The Finite Field Distance Problem (Carus Mathematical Monographs) (The Carus Mathematical Monographs, 37) 1470460319, 9781470460310

Table of contents :
Cover
Title page
Copyright
Contents
Preface
Acknowledgments
Chapter 1. Background
1.1. Equivalence relations and the pigeonhole principle
1.2. Algebra and finite fields
1.3. Basic inequalities
1.4. Notation
1.5. Exercises: Chapter 1
Chapter 2. The distance problem
2.1. Introduction to the distance problem
2.2. Falconer distance problem
2.3. Finite field distance problem
2.4. Exercises: Chapter 2
Chapter 3. The Iosevich-Rudnev bound
3.1. Counting-method
3.2. The 𝐿²-method
3.3. Finite field spherical averages
3.4. Size and decay estimates for spheres
3.5. Finite field counterexample
3.6. Relations to the Falconer problem
3.7. Exercises: Chapter 3
Chapter 4. Wolff’s exponent
4.1. Introduction
4.2. Proof of 𝐿² estimate for 𝜈(𝑡)
4.3. Restriction and extension theory
4.4. Exercises: Chapter 4
Chapter 5. Rings and generalized distances
5.1. Distances in finite rings
5.2. Distances between two sets
5.3. Generalized distances
5.4. Pinned distances
5.5. Exercises: Chapter 5
Chapter 6. Configurations and group actions
6.1. Finite configurations
6.2. The Elekes-Sharir framework
6.3. Triangles: The enquote {7/4} bound
6.4. Triangles: The enquote {8/5} bound
6.5. Distance graph
6.6. Exercises: Chapter 6
Chapter 7. Combinatorics in finite fields
7.1. Incidence theory
7.2. Sum-product phenomena
7.3. Kakeya conjecture
7.4. Waring’s theorem
7.5. Roth’s theorem and the cap-set problem
7.6. The spectral gap theorem
7.7. Exercises: Chapter 7
Bibliography
Index
Back Cover

Citation preview

AMS / MAA THE CARUS MATHEMATICAL MONOGRAPHS

VOL 37

The Finite Field Distance Problem

David J. Covert

10.1090/car/037

The Finite Field Distance Problem

AMS/MAA

THE CARUS MATHEMATICAL MONOGRAPHS

VOL 37

The Finite Field Distance Problem David Covert

2019–2020 Editorial Committee Bruce P. Palka, Editor Francis Bonahon Steven J. Miller

Annalisa Crannel Carol Schumacher

Alex Iosevich Henry Sagerman

2020 Mathematics Subject Classification. Primary 05-02, 52-02, 52C10, 11E08, 11B13, 11T23. Cover image © 2018 by William Banks For additional information and updates on this book, visit www.ams.org/bookpages/car-37 Library of Congress Cataloging-in-Publication Data Names: Covert, David J., 1984– author. Title: The finite field distance problem / David J. Covert. Description: Providence, Rhode Island: MAA Press, an imprint of the American Mathematical Society, [2021] | Series: The Carus mathematical monographs, 0069-0813; volume 37 | Includes bibliographical references and index. Identifiers: LCCN 2020055010 | ISBN 9781470460310 (paperback) | (ebook) Subjects: LCSH: Finite fields (Algebra) | Combinatorial analysis. | Commutative rings. | AMS: Convex and discrete geometry – Research exposition (monographs, survey articles). | Number theory – Forms and linear algebraic groups – Quadratic forms over local rings and fields. | Number theory – Sequences and sets – Additive bases, including sumsets. | Number theory – Finite fields and commutative rings (number-theoretic aspects) – Exponential sums. Classification: LCC QA247.3 .C68 2021 | DDC 512/.3–dc23 LC record available at https://lccn.loc.gov/2020055010 Copying and reprinting. Individual readers of this publication, and nonprofit libraries acting for them, are permitted to make fair use of the material, such as to copy select pages for use in teaching or research. Permission is granted to quote brief passages from this publication in reviews, provided the customary acknowledgment of the source is given. Republication, systematic copying, or multiple reproduction of any material in this publication is permitted only under license from the American Mathematical Society. Requests for permission to reuse portions of AMS publication content are handled by the Copyright Clearance Center. For more information, please visit www.ams.org/publications/ pubpermissions. Send requests for translation rights and licensed reprints to reprint-permission@ ams.org. © 2021 by the American Mathematical Society. All rights reserved. The American Mathematical Society retains all rights except those granted to the United States Government. Printed in the United States of America. ∞ The paper used in this book is acid-free and falls within the guidelines ⃝

established to ensure permanence and durability. Visit the AMS home page at https://www.ams.org/ 10 9 8 7 6 5 4 3 2 1

26 25 24 23 22 21

To Patrick and Leo

Contents Preface

ix

Acknowledgments

xi

Chapter 1 Background 1.1 Equivalence relations and the pigeonhole principle 1.2 Algebra and finite fields 1.3 Basic inequalities 1.4 Notation 1.5 Exercises: Chapter 1

1 1 8 16 17 18

Chapter 2 The distance problem 2.1 Introduction to the distance problem 2.2 Falconer distance problem 2.3 Finite field distance problem 2.4 Exercises: Chapter 2

21 21 25 28 32

Chapter 3 The Iosevich-Rudnev bound 3.1 Counting-method 3.2 The 𝐿2 -method 3.3 Finite field spherical averages 3.4 Size and decay estimates for spheres 3.5 Finite field counterexample 3.6 Relations to the Falconer problem 3.7 Exercises: Chapter 3

35 35 41 42 52 57 59 60

Chapter 4 Wolff’s exponent 4.1 Introduction 4.2 Proof of 𝐿2 estimate for 𝜈(𝑡)

63 63 65 vii

viii 4.3 Restriction and extension theory 4.4 Exercises: Chapter 4

Contents 70 73

Chapter 5 Rings and generalized distances 5.1 Distances in finite rings 5.2 Distances between two sets 5.3 Generalized distances 5.4 Pinned distances 5.5 Exercises: Chapter 5

75 75 91 94 98 103

Chapter 6 Configurations and group actions 6.1 Finite configurations 6.2 The Elekes-Sharir framework 6.3 Triangles: The “7/4” bound 6.4 Triangles: The “8/5” bound 6.5 Distance graph 6.6 Exercises: Chapter 6

105 105 111 113 117 120 126

Chapter 7 Combinatorics in finite fields 7.1 Incidence theory 7.2 Sum-product phenomena 7.3 Kakeya conjecture 7.4 Waring’s theorem 7.5 Roth’s theorem and the cap-set problem 7.6 The spectral gap theorem 7.7 Exercises: Chapter 7

129 129 137 146 152 155 161 166

Bibliography

169

Index

179

Preface This book is meant to be a succinct, self-contained, but thorough introduction to the state-of-the-art of the socalled Erdős-Falconer Distance Problem, also known as the finite field distance problem. This problem has the fortune of being new, easily understood, and yet it remains unsolved, making this topic ideal for study by advanced undergraduates, graduate students, and research mathematicians alike. The book is structured as follows. In Chapter 1 we start with background material such as a review of properties of finite fields and some basic ideas in combinatorics. In Chapter 2 we will outline the original distance problem by Erdős, a continuous version of the problem by Falconer, and a finite field distance problem which will be the core topic of the book. Chapter 3 will be spent proving the first explicit result on the Erdős-Falconer distance problem in great detail. Chapter 4 will introduce the topic of restriction theory from harmonic analysis, and we will discuss the role that it plays in proving results on the size of distance sets. Chapter 5 will include many of the generalizations of the distance problem including an extension of the problem to finite rings. Chapter 6 will outline a grouptheoretic approach to the distance problem pioneered by Elekes and Sharir ([43]) and ultimately used by Guth and Katz ([70]) to establish the Erdős distance problem in the plane. We will discuss what implications ix

x

Preface these ideas will have in regards to the finite field distance problem. The final chapter (Chapter 7) will be devoted to other topics in the area including incidence theory, sum-product phenomena, the Kakeya conjecture, the cap set problem, and Waring’s problem. The goal of this final chapter is to introduce the non-expert to a handful of research problems in the area, some of which are still active areas of research, while others only having been resolved very recently. We provide enough background in finite fields to allow the reader to understand the material we discuss here. Otherwise, we provide only a cursory introduction to each topic, leaving many resources and references for the motivated reader.

Acknowledgments This book would not have been possible without a long list of supportive colleagues, friends, and family. A special thanks is due to Alex Iosevich—my mentor and friend who taught me much of what I know about combinatorics and analysis. Many thanks also to Jonathan Pakianathan, Bill Banks (for sharing both his mathematical expertise as well as his painting for the book cover), A. Prabhakar Rao, Ravindra Girivaru, Derrick Hart, Doowon Koh, Steve Senger, Krystal Taylor, Jeremy Chapman, Yeşim Demiroğlu-Karabulat, and many others. I owe a debt of gratitude to the editors at the AMS Book Program for their encouragement and detailed feedback on earlier drafts of this book. Thanks to my supportive and encouraging family: to my parents for their unwavering support, to my brother who inspired my love of math when I was small and whose friendship I value highly, to my sister who has always stood by my side, and to my wife Laura and sons Patrick and Leo who bring much love, meaning, and entertainment to my life.

xi

10.1090/car/037/01

Chapter

1

Background Before we discuss anything related to the distance problem, we first discuss some background material. Particularly, we will lay out some of the basic ideas in combinatorics and abstract algebra. Readers familiar with notions like the pigeonhole principle, the integers modulo 𝑛, and the construction of finite fields can safely skip to the next chapter.

1.1 Equivalence relations and the pigeonhole principle The pigeonhole principle states that if you were to place 𝑚 objects each into one of 𝑘 possible bins, then at least one bin must contain at least 𝑚 ⌈ 𝑘 ⌉ objects in it. Here ⌈𝑥⌉ is the ceiling of 𝑥 (meaning ⌈𝑥⌉ is the smallest integer 𝑛 that is greater than or equal to 𝑥). Thus, if I put 10 objects into 10 one of 3 possible bins, then at least one bin must contain at least ⌈ 3 ⌉ = 4 objects (see Figure 1.1). This obvious-sounding theorem is surprisingly useful and therefore surprisingly pervasive in combinatorics. This problem can be used both to prove the whimsical result that in the city of Saint Louis, Missouri, at least 3 people have exactly the same number of hairs on their head (humans having around 150, 000 hairs on their head, and there being around 318, 000 people living in the city limits of Saint Louis), and yet the pigeonhole principle can also be used to prove that if a real number can be well-approximated by many rational numbers, then that real number must be irrational (see Dirichlet’s approximation theorem). Frequently we will use the pigeonhole principle to get quantitive results such as 1

2

Chapter 1. Background

Figure 1.1. Three bins containing a total of 10 objects. Note that if half of the objects were in 1 bin and the other half of the objects were in another bin (so that the third bin was empty), the statement would still be true: at least one bin contains at least 4 objects!

the following: If 𝑥1 , 𝑥2 , . . . , 𝑥𝑛 are positive real numbers, and if 𝑥1 ⋅ 𝑥2 ⋅ ⋯ ⋅ 𝑥𝑛 ≥ 𝐶, then max(𝑥𝑖 ) ≥ 𝐶 1/𝑛 where the maximum is taken over 𝑖 ∈ {1, . . . , 𝑛} (see Exercise 1.3). Another elementary notion that we will use is that of equivalence. An equivalence relation on a set 𝑆 is a subset 𝑅 ⊆ 𝑆 × 𝑆 such that (𝑥, 𝑥) ∈ 𝑅 for all 𝑥 ∈ 𝑆, if (𝑥, 𝑦) ∈ 𝑅, then (𝑦, 𝑥) ∈ 𝑅, and if (𝑥, 𝑦) ∈ 𝑅 and (𝑦, 𝑧) ∈ 𝑅, then (𝑥, 𝑧) ∈ 𝑅. These properties are called reflexivity, symmetry, and transitivity, respectively. We will examine some examples of equivalence relations below. Furthermore, if 𝑅 is an equivalence relation on 𝑆, then for each element 𝑥 ∈ 𝑆, we may consider the equivalence class [𝑥] = {𝑦 ∈ 𝑆 ∶ (𝑥, 𝑦) ∈ 𝑅}.

(1.1)

The set of all equivalence classes is called the quotient set of 𝑆 mod 𝑅, and it will be denoted 𝑆/𝑅. The first example of an equivalence relation we should discuss is that of the integers modulo 𝑛. If 𝑎 and 𝑏 are integers, then we say that 𝑎 divides 𝑏 (written 𝑎 ∣ 𝑏) if 𝑏 = 𝑎𝑘 for some 𝑘 ∈ ℤ. Thus, 4 ∣ 12 as 12 = 4(3), but 5 ∤ 12. For a fixed integer 𝑛 ≥ 2, we say that 𝑥 is congruent to 𝑦 modulo 𝑛, written 𝑥 ≡ 𝑦 (mod 𝑛), if 𝑛 ∣ (𝑥 − 𝑦). It can be checked (this is code for you should check) that 𝑅 = {(𝑥, 𝑦) ∈ ℤ × ℤ ∶ 𝑥 ≡ 𝑦

(mod 𝑛)}

is an equivalence relation. Let’s work with 𝑛 = 5 for concreteness, so 𝑅 = {(𝑥, 𝑦) ∈ ℤ × ℤ ∶ 𝑥 ≡ 𝑦

(mod 5)}

= {(𝑥, 𝑦) ∈ ℤ × ℤ ∶ 5 ∣ (𝑥 − 𝑦)} = {(𝑥, 𝑦) ∈ ℤ × ℤ ∶ 𝑥 = 𝑦 + 5𝑘 for some integer 𝑘}.

1.1. Equivalence relations and the pigeonhole principle

3

Now, let’s look at the equivalence class [1]. By definition (1.1) we have [1] = {𝑦 ∈ ℤ ∶ 1 ≡ 𝑦

(mod 5)}

= {𝑦 ∈ ℤ ∶ 5 ∣ (1 − 𝑦)} = {. . . , −9, −4, 1, 6, 11, . . . }. Notice that there are 5 possible equivalence classes of this equivalence relation: [0], [1], [2], [3], and [4], corresponding to the 5 possible remainders when dividing an integer by 5. Working mod 5 can also be thought of as doing arithmetic along a circle rather than along a line.

Addition by some integer 𝑛 corresponds to moving 𝑛 additional places along this circle. In this setting 4 + 2 = 1, since if we start at 4 and move 2 units in the clockwise direction, we land on the number 1. We could also write this as 4 + 2 ≡ 1 (mod 5), which follows by the definition. Multiplication of 𝑥 by some integer 𝑛 then corresponds to moving 𝑥 units 𝑛 times (or equivalently, moving 𝑛 units 𝑥 times) starting at the position 0. For example, 4 ⋅ 2 = 3 since if we move four units 2 times (or if we move 2 units 4 times) we land on the number 3. It should now be clear why modular arithmetic is sometimes referred to as clock arithmetic. Changing the modulus from 5 to some other integer 𝑛 requires making a whole new clock. It is convenient to take the integers mod 𝑛, and turn them into a set ℤ𝑛 , where addition and multiplication are performed mod 𝑛. In ℤ5 , we have 4+2 = 1 and 4⋅2 = 3 as we saw above. The set ℤ𝑛 is often called the

4

Chapter 1. Background

ring of integers mod 𝑛, and this is language which we will justify shortly. Of special importance to us is the set of integers modulo a prime 𝑝. When 𝑝 is prime, we write ℤ𝑝 = 𝔽𝑝 . It can be shown (see Exercise 1.12) that the set 𝔽𝑝 has the very nice property that every nonzero element has a multiplicative inverse. Thus for all 𝑥 ∈ 𝔽𝑝 ⧵ {0}, there exists 𝑦 ∈ 𝔽𝑝 such that 𝑥𝑦 = 1 in 𝔽𝑝 . Consequently 𝔽𝑝 has the zero product property: For 𝑎, 𝑏 ∈ 𝔽𝑝 , if 𝑎𝑏 = 0, then either 𝑎 = 0 or 𝑏 = 0. These sets 𝔽𝑝 are called prime order finite fields, and fields will be described in detail below. That these sets are fields is why we write them as 𝔽𝑝 . Another important equivalence relation is the notion of a projective plane. For example let 𝐸 = ℝ3 ⧵{(0, 0, 0)}, and define the relation 𝑅 ⊆ 𝐸× 𝐸, where 𝑥, 𝑦 ∈ 𝔼 are related if 𝑥 and 𝑦 lie on the same line through the origin. One can easily check (and you should!) that this is an equivalence relation (see Exercise 1.1). Its resulting quotient set is called the real projective plane 𝑃ℝ2 . The same projective plane relation can be defined over other sets, such as 𝔽𝑝 for primes 𝑝, though we need to be more careful with what we mean by a line. For a set 𝑆, we let 𝑆 𝑑 be the 𝑑-fold Cartesian project of 𝑆 on itself. That is, 𝑆𝑑 = 𝑆 ×⎵⋯ ×⏟ 𝑆 = {(𝑥1 , 𝑥2 , . . . , 𝑥𝑑 ) ∶ 𝑥1 , 𝑥2 , . . . , 𝑥𝑑 ∈ 𝑆}. ⏟⎵ ⏟⎵⎵ 𝑑−times

Much of the setting of this book will take place in settings like 𝔽𝑝𝑑 . Formally, 𝔽𝑝𝑑 is the 𝑑-dimensional vector space over 𝔽𝑝 , but really we only need to think about 𝔽𝑝𝑑 as the set of 𝑑-tuples of elements in 𝔽𝑝 . For example, thinking of 𝔽3 as {0, 1, 2}, we have 𝔽23 = {(0, 0); (0, 1); (0, 2); (1, 0); (1, 1); (1, 2); (2, 0); (2, 1); (2, 2)}. It can help to visualize these sets as grids of points with integer coordinates (see Figure 1.2). At this point we are ready to discuss some geometry in 𝔽𝑝𝑑 . A line in 2 𝔽𝑝 is a set of the form {𝑡𝑥 + 𝑦 ∶ 𝑡 ∈ 𝔽𝑝 } where 𝑥, 𝑦 ∈ 𝔽𝑝2 and 𝑥 ≠ (0, 0). Notice that in 𝔽𝑝2 , every line has exactly 𝑝 points. For example the line in 𝔽23 between (0, 1) and (1, 2) is given by: ℓ1 = {𝑡(1, 1) + (0, 1) ∶ 𝑡 ∈ 𝔽3 } = {(0, 1), (1, 2), (2, 0)}.

(1.2)

See Figure 1.3. Notice also that (1, 2) − (0, 1) = (1, 1) is the direction or slope of this line. This concept of direction is well defined since

1.1. Equivalence relations and the pigeonhole principle

5

Figure 1.2. Representations of 𝔽23 with and without axes

Figure 1.3. The line ℓ1 above

𝑡(1, 1)+(0, 1)−(𝑠(1, 1) + (0, 1)) = (𝑡−𝑠)(1, 1) is a multiple of (1, 1). Thus, (1, 1) and (2, 2) give the same direction of a line since they are nonzero multiples of one another. There are 4 “directions” in which a line could travel 𝔽23 . We will cover this idea of directions in more detail in Section 7.3. As another example, consider the unique line1 ℓ2 in 𝔽213 containing the points (0, 5) and (3, 3) (see Figure 1.4). A similar construction defines lines in higher dimensions, though they are more difficult to visualize. A line through 𝑦 ∈ 𝔽𝑝𝑑 in direction 1 It is not immediately obvious that there is a unique line between two points, so try to prove it!

6

Chapter 1. Background

Figure 1.4. The line ℓ2 in 𝔽213 through (0, 5) and (3, 3)

𝑥 ≠ (0, . . . , 0) is the set of points given by ℓ = {𝑡𝑥 + 𝑦 ∶ 𝑡 ∈ 𝔽𝑝 }. Again, lines in 𝔽𝑝𝑑 contain exactly 𝑝 points. For example, the unique line through the points (1, 0, 0) and (2, 2, 1) in 𝔽35 (that is, the line through (1, 0, 0) in the direction (1, 2, 1)) is given by ℓ3 = {𝑡(1, 2, 1) + (1, 0, 0) ∶ 𝑡 ∈ 𝔽5 } = {(1, 0, 0); (2, 2, 1); (3, 4, 2); (4, 1, 3); (0, 3, 4)}. In 𝔽𝑝𝑑 , a sphere2 of “radius” 𝑡 centered at the point (𝑎1 , . . . , 𝑎𝑑 ) ∈ 𝔽𝑑𝑞 refers to the set 𝑆 𝑡 = {(𝑥1 , . . . , 𝑥𝑑 ) ∈ 𝔽𝑑𝑞 ∶ (𝑥1 − 𝑎1 )2 + . . . (𝑥𝑑 − 𝑎𝑑 )2 = 𝑡} . 2 We use the term sphere to describe the set 𝑆 even in the case 𝑑 = 2, when it might 𝑡 be more accurate to call the set a circle.

1.1. Equivalence relations and the pigeonhole principle

7

Note that we do not take square roots since not all elements in a finite field are squares. We will typically work with spheres centered at the origin. As an example, the sphere 𝑆 1 in 𝔽25 centered at the origin consists of the four points 𝑆 1 = {(1, 0); (0, 1); (4, 0); (0, 4)} (see Figure 1.5).

Figure 1.5. The sphere 𝑆 1 ⊆ 𝔽25

Figure 1.6 shows the spheres 𝑆 1 and 𝑆 2 in 𝔽211 (both centered at the origin), both of which containing 12 points. It turns out that all spheres with nonzero radius have the same number of points in 𝔽𝑝2 .

Figure 1.6. The spheres 𝑆 1 (left) and 𝑆 2 (right) in 𝔽211

There are some quirks here. In Euclidean space, a sphere of radius 0 is simply a point. In finite fields, we can have spheres with radius 0

8

Chapter 1. Background

containing many points. For example, the sphere 𝑆 0 in 𝔽25 centered at the origin contains the 9 points 𝑆 0 = {(0, 0); (1, 2); (1, 3); (2, 1); (3, 1); (2, 4); (3, 4); (4, 2); (4, 3)}.

Figure 1.7. The sphere 𝑆 0 in 𝔽211 with and without the (Euclidean) circle (𝑥 − 5/2)2 + (𝑦 − 5/2)2 = 5/2 superimposed

In 𝔽35 , the sphere 𝑆 0 contains 30 points (see Figure 1.8). We will give an exact formula for the number of points in a sphere in Chapter 3. It is instructive to consider and illustrate your own examples of lines and sphere in the sets 𝔽𝑝𝑑 . We will delve deeper into these types of geometrical objects later in the text, but at this point we can construct projective planes over some finite fields. Let 𝑝 be prime, suppose 𝐸 = 𝔽𝑝3 ⧵ {(0, 0, 0)}, and define 𝑅 ⊆ 𝐸 × 𝐸 where 𝑥 ∈ 𝐸 is related to 𝑦 ∈ 𝐸 if 𝑦 lies on the line ℓ𝑥 = {𝑡𝑥 ∶ 𝑡 ∈ 𝔽𝑝 }. Notice that ℓ𝑥 is really just the line containing 𝑥 and the origin (0, 0, 0) This relation is again an equivalence relation, and the resulting quotient set is called the finite projective plane 𝑃𝐺(2, 𝑝). Perhaps the best known of these projective planes is the Fano plane 𝑃𝐺(2, 2) over 𝔽2 (see Figure 1.9 below). There is much more we could say about these finite projective planes as they are interesting in their own right, but instead we refer the interested reader to [22].

1.2 Algebra and finite fields Given a set 𝑆, a binary operation on 𝑆 is an operation acting on pairs of elements of 𝑆. That is, a binary operation ∗ is any function 𝑆 × 𝑆 → 𝑋

1.2. Algebra and finite fields

Figure 1.8. The sphere 𝑆 0 in 𝔽35 superimposed on the (Euclidean) sphere (𝑥 − 5/2)2 + (𝑦 − 5/2)2 + (𝑧 − 5/2)2 = 19/4.

Figure 1.9. The Fano plane. Image taken from [155].

9

10

Chapter 1. Background

for some set 𝑋 Thus 𝑥 ∗ 𝑦 is well defined for every pair (𝑥, 𝑦) ∈ 𝑆 × 𝑆. Addition, multiplication, and subtraction are all binary operations on ℝ, for example.

1.2.1 Rings A ring (𝑅, +, ∗) is a set 𝑅 together with two binary operations + and ∗ satisfying a certain set of properties (which we list below). Rings typically are required to have a unit element 1 ∈ 𝑅 such that 1∗𝑎 = 𝑎∗1 = 𝑎 for all 𝑎 ∈ 𝑅. Table 1.1. Addition and multiplication modulo 5

+ 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

∗ 0 1 2 3 4

0 0 0 0 0 0

1 0 1 2 3 4

2 0 2 4 1 3

3 0 3 1 1 2

4 0 4 3 2 1

Definition 1.2.1. A set (𝑅, +, ∗) is a ring if it satisfies all of the following conditions: •

Closure of addition and multiplication: 𝑎+𝑏∈𝑅

•

and

𝑎∗𝑏∈𝑅

Additive Associativity: 𝑎 + (𝑏 + 𝑐) = (𝑎 + 𝑏) + 𝑐

•

for all 𝑎, 𝑏 ∈ 𝑅.

Additive Identity: There exists 0 ∈ 𝑅 such that 𝑎+0=0+𝑎=𝑎

•

for all 𝑎, 𝑏, 𝑐 ∈ 𝑅.

Additive Commutativity: 𝑎+𝑏=𝑏+𝑎

•

for all 𝑎, 𝑏 ∈ 𝑅.

for all 𝑎 ∈ 𝑅.

Additive Inverses: For every 𝑎 ∈ 𝑅, there exists 𝑏 ∈ 𝑅 such that 𝑎+𝑏=0

(we will write 𝑏 = −𝑎).

1.2. Algebra and finite fields •

•

11

Left and Right Distributivity: 𝑎 ∗ (𝑏 + 𝑐) = 𝑎 ∗ 𝑏 + 𝑎 ∗ 𝑐

for all 𝑎, 𝑏, 𝑐 ∈ 𝑅

(𝑎 + 𝑏) ∗ 𝑐 = 𝑎 ∗ 𝑐 + 𝑏 ∗ 𝑐

for all 𝑎, 𝑏, 𝑐 ∈ 𝑅.

Multiplicative Associativity: 𝑎 ∗ (𝑏 ∗ 𝑐) = (𝑎 ∗ 𝑏) ∗ 𝑐

•

for all 𝑎, 𝑏, 𝑐 ∈ 𝑅.

Multiplicative Identity: There exists 1 ∈ 𝑅 so that 1∗𝑎=𝑎∗1=𝑎

for all 𝑎 ∈ 𝑅.

The canonical example of a ring (and one that will be particularly important in this text) is the set (ℤ𝑛 , +, ∗) where ℤ𝑛 = {0, 1, 2, . . . , 𝑛 − 1} is the set of integers mod 𝑛, and where the addition and multiplication are performed mod 𝑛. For the remainder of the text, we will frequently refer to the ring (𝑅, +, ∗) as simply the set 𝑅 if the operations of addition and multiplication are easily understood. Recall from the previous section that formally the elements of ℤ𝑛 are the equivalence classes [0], [1], . . . , [𝑛 − 1] of the equivalence relation {(𝑥, 𝑦) ∈ ℤ × ℤ ∶ 𝑥 ≡ 𝑦

(mod 𝑛)},

though it is more convenient to write 𝑥 rather than [𝑥], so if we are working in ℤ5 we would write 2 ⋅ 3 = 1 rather than [2] ∗ [3] = [1]. There are many other examples of rings including the integers ℤ, the rational numbers ℚ, the real numbers ℝ, and the complex numbers ℂ all taken with the standard operations of addition and multiplication. Note that the set of natural numbers ℕ is not a ring since elements of ℕ do not have additive inverses. Other examples include sets such as 𝑀𝑛×𝑛 (ℚ) which is the ring of 𝑛×𝑛 matrices with rational entries, where + denotes the usual sum of matrices and ∗ stands for matrix multiplication. What are the additive and multiplicative identity in 𝑀𝑛×𝑛 (ℚ)? Given any ring 𝑅, we define 𝑅[𝑥] to be the set of all polynomials in variable 𝑥 where the coefficients are taken from some ring (𝑅, +, ∗). Here, function addition is the operation (𝑓 + 𝑔)(𝑥) = 𝑓(𝑥) + 𝑔(𝑥) and function multiplication is (𝑓 ⋅ 𝑔)(𝑥) = 𝑓(𝑥) ∗ 𝑔(𝑥), where ∗ is multiplication in the ring 𝑅. One can (and should) check that 𝑅[𝑥] is indeed a ring. For example, ℝ[𝑥] is the ring of polynomials with real coefficients.

12

Chapter 1. Background

Many more examples of rings exist, and we even encounter some more examples throughout the text.

1.2.2 Groups Another fundamental concept in abstract algebra is that of a group. We say that a set 𝐺 with a binary operation ∗ is a group if (𝐺, ∗) satisfies the following: •

Closure: 𝑎∗𝑏∈𝐺

•

for all 𝑎, 𝑏 ∈ 𝐺.

Associativity: 𝑎 ∗ (𝑏 ∗ 𝑐) = (𝑎 ∗ 𝑏) ∗ 𝑐

•

Identity: there exists 𝑒 ∈ 𝐺 such that 𝑎∗𝑒=𝑒∗𝑎=𝑒

•

for all 𝑎, 𝑏, 𝑐 ∈ 𝐺

for all 𝑎 ∈ 𝐺.

Existence of inverses: For every 𝑎 ∈ 𝐺 there exists 𝑏 ∈ 𝐺 such that 𝑎 ∗ 𝑏 = 𝑏 ∗ 𝑎 = 𝑒.

We often write 𝑏 = 𝑎−1 as 𝑏 = −𝑎 if we are thinking of the group operation on 𝐺 as being a type of addition. There are numerous examples of groups with which you are (probably) already familiar. For example (ℤ𝑛 , +) the group of integers modulo 𝑛 under addition is a group. Another example is the set ℤ×𝑛 = {𝑥 ∈ ℤ𝑛 ∶ gcd(𝑥, 𝑛) = 1}. ℤ×𝑛 is called the multiplicative subgroup of ℤ𝑛 , and it consists of all the elements in ℤ𝑛 which have inverses with respect to multiplication (Exercise 1.12 strikes again!). For example, ℤ×10 = {1, 3, 7, 9}, and for any prime 𝑝 we have 𝔽𝑝× = {1, 2, . . . , 𝑝 − 1} = 𝔽𝑝 ⧵ {0}.

(1.3)

1.2. Algebra and finite fields

13

Other examples of groups include the group of invertible matrices in 𝑀𝑛×𝑛 (ℝ) together with matrix multiplication (called the general linear group): 𝐺𝐿𝑛 (ℝ) = {𝑀 ∈ 𝑀𝑛×𝑛 (ℝ) ∶ det(𝑀) ≠ 0}, the orthogonal group: 𝑂𝑛 (ℝ) = {𝑀 ∈ 𝑀𝑛×𝑛 (ℝ) ∶ 𝑀 −1 = 𝑀 𝑇 }, and the special orthogonal group: 𝑂𝑛 (ℝ) = {𝑀 ∈ 𝑀𝑛×𝑛 (ℝ) ∶ 𝑀 −1 = 𝑀 𝑇 and det(𝑀) = 1}, where 𝑀 𝑇 denote the matrix transpose of the matrix 𝑀.

1.2.3 Fields A field is a ring (𝑅, +, ∗) where every nonzero element in 𝑅 has a multiplicative inverse under the operation ∗. Examples of fields include the rational numbers ℚ, the real numbers ℝ, and the complex numbers ℂ Notice that ℤ is not a field as most integers do not have multiplicative inverses in the integers. Furthermore by (1.3), the ring (ℤ𝑛 , +, ×) is a field if and only if 𝑛 is prime, and recall this is why we write ℤ𝑝 = 𝔽𝑝 . There are numerous facts about fields which will be useful throughout the text. Proposition 1.2.2. If 𝔽 is a field and if 𝑥, 𝑦 ∈ 𝔽 satisfy 𝑥𝑦 = 0, then either 𝑥 = 0 or 𝑦 = 0. Proposition 1.2.3. Let 𝔽 be a field, and let 𝑝(𝑥) ∈ 𝔽[𝑥] be a polynomial in 𝑥 with coefficients from 𝔽 and with degree deg(𝑝) = 𝑑. Then 𝑝(𝑥) = 0 has at most 𝑑 solutions in 𝔽. The proof of the first proposition is left as an exercise (see Exercise 1.6). We refer the reader to the excellent3 book [104] for proof of the last proposition and for more facts about fields (and finite fields in particular). 3 It really cannot be overstated: The more time you spend reading Finite Fields by Lidl and Niederetter ([104]), the better!

14

Chapter 1. Background

A field is said to have characteristic 𝑛 if 𝑛 is the smallest positive integer such that 𝑛𝑥 = 0 for all 𝑥 ∈ 𝑅, and where 0 ∈ 𝑅 is the additive identity in 𝑅. If no such positive 𝑛 exists (such as for ℂ, ℝ, and ℚ), then we say these fields have characteristic zero. Furthermore, the field 𝔽𝑝 has characteristic 𝑝. It turns out that all finite fields will have positive characteristic, and there also exist some fields of infinite order with positive characteristic (such as the 𝑝-adic integers). It also turns out that any positive characteristic must be prime (see Exercise 1.7). The order of a finite field is simply the number of elements contained in that finite field. The finite fields of prime order are the easiest to comprehend and work with computationally, but they are far from the only finite fields. It turns out that there exist finite fields of order 𝑝𝑟 for every prime 𝑝 and every positive integer 𝑟. We will provide explicit methods of constructing such finite fields below. If 𝐾 and 𝐿 are two fields with the same operations, and if 𝐾 ⊆ 𝐿, then we say that 𝐾 is a subfield of 𝐿 (or equivalently that 𝐿 is an extension field of 𝐾). If a field 𝐾 ⊊ 𝐿 is a proper subset, we call 𝐾 a proper subfield of 𝐾 (or 𝐿 a proper extension field of 𝐾). We will frequently talk about polynomials with coefficients in a field 𝔽. Recall that the set of such polynomials is denoted 𝔽[𝑥]. Given 𝑓 ∈ 𝔽[𝑥], the splitting field of 𝑓(𝑥) is the smallest extension field of 𝐹 over which 𝑓(𝑥) splits into linear factors. It can be shown that the splitting field of a polynomial always exists and is unique ([104]). Example 1.2.4. The function 𝑓 ∈ ℝ[𝑥] given by 𝑓(𝑥) = 𝑥2 + 1 has splitting field ℂ. Example 1.2.5. The splitting field of 𝑓 ∈ 𝔽5 [𝑥] given by 𝑓(𝑥) = 𝑥2 + 1 is simply 𝔽5 since 𝑓(𝑥) = 𝑥2 + 1 = (𝑥 + 3)(𝑥 + 2). Example 1.2.6. The polynomial 𝑓 ∈ 𝔽3 [𝑥] given by 𝑓(𝑥) = 𝑥2 + 1 has splitting field 𝔽3 [𝛼] ≔ {𝑥 + 𝛼𝑦 ∶ 𝑥, 𝑦 ∈ 𝔽3 and 𝛼2 = −1}. This set 𝔽3 [𝛼] has nine elements: 𝔽3 [𝛼] = {0, 1, 2, 𝛼, 𝛼 + 1, 𝛼 + 2, 2𝛼, 2𝛼 + 1, 2𝛼 + 2}.

1.2. Algebra and finite fields

15

We will show that 𝔽3 [𝛼] is itself a field with characteristic 3, having 𝔽3 as a subfield, and having order 9. If we can show this set is a field, it follows immediately that 𝔽3 is a subfield, that the field has characteristic 3, and the field has order 9. Now the simplest way to show this set is a field is to notice that each nonzero element has a multiplicative inverse: 1⋅1=1 2⋅2=1 𝛼 ⋅ 2𝛼 = 1 (𝛼 + 1) ⋅ (𝛼 + 2) = 1 (2𝛼 + 1) ⋅ (2𝛼 + 2) = 1. It remains to show that 𝔽3 [𝛼] splits 𝑓(𝑥) = 𝑥2 + 1, but this follows immediately from 𝑥2 + 1 = (𝑥 + 𝛼)(𝑥 + 2𝛼). It turns out that the splitting field of the function 𝑓 ∈ 𝔽𝑝 [𝑥] given by 𝑟 𝑓(𝑥) = 𝑥𝑝 − 𝑥 is a field with 𝑝𝑟 elements ([104]). In particular, there exists a finite field of every prime power order. Since any two fields of the same order are isomorphic ([104]), we can refer to the finite field 𝔽𝑝𝑟 . From this we can clearly see some immediate consequences. Recalling that the characteristic of a finite field must be prime, it follows that if 𝑛 is not some power of a prime, then there is no field of order 𝑛. If 𝑞 = 𝑝𝑟 is a power of a prime, then there exists a field of order 𝑝𝑟 and the field is unique up to isomorphism. Every field of order 𝑝𝑟 has 𝔽𝑝 as a subfield. We can view the finite field 𝔽𝑝𝑟 as an 𝑟-dimensional vector space over its prime field 𝔽𝑝 . The field 𝔽𝑝𝑛 is a subfield of 𝔽𝑝𝑚 if and only if 𝑛 ∣ 𝑚. In particular if 𝔽𝑞 is a finite field then its proper subfields must have order √𝑞 or smaller. This simple and seemingly insignificant point will be relevant throughout the text. If 𝔽𝑝𝑛 is a subfield of 𝔽𝑝𝑚 , then we can view 𝔽𝑝𝑚 as a vector space over 𝔽𝑝𝑛 . One final point bears repeating: The fields whose orders are prime powers but not primes such as 𝔽4 , 𝔽8 , and 𝔽9 are not isomorphic to the rings ℤ4 , ℤ8 , and ℤ9 . The additive group (𝔽𝑞 , +) is not cyclic unless 𝑞 is prime. However, for every finite field 𝔽𝑞 , the multiplicative subgroup 𝔽∗𝑞 = 𝔽𝑞 ⧵ {0} is cyclic ([104]).

16

Chapter 1. Background

1.3 Basic inequalities Most of the inequalities we will use throughout this book will be straightforward, and we will discuss them as we go along. However, a few inequalities are so ubiquitous that is worth recording them here. We will only state these inequalities, and we leave it to the reader to verify them. The first inequality we will encounter with great regularity is the triangle inequality: Proposition 1.3.1. Suppose that 𝑓(𝑡) is any real valued function. Then | | |∑ 𝑓(𝑡)| ≤ ∑ |𝑓(𝑡)| . | | | 𝑡 | 𝑡 Perhaps the most useful inequality for us will be the Cauchy-Schwarz inequality (also called the Cauchy-Bunyakovsky-Schwarz inequality). This inequality takes many forms (for example, it is valid over any inner product space), but we will frequently used the Cauchy-Schwarz inequality for sums as follows. Proposition 1.3.2 (Cauchy-Schwarz). Suppose that 𝑓(𝑡) and 𝑔(𝑡) are nonnegative and real-valued functions. Then, 1/2 2

∑ 𝑓(𝑡)𝑔(𝑡) ≤ (∑ [𝑓(𝑡)] ) 𝑡

𝑡

1/2 2

.

(∑ [𝑔(𝑡)] ) 𝑡

The proof uses nothing more than the fact that 2𝑥𝑦 ≤ 𝑥2 + 𝑦2 for all real numbers 𝑥 and 𝑦. A generalization of the Cauchy-Schwarz inequality is Hölder’s inequality: Proposition 1.3.3 (Hölder’s inequality). Suppose that 𝑓(𝑡) and 𝑔(𝑡) are nonnegative and real-valued functions. Suppose that 𝑝 and 𝑞 are both pos1 1 itive real numbers satisfying 𝑝 + 𝑞 = 1. Then, 1/𝑝 𝑝

∑ 𝑓(𝑡)𝑔(𝑡) ≤ (∑ [𝑓(𝑡)] ) 𝑡

𝑡

1/𝑞 𝑞

(∑ [𝑔(𝑡)] ) 𝑡

.

1.4. Notation

17

The values 𝑝 and 𝑞 are often called Hölder conjugates in this context. The proof of this inequality uses Young’s inequality: If 𝑎 and 𝑏 are nonnegative real numbers, and if 𝑝 and 𝑞 are Hölder conjugates, then 𝑎𝑝 𝑏𝑞 + ≥ 𝑎𝑏. 𝑝 𝑞 Notice that the Cauchy-Schwarz inequality is simply Hölder’s inequality in the case 𝑝 = 𝑞 = 2. Finally, we have the following trivial but useful inequality which we can view as Hölder’s inequality in the case that 𝑝 = ∞ and 𝑞 = 1: Proposition 1.3.4. Suppose that 𝑓(𝑡) and 𝑔(𝑡) are nonnegative real-valued functions. Then, ∑ 𝑓(𝑡)𝑔(𝑡) ≤ (max 𝑓(𝑡)) ∑ 𝑔(𝑡). 𝑡

𝑡

𝑡

1.4 Notation We are almost ready to dive into the substance of the text. Before we do, we first provide a short list of some notation and conventions. •

𝑞 will always denote a power of an odd prime, unless explicitly stated otherwise.

•

𝔽𝑞 will denote the finite field with 𝑞 elements, where 𝑞 is always assumed to be odd. 𝔽𝑑𝑞 is the 𝑑-dimensional vector space over 𝔽𝑞 .

•

|𝐸| will denote the cardinality of a finite set 𝐸. For 𝑥 ∈ ℝ𝑛 , we may also use |𝑥| to denote the vector’s length (i.e. Euclidean norm). For a complex number 𝑧 ∈ ℂ, we frequently rely on the trivial observation that |𝑧|2 = 𝑧𝑧, where 𝑧 is the complex conjugation of 𝑧. The notation | ⋅ | should always be clear from the context.

•

The constant 𝑐 will often stand for a generic (positive) constant which may change from one occurrence to the next. We may add subscripts to particular constants to emphasize dependence on particular parameters.

•

The quantity log 𝑥 always denotes the natural logarithm.

18

Chapter 1. Background

•

Given a ring 𝑅, 𝑅× denotes the set of units in 𝑅, and 𝑅∗ denotes the nonzero elements in 𝑅. Thus, 𝑅 is a field if and only if 𝑅∗ = 𝑅× .

•

We use 𝑓(𝑥) ≪ 𝑔(𝑥) (or 𝑔(𝑥) ≫ 𝑓(𝑥)) to imply the existence of a constant 𝑐 > 0 such that 𝑓(𝑥) ≤ 𝑐𝑔(𝑥). Some authors use 𝑓(𝑥) = 𝑂(𝑔(𝑥)) interchangeably with 𝑓(𝑥) ≪ 𝑔(𝑥). If we wish to emphasize that the implicit constants depend on some parameter 𝑡, then we may write 𝑓(𝑥) ≪𝑡 𝑔(𝑥).

•

𝑓(𝑥) ≍ 𝑔(𝑥) signifies that 𝑓(𝑥) ≪ 𝑔(𝑥) ≪ 𝑓(𝑥).

•

𝑓(𝑥) = 𝑜(𝑔(𝑥)) means that lim𝑥→∞ 𝑓(𝑥)/𝑔(𝑥) = 0 .

•

𝑓(𝑥) ∼ 𝑔(𝑥) means 𝑓(𝑥) = 𝑔(𝑥)(1 + 𝑜(1)).

1.5 Exercises: Chapter 1 Exercise 1.1. Prove that the equivalence relation defining the projective plane 𝑃ℝ2 is indeed an equivalence relation. Notice that a projective plane can be described in such a manner over any field 𝔽. Exercise 1.2. Construct addition and multiplication tables for ℤ6 . Exercise 1.3. Prove that if 𝑥1 + ⋯ + 𝑥𝑛 ≥ 𝐶, there exists 𝑖 ∈ {1, . . . , 𝑛} such that 𝑥𝑖 ≥ 𝐶/𝑛. Exercise 1.4. Write down the 9 × 9 multiplication table for 𝔽9 , using the construction for 𝔽3 [𝛼] we discussed in Example 1.2.6. Exercise 1.5. Prove that if 𝔽𝑑𝑞 is the 𝑑-dimensional vector space over the finite field with 𝑞 elements, then 𝔽𝑑𝑞 contains 𝑞𝑑 vectors. Write out all nine vectors from the vector space 𝔽23 and all twenty-five vectors in 𝔽25 . Exercise 1.6. Prove that if 𝔽 is a field, and if 𝑥, 𝑦 ∈ 𝔽 satisfy 𝑥𝑦 = 0, then either 𝑥 = 0 or 𝑦 = 0. Exercise 1.7. Prove that if a field has positive characteristic, then that characteristic must be prime.

1.5. Exercises: Chapter 1

19

Exercise 1.8. Prove that 𝑥2 ≡ 1 (mod 𝑝) has at most two solutions for all primes 𝑝, while 𝑥2 ≡ 1 (mod 8) has 4 unique solutions (mod 8). Is this consistent with what we know about fields? Hint: Yes, it is! Exercise 1.9. Show that ℤ8 is not a field. Use the splitting field construction to construct a field of order 8. Exercise 1.10. Euclid’s Algorithm gives a method for computing gcd(𝑎, 𝑏) for two positive integers 𝑎 and 𝑏. Look up this algorithm, and compute gcd(2020, 7) using the algorithm. Exercise 1.11. Use Euclid’s Algorithm (see Exercise 1.10) to prove Bézout’s Little Theorem: If 𝑎 and 𝑏 are relatively prime, then there exist integers 𝑥 and 𝑦 (called the Bézout coefficients of 𝑎 and 𝑏) such that 𝑎𝑥 + 𝑏𝑦 = 1. Exercise 1.12. Use Bézout’s Little Theorem (see previous exercise) to prove that an element 𝑎 ∈ ℤ𝑛 has a multiplicative inverse if and only if gcd(𝑎, 𝑛) = 1. That is, the congruence 𝑎𝑥 ≡ 1

(mod 𝑛)

has a solution 𝑥 ∈ ℤ𝑛 if and only if gcd(𝑎, 𝑛) = 1. Hint: Write 1 = 𝑎𝑥+𝑛𝑦 since gcd(𝑎, 𝑛) = 1, and recall that equivalence relations are reflexive. Exercise 1.13. Using your calculation from Exercise 1.10, find the Bézout coefficients of 2020 and 7. Then, use Exercise 1.12 to compute the multiplicative inverse of 7 (mod 2020). Exercise 1.14. Show that 5 ⋅ 2−1 ≡ 0 (mod 5) and 19 ⋅ 4−1 ≡ 1 (mod 5). Compare this to the circle and sphere superimposed in Figures 1.7 and 1.8.

10.1090/car/037/02

Chapter

2

The distance problem 2.1 Introduction to the distance problem Let’s play a game. Your goal is to draw three points on a blank sheet of paper, and then connect each pair of dots with a line, trying to minimize the number of distinct distances. How many distinct distances did you draw? If you picked three random points, then most likely you will have 3 distinct distances. If you drew an isosceles triangle, with two sides being the same length, then your drawing determined only two distances. Better still, if you drew an equilateral triangle so that all sides of the triangle are equal, then you win as you have three points determining only one distance, which is clearly the best you can do. Now, what happens if you have to select four points? What is the minimum number of distinct distances you achieve here? It turns out that a square will determine only two distances (the edge of the square and the diagonal), and this is best possible. Now what if you draw 5 points? 10 points? 100 points?

The complexity of the problem increases dramatically even for a small number of points (try this for 10 points for example). Therefore, rather than trying to find a function which would give the minimum of distances for exactly 𝑛 points, it is more productive to ask about the longterm behavior of such a quantity: Roughly how many distances do we 21

22

Chapter 2. The distance problem

achieve if we were to draw 𝑛 points on a sheet of a paper, when 𝑛 is very large? This question was originally posed by Paul Erdős in 1946 ([49]). Suppose 𝑓(𝑛) denotes the minimum number of distances determined by a set of 𝑛 points in ℝ2 , so that 𝑓(3) = 1 as an equilateral triangle determines a single distinct distance, 𝑓(4) = 𝑓(5) = 2 as shown above with the square and pentagon, and so on. What is the best asymptotic lower bound for 𝑓(𝑛), if we let 𝑛 go to infinity? More precisely, given a set 𝐸 ⊆ ℝ2 of cardinality |𝐸| = 𝑛, define Δ(𝐸) = {|𝑥 − 𝑦| ∶ 𝑥, 𝑦 ∈ 𝐸} ⊆ ℝ,

(2.1)

where |𝑥| = √𝑥12 + ⋯ + 𝑥𝑑2 denotes the standard (Euclidean) distance. Then, the Erdős distance problem asks one for a lower bound on the quantity 𝑓(𝑛) = min{|Δ(𝐸)| ∶ 𝐸 ⊆ ℝ2 and |𝐸| = 𝑛}. In [49], Erdős showed that one always has the bound 𝑛 √𝑛 ≪ 𝑓(𝑛) ≪ . √log 𝑛

(2.2)

The upper bound in (2.2) came by considering the set 𝐸 = {(𝑥, 𝑦) ∈ ℤ × ℤ ∶ 0 ≤ 𝑥, 𝑦 ≤ √𝑛} when 𝑛 is a square. It is well known ([90]) that if 𝐴(𝑥) denotes the set of nonnegative integers less than or equal to 𝑥 which are sums of two squares, then |𝐴(𝑥)| ∼ 𝛾𝑥/√log 𝑥, where 𝛾 = 0.7642 . . . is the LandauRamanjuan constant. It follows that |Δ(𝐸)| ≤ 2|𝐴(𝑛)|, so that |Δ(𝐸)| ≤

2𝛾𝑛 √log 2𝑛

≍

𝑛 √log 𝑛

.

(2.3)

The lower bound was achieved by some clever combinatorial reasoning. Essentially, Erdős’ argument involved only elementary notions like the fact that two distinct circles intersect in at most 2 points, coupled with the pigeonhole principle. Details of this argument can be found in Section 6.2, and a thorough and intuitive description of these results and the progress thereafter are the subject of the excellent book The Erdős distance problem by Garibaldi, Iosevich, and Senger ([61]). Erdős’ bounds from (2.2) led him to conjecture ([15]) that 𝑓(𝑛) ≫ 𝑛1−𝑜(1) for all sets 𝐸 ⊆ ℝ2 with cardinality 𝑛. The lower bound was

2.1. Introduction to the distance problem

23

steadily improved ([18, 19, 98, 112, 143, 146, 151]) until more than sixty years after Erdős first wrote about this problem, a breakthrough was achieved by Guth and Katz ([70]). They showed that 𝑛 , 𝑓(𝑛) ≫ (2.4) log 𝑛 a very near optimal bound which established the Erdős distance conjecture! Later (in Section 6.2) we will provide a brief outline of their proof which follows the so-called Elekes-Sharir framework ([43]), and we provide some related insights by applying analogues of their ideas to the finite field problem which we will describe below. The Erdős distinct distance problem has been considered in higher dimensions as well. Let 𝑓𝑑 (𝑛) denote the minimum number of Euclidean distances determined by a set of 𝑛 points in ℝ𝑑 . The integer lattice example giving the upper bound for 𝑑 = 2 shows that when 𝑑 ≥ 3, we can hope for 𝑓𝑑 (𝑛) ≪ 𝑛2/𝑑 , and no better. More precisely, let 𝐸 = {(𝑥1 , . . . , 𝑥𝑑 ) ∈ ℤ𝑑 ∶ 1 ≤ 𝑥𝑖 ≤ 𝑛1/𝑑 for all 𝑖 = 1, . . . , 𝑑} for some integer 𝑛 which is a perfect 𝑑th power. Legendre showed that if 𝐵(𝑥) denotes the set of positive integers which are the sum of three 5 squares, then 𝐵(𝑥) satisfies 𝐵(𝑥) ∼ 6 𝑥 (see Exercise 2.2). Additionally, every integer is the sum of at most four squares by Lagrange’s four-square theorem. Hence, 𝑓𝑑 (𝑛) ≤ 𝑐 𝑑 𝑛2/𝑑 ≪ 𝑛2/𝑑 for 𝑑 ≥ 3. It is further conjectured that 𝑓𝑑 (𝑛) ≫ 𝑛2/𝑑 with no logarithmic loss whenever 𝑑 ≥ 3. Erdős’ bound 𝑓(𝑛) ≫ 𝑛1/2 can be adapted to higher dimensions to show that 𝑓𝑑 (𝑛) ≫ 𝑛1/𝑑 . The best result in general higher dimensions belongs to Solymosi and Vu ([144]) who showed that 2

𝑓𝑑 (𝑛) ≫ 𝑛 𝑑

−

2 𝑑2 +2𝑑

.

2.1.1 Erdős unit-distance problem Erdős also considered a second distance problem. Given a set of 𝑛 points, how many times is it possible for a single distance to occur? Formally, we define 𝑔(𝑛) = max |{‖𝑥 − 𝑦‖ = 1 ∶ 𝑥, 𝑦 ∈ 𝐸}| . |𝐸|=𝑛

𝐸⊆ℝ2

24

Chapter 2. The distance problem

Note that given a point set 𝐸, we can simply scale the set so that the most frequent distance achieved is 1, and hence there is no harm in replacing “most popular distance” with the distance 1. Now, Erdős showed that we must have the bound 𝑔(𝑛) ≫ 𝑛1+𝑐/ log log 𝑛 , for some 𝑐 > 0 by again considering the set {(𝑥, 𝑦) ∈ ℤ × ℤ ∶ 1 ≤ 𝑥, 𝑦 ≤ √𝑛} and using known results about the distribution of primes of the form 𝑝 = 4𝑚 + 1 ([48]), since such primes can always be written as the sum of squares (see Exercise 2.4). He conjectured ([15]) that 𝑔(𝑛) ≪ 𝑛1+𝑜(1) , but the best bound established thus far is 𝑔(𝑛) ≪ 𝑛4/3 ,

(2.5)

due to the work of Spencer, Szemerédi, and Trotter ([140]), which relied on the work of Szemerédi and Trotter on incidences: Theorem 2.1.1 ([148]). Let 𝑃 ⊆ ℝ2 be a set of points, and let 𝐿 be a set of lines in ℝ2 . An incidence is a pair (𝑝, ℓ) such that 𝑝 ∈ ℓ. The number of incidences determined by the sets 𝑃 and 𝐿 is then 𝐼(𝑃, 𝐿) = |{(𝑝, ℓ) ∈ 𝑃 × 𝐿 ∶ 𝑝 ∈ ℓ}|. Finally we have 𝐼(𝑃, 𝐿) ≪ (|𝑃|2/3 |𝐿|2/3 + |𝑃| + |𝐿|). See Section 7.1 for much more information on incidence theory and its applications to problems in additive combinatorial problems in particular. Really, the result of Spencer, Szemerédi, and Trotter relied on an adaptation of the work of Szemerédi and Trotter where lines were replaced by curves with no self-intersections. This was then applied to circles in the plane. This remarkable theorem of Szemerédi and Trotter has many applications to geometric combinatorics including a result on the sum-product problem ([41]) and the unit-distance problem ([140]). We explore incidence theory in more detail in Chapter 7.

2.2. Falconer distance problem

25

The unit-distance problem is only interesting in dimensions 3 and lower due to some arithmetic obstructions ([89]). In ℝ3 , we have that 𝑓(𝑛) ≪ 𝑛2/3 , so that 𝑔3 (𝑛) ≫ 𝑛4/3 by the pigeonhole principle, where 𝑔3 (𝑛) = max |{‖𝑥 − 𝑦‖ = 1 ∶ 𝑥, 𝑦 ∈ 𝐸}| . |𝐸|=𝑛

𝐸⊆ℝ3

Until recently, the best known bound was 𝑔3 (𝑛) ≪ 𝑛3/2 , but now the world record belongs to Joshua Zahl ([160]), as he has demonstrated that 3

1

𝑔3 (𝑛) ≪ 𝑛 2 − 394 +𝑜(1) . The unit distance problem remains one of the most important unsolved problems in combinatorics. See, for example, [1, 15, 93, 108, 117] and the references therein for current results on the problem and its variants.

2.2 Falconer distance problem In 1986, Falconer ([52]) considered a continuous version of the distance problem. While the Falconer distance problem is very similar to that of the Erdős distance problem, the Falconer distance problem requires a more technical setup, so these ideas should not be taken too seriously, especially if the notions of measure and fractal dimension are new to you. There are two main ideas necessary to understand the gist of the Falconer distance problem. The first idea is that of (Lebesgue) measure. The measure of a set 𝐸 ⊆ ℝ𝑑 is the quantity that mathematicians use which can loosely be thought of as length in ℝ, area in ℝ2 , volume in ℝ3 , and so on. Also, a set 𝐸 ⊆ ℝ𝑑 has measure 0 if for any 𝜖 > 0, the set can be covered by a countable (or finite) collection of open balls whose union has volume 𝜖. The only fact concerning measure which we will require here is that any countable set 𝐸 ⊆ ℝ𝑑 has measure 0 (see Exercise 2.5). The second idea to convey is that of Hausdorff dimension. We loosely describe dimension in the following intuitive way. If you were to take a line segment of length 1 and scale down the line segment by 1/2, you would unsurprisingly (and perhaps uninterestingly) get a line segment of length 1/2 = (1/2)1 . If you take a square of area 1, and scale down

26

Chapter 2. The distance problem

each side-length to half its size, the resulting square would be of area 1/4 = (1/2)2 . If you take a cube of area 1 and scale down each side to half its original size, you would make a cube with volume 1/8 = (1/2)3 . You can think of those powers in the terms (1/2)1 for the line segment, (1/2)2 for the square, and (1/2)3 for the cube as the reason why the line segment is 1-dimensional, the square is 2-dimensional, and the cube is 3-dimensional.

Now let’s consider the strange and wonderful fractal that is the Cantor set. The Cantor set 𝒞 is defined as follows: Start with the unit interval [0, 1] ⊆ ℝ, and remove the middle third (1/3, 2/3). You are left with only the union of two intervals [0, 1/3] ∪ [2/3, 1]. Now from each of those intervals, remove the middle third, to be left with a union of 4 intervals, each of length 1/9. Taking a limit of this process gives a fractal called the Cantor set1 :

Notice that if you scale down the Cantor set by a factor of 3, you have split the Cantor set into two identical copies of itself, each 1/3 the size of the original. This means that the dimension 𝑠 of the Cantor set should satisfy (1/3)𝑠 = 1/2. In this sense the Cantor set 𝒞 has fractal dimension ln(2)/ ln(3) ≈ 0.63093 . . . . 1 Image

taken from [156].

2.2. Falconer distance problem

27

While Hausdorff dimension is the notion of fractal dimension used in the Falconer distance problem, there are many other notions of fractal dimension, particularly Minkowski dimension, box-counting dimension, and packing dimension. We will write dim𝐻 (𝐸) to denote the Hausdorff dimension of the set 𝐸. See Section 7.3 for an introduction to the Minkowski dimension, and the interested reader should see [110] for an excellent treatise on fractal geometric including rigorous definitions of measure, fractal dimension, and much more. Let Δ(𝐸) be the distance set defined as before. Falconer showed that if 𝐸 ⊆ ℝ𝑑 is compact and the set 𝐸 has Hausdorff dimension dim𝐻 (𝐸) > 𝑑+1 , then the set Δ(𝐸) would have positive Lebesgue measure. Further2 𝑑

more, he showed that for any 𝑠 < 2 , there exists a set 𝐸 ⊆ ℝ𝑑 such that dim𝐻 (𝐸) = 𝑠, and yet the distance set Δ(𝐸) has Lebesgue measure zero. This led him to conjecture that if a set 𝐸 ⊆ ℝ𝑑 is compact with 𝑑 dim𝐻 (𝐸) > 2 , then the distance set Δ(𝐸) would have positive measure. The details of Falconer’s construction can be found in [53] or [110], for example. The Falconer distance problem can be thought of as a generalization of the Steinhaus Theorem ([53]), which states that if 𝐴 ⊆ ℝ is of positive (Lebesgue) measure, then the set 𝐴 − 𝐴 = {𝑎 − 𝑎′ ∶ 𝑎, 𝑎′ ∈ 𝐴} contains some interval of the form [0, 𝜀) for some 𝜀 > 0. The Falconer distance problem has received considerable attention, most notably from Wolff, Bourgain, and Erdoğan. For a long while, the best known result in the plane was due to Wolff ([157]) as he demonstrated that Δ(𝐸) has positive Lebesgue measure when dim𝐻 (𝐸) > 4/3 for all compact sets 𝐸 ⊆ ℝ2 , hence why we refer to the results in Chapter 4 as “Wolff’s Exponent.” Interestingly, Wolff’s bound was improved by Guth, Iosevich, Ou, and Wang ([69]) who demonstrated that if 𝐸 ⊆ ℝ2 is compact and has dim𝐻 (𝐸) > 5/4, then Δ(𝐸) has positive Lebesgue measure2 . If 𝐸 ⊆ ℝ2 is compact and also 𝑠-Ahlfors-David regular for some 𝑠 ≥ 1, then Orponen ([118]) has shown that Δ(𝐸) has packing dimension dim𝑝 (Δ(𝐸)) = 1. In higher dimensions the best known results currently belong to Du, Guth, Ou, Wang, Wilson, Zhang ([37]) for 𝑑 = 3 and Du and Zhang ([38] for 𝑑 ≥ 4. They have shown that if 𝐸 ⊆ ℝ3 satisfies dim𝐻 (𝐸) > 9/5 and if 𝐸 ⊆ ℝ𝑑 2 They actually proved a stronger so-called pinned-distance result, the likes of which we discuss in Section 5.4.

28

Chapter 2. The distance problem 𝑑

1

1

(for 𝑑 ≥ 4) satisfies dim𝐻 (𝐸) > 2 + 4 + 8𝑑−4 , then Δ(𝐸) has positive Lebesgue measure in each case. See [47, 110, 157] for a more detailed treatise of the subject.

2.3 Finite field distance problem Finite fields have long been used as an uncomplicated setting in which one could play with Euclidean problems in an environment with fewer technical difficulties. To this end the distance problem was considered in this finite setting in order to gain some insight into the Euclidean distances problems discussed above. A finite field version of the distance problem will be described below. We will always require the parameter 𝑞 to be odd unless explicitly stated otherwise. Let 𝔽𝑑𝑞 denote the 𝑑-dimensional vector space over the finite field with 𝑞 elements. Recall that 𝑞 must be the power of an odd prime3 . For 𝑥 = (𝑥1 , . . . , 𝑥𝑑 ) ∈ 𝔽𝑑𝑞 , we set ‖𝑥‖ = 𝑥 ⋅ 𝑥 = 𝑥𝑡 𝑥 = 𝑥12 + ⋯ + 𝑥𝑑2 ∈ 𝔽𝑞 , viewing 𝑥 as a column vector. The quantity ‖𝑥‖ is clearly not a norm in any analytic sense, though it should be observed that the function 𝑓(𝑥) = ‖𝑥‖ is preserved by rigid motions. More precisely, let 𝑂 𝑑 (𝔽𝑞 ) denote the set of 𝑑 × 𝑑 orthogonal matrices with entries in 𝔽𝑞 . Then, it is easy to check that ‖𝑥‖ = ‖𝑂𝑥‖ for any 𝑂 ∈ 𝑂 𝑑 (𝔽𝑞 ). As before, for 𝐸 ⊆ 𝔽𝑑𝑞 we define the distance set of 𝐸 as Δ(𝐸) = {‖𝑥 − 𝑦‖ ∶ 𝑥, 𝑦 ∈ 𝐸} ⊆ 𝔽𝑞 . This notion of distance is also preserved by rigid motions in 𝔽𝑑𝑞 . Example 2.3.1. Consider the set 𝔽213 consisting of the five points 𝐸 = {(4, 2); (4, 3); (5, 4); (7, 4); (3, 6)}. You can check that Δ(𝐸) = {0, 1, 2, 4, 5, 7, 8, 10} ⊆ 𝔽13 (see Figure 2.1). The first result on distances in finite fields came from Bourgain, Katz, and Tao ([14]) who showed the following. 3 If finite fields are new mathematical objects to you, then it may be helpful to first consider the case when 𝑞 is prime, as 𝔽𝑞 is indeed a finite field when 𝑞 is prime.

2.3. Finite field distance problem

29

Figure 2.1. The set 𝐸 from Example 2.3.1 represented by the set of integer points (𝑥, 𝑦) where 0 ≤ 𝑥, 𝑦 ≤ 12

Theorem 2.3.2. Suppose that 𝐸 ⊆ 𝔽𝑝2 , where 𝑝 ≡ 3 (mod 4) is a prime. Suppose that 𝐸 has cardinality |𝐸| = 𝑝𝛼 for some 0 < 𝛼 < 2. Then, there exists a quantity 𝜖 = 𝜖(𝛼) such that 1

|Δ(𝐸)| ≫ |𝐸| 2 +𝜖 . Note that if 𝛼 = 2, then |Δ(𝐸)| = |𝐸|1/2 , and no better. Notice also that if 𝑝 ≡ 1 (mod 4), then there exists an element 𝑖 ∈ 𝔽𝑝 such that 𝑖2 = −1. Hence, we can construct a set 𝐸 = {(𝑥, 𝑖𝑥) ∶ 𝑥 ∈ 𝔽𝑝 } such that Δ(𝐸) = {0}, and yet |𝐸| = 𝑝. This finite field distance problem was rephrased and improved upon by Iosevich and Rudnev ([87]). Rather than showing an Erdős-style bound of |Δ(𝐸)| > |𝐸|𝛽 , it is more fruitful to ask how large a set 𝐸 ⊆ 𝔽𝑑𝑞 needs to be in order to ensure that Δ(𝐸) = 𝔽𝑞 , or more modestly that |Δ(𝐸)| ≫ 𝑞. Therefore the finite field distance problem has flavors of both the Erdős distance problem and the Falconer distance problem, and is henceforth referred to as the Erdős-Falconer distance problem. Before we state the current conjecture, we first detail what progress has been made on this problem. Theorem 2.3.3 ([87]). Let 𝐸 ⊆ 𝔽𝑑𝑞 with |𝐸| > 2𝑞

𝑑+1 2

. Then, Δ(𝐸) = 𝔽𝑞 .

30

Chapter 2. The distance problem We will prove this theorem in Chapter 3. In the same chapter, we will 𝑑+1

give proofs of the weaker result |Δ(𝐸)| ≫ 𝑞 when |𝐸| ≫ 𝑞 2 , which turns out to have many applications to generalizations of the distance problem. Regarding counterexamples, notice that if 𝑞 = 𝑝2 , and since 𝔽𝑞 contains a subfield isomorphic to 𝔽𝑝 , then we have found a set 𝐸 ⊆ 𝔽𝑑𝑞 with |𝐸| = 𝑝𝑑 = 𝑞𝑑/2 , and yet |Δ(𝐸)| = 𝑝 = √𝑞, and no better. Even worse, adapting a counterexample from above, note that in any even dimension 𝑑 = 2𝑘, if −1 is a square in 𝔽𝑞 , then we could construct 𝐸 = {(𝑥1 , 𝑖𝑥1 , 𝑥2 , 𝑖𝑥2 , . . . , 𝑥𝑘 , 𝑖𝑥𝑘 ) ∶ 𝑥𝑖 ∈ 𝔽𝑞 } which gives a set of size |𝐸| = 𝑞𝑑/2 with Δ(𝐸) = {0}. Using these examples and the Falconer distance conjecture as a guide, it would be reasonable to assume that if |𝐸| ≥ 𝐶𝑞𝑑/2 for a sufficiently large constant 𝐶, then one achieves 𝔽𝑞 = Δ(𝐸). However, it was shown in [76] that the exponent

𝑑+1 2

is best possible at least in odd dimensions. That is, given 𝑑+1

where 𝑑 ≥ 3 is odd, we can construct 𝐸 ⊆ 𝔽𝑑𝑞 such that |𝐸| ≍ 𝑞 2 −𝜖 and yet |Δ(𝐸)| ≪ 𝑞1−𝜖 . In the Euclidean setting point-sets behave more or less the same regardless of the parity of the dimension of the ambient space. In finite fields, however, the parity of the dimension will play a much larger role. We will construct the above counterexample in Section 3.5. Finally, some partial improvement to Theorem 2.3.3 is known but only in the case 𝑑 = 2. Chapman, Erdoğan, Hart, Iosevich, and Koh ([17]), showed that if 𝐸 ⊆ 𝔽2𝑞 where 𝑞 ≡ 3 (mod 4) and |𝐸| ≥ 𝑞4/3 , then |Δ(𝐸)| ≥ 𝑐𝑞 for some 0 < 𝑐 ≤ 1. A further paper ([7]) was able to extend this result to the case 𝑞 ≡ 1 (mod 4). Combining these results, we have: 𝔽𝑑𝑞 ,

Theorem 2.3.4. Let 𝐸 ⊆ 𝔽2𝑞 such that |𝐸| ≥ 𝑞4/3 for a sufficiently large constant 𝐶. Then, there exists an absolute constant 𝑐 ∈ (0, 1] such that |Δ(𝐸)| ≥ 𝑐𝑞. While this shows that the distance set is large when 𝐸 is of sufficiently large cardinality (in fact we show that at the very least the distance set contains a positive proportion of the field 𝔽𝑞 ), we still do not know which elements of 𝔽∗𝑞 are in the distance set. Thus, if we care whether or not 1 ∈ Δ(𝐸), for example, then we must require that |𝐸| > 2𝑞

𝑑+1 2

.

2.3. Finite field distance problem

31

In an interesting twist Murphy and Petridis ([114]) demonstrated that the 4/3 threshold is best possible in 𝔽2𝑞 , at least for arbitrary finite fields when you want to guarantee that all elements are in the distance set Δ(𝐸). Specifically they construct a family of sets 𝐸 ⊆ 𝔽2𝑞 with |𝐸| = 𝑞4/3 such that |Δ(𝐸)| < 𝑞. Other results are known to hold for certain sets 𝐸 (such as Salem sets and subsets of spheres), and we collect such results in later chapters (see Subsection 3.3.1 and Theorem 5.3.6). Putting all of the results together, we now state two versions of the Erdős-Falconer distance conjecture. Conjecture 2.3.5 (Strong Erdős-Falconer Distance Conjecture). Let 𝐸 ⊆ 𝔽𝑑𝑞 , 𝑑 ≥ 4 even, be such that |𝐸| ≥ 𝐶𝑞𝑑/2 for a sufficiently large constant 𝐶. Then, Δ(𝐸) = 𝔽𝑞 . Conjecture 2.3.6 (Weak Erdős-Falconer Distance Conjecture). Let 𝐸 ⊆ 𝔽𝑑𝑞 , 𝑑 ≥ 2 even, be such that |𝐸| ≥ 𝐶𝑞𝑑/2 for a sufficiently large constant 𝐶. Then, |Δ(𝐸)| ≥ 𝑐𝑞 for some 0 < 𝑐 ≤ 1. In the following chapters, we will recount the progress made on the strong and weak versions of the Erdős-Falconer conjectures and its many variants thus far. Before we prove any finite field distance result, I want to emphasize the importance and utility of studying finite field analogues of classical (Euclidean) problems. While we have not seen all of these problems just yet, I will mention a strange intertwining of results between both the finite field and Euclidean worlds. The Kakeya conjecture (see Section 7.3) asserts that any set containing a line in every direction must be appropriately large. The finite field analogue, long thought to be a difficult analytical problem, turned out to have an easy (one paragraph) algebraic solution using the so-called algebraic method (see Theorem 7.3.9). This algebraic method turned out not to be useful for the original Kakeya problem, though it did inspire G. Elekes and M. Sharir to recast the Erdős distance problem in terms of algebraic operations (see Section 6.2). This recasting of the problem allowed for the aforementioned solution of the Erdős distance problem in dimension 𝑑 = 2 by GuthKatz, but it did not yield any gains in higher dimensions due to some annoying algebraic obstructions there. The progress made on the Erdős

32

Chapter 2. The distance problem

distance problem further helped to improve the Falconer distance problem in dimension 𝑑 = 2, but also did not yield any new results in higher dimensions. The Erdős distance problem remains open for 𝑑 ≥ 3, the Erdős-unit distance problem remains open in dimensions 2 and 3, the Falconer distance problem remains open in all dimensions 𝑑 ≥ 2, and the finite field distance problem is open for all even dimensions 𝑑 ≥ 2. At any step in this strange winding road of progress, it was not obvious that a solution to some problem was going to assist with the solution to another, but undoubtedly, finite fields have played a key role in better understanding some combinatorial problems, even in the Euclidean setting.

2.4 Exercises: Chapter 2 Exercise 2.1. Find a set 𝐸 ⊆ 𝔽23 such that |𝐸| = 3 and 1 ∉ Δ(𝐸). On the other hand, prove that any set 𝐸 ⊆ 𝔽23 with cardinality |𝐸| ≥ 4 satisfies Δ(𝐸) = 𝔽3 . Moreover, find a set 𝐸 ⊆ 𝔽25 such that |𝐸| = 10, and yet 1 ∉ Δ(𝐸). Then, prove that if 𝐸 ⊆ 𝔽25 has cardinality |𝐸| ≥ 11, then Δ(𝐸) = 𝔽5 . Exercise 2.2. Let 𝐵 = {𝑛 ∈ ℤ+ ∶ 𝑛 = 𝑥2 + 𝑦2 + 𝑧2 for some 𝑥, 𝑦, 𝑧 ∈ ℤ}, and consider its counting function 𝐵(𝑥) = {𝑛 ∈ 𝐵 ∶ 𝑛 ≤ 𝑥}. Provide a counting argument for: 𝐵(𝑥) 5 lim = . 6 𝑥→∞ 𝑥 Hint: Use the following heuristic: Legendre’s 3 squares theorem states that an integer is not a sum of three squares if and only if it is of the form 4𝑎 (8𝑏 + 7) for nonnegative integers 𝑎 and 𝑏. Finally, notice ∞

1 1 = . 𝑘 6 8 ⋅ 4 𝑘=0 ∑

Exercise 2.3. Show that if you take 𝐸 ⊆ 𝔽𝑑2 to be the set of vertices which have an even number of zero components, then |𝐸| = 2𝑑−1 and yet Δ(𝐸) = {0}. However, using the pigeonhole-principle, show that if |𝐸| > 2𝑑−1 , then Δ(𝐸) = 𝔽2 . This is why we emphasize that in the above theorems (and the text throughout), 𝑞 is assumed to be odd.

2.4. Exercises: Chapter 2

33

Exercise 2.4. Look up and write down any proof of Fermat’s Theorem on sums of two squares which states that an odd prime 𝑝 is a sum of two squares if and only if 𝑝 ≡ 1 (mod 4). Exercise 2.5. Prove that any countable set in ℝ𝑑 has measure 0, using the definitions laid out in Section 2.2. Exercise 2.6. Prove that every element in 𝔽𝑞 is the sum of two squares using the pigeonhole principle. Conclude that the square lattice construction in (2.3) does not yield a logarithmic loss in 𝔽2𝑞 .

10.1090/car/037/03

Chapter

3

The Iosevich-Rudnev bound 3.1 Counting-method The goal of this chapter is to understand two different proofs of Theorem 2.3.3. The first proof will show that every element from 𝔽𝑞 is in the distance set of 𝐸 so long as |𝐸| ≥ 4𝑞

𝑑+1 2

. The second proof will simply show

𝑑+1 2

that if |𝐸| ≥ 𝐶𝑞 for a sufficiently large constant 𝐶, then there exists a constant 𝑐 ∈ (0, 1] such that the distance set satisfies |Δ(𝐸)| ≥ 𝑐𝑞. Both proofs involve studying the counting function 𝜈(𝑡) = |{(𝑥, 𝑦) ∈ 𝐸 × 𝐸 ∶ ‖𝑥 − 𝑦‖ = 𝑡}| which counts the number of pairs of points which are distance 𝑡 apart from one another. Note that 𝑡 ∈ Δ(𝐸) if and only if 𝜈(𝑡) > 0. Thus, we aim to show simply that 𝜈(𝑡) > 0 for all 𝑡 ∈ 𝔽∗𝑞 , since 𝜈(0) > 0 trivially when 𝐸 ≠ ∅ as ‖𝑥 − 𝑥‖ = 0 for all 𝑥 ∈ 𝐸. Our first step is to convert 𝜈(𝑡) into a sum 𝜈(𝑡) =

∑

1,

(3.1)

𝑥,𝑦∈𝐸

‖𝑥−𝑦‖=𝑡

and we will study this sum in detail. We next use a technique of analytic number theory to turn this sum into an exponential sum. Recall we are working in the finite field 𝔽𝑞 , where 𝑞 = 𝑝𝑟 for some odd prime 𝑝. A group character is defined on any abelian group 𝐺 as a group homomorphism from that group 𝐺 to the set of complex numbers of modulus 1. In addition to additive characters on the (additive) group 35

36

Chapter 3. The Iosevich-Rudnev bound

(𝔽𝑞 , +), we will later encounter so-called multiplicative characters on the group (𝔽∗𝑞 , ⋅). The canonical additive character of 𝔽𝑑𝑞 is the function 𝜒 ∶ 𝔽𝑞 → ℂ defined by 𝜒(𝑘) = 𝑒2𝜋𝑖 Tr(𝑘)/𝑝 where Tr ∶ 𝔽𝑝𝑟 → 𝔽𝑝 is the linear function (called the field trace) given 2 𝑟−1 by Tr(𝑘) = 𝑘 + 𝑘𝑝 + 𝑘𝑝 + ⋯ + 𝑘𝑝 . Notice that when 𝑟 = 1 (that is, when 𝑞 itself is prime), Tr(𝑘) = 𝑘, so that 𝜒(𝑘) = 𝑒2𝜋𝑖𝑘/𝑞 . Example 3.1.1. For the finite field 𝔽5 , the canonical additive character is the function 𝜒 ∶ 𝔽5 → ℂ given by 𝜒(𝑥) = 𝑒2𝜋𝑖𝑥/5 . Notice that for 𝑥 ∈ 𝔽5 , any integer congruent to 𝑥 (mod 5) will yield the same value for 𝜒(𝑥). Thus, 𝜒(2) = 𝑒2𝜋𝑖⋅2/5 = 𝑒2𝜋𝑖⋅7/5 = 𝜒(7) = 𝑒4𝜋𝑖/5 . Example 3.1.2. Consider the finite field 𝔽9 as constructed in Example 1.2.6. We had 𝔽9 = 𝔽3 [𝛼] = {𝑥 + 𝛼𝑦 ∶ 𝑥, 𝑦 ∈ 𝔽3 and 𝛼2 = −1}. Now, the additive character on 𝔽9 is 𝜒(𝑥) = 𝑒2𝜋𝑖 Tr(𝑥)/3 where Tr ∶ 𝔽9 → 𝔽3 is defined as Tr(𝑥) = 𝑥 + 𝑥3 . One can compute Tr(0) = 0, Tr(1) = 2, Tr(2) = 1, and Tr(𝛼) = 𝛼(𝛼2 + 1) = 0. One can also compute (or show by linearity) that Tr(𝛼 + 1) = Tr(𝛼) + Tr(1) = 2, for example. It then follows that 𝜒(𝛼 + 1) = 𝑒2𝜋𝑖⋅Tr(𝛼+1)/3 = 𝑒2𝜋𝑖⋅2/3 = 𝑒4𝜋𝑖/3 = 𝜒(𝛼)𝜒(1). The linearity of the field trace is what allows 𝜒 to be a homomorphism (see Exercise 3.1). All additive characters on 𝔽𝑞 are of the form 𝜒𝑗 (𝑘) = 𝜒(𝑗𝑘) = 𝑒2𝜋𝑖𝑗 Tr(𝑘)/𝑝 for 𝑗 ∈ 𝔽𝑞 . The character 𝜒0 (𝑘) ≡ 1 is called the trivial character on 𝔽𝑞 , and all other characters are called nontrivial. All results mentioned in this book will be independent of which character is used (so long as we avoid the trivial character), so we will simply use the canonical additive character 𝜒(⋅) = 𝜒1 (⋅) throughout the text. For more information on characters in finite fields, see [104].

3.1. Counting-method

37

Characters satisfy the following orthogonality principle: 𝑞−𝑑 ∑ 𝜒(𝑥 ⋅ 𝑚) = { 𝑥∈𝔽𝑑 𝑞

1 0

𝑚=0 𝑚≠0

where for 𝑥, 𝑚 ∈ 𝔽𝑑𝑞 , the quantity 𝑥 ⋅ 𝑚 = 𝑥1 𝑚1 + . . . 𝑥𝑑 𝑚𝑑 denotes the standard inner product. In particular if 𝑑 = 1 we get ∑ 𝜒(𝑎𝑡) = { 𝑡∈𝔽𝑞

𝑞 0

𝑎=0 otherwise.

(3.2)

We denote the sphere of radius 𝑡 by 𝑆 𝑡 = {𝑥 ∈ 𝔽𝑑𝑞 ∶ ‖𝑥‖ = 𝑡}. We will see shortly that the sphere 𝑆 𝑡 behaves like a Euclidean sphere in ℝ𝑑 in many ways. For example two distinct spheres in 𝔽𝑑𝑞 intersect in at most 2 points, the sphere 𝑆 𝑡 is (𝑑 − 1)-dimensional, and the sphere 𝑆 𝑡 even satisfies the same so-called stationary phase estimates as Euclidean spheres. Spheres in 𝔽𝑑𝑞 can exhibit some strange behaviors, such as the existence of nontrivial spheres of radius 0 (as we saw in Chapter 1). For a set 𝐸, we will write 𝐸(⋅) to denote its characteristic function, so that we would write 𝐸(𝑥) = {

1 0

𝑥∈𝐸 𝑥 ∉ 𝐸.

We will use this characteristic function notation frequently, especially for spheres 𝑆 𝑡 . Orthogonality will allow us to turn the sum (3.1) into the following sum: 𝜈(𝑡) = ∑ 𝐸(𝑥)𝐸(𝑦)𝑆 𝑡 (𝑥 − 𝑦). (3.3) 𝑥,𝑦

We will study this sum by applying some techniques from harmonic analysis. For a function 𝑓 ∶ 𝔽𝑑𝑞 → ℂ, we define the normalized Fourier transform of 𝑓 as ˆ 𝑓(𝑚) = 𝑞−𝑑 ∑ 𝑓(𝑥)𝜒(−𝑥 ⋅ 𝑚), 𝑥∈𝔽𝑑 𝑞

where 𝜒 is the canonical additive character of 𝔽𝑞 discussed above, and where 𝑥 ⋅ 𝑚 again denotes the standard dot-product for 𝑥, 𝑚 ∈ 𝔽𝑑𝑞 . It turns that if 𝜒 is a nontrivial character on 𝔽, then all the characters on

38

Chapter 3. The Iosevich-Rudnev bound

𝔽𝑑𝑞 are of the form 𝜒(𝑥 ⋅ 𝑚) for some 𝑚 ∈ 𝔽𝑑𝑞 . This finite field Fourier transform is analogous to the Fourier transform often used in harmonic analysis, signal processing, modeling using PDEs, and engineering: ˆ 𝑓(𝑚) = ∫ 𝑓(𝑥)𝑒−𝜋𝑖𝑚 𝑓(𝑥 ⋅ 𝑚)𝑑𝑥. ℝ𝑑

The orthogonality of characters in finite fields leads us to the following results: Proposition 3.1.3 (Inversion). Let 𝑓 ∶ 𝔽𝑑𝑞 → ℂ. Then, ˆ 𝑓(𝑥) = ∑ 𝑓(𝑚)𝜒(𝑥 ⋅ 𝑚). 𝑚∈𝔽𝑑 𝑞

Proposition 3.1.4 (Plancherel). For a function 𝑓 ∶ 𝔽𝑑𝑞 → ℂ, we have 2

ˆ || = 𝑞−𝑑 ∑ |𝑓(𝑥)|2 . ∑ ||𝑓(𝑚) 𝑚∈𝔽𝑑 𝑞

𝑥∈𝔽𝑑 𝑞

We leave the proofs of these elementary Propositions to the reader (see Exercises 3.2 and 3.3). We are now ready to prove Theorem 2.3.3 for the first time. Applying inversion and Plancherel to the sum (3.3), and using that 𝜒 is a homomorphism, we see that 𝜈(𝑡) = ∑ 𝐸(𝑥)𝐸(𝑦)𝑆 𝑡 (𝑥 − 𝑦) 𝑥,𝑦∈𝔽𝑑 𝑞

=

𝐸(𝑥)𝐸(𝑦)𝑆ˆ𝑡 (𝑚)𝜒((𝑥 − 𝑦) ⋅ 𝑚)

∑ 𝑥,𝑦,𝑚∈𝔽𝑑 𝑞

= ∑ (∑ 𝐸(𝑦)𝜒(−𝑦 ⋅ 𝑚)) (∑ 𝐸(𝑥)𝜒(−𝑥 ⋅ 𝑚))𝑆ˆ𝑡 (𝑚). 𝑚∈𝔽𝑑 𝑞

𝑦

𝑥

Applying the definition of the Fourier transform, this becomes

ˆ 𝑆ˆ𝑡 (𝑚) ˆ 𝜈(𝑡) = ∑ 𝑞𝑑 𝐸(𝑚) ⋅ 𝑞𝑑 𝐸(𝑚) 𝑚 2

ˆ || 𝑆 𝑡̂ (𝑚). = 𝑞2𝑑 ∑ ||𝐸(𝑚) 𝑚

3.1. Counting-method

39

In our first step, we applied inversion on the sphere, and then in the next step, we summed in 𝑥 and 𝑦 to get the Fourier transform of the characteristic function of the set 𝐸. This establishes 2

ˆ || 𝑆 𝑡̂ (𝑚). 𝜈(𝑡) = 𝑞2𝑑 ∑ ||𝐸(𝑚)

(3.4)

𝑚

Now, 2

ˆ || 𝑆 𝑡̂ (0) + 𝑞2𝑑 𝜈(𝑡) = 𝑞2𝑑 ||𝐸(0)

2

ˆ || 𝑆 𝑡̂ (𝑚) ≔ 𝑀 + 𝑅𝑡 , (3.5) ||𝐸(𝑚)

∑ 𝑚≠(0,. . .,0)

where 𝑀 is the term corresponding to 𝑚 = (0, . . . , 0) and where 𝑅𝑡 is the sum of the nonzero values of 𝑚. Notice that by definition, we have ˆ . . . , 0) = 𝑞−𝑑 ∑ 𝐸(𝑥) = 𝑞−𝑑 |𝐸|, 𝐸(0, 𝑥∈𝔽𝑑 𝑞

and likewise 𝑆ˆ𝑡 (0, . . . , 0) = 𝑞−𝑑 |𝑆 𝑡 |. We next need to determine the cardinality of the sphere 𝑆 𝑡 ⊆ 𝔽𝑑𝑞 . Proposition 3.1.5. Let 𝑆 𝑡 ⊆ 𝔽𝑑𝑞 be as above. Then for 𝑑 ≥ 2, we have |𝑆 𝑡 | = {

𝑞𝑑−1 + 𝑞

𝑑−1 2

𝜂 ((−1)

𝑡)

𝑑 is odd

𝑞𝑑−1 + 𝑞

𝑑−2 2

𝜂 ((−1)𝑑/2 ) (𝛿(𝑡)𝑞 − 1)

𝑑 is even.

𝑑−1 2

Here we are using the notation 𝜂(𝑥) = 1 if 𝑥 ≠ 0 is a square in 𝔽𝑞 , 𝜂(𝑥) = −1 if 𝑥 is not a square in 𝔽𝑞 , and 𝜂(0) = 0. Also the quantity 𝛿 ∶ 𝔽𝑞 → {0, 1} is the 1-dimensional delta function 𝛿(𝑡) = {

1 0

𝑡=0 𝑡 ≠ 0.

In particular |𝑆 𝑡 | ≤ 2𝑞𝑑−1 independent of 𝑡, and |𝑆 𝑡 | = 𝑞𝑑−1 (1 + 𝑜(1))

when either 𝑡 ≠ 0 or 𝑑 ≥ 3.

In this sense, the sphere 𝑆 𝑡 ⊆ 𝔽𝑑𝑞 is (𝑑 − 1)-dimensional. Applying these bounds to the main term of our sum (3.5) we see that 𝑀 = 𝑞−𝑑 |𝐸|2 |𝑆 𝑡 | = 𝑞−𝑑 |𝐸|2 𝑞𝑑−1 (1 + 𝑜(1)) =

|𝐸|2 (1 + 𝑜(1)), 𝑞

(3.6)

by Proposition 3.1.5. Finally, we estimate the remainder, though first we need to more closely examine the quantity 𝑆 𝑡̂ (𝑚).

40

Chapter 3. The Iosevich-Rudnev bound

Proposition 3.1.6. Let 𝑆 𝑡 (⋅) denote the characteristic function of the set 𝑆 𝑡 , and let 𝛿(𝑚) = {

1 0

𝑚 = (0, . . . , 0) 𝑚 ≠ (0, . . . , 0).

Then, 𝑆 𝑡̂ (𝑚) =

𝑑+2 ‖𝑚‖ 𝛿(𝑚) + 𝜀𝑑𝑞 𝑞− 2 ∑ 𝜒 (−𝑡𝑠 − ) 𝑞 4𝑠 𝑠≠0

where for odd prime powers 𝑞 = 𝑝ℓ , (−1)ℓ−1

𝑝≡1

(mod 4)

(−1)ℓ−1 𝑖ℓ

𝑝≡3

(mod 4).

𝜀𝑞 = { In particular for 𝑚 ≠ (0, . . . , 0) we have |𝑆 0̂ (𝑚)| ≤ 𝑞−𝑑/2 , while for 𝑡 ≠ 0 and 𝑚 ≠ (0, . . . , 0): ||𝑆ˆ𝑡 (𝑚)|| ≤ 2𝑞−

𝑑+1 2

.

We will delay the proofs Propositions 3.1.5 and 3.1.6 until Section 3.4 later in the chapter. For now, just implementing Proposition 3.1.6, we see that the remainder term from (3.5) satisfies |𝑅𝑡 | ≤ 𝑞2𝑑

2

ˆ || ||𝑆ˆ𝑡 (𝑚)|| ||𝐸(𝑚)

∑ 𝑚≠(0,. . .,0)

≤ max |𝑆 𝑡̂ (𝑚)| 𝑞2𝑑 𝑡≠0,𝑚≠0⃗

≤ 2𝑞

−

𝑑+1 2

ˆ || ||𝐸(𝑚)

∑

2

𝑚≠(0,. . .,0)

𝑞2𝑑

∑

2

ˆ || . ||𝐸(𝑚)

𝑚≠(0,. . .,0)

We can bound the sum over all nonzero 𝑚 to that of all 𝑚 yielding |𝑅𝑡 | ≤ 2𝑞−

𝑑+1 2

2

ˆ || . 𝑞2𝑑 ∑ ||𝐸(𝑚) 𝑚∈𝔽𝑑 𝑞

3.2. The 𝐿2 -method

41

Applying Plancherel’s identity (Proposition 3.1.4), we see that |𝑅𝑡 | ≤ 2𝑞−

𝑑+1 2

2

𝑞2𝑑 𝑞−𝑑 ∑ |𝐸(𝑥)| 𝑥

= 2𝑞

𝑑−1 2

|𝐸|.

Putting together all of our estimates, we have seen that for 𝑡 ≠ 0, we have the estimate |𝐸|2 (1 + 𝑜(1)) + 𝑅𝑡 , 𝑞

𝜈(𝑡) =

where |𝑅𝑡 | ≤ 2𝑞

𝑑−1 2

(3.7)

|𝐸|. It follows that 𝜈(𝑡) > 0 whenever 𝑑−1 |𝐸|2 (1 + 𝑜(1)) > 2𝑞 2 |𝐸|, 𝑞

which is certain to happen if |𝐸| ≥ 4𝑞 Theorem 2.3.3.

𝑑+1 2

. This finishes our first proof of

Note 3.1.7. In the proof, we used only that |𝑆 𝑡 | = 𝑞𝑑−1 (1 + 𝑜(1)). One can go more carefully through the previous argument using the precise values for the size of the sphere, and it turns out that 𝜈(𝑡) > 0 whenever |𝐸| > 2𝑞

𝑑+1 2

(see Exercise 3.16).

3.2 The 𝐿2 -method As before set 𝜈(𝑡) = |{(𝑥, 𝑦) ∈ 𝐸 × 𝐸 ∶ ‖𝑥 − 𝑦‖ = 𝑡}|, and we note that ∑ 𝜈(𝑡) = |𝐸|2 . 𝑡

42

Chapter 3. The Iosevich-Rudnev bound

Applying Cauchy-Schwarz to this sum, and letting 1∆(𝐸) (⋅) denote the characteristic function of the distance set Δ(𝐸), we see that 2

2

4

|𝐸| = ( ∑ 𝜈(𝑡)) = ( ∑ 1∆(𝐸) (𝑡) ⋅ 𝜈(𝑡)) 𝑡∈𝔽𝑞

𝑡∈𝔽𝑞

≤ ∑ 12∆(𝐸) (𝑡) ⋅ ∑ 𝜈2 (𝑡) 𝑡∈𝔽𝑞

𝑡∈𝔽𝑞 2

= |Δ(𝐸)| ∑ 𝜈 (𝑡) 𝑡∈𝔽𝑞

which immediately yields |Δ(𝐸)| ≥

|𝐸|4 . ∑𝑡 𝜈2 (𝑡)

(3.8)

This method has some flexibility. For example we could also write: 2

2

4

|𝐸| = (∑ 𝜈(𝑡) + 𝜈(0)) ≤ 2 (∑ 𝜈(𝑡)) + 2𝜈(0)2 𝑡≠0

𝑡≠0

≤ 2|Δ(𝐸)| ∑ 𝜈2 (𝑡) + 2𝜈2 (0) 𝑡≠0

which implies that |Δ(𝐸)| ≥

1 |𝐸|4 2

− 𝜈2 (0)

∑𝑡≠0 𝜈2 (𝑡)

.

(3.9)

Note that we used the trivial inequality (𝑎 + 𝑏)2 ≤ 2(𝑎2 + 𝑏2 ). We will choose whether we need (3.8) or (3.9) or other variants of this idea depending on each individual application of the method. Thus all of the work involved with showing that a set 𝐸 has a large distance set is in finding an upper bound on the quantities ∑𝑡 𝜈2 (𝑡) and ∑𝑡≠0 𝜈2 (𝑡) (hence why we call this the “𝐿2 -method”).

3.3 Finite field spherical averages We next turn to studying a finite field quantity involving so-called spherical averages.

3.3. Finite field spherical averages

43

Theorem 3.3.1. Let 2

2

ˆ || = ∑ ||𝐸(𝑚) ˆ || 𝑆 𝑡 (𝑚), 𝜎𝐸 (𝑡) = ∑ ||𝐸(𝑚) ‖𝑚‖=𝑡

𝑚∈𝔽𝑑 𝑞

and define 𝑀𝐸 (𝑞) =

𝑞3𝑑+1 ∑ 𝜎2 (𝑡). |𝐸|4 𝑡∈𝔽∗ 𝐸 𝑞

𝑑 2

Then for |𝐸| ≥ 4𝑞 , we have |Δ(𝐸)| ≥ min(𝑐𝑞,

𝑞 ). 𝑀𝐸 (𝑞)

In particular for sets 𝐸 satisfying |𝐸| ≥ 4𝑞𝑑/2 , we will have |Δ(𝐸)| ≫ 𝑞 whenever 𝑀𝐸 (𝑞) ≫ 1. The quantity 𝜎𝐸 (𝑡) is called a spherical average since the sum is taken over the sphere of radius 𝑡. We will follow the main idea of the original proof from [87]. Recall that we already know (see Note 3.1.7) that if |𝐸| > 2𝑞

𝑑+1 2

, then Δ(𝐸) = 𝔽𝑞 , so we may assume that the cardinality of 𝑑

𝐸 satisfies 4𝑞 2 ≤ |𝐸| ≤ 2𝑞 𝜈(0) =

𝑑+1 2

. Note that

2 |𝐸|2 |𝑆 0 | ˆ0 (𝑚) ˆ || 𝑆 ||𝐸(𝑚) ∑ + 𝑞2𝑑 𝑞𝑑 𝑚≠(0,. . .,0)

so that |𝜈(0)| ≤ 2 since |𝐸| ≤ 2𝑞

𝑑+1 2

𝑑 𝑑 |𝐸|2 + 𝑞 2 |𝐸| ≤ 2𝑞 2 |𝐸| 𝑞

. Hence 𝜈2 (0) ≤ 4𝑞𝑑 |𝐸|2 ≤

1 4 |𝐸| 4

𝑑

when |𝐸| ≥ 4𝑞 2 . Note that the constant 4 can be improved though we make no attempt to do so here. Combing this with (3.9) we have established that |𝐸|4 . |Δ(𝐸)| ≥ (3.10) 4 ∑𝑡≠0 𝜈2 (𝑡) To prove Theorem 3.3.1, we start with some preliminary lemmas.

44

Chapter 3. The Iosevich-Rudnev bound

Proposition 3.3.2. For 𝑚 ∈ 𝔽𝑑𝑞 and 𝑘 ∈ 𝔽𝑞 , define 𝐺(𝑚, 𝑘) = ∑ 𝜒(𝑘‖𝑥‖ + 𝑚 ⋅ 𝑥). 𝑥∈𝔽𝑑 𝑞

Then 𝜀𝑑𝑞 𝑞𝑑/2 𝜒 (‖𝑚‖(4𝑘)−1 )

𝑘≠0

𝛿(𝑚)𝑞𝑑

𝑘=0

𝐺(𝑚, 𝑘) = {

where 𝜀𝑞 ∈ {±1, ±𝑖} is independent of 𝑘 and 𝑚, and where 𝛿 is the delta function from above. Proof. The proof follows from the orthogonality of characters when 𝑘 = 0. When 𝑘 ≠ 0, the result follows by completing the square and using the well known value of the so-called Gauss sum: 𝑔𝑘 ≔ ∑ 𝜒(𝑘𝑥2 ) = 𝜀𝑞 √𝑞.

(3.11)

𝑥∈𝔽𝑞

Our first step is to break this sum into the sums of the components of the vector 𝑥 = (𝑥1 , . . . , 𝑥𝑑 ): 𝐺(𝑚, 𝑘) = ∑ 𝜒(𝑘‖𝑥‖ + 𝑚 ⋅ 𝑥) 𝑥∈𝔽𝑑 𝑞

= ∑ 𝜒(𝑘𝑥12 + ⋯ + 𝑘𝑥𝑑2 + 𝑚1 𝑥1 + ⋯ + 𝑚𝑑 𝑥𝑑 ) 𝑥∈𝔽𝑑 𝑞 𝑑

= ∏ ∑ 𝜒(𝑘𝑥𝑖2 + 𝑚𝑖 𝑥𝑖 ). 𝑖=1 𝑥𝑖 ∈𝔽𝑞

Completing the square is only possible in fields of odd order since we require 2 ∈ 𝔽𝑞 to be invertible. To simplify notation we will write write 𝑎 to denote the quantity 𝑎𝑏−1 where 𝑏−1 is the multiplicative inverse of 𝑏 𝑏 in the underlying field. For example in 𝔽11 , we have

1 2

= 6. Recall that

3.3. Finite field spherical averages

45

we are assuming 𝑘 ≠ 0. Now in any field of odd order we have 𝑚 𝑘𝑥𝑖2 + 𝑚𝑖 𝑥𝑖 = 𝑘 (𝑥𝑖2 + 𝑖 𝑥𝑖 ) 𝑘 𝑚2 𝑚2 𝑚 = 𝑘 (𝑥𝑖2 + 𝑖 𝑥𝑖 + 𝑖2 − 𝑖2 ) 𝑘 4𝑘 4𝑘 𝑚𝑖 2 𝑚2𝑖 . ) − 𝑘 4𝑘 Thus our expression for 𝐺(𝑚, 𝑘) becomes = 𝑘 (𝑥𝑖 +

𝑑

𝐺(𝑚, 𝑘) = ∏ ∑ 𝜒 (𝑘 (𝑥𝑖 + 𝑖=1 𝑥𝑖 ∈𝔽𝑞 𝑑

= ∏ 𝜒 (− 𝑖=1

𝑚2 𝑚𝑖 2 ) ) 𝜒 (− 𝑖 ) 𝑘 4𝑘

𝑚2𝑖 𝑚 2 ) ∑ 𝜒 (𝑘 (𝑥𝑖 + 𝑖 ) ) 4𝑘 𝑥 ∈𝔽 𝑘 𝑞

𝑖

𝑑

= 𝜒(‖𝑚‖/4𝑘)(𝑔𝑘 ) . Applying (3.11), the expression for 𝐺(𝑚, 𝑘) follows. To establish (3.11), we first write |𝑔𝑘 |2 = ∑ 𝜒(𝑘(𝑥2 − 𝑦2 )) 𝑥,𝑦∈𝔽𝑞

= ∑ 𝜒(𝑘𝑡)𝐻(𝑡) 𝑡∈𝔽𝑞

where 𝐻(𝑡) = |{(𝑥, 𝑦) ∈ 𝔽𝑞 ∶ 𝑥2 − 𝑦2 = 𝑡}|. A simple counting argument (see Exercise 3.4) shows that 𝐻(0) = 2𝑞 − 1, while 𝐻(𝑡) = 𝑞 − 1 for 𝑡 ≠ 0. It follows that |𝑔𝑘 |2 = (2𝑞 − 1) + (𝑞 − 1) ∑ 𝜒(𝑘𝑡) = 𝑞.

(3.12)

𝑡≠0

Next, note that 𝑔𝑘 = ∑ 𝜒(𝑘𝑥2 ) = ∑ 𝜒(𝑧)𝜆(𝑧), 𝑥∈𝔽𝑞

𝑧∈𝔽𝑞

where 𝜆(𝑧) = |{𝑥 ∈ 𝔽𝑞 ∶ 𝑘𝑥2 = 𝑧}| = 1 + 𝜂(𝑧𝑘−1 ) = 1 + 𝜂(𝑘)𝜂(𝑧), since 𝜂(𝑘−1 ) = 𝜂(𝑘) for 𝑘 ≠ 0. This shows that 𝑔𝑘 = ∑ 𝜒(𝑧)(1 + 𝜂(𝑘)𝜂(𝑧)) = 𝜂(𝑘) ∑ 𝜒(𝑥)𝜂(𝑥). 𝑧∈𝔽𝑞

𝑥∈𝔽𝑞

(3.13)

46

Chapter 3. The Iosevich-Rudnev bound

Using (3.13), we obtain 𝑔𝑘 = 𝑔−𝑘 = 𝜂(−𝑘) ∑ 𝜒(𝑥)𝜂(𝑥) = 𝜂(−1)𝑔𝑘 .

(3.14)

𝑥∈𝔽𝑞

Putting together (3.12) and (3.14), we see that 𝑔2𝑘 = 𝑔𝑘 𝜂(−1)𝑔𝑘 = 𝜂(−1)|𝑔𝑘 |2 = 𝜂(−1)𝑞, and the result follows by taking square roots. It should be noted that precisely determining the signs of the quantity 𝜀𝑞 is much more delicate than one first imagines, and such details can be found, for example, in [104]. Proposition 3.3.3. Fix 𝑥 ∈ 𝔽𝑑𝑞 . Consider 𝑢𝑥 ∶ 𝔽𝑞 → ℂ given by 𝑢𝑥 (𝑗) = 𝑆𝑗 (𝑥). Then, 1 𝑢𝑥̂ (𝑘) = 𝜒(−𝑘‖𝑥‖). 𝑞 The proof is again straight-forward though it uses the orthogonality relation 𝑢𝑥 (𝑗) = 𝑞−1 ∑ 𝜒(−𝑠(‖𝑥‖ − 𝑗)). 𝑠

The details are left as an exercise (Exercise 3.11). Now, we put together these estimates. Recall (see (3.4)) that for 𝑡 ∈ 𝔽𝑞 we have 2

ˆ || 𝑆 𝑡̂ (𝑚). 𝜈(𝑡) = 𝑞2𝑑 ∑ ||𝐸(𝑚) 𝑚∈𝔽𝑑 𝑞

Taking the (1-dimensional) Fourier transform of 𝜈 and applying the above expression for 𝜈 gives 𝜈(𝑘) ̂ = 𝑞−1 ∑ 𝜈(𝑗)𝜒(−𝑗𝑘) 𝑗∈𝔽𝑞 2

ˆ || 𝑆𝑗̂ (𝑚)𝜒(−𝑗𝑘) = 𝑞2𝑑−1 ∑ ∑ ||𝐸(𝑚) 𝑗∈𝔽𝑞 𝑚∈𝔽𝑑 𝑞

= 𝑞𝑑−1 ∑

2

ˆ || 𝑆𝑗 (𝑥)𝜒(−𝑚 ⋅ 𝑥)𝜒(−𝑗𝑘) ∑ ||𝐸(𝑚)

𝑗∈𝔽𝑞 𝑚,𝑥∈𝔽𝑑 𝑞

3.3. Finite field spherical averages

47

where in the last step we simply applied the definition of the (𝑑-dimensional) Fourier transform of the indicator function 𝑆 𝑡 . Applying orthogonality we have 𝜈(𝑘) ̂ = 𝑞𝑑−2 ∑

2

ˆ || 𝜒(−𝑚 ⋅ 𝑥)𝜒(𝑠(‖𝑥‖ − 𝑗))𝜒(−𝑗𝑘) ∑ ||𝐸(𝑚)

𝑠,𝑗∈𝔽𝑞 𝑚,𝑥∈𝔽𝑑 𝑞 2

ˆ || 𝜒(−𝑚 ⋅ 𝑥 + 𝑠‖𝑥‖). = 𝑞𝑑−2 ∑ [ ∑ 𝜒(−𝑗(𝑘 + 𝑠))] ∑ ||𝐸(𝑚) 𝑠∈𝔽𝑞

𝑗∈𝔽𝑞

𝑚,𝑥∈𝔽𝑑 𝑞

Orthogonality implies that the sum inside the brackets is 0 unless 𝑠 = −𝑘. Hence 2

ˆ || 𝜒(−𝑚 ⋅ 𝑥)𝜒(−𝑘‖𝑥‖) 𝜈(𝑘) ̂ = 𝑞𝑑−1 ∑ ||𝐸(𝑚) 𝑚,𝑥∈𝔽𝑑 𝑞 2

ˆ || 𝐺(−𝑚, −𝑘) = 𝑞𝑑−1 ∑ ||𝐸(𝑚) 𝑚∈𝔽𝑑 𝑞

where 𝐺(−𝑚, −𝑘) is the sum appearing in Proposition 3.3.2. Applying our bound for 𝐺(−𝑚 − 𝑘), we see that 2

3

ˆ || 𝜒 (− 𝜈(𝑘) = 𝜀𝑑𝑞 𝑞 2 𝑑−1 ∑ ||𝐸(𝑚) 𝑚∈𝔽𝑑 𝑞

‖𝑚‖ ) 4𝑘

(3.15)

so long as 𝑘 = 0. Note that 𝜈(0) ̂ = |𝐸|2 /𝑞. Applying the above expression for 𝜈(𝑘), ̂ we have that 2

∑ |𝜈(𝑘)| ̂ = 𝑞3𝑑−2 𝑘≠0

∑

2

2

ˆ || ||𝐸(𝑚 ˆ ′ )|| ∑ 𝜒 ( ||𝐸(𝑚) 𝑘≠0

𝑚,𝑚′ ∈𝔽𝑑 𝑞

= 𝑞3𝑑−2

∑ 𝑚,𝑚′ ∈𝔽𝑑 𝑞

2

2

‖𝑚′ ‖ − ‖𝑚‖ ) 4𝑘

ˆ || ||𝐸(𝑚 ˆ ′ )|| ∑ 𝜒 (𝑘 (‖𝑚′ ‖ − ‖𝑚‖)) ||𝐸(𝑚) 𝑘≠0

48

Chapter 3. The Iosevich-Rudnev bound

by the change of variables 𝑘 ↦ 1/(4𝑘). Thus, 2

∑ |𝜈(𝑘)| ̂ = 𝑞3𝑑−2 𝑘≠0

2

2

ˆ || ||𝐸(𝑚 ˆ ′ )|| ∑ 𝜒 (𝑘 (‖𝑚′ ‖ − ‖𝑚‖)) ||𝐸(𝑚)

∑

𝑘≠0

𝑚,𝑚′ ∈𝔽𝑑 𝑞

= 𝑞3𝑑−2

2

2

2

2

ˆ ′ )|| ∑ 𝜒 (𝑘 (‖𝑚′ ‖ − ‖𝑚‖)) ˆ || ||𝐸(𝑚 ||𝐸(𝑚)

∑

𝑘

𝑚,𝑚′ ∈𝔽𝑑 𝑞

− 𝑞3𝑑−2

ˆ || ||𝐸(𝑚 ˆ ′ )|| . ||𝐸(𝑚)

∑ 𝑚,𝑚′ ∈𝔽𝑑 𝑞

By Plancherel the second sum is simply 𝑞3𝑑−2

∑

2

2

2

ˆ || ||𝐸(𝑚 ˆ ′ )|| = 𝑞3𝑑−2 (|𝐸|/𝑞𝑑 ) = |𝐸|2 𝑞𝑑−2 . ||𝐸(𝑚)

𝑚,𝑚′ ∈𝔽𝑑 𝑞

Applying orthogonality yet again, we see that ∑ 𝜒 (𝑘 (‖𝑚′ ‖ − ‖𝑚‖)) = 𝑞 𝑘∈𝔽𝑞

so long as ‖𝑚‖ = ‖𝑚′ ‖, and the sum is zero otherwise. Thus, 2

∑ |𝜈(𝑘)| ̂ = −|𝐸|2 𝑞𝑑−2 + 𝑞3𝑑−1

2

2

ˆ ′ )|| . ˆ || ||𝐸(𝑚 ||𝐸(𝑚)

∑ ‖𝑚‖=‖𝑚′ ‖

𝑘≠0

Now, recall that the spherical average is given by 2

ˆ || . 𝜎𝐸 (𝑡) = ∑ ||𝐸(𝑚) ‖𝑚‖=𝑡

In particular, ∑ 𝜎𝐸2 (𝑡) = ∑

2

2

ˆ || ||𝐸(𝑚 ˆ ′ )|| ||𝐸(𝑚)

∑

𝑡∈𝔽𝑞 ‖𝑚‖=𝑡,‖𝑚′ ‖=𝑡

𝑡∈𝔽𝑞

=

∑ ‖𝑚‖=‖𝑚′ ‖

2

2

ˆ || ||𝐸(𝑚 ˆ ′ )|| . ||𝐸(𝑚)

(3.16)

3.3. Finite field spherical averages

49

Plugging this in yields 2

∑ |𝜈(𝑘)| ̂ = 𝑞3𝑑−1 ∑ 𝜎𝐸2 (𝑡) − 𝑞𝑑−2 |𝐸|2 𝑘≠0

𝑡

=𝑞

3𝑑−1

∑ 𝜎𝐸2 (𝑡) − 𝑞𝑑−2 |𝐸|2 + 𝑞3𝑑−1 𝜎𝐸2 (0) 𝑡≠0

|𝐸|4 = 2 𝑀𝐸 (𝑞) − 𝑞𝑑−2 |𝐸|2 + 𝑞3𝑑−1 𝜎𝐸2 (0) 𝑞 where 𝑀𝐸 (𝑞) is the quantity appearing in Theorem 3.3.1. Using Plancherel once again: 2

∑ 𝜈2 (𝑡) = 𝑞 ∑ |𝜈(𝑘)| ̂ 𝑡∈𝔽𝑞

𝑘∈𝔽𝑞

= 𝑞( =

|𝐸|4 2 + ∑ |𝜈(𝑘)| ̂ ) 𝑞2 𝑘≠0

|𝐸|4 |𝐸|4 + 𝑀𝐸 (𝑞) − 𝑞𝑑−1 |𝐸|2 + 𝑞3𝑑 𝜎𝐸2 (0). 𝑞 𝑞

Peeling off the term 𝜈(0), we see that ∑ 𝜈2 (𝑗) = 𝑗≠0

|𝐸|4 |𝐸|4 − |𝐸|2 𝑞𝑑−1 + 𝑞3𝑑 𝜎𝐸2 (0) − 𝜈2 (0) + 𝑀𝐸 (𝑞). 𝑞 𝑞

Finally we need to establish that −|𝐸|2 𝑞𝑑−1 + 𝑞3𝑑 𝜎𝐸2 (0) − 𝜈2 (0) ≪

|𝐸|4 , 𝑞

which follows immediately if 𝑞3𝑑 𝜎𝐸2 (0) ≪

|𝐸|4 . 𝑞

To this end, note that 2

ˆ || 𝑆 0 (𝑚) 𝜎𝐸 (0) = ∑ ||𝐸(𝑚) 𝑚∈𝔽𝑑 𝑞

= 𝑞−2𝑑

∑

𝐸(𝑥)𝐸(𝑦)𝜒(−𝑥 ⋅ 𝑚)𝜒(𝑦 ⋅ 𝑚)𝑆 0 (𝑥)

𝑥,𝑦,𝑚∈𝔽𝑑 𝑞

= 𝑞𝑑 ∑ 𝐸(𝑥)𝐸(𝑦)𝑆 0̂ (𝑥 − 𝑦). 𝑥,𝑦∈𝔽𝑑 𝑞

(3.17)

50

Chapter 3. The Iosevich-Rudnev bound Next, by Proposition 3.1.6, we have 𝑆 0̂ (𝑥 − 𝑦) = 𝑞−1 𝛿(𝑥 − 𝑦) + 𝜀𝑑𝑞 𝑞−

𝑑+2 2

∑ 𝜒 (− 𝑠≠0

𝑑+2 − 2

= 𝑞−1 𝛿(𝑥 − 𝑦) + 𝜀𝑑𝑞 𝑞

‖𝑥 − 𝑦‖ ) 4𝑠

∑ 𝜒 (𝑠‖𝑥 − 𝑦‖) 𝑠≠0

= 𝑞−1 𝛿(𝑥 − 𝑦) + 𝜀𝑑𝑞 𝑞−

𝑑+2 2

∑ 𝜒 (𝑠‖𝑥 − 𝑦‖) − 𝜀𝑑𝑞 𝑞−

𝑑+2 2

𝑠∈𝔽𝑞 𝑑+2 − 2

= 𝑞−1 𝛿(𝑥 − 𝑦) + 𝜀𝑑𝑞 𝑞

⋅ 𝑞𝛿(‖𝑥 − 𝑦‖) − 𝜀𝑑𝑞 𝑞−

𝑑+2 2

by orthogonality. Note that 𝛿(𝑥 − 𝑦) is the 𝑑-dimensional delta function, while 𝛿(‖𝑥 − 𝑦‖) denotes the one-dimensional delta function. Our expression for 𝜎𝐸 (0) is thus 𝜎𝐸 (0) = 𝐼 + 𝐼𝐼 + 𝐼𝐼𝐼, where 𝐼 = 𝑞−𝑑−1 ∑ 𝛿(𝑥 − 𝑦) = 𝑞−𝑑−1 |𝐸|, 𝑥,𝑦∈𝔽𝑑 𝑞

𝐼𝐼 = 𝜀𝑑𝑞 𝑞−

3𝑑 2

∑

3𝑑

𝐸(𝑥)𝐸(𝑦) = 𝜀𝑑𝑞 𝑞− 2 𝜈(0), and

𝑥,𝑦∈𝔽𝑑 𝑞

‖𝑥−𝑦‖=0

𝐼𝐼𝐼 = 𝜀𝑑𝑞 𝑞−

3𝑑+2 2

∑ 𝐸(𝑥)𝐸(𝑦) = 𝜀𝑑𝑞 𝑞−

3𝑑+2 2

|𝐸|2 .

𝑥,𝑦∈𝔽𝑑 𝑞

Note that for each expression we have |𝐼| = 𝑞−𝑑−1 |𝐸| ≤ |𝐸|2 𝑞− 3𝑑

|𝐼𝐼| = 𝑞− 2 𝜈(0) ≤ |𝐸|2 𝑞− |𝐼𝐼𝐼| = 𝑞

3𝑑+2 − 2

|𝐸|2 ≤ |𝐸|2 𝑞

3𝑑+1 2 3𝑑+1 2

, and

3𝑑+1 − 2

,

using the trivial bound 𝜈(0) ≤ |𝐸|2 and recalling that we are assuming 4𝑞𝑑/2 ≤ |𝐸| ≤ 2𝑞

𝑑+1 2

. Finally it follows that

𝑞3𝑑 𝜎𝐸2 (0) ≪ 𝑞3𝑑 (|𝐸|2 𝑞−

3𝑑+1 2

2

) =

|𝐸|4 , 𝑞

3.3. Finite field spherical averages

51

which implies the following bound for (3.17): ∑ 𝜈2 (𝑗) ≪ 𝑗≠0

|𝐸|4 |𝐸|4 + 𝑀𝐸 (𝑞). 𝑞 𝑞

By (3.10) we have achieved the lower bound |Δ(𝐸)| ≫

|𝐸|4 |𝐸|4 𝑞

𝑀𝐸 (𝑞) +

|𝐸|4 𝑞

≫

𝑞 . 𝑀𝐸 (𝑞) + 1

This completes the proof of Theorem 3.3.1 since if 𝑀𝐸 (𝑞) ≪ 1, then |Δ(𝐸)| ≫ 𝑞.

3.3.1 Distance results for Salem sets In general, Salem sets are sets whose characteristic functions have maximal Fourier decay. Suppose that for 𝑚 ≠ (0, . . . , 0) we have some bound of the form ˆ || ≪ 𝑞𝛼 |𝐸|𝛽 . ||𝐸(𝑚) It follows that 2

ˆ || ≪ 𝑞𝑑+2𝛼 |𝐸|2𝛽 + 𝑞−𝑑 |𝐸| = ∑ ||𝐸(𝑚) 𝑚

|𝐸|2 𝑞2𝑑

which implies that 𝑞𝑑+2𝛼 |𝐸|2𝛽 ≫

|𝐸| |𝐸|2 |𝐸| − 2𝑑 ≥ 𝑑 . 𝑞𝑑 𝑞 𝑞

In general the best we can hope for is 𝛽 = 1/2 and 𝛼 = −𝑑. This leads us to the following definition. Definition 3.3.4. A set 𝐸 ⊆ 𝔽𝑑𝑞 is a Salem set if the bound ˆ || ≪ 𝑞−𝑑 √|𝐸| ||𝐸(𝑚) holds for all 𝑚 ∈ 𝔽𝑑𝑞 ⧵ {(0, . . . , 0)}. Theorem 3.3.5. If 𝐸 ⊆ 𝔽𝑑𝑞 is a Salem set with |𝐸| ≥ 𝐶𝑞𝑑/2 for a sufficiently large constant 𝐶, then |Δ(𝐸)| ≫ 𝑞.

52

Chapter 3. The Iosevich-Rudnev bound

Proof. It is enough to show 𝑀𝐸 (𝑞) ≪ 1 by Theorem 3.3.1. Now, 2

ˆ || 𝑆 𝑡 (𝑚) 𝜎𝐸 (𝑡) = ∑ ||𝐸(𝑚) 𝑚∈𝔽𝑑 𝑞

≪ 𝑞−2𝑑 |𝐸| ∑ 𝑆 𝑡 (𝑚) 𝑚∈𝔽𝑑 𝑞

= 𝑞−2𝑑 |𝐸||𝑆 𝑡 |. Hence, 𝑀𝐸 (𝑞) =

𝑞3𝑑+1 ∑ 𝜎2 (𝑡) |𝐸|4 𝑡≠0 𝐸

≪ 𝑞3𝑑+1 |𝐸|4 ∑ 𝑞−4𝑑 |𝐸|2 |𝑆 𝑡 |2 𝑡≠0 3𝑑+1

𝑞 ⋅ 𝑞−4𝑑 |𝐸|2 ⋅ 𝑞2𝑑−1 |𝐸|4 ≪1 ≪

when |𝐸| ≫ 𝑞𝑑/2 as claimed.

3.4 Size and decay estimates for spheres We have proved Theorem 2.3.3 save for the proofs of Propositions 3.1.5 and 3.1.6, which we prove in detail now.

3.4.1 Proof of Proposition 3.1.5 We start with a few preliminaries. In what follows, we let 𝑁(𝑓(𝑥) = 𝑡) denote the number of solutions to the equation 𝑓(𝑥) = 𝑡. Recall that the function 𝛿 ∶ 𝔽𝑞 → {0, 1} is given by 𝛿(𝑡) = {

1 0

𝑡=0 𝑡≠0

and 𝜂 ∶ 𝔽𝑞 → {−1, 0, 1} is the quadratic character of 𝔽𝑞 : 1 𝜂(𝑡) = { 0 −1

𝑡 is a nonzero square 𝑡=0 𝑡 is not a square.

3.4. Size and decay estimates for spheres

53

𝑡

If 𝑞 is prime, then 𝜂(𝑡) = ( 𝑝 ) is called the Legendre symbol. The Legendre symbol ıis a multiplicative character on the (multiplicative) group 𝔽∗𝑞 with the added condition that 𝜂(0) = 0. Note that this implies that for 𝑥, 𝑦 ∈ 𝔽∗𝑞 , we have 𝜂(𝑥𝑦) = 𝜂(𝑥)𝜂(𝑦) and that ∑ 𝜂(𝑐) = ∑ 𝜂(𝑐) = 0. 𝑐∈𝔽𝑞

𝑐∈𝔽𝑞 ∗

See (3.19) below. We further define 𝜇(𝑡) = 𝛿(𝑡)𝑞 − 1 = {

𝑞−1 −1

𝑡=0 𝑡 ≠ 0.

In a few instances, we will complete the square, so recall that we must assume that 𝑞 ≥ 3 is odd. We next lay out a series of simple but useful equalities involving the above functions. Proposition 3.4.1. Let 𝑐, 𝑡, 𝑎 ∈ 𝔽𝑞 . Then the following hold: 2 𝑁(𝑥 = 𝑐) = 1 + 𝜂(𝑐) = { 1 0 2

𝑐 is a nonzero square 𝑐=0 𝑐 is not a square

∑ 𝜂(𝑐) = 0

(3.18)

(3.19)

𝑐∈𝔽𝑞

∑ 𝜇(𝑡) = 0

(3.20)

𝑡∈𝔽𝑞

∑ 𝛿(𝑐) = 1

(3.21)

𝑐∈𝔽𝑞

(𝜇 ∗ 𝜇)(𝑡) = ∑ 𝜇(𝑐)𝜇(𝑡 − 𝑐) = 𝑞 ⋅ 𝜇(𝑡)

(3.22)

𝑐∈𝔽𝑞

∑ 𝜂(𝑐2 + 4𝑎𝑐) = 𝜇(𝑎)

(3.23)

𝑐∈𝔽𝑞

∑ 𝜂(𝑡𝑐 − 𝑐2 ) = 𝜂(−1)𝜇(𝑡)

(3.24)

𝑐∈𝔽𝑞

∑ 𝜇(𝑐)𝜂(𝑡 − 𝑐) = 𝑞𝜂(𝑡). 𝑐∈𝔽𝑞

(3.25)

54

Chapter 3. The Iosevich-Rudnev bound

Proof. We will leave (3.18), (3.20), (3.21), and (3.22) as exercises. Note that (3.24) follows from (3.23). Now, (3.19) follows by orthogonality as 𝜂 is a character over the group 𝔽∗𝑞 . Alternatively, ∑ 𝜂(𝑐) = ∑ (1 + 𝜂(𝑐) − 1) = ∑ (𝑁(𝑥2 = 𝑐) − 1) = 𝑞 − 𝑞 = 0. 𝑐∈𝔽𝑞

𝑐∈𝔽𝑞

𝑐∈𝔽𝑞

The equality ∑ 𝑁(𝑥2 = 𝑐) = 𝑞 𝑐∈𝔽𝑞

follows from the fact that if 𝑐 is a nonzero square, then 𝑥2 = 𝑐 has two distinct solutions, say 𝑎 and −𝑎. Thus, each nonzero element in 𝔽∗𝑞 is a solution to some equation 𝑥2 = 𝑐 for a unique 𝑐. It follows that there 𝑞−1 𝑞−1 are 2 values 𝑐 such that 𝑥2 = 𝑐 has two solutions. In other words, 2 elements in 𝔽𝑞 are nonzero squares. The rest now follows easily. Finally, we prove (3.23). Now, ∑ 𝜂(𝑐2 + 4𝑡𝑐) = ∑ 𝜂((𝑐 + 2𝑡)2 − 4𝑡2 ) 𝑐∈𝔽𝑞

𝑐∈𝔽𝑞

= ∑ 𝜂(𝑐2 − 4𝑡2 ) 𝑐∈𝔽𝑞

= ∑ [1 + 𝜂(𝑐2 − 4𝑡2 ) − 1] 𝑐∈𝔽𝑞

= 𝑁(𝑥2 = 𝑐2 − 4𝑡2 ) − 𝑞. In Exercise 3.4, you are asked to show that 𝑁(𝑥2 − 𝑐2 = 𝑡) = 𝑞 + 𝜇(𝑡). This completes the proof. We are finally ready to prove Proposition 3.1.5. When 𝑑 = 2, by (3.18) we have |𝑆 𝑡 | = ∑ 𝑁(𝑥12 = 𝑎)𝑁(𝑥22 = 𝑏) 𝑎+𝑏=𝑡

= ∑ 𝑁(𝑥12 = 𝑐)𝑁(𝑥22 = 𝑡 − 𝑐) 𝑐∈𝔽𝑞

= ∑ [1 + 𝜂(𝑐)][1 + 𝜂(𝑡 − 𝑐)]. 𝑐∈𝔽𝑞

3.4. Size and decay estimates for spheres

55

Now because 𝜂 is multiplicative we have [1 + 𝜂(𝑐)][1 + 𝜂(𝑡 − 𝑐)] = 1 + 𝜂(𝑐) + 𝜂(𝑡 − 𝑐) + 𝜂(𝑡𝑐 − 𝑐2 ). Note that by (3.19), the sum over 𝜂(𝑐) and 𝜂(𝑡 − 𝑐) is zero. Hence |𝑆 𝑡 | = ∑ [1 + 𝜂(𝑡𝑐 − 𝑐2 )] 𝑐∈𝔽𝑞

= 𝑞 + ∑ 𝜂(𝑡𝑐 − 𝑐2 ) 𝑐∈𝔽𝑞

= 𝑞 + 𝜂(−1)𝜇(𝑡), where the last line followed from (3.24). For even values 𝑑 ≥ 2 we apply induction with the case 𝑑 = 2 above being our base case. For 𝑑 ≥ 4 we have 2 2 |𝑆 𝑡 | = ∑ 𝑁(𝑥12 + ⋯ + 𝑥𝑑−2 = 𝑐)𝑁(𝑥𝑑−1 + 𝑥𝑑2 = 𝑡 − 𝑐) 𝑐∈𝔽𝑞

= ∑ (𝑞𝑑−3 + 𝜇(𝑐)𝜂 ((−1)

𝑑−2 2

)𝑞

𝑑−4 2

) ⋅ (𝑞 + 𝜂(−1)𝜇(𝑡 − 𝑐)) ,

𝑐∈𝔽𝑞

since 2 𝑁(𝑥12 + ⋯ + 𝑥𝑑−2 = 𝑐) = 𝑞𝑑−3 + 𝜇(𝑐)𝜂 ((−1)

𝑑−2 2

)𝑞

𝑑−4 2

by the inductive hypothesis and 2 𝑁(𝑥𝑑−1 + 𝑥𝑑2 = 𝑡 − 𝑐) = 𝑞 + 𝜂(−1)𝜇(𝑡 − 𝑐)

by the base case. It follows that |𝑆 𝑡 | = ∑ (𝑞𝑑−2 + 𝜇(𝑐)𝜂 ((−1)

𝑑−2 2

)𝑞

𝑑−4 2

𝜂(−1)𝜇(𝑡 − 𝑐))

𝑐∈𝔽𝑞 𝑑

= 𝑞𝑑−1 + 𝜂 ((−1) 2 ) 𝑞

𝑑−4 2

∑ 𝜇(𝑐)𝜇(𝑡 − 𝑐) 𝑐∈𝔽𝑞

𝑑

= 𝑞𝑑−1 + 𝜂 ((−1) 2 ) 𝑞

𝑑−2 2

𝜇(𝑡)

56

Chapter 3. The Iosevich-Rudnev bound

where the last line follows from (3.19). This established Proposition 3.1.5 when 𝑑 ≥ 2 is even. When 𝑑 ≥ 3 is odd, 2 |𝑆 𝑡 | = ∑ 𝑁(𝑥12 + . . . 𝑥𝑑−1 = 𝑐)𝑁(𝑥𝑑2 = 𝑡 − 𝑐) 𝑐∈𝔽𝑞

= ∑ [𝑞𝑑−2 + 𝜂 ((−1)

𝑑−1 2

)𝑞

𝑑−3 2

𝜇(𝑐)] (1 + 𝜂(𝑡 − 𝑐)) .

𝑐∈𝔽𝑞

Distributing the quantity in the sum, and applying (3.20) and (3.19) (and a change of variables), we see that |𝑆 𝑡 | = 𝑞𝑑−1 + 𝑞

𝑑−3 2

𝜂 ((−1)

𝑑−1 2

𝜂 ((−1)

𝑑−1 2

) ∑ 𝜇(𝑐)𝜂(𝑡 − 𝑐) 𝑐∈𝔽𝑞

= 𝑞𝑑−1 + 𝑞

𝑑−1 2

𝑡) ,

by (3.25). This completes the proof of Proposition 3.1.5.

3.4.2 Proof of Proposition 3.1.6 Let’s jump right in. First, write 𝑆 𝑡̂ (𝑚) = 𝑞−𝑑 ∑ 𝑆 𝑡 (𝑥)𝜒(−𝑥 ⋅ 𝑚) 𝑥∈𝔽𝑑 𝑞

= 𝑞−𝑑−1 ∑ ∑ 𝜒(𝑠‖𝑥‖ − 𝑠𝑡)𝜒(−𝑥 ⋅ 𝑚) 𝑠∈𝔽𝑞 𝑥∈𝔽𝑑 𝑞

=

𝛿(𝑚) + ∑ ∑ 𝜒(𝑠‖𝑥‖ − 𝑠𝑡)𝜒(−𝑥 ⋅ 𝑚). 𝑞 𝑑 𝑠≠0 𝑥∈𝔽𝑞

Breaking apart the sum in 𝑥 = (𝑥1 , . . . , 𝑥𝑑 ) into the sum of its component pieces, we see that 𝑑

𝑆 𝑡̂ (𝑚) =

𝛿(𝑚) + 𝑞−𝑑−1 ∑ 𝜒(−𝑠𝑡) ∏ ∑ 𝜒(𝑠𝑥𝑖2 − 𝑥𝑖 𝑚𝑖 ) 𝑞 𝑠≠0 𝑖=1 𝑥 ∈𝔽 𝑖

𝑞

𝑑

=

𝑚 2 𝑚2 𝛿(𝑚) + 𝑞−𝑑−1 ∑ 𝜒(−𝑠𝑡) ∏ ∑ 𝜒 (𝑠 (𝑥𝑖 − 𝑖 ) − 𝑖 ) 𝑞 2𝑠 4𝑠 𝑠≠0 𝑖=1 𝑥 ∈𝔽 𝑖

𝑞

𝑑

=

𝑚2 𝛿(𝑚) + 𝑞−𝑑−1 ∑ 𝜒(−𝑠𝑡) ∏ ∑ 𝜒(𝑠𝑥𝑖2 )𝜒 (− 𝑖 ) . 𝑞 4𝑠 𝑠≠0 𝑖=1 𝑥 ∈𝔽 𝑖

𝑞

3.5. Finite field counterexample

57

Applying (3.11) yields 𝑆 𝑡̂ (𝑚) = =

𝑚21 + ⋯ + 𝑚2𝑑 𝛿(𝑚) + 𝑞−𝑑−1 𝜀𝑑𝑞 𝑞𝑑/2 ∑ 𝜒 (−𝑠𝑡 − ) 𝑞 4𝑠 𝑠≠0 𝑑+2 ‖𝑚‖ 𝛿(𝑚) + 𝜀𝑑𝑞 𝑞− 2 ∑ 𝜒 (−𝑠𝑡 − ). 𝑞 4𝑠 𝑠≠0

𝑎

Notice that for 𝑎, 𝑏 ∈ 𝔽∗𝑞 we are using the notation 𝑏 to mean 𝑎𝑏−1 . In the sum above, the exact value of 𝜀𝑞 is surprisingly difficult to compute, but it turns out (see [90, 104]) that if 𝑞 = 𝑝ℓ for an odd prime 𝑝, then (−1)ℓ−1

𝑝≡1

(mod 4)

(−1)ℓ−1 𝑖ℓ

𝑝≡3

(mod 4).

𝜀𝑞 = { The two inequalities in Proposition 3.1.6 will follow from the following deep result of Weil ([154]) on so-called Kloosterman sums, finishing our proof of the proposition. Theorem 3.4.2. Define 𝐾(𝑎, 𝑏) = ∑ 𝜒 (𝑎𝑠 + 𝑏𝑠−1 ) . 𝑠≠0

Then, |𝐾(𝑎, 𝑏)| ≤ 2√𝑞 if 𝑎𝑏 ≠ 0. Note that 𝐾(0, 0) = 𝑞 − 1, while 𝐾(0, 𝑡) = 𝐾(𝑡, 0) = −1 for 𝑡 ≠ 0. The proof of Theorem 3.4.2 is beyond the scope of this book. For an elementary1 proof, see [136].

3.5 Finite field counterexample In this section we prove the following counterexample to the ErdősFalconer distance conjecture constructed in [76]. 𝑑+1

Theorem 3.5.1. Let 𝑑 ≥ 3 be odd. Then the exponent 2 in Theorem 2.3.3 is best possible. In particular for any 𝜀 > 0, there exists a set 𝐸 ⊆ 𝔽𝑑𝑞 such that |𝐸| ≍ 𝑞

𝑑+1 −𝜀 2

and yet the distance set satisfies |Δ(𝐸)| ≪ 𝑞1−𝜀 .

1 An elementary proof is a proof which only relies on basic techniques. It does not mean that the proof is easy to follow! Schmidt’s approach to the proof (following work of Stepanov and Thue) is both elegant and intricate.

58

Chapter 3. The Iosevich-Rudnev bound

We will call 𝑣 ∈ 𝔽𝑑𝑞 a null-vector if 𝑣 ≠ (0, . . . , 0) and yet 𝑣 ⋅ 𝑣 = 0. We start with a preliminary lemma. Proposition 3.5.2. Let 𝑑 ≥ 4 be even. Then, there exist at least 𝑑/2 mutually orthogonal null vectors in 𝔽𝑑𝑞 . Proof. If 𝑞 ≡ 1 (mod 4), then there exists an element 𝑖 ∈ 𝔽𝑞 such that 𝑖2 = −1. Then, let 𝑣𝑗 = (0, . . . , 0, 1, 𝑖, . . . , 0), where there is a 1 in the (2𝑗−1)-st coordinate, and an 𝑖 in the 2𝑗-th coordinate. This produces 𝑑/2 mutually orthogonal null vectors in 𝔽𝑑𝑞 . If 𝑞 ≡ 3 (mod 4), we will need to be more creative. Suppose that 𝑑 ≡ 0 (mod 4). Then we will take 𝑣 1 = (𝑎, 𝑏, 𝑐, 0, . . . , 0) and 𝑣 2 = (0, −𝑐, 𝑏, 𝑎, 0, . . . , 0). Note that 𝑣 1 ⋅ 𝑣 2 = 0 for all 𝑎, 𝑏, 𝑐 ∈ 𝔽𝑞 . Next, we take 𝑣 3 = (0, 0, 0, 0, 𝑎, 𝑏, 𝑐, 0, . . . , 0) and 𝑣 4 = (0, 0, 0, 0, 0, −𝑏, −𝑐, 𝑎, 0 . . . , 0), and so on. Note that there always exists a triple (𝑎, 𝑏, 𝑐) ∈ 𝔽3𝑞 such that 𝑎2 + 𝑏2 + 𝑐2 = 0 and 𝑎𝑏𝑐 ≠ 0 (see Exercise 3.13). This provides 𝑑/2 mutually orthogonal null vectors when 𝑑 ≡ 0 (mod 4). It remains to treat the case 𝑑 ≡ 2 (mod 4). We first handle 𝑑 = 6 by simply taking 𝑣 1 = (𝑎, 𝑏, 𝑐, 0, 0, 0) 𝑣 2 = (−𝑏, 𝑎, 0, 𝑐, 0, 0) 𝑣 3 = (0, −𝑐, 𝑏, 𝑎, 0, 0). All higher dimensions 𝑑 ≡ 2 (mod 4) can be handled similarly. For example when 𝑑 = 10, we may take 𝑣 1 = (𝑎, 𝑏, 𝑐, 0, 0, 0, 0, 0, 0, 0) 𝑣 2 = (−𝑏, 𝑎, 0, 𝑐, 0, 0, 0, 0, 0, 0) 𝑣 3 = (0, −𝑐, 𝑏, 𝑎, 0, 0, 0, 0, 0, 0) 𝑣 4 = (0, 0, 0, 0, 𝑎, 𝑏, 𝑐, 0, 0, 0) 𝑣 5 = (0, 0, 0, 0, 0, −𝑐, 𝑏, 0, 0, 0). This completes the proof of Proposition 3.5.2. Now let 𝑑 ≥ 3 be odd, so that 𝑑 = 2𝑘 + 1 for some 𝑘 ≥ 1. We have shown that there exist at least 𝑘 mutually orthogonal null-vectors whose

3.6. Relations to the Falconer problem

59

last coordinate is 0. Suppose we take 𝐴 ⊆ 𝔽𝑞 to be any arithmetic pro𝑞+1 gression of length 𝑛 ≤ 2 , and let 𝑒 𝑑 = (0, . . . , 0, 1). Finally, put 𝐸 = {𝑡 𝑖 𝑣 𝑖 + 𝑎𝑒 𝑑 ∶ 𝑎 ∈ 𝐴, and 𝑖 = 1, . . . , 𝑘, where 𝑡 𝑖 runs through 𝔽𝑞 }. First, note that |𝐸| = 𝑞𝑘 ⋅𝑛, since for each of the 𝑘 vectors 𝑣 𝑖 , the quantity 𝑡 𝑖 runs through 𝔽𝑞 . Next, note that since 𝑣 𝑖 ⋅ 𝑣𝑗 = 0 for all 1 ≤ 𝑖 ≤ 𝑗 ≤ 𝑘, it follows that for 𝑥, 𝑦 ∈ 𝐸 we have ‖𝑥 − 𝑦‖ = ‖𝑡1 𝑣 1 + 𝑎𝑒 𝑑 − 𝑡2 𝑣 2 − 𝑎′ 𝑒 𝑑 ‖ = (𝑎 − 𝑎′ )2 ,

(3.26)

by a direct calculation (see Exercise 3.10), and thus |Δ(𝐸)| ≤ 2𝑛 − 1. We leave the remainder of the proof of Theorem 3.5.1 as an exercise (Exercise 3.14).

3.6 Relations to the Falconer problem Finally we should mention that the approaches to the Erdős-Falconer distance problem have been heavily influenced by some of the approaches to the Falconer distance problem, and it will be instructive to (very) briefly outline the strategies employed in the Falconer distance problem, though one should not take the following integrals and inequalities too seriously. Kenneth Falconer’s original proof of his result concerning the Falconer distance problem centered around the observation that 𝜇 × 𝜇 {(𝑥, 𝑦) ∈ ℝ𝑑 × ℝ𝑑 ∶ 1 ≤ |𝑥 − 𝑦| ≤ 1 + 𝜖} ≪ 𝜖, for an appropriate probability measure2 𝜇, and this estimate in turn utilized the stationary phase estimate |1̂𝑆 (𝜉)| ≪ |𝜉|−

𝑑+1 2

,

(3.27)

where 1𝑆 denotes the characteristic equation of the sphere 𝑆 𝑑−1 ⊆ ℝ𝑑 . This idea essentially reduces to computing the number of times a distance is repeated, and this was our first line of attack (from Section 3.1). 2 Loosely, a probability measure on a set 𝐸 is a measure such that 𝜇(𝐸) = 1. These particular measures used by Falconer are called Frostman measures.

60

Chapter 3. The Iosevich-Rudnev bound

Furthermore, Mattila showed that in order to prove Falconer’s conjecture it is enough to get a bound of the form ∞

2

2

∫ (∫ ||𝜇 ˆ(𝑡𝑒𝑖𝜃 )|| 𝑑𝜃) 𝑡 𝑑𝑡 < ∞

(3.28)

1

where 𝜇 is again the Frostman measure ([109]). The integral (3.28) will be referred to simply as 𝑀(𝜇). Again, do not worry at all about understanding this integral, perhaps other than to notice that an 𝐿2 -bound on this integral gives a distance set result. This was the motivation for having examined the finite field Mattila sum 𝑀𝐸 (𝑞) from Section 3.3. The ubiquity of the connections between continuous problems and finite field analogues can be surprising, and they will continue to be highlighted throughout the remainder of the book.

3.7 Exercises: Chapter 3 Exercise 3.1. Suppose 𝑞 = 𝑝𝑟 , and let 𝑇𝑟 ∶ 𝔽𝑞 → 𝔽𝑝 be the field trace, i.e., the function defined by 2

Tr(𝑥) = 𝑥 + 𝑥𝑝 + 𝑥𝑝 + ⋯ + 𝑥𝑝

𝑟−1

.

Prove that 𝜒(𝑥) = 𝑒2𝜋𝑖 Tr(𝑥)/𝑝 is a group homomorphism from 𝔽𝑞 to the set ℂ∗ = ℂ ⧵ {0}, thereby establishing that 𝜒(𝑥) is a character on 𝔽𝑞 . Exercise 3.2. Prove Proposition 3.1.3. That is, prove that we have ˆ ⋅ 𝑚) 𝑓(𝑥) = ∑ 𝑓(𝑚)𝜒(𝑥 𝑚∈𝔽𝑑 𝑞

for any function 𝑓 ∶ 𝔽𝑑𝑞 → ℂ. Exercise 3.3. Prove that we have ˆ 𝑔(𝑚) ∑ 𝑓(𝑚) ̂ = 𝑞−𝑑 ∑ 𝑓(𝑥)𝑔(𝑥) 𝑚∈𝔽𝑑 𝑞

𝑥∈𝔽𝑑 𝑞

for all functions 𝑓, 𝑔 ∶ 𝔽𝑑𝑞 → ℂ. This establishes Proposition 3.1.4. Exercise 3.4. Let 𝐻(𝑡) = |{(𝑥, 𝑦) ∈ 𝔽𝑞 ∶ 𝑥2 − 𝑦2 = 𝑡}|. Show that 𝐻(𝑡) = {

2𝑞 − 1 𝑞−1

𝑡=0 𝑡 ≠ 0.

3.7. Exercises: Chapter 3

61

Exercise 3.5. Prove (3.18) by showing that the number of solutions to the equation 𝑥2 = 𝑐 is given by 1 + 𝜂(𝑐), where 𝜂 is the quadratic character of 𝔽𝑞 . Exercise 3.6. Prove equation (3.20). Exercise 3.7. Verify that ∑ 𝛿(𝑐 + 𝑎) = 1 𝑐∈𝔽𝑞

holds for any 𝑎 ∈ 𝔽𝑞 by a direct calculation. This generalizes (3.21). Exercise 3.8. Prove that (𝜇 ∗ 𝜇)(𝑡) = ∑𝑐 𝜇(𝑐)𝜇(𝑡 − 𝑐) = 𝑞 ⋅ 𝜇(𝑡), verifying (3.22). Exercise 3.9. Prove (3.25). That is, show that ∑ 𝜇(𝑐)𝜂(𝑡 − 𝑐) = 𝑞𝜂(𝑡). 𝑐∈𝔽𝑞

Exercise 3.10. Verify equation (3.26) by a direct calculation. Exercise 3.11. Prove Proposition 3.3.3. More precisely, for fixed 𝑥 ∈ 𝔽𝑑𝑞 , define 𝑢𝑥 ∶ 𝔽𝑞 → ℂ given by 𝑢𝑥 (𝑗) = 𝑆𝑗 (𝑥). Show that we have 𝑢𝑥̂ (𝑘) =

1 𝜒(−𝑘‖𝑥‖). 𝑞

Exercise 3.12. Prove the following generalization of (3.7). Show for functions 𝑓, 𝑔 ∶ 𝔽𝑑𝑞 → ℂ, that ∑ 𝑓(𝑥)𝑔(𝑦)𝑆 𝑡 (𝑥 − 𝑦) = 𝑡∈𝔽𝑞

where |𝑅(𝑓, 𝑔)| ≤ 2𝑞

𝑑−1 2

|𝑆 𝑡 | ‖𝑓‖1 ‖𝑔‖1 + 𝑅(𝑓, 𝑔), 𝑞𝑑

‖𝑓‖2 ‖𝑔‖2 . Here, we are writing 1/𝑝

‖𝑓‖𝑝 = ( ∑ |𝑓(𝑥)|𝑝 )

.

𝑥∈𝔽𝑞

Exercise 3.13. Show that when 𝑞 ≡ 3 (mod 4), the equation 𝑎2 + 𝑏2 + 𝑐2 = 0 has a nontrivial solution (𝑎, 𝑏, 𝑐) ∈ 𝔽∗𝑞 × 𝔽∗𝑞 × 𝔽∗𝑞 .

62

Chapter 3. The Iosevich-Rudnev bound

Exercise 3.14. Finish the proof of Theorem 3.5.1. More precisely, given any 𝜀 ∈ (0, 1), construct a set 𝐸 such that |𝐸| ≍ 𝑞 𝑞1−𝜀 .

𝑑+1 −𝜀 2

and yet |Δ(𝐸)| ≪

Exercise 3.15. Show that two distinct spheres in 𝔽2𝑞 intersect in at most two points. What can we say about the intersection of spheres in higher dimensions? Exercise 3.16. Strengthen our proof of Theorem 2.3.3. Show that if 𝐸 ⊆ 𝔽𝑑𝑞 satisfies |𝐸| > 2𝑞 for |𝑆 𝑡 | and 𝑆 𝑡̂ (𝑚).

𝑑+1 2

, then Δ(𝐸) = 𝔽𝑞 by using more precise estimates

10.1090/car/037/04

Chapter

4

Wolff’s exponent 4.1 Introduction We previously mentioned that Tom Wolff gave a “4/3-result” toward the Falconer distance problem (see Section 2.2). In this chapter, we prove a 4/3-result toward the finite field distance problem. Namely we will discuss the following improvement to the distance problem in two dimensions, first proved in [17]. Theorem 4.1.1. Suppose that 𝐸 ⊆ 𝔽2𝑞 with |𝐸| ≥ 𝑞4/3 , where 𝑞 is sufficiently large. Then there exists a constant 𝑐 > 0 such that |Δ(𝐸)| ≥ 𝑐𝑞. Proof of Theorem 4.1.1. We will use the 𝐿2 -methods outlined in Section 3.2. Whether we use (3.8) or (3.9) will depend on whether 𝑞 ≡ 1 (mod 4) or 𝑞 ≡ 3 (mod 4). When 𝑞 ≡ 3 (mod 4), we will utilize (3.8): |Δ(𝐸)| ≥

|𝐸|4 , ∑𝑡 𝜈(𝑡)2

since in this case there do not exist nontrivial circles of radius zero in two dimensions. When 𝑞 ≡ 1 (mod 4), we will need to account for such isotropic spheres. Theorem 4.1.1 will follow from the following upper bound, which we achieve when 𝑞 ≡ 3 (mod 4) and |𝐸| ≥ 𝑞4/3 : ∑ 𝜈(𝑡)2 ≤ (1 + √3)𝑞−1 |𝐸|3 .

(4.1)

𝑡

Thus it remains to prove (4.1). First we apply the following result. 63

64

Chapter 4. Wolff’s exponent

Theorem 4.1.2. For all odd values 𝑞 and for 𝐸 ⊆ 𝔽2𝑞 , we have the equality 2

∑ 𝜈(𝑡)2 = 𝑞6 ∑ (𝜎𝐸 (𝑡)) + Γ(𝑞, 𝐸), 𝑡∈𝔽𝑞

𝑡∈𝔽𝑞

where Γ(𝑞, 𝐸) =

|𝐸|4 − 𝑞|𝐸|2 , 𝑞

and where

2

ˆ || 𝜎𝐸 (𝑡) = ∑ ||𝐸(𝑚) 𝑚∈𝑆𝑡 1

is the circular average of 𝐸 which we encountered in the previous chapter. We will prove Theorem 4.1.2 shortly. The bound (4.1) will follow from combining Theorem 4.1.2 with the following remarkable bound derived by Iosevich and Koh ([83]). Theorem 4.1.3. If 𝐸 ⊆ 𝔽2𝑞 , then we have that 2

ˆ || ≤ √3𝑞−3 |𝐸|3/2 . max ∑ ||𝐸(𝑚) 𝑡≠0

‖𝑚‖=𝑡

The proof of Theorem 4.1.3 lies just beyond the scope of the book. The motivated reader is encouraged to look up the original source. However, we will next show that the combination of Theorems 4.1.2 and 4.1.3 is sufficient to achieve (4.1). We start by pulling out the term 𝑚 = 0,⃗ and then used Proposition 1.3.4. Thus, 4

2

ˆ 0)|| + 𝑞6 (max 𝜎𝐸 (𝑡)) ∑ ||𝐸(𝑚) ˆ || + Γ(𝑞, 𝐸) ∑ 𝜈(𝑡)2 ≤ 𝑞6 ||𝐸(0, 𝑡≠0

𝑡∈𝔽𝑞

𝑚≠(0,0)

≤ 𝑞6 𝑞−8 |𝐸|4 + 𝑞6 √3𝑞−3 |𝐸|3/2 ⋅ (𝑞−2 |𝐸| − 𝑞−4 |𝐸|2 ) + Γ(𝑞, 𝐸) = 𝑞−2 |𝐸|4 + √3𝑞|𝐸|5/2 − √3𝑞−1 |𝐸|5/2 + 𝑞−1 |𝐸|4 − 𝑞|𝐸|2 ≤ 𝑞−1 |𝐸|4 + √3𝑞|𝐸|5/2 ≤ (1 + √3)𝑞−1 |𝐸|4 . 1 We previously called this quantity the spherical average, though we use circles to refer to 2-dimensional spheres.

4.2. Proof of 𝐿2 estimate for 𝜈(𝑡)

65

In the final steps we grouped everything together and used that |𝐸| ≥ 𝑞4/3 . Putting everything together, we have shown that when 𝑞 ≡ 3 (mod 4) and 𝐸 ⊆ 𝔽2𝑞 satisfies |𝐸| ≥ 𝑞4/3 , then |Δ(𝐸)| ≥

1 1 + √3

𝑞.

When 𝑞 ≡ 1 (mod 4), the proof of Theorem 4.1.1 is similar, though we must use the inequality (3.9) since in this case, there exist nontrivial circles of radius 0. We leave it to the motivated reader to complete the proof in the case 𝑞 ≡ 1 (mod 4), using the original paper [17] as a guide. When 𝑞 ≡ 1 (mod 4), the authors ([17]) actually showed that if |𝐸| ≥ 𝑞4/3 , then |Δ(𝐸)| ≥ 𝑐𝑞 ⋅ 𝑞, where 2

𝑐𝑞 =

(1 − 2/𝑞)

1 + √3 − √3/𝑞2/3

Notice that 𝑐𝑞 is increasing if 𝑞 ≥ 2, 𝑐𝑞 → 𝑐𝑞 ≥

1 3

. 1

1+√3

as 𝑞 → ∞, and that

if 𝑞 ≥ 25.

4.2 Proof of 𝐿2 estimate for 𝜈(𝑡) We still need to prove Theorem 4.1.2. Recall that 𝜈(𝑡) is our counting function: 𝜈(𝑡) = |{(𝑥, 𝑦) ∈ 𝐸 × 𝐸 ∶ ‖𝑥 − 𝑦|| = 𝑡}|. We start by applying our bound (3.4): 2

ˆ || 𝑆 𝑡̂ (𝑚) 𝜈(𝑡) = 𝑞4 ∑ ||𝐸(𝑚) 𝑚∈𝔽2𝑞

so that 𝜈2 (𝑡) = 𝑞8

∑

2

2

ˆ || ||𝐸(𝑚 ˆ ′ )|| 𝑆 𝑡̂ (𝑚)𝑆 𝑡̂ (𝑚′ ) ||𝐸(𝑚)

𝑚,𝑚′ ∈𝔽2𝑞

= 𝐴(𝑡) + 𝐵(𝑡) + 𝐶(𝑡)

66

Chapter 4. Wolff’s exponent

where 𝐴(𝑡) = 𝑞−4 |𝐸|4 |𝑆 𝑡 |2 𝐵(𝑡) = 2𝑞8

∑

2

2

ˆ || ||𝐸(0, ˆ 0)|| 𝑆 𝑡̂ (𝑚)𝑆 𝑡̂ (0, 0), and ||𝐸(𝑚)

𝑚∈𝔽2𝑞 ⧵{(0,0)}

𝐶(𝑡) = 𝑞8

2

2

ˆ ′ )|| 𝑆 𝑡̂ (𝑚)𝑆 𝑡̂ (𝑚′ ). ˆ || ||𝐸(𝑚 ||𝐸(𝑚)

∑ 𝑚,𝑚′ ∈𝔽2𝑞 ⧵{(0,0)}

Note that 𝐵(𝑡) contains both the sums when 𝑚 = (0, 0) and 𝑚′ ≠ (0, 0) as well as 𝑚′ = (0, 0) and 𝑚 ≠ (0, 0), so we can combine the sums by symmetry. Hence, ∑ 𝜈(𝑡)2 = ∑ [𝐴(𝑡) + 𝐵(𝑡) + 𝐶(𝑡)] . 𝑡∈𝔽𝑞

𝑡∈𝔽𝑞

The easiest term to handle is ∑ 𝐴(𝑡) = 𝑞−4 |𝐸|4 ∑ |𝑆 𝑡 |2 . 𝑡∈𝔽𝑞

𝑡∈𝔽𝑞

We will need the following averaging results for the sphere, whose proofs we delay until the end of the chapter. Proposition 4.2.1. Let 𝑆 𝑡 ⊆ 𝔽𝑑𝑞 . Then we have ∑ |𝑆 𝑡 |2 = 𝑞2𝑑−1 + 𝑞𝑑 − 𝑞𝑑−1 . 𝑡∈𝔽𝑞

Moreover, if 𝑚, 𝑚′ ≠ 0,⃗ then we have −𝑑−1 ∑ 𝑆 𝑡̂ (𝑚)𝑆 𝑡̂ (𝑚′ ) = 𝜀2𝑑 [−1 + 𝑞𝛿(‖𝑚‖ − ‖𝑚′ ‖)] 𝑞 𝑞 𝑡

where 𝛿(⋅) is the 1-dimensional delta function. If 𝑑 ≥ 2 is even and 𝑚 ≠ 0,⃗ then ∑ |𝑆 𝑡 |𝑆 𝑡̂ (𝑚) = 𝛿(‖𝑚‖) − 𝑞−1 . 𝑡∈𝔽𝑞

Note 4.2.2. For reference recall that (−1)ℓ−1

𝑞 = 𝑝ℓ , 𝑝 ≡ 1

(mod 4)

(−1)ℓ−1 𝑖ℓ

𝑞 = 𝑝ℓ , 𝑝 ≡ 3

(mod 4).

𝜀𝑞 = {

4.2. Proof of 𝐿2 estimate for 𝜈(𝑡)

67

In particular notice that 𝜀4𝑞 = 1 for all 𝑞, and 1

𝑞≡1

(mod 4)

−1

𝑞≡3

(mod 4).

𝜀2𝑞 = { From Proposition 4.2.1 it follows that ∑ 𝐴(𝑡) = |𝐸|4 (𝑞−1 + 𝑞−2 − 𝑞−3 ) .

(4.2)

𝑡

By Proposition 4.2.1 if 𝑑 = 2 and 𝑚 ≠ (0, 0), then we have ∑ |𝑆 𝑡 |𝑆 𝑡̂ (𝑚) = 𝛿(‖𝑚‖) − 𝑞−1 . 𝑡∈𝔽𝑑 𝑞

We are ready to handle the second sum. By the results above we have 2

ˆ || 𝑆 𝑡̂ (𝑚) ∑ 𝐵(𝑡) = 2𝑞8 𝑞−6 |𝐸|2 ∑ |𝑆 𝑡 | ∑ ||𝐸(𝑚) 𝑡

𝑡∈𝔽𝑞

𝑚≠(0,0) 2

ˆ || ∑ |𝑆 𝑡 |𝑆 𝑡̂ (𝑚) ∑ ||𝐸(𝑚)

= 2𝑞2 |𝐸|2

𝑡

𝑚≠(0,0)

= 2𝑞2 |𝐸|2

2

ˆ || − 2𝑞|𝐸|2 (𝑞−2 |𝐸|4 − 𝑞−4 ) ||𝐸(𝑚)

∑ 𝑚≠(0,0),‖𝑚‖=0

2

ˆ || − 2𝑞−2 |𝐸|4 + 2𝑞−3 |𝐸|4 − 2𝑞−1 |𝐸|3 . = 2𝑞2 |𝐸|2 ∑ ||𝐸(𝑚) ‖𝑚‖=0

Finally, we examine ∑ 𝐶(𝑡) = 𝑞8 ∑

∑

2

2

ˆ ′ )|| 𝑆 𝑡̂ (𝑚)𝑆 𝑡̂ (𝑚′ ). ˆ || ||𝐸(𝑚 ||𝐸(𝑚)

𝑡∈𝔽𝑞 𝑚,𝑚′ ∈𝔽2𝑞 ⧵{(0,0)}

𝑡∈𝔽𝑞

From Proposition 4.2.1 and Note 4.2.2, we have ∑ 𝑆 𝑡̂ (𝑚)𝑆 𝑡̂ (𝑚′ ) = 𝑞−2 𝛿(‖𝑚‖ − ‖𝑚′ ‖) − 𝑞−3 . 𝑡

68

Chapter 4. Wolff’s exponent

Hence, ∑ 𝐶(𝑡) = 𝑞5 𝑡∈𝔽𝑞

2

2

ˆ || ||𝐸(𝑚 ˆ ′ )|| (−1 + ∑ 𝜒 (𝑠(‖𝑚‖ − ‖𝑚′ ‖))) ∑ ||𝐸(𝑚) 𝑠

𝑚,𝑚′ ≠0⃗

2

= 𝑞6

2

2

2

ˆ || ||𝐸(𝑚 ˆ ′ )|| − 𝑞5 ( ∑ ||𝐸(𝑚) ˆ || ) . ||𝐸(𝑚)

∑ 𝑚,𝑚′ ≠0⃗ ‖𝑚‖=‖𝑚′ ‖

𝑚≠0⃗

For the first sum, we use the same expression we derived in (3.16). For the second sum, we add and subtract in the term corresponding to 𝑚 = 0⃗ and apply Plancherel to see that 2

⎛ ⎞ 2⎟ ˆ ∑ 𝐶(𝑡) = 𝑞6 ∑ ⎜ ∑ |𝐸(𝑚)| − 𝑞5 (𝑞−2 |𝐸| − 𝑞−4 |𝐸|2 )2 ⎟ 𝑡∈𝔽𝑞 𝑡∈𝔽𝑞 ⎜‖𝑚‖=𝑡 ⎝ 𝑚≠0⃗ ⎠ 2

= 𝑞6 (

2

2 ˆ |𝐸(𝑚)| ) + 𝑞6 ∑ (𝜎𝐸 (𝑡)) + Λ(𝑞, 𝐸)

∑

𝑡∈𝔽∗𝑞

𝑚∈𝑆0 ⧵{(0,0)} 2

2

= 𝑞6 (𝜎𝐸 (0) − 𝑞−4 |𝐸|2 ) + 𝑞6 ∑ (𝜎𝐸 (𝑡)) + Λ(𝑞, 𝐸) 𝑡∈𝔽∗𝑞 2

= 𝑞6 ∑ (𝜎𝐸 (𝑡)) − 2𝑞2 |𝐸|2 𝜎𝐸 (0) + 𝑞−2 |𝐸|4 + Λ(𝑞, 𝐸), 𝑡∈𝔽𝑞

where Λ(𝑞, 𝐸) = −𝑞|𝐸|2 +2𝑞−1 |𝐸|3 −𝑞−3 |𝐸|4 . Putting together the above estimates, finally gives us 2 2

ˆ || ) + ∑ 𝜈(𝑡) = 𝑞 ∑ ( ∑ ||𝐸(𝑚) 2

𝑡∈𝔽𝑞

6

𝑡∈𝔽𝑞

‖𝑚‖=𝑡

|𝐸|4 − 𝑞|𝐸|2 . 𝑞

This completes the proof of Theorem 4.1.2.

4.2.1 Proof of Proposition 4.2.1 We will prove Proposition 4.2.1 only in the case 𝑑 = 2. We leave the general case as an exercise. Let 𝑆 𝑡 ⊆ 𝔽𝑑𝑞 . Recall that 𝜂(𝑥) = 1 if 𝑥 ∈ 𝔽∗𝑞 is a nonzero square, 𝜂(𝑥) = −1 if 𝑥 ∈ 𝔽∗𝑞 is not a square, and 𝜂(0) = 0.

4.2. Proof of 𝐿2 estimate for 𝜈(𝑡)

69

Moreover, recall that 𝜇(0) = 𝑞 − 1 and 𝜇(𝑡) = −1 for 𝑡 ≠ 0. In particular, we will use that ∑ 𝜇(𝑡) = 0. 𝑡∈𝔽𝑞

Now we have that ∑ |𝑆 𝑡 |2 = ∑(𝑞 + 𝜂(−1)𝜇(𝑡))2 𝑡

𝑡

= ∑(𝑞2 + 𝑞𝜂(−1)𝜇(𝑡) + 𝜇(𝑡)2 ) 𝑡

= 𝑞3 + 𝑞𝜂(−1) ∑ 𝜇(𝑡) + (𝑞 − 1)2 + 𝑞 − 1 𝑡

= 𝑞3 + 𝑞2 − 𝑞. Next, we look at ∑ 𝑆 𝑡̂ (𝑚)𝑆 𝑡̂ (𝑚′ ). 𝑡

Recall from Section 3.4.2 that if 𝑑 = 2 and 𝑚 ≠ (0, 0), we have 𝑆 𝑡̂ (𝑚) = 𝜀2𝑞 𝑞−2 ∑ 𝜒 (𝑠𝑡 + 𝑠≠0

‖𝑚‖ ). 4𝑠

Note that 𝜀𝑞 ∈ {±1, ±𝑖}, so that 𝜀4𝑞 = 1. Hence for 𝑑 = 2 and 𝑚, 𝑚′ ≠ (0, 0), we have ∑ 𝑆 𝑡̂ (𝑚)𝑆 𝑡̂ (𝑚′ ) = 𝜀4𝑞 𝑞−4 ∑ 𝜒 (𝑠𝑡 + 𝑡

𝑠≠0

‖𝑚‖ ‖𝑚′ ‖ ) ∑ 𝜒 (−𝑠′ 𝑡 − ) 4𝑠 𝑠′ ≠0 4𝑠′

= 𝑞−4 ∑ ∑ 𝜒(𝑡(𝑠 − 𝑠′ ))𝜒 ( 𝑠,𝑠′ ≠0 𝑡

= 𝑞−3 ∑ 𝜒 ( 𝑠≠0

‖𝑚‖ ‖𝑚′ ‖ − ) 4𝑠 4𝑠′

‖𝑚‖ ‖𝑚′ ‖ − ). 4𝑠 4𝑠

Again applying a change of variables, we have ∑ 𝑆 𝑡̂ (𝑚)𝑆 𝑡̂ (𝑚′ ) = 𝑞−3 ∑ 𝜒(𝑠(‖𝑚‖ − ‖𝑚′ ‖)) 𝑡

𝑠≠0

= 𝑞−3 [−1 + ∑ 𝜒(𝑠(‖𝑚‖ − ‖𝑚′ ‖))] 𝑠

= −𝑞

−3

−2

+ 𝑞 𝛿(‖𝑚‖ − ‖𝑚′ ‖),

70

Chapter 4. Wolff’s exponent

where 𝛿(⋅) is the (1-dimensional) delta function. Finally we examine ∑ |𝑆 𝑡 |𝑆 𝑡̂ (𝑚). 𝑡∈𝔽𝑞

For 𝑚 ≠ (0, 0) we have ∑ |𝑆 𝑡 |𝑆 𝑡̂ (𝑚) = 𝜀2𝑞 𝑞−2 ∑ 𝜒 (− 𝑡

𝑠≠0

‖𝑚‖ ) ⋅ ∑ (𝑞 + 𝜂(−1)𝜇(𝑡)) 𝜒(−𝑠𝑡) 4𝑠 𝑡

= 𝜀2𝑞 𝑞−2 𝜂(−1) ∑ 𝜒 (− 𝑠≠0

‖𝑚‖ ) ∑ (𝜇(𝑡)𝜒(−𝑠𝑡)) . 4𝑠 𝑡

Applying the definition of 𝜇 and using orthogonality multiple times (along with a change of variables), we see that ∑ |𝑆 𝑡 |𝑆 𝑡̂ (𝑚) = 𝑞−2 ∑ 𝜒 (− 𝑡

𝑠≠0

= 𝑞−2 ∑ 𝜒 (− 𝑠≠0

= 𝑞−1 ∑ 𝜒 (− 𝑠≠0

‖𝑚‖ ) [𝑞 − 1 − ∑ 𝜒(−𝑠𝑡)] 4𝑠 𝑡≠0 ‖𝑚‖ ) [𝑞 − ∑ (𝜒(−𝑠𝑡))] 4𝑠 𝑡 ‖𝑚‖ ) 4𝑠

= 𝑞−1 ∑ 𝜒 (𝑠‖𝑚‖) 𝑠≠0

1 = 𝛿(‖𝑚‖) − . 𝑞

4.3 Restriction and extension theory As before it is instructive to try to understand the connections between the continuous restriction problem and the finite field analogue. This section will introduce us to these ideas, but the standard disclaimer applies: Do not take this section too seriously, and it is okay to skip this section if you would like. Before we delve into restriction theory, we first want to establish the notation used in restriction and extension theory.

4.3. Restriction and extension theory

71

4.3.1 Fourier transform notation for finite fields It will be helpful to discuss some notation at this time. For 𝑓 ∶ 𝔽𝑑𝑞 → ℂ, ˆ 𝑓, ˜ and 𝑓∨ for the normalized Fourier transform, the nonwe write 𝑓, normalized Fourier transform, and the inverse Fourier transform, respectively. More precisely, if 𝜒 is the canonical additive character over 𝔽𝑞 , then we define ˆ 𝑓(𝑚) = 𝑞−𝑑 ∑ 𝑓(𝑥)𝜒(−𝑚 ⋅ 𝑥), 𝑥∈𝔽𝑑 𝑞

˜ ˆ 𝑓(𝑚) = ∑ 𝑓(𝑥)𝜒(−𝑚 ⋅ 𝑥) = 𝑞𝑑 𝑓(𝑚), 𝑥∈𝔽𝑑 𝑞

and ˆ ˜ 𝑓∨ (𝑚) = ∑ 𝑓(𝑥)𝜒(𝑚 ⋅ 𝑥) = 𝑓(−𝑚) = 𝑞𝑑 𝑓(𝑚). 𝑥∈𝔽𝑑 𝑞

We will write

1/𝑟 𝑟

‖𝑓‖𝐿𝑟 (𝔽𝑑𝑞 ,𝑑𝑚) = ( ∑ |𝑓(𝑚)| ) 𝑚∈𝔽𝑑 𝑞

while

1/𝑟

‖𝑓‖𝐿𝑟 (𝔽𝑑𝑞 ,𝑑𝑥) = (𝑞−𝑑 ∑ |𝑓(𝑥)|𝑟 ) 𝑥∈𝔽𝑑 𝑞

and ‖𝑓‖𝐿𝑟 (𝑆,𝑑𝜍)

1 ∑ |𝑓(𝑥)|𝑟 ) =( |𝑆| 𝑥∈𝑆

1/𝑟

where 𝑆 ⊆ 𝔽𝑑𝑞 is any subset sphere of some given radius, and where 𝑑𝜎 denotes the surface measure of 𝑆 which we can think of as the function 𝑞𝑑 𝑑𝜎(𝑥) = |𝑆| 𝑆(𝑥), again using 𝑆(𝑥) as the characteristic function of the set 𝑆. Finally for a set 𝐸 ⊆ 𝔽𝑑𝑞 , we will write ‖𝑓‖𝐿∞ (𝐸,𝑑𝑥) = 𝑞−𝑑 max |𝑓(𝑥)|. 𝑥∈𝐸

This notation is economical as we can now write the Plancherel identity as ‖𝑓ˆ‖ = ‖𝑓‖𝐿2 (𝔽2𝑞 ,𝑑𝑥) . ‖ ‖ 2 𝐿2 (𝔽𝑞 ,𝑑𝑚)

72

Chapter 4. Wolff’s exponent

We further define the (additive) convolution of 𝑓 and 𝑔 as 𝑓 ∗ 𝑔(𝑥) = ∑ 𝑓(𝑦)𝑔(𝑥 − 𝑦). 𝑦∈𝔽𝑑 𝑞

It follows easily that ˜ ˜ 𝑔(𝑚) 𝑓 ∗ 𝑔(𝑚) = 𝑓(𝑚) ̃

(4.3)

˜ = 𝑓˜ ∗ 𝑔.̃ 𝑓𝑔

(4.4)

and We leave these last two equalities as an exercise (see Exercise 4.2). The interested reader can consult the Proof of Lemma 4.4 in [17] for a more ˆ 𝑓, ˜ and 𝑓∨ along with some of the in-depth discussion of the objects 𝑓, underlying mathematics.

4.3.2 Restriction theory The finite field restriction problem refers to establishing the following inequality. We use the notations developed in Section 4.3.1. Problem 4.3.1 (Finite Field Restriction Problem). Determine the values 1 ≤ 𝑝, 𝑟 ≤ ∞ such that there exists a constant 𝐶𝑝,𝑟,𝑑 so that for all functions 𝑓 ∶ 𝔽𝑑𝑞 → ℂ, we have ˜ 𝐿𝑟 (𝑆,𝑑𝜍) ≪𝑝,𝑟,𝑑 ‖𝑓‖ 𝑝 𝑑 ‖𝑓‖ 𝐿 (𝔽𝑞 ,𝑑𝑚) .

(4.5)

By duality this is equivalent to asking for ‖(𝑓𝑑𝜎)∨ ‖𝐿𝑝′ (𝔽𝑑𝑞 ,𝑑𝑚) ≪𝑝,𝑟,𝑑 ‖𝑔‖𝐿𝑟′ (𝑆,𝑑𝜍) .

(4.6)

We will use the notation 𝑅(𝑟 → 𝑝) and 𝑅∗ (𝑝′ → 𝑟′ ) to denote the best constants appearing in (4.5) and (4.6), respectively. The restriction problem has been intricately and intimately tied to many problems in geometric combinatorics including the distance problem. We next note some straight-forward estimates. Proposition 4.3.2. We have 𝑅∗ (𝑝1 → 𝑟) ≤ 𝑅∗ (𝑝2 → 𝑟)

for

𝑝2 ≤ 𝑝1

𝑅∗ (𝑝 → 𝑟1 ) ≤ 𝑅∗ (𝑝 → 𝑟2 )

for

𝑟2 ≤ 𝑟1 .

and

4.4. Exercises: Chapter 4

73

Furthermore, we have 𝑅∗ (𝑝 → ∞) = 1

for

1 ≤ 𝑝 ≤ ∞,

(4.7)

𝑅∗ (𝑝 → 2) ∼ √𝑞

for

2 ≤ 𝑝 ≤ ∞.

(4.8)

and

We will leave the proof of this proposition as an exercise (see Exercise 4.1). There are many results in restriction theory which have been proven in the setting of finite fields ([84, 85, 105–107, 111]). Theorem 4.1.3 can be and should be properly viewed as a restriction inequality in the case 𝑓(𝑥) = 𝐸(𝑥) for some set 𝐸. It is instructive to compare the finite field restriction problem with its Euclidean analogue, and we encourage the reader to do so. Indeed the Euclidean restriction problem has a rich history, but we leave it to the interested reader to pursue the Euclidean restriction problem on their own time (see, for example, [102, 149]).

4.4 Exercises: Chapter 4 Exercise 4.1. Prove the four parts of Proposition 4.3.2. Exercise 4.2. Verify equations (4.3) and (4.4). Exercise 4.3. Find a formula for 𝜈(0) when 𝑞 ≡ 3 (mod 4). Exercise 4.4. Prove Proposition 4.2.1 for general dimensions 𝑑. Exercise 4.5. Prove that ∑ |𝑆 𝑡̂ (𝑚)|2 = 𝑞−𝑑 − 𝑞−𝑑−1 + 𝛿(𝑚)𝑞−1 , 𝑡∈𝔽𝑞

where 𝛿(⋅) is the 𝑑-dimensional delta function. Exercise 4.6. Let 𝑝 and 𝑞 be conjugate exponents with 1 < 𝑝, 𝑞 < ∞. 1 1 That is, for 1 < 𝑝, 𝑞 < ∞, assume that 𝑝 + 𝑞 = 1. Show that |Δ(𝐸)| ≥

|𝐸|2𝑞 𝑞/𝑝

(∑𝑡 𝜈(𝑡)𝑝 )

.

74

Chapter 4. Wolff’s exponent

Exercise 4.7. Prove the following generalization of Theorem 4.1.2. For even dimensions 𝑑 ≥ 2 show that: 2 2

−1

4

∑ 𝜈(𝑡) = 𝑞 |𝐸| + 𝑞 𝑡

3𝑑

2

ˆ || ) − 𝑞𝑑−1 |𝐸|2 . ∑ ( ∑ ||𝐸(𝑚) 𝑡

‖𝑚‖=𝑡

Exercise 4.8. Find an explicit value for ∑𝑡 𝜈(𝑡)2 for odd dimensions 𝑑 ≥ 3.

10.1090/car/037/05

Chapter

5

Rings and generalized distances Next we turn our attention to studying generalizations of the distance problem. We will examine some methods by which to generalize the problem: we can either vary the ambient space of the underlying problem (which brought finite field models to light in the first place), or we can generalize our notion of a distance. We could also think of distance set problems in the more general category of finite point configurations, though we take up this topic in Chapter 6.

5.1 Distances in finite rings The first generalization we will discuss involves moving from the finite field 𝔽𝑞 to the cyclic ring ℤ𝑞 ≔ ℤ/𝑞ℤ. We use the notation ℤ𝑞 to denote this ring for maximal clarity. For a finite field 𝔽𝑞 to be defined, 𝑞 must be a power of a prime. In the case of the ring of integers modulo 𝑞, ℤ𝑞 is meaningfully defined for all integers 𝑞 ≥ 2. Recall that ℤ×𝑞 is the set of units in ℤ𝑞 , so that ℤ×𝑞 = {𝑥 ∈ ℤ ∶ 1 ≤ 𝑥 ≤ 𝑞 and gcd(𝑥, 𝑞) = 1}1 . In this setting of integers mod 𝑞, the distance problem can be worded similarly to the finite field case. First, ℤ𝑑𝑞 = ℤ𝑞 ×⋯×ℤ𝑞 is the free module 1 Again we are conflating integers with the set of equivalence classes mod 𝑞. Technically, the set of units modulo 𝑞 refers to the set of equivalence classes defined by the equivalence relation which we used to define the integers mod 𝑞, such that the representatives of those equivalence classes are relatively prime to 𝑞. To enumerate this technically correct definition is much too burdensome to be helpful, hence why we will continue to conflate the set of integers and their equivalence classes.

75

76

Chapter 5. Rings and generalized distances

of rank 𝑑 over the set of integers modulo 𝑞. For 𝑥 = (𝑥1 , . . . , 𝑥𝑑 ) ∈ ℤ𝑑𝑞 , we define ‖𝑥‖ = 𝑥12 + ⋯ + 𝑥𝑑2 ∈ ℤ𝑞 , and informally we refer to ‖𝑥‖ as the “length” or “norm” of 𝑥. Again, the function 𝑓(𝑥) = ‖𝑥‖ is not a norm in any (analytic) sense, though it behaves like a norm in an algebraic sense. Indeed, ‖𝑂𝑥‖ = ‖𝑥‖ for any 𝑥 ∈ ℤ𝑑𝑞 and 𝑂 ∈ 𝑂 𝑑 (ℤ𝑞 ). Here 𝑂 𝑑 (ℤ𝑞 ) = {𝑂 ∈ 𝑀𝑎𝑡 𝑑 (ℤ𝑞 ) ∶ 𝑂𝑇 𝑂 = 𝐼} is the set of 𝑑 × 𝑑 orthogonal matrices with entries in ℤ𝑞 . The distance set of 𝐸 ⊆ ℤ𝑑𝑞 is defined analogously: Δ(𝐸) = {‖𝑥 − 𝑦‖ ∶ 𝑥, 𝑦 ∈ 𝐸}. The Erdős-Falconer distance problem in ℤ𝑑𝑞 is as follows. Problem 5.1.1. Let 𝐸 ⊆ ℤ𝑑𝑞 . Find the minimal value of 𝛼 such that there exist positive constants 𝐶, 𝑐 (independent of 𝑞) so that if |𝐸| ≥ 𝐶𝑞𝛼 then |Δ(𝐸)| ≥ 𝑐𝑞. The first such result was achieved in [26]. Theorem 5.1.2. Suppose that 𝑞 = 𝑝ℓ is a power of an odd prime. If 𝐸 ⊆ ℤ𝑑𝑞 has cardinality |𝐸| ≫ ℓ(ℓ + 1)𝑞

(2ℓ−1)𝑑+1 2ℓ

, then Δ(𝐸) ⊇ ℤ×𝑞 ∪ {0}.

Remark 5.1.3. Note that when ℓ = 1, ℤ𝑝 = 𝔽𝑝 is a field and the results in Theorem 5.1.2 match those of Theorem 2.3.3. Also, note |ℤ×𝑞 | = 2 𝑝ℓ − 𝑝ℓ−1 . Thus, we have shown that |Δ(𝐸)| ≥ 3 𝑞 for sets 𝐸 ⊆ ℤ𝑞 of sufficiently large cardinality. Other than Theorem 5.1.2, only one other such result is known ([24]). Theorem 5.1.4. Suppose that 𝑑 ≥ 3, and 𝑞 has prime factorization 𝑞 = 𝛼 𝛼 𝑝1 1 . . . 𝑝𝑘 𝑘 , where 2 < 𝑝1 < ⋯ < 𝑝 𝑘 . Then if 𝐸 ⊆ ℤ𝑑𝑞 has cardinality −

|𝐸| ≫ 𝜏(𝑞)𝑞𝑑 𝑝1 divisors of 𝑞.

𝑑−2 2

, then Δ(𝐸) = ℤ𝑞 . Here, 𝜏(𝑞) is the number of positive

Corollary 5.1.5. Let 𝑑 ≥ 3, and assume 𝑞 = 𝑝ℓ is the power of an odd prime. If 𝐸 ⊆ ℤ𝑑𝑞 has cardinality |𝐸| ≫ ℓ𝑞

(2ℓ−1)𝑑+2 2ℓ

, then Δ(𝐸) = ℤ𝑞 .

5.1. Distances in finite rings

77

The proof of Theorem 5.1.2 will mimic the proof of Theorem 2.3.3, though we will have more arithmetic difficulties with which to contend such as the presence of zero divisors. In turn, the proof of Theorem 5.1.4 will follow the outline to Theorem 5.1.2, though much more care is needed in evaluating the corresponding exponential sums.

5.1.1 Fourier analysis over ℤ𝑑𝑞 Just as in the finite field setting, for a function 𝑓 ∶ ℤ𝑑𝑞 → ℂ, we define the Fourier transform as ˆ 𝑓(𝑚) = 𝑞−𝑑 ∑ 𝜒(−𝑥 ⋅ 𝑚)𝑓(𝑥). 𝑥∈ℤ𝑑 𝑞

Here, 𝜒(𝑥) = 𝑒2𝜋𝑖𝑥/𝑞 is the canonical additive character in 𝔽𝑞 . Note that 𝜒 is well defined since 𝜒 is periodic with period 𝑞, and hence evaluating 𝜒 at any two elements in some equivalence class will yield the same output. Just as was the case in finite fields, we have orthogonality, Plancherel and inversion identities: Proposition 5.1.6. We have 𝑞−𝑑 ∑ 𝜒(𝑥 ⋅ 𝑚) = { 𝑥∈ℤ𝑑 𝑞

1 0

𝑚 ≠ (0, . . . , 0) 𝑚 = (0, . . . , 0).

Moreover, let 𝑓 and 𝑔 be functions mapping ℤ𝑑𝑞 to the complex numbers ℂ. Then, ˆ 𝑔(𝑚) ∑ 𝑓(𝑚) ̂ = 𝑞−𝑑 ∑ 𝑓(𝑥)𝑔(𝑥) 𝑚∈ℤ𝑑 𝑞

𝑥∈ℤ𝑑 𝑞

ˆ 𝑓(𝑥) = ∑ 𝑓(𝑚)𝜒(𝑥 ⋅ 𝑚). 𝑥∈ℤ𝑑 𝑞

Lemma 5.1.7. Let 𝑆 𝑡 = {𝑥 ∈ ℤ𝑑𝑞 ∶ ‖𝑥‖ = 𝑡} denote the sphere of radius 𝑡 in ℤ𝑑𝑞 . Suppose that 𝑞 is odd. If 𝑡 ∈ ℤ×𝑞 is a unit and 𝑑 = 2, or if 𝑑 ≥ 3 and 𝑡 ∈ ℤ𝑞 is arbitrary, then we have |𝑆 𝑡 | = 𝑞𝑑−1 (1 + 𝑜(1)). For all dimensions 𝑑 ≥ 2 and all radii 𝑡 ∈ ℤ𝑞 , we have |𝑆 𝑡 | ≪ 𝑞𝑑−1 .

78

Chapter 5. Rings and generalized distances

Finally, we will need the following estimates on the Fourier transform of the characteristics function of the sphere: Lemma 5.1.8. Let 𝑆 𝑡 denote the indicator function of the sphere of radius 𝑡. First, suppose that 𝑞 = 𝑝ℓ is a power of an odd prime. Then, for 𝑑 ≥ 2, 𝑡 ∈ ℤ×𝑞 , and 𝑚 ≠ (0, . . . , 0) we have |𝑆 𝑡̂ (𝑚)| ≤ ℓ(ℓ + 1)𝑞−

𝑑+2ℓ−1 2ℓ

𝛼

. 𝛼

If 𝑑 > 2, and 𝑞 has the prime factorization 𝑞 = 𝑝1 1 . . . 𝑝𝑘 𝑘 where 2 < 𝑝1 < ⋯ < 𝑝 𝑘 , then for 𝑚 ≠ (0, . . . , 0), we have |𝑆 𝑡̂ (𝑚)| ≤

𝜏(𝑞) − 𝑑−2 𝑝 2 𝑞 1

for all 𝑡 ∈ ℤ𝑞 . We are now primed to prove Theorems 5.1.2 and 5.1.4. As usual for 𝐸 ⊆ ℤ𝑑𝑞 we define 𝜈(𝑡) = |{(𝑥, 𝑦) ∈ 𝐸 × 𝐸 ∶ ‖𝑥 − 𝑦| = 𝑡}|. For each unit 𝑡 ∈ ℤ×𝑞 , our goal will be to show that 𝜈(𝑡) > 0 which implies the existence of a pair of elements 𝑥 and 𝑦 in 𝐸 such that ‖𝑥 − 𝑦‖ = 𝑡, and thus 𝑡 ∈ Δ(𝐸). First, suppose that 𝑞 = 𝑝ℓ is the power of an odd prime. We again use the set to denote its own characteristic (or indicator) function. For example: 1

𝑥∈𝐸

0

𝑥 ∉ 𝐸.

𝐸(𝑥) = { Thus, we can write 𝜈(𝑡) as: 𝜈(𝑡) = ∑ 𝐸(𝑥)𝐸(𝑦)𝑆 𝑡 (𝑥 − 𝑦) 𝑥,𝑦

= ∑ 𝐸(𝑥)𝐸(𝑦)𝑆ˆ𝑡 (𝑚)𝜒((𝑥 − 𝑦) ⋅ 𝑚) 𝑥,𝑦,𝑚

= ∑ (∑ 𝐸(𝑦)𝜒(−𝑦 ⋅ 𝑚)) (∑ 𝐸(𝑥)𝜒(−𝑥 ⋅ 𝑚))𝑆ˆ𝑡 (𝑚). 𝑚

𝑦

𝑥

5.1. Distances in finite rings

79

Applying the definition of the Fourier transform, this becomes

ˆ ˆ 𝑆ˆ𝑡 (𝑚) ⋅ 𝑞𝑑 𝐸(𝑚) 𝜈(𝑡) = ∑ 𝑞𝑑 𝐸(𝑚) 𝑚 2

ˆ || 𝑆 𝑡̂ (𝑚) = 𝑞2𝑑 ∑ ||𝐸(𝑚) 𝑚

Just as before, if we separate the zero vector term 𝑚 = (0, . . . , 0) from the rest of the sum, we see that

𝜈(𝑡) = 𝑞−𝑑 |𝐸|2 |𝑆 𝑡 | + 𝑞2𝑑

2

ˆ || 𝑆 𝑡̂ (𝑚). ||𝐸(𝑚)

∑ 𝑚≠(0,. . .,0)

≔ 𝑀 + 𝑅𝑡 . By Lemma 5.1.7, since 𝑡 ∈ ℤ×𝑞 , we have |𝑆 𝑡 | = 𝑞𝑑−1 (1 + 𝑜(1)). Hence the main term satisfies 𝑀 = 𝑞−𝑑 |𝐸|2 |𝑆 𝑡 | = 𝑞−1 |𝐸|2 (1 + 𝑜(1)). By Lemma 5.1.8, since 𝑚 ≠ (0, . . . , 0) and again using the fact that 𝑡 ∈ ℤ×𝑞 , we see that |𝑆 𝑡̂ (𝑚)| ≤ ℓ(ℓ + 1)𝑞−

𝑑+2ℓ−1 2ℓ

.

Thus, the modulus of the remainder term satisfies |𝑅𝑡 | ≤ 𝑞2𝑑

max

𝑚≠(0,. . .,0)

|𝑆 𝑡̂ (𝑚)|

≤ 𝑞2𝑑 ⋅ ℓ(ℓ + 1)𝑞

∑

2

ˆ || ||𝐸(𝑚)

𝑚≠(0,. . .,0) 𝑑+2ℓ−1 − 2ℓ

2

ˆ || ∑ ||𝐸(𝑚) 𝑚∈𝔽𝑑 𝑞

80

Chapter 5. Rings and generalized distances

where we put back in the term 𝑚 = (0, . . . , 0). We can finish our estimate by applying Plancherel: |𝑅𝑡 | ≤ 𝑞2𝑑 ⋅ ℓ(ℓ + 1)𝑞−

𝑑+2ℓ−1 2ℓ

𝑞−𝑑 ∑ 𝐸(𝑥)2 𝑥∈ℤ𝑑 𝑞

= 𝑞2𝑑 ⋅ ℓ(ℓ + 1)𝑞− = ℓ(ℓ + 1)𝑞−

𝑑+2ℓ−1 2ℓ

𝑑+2ℓ−1 2ℓ

𝑞−𝑑 |𝐸|

𝑞𝑑 |𝐸|.

We have demonstrated that 𝜈(𝑡) =

|𝐸|2 (1 + 𝑜(1)) + 𝑅𝑡 𝑞

where the Remainder term 𝑅𝑡 satisfies |𝑅𝑡 | ≤ ℓ(ℓ + 1)𝑞−

𝑑+2ℓ−1 2ℓ

𝑞𝑑 |𝐸|,

for any 𝑡 ∈ ℤ×𝑞 . It follows that if 𝑑+2ℓ−1 |𝐸|2 (1 + 𝑜(1)) > ℓ(ℓ + 1)𝑞− 2ℓ 𝑞𝑑 |𝐸|, 𝑞

(5.1) (2ℓ−1)𝑑+1

then 𝜈(𝑡) > 0, and this inequality (5.1) holds if |𝐸| ≥ 𝐶ℓ(ℓ + 1)𝑞 2ℓ for a sufficiently large constant 𝐶. This establishes Theorem 5.1.2. 𝛼 𝛼 Now, assume that 𝑞 has prime decomposition 𝑞 = 𝑝1 1 . . . 𝑝𝑘 𝑘 for 𝑑−1 primes 2 < 𝑝1 < ⋯ < 𝑝 𝑘 . We again have |𝑆 𝑡 | = 𝑞 (1 + 𝑜(1)), but the decay estimate is changed. For 𝑚 ≠ (0, . . . , 0), we have the bound −

𝑑−2

|𝑆 𝑡̂ (𝑚)| ≤ 𝑞−1 𝜏(𝑞)𝑝1 2 . Following the same outline as the proof of the previous theorem, write 𝜈(𝑡) = 𝑞−1 |𝐸|2 (1 + 𝑜(1)) + 𝑅𝑡 , −

where |𝑅𝑡 | ≤ 𝜏(𝑞)𝑞𝑑−1 |𝐸|𝑝1

𝑑−2 2

. Hence, 𝜈(𝑡) > 0 whenever −

|𝐸| ≥ 𝐶𝜏(𝑞)𝑞𝑑 𝑝1

𝑑−2 2

for a sufficiently large constant 𝐶, which establishes Theorem 5.1.4. It remains to prove the Lemmas about the sphere 𝑆 𝑡 .

5.1. Distances in finite rings

81

5.1.2 Gauss sums over ℤ𝑛 Before we prove Lemmas 5.1.7 and 5.1.8, we first provide background in character theory over ℤ𝑛 as well as some well-known results on quadratic Gauss sums over ℤ𝑛 . First we consider the additive group ℤ𝑛 . The canonical (additive) character on this group is the function 𝜒(𝑥) = 𝑒2𝜋𝑖𝑥/𝑛 . This function is periodic with period 𝑛, so we can view this function as taking inputs from ℤ, though it is also well-defined on ℤ𝑛 . All other characters on ℤ𝑛 are of the form 𝜒𝑎 (𝑥) ≔ 𝜒(𝑎𝑥) = 𝑒2𝜋𝑖𝑎𝑥/𝑛 for 𝑎 ∈ ℤ𝑛 . Thus if 𝑎 = 0, then we achieve the trivial character: 𝜒0 (𝑥) ≡ 1. Proposition 5.1.6 yields an orthogonality relation for the canonical additive character: 𝑛−1 ∑ 𝜒(𝑚𝑥) = { 𝑥∈ℤ𝑛

1 0

𝑚≡0 𝑚≢0

(mod 𝑛) (mod 𝑛)

It is important to note that all characters on ℤ𝑑𝑛 are of the form 𝜒(𝛼 ⋅ 𝑥), where 𝜒(𝑥) = 𝑒2𝜋𝑖𝑥/𝑛 , and where 𝛼, 𝑥 ∈ ℤ𝑑𝑛 with 𝛼 ⋅ .𝑥 denoting the usual dot product: If 𝛼 = (𝛼1 , . . . , 𝛼𝑑 ) ∈ ℤ𝑑𝑛 and 𝑥 = (𝑥1 , . . . , 𝑥𝑑 ) ∈ ℤ𝑑𝑛 , then 𝛼 ⋅ 𝑥 = 𝛼1 𝑥1 + ⋯ + 𝛼𝑑 𝑥𝑑 . Next, we consider the multiplicative subgroup ℤ×𝑛 . Recall that 𝑥 ∈ ℤ𝑛 is a unit if and only if gcd(𝑥, 𝑛) = 1. The multiplicative characters on ℤ𝑛 (i.e., the characters on the multiplicative subgroup ℤ×𝑛 ) are functions 𝜓 ∶ ℤ×𝑛 → ℂ such that 𝜓 is not identically 0, 𝜓 is totally multiplicative (meaning that 𝜓(𝑥𝑦) = 𝜓(𝑥)𝜓(𝑦)), and 𝜓(𝑥 + 𝑛) = 𝜓(𝑥) for all 𝑥, so that 𝜓 is periodic with period 𝑛. We can extend these functions to be defined on all of ℤ𝑛 (and thus on all of ℤ) by adding the condition that 𝜓(𝑥) = 0 if gcd(𝑥, 𝑛) ≠ 1. These characters are called Dirichlet characters mod 𝑛. While there is much to say and learn about Dirichlet characters, we focus only on the Jacobi symbol, which is the one and only Dirichlet character that I know that is relevant to our study of distance sets2 . Recall that we 2 I cannot say for sure whether other Dirichlet characters are irrelevant to the study of distance sets since we cannot fully solve the problem!

82

Chapter 5. Rings and generalized distances

have already seen the Legendre symbol modulo an odd prime 𝑝: 1 𝑥 ( )={ 0 𝑝 −1

𝑥 is congruent to a nonzero square mod 𝑝 𝑥 ≡ 0 (mod 𝑝) 𝑥 is not congruent to a nonzero square mod 𝑝.

The Jacobi symbol is defined as a product of Legendre symbols, and it is defined only for odd integers 𝑛 ≥ 1. Definition 5.1.9. Let 𝑛 ≥ 3 be an odd integer with prime factorization 𝑒 𝑒 𝑛 = 𝑝11 . . . 𝑝𝑘𝑘 . Then the Jacobi symbol is the function 𝑒

𝑒

𝑥 1 𝑥 𝑥 𝑘 ( ) = ( ) ... ( ) . 𝑛 𝑝1 𝑝𝑘 𝑥

The Jacobi symbol also follows the convention that ( 1 ) = 1 for all integers 𝑥. We leave it to the reader to verify that this truly defines a Dirichlet character. We are now ready to discuss quadratic Gauss sums over the integers modulo 𝑛. Definition 5.1.10 (Quadratic Gauss sums). For a fixed integer 𝑛 ≥ 2 and for integers 𝑎 and 𝑏, we denote by 𝐺(𝑎, 𝑏, 𝑛) the following sum: 𝐺(𝑎, 𝑏, 𝑛) ≔ ∑ 𝜒(𝑎𝑥2 + 𝑏𝑥) 𝑥∈ℤ𝑛

where 𝜒(𝑥) = 𝑒2𝜋𝑖𝑥/𝑛 . For convenience, we set 𝐺(𝑎, 0, 𝑛) = 𝐺(𝑎, 𝑛). Since 𝑛 is fixed, the quantity 𝐺(𝑎, 𝑏, 𝑛) can be viewed as a function in the variables 𝑎 and 𝑏, but keeping track of the value 𝑛 will turn out to be important for us, so we include this value in the definition above. Note also that 𝐺(𝑎, 𝑏, 𝑛) = 𝐺(𝑎′ , 𝑏′ , 𝑛) if 𝑎 ≡ 𝑎′ (mod 𝑛) and 𝑏 ≡ 𝑏′ (mod 𝑛). This is because we can view 𝜒 as a function over ℤ𝑛 , as well as a function over ℤ. This slight abuse of notation should not cause any confusion. For the same reason, we may think of 𝐺(𝑎, 𝑏, 𝑛) as taking values from ℤ𝑛 or ℤ, since only the equivalence class of the integers 𝑎 and 𝑏 is relevant for this sum. Proposition 5.1.11 ([90]). Fix an odd positive integer 𝑛 ≥ 3, and let 𝐺(𝑎, 𝑏, 𝑛) denote the Gauss sum above, keeping in mind that we write

5.1. Distances in finite rings

83

𝐺(𝑎, 𝑛) = 𝐺(𝑎, 0, 𝑛). For any positive integer 𝑎 with gcd(𝑎, 𝑛) = 1, we have 𝑎 𝐺(𝑎, 𝑛) = 𝜀𝑛 ( ) √𝑛 𝑛 ⋅

where ( 𝑛 ) denotes the Jacobi symbol mod 𝑛, and 𝜀𝑛 = {

1 𝑖

𝑛 ≡ 1 (mod 4) 𝑛 ≡ 3 (mod 4).

Furthermore, for general values of 𝑎 ∈ ℤ𝑛 , we have 𝑎

𝐺(𝑎, 𝑏, 𝑛) = {

𝑏

𝑛

gcd(𝑎, 𝑛) ⋅ 𝐺 ( gcd(𝑎,𝑛) , gcd(𝑎,𝑛) , gcd(𝑎,𝑛) ) 0

gcd(𝑎, 𝑛) ∣ 𝑏 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒.

Definition 5.1.12 (Generalized Gauss Sum). For a fixed odd integer 𝑛 ≥ 3, we let 𝜓(𝑥) denote a Dirichlet character modulo 𝑛, and we let 𝜒𝑎 (𝑥) = 𝑒2𝜋𝑖𝑎𝑥/𝑛 be an additive character on ℤ𝑛 . Then, we set 𝜏(𝜓, 𝜒𝑎 ) = ∑ 𝜓(𝑥)𝜒𝑎 (𝑥). 𝑥∈ℤ𝑛

When 𝑎 = 1, it is customary to write 𝜏(𝜂, 𝜒1 ) = 𝜏(𝜓). Proposition 5.1.13 ([90]). If 𝜂(𝑥) denotes the Jacobi symbol mod 𝑛, then we have the bound |𝜏(𝜂, 𝜒𝑎 )| ≤ √𝑛. 𝑥

In particular, if 𝜂(𝑥) = ( 𝑝 ) is the Legendre symbol mod 𝑝, and if 𝜒𝑎 (𝑥) = 𝑒2𝜋𝑖𝑎𝑥/𝑝 , then we have the same bound: |𝜏(𝜂, 𝜒𝑎 )| ≤ √𝑝.

5.1.3 Proof of Lemma 5.1.7 We are ready to prove the estimates on the cardinality of the sphere as well as the bound on the Fourier transforms of the characteristic function of the sphere. We first prove the Lemma in the case that 𝑞 = 𝑝ℓ is a power of an odd prime. As before, put 𝜒(𝑥) = 𝑒2𝜋𝑖𝑥/𝑞 , and suppose 𝑗 is a fixed unit in ℤ𝑞 .

84

Chapter 5. Rings and generalized distances

Then by orthogonality (see Proposition 5.1.6), we have |𝑆𝑗 | = ∑ 𝑆𝑗 (𝑥) = 𝑞−1 ∑ ∑ 𝜒(𝑠(‖𝑥‖ − 𝑗)) 𝑠∈ℤ𝑞 𝑥∈ℤ𝑑 𝑞

𝑥∈ℤ𝑑 𝑞

= 𝑞−1 ∑ ∑ 𝜒 (𝑠 (𝑥12 + ⋯ + 𝑥𝑑2 − 𝑗)) 𝑠∈ℤ𝑞 𝑥∈ℤ𝑑 𝑞

= 𝑞−1 ∑ ∑ 𝜒 (𝑠𝑥12 ) . . . 𝜒 (𝑠𝑥𝑑2 ) 𝜒(−𝑠𝑗) 𝑠∈ℤ𝑞 𝑥∈ℤ𝑑 𝑞

= 𝑞−1 (𝑇∞ + 𝑇0 + ⋯ + 𝑇ℓ−1 ) , where ∑

𝑇𝑟 =

∑ 𝜒(𝑠𝑥12 ) . . . 𝜒(𝑠𝑥𝑑2 )𝜒(−𝑠𝑗)

0≤𝑠≤𝑞−1 𝑥∈ℤ𝑑 𝑞 𝑣𝑎𝑙𝑝 (𝑠)=𝑟 𝑑

∑

=

0≤𝑠≤𝑞−1 𝑣𝑎𝑙𝑝 (𝑠)=𝑟

∑

=

2

( ∑ 𝜒(𝑠𝑥 )) 𝜒(−𝑠𝑗) 𝑥∈ℤ𝑞

(𝐺(𝑠, 𝑞))𝑑 𝜒(−𝑠𝑗).

0≤𝑠≤𝑞−1 𝑣𝑎𝑙𝑝 (𝑠)=𝑟

Recall that for an integer 𝑥, we write 𝑣𝑎𝑙𝑝 (𝑥) = 𝑟 if 𝑝𝑟 is the largest power of the prime 𝑝 that divides 𝑥, and 𝑣𝑎𝑙𝑝 (0) = ∞. The term 𝑇∞ refers to the sum when 𝑠 = 0, so that 𝑇∞ = 𝑞𝑑 = 𝑝ℓ𝑑 . For 0 ≤ 𝑟 ≤ ℓ − 1, note that if 0 ≤ 𝑠 ≤ 𝑞 − 1 and 𝑣𝑎𝑙𝑝 (𝑠) = 𝑟, then 𝑠 can be written in the form 𝑠 = 𝑝𝑟 𝑠′ , where gcd(𝑠′ , 𝑝) = 1 and 0 ≤ 𝑠′ ≤ 𝑝ℓ−𝑟 is uniquely determined3 . Using this fact, along with Proposition 5.1.11, we see that 𝑇𝑟 = 𝑝𝑟𝑑

∑ (𝐺(𝑠, 𝑝ℓ−𝑟 ))𝑑 𝜒(−𝑠𝑗) 𝑠∈ℤ× 𝑝ℓ−𝑟 𝑑

= 𝑝𝑟𝑑 𝜀𝑝𝑑ℓ−𝑟 (𝑝ℓ−𝑟 ) 2

∑ 𝜂 (𝑠)

𝑑(ℓ−𝑟)

𝜒(−𝑠𝑗)

𝑠∈ℤ× 𝑝ℓ−𝑟

𝑠

where 𝜂(𝑠) = ( 𝑝 ) is the Legendre symbol mod 𝑝. 3 Note that the condition that 𝑟 is finite precludes the possibility of having 𝑠 = 0, but the statement is still true including 0 in the range of 𝑠.

5.1. Distances in finite rings

85

If 𝑑(ℓ − 𝑟) is even, we see that 𝑑

𝑑

𝑇𝑟 = 𝑝ℓ 2 +𝑟 2 𝜀𝑝𝑑ℓ−𝑟

𝑑

∑ 𝜒(−𝑠𝑗) 𝑠∈ℤ× 𝑝ℓ−𝑟

𝑑

= 𝑝ℓ 2 +𝑟 2 𝜀𝑝𝑑ℓ−𝑟 ( ∑ 𝜒(−𝑠𝑗) − 𝑠∈ℤ𝑝ℓ−𝑟 𝑑

𝑑

= −𝑝ℓ 2 +𝑟 2 𝜀𝑝𝑑ℓ−𝑟

∑

∑

𝜒(−𝑠𝑗))

𝑠∈𝑝ℤ𝑝ℓ−𝑟−1

𝜒(−𝑠𝑗),

𝑠∈𝑝ℤ𝑝ℓ−𝑟−1 𝑑

so that |𝑇𝑟 | ≤ 𝑝(ℓ+𝑟) 2 (𝑝ℓ−𝑟−1 ) = 𝑝 using the notation

ℓ(

𝑑+2 𝑑−2 )+𝑟( 2 )−1 2

. Note that we are

𝑝ℤ𝑝𝛼 = {𝑝𝑥 ∶ 𝑥 ∈ ℤ and 0 ≤ 𝑥 ≤ 𝑝𝛼 − 1} . We are tacitly identifying the ring ℤ𝑛 with the set of integers {0, . . . , 𝑛−1} which represent its equivalence classes. In doing so, we can think of 𝑝ℤ𝑝𝛼 as the set of non-units in the ring ℤ𝑝𝛼+1 . In other words, we identify 𝑝ℤ𝑝𝛼 = ℤ𝑝𝛼+1 ⧵ ℤ𝑝×𝛼+1 , which is equivalent to viewing ℤ𝑝𝛼+1 = ℤ𝑝×𝛼+1 ∪ 𝑝ℤ𝑝𝛼 . Now if 𝑑(ℓ − 𝑟) is odd, then, 𝑑

𝑇𝑟 = 𝑝(ℓ+𝑟) 2 𝜀𝑝𝑑ℓ−𝑟

∑ 𝜂(𝑠)𝜒(−𝑠𝑗) 𝑠∈ℤ× 𝑝ℓ−𝑟

𝑑

= 𝑝(ℓ+𝑟) 2 𝜀𝑝𝑑ℓ−𝑟 ( ∑ 𝜂(𝑠)𝜒(−𝑠𝑗) − 𝑠∈ℤ𝑝ℓ−𝑟

∑

𝜂(𝑠)𝜒(−𝑠𝑗)) .

𝑠∈𝑝ℤ𝑝ℓ−𝑟−1

⏟⎵⎵⎵⎵⎵⏟⎵⎵⎵⎵⎵⏟

⏟⎵⎵⎵⎵⎵⎵⏟⎵⎵⎵⎵⎵⎵⏟

𝜏(𝜂,𝜒−𝑗 )

𝑅

Proposition 5.1.13 gives |𝜏(𝜂, 𝜒−𝑗 )| ≤ √𝑝ℓ−𝑟 , while the triangle inequality yields the crude bound | | ∑ 𝜂(𝑠)𝜒(−𝑠𝑗)|| ≤ |𝜂(𝑠)𝜒(−𝑠𝑗)| ≤ 𝑝ℓ−𝑟−1 . |𝑅| = || ∑ 𝑠∈𝑝ℤ 𝑠∈𝑝ℤ | | 𝑝ℓ−𝑟−1 𝑝ℓ−𝑟−1

86

Chapter 5. Rings and generalized distances

When 𝑑(ℓ − 𝑟) is odd, it follows that 𝑑

1

|𝑇𝑟 | ≤ 𝑝(ℓ+𝑟) 2 (𝑝(ℓ−𝑟) 2 + 𝑝ℓ−𝑟−1 ) . ℓ−𝑟 2 𝑑+2 𝑑−2 ℓ 2 +𝑟 2 −1

Noting that

≤ ℓ − 𝑟 − 1 for 𝑟 ≤ ℓ − 2, we have shown that |𝑇𝑟 | ≤

2𝑝 when 0 ≤ 𝑟 ≤ ℓ − 2, and |𝑇ℓ−1 | ≤ 2𝑝ℓ𝑑− our estimates show: 𝑑

𝑝ℓ𝑑− 2

|𝑇𝑟 | ≤ |𝑇ℓ−1 | ≤ {

ℓ𝑑−

2𝑝

𝑑−1 2

. Altogether,

𝑑(ℓ − 1) is even 𝑑−1 2

𝑑(ℓ − 1) is odd.

Thus, we have |𝑆𝑗 | = 𝑞𝑑−1 + 𝑞−1 (𝑇0 + ⋯ + 𝑇ℓ−1 ), where 𝑑

ℓ−1

|𝑇0 + ⋯ + 𝑇ℓ−1 | ≤ ∑ |𝑇𝑟 | ≤ { 𝑟=0

ℓ𝑝ℓ𝑑− 2 2ℓ𝑝

ℓ𝑑−

𝑑(ℓ − 1) is even 𝑑−1 2

𝑑(ℓ − 1) is odd.

(5.2)

Putting everything together, and recalling that we set 𝑞 = 𝑝ℓ , we have ℓ−1

|𝑆𝑗 | = 𝑝ℓ(𝑑−1) + 𝑂 (𝑞−1 ∑ |𝑇𝑟 |)

(5.3)

𝑟=0 𝑑

=𝑞

𝑑−1

+ 𝑂 ({

ℓ𝑝ℓ(𝑑−1)− 2 ℓ𝑝

ℓ(𝑑−1)−

𝑑(ℓ − 1) is even

𝑑−1 2

𝑑(ℓ − 1) is odd

})

= 𝑝ℓ(𝑑−1) (1 + 𝑜(1)).

(5.4)

The general case follows from the Chinese Remainder Theorem. Recall ℓ ℓ that if 𝑞 = 𝑝11 . . . 𝑝𝑘𝑘 , then ℤ𝑞 ≅ ℤ𝑝ℓ1 × ⋯ × ℤ 1

ℓ

𝑝𝑘𝑘

and

ℤ×𝑞 ≅ ℤ×ℓ1 × ⋯ × ℤ×ℓ𝑘 . 𝑝1

𝑝𝑘

(5.5)

See Exercise 5.6 for more details. Write 𝑡 ∈ ℤ×𝑞 as 𝑡 = (𝑡1 , . . . , 𝑡 𝑘 ), where 𝑡𝑟 ∈ ℤ×ℓ𝑟 . To find the number of solutions to the equation ‖𝑥‖ = 𝑡 in 𝑝𝑟 ℤ𝑞 , one must solve the equation ‖𝑥‖ = 𝑡𝑟 in each component ℤ𝑝ℓ𝑟 . The 𝑟 number of solutions in ℤ𝑞 is then the product of the number of solutions in ℤ𝑝ℓ𝑟 . Hence: 𝑟

𝑘

|𝑆 𝑡 | = ∏ |𝑆 𝑡𝑟 | = 𝑞𝑑−1 (1 + 𝑜(1)), 𝑟=1

which completes the proof of Lemma 5.1.7.

5.1. Distances in finite rings

87

5.1.4 Fourier decay of the sphere Next we prove Lemma 5.1.8. We first require 𝑞 = 𝑝ℓ to be a power of an odd prime. For 𝑚 ∈ ℤ𝑑𝑞 ⧵ {(0, . . . , 0)}, by applying the definition of the Fourier transform, and using orthogonality, we see that 𝑆𝑗̂ (𝑚) = 𝑞−𝑑 ∑ 𝑆𝑗 (𝑥)𝜒(−𝑥 ⋅ 𝑚) 𝑥∈ℤ𝑑 𝑞

= 𝑞−𝑑−1 ∑ ∑ 𝜒((𝑥12 + ⋯ + 𝑥𝑑2 − 𝑗)𝑡)𝜒(−𝑚 ⋅ 𝑥) 𝑡∈ℤ𝑞 𝑥∈ℤ𝑑 𝑞

= 𝑞−𝑑−1 ∑

∑

𝑥∈ℤ𝑑 𝑞

𝜒((𝑥12 + ⋯ + 𝑥𝑑2 − 𝑗)𝑡)𝜒(−𝑚 ⋅ 𝑥)

𝑡∈ℤ𝑞 ⧵{0}

since the term corresponding to 𝑡 = 0 is 0 by Proposition 5.1.6 as 𝑚 ≠ (0, . . . , 0). Thus, 𝑑

𝑆𝑗̂ (𝑚) = 𝑞−𝑑−1 ∑ 𝜒(−𝑗𝑡) ∏ ( ∑ 𝜒(𝑥𝑖2 𝑡 − 𝑚𝑟 𝑥𝑖 )) 𝑡≠0

𝑟=1

𝑥𝑖 ∈ℤ𝑞

𝑑

= 𝑞−𝑑−1 ∑ 𝜒(−𝑗𝑡) ∏ 𝐺(𝑡, −𝑚𝑟 , 𝑞) 𝑡≠0

(5.6)

𝑟=1

where we regrouped our terms and recognized the last sums as quadratic Gauss sums. By Proposition 5.1.11, 𝐺(𝑡, −𝑚𝑟 , 𝑞) = 0, unless 𝑚𝑟 ≡ 0 (mod gcd(𝑡, 𝑞)). Note also that 𝑣𝑎𝑙𝑝 (𝑡) = 𝜈 implies that gcd(𝑡, 𝑞) = 𝑝𝜈 . Now, we partition ℓ−1

𝑆𝑗̂ (𝑚) = ∑ 𝑆𝑗𝜈̂ (𝑚), 𝜈=0

where applying the last equality in Proposition 5.1.11 yields 𝑆𝑗𝜈̂ (𝑚) ≔ 𝑆𝑗̂ (𝑚)|𝑣𝑎𝑙

𝑝 (𝑡)=𝜈

𝑑

= 𝑞−𝑑−1

∑ 𝑣𝑎𝑙𝑝 (𝑡)=𝜈

𝜒(−𝑗𝑡)𝑝𝜈𝑑 ∏ 𝐺 ( 𝑟=1

𝑡 −𝑚𝑟 ℓ−𝜈 , ,𝑝 ). 𝑝𝜈 𝑝𝜈

For convenience, we put 𝑢 = 𝑡/𝑝𝜈 and 𝜇𝑟 = 𝑚𝑟 /𝑝𝜈 . We note that 𝑢 is a unit and 𝜇𝑟 are integers, since 𝑚𝑟 ≡ 0 (mod 𝑝𝜈 ) (as otherwise the quadratic Gauss sum vanishes). Again, we write 𝜇 = (𝜇1 , . . . , 𝜇𝑑 ) and

88

Chapter 5. Rings and generalized distances 𝑑

‖𝜇‖ = ∑𝑟=1

𝑚2𝑟 , 𝑝2𝜈

𝑆𝑗𝜈̂ (𝑚) = 𝑞−𝑑−1

⋅

and ( 𝑝𝑚 ) denotes the Jacobi symbol. Hence, 𝑑

𝑝𝜈𝑑 ⋅ 𝑝 2 (ℓ−𝜈) 𝜀𝑝𝑑ℓ−𝜈 ⋅ 𝜒 (−𝑗𝑢 −

∑ 𝑣𝑎𝑙𝑝 (𝑡)=𝜈 ℓ−1

𝑑

= 𝑞−𝑑−1 ∑ 𝑝 2 (ℓ+𝜈) 𝜀𝑝𝑑ℓ−𝜈 𝜈=0

∑

𝜒 (−

ᵆ∈ℤ× 𝑝ℓ−𝜈

𝑑 ‖𝜇‖ 𝑢 ) ⋅ ( ℓ−𝜈 ) 4𝑢 𝑝

(ℓ−𝜈)𝑑 ‖𝜇‖ 𝑢 − 𝑗𝑢) ( ) . 4𝑢 𝑝

To finish the argument, we claim that we have the bound | 𝛾𝑑 | 𝛾 | | 𝑢 | ∑ 𝜒 (𝑎𝑢−1 + 𝑏𝑢) ( ) | ≤ (𝛾 + 1)𝑝 2 , 𝑝 | |ᵆ∈ℤ× | 𝑝𝛾 |

(5.7)

where 𝑎 ∈ ℤ𝑝𝛾 is arbitrary, 𝑏 ∈ ℤ𝑝𝛾 is a unit, and 𝛾 is a positive integer. Accepting the claim for the moment, we see that ℓ−1

𝑑

|𝑆𝑗̂ (𝑚)| ≤ 𝑞−𝑑−1 ∑ (ℓ − 𝜈 + 1)𝑝 2 (ℓ+𝜈) 𝑝

ℓ−𝜈 2

𝜈=0

≤ (ℓ + 1)𝑞−𝑑−1 𝑝

(𝑑+1)ℓ 2

≤ (ℓ + 1)𝑞−𝑑−1 𝑞

(𝑑+1)ℓ 2ℓ

ℓ−1 (

∑𝑝

𝑑−1 )𝜈 2

𝜈=0

≤ ℓ(ℓ + 1)𝑞−

1

ℓ𝑞 ℓ

((

𝑑−1 )(ℓ−1)) 2

𝑑+2ℓ−1 2ℓ

from which the result follows. It remains to justify (5.7). For convenience, we define the Salié sum (or twisted Kloosterman sum) as 𝑥 𝑆(𝑎, 𝑏, 𝑛) = ∑ 𝜒 (𝑎𝑥−1 + 𝑏𝑥) ( ) , 𝑛 𝑥∈ℤ× 𝑛

and we define the Kloosterman sum as 𝐾(𝑎, 𝑏, 𝑛) = ∑ 𝜒 (𝑎𝑥−1 + 𝑏𝑥) . 𝑥∈ℤ× 𝑛

H. Salié ([132]) gave the bound |𝑆(𝑎, 𝑏, 𝑝)| ≤ 2√𝑝 for gcd(𝑎, 𝑏, 𝑝) = 1 and 𝑝 an odd prime. Note that when 𝑞 = 𝑝𝛾 and 𝛾 is even, we have 𝑆(𝑎, 𝑏, 𝑞) = 𝐾(𝑎, 𝑏, 𝑞). A. Weil ([154]) provided the well known bound |𝐾(𝑎, 𝑏, 𝑞)| ≤ 𝜏(𝑞) gcd(𝑎, 𝑏, 𝑞)1/2 𝑞1/2 , where 𝜏(𝑞) denotes the number of

5.1. Distances in finite rings

89

positive divisors of 𝑞. If 𝛾𝑑 is even, then the sum (5.7) is reduced to the Kloosterman sum 𝐾(𝑎, 𝑏, 𝑞) which has size |𝐾(𝑎, 𝑏, 𝑞)| ≤ (𝛾 + 1)𝑝𝛾/2 , since gcd(𝑎, 𝑏, 𝑞) = 1 because 𝑏 is a unit (mod 𝑞). Finally, when 𝛾 and 𝑑 are odd we invoke the bound | | | ∑ 𝜒 (𝑎𝑥−1 + 𝑏𝑥) ( 𝑥 )| ≤ (𝛾 + 1)𝑝𝛾/2 | 𝑝 | |𝑥∈ℤ𝑝𝛾 |

where 𝛾 is odd ([134]). Next we turn our attention to the case when 𝑞 has prime factorization 𝛼 𝛼 𝑞 = 𝑝1 1 . . . 𝑝𝑘 𝑘 . We start with the equation (5.6):

𝑑

𝑆𝑗̂ (𝑚) = 𝑞−𝑑−1 ∑ 𝜒(−𝑗𝑡) ∏ 𝐺(𝑡, −𝑚𝑟 , 𝑞)𝑠. 𝑡≠0

𝛽

𝑟=1

𝛽

Our first step is to write 𝑡 = 𝑝1 1 . . . 𝑝𝑘𝑘 𝑢 where 𝑢 ∈ ℤ×𝑞′ is uniquely de𝛼 −𝛽

𝛼 −𝛽

termined for modulus 𝑞′ = 𝑝1 1 1 . . . 𝑝𝑘 𝑘 𝑘 and 𝛽𝑟 ≥ 0. Note that 𝛽𝑟 < 𝛼𝑟 for some 𝑟 since 𝑡 ≠ 0. We will use the notation ∑𝛽 to denote the sum over all such (𝛽1 , . . . , 𝛽 𝑘 ). For 𝑚 = (𝑚1 , . . . , 𝑚𝑑 ) and 𝛽 = (𝛽1 , . . . , 𝛽 𝑘 ), we 𝛽 𝛽 define 𝜆𝑚,𝛽 to be 1 if 𝑝1 1 . . . 𝑝𝑘𝑘 ∣ 𝑚𝑟 for all 𝑟, and zero otherwise. When 𝑚𝑟 𝜆𝑚,𝛽 = 1, we put 𝜇𝑟 = 𝛽1 𝛽𝑘 . For notational convenience, we will write 𝑋𝛽 =

𝛽 𝑝1 1

𝛽 . . . 𝑝𝑘𝑘 .

𝑝1 . . .𝑝𝑘

Applying Proposition 5.1.11,

𝑑

𝑆 𝑡̂ (𝑚) = 𝑞−𝑑−1 ∑ ∑ 𝜒(−𝑋𝛽 𝑢𝑡) ∏ 𝐺 (𝑋𝛽 𝑢, −𝑚𝑟 , 𝑞) 𝛽 ᵆ∈ℤ×′ 𝑞

𝑟=1 𝑑

= 𝜆𝑚,𝛽 𝑞−𝑑−1 ∑ ∑ 𝑋𝛽𝑑 𝜒(−𝑋𝛽 𝑢𝑡) ∏ 𝐺(𝑢, −𝜇𝑟 , 𝑞′ ). 𝛽 ᵆ∈ℤ×′ 𝑞

𝑟=1

90

Chapter 5. Rings and generalized distances

Noting that by completing the square we have 𝐺(𝑢, −𝜇𝑟 , 𝑞′ ) = ∑ 𝜒 (𝑢𝑥2 − 𝜇𝑟 𝑥) 𝑥∈ℤ𝑞′

= ∑ 𝜒 (𝑢 (𝑥 − 𝑥∈ℤ𝑞′

𝜇𝑟 2 −𝜇2 ) )𝜒( 𝑟) 2𝑢 4𝑢

= ∑ 𝜒 (𝑢𝑥2 ) 𝜒 ( 𝑥∈ℤ𝑞′

−𝜇2𝑟 ) 4𝑢

where we used a change of variables in the last line. Taking the product of these terms and applying Proposition 5.1.11 yields 𝑑

𝑆 𝑡̂ (𝑚) = 𝜆𝑚,𝛽 𝑞−𝑑−1 ∑ ∑ 𝑋𝛽𝑑 𝜒(−𝑋𝛽 𝑢𝑡) ∏ 𝐺(𝑢, −𝜇𝑟 , 𝑞′ ) 𝛽 ᵆ∈ℤ×′ 𝑞

𝑟=1 𝑑

= 𝜆𝑚,𝛽 𝑞−𝑑−1 ∑ 𝑋𝛽𝑑 (𝑞′ ) 2 𝜀𝑑𝑞′ ∑ 𝜒 (−𝑋𝛽 𝑢𝑡 − ᵆ∈ℤ× 𝑞′

𝛽

‖𝜇‖ 𝑢 𝑑 )( ′) . 4𝑢 𝑞

We will apply the trivial bound to the sum in 𝑢 ∈ ℤ×𝑞′ : | | | ‖𝜇‖ 𝑢 𝑑 | 𝛽 𝛼 −𝛽 𝛽 𝛼 −𝛽 | ∑ 𝜒 (−𝑝1 1 . . . 𝑝𝑘𝑘 𝑢𝑡 − ) ( ′ ) | ≤ 𝑞′ = 𝑝1 1 1 . . . 𝑝𝑘 𝑘 𝑘 |ᵆ∈ℤ× 4𝑢 𝑞 | | | 𝑞′ to see that (

|𝑆 𝑡̂ (𝑚)| ≤ 𝑞−𝑑−1 ∑ 𝑝1

𝛼1 +𝛽1 2

)𝑑

(

. . . 𝑝𝑘

𝛼𝑘 +𝛽𝑘 2

)𝑑 𝛼 −𝛽 𝑝1 1 1

𝛽 𝑘

(

= 𝑞−𝑑−1 ∑ ∏ 𝑝𝑖 𝛽 𝑟=1

𝛼𝑟 +𝛽𝑟 2

)𝑑+𝛼𝑟 −𝛽𝑟

.

𝛼 −𝛽𝑘

. . . 𝑝𝑘 𝑘

5.2. Distances between two sets

91

Writing 𝛽𝑟 = 𝛼𝑟 − 𝜖𝑟 , we have 𝑘

(

|𝑆 𝑡̂ (𝑚)| ≤ 𝑞−𝑑−1 ∑ ∏ 𝑝𝑖

2𝛼𝑟 −𝜖𝑟 2

)𝑑+𝜖𝑟

𝜖𝑟 2

)𝑑+𝜖𝑟

𝛽 𝑟=1 𝑘

(𝛼𝑟 −

= 𝑞−𝑑−1 ∑ ∏ 𝑝𝑖 𝛽 𝑟=1 𝑘

−

= 𝑞−𝑑−1 ∑ 𝑞𝑑 ∏ 𝑝𝑖 𝛽

𝑑𝜖𝑟 −2 2

𝑟=1 𝑑−2 − 2

≤ 𝑞−1 𝜏(𝑞)𝑝𝑖

for some 𝑝𝑟 since 𝜖𝑟 > 0 for at least one value 𝑟 as 𝑡 ≠ 0. The largest value obtained by the quantity occurs when 𝜖1 = 1 and 𝜖𝑟 = 0 for 𝑟 > 1. Hence, −

|𝑆 𝑡̂ (𝑚)| ≤ 𝑞−1 𝜏(𝑞)𝑝1

𝑑−2 2

.

This completes the proof of Proposition 5.1.8.

5.2 Distances between two sets Our next generalization consists of examining distances determined by pairs of sets. We put Δ(𝐸, 𝐹) = {‖𝑥 − 𝑦‖ ∶ 𝑥 ∈ 𝐸, 𝑦 ∈ 𝐹}, where 𝐸, 𝐹 ⊆ 𝔽𝑑𝑞 or 𝐸, 𝐹 ⊆ ℤ𝑑𝑞 . The techniques used in Section 3.1 show that Δ(𝐸, 𝐹) ⊇ 𝔽∗𝑞 whenever |𝐸||𝐹| ≥ 𝐶𝑞𝑑+1 for a sufficiently large constant 𝐶 (see Exercise 5.9). Note that 0 ∈ Δ(𝐸, 𝐹) if 𝐸 ∩ 𝐹 ≠ ∅, though this need not be guaranteed by the size condition on |𝐸||𝐹|. Notice that the exponent 𝑑 + 1 cannot be improved unconditionally because of the previously discussed counterexamples in the finite field setting. However, some progress has been made on the problem. The first result in this direction was given by Koh and Shen ([99]). Theorem 5.2.1. Suppose that 𝐸, 𝐹 ⊆ 𝔽2𝑞 are such that |𝐸||𝐹| ≫ 𝑞8/3 . Then, there exists a constant 0 < 𝑐 ≤ 1 such that |Δ(𝐸, 𝐹)| ≥ 𝑐𝑞.

92

Chapter 5. Rings and generalized distances

We will delve into some of these ideas later in the section, though we leave it to the interested reader to fill in any missing details. This is the only result (out of Theorems 5.2.1, 5.2.2, and 5.2.3) that allows 𝐸 and 𝐹 to be arbitrary sets, and hence it is the only result that is a true generalization of the Erdős-Falconer distance problem for arbitrary sets. Up to logarithms, the problem was settled by Dietmann [36] in all dimensions, albeit for sets 𝐸 and 𝐹 of different sizes. Because of this restriction, these results will not push forward any progress on the finite field distance problem. Theorem 5.2.2. Suppose that 𝐸, 𝐹 ⊆ 𝔽𝑑𝑞 are such that |𝐸||𝐹| ≫ 𝑞𝑑 log 𝑞 where we have |𝐹| ≫ 𝑞 such that

𝑑+1 2

log 𝑞. Then there exists a constant 0 < 𝑐 ≤ 1 |Δ(𝐸, 𝐹)| ≥ 𝑐𝑞.

The logarithmic factors were removed in [100], thus completely settling the conjecture in the general case: Theorem 5.2.3. Suppose that 𝐸, 𝐹 ⊆ 𝔽𝑑𝑞 such that |𝐸||𝐹| ≫ 𝑞𝑑 where we have |𝐹| ≫ 𝑞

𝑑+1 2

. Then there exists a constant 0 < 𝑐 ≤ 1 such that |Δ(𝐸, 𝐹)| ≥ 𝑐𝑞.

Proof. We will sketch the proof of Theorem 5.2.3 as we have already developed the necessary machinery to do so. We need to recall a few facts from earlier chapters. First and foremost we will be using 𝜈(𝑡) = {(𝑥, 𝑦) ∈ 𝐸 × 𝐹 ∶ ‖𝑥 − 𝑦‖ = 𝑡}. There is no harm in using the same notation as in the single set case (𝐸 = 𝐹) since no confusion should occur. Proposition 5.2.4. Let 𝐸, 𝐹 ⊆ 𝔽𝑑𝑞 . Then, |Δ(𝐸, 𝐹)| ≥

|𝐸|2 |𝐹|2 ∑𝑡 𝜈(𝑡)2

(5.8)

and 2

|Δ(𝐸, 𝐹)| ≥

(|𝐸||𝐹| − 𝜈(0)) . ∑𝑡≠0 𝜈(𝑡)2

(5.9)

5.2. Distances between two sets

93

You have probably already guessed that we will use equation (5.8) when 𝑞 ≡ 3 (mod 4) and thus we need (5.9) in the case 𝑞 ≡ 1 (mod 4). Also, recall that 𝜎𝐸 (𝑡) denotes the spherical average: 2

ˆ || . 𝜎𝐸 (𝑡) = ∑ ||𝐸(𝑚) ‖𝑚‖=𝑡

Proposition 5.2.5. Let 𝐸, 𝐹 ⊆ 𝔽𝑑𝑞 . Then ∑ 𝜈(𝑡)2 ≤ 𝑡

|𝐸|2 |𝐹|2 + 𝑞2𝑑 |𝐹| max(𝜎𝐸 (𝑡)) 𝑞 𝑡∈𝔽𝑞

(5.10)

and 2

| |𝐸|2 |𝐹|2 3𝑑 || ˆ 𝐹(𝑚) ˆ || +𝑞2𝑑 |𝐹| max(𝜎𝐸 (𝑡)). (5.11) ∑ 𝜈(𝑡) ≤ +𝑞 | ∑ 𝐸(𝑚) 𝑞 𝑡≠0 |‖𝑚‖=0 | 𝑡 2

Proposition 5.2.6. For 𝑑 ≥ 2 even and 𝐸, 𝐹 ⊆ 𝔽𝑑𝑞 with |𝐸||𝐹| ≥ 16𝑞𝑑 we have 1 2 |𝐸||𝐹|, (5.12) (|𝐸||𝐹| − 𝜈(0)) ≥ 36 and 2 2 2 | | 3𝑑 | | ˆ 𝐹(𝑚) ˆ | ≤ |𝐸| |𝐹| + 𝜈(0)2 . (5.13) 𝑞 | ∑ 𝐸(𝑚) 𝑞 | |‖𝑚‖=0 Finally we have estimates on the maximal spherical average, which we have already seen play a major role in Theorem 4.1.1. Proposition 5.2.7. Let 𝐸 ⊆ 𝔽𝑑𝑞 . If 𝑑 ≥ 3 is odd then max (𝜎𝐸 (𝑡)) ≤ min(𝑞−𝑑 |𝐸|, 2𝑞−𝑑−1 |𝐸| + 2𝑞−

3𝑑+1 2

𝑡∈𝔽𝑞

|𝐸|2 ) .

(5.14)

If 𝑑 ≥ 4 is even, then we have the same bound away from the 0-sphere: max(𝜎𝐸 (𝑡)) ≤ min(𝑞−𝑑 |𝐸|, 2𝑞−𝑑−1 |𝐸| + 2𝑞− 𝑡≠0

3𝑑+1 2

|𝐸|2 ) .

When 𝑑 = 2 we have a superior bound (see Theorem 4.1.3): max (𝜎𝐸 (𝑡)) ≤ 3𝑞−3 |𝐸|3/2 . 𝑡≠0

(5.15)

94

Chapter 5. Rings and generalized distances

Putting together Propositions 5.2.4, 5.2.5, 5.2.6, and 5.2.7, we see that if |𝐸||𝐹| ≫ 𝑞𝑑 and |𝐸| ≤ |𝐹|, then

|Δ(𝐸, 𝐹)| ≫

⎧ min(𝑞, |𝐹| 𝑑−1 ) ⎪ 𝑞 2

𝑑≥2

⎨ ⎪ |𝐸|1/2 |𝐹| ⎩ min(𝑞, 𝑞 )

𝑑 = 2.

Theorem 5.2.3 immediately follows from this inequality.

5.3 Generalized distances Instead of using two different sets, we could simply change what we mean by a distance. For example Koh and Iosevich ([82]) studied the set Δ𝑛 (𝐸) ≔ {(𝑥1 − 𝑦1 )𝑛 + ⋯ + (𝑥𝑑 − 𝑦 𝑑 )𝑛 ∶ 𝑥, 𝑦 ∈ 𝐸} for 𝐸 ⊆ 𝔽𝑑𝑞 . When 𝑛 = 2, we get the familiar finite field distance set. Notice that when 𝑛 = 3, rather than utilizing the well-known Gauss sums, the so-called Kummer sums come into play. They demonstrated the following result. Theorem 5.3.1. Let 𝐸 ⊆ 𝔽𝑑𝑞 be such that |𝐸| ≫ 𝑞 (mod 3) is prime. Then,

𝑑+1 2

, where 𝑞 ≡ 1

Δ3 (𝐸) = 𝔽𝑞 . If 𝑑 = 2, then under the same conditions on 𝐸 and 𝑞 we have Δ𝑛 (𝐸) = 𝔽𝑞 for all 𝑛 ≥ 2. Furthermore, if 𝐸 ⊆ 𝔽𝑑𝑞 and 𝐹 ⊆ 𝔽𝑑𝑞 satisfy |𝐸||𝐹| ≫ 𝑞𝑑+1 , then Δ3 (𝐸, 𝐹) ≔ {(𝑥1 − 𝑦1 )3 + ⋯ + (𝑥𝑑 − 𝑦 𝑑 )3 ∶ 𝑥 ∈ 𝐸, 𝑦 ∈ 𝐹} satisfies Δ3 (𝐸, 𝐹) ⊇ 𝔽∗𝑞 . An alternative way to generalize the distance set was undertaken in [27]. The authors studied the so-called 𝑘-resultant set Δ𝑘 (𝐸) ≔ {‖𝑥1⃗ − ⋯ − 𝑥𝑘⃗ ‖ ∶ 𝑥1⃗ , . . . , 𝑥𝑘⃗ ∈ 𝐸}.

(5.16)

5.3. Generalized distances

95

Recall that for 𝑥⃗ = (𝑥1 , . . . , 𝑥𝑑 ) ∈ 𝔽𝑑𝑞 , we have ‖𝑥‖ = 𝑥12 + ⋯ + 𝑥𝑑2 . The results obtained were independent of the sign ± between the vectors 𝑥𝑖⃗ , so the same conclusion could apply to sets of the form {‖𝑥1⃗ ± ⋯ ± 𝑥𝑘⃗ ‖ ∶ 𝑥1⃗ , . . . , 𝑥𝑘⃗ ∈ 𝐸}, though we will stick with our original definition (5.16). Note that the usual finite-field distance set is then Δ2 (𝐸) = Δ(𝐸). The results are as follows. Theorem 5.3.2. Let 𝐸 ⊆ 𝔽𝑑𝑞 . Then there exists a constant 𝑐 ∈ (0, 1] such that |Δ3 (𝐸)| ≥ 𝑐𝑞 when |𝐸| ≫ 𝑞 |Δ4 (𝐸)| ≥ 𝑐𝑞 when |𝐸| ≫ 𝑞

𝑑+1 1 − 6𝑑+2 2

1 𝑑+1 − 6𝑑+2 2

for 𝑑 = 4 and 𝑑 = 6, and

for even 𝑑 ≥ 8.

Proof. We will provide a sketch of a proof of Theorem 5.3.2, though the details get somewhat technical, so we instruct the interested reader to comb through the original source as much as possible. We let 𝜈 𝑘 (𝑡) be our counting function: 𝜈 𝑘 (𝑡) = |{(𝑥1⃗ , . . . , 𝑥𝑘⃗ ) ∈ 𝐸 𝑘 ∶ ‖𝑥1⃗ − ⋯ − 𝑥𝑘⃗ ‖ = 𝑡}|. Then we have |Δ𝑘 (𝐸)| ≥

(|𝐸|𝑘 − 𝜈 𝑘 (0))2 ∑𝑡∈𝔽∗ 𝜈2𝑘 (𝑡)

(5.17)

𝑞

so as in the previous chapters we need a good upper bound on ∑ 𝜈2𝑘 (𝑡).

𝑡∈𝔽∗𝑞

The authors were able to obtain the estimate 2

2

−1

∑ 𝜈 𝑘 (𝑡) ≤ 𝑞 |𝐸|

2𝑘

𝑡∈𝔽𝑞

+𝑞

𝑑(2𝑘−1)

| 𝑘| | . ˆ ∑ || ∑ (𝐸(𝑣)) | | 𝑟∈𝔽𝑞 |‖𝑣‖=𝑟

Additionally, we use some restriction estimates. Proposition 5.3.3. If 𝑑 ≥ 4 is even, then for 𝑘 > 4 − 𝑘

ˆ || ) ≪ 𝑞−𝑑𝑘+𝑑−1 |𝐸| max ( ∑ ||𝐸(𝑚) 𝑡≠0

‖𝑚‖=𝑡

24 , 3𝑑+4

we have

3𝑘𝑑−3𝑑+4𝑘+2 3𝑑+4

.

96

Chapter 5. Rings and generalized distances

Note that this proposition is meaningful for 𝑘 = 3 if 𝑑 = 4 or 𝑑 = 6, and it is meaningful for 𝑘 ≥ 4 and in even dimensions 𝑑 ≥ 8. We have one further proposition before we assemble the pieces. Proposition 5.3.4. If 𝑑 ≥ 2 is even, 𝑘 ≥ 2, and 𝐸 ⊆ 𝔽𝑑𝑞 satisfies |𝐸| ≥ 3𝑞𝑑/2 , then |𝐸|𝑘 |𝐸|𝑘 − 𝜈 𝑘 (0)2 ≥ . 3 Furthermore, we have 2

𝑞

2𝑘𝑑−𝑑

| 𝑘| | − 𝜈 (0)2 ≤ 4𝑞−1 |𝐸|2𝑘 | ∑ (𝐸(𝑚)) ˆ 𝑘 | | | |‖𝑚‖=0

and 𝑘

ˆ || ) . ∑ 𝜈 𝑘 (𝑡)2 ≪ 𝑞−1 |𝐸|2𝑘 + 𝑞𝑑𝑘 |𝐸|𝑘−1 max ( ∑ ||𝐸(𝑚) 𝑟≠0

𝑡≠0

‖𝑚‖=𝑟

Putting everything together and using (5.17) we have |Δ𝑘 (𝐸)| ≫

(|𝐸|𝑘 − 𝜈 𝑘 (0))2 𝑘

ˆ || ) 𝑞−1 |𝐸|2𝑘 + 𝑞𝑑𝑘 |𝐸|𝑘−1 max𝑟≠0 (∑‖𝑚‖=𝑟 ||𝐸(𝑚) |𝐸|2𝑘

≫

𝑘

.

ˆ || ) 𝑞−1 |𝐸|2𝑘 + 𝑞𝑑𝑘 |𝐸|𝑘−1 max𝑟≠0 (∑‖𝑚‖=𝑟 ||𝐸(𝑚) Depending on which term in the denominator is the dominant term, and using Proposition 5.3.3, we see that ⎛ ⎞ |𝐸|𝑘+1 |Δ𝑘 (𝐸)| ≫ min ⎜𝑞, ⎟ 𝑘 ⎜ 𝑑𝑘 max ˆ || ) ⎟ ||𝐸(𝑚) 𝑞 (∑ 𝑟≠0 ‖𝑚‖=𝑟 ⎝ ⎠ ≫ min(𝑞,

|𝐸|𝑘+1 𝑞𝑑−1 |𝐸|

3𝑑𝑘−3𝑑+4𝑘+2 3𝑑+4

)

≫𝑞 whenever |𝐸| ≫ 𝑞

𝑑+1 1 − 6𝑑+2 2

.

We have a weaker result in even dimensions 𝑑 ≥ 8 concerning Δ3 (𝐸).

5.3. Generalized distances

97

Theorem 5.3.5 ([29]). Let 𝑑 ≥ 8 be even, and assume 𝐸 ⊆ 𝔽𝑑𝑞 . Then we are guaranteed that |Δ3 (𝐸)| ≥ 𝑐𝑞, so long as |𝐸| ≫ 𝑞

𝑑+1 1 − 9𝑑+18 2

.

The above results are intriguing in that for the first time in even dimensions 𝑑 ≥ 4, we are able to dip below the exponent threshold of 𝑑+1 implying that there is not some algebraic obstruction as in the odd2 𝑑+1

dimensional case. This gives some hope that the exponent 2 in higher even dimensions can be improved in the Erdős-Falconer distance prob𝑑+1 lem. It should be noted that we are able to dip below the previous 2 threshold because, for example, one can think of the 3-resultant set Δ3 (𝐸) = {‖𝑥 − 𝑦 − 𝑧‖ ∶ 𝑥, 𝑦, 𝑧 ∈ 𝐸} = {‖𝑥 − 𝑤‖ ∶ 𝑥 ∈ 𝐸, 𝑤 ∈ 𝐸 + 𝐸} = Δ(𝐸, 𝐸 + 𝐸). Now for some sets 𝐸 ⊆ 𝔽𝑑𝑞 , it may be that 𝐸+𝐸 is sufficiently close to 𝐸 (if 𝐸 has a lattice structure, for example), but we were able to overcome this obstacle by directly appealing to the applications of restriction theory, rather than viewing the 3-resultant distance set as a distance set result between two different sets. There has also been progress on the 𝑘-resultant modulus problem for subsets of algebraic varieties. For a polynomial 𝑄 ∈ 𝔽𝑞 [𝑥1 , . . . , 𝑥𝑑 ], we call 𝑉 = {𝑥 ∈ 𝔽𝑑𝑞 ∶ 𝑄(𝑥) = 0} a regular variety if |𝑉| ≍ 𝑞𝑑−1 and 𝑑+1

ˆ ||𝑉(𝑚) || ≪ 𝑞− 2 when 𝑚 ≠ (0, . . . , 0). A regular variety 𝑉 ⊆ 𝔽2𝑞 will be called nondegenerate if 𝑄(𝑥) does not contain any linear factor. Theorem 5.3.6 ([28]). Let 𝑉 ⊆ 𝔽𝑑𝑞 be a regular variety, 𝐸 ⊆ 𝑉, and 𝑘 ≥ 3. 𝑑−1

1

If |𝐸| ≫ 𝑞 2 + 𝑘−1 then Δ𝑘 (𝐸) ⊇ 𝔽∗𝑞 when 𝑑 ≥ 2 is even and Δ𝑘 (𝐸) = 𝔽𝑞 when 𝑑 ≥ 3 is odd. Many of these results can also be extended to include some of the generalizations previously discussed in this chapter. For example we considered the multiset case: Theorem 5.3.7. For 𝐸1 , 𝐸2 , 𝐸3 ⊆ 𝔽𝑑𝑞 , define Δ3 (𝐸1 , 𝐸2 , 𝐸3 ) = {‖𝑥1⃗ + 𝑥2⃗ + 𝑥3⃗ ‖ ∶ 𝑥𝑖⃗ ∈ 𝐸𝑖 , for 𝑖 = 1, 2, 3}.

98

Chapter 5. Rings and generalized distances

Then, |Δ3 (𝐸)| ≫ 𝑞 whenever |𝐸1 ||𝐸2 ||𝐸3 | ≫ 𝑞 𝑑 = 6.

3(

𝑑+1 1 − 6𝑑+2 ) 2

for 𝑑 = 4 and

Similar results hold for 𝑑 ≥ 8, and for the 3-resultant set when 𝑑 = 8. Further generalizations of Theorem 5.3.6 were achieved by Hieu-Pham ([79]). Theorem 5.3.8. Let 𝑄 be a non-degenerate quadratic form on 𝔽𝑑𝑞 , and assume that 𝑉 ⊆ 𝔽𝑑𝑞 is a regular variety, with 𝐸 ⊆ 𝑉, and 𝑘 ≥ 3. If |𝐸| ≫ 𝑞

𝑑−1 1 + 𝑘+1 2

, then for 𝑡 ≠ 0, we have

|{(𝑥1 , . . . , 𝑥𝑘 ) ∶ 𝑄(𝑥1 , . . . , 𝑥𝑘 ) = 𝑡}| = (1 − 𝑜(1))

|𝐸|𝑘 . 𝑞

5.4 Pinned distances Instead of looking at the entire set of distances, sometimes it is sufficient to look at the set of distances where one endpoint is fixed or pinned. One of the first results to this end was a continuous result by Peres and Schlag ([121]), where they prove a pinned version of Falconer’s result. Specifically they proved: Theorem 5.4.1. Let 𝐸 ⊆ ℝ𝑑 have dim𝐻 (𝐸) > point 𝑦 ∈ 𝐸 such that the pinned distance set

𝑑+1 . 2

Then, there exists a

{|𝑥 − 𝑦| ∶ 𝑥 ∈ 𝐸} has positive 1-dimensional Lebesgue measure. Recall that we have intuitively (and informally) defined the notions of Hausdorff dimension and Lebesgue measure in Section 2.2. Guth, Iosevich, Ou, and Wang ([69]) have improved this bound in two dimensions, thereby improving the best known bound for the “unpinned” Falconer distance problem in ℝ2 as well. Theorem 5.4.2. Let 𝐸 ⊆ ℝ2 be compact with dim𝐻 (𝐸) > 5/4. Then, there exists a point 𝑦 ∈ 𝐸 such that the pinned distance set {|𝑥 − 𝑦| ∶ 𝑥 ∈ 𝐸} has positive 1-dimensional Lebesgue measure.

5.4. Pinned distances

99

Naturally, we consider the situation in finite fields. For 𝑦 ∈ 𝐸, define the set of pinned distances to be Δ𝑦 (𝐸) = {‖𝑥 − 𝑦‖ ∶ 𝑥 ∈ 𝐸}. It is not necessary to specify that 𝑦 ∈ 𝐸 in order for the pinned distance set Δ𝑦 (𝐸) to be studied, but notice that if 𝑦 ∈ 𝐸, then Δ𝑦 (𝐸) ⊆ Δ(𝐸), so requiring 𝑦 ∈ 𝐸 allows us to turned pinned distance results into generalizations of classical distance problem results. We first state and prove a simple pinned distance result ([17]). Theorem 5.4.3. Let 𝐸 ⊆ 𝔽𝑑𝑞 be such that |𝐸| ≫ 𝑞 1 𝑦 ∈ 𝐸 such that |Δ𝑦 (𝐸)| ≥ 2 𝑞.

𝑑+1 2

. Then there exists

Proof of Theorem 5.4.3. This proof will be similar to the proof discussed in “𝐿2 -method” discussed in Section 3.2. Just like in the proof of Theorem 2.3.3, we let 𝜈𝑦 (𝑡) = |{𝑥 ∈ 𝐸 ∶ ‖𝑥 − 𝑦‖ = 𝑡}|, so that 𝜈𝑦 (𝑡) denotes the number of points 𝑥 ∈ 𝐸 which are at distance 𝑡 from the pinned value 𝑦. We first note that by changing the order of summation, we obtain: ∑ ∑ 𝜈𝑦 (𝑡) = |𝐸|2 . 𝑦∈𝐸 𝑡∈𝔽𝑑 𝑞

Applying Cauchy-Schwarz in 𝑦, and using 1∆𝑦 (𝐸) (⋅) to denote the characteristic function of Δ𝑦 (𝐸), we see that 2

|𝐸|4 = ( ∑ ∑ 𝜈𝑦 (𝑡)) 𝑦∈𝐸 𝑡∈𝔽𝑞 2

= ( ∑ ∑ 1∆𝑦 (𝐸) (𝑡)𝜈𝑦 (𝑡)) 𝑦∈𝐸 𝑡∈𝔽𝑞

≤ ∑ ∑ 1∆𝑦 (𝐸) (𝑡)2 ∑ 𝜈𝑦 (𝑡)2 𝑡∈𝔽𝑞 𝑦∈𝐸

𝑦∈𝐸

= ∑ |Δ𝑦 (𝐸)| ⋅ ∑ ∑ 𝜈𝑦 (𝑡)2 . 𝑦∈𝐸

𝑡∈𝔽𝑞 𝑦∈𝐸

100

Chapter 5. Rings and generalized distances

We can conclude that ∑ |Δ𝑦 (𝐸)| ≥ 𝑦∈𝐸

∑𝑡∈𝔽

𝑞

|𝐸|4 , ∑𝑦∈𝐸 𝜈𝑦 (𝑡)2

so we need only to find a suitable upper bound for ∑𝑡,𝑦 𝜈𝑦 (𝑡)2 . To this end notice that 𝜈𝑦 (𝑡) = ∑ 𝐸(𝑥). 𝑥∈𝔽𝑑 𝑞

‖𝑥−𝑦‖=𝑡

Squaring gives us 𝜈𝑦 (𝑡)2 =

𝐸(𝑥)𝐸(𝑥′ ).

∑ 𝑥,𝑥′ ∈𝔽𝑑 𝑞 ‖𝑥−𝑦‖=‖𝑥′ −𝑦‖=𝑡

Summing over 𝑡 ∈ 𝔽𝑞 and 𝑦 ∈ 𝐸 then leads us to ∑ ∑ 𝜈𝑦 (𝑡)2 = ∑ 𝑡∈𝔽𝑞 𝑦∈𝐸

∑

𝑡∈𝔽𝑞

𝐸(𝑥)𝐸(𝑥′ )𝐸(𝑦)

𝑥,𝑥′ ,𝑦∈𝔽𝑑 𝑞

‖𝑥−𝑦‖=‖𝑥′ −𝑦‖=𝑡

∑

=

𝐸(𝑥)𝐸(𝑥′ )𝐸(𝑦).

𝑥,𝑥′ ,𝑦∈𝔽𝑑 𝑞 ‖𝑥−𝑦‖=‖𝑥′ −𝑦‖

Applying orthogonality in dimension 𝑑 = 1, we see that ∑𝑡∈𝔽 ∑𝑦∈𝐸 𝜈𝑦 (𝑡)2 𝑞 equals the quantity 𝑞−1 ∑

∑

𝜒(𝑠(‖𝑥 − 𝑦‖ − ‖𝑥′ − 𝑦‖))𝐸(𝑥)𝐸(𝑥′ )𝐸(𝑦).

(5.18)

𝑠∈𝔽𝑞 𝑥,𝑥′ ,𝑦∈𝔽𝑑 𝑞

A quick calculation (see Exercise 5.1) shows that (5.18) becomes ∑ ∑ 𝜈𝑦 (𝑡)2 = 𝑞−1 ∑

∑

𝜒(𝑠(‖𝑥‖ − 2𝑥 ⋅ 𝑦))𝜒(−𝑠(‖𝑥′ ‖ − 2𝑥′ ⋅ 𝑦))

𝑠∈𝔽𝑞 𝑥,𝑥′ ,𝑦∈𝐸

𝑡∈𝔽𝑞 𝑦∈𝐸

2

=𝑞

−1

| | ∑ ∑ || ∑ 𝜒(𝑠(‖𝑥‖ − 2𝑥 ⋅ 𝑦))|| | 𝑠∈𝔽𝑞 𝑦∈𝐸 |𝑥∈𝐸

= 𝐼𝑠=0 + 𝐼𝐼𝑠≠0 , where 2

| | 𝐼𝑠=0 = 𝑞−1 ∑ ∑ || ∑ 𝜒(𝑠(‖𝑥‖ − 2𝑥 ⋅ 𝑦))|| = 𝑞−1 |𝐸|3 | 𝑠=0 𝑦∈𝐸 |𝑥∈𝐸

5.4. Pinned distances

101

and

2

𝐼𝐼𝑠≠0 = 𝑞

−1

| | ∑ ∑ || ∑ 𝜒(𝑠(‖𝑥‖ − 2𝑥 ⋅ 𝑦))|| . | 𝑠≠0 𝑦∈𝐸 |𝑥∈𝐸

We finally need only to bound the last sum 𝐼𝐼𝑠≠0 . We first extend the sum in 𝑦 ∈ 𝐸 to a sum in 𝑦 ∈ 𝔽𝑑𝑞 : 2

| | 𝐼𝐼𝑠≠0 = 𝑞−1 ∑ ∑ || ∑ 𝜒(𝑠(‖𝑥‖ − 2𝑥 ⋅ 𝑦))|| | 𝑠≠0 𝑦∈𝐸 |𝑥∈𝐸

2

≤𝑞

−1

| | ∑ ∑ || ∑ 𝜒(𝑠(‖𝑥‖ − 2𝑥 ⋅ 𝑦))|| | 𝑠≠0 𝑦∈𝔽𝑑 |𝑥∈𝐸 𝑞

=𝑞

−1

∑ 𝜒(𝑠(‖𝑥‖ − ‖𝑥′ ‖))𝜒(2𝑠𝑦 ⋅ (𝑥 − 𝑥′ )).

∑ ∑

𝑠≠0 𝑦∈𝔽𝑑 𝑥,𝑥′ ∈𝐸 𝑞

Summing in 𝑦 ∈ 𝔽𝑑𝑞 and applying orthogonality yields 𝐼𝐼𝑠≠0 ≤ 𝑞𝑑−1 ∑ 𝑠≠0

∑

𝜒(𝑠(‖𝑥‖ − ‖𝑥′ ‖))

𝑥,𝑥′ ∈𝐸 ‖𝑥‖=‖𝑥′ ‖

= 𝑞𝑑−1 (𝑞 − 1)|𝐸| ≤ 𝑞𝑑 |𝐸|. Notice that 𝐼𝑠=0 ≥ 𝐼𝐼𝑠≠0 whenever 𝑞−1 |𝐸|3 ≥ 𝑞𝑑 |𝐸|. It follows that ∑ ∑ 𝜈𝑦 (𝑡)2 = 𝐼𝑠=0 + 𝐼𝐼𝑠≠0 ≤ 2𝑞−1 |𝐸|3 𝑡∈𝔽𝑞 𝑦∈𝐸

whenever |𝐸| ≥ 𝑞

𝑑+1 2

. Putting together all of the pieces, we see that

∑ |Δ𝑦 (𝐸)| ≥ 𝑦∈𝐸

so long as |𝐸| ≥ 𝑞

∑𝑡∈𝔽

𝑞

𝑑+1 2

|𝐸|4 |𝐸|4 ≥ 2𝑞−1 |𝐸|3 ∑𝑦∈𝐸 𝜈𝑦 (𝑡)2

. Thus, in this range of 𝐸, we have 1 1 ∑ |Δ (𝐸)| ≥ 𝑞. |𝐸| 𝑦∈𝐸 𝑦 2

Theorem 5.4.3 follows from the pigeonhole principle. That is, there must 1 exist a 𝑦 ∈ 𝐸 such that |Δ𝑦 (𝐸)| ≥ 2 𝑞.

102

Chapter 5. Rings and generalized distances

Note that the proof of the above result was from [17], and if you more carefully analyze the final bound: 1 1 ∑ |Δ (𝐸)| ≥ 𝑞, |𝐸| 𝑦∈𝐸 𝑦 2 you achieve the following. 𝑑+1

Theorem 5.4.4. Let 𝐸 ⊆ 𝔽𝑑𝑞 , and suppose |𝐸| ≫ 𝑞 2 . Then there exists a subset 𝐸 ′ ⊆ 𝐸 such that |𝐸 ′ | ≫ |𝐸|, and for all 𝑦 ∈ 𝐸 ′ , we have |Δ𝑦 (𝐸)| ≥

1 𝑞. 2

When 𝑑 = 2, this result was improved in [71]. Theorem 5.4.5. Let 𝐸 ⊆ 𝔽2𝑞 , and suppose |𝐸| ≫ 𝑞4/3 . Then there exists a subset 𝐸 ′ ⊆ 𝐸 such that |𝐸 ′ | ≫ |𝐸|, and for all 𝑦 ∈ 𝐸 ′ , we have |Δ𝑦 (𝐸)| ≫ 𝑞. Interestingly, Theorem 5.4.5 was proved used a geometric argument involving perpendicular bisectors and point-line incidences in 𝔽𝑑𝑞 . It should be noted that Theorem 5.4.5 is a strengthening of Theorem 4.1.1. Finally the following was obtained in [122] for product sets in 𝔽𝑝2 . Theorem 5.4.6. Let 𝐸 = 𝐴 × 𝐴 ⊆ 𝔽𝑝2 , where 𝑝 is prime. Then, there exists 𝑦 ∈ 𝐸 such that |Δ𝑦 (𝐸)| ≫ min{𝑝, |𝐸|3/4 }. Note that Theorems 5.4.5 and 5.4.6 both give the same bound: |𝐸| ≫ 𝑞4/3 ⟹ |Δ𝑦 (𝐸)| ≫ 𝑞. Furthermore Theorem 5.4.5 holds for all sets 𝐸 ⊂ 𝔽2𝑞 for a positive proportion of 𝑥 ∈ 𝐸, while Theorem 5.4.6 holds only for product sets in 𝔽𝑝2 for 𝑝 a prime, and for at least one 𝑦 ∈ 𝐸. However, Theorem 5.4.6 does give a quantitative upper bound for |Δ𝑦 (𝐸)| when 𝐸 = 𝐴 × 𝐴 has small cardinality.

5.5. Exercises: Chapter 5

103

5.5 Exercises: Chapter 5 Exercise 5.1. Verify that for 𝑥, 𝑦, 𝑧 ∈ 𝔽𝑑𝑞 , we have ‖𝑥 − 𝑦‖ − ‖𝑧 − 𝑦‖ = (‖𝑥‖ − 2𝑥 ⋅ 𝑦) − (‖𝑧‖ − 2𝑧 ⋅ 𝑦). Exercise 5.2. Complete the square in ℤ𝑛 . That is if gcd(2𝑎, 𝑛), then find all 𝑥 ∈ ℤ𝑛 such that 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0. Exercise 5.3. Suppose that 𝑎 ∈ ℤ×𝑛 , where 𝑛 ≥ 3 is odd. Prove that 𝐺(𝑎, 𝑏, 𝑛) = 𝐺(𝑎, 𝑛)𝜒(−𝑏2 /4𝑎). Exercise 5.4. Prove that |𝐺(𝑎, 𝑛)|2 ≤ 𝑛 for all 𝑎 ∈ ℤ𝑞 . Exercise 5.5. Suppose 𝜓 is a Dirichlet character mod 𝑞 and gcd(𝑎, 𝑞) = 1. Show that 𝜏(𝜓, 𝜒𝑎 ) = 𝜓(𝑎)𝜏(𝜓). Exercise 5.6 (Chinese Remainder Theorem). Suppose that 𝑎 and 𝑏 are positive integers, each greater than or equal to 2. Prove that if gcd(𝑎, 𝑏) = 1, then we have the following ring isomorphisms: ℤ𝑎𝑏 ≅ ℤ𝑎 × ℤ𝑏 and ℤ×𝑎𝑏 ≅ ℤ×𝑎 × ℤ×𝑏 . Exercise 5.7. Prove Theorem 5.2.1. That is, using the techniques laid out in the proof of Theorem 4.1.1, prove that if 𝐸, 𝐹 ⊆ 𝔽2𝑞 and |𝐸||𝐹| ≫ 𝑞8/3 , then |Δ(𝐸, 𝐹)| ≫ 𝑞. Exercise 5.8. Prove the following elementary interpolation theorem4 : Let (𝑎𝑗 ) be a sequence of positive real numbers, and suppose that 𝑁

∑ 𝑎𝑗 ≤ 𝐴 𝑗=1 4 First shown to me by Alex Iosevich, and it appears in his fantastic book A View From the Top ([81]).

104

Chapter 5. Rings and generalized distances

while

1/2

𝑁

(∑

𝑎𝑗2 )

≤ 𝐵.

𝑗=1

Then we have 𝑁

(∑

1/𝑝 𝑝 𝑎𝑗 )

2

≤ 𝐴𝑝

2

−1 2− 𝑝

𝐵

𝑗=1

for all 1 ≤ 𝑝 ≤ 2. Exercise 5.9. Show that if the sets 𝐸, 𝐹 ⊆ 𝔽𝑑𝑞 satisfy |𝐸||𝐹| ≫ 𝑞𝑑+1 , then Δ(𝐸, 𝐹) ⊇ 𝔽∗𝑞 . When must 0 ∈ Δ(𝐸, 𝐹)?

10.1090/car/037/06

Chapter

6

Configurations and group actions 6.1 Finite configurations In the previous chapters we were working with the distance set Δ(𝐸) = {‖𝑥 − 𝑦‖ ∶ 𝑥, 𝑦 ∈ 𝐸} and its vertex set 𝐷𝑡 (𝐸) = {(𝑥, 𝑦) ∈ 𝐸 × 𝐸 ∶ ‖𝑥 − 𝑦‖ = 𝑡}, where 𝐸 ⊆ 𝔽𝑑𝑞 . It is instructive to think of distances as 2-point configurations. In this chapter, we study 𝑛-point configurations, but some extra care is required.1 All of these problems can be extended in the same ways as the distance problem (working over different spaces, using different notions of distance, etc.), but we will concentrate on configurations over finite fields as that setting is the simplest and cleanest. As a first glimpse let’s consider an example. Let 𝐸 ⊆ 𝔽2𝑞 , let ℓ1 , ℓ2 , ℓ3 ∈ ∗ 𝔽𝑞 , and consider the set {(𝑥, 𝑦, 𝑧) ∶ ‖𝑥 − 𝑦‖ = ℓ1 , ‖𝑦 − 𝑧‖ = ℓ2 , ‖𝑥 − 𝑧‖ = ℓ3 } of all 3-point configurations with side-lengths ℓ1 , ℓ2 , and ℓ3 , respectively. It is sometimes convenient to assume that ℓ𝑖 ≠ 0 so that we can avoid 1 At this point in the text, we cease to provide as elementary an overview as possible, opting instead to deliver a succinct history of the development of this area as it is expansive. We invite the reader to read the original sources whenever possible as we only provide the gist of each proof where appropriate.

105

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triangles containing isotropic lines (i.e., nontrivial lines with length 0), specifically when 𝑞 ≡ 1 (mod 4).

Figure 6.1. The nondegenerate triangle (𝐴, 𝐵, 𝐶) with distances ℓ1 , ℓ2 , ℓ3 ; the degenerate triangle (𝐷, 𝐸, 𝐹).

We will refer to a 3-point configuration as a “triangle” though it should be clear we are only ever discussing the vertex set of a triangle and not the edges between the vertices (see Figure 6.1). We will say that two triangles are equivalent if one is a rotated, shifted copy of the other2 . We say that a triangle is nondegenerate if there is no line containing all three vertices of a triangle. It turns out that this set of equivalence classes of triangles is 3-dimensional3 . Already a natural question should spring to mind: Question 6.1.1. Let 𝐸 ⊆ 𝔽2𝑞 . What is the minimum power 𝛽 such that there exists a constant 𝑐 > 0 so that |{(‖𝑥 − 𝑦‖, ‖𝑦 − 𝑧‖, ‖𝑥 − 𝑦‖) ∶ 𝑥, 𝑦, 𝑧 ∈ 𝐸}| ≥ 𝑐𝑞3 , whenever |𝐸| ≫ 𝑞𝛽 ? Before we can answer this question, we must now rigorously define finite point configurations in 𝔽𝑑𝑞 . A 𝑘-simplex will refer to a set of (𝑘+1)points in 𝔽𝑑𝑞 , so that a triangle (or more precisely, the vertices of a triangle) forms a 2-simplex, and a distance is a 1-simplex. A priori there 2 This 3 This

language will be justified shortly. is not as obvious as one may think!

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107

is no relationship between 𝑘 and 𝑑. We say that two 𝑘-simplices Δ1 = {𝑥0 , . . . , 𝑥𝑘 } ⊆ 𝔽𝑑𝑞 and Δ2 = {𝑦0 , . . . , 𝑦 𝑘 } ⊆ 𝔽𝑑𝑞 are equivalent (Δ1 ∼ Δ2 ) if there exists a translation vector 𝜏 ∈ 𝔽𝑑𝑞 and an orthogonal matrix 𝑂 ∈ 𝑂 𝑑 (𝔽𝑞 ) such that 𝑥𝑖 = 𝑂𝑦 𝑖 + 𝜏, for all 𝑖 = 0, . . . , 𝑘, where 𝑥𝑖 , 𝑦 𝑖 , and 𝜏 are viewed as column vectors. Just like in the Euclidean setting, a matrix 𝑂 is orthogonal if 𝑂𝑇 = 𝑂−1 . It is easy to check that this defines an equivalence relation (see Exercise 6.3), so we can now genuinely say that Δ1 and Δ2 are equivalent. We frequently abuse notation and refer to the linear transformation determined by 𝑂 as simply 𝑂 itself. Now, we say a 𝑘-simplex Δ is nondegenerate if the vertices 𝑥0 , . . . , 𝑥𝑘 are affinely independent (i.e., the set of points 𝑥1 − 𝑥0 , . . . , 𝑥𝑘 − 𝑥0 are linearly independent). Note that a 𝑘simplex in 𝔽𝑑𝑞 must be degenerate if 𝑘 > 𝑑, so we will only concentrate on the case 𝑘 ≤ 𝑑 with the case 𝑘 = 𝑑 (such as triangles in 𝔽2𝑞 ) perhaps being the most compelling. Let ∼ denote the equivalence of simplices in 𝔽𝑑𝑞 under translations and rotations defined above, and consider 𝑇𝑘 (𝐸) = 𝐸 𝑘+1 / ∼, the resulting quotient set of equivalence classes. The following proposition will give the dimensionality of the set of simplices. Proposition 6.1.2. Suppose that Δ1 = {𝑥0 , . . . , 𝑥𝑘 } and Δ2 = {𝑦0 , . . . , 𝑦 𝑘 } are two nondegenerate simplices in 𝔽𝑑𝑞 . Then Δ1 ∼ Δ2 if and only if you can order the simplices Δ1 and Δ2 such that ‖𝑥𝑖 − 𝑥𝑗 ‖ = ‖𝑦 𝑖 − 𝑦𝑗 ‖ for all pairs 0 ≤ 𝑖 < 𝑗 ≤ 𝑘. Proof. Exercise. (See Exercise 6.4.) Note that we can encode any nondegenerate 𝑘-simplex as a (𝑘 + 1) × (𝑘 + 1) symmetric matrix. For example, the set of triangles {𝑥, 𝑦, 𝑧} with sides ‖𝑥 − 𝑦‖ = 𝑎, ‖𝑥 − 𝑧‖ = 𝑏, ‖𝑦 − 𝑧‖ = 𝑐 can be encoded as a 3 × 3 matrix 0 𝑎 𝑏 [𝑎 0 𝑐 ] . 𝑏 𝑐 0

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Different values of 𝑎, 𝑏, and 𝑐 will lead to different triangles. A straightforward count shows that the set of all (possibly degenerate) triangles is 3-dimensional. Generalizing this, we see that the set of nondegenerate 𝑘-simplices is (𝑘+1 )-dimensional. When 𝑘 ≤ 𝑑, the set of degenerate 2 simplices is much smaller than the nondegenerate simplices (see Exercise 6.6). We are now ready to state some results. We will start with the earliest results and work our way forward chronologically. In [25] the authors studied 𝑑-simplices in 𝔽𝑑𝑞 . The main result was the following: Theorem 6.1.3. Let 𝐸 ⊆ 𝔽𝑑𝑞 satisfy |𝐸| ≥ 𝜌𝑞𝑑 , where 𝑞−1/2 ≪ 𝜌 ≤ 1. Then the set of 𝑑-simplices satisfies |𝑇𝑑 (𝐸)| ≫ 𝜌𝑑−1 𝑞(

𝑑+1 ) 2

.

Remark 6.1.4. The proof of Theorem 6.1.3 essentially amounts to estimating |{(𝑥, 𝑥1 , . . . , 𝑥𝑑 ) ∈ 𝐸 𝑑+1 ∶ ‖𝑥 − 𝑥𝑖 ‖ = 1}| , and then employing some combinatorial reasoning. The original bound on that quantity led to the range 𝑞−1/2 ≪ 𝜌 ≤ 1. The range was later 𝑑−1

improved to 𝑞− 2 ≪ 𝜌 ≤ 1 in a subsequent paper ([6]), though better results are now known. In [73] Hart and Iosevich gave the first bound on general 𝑘-simplices in 𝔽𝑑𝑞 albeit for 𝑘 ≪ √𝑑: Theorem 6.1.5. Fix 𝑘, and let 𝑑 > (𝑘+1 ). Let 𝐸 ⊆ 𝔽𝑑𝑞 . Then 2 𝑘+1 ) 2

|𝑇𝑘 (𝐸)| ≫ 𝑞( 𝑑𝑘

𝑘

so long as |𝐸| ≫ 𝑞 𝑘+1 + 2 . The main thrust of the argument was an induction argument used to “bootstrap” from (𝑘 − 1)-simplices to 𝑘-simplices. Theorem 6.1.6. We define a 𝑘-simplex recursively: Let 𝑇 = (𝑡 𝑖,𝑗 ) denote a 𝑑 × 𝑑 symmetric matrix with entries in 𝔽∗𝑞 . Let 𝑟 𝑘 = {𝑡 𝑘,1 , . . . , 𝑡 𝑘,𝑘 }

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109

denote the first 𝑘 values in the 𝑘th row, and finally let ℓ𝑘 denote the set of values on or under the main diagonal in the first 𝑘 rows of the matrix, so ℓ𝑘 = {𝑡1,1 ; 𝑡2,1 ; 𝑡2,2 ; . . . ; 𝑡 𝑘,1 ; . . . ; 𝑡 𝑘,𝑘 } = 𝑟1 ∪ ⋯ ∪ 𝑟 𝑘 . Now, we write 𝒯ℓ1 = {(𝑥0 , 𝑥1 ) ∈ 𝐸×𝐸 ∶ ‖𝑥0 −𝑥1 ‖ = 𝑡1,1 }, so that is the pair of vertices in 𝐸 who are distance 𝑡1,1 apart. Then, 𝒯ℓ2 is the set of triples in 𝐸 whose vertices form triangles with side lengths 𝑡1,1 ; 𝑡2,1 ; 𝑡2,2 . 𝑇ℓ3 is the set of 6 = (3+1 ) vertices in 𝐸 determining distances 𝑡1,1 ; 𝑡2,1 ; 𝑡2,2 ; 𝑡3,1 ; 𝑡3,2 ; 𝑡3,3 , 2 and so on for larger simplices. Formally, we have 𝒯ℓ𝑘 ={(𝑥0 , . . . , 𝑥𝑘−1 , 𝑥𝑘 ) ∈ 𝒯ℓ𝑘−1 × 𝐸 ∶ ‖𝑥𝑖 − 𝑥𝑘 ‖ ∈ 𝑟 𝑘 }. Note that varying the values in the matrix 𝑇 will change the vertex set, though just like the sphere of varying radii having the same cardinality, 𝑘+1 the values in 𝑇 will not change our estimates. Then for all ℓ𝑘 ∈ (𝔽∗𝑞 )( 2 ) , we have |𝐸|𝑘+1 |𝒯ℓ𝑘 | = 𝑘+1 (1 + 𝑜(1)) 𝑞( 2 ) 𝑘𝑑

𝑘

when |𝐸| ≫ 𝑞 𝑘+1 + 2 . In particular |𝒯ℓ𝑘 | > 0 for 𝐸 in the same range. The proof that |𝒯ℓ1 | =

|𝐸|2 (1 + 𝑜(1)) 𝑞

𝑑+1

when |𝐸| ≫ 𝑞 2 is the estimate we derived in (3.6). For the inductive step, the authors write |𝑇ℓ𝑘 | =

∑

∑

𝒯ℓ𝑘−1 (𝑥0 , . . . , 𝑥𝑘−1 )𝐸(𝑥𝑘 ),

𝑥0 ,. . .,𝑥𝑘 ‖𝑥0 −𝑥𝑘 ‖=𝑡1,𝑘 ;. . .;‖𝑥𝑘−1 −𝑥𝑘 ‖=𝑡𝑘,𝑘

and they apply the same Fourier Analytic techniques that were applied in the distance problem. Here 𝒯ℓ𝑗 (⋅) is the characteristic function of the set 𝑇ℓ𝑗 . An improvement to Theorem 6.1.5 was given in [17]. Theorem 6.1.7. Fix 𝑘 ≤ 𝑑 − 1, and suppose that 𝐸 ⊆ 𝔽𝑑𝑞 satisfies |𝐸| ≫ 𝑘+𝑑

𝑞 2 . Then 𝐸 determines a positive proportion of the 𝑘-simplices. That is, for sets 𝐸 of sufficiently large cardinality we have |𝑇𝑘 (𝐸)| ≫ 𝑞(

𝑘+1 ) 2

.

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The proof of Theorem 6.1.7 involves aspects of the previous proofs: both bootstrapping and estimating the number of so-called 𝑘-stars: |{(𝑥, 𝑥1 , . . . , 𝑥𝑘 ) ∈ 𝐸 𝑑+1 ∶ ‖𝑥 − 𝑥𝑖 ‖ = 𝑡 𝑖 }| where 𝑡 𝑖 ∈ 𝔽∗𝑞 for 𝑖 = 1, . . . , 𝑘. The authors were essentially able to prove that |{(𝑥, 𝑥1 , . . . , 𝑥𝑘 ) ∈ 𝐸 𝑑+1 ∶ ‖𝑥 − 𝑥𝑖 ‖ = 𝑡 𝑖 }| > 0 (6.1) 𝑑+𝑘

for all choices 𝑡 𝑖 ∈ 𝔽∗𝑞 for 𝑖 = 1, . . . , 𝑘, whenever |𝐸| ≫ 𝑞 2 . The pigeonhole principle then allows us to move from 𝑘-stars to 𝑘-simplices. Despite not being able to improve the 𝑘-simplex bound |𝐸| ≫ 𝑞 inequality (6.1) was extended to sets of smaller cardinality ([6]):

𝑑+𝑘 2

, the

Proposition 6.1.8. Let 𝐸 ⊆ 𝔽𝑑𝑞 , and let 𝑡1 , . . . , 𝑡 𝑘 ∈ 𝔽∗𝑞 . Then |{(𝑥, 𝑥1 , . . . , 𝑥𝑘 ) ∈ 𝐸 𝑑+1 ∶ ‖𝑥 − 𝑥𝑖 ‖ = 𝑡 𝑖 }| > 0 for all 𝑘
0, then max(𝑎, 𝑏) ≥ √𝑎𝑏. Exercise 6.2. Let 𝐺 denote the 1-distance graph of the entire finite field 𝔽𝑑𝑞 . Prove that the chromatic number of the distance graph satisfies 10 𝜒(𝐺) ≫ 𝑞

𝑑−1 2

.

Exercise 6.3. Fix 𝑡 ∈ 𝔽∗𝑞 . Prove that 𝐺 𝑡 is simple, undirected, connected, and regular. What other properties does 𝐺 𝑡 have? Exercise 6.4. Let Δ1 = {𝑣 0 , . . . 𝑣 𝑘 } and Δ2 = {𝑤 0 , . . . , 𝑤 𝑘 } be two nondegenerate simplices in 𝔽𝑑𝑞 . Prove that if ‖𝑣 𝑖 − 𝑣𝑗 ‖ = ‖𝑤 𝑖 − 𝑤𝑗 ‖ for all 0 ≤ 𝑖 < 𝑗 ≤ 𝑘, then Δ1 ∼ Δ2 . Exercise 6.5. If Δ1 is nondegenerate and Δ1 ∼ Δ2 , then Δ2 is nondegenerate. Exercise 6.6. Fix 𝑘 and 𝑑 such that 𝑘 ≤ 𝑑. Let 𝐴 ⊆ 𝑇𝑘 (𝑑) be the set of 𝑘+1 nondegenerate simplices. Prove that 𝐴 = 𝑜 (𝑞( 2 ) ) . 10 I had once thought that 𝜒(𝐺) should be uniformly bounded independent of 𝑞, which turned out to be absurdly false!

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127

Exercise 6.7. Let 𝐺𝐿2 (𝔽𝑞 ) denote the set of invertible 2 × 2-matrices with entries in 𝔽𝑞 . Show that |𝐺𝐿2 (𝔽𝑞 )| = (𝑞2 − 1)(𝑞2 − 𝑞). What is the cardinality of 𝐺𝐿𝑛 (𝔽𝑞 )? Exercise 6.8. Let 𝑂(2; 𝑞) denote the set of 2 × 2-orthogonal matrices with entries in 𝔽𝑞 . That is, 𝑂(2; 𝑞) = {𝐴 ∈ 𝑀2 (𝔽𝑞 ) ∶ 𝐴𝐴𝑇 = 𝐼} . Prove that |𝑂(2; 𝑞)| = {

2𝑞 − 2 2𝑞 + 2

𝑞 ≡ 1 (mod 4) 𝑞 ≡ 3 (mod 4).

Hint: Start by showing that |𝑆𝑂(2; 𝑞)| = |𝑆 1 |, where 𝑆𝑂(2; 𝑞) = {𝐴 ∈ 𝑂(2; 𝑞) ∶ det(𝐴) = 1}. Exercise 6.9. Let Graff(𝑘, 𝑑) denote the set of 𝑘-dimensional affine planes in 𝔽𝑑𝑞 . Let (𝑥, ℎ) ∈ 𝔽𝑑𝑞 × Graff(𝑑 − 1, 𝑑). We say that (𝑥, ℎ) is equivalent to (𝑥′ , ℎ′ ) if after translating 𝑥 to 𝑥′ , there exist an orthogonal matrix 𝑂 ∈ 𝑂 𝑑 (𝔽𝑞 ) that maps ℎ to ℎ′ . Prove that this is indeed an equivalence relation. Let Δ(𝐸, 𝐹) denote the quotient set of this equivalence relation. Exercise 6.10 ([10, 124]). Prove that if 𝐸 ⊆ 𝔽𝑑𝑞 , and 𝐹 is a set of non1 degenerate11 planes in Graff(𝑑 − 1, 𝑑), then |Δ(𝐸, 𝐹)| ≥ 2 𝑞, so long as 1

|𝐸||𝐹| > 𝑞𝑑+1 . If 𝑑 = 2, then |Δ(𝐸, 𝐹) ≥ 2 𝑞 when |𝐸||𝐹| > 𝑞8/3 .

11 Here,

non-degenerate means that the normal vector to the plane is nonisotropic.

10.1090/car/037/07

Chapter

7

Combinatorics in finite fields While I am primarily captivated by the study of the distance set problem and its variants in Euclidean and discrete settings, there are many other problems in geometric combinatorics that are worth exploring. We give only a brief and terse introduction to each topic. This list is not meant to be exhaustive, but these are the problems that I personally find most captivating in the area. The motivated reader should explore topics as deeply as possible, and we provide sufficient resources for one to do so.

7.1 Incidence theory Let 𝔽𝑑 denote the 𝑑-dimensional vector space over some field 𝔽. Given a set 𝑃 ⊆ 𝔽𝑑 of points and set Γ of curves in 𝔽𝑑 , we define an incidence as a pair (𝑝, 𝛾) ∈ 𝑃 × Γ such that 𝑝 ∈ 𝛾 (see Figure 7.1). A major problem in geometric combinatorics is to determine the number of incidences between finite point-sets and finite sets of lines. As geometric as this problem feels, counting incidences turns out to have major combinatorial and number-theoretic implications. For example, we have already seen that bounding the number of incidences of points and lines in ℝ2 can be used to give a nontrivial1 bound on the Erdős unit-distance problem (see (2.5)). Recall that the main driver of this result was an incidence result of Szemerédi and Trotter which we restate here for convenience. Theorem 7.1.1 (Szemerédi-Trotter, Version 1). Let 𝑃 ⊆ ℝ2 be a finite point set, and suppose 𝐿 is a finite set of lines in the plane ℝ2 . Then the 1 In

fact, this method gives the current world record!

129

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Chapter 7. Combinatorics in finite fields

Figure 7.1. A set of points 8 points and 6 lines determining 11 incidences.

number of incidences 𝐼(𝑃, 𝐿) satisfies the bound 𝐼(𝑃, 𝐿) ≪ |𝑃|2/3 |𝐿|2/3 + |𝑃| + |𝐿|. In particular if we consider a set of 𝑛 points 𝑃 and 𝑛 lines 𝐿 in ℝ2 , then we have the bound 𝐼(𝑃, 𝐿) ≪ 𝑛4/3 . The Szemerédi-Trotter bound can also be worded as follows: Theorem 7.1.2 (Szemerédi-Trotter, Version 2). Let 𝐿 be a finite set of lines in ℝ2 . Let 𝑋𝑘 denote the set of points in ℝ2 that lie in at least 𝑘 lines. Then |𝑋𝑘 | ≪ |𝐿|2 𝑘−3 + |𝐿|𝑘−1 . The Szemerédi-Trotter bound is sharp up to constants, at least when 𝑃 and 𝐿 are comparable in cardinality. Example 7.1.3. Let 𝑃 = {(𝑥, 𝑦) ∈ ℤ2 ∶ 1 ≤ 𝑥 ≤ 𝑛, 1 ≤ 𝑦 ≤ 2𝑛2 } and 𝐿 = {(𝑡, 𝑎𝑡 + 𝑏) ∶ 𝑎, 𝑏 ∈ ℤ, 1 ≤ 𝑎 ≤ 𝑛, 1 ≤ 𝑏 ≤ 𝑛2 }. Then |𝑃| ≍ |𝐿| ≍ 𝑛3 , and 𝐼(𝑃, 𝐿) ≍ 𝑛4 . There are numerous incidence results for higher dimensions ([144], for example), though none have proven as useful (so far) as SzemerédiTrotter. My favorite proof of the Szemerédi-Trotter theorem is a clever

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131

graph-theoretic proof originally given by Székely ([146]), and we outline that proof now.

7.1.1 Proof of Szemerédi-Trotter (Theorem 7.1.1) We refer the reader unfamiliar with graph theory to Section 6.5 for a terse introduction to the necessary terminology. We assume all graphs are undirected, finite (meaning that the graph has a finite number of vertices), and simple. A cycle is a path that starts and ends on the same vertex, and which traverses no other vertex more than once. Recall that a graph is planar if its crossing number satisfies cr(𝐺) = 0. That is, a graph is planar if and only if it can be drawn in the plane so that no edges intersect. For a planar drawing of a planar graph, there is a well-defined number of faces or regions bounded by edges, including the outer unbounded region (see Figure 7.2).

Figure 7.2. A planar graph with 5 vertices, 6 edges, and 3 faces (including the unbounded face).

Proposition 7.1.4. Let 𝐺 = (𝑉, 𝐸) be planar, finite, simple, and connected. If 𝐺 contains at least 3 vertices, then the number of faces |𝐹| satisfies 3|𝐹| ≤ 2|𝐸|. Proof. If 𝐺 has 3 vertices and 2 edges, then 𝐺 has 1 face, in which case 3(1) ≤ 2(2). Otherwise, each face is determined by at least 3 edges, and each edge corresponds to at most 2 faces. The result follows.

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Chapter 7. Combinatorics in finite fields

Proposition 7.1.5. Let 𝐺 = (𝑉, 𝐸) be planar, finite, simple, and connected. If 𝐺 contains at least 3 vertices, then |𝐸| ≤ 3|𝑉|. Proof. This follows from the previous proposition together with the wellknown Euler formula on planar graphs: A planar graph with |𝐸| edges, |𝑉| vertices, and |𝐹| faces satisfies the formula |𝐹| + |𝑉| − |𝐸| = 2.

Proposition 7.1.6. Let 𝐺 be a finite, simple, connected graph. Then cr(𝐺) > |𝐸| − 3|𝑉|. Proof. Suppose for a contradiction that cr(𝐺) ≤ |𝐸| − 3|𝑉|. Then the planar subgraph 𝐺 ′ obtained by removing the cr(𝐺) ≤ |𝐸| − 3|𝑉| overlapping edges would have more than |𝐸| − (|𝐸| − 3|𝑉|) = 3|𝑉| edges, contradicting Proposition 7.1.5. The key ingredient is the following result colloquially referred to as the Crossing Number Inequality. Theorem 7.1.7. Let 𝐺 = (𝑉, 𝐸) be a finite simple graph, and suppose that |𝐸| ≥ 4|𝑉|. Then |𝐸|3 cr(𝐺) ≫ . |𝑉|2 We now prove Szemerédi-Trotter via the Crossing Number Inequality. Let 𝑃 be a finite point set in which we will make the vertices of our graph 𝐺. Now consider a finite set of lines 𝐿. We say that two points 𝑥, 𝑦 ∈ 𝑃 are consecutive on some line ℓ ∈ 𝐿 if there is no third point 𝑧 on the line segment 𝑥𝑦. Finally, we connect two vertices 𝑥, 𝑦 ∈ 𝑃 with an edge if and only if 𝑥 and 𝑦 are consecutive on some line. We have 𝐼(𝑃, 𝐿) = |𝐸| + |𝐿|, by noticing that any line containing 𝑛 points will determine 𝑛 − 1 edges to the graph, and then adding this over all lines. In particular, |𝐸| = 𝐼(𝑃, 𝐿) − |𝐿|. If |𝐸| ≥ 4|𝑃|, we apply the Crossing Number Inequality to see that (𝐼(𝑃, 𝐿) − |𝐿|)3 . cr(𝐺) ≫ |𝑃|2

7.1. Incidence theory

133

On the other hand, notice that cr(𝐺) ≤ (|𝐿| ) ≍ |𝐿|2 , since each crossing 2 comes from a pair of lines. It follows that (𝐼(𝑃, 𝐿) − |𝐿|)3 ≪ |𝐿|2 , |𝑃|2 and it follows that 𝐼(𝑃, 𝐿) ≪ |𝐿|2/3 |𝑃|2/3 + |𝐿|. If |𝐸| < 4|𝑃|, then 𝐼(𝑃, 𝐿) − |𝐿| < 4|𝑃|, so that 𝐼(𝑃, 𝐿) < 4|𝑃| + |𝐿|. This completes the proof of Theorem 7.1.1. Notice that the only fact about lines we needed in the proof is that two distinct lines can intersect in at most a finite number of points. Thus, the Szemerédi-Trotter Theorem can be extended to any set of curves which do not intersect themselves in more than a constant number of points.

7.1.2 Incidences over finite fields We are now ready to discuss incidences in finite fields. At this point we should be comfortable with points and lines in 𝔽𝑑𝑞 , and incidences in 𝔽𝑑𝑞 again refer to ordered pairs of points 𝑝 and lines ℓ such that 𝑝 ∈ ℓ. We first note that nothing close to as strong as the Szemerédi-Trotter can hold in finite fields. Example 7.1.8. Let 𝑃 = 𝔽2𝑞 , and consider the set of all lines 𝐿 in 𝔽2𝑞 . Note that |𝑃| = 𝑞2 , while |𝐿| = 𝑞2 + 𝑞 ≍ 𝑞2 = |𝑃|. Then 𝐼(𝑃, 𝐿) ≍ 𝑞3 = (𝑞2 )3/2 , and thus the exponent 3/2 is best possible in finite fields. The part of the proof of Szemerédi-Trotter that breaks down in the finite field setting is the notion of “consecutiveness,” which was the condition we used to determine the edge set in our graph. Thus an alternate approach is needed. Proposition 7.1.9. Suppose that we have a set of points 𝑃 ⊆ 𝔽2𝑞 and a set of lines 𝐿 in 𝔽2𝑞 . Then the number of incidences 𝐼 between the set of points 𝑃 and the set of lines 𝐿 satisfies 𝐼 ≤ |𝑃| + |𝑃|1/2 |𝐿| and 𝐼 ≤ |𝐿| + |𝐿|1/2 |𝑃|. In particular, if we have a set of 𝑁 points and 𝑁 lines in 𝔽2𝑞 , then, 𝐼 ≪ 𝑁 3/2 .

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Proof. We prove only the first inequality. We will use Iverson-bracket notation: 1 𝑥∈ℓ [𝑥 ∈ ℓ] = { 0 otherwise. Let 𝑓(ℓ) denote the number of points on the line ℓ, so that 𝑓(ℓ) = ∑ [𝑥 ∈ ℓ]. 𝑥∈𝑃

Thus we have 𝐼 = ∑ 𝑓(ℓ) = ∑ ∑ [𝑥 ∈ ℓ]. ℓ∈𝐿

ℓ∈𝐿 𝑝∈𝑃

By Cauchy-Schwarz, we have 2 2

𝐼 = ( ∑ 𝑓(ℓ)) ≤ ∑ 12 ∑ 𝑓(ℓ)2 = |𝐿| ∑ 𝑓(ℓ)2 . ℓ∈𝐿

ℓ∈𝐿

ℓ∈𝐿

(7.1)

ℓ∈𝐿

Now we see that ∑ 𝑓(ℓ)2 = ∑ ∑ [𝑥 ∈ ℓ] ∑ [𝑦 ∈ ℓ] ℓ∈𝐿

ℓ∈𝐿 𝑥∈𝑃

𝑦∈𝑃

= ∑ ∑ [𝑥 ∈ ℓ]2 + ∑ ∑ [𝑥 ∈ ℓ][𝑦 ∈ ℓ] ℓ∈𝐿 𝑥∈𝑃

ℓ∈𝐿 𝑥,𝑦∈𝑃 𝑥≠𝑦

where the first sum handles the situation when 𝑥 = 𝑦. Clearly, ∑ ∑ [𝑥 ∈ ℓ]2 = ∑ ∑ [𝑥 ∈ ℓ] = 𝐼. ℓ∈𝐿 𝑥∈𝑃

ℓ∈𝐿 𝑥∈𝑃

Also, since there is at most one line that passes through the distinct points 𝑥 and 𝑦, it follows that ∑ ∑ [𝑥 ∈ ℓ][𝑦 ∈ ℓ] ≤ |𝑃|2 . ℓ∈𝐿 𝑥,𝑦∈𝑃 𝑥≠𝑦

Combining these results with (7.1), we see that 𝐼 2 ≤ |𝐿| ⋅ (𝐼 + |𝑃|2 ). Completing the square in 𝐼 yields 2

(𝐼 −

|𝐿|2 |𝐿| . ) ≤ |𝐿||𝑃|2 + 2 4

7.1. Incidence theory

135

Taking square roots and applying the inequality √𝑥 + 𝑦 ≤ √𝑥 + √𝑦 (which is trivially valid for all nonnegative 𝑥 and 𝑦), we finally see that 𝐼 ≤ |𝐿|1/2 |𝑃| + |𝐿|. The second inequality follows similarly, and it uses only the fact that two distinct lines can have at most one common point. As before the proof can be modified slightly to accommodate other curves (such as circles, arcs, etc.) in the plane so long as the curves do not self-intersect too often. We have already seen numerous (often ingenious) applications of incidence theory used to solve many combinatorial problems. In the next section we will explore my favorite such application: a connection discovered by G. Elekes ([41]) between incidence theory and the so-called sum-product problem of Erdős and Szemerédi. Other applications include partial solutions to the unit-distance problem (see (2.5)) as we discussed above. We will spend the rest of the section detailing the most recent incidence bounds in finite fields and other fields. These bounds will be useful in turn in later sections. The first nontrivial incidence result for finite fields was given by Bourgain, Katz, and Tao ([14]). In what follows, 𝑃𝔽3𝑞 can be viewed as the plane 𝔽2𝑞 plus the line at infinity2 . Theorem 7.1.10. Let 𝑝 be prime, and consider a set of points 𝑃 and a set of lines 𝐿 in 𝑃𝔽𝑝3 . Let 𝛼 ∈ (0, 2), and set 𝑁 = 𝑝𝛼 . If |𝑃|, |𝐿| ≤ 𝑁, then 𝐼(𝑃, 𝐿) ≪𝜀 𝑁 3/2−𝜀 for some 𝜀 > 0. The relation between 𝜀 and 𝛼 was nontransparent from their proof. However an explicit estimate was first given Helfgott and Rudnev ([77]). Theorem 7.1.11. Let 𝑝 be prime, and consider a set of points 𝑃 and a set of lines 𝐿 in 𝑃𝔽𝑝3 . If |𝑃| = |𝐿| = 𝑁, where 𝑁 < 𝑝, then 3

1

𝐼(𝑃, 𝐿) ≪ 𝑁 2 − 10678 . 2 Really,

details.

𝑃𝔽3𝑞 is the set 𝔽3𝑞 ⧵ {(0, 0, 0)} quotiented out by dilations. See [14] for more

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This result was later improved by Jones ([91]). Theorem 7.1.12. Let 𝑝 be prime, and consider a set of points 𝑃 and a set of lines 𝐿 in 𝑃𝔽𝑝3 . If |𝑃| = |𝐿| = 𝑁, where 𝑁 < 𝑝, then 3

1

𝐼(𝑃, 𝐿) ≪ 𝑁 2 − 806 +𝑜(1) . This result was further improved numerous times, with the best known results now belonging to Stevens-de Zeeuw ([145]). Theorem 7.1.13. Let 𝑞 = 𝑝𝑟 be the power of a prime, and consider a set of points 𝑃 and a set of lines 𝐿 in 𝑃𝔽3𝑞 . If |𝑃| = |𝐿| = 𝑁, where 𝑁 < 𝑝15/11 , then 3 1 𝐼(𝑃, 𝐿) ≪ 𝑁 2 − 30 +𝑜(1) . Theorem 7.1.13 is only a corollary of the work of Stevens-de Zeeuw, and we encourage the reader to read their paper as it contains many important mathematical ideas, along with a terse recent history of incidence results over general (not necessarily finite) fields. Again many incidence results hold for sets of curves other than lines, so long as the curves do not self-intersect, and as long as we can bound the number of curves on which a single point may lie. Next we explore incidences in higher dimensions and with general curves. In what follows, a plane in 𝔽3𝑞 is a set of the form {𝑥⃗ + 𝑡𝑦 ⃗ + 𝑠𝑧 ⃗ ∶ 𝑡, 𝑠 ∈ 𝔽𝑞 }, where 𝑥,⃗ 𝑦,⃗ 𝑧 ⃗ ∈ 𝔽3𝑞 , and where 𝑦 ⃗ and 𝑧 ⃗ are nonzero. Theorem 7.1.14 ([130]). Let 𝑞 = 𝑝𝑟 be a power of an odd prime, and consider a set 𝑃 points and a set 𝑋 planes in 𝔽3𝑞 where |𝑃| = 𝑚 and |𝑋| = 𝑛. Assume that every line in 𝔽3𝑞 contains 𝑘 or fewer points of 𝑃. Assume also that 𝑚 ≥ 𝑛 and 𝑛 ≪ 𝑝2 . Then we have the bound 𝐼(𝑃, 𝑋) ≪ 𝑚√𝑛 + 𝑚𝑘. Theorem 7.1.14 served as the basis for the proof of Theorem 7.1.13 along with other important results (such as [131]). Theorem 7.1.15 (Cilleruelo, Iosevich, Lund, Roche-Newton, Rudnev, [20]). Let 𝑃 ⊆ 𝔽𝑑𝑞 be a set of points, and 𝒮 a set of spheres in 𝔽𝑑𝑞 . Then |𝑃||𝒮| | | |𝐼(𝑃, 𝒮) − | < |𝑃||𝒮|𝑞𝑑 . | 𝑞 | √

7.2. Sum-product phenomena

137

Theorem 7.1.15 was used to establish some pinned-distance results in 𝔽𝑑𝑞 , and it was also used to establish an analogue of Beck’s Theorem for Circles in 𝔽2𝑞 . The Stephens-De Zeeuw bound (Theorem 7.1.13) and a related result ([131]) were used to prove the following variant (a so-called “small set” variant) of the Finite Field Distance Problem. Theorem 7.1.16 ([86]). Let 𝐴 ⊆ 𝔽𝑝2 , where |𝐴| ≤ 𝑝7/6 , and 𝑝 ≡ 3 (mod 4). Then, 1

149

|Δ(𝐴)| ≫ |𝐴| 2 + 4214 . Finally, there has also been an incidence result over cyclic rings. Theorem 7.1.17 ([125]). Let ℤ𝑚 be the ring of integers modulo 𝑚, and let 𝛾(𝑚), 𝜙(𝑚), and 𝜏(𝑚) denote the smallest prime factor of 𝑚, the Euler totient function, and the number of positive divisors of 𝑚, respectively. Then for a collection of points 𝑃 and lines 𝐿 in ℤ2𝑚 , we have 𝐼(𝑃, 𝐿) ≤

|𝑃||𝐿| 2𝜏(𝑚)𝑚2 |𝑃||𝐿| . + 𝑚 𝜙(𝑚) √ 𝛾(𝑚)

7.2 Sum-product phenomena Given a subset 𝐴 of a ring (𝑅, +, ⋅), we define the sum set, product set, difference set, and quotient set3 of 𝐴 as 𝐴 + 𝐴 = {𝑎 + 𝑏 ∶ 𝑎, 𝑏 ∈ 𝐴}, 𝐴 ⋅ 𝐴 = {𝑎 ⋅ 𝑏 ∶ 𝑎, 𝑏 ∈ 𝐴}, 𝐴 − 𝐴 = {𝑎 − 𝑏 ∶ 𝑎, 𝑏 ∈ 𝐴}, 𝐴/𝐴 = {𝑎𝑏−1 ∶ 𝑎, 𝑏 ∈ 𝐴}, respectively. If 𝐴 ⊆ ℝ and |𝐴| = 𝑘, then we have the trivial bounds 2𝑘 − 1 ≤ |𝐴 + 𝐴|, |𝐴 ⋅ 𝐴| ≤ (

𝑘+1 ). 2

3 In order for the quotient set of 𝐴 to be well defined, one needs to ensure that all elements of 𝐴 are invertible.

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As an example consider the set 𝐴 = {𝑎, 𝑎 + 𝑑, . . . , 𝑎 + (𝑘 − 1)𝑑}, an arithmetic progression of length 𝑘. Then 𝐴 + 𝐴 = {2𝑎, 2𝑎 + 𝑑, . . . , 2(𝑎 + (𝑘 − 1)𝑑)}, so that |𝐴 + 𝐴| = 2𝑘 − 1. On the other hand, it has been shown by Ford ([56]) that 𝑘2 |𝐴 ⋅ 𝐴| ≍ , (log 𝑘)𝛿 (log log 𝑘)3/2 1+log log 2

= 0.086071 . . . .4 Thus while the sum set is where 𝛿 = 1 − log 2 small, the product set is large. This result of Ford also settled the socalled Multiplication Table Problem. Given an 𝑛×𝑛 multiplication table, how many unique entries exist on that table? This is really asking for the exact size of the product set 𝐴 ⋅ 𝐴, where 𝐴 = {1, . . . , 𝑛}. On the other hand, if 𝐴 = {𝑎, 𝑎𝑟, 𝑎𝑟2 , . . . 𝑎𝑟𝑘−1 } is a geometric progression, then we have |𝐴 ⋅ 𝐴| = 2𝑘 − 1, while |𝐴 + 𝐴| = (𝑘+1 ). Thus in these 2 cases, if either the sum set or the product set is small, then the other is large. Erdős and Szemerédi5 conjectured the following: Conjecture 7.2.1 (Sum-Product Conjecture, [51]). Let 𝐴 ⊆ ℤ, with |𝐴| = 𝑘. Then for all 𝜀 > 0, we have max(|𝐴 + 𝐴|, |𝐴 ⋅ 𝐴|) ≫𝜀 𝑘2−𝜀 . In the same paper Erdős and Szemerédi proved the following: Theorem 7.2.2. Let 𝐴 ⊆ ℤ, with |𝐴| = 𝑘. Then there exists 𝜀 > 0 such that max(|𝐴 + 𝐴|, |𝐴 ⋅ 𝐴|) ≫ 𝑘1+𝜀 . Notice that the bound max(|𝐴 + 𝐴|, |𝐴 ⋅ 𝐴|) ≫ 𝑘1+𝜀 is exponentially better than the trivial bound max(|𝐴 + 𝐴|, |𝐴 ⋅ 𝐴|) ≫ 𝑘, but no attempt 4 Ford’s result postdates the Erdős-Szemerédi Problem, but Erdős already knew ([50]) that |𝐴 ⋅ 𝐴| ≍ 𝑘2 /(log 𝑘)𝛿+𝑜(1) where 𝛿 = 0.086 . . . is the value appearing in Ford’s result. 5 Mel Nathanson once told me that this is really only Erdős’ conjecture (and not Szemerédi’s). However, due to Erdős’ prolific career, it is helpful to append someone else’s name to the conjecture in order to help distinguish this conjecture of Erdős from his many others. For this reason and since historically this has been referred to as the ErdősSzemerédi Problem, we choose to retain both names.

7.2. Sum-product phenomena

139

at an explicit value of 𝜀 > 0 was made until Nathanson’s work ([115]), where he showed that every set 𝐴 ⊆ ℤ with |𝐴| = 𝑘 must satisfy 1

max(|𝐴 + 𝐴|, |𝐴 ⋅ 𝐴|) ≫ 𝑘1+ 31 . Ford ([55]) refined Nathanson’s work to achieve the bound 1

max(|𝐴 + 𝐴|, |𝐴 ⋅ 𝐴|) ≫ 𝑘1+ 15 . Around the same time, Elekes ([41]) changed the way we think about the sum-product problem by connecting two seemingly disparate areas of mathematics in number theory and extremal geometry. Theorem 7.2.3. Let 𝐴 ⊆ ℝ with |𝐴| = 𝑘. Then |𝐴 + 𝐴||𝐴 ⋅ 𝐴| ≫ 𝑘5/2 . In particular, max(|𝐴 + 𝐴|, |𝐴 ⋅ 𝐴|) ≫ 𝑘5/4 . Proof. We may assume that 0 ∉ 𝐴 without affecting our asymptotic bounds. Consider the set of points 𝑃 = (𝐴 ⋅ 𝐴) × (𝐴 + 𝐴) and the set of lines 𝐿 = {𝑚𝑥 + 𝑏 ∶ 𝑏 ∈ 𝐴,

1 ∈ 𝐴} . 𝑚

Then |𝑃| = |𝐴 + 𝐴||𝐴 ⋅ 𝐴| while |𝐿| = 𝑘2 . Note that each line is incident to 𝑘 points in 𝑃. Hence, 𝐼(𝑃, 𝐿) = 𝑘3 . Thus, 𝑘3 ≪ (|𝐴 ⋅ 𝐴||𝐴 + 𝐴|)2/3 (𝑘2 )2/3 by the Szemerédi-Trotter inequality. The result follows. More than simply improving the Sum-Product results, Elekes’ short proof drastically changed our viewpoint of the sum-product problem, and efforts to improve on the sum-product conjecture ([101, 138, 141, 142]) have relied on incidence theory ever since. Note also that Elekes’ result works over any set of reals 𝐴 ⊆ ℝ, so that 𝐴 need not contain only integers. Unfortunately, there is a natural boundary of |𝐴 + 𝐴||𝐴 ⋅ 𝐴| ≫ 𝑘3 , so that 3/2 is the optimal exponent we could achieve (for the sumproduct problem) with this method. It is known ([138]) that max(|𝐴 + 𝐴|, |𝐴 ⋅ 𝐴|) ≫ |𝐴|4/3+5/5277 for all 𝐴 ⊆ ℝ.

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Interestingly, there are numerous results that state when either the sum set or the product set is small, then the other must be large. Results of this type, specifically showing that the product set is large when the sum set is small, are sometimes called Freiman-type results, as the first such result was the work of Gregory Freiman ([57]). Theorem 7.2.4 (Freiman’s Theorem). Let 𝑏 denote a positive integer, and suppose 𝐴 ⊆ ℤ, where |𝐴| = 𝑘. If 𝑘 ≥ 3, and if |𝐴+𝐴| = 2𝑘−1+𝑏 ≤ 3𝑘−4, then 𝐴 is a subset of an arithmetic progression of length 𝑘 + 𝑏. This result is highly combinatorial in nature, and improvements have long been sought by many mathematicians ([11, 133]). Nathanson and Tenenbaum ([116]) were the first to explicitly evaluate the size of the product set when the sum set is small. Theorem 7.2.5. Let 𝐴 ⊆ ℤ, with |𝐴| = 𝑘. If |𝐴 + 𝐴| ≤ 3𝑘 − 4, then |𝐴 ⋅ 𝐴| ≥ 𝑘2 /(log 𝑘)2 if 𝑘 is sufficiently large. A quantitive version of the “small sum set, large product set” phenomenon was also given by Elekes and Ruzsa ([42]). Theorem 7.2.6. Let 𝐴 ⊆ ℝ, where |𝐴| = 𝑘 ≥ 2. Then, |𝐴 + 𝐴|4 |𝐴 ⋅ 𝐴| ≫

𝑘6 . log 𝑘

In particular, if |𝐴 + 𝐴| ≪ 𝑘, then |𝐴 ⋅ 𝐴| ≫

𝑘2 . log 𝑘

Note that Theorem 7.2.6 is sharp in the sense that 𝑘6 cannot be replaced by any smaller power of 𝑘. Nonetheless, their result only gives an inferior sum-product estimate of max(|𝐴 + 𝐴|, |𝐴 ⋅ 𝐴|) ≫ 𝑘6/5−𝜀 for any 𝜀 > 0. On the other hand, Elekes’ Theorem (Theorem 7.2.3) was not a sharp inequality, and yet it yielded a superior sum-product estimate. The “small product set implies large sum set” results are more difficult. Nonetheless Chang ([16]) gave the following result: Theorem 7.2.7. Let 𝐴 ⊆ ℤ, with |𝐴| = 𝑘. If |𝐴 ⋅ 𝐴| < 𝛼𝑘, then |𝐴 + 𝐴| > 36−𝛼 𝑘2 . Before we explore some questions related to the Sum-Product Problem, we first review some of the results achieved and variants considered

7.2. Sum-product phenomena

141

towards a better understanding of the sum-product phenomena. For example the following was achieved by Iosevich, Roche-Newton, and Rudnev ([88]). The following relied heavily on the Guth-Katz distance theorem (see (2.4)): Theorem 7.2.8. Let 𝐴 ⊆ ℝ with |𝐴| = 𝑘. Then, |𝐴 ⋅ 𝐴 + 𝐴 ⋅ 𝐴 + 𝐴 ⋅ 𝐴| ≫ 𝑘2 / log 𝑘.

7.2.1 Sum-product phenomena in rings It is time to delve into other analogues of the problem. Indeed, the sum set 𝐴 + 𝐴 and product set 𝐴 ⋅ 𝐴 are well defined when 𝐴 ⊆ 𝑅, where 𝑅 is any ring. Naturally, we will primarily consider the case 𝑅 = 𝔽𝑞 , where 𝑞 = 𝑝ℓ for an odd prime 𝑝. There are numerous idiosyncrasies for the sum-product problem over finite fields. For example, if 𝐴 ⊆ 𝔽𝑞 is a subring, then the sum set and product set satisfy 𝐴 + 𝐴 = 𝐴 ⋅ 𝐴 = 𝐴. Thus we must not allow 𝐴 to be “too close” to being a subring of 𝑅. For subsets of finite rings 𝐴 ⊆ 𝑅, we have the trivial bounds |𝐴| ≤ max(|𝐴 + 𝐴|, |𝐴 ⋅ 𝐴|) ≤ min(|𝑅|, (

|𝐴| + 1 )) . 2

Fortunately, there are many methods available to us to ensure that 𝐴 is not close to being a subring, including cardinality conditions (supposing 𝑞1/2 ≪ |𝐴| ≪ 𝑞1−𝜀 for some 𝜀 > 0, for instance). The first pure sum-product result in finite fields was work of Bourgain, Katz, and Tao ([14]). Theorem 7.2.9. Let 𝐴 ⊆ 𝔽𝑝 for a prime 𝑝, where |𝐴| ≪ 𝑝1−𝜀 for some 𝜀 > 0. Then there exists 𝛿 = 𝛿(𝜀) > 0 such that max(|𝐴 + 𝐴|, |𝐴 ⋅ 𝐴|) ≫ |𝐴|1+𝛿 . Note 7.2.10. Theorem 7.2.9 does not establish an explicit relationship between 𝜀 and 𝛿. Also, the condition |𝐴| ≪ 𝑝1−𝜀 was sufficient in this case as the prime field 𝔽𝑝 has only the trivial subrings: itself and {0}. The original statement of Theorem 7.2.9 also assumed that |𝐴| ≫ 𝑝𝜀 , though this condition was later shown ([13]) to be unnecessary.

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Chapter 7. Combinatorics in finite fields

Hart, Iosevich, and Solymosi ([75]) established the first explicit sumproduct bound. Theorem 7.2.11. Let 𝐴 ⊆ 𝔽𝑞 with 𝑞1/2 ≪ |𝐴| ≪ 𝑞7/10 . Then, max(|𝐴 + 𝐴|, |𝐴 ⋅ 𝐴|) ≫ 𝑞−1/4 |𝐴|3/2 . In particular if |𝐴| ≍ 𝑞7/10 , then max(|𝐴 + 𝐴|, |𝐴 ⋅ 𝐴|) ≫ |𝐴|8/7 . Additionally, if |𝐴| ≫ 𝑞3/4 and |𝐴 + 𝐴| ≪ |𝐴|, then |𝐴 ⋅ 𝐴| ≫ 𝑞. The proof of Theorem 7.2.11 essentially amounts to using the following incidence bound for hyperbolas. Theorem 7.2.12 ([75]). Let 𝐸, 𝐹 ⊆ 𝔽2𝑞 , and 𝑗 ∈ 𝔽∗𝑞 . Then |{(𝑥, 𝑦) ∈ 𝐸 × 𝐹 ∶ (𝑥1 − 𝑦1 )(𝑥2 − 𝑦2 ) = 𝑗}| ≪ 𝑞−1 |𝐸||𝐹| + 𝑞1/2 √|𝐸||𝐹|. Additional results exist for small subsets of finite fields. Theorem 7.2.13 (Garaev, [59]). Let 𝐴 ⊆ 𝔽𝑝 , where 𝑝 is prime. Then for sets 𝐴 with |𝐴| < 𝑝7/13 /(log 𝑝)4/13 , max(|𝐴 + 𝐴|, |𝐴 ⋅ 𝐴|) ≫ |𝐴|15/14+𝑜(1) . Garaev’s inequality has been improved many times ([12, 95, 96]) with the current best result belonging to Roche-Newton, Rudnev, and Shkredov ([127]). Theorem 7.2.14. Let 𝐴 ⊆ 𝔽𝑞 , where 𝑞 = 𝑝ℓ for a prime 𝑝. Then we have max(|𝐴 + 𝐴|, |𝐴 ⋅ 𝐴|) ≫ |𝐴|6/5 , so long as |𝐴| < 𝑝5/8 . A similar result holds if 𝐴 + 𝐴 is replaced by 𝐴 − 𝐴. In arbitrary finite fields, the strongest known result is the following result by Jones and Roche-Newton ([92]). Theorem 7.2.15. Let 𝐴 ⊆ 𝔽𝑞 , where 𝐴 is such that |𝐴 ∩ 𝑐𝐺| ≤ |𝐺|1/2 for all subfields 𝐺 ⊆ 𝔽𝑞 and for all 𝑐 ∈ 𝔽𝑞 . Then max(|𝐴 + 𝐴|, |𝐴 ⋅ 𝐴|) ≫ |𝐴|12/11−𝑜(1) . Note 7.2.16. Again the condition that |𝐴∩𝑐𝐺| ≤ |𝐺|1/2 in Theorem 7.2.15 ensures that 𝐴 is sufficiently far from being a subring.

7.2. Sum-product phenomena

143

Another such result concerns the set 𝐴(𝐴 + 1) = {𝑎(𝑏 + 1) ∶ 𝑎, 𝑏 ∈ 𝐴}. Theorem 7.2.17 ([92]). Let 𝐴 ⊆ 𝔽𝑝 , where 𝑝 is prime. Then for sets 𝐴 with |𝐴| < 𝑝1/2 , |𝐴(𝐴 + 1)| ≫ |𝐴|57/56−𝑜(1) . This result was an improvement on [60], and both papers also included results for sets of real numbers 𝐴 ⊆ ℝ. Theorem 7.2.18 ([127]). Let 𝐴 ⊆ 𝔽𝑞 with 𝑞 = 𝑝ℓ for a prime 𝑝. For sets 𝐴 with |𝐴| ≪ 𝑝18/35 , max(|𝐴 + 𝐴 + 𝐴|, |𝐴 ⋅ 𝐴|) ≫ |𝐴|16/13 . We mention one more analogue of the sum-product phenomena in finite fields. We define 𝐴 ⋅ 𝐴 + 𝐴 ⋅ 𝐴 = {𝑎𝑏 + 𝑐𝑑 ∶ 𝑎, 𝑏, 𝑐, 𝑑 ∈ 𝐴}. It is convenient to denote 𝐴 + 𝐴 = 𝐴2 , and 𝐴 + 𝐴 = 2𝐴. Thus we would write 𝐴 ⋅ 𝐴 + 𝐴 ⋅ 𝐴 = 2𝐴2 . Likewise we define 𝑑⎵ 𝑘𝐴𝑑 = 𝐴 +⎵⏟⎵ ⋯+ 𝐴𝑑 , ⏟⎵ ⎵⎵⏟ 𝑘−times

where 𝐴𝑑 = {𝑎1 . . . 𝑎𝑑 ∶ 𝑎1 , . . . , 𝑎𝑑 ∈ 𝐴} is the 𝑑-fold product set. In this context the sum-product problem can then be framed similarly to the Erdős-Falconer distance problem. Problem 7.2.19 (Dot-Product Problem). Let 𝐴 ⊆ 𝔽𝑞 . Find the minimum exponent 𝛼 ∈ (0, 1) such that we are guaranteed |𝑘𝐴𝑑 | ≫ 𝑞 whenever |𝐴| ≫ 𝑞𝛼 . In general 𝛼 > 1/2 as otherwise 𝐴 could be a subring of 𝔽𝑞 . 1

1

Theorem 7.2.20. ([74]) Let 𝐴 ⊆ 𝔽𝑞 with |𝐴| > 𝑞 2 + 2𝑑 . Then 𝔽∗𝑞 ⊆ 𝑑𝐴2 .

144

Chapter 7. Combinatorics in finite fields

Proof. For 𝐸 ⊆ 𝔽𝑑𝑞 , we define Π(𝐸) = {𝑥 ⋅ 𝑦 ∶ 𝑥, 𝑦 ∈ 𝐸}, where 𝑥 ⋅ 𝑦 = 𝑥1 𝑦1 + ⋯ + 𝑥𝑑 𝑦 𝑑 denotes the standard inner product. Let 𝜈(𝑗) = |{(𝑥, 𝑦) ∈ 𝐸 × 𝐸 ∶ 𝑥 ⋅ 𝑦 = 𝑗}|. We will show that 𝑗 ∈ Π(𝐸) for all 𝑗 ∈ 𝔽∗𝑞 so long as 𝐸 is sufficiently large, which is more than what we need to show. We will make the connection between these results at the end of the proof. As usual, we use the exponential sums method:

𝜈(𝑗) = 𝑞−1 ∑ ∑ 𝜒(𝑠 ⋅ (𝑥 ⋅ 𝑦 − 𝑗)) 𝑠∈𝔽𝑞 𝑥,𝑦∈𝐸

=𝑞

−1

∑ ∑ 𝜒(𝑠(𝑥 ⋅ 𝑦 − 𝑗)) + 𝑞−1 ∑ ∑ 𝜒(𝑠(𝑥 ⋅ 𝑦 − 𝑗)) 𝑠≠0 𝑥,𝑦∈𝐸

𝑠=0 𝑥,𝑦∈𝐸 −1

2

= 𝑞 |𝐸| + 𝑅𝑗 . Note that by rearranging our sums and applying the triangle inequality, we have | | 𝑅𝑗 = 𝑞−1 ∑ ∑ ∑ 𝜒(𝑠(𝑥 ⋅ 𝑦 − 𝑗)) ≤ 𝑞−1 ∑ || ∑ ∑ 𝜒(𝑠(𝑥 ⋅ 𝑦 − 𝑗))|| . | 𝑥∈𝐸 𝑦∈𝐸 𝑠≠0 𝑥∈𝐸 |𝑦∈𝐸 𝑠≠0 Squaring 𝑅𝑗 and applying the Cauchy-Schwarz inequality yields 2

|𝑅𝑗 | ≤ 𝑞

−2

| | ( ∑ 1 ⋅ || ∑ ∑ 𝜒(𝑠(𝑥 ⋅ 𝑦 − 𝑗))||) | |𝑦∈𝐸 𝑠≠0 𝑥∈𝐸

≤𝑞

−2

| | ( ∑ 1 ) ( ∑ || ∑ ∑ 𝜒(𝑠(𝑥 ⋅ 𝑦 − 𝑗))|| ) | 𝑥∈𝐸 𝑥∈𝐸 |𝑦∈𝐸 𝑠≠0

2

2

2

2

| | ≤ 𝑞 |𝐸| ∑ || ∑ ∑ 𝜒(𝑠(𝑥 ⋅ 𝑦 − 𝑗))|| . |𝑦∈𝐸 𝑠≠0 | 𝑥∈𝔽𝑑 𝑞 −2

7.2. Sum-product phenomena

145

Using only the fact that a complex number 𝑧 = 𝑥 + 𝑖𝑦 satisfies |𝑧| = 𝑧𝑧 = (𝑥 + 𝑖𝑦)(𝑥 − 𝑖𝑦), we see that 2

| | ∑ || ∑ ∑ 𝜒(𝑠(𝑥 ⋅ 𝑦 − 𝑗))|| |𝑦∈𝐸 𝑠≠0 | 𝑥∈𝔽𝑑 𝑞 = ∑

∑

∑ 𝜒(𝑥 ⋅ (𝑠𝑦 − 𝑠′ 𝑦′ ))𝜒(𝑗(𝑠′ − 𝑠))

𝑠,𝑠′ ≠0 𝑦,𝑦′ ∈𝐸 𝑥∈𝔽𝑑 𝑞

= ∑ 𝜒(𝑗(𝑠′ − 𝑠)), 𝑠,𝑠′ ≠0 𝑦,𝑦∈𝐸 𝑠𝑦=𝑠′ 𝑦′

where we regrouped and then applied orthogonality. Note that 𝑠, 𝑠′ ∈ 𝔽∗𝑞 . Thus we substitute 𝑎 = 𝑠/𝑠′ and 𝑏 = 𝑠′ to see that |𝑅𝑗 |2 ≤ 𝑞𝑑−2 |𝐸| ∑ 𝜒(𝑗𝑏(1 − 𝑎)) ≔ 𝐼 + 𝐼𝐼, 𝑎,𝑏∈𝔽∗𝑞 𝑦,𝑦′ ∈𝐸 𝑎𝑦=𝑦′

where 𝐼 = 𝑞𝑑−2 |𝐸| ∑ 1 = |𝐸|2 𝑞𝑑−2 (𝑞 − 1) ≤ |𝐸|2 𝑞𝑑−1 , 𝑏∈𝔽∗𝑞 𝑦,𝑦′ ∈𝐸 𝑦=𝑦′

and where 𝐼𝐼 = 𝑞𝑑−2 |𝐸|

𝜒(𝑗𝑏(1 − 𝑎)) − 𝑞𝑑−2 |𝐸| ∑ 1

∑

𝑎≠0,1 𝑦,𝑦′ ∈𝐸 𝑎𝑦=𝑦′

𝑏∈𝔽𝑞 ;𝑎≠0,1 𝑦,𝑦′ ∈𝐸 𝑎𝑦=𝑦′

= −𝑞𝑑−2 |𝐸| ∑ 1. 𝑎≠0,1 𝑦,𝑦′ ∈𝐸 𝑎𝑦=𝑦′

Hence 𝐼𝐼 < 0. This shows that |𝑅𝑗 |2 ≤ 𝐼 + 𝐼𝐼 ≤ 𝐼 ≤ 𝑞𝑑−1 |𝐸|2 , and hence 𝜈(𝑗) = 𝑞−1 |𝐸|2 + 𝑅𝑗 , where |𝑅𝑗 | ≤ 𝑞

𝑑−1 2

|𝐸|.

146

Chapter 7. Combinatorics in finite fields 𝑑+1

Therefore, 𝜈(𝑗) > 0 (ensuring 𝑗 ∈ 𝑃𝑖(𝐸)) whenever |𝐸| > 𝑞 2 . In particular if 𝐸 = 𝐴 × ⋯ × 𝐴 is the 𝑑-fold Cartesian product of a set 𝐴 ⊆ 𝔽𝑞 , then Π(𝐸) = 𝑑𝐴2 . It follows that if |𝐴| > 𝑞

𝑑+1 2𝑑

, then 𝔽∗𝑞 ⊆ 𝑑𝐴2 .

In [26], Theorem 7.2.20 was extended to the case ℤ𝑞 . Theorem 7.2.21. Let 𝑞 = 𝑝ℓ for a prime 𝑝, and suppose 𝐴 ⊆ ℤ𝑞 satisfies |𝐴| > 𝑞

(2ℓ−1)𝑑+1 2ℓ𝑑

. Then, ℤ×𝑞 ⊆ 𝑑𝐴2 .

As before, the proof of Theorem 7.2.21 mimicked the proof of Theorem 7.2.20 in spirit, though some of the technical difficulties were greater.

7.3 Kakeya conjecture Another problem in the “big sets must be interesting” category is the Kakeya Conjecture. A Kakeya-Besicovith set (which for convience we will call simply a Kakeya set) is a compact set 𝐸 ⊆ ℝ2 containing a line segment of length 1 in every direction. Of course the first such set that springs to mind is a circle with diameter 1. Is there another set with a smaller area? It turns out that an equilateral triangle with sidelength 1 is also a Kakeya set, and it has a smaller area. A deltoid of diameter 1 has area smaller still. How small can the area be of such a Kakeya set?

Theorem 7.3.1 (Besicovitch, [8]). There exist Kakeya sets with measure 06 . 6 See

Section 2.2 for what it means for a set to have measure 0.

7.3. Kakeya conjecture

147

This is far from the end of the story as there is much more to consider. What happens in higher dimensions? Do there exist Kakeya sets in ℝ𝑑 of measure zero? What do they look like? The so-called Kakeya conjecture is simple to state, given that we remember the Hausdorff dimension: Conjecture 7.3.2. Let 𝐸 ⊆ ℝ𝑑 be a Kakeya set. Then dim𝐻 (𝐸) = 𝑑. Again, one should not worry too much about the notion of Hausdorff dimension if this concept is foreign to you. Simply put, the Kakeya conjecture asserts that Kakeya sets must all be large in terms of their (Hausdorff) dimension. Roy Davies ([32]) showed that a Kakeya set 𝐸 ⊆ ℝ2 satisfies dim𝐻 (𝐸) = 2, but the conjecture is unsolved in dimensions 𝑑 ≥ 3. The best known bounds for Kakeya sets in higher dimensions involve the Minkowski dimension, which is closely related to the Hausdorff dimension, but it is much easier to define. For a compact set 𝐸 ⊆ ℝ𝑑 , let 𝑁 𝛿 (𝐸) denote the minimum number of 𝛿-balls7 needed to cover 𝐸. Then the upper Minkowski dimension is log(𝑁 𝛿 (𝐸)) dim𝑀 (𝐸) = lim sup , log(1/𝛿) + 𝛿→0 while the lower Minkowski dimension is log(𝑁 𝛿 (𝐸)) , dim𝑀 (𝐸) = lim inf + log(1/𝛿) 𝛿→0 and thus the Minkowski dimension is dim𝑀 (𝐸) = lim+ 𝛿→0

log(𝑁 𝛿 (𝐸)) , log(1/𝛿)

provided the limit exists. Note 7.3.3. There are many sets for which the Hausdorff dimension and the Minkowski dimension are equal, but in general one has the inequality 0 ≤ dim𝐻 (𝐸) ≤ dim𝑀 (𝐸) ≤ dim𝑀 (𝐸) ≤ 𝑑 7A

𝛿-ball is simply a ball with radius 𝛿.

148

Chapter 7. Combinatorics in finite fields

for all sets 𝐸 ⊆ ℝ𝑑 . Thus the condition that dim𝐻 𝐸 = 𝑑 is stronger than the condition dim𝑀 𝐸 = 𝑑. The best known bounds for the Minkowski dimension of a Kakeya set in ℝ𝑑 are as follows ([94, 97, 103]). Theorem 7.3.4. Let 𝐸 ⊆ ℝ𝑑 be a Kakeya set. Then, 5/2 + 10−10 ⎧ ⎪ ⎪ 3 + 10−10 ⎪ dim𝑀 (𝐸) ≥

⎨ ⎪ (2 − √2)(𝑑 − 4) + 3 ⎪ ⎪ 𝑑+𝑟−1 ⎩ 𝑟

𝑑=3 𝑑=4 5 ≤ 𝑑 ≤ 23 𝑑 ≥ 24.

Here, 𝑟 = 1.675 . . . is the root of 𝑥3 = 4𝑥 − 2 with 𝑟 > 1. The following bounds for the Hausdorff dimension of Kakeya sets are also known ([97, 158]). Theorem 7.3.5. Let 𝐸 ⊆ ℝ𝑑 be a Kakeya set. Then dim𝐻 (𝐸) ≥ {

𝑑+2 2

(2 − √2)(𝑑 − 4) + 3

𝑑 = 3, 4 𝑑 ≥ 5.

The proofs of these results are very technical in nature (notice that we haven’t even defined the Hausdorff dimension!), and so we will not give the proofs here. The interested reader should consult the original sources for such proofs. However, Tom Wolff ([157]) devised a finite field analogue of the Kakeya conjecture, and we will explore that analogue in detail below. Even in finite fields, a Kakeya set refers to a set containing a line in every direction. We first recall some basic facts about lines in finite fields, and we define directions in finite fields. Recall a line in 𝔽𝑑𝑞 is a set of the form ℓ = {𝑎 + 𝑡𝑏 ∶ 𝑡 ∈ 𝔽𝑞 } ⊆ 𝔽𝑑𝑞 , where 𝑎, 𝑏 ∈ 𝔽𝑑𝑞 and 𝑏 ≠ 0.⃗ Notice that every line contains precisely 𝑞 points. Now, we say two lines ℓ1 and ℓ2 in 𝔽𝑑𝑞 are in the same direction

7.3. Kakeya conjecture

149

if there exists a vector 𝑥 ∈ 𝔽𝑑𝑞 such that ℓ1 + 𝑥 = {𝑎 + 𝑥 + 𝑡𝑏 ∶ 𝑡 ∈ 𝔽𝑞 } satisfies ℓ1 + 𝑥 = ℓ2 . “Being in the same direction” turns out to be an equivalence relation on the set of all lines. Directions are the resulting set of equivalence classes. Moreover, the number of directions in 𝔽𝑑𝑞 is given by 𝑞𝑑 − 1 , 𝑞−1 since given any point, there are 𝑞𝑑 − 1 other points in 𝔽𝑑𝑞 through which we can draw a line, and there are exactly 𝑞 points on a line (which means there are 𝑞 − 1 other points on a line if we exclude our starting point). Thus, there are exactly 𝑞 + 1 directions in 𝔽2𝑞 and approximately 𝑞𝑑−1 directions in 𝔽𝑑𝑞 . We are now ready to state the Finite Field Kakeya Conjecture. Recall that the notion of dimension in ℝ𝑑 can be replaced by cardinality in 𝔽𝑑𝑞 . Loosely speaking, if 𝐸 ⊆ ℝ𝑑 has some algebraic or geometric property whenever dim𝐻 (𝐸) > 𝛼, then one should expect8 the same for sets 𝐸 ⊆ 𝔽𝑑𝑞 with |𝐸| ≫ 𝑞𝛼 . Conjecture 7.3.6. Let 𝐸 ⊆ 𝔽𝑑𝑞 be a set containing a line in every direction. ⃗ there More precisely, suppose that 𝐸 ⊆ 𝔽𝑑𝑞 is such that for all 𝑏 ∈ 𝔽𝑑𝑞 ⧵ {0}, 𝑑 𝑑 exists 𝑎 ∈ 𝔽𝑞 such that 𝑎 + 𝑡𝑏 ∈ 𝐸 for all 𝑡 ∈ 𝔽𝑞 . Then |𝐸| ≫𝑑 𝑞 . This conjecture has been resolved in the affirmative ([40]), but before we discuss this result, it is instructive to consider Wolff’s results. Theorem 7.3.7 (Wolff, [158]). Let 𝐸 ⊆ 𝔽𝑑𝑞 be a Kakeya set. Then |𝐸| ≫𝑑 𝑞 In particular a Kakeya set 𝐸 ⊆

𝔽2𝑞

𝑑+2 2

.

satisfies |𝐸| ≫ 𝑞2 .

Proof. We only prove the result in the case that 𝑑 = 2, and in fact, we prove a strengthened version of the result: Lemma 7.3.8. Let 𝐸 ⊆ 𝔽2𝑞 satisfy the condition that 𝐸 contains at least 𝑞/2 points on a line in a set of lines having 𝑚 different directions. Then |𝐸| ≫ 𝑚𝑞. 8 This is a vast overgeneralization, but the number of instances of this duality transcends mere coincidence.

150

Chapter 7. Combinatorics in finite fields

We first prove the lemma. Let {ℓ1 , . . . , ℓ𝑚 } be the set of 𝑚 lines. Since |𝐸 ∩ ℓ| ≥ 𝑞/2 for all lines ℓ, it follows that 𝑚

1 𝑚𝑞 ≤ ∑ |𝐸 ∩ ℓ𝑗 |. 2 𝑗=1 Squaring our previous inequality, we see that 2

𝑚 2

(𝑚𝑞) ≪ ( ∑ |𝐸 ∩ ℓ𝑗 |) 𝑗=1 𝑚

𝑚

= ∑ ∑ |𝐸 ∩ ℓ𝑗 ||𝐸 ∩ ℓ𝑘 | 𝑗=1 𝑘=1 𝑚

𝑚

= ∑ ∑ ∑ 𝜒ℓ𝑗 (𝑥)𝜒ℓ𝑘 (𝑥) 𝑥∈𝐸 𝑗=1 𝑘=1 𝑚

𝑚

≤ |𝐸| ∑ ∑ |ℓ𝑗 ∩ ℓ𝑘 |, 𝑗=1 𝑘=1

where 𝜒ℓ is the characteristic function of the line ℓ so that 𝜒ℓ (𝑥) = 1 if 𝑥 ∈ ℓ and 0 otherwise. Using that two lines intersect in at most 1 point, we see that 𝑚

(𝑚𝑞)2 ≪ |𝐸| ( ∑ |ℓ𝑗 | + ∑ |ℓ𝑗 ∩ ℓ𝑘 |) 𝑗=1

𝑗≠𝑘

≤ |𝐸|(𝑚𝑞 + 𝑚(𝑚 − 1)) from which the Lemma follows, since there are 𝑚 = 𝑞 + 1 directions when 𝑑 = 2. Thus we have shown that |𝐸| ≫ 𝑞2 for Kakeya sets 𝐸 ⊆ 𝔽2𝑞 , and this proves Theorem 7.3.7 when 𝑑 = 2. Wolff’s bounds were steadily improved upon by numerous authors ([14, 111, 128, 149]), whose proofs were quite difficult and typically involved some harmonic analysis, additive combinatorics, and even some algebraic geometry. It was widely believed that the finite field Kakeya problem was nearly as difficult as the Euclidean version until Zeev Dvir shocked the harmonic analysis world with a full (algebraic) proof of the

7.3. Kakeya conjecture

151

finite field Kakeya conjecture, whose difficulty even some high school students9 would understand. Theorem 7.3.9 (Dvir, [40]). Let 𝐸 ⊆ 𝔽𝑑𝑞 be a Kakeya set. Then |𝐸| ≫𝑑 𝑞𝑑 . Proof. We need the following two Propositions10 . Proposition 7.3.10 (Paramater Counting Argument). Suppose that 𝐸 ⊆ 𝔽𝑑𝑞 . Then there exists a nonzero polynomial 𝑃 in 𝑑 variables such that 𝑃(𝑥) = 0 for all 𝑥 ∈ 𝐸 and such that deg(𝑃) ≤ 𝑑|𝐸|1/𝑑 . Proposition 7.3.11 (Vanishing Proposition). Let 𝑉 𝐷,𝑑 be the vector space of polynomials over 𝔽𝑞 in 𝑑 variables with degree at most 𝐷. If 𝑃 ∈ 𝑉 𝐷,𝑑 vanishes on a set 𝐸 where |𝐸| ≥ 𝐷 + 1, then 𝑃 ≡ 0 is the zero polynomial. Note 7.3.12. Analogues of these two preceding Propositions hold over any field, though we have only stated them for finite fields. Suppose that 𝐸 ⊆ 𝔽𝑑𝑞 is a Kakeya set and that |𝐸| ≤ (10𝑑)−𝑑 𝑞𝑑 . By Proposition 7.3.10 there must exist a nonzero polynomial 𝑃 in 𝑑 variables that vanishes on 𝐸 and has degree 𝐷 ≔ deg(𝑃) ≤ 𝑑|𝐸|1/𝑑 < 𝑞. We write 𝑃 = 𝑃𝐷 + 𝑅, where 𝑃𝐷 is nonzero and homogeneous of degree 𝐷 and where deg(𝑅) < 𝐷. Let 𝑎 ∈ 𝔽𝑑𝑞 ⧵ {(0, . . . , 0)}, and choose 𝑏 ∈ 𝔽𝑑𝑞 ⧵ {(0, . . . , 0)} so that {𝑎 + 𝑡𝑏 ∶ 𝑡 ∈ 𝔽𝑞 } ⊆ 𝐸. Then the polynomial 𝑄(𝑡) = 𝑃(𝑎𝑡 + 𝑏) = 0 for all 𝑡 ∈ 𝔽𝑞 . Note that deg(𝑄) ≤ 𝐷 < 𝑞. By Proposition 7.3.11, 𝑄 must be the zero polynomial, so that every coefficient of 𝑄 must be zero. But the 𝑡𝐷 coefficient in 𝑄 is nothing else but 𝑃𝐷 (𝑎), and hence 𝑃𝐷 (𝑎) = 0 for all 𝑎 ∈ 𝔽𝑑𝑞 ⧵ {(0, . . . , 0)}. Of course 𝑃𝐷 (0, . . . , 0) = 0, since 𝑃𝐷 is homogeneous. Thus, 𝑃𝐷 also vanishes at every point in 𝔽𝑑𝑞 . Since 𝐷 < 𝑞, 𝑃𝐷 is the zero polynomial, which is a contradiction. Thus, we must have |𝐸| ≥ (10𝑑)−𝑑 𝑞𝑑 for Kakeya sets 𝐸 ⊆ 𝔽𝑑𝑞 . 9 At a conference I once attended, Nets Katz was giving a detailed talk of his work on the “Joints” problem (this was the joint work with Larry Guth that eventually lead to the solution of the Erdős distance problem), describing most of the steps of the proof as “high school mathematics”. Halfway through the lecture, another mathematician stood up and shouted, “Where the hell did you go to high school?” 10 The following proof appears in Larry Guth’s excellent book Polynomial Methods in Combinatorics ([68]).

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And the story in finite fields is done! Despite being a short proof, the proof remains mysterious to me (and I hope to you too). Why did the proof work? Does any modification of the proof apply to the Euclidean Kakeya Conjecture? What is it about the Kakeya problem that lent itself to such an easy solution? What other combinatorial problems can be solved with the polynomial method? What happens in ℤ𝑑𝑛 ? Note that lines in ℤ𝑑𝑛 are defined similarly to the finite field case: for 𝑎, 𝑏 ∈ ℤ𝑑𝑛 (and 𝑏 ≠ (0, . . . , 0)) we define ℓ = {𝑎 + 𝑏𝑡 ∶ 𝑡 ∈ ℤ𝑛 }. Note that lines need not have 𝑛 points anymore (such as when gcd(𝑏, 𝑛) > 1), so this problem is even more interesting (and difficult)! There has been some progress on the analogue for the Kakeya problem for the modules over the ring ℤ𝑛 , at least in the case 𝑑 = 2 when 𝑛 is the power of a prime. Theorem 7.3.13 (Dummit and Hablicsek, [39]). Let 𝑝 be a prime and suppose that 𝐸 ⊆ ℤ𝑝2𝑘 is a Kakeya set. Then |𝐸| ≥

𝑝2𝑘 . 2𝑘

There are various other results in the area (see [45], for example), but this area is ripe for further exploration.

7.4 Waring’s theorem The classical Waring problem, dating back to the late 18th century, asks one to find the minimum number 𝑚 so that for a fixed 𝑘 ≥ 2, every nonnegative integer 𝑛 can be written as a sum of 𝑚 𝑘th powers. We use 𝑔(𝑘) to denote the minimum such value of 𝑚. Lagrange showed that 𝑔(2) ≤ 4, which together with the trivial observation that 7 cannot be written as a sum of 3 squares, shows that 𝑔(2) = 4. That 𝑔(𝑘) exists for all 𝑘 ≥ 2 was shown by Hilbert ([80]), and an (almost) exact evaluation of 𝑔(𝑘) is known (see [153] for details): It is known that 𝑘

𝑔(𝑘) = 2𝑘 + ⌊(3/2) ⌋ − 2 so long as 𝑘

2𝑘 {(3/2)𝑘 } + ⌊(3/2) ⌋ ≤ 2𝑘 ,

7.4. Waring’s theorem

153

and 𝑘

𝑘

𝑔(𝑘) = 2𝑘 + ⌊(3/2) ⌋ + ⌊(4/3) ⌋ − 𝐴 otherwise, where 𝐴 is either 2 or 3 depending on whether or not 𝑘 satisfies a certain inequality. Here ⌊⋅⌋ is the usual floor function and {𝑥} = 𝑥 − ⌊𝑥⌋ is the fractional part of 𝑥. The modern formulation of Waring’s problem is to fix an integer 𝑘 ≥ 2 and determine the minimal integer 𝑚 such that every sufficiently large integer can be written as the sum of 𝑚 𝑘th powers. As is customary, we use 𝐺(𝑘) to denote the minimal such integer. Only two values of the function are known precisely: 𝐺(2) = 4 is a corollary of Lagrange’s four squares theorem (and the trivial fact that, if 𝑛 ≡ 7 (mod 8), then 𝑛 is not the sum of three squares), and 𝐺(4) = 16 is work of Davenport ([31]). We know only that 𝐺(3) ∈ {4, 5, 6, 7}, for example ([30]). Finding upper bounds for 𝐺(𝑘) for certain values of 𝑘 and asymptotic bounds are areas of active research. The best known general asymptotic bound for 𝐺(𝑘) is work of Wooley ([159]): 𝐺(𝑘) ≤ 𝑘(log 𝑘 + log log 𝑘 + 2 + 𝑂(log log 𝑘/ log 𝑘)). Let 𝐺 1 (𝑘) be the smallest integer 𝑚 such that almost all11 integers 𝑛 are the sum of 𝑚 𝑘th powers. Even with the weakening of the definition the precise values of 𝐺 1 (𝑘) are only known for six values of 𝑘. Of course Lagrange’s four squares theorem and the observation that if 𝑥 ≡ 7 (mod 8), then 𝑥 is not the sum of three squares implies that 𝐺 1 (2) = 4. The other known values of the function 𝐺 1 are 𝐺 1 (3) = 4, 𝐺 1 (4) = 15, 𝐺 1 (8) = 32, 𝐺 1 (16) = 64, and 𝐺 1 (32) = 128 ([153]). If 𝑆 is a subset of the ring and 𝑘 is a positive integer, we let Γ(𝑘, 𝑆) denote the smallest integer 𝑚 such that for all 𝑟 ∈ 𝑆, there exists a solution {𝑟1 , . . . , 𝑟𝑚 } to the equation 𝑘 𝑟 = 𝑟1𝑘 + ⋯ + 𝑟𝑚 .

11 More

precisely if 𝐸 ⊆ ℕ is the exceptional set, then lim

𝑛→∞

|𝐸 ∩ {1, . . . , 𝑛}| = 0. 𝑛

(7.2)

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In the case that no such minimum exists, then we put Γ(𝑘, 𝑆) = +∞. Thus, the classical version of Waring’s problem is determining 𝑔(𝑘) = Γ(𝑘, ℤ+ ). A natural case to consider beyond the above versions of Waring’s problem is when 𝑆 = ℤ𝑛 , the ring of integers mod 𝑛. For example we know ([139]) that Γ(2, ℤ𝑛 ) ≤ 2 if and only if 𝑛 has the property that, if a prime 𝑝 satisfies 𝑝2 ∣ 𝑛, then 𝑝 ≡ 1 (mod 4). Furthermore, we have Γ(2, ℤ𝑛 ) ≤ 3 if and only if 8 ∤ 𝑛, and of course Γ(2, ℤ𝑛 ) ≤ 4 for all 𝑛 by Lagrange’s four squares theorem. We further set Γ(𝑘) = max Γ(𝑘, ℤ𝑛 ). ℤ𝑛

Note that this quantity Γ(𝑘) exists for each 𝑘 (Exercise 7.9). Of course we have Γ(2) = 4, but cubes are more interesting as it turns out that the following holds: Proposition 7.4.1. Γ(3, ℤ𝑛 ) ≤ 2 if and only if 7 ∤ 𝑛 and 9 ∤ 𝑛, and Γ(3, ℤ𝑛 ) ≤ 3 if and only if 9 ∤ 𝑛. Finally Γ(3) = 4. The first to calculate Γ(𝑘) for small values of 𝑘 were Hardy and Littlewood ([72]), who calculated Γ(𝑘) for 2 ≤ 𝑘 ≤ 100, with a followup and some corrections from Sekigawa-Koyama ([137]) who calculated Γ(𝑘) for 2 ≤ 𝑘 ≤ 200. For example it turns out that Γ(4) = 15, Γ(5) = 5, and Γ(6) = 9. Much remains unknown about Γ(𝑘) despite it being evaluated for small values of 𝑘. See [46] for an evaluation of (a slight variation of) Γ(𝑘) and some related problems. The main tools used in this context are the Chinese Remainder Theorem (Exercise 5.6) and an elementary version of Hensel’s Lemma: Lemma 7.4.2 (Hensel’s Lemma). Let 𝑓(𝑥) be a polynomial with integer coefficients, let 𝑝 be prime, and let 𝑛 be a positive integer. If 𝑎 ∈ ℤ satisfies 𝑓(𝑎) ≡ 0 (mod 𝑝𝑛 ) and if 𝑓′ (𝑎) ≢ 0 (mod 𝑝), then there exists 𝑏 ∈ ℤ such that 𝑓(𝑏) ≡ 0 (mod 𝑝𝑛+1 ) and 𝑎 ≡ 𝑏 (mod 𝑝). More robust and technical (and more useful) versions of Hensel’s Lemma exist (see [126], for example).

7.5. Roth’s theorem and the cap-set problem

155

7.4.1 Waring’s problem for finite fields The finite field version of the problem is much the same as that of the ring of integers modulo 𝑛, except that Γ(𝑘, 𝔽𝑞 ) need not always exist. For example, the cubes in 𝔽4 are {0, 1}, so that every element outside of the subfield {0, 1} cannot be written as a sum of any number of cubes. Thus Γ(3, 𝔽4 ) = +∞. We start by noting that Γ(𝑘, 𝔽𝑝𝑛 ) exists if and only if for all 𝑑 ∣ 𝑛 where 𝑝𝑛 −1

𝑑 ≠ 𝑛, then 𝑝𝑑 −1 ∤ 𝑘 ([9]), and we have already seen that Γ(𝑘, 𝔽𝑝 ) exists for all primes 𝑝. We summarize some of the basic known results in the following theorem. Theorem 7.4.3. Let 𝑞 = 𝑝𝑛 , and assume that 𝑘 and 𝑞 are such that Γ(𝑘, 𝔽𝑞 ) exists. Put 𝑑 = gcd(𝑘, 𝑝 − 1). Then: Γ(𝑑, 𝔽𝑝 ) = Γ(𝑘, 𝔽𝑝 )

(7.3)

Γ(𝑘, 𝔽𝑞 ) ≤ 𝑘

(7.4)

Γ(𝑘, 𝔽𝑞 ) = 1 for all 𝔽𝑞 ⟺ 𝑘 = 1

(7.5)

Γ(2, 𝔽𝑞 ) = 2

(7.6)

Γ(𝑘, 𝔽𝑞 ) ≤ 𝑠 Γ(𝑘, 𝔽𝑞 ) ≪ 𝑘

if 𝑞 > 𝑘 1/2

2𝑠 𝑠−1

.

(7.7) (7.8)

The result Γ(2, 𝔽𝑞 ) = 2 follows from the pigeonhole principle. The bound Γ(𝑘, 𝔽𝑞 ) ≪ 𝑘1/2 is known as the second Heilbronn conjecture, and it was established by Cipra, Cochrane, and Pinner ([21]) in 2006. Analogous to what we did for the ring of integers mod 𝑛, set Γ∗ (𝑘) = max Γ(𝑘, 𝔽𝑞 ) where the maximum is taken over all prime powers 𝑞 such that Γ(𝑘, 𝔽𝑞 ) exists. For example it is known ([33]) that Γ∗ (4) = 5, Γ∗ (5) = 5, and Γ∗ (6) = 6. What happens to Γ∗ (𝑘) as 𝑘 → ∞?

7.5 Roth’s theorem and the cap-set problem Often cited as one of Martin Gardner’s main influences, Henry Dudeney was a mathematician and author who specialized in creating mathematical games. My favorite of his games is his aptly titled “no-three-in-line”

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Chapter 7. Combinatorics in finite fields

game: Find the maximum number of points, 𝑀(𝑛), on an 𝑛 × 𝑛 grid, so that no three points are collinear.

Figure 7.3. A set of 20 points on a 10 × 10 grid so that no line drawn contains any 3 points.

Note that Figure 7.3 shows that 𝑀(10) ≥ 20. The pigeonhole principle implies that 𝑀(𝑛) ≤ 2𝑛 in general (see Exercise 7.15). The bound 𝑀(𝑛) = 2𝑛 has been achieved for some small values of 𝑛, but it has been 𝜋𝑛 conjectured ([119]) that 𝑀(𝑛) ≪ + 𝑜(𝑛) in general. This problem is √3

surprisingly difficult even for some small values of 𝑛, and the reader is highly encouraged to find a solution for a 9 × 9 grid. One of the earliest and best-known problems in Ramsey theory is the following. Consider any 9 numbers, say {1, . . . , 9}. Now assign to each element in {1, . . . , 9} one of two colors, say red and blue. A natural question which arises is whether such a coloring must necessarily contain a 3-term arithmetic progression. For example: 1𝐵 2𝑅 3𝑅 4𝐵 5𝐵 6𝑅 7𝑅 8 𝐵 9𝑅 ⟶ 1𝐵 2𝑅 3𝑅 4𝐵 5𝐵 6𝑅 7𝑅 8 𝐵 9𝑅

7.5. Roth’s theorem and the cap-set problem

157

and 1𝐵 2𝐵 3𝑅 4𝑅 5𝐵 6𝐵 7𝑅 8 𝐵 9𝑅 ⟶ 1𝐵 2𝐵 3𝑅 4𝑅 5𝐵 6𝐵 7𝑅 8 𝐵 9𝑅 . Is there a way to color these 9 integers using only blue or red so that no three-term arithmetic progression has the same color? It turns out there is no such coloring, though in order to prove this, you must show that every coloring of {1, . . . , , 9} (of which there are 512 possible colorings) cannot be so colored. Indeed there is nothing special about having 9 integers, 2 colors, or even that the arithmetic progression has 3-terms. In 1927, Dutch mathematician Bartel Leendert van der Waerden ([152]) proved the following general result. Theorem 7.5.1 (van der Waerden’s Theorem). Fix integers 𝑘 ≥ 3 and 𝑟 ≥ 2. Then there exists a constant 𝑊(𝑟, 𝑘) such that for 𝑛 ≥ 𝑊(𝑟, 𝑘) if the set {1, . . . , 𝑛} is 𝑟-colored in the sense that each element in {1, . . . , 𝑛} corresponds to a color {𝐶1 , . . . , 𝐶𝑟 }, then there must exist at least one subset of {1, . . . , 𝑛} containing an arithmetic progression of length 𝑘, where all elements in the arithmetic progression have the same color. In our example above, we have 𝑊(2, 3) = 9. The constant 𝑊(𝑟, 𝑘) is called van der Waerden’s constant in his honor. Of course lower bounds on 𝑊(𝑟, 𝑘) can be exhibited by examples, but upper bounds are much more difficult as you have to show that any sufficiently large set of integers cannot be colored using fewer colors. For general 𝑟 and 𝑘, Gowers ([63]) gave the terrifying-looking bound 𝑟2

𝑊(𝑟, 𝑘) ≤ 22

2𝑘+9

.

It is remarkable that any finite bound can be achieved for 𝑊(𝑟, 𝑘), though 101233

Gowers’ bound only shows that 𝑊(2, 3) ≤ 1010 . Yikes! Ron Graham has conjectured ([67]) that we should have the more modest bound 2

𝑊(2, 𝑘) ≤ 2𝑘 . A problem closely related to van der Waerden’s theorem is Roth’s theorem. Consider any subset of integers 𝐴 ⊆ {1, 2, 3, . . . } with positive upper density in the sense that lim sup 𝑛→∞

|𝐴 ∩ [1, 𝑛]| > 0. 𝑛

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Chapter 7. Combinatorics in finite fields

Does 𝐴 necessarily contain 3-term arithmetic progressions? What about 4-term arithmetic progressions, or even 𝑘-term arithmetic progressions? Roth showed this must be the case for 𝑘 = 3 ([129]), and Szemerédi gave an affirmative answer for every 𝑘 ≥ 4 in his groundbreaking work [147]. For convenience we state these results together in the following theorem. Theorem 7.5.2 (Roth-Szemerédi). Let 𝐴 ⊆ ℤ have positive upper density. Then 𝐴 necessarily contains a 𝑘-term arithmetic progression for any 𝑘 ≥ 3. Szemerédi’s Theorem is a very deep and difficult result. Famously, Erdős called Szemerédi’s proof a “masterpiece of combinatorial reasoning.” There are many different proofs of Szemerédi’s Theorem (at least 10 different proofs)12 . Not only is Szemerédi’s Theorem important, but even some proofs of Szemerédi’s Theorem are important in their own right (in particular, Gowers’ proof ([63])). Interestingly, Szemerédi’s Theorem has been extended to the primes by Green and Tao. Theorem 7.5.3 (Green-Tao, [65]). Let 𝒫 = {2, 3, 5, . . . } be the primes, and as usual let 𝜋(𝑛) = |𝒫∩{1, . . . , 𝑛}| be the prime counting function. If 𝐴 ⊆ 𝒫 has positive upper density in the primes in the sense that lim sup 𝑛→∞

|𝐴 ∩ {1, . . . , 𝑛}| >0 𝜋(𝑛)

then 𝐴 contains some 𝑘-term arithmetic progression for each 𝑘 ≥ 3. Notice that the primes 𝒫 have zero density in the positive integers, and yet the primes still contain arbitrarily long arithmetic progressions. Improved bounds for Roth’s Theorem have long been sought, with the best known bounds belonging to Thomas Bloom ([11]).

12 The shortest proof of Szemerédi’s Theorem is six pages and belongs to Terry Tao and Ben Green ([65]), though as Terry notes on his blog “Technically, one might call this the shortest proof of Szemerédi’s theorem in the literature (and would be something like the sixteenth such genuinely distinct proof, by our count), but that would be cheating quite a bit, primarily due to the fact that it assumes the inverse conjecture for the Gowers norm, our current proof of which is checking in at about 100 pages. . . ” ([150]).

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159

Theorem 7.5.4. Let 𝐴 ⊆ {1, . . . , 𝑁} be a set not containing any 3-term arithmetic progression. Then, |𝐴| ≪

𝑁 (log log 𝑁)4 . log 𝑁

It has been said ([64]) that it is “by no means unreasonable to think that some of the methods introduced in [Bloom’s] paper could be useful in proving a bound |𝐴| ≪ 𝑁(log 𝑁)−1−𝑐 which would be strong enough to conclude the first nontrivial case 𝑘 = 3 1 of Erdős’ celebrated conjecture that, if 𝐴 ⊆ ℕ is a set with ∑𝑎∈𝐴 𝑎 = ∞, then 𝐴 contains infinitely many arithmetic progressions of length 𝑘”.13 We should note that the Green-Tao Theorem is the first theorem exhibiting a set of zero density (so that Szeméredi’s Theorem does not apply) whose sum of reciprocals diverges and which contains arbitrarily long arithmetic progressions. We return to Dudeney’s No-Three-In-Line Problem from the start of the section. How would the results change if instead of considering a 10 × 10 grid, we worked over ℤ210 ? This would mean identifying the left and right edges and identifying the top and bottom edges, so that we are now playing the game not on a plane, but on a torus. This version of Dudeney’s problem has actually been well studied and is called the cap-set problem. Here a “cap” refers to a 3-term arithmetic progression meaning a set of the forn {𝑥, 𝑦, 𝑥 + 𝑦}. Problem 7.5.5 (Cap-Set Problem). What is the largest subset 𝐴 ⊆ 𝔽𝑑3 so that 𝐴 does not contain any 3-term arithmetic progressions? Just like the Erdős-distance problem, it is too much to ask for a precise bound on such sets 𝐴 for varying 𝑑, but we would like to better understand the asymptotics. There is nothing special about 𝔽3 , and indeed we can consider this question over any finite field 𝔽𝑞 or even any ring ℤ𝑛 . 13 Excitingly, Thomas Bloom and Olof Sisack have recently submitted a paper claiming to give this bound, establishing this remarkable conjecture of Erdős for three-term arithmetic progressions!

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Chapter 7. Combinatorics in finite fields

If you want to get an intuitive handle on such a problem, I suggest playing a card game called Set®. In this game the deck consists of 81 cards, each card having one of three colors, one of three shapes, one of three numbers, and one of three shadings. One must form a “set” which consists of three cards all with three of the four same identical properties, or each characteristic being different. Thus, “1 red filled diamond, 2 red filled diamonds, and 3 red filled diamonds” form a set as they all share the same shape, color, and shading. Likewise, “1 red filled diamond, 2 green shaded circles, and 3 purple blank squiggles” also form a set as each card has properties different from every other. This game is really just the cap-set problem for 𝔽43 . Again a natural question arises. How many cards can you have such that you cannot form a set? Pelegrino ([120]) showed that it is possible to have 20 cards with no sets, but any collection of 21 cards must contain a set, thus settling the question. In general what is the largest subset of 𝔽𝑝𝑑 containing no three-term arithmetic progressions? More precisely define 𝑟3 (𝑑) to be the largest cardinality of a set 𝐴 ⊆ ℤ𝑑𝑛 so that 𝐴 does not contain any 3-term arithmetic progressions. When 𝑑 = 1, this is very similar to Roth’s Theorem discussed above (Theorem 7.5.2). For example, consider the cap-set problem in 𝐴 ⊆ ℤ10 . Then 𝐴 = {1, 2, 4, 9} does not contain any APs of length 3, which shows that 𝑟3 (𝐴) ≥ 4. Again upper bounds are more difficult to achieve. Notice that the set 𝐴 = {1, 2, 4, 8} may at first glance look like a cap set in ℤ10 , but it is not, as {4, 8, 2} is a 3-term AP in ℤ10 . Can you find any cap sets in ℤ10 consisting of 5 or more elements? The cap-set problem was recently solved by Ellenberg and Gijswijt ([44]) wherein they proved the following. Theorem 7.5.6. Suppose that 𝐴 ⊆ ℤ𝑛3 is a set not containing any 3-term arithmetic progressions. Then |𝐴| = 𝑜((2.756)𝑛 ). This Theorem has an interesting history (see [4], for example). In a series of papers, it was shown that 𝑟3 (𝑑) = 𝑜(3𝑛 ), and then 𝑟3 (𝑑) = 𝑂(3𝑛 /𝑛). In a much praised paper, Bateman and Katz ([4]) showed that

7.6. The spectral gap theorem

161

𝑟3 (𝑑) = 𝑂(3𝑛 /𝑛1+𝜀 ) for some 𝜀 > 0. Thus, Ellenberg’s and Gijswijt’s result that 𝑟3 (𝑑) = 𝑜(2.756 . . . )𝑛 was quite a breakthrough! We should note that Ellenberg first posted a solo-authored paper of this result, though Dion Gijswijt happened to prove the same result independently around the same time. Thus, their proofs, which both utilized the Polynomial Method along with similar ideas from a paper of Croot, Lev, and Pach ([23]) on cap-sets in ℤ𝑛4 , were combined and submitted as a joint paper. Truly, math is a social enterprise!

7.6 The spectral gap theorem Algebraic graph theory is a subject which aims to bring the methods of algebra (specifically group theory and linear algebra) to the study of graphs. The Spectral Gap Theorem is an oft underused and undervalued tool from algebraic graph theory. It has been used to study a wide variety of problems in combinatorics, including the Erdős-distance problem, Waring’s problem, and the sum-product problem. We explore some of these results below, but first we describe the Spectral Gap Theorem and the related notion of Cayley Graphs. The various terms that appear in the statement of the theorem will be explained in the course of giving its proof. Theorem 7.6.1 (Spectral Gap Theorem for Symmetric Graphs). Let 𝐺 be a 𝑑-regular simple graph with 𝑛 eigenvalues 𝜆1 ≥ ⋯ ≥ 𝜆𝑛 . Let 𝑛 𝑛∗ = ( max |𝜆𝑖 |) , 𝑑 2≤𝑖≤𝑛 and let 𝑋, 𝑌 ⊆ 𝑉(𝐺) be subsets of vertices. If √|𝑋||𝑌 | > 𝑛∗ then there exists an 𝑋 − 𝑌 edge in 𝐺 (that is, there exists an edge incident to a vertex in 𝑋 and a vertex in 𝑌 ). In particular if |𝑋| > 𝑛∗ , then there exist two distinct vertices 𝑥, 𝑦 ∈ 𝑋 such that there is an edge from 𝑥 to 𝑦. Proof. 14 Let 𝔸 = (𝑎𝑖𝑗 ) be the 0 − 1 matrix, called the adjacency matrix, where 𝑎𝑖𝑗 = 1 if there exists an edge from 𝑥𝑖 to 𝑥𝑗 , and 𝑎𝑖𝑗 = 0 otherwise. Since 𝐺 is undirected, the matrix 𝔸 is symmetric, and hence 𝔸 has real eigenvalues. Let 𝑣 1 , . . . , 𝑣 𝑛 be orthonormal eigenvectors of 𝔸 14 This

proof was first shown to me by Jonathan Pakianathan.

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with eigenvalues 𝜆1 ≥ 𝜆2 ≥ ⋯ ≥ 𝜆𝑛 . It turns out (see Exercise 7.16) that the degree 𝑑 = 𝜆1 is the largest eigenvalue and corresponds to the eigenvector (1, . . . , 1). Any real 𝑛-dimensional vector 𝑣 can be written 𝑛 𝑣 = ∑𝑗=1 ⟨𝑣, 𝑣𝑗 ⟩𝑣𝑗 where ⟨⋅, ⋅⟩ is the dot product. Dotting this equation 𝑛

with itself yields the Plancherel identity ⟨𝑣, 𝑣⟩ = ∑𝑗=1 ⟨𝑣, 𝑣𝑗 ⟩2 . Let 1𝑋 be the 0 − 1 vector whose 𝑖th entry is 1 when 𝑣 𝑖 ∈ 𝑋 and zero otherwise. Define 1𝑌 similarly. The number of directed 𝑋 − 𝑌 edges is given by 𝑛 𝑇 1𝑋 𝔸1𝑌 = ⟨1𝑋 , 𝔸1𝑌 ⟩ = ∑ 𝜆𝑗 ⟨1𝑋 , 𝑣𝑗 ⟩⟨1𝑌 , 𝑣𝑗 ⟩. 𝑗=1

Note that ⟨1𝑋 , 𝑣 1 ⟩ = |𝑋|/√𝑛. Hence, peeling off the 𝑗 = 1 term, we see that: 𝑛 |𝑋||𝑌 |𝑑 𝑇 1𝑋 𝔸1𝑌 = + ∑ 𝜆𝑗 ⟨1𝑋 , 𝑣𝑗 ⟩⟨1𝑌 , 𝑣𝑗 ⟩. 𝑛 𝑗=2 If we let 𝑛

𝑅 = ∑ 𝜆𝑗 ⟨1𝑋 , 𝑣𝑗 ⟩⟨1𝑌 , 𝑣𝑗 ⟩ 𝑗=2

then by Cauchy-Schwarz we have 1/2

| | |𝑅| ≤ max |𝜆𝑗 | ||∑⟨1𝑋 , 𝑣𝑗 ⟩2 || 2≤𝑗≤𝑛 |𝑗 |

1/2

| | |∑⟨1 , 𝑣 ⟩2 | 𝑌 𝑗 | | |𝑗 |

.

Thus by Plancherel, we see that |𝑅| ≤ max |𝜆𝑗 |√|𝑋||𝑌 |. 2≤𝑗≤𝑛

𝑑|𝑋||𝑌 |

Since the number of 𝑋 −𝑌 edges is 𝑛 +𝑅, then there will be an 𝑋 −𝑌 edge so long as 𝑑|𝑋||𝑌 | > max |𝜆𝑗 |√|𝑋||𝑌 |. 𝑛 2≤𝑗≤𝑛 The theorem follows immediately. The Spectral Gap Theorem has an analogue when 𝐺 is directed, though in general the eigenvalues of 𝐺 are complex. We refer the interested reader to [2], for example.

7.6. The spectral gap theorem

163

We next introduce the notion of Cayley graphs and explore a Spectral Gap Theorem for Cayley digraphs (directed graphs). Let 𝐺 denote an additive abelian group, and suppose that 𝑆 ⊆ 𝐺 is a subset. We then consider the Cayley graph 𝒢(𝑉, 𝐸) = 𝐶𝑎𝑦(𝐺, 𝑆) where 𝑉 = 𝐺 and where two vertices 𝑥, 𝑦 ∈ 𝐺 are connected if and only if 𝑦 − 𝑥 ∈ 𝑆. Note that if 𝑆 is a symmetric set (that is, if 𝑥 ∈ 𝑆, then −𝑥 ∈ 𝑆), the Cayley graph is undirected (possibly with loops if we allow 0 ∈ 𝑆). In general the Cayley graph is a digraph. First we describe the Spectral Gap Theorem in this context. Theorem 7.6.2. Let 𝐶𝑎𝑦(𝐺, 𝑆) be the Cayley digraph of the abelian group 𝐺 over the set 𝑆 ⊆ 𝐺. Let {𝜒1 , . . . , 𝜒𝑛 } denote the set of additive characters on 𝐺, where 𝜒1 ≡ 1 denotes the trivial character. Define 𝑛∗ =

| | 𝑛 max || ∑ 𝜒𝑖 (𝑠)|| . |𝑆| 2≤𝑖≤𝑛 |𝑠∈𝑆 |

Let 𝑋 and 𝑌 be sets of vertices from the Cayley graph. Then 𝐶𝑎𝑦(𝐺, 𝑆) has an 𝑋 − 𝑌 edge if the geometric mean √|𝑋||𝑌 | > 𝑛∗ ; if |𝑋| > 𝑛∗ , then there exist two distinct vertices 𝑥, 𝑦 ∈ 𝑋 such that there is an edge from 𝑥 to 𝑦. We note that 𝐶𝑎𝑦(𝐺, 𝑆) is connected if and only if 𝑆 is a generating set (i.e., a subset whose elements and inverses span the group 𝐺). Cayley digraphs have a beautiful correspondence between the characters of the underling group 𝐺 and the eigenvalues of the adjacency matrix of 𝐶𝑎𝑦(𝐺, 𝑆). Proposition 7.6.3. Let 𝐶𝑎𝑦(𝐺, 𝑆) be the Cayley graph of the abelian group 𝐺 with respect to the set 𝑆. Let Λ be the eigenvalues of the adjacency matrix of 𝐶𝑎𝑦(𝐺, 𝑆), and let 𝑋 denote the characters of 𝐺. Then each eigenvalue 𝜆 ∈ Λ corresponds to a sum of the form 𝜆 = ∑ 𝜒(𝑠), 𝑠∈𝑆

for some 𝜒 ∈ 𝑋. A priori, it is not clear that the number of eigenvalues of 𝐶𝑎𝑦(𝐺, 𝑆) and the number of characters on 𝐺 should be equal in number. However this does occur since we have |𝐺|-many eigenvectors, and they are

164

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all distinct. In particular, the adjacency matrix of 𝐶𝑎𝑦(𝐺, 𝑆) is always diagonalizable. See [34] for a nice survey of these and similar results.

7.6.1 Applications of the spectral gap theorem Finally we describe some problems15 which employ the malleable Spectral Gap Theorem in various settings. When studying Waring’s problem for a fixed integer 𝑘 ≥ 2 and for a fixed ring 𝑅, we define 𝐺(𝑉, 𝐸) to be the graph with vertex set 𝑉 = 𝑅, where two vertices 𝑥 and 𝑦 are connected by an edge if and only if 𝑥 − 𝑦 is a 𝑘th power in 𝑅. Note that this graph need not be symmetric. In particular its symmetry will depend on whether or not −1 is a 𝑘th power in 𝑅. This same choice of graph also gives a result analogous to the Furstenberg-Sarkőzy Theorem on squares ([58, 135]). We first recall the classical theorem. Theorem 7.6.4. Let 𝐸 ⊆ ℤ+ have positive natural density in the sense that |𝐸 ∩ {1, . . . 𝑛}| lim > 0. 𝑛 𝑛→∞ Then 𝐸 contains two distinct elements 𝑥, 𝑦 ∈ 𝐸 such that 𝑥 − 𝑦 is a square. A finite field analogue was obtained by Demiroğlu ([33]): Theorem 7.6.5. Let 𝐸 ⊆ 𝔽𝑞 satisfy |𝐸| >

𝑞𝑘 . √𝑞−1

Then 𝐸 contains two

distinct elements 𝑥 and 𝑦 such that 𝑥 − 𝑦 is a 𝑘th power in 𝔽𝑞 . Other similar results include the following. Theorem 7.6.6. Let 𝐸 ⊆ 𝑀2 (𝔽𝑞 ) be such that |𝐸| ≫ 𝑞5/2 . Then for all 𝑡 ∈ 𝔽∗𝑞 , there exist matrices 𝐴 and 𝐵 in 𝐸 such that det(𝐴 − 𝐵) = 𝑡. The graph used here is the natural one: Fix 𝑡 ∈ 𝔽∗𝑞 . The vertex set is 𝑀2 (𝔽𝑞 ), and two matrices (i.e., vertices) 𝐴 and 𝐵 are connected if and only if 𝐵 − 𝐴 has determinant 𝑡. Note that we must have |𝐸| ≫ 𝑞2 as the set 𝑎 𝑏 𝐹 = {[ ] ∶ 𝑎, 𝑏 ∈ 𝔽𝑞 } 0 0 15 Many of these results are work of my former graduate student Yeşim Demiroğlu Karabulat.

7.6. The spectral gap theorem

165

has cardinality 𝑞2 , and yet the difference of any two matrices from 𝐹 has determinant zero. An immediate corollary is the following: 4 Corollary 7.6.7. Let 𝐴, 𝐵, 𝐶, 𝐷 ⊆ 𝔽𝑞 be such that √|𝐴||𝐵||𝐶||𝐷| ≫ 𝑞3/4 . Then (𝐴 − 𝐵)(𝐶 − 𝐷) = 𝔽∗𝑞 .

In particular if |𝐴| ≫ 𝑞3/4 , then (𝐴 − 𝐴)(𝐴 − 𝐴) = 𝔽𝑞 . This theorem on sums and products is just shy of the following remarkable result of Murphy and Petridis ([113]). Just like the finite field distance problem, you are able to get away with less if you only want a positive proportion of 𝔽𝑞 rather than the entire field. Theorem 7.6.8. Let 𝐴 ⊆ 𝔽𝑞 with |𝐴| > 𝑞2/3−𝛿 for any 𝛿 < 1/13, 542. 𝑞 Then |(𝐴 − 𝐴)(𝐴 − 𝐴)| > 2 . Another result achieved using the Spectral Gap Theorem is as follows. Theorem 7.6.9. Every matrix 𝐴 ∈ 𝑀𝑛 (𝔽𝑞 ) is a sum of 2 invertible matrices except in the case that 𝑀𝑛 (𝔽𝑞 ) = 𝔽2 (that is if 𝑛 = 1 and 𝑞 = 2). Moreover if 𝑛 = 2, then every matrix in 𝑀2 (𝔽𝑞 ) is the sum of two matrices of determinant 1. Theorem 7.6.9 along with The Artin-Weddenburn Theorem can be used to reconstruct the following classic result of Henriksen ([78]): Theorem 7.6.10. Let 𝑅 be a finite ring (with identity) of odd order. Then every element of 𝑅 is the sum of two units of 𝑅. More results using the Spectral Gap Theorem appear in the literature, but hopefully this gives an idea of exactly how flexible and hence how useful the Spectral Gap Theorem can be. There is so much more we could explore in combinatorics in finite fields and beyond, but I strongly recommend you start looking up some of the references in this book and dig into some of these problems, especially the unsolved problems (including the finite field distance problem)! There is much work left to do. As Richard Feynman ([54]) once said, “Some new ideas are here needed!”

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7.7 Exercises: Chapter 7 Exercise 7.1. Prove that we have the trivial incidence bound 𝐼(𝑃, 𝐿) ≤ |𝐿||𝑃| for any set of points and lines over any field 𝔽. Exercise 7.2. Show that taking 𝑃 = 𝔽2𝑞 and 𝐿 the set of all possible lines contradicts the Szemerédi-Trotter Theorem thereby showing that Theorem 7.1.1 does not hold in finite fields. Exercise 7.3. Let 𝐴 ⊆ ℤ, and notice that since addition is commutative and since subtraction is not commutative one would expect to have |𝐴 − 𝐴| ≥ |𝐴 + 𝐴| for most sets 𝐴. Nonetheless find a set 𝐴 ⊆ ℤ such that |𝐴 − 𝐴| < |𝐴 + 𝐴|. Such sets are called MSTD sets (More Sums Than Differences). Exercise 7.4. Prove the following analogue of the Kakeya conjecture for circles: If 𝐸 ⊆ 𝔽𝑑𝑞 contains a sphere of radius 𝑡 for all 𝑡 ∈ 𝔽∗𝑞 , then |𝐸| ≫𝑑 𝑞𝑑 . Exercise 7.5. ([41]) Let 𝐸 + (𝐴) denote the additive energy of 𝐴 ⊆ ℝ: 𝐸 + (𝐴) = {(𝑎, 𝑏, 𝑐, 𝑑) ∈ 𝐴 × 𝐴 × 𝐴 × 𝐴 ∶ 𝑎 + 𝑏 = 𝑐 + 𝑑}. Use Cauchy-Schwarz to prove |𝐴|4 ≤ 𝐸 + (𝐴)|𝐴 + 𝐴|. (Compare to Lemma 6.2.2.) Exercise 7.6. ([3]) Prove that |𝐴/𝐴 + 𝐴| ≫ |𝐴|3/2 for all sets 𝐴 ⊆ ℝ ⧵ {0}. Exercise 7.7. Recall that the Minkowski dimension of a set 𝐸 ⊆ ℝ𝑑 is given by log 𝑁𝜖 , dim𝑀 (𝐸) = lim+ 𝜖→0 log(1/𝜖) where 𝑁𝜖 is the number of 𝜖-balls needed to cover the set 𝐸. Show that the 1 set 𝐸 = {1/𝑛}𝑛∈ℤ+ has Minkowski dimension dim𝑀 (𝐸) = 2 . Exercise 7.8. Adapt the previous example to find a set 𝐸 ⊆ ℝ with Minkowski dimension dim𝑀 (𝐸) = 𝛼 for any 𝛼 ∈ [0, 1]. Exercise 7.9. Prove that Γ(𝑘) exists for all 𝑘 ≥ 2.

7.7. Exercises: Chapter 7

167

Exercise 7.10. Prove that Γ(𝑘, ℤ𝑛 ) = 1 if and only if gcd(𝑘, 𝑛 − 1) = 1. Conclude that (7.5) holds: Γ(𝑘, 𝔽𝑞 ) = 1 for all finite fields 𝔽𝑞 if and only if 𝑘 = 1. Exercise 7.11. Prove that (7.6) holds using the pigeonhole principle. 𝑒

𝑒

Exercise 7.12. Let 𝑛 = 𝑝11 . . . 𝑝ℓℓ be the prime factorization of 𝑛. Then, Γ(𝑘, ℤ𝑛 ) ≤ 𝑚 if and only if Γ(𝑘, ℤ𝑝𝑒𝑖 ) ≤ 𝑚 for 𝑖 = 1, . . . , ℓ. 𝑖

Exercise 7.13. Prove that if 𝑛 and 𝑛′ are positive integers such that 𝑛 ∣ 𝑛′ , and if Γ(𝑘, 𝑛) > 𝑚, then Γ(𝑘, 𝑛′ ) > 𝑚. Exercise 7.14. Let 𝑅∗𝑘 = {𝑥𝑘 ∶ 𝑥 ∈ ℤ𝑝ℓ } ⧵ {0}. Then, 𝑅∗𝑘 ⊆ ℤ𝑝×ℓ if 𝑘 ≥ ℓ. That is, if 𝑘 ≥ ℓ then all the nonzero 𝑘th power residues in ℤ𝑝ℓ are units. Exercise 7.15. Let 𝑀(𝑛) be as in the “No-three-in-line” game (see Section 7.5). Prove that 𝑀(𝑛) ≤ 2𝑛 for all 𝑛. Exercise 7.16. Let 𝐺 be a regular (undirected) graph of degree 𝑑 (meaning every vertex has degree 𝑑). Prove that if the spectrum of the adjacency matrix of 𝐺 is {𝜆1 , . . . , 𝜆𝑘 }, then we have max |𝜆𝑖 | = 𝑑.

1≤𝑖≤𝑘

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Index

Additive character, 35 Additive energy, 166 Algebraic method, 31 Arithmetic progression, 138 Bézout’s little theorem, 19 Binary operation, 8 Cap-set problem, 159 Cauchy-Schwarz inequality, 16 Cayley graph, 163 Character, 35, 81 Characteristic (field), 14 Characteristic function, 37 Chinese remainder theorem, 86, 103, 154 Completing the square, 44, 103 Convolution, 72 Direction (of line), 4 Distance graph, 121 Distance set Euclidean, 22 finite field, 28 integers mod 𝑛, 76 Dot-product problem, 143 Dvir’s theorem, 151

Elekes-Sharir framework, 23, 111 Energy set, 112 Equivalence class, 2 Equivalence relation, 2 Erdős distance problem, 22 Erdős unit-distance problem, 23 Erdős-Falconer distance problem finite field, 31 integer mod 𝑛, 76 Euclid’s algorithm, 19 Extension field, 14 Falconer distance problem, 25 Fano plane, 8 Fermat’s two-squares theorem, 33 Field, 13 Field trace, 36 Finite configuration, 105 Finite field distance problem, 29 Finite field restriction problem, 72 Fourier transform 179

180 finite field, 37 integer mod 𝑛, 77 Freiman’s theorem, 140 Gauss sum, 44, 82, 83 General linear group, 13, 127 Graph theory, 120 adjacent, 121 chromatic number, 121, 126 connected, 121 crossing number, 121 degree, 121 directed graph, 121 finite graph, 121 incident, 121 loop, 121 multigraph, 121 path, 121 regular graph, 121 simple graph, 121 undirected graph, 121 Group, 12 Hölder’s inequality, 16 Hausdorff dimension, 25, 147 Hensel’s lemma, 154 Incidence, 24 Incidence theory, 129 Integers mod 𝑛, 2 Interpolation, 103 Inversion theorem finite field, 38 integers mod 𝑛, 77 𝑘-resultant set, 94 𝑘-simplex, 106

Index 𝑘-star, 125 Kakeya conjecture, 31, 146 Kloosterman sum, 57, 88 Kummer sum, 94 Lagrange’s four-square theorem, 23 Lebesgue measure, 25 Legendre symbol, 53 Line, 4, 5 Minkowski dimension, 147 MSTD set, 166 Multiplication table problem, 138 multiplicative character, 36 Multiplicative subgroup of the integers mod 𝑛, 12 Nondegenerate, 107 Null vector, 58 Order (of a Finite Field), 14 Orthogonal group, 13, 127 Orthogonality finite field, 37 integers mod 𝑛, 77 Parameter counting argument, 151 Pigeonhole principle, 1 Pinned distance set, 98 Plancherel’s theorem finite field, 38 integers mod 𝑛, 77 Polynomial ring, 11 Product set, 137 Projective plane, 4

Index Quotient set, 2, 137 Ramsey theory, 156 Regular variety, 97 Restriction theory, 70 Ring, 10 Roth’s theorem, 157 Salem set, 51 Salié sum, 88 Semerédi’s theorem, 158 Special Euclidean group, 112 Special orthogonal group, 13 Spectral gap theorem, 161 Speherical average, 42 Sphere of radius 𝑡 finite field, 6, 37

181 integer mod 𝑛, 77 Splitting field, 14 Steinhaus theorem, 27 Subfield, 14 Sum set, 137 Sum-product problem, 24, 135, 138 Szemerédi-Trotter theorem, 24, 129, 130 Triangle, 106 Triangle inequality, 16 van der Waerden’s theorem, 157 Vanishing proposition, 151 Waring’s theorem, 152

AMS / MAA THE CARUS MATHEMATICAL MONOGRAPHS

Erdo˝s asked how many distinct distances must there be in a set of n points in the plane. Falconer asked a continuous analogue, essentially asking what is the minimal Hausdorff dimension required of a compact set in order to guarantee that the set of distinct distances has positive Lebesgue measure in R . The finite field distance problem poses the analogous question in a vector space over a finite field. The problem is relatively new but remains tantalizingly out of reach. This book provides an accessible, exciting summary of known results. The tools used range over combinatorics, number theory, analysis, and algebra. The intended audience is graduate students and advanced undergraduates interested in investigating the unknown dimensions of the problem. Results available until now only in the research literature are clearly explained and beautifully motivated. A concluding chapter opens up connections to related topics in combinatorics and number theory: incidence theory, sum-product phenomena, Waring’s problem, and the Kakeya conjecture.

For additional information and updates on this book, visit www.ams.org/bookpages/car-37

CAR/37