedia.org/w/index.php?curid¼1017771/ Made available under CC BY-SA 3.0 license Page 138, Figure 99: From left to right “Tetrahedron,” “Hexahedron,” “Dodecahedron,” “Cuboctahedron” by Robert Webb, Rendered with Robert Webb’s Great Stella software, http://www.software3d.com/Stella.html// Reproduced with permission Page 138, Figure 102: “Human anatomy planes” by YassineMrabet, accessed at https:// commons.wikimedia.org/wiki/File:Human_anatomy_planes.svg/ Reproduced and modified with permission Page 152, Figure 118: Flail illustration by Robo Blazek, accessed at https://commons. wikimedia.org/wiki/File:Rincak.jpg/ Public domain Page 152, Figure 119: Golf ball image, accessed at freedigitalphotos.net/ Reproduced with permission Page 153, Figure 120 (left): Epcot Center photograph by Paul Brennan, accessed at https://www.publicdomainpictures.net/en/view-image.php?image¼90140& picture¼spaceship-earth-sphere/ Public domain

255 Image Credits

Page 153, Figure 120 (right): Goats beard photograph by Ken Tapp, reproduced with permission Page 155, Figure 122 (left): Illustration of Th symmetry by Tamfang, accessed at https:// en.wikipedia.org/wiki/File:Volleyball_seams_diagram.png/ Public domain Page 155, Figure 123: “Alice stepping through the looking glass” from Through the Looking Glass and What Alice Found There by Lewis Carroll, illustrated by Sir John Tenniel, accessed at https://es.wikipedia.org/wiki/Archivo:Aliceroom3.jpg/ Public domain Page 156, Figure 124: Amino Acid Chirality by NASA, accessed at https://en.wikipedia. org/wiki/File:Chirality_with_hands.jpg/ Public domain Page 159, Figure 128: SN2006gy Supernova by NASA, accessed at https://upload. wikimedia.org/wikipedia/commons/2/25/Bright_Supernova_Sn2006gy.jpg/ Public domain Page 160, Figure 129: “Dual tetrahedron” by Robert Webb, Rendered with Robert Webb’s Great Stella software, http://www.software3d.com/Stella.html// Reproduced with permission Page 163, Figure 133: Image of volleyball, accessed at https://commons.wikimedia.org/ wiki/File:Volley-ball_andrea_bianc_01.svg/ Public domain Page 165, Object E: “Pentagram pyramid” by Robert Webb, Rendered with Robert Webb’s Great Stella software, http://www.software3d.com/Stella.html// Reproduced with permission Page 165, Object G: “Illustration of a triple torus” by Oleg Alexandrov, accessed at https://commons.wikimedia.org/wiki/File:Triple_torus_illustration.png/ Public domain Page 170, Figure 142: “Illustration of a typical member of each of 7 infinite families of 3D point groups” by AndrewKepert, accessed at https://commons.wikimedia.org/ wiki/File:Uniaxial.png/ Made available under CC BY-SA 3.0 license, reproduced and modified with permission Page 176, Figure 153: Photo by Karl Horton, accessed at https://www.flickr.com/ photos/karlhorton/917587002/sizes/z/in/photostream// Made available under CC BY-SA 2.0 license Page 178, Figure 154: “Dual Cube-Octahedron” by Cinabrium, accessed at https://en. wikipedia.org/wiki/File:Dual_Cube-Octahedron.jpg/ Reproduced with permission Page 179: Images of polyhedra by ATRACTOR, accessed at https://www.atractor.pt/ mat/Polied/index-_en.html/ Reproduced and modified with permission Page 184, Figure 162 (left): “Torus illustration” by Oleg Alexandrov, accessed at https:// commons.wikimedia.org/wiki/File:Torus_illustration.png/ Public domain Page 184, Figure 162 (middle): “Double torus illustration” by Oleg Alexandrov, accessed at https://commons.wikimedia.org/wiki/File:Torus_illustration.png/ Public domain Page 186, Figure 164: “Kepler’s Platonic solid model of the Solar system” by Johannes Kepler, Mysterium Cosmographicum, Tübingen (1596), accessed at https://en. wikipedia.org/wiki/File:Kepler-solar-system-1.png / Public domain Page 227: Photo of Georg Cantor, accessed at https://i12bent.tumblr.com/post/ 3622180726/georg-cantor-german-mathematician-and-philosopher/ Public domain

257

Index

A Algebraic operation, 35 Alternating group, 120 Associative, 35 Asymmetric, 12 3D object, 136 B Beveled polygon, 142 Border pattern, 15 classification of, 61, 62 Bounded object in space, 136 object in the plane, 15 Bubble Theorem, 191–193 C Cantor’s Theorem, 226 Cayley table, 31 Center Point Theorem in Euclidean space, 246 3D, 139 2D, 16 Centrally symmetric, 158 Chiral, 153 Circle Theorem, 194, 196, 197 Classification of plane rigid motions version 1, 13–14 version 2, 44 Commutative group, 35 Commute, 105 Compose, see Composition

Composition of permutations, 114 of rigid motions of space, 136 of rigid motions of the plane, 29 Connected planar graph, 180 Contrapositive, 17 Converse, 89 Corollary, 159 Cosine, 235 Countable, 223 Counterexample, 45 Cube, 147 Cycle notation, 114 Cyclic group, 38 better notation for, 87, 88 D Da Vinci’s Theorem, 58–60 Decomposable, 48 Determinant, 244 Dihedral group, 38 Disprove, 45 Distance formula, 233 Dodecahedron, 149 Dot product, 236 Duality, 177 E Equivalence class, 69 Equivalence relation, 69 Essentially two-dimensional, 143 Euclidean space, 231

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Index

Euclid’s Theorem, 217–219 Euler characteristic, 185 Euler’s Formula, 180 Even-or-Odd Theorem, 119–121 Even permutation, 118 F Fixed point, 11 Full-vs-Proper Theorem 3D, 156–158 2D, 105, 106 G Generalization, 106 Generated subgroup, 99 Glide reflection, 30 Graph, 180 Group, 35 I Icosahedron, 172 Identity of a group, 35 rigid motion of space, 136 rigid motion of the plane, 12 If and only if, 89 Improper rigid motion of Euclidean space, 250 rigid motion of space, 138 rigid motion of the plane, 12 Indecomposable, 48 Indirect proof, see Proof by contradiction Inductive proof, 113, 127 Integer, 203 Inverse of a member of a group, 35 Inversion, 158 Irrational numbers, 203 Isomorphic, see Isomorphism Isomorphism, 82 Isoperimetric Theorem, 199 M Matrix, 243 Minimal surface, 193 N Natural numbers, 202 Norm, 236

O Object in Euclidean space, 241 3D, 135 2D, 4 Octahedron, 172 Odd permutation, 118 Order of a member of a group, 101 of a rotation, 12 of a rotation axis, 137 Oriented, 14 Oriented polygon, 6 Origami, 166 Orthogonal matrix, 249 Overcrowded, 173 P Permutation group, 112 Perpendicular, 237, 238 Plane, 4 Platonic solid, 171 Plato’s Theorem, 175, 176 Polygon, 5 Prime factorization, 202 Prime number, 202 Prime Number Theorem, 219, 220 Product group, 101 Proof, 16 Proof by contradiction, 197 Proof by induction, 127 Proper rigid motion of Euclidean space, 250 rigid motion of space, 138 rigid motion of the plane, 12 Proper symmetry group of a 3D object, 138 of a 2D object, 96 Properly rigidly equivalent, 139 Pythagorean Theorem, 232 R Rational numbers, 203 construction of, 211 Real Number Redundancy Rule, 206, 207 Real numbers, 205 construction of, 211 Reflection, 2 of space, 137, 138 Reflexive, 69 Regular polygon, 5 Regular polyhedron, see Platonic solid

259 Index

Rigid motion of Euclidean space, 241 of the plane, 9 of space, 136 Rigidly equivalent 2D objects, 56 3D objects, 139 Rigid Motion Detector Theorem, 42 Rotation, 2 of space, 136, 137 S Same size, 221–223 Set, 35 Shear, 65 Sine, 235 Soccer ball, 169, 187 Space, 135 Squeeze, 64 Stellated, 168 Subgroup, 95 Sudoku Theorem, 37 Swap, 117 Symmetric, 69 Symmetry of an object in Euclidean space, 241 of a 3D object, 136 of a 2D object, 10 Symmetry group of a 3D object, 138 of a 2D object, 37

T Tetrahedron, 145 Theorem, 13 Thick polygon, 142 Topology, 184 Transitive, 69 Translation, 7 of space, 136 U Unbounded object in space, 136 object in the plane, 15 V Volleyball, 154 W Wallpaper pattern, 15 classification of, 63, 65, 66 Well-defined, 124 Z Zero-or-Equal Theorem 3D, 145 2D, 39