Additive combinatorics

Terence Tao, Van Vu

Contents

Prologue page 6 1 Existence proofs 14 1.1 The first moment method 15 1.2 The second moment method 19 1.3 The exponential moment method 23 1.4 Correlation inequalities 33 1.5 The Lovász local lemma 38 1.6 Janson’s inequality 42 1.7 Concentration of polynomials 48 1.8 Thin bases of higher order 53 1.9 Thin Waring bases 58 1.10 Appendix: the distribution of the primes 61 2 Sum set estimates 67 2.1 Sum sets 70 2.2 Doubling constants 73 2.3 Ruzsa distance and additive energy 76 2.4 Covering lemmas 87 2.5 The Balog-Szemerédi-Gowers theorem 97 2.6 Symmetry sets and imbalanced partial sumsets 102 2.7 Non-commutative analogues 111 2.8 Elementary sum-product estimates 117 3 Additive geometry 129 3.1 Additive groups 130 3.2 Progressions 137 3.3 Convex bodies 140 3.4 The Brunn-Minkowski inequality 145 3.5 Intersecting a convex set with a lattice 148 3.6 Progressions and proper progressions 162 4 Fourier analytic methods 168 4.1 Basic theory 169 4.2 L^{p} theory 176 4.3 Linear bias 180 4.4 Bohr sets 186 4.5 Λ(p) constants, Sidon sets, and dissociated sets 193 4.6 The spectrum of an additive set 202 4.7 Progressions in sumsets 211 5 Inverse sumset theorems 221 5.1 Minimal size of sumsets and the Dyson e-transform 221 5.2 Sum sets in vector spaces 234 5.3 Freiman homomorphisms 244 5.4 Torsion and torsion-free inverse theorems 253 5.5 Universal ambient groups 258 5.6 Freiman’s theorem in an arbitrary group 265 6 Graph theoretic methods 273 6.1 Basic Notions 274 6.2 Independent sets, sum-free subsets, and Sidon sets 275 6.3 Ramsey theory 282 6.4 Proof of the Balog-Szemerédi-Gowers theorem 290 6.5 Plünnecke’s theorem 297 7 The Littlewood-Offord problem 307 7.1 The combinatorial approach 308 7.2 The Fourier-analytic approach 312 7.3 The Esséen concentration inequality 322 7.4 Inverse Littlewood-Offord results 324 7.5 Random Bernoulli matrices 329 7.6 The quadratic Littlewood-Offord problem 335 8 Incidence geometry 339 8.1 The crossing number of a graph 339 8.2 The Szemerédi-Trotter theorem 342 8.3 The sum-product problem in R 346 8.4 Cell decompositions and the distinct distances problem 351 8.5 The sum-product problem in other fields 357 9 Algebraic methods 361 9.1 The combinatorial Nullstellensatz 362 9.2 Restricted sum sets 365

Contents 5

9.3 Snevily’s conjecture 375 9.4 Finite fields 379 9.5 Davenport’s problem 384 9.6 Kemnitz’s conjecture 388 9.7 Stepanov’s method 390 9.8 Cyclotomic fields, and the uncertainty principle 397

10 Szemerédi’s theorem for k = 3 405 10.1 General strategy 409 10.2 The small torsion case 414 10.3 The integer case 424 10.4 Quantitative bounds 427 10.5 An ergodic argument 435 10.6 The Szemerédi regularity lemma 445 10.7 Szemerédi’s argument 451

11 Szemerédi’s theorem for k > 3 454 11.1 Gowers uniformity norms 458 11.2 Hard obstructions to uniformity 465 11.3 Proof of Theorem 11.6 474 11.4 Soft obstructions to uniformity 482 11.5 The infinitary ergodic approach 490 11.6 The hypergraph approach 497 11.7 Arithmetic progressions in the primes 507

12 Long arithmetic progressions in sumsets 515 12.1 Proof of Theorem 12.3 519 12.2 Generalizations and variants 523 12.3 Complete and subcomplete sequences 525 12.4 Proof of Theorem 12.15 528 12.5 Further applications 530 Bibliography 534

Prologue

This book arose out of lecture notes developed by us while teaching courses on additive combinatorics at the University of California, Los Angeles and the University of California, San Diego. Arithmetic combinatorics is currently a highly active area of research, not least of all because of its applications to additive number theory. One remarkable feature of the field is the use of tools from many diverse fields of mathematics, including elementary combinatorics, harmonic analysis, convex geometry, incidence geometry, graph theory, probability, algebraic geometry and ergodic theory; this wealth of perspectives makes arithmetic combinatorics a rich, fascinating, and multi-fascinated subject. There are still many major problems left in the field, and it seems likely that many of these will require a combination of several of the tools mentioned above in order to solve them.

The main purpose of this book is to gather all these diverse tools in one location, present them in a self-contained and introductory manner, and illustrate their application to problems in arithmetic combinatorics. Many aspects of this material have already been covered in other papers and texts (and in particular several earlier books [161], [246], [110] have focused on some of the aspects of additive combinatorics), but this book attempts to present as many perspectives and techniques as possible in a unified setting.

Arithmetic combinatorics is largely concerned with the additive (and occasionally multiplicative) structure of sets. To clarify what we mean by “additive structure”, let us introduce the following definitions.

Definition 0.1 An additive group is any abelian group Z with group operation +. Note that we can define a multiplication operation nx ∈ Z whenever n ∈ Z and x ∈ Z in the usual manner, thus 3x = x + x + x, −2x = −x − x, etc. An additive set is a pair (A,Z), where Z is an additive group, and A is a finite non-empty subset of Z. We often abbreviate an additive set (A,Z) simply as A, and refer to Z as the ambient group of the additive set. If A,B are additive sets in Z, we define the sum set

A+B := {a+b : a ∈ A,b ∈ B}

and difference set

A−B := {a−b : a ∈ A,b ∈ B}.

Also, we define the iterated sumset kA for k ∈ Z^{+} by

kA := {a_{1} +...+a_{k} : a_{1},...,a_{k} ∈ A}.

We caution that the sumset kA is usually distinct from the dilation k ·A of A, defined by

k ·A := {ka : a ∈ A}.

For us, typical examples of additive groups Z will be the integers Z, a cyclic group Z_{N}, or a finite field geometry F_{p}^{n}. As the notation suggests, we will eventually be viewing additive sets as “intrinsic” objects, which can be embedded inside any number of different ambient groups; this is somewhat similar to how a manifold can be thought of intrinsically, or alternatively can be embedded into an ambient space. To make these ideas rigorous we will need to develop the theory of Freiman homomorphisms, but we will defer this to Section 5.3.

Additive sets may have a large or small amount of additive structure. A good example of a set with little additive structure would be a randomly chosen subset A of a finite additive group Z with some fixed cardinality. At the other extreme, examples of sets with very strong additive structure would include arithmetic progressions

a+[0,N)·r := {a,a+r,...,a+(N −1)r}

where a,r ∈ Z and N ∈ Z_{+}, d-dimensional generalized arithmetic progressions

a+[0,N)·v := {a+n_{1}v_{1} +...+n_{d}v_{d} : 0 ≤ n_{j} < N_{j} for all 1 ≤ j ≤ d}

where a ∈ Z, v = (v_{1},...,v_{d}) ∈ Z^{d}, and N = (N_{1},...,N_{d}) ∈ Z^{+}, or d-dimensional cubes

a+{0,1}^{d} ·v = {a+ϵ_{1}v_{1} +...+ϵ_{d}v_{d} : ϵ_{1},...,ϵ_{d} ∈ {0,1}}.

A fundamental task in this subject is to give some quantitative measures of additive structure in a set, and then investigate to what extent these measures are equivalent to each other. For example, one could try to quantify each of the following informal statements as being some version of the assertion “A has additive structure”:

• A+A is small.

• A−A is small.

• A−A can be covered by a small number of translates of A.

• kA is small for any fixed k.

• There are many quadruples (a_{1},a_{2},a_{3},a_{4}) ∈ A×A×A×A such that a_{1} +a_{2} = a_{3} +a_{4}.

• There are many quadruples (a_{1},a_{2},a_{3},a_{4}) ∈ A×A×A×A such that a_{1} −a_{2} = a_{3} −a_{4}.

• The convolution 1_{A} ∗1_{A} is highly concentrated.

• The subset sums F S(A) := {∑_{a∈B} a : B ⊆ A} have high multiplicity.

• The Fourier transform ˆ1_{A} is highly concentrated.

• The Fourier transform ˆ1_{A} is highly concentrated in a cube.

• A has a large intersection with a generalized arithmetic progression, of size comparable to A.

• A is contained in a generalized arithmetic progression, of size comparable to A.

• A (or perhaps A−A, or 2A−2A) contains a large generalized arithmetic progression.

The reader is invited to investigate to what extent these informal statements are true for sets such as progressions and cubes, and false for sets such as random sets. As it turns out, once one makes the above assertions more quantitative, there are a number of deep and important equivalences between them; indeed, to oversimplify tremendously, all of the above criteria for additive structure are “essentially” equivalent. There is also a similar heuristic to quantify what it would mean for two additive sets A,B of comparable size to have a large amount of “shared additive structure” (e.g. A and B are progressions with the same step size v); we invite the reader to devise analogues of the above criteria to capture this concept.

To make the above heuristics precise and rigorous will require some work, and in fact will occupy large parts of Chapters 2, 3, 4, 5, 6. In deriving these basic tools of the field, we shall need to develop and combine techniques from elementary combinatorics, additive geometry, harmonic analysis, and graph theory; many of these methods are of independent interest in their own right, and so we have devoted some space to treating them in detail.

Of course, a “typical” additive set will most likely behave like a random additive set, which one expects to have very little additive structure. Nevertheless, it is a deep and surprising fact that as long as an additive set is dense enough in its ambient group, it will always have some level of additive structure. The most famous example of this principle is Szemerédi’s theorem, which asserts that every subset of the integers of positive upper density will contain arbitrarily long arithmetic progressions; we shall devote all of Chapter 11 to this beautiful and important theorem. A variant of this fact is the very recent Green-Tao theorem, which asserts that every subset of the prime numbers of positive upper relative density also contains arbitrarily long arithmetic progressions; in particular, the primes themselves have this property. If one starts with an even sparser set A than the primes, then it is not yet known whether A will necessarily contain long progressions; however, if one forms sum sets such as A+A, A+A+A, 2A−2A, F S(A) then these sets contain extraordinarily long arithmetic progressions (see in particular Section 4.7 and Chapter 12). This basic principle - that sumsets have much more additive structure than general sets - is closely connected to the equivalences between the various types of additive structure mentioned previously; indeed results of the former type can be used to deduce results of the latter type, and conversely.

We now describe some other topics covered in this text. In Chapter 1 we recall the simple yet powerful probabilistic method, which is very useful in additive combinatorics for constructing sets with certain desirable properties (e.g. thin additive bases of the integers), and provides an important conceptual framework that complements more classical deterministic approaches to such constructions. In Chapter 6 we present some ways in which graph theory interacts with additive combinatorics, for instance in the theory of sum-free sets, or via Ramsey theory. Graph theory is also decisive in establishing two important results in the theory of sum sets, the Balog-Szemerédi-Gowers theorem and the Plünnecke inequalities. Two other important tools from graph theory, namely the crossing number inequality and the Szemerédi regularity lemma, will also be covered in Chapter 8 and Sections 10.6, 11.6 respectively. In Chapter 7 we view sum sets from the theory of random walks, and give some classical and recent results concerning the distribution of these sum sets, and in particular on recent applications to random matrices. Last, but not least, in Chapter 9 we describe some algebraic methods, notably the combinatorial Nullstellensatz and Chevalley-Waring type methods, which have led to several deep arithmetical results (often with very sharp bounds) not obtainable by other means.

Acknowledgements

The authors would like to thank Robin Chapman, Michael Cowling, Andrew Granville, Ben Green, Timothy Gowers, Roy Meschulam, Melvyn Nathanson, Imre Ruzsa, Roman Sasyk, and Benny Sudakov for helpful comments and corrections, and to Australian National University and the University of Edinburgh for their hospitality while portions of this book were being written. Parts of this work were inspired by the lecture notes of Ben Green [138], the expository article of Imre Ruzsa [284], and the book by Melvyn Nathanson [246]. TT is also particularly indebted to Roman Sasyk and Hillel Furstenberg for explaining the ergodic theory proof of Szemerédi’s theorem.

General notation

The following general notational conventions will be used throughout the book.

Sets and functions For any set A, we use

A^{d} := A×...×A = {(a_{1},...,a_{d}) : a_{1},...,a_{d} ∈ A}

to denote the Cartesian product of d copies of A, thus for instance Z^{d} is the d-dimensional integer lattice. We shall occasionally denote A^{d} by A^{⊕d}, in order to distinguish this Cartesian product from the d-fold product set A^{·d} = A·...·A of A, or the d-fold powers A^{∧}d := {a^{d} : a ∈ A} of A.

If A,B are sets, we use A\B := {a ∈ A : a ∉ B} to denote the settheoretic difference of A and B, and B^{A} to denote the space of functions f : A → B from A to B. We also use 2^{A} := {B : B ⊂ A} to denote the power set of A. We use |A| to denote the cardinality of A. (We shall also use |x| to denote the magnitude of a real or complex number x, and |v| = \sqrt{v_{1}^{2} +...+v_{d}^{2}} to denote the magnitude of a vector v = (v_{1},...,v_{d}) in a Euclidean space R^{d}. The meaning of the absolute value signs should be clear from context in all cases.)

If A ⊂ Z, we use 1_{A} : Z → {0,1} to denote the indicator function of A, thus 1_{A}(x) = 1 when x ∈ A and 1_{A}(x) = 0 otherwise. Similarly if P is a property, we let I(P) denote the quantity 1 if P holds and 0 otherwise, thus for instance 1_{A}(x) = I(x ∈ A).

We use \binom{n}{k} = \frac{n!}{k!(n−k)!} to denote the number of k-element subsets of an n-element set. In particular we have the natural convention that \binom{n}{k} = 0 if k > n or k < 0.

Number systems We shall rely frequently on the integers Z, the positive integers Z^{+} := {1,2,...}, the natural numbers N := Z_{≥0} = {0,1,...}, the reals R, the positive reals R^{+} := {x ∈ R : x > 0}, the non-negative reals R_{≥0} := {x ∈ R : x ≥ 0}, and the complex numbers C, as well as the circle group R/Z := {x+Z : x ∈ R}.

For any natural number N ∈ N, we use Z_{N} := Z/NZ to denote the cyclic group of order N, and use n ↦ nmod N to denote the canonical projection from Z to Z_{N}. If q is a prime power, we use F_{q} to denote the finite field of order q (see Section 9.4). In particular if p is a prime then F_{p} is identifiable with Z_{p}.

If x is a real number, we use ⌊x⌋ to denote the greatest integer less than or equal to x.

Landau asymptotic notation Let n be a positive variable (usually taking values on N, Z^{+}, R_{≥0}, or R^{+}, and often assumed to be large) and let f(n) and g(n) be real-valued functions of n.

• g(n) = O(f(n)) means that f is non-negative, and there is a positive constant C such that |g(n)| ≤ Cf(n) for all n.

• g(n) = Ω(f(n)) means that f,g are non-negative, and there is a positive constant c such that g(n) ≥ cf(n) for all sufficiently large n.

• g(n) = Θ(f(n)) means that f,g are non-negative and both g(n) = O(f(n)) and g(n) = Ω(f(n)) hold, namely, there are positive constants c and C such that cf(n) ≥ g(n) ≥ Cf(n) for all n.

• g(n) = o_{n→∞}(f(n)) means that f is non-negative and g(n) = O(a(n)f(n)) for some a(n) which tends to zero as n → ∞; if f is strictly positive, this is equivalent to lim_{n→∞} g(n)/f(n) = 0.

• g(n) = ω_{n→∞}(f(n)) means that f,g are non-negative and f(n) = o_{n→∞}(g(n)).

In most cases the asymptotic variable n will be clear from context, and we shall simply write o_{n→∞}(f(n)) as o(f(n)), and similarly write ω_{n→∞}(f(n)) as ω(f(n)). In some cases the constants c,C and the decaying function a(n) will depend on some other parameters, in which case we indicate this by subscripts. Thus for instance g(n) = O_{k}(f(n)) would mean that g(n) ≤ C_{k}f(n) for all n, where C_{k} depend on the parameter k; similarly, g(n) = o_{n→∞;k}(f(n)) would mean that g(n) = O(a_{k}(n)f(n)) for some a_{k}(n) which tends to zero as n → ∞ for each fixed k.

The notation g(n) = Õ(f(n)) has been used widely in the combinatorics and theoretical computer science community in recent years; g(n) = Õ(f(n)) means that there is a constant c such that g(n) ≤ f(n)log^{c} n for all sufficiently large n. We can define, in a similar manner, Ω̃ and Θ̃, though this notation will only be used occasionally here. Here and throughout the rest of the book, log shall denote the natural logarithm unless specified by subscripts, thus log_{x} y = \frac{log y}{log x}.

Progressions We have already encountered the concept of a generalized arithmetic progression. We now make this concept more precise.

Definition 0.2 (Progressions) For any integers a ≤ b, we let [a,b] denote the discrete closed interval [a,b] := {n ∈ Z : a ≤ n ≤ b}; similarly define the half-open discrete interval [a,b), etc. More generally, if a = (a_{1},...,a_{d}) and b = (b_{1},...,b_{d}) are elements of Z^{d} such that a_{j} ≤ b_{j}, we define the discrete box

[a,b] := {(n_{1},...,n_{d}) ∈ Z^{d} : a_{j} ≤ n_{j} ≤ b_{j} for all 1 ≤ j ≤ d},

and similarly

[a,b) := {(n_{1},...,n_{d}) ∈ Z^{d} : a_{j} ≤ n_{j} < b_{j} for all 1 ≤ j ≤ d},

etc. If Z is an additive group, we define a generalized arithmetic progression (or just progression for short) in Z to be any set^{1} of the form

1 Strictly speaking, this is an abuse of notation; the arithmetic progression should really be the sextuple (P,d,N,a,v,Z), because the set P alone does not always

P = a + [0,N] · v, where a ∈ Z, [0,N] ⊂ Z^{d} is a discrete box, v = (v_{1},...,v_{d}) ∈ Z^{d}, · : Z^{d} ×Z^{d} → Z is the dot product

(n_{1},...,n_{d})·(v_{1},...,v_{d}) := n_{1}v_{1} +...n_{d}v_{d}

and [0,N]·v := {n·v : n ∈ [0,N]}. In other words,

P = {a+n_{1}v_{1} +...+n_{d}v_{d} : 0 ≤ n_{j} ≤ N_{j} for all 1 ≤ j ≤ d}.

We call a the base point of P, v = (v_{1},...,v_{d}) the basis vectors of P, N the dimensions of P, d the dimension or rank of P, and vol(P) := |[0,N]| = ∏_{j=1}^{d}(N_{j} + 1) the volume of P. We say that the progression P is proper if the map n ↦ n·v is injective on [0,N], or equivalently if the cardinality of P is equal to its volume (as opposed to being strictly smaller than the volume, which can occur if the basis vectors are linearly dependent over Z). We say that P is symmetric if −P = P, for instance [−N,N]·v = −N ·v +[0,2N]·v is a symmetric progression.

uniquely determine the base point, step, ambient space or even length (if the the progression is improper) of the progression P. However as it would be cumbersome to use this sextuple continually, we shall usually just use P to denote the progression.

1

Existence proofs

In additive number theory, one frequently faces the problem of showing that a set A contains a subset B with a certain property P. A very powerful tool for such a problem is Erdös’ probabilistic method. In order to show that such a subset B exists, it suffices to prove that a properly defined random subset of A satisfies P with positive probability. The power of the random method has been justified by the fact that in most problems solved using this approach, it seems impossible to come up with a deterministically constructive proof of comparable simplicity.

In this chapter we are going to present several basic probabilistic tools together with some representative applications of the probabilistic method, particularly with regard to additive bases and the primes. We shall require several standard facts about the distribution of primes P = {2,3,5,...}; so as not to disrupt the flow of the chapter we have placed these facts in an appendix (Section 1.10).

Notation. We assume the existence of some sample space (usually this will be finite). If E is an event in this sample space, we use P(E) to denote the probability of E, and I(E) to denote the indicator function (thus I(E) = 1 if E occurs and 0 otherwise). If E,F are events, we use E ∧F to denote the event that E,F both hold, E ∨F to denote the event that at least one of E,F hold, and Ē to denote the event that E does not hold. In this chapter all random variables will be assumed to be real valued (and usually denoted as X or Y) or set-valued (and usually denoted as B). If X is a real-valued random variable, we use

E(X) := ∑_{x}xP(X = x)

to denote the expectation of X, and

Var(X) := E(|X −E(X)|^{2}) = E(|X|^{2})−E(|X|)^{2}

to denote the variance. Thus for instance

E(I(E)) = P(E); Var(I(E)) = P(E)−P(E)^{2}.  (1.1)

If F is an event of non-zero probability, we define the conditional probability of another event E with respect to F by Bayes’ formula

P(E|F) := \frac{P(E ∧F)}{P(F)}

and similarly the conditional expectation of a random variable X by

E(X|F) := \frac{E(XI(F))}{E(I(F))} = ∑_{x}xP(X = x|F).

A random variable is boolean if it takes values in {0,1}, or equivalently if it is an indicator function I(E) for some event E.

1.1 The first moment method

The simplest instance of the probabilistic method is the first moment method, which seeks to control the distribution of a random variable X in terms of its expectation E(X). Firstly, we make the trivial observation (essentially the pigeonhole principle) that X ≤ E(X) with positive probability, and X ≥ E(X) with positive probability. A slight variant of this is

Theorem 1.1 (Markov’s inequality) Let X be a non-negative random variable. Then for any positive real λ > 0

P(X ≥ λ) ≤ \frac{E(X)}{λ}.  (1.2)

Proof Start with the trivial inequality X ≥ λI(X ≥ λ) and take expectations of both sides.

Informally, this inequality asserts that X = O(E(X)) with high probability; for instance, X ≤ 10E(X) with probability at least 0.9. Note that this is only an upper tail estimate; it gives an upper bound for how likely X is to be much larger than E(X), but does not control how likely X is to be much less than E(X). Indeed, if all one knows is the expectation E(X), it is easy to see that X could be as small as zero with probability arbitrarily close to 1, so the first moment method cannot give any non-trivial lower tail estimate. Later on we shall introduce more refined methods, such as the second moment method, that give further upper and lower tail estimates.

To apply the first moment method, we of course need to compute the expectations of random variables. A fundamental tool in doing so is linearity of expectation, which asserts that

E(c_{1}X_{1} +···+c_{n}X_{n}) = c_{1}E(X_{1})+···+c_{n}E(X_{n})  (1.3)

whenever X_{1},...,X_{n} are random variables and c_{1},...,c_{n} be real numbers. The power of this principle comes from there being no restriction on the independence or dependence between the X_{i}’s. A very typical application of (1.3) is in estimating the size |B| of a subset B of a given set A, where B is generated in some random manner. From the obvious identity

|B| = ∑_{a∈A}I(a ∈ B)

and (1.3), (1.1) we see that

E(|B|) = ∑_{a∈A}P(a ∈ B).  (1.4)

Again, we emphasize that the events a ∈ B do not need to be independent in order for (1.4) to apply.

A weaker (but still useful) version of the linearity of expectation principle is the union bound

P(E_{1} ∨...∨E_{n}) ≤ P(E_{1})+...+P(E_{n})  (1.5)

for arbitrary events E_{1},...,E_{n} (compare this with (1.3) with X_{i} := I(E_{i}) and c_{i} := 1). This trivial bound is still useful, especially in the case when the events E_{1},...,E_{n} are rare and not too strongly correlated (see Exercise 3). A related estimate is as follows.

Lemma 1.2 (Borel-Cantelli lemma) Let E_{1},E_{2},... be a sequence of events (possibly infinite or dependent), such that ∑_{n} P(E_{n}) < ∞. Then for any integer M, we have

P(Fewer than M of the events E_{1},E_{2},... hold) ≥ 1− \frac{∑_{n} P(E_{n})}{M}.

In particular, with probability 1 at most finitely many of the events E_{1},E_{2},... hold.

Another useful way of phrasing the Borel-Cantelli lemma is that if F_{1},F_{2},... are events such that ∑_{n}(1−P(F_{n})) < ∞, then with probability n all but finitely many of the events F_{n} hold.

Proof By monotone convergence it suffices to prove the claim when there are only finitely many events. From (1.3) we have E(∑_{n} I(E_{n})) = ∑_{n} P(E_{n}). If one now applies Markov’s inequality with λ = M, the claim follows.

1.1.1 Sum-free sets We now apply the first moment method to the theory of sum-free sets. An additive set A is called sum-free iff it does not contain three elements x,y,z such that x+y = z; equivalently, A is sum-free iff A∩2A = ∅.

Theorem 1.3 Let A be an additive set of non-zero integers. Then A contains a sum-free subset B of size |B| > |A|/3.

Proof Choose a prime number p = 3k +2, where k is sufficiently large so that A ⊂ [−p/2,p/2]\{0}. We can thus view A as a subset of the cyclic group Z_{p} rather than the integers Z, and observe that a subset B of A will be sum-free in Z_{p} if and only if^{1} it is sum-free in Z.

Now choose a random number x ∈ Z_{p}\{0} uniformly, and form the random set

B := A∩(x·[k +1,2k +1]) = {a ∈ A : x^{−1}a ∈ {k +1,...,2k +1}}.

Since [k + 1,2k + 1] is sum-free in Z_{p}, we see that x · [k + 1,2k + 1] is also, and thus B is a sum-free subset of A. We would like to show that |B| > |A|/3 with positive probability; by the first moment method it suffices to show that E(|B|) > |A|/3. From (1.4) we have

E(|B|) = ∑_{a∈A}P(a ∈ B) = ∑_{a∈A}P(x^{−1}a ∈ [k +1,2k +1]).

If a ∈ A, then a is an invertible element of Z_{p}, and thus x^{−1}a is uniformly distributed in Z_{p}\\{0\}. Since |[k + 1,2k + 1]| > \frac{p−1}{3}, we conclude that P(x^{−1}a ∈ [k +1,2k +1]) > \frac{1}{3} for all a ∈ A. Thus we have E(|B|) > \frac{|A|}{3} as desired.

1 This trick can be placed in a more systematic context using the theory of Freiman homomorphisms: see Section 5.3.

Theorem 1.3 was proved by Erdös in 1965 [83]. Several years later, Bourgain (private communication) used harmonic analysis arguments to improve the bound slightly. It is surprising that the following question is open:

Question 1.4 Can one replace n/3 by n/3+10?

Alon and Kleiman [10] considered the case of more general additive sets (not necessarily in Z). They showed that in this case A always contains a sum-free subset of 2|A|/7 elements and the constant 2/7 is best possible.

1.1.1 If X is a non-negative random variable, establish the identity

E(X) = ∫_{0}^{∞} P(X > λ) dλ  (1.6)

and more generally for any 0 < p < ∞

E(X^{p}) = p∫_{0}^{∞} λ^{p−1}P(X > λ) dλ  (1.7)

Thus the probability distribution function P(X > λ) controls all the moments of X.

1.1.2 When does equality hold in Markov’s inequality? 1.1.3 If E_{1},...,E_{n} are arbitrary probabilistic events, establish the

lower bound

P(E_{1} ∨...∨E_{n}) ≥ ∑_{i=1}^{n}P(E_{i})− ∑_{1≤i<j≤n}P(E_{i} ∧E_{j});

this bound should be compared with (1.5), and can be thought of as a variant of the second moment method which we discuss in the next section. (Hint: consider the random variable ∑_{i=1}^{n} I(E_{i}) − ∑_{1≤i<j≤n} I(E_{i})I(E_{j}).) More generally, establish the Bonferroni inequalities

P(E_{1} ∨...∨E_{n}) ≥ ∑_{A⊂[1,n]:1≤|A|≤k}(−1)^{k}P( ⋀_{i∈A}E_{i})

when k is even, and

P(E_{1} ∨...∨E_{n}) ≤ ∑_{A⊂[1,n]:1≤|A|≤k}(−1)^{k}P( ⋀_{i∈A}E_{i})

when k is odd.

1.1.4 Let X be a non-negative random variable. Establish the popularity principle E(XI(X > \frac{1}{2}E(X))) ≥ \frac{1}{2}E(X). In particular, if X is bounded by some constant M, then P(X > \frac{1}{2}E(X)) ≥ \frac{1}{2M} E(X). Thus while there is in general no lower tail estimate on the event X ≤ \frac{1}{2}E(X), we can say that the majority of the expectation of X is generated outside of this tail event, which does lead to a lower tail estimate if X is bounded.

1.1.5 Let A,B be non-empty subsets of a finite additive group Z. Show that there exists an x ∈ Z such that

1− \frac{|A∩(B +x)|}{|Z|} ≤ (1− \frac{|A|}{|Z|})(1− \frac{|B|}{|Z|})

and a y ∈ Z such that

1− \frac{|A∩(B +y)|}{|Z|} ≥ (1− \frac{|A|}{|Z|})(1− \frac{|B|}{|Z|}).

1.1.6 Consider a set A as above. Show that there exists a subset \{v_{1},...,v_{d}\} of Z with d = O(log \frac{|Z|}{|A|}) such that

|A+[0,1]^{d} ·(v_{1},...,v_{d})| ≥ |Z|/2.

1.1.7 Consider a set A as above. Show that there exists a subset \{v_{1},...,v_{d}\} of Z with d := O(log \frac{|Z|}{|A|} + loglog(10 + |Z|)) such that

A+[0,1]^{d} ·(v_{1},...,v_{d}) = Z.

1.2 The second moment method

The first moment method allows one to control the order of magnitude of a random variable X by its expectation E(X). In many cases, this control is insufficient, and one needs to also establish that X usually does not deviate too greatly from its expected value. These types of estimates are known as large deviation inequalities, and are a fundamental set of tools in the subject. They can be significantly more powerful than the first moment method, but often require some assumptions concerning independence or approximate independence.

The simplest such large deviation inequality is Chebyshev’s inequality, which controls the deviation in terms of the variance Var(X):

Theorem 1.5 (Chebyshev’s inequality) Let X be a random variable. Then for any positive λ

P(|X −E(X)| > λVar(X)^{1/2}) ≤ \frac{1}{λ^{2}} .  (1.8)

Proof We may assume Var(X) > 0 as the case Var(X) = 0 is trivial. From Markov’s inequality we have

P(|X −E(X)|^{2} > λ^{2}Var(X)) ≤ \frac{E(|X −E(X)|^{2})}{λ^{2}Var(X)} = \frac{1}{λ^{2}}

and the claim follows.

Thus Chebyschev’s inequality asserts that X = E(X)+O(Var(X)^{1/2}) with high probability, while in the converse direction it is clear that |X − E(X)| ≥ Var(X)^{1/2} with positive probability. The application of these facts is referred to as the second moment method. Note that Chebyshev’s inequality provides both upper tail and lower tail bounds on X, with the tail decaying like 1/λ^{2} rather than 1/λ. Thus the second moment method tends to give better distributional control than the first moment method. The downside is that the second moment method requires computing the variance, which is often trickier than computing the expectation.

Assume that X = X_{1} + ...X_{n}, where X_{i}’s are random variables. In view of (1.3), one might wonder whether

Var(X) = Var(X_{1})+···+Var(X_{n}).  (1.9)

This equality holds in the special case when the X_{i}’s are pairwise independent (and in particular when the X_{i} are jointly independent), but does not hold in general. For arbitrary X_{i}’s, we instead have

Var(X) = ∑_{i=1}^{n}Var(X_{i})+ ∑_{i,j∈[1,n]:i≠j} Cov(X_{i},X_{j}),  (1.10)

where the covariance Cov(X_{i},X_{j}) is defined as

Cov(X_{i},X_{j}) := E((X_{i}−E(X_{i}))(X_{j} −E(X_{j})) = E(X_{i}X_{j})−E(X_{i})E(X_{j}).

Applying (1.9) to the special case when X = |B|, where B is some randomly generated subset of a set A, we see from (1.1) that if the events a ∈ B are pairwise independent for all a ∈ A, then

Var(|B|) = ∑_{a∈A}P(a ∈ B)−P(a ∈ B)^{2}  (1.11)

and in particular we see from (1.4) that

Var(|B|) ≤ E(|B|).  (1.12)

In the case when the events a ∈ B are not pairwise independent, we must replace (1.11) by the more complicated identity

\begin{aligned}Var(|B|) & = ∑_{a∈A}P(a ∈ B)−P(a ∈ B)^{2} \\ & + ∑_{a,a′∈A:a≠a′} Cov(I(a ∈ B),I(a′ ∈ B)). \end{aligned}  (1.13)

1.2.1 The number of prime divisors Now we present a nice application of the second moment method to classical number theory. To this end, let^{1}

ν(n) := ∑_{p≤n}I(p|x)

denote the number of prime divisors of n. This function is among the most studied objects in classical number theory. Hardy and Ramanujan in the 1920s showed that “almost” all n have about loglogn prime divisors. We give a very simple proof of this result, found by Turán in 1934 [351].

Theorem 1.6 Let ω(n) tend to infinity arbitrarily slowly. Then

\{x ∈ [1,n] : |ν(x)−loglogx| > ω(n)\sqrt{loglogn}\} = o(n).  (1.14)

Informally speaking, this result asserts that for a “generic” integer x, we have ν(x) = loglogx+O(\sqrt{loglogx}) with high probability.

Proof Let x be chosen uniformly at random from the interval {1,2,...,n}. Our task is now to show that

P(|ν(x)−lnlogx| > ω(n)\sqrt{loglogn}) = o(1).

Due to a technical reason, instead of ν(x) we shall consider the related set

B := {p prime : p ≤ n^{1/10},p|x}.

1 We shall adopt the convention that whenever a summation is over the index p, then p is understood to be prime.

Since x cannot have 10 different prime divisors larger than n^{1/10}, it follows that |B|−10 ≤ ν(x) ≤ |B|. Thus, to prove (1.14), it suffices to show

P(||B|−loglogn| ≥ ω(n)\sqrt{lnlogn}) = o(1).

In light of Chebyshev’s inequality, this will follow from the following expectation and variance estimates:

E(|B|),Var(|B|) = loglogn+O(1).

It remains to verify the expectation and variance estimate. From linearity of expectation (1.4) we have

E(|B|) = ∑_{p≤n^{1/10}} P(p|x)

while from the variance identity (1.13) we have

Var(|B|) = ∑_{p≤n^{1/10}} P(p|x)−P(p|x)^{2} + ∑_{p,q≤n^{1/10}:p≠q} Cov(I(p|x),I(q|x)).

Observe from the Chinese remainder theorem that I(p|x)I(q|x) = I(pq|x). Since I(d|x) = \frac{1}{d} +O(\frac{1}{n}) for any d ≥ 1, we conclude that

P(p|x) = \frac{1}{p} +O(\frac{1}{n})

and

Cov(I(p|x),I(q|x)) = \frac{1}{pq} +O(\frac{1}{n})−(\frac{1}{p} +O(\frac{1}{n}))(\frac{1}{q} +O(\frac{1}{n})) = O(\frac{1}{n}).

We thus conclude that

E(|B|) = ∑_{p≤n^{1/10}} \frac{1}{p} +O(n^{−9/10})

and

Var(|B|) = ∑_{p≤n^{1/10}} \frac{1}{p} − \frac{1}{p^{2}} +O(n^{−8/10}).

The expectation and variance estimates now follow from Mertens’ theorem (see Proposition 1.51) and the convergence of the sum ∑_{k} \frac{1}{k^{2}} .

1.2.1 When does equality hold in Chebyshev’s inequality? 1.2.2 If X and Y are two random variables, verify the Cauchy-Schwarz

inequality |Cov(X,Y)| ≤ Var(X)^{1/2}Var(Y)^{1/2} and the triangle inequality Var(X + Y)^{1/2} ≤ Var(X)^{1/2} + Var(Y)^{1/2}. When does equality occur?

1.2.3 Prove (1.10). 1.2.4 If ϕ : R → R is a convex function and X is a random variable,

verify Jensen’s inequality E(ϕ(X)) ≤ ϕ(E(X)). If ϕ is strictly convex, when does equality occur?

1.2.5 Generalize Chebyschev’s inequality using higher moments E(|X −

E(X)|^{p}) instead of the variance. 1.2.6 By obtaining an upper bound on the fourth moment, improve

Theorem 1.6 to

\frac{1}{N} |\{x ∈ [1,N] : |ν(x)−loglogN| > K\sqrt{loglogN}| = O(K^{−4}).

Can you generalize this to obtain a bound of O_{m}(K^{−m}) for any even integer m ≥ 2, where the constant in the O() notation is allowed to depend on m?

1.3 The exponential moment method

Chebyshev’s inequality shows that if one has control of the second moment Var(X) = E(|X −E(X)|^{2}), then a random variable X takes the value E(X)+O(λVar(X)^{1/2}) with probability 1−O(λ^{−2}). If one uses higher moments, one can obtain better decay of the tail probability than O(λ^{−2}). In particular, if one can control exponential moments^{1} such as E(e^{tX}) for some real parameter t, then one can obtain exponential decay in upper and lower tail probabilities, since Markov’s inequality yields

P(X ≥ λ) = P(e^{tX} ≥ e^{tλ}) ≤ \frac{E(e^{tX})}{e^{tλ}}  (1.15)

for t > 0 and λ ∈ R, and similarly

P(X ≤ −λ) = P(e^{−tX} ≥ e^{tλ}) ≤ \frac{E(e^{−tX})}{e^{tλ}}  (1.16)

for the same range of t,λ. The quantity E(e^{tX}) is known as an exponential moment of X, and the function t ↦ E(e^{tX}) is known as the moment generating function, thanks to the Taylor expansion

E(e^{tX}) = 1+tE(X)+ \frac{t^{2}}{2!} E(X^{2})+ \frac{t^{3}}{3!} E(X^{3})+...

The application of (1.15) or (1.16) is known as the exponential moment

1 To avoid questions of integrability or measurability, let us assume for sake of discussion that the random variable X here only takes finitely many values; this is the case of importance in combinatorial applications.

method. Of course, to use it effectively one needs to be able to compute the exponential moments E(e^{tX}). A preliminary tool for doing so is

Lemma 1.7 Let X be a random variable with |X| ≤ 1 and E(X) = 0. Then for any −1 ≤ t ≤ 1 we have E(e^{tX}) ≤ exp(t^{2}Var(X)).

Proof Since |tX| ≤ 1, a simple comparison of Taylor series gives the inequality

e^{tX} ≤ 1+tX +t^{2}X^{2}.

Taking expectations of both sides and using linearity of expectation and the hypothesis E(X) = 0 we obtain

E(e^{tX}) ≤ 1+t^{2}Var(X) ≤ exp(t^{2}Var(X))

as desired.

This lemma by itself is not terribly effective as it requires both X and t to be bounded. However the power of this lemma can be amplified considerably when applied to random variables X which are sums of bounded random variables, X = X_{1}+...+X_{n}, provided that we have the very strong assumption of joint independence between the X_{1},...,X_{n}. More precisely, we have

Theorem 1.8 (Chernoff’s inequality) Assume that X_{1},...,X_{n} are jointly independent random variables where |X_{i} −E(X_{i})| ≤ 1 for all i. Set X := X_{1} +···+X_{n} and let σ := \sqrt{Var(X)} be the standard deviation of X. Then for any λ > 0

P(|X −E(X)| ≥ λσ) ≤ 2max(e^{−λ^{2}/4},e^{−λσ/2}).  (1.17)

Informally speaking, this inequality asserts that X = E(X)+O(Var(X)^{1/2}) with high probability, and X = E(X) + O(ln^{1/2} nVar(X)^{1/2}) with extremely high probability (1 − O(n^{−C}) for some large C). The bound in Chernoff’s theorem provides a huge improvement over Chebyschev’s inequality when λ is large. However the joint independence of the X_{i} is essential (Exercise 8). Later on we shall develop several variants of Chernoff’s inequality in which there is some limited interaction between the X_{i}.

Proof By subtracting a constant from each of the X_{i} we may normalize E(X_{i}) = 0 for each i. Observe that P(|X| ≥ λσ) = P(X ≥ λσ)+P(X ≤ −λσ). By symmetry, it thus suffices to prove that

P(X ≥ λσ) ≤ e^{−tλσ/2}  (1.18)

where t := min(λ/2σ,1).

Applying (1.15) we have

P(X ≥ λσ) ≤ e^{−tλσ}E(e^{tX_{1}} ...e^{tX_{n}}).

Since the X_{i} are jointly independent, so are the e^{tX_{i}}. Using this and Lemma 1.7 we obtain

E(e^{tX_{1}} ...e^{tX_{n}}) = E(e^{tX_{1}})...E(e^{tX_{n})} ≤ exp(t^{2}Var(X_{1}))...exp(t^{2}Var(X_{n})).

On the other hand, from (1.9) we have

Var(X_{1})+...+Var(X_{n}) = σ^{2}.

Putting all this together, we obtain

P(X ≥ λσ) ≤ e^{−tλσ}e^{t^{2}σ^{2}}.

Since t ≤ λ/2σ, the claim follows.

Now let us consider a special, but important case when X_{i}’s are independent boolean variables.

Corollary 1.9 Let X = t_{1}+···+t_{n} where the t_{i} are independent boolean random variables. Then for any ϵ > 0

P(|X −E(X)| ≥ ϵE(X)) ≤ 2e^{− min(ϵ^{2}/4,ϵ/2)E(X)}.  (1.19)

Applying this with ϵ = 1/2 (for instance), we conclude in particular that

P(X = Θ(E(X))) ≥ 1−2e^{−E(X)/16}.  (1.20)

Proof From (1.1) we have that |t_{i} − E(t_{i})| ≤ 1 and Var(t_{i}) ≤ E(t_{i}). Summing this using (1.3), (1.9), we conclude that Var(X) ≤ E(X) (cf. (1.12)). The claim now follows from Theorem 1.8 with λ := ϵE(X).

As an immediate consequence of Corollary 1.9 and (1.4) we obtain the following concentration of measure property for the distribution of certain types of random sets.

Corollary 1.10 Let A be a set (possibly infinite), and let B ⊂ A be a random subset of A with the property that the events a ∈ B are independent for every a ∈ A. Then for any ϵ > 0 and any finite A′ ⊆ A we have

P(||B ∩A′|− ∑_{a∈A′} p_{a}| ≥ ϵ ∑_{a∈A′} p_{a}) ≤ 2e^{− min(ϵ^{2}/4,ϵ/2) ∑_{a∈A′} p_{a}}

where p_{a} := P(a ∈ B). In particular

P(|B ∩A′| = Θ( ∑_{a∈A′} p_{a})) ≥ 1−2e^{− ∑_{a∈A′} p_{a}/16}.

1.3.1 Sidon’s problem on thin bases

We now apply Chernoff’s inequality to the study of thin bases in additive combinatorics.

Definition 1.11 (Bases) Let B ⊂ N be an (infinite) set of natural numbers, and let k ∈ Z_{+}. We define the counting function r_{k,B}(n) for any n ∈ N as

r_{k,B}(n) := |{(b_{1},...,b_{k}) ∈ B^{k} : b_{1} +...+b_{k} = n}|.

We say that B is a basis of order k if every sufficiently large positive integer can be represented as sum of k (not necessarily distinct) elements of B, or equivalently if r_{k,B}(n) ≥ 1 for all sufficiently large n. Alternatively, B is a basis of order k if and only if N\kB is finite.

Examples 1.12 The squares N^{∧}2 = {0,1,4,9,...} are known to be a basis of order 4 (Legendre’s theorem), while the primes P = {2,3,5,7,...} are conjectured to be a basis of order 3 (Goldbach’s conjecture) and are known to be a basis of order 4 (Vinogradov’s theorem). Furthermore, for any k ≥ 1, the k^{th} powers N^{∧}k = {0^{k},1^{k},2^{k},...} are known to be a basis of order C(k) for some finite C(k) (Waring’s conjecture, first proven by Hilbert). Indeed in this case, the powerful Hardy-Littlewood circle method yields the stronger result that r_{m,N^{∧}k}(n) = Θ_{m,k}(n^{\frac{m}{k} −1}) for all large n, if m is sufficiently large depending on k (see for instance [361] for a discussion). On the other hand, the powers of k k^{∧}N = {k^{0},k^{1},k^{2},...} and the infinite progression k · N = {0,k,2k,...} are not bases of any order when k > 1.

The function r_{k,B} is closely related to the density of the set B. Indeed, we have the easy inequalities

∑_{n≤N} r_{k,B}(n) ≤ |B ∩[0,N]|^{k} ≤ ∑_{n≤kN} r_{k,B}(n)  (1.21)

for any N ≥ 1; this reflects the obvious fact that if n = b_{1} +...+b_{k} is a decomposition of a natural number n into k natural numbers b_{1},...,b_{k}, then n ≤ N implies that b_{1},...,b_{k} ∈ [0,N], and conversely b_{1},...,b_{k} ∈ [0,N] implies n ≤ kN. In particular if B is a basis of order k then

|B ∩[0,N]| = Ω(N^{1/k}).  (1.22)

Let us say that a basis B of order k is thin if r_{k,B}(n) = O(logn) for all large n. This would mean that |B ∩ [0,N]| = N^{1/k+o_{k}(1)}, thus the basis B would be nearly as “thin” as possible given (1.22). In the 1930s, Sidon asked the question of whether thin bases actually exist (or more generally, any basis which is “high quality” in the sense that r_{k,B}(n) = n^{o(1)} for all n. As Erdös recalled in one of his memoirs, he thought he could provide an answer within a few days. It took a little bit longer. In 1956, Erdös [89] positively answered Sidon’s question.

Theorem 1.13 There exists a basis B ⊂ Z^{+} of order 2 so that r_{2,B}(n) = Θ(logn) for every sufficiently large n. In particular, there exists a thin basis of order 2.

Remark 1.14 A very old, but still unsolved conjecture of Erdös and Turán [94] states that if B ⊂ N is a basis of order 2, then limsup_{n→∞} r_{2,B}(n) = ∞. In fact, Erdös later conjectured that limsup_{n→∞} r_{2,B}(n)/logn > 0 (so that the thin basis constructed above is essentially as thin as possible) Nothing is known concerning these conjectures (though see Exercise 10 for a much weaker result).

Proof Define^{1} a set B ⊂ Z^{+} randomly by requiring the events n ∈ B (for n ∈ Z^{+}) to be jointly independent with probability

P(n ∈ B) = max(C\sqrt{\frac{logn}{n}},1)

1 Strictly speaking, to make this argument rigourous one needs an infinite probability space such as Wiener space, which in turn requires a certain amount of measure theory to construct. One can avoid this by proving a “finitary” version of Theorem 1.13 to provide a thin basis for an interval [1,N] for all sufficiently large N, and then gluing those bases together; we leave the details to the interested reader. A similar remark applies to other random subsets of Z^{+} which we shall construct later in this chapter.

where C > 0 is a large constant to be chosen later. We now show that r_{2,B}(n) = Θ(logn) for all sufficiently large n with positive probability (indeed, it is true with probability 1). Writing

r_{2,B}(n) = ∑_{i+j=n}I(i ∈ B)I(j ∈ B) = Θ( ∑_{1≤i<n/2}I(i ∈ B)I(n−i ∈ B))+O(1)

we see that it suffices to show that the probability

P( ∑_{1≤i<n/2}I(i ∈ B)I(n−i ∈ B) = Θ(logn) for all but finitely many n)

is positive (if the constants in the Θ() notation are chosen appropriately). By the Borel-Cantelli lemma (Lemma 1.2) and the convergence of ∑_{n=1}^{∞} \frac{1}{n^{2}} , it thus suffices to show that

P( ∑_{1≤i<n/2}I(i ∈ B)I(n−i ∈ B) = Θ(logn)) = 1−O(\frac{1}{n^{2}} )

for all large n.

By linearity of expectation (1.3), we have for n > 1

E( ∑_{1≤i<n/2}I(i ∈ B)I(n−i ∈ B)) = ∑_{1≤i<n/2}C^{2}\sqrt{\frac{logi}{i}}\sqrt{\frac{log(n−i)}{n−i}} +O_{C}(1)

= Θ(C^{2} \frac{ln^{1/2} n}{n^{1/2}} ∑_{1≤i<n/2} \frac{ln^{1/2} i}{i^{1/2}})+O_{C}(1)

= Θ(C^{2} logn)+O_{C}(1).

In particular, by choosing C large enough, we may take

E( ∑_{1≤i<n/2}I(i ∈ B)I(n−i ∈ B)) ≥ 32logn

for all n > 1.

Observe that the restriction i < n/2 ensures that the boolean random variables I(i ∈ B)I(n−i ∈ B) are jointly independent. If we now apply Corollary 1.9 with ϵ := 1/2, we conclude that

P( ∑_{1≤i<n/2}I(i ∈ B)I(n−i ∈ B) ≤ 16logn) ≤ 2/n^{2}.

and

P( ∑_{1≤i<n/2}I(i ∈ B)I(n−i ∈ B) ≤ 48logn) ≤ 2/n^{2}.

The claim follows.

It is quite natural to ask whether Theorem 1.13 can be generalized to arbitrary k. Using the above approach, in order to obtain a basis B such that r_{k,B}(n) = Θ(logn), we should set P(n ∈ B) = cn^{1/k−1} ln^{1/k} n for all sufficiently large n. As before, we have

r_{k,B}(n) = ∑_{x_{1}+···+x_{k}=n}I(x_{1} ∈ B)...I(x_{k} ∈ B).  (1.23)

Although r_{k,B}(n) does have the right expectation Θ(logn), we face a major problem: the products I(x_{1} ∈ B)...I(x_{k} ∈ B) with k > 2 are no longer independent. In fact, a typical number x appears in quite many (Ω(n^{k−2})) solutions of x_{1} +···+x_{k} = n. This dashes the hope that one can use Theorem 1.8 to conclude the argument.

It took a long time to overcome this problem of dependency. In 1990, Erdös and Tetali [93] successfully generalized Theorem 1.13 for arbitrary k:

Theorem 1.15 For any fixed k, there is a subset B ⊂ N such that r_{k,B}(n) = Θ(logn) for all sufficiently large n. In particular, there exists a thin basis of order k for any k.

We shall discuss this theorem later in a later section. Let us now turn instead to another application.

1.3.2 Complementary bases

Given a set A ⊂ N and an integer k ≥ 1, a set B ⊂ N is a complementary basis of order k of A if every sufficiently large natural number can be written as a sum of an element in A and k elements in B (not necessarily distinct), or equivalently if N\(A+kB) is finite.

As in the theory of bases, it is convenient to introduce the counting function

r_{A+B+...+B}(n) := |{(a,b_{1},...,b_{k}) ∈ A×B^{k} : n = a+b_{1} +...+b_{k}}|

and observe (analogously to (1.21)) that

∑_{n≤N} r_{A+B+...+B}(n) ≤ |A∩[0,N]||X ∩[0,N]| ≤ ∑_{n≤(k+1)N} r_{A+B+...+B}(n).

Now consider the set P = {2,3,5,...} of primes, and let B be a complementary basis for P of order 1. Recall that |P ∩[0,N]| = Θ(n/logn)

(Exercise 4 from Section 1.10). From the preceding inequality we thus have the lower bound

|B ∩[0,n]| = Ω(logn)

for all large n. It is not known whether this bound can actually be attained. However, Erdös shown that P has a complementary base of size O(log^{2} n) [89, 163]:

Theorem 1.16 P has a complementary base B ⊂ Z^{+} of order 1 such that |B ∩[0,n]| = O(log^{2} n) for all sufficiently large n.

Proof Again B is created in a random manner, setting the events n ∈ B to be jointly independent with probability

P(n ∈ B) = max(C \frac{logn}{n},1)

for some large constant C. From Corollary 1.10 we have

P(|B ∩[0,n]| < 10C log^{2} n) = O(\frac{1}{n^{2}} )

(say) for each n, and hence by the Borel-Cantelli lemma (Lemma 1.2) we have with probability 1 that |B ∩[0,n]| = O(log^{2} n) for all sufficiently large n. Thus it suffices to show that with probability 1, r_{P +B}(n) > 0 for all sufficiently large n. By the Borel-Cantelli lemma again, it will suffice to show that

P(r_{P +B}(n) < 10log^{2} n) = 1−O(\frac{1}{n^{2}} )

for all large n. To show this, we write r_{P +B}(n) = |B ∩(n−P)|. From linearity of expectation (1.4) we have

E(|B ∩(n−P)|) = C ∑_{p∈P ∩[1,n)} \frac{log(n−p)}{n−p} +O_{C}(1).

We now use the estimate

∑_{p∈P ∩[1,n)} \frac{log(n−p)}{n−p} = Ω(logn)

for all sufficiently large n (see Proposition 1.54 in the Appendix); if we choose C large enough, we thus conclude that

E(|B ∩(n−P)|) > 8logn

for all sufficiently large n. From Corollary 1.10 (or Corollary 1.8), the desired claim follows.

1.3.1 Let ε be the uniform distribution on {−1,+1}, and let ε_{1},...,ε_{n} be independent trials of ε. For any λ > 0, prove the reflection principle

P( max_{1≤j≤n} ∑_{i=1}^{j}ε_{i} ≥ λ) = 2P(∑_{i=1}^{n}ε_{i} ≥ λ).

Hint: Let A ⊂ {−1,1}^{n} be the set of n-tuples (ε_{1},...,ε_{n}) such that ∑_{i=1}^{n} ε_{i} ≥ λ, and let B ⊂ {−1,1}^{n} be the set of n-tuples (ε_{1},...,ε_{n}) such that ∑_{i=1}^{n} ε_{i} < λ but ∑_{i=1}^{j} ε_{i} ≥ λ for some 1 ≤ j < n. Create a “reflection map” which exhibits a bijection between A and B.

1.3.2 With the same notation as the previous exercise, show that

P( max_{1≤j≤n} ∑_{i=1}^{j}a_{i}ε_{i} ≥ λ) ≤ 2P(∑_{i=1}^{n}a_{i}ε_{i} ≥ λ)

for all non-negative real numbers a_{1},...,a_{n}. 1.3.3 By considering the case when X_{1},...,X_{n} ∈ {−1,1} are inde-

pendent variables taking +1 and −1 with equal probability 1/2, show that Theorem 1.8 cannot be improved except for the constant in the exponent.

1.3.4 Let the hypotheses be as in Theorem 1.8, but with the X_{i} complex-valued instead of real-valued. Show that

E(|X −E(X)| ≥ λσ) ≤ 4max(e^{−λ^{2}/8},e^{λσ/2\sqrt{2}})

for all λ > 0. (Hint: if |z| ≥ λσ, then either |Re(z)| ≥ \frac{1}{\sqrt{2}}λσ or |Im(z)| ≥ \frac{1}{\sqrt{2}}λσ.) The constants here can be improved slightly.

1.3.5 (Hoeffding’s inequality) Let X_{1},...,X_{n} be jointly independent random variables, taking finitely many values, with a_{i} ≤ X_{i} ≤ b_{i} for all i and some real numbers a_{i} < b_{i}. Let X := X_{1} +...+X_{n}. Using the exponential moment method, show that

P(|X −E(X)| ≥ λ(∑_{i=1}^{n}|b_{i} −a_{i}|^{2})^{1/2}) ≤ 2e^{−2λ^{2}}.

1.3.6 (Azuma’s inequality) Let X_{1},...,X_{n} are random variables taking finitely many values with |X_{i}| ≤ 1 for all i. We do not assume that the X_{i} are jointly independent, however we do require that the X_{i} form a martingale difference sequence, by which we mean that E(X_{i}|X_{1} = x_{1},...,X_{i−1} = x_{i−1}) = 0 for all 1 ≤ i ≤ n and all x_{1},...,x_{i−1}. Using the exponential moment method, establish the large deviation inequality

P(|X_{1} +...+X_{n}| ≥ λ\sqrt{n}) ≤ 2e^{−λ^{2}/2}.  (1.24)

1.3.7 Let n be a sufficiently large integer, and color each of the elements in [1,n] red or blue, uniformly and independently at random (so each element is red with probability 1/2 and blue with probability 1/2). Show that the following statements hold with probability at least 0.9:

(a) There is a red arithmetic progression of length at least \frac{log n}{10}.

(b) There is no monochromatic arithmetic progression of length exceeding 10logn.

(c) The number of red elements and the number of blue elements in [1,n] differ by O(n^{1/2}).

(d) In every arithmetic progression in [1,n], the number of red elements and the number of blue elements differ by O(n^{1/2} log^{1/2} n).

1.3.8 Let us color the elements of [1,n] red or blue as in the preceding exercise. For each A ⊂ [1,n], let t_{A} denote the parity of the red elements in A, thus t_{A} = 1 if there are an odd number of red elements in A, and t_{A} = 0 otherwise. Let X = ∑_{A⊆[1,n]} t_{A}. Show that the t_{A} are pairwise (but not necessarily jointly) independent, that E(X) = 2^{n−1}, and that Var(X) = 2^{n−2}. Furthermore, show that P(X = 0) = 2^{−n}. This shows that Chernoff’s inequality can fail dramatically if one only assumes pairwise independence instead of joint independence (though Chebyshev’s inequality is of course still valid in this case).

1.3.9 For any k ≥ 1, find a basis B ⊂ N of order k such that |B ∩ [0,n]| = Θ_{k}(n^{1/k}) for all large n. (This can be done constructively, without recourse to the probabilistic method, for instance by taking advantage of the base k representation of the integers.)

1.3.10 Prove that there does not exist positive integers k,m ≥ 1, and a set B ⊂ N such that r_{k,B}(n) = m for all sufficiently large n; thus a base of order k cannot be perfectly regular. (Hint: consider the complex-analytic function ∑_{n∈B} z^{n}, defined for |z| < 1, and compute the k^{th} power of this function. It is rather challenging to find an elementary proof of this fact that does not use complex analysis, or the closely related tools of Fourier analysis.)

1.3.11 With the hypotheses of Theorem 1.8, establish the moment es-

timates

E(|X|^{p})^{1/p} = O(\sqrt{p}σ +p)

for all p ≥ 1. 1.3.12 With the hypotheses of Corollary 1.9, establish the inequality

E(\binom{X}{n}) ≤ \frac{1}{n!}E(X)^{n}

for all n ∈ N. (Hint: expand \binom{X}{n} as ∑_{i_{1}<...<i_{n}} t_{i_{1}} ...t_{i_{n}}). Use this (and Stirling’s formula (1.52)) to derive an inequality similar to that in Corollary 1.9 in the case ϵ > 1. For a generalization of this inequality, see Lemma 1.40 below.

1.4 Correlation inequalities

Chernoff’s inequality is useful for controlling quantities of the form t_{1} +...+t_{n} where t_{1},...,t_{n} are independent variables. In many applications, however, one needs to instead control more complicated polynomial expressions of t_{1},...,t_{n}, such as monotone quantities.

Definition 1.17 (Monotone increasing variables) Let t_{1},...,t_{n} be independent boolean random variables. A random variable X = X(t_{1},...,t_{n}) is monotone increasing if we have

X(t_{1},...,t_{n}) ≥ X(t′_{1},...,t′_{n}) whenever t_{i} ≥ t′_{i} for all 1 ≤ i ≤ n

or equivalently if X is monotone increasing in each of the variables t_{i} separately. We call X monotone decreasing if −X is monotone increasing. We say that an event A is monotone increasing (resp. decreasing) if the indicator I(A) is monotone increasing (resp. decreasing).

Example 1.18 If P (t_{1},...,t_{n}) is any polynomial of t_{1},...,t_{n} with nonnegative coefficients, then P is monotone increasing and −P is monotone decreasing, and the event P (t_{1},...,t_{n}) ≥ k is monotone increasing for any fixed k.

It is reasonable to think that any two increasing (resp. decreasing) variables or events are, in some way, positively correlated; intuitively, if both X and Y are monotone increasing (resp. decreasing), then the event that X is large (resp. small) should boost up the chance that Y is also large (resp. small). This intuition was materialized by Fortuin, Kasteleyn and Ginibre [100], motivated by problems in statistical mechanics:

Theorem 1.19 (FKG inequality) Let n ≥ 0, and let X and Y be two monotone increasing variables. Then

E(XY) ≥ E(X)E(Y)

or equivalently

Cov(X,Y) ≥ 0.

The same inequality holds for the case both X and Y are monotone decreasing.

Proof By replacing X,Y with −X,−Y if necessary, we may assume that X and Y are both monotone increasing.

We use induction on n. The base case n = 0 is trivial since in this case X and Y are deterministic. Now assume inductively that n ≥ 1 and the claim has already been proven for n−1. We may assume that P(t_{n} = 0) and P(t_{n} = 1) are non-zero since otherwise the claim follows immediately from the induction hypothesis. Observe that the covariance Cov(X,Y) is unaffected if we shift X and Y by constants. Thus we may normalize

E(X|t_{n} = 0) = E(Y |t_{n} = 0) = 0  (1.25)

where E(X|t_{n} = 0) denotes the conditional expectation of X relative to the event t_{n} = 0. By monotonicity of X,Y in the t_{n} variable and the joint independence of the t_{i} we then have

E(X|t_{n} = 1),E(Y |t_{n} = 1) ≥ 0.  (1.26)

Observe that conditioning on the event t_{n} = 0, the random variables X,Y are monotone increasing functions of t_{1},...,t_{n−1}. Thus by the induction hypothesis

E(XY |t_{n} = 0) ≥ E(X|t_{n} = 0)E(Y |t_{n} = 0) = 0

and similarly

E(XY |t_{n} = 1) ≥ E(X|t_{n} = 1)E(Y |t_{n} = 1).

By Bayes’ formula we thus have

E(XY) = E(XY |t_{n} = 0)P(t_{n} = 0)+E(XY |t_{n} = 1)P(t_{n} = 1)
≥ E(X|t_{n} = 1)E(Y |t_{n} = 1)P(t_{n} = 1).

On the other hand, from (1.25) and another application of Bayes’ formula we have

E(X)E(Y) = E(X|t_{n} = 1)P(t_{n} = 1)E(Y |t_{n} = 1)P(t_{n} = 1).

Since P(t_{n} = 1) ≤ 1, the claim now follows from (1.26).

From (1.1) and an easy induction we have an immediate corollary to Theorem 1.19:

Corollary 1.20 Let A and B be two increasing events, then

P(A∧B) ≥ P(A)P(B).

More generally, if A_{1},...,A_{k} are increasing events, then

P(A_{1} ∧...∧A_{k}) ≥ P(A_{1})...P(A_{k}).

1.4.1 Asymptotic complementary bases Now we are going to use the FKG inequality to prove a result of Ruzsa [280] concerning asymptotic complementary bases.

Definition 1.21 (Asymptotic complementary bases) Let A ⊂ N be a set of natural numbers and k ≥ 1. We define the lower density \underline{σ}(A) and upper density \overline{σ}(A) of A to be the numbers

\underline{σ}(A) := liminf_{n→∞} \frac{|A∩[0,n)|}{n};\overline{σ}(A) := limsup_{n→∞} \frac{|A∩[0,n)|}{n}.

If ϵ > 0 and X ⊂ N, we say that X is a (1 − ε)-complementary base of order k of A if \underline{σ}(A + kX) ≥ 1 − ε, and that X is an asymptotic complementary base of order k of A if \underline{σ}(A+kX) = 1.

Theorem 1.22 [280] Let P = {2,3,5,...} be the primes. For any 0 < ϵ < 1, there is an (1−ϵ)-complementary base X ⊂ Z^{+} of order 1 of P with |X ∩[1,n]| = O_{ϵ}(logn) for all large n.

It follows that (the proof is left as an exercise)

Corollary 1.23 For any function ω(n) tending to infinity with n, there is an asymptotic complementary base X ∈ Z^{+} of order 1 of P with |X ∩[1,n]| ≤ ω(n)logn for all large n.

Corollary 1.23 improves an earlier result of Kolountzakis [204], and should also be compared with Theorem 1.16 (note that every complementary basis is automatically an asymptotic complementary basis). Since P has density Θ(n/logn), it is clear that an asymptotic complementary base of P should have density Ω(logn). Thus, Corollary 1.23 is nearly best possible.

Proof [of Theorem 1.22] The theorem follows from the following finite statement.

Lemma 1.24 For every ε > 0, and all natural numbers n which are sufficiently large depending on ε, there exists a set B ⊂ [n^{2/3},2n^{2/3}] with |B| = O_{ε}(logn) such that

|[1,x]\(P +B)| ≤ εx,  (1.27)

for all n^{3/4} ≤ x ≤ n.

The deduction of Theorem 1.22 from Lemma 1.24 is straightforward and is left as an exercise. To prove Lemma 1.24, we use the probabilistic method. We choose B ⊂ [n^{2/3},2n^{2/3}] randomly, by letting the events l ∈ B with l ∈ [n^{2/3},2n^{2/3}] be jointly independent with probability

P(l ∈ B) = \frac{K logn}{n^{2/3}}

where K = K_{ε} is a large constant to be chosen later. From Corollary 1.10 we have

P(|B| ≥ \frac{K}{100} logn) < \frac{1}{n}  (1.28)

(say).

Now let J := ⌊\frac{1}{4} log_{2} n⌋. If j ∈ [0,J], we say that j is good if

|[2n^{2/3},n/2^{j}]\(P +B)| ≤ \frac{ε}{2} \frac{n}{2^{j}} .

It is easy to verify that if all the elements of [0,J] are good, then (1.27) holds (recall that we assume n large depending on ε). In view of (1.28), it thus suffices to show that

P(j is good for all j ∈ [0,J]) ≥ \frac{1}{n}.  (1.29)

Let us first estimate the probability that a single j ∈ [0,J] is good. Fixing j ∈ [0,J], we observe for each m ∈ [2n^{2/3},n/2^{j}] that

P(m ∉ P +B) = P(m−p ∉ B for all p ∈ m−[n^{2/3},2n^{2/3}])

= ∏_{p∈m−[n^{2/3},2n^{2/3}]}P(m−p ∉ B)

= ∏_{p∈m−[n^{2/3},2n^{2/3}]}(1− \frac{K logn}{n^{2/3}})

≤ exp(−|P ∩m−[n^{2/3},2n^{2/3}]|\frac{K logn}{n^{2/3}})

where we have used the independence of the events l ∈ B. By Lemma 1.53, we conclude

P(m ∉ P +B) ≤ exp(−Ω(K)).

Summing this over m ∈ [1,n/2^{j}] and using linearity of expectation (1.4), we conclude

E(|[2n^{2/3},n/2^{j}]\(P +B)|) ≤ exp(−Ω(K))\frac{n}{2^{j}} .

If we choose K sufficiently large depending on ε, we thus see from Markov’s inequality that

P(j is good ) ≥ \frac{1}{2}.

Now we come to the final and most important observation: For any fixed j, the event that x_{j} is good is a monotone increasing random variable, with respect to indicator variables t_{l} := I(l ∈ B). Thus, by Corollary 1.20,

P(j is good for all j ∈ [0,J]) ≥ ∏_{j∈[0,j]}P(j is good)
≥ 2^{−J−1}.

Since J = \frac{1}{4} ln_{2} n+O(1) and n is assumed to be large, the claim (1.29) follows.

1.4.1 Deduce Theorem 1.22 from Lemma 1.24. 1.4.2 Deduce Corollary 1.23 from Theorem 1.22. 1.4.3 Let the notation and assumptions be as in Theorem 1.19. Sup-

pose that each of the independent variables t_{1},...,t_{n} attain the values 0 and 1 with positive probability. Show that equality holds in Theorem 1.19 if and only if X and Y depend on disjoint subsets of the random variables t_{1},...,t_{n}.

1.5 The Lovász local lemma

Let (A_{i})_{i∈V} be a finite collection of events in a probabilistic space; we will later view the index set V as the vertex set of a graph. In many situations, it is desirable to show that there is a chance that the complementary events (Ā_{i})_{i∈V} hold simultaneously, i.e. that P(⋀_{i∈V} Ā_{i}) > 0. This is particularly useful when the A_{i} are bad events that we would like to avoid.

If the A_{i} are mutually independent, then the problem is trivial, as we have

P( ⋀_{i∈V} Ā_{i}) = ∏_{i∈V} P(Ā_{i}) = ∏_{v∈V} (1−P(A_{i})),  (1.30)

which is positive if P(A_{i}) are all strictly less than one. On the other hand, mutual independence is a very strong assumption which rarely holds.

One may expect that something similar to (1.30) is still true if we allow a sufficiently “local” dependence among the A_{i}’s, so that we still have good control on P(A_{i}) even after conditioning on most of the events Ā_{j}. This is indeed possible, as shown by Lovász in 1975 in a joint paper with Erdös [90]. We present a modern version of this lemma as follows.

Lemma 1.25 (Lovász local lemma) Let V be a finite set, and for each i ∈ V let A_{i} be a probabilistic event. Assume that there is a directed graph G(V,E) (without loops) on the vertex set V (which is known as the dependency graph of the A_{i}) and a sequence of numbers 0 ≤ x_{i} < 1 for each i ∈ V such that the estimate

P(A_{i}| ⋀_{j∈S} Ā_{j}) ≤ x_{i} ∏_{(i,j)∈E}(1−x_{j})  (1.31)

holds whenever i ∈ V and S ⊆ V \{i} is such that ⋀_{j∈S} Ā_{j} has non-zero probability and (i,j) ∉ E for all j ∈ S. Then for any disjoint S,S′ ⊆ V we have

P(∧_{i∈S}Ā_{i}|∧_{i∈S′} Ā_{i}) ≥ ∏_{i∈S}(1−x_{i}) > 0.  (1.32)

In particular we have

P(∧_{i∈V} Ā_{i}) ≥ ∏_{i∈V}(1−x_{i}) > 0.

The graph G is usually referred to as the dependency graph of the A_{i}. Note that (1.31) will hold if we have

P(A_{i}) ≤ x_{i} ∏_{(i,j)∈E}(1−x_{j})

and each A_{i} is mutually independent to all of the A_{j} with (i,j) ∉ E and j ≠ i. This was in fact the hypothesis stated in the original formulation of the lemma. However, there are situations where these rather strong mutual independence hypotheses are not available and one needs the full strength of Lemma 1.25. Alon and Spencer’s book [12]Chapter 5 contains many interesting applications.

Proof [of Lemma 1.25] We shall induct on the total cardinality |S|+|S′|. If |S| + |S′| = 0 then S,S′ are empty, and the claim (1.32) is trivial. Now assume inductively that |S|+|S′| ≥ 1, and the claim has already been proven for smaller values of |S|+|S′|. Note that the case |S| = 0 is trivial. To establish the claim for |S| ≥ 1, it suffices to do so for the case |S| = 1. Indeed, if |S| ≥ 1, then we can split S = {j}∪(S\{j}) for some j ∈ S. From Bayes’ identity we have

P(∧_{i∈S}Ā_{i}|∧_{i∈S′} Ā_{i}) = P(Ā_{i}|∧_{i∈S′∪S\{j}} Ā_{i})P(∧_{i∈S\{j}}Ā_{i}|∧_{i∈S′} Ā_{i})

and the claim (1.32) then follows by applying the induction hypothesis to estimate the second factor.

Thus it remains to verify the |S| = 1 case of (1.32). Writing S = {i}, we reduce to showing that

P(A_{i}|∧_{j∈S′} Ā_{j}) ≥ x_{i}.

We split S′ = ∪S_{1} ∪S_{2} where S_{1} := {j ∈ S|(i,j) ∈ E} are those indices j which are adjacent to i in the dependency graph, and S_{2} := S′\S_{1}. From Bayes’ identity again we have

P(A_{i}|∧_{j∈S′} Ā_{j}) = \frac{P(A_{i},∧_{j∈S_{1}} Ā_{j}|∧_{j∈S_{2}} Ā_{j})}{P(∧_{j∈S_{1}} Ā_{j}|∧_{j∈S_{2}} Ā_{j})}.

Note that by induction hypothesis, ⋀_{j∈S_{2}} Ā_{j} occurs with positive probability. From (1.31) we have

P(A_{i},∧_{j∈S_{1}} Ā_{j}|∧_{j∈S_{2}} Ā_{j}) ≤ P(A_{i}|∧_{j∈S_{2}} Ā_{j}) ≤ x_{i} ∏_{j∈V:(i,j)∈E}(1−x_{j}).

On the other hand, from the induction hypothesis (since |S_{1}| + |S_{2}| < 1+|S′|) we have

P(∧_{j∈S_{1}} Ā_{j}|∧_{j∈S_{2}} Ā_{j}) ≥ ∏_{j∈S_{1}}(1−x_{j}) ≥ ∏_{j∈V:(i,j)∈E}(1−x_{j}).

Combining the two, we obtain the claim.

In practice, the following corollary of Lemma 1.25 is sometimes easier to apply.

Corollary 1.26 Let d ≥ 1 and 0 < p < 1 be numbers such that

p ≤ \frac{1}{e(d+1)},

where e = 2.718... is the base of the natural logarithm. Let V be a finite set, and for each i ∈ V let A_{i} be a probabilistic event with P(A_{i}) ≤ p. Assume also that each A_{i} is mutually independent of all but at most d of the other events A_{j}. Then

P(∧_{i∈V} Ā_{i}) ≥ (1−p)^{|V |} > 0.

If d = 0, then Corollary 1.26 follows from (1.30). For d ≥ 1, the corollary follows from Lemma 1.25 by setting x_{i} = \frac{1}{d+1} and using the fact that (1− \frac{1}{d+1})^{d} > \frac{1}{e}. The constant e is best possible as shown by Shearer.

1.5.1 Colorings of the real line We now give an application of Corollary 1.26. This is the original result from the paper [90] of Erdös and Lovász, which motivated the development of the local lemma.

Let us use k colors [1,k] to color the real numbers. (Thus, a coloring is a map from R to [1,k].) A subset T of R is called colorful if it contains all k colors.

Theorem 1.27 Let m and k be two positive integers satisfying

e(m(m−1)+1)k(1− \frac{1}{k})^{m} ≤ 1.  (1.33)

Then for any set S of real numbers with |S| = m, and any set X ⊂ R (possibly infinite), there is a k-coloring of R such that the translates x+S of S are colorful for every x ∈ X.

Proof We first prove this theorem in the special case when X is finite, and then use a compactness argument to handle the general case (of course, the theorem is strongest when X = R). The point is that the bound (1.33) does not depend on the cardinality of X.

Fix X to be finite, thus X +S is also finite. Note that we only need to color the real numbers in X +S, as the real numbers outside of X + S are irrelevant. For each element y in X + S, we color it randomly and independently: y receives each of the colors in [1,k] with the same probability 1/k. Let A_{x} be the event that the translate x + S is not colorful. We need to show that

P(∧_{x∈X} Ā_{x}) > 0.

In order to apply Corollary 1.26, we first estimate P(A_{x}). If x_{S} is not colorful, then at least one color is missing. The probability that a particular color (say 1) is missing is (1− \frac{1}{k})^{|x+S|} = (1− \frac{1}{k})^{m}. As there are k colors, we conclude

P(A_{x}) ≤ k(1− \frac{1}{k})^{m}.

(In fact we have a strict inequality as there is a positive chance that more than one colors are missing.) Next, observe that if two translates x+S and x′+S are disjoint, then the events A_{x} and A_{x′} are independent. On the other hand, x+S and x′ ∈ S intersects if and only if there are two elements s_{1},s_{2} ∈ S such that x+s_{1} = x′ +s_{2}. It follows that x′ = x+(s_{1} −s_{2}). Since that number of (ordered) pairs (s_{1},s_{2}) with s_{1} ≠ s_{2} and s_{1},s_{2} ∈ S is m(m − 1), we conclude that each A_{x} is independent from all but at most m(m − 1) events A_{x′}. Set p = k(1 − \frac{1}{k})^{m} and d = m(m − 1). The condition (1.33) guarantees that the condition of Corollary 1.26 is met and this corollary implies that P(∧_{x∈X} Ā_{x}) > 0, as desired.

A routine way to pass from a finite statement to an infinite one is to use a compactness argument and that is what we do next. The space of colorings of R can be identified with the product space [1,k]^{R}, which is compact in the product topology by Tychonoff’s theorem. In this product space, for each x ∈ R we set K_{x} as the set of all k-colorings such that x+S is colorful. It is easy to see that each K_{x} is closed. The finite statement proved above asserts that any finite collection of K_{x}’s has a non-empty intersection. It follows, by compactness, that all K_{x}, x ∈ R, has a non-empty intersection. Any element in this intersection is a coloring desired by the theorem.

1.5.1 Show that there exists a positive constant c such that the following holds. For every sufficiently large n, there is a graph on n points which does not contain the following two objects: a triangle and an independent set of size c\sqrt{n}logn.

1.6 Janson’s inequality

Let t_{1},...,t_{n} be jointly independent boolean random variables. In Corollary 1.9 we established a large deviation inequality for the polynomial t_{1} +...+t_{n}. In many applications, it is also of interest to obtain large deviation inequalities for more general polynomials P (t_{1},...,t_{n}) of the boolean variables t_{1},...,t_{n}. One particularly important case is that of a boolean polynomial

X := ∑_{A∈A} ∏_{j∈A}t_{j}

where A is some collection of non-empty subsets of [1,n]. Observe that boolean polynomials are automatically positive and monotone increasing, and hence any two boolean polynomials are positively correlated via the FKG inequality (Theorem 1.19). More generally, if X and Y are boolean polynomials, then f(X) and f(Y) will be positively correlated whenever f is a monotone increasing or decreasing function. In particular, we see that

E(e^{−s(X+Y)}) ≥ E(e^{−sX})E(e^{−sY})  (1.34)

for any real number s. Using this fact, the exponential moment method, and some additional convexity arguments, Janson [181] derived a powerful bound for the lower tail probability P(X ≤ E(X)−T):

Theorem 1.28 (Janson’s inequality) Let t_{1},...,t_{n},A,X be as above. Then for any 0 ≤ T ≤ E(X) we have the lower tail estimate

P(X ≤ E(X)−T) ≤ exp(−\frac{T^{2}}{2Δ})

where

Δ = ∑_{A,B∈A:A∩B≠∅}E( ∏_{j∈A∪B} t_{j}).

In particular, we have

P(X = 0) ≤ exp(−\frac{E(X)^{2}}{2Δ} ).

Remark 1.29 Informally, Janson’s inequality asserts that if Δ = O(E(X)^{1/2}), then X = Ω(E(X)) with large probability. In the case where A is just the collection of singletons {1},...,{n}, then X = t_{1} +...+t_{n}, Δ = E(X), and the above claim is then essentially (the lower half of) Corollary 1.9.

The quantity Δ is somewhat inconvenient to work with directly. Using the independence of the t_{j}, one can rewrite it as

Δ = ∑_{A∈A}E(∏_{j∈A}t_{j}) ∑_{B∈A:A∩B≠∅}E( ∏_{j∈B\A}t_{j}).

Since E(X) = ∑_{A∈A} E(∏_{j∈A} t_{j}), we thus have

Δ ≤ E(X) sup_{A∈A} ∑_{B∈A:A∩B≠∅}E( ∏_{j∈B\A}t_{j}).  (1.35)

We record a particular consequence of this estimate concerning quadratic boolean polynomials that we shall use shortly.

Corollary 1.30 Let t_{1},...,t_{n} be as above, and let X = ∑_{1≤i≤j≤n:i∼j} t_{i}t_{j}, where i ∼ j is some symmetric relation on [1,n]. Then we have

P(X = 0) ≤ exp(−\frac{E(X)}{2+4sup_{i} ∑_{j:i∼j} E(t_{j})}).

Proof We take A := {{i,j} : i ∼ j}. For any A ∈ A, it is easy to verify that

∑_{B∈A:A∩B≠∅}E( ∏_{j∈B\A}t_{j}) ≤ 1+2sup_{i} ∑_{j:i∼j} E(t_{j})

and so the claim follows from (1.35) and Theorem 1.28.

Before presenting the proof of Theorem 1.28, let us give an application. This application again concerns complementary bases of primes, but this time of order 2 rather than 1. The following result (which should be compared with Theorems 1.16 and 1.22) in the case k = 2 was recently proved by Vu [358].

Theorem 1.31 For any k ≥ 2, P has a complementary base B ∈ Z^{+} of order k with |B ∩[1,n]| = O(logn) for all large n.

Proof It suffices to establish the claim when k = 2. To construct B shall use the random method. More precisely, we let B ⊂ Z^{+} be a random set with the events n ∈ B being independent with probability

P(n ∈ B) = min(\frac{c}{n},1)

for all n ∈ Z^{+}, where c is a positive constant to be determined. As before, we will not discuss the measure-theoretic issues associated with requiring infinitely many independent random variables, as they can be dealt with by a suitable finitization of this argument. Let t_{n} be the boolean random variable t_{n} := I(n ∈ B). By Corollary 1.10 we have

P(|B ∩[1,m]| ≤ 10clogm) = 1−O(\frac{1}{m^{2}} )

for all large m, and hence by the Borel-Cantelli lemma (Lemma 1.2) we have with probability 1 that

|B ∩[1,m]| = O_{c}(logm) for all sufficiently large m > 1.  (1.36)

Now for each n ∈ Z^{+}, consider the counting function

r_{P +B+B}(n) = |{(p,i,j) ∈ P ×B ×B : n = p+i+j}|

= ∑_{p<n} ∑_{i+j=n−p}t_{i}t_{j}.

This is of course a random variable for each n. In view of (1.36), it will suffice to show that with probability 1, we have r_{P +B+B}(n) ≠ 0 for all but finitely many n. From the Borel-Cantelli lemma, it thus suffices to show that

P(r_{P +B+B}(n) = 0) = O(\frac{1}{n^{2}} )

for all large n, if c is chosen large enough.

Fix n to be large. It will be convenient to work with a reduced version of r_{P +B+B}(n), namely the boolean polynomial

Y_{n} := ∑_{i>j≥n^{2/3}:i+j∈n−P} t_{i}t_{j}.

Clearly we have Y_{n} ≤ r_{P +B+B}(n), and so it suffices to show that

P(Y_{n} = 0) = O(\frac{1}{n^{2}} ).

We now apply Corollary 1.30 (using the relation i ∼ j if i ≠ j and i+j ∈ n−P) to give

P(Y_{n} = 0) ≤ exp(−\frac{E(Y_{n})}{2+4sup_{i≥n^{2/3}} ∑_{j≥n^{2/3}:i+j∈n−P} E(t_{j})}).

By construction of the t_{j}, and Proposition 1.54 in the appendix, we have for any i ≥ n^{2/3}

∑_{j≥n^{2/3}:i+j∈n−P} E(t_{j}) = ∑_{p≤n−i−n^{2/3}} max(\frac{c}{n−i−p},1)
= O(c).

On the other hand, from linearity of expectation (1.3) and independence we have

E(Y_{n}) = ∑_{i>j≥n^{2/3}:i+j∈n−P} E(t_{i}t_{j})

= ∑_{i>j≥n^{2/3}:i+j∈n−P} \frac{c^{2}}{ij}

= c^{2} ∑_{p≤n−2n^{2/3}} ∑_{i>j≥n^{2/3}:i+j=n−p} \frac{1}{ij}

= c^{2} ∑_{p≤n−2n^{2/3}} Ω(\frac{log(n−p)}{n−p})

= Ω(c^{2} logn)

where in the last line we used Proposition 1.54 from the appendix. Putting all of these estimates together we obtain

P(Y_{n} ≥ 0) ≤ exp(−Ω(clogn))

and the claim follows by choosing c to be suitably large.

Now we are going to prove Theorem 1.28.

Proof [of Theorem 1.28] We shall use the exponential moment method. By a limiting argument we may assume that P(t_{j} = 0),P(t_{j} = 1) > 0 for all j. We introduce the moment generating function F (t) := e^{−tX} for any t > 0. By (1.16) we have

P(X ≤ E(X)−T) ≤ \frac{F (t)}{e^{−t(E(X)−T)}} .

Taking logarithms, we see that it will suffice to establish the inequality

logF (t)+t(E(X)−T) ≤ −\frac{T^{2}}{2Δ}

for some t > 0. Unlike the situation in Theorem 1.8, the summands in X are not necessarily independent, so we cannot factorize F (t) = E(e^{−tX}) easily. Janson found a beautiful argument to get around this difficulty. Since F (0) = 0, we see from the fundamental theorem of calculus that

logF (t) = ∫_{0}^{t} \frac{F′(s)}{F (s)} ds.

Direct calculation shows that

F′(s) = −E(Xe^{−sX})

= − ∑_{A∈A}E(e^{−sX} ∏_{j∈A}t_{j})

= − ∑_{A∈A}E(e^{−sX}|E_{A})P(E_{A})

where E_{A} is the event that t_{j} = 1 for all j ∈ A. Thus it suffices to show that

∑_{A∈A}P(E_{A})∫_{0}^{t} \frac{E(e^{−sX}|E_{A})}{F (s)} ds−t(E(X)−T) ≥ \frac{T^{2}}{2Δ}

for some t > 0.

We now exploit the fact that some of the factors of e^{−sX} are independent of E_{A}. For each A ∈ A, we split X = Y_{A} + Z_{A}, are the boolean polynomials

Y_{A} := ∑_{B∈A:A∩B≠∅} ∏_{j∈B} t_{j}; Z_{A} = ∑_{B∈A:A∩B=∅} ∏_{j∈B} t_{j}.

By (1.34) (conditioning on the variables in E_{A}), we conclude

E(e^{−sX}|E_{A}) ≥ E(e^{−sY_{A}}|E_{A})E(e^{−sZ_{A}}|E_{A}).

On the other hand, Z_{A} is independent from E_{A} and is bounded from above by X, thus

E(e^{−sZ_{A}}|E_{A}) = E(e^{−sZ_{A}}) ≥ E(e^{−sX}) = F (s).

Combining all these estimates, we reduce to showing that

∑_{A∈A}P(E_{A})∫_{0}^{t} E(e^{−sY_{A}}|E_{A}) ds−t(E(X)−T) ≥ \frac{T^{2}}{2Δ}

for some t > 0.

Next, we exploit the convexity of the function x ↦ e^{−sx} via Jensen’s inequality (Exercise 4), concluding that

E(e^{−sY_{A}}|E_{A}) ≥ e^{−sE(Y_{A}|E_{A})}.

From linearity of expectation we have ∑_{A∈A} P(E_{A}) = E(X), and so another application of Jensen’s inequality gives

∑_{A∈A}P(E_{A})e^{−sE(Y_{A}|E_{A})} ≥ E(X)e^{−s ∑_{A∈A} \frac{P(E_{A})}{E(X)} E(Y_{A}|E_{A})}.

On the other hand, from Bayes formula we have

∑_{A∈A}P(E_{A})E(Y_{A}|E_{A}) = ∑_{A∈A} ∑_{B∈A:A∩B≠∅}E(I(E_{A}) ∏_{j∈B} t_{j}) = Δ.

We thus have

∑_{A∈A}P(E_{A})∫_{0}^{t} E(e^{−sY_{A}}|E_{A}) ds−t(E(X)−T)  (1.37)

≥ E(X)∫_{0}^{t} e^{−sΔ/E(X)} ds−t(E(X)−T)
= \frac{E(X)^{2}}{Δ} (1−e^{−tΔ/E(X)})−t(E(X)−T).  (1.38)

If we set t := T/Δ, then tΔ/E(X) = T/E(X) ≤ 1, and we have

1−e^{−tΔ/E(X)} = 1−e^{−T/E(X)}
≥ T/E(X)−T^{2}/2E(X)^{2}

and hence

∑_{A∈A}P(E_{A})∫_{0}^{t} E(e^{−sY_{A}}|E_{A}) ds−t(E(X)−T)

≥ \frac{T E(X)}{Δ} − \frac{T^{2}}{2Δ} − \frac{T}{Δ}(E(X)−T)

= \frac{T^{2}}{2Δ}

as desired.

Remark 1.32 t = T/Δ is a convenient choice, but might not be the optimal one. One can have a slightly better bound by optimizing the right hand side of (1.38) over t.

Remark 1.33 The proof of Janson’s inequality is not symmetric. In other words, it cannot be extended to give a bound for the upper tail probability P(X ≥ μ+T). This probability will be addressed in the next section.

1.6.1 By refining the argument, show that the complementary base B constructed in the proof of Theorem 1.31 has (with high probability) the property that r_{P +B+B}(n) = Ω(logn) for all sufficiently large n.

1.6.2 Define a random graph G(n,p) on the vertex set [1,n] as follows. For each pair i,j (1 ≤ i < j ≤ n) draw an edge between i and j with probability p, independently.

(a) Prove that if p = o(n^{−1}), then with probability 1 − o(1), G(n,p) does not contain a triangle.

(b) Assume that p = n^{−1+ϵ} for some small positive constant ϵ. Bound the probability that G does not contain a triangle.

1.6.3 Prove that for any k ≥ 2 there is a basis B of order k with with |B ∩[1,n]| = O(n^{1/2} log^{1/k} n) for all large n.

1.7 Concentration of polynomials

In previous sections, we often considered a polynomial quantity Y = Y (t_{1},...,t_{n}) of n independent random variables t_{1},...,t_{n}, and wished to control the distribution of these variables. For instance Chernoff’s inequality shows that the polynomial t_{1} +...+t_{n} is concentrated around its mean, while Janson’s inequality shows that certain boolean polynomials (especially those of low degree) are usually at least as large as the mean.

In this section, we present some further results of this type, that assert that certain polynomials with small degrees are strongly concentrated. These results can be seen as generalizing Chernoff’s bound, and also provide (in certain cases) the missing half (upper tail bound) of Janson’s inequality.

To motivate the results, let us first give a classical result which works for any function Y (not just a polynomial) provided that the Lipschitz constant of Y is small.

Lemma 1.34 (Lipschitz concentration inequality) Let Y : {0,1}^{n} → R be a function such that such that |Y (t)−Y (t′)| ≤ K whenever t,t′ ∈ {0,1}^{n} differ in only one co-ordinate. Then if t_{1},...,t_{n} are independent boolean variables, we have

P(|Y (t_{1},...,t_{n})−E(Y (t_{1},...,t_{n}))| ≥ λK\sqrt{n}) ≤ 2e^{−λ^{2}/2}

for all λ > 0.

Remark 1.35 This inequality asserts that if each t_{i} can only influence the random variable Y (t_{1},...,t_{n}) by O(K) at most, then Y (t_{1},...,t_{n}) itself is concentrated in an interval of length O(K\sqrt{n}) around its mean. It should be compared with Hoeffding’s inequality, which deals with the case Y (t_{1},...,t_{n}) := t_{1} +...+t_{n}, and also with Corollary 1.30.

Proof By dividing Y by K we may renormalize K = 1. Introduce the partially conditioned random variables Y_{0},Y_{1}(t_{1}),...,Y_{n}(t_{1},...,t_{n}) = Y (t_{1},...,t_{n}) by Y_{j}(t_{1},...,t_{j}) := E(Y |t_{1},...,t_{j}), thus Y_{j} is the conditional expectation of Y with the first j boolean variables t_{j} fixed. In particular Y_{0} = E(Y) and Y_{n} = Y (t_{1},...,t_{n}). We can thus write

Y (t_{1},...,t_{n})−E(Y (t_{1},...,t_{n})) = X_{1} +...+X_{n}

where X_{j} := Y_{j} − Y_{j−1}. One then easily verifies that |X_{j}| ≤ 1 and X_{1},...,X_{n} form a martingale difference sequence in the sense of Exercise 6. The claim then follows from Azuma’s inequality (1.24).

The above lemma is very useful when one has uniform Lipschitz control on Y, for instance if Y = Y (t_{1},...,t_{n}) is a polynomial for which the partial derivatives \frac{∂Y}{∂t_{i}} are small for all t_{1},...,t_{n} in the unit cube. However in many applications (especially to thin bases), these partial derivatives will only be small on the average. Fortunately there are analogues of the above lemma which apply in this case, though they also require some average control on higher derivatives of Y. To state the results we need some notation. Let Y = Y (t_{1},...,t_{n}) be a polynomial of n real variables. We say that Y is totally positive if all of its coefficients are non-negative, and furthermore that Y is regular if all the coefficients are between zero and one. We also say that Y is simplified if all of its monomials are square-free (i.e. do not contain any factor of t_{i}^{2}), and homogeneous if all the monomials have the same degree. Thus for instance a boolean polynomial is automatically regular and simplified, though not necessarily homogeneous. Given any multi-index α = (α_{1},...,α_{n}) ∈ Z_{+}^{n}, we define the partial derivative ∂^{α}Y as

∂^{α}Y := (\frac{∂}{∂t_{1}} )^{α_{1}} ...(\frac{∂}{∂t_{n}})^{α_{n}}Y (t_{1},...,t_{n}),

and denote the order of α as |α| := α_{1}+...+α_{n}. For any order d ≥ 0, we denote E_{d}(Y) := max_{α:|α|=d} E(∂^{α}Y), thus for instance E_{0}(Y) = E(Y), and E_{d}(Y) = 0 if d exceeds the degree of Y. These quantities are vaguely reminiscent of Sobolev norms for the random variable Y. We also define E_{≥d}(Y) := max_{d′≥d} E_{d′}(Y).

The following result is due to Kim and Vu [193].

Theorem 1.36 Let k ≥ 1, and let Y = Y (t_{1},...,t_{n}) be a totally positive polynomial of n independent boolean variables t_{1},...,t_{n}. Then there exists a constant C_{k} > 0 depending only on k such that

P(|Y −E(Y)| ≥ C_{k}λ^{k−1/2}\sqrt{E_{≥0}(Y)E_{≥1}(Y)}) = O_{k}(e^{−λ/4+(k−1) log n})

for all λ > 0.

Informally Theorem 1.36 asserts that when the derivatives of Y are smaller on average than Y itself, and the degree of Y is small, then Y is concentrated around its mean, and in fact we have Y = (1 + O_{k}(\sqrt{\frac{E_{≥1}(Y)}{E_{≥0}(Y)}} log^{k−1/2} n)E(Y) with high probability.

In applications in additive number theory, we frequently deal with the case when Y is roughly of size logn. In this case, the error term e^{(k−1) log n} renders Theorem 1.36 ineffective. We, however, have a variant which is designed to handle this case:

Theorem 1.37 [360] Let k,n ≥ 1 and β,γ,ϵ > 0. If Y = Y (t_{1},...,t_{n}) is a regular polynomial (not necessarily simplified) of n independent boolean variables t_{1},...,t_{n}, which is homogeneous of degree k and obeys the expectation bounds

Qlogn ≤ E(Y) ≤ n/Q; E_{1}(Y),...,E_{k−1}(Y) ≤ n^{−γ}

for some sufficiently large Q = Q(k,ϵ,α,β) (independent of n), then

P(|Y −E(Y)| ≥ ϵE(Y)) ≤ n^{−β}.

In the next section, we will use this theorem to prove Theorem 1.15. The next theorem deals with the case when the expectation of Y is

less than one. In this case it is convenient to remove the constant term from any derivative of Y which appears. More precisely, introduce the renormalized derivative ∂_{∗}^{α}Y (t) := ∂^{α}Y (t)−∂^{α}Y (0).

Theorem 1.38 Let Y = Y (t_{1},...,t_{n}) be a simplified regular polynomial of n independent boolean variables (not necessarily homogeneous) such that E(Y) < 1 and E(∂_{∗}^{α}Y) ≤ n^{−γ} for some γ > 0 and all non-zero α. Then for any β > 0 we have the bound P(Y ≥ K_{β,γ}) < n^{−β} for some K_{β,γ} which is independent of n and Y.

The proof of Theorem 1.36 relies on the so-called “divide and conquer martingale” technique, together with the exponential moment method. It is not too technical but requires lots of introduction. We thus skip it and refer the reader to [193]. The proof of Theorem 1.37 is more complicated. Beside the above mentioned martingale technique, it also requires some non-trivial combinatorial considerations. Theorem 1.38 is a by-product of this proof (for details see [360]). These theorems have a wide range of applications in several areas and we refer the reader to [359] for a survey.

1.7.1 B_{h}[g] sets Let us conclude this section by an application of Theorem 1.38. A set A ⊂ N is called a B_{h}[g] set if for any positive integer m, the equation m = x_{1} +···+x_{h}, x_{1} ≤ x_{2} ≤ ··· ≤ x_{h}, x_{i} ∈ A, has at most g solutions; up to a factor of h!, this is equivalent to requiring r_{h,A}(m) to be bounded by g for all m. B_{h}[g] sequences were studied by Erdös and Turán in [94]. From (1.21) we see that if A is a B_{h}[g] set, then |A∩[0,n]| = O_{h,g}(n^{1/h}) for all n. In the converse direction, Erdös and Turán proved

Theorem 1.39 For any h ≥ 1, there exists a set A ⊂ Z^{+} with |A ∩ [0,n]| = Ω_{h}(n^{1/h}) for all large n, which is a B_{h}[g] sequence for some g = g_{h} (or in other words, r_{h,A}(n) is uniformly bounded in n).

Proof By using Theorem 1.38 we can give a short proof of this theorem. As before, we construct A randomly, letting the events n ∈ A be independent with probability P(n ∈ A) = cn^{1/h−1} for some small constant 0 < c < 1. A simple application of Corollary 1.9 and the Borel-Cantelli lemma also gives |A ∩ [0,n]| = Ω_{c,h}(n^{1/h}) for all but finitely many n with probability 1. Thus it will suffice to show that A is a B_{h}[g] set with probability 1 (perhaps after removing finitely many elements), for some suitably large g = g_{h} depending only on g.

Let t_{n} denote the indicator variables t_{n} := I(n ∈ A). For each m, we observe that the random variable

Y_{m} = Y_{m}(t_{1},...,t_{m}) = ∑_{n_{1}≤...≤n_{h}:n_{1}+···+n_{h}=m}t_{n_{1}} ...t_{n_{h}}

will become a regular polynomial of degree h in the t_{1},...,t_{m} once we use the identity t_{i}^{a} = t_{i} for a = 2,3,... to make the monomials square-free. To show that A is a B_{h}[g] set after removing finitely many elements, it will suffice to show that Y_{m} ≤ g for all but finitely many m; by the Borel-Cantelli lemma, it will thus suffice to establish the upper tail estimate

P(Y_{m} > g) ≤ m^{−2}

for all large m. From linearity of expectation and independence we have

E(Y_{m}) = ∑_{n_{1}≤...≤n_{h}:n_{1}+···+n_{h}=m}c^{h}n_{1}^{1/h−1} ...n_{h}^{1/h−1}

≤ c^{h}O_{h}(m^{1/h−1} ∑_{n_{1},...,n_{h−1}≤m}n_{1}^{1/h−1} ...n_{h−1}^{1/h−1})

≤ c^{h}m^{1/h−1}O_{h}(( ∑_{n≤m}n^{1/h−1})^{h−1})

≤ O_{h}(c^{h}).

In particular P(Y_{m}) < 1 if c is small enough. This already gives some non-trivial bound on P(Y_{m} > g) from Markov’s inequality, but does not give the required decay in m. However, a similar computation to the above (which we leave as an exercise) establishes that P(∂_{∗}^{α}Y_{m}) = O_{h}(m^{−1/h}) for all non-zero α. The claim now follows from Theorem 1.38.

The study of B_{h}[g] sequences is a popular topic in additive combinatorics. A detailed discussion of this topic is beyond the scope of our book. Let us, however, mention one new result of Cilleruelo, Ruzsa and Trujillo from [61]. Many other recent results can be found in [61, 182, 203, 60, 139, 261].

Let A ⊂ [1,N] be a B_{h}[g] set. A simple counting argument (related to (1.21)) gives \binom{|A|+h−1}{h} ≤ ghN, which in turn yields the trivial bound |A| ≤ (ghh!N)^{1/h}. Cilleruelo, Ruzsa and Trujillo gave the first nontrivial bounds for the case g ≥ 2. They prove that |A| ≤ 1.864(gN)^{1/2}+1 when h = 2, and that

F_{h}(g,N) ≤ (1+cos^{h}(π/h))^{−1/h}(hh!gN)^{1/h}

when h > 2. The proofs made use of harmonic analysis methods via the consideration of the trigonometric polynomials f(t) = ∑_{a∈X} e^{iat}. The authors also constructed sets to establish for any g, the existence of a B_{2}[g] set A ⊂ [1,N] with

|A| ≥ (\frac{g +[g/2]}{\sqrt{g +2[g/2]}} +o_{g}(1))N^{1/2}.

1.7.1 Consider the random graph G(n,p) defined in Exercise 2, and set p := n^{−1+ϵ}. Let Y be the number of triangles in G(n,p). Give an upper bound and a lower bound for

P(Y ≥ \frac{3}{2}E(Y)).

1.7.2 Verify the bound P(∂_{∗}^{α}Y_{m}) = O_{h}(n^{−1/h}) claimed in Theorem 1.39.

1.8 Thin bases of higher order

We now return to the study of thin bases B and their associated counting functions r_{k,B}(n), initiated in Section 1.3. However, in this section we can use Theorem 1.37 to present a proof of Theorem 1.15, which asserted for each k ≥ 1 the existence of a base B of order k with r_{k,B}(n) = O_{k}(logn) for all large n. This was proven in the k = 2 case (see Theorem 1.13) using Chernoff’s inequality, but that method does not directly apply for higher k because r_{k,B}(n) cannot be easily expressed as the sum of independent random variables.

We begin with a simple lemma on boolean polynomials that shows that if E(X) is not too large, then at most points (t_{1},...,t_{n}) of the sample space, the polynomial X does not contain too many independent terms (cf. Exercise 12).

Lemma 1.40 Let X = ∑_{A∈A} ∏_{j∈A} t_{j} be a boolean polynomial of n independent boolean variables t_{1},...,t_{n}, let B ⊆ [1,n] be the random set B := {j ∈ [1,n] : t_{j} = 1}, and let D ∈ N be the random variable, defined as the largest number of disjoint sets in A which are contained in B. Then for any integer K ≥ 1 we have

P(D ≥ K) ≤ \frac{E(X)^{K}}{K!} .

Proof Observe that

I(D ≥ K) ≤ \frac{1}{K!} ∑_{A_{1},...,A_{K}∈A, disjoint} ∏_{j∈A_{1}} t_{j} ... ∏_{j∈A_{k}} t_{j}.

Taking expectations of both sides and using linearity of expectation (1.3) followed by independence, we conclude

P(D ≥ K) ≤ \frac{1}{K!} ∑_{A_{1},...,A_{K}∈A}E( ∏_{j∈A_{1}} t_{j})...E( ∏_{j∈A_{k}} t_{j}).

But by linearity of expectation again, the left-hand side is just E(X)^{K}/K!, and the claim follows.

This lemma is particularly useful when combined with the sunflower lemma of Erdös and Rado [92]. A collection of sets A_{1},...,A_{l} forms a sunflower if the pairwise intersections A_{i} ∩A_{j} for i = j are all the same (the A_{i} are called the petals of the flower).

Lemma 1.41 (Sunflower lemma) If B is a collection of sets, each of size at most k, and |B| > (l − 1)^{k}k!, then A contains l sets forming a sunflower.

This lemma can be proven by elementary combinatorics and is left as an exercise. It has the following consequence for the counting function r_{k,B}(n).

Corollary 1.42 Let B ⊂ Z^{+} and k ≥ 2, and for each n ∈ Z^{+} let D_{k,n} be the largest number of disjoint multisets {x_{1},...,x_{k}} of elements of B which sum to n. Then

r_{k,B}(n) ≤ k^{k} max(D_{k,n},( sup_{m<n}r_{k−1,B}(m)−1)^{k}k!).

Proof Fix n, and consider the collection B of sets which arise from taking the multisets {x_{1},...,x_{k}} of elements of B which sum to n and removing redundant elements. Clearly r_{k,B}(n) ≤ k^{k}|B|. Also observe that any sunflower in B has cardinality at most D_{k,n} (if the petals are disjoint) or sup_{m<n} r_{k−1,B}(m) (if the petals are not disjoint); the latter follows by taking one of the elements in the common intersection of the sunflower and removing it once from each of the associated multisets. The claim then follows from the sunflower lemma.

Using the above methods, we can now give a preliminary result towards proving Theorem 1.15.

Proposition 1.43 Let k ≥ 2, and let B ⊂ Z^{+} be a random subset of Z^{+}, defined by letting x ∈ B be independent with probability

P(x ∈ B) = min(Cx^{1/k−1} log^{1/k} x,1)

for some positive constant C > 1. Then with probability 1, we have sup_{n} r_{k′,B}(n) = O_{C,k,k′,B}(1) for all 1 ≤ k′ < k.

Proof We induct on d. The case d = 1 is obvious. Now suppose that 1 < k′ < k and the claim has already been proven for k′ −1. Applying Corollary 1.42, we conclude that with probability 1 that

r_{k′,B}(n) = O_{C,k,k′,B}(D_{k′,n} +1).  (1.39)

On the other hand, if we apply Lemma 1.40 with t_{x} := I(x ∈ B) for 1 ≤ x ≤ n, and A = A_{n} equal to all the sets which arise from the multisets {x_{1},...,x_{k′}} which sum to n, we observe that

P(D_{k′,n} ≥ K) ≤ \frac{E(∑_{A∈A_{n}} ∏_{j∈A} t_{j})^{K}}{K!}

for any K ∈ Z^{+}. On the other hand, from linearity of expectation (1.3) and independence we have

E( ∑_{A∈A_{n}} ∏_{j∈A}t_{j}) = ∑_{A∈A_{n}} ∏_{j∈A}min(Cj^{1/k−1} log^{1/k} j,1)

≤ O_{C,k,k′}( ∑_{j_{1}≤...≤j_{k′}:j_{1}+...+j_{k′} =n}j_{1}^{1/k−1} ...j_{k′}^{1/k−1})logn

≤ O_{C,k,k′}( ∑_{j_{1},...,j_{k′−1}∈[1,n]}j_{1}^{1/k−1} ...j_{k′−1}^{1/k−1})n^{1/k−1} logn

= O_{C,k,k′}(( ∑_{j∈[1,n]}j^{1/k−1})^{k′−1})n^{1/k−1} logn

= O_{C,k,k′}(n^{k′/k−1} logn).

Since k′ < k, we thus see that by choosing K depending on k sufficiently large (e.g. K = 2k +1) then we have

P(D_{k′,n} ≥ K) = O_{C,k,k′,K}(\frac{1}{n^{2}} ).

Applying the Borel-Cantelli lemma (Lemma 1.2) we see that with probability 1, we have D_{k′,n} < K for all but finitely many n. Combining this with (1.39) we obtain the claim.

Now we prove Theorem 1.15. It will suffice to show that

Proposition 1.44 Let k ≥ 2, and let B ⊂ Z^{+} be a random subset of Z^{+}, defined by letting x ∈ B be independent with probability

P(x ∈ B) = min(Cx^{1/k−1} log^{1/k} x,1)

for some positive constant C > 1. If C is sufficiently large depending on k, then with probability 1, we have r_{k,B}(n) = Θ_{C,k}(logn) for all but finitely many n. In particular, B is a thin basis of order k with probability 1.

Proof We shall estimate r_{k,B}(n) in terms of two related expressions,

R(n) := {(x_{1},...,x_{k}) ∈ B : x_{1} +...+x_{k} = n;n^{0.1} < x_{1} < x_{2} < ... < x_{k}} (1.40)
E(n) := {(x_{1},...,x_{k}) ∈ B : x_{1} +...+x_{k} = n;x_{1} = x_{2} or x_{1} ≤ n^{0.1}}. (1.41)

It is clear (using the symmetry of x_{1} +...+x_{k} under permutations) that

k!R(n) ≤ r_{k,B}(n) ≤ k!R(n)+k^{2}E(n).

We view R(n) as the main term and E(n) as the error term; this reflects the intuitive fact that for most representations n = x_{1} + ... + x_{k}, the x_{i} will be distinct and comparable in magnitude to n. It will suffice to show that with probability 1 we have

E(n) = O_{C,k,B}(1); R(n) = Θ_{C,k,B}(logn)

for all but finitely many n.

Let us deal first with the error term E(n). We argue as in the proof of Proposition 1.43. Let A_{n} denote those sets which arise from the multisets {x_{1},...,x_{k}} with x_{1} +...+x_{k} = n and either x_{1} = x_{2} or x_{1} ≤ n^{0.1}. By arguing as in Corollary 1.42, we have

E(n) ≤ k^{k} max(D_{n},( sup_{m<n}r_{k−1,B}(m)−1)^{k}k!)

where D_{n} is the largest number of disjoint sets that one can find in A_{n}. Applying Proposition 1.43, we conclude that

E(n) = O_{C,k,B}(D_{n} +1)

with probability 1. On the other hand, from Lemma 1.40 we for any K that

P(D_{n} ≥ K) ≤ \frac{E(∑_{A∈A_{n}} ∏_{j∈A} t_{j})^{K}}{K!} .

By arguing as in Proposition 1.43, one can establish

E( ∑_{A∈A_{n}} ∏_{j∈A}t_{j}) ≤ O_{k}(n^{−1/k}n^{0.1k} logn)

and thus for a suitably large constant K depending only on k

P(D_{n} ≥ K) = O_{k}(1/n^{2}).

From the Borel-Cantelli lemma we conclude that with probability 1, we have

E(n) = O_{C,k,B}(1)

for all but finitely many n, and so the contribution of E(n) is negligible.

Now we estimate the main term R(n). Observe that we can write R(n) as a homogeneous boolean polynomial Y = Y (t_{1},...,t_{n}) of degree k; more explicitly, we have

Y (t_{1},...,t_{n}) = ∑_{A∈A′_{n}} ∏_{j∈A}t_{j}

where A′_{n} is the collection of all sets {x_{1},...,x_{k}} where x_{1} +...+x_{k} = n and n^{0.1} < x_{1} < x_{2} < ... < x_{k}. Repeating the computations in Proposition 1.43 we see that

E(Y) = Θ_{k}(C logn)

when n is sufficiently large depending on C,k. To conclude the proof it would thus suffice by the Borel-Cantelli lemma to establish the large deviation inequality

P(|Y −E(Y)| > \frac{1}{2}E(Y)) = O_{C,k}(\frac{1}{n^{2}} )

for all large n. Applying Theorem 1.37 (and choosing C sufficiently large), we see that it suffices to show the derivative estimates

E_{1}(R(n)),...,E_{k−1}(Y) ≤ n^{−γ}

for all large n and some γ > 0. In other words, we need to establish

E((\frac{∂}{∂t_{1}} )^{α_{1}} ...(\frac{∂}{∂t_{n}})^{α_{n}}Y (t_{1},...,t_{n})) ≤ n^{−γ}

whenever n is large and 1 ≤ α_{1} +...+α_{n} ≤ k −1. From the definition of A′_{n} we see that we may take α_{j} = 0 for all j ≤ n^{0.1}, and all the other α_{j} equal to zero or one, since the above partial derivative vanishes otherwise. One can then compute the partial derivative and reduce to showing that

E( ∑_{A∈A′_{n}:A⊃A_{0}} ∏_{j∈A\A_{0}} t_{j}) ≤ n^{−γ}

whenever A_{0} is any subset of [n^{0.1},n] of cardinality 1 ≤ |A_{0}| ≤ k−1 (this is the set of indices where α_{j} = 1). Applying linearity of expectation and independence, and noting that j ∈ [n^{0.1},n] for all j ∈ A\A_{0}, we conclude that

E( ∑_{A∈A′_{n}:A⊃A_{0}} ∏_{j∈A\A_{0}} t_{j}) ≤ ∑_{A∈CA′_{n}:A⊃A_{0}} O_{C,k}(n^{1/k−1} log^{1/k} n)^{k−|A_{0}|}
≤ O_{k}(n^{k−|A_{0}|−1})O_{C,k}(n^{1/k−1} log^{1/k} n)^{k−|A_{0}|}
≤ O_{C,k}(n^{−1/k} logn)

and the claim follows for large n.

Remark 1.45 The proof above is from [360] and is based on the proof of Theorem 1.48 in [361]. The original proof in [94] was different and did not use Theorem 1.37.

1.8.1 Let A ∈ Z^{+} be a set of n different integers. Prove that A contains a subset B of cardinality Ω(logn) with the following property. No two elements of B add up to an element of A (thus r_{2,B}(m) vanishes for all m ∈ A, or equivalently A∩2B = ∅).

1.8.2 Prove Lemma 1.41. (Hint: first use the pigeonhole principle to show that if |B| > (l−1)k, then either B contains l disjoint sets, or that there exist at least |B|/(l −1)k sets in B which all have a common element x_{0}. Then use induction on k.)

1.9 Thin Waring bases

Recall that a thin basis of order k is a set B ⊂ N such that r_{k,B}(n) = O(logn) for all large n. Theorem 1.15, proved above, asserts that ℕ contains a thin basis of any order. Given the abundance of classical bases such as the squares and primes, it is then natural to pose the following question:

Question 1.46 Let A be any fixed basis of order k. Does A contain a thin subbasis B?

Note that Sidon’s original question can be viewed as the k = 2, A = N case of this question. From (1.21) we know that a thin basis B enjoys the bounds

|B ∩[0,N]| = Ω_{k}(N^{1/k}); |B ∩[0,N]| = O_{k}(N^{1/k} log^{1/k} N)

for all large N. Thus we can consider the following weaker version of Question 1.46:

Question 1.47 Let A be any fixed basis of order k. Does A contain a subbasis B with |B ∩[0,N]| = O_{k}(N^{1/k} log^{1/k} N) for all large N?

Question 1.47 has been investigated intensively for the Waring bases N^{∧}r = {0^{r},1^{r},2^{r},...}, especially when r = 2 [87, 55, 369, 370, 366, 318]. For these bases it is known that if k is sufficiently large depending on k, then N^{∧}r is a basis of order k, and furthermore that

r_{k,N^{∧}r}(n) = Θ_{k,r}(n^{\frac{k}{r} −1})  (1.42)

(note that this is consistent with (1.21)).

Choi, Erdös and Nathanson proved in [55] that N^{∧}2, the set of squares, contains a subbasis B of order 4, with |B ∩[0,N]| = O_{ε}(N^{1/3} +ε) for all N > 1 and all ε > 0. This was generalized by Zöllner [369, 370], who showed that for any k ≥ 4 there was a subbasis B ⊂ N^{∧}2 of order k with |B ∩[0,N]| = O_{k,ε}(N^{1/k+ε}) for any ε > 0 and N > 1. This bound was then sharpened further to |B ∩[0,N]| = O_{k}(N^{1/k} log^{1/k} N); from (1.21) we know that this is sharp except for the logarithmic factor. A short proof of Wirsing’s result for the case k = 4 was given by Spencer in [318]. For r ≥ 3, much less was known. In 1980, Nathanson [248] proved that N^{∧}r contains a subbasis of some order with density o(N^{1/r}). In the same paper, he posed a special case of Question 1.47, when A = N^{∧}r.

In a recent paper, Vu [361] positively answered Question 1.46 (and hence Question 1.47) for the case A = N^{∧}r for any r ≥ 1:

Theorem 1.48 For any fixed r there is an integer k_{0} such that the following holds. For any k ≥ k_{0}, the set N^{∧}r of all rth powers contains a thin basis B of order k. In particular, from (1.21) we have |B∩[0,n)| = O_{k}(N^{1/k} log^{1/k} N) for all large N.

Remark 1.49 The sharp concentration result in Theorem 1.37 was first developed in order to prove Theorem 1.48.

Just as Theorem 1.15 followed from Proposition 1.44, Theorem 1.48 is an immediate consequence of

Proposition 1.50 Let k,r ≥ 2, and let B be a random subset of (Z^{+})^{∧}r, defined by letting x^{r} ∈ B be independent with probability

P(x^{r} ∈ B) = min(Cx^{\frac{r}{k} −1} log^{1/k} x,1)

for some positive constant C > 1. If k is sufficiently large depending on r, and C is sufficiently large depending on k,r, then with probability 1 we have r_{k,B}(n) = Θ_{C,k,r,B}(logn) for all but finitely many n. In particular, B is a thin basis of order k with probability 1.

Proof (Sketch) As in the proof of Proposition 1.44, it suffices to show that with probability 1 we have

E(n) = O_{C,k,r,B}(1); R(n) = Θ_{C,k,r,B}(logn)

for all but finitely many n, where R(n) and E(n) were defined in (1.40), (1.41). The contribution of E(n) can be dealt with by similar arguments to the previous section and is left as an exercise, so we focus on R(n). As before we can write R(n) as a boolean polynomial Y_{n} = Y_{n}(t_{1},...,t_{m}), where m = ⌊n^{1/k}⌋, t_{x} = I(x^{r} ∈ B), and

Y_{n} = ∑_{A∈A_{n}} ∏_{x∈A}t_{x}

where A_{n} is the collection of sets {x_{1},...,x_{k}} of positive integers with x_{1}^{r} + ... + x_{k}^{r} = n and n^{0.1} < x_{1}^{r} < ... < x_{k}^{r}. Given the framework presented in the last section, the substantial difficulty remaining is to estimate the expectations of Y_{n} and its partial derivatives. In the following, we shall focus on the expectation of Y_{n}, establishing in particular that

E(Y_{n}) = Θ_{k,r}(C^{k} logn).

This is the main estimate, and the remainder of the argument proceeds as in Proposition 1.44. Notice that

E(Y_{n}) = C^{k} ∑_{x_{1}<...<x_{k}:\{x_{1},...,x_{k}\}∈A_{n}} ∏_{j=1}^{k}x_{j}^{\frac{r}{k} −1} log^{1/k} x_{j};

since all the x_{j} range between n^{1/10r} and n^{1/r}, it thus suffices to show that

∑_{x_{1}<...<x_{k}:\{x_{1},...,x_{k}\}∈A_{n}} x_{1}^{\frac{r}{k} −1} ...x_{k}^{\frac{r}{k} −1} = Θ_{r,k}(1).  (1.43)

This bound implies, but is a little bit stronger than, the standard bound (1.42), as the estimate also asserts some improved bound on the counting function r_{k,N^{∧}r}(n) when one or more of the summands are restricted to be small (so that the corresponding weight x^{\frac{r}{k} −1} is large.

The proof of (1.43) is a standard but lengthy application of the Hardy-Littlewood circle method, and is beyond the scope of this book. The reader may consult [361] for the full proof.

Wooley [364] shown that one can set k_{0} = O(r logr). This is (up to a constant factor) also the best current bound for k in (1.42). His proof also relies on Theorem 1.37, but the number theoretic part is different.

1.9.1 In the proof of Proposition 1.50, verify that with probability 1 one has E(n) = O_{C,k,r,B}(1) for all but finitely many n.

1.10 Appendix: the distribution of the primes

Several results in this chapter relied on facts concerning the distribution of the primes

P = {2,3,5,...}.

The distribution of this set is of course a very well-studied subject in analytic number theory, with one of the fundamental results being the prime number theorem

|P ∩[1,n]| = (1+o(1))\frac{n}{logn}.  (1.44)

An equivalent formulation is that if p_{k} denotes the k^{th} prime, then p_{k} = (1+o(1))k logk. The famous Riemann hypothesis, which is still unsolved, is equivalent to the stronger statement that

|P ∩[1,n]| = \frac{n}{logn} +O_{ε}(n^{1/2+ε})  (1.45)

for any ε > 0, or equivalently that p_{k} = k logk + O_{ε}(k^{1/2+ε}) for any ε > 0.

The prime number theorem is rather deep and will not be proven here. In this appendix we present some related results, most of which have surprisingly elementary and beautiful proofs. As they are numbertheoretical rather than probabilistic in nature we have chosen to place these results in an appendix to this chapter.

We begin with some classical estimates of Chebyshev and Mertens). As is customary, when summing over a variable p, p is understood to denote a prime.

Proposition 1.51 (Elementary prime number estimates) Let n ≥ 1 be an integer. Then we have the estimates

∑_{p≤n}logp = O(n)  (1.46)
∑_{p≤n} \frac{logp}{p} = logn+O(1)  (1.47)
∑_{p≤n} \frac{1}{p} = loglogn+O(1).  (1.48)

Remark 1.52 With the prime number theorem, we can improve (1.46) to ∑_{p≤n} logp = (1 + o(1))n, but it is not necessary to do so for our applications here.

Proof We first prove (1.46). Without loss of generality we may take n to be a power of two. Consider the binomial \binom{2n}{n}. From Pascal’s formula we know that \binom{2n}{n} ≤ 4^{n}. On the other hand, it is clear that every prime between n and 2n will divide \binom{2n}{n}. Thus

∏_{n<p≤2n}p ≤ 4^{n}.

Taking logarithms we conclude

∑_{n<p≤2n}logp = O(n).

Applying this bound to n/2, n/4, and so forth, and then summing the geometric series, the claim (1.46) follows.

Now we prove (1.47). This is a similar argument but based around the factorial n! instead of \binom{2n}{n}. Observe that the only primes dividing n! are those less than or equal to n. For each prime p ≤ n, there are ⌊n/p⌋ numbers (between 1 and n) divisible by p, ⌊n/p^{2}⌋ numbers (between 1 and n) divisible by p^{2} and so on. Thus

n! = ∏_{p≤n}p^{⌊n/p⌋+⌊n/p^{2}⌋+...}.  (1.49)

Taking the logarithm of both sides and applying Stirling’s formula (Exercise 1) we obtain

nlogn+O(n) = ∑_{p≤n}(⌊n/p⌋+⌊n/p^{2}⌋+...)logp.

Since

⌊n/p⌋+⌊n/p^{2}⌋+··· = \frac{n}{p} +O(1)+O(\frac{n}{p^{2}} ),

we conclude after some rearranging

∑_{p≤n}n\frac{logp}{p} = nlogn+O(n)+ ∑_{p≤n}O(logp)+ ∑_{p≤n}O(\frac{nlogp}{p^{2}}).

Since ∑_{k} \frac{log k}{k^{2}} is convergent, the last term is O(n). The claim now follows from (1.46).

We shall deduce (1.48) from (1.47) using Abel’s summation technique, rewriting one partial sum over primes as an average of others. Observe from the fundamental theorem of calculus that

\frac{1}{p} = \frac{logp}{p} \frac{1}{logp}

= \frac{logp}{p} ∫_{1}^{∞} I(t > p)\frac{dt}{tlog^{2} t}

and hence

∑_{p≤n} \frac{1}{p} = ∑_{p≤n} \frac{logp}{p} ∫_{1}^{∞} I(t > p)\frac{dt}{tlog^{2} t}.

Swapping the sum and integral, we obtain

∑_{p≤n} \frac{1}{p} = ∫_{1}^{∞}(∑_{p≤t} \frac{logp}{p})\frac{dt}{tlog^{2} t}.

Applying (1.47), we obtain

∑_{p≤n} \frac{1}{p} = ∫_{1}^{∞}(logt+O(1))\frac{dt}{tlog^{2} t}.

Since logt\frac{t}{log^{2} t} is the antiderivative of loglogt, and \frac{1}{t log^{2} t} is absolutely convergent, the claim follows.

We now turn to a deeper fact concerning the distribution of primes in intervals.

Theorem 1.53 For all sufficiently large n, we have |P ∩ [n − x,n)| = Θ(\frac{x}{log n}) for all n^{2/3} < x < n.

Results of this type first appeared by Hoheisel [174]; the result as claimed is due to Ingham [179]. Note that this theorem follows immediately from the Riemann hypothesis (1.45). However, this theorem can be proven without using the Riemann hypothesis, instead using some weaker (but still very non-trivial) facts on the distribution of zeroes of the Riemann zeta function: see [163]. We remark that if one only seeks the upper bound on |P ∩[n−x,n)| then one can use relatively elementary sieve theory methods to establish the claim. The constant 2/3 has been lowered (for instance to 7/12, see [178], [170]). However, for the applications here, any exponent less than 1 will suffice.

We now combine this theorem with the Abel summation method to establish some further estimates on sums involving primes.

Proposition 1.54 Let n be a large integer. Then we have the estimates

∑_{p∈P ∩[1,n−n^{2/3})} \frac{1}{n−p} = Θ(1)  (1.50)

∑_{p∈P ∩[1,n−n^{2/3})} \frac{log(n−p)}{n−p} = Θ(logn).  (1.51)

Proof We begin by proving (1.50). From the fundamental theorem of calculus we have

\frac{1}{n−p} = ∫_{1}^{∞} 1_{p∈[n−x,n−n^{2/3})} \frac{1}{x^{2}} dx

for all p ∈ P ∩[1,n−n^{2/3}), and hence

∑_{p∈P ∩[1,n−n^{2/3})} \frac{1}{n−p} = ∫_{1}^{∞} |P ∩[n−x,n−n^{2/3})| \frac{dx}{x^{2}}.

The integrand vanishes when x ≤ n^{2/3}. When n^{2/3} < x ≤ 2n^{2/3}, Theorem 1.53 shows that the integrand is O(\frac{1}{n^{2/3} log n}, while for x ≥ n^{2/3} another application of Theorem 1.53 shows that the integrand is Θ(\frac{1}{x log n}) when x ≤ n and Θ(\frac{n}{x^{2} log n}) when x > n. Putting all these estimates together we obtain (1.50). The estimate (1.51) then follows immediately from (1.50) since log(n − p) = Θ(logn) when p ∈ [1,n − n^{2/3}].

1.10.1 By approximating the sum ∑_{m=1}^{n} logm by the integral ∫_{1}^{n} logx dx, prove Stirling’s formula

logn! = nlogn−n+O(logn)  (1.52)

for all n > 1.

1.10.2 Using Proposition 1.51, show that there is a constant c so that there is always a prime between n and cn for every positive integer n.

1.10.3 By being more careful in the proof of (1.46), show that

∑_{p<n}logp ≤ 2nlog2+O(n^{1/2})

and

∑_{n≤p<2n}logp+ ∑_{p≤2n/3}logp ≥ 2nlog2−O(n^{1/2}),

and conclude Bertrand’s postulate, namely that for every sufficiently large integer n there exists a prime between n and 2n. (This argument is due to Ramanujan. Bertrand’s postulate in fact holds for all integers n, as the case of small n can be verified directly.)

1.10.4 Without using the prime number theorem, prove that |P ∩[1,n]| = Θ(\frac{n}{log n}); this is known as Chebyschev’s theorem. This theorem is of course superceded by the prime number theorem π(n) = (1+o(1))\frac{n}{log n}, but has the advantage of having a short elementary proof.

1.10.5 Prove that p_{k} = Θ(k logk), where p_{k} denotes the k^{th} prime. Again, this is superceded by the prime number theorem p_{k} = (1+o(1))k logk.

1.10.6 Define the von Mangoldt function Λ : Z^{+} → R by setting Λ(n) := logp if n > 1 is a power of a prime p, and Λ(n) = 0 otherwise. Show that

∑_{d|n}Λ(d) = logn  (1.53)

for all integers n ≥ 1. Use this to prove that

(∑_{n=1}^{∞} \frac{Λ(n)}{n^{s}})(∑_{n=1}^{∞} \frac{1}{n^{s}}) = ∑_{n=1}^{∞} \frac{logn}{n^{s}}

for all real numbers s > 1. Also, use (1.53) to give an alternate proof of (1.49).

1.10.7 Using the preceding exercise, show that

∑_{n=1}^{∞} \frac{logp}{p^{s}} = \frac{1}{s−1} +O(1)

for all s > 1; integrate this to conclude

∑_{n=1}^{∞} \frac{1}{p^{s}} = log \frac{1}{s−1} +O(1)  (1.54)

for all s > 1. Show that these estimates can also be deduced from Proposition 1.51 via Abel’s method. Conversely, use (1.54) and (1.46) to give an alternate proof of (1.48).

1.10.8 Using Abel’s summation method, show that the prime number theorem π(x) = (1+o(1))\frac{x}{log x} is equivalent to the estimate ∑_{n≤x} Λ(n) = (1+o(1))x.

1.10.9 By being more careful in the proof of (1.48), show that

∑_{p<n} \frac{1}{p} = loglogn+C +O(\frac{1}{logn})

for some absolute constant C. Use this to deduce Merten’s theorem

∏_{p<n}(1− \frac{1}{p}) = (1+o(1))\frac{C′}{logn}  (1.55)

for some other absolute constant C′ and all n > 1. (In fact one has C′ = e^{−γ}, where γ = 0.577... is Euler’s constant.)

2

Sum set estimates

Many classical problems in additive number theory revolve around the study of these sum sets for specific sets A, B (though one typically works with infinite sets rather than finite ones). For instance, if N^{∧}2 := {0,1,4,9,16,...} is the set of square numbers, then it is a famous theorem of Lagrange that 4N^{∧}2 = N, i.e. every natural number is the sum of four squares; if P := {2,3,5,7,11,...} is the set of prime numbers, then it is a famous theorem of Vinogradov that (2·N+1)\3P is finite (i.e. every sufficiently large odd number is the sum of three primes); in fact it is conjectured that this set consists only of 1, 3, and 5. The corresponding result for (2 · N)\2P remains open; the infamous Goldbach conjecture asserts that 2P contains every even integer greater than 2, but this conjecture remains far from resolution.

In this text, we shall not focus on these types of problems, which rely heavily on the specific number-theoretic structure of the sets involved. Instead, we shall focus instead on the analysis of sumsets A + B and related objects for more general sets A,B. To simplify the discussion we shall focus primarily on additive sets A,B, which are finite and nonempty subsets of an additive group such as Z; thus our theory will not cover infinite sets such as the squares N^{∧}2 or the primes P directly, although one can certainly use this theory to analyze those sets simply by considering finite truncations, say to an interval [0,N].

A fundamental problem in this field is the inverse sumset problem: if A + B or A − B is small, what can one say about A and B? A more specific question is as follows: if A is a finite non-empty subset of integers such that |A+A| = K|A| for some small number K, what can one say about A? Here and in the rest of the text we use |A| to denote the cardinality of a finite set A. The number K := |A+A|/|A| is referred to as the doubling constant of A. It is easy to see that this constant is at least 1, but it can be much larger; for instance, if A is a geometric progression such as A = 2^{∧}[0,N) = {1,2,2^{2},...,2^{N−1}} then one can easily verify that K = (N +1)/2, so the doubling constant can be arbitrarily large; indeed for “generic” sets A we will have K = (|A|+1)/2.

At the other extreme, if A is an arithmetic progression A := a+[0,N)· r = {a,a + r,a + 2r,...,a + (N − 1)r} of length N then one can check that A has doubling constant K = 2− \frac{1}{N}. Thus arithmetic progressions are examples of sets with small doubling constant. One can perturb this example to produce a number of other examples of sets with small doubling constant; for instance if A is the above arithmetic progression, and we let A′ be a subset of A of cardinality N/2 (say), then one can easily check that A′ has doubling constant at most 4. Another example is if one adds an arbitrary integer n to A, then the set A∪{n} also has doubling constant at most 4.

One can generalize the concept of an arithmetic progression, to create more sets with small doubling constant. Consider the set

A := a+[0,(N_{1},N_{2}))·(v_{1},v_{2}) = {a+n_{1}v_{1}+n_{2}v_{2} : 0 ≤ n_{1} < N_{1};0 ≤ n_{2} < N_{2}},

where a,v_{1},v_{2} are integers, N_{1},N_{2} are positive integers and n_{1},n_{2} are understood to lie in the integers; this is an example of a generalized arithmetic progression of rank 2. One can verify that such sets have a doubling constant of at most 4. Note that such sets can look quite different from an ordinary arithmetic progression if N_{1},N_{2} are large and v_{1},v_{2} are very widely separated.

We have just remarked that generalized arithmetic progressions have small doubling constant. One of the fundamental theorems in this subject is Freiman’s theorem, which asserts a partial converse to this claim; indeed, it asserts that any finite subset of the integers with small doubling constant can be efficiently contained in a generalized arithmetic progression (of bounded rank). This theorem is very useful, but is rather deep, and we will defer its proof to Section 5.4. It also has the drawback that some of the constants in this theorem depend exponentially on the doubling constant K. As such, it tends to only be useful in contexts where the doubling constant K is of the order of log|A| or smaller.

Roughly speaking, one can classify results in inverse sum set theory by the range of K for which the results are non-trivial. The case K = 1 is group theory (see Proposition 2.7). When K is very small, e.g. K < 2 or K < 3, we have a complete characterization of the inverse problem, characterizing A in terms of groups and arithmetic progressions. When K = O(log|A|), the best result is Freiman’s theorem, which characterizes A in terms of generalized arithmetic progressions. When K = O(|A|^{ε}) for some small ε, we have Proposition 2.25 (as well as many of the other results in this chapter), which characterizes A in terms of approximate groups. In the remaining cases |A|^{ε} ≪ K ≤ |A|, some of the estimates here are still useful, but our understanding is still quite poor.

We will not prove Freiman’s theorem in this chapter. However, we shall develop the more elementary theory of sum set estimates, which can be used as substitutes for Freiman’s theorem in some cases and are also of interest in their own right; this theory will also be needed in the proof of Freiman’s theorem later on. These estimates are obtained by very simple combinatorial considerations, and rely on simple arithmetic facts such as a−c = (a−b)+(b−c) and a+b = a′+b′ ⟺ a−b′ = a′−b. Because of the simplicity of the techniques used here, the results in this section are quite general, being applicable to any additive group and even to a large extent to non-abelian groups (see Section 2.7); we will wait until Chapter 5 until developing sum set estimates which exploit the specific structure of the ambient group. Also, the bounds obtained here are fairly reasonable, for instance the dependence of constants on the doubling constant K is only polynomial in all the results in this section (in contrast to the exponential dependence on K in Freiman’s theorem). In some cases, though, the results in this section will be superceded by more precise results proven using advanced techniques, which we will address in later sections; for instance, in Section 6.5 we shall develop the theory of Plünnecke inequalities, which give more precise control on iterated sum sets and also handle the case when A and B have very different sizes, a case which is not treated efficiently by the tools in this section.

There are a large number of results in this chapter, but we point out a couple of specific results proven here which have a very large number of applications. The first is Ruzsa’s triangle inequality, Lemma 2.6, which allows us to define a metric on the space of additive sets which measures how small their sumsets are. Then there is Proposition 2.11, which links the size of |A+B| and |A−B| for arbitrary additive sets A,B. This generalizes to the iterated sum set estimates in Corollary 2.22 and Corollary 2.23. Another very useful class of tools are the covering lemmas - Ruzsa’s covering lemma (Lemma 2.13), Green-Ruzsa’s covering lemma (Lemma 2.16), and Chang’s covering lemma (Lemma 5.31), which gives conditions under which one set A can be efficiently covered by translates of another B. These results are collected together in Proposition 2.25 and Proposition 2.26, which characterize sets with small sum set in terms of approximate groups. Last, but certainly not least, there is the Balog-Szemerédi-Gowers theorem, which generalizes the previous results to the setting when one has only partial information on a sum set (or equivalently, one only controls the “additive energy” between two sets); see Theorem 2.28 and Theorem 2.30. We also develop an asymmetric version of this theorem in Section 2.6.

2.1 Sum sets

We now systematically study the sum sets A + B and difference sets A − B of two additive sets A,B in an ambient group Z as defined in Definition 0.1, as well as the iterated sum sets nA. We should caution the reader that the iterated sum set nA is in general not the same as the dilate n · A := {n · a : a ∈ A} though we do have the inclusion n · A ⊆ nA. Similarly the difference set A − B should not be confused with the set-theoretic difference A\B := {x ∈ A : x ∉ B}. We also write A+x = A+{x} for the translate of A by an element x ∈ Z.

Since addition of group elements is associative and commutative, one can easily verify the same is true for addition of sets. We should caution however that the sum set operation is not invertible: for instance, A+ B −B contains A but is generally not equal to A. Similarly, when n and m have opposing signs, then nA + mA will contain (n + m)A but will generally be larger.

A very fundamental question in this topic is the following: under what conditions is A + B “small”, and under what conditions is it “large”? More precisely, we will be interested in the cardinality |A + B| of the sum set A+B. We have the following trivial estimates:

Lemma 2.1 (Trivial sum set estimates) Let A,B be additive sets with common ambient group Z, and let x ∈ Z. Then we have the identities |A+x| = |−A| = |A|, the inequalities

max(|A|,|B|) ≤ |A+B|,|A−B| ≤ |A||B|  (2.1)

and the inequalities

|A| ≤ |A+A| ≤ \frac{|A|(|A|+1)}{2}.  (2.2)

More generally, for any integer n ≥ 1, we have |(n+1)A| ≥ |nA| and

|nA| ≤ \binom{|A|+n−1}{n} = \frac{|A|(|A|+1)...(|A|+n−1)}{n!}.  (2.3)

We remark that the lower bound in (2.1) can be improved for specific groups Z, or when A and B have large “dimension”; see Theorem 3.16, Lemma 5.3, Theorem 5.17, Corollary 5.13, Theorem 5.4.

Proof We shall just prove (2.3), as all the other inequalities either follow from this inequality or are trivial. We argue by induction on |A|. If |A| = 1 then both sides of (2.3) are equal to 1. If |A| > 1, then we can write A = B ∪{x} where B is a non-empty set with |B| = |A|−1. Then

nA = ⋃_{j=0}^{n}(jB +(n−j)·x)

and hence by the induction hypothesis and Pascal’s triangle identity

|nA| ≤ ∑_{j=0}^{n}|jB| ≤ ∑_{j=0}^{n} \binom{|A|−1+j −1}{j} = \binom{|A|+n−1}{n}

as claimed.

Observe from the above facts that the magnitude of sum sets such as A+B, A−B, kA are unaffected if one translates A or B by an arbitrary amount. This gives much of the theory of sum sets a “translationinvariant” or “affine” flavor. We will sometimes take advantage of this translation invariance to normalize one of the sets, for instance to contain the origin.

For “generic” additive sets A and B, the cardinalities of the sum sets considered in Lemma 2.1 are much more likely to closer to the upper bounds listed above than the lower bounds, see for instance Exercise 1. This suggests that the lower bounds are only attainable, or close to being attainable, when the sets A and B have a considerable amount of structure; we shall develop this theme in the remainder of this chapter, by introducing tools such as doubling constants, Ruzsa distance and diameter, additive energy, and K-approximate groups to quantify some of these notions of “structure”. For now, we at least settle the question of when the lower bound in (2.1) is attained.

Proposition 2.2 (Exact inverse sumset theorem) Suppose that A,B are additive sets with common ambient group Z. Then the following are equivalent.

• |A+B| = |A|.
• |A−B| = |A|.

• |A + nB − mB| = |A| for at least one pair of integers (n,m) ≠ (0,0).

• |A+nB −mB| = |A| for all integers n,m.

• Then there exists a finite subgroup G of Z such that B is contained in a coset of G, and A is the union of cosets of G.

Proof We shall just show that the first claim implies the fifth; the remaining claims are either similar or easy and are left to the exercises. By translating B if necessary we may assume that B contains 0. Then A+B ⊃ {0}+A = A, but since |A+B| = |A| we thus have A+B = A. In particular A+b = A for all b ∈ B. Thus if we define the symmetry group Sym_{1}(A) (also known as the period of A) to be the set Sym_{1}(A) := {h ∈ Z : A + h = A}, then we have B ⊆ Sym_{1}(A). We leave as an exercise to the reader the verification that Sym_{1}(A) is a finite group, and A is the union of cosets of Sym_{1}(A); the claim then follows by setting G := Sym_{1}(A).

We shall study the symmetry group Sym_{1}(A), as well as the more general symmetry sets Sym_{α}(A), more systematically in Section 2.6.

As for when the upper bound is attained, we do not have as explicit a description, but we can give a number of equivalent formulations of the condition.

Proposition 2.3 Suppose that A,B are additive sets with common ambient group Z. Then the following are equivalent.

• |A+B| = |A||B|.

• |A−B| = |A||B|.

• |{(a,a′,b,b′) ∈ A×A×B ×B : a+b = a′ +b′}| = |A||B|.

• |{(a,a′,b,b′) ∈ A×A×B ×B : a−b = a′ −b′}| = |A||B|.

• |A∩(x−B)| = 1 for all x ∈ A+B.

• |A∩(B +y)| = 1 for all y ∈ A−B.

• (A−A)∩(B −B) = {0}.

We leave the easy proof of this Proposition to the exercises. For a partial generalization of this proposition, see Corollary 2.10 below.

In Proposition 2.2 and Proposition 2.3, the sets A + B and A − B have the same size (see also Exercise 6). However, this is not true in general. A basic example is the set A = {0,1,3} ⊂ Z, then A + A = {0,1,2,3,4,6} has six elements and A − A = {−3,−2,−1,0,1,2,3} has seven elements. More generally, if A = {0,1,3}^{d} ⊂ Z, then A + A has 6^{d} elements and A − A has 7^{d}. Thus A − A can be larger than A + A by an arbitrarily large amount. In the converse direction, the set A := {(0,0),(1,0),(2,0),(3,1),(4,0),(5,1),(6,1),(7,0),(8,1),(9,1)} ∈ Z_{10} ×Z_{2} is such that A + A = Z_{10} × Z_{2} has 20 elements, but A − A = Z_{10} × Z_{2}\{(0,1)} has only 19 elements; one can raise this example to the power d as before. Despite these examples, however, there are still several relationships between the size of |A+A| and |A−A|; see in particular (2.12) below.

2.1.1 Let N,M ≥ 1 be integers, and let A and B to be sets of cardinality N and M respectively chosen uniformly at random from the real interval {x ∈ R : 0 ≤ x ≤ 1}. Show that with probability 1 we have |A+B| = |A||B| and |nA| = \binom{|A|+n−1}{n} for all n ≥ 1.

2.1.2 Prove the remaining claims in Proposition 2.2. 2.1.3 Let A be an additive set. Show that A is a group if and only if

2A = A. 2.1.4 Prove (2.3). 2.1.5 [276] Find an additive set A of integers such that |A − A| <

|A+A|. (Hint: there are several ways to proceed. One way is to tile the lattice Z^{2} with the Z_{10} ×Z_{2} example given above, and somehow truncate and then project this back to Z).

2.1.6 Let A,B be additive sets in a finite additive group Z, such that |A| + |B| > |Z|. Prove that A + B = A − B = Z. Give an example to show that the condition |A| + |B| > |Z| cannot be improved.

2.1.7 Show that for any additive set A, the symmetry group Sym_{1}(A) of A as defined in the proof of Proposition 2.2 is a finite group contained in A−A, obeys the identity A = A+Sym_{1}(A), and that A is the union of cosets of Sym_{1}(A). (We shall define a more general notion of symmetry sets Sym_{α}(A) of an additive set in Section 2.6.)

2.1.8 Let d ≥ 1. Give an example of an additive set A of integers such that |A+A| = 6^{d} and |A−A| = 7^{d}.

2.2 Doubling constants

The traditional way to measure the additive structure inside an additive set A is via doubling constants Dbl[A], which we now define. We will shortly develop three other measures of additive structure, namely the Ruzsa diameter d(A,A), the additive energy E(A,A), and the concept of a K-approximate group, which are also useful, and are closely related to the doubling constant.

Definition 2.4 (Doubling constant) For an additive set A, the doubling constant Dbl[A] is defined to be the quantity

Dbl[A] := \frac{|2A|}{|A|} = \frac{|A+A|}{|A|}.

From (2.2) we thus have the bounds

1 ≤ Dbl[A] ≤ \frac{|A|+1}{2}.

The upper bound here is quite easy to attain; for instance if A = 2^{∧}[0,N) = \{1,2,2^{2},...,2^{N−1}\} ⊂ Z, then |A| = N and |A+A| = \frac{N(N+1)}{2}, hence Dbl[A] = \frac{N+1}{2}. In the converse direction, Proposition 2.2 shows that Dbl[A] = 1 if and only if A is a coset of a group; we shall elaborate upon this in Proposition 2.7 below.

An additive set A with the maximal value of doubling constant Dbl[A] = (|A|+1)/2 is known as a Sidon set or a B_{2} set. Informally, this means that all the pairwise sums of A are distinct, excluding the trivial equalities coming from the identity a + b = b + a; see Exercise 1. We will revisit Sidon sets in Section 4.5. There are various senses in which this behavior is “generic”; for instance, if A is a set of N real numbers chosen uniformly at random from the unit interval {x ∈ R : 0 ≤ x ≤ 1}, then we see from Exercise 1 from Section 2.1 that A is a Sidon set with probability 1, and so |A + A| = \frac{N(N+1)}{2}; the point is that if \{a,b\} ≠ \{c,d\} then a + b and c + d will “generically” be distinct. A more interesting question is to understand the conditions under the doubling constant Dbl[A] can be small.

Specializing Proposition 2.2 to the case B = A, we see that Dbl[A] = 1 if and only if A is the coset of a finite subgroup G of Z. We thus expect that if A has a doubling constant which is small, but not actually equal to 1, then it should behave “approximately” like a group (up to translations); we shall see several manifestations of this heuristic throughout this book, when we develop more tools with which to analyze the doubling constant. Indeed, the study of sets of small doubling constant can be thought of as a kind of “approximate group theory”, with the inverse sumset theorems of Chapter 5 then being analogous to a classification theorem for groups.

The study of sets with close to maximal doubling appears to be hopeless at present. A probabilistic construction of Ruzsa [278] shows that there exist large additive sets A with |A−A| very close to the maximal value of |A|^{2}, but |A+A| < |A|^{2−c} for some explicit absolute constant c > 0; and similarly with the roles of A−A and A+A reversed.

2.2.1 Let A be an additive set. Show that A is a Sidon set if and only

if, for any a,b,c,d ∈ A, that a+b ≠ c+d unless {a,b} = {c,d}. 2.2.2 Let Z be an additive group, let a,r ∈ Z, and let N ≥ 1 be an

integer. Let P = {a,a + r,...,a + (N − 1)r} be an arithmetic progression in Z. Show that Dbl[P] ≤ 2 − \frac{1}{N}, with equality if and only if ord(r) ≥ 2N, where ord(r) is the order of the group element r in Z.

2.2.3 If ϕ : Z′ → Z is a surjective group homomorphism whose kernel ker(ϕ) := ϕ^{−1}({0}) is finite, and A is an additive set in Z, show that Dbl[ϕ^{−1}(A)] = Dbl[A].

2.2.4 If A, A′ are additive sets in Z,Z′ respectively, show tha Dbl[A× A′] = Dbl[A]Dbl[A′]. In particular Dbl[A^{d}] = Dbl[A]^{d} for all d ≥ 1.

2.2.5 Let A be any additive set. Show that a non-empty subset of A can have doubling constant at most \sqrt{Dbl[A]|A|/2}. Give examples that show that this bound cannot be improved except by an absolute constant.

2.2.6 [96] Let A be any additive set. Show that a Sidon set contained in A can have cardinality at most \sqrt{2Dbl[A]|A|}. (Thus sets with small doubling constant cannot contain very large Sidon sets.)

2.2.7 [281] Let p be a prime, let θ ∈ Z_{p}\0 be a multiplicative generator of Z_{p}, and let Z := Z_{p−1} × Z_{p}. Let A ⊂ Z be the set A := {(t,θ^{t}) : t = 1,...,p − 1}. Show that A is a Sidon set, and compare this to Exercise 6. Modify this construction to give an example of a Sidon set A ⊂ [0,N] for a large integer N such that |A| is comparable to N^{1/2}. A similar example can be given by using the discrete parabola {(t,t^{2}) : t ∈ Z_{p}} in Z_{p} ×Z_{p}. For a survey of other constructions of Sidon sets, see [253].

2.2.8 Let N be a large integer. Give examples of finite non-empty sets A,B of integers such that |A| = |B| = N and Dbl[A],Dbl[B] ≤ 2, but Dbl[A ∪ B] ≥ \frac{N}{2} . This example shows that doubling constants can behave very badly under set union (see however Exercise 16 in Section 2.3 below). On the other hand, establish the inequality Dbl[A ∪ B] ≥ Dbl[A] + |B|; thus adding a small set to A will not significantly affect the doubling constant.

2.2.9 Let N be a large integer. Give examples of finite non-empty sets A,B of integers such that |A| = |B| = N and Dbl[A],Dbl[B] ≤ 10, but Dbl[A ∩ B] ≥ \frac{1}{10}N^{1/2}. (Hint: concatenate a Sidon set with an arithmetic progression.) Compare this result against Exercise 6. This example shows that doubling constants can behave badly under set intersection (but see Exercise 7 in Section 2.4 below).

2.2.10 Let A be an additive set in Z, and let π : Z → Z′ be a group homomorphism. Show by example that Dbl[π(A)] is not necessarily less than or equal to Dbl[A]. (Hint: this is surprisingly delicate. One way is to start with an additive set C ⊂ Z_{0} with |2C| > |C − C|, and consider the additive set A := (−C)^{n} × {0} × G ∪ C^{n} ∪ C^{m} ∪ {0} in Z_{0}^{n} × Z_{0}^{m} × G, where n,m ≥ 1 and G is a finite group such that |C|^{m} < |G|, and such that n is large compared to m.) See however Exercise 7 and Exercise 17.

2.2.11 Let A be an additive set in Z, and let G be a subgroup of Z. Show by example that Dbl[A+G] is not necessarily less than or equal to Dbl[A]. (Hint: use the previous exercise.)

2.3 Ruzsa distance and additive energy

The doubling constant measures the amount of internal additive structure of a single additive set A. We now introduce two useful quantities measuring the amount of common additive structure between two additive sets A,B - the Ruzsa distance and the additive energy.

Definition 2.5 (Ruzsa distance) Let A and B be two additive sets with a common ambient group Z. We define the Ruzsa distance d(A,B) between these two sets to be the quantity

d(A,B) := log \frac{|A−B|}{|A|^{1/2}|B|^{1/2}} .

We refer to the quantity d(A,A) := log \frac{|A−A|}{|A|} as the Ruzsa diameter of A.

We now justify the terminology “Ruzsa distance”.

Lemma 2.6 (Ruzsa triangle inequality) [284] The Ruzsa distance d(A,B) is non-negative, symmetric, and obeys the triangle inequality

d(A,C) ≤ d(A,B)+d(B,C)

for all additive sets A,B,C with common ambient group Z.

Proof The non-negativity follows from (2.1). The symmetry follows since B −A = −(A−B). Now we prove the triangle inequality. Using the definition of Ruzsa distance and some algebra, we reduce to showing that

|A−C| ≤ \frac{|A−B||B −C|}{|B|}.

Consider the addition map + : (A − B) × (B − C) → Z defined by +(x,y) := x+y. From the identity

a−c = (a−b)+(b−c) = f(a−b,b−c)

we see that every element a−c in A−C has at least |B| distinct representations of the form x+y with (x,y) ∈ (A−B)×(B −C). The claim then follows.

For an approximate version of this inequality in which one replaces complete difference sets with nearly complete difference sets (using at least 75% of the differences), see Exercise 4 in Section 2.5.

The Ruzsa distance thus satisfies all the axioms of a metric except one; we do not have that d(A,A) = 0 for all sets A (also, we have d(G+x,G+y) whenever G+x, G+y are cosets of a group G). Indeed we have a precise characterization on when this Ruzsa diameter vanishes:

Proposition 2.7 (Exact inverse theorem for Ruzsa diameter) Suppose that (A,Z) is an additive set. Then the following are equivalent:

• Dbl[A] = 1.

• d(A,A) = 0.

• d(A,B) = 0 for at least one additive set B.

• |A+A| = |A|.

• |A−A| = |A|.

• |nA−mA| = |A| for at least one pair of integers n,m with |n|+ |m| ≥ 2.

• |nA−mA| = |A| for all integers n,m.

• A is a coset of a finite subgroup G of Z.

Proof Apply Proposition 2.2 and the Ruzsa triangle inequality.

Later on in this chapter we shall generalize this proposition to the case when the Ruzsa diameter, Ruzsa distance or doubling constant are a little larger than 0, 0, or 1 respectively, but still fairly small; see Proposition 2.25.

Despite the non-vanishing of the Ruzsa diameter in general, it is still a useful heuristic to view the Ruzsa distance as behaving like a metric^{1}. Now we relate the Ruzsa diameter to the doubling constant. From the definition of Ruzsa distance and doubling constant we have the identity

d(A,−A) = logDbl[A].  (2.4)

In particular, from Lemma 2.6 we can bound the Ruzsa diameter of a set by its doubling constant

d(A,A) ≤ 2logDbl[A]

and hence we obtain the difference set estimate

|A−A| ≤ Dbl[A]^{2}|A| = \frac{|A+A|^{2}}{|A|} .  (2.5)

A similar argument gives the more general estimate

|B −B| ≤ \frac{|A+B|^{2}}{|A|}  (2.6)

for any two additive sets A,B with common ambient group Z.

It turns out that we can conversely bound the doubling constant of a set by its Ruzsa diameter; see (2.11) below.

There are many further inequalities relating the Ruzsa distance of various sets, most of which we leave to the exercises. We will however mention one useful inequality here, namely that

d(A,C) ≤ d(A,B)+ \frac{1}{2} log \frac{|B|}{|C|}  (2.7)

whenever C ⊆ B; this follows immediately from the estimate |A−C| ≤ |A−B|, and shows that the Ruzsa distance d(A,B) is stable under refinement of one or both of the sets A,B.

1 One could artificially convert the Ruzsa distance into a genuine metric by identifying A with A + x for all x, and redefining d(A,A) to be zero, or alternatively by introducing the metric space X := {A×{j} : A ⊆ Z;0 < |A| < ∞;j ∈ {1,2}} consisting of two copies of each finite non-empty subset of Z (again identifying A with its translations), with the metric d_{X}(A × {j},B × {k}) defined to equal d(A,B) if A×{j} ≠ B ×{k} and equal to 0 otherwise. However there appears to be no significant advantage in working in such an artificial setting.

Having introduced the Ruzsa distance, we now turn to the closely related notion of additive energy E(A,B) between two additive sets.

Definition 2.8 (Additive energy) If A and B are two additive sets with ambient group Z, we define the additive energy E(A,B) between A and B to be the quantity

E(A,B) := |{(a,a′,b,b′) ∈ A×A×B ×B : a+b = a′ +b′}|.

We observe the trivial bounds

|A||B| ≤ E(A,B) ≤ |A||B|min(|A||B|).  (2.8)

The lower bound follows since a+b = a′ +b′ whenever (a,b) = (a′,b′). To see the upper bound, observe that if one fixes a,a′,b, then b′ = a + a′ − b is completely determined, and hence E(A,B) ≤ |A|^{2}|B|. A similar argument gives E(A,B) ≤ |A||B|^{2}. Note that Corollary 2.3 addresses the case when E(A,B) = |A||B|.

We will analyze the additive energy more comprehensively in Section 4.2, when we have developed the machinery of Fourier transforms, and in Section 2.5, when we have developed the Balog-Szemerédi-Gowers theorem. For now we concentrate on the elementary properties of this energy. We first observe the symmetry property E(A,B) = E(B,A) and the translation invariance property E(A + x,B + y) = E(A,B) for all x,y ∈ Z. From the trivial observation

a+b = a′ +b′ ⟺ a−b′ = a′ −b

we also see that E(A,B) = E(A,−B), and similarly if we reflect A to −A.

The additive energy reflects the extent to which A intersects with translates of B or −B, as the following simple identities show:

Lemma 2.9 Let A,B be additive sets with ambient group Z. Then we have the identities

|A||B| = ∑_{x∈A+B} |A∩(x−B)| = ∑_{y∈A−B} |A∩(B +y)|

and

E(A,B) = ∑_{x∈A+B} |A∩(x−B)|^{2} = ∑_{y∈A−B} |A∩(B+y)|^{2} = ∑_{z∈(A−A)∩(B−B)}|A∩(z+A)||B∩(z+B)|.

In particular, if we let r_{A+B}(n) denote the number of representations of n as a + b for some a ∈ A and b ∈ B, and define r_{A−B}(n) similarly, then we have

|A||B| = ∑_{n}r_{A+B}(n) = ∑_{n}r_{A−B}(n); E(A,B) = ∑_{n}r_{A+B}(n)^{2} = ∑_{n}r_{A−B}(n)^{2}.

Proof A simple counting argument yields

|A||B| = ∑_{x∈A+B} |{(a,b) ∈ A×B : a+b = x}| = ∑_{x∈A+B} |A∩(x−B)|;

By replacing B with −B we similarly obtain |A||B| = ∑_{y∈A−B} |A∩(B+ y)|. This gives the first set of identities. For the second set of identities, we compute

∑_{x∈A+B} |A∩(x−B)|^{2} = ∑_{x∈A+B} |{(a,b) ∈ A×B : a+b = x}|^{2}

= ∑_{x∈A+B} |{(a,a′,b,b′) ∈ A×A×B ×B : a+b = a′ +b′ = x}|

= |{(a,a′,b,b′) ∈ A×A×B ×B : a+b = a′ +b′}| = |{(a,a′,b,b′) ∈ A×A×B ×B : a−b′ = a′ −b}|

= ∑_{y∈A−B} |{(a,b′) ∈ A×B : a−b′ = a′ −b}|^{2}

= ∑_{y∈A−B} |A∩(B +y)|^{2}

and

∑_{z∈(A−A)∩(B−B)}|A∩(z +A)||B ∩(z +B)|

= ∑_{z∈(A−A)∩(B−B)}|{(a,a′,b,b′) ∈ A×A×B ×B : z = a−a′ = b′ −b}|

= |{(a,a′,b,b′) ∈ A×A×B ×B : a−a′ = b′ −b}| = |{(a,a′,b,b′) ∈ A×A×B ×B : a+b = a′ +b′}|

and the claims follow from the definition of E(A,B). The last identity follows since r_{A+B}(n) = |A∩(n−B)| and r_{A−B}(n) = |A∩(B +n)|.

As a consequence of this Lemma we have the following inequalities, which assert that pairs of sets with small Ruzsa distance have large additive energy, and pairs with large additive energy have large intersection (after translating and possibly reflecting one of the sets).

Corollary 2.10 Let A,B be additive sets with ambient group Z. Then there exists x ∈ A+B and y ∈ A−B such that

\begin{aligned}|(A−A)∩(B −B)| & ≥ |A∩(x−B)|,|A∩(B +y)| \\ & ≥ \frac{E(A,B)}{|A||B|} \\ & ≥ e^{−d(A,±B)}|A|^{1/2}|B|^{1/2} \\ & = \frac{|A||B|}{|A∓B|} \end{aligned}  (2.9)

for either choice of sign ±. Finally we have the Cauchy-Schwarz inequality

E(A,B) ≤ E(A,A)^{1/2}E(B,B)^{1/2}.  (2.10)

This Corollary should be compared with Proposition 2.3.

Proof From Lemma 2.9 and Cauchy-Schwarz we have

E(A,B) ≥ \frac{|A||B|}{|A±B|} = e^{−d(A,∓B}|A|^{1/2}|B|^{1/2}.

To prove the remaining estimates in (2.9), observe that if z ∈ A∩(x−B), then A∩(x−B) ⊂ z +((A−A)∩(B −B)), hence |A∩(x−B)| ≤ |(A− A)∩(B −B)|, and similarly |A∩(B +y)| ≤ |(A−A)∩(B −B)|. Finally, (2.10) follows from the final formula for E(A,B) in Lemma 2.9 and the Cauchy-Schwarz inequality.

We can use this Corollary to obtain a useful relationship between d(A,B) and d(A,−B):

Proposition 2.11 Let A,B be additive sets with ambient group Z, and let x ∈ A+B. Then we have

|A+B| ≤ \frac{|A−B|^{2}}{|A∩(x−B)|} ≤ \frac{|A−B|^{2}|A||B|}{E(A,B)} ≤ \frac{|A−B|^{3}}{|A||B|} .  (2.11)

In particular we have

d(A,−B) ≤ 3d(A,B)

Proof By translating B if necessary we may assume x = 0. From the Ruzsa triangle inequality and (2.7) we have

d(A,−B) ≤ d(A,−A∩B)+d(−A∩B,−B)

≤ d(A,B)+ \frac{1}{2} log \frac{|B|}{|−A∩B|} +d(−A,−B)+ \frac{1}{2} log \frac{|−A|}{|−A∩B|}

= 2d(A,B)+ \frac{1}{2} log \frac{|A||B|}{|A∩(−B)|^{2}} .

Expanding out the definition of Ruzsa distance, we obtain the first inequality in (2.11). The remaining two inequalities in (2.11), then follow from (2.9), and the final inequality d(A,−B) ≤ 3d(A,B) then follows from the definition of Ruzsa distance.

From (2.11) and and (2.5) we obtain the inequalities

Dbl[A]^{1/3}|A| ≤ |A−A| ≤ Dbl[A]^{2}|A|  (2.12)

which were first observed in [276]. Equivalently, we have

e^{d(A,A)/2} ≤ Dbl[A] ≤ e^{3d(A,A)}.  (2.13)

Thus an additive set has small doubling constant if and only if its Ruzsa diameter is small. It is not known whether the lower bound is best possible. However, the upper bound can be improved to Dbl[A] ≤ e^{2d(A,A)} using Plünnecke inequalities; see Exercise 15 in Section 6.5.

We now show how the Ruzsa distance can be used to control iterated sum sets. We begin with a lemma which controls iterated sum sets of “most” of A+B.

Lemma 2.12 Let A and B be additive sets in a common ambient group. Then there exists S ⊂ A+B such that

|{(a,b) ∈ A×B : a+b ∈ S}| ≥ |A||B|/2  (2.14)

and

|S| ≥ max(|A|,|B|)/2  (2.15)

such that

|A+B +nS| ≤ \frac{2^{n}|A+B|^{2n+1}}{|A|^{n}|B|^{n}}  (2.16)

for all integers n ≥ 0.

Proof If we define S to be the set of all x ∈ A+B such that

|\{(a,b) ∈ A×B : a+b = x\}| ≥ \frac{|A||B|}{2|A+B|}

then we have

|\{(a,b) ∈ A×B : a+b ∈ (A+B)\S\}| < |A+B|\frac{|A||B|}{2|A+B|}

which gives (2.14). Since each element of S has at most min(|A|,|B|) representations of the form a+b, we deduce (2.15) from (2.14).

Now we prove (2.16). A typical element of A+B +nS can be written as

a_{0} +s_{1} +s_{2} +...+s_{n} +b_{n+1}

where a_{0} ∈ A, b_{n+1} ∈ B, and s_{1},...,s_{n} ∈ S. By definition of S, we can expand this in at least (\frac{|A||B|}{2|A+B|})^{n} different ways as

a_{0} +(b_{1} +a_{1})+(b_{2} +a_{2})+...+(b_{n} +a_{n})+b_{n+1}

where b_{i} ∈ B, a_{i} ∈ A, and b_{i} +a_{i} = s_{i} for all 1 ≤ i ≤ n. We regroup this as the sum of n+1 elements from A+B,

(a_{0} +b_{1})+(a_{1} +b_{2})+...+(a_{n} +b_{n+1})

and observe that for fixed a_{0},s_{1},...,s_{n},b_{n+1}, the quantities a_{0} +b_{1},a_{1} + b_{2},...,a_{n}+b_{n+1} completely determine all the variables a_{0},...,a_{n},b_{1},...,b_{n+1}. Thus we have shown that every element of A + B + nS has at least (\frac{|A||B|}{2|A+B|})^{n} representations of the form t_{0} +...+t_{n} where each t_{i} ∈ A+B. The claim then follows.

This result can then be used, together with the Ruzsa triangle inequality to deduce control on iterated sum sets of A and B; see Exercise 9. However we will pursue an approach that gives slightly better bounds in the next section (and an even better result will be developed in Section 6.5).

2.3.1 If ϕ : Z′ → Z is a surjective group homomorphism whose kernel ker(ϕ) := ϕ^{−1}({0}) is finite, and A,B are additive sets in Z, show that d(ϕ^{−1}(A),ϕ^{−1}(B)) = d(A,B). Also show that d(A+ x,B +y) = d(A,B) for any x,y ∈ Z.

2.3.2 If A,B,C,D are additive sets in Z, show that

d(A,B)− \frac{1}{2} log|C||D| ≤ d(A+C,B +D) ≤ d(A,B)+log|C +D|

and

d(A,B ∪C) ≤ max(d(A,B),d(A,C))+log2.

If A′,B′ are additive sets in Z′, show that

d(A×A′,B ×B′) = d(A,B)+d(A′,B′).

2.3.3 Let A,B be additive sets with common ambient group. Show that d(A,B) ≤ \frac{1}{2} log|A|+ \frac{1}{2} log|B|, and that d(A,B) = \frac{1}{2} log|A|+ \frac{1}{2} log|B| if and only if d(A,−B) = \frac{1}{2} log|A|+ \frac{1}{2} log|B|.

2.3.4 Show that for any n ≥ 1, there exists an additive set A such that |A| = 4^{n}, |A + A| = 10^{n}, and |A + A − A| = 28^{n}. Thus it is not possible to obtain an estimate of the form |2A−A| = O(Dbl^{2}[A]|A|).

2.3.5 Let A,B be additive sets with common ambient group. Show that e^{−2d(A,B)}|A| ≤ |B| ≤ e^{2d(A,B)}|A|. Thus sets which are close in the Ruzsa distance are necessarily close in cardinality also. Of course the converse is far from true.

2.3.6 Let A,B be additive sets with common ambient group Z. Show that d(A,B) = 0 if and only if A,B are cosets of the same finite subgroup G of Z. (We shall generalize this result later; see Proposition 2.26).

2.3.7 Let A be an additive set in an additive group Z, and let G be a finite subgroup of Z. Show that Dbl[A+G] ≤ \frac{|3A|}{|A|}. (Hint: apply the Ruzsa triangle inequality to 2A, −A, and G.) Conclude that if π : Z → Z′ is a group homomorphism then Dbl[π(A)] ≤ \frac{|3A|}{|A|}. One cannot replace the tripling constant \frac{|3A|}{|A|} with the doubling constant; see Exercise 10. See however Exercise 17.

2.3.8 Let K be a large integer, and let A = B = {e_{1},...,e_{K}} be the standard basis of Z^{K}. Show that if S is any subset of A + B obeying (2.14) then

|A+B +nS| = Ω_{n}(\frac{|A+B|^{2n+1}}{|A|^{n}|B|^{n}})

where we are using the Landau notation Ω(). This shows that Lemma 2.12 cannot be significantly improved (except possibly in improving the bound (2.15)).

2.3.9 Let A,B be additive sets with common ambient group such that |A + B| ≤ K|A|^{1/2}|B|^{1/2} for some K ≥ 1. Using (2.12), and many applications of the Ruzsa triangle inequality, establish the estimate

|n_{1}A−n_{2}A+n_{3}B−n_{4}B| = O_{n_{1},n_{2},n_{3},n_{4}}(K^{O_{n_{1},n_{2},n_{3},n_{4}} (1)}|A|^{1/2}|B|^{1/2})

for all integers n_{1},n_{2},n_{3},n_{4}. In particular, establish the bounds

d(n_{1}A−n_{2}A+n_{3}B−n_{4}B,n_{5}A−n_{6}A+n_{7}B−n_{8}B) ≤ O_{n_{1},...,n_{8}}(1+d(A,B))

for all integers n_{1},...,n_{8}. We shall improve this bound slightly in Corollary 2.22 and Corollary 2.23; see also Corollary 2.18 for the “tensor power trick” that can eliminate lower order terms such as the implicit constant preceding the K^{O_{n_{1},n_{2},n_{3},n_{4}} (1)} factor.

2.3.10 Let G and H be subgroups of Z. Show that

d(G,H) = log \frac{|G|^{1/2}|H|^{1/2}}{|G∩H|}

Conclude that d(G,H) = d(G,G+H)+d(G+H,H) = d(G,G∩ H) + d(G ∩ H,H). Also, if K is another subgroup of Z, prove the contractivity properties d(G + K,H + K) ≤ d(G,H) and d(G∩K,H ∩K) ≤ d(G,H). Note that the Ruzsa distance, when restricted to subgroups of Z, is indeed a genuine metric, thanks to Proposition 2.7. See also Exercise 7 and 8 in Section 2.4 below.

2.3.11 Let A be an additive set. Show that

Dbl[A∪(−A)] ≤ 2Dbl[A]+Dbl[A]^{2}.

Thus a set with small doubling can be contained in a symmetric set (i.e. a set B such that −B = B) with small doubling which has at most twice the cardinality.

2.3.12 [113],[276] Let A be an additive set. Prove the inequality |A− A| ≤ |A+A|^{3/2}. (Hint: use (2.12) and (2.1)). The corresponding inequality |A+A| ≤ |A−A|^{3/2} is also true, but requires the Plünnecke inequalities to prove it; see Exercise 15 from Section 6.5.

2.3.13 [26] Let A be an additive set. Show that there exists an element x ∈ A − A such that the set F := A ∩ (x − A) has size |F | ≥ |A|/Dbl[A] and doubling constant Dbl[F] ≤ Dbl[A]^{2}. Thus every additive set A of small doubling contains a large symmetric subset F of small doubling, though the set F may be symmetric around a non-zero origin x/2.

2.3.14 Let A,B be additive sets with common ambient group Z. Show that Dbl[A] ≤ e^{4d(A,B)}. Thus only sets with small doubling constant can be close to other sets in the Ruzsa metric.

2.3.15 Let A,B be additive sets with common ambient group Z. Show that Dbl[A∪B] ≤ e^{d(A,B)} +2e^{4d(A,B)}. Thus a pair of sets which are close in the Ruzsa metric can be contained in a slightly larger set with small doubling. In the converse direction, establish the estimate

d(A,B) ≤ logDbl[A∪B]+ \frac{1}{2} log \frac{|A∪B|}{|A|} + \frac{1}{2} log \frac{|A∪B|}{|B|}.

2.3.16 Let A, B be additive sets with common ambient group Z, such that Dbl[A],Dbl[B] ≤ K for some K ≥ 1, and such that A∩B is non-empty. Show that

Dbl[A∪B] ≤ 2K +K^{3} \frac{min(|A|,|B|)}{|A∩B|}.

Thus the union of sets with small doubling remains small doubling provided that those two sets had substantial intersection.

2.3.17 [39], [40] Let K ≥ 1, and let A_{1},A_{2},A_{3} be additive sets with common ambient group Z, such that

|A_{j} ∩A_{3}| ≥ \frac{1}{K} |A_{j}| and |A_{j} +A_{j}| ≤ K|A_{j}|

for all j = 1,2,3. Prove that |A_{1} +A_{2}| ≤ K^{6}|A_{3}|. (Hint: apply the Ruzsa triangle inequality to A_{1}, −(A_{1} ∩ A_{3}), A_{2} ∩ A_{3}, and −A_{2}).

2.3.18 Suppose that A and B are subgroups of Z, and let x = y = 0.

Show that all the inequalities in (2.9) are in fact equalities. 2.3.19 Let A,B,C be additive sets in an ambient group Z. Show that

max(E(A,B),E(A,C)) ≤ E(A,B∪C)^{1/2} ≤ E(A,B)^{1/2}+E(A,C)^{1/2}.

(Hint: use Lemma 2.9 and the triangle inequality for the l^{2} norm).

2.3.20 Let A,B,C be additive sets in an ambient group Z with |A| = |B| = |C| = N. Give examples of such sets where E(A,B) and E(A,C) are comparable to N^{2} and E(B,C) are comparable to N^{3}, or where E(A,B) and E(A,C) are comparable to N^{3} and E(B,C) are comparable to N^{2}. These examples show that there is no hope of any useful “triangle inequality” connecting E(A,B), E(B,C), and E(A,C).

2.3.21 Suppose A,B are additive sets in an ambient group Z. Show that E(A,B) = |A|^{2}|B| holds if and only if |A+B| = |B|. One can thus use Proposition 2.2 to determine when the upper bound in (2.8) is obtained. Conclude in particular that E(A,B) = |A|^{3/2}|B|^{3/2} if and only if d(A,B) = 0, which in turn occurs if and only if A and B are cosets of the same finite group G.

2.3.22 Give an example of an additive set A ⊂ Z of cardinality |A| = N such that E(A,A) ≥ \frac{1}{100}N^{3} but d(A,A) ≥ \frac{1}{100} logN. Compare this with (2.9) (and with Corollary 2.30 below).

2.4 Covering lemmas

We now collect a number of covering lemmas, which roughly speaking have the following flavor: if A and B have similar additive structure (for instance, if their Ruzsa distance is small) then one can cover A by translates of B (or some modification of B).

Lemma 2.13 (Ruzsa’s covering lemma) [287] For any additive sets A,B with common additive group Z, there exists an additive set X_{+} ⊆ B with

B ⊆ A−A+X_{+}; |X_{+}| ≤ \frac{|A+B|}{|A|}; |A+X_{+}| = |A||X_{+}|

and similarly there exists an additive set X_{−} ⊆ B with

B ⊆ A−A+X_{−}; |X_{+}| ≤ \frac{|A−B|}{|A|}; |A−X_{−}| = |A||X_{−}|.

In particular, B can be covered by min(\frac{|A+B|}{|A|}, \frac{|A−B|}{|A|}) translates of A−A.

Remark 2.14 One useful side-benefit of this covering lemma is that there exist at least \frac{|B|}{|A−A|} disjoint translates A+b of A with b ∈ B, as can be seen by restricting b to X_{+}.

Proof It suffices to prove the claim concerning A+B, since the claim concerning A − B follows by replacing B with −B and X_{+} with −X_{−} (note that A−A is symmetric around the origin). Consider the family {A+b : b ∈ B} of translates of A by elements of B. All of these translates have volume |A| and are contained inside A + B. Thus if we take a maximal disjoint sub-family of these translates, i.e. {A + x : x ∈ X_{+}} for some X_{+} ⊆ B, then X_{+} can have cardinality at most \frac{|A+B|}{|A|}. Also we have |A + X_{+}| = |A||X_{+}| by construction. Now for any element b ∈ B, we see that A + b cannot be disjoint from every member of {A+x : x ∈ X_{+}} as this would contradict the maximality of X_{+}. Thus A + b must intersect A + X_{+}, which implies that b is in A − A + X_{+}. Since b ∈ B was arbitrary, we thus have B ⊆ A−A+X_{+} and the claim follows.

Covering lemmas such as the one above are convenient for a number of reasons. Firstly, they allow for easy computation of iterated sumsets. For instance, if one knows that

A+B ⊆ A+X

then one can immediately deduce that

A+nB ⊆ A+nX for all n ≥ 0;

this is advantageous if X is substantially smaller than B. Also they are stable under Freiman homomorphisms (whereas doubling constants are only preserved by Freiman isomorphisms); see Exercise 13 in Section 5.3.

Remark 2.15 Observe that we are covering B by A−A rather than by A. This reflects the fact that A−A is a “smoother” set than A, and tends to contain fewer “holes” that would render it unsuitable for covering other sets. Later on we shall see that higher order sum-difference sets such as 2A − 2A are even smoother, in that they tend to contain very large arithmetic progressions; see Section 4.7 and Chapter 12 for further discussion.

One can modify Ruzsa’s covering lemma in a number of ways. For instance, one can ensure the covering of B by translates of A − A has very high multiplicity (at the cost of increasing the number of covers by a factor of 2).

Lemma 2.16 (Green-Ruzsa covering lemma) [148] Let A and B be additive sets with common ambient group. Then there exists an additive set X ⊆ B with |X| ≤ 2\frac{|A+B|}{|A|} −1 such that for every y ∈ B there are at least |A|/2 triplets (x,a,a′) ∈ X × A × A such that x + a − a′ = y. More informally, A − A + X covers B with multiplicity at least |A|/2. Furthermore, we have

B −B ⊆ A−A+X −X.

Similar claims hold if \frac{|A+B|}{|A|} is replaced by \frac{|A−B|}{|A|}.

Proof Again it suffices to prove the claim for \frac{|A+B|}{|A|}. We perform the following algorithm. Initialize X to be the empty set, so that X +A−A is also the empty set. We now run the following loop. If we cannot find any element y in B which is “sufficiently disjoint from X + A − A” in the sense that |(y +A)∩(X +A)| ≤ |A|/2, we terminate the algorithm. Otherwise, if there is such an element y, we add it to X, and then repeat the algorithm.

Every time we add an element to X, the size of |X +A| increases by at least |A|/2, by construction, and at the first stage it increases by |A|. However, X + A must always lie within the set B + A. Thus this algorithm terminates after at most \frac{2|A+B|}{|A|} −1 steps.

Now let y be any element of B. By construction, we have |(y +A)∩ (X +A)| > |A|/2, and hence y has at least |A|/2 representations of the form x+a−a′ for some (x,a,a′) ∈ X ×A×A′, as desired.

Finally, if y and y′ are two elements of B, then we have

|{a ∈ A : y +a ∈ X +A}| = |(y +A)∩(X +A)| > |A|/2

and similarly we have |{a ∈ A : y′ + A ∈ X + A}| > |A|/2. Thus by the pigeonhole principle there exists a ∈ A such that y +a ∈ X +A and y′ +a ∈ X +A, thus y −y′ = (y +a)−(y′ +a) ∈ X +A−(X +A) = A− A+X −X. Since y,y′ ∈ B is arbitrary, we have B −B ⊆ A−A+X −X as claimed.

In Section 5.4 we develop yet another covering lemma (Lemma 5.31), in which the covering set X is not arbitrary, but is in fact a cube.

We now give an application of the Green-Ruzsa covering lemma, namely a variant of (2.6) which controls quadruple sums rather than double sums.

Proposition 2.17 Let A,B be additive sets in an ambient group Z. Then

|2B −2B| ≤ 16\frac{|A+B|^{4}|A−A|}{|A|^{4}}.

Proof Applying the Green-Ruzsa covering lemma, and reflecting, we may find a set X of cardinality |X| ≤ 2\frac{|A+B|}{|A|} such that A − A + X covers B with multiplicity at least |A|/2.

Now let z be any element of B −B. By definition, we have z = b_{1} −b_{2} for some b_{1},b_{2} ∈ A. By construction of X, we can find at least |A|/2 triplets (x,a_{1},a_{2}) ∈ X ×A×A such that b_{2} = x+a_{1} −a_{2}, and thus

|{(x,a_{1},a_{2}) ∈ X ×A×A : z = b_{1} −a_{1} +a_{2} −x}| ≥ |A|/2.

Making the change of variables c := b_{1} +a_{2} ∈ A+B, we conclude that

|{(x,c,a_{1}) ∈ X ×(A+B)×A : z = c−a_{1} −x}| ≥ |A|/2.

Similarly, if z′ is another element of B −B, we have

|{(x′,c′,a′_{1}) ∈ X ×(A+B)×A : z′ = c′ −a′_{1} −x}| ≥ |A|/2,

and hence

|{(x,x′,c,c′,a_{1},a′_{1}) ∈ X ×X ×(A+B)×(A+B)×A×A :
z = c−a_{1} −x;z′ = c′ −a′_{1} −x′}| ≥ |A|^{2}/4.

Now write d := a_{1} −a′_{1} ∈ A−A, and observe that if z = c−a_{1} −x and z′ = c′ −a′_{1} −x′ then

z −z′ = c−c′ −d−x+x′.

Also, if one fixes z,z′,c,c′,d,x,x′, then a_{1} and a′_{1} are determined by the equations a_{1} = c−x−z, a′_{1} = c′ −x′ −z′. Thus we have

|{(x,x′,c,c′,d) ∈ X ×X ×(A+B)×(A+B)×(A−A) :
z −z′ = c−c′ −d−x+x′}| ≥ |A|^{2}/4.

Note that z −z′ is an arbitrary element of (B −B)−(B −B) = 2B −2B. Thus we have shown that an arbitrary element of 2B −2B has at least |A|^{2}/4 representations of the form c−c′−d−x+x′ where (x,x′,c,c′,d) ∈ X ×X ×(A+B)×(A+B)×(A−A). The claim then follows since |X| ≤ 2\frac{|A+B|}{|A|}.

We can eliminate the factor of 16 by the following elegant “tensor power trick” of Ruzsa[284]:

Corollary 2.18 [284] Let A,B be additive sets in an ambient group Z. Then

|2B −2B| ≤ \frac{|A+B|^{4}|A−A|}{|A|^{4}}.

Proof Fix A,B, and let M be a large integer parameter. We consider the M-fold Cartesian product A^{⊕M} := A×...×A, which is a subset of the additive group Z^{⊕M} := Z ⊕...⊕Z; similarly consider B^{⊕M}. Then one easily verifies

2B^{⊕M} −2B^{⊕M} = (2B −2B)^{⊕M};
A^{⊕M} +B^{⊕M} = (A+B)^{⊕M};
A^{⊕M} −A^{⊕M} = (A−A)^{⊕M}.

Thus by applying Lemma 2.17 with A,B replaced by A^{⊕M}, B^{⊕M} we obtain

|2B −2B|^{M} ≤ 16\frac{|A+B|^{4M}|A−A|^{M}}{|A|^{4M}} .

Taking M^{th} roots of both sides and letting M → ∞, we obtain the result.

Specializing Corollary 2.18 to the case B := −A, we see in particular that

Corollary 2.19 Let A be an additive set. Then

|2A−2A| ≤ \frac{|A−A|^{5}}{|A|^{4}}

or in other words

d(A−A,A−A) ≤ 4d(A,A).

Remark 2.20 One can improve these estimates slightly by using the machinery of Plünnecke inequalities; see Corollary 6.27.

Combining Corollary 2.18 with the Ruzsa covering lemma (Lemma 2.13), and setting K := e^{d(A,A)}, we obtain

Corollary 2.21 For any additive set A, 2A − A can be covered by e^{5d(A,A)} translates of A−A.

An easy induction argument then shows that

mA−nA can be covered by e^{5(m+n−2)d(A,A)} translates of A−A

(2.17)

for all m,n ≥ 1. In particular we have

|mA−nA| ≤ e^{5(m+n−1)d(A,A)}|A| for all m,n ≥ 1.  (2.18)

From this we obtain Corollary 2.22 (Symmetric sum set estimates, preliminary version) Let A be an additive set. Then we have the estimates

d(n_{1}A−n_{2}A,n_{3}A−n_{4}A) ≤ 5(n_{1} +n_{2} +n_{3} +n_{4})d(A,A)

for any non-negative integers n_{1},n_{2},n_{3},n_{4}. (The constant 5 is not best possible; we will improve it later).

Thus if A has small Ruzsa diameter, then in fact all iterated sum sets of A are close to each other in the Ruzsa metric. Combining this with (2.5) we obtain

Dbl[n_{1}A−n_{2}A] ≤ Dbl[A]^{20(n_{1}+n_{2})}

for all non-negative integers n_{1},n_{2}. The factor of 20 is not best possible; we shall obtain improvements to this constant later when we develop the machinery of Plünnecke inequalities in Section 6.5. However, the linear growth in n_{1} and n_{2} is necessary; see Exercise 9.

By combining the above corollary with the Ruzsa triangle inequality one can obtain similar estimates for pairs of sets:

Corollary 2.23 (Asymmetric sum set estimates, preliminary version) Let A,B be additive sets with common ambient group Z. Then we have the estimates

d(n_{1}A−n_{2}A+n_{3}B −n_{4}B,n_{5}A−n_{6}A+n_{7}B −n_{8}B)
= O((n_{1} +...+n_{8})d(A,B))

for any n_{1},...,n_{8} ∈ N.

The proof is left as an exercise.

We can use the above machinery to place additive sets with small Ruzsa diameter or doubling constant inside a more structured set, namely an “approximate group”.

Definition 2.24 (Approximate groups) Let K ≥ 1. A set H is said to be an K-approximate group if it is symmetric (so H = −H), contains the origin, and H +H can be covered by at most K translates of H.

Observe that a 1-approximate group is necessarily a finite group, and conversely every finite group is a 1-approximate group.

We can summarize many of the preceding results by giving the following partial generalization of Proposition 2.7.

Proposition 2.25 Let A be an additive set and let K ≥ 1. Then the following statements are equivalent up to constants, in the sense that if the j^{th} property holds for some absolute constant C_{j}, then the k^{th} property will also hold for some absolute constant C_{k} depending on C_{j}:

(i) Dbl[A] ≤ K^{C_{1}}.

(ii) d(A,A) ≤ C_{2} logK.

(iii) d(A,B) ≤ C_{3} logK for at least one additive set B.

(iv) |A+A| ≤ K^{C_{4}}|A|.

(v) |A−A| ≤ K^{C_{5}}|A|.

(vi) |nA−mA| ≤ K^{C_{6}(|n|+|m|)}|A| for all integers n,m. (vii) There exists a K^{C_{7}}-approximate group H such that A ⊆ x + H

for all x ∈ A, and furthermore |A| ≥ K^{−C_{7}}|H|.

Proof The equivalence of the first five properties follows from the Ruzsa triangle inequality and (2.13). The equivalence of the sixth property with (say) the second follows from Corollary 2.23. To see that the seventh property implies (say) the fourth, observe that

|A+A| ≤ |H +H| ≤ K^{C_{7}}|H| ≤ K^{2C_{7}}|A|.

It remains to deduce the seventh property from (say) the sixth. But this follows by taking H := A−A and applying the Ruzsa covering lemma (Lemma 2.13).

Thus, in a qualitative sense, we have reduced the study of additive sets with small Ruzsa diameter or doubling constant to the study of approximate groups, or precisely to the study of dense subsets of translates of approximate groups. This is a fairly satisfactory state of affairs, except for the fact that we do not have a good characterization of which sets are approximate groups. The well known structure theorem for finite groups (see Corollary 3.8 below) asserts that every finite group is the product of finite cyclic groups; we shall eventually be able to obtain a somewhat similar characterization of approximate groups, showing that they are efficiently contained in a generalized arithmetic progression. For some other properties of approximate groups, see the exercises below.

There is an asymmetric counterpart to Proposition 2.25, whose proof we leave as an exercise.

Proposition 2.26 Let A,B be additive sets in an ambient group Z, and let K ≥ 1. Then the following statements are equivalent up to constants, in the sense that if the j^{th} property holds for some absolute constant C_{j}, then the k^{th} property will also hold for some absolute constant C_{k} depending on C_{j}:

(i) d(A,B) ≤ C_{1} logK.

(ii) d(A,−B) ≤ C_{2} logK.

(iii) |A+B| ≤ K^{C_{3}} min(|A|,|B|).

(iv) |A−B| ≤ K^{C_{4}} min(|A|,|B|).

(v) |n_{1}A − n_{2}A + n_{3}B − n_{4}B| ≤ K^{C_{5}(n_{1}+n_{2}+n_{3}+n_{4})}|A| for all nonnegative integers n_{1},n_{2},n_{3},n_{4}.

(vi) Dbl[A],Dbl[B] ≤ K^{C_{6}}, and there exists x ∈ Z such that |A∩(B +

x)| ≥ K^{−C_{6}}|A|^{1/2}|B|^{1/2}. (vii) Dbl[A],Dbl[B] ≤ K^{C_{7}}, and E(A,B) ≥ K^{−C_{7}}|A|^{3/2}|B|^{3/2}. (viii) There exists a K^{C_{8}}-approximate group H such that A ⊆ H + a

and B ⊆ H +b for all a ∈ A, b ∈ B, and furthermore |A|,|B| ≥ K^{−C_{8}}|H|.

Observe that Exercise 6 from Section 2.3 is essentially the K = 1 case of this Proposition.

Proposition 2.26 gives a satisfactory characterization of pairs of sets with small Ruzsa distance, in terms of approximate groups, provided that one is willing to lose some absolute constants in the exponents. Note however that it is restricted to treating those sets A, B which are comparable in magnitude up to powers of K (cf. Exercise 5 from Section 2.3). A partial analogue of this proposition exists in the case when A and B are very different in magnitude, but the theory here is not as satisfactory; see Section 2.6.

2.4.1 Let Z be a finite additive group, and let A be a random subset of Z such that the events a ∈ A are independent with probability 3/4 for all a ∈ Z. Show that with probability 1 − o_{|Z|→∞}(1), that |A| > |Z|/2 (so in particular A+A = A−A = Z, thanks to Exercise 6), but that it is not possible to cover Z using fewer than \frac{1}{10} log|Z| translates of A. (Hint: if X is an additive set with |X| ≤ \frac{1}{10} log|Z|, use Lemma 2.13 to find an additive set Y with |Y | = Θ(|Z|/log^{2} |Z|) such that the translates b + X are disjoint for all b. Use this and independence to obtain an upper bound for the probability that A+X = Z. Now take the union bound over all choices of X.) This shows that we cannot replace A−A by A in Lemma 2.13 without admitting some sort of logarithmic loss.

2.4.2 Let A be an additive set in a group Z, and let ϕ : Z → Z′ be a group homomorphism. Establish the inequalities

|A| ≤ |ϕ(A)| sup_{x∈Z′} |A∩ϕ^{−1}(x)| ≤ |2A|.

(Hint: use the Ruzsa covering lemma to cover A by translates of a subset of ϕ^{−1}(0).) In particular equality is attained in both inequalities when A is the coset of a group.

2.4.3 Prove Corollary 2.23. What value of the implicit constant in

the O() notation do you get? 2.4.4 Let A be an additive set such that |2A−2A| < 2|A|. Conclude

that A−A is a group. (Hint: use Lemma 2.13). From this and Corollary 2.18 we see that if |A−A| < 2^{1/5}|A|, then A−A is a group. The constant 2^{1/5} can be improved to \frac{3}{2}; see Exercise 5 in Section 2.6 below.

2.4.5 Let G be a K-approximate group for some integer K ≥ 1. Show that |nG| ≤ \binom{K+n−1}{n}|G| for all integers n ≥ 1. Conclude in particular the bounds

|nG| ≤ min(K^{n},n^{K−1})|G| for all n ≥ 1;

thus the numbers |nG| grow exponentially in n for n ≤ K but settle down to become polynomial growth for n > K. In fact for any additive set, |nA| is a polynomial in n for sufficiently large n; see [250] for a proof of this fact and some further discussion.

2.4.6 Let A be an additive set with doubling constant Dbl[A] = K for some K ≥ 1. Show that

|nA| ≤ min(K^{Cn},n^{K^{C}−1})|A|

for all n ≥ 1 and some absolute constant C > 0. (Note that if K is very close to 1, then one can use Exercise 4 to obtain a much stronger bound.)

2.4.7 Let G be a K-approximate group in an ambient group Z, and let H be a K′-approximate group in Z. Show that G + H is a KK′-approximate group. Show that 2G ∩ 2H is a (KK′)^{3}- approximate group. (Hint: first show that (2G ∩ 2H) − (2G ∩ 2H) ⊂ (G+X)∩(H +Y) for some X,Y of cardinality at most K^{3} and (K′)^{3} respectively, and then show that each set of the form (G+x)∩(H +y) is contained in a translate of 2G∩2H). Modify Exercise 9 from Section 2.2 to show that this type of statement fails quite badly if the set 2G ∩ 2H is replaced by G∩H. Also, establish the cardinality bounds

\frac{|G||H|}{|G+H|} ≤ |2G∩2H| ≤ \frac{1}{(KK′)^{3}} \frac{|G||H|}{|G+H|}.

(Hint: use (2.9) for the lower bound, and the Ruzsa triangle inequality for the upper bound). Conclude the estimates

d(G,H) ≤ d(G,G+H)+d(G+H,H) ≤ d(G,H)+logKK′

and

d(G,H) ≤ d(G,2G∩2H)+d(2G∩2H,H) ≤ d(G,H)+3logKK′,

and compare this with Exercise 10 from Section 2.3. 2.4.8 For each j = 1,2,3, let G_{j} be a K_{j}-approximate group in an

ambient group Z. Using the Ruzsa triangle inequality, show that

|G_{1} +G_{2} +G_{3}| ≤ K_{2} \frac{|G_{1} +G_{2}||G_{2} +G_{3}|}{|G_{2}|}.

Conclude that

d(G_{1} +G_{2},G_{1} +G_{2} +G_{3}) ≤ d(G_{2},G_{2} +G_{3})+logK_{1}K_{2}.

Similarly for permutations. Conclude from this and the preceding exercise that

d(G_{1},G_{2}) ≤ d(G_{1} +G_{3},G_{2} +G_{3})+2logK_{1}K_{2}K_{3}

and compare this with Exercise 10 from Section 2.3. (A corresponding statement exists for intersections but is somewhat tricky).

2.4.9 For any integers K,n_{1},n_{2} ≥ 1, give an example of an additive

set A with Dbl[A] = K and Dbl[n_{1}A−n_{2}A] = Ω_{n_{1},n_{2}}(K^{n_{1}+n_{2}}). 2.4.10 Let A, B be additive sets in a common ambient group Z. Show

that Dbl[A+B] ≤ (Dbl[A]Dbl[B])^{C} where C ≥ 1 is an absolute constant. (Hint: use Proposition 2.25 to place A and B inside translates of approximate groups. To obtain lower bounds on |A+B|, use the inequality

|A+B| ≥ \frac{|A||B|}{|(A−A)∩(B −B)}

from (2.9).)

2.4.11 Prove Proposition 11. (Hint: to construct the approximate

group H, one possible choice is H = A−A+B −B.) 2.4.12 Try to improve upon the constant 5 in (2.18), by using the

Ruzsa triangle inequality instead of the Ruzsa covering lemma. This exercise demonstrates that the triangle inequality is slightly sharper than the covering lemma when one wants cardinality bounds, but the covering lemmas of course give much more information than just cardinality.

2.4.13 [199] Let A,B be additive sets in an ambient group Z, and let G be the group generated by A. Show that there exists an additive set B′ ∈ B such that B′ is contained in a coset of G, and such that |A+B′| ≤ \frac{|B′|}{|B|}|A+B|.

2.4.14 Let A,B,A′,B′ be additive sets with common ambient group Z. Establish the inequality d(A + A′,B + B′) = O(d(A,B) + d(A′,B′)). (Hint: argue as in Exercise 10.) Conclude that if ϕ : Z → Z′ is a group homomorphism, then d(ϕ(A),ϕ(B)) = O(d(A,B)). Thus group homomorphisms are “Lipschitz” with respect to the Ruzsa distance.

2.5 The Balog-Szemerédi-Gowers theorem

In the previous sections we have only considered complete sum sets A+ B and complete difference sets A − B. In many applications one only controls a partial collection of sums and differences. Fortunately, there is a very useful tool, the Balog-Szemerédi-Gowers theorem, which allows one to pass from control of partial sum and difference sets to control of complete sum and difference sets (after refining the sets slightly). We begin with some notation.

Definition 2.27 (Partial sum sets) If A,B are additive sets with common ambient group Z, and G is a subset of A × B, we define the partial sum set

A+_{G} B := {a+b : (a,b) ∈ G}

and the partial difference set

A−_{G} B := {a−b : (a,b) ∈ G}.

One may like to think of G as a bipartite graph connecting A and B. Note that when G = A×B is complete, then the notion of partial sum set and partial difference set collapse to just the complete sum set and difference set.

Partial sum sets and partial difference sets are not as nice to work with algebraically as complete sum sets. In particular, the above machinery of sum set estimates do not directly yield any conclusion if one only assumes that the cardinality |A+_{G} B| of a partial sum set is small. Note that even when G is very large, it is possible for |A+_{G} B| to be small while |A+B| is large; see exercises. Fortunately, the Balog-Szemerédi-Gowers theorem, which we will present shortly, does allow us to conclude information on complete sum sets from information on partial sum sets, if we are willing to refine A and B by a small factor (i.e. replace A and B by subsets A′ and B′ which are only slightly smaller than A and B). The first result in this direction was by Balog and Szemerédi [16]; the argument was then simplified and improved by Gowers [131] (with a slight refinement by Bourgain [37]), in particular with dependence of constants that are only polynomial in nature; see Theorem 2.28. A modern formulation of the theorem is as follows.

Theorem 2.28 (Balog-Szemerédi-Gowers theorem) [131], [37] Let A,B be additive sets in an ambient group Z, and let G ⊆ A×B be such that

|G| ≥ |A||B|/K and |A+_{G} B| ≤ K′|A|^{1/2}|B|^{1/2}

for some K ≥ 1 and K′ > 0. Then there exists subsets A′ ⊆ A, B′ ⊆ B such that

|A′| ≥ \frac{|A|}{4\sqrt{2}K}  (2.19)
|B′| ≥ \frac{|B|}{4K}  (2.20)
|A′ +B′| ≤ 2^{12}K^{4}(K′)^{3}|A|^{1/2}|B|^{1/2}  (2.21)

In particular we have

d(A′,−B′) ≤ 5logK +3logK′ +O(1).

The proof of this theorem is graph theoretical. It is elementary, but a little lengthy and so we postpone it to Section 6.4. One can of course combine this theorem with Corollary 2.23 and Proposition 2.25 to gain more information on the iterated sum and difference sets of A′′ and B′′. It is likely that the factor of 2^{12}K^{4}(K′)^{3} in (2.21) can be improved.

However, the bounds (2.19), (2.20) cannot be significantly improved; see exercises.

To apply the Balog-Szemerédi theorem, it is convenient to introduce the following lemma connecting large additive energy to small partial sum sets or small partial difference sets.

Lemma 2.29 Let A, B be additive sets in an ambient group Z, and let G be a non-empty subset of A×B. Then

E(A,B) ≥ \frac{|G|^{2}}{|A+_{G} B|}, \frac{|G|^{2}}{|A−_{G} B|}.

Conversely, if E(A,B) ≥ |A|^{3/2}|B|^{3/2}/K for some K ≥ 1, then there exists G ⊆ A×B such that

|G| ≥ |A||B|/2K; |A+_{G} B| ≤ 2K|A|^{1/2}|B|^{1/2}.

and similarly there exists H ⊆ A×B such that

|H| ≥ |A||B|/2K; |A−_{H} B| ≤ 2K|A|^{1/2}|B|^{1/2}.

Proof Observe that

∑_{x∈A+_{G}B} |{(a,b) ∈ G : a+b = x}| = |G|

and hence by Cauchy-Schwarz

∑_{x∈A+_{G}B} |{(a,b) ∈ G : a+b = x}|^{2} ≥ |G|^{2}/|A+_{G} B|.

But the left-hand side is equal to

|{(a,a′,b,b′) ∈ A×A×B ×B : a+b = a′ +b′;(a,b),(a′,b′) ∈ G}|

which was less than E(A,B). This proves that E(A,B) ≥ \frac{|G|^{2}}{|A+_{G}B|}; using the symmetry E(A,B) = E(A,−B) we thus also obtain E(A,B) ≥ \frac{|G|^{2}}{|A−_{G}B|}.

Now assume E(A,B) ≥ |A|^{3/2}|B|^{3/2}/K. Then by Lemma 2.9 we have

∑_{x∈A+B} |A∩(x−B)|^{2} ≥ |A|^{3/2}|B|^{3/2}/K.

If we set S := {x ∈ A+B : |A∩(x−B)| ≥ |A|^{1/2}|B|^{1/2}/2K}, we then have (by Lemma 2.9 again)

∑_{x∈S} |A∩(x−B)|^{2} ≥ |A|^{3/2}|B|^{3/2}/K−|A||B||A|^{1/2}|B|^{1/2}/2K = |A|^{3/2}|B|^{3/2}/2K.

Now observe from Lemma 2.9 again that

|S||A|^{1/2}|B|^{1/2}/2K ≤ ∑_{x∈S} |A∩(x−B)| ≤ |A||B|

and hence

|S| ≤ 2K|A|^{1/2}|B|^{1/2}.

Now let G := {(a,b) ∈ A×B : a+b ∈ S}, then clearly A+_{G} B ⊆ S and hence

|A+_{G} B| ≤ 2K|A|^{1/2}|B|^{1/2}.

Furthermore we have

|G| = ∑_{x∈S} |{(a,b) ∈ A×B : a+b = x}|

= ∑_{x∈S} |A∩(x−B)|

≥ ∑_{x∈S} \frac{|A∩(x−B)|^{2}}{|A|^{1/2}|x−B|^{1/2}}

≥ \frac{|A|^{3/2}|B|^{3/2}/2K}{|A|^{1/2}|B|^{1/2}}

= |A||B|/2K.

This gives the desired set G. The construction of H follows by using the symmetry E(A,B) = E(A,−B).

Combining this Lemma with the Balog-Szemerédi-Gowers theorem, we can obtain a characterization of pairs of sets with large additive energy.

Theorem 2.30 (Balog-Szemerédi-Gowers theorem, alternate version) Let A,B be additive sets in an ambient group Z, and let K ≥ 1. Then the following statements are equivalent up to constants, in the sense that if the j^{th} property holds for some absolute constant C_{j}, then the k^{th} property will also hold for some absolute constant C_{k} depending on C_{j}:

(i) E(A,B) ≥ K^{−C_{1}}|A|^{3/2}|B|^{3/2}.

(ii) There exists G ⊂ A×B such that |G| ≥ K^{−C_{2}}|A||B| and |A+_{G}

B| ≤ K^{C_{2}}|A|^{1/2}|B|^{1/2}. (iii) There exists G ⊂ A×B such that |G| ≥ K^{−C_{3}}|A||B| and |A−_{G}

B| ≤ K^{C_{3}}|A|^{1/2}|B|^{1/2}.

(iv) There exists subsets A′ ⊆ A, B′ ⊆ B with |A′| ≥ K^{−C_{4}}|A|,

|B′| ≥ K^{−C_{4}}|B|, and d(A′,B′) ≤ C_{4} logK.

(v) There exists a K^{C_{5}}-approximate group H and x,y ∈ Z such that |A∩(H +x)|,|B ∩(H +y)| ≥ K^{−C_{5}}|H| and |A|,|B| ≤ K^{C_{5}}|H|.

We leave the proof of this theorem to the exercises. Theorem 2.30 should be compared with Exercise 21 from Section 2.3, which is the K = 1 case of this Theorem. As with Proposition 2.26, this Theorem is restricted to sets A,B which are close in cardinality (see exercises). We shall address the question of sets A,B of widely differing cardinalities in the next section.

2.5.1 Let A,B be additive sets with common ambient group Z such that E(A,B) ≥ K^{−1}|A|^{3/2}|B|^{3/2}. Show that K^{−2}|A| ≤ |B| ≤ K^{2}|A|, and show by means of an example that these bounds cannot be improved.

2.5.2 Give an example of an additive set A ⊂ Z of cardinality N, and a set G ⊂ A×A of cardinality N^{2}/4, such that |A+_{G} A| ≤ N but |A + A| ≥ N^{2}/8. (Hint: concatenate a Sidon set with an arithmetic progression).

2.5.3 Let N ≫ K ≫ 1 be large integers, with N a multiple of K. Give an example of sets A,B ⊂ Z of cardinality |A| = |B| = N and a subset G ⊂ A × B of cardinality |G| = |A||B|/K with the property that |A +_{G} B| ≤ 2N, but such that |A′′ + B′′| ≥ N^{2}/K^{2} whenever A′′ ⊂ A and B′′ ⊂ B is such that |A′′| ≥ 2|A|/K. (Hint: take B to be a long progression, and take A to be a short progression concatenated with some generic integers.) This shows that the conditions (2.19), (2.20) in Theorem 2.28 cannot be significantly improved.

2.5.4 Let A,B,C be additive sets in an ambient group Z, let 0 < ε < 1/4, and let G ⊂ A × B,H ⊂ B × C be such that |G| ≥ (1 − ε)|A||B| and |H| ≥ (1 − ε)|B||C|. Show that there exists subsets A′ ⊆ A and C′ ⊆ C with |A′| ≥ (1 − ε^{1/2})|A| and |C′| ≥ (1−ε^{1/2})|C| such that |A′ −C′| ≤ \frac{|A−_{G}B||B−_{H}C|}{(1−2ε^{1/2})|B|}. (Hint: show that at most ε^{1/2}|B| elements of B have a G-degree of less than (1 − ε^{1/2})|A|, and similarly at most ε^{1/2}|B| elements have a H-degree of less than (1−ε^{1/2})|C|.) This result is can be used as a substitute for the Balog-Szemerédi-Gowers in the case when the graph G is extremely dense; it has the advantage that it does not require A,B,C to be comparable in size and it does not lose any constants in the limit ε → 0; indeed it collapses to Ruzsa’s triangle inequality in that limit.

2.5.5 Prove Proposition 2.30. (Hint: for K large, e.g. K ≥ 1.1, one can use the Balog-Szemerédi-Gowers theorem and Proposition 2.26. For K small, e.g. 1 ≤ K < 1.1, one can use Exercise 4 as a substitute for the Balog-Szemerédi-Gowers theorem.)

2.6 Symmetry sets and imbalanced partial sumsets The Balog-Szemerédi-Gowers theorem is a very powerful tool when studying two additive sets A,B with additive energy E(A,B) close to |A|^{3/2}|B|^{3/2}, however from (2.8) we see that this situation only occurs when |A| and |B| are comparable in size. This leaves open the question of what happens in the case |A| ≫ |B| (say) and E(A,B) is close to the upper bound of |A||B|^{2} given by (2.8). A special sub-case of this (thanks to (2.9)) is the case when |A + B| or |A − B| is comparable to |A|. Note that Proposition 2.2 already gives an answer to this question in the extreme case when |A + B| = |A| or |A − B| = |A| (or equivalently if E(A,B) = |A||B|^{2}; see Exercise 21 in Section 2.3). However, an example of Ruzsa [284] shows that things become bad when |A| and |B| are very widely separated, see exercises.

If however we are prepared to endure logarithmic-type losses in the ratio |A|/|B| (or more precisely losses of the form (|A|/|B|)^{ε} where ε can be chosen to be small), then one can recover a reasonable theory. In analogy with Proposition 2.2, one expects that if |A+B| is comparable to |A|, or if E(A,B) is close to |A||B|^{2}, then there should be an approximate group H such that A is approximately the union of translates of H, and B is approximately contained in a single translate of H. To achieve this will be the main objective of this section.

In the extreme case when |A + B| = |A| or E(A,B) = |A||B|^{2}, the approximate group H was in fact an exact group and in the proof of Proposition 2.2 it was constructed as the symmetry group Sym_{1}(A) of the larger additive set A. In the general case this symmetry group is likely to be trivial. However, a more general notion is still useful.

Definition 2.31 (Symmetry sets) Let (A,Z) be an additive set. For any non-negative real number α ≥ 0, define the symmetry set Sym_{α}(A) ⊆ Z at threshold α to be the set

Sym_{α}(A) := {h ∈ Z : |A∩(A+h)| ≥ α|A|}.

Note that Sym_{1}(A) = {h ∈ Z : A + h = A} is the same symmetry group applied in the proof of Proposition 2.2. The other symmetry sets are not groups in general, but nevertheless they are still symmetric (so −Sym_{α}(A) = Sym_{α}(A)) and contain the origin, and they obey the nesting property Sym_{α}(A) ⊆ Sym_{β}(A). It is also clear that Sym_{α}(A) ⊆ A − A for all 0 < α ≤ 1. Note that as Sym_{α}(A) is empty for α > 1 and equal to all of Z for α ≤ 0, we shall mostly restrict ourselves to the non-trivial region where 0 < α ≤ 1.

We now relate the size of these symmetry sets to the additive energy. From (2.9) we have

E(A,A) = ∑_{h∈A−A}|A∩(A+h)|^{2}

and hence for any 0 < α ≤ 1 and the crude bounds |A∩(A+h)| ≤ |A| when h ∈ Sym_{α}(A) and |A ∩ (A + h)| ≤ α|A| when h ∉ Sym_{α}(A), we have

α^{2}|A|^{2}|Sym_{α}(A)| ≤ E(A,A) ≤ α^{2}|A|^{2}|A−A|+|A|^{2}|Sym_{α}(A)|,

which indicates that Sym_{α}(A) should be large whenever the energy is large. In particular, from (2.8) we have

|Sym_{α}(A)| ≤ |A|/α^{2}.  (2.22)

Now let A,B be additive sets in an additive group Z. From Lemma 2.9 again, we have

E(A,B) = ∑_{b,b′∈B} |A∩(A+b−b′)|

and hence for any 0 < α ≤ 1 we have

E(A,B) ≤ |B|^{2}α|A|+|A||{(b,b′) ∈ B : b−b′ ∈ Sym_{α}(A)}|.

In particular, if E(A,B) ≥ 2α|A||B|^{2}, then we conclude that there is a set G ⊂ B ×B of cardinality |G| ≥ α|B|^{2} such that

B −_{G} B ⊆ Sym_{α}(A).  (2.23)

At first glance it seems that one may now be able to apply the symmetric Balog-Szemerédi-Gowers theorem. However, the fact that A is much larger than B means that B −_{G} B may be much larger than B (compare (2.23) to (2.22)). To get around this difficulty we need to iterate this construction, and exploit the fact that Sym_{α}(A) behaves like a group. This is already clear when α = 1, when Sym_{1}(A) is indeed a genuine group; the following lemma shows that this behavior persists in an approximate sense for α less than 1.

Lemma 2.32 Let A be an additive set. Then we have

Sym_{1−ε}(A)+Sym_{1−ε′}(A) ⊆ Sym_{1−ε−ε′}(A)  (2.24)

whenever ε,ε′ > 1. Furthermore, if 0 < α ≤ 1 and S ⊆ Sym_{α}(A) is a non-empty set, then there exists a set G ⊆ |S|^{2} with

|G| ≥ α^{2}|S|^{2}/2  (2.25)

such that

S −_{G} S ⊆ Sym_{α^{2}/2}(A).  (2.26)

Proof To verify the first claim, observe that if x ∈ Sym_{1−ε}(A) and y ∈ Sym_{1−ε′}(A) then

|(A+x)\A| = |A|−|A∩(A+x)| ≤ ε|A|

and

|(A+x)\(A+x+y)| = |A|−|A∩(A+y)| ≤ ε′|A|

and hence

|A∩(A+x+y)| ≥ |(A+x)∩A∩(A+x+y)| ≥ (1−ε−ε′)|A|

which proves (2.24).

Now we prove the second claim. By definition of S, we see that for each x ∈ S there exists at least α|A| elements a ∈ A such that a+x ∈ A. Summing this over all x we see that

∑_{a∈A}|{x ∈ S : a+x ∈ A}| ≥ α|A||S|.

Applying Cauchy-Schwarz we conclude that

∑_{x,y∈S×S} |{a ∈ A : a+x,a+y ∈ A}| = ∑_{a∈A}|{x ∈ S : a+x ∈ A}|^{2} ≥ α^{2}|A||S|^{2}.

If we set G ⊆ S ×S to be all the pairs (x,y) such that

|{a ∈ A : a+x,a+y ∈ A}| ≥ α^{2}|A|/2

then we have

|A||G| ≥ ∑_{(x,y)∈G}|\{a ∈ A : a+x,a+y ∈ A\}| ≥ α^{2}|A||S|^{2} − \frac{α^{2}|A|}{2}|S|^{2}

which gives (2.25). Also, if (x,y) ∈ G then |A∩(A+x−y)| ≥ α^{2}|A|/2 by definition of G, which gives (2.26).

Before we proceed with the main theorem, we need a technical lemma that uniformizes the size of the fibers {(a,a′) ∈ G : a−a′ = x} of A−_{G}A.

Lemma 2.33 (Dyadic pigeonhole principle) Let A be an additive set, and let G ⊂ A×A be such that |G| ≥ α|A|^{2} and |A−_{G} A| ≤ L|A| for some 0 < α < 1 and L ≥ 1. Then there exists a subset G′ of G with

|G′| = Ω(\frac{α}{1+log \frac{1}{α} +logL}|A|^{2})

and

|\{(a,a′) ∈ G′ : a−a′ = x\}| ≥ \frac{|G′|}{2|A−_{G′} A|}

for all x ∈ A−_{G′} A.

It is important to note that the dependence on L only enters in a logarithmic manner.

Proof Let D be the set of all x such that

|\{(a,a′) ∈ G̃ : a−a′ = x\}| ≥ \frac{α|A|^{2}}{2L|A|} = \frac{α}{2L}|A|

(thus D is the set of “popular differences”) and set G̃ to be the pairs (a,a′) in G such that a−a′ ∈ D. Then we have |G\G̃| ≤ \frac{α}{2L}|A||A−_{G}A| ≤ α|A|^{2}/2, and hence |G̃| ≥ α|A|^{2}/2. On the other hand, we have the crude upper bound

|{(a,a′) ∈ G̃ : a−a′ = x}| ≤ ∑_{a′∈A}|{a ∈ A : a = x+a′}| ≤ |A|.

Thus if we let M be the least integer such that 2^{−M} < \frac{α}{2L}, we can partition G̃ = G_{1} ∪ ... ∪ G_{M} where G_{m} := {(a,a′) ∈ G̃ : a − a′ ∈ D_{m}} and

D_{m} := {x ∈ A−G̃A : 2^{−m}|A| < |{(a,a′) ∈ G̃ : a−a′ = x}| ≤ 2^{−m+1}|A|}.

By the pigeonhole principle, there exists 1 ≤ m ≤ M such that

|G_{m}| ≥ \frac{1}{M} |G| ≥ \frac{α}{C(1+log \frac{1}{α} +logL)}|A|^{2}.

By definition of D_{m}, we have

\frac{|G_{m}|}{2^{−m+1}|A|} ≤ |D_{m}| ≤ \frac{|G_{m}|}{2^{−m}|A|};

since D_{m} = A−_{G_{m}} A, we thus see that

|\{(a,a′) ∈ G′ : a−a′ = x\}| ≥ 2^{−m}|A| ≥ \frac{|G′|}{2|A−_{G′} A|}

for all x ∈ A−_{G_{m}} A. The claim then follows by setting G′ := G_{m}.

Now we give the main theorem of this section.

Theorem 2.34 (Asymmetric Balog-Szemerédi-Gowers theorem) Let A,B be additive sets in an additive group Z such that E(A,B) ≥ 2α|A||B|^{2} and |A| ≤ LB for some L ≥ 1 and 0 < α ≤ 1. Let ε > 0. Then there exists a O_{ε}(α^{−O_{ε}(1)}L^{ε}-approximate group H in Z, an additive set X in Z of cardinality |X| = O_{ε}(K^{O_{ε}(1)}L^{ε}|A|/|H|) such that |A ∩ (X + H)| = Ω_{ε}(α^{O_{ε}(1)}L^{−ε}|A|), and an x ∈ Z such that |B ∩ (x + H)| = Ω_{ε}(α^{O_{ε}(1)}L^{ε}|B|).

Observe in the converse direction that if the conclusions of this theorem are true, then E(A,B) = Ω_{ε}(α^{O_{ε}(1)}L^{−O(ε)}|A||B|^{2}) (Exercise 3. Thus this theorem is sharp up to polynomial losses in α and L^{ε}, where ε can be made arbitrary small; the example in Exercise 1 can be adapted to show that this loss is necessary (Exercise 2).

Proof A direct application of Theorem 2.30 will lose far too many powers of L. The trick is to embed B in a long increasing sequence of sets B_{0},B_{1},B_{2},..., with each B_{j} being (roughly speaking) a partial difference set of the previous one, and use the pigeonhole principle to show that at some stage the ratio |B_{j+1}|/|B_{j}| is bounded by a small power of L. One can then apply Theorem 2.30 with acceptable losses and conclude the theorem. (This method of proof is inspired by a similar argument in [39].)

We turn to the details. It will be convenient to use a variant of the Landau O() and Ω() notation which can absorb factors of α and logL (which we think of as being relatively close to 1). If X,Y are nonnegative quantities and j is a parameter, let us say that X = Õ_{j}(Y) or Y = Ω̃_{j}(X) if one has an estimate of the form

X ≤ C(j)α^{−C(j)}Y log^{C(j)} L

for some C(j) > 0 depending only on j.

Let J = J(ε) ≫ 1 be a large integer to be chosen later. Let 1 > α_{1} > ... > α_{J+1} > 0 be the sequence defined recursively by α_{1} := α and α_{j+1} := α_{j}^{2}/2 for all 1 ≤ j ≤ J. From induction we see that α_{j} = Ω̃_{j}(1).We claim that we can find a sequence B_{0},B_{1},...,B_{J},B_{J+1} of additive sets in Z with the following properties:

• B_{0} = B, and for all 1 ≤ j ≤ J +1 we have

B_{j} ⊆ Sym_{α_{j}}(A).  (2.27)

• For all 0 ≤ j ≤ J +1, we have

α_{j}^{−2}L|B| ≥ |B_{j}| = Ω̃_{j}(|B|).  (2.28)

• For all 0 ≤ j ≤ J, there exists G_{j} ⊆ B_{j} ×B_{j} such that

|G_{j}| = Ω̃_{j}(|B_{j}|^{2})  (2.29)

and

B_{j+1} = B_{j} −_{G_{j}} B_{j}.  (2.30)

Furthermore, for all x ∈ B_{j+1} we have

|\{(b,b′) ∈ G_{j} : b−b′ = x\}| = Ω̃_{j}(\frac{|B_{j}|^{2}}{|B_{j+1}|}).  (2.31)

We construct the B_{j} as follows. We set B_{0} := B. From (2.23) followed by Lemma 2.33 we can construct G_{0} ⊆ B_{0} ×B_{0} and B_{1} := B_{0} −_{G_{0}} B_{0} obeying (2.27), (2.29), (2.30), (2.31). Since each element in B_{0} −_{G_{0}} B_{0} can be represented as a difference of a pair in G in at most |B_{0}| ways, we have

|B_{1}| = |B_{0} −_{G_{0}} B_{0}| ≥ |G_{0}|/|B_{0}| = Ω̃_{j}(|B|)

which is the lower bound in (2.28); the upper bound follows from (2.27) and (2.22).

Next, suppose inductively that B_{j} ⊆ Sym_{α_{j}}(A) has already been chosen for some 1 ≤ j ≤ J. Applying Lemma 2.32 (with S := B_{j}) followed by Lemma 2.33, and using the cardinality bounds already obtained in (2.28) and the construction α_{j+1}^{2} := α_{j}^{2}/2 of the α_{j}, we can thus find G_{j} ⊆ B_{j} × B_{j} and B_{j+1} := B_{j} −_{G_{j}} B_{j} obeying (2.27), (2.29), (2.30), (2.31). This closes the induction and so we can construct the B_{j} for all 0 ≤ j ≤ J +1, and similarly obtain the G_{j} for all 1 ≤ j ≤ J.

Now for the crucial step (which explains why we iterated the above procedure so many times). From (2.28) and the pigeonhole principle, there exists 1 ≤ j ≤ J such that

|B_{j+1}| = Õ_{J}(L^{O(1/J)}|B_{j}|);

the point is that we have managed to replace L by the substantially smaller quantity L^{O(1/J)}. If we now apply (2.30), (2.29), and Theorem 2.30, we can thus find a Õ_{J}(L^{O(1/J)})-approximate group H of cardinality

|H| = Õ_{J}(L^{O(1/J)}|B_{j}|)  (2.32)

and an x_{j} ∈ Z such that

|B_{j} ∩(H +x_{j})| = Ω̃_{J}(L^{−C_{0}/J}|B_{j}|)  (2.33)

for some absolute constant C_{0}. It remains to relate H to B and to A. We begin with B. From (2.33) and (2.31) (with j replaced by j −1) we have

|{(b,b′) ∈ G_{j−1} : b−b′ ∈ B_{j} ∩(H +x_{j})}| = Ω̃_{J}(L^{−C_{0}/J}|B_{j−1}|^{2})

so in particular

|{(b,b′) ∈ B_{j−1} ×B_{j−1} : b ∈ H +x_{j} +b′}| = Ω̃_{J}(L^{−C_{0}/J}|B_{j−1}|^{2})

Thus by the pigeonhole principle, there exists a b′ such that

|{b ∈ B_{j−1} : b−b′ ∈ H +x_{j} +b′}| = Ω̃_{J}(L^{−C_{0}/J}|B_{j−1}|).

Thus if we set x_{j−1} := x_{j} +b′ then we have

|B_{j−1} ∩(H +x_{j−1})| = Ω̃_{J}(L^{−C_{0}/J)}|B_{j−1}|).  (2.34)

We now repeat this argument with j replaced by j−1 and (2.33) replaced by (2.34). Iterating this at most J times, we eventually locate an x = x_{0} ∈ Z such that

|B ∩(H +x))| = Ω̃_{J}(L^{−C_{0}/J}|B|),

which gives the desired control on B if J is sufficiently large depending on ε.

It remains to control A. From (2.33), (2.32) and (2.27) we have

|{y ∈ H +x_{j} : y ∈ Sym_{α_{j}}(A)}| = Ω̃_{J}(L^{−O(1/J)}|H|)

and thus by definition of Sym_{α_{j}}(A) and α_{j}

|{(a,y) ∈ A×(H +x_{j}) : a+y ∈ A}| = Ω̃_{J}(L^{−O(1/J)}|H||A|).

We rewrite this as

∑_{x∈x_{j}+A}|A∩(H +x)| = Ω̃_{J}(L^{−O(1/J)}|H||A|).

We can therefore find a subset X_{0} of x_{j} +A with

|X_{0}| = Ω̃_{J}(L^{−O(1/J)}|A|)  (2.35)

such that

|A∩(H +x)| = Ω̃_{J}(L^{−O(1/J)}|H|) for all x ∈ X_{0}.

Now we use an argument similar to that used to prove Ruzsa’s covering lemma (Lemma 2.13). Let X be a subset of X_{0} such that the sets {H +x : x ∈ X} are all disjoint, and which is maximal with respect to set inclusion. Then we have

|A∩(H +X)| = ∑_{x∈X} |A∩(H +x)| = Ω̃_{J}(L^{−O(1/J)}|H||X|). (2.36)

On the other hand, if y ∈ X_{0} then by maximality of X there exists x ∈ X such that x+H intersects y +H. In other words, X_{0} is covered by X +H −H, and hence (since H is a Õ(L^{O(1/J)})-approximate group)

|X_{0}| ≤ |X||H −H| = Õ(|X|L^{O(1/J)}|H|).  (2.37)

Combining (2.35), (2.36), (2.37) we see that X obeys all the desired properties, if J is chosen sufficiently small depending on ε.

The above theorem can also be put in a form resembling Theorem 2.28:

Corollary 2.35 Let A,B be additive sets with common ambient group such that E(A,B) ≥ 2α|A||B|^{2} and |A| ≤ LB for some L ≥ 1 and 0 < α ≤ 1. Let ε > 0. Then there exists subsets A′ ⊆ A and B′ ⊆ B such that

|A′| = Ω_{ε}(α^{O_{ε}(1)}L^{−ε}|A|)
|B′| = Ω_{ε}(α^{O_{ε}(1)}L^{−ε}|B|)
|A′ +nB′ −mB′| = O_{ε}(α^{−O_{ε}(1)}L^{ε})^{n+m}|A|

for all integers n,m ≥ 0.

Proof Apply Theorem 2.34 and set A′ := A∩(X +H) and B′ := B ∩ (x+H).

Because of (2.9), the above results give some partial results concerning the situation when |A+B| ≤ K|A| and |A| is much larger than B, but these results will be rather weak. We will give a better result concerning this problem in Section 6.5, once we develop the Plünnecke inequalities.

2.6.1 [284] Let n be a large integer, and let Z := Z^{2n}. Let A be the additive set

A := {(x_{1},x_{2},...,x_{2n}) ∈ Z^{2n} : x_{1} +...+x_{n} = n;x_{1},...,x_{n} ≥ 0}

and let B := {e_{1},...,e_{2n}}. Show that |B| = 2n, that |A| = (27/4)^{n+o(1)}, that |A + B| = O(|A|), but that |A − B| ≥ n|A|. (You may find Stirling’s formula (1.52) to be useful.)

2.6.2 Modify Exercise 1 to show that one cannot take ε = 0 in Theo-

rem 2.34. 2.6.3 Let A,B be additive sets and let ε > 0, 0 < α < 1, and L ≥

1 be such that the conclusions of Theorem 2.34 are satisfied. Conclude that E(A,B) = Ω_{ε}(α^{O_{ε}(1)}L^{−O(ε)}|A||B|^{2}).

2.6.4 Let A be an additive set. By modifying the proof of Lemma 2.12, establish the inequality

|A−A+nSym_{α}(A)| ≤ \frac{e^{(n+1)d(A,A)}}{α^{n}} |A|

for all integers n ≥ 0 and all 0 < α < 1. 2.6.5 [210] Let A be an additive set such that A−A is not a group.

Show that there exists h ∈ A−A such that 1 ≤ |A∩(A+h)| ≤ |A|/2. (Hint: argue by contradiction, and analyze Sym_{α}(A) for some α slightly greater than 1/2). Conclude in particular that if |A−A| < \frac{3}{2}|A|, then A−A is a group. Note that the example A = \{0,1\} ⊂ Z shows that the constant \frac{3}{2} cannot be improved; one can also make this example larger, for instance by taking the Cartesian product of {0,1} with a finite group. For a more refined estimate on A−A, see Theorem 5.5 and Corollary 5.6.

2.6.6 Let A,B be additive sets with common ambient group such that |A + B| ≤ K|A| and |A| ≤ L|B| for some K,L ≥ 1. Let ε > 0. Show that there exists a O_{ε}(K^{O_{ε}(1)}L^{ε})-approximate group H such that B is contained in a translate of H, and that A is contained in at most O_{ε}(K^{O_{ε}(1)}L^{ε}|A|/|H|) translates of H; compare this with Proposition 2.2. (Hint: Apply Theorem 2.34 and the Ruzsa covering lemma (Lemma 2.13).

2.6.7 Let A be an additive set, and let B be a subset of A such that |B| ≥ (1−ε)|A| for some 0 < ε < 1. Prove that

Sym_{α/(1−ε)}(B) ⊆ Sym_{α}(A) ⊆ Sym_{(α−2ε)/(1−ε)}(B)

for every α ∈ R. 2.6.8 Let A be an additive set. Refine (2.22) slightly to

|Sym_{α}(A)| ≤ 1+ \frac{|A|(|A|−1)}{α} for all α > 0.

2.6.9 [335] Let A,B be an additive sets in Z, such that B consists entirely of positive numbers. Show that there exists b ∈ B such that

|A∩(A+b)| < \frac{|A|−1}{|B|} \frac{|A|}{2}.

(Hint: use Exercise 8, and exploit the fact that only half of the elements of Sym_{α}(A)\{0} are positive.

2.6.10 [43] Let A be an additive set such that |A+A| ≤ K|A| for some K ≥ 1. Let G be the group generated by Sym_{\frac{2}{9K}} (A). Show that there exists a coset x + G of G such that |A ∩ (x + G)| ≥ |A|/3. (Hint: suppose for contradiction that |A ∩ (x + G)| < |A|/3 for all x. Use the greedy algorithm to partition A = A′ ∪A′′ where |A|/3 ≤ |A′|,|A′′| ≤ 2|A|/3 and such that A′ −A′′ is disjoint from G (and thus disjoint from Sym_{\frac{2}{9K}} (A)). Use this to obtain an upper bound on E(A′,A′′) and use (2.9) to obtain a contradiction.)

2.7 Non-commutative analogues

Many of the above arguments carry over to the non-commutative setting, though one of course now needs to take care with the ordering of multiplication. We sketch some of the main points here and leave the details as exercises.

Definition 2.36 A multiplicative group is any group G (not necessarily abelian) with group operation ·, with inversion operation x ↦ x^{−1}, and identity element 1. An multiplicative set is a pair (A,G), where G is a multiplicative group, and A is a finite non-empty subset of G. We often abbreviate a multiplicative set (A,G) simply as A, and refer to G as the ambient group.

If A and B are multiplicative sets with common ambient group G, we define their product set

A·B := {ab : a ∈ A,b ∈ B}

and the inverse set

A^{−1} := {a^{−1} : a ∈ A}.

We also define right translates A·x and left translates x·A for x ∈ G in the usual manner. Note that x·A ≠ A·x and A·B ≠ B ·A in general, although we do have |A| = |x · A| = |A · x| = |A^{−1}|. We also define iterated product sets A^{·n} := A · ... · A for n ≥ 1, with the conventions that A^{·0} := {1} and A^{·−n} := (A^{·n})^{−1} = (A^{−1})^{·n}.

We remark that A·B and B·A may have widely different cardinalities; for instance if H is a finite subgroup of G and x is an element of G that does not lie in the normalizer N(H) := {x ∈ G : xH = Hx} of H, then H ·(x·H) and (x·H)·H can have very different cardinalities. However, we still have the analogue of (2.1):

max(|A|,|B|) ≤ |A·B|,|B ·A| ≤ |A||B|;

see exercises.

We define the (left-invariant) Ruzsa distance d(A,B) between two multiplicative sets

d(A,B) := log \frac{|A·B^{−1}|}{|A|^{1/2}|B|^{1/2}} .

This distance still obeys the Ruzsa triangle inequality, mainly thanks to the identity (ab^{−1})(bc^{−1}) = ac^{−1}. It is left-invariant in each variable, thus d(x · A,B) = d(A,x · B) = d(A,B), and is jointly right-invariant, d(A · x,B · x) = d(A,B), but is not separately right-invariant in each variable. Also it is not reflection invariant; the metric d^{∗}(A,B) := d(A^{−1},B^{−1}) is the right-invariant Ruzsa distance, which we will not use here.

We can characterize when d(A,B) is zero:

Proposition 2.37 Let A,B be multiplicative sets in an ambient group G. Then d(A,B) = 0 if and only if A and B are both left cosets of the same finite subgroup H, thus A = x·H and B = y·H for some x,y ∈ G.

We leave the proof as an exercise. Observe that d(A,B) = 0 does not necessarily imply that A or B has small doubling; if x or y lie outside the normalizer of H then A^{2} or B^{2} can be significantly larger than A or B. Similarly we see that d(A,B) = 0 does not imply that d(A,B^{−1}) = 0. So there does not appear to be an analogue of Proposition 2.11. Furthermore, if A has small doubling, e.g. |A · A| ≤ K|A|, this does not necessarily imply that the iterated product sets of A are also small. Consider for instance A := H ∪{x} where H is a subgroup of G and x lies outside the normalizer N(H) of H. Then A·A = H ∪(x·H)∪(H · x)∪{x^{2}}, so |A·A| ≤ 3|A|−2; but A·A·A contains H ·x·H which can be as large as |H|^{2} = (|A|−1)^{2}.

However, we can still obtain a reasonable characterization of sets of small doubling, or pairs of sets with small Ruzsa distance, by passing to subsets and using Balog-Szemerédi-Gowers techniques. Define the multiplicative energy E(A,B) between two multiplicative sets A,B with common ambient group to be

E(A,B) := |{(a,a′,b,b′) ∈ A×A×B ×B : ab = a′b′}|.  (2.38)

A significant difficulty here is that E(A,B) obeys far fewer symmetries in the non-commutative case than in the commutative case; indeed, the only symmetry available is that E(A,B) = E(B^{−1},A^{−1}). However in the case when B = A^{−1} we have a crucial additional identity E(A,A^{−1}) = E(A^{−1},A) (see exercises), which can be thought of as a very weak, restricted form of commutativity.

There is a non-commutative Balog-Szemerédi-Gowers theorem:

Lemma 2.38 (Multiplicative Balog-Szemerédi-Gowers theorem) Let A,B be multiplicative sets in an ambient group Z, let K,K′,ε > 0, and let G ⊂ A×B be such that

|G| ≥ |A||B|/K and |A·_{G} B| ≤ K′|A|^{1/2}|B|^{1/2}.

Then there exists a subset A′ ⊂ A such that

|A′| ≥ \frac{|A|}{\sqrt{2}K}

and a set G′ ⊆ A′ ×(A′)^{−1} such that

|G′| ≥ (1−ε)|A′|^{2}

with the property that

|A′ ·_{G′} (A′)^{−1}| ≤ \frac{2(KK′)^{2}}{ε} |A|.

We defer the proof of this Lemma to Section 6.4. This can then be used to deduce the following corollary (see Exercise 10):

Corollary 2.39 Let A,B be multiplicative sets in an ambient group Z such that E(A,B) ≥ |A|^{3/2}|B|^{3/2}/K for some K > 1. Then there exists a subset A′ ⊂ A such that |A′| = Ω(K^{−O(1)}|A|) and |A′ · (A′)^{−1}| = O(K^{O(1)}|A|) for some absolute constant C.

Combining this with the identity E(A,A^{−1}) = E(A^{−1},A) we obtain the following weak commutativity property between A and A^{−1}:

Corollary 2.40 Let A be a multiplicative set such that d(A^{−1},A^{−1}) ≤ logK for some K ≥ 1. Then there exists a subset A′ ⊂ A such that |A′| = Ω(K^{−O(1)}|A|) and |A′ ·(A′)^{−1}| = O(K^{O(1)}|A|).

To use this conclusion, we use the following variant of Lemma 2.12, whose proof we leave as an exercise.

Lemma 2.41 Let A be a multiplicative set. Then there exists S ⊂ A^{−1} ·A such that |S| ≥ |A|/2 and

|A·S^{·n} ·A^{−1}| ≤ \frac{2^{n}|A·A^{−1}|^{n+1}|A^{−1} ·A|^{n}}{|A|^{2n}}

for all integers n ≥ 0.

Combining this result with Corollary 2.40 and the Ruzsa triangle inequality gives

Corollary 2.42 Let A be a multiplicative set such that d(A,B) ≤ logK for some K ≥ 1. Then there exists a set S such that |S| ≥ Ω(K^{−O(1)}|A|) and

|A·S^{·n} ·A^{−1}| ≤ O(K)^{O(1+n)}|A|

for all integers n ≥ 0.

From this, the Ruzsa triangle inequality, and the Ruzsa covering lemma, one can now obtain a characterization of when the Ruzsa distance is small or when the multiplicative energy is large. Define a multiplicative K-approximate group to be any multiplicative set H which is symmetric (so H = H^{−1}), contains the identity, and is such that there exists a set X of cardinality |X| ≤ K such that we have the inclusions

H ·H ⊆ X ·H ⊆ H ·X ·X; H ·H ⊆ H ·X ⊆ X ·X ·H.

Theorem 2.43 Let A,B be multiplicative sets in a group G, and let K ≥ 1. Then the following statements are equivalent up to constants, in the sense that if the j^{th} property holds for some absolute constant C_{j}, then the k^{th} property will also hold for some absolute constant C_{k} depending on C_{j}:

(i) d(A,B) ≤ C_{1}(1+logK).

(ii) There exists a C_{2}K^{C_{2}}-approximate group H such that |H| ≤ C_{2}K^{C_{2}}|A|, A ⊂ X · H and B ⊂ Y · H for some multiplicative sets X,Y of cardinality at most C_{2}K^{C_{2}}.

Theorem 2.44 Let A,B be multiplicative sets in a group G, and let K ≥ 1. Then the following statements are equivalent up to constants, in the sense that if the j^{th} property holds for some absolute constant C_{j}, then the k^{th} property will also hold for some absolute constant C_{k} depending on C_{j}:

(i) E(A,B) ≥ C_{1}^{−1}K^{−C_{1}}|A|^{3/2}|B|^{3/2}.

(ii) There exists a subset G ⊂ A·B with |G| ≥ C_{2}^{−1}K^{−C_{2}}|A||B| such

that |A·_{G} B| ≤ C_{2}K^{C_{2}}|A|^{1/2}|B|^{1/2}. (iii) There exists a C_{3}K^{C_{3}}-approximate group H and x,y ∈ G such

that |H| ≤ C_{3}K^{C_{3}}|A|^{1/2}|B|^{1/2} and

|A∩(x·H)|,|B ∩(H ·y)| ≥ C_{3}^{−1}K^{−C_{3}}|H|.

We leave the proofs of these theorems to the exercises. Despite these characterizations, there is much left to be done in the study of product sets in non-commutative groups; for instance no analogue of Freiman’s theorem is currently known, except in the case of very small doubling (at most 2) in torsion-free abelian groups; see [165].

2.7.1 Prove a multiplicative version of Lemma 2.1. 2.7.2 Prove a multiplicative version of Lemma 2.6. 2.7.3 Prove Proposition 2.37. 2.7.4 Let (A,G) be a multiplicative set. Prove that |A · A| = |A| if

and only if A is a normal coset of H, i.e. A = x·H = H ·x for some x ∈ N(H).

2.7.5 Let A be a multiplicative set, and define Dbl[A] := |A·A|/|A|. Using the Ruzsa triangle inequality, establish the inequality

Dbl[A∪A^{−1}] ≤ 2Dbl[A]+2Dbl[A]^{2}.

Thus any multiplicative set of small doubling can be placed inside a symmetric set of at most twice the cardinality and not much larger doubling.

2.7.6 Let A be a symmetric multiplicative set, so A = A^{−1}, and let Dbl_{n}[A] denote the n-fold doubling numbers |A^{n}|/|A|. Using the Ruzsa triangle inequality, show that Dbl_{m+n−2}[A] ≤ Dbl_{m}[A]Dbl_{n}[A] for all m,n ≥ 2.

2.7.7 Let A be a multiplicative set (not necessarily symmetric), and suppose that |A·A·A| ≤ K|A| for some K ≥ 1. Using the Ruzsa triangle inequality, show that for each integer n ≥ 1 there exists a constant C_{n} depending only on n such that

|(A∪A^{−1})^{n}| ≤ C_{n}K^{C_{n}}|A|.

(Hint: first establish this for n = 2,3 using many applications of the Ruzsa triangle inequality, and then use Exercise 6).

2.7.8 Let A and B be multiplicative sets. Establish the identities E(A,B) = E(B^{−1},A^{−1}) and E(A,A^{−1}) = E(A^{−1},A), and the inequality E(A,B) ≥ \frac{|A|^{2}|B|^{2}}{|A·B|}.

2.7.9 Let A,B,C be additive sets in an ambient group Z, let 0 < ε < 1/4, and let G ⊂ A × B^{−1},H ⊂ B × C^{−1} be such that |G| ≥ (1 − ε)|A||B| and |H| ≥ (1 − ε)|B||C|. By modifying the solution of Exercise 4 in Section 2.5, show that there exists subsets A′ ⊆ A and C′ ⊆ C with |A′| ≥ (1 − ε^{1/2})|A| and |C′| ≥ (1−ε^{1/2})|C| such that |A′ ·(C′)^{−1}| ≤ \frac{|A·_{G}B^{−1}||B·_{H}C^{−1}|}{(1−2ε^{1/2})|B|}.

2.7.10 Prove Corollary 2.39. (Hint: first develop a multiplicative analogue of Lemma 2.29 to establish a large set G ⊂ A × B such that A ·_{G} B is small. Then apply Lemma 2.38 with ε = 1/9 (say), and then apply Exercise 9.)

2.7.11 Let A be a multiplicative set such that |A · A^{−1}| ≤ K|A| and |A^{−1} ·A| ≤ K|A|. Show that there exists a subset Ã of A such that |Ã| ≥ |A|/2K and

|Ã·Ã^{−1} ·...·Ã^{(−1)^{n+1}}| ≤ 2^{n−2}K^{2n−3}|A|

for all n ≥ 2, where the product consists of n factors alternating between Ã and Ã^{−1}.

2.7.12 If A and B are multiplicative sets in a group G, show that there exist sets X_{1},X_{2} ⊆ A such that |X_{1}| ≤ \frac{|A·B|}{|B|}, |X_{2}| ≤ \frac{|B·A|}{|B|}, and A ⊆ X_{1} ·B ·B^{−1} and A ⊆ B^{−1} ·B ·X_{2}, by modifying the proof of Lemma 2.13.

2.7.13 Prove Lemma 2.41. 2.7.14 Show that the direct analogue of Proposition 2.17 fails in the

non-commutative case, even when A = B = A^{−1}.

2.7.15 Let A,B be multiplicative sets in an ambient group G, and let Ã be the set

Ã := \{a ∈ A : |\{(a′,b,b′) ∈ A×B×B : a = a′b′b^{−1}\}| ≥ \frac{|A||B|^{2}}{2|A·B|}\}.

Establish the bounds

|Ã| ≥ \frac{|A|^{2}}{2|A·B|}

and

|A·Ã^{−1} ·Ã·A^{−1}| ≤ 4\frac{|A·B|^{4}|A^{−1} ·A|}{|A|^{4}}.

Compare this against Exercise 14. Hint: if x := a_{1}a_{2}^{−1}a_{3}a_{4}^{−1} be a typical element of A · Ã^{−1} · Ã· A^{−1}, obtain at least (\frac{|A||B|^{2}}{2|A·B|})^{2} representations of the form

x = [a_{1}b_{2}](b′_{2})^{−1}[(a′_{2})^{−1}a′_{3}]b′_{3}[a_{4}b_{2}]^{−1}

where a_{1}b_{2},a_{4}b_{2} ∈ A·B, b′_{2},b′_{3} ∈ B, and (a′_{2})^{−1}a′_{3} ∈ A^{−1} ·A. 2.7.16 Prove Theorem 2.43. (Hint: the backwards implication is easy.

For the forward implication, first use Corollary 2.42 to obtain the set S, and set H := S^{−1} ∪S. Use Exercise 7, Exercise 12, and the Ruzsa triangle inequality several times. Alternatively one can proceed using Exercise 15.)

2.7.17 Prove Theorem 2.44.

2.8 Elementary sum-product estimates

We now discuss some results concerning the sum-set and product-set of a subset A of a commutative ring Z, thus combining both the additive and multiplicative theory of the preceding sections (but keeping the multiplication commutative, for simplicity). The question here is to analyze the extent to which a set A can be approximately closed under addition and multiplication simultaneously. Of course, one way that this can happen is if A is a subring of Z; it appears that up to trivial changes (such as removing some elements, adding a small number of new elements, or dilating the set) that this is essentially the only such example, although we currently only have a satisfactory and complete formalization of this principle when Z is a field (Theorem 2.51). In some ways the theory here is in fact easier than the sum set theory, because one can exploit two rather different structures arising from the smallness of A + A and the smallness of A · A to obtain a conclusion. As in the rest of this chapter, our discussion is for general fields, with a particular emphasis on the finite field Z_{p}. We remark that for the field R much better results are known, see Sections 8.3, 8.5.

In this section Z will always denote a commutative ring, and Z^{×} will denote the elements of Z which are not zero-divisors; these form a multiplicative cancellative monoid in Z. The situation is significantly better understood in the case that Z is a field (see in particular Theorem 2.51 below); in such cases we shall emphasize this by writing the field as F instead of Z; note that F^{×} is simply F \{0} in this case. A fundamental concept in that setting is that of a quotient set, which is the arithmetic equivalent of the concept of a quotient field of a division ring.

Definition 2.45 (Quotient set) Let A be a finite subset of a field F such that |A| ≥ 2. Then the quotient set Q[A] of A is defined to be

Q[A] := \frac{A−A}{(A−A)\0} := \{\frac{a−b}{c−d} : a,b,c,d ∈ A;c ≠ d\}.

We also set Q[A]^{×} := Q[A]\0 to be the invertible elements in Q[A].

Observe that Q[A] contains both 0 and 1, and is symmetric under both additive and multiplicative inversion, thus Q[A] = −Q[A] and Q[A]^{×} = (Q[A]^{×})^{−1}. It is also invariant under translations and dilations of A, thus Q[A] = Q[A+x] = Q[λ·A] for all x ∈ F and λ ∈ F^{×}. Geometrically, Q[A] can be viewed as the set of slopes of lines connecting points in A×A.

The relevance of the quotient set to sum-product estimates lies in the trivial but fundamental observation:

Lemma 2.46 Let A be a finite subset of a field F such that |A| ≥ 2, and let x ∈ F. Then |A+x·A| = |A|^{2} if and only if x ∉ Q[A].

Proof We have |A+x·A| = |A|^{2} if and only if the map (a,b) ↦ a+xb is injective on A×A, which is true if and only if a+xb ≠ c+xd for all distinct (a,b),(c,d) ∈ A × A, which after some algebra is equivalent to asserting that x ∉ Q[A].

This has an immediate corollary:

Corollary 2.47 If A is a subset of a finite field F such that |A| > |F |^{1/2}, then Q[A] = F.

Note that the condition |A| > |F |^{1/2} is absolutely sharp, as can be seen by considering the case when A is a subfield of F of index 2.

Lemma 2.46 has another important consequence: it gives a criterion under which Q[A] is a subfield of F.

Corollary 2.48 Let A be a finite subfield of a field F such that |A| ≥ 2 and

|A+Q[A]·Q[A]·A|,|A+(Q[A]+Q[A])·A| < |A|^{2}.

Then Q[A] is a subfield of F.

This corollary may be compared with Exercise 5 from Section 2.6.

Proof From Lemma 2.46 and the hypotheses we see that Q[A]·Q[A] ⊆ Q[A] and Q[A] + Q[A] ⊆ Q[A]. In particular Q[A]^{×} · Q[A]^{×} = Q[A]^{×}. Since Q[A] is finite and contains 0,1, we see from Proposition 2.7 that Q[A] is an additive group, and similarly from the multiplicative version of this Proposition we see that Q[A]^{×} is a multiplicative group. The claim follows.

In order to use this corollary, one needs to control rational expressions of A such as A+Q[A]·Q[A]·A. In analogy with sum set estimates such as Corollary 2.22, one might first expect that once |A + A| ≤ K|A| and |A · A| ≤ K|A|, then all polynomial or rational expressions of A are controlled in cardinality by CK^{C}|A|. This however is not the case, even if one normalizes A to contain 0 and 1. To see this, consider A = G∪{x} where G is a subfield of F and x ∉ G. Then one easily verifies |A+A|,|A·A| < 2|A| but |A·A+A·A| ≥ (|A|−1)^{2}, since A·A+A·A contains G+x·G, which has size |G|^{2} by Lemma 2.46. This example is similar to one appearing in the preceding section, and it is resolved in a similar way, namely by passing from A to a subset of A.

Lemma 2.49 (Katz-Tao lemma) [190], , [40] Let Z be a commutative ring, and let A ⊆ Z^{∗} be a finite non-empty subset such that |A+A| ≤ K|A| and |A·A| ≤ K|A| for some K ≥ 1. Then there exists a subset A′ of A such that |A′| ≥ |A|/2K −1 and |A′ ·A′ −A′ ·A′| = O(K^{O(1)}|A′|).

Note that this lemma works in arbitrary commutative rings, not just in fields. The requirement that none of the elements of A be zero divisors is not serious in the case of a field, since one can simply remove the origin 0 from A if necessary, but is a non-trivial requirement in other commutative rings.

Proof We use an argument from [40]. We may assume that A > 10K (for instance) since the claim is trivial otherwise. Consider the dilates {a·A : a ∈ A} of A. Since a ∈ Z^{∗}, a·A has the same cardinality as A. In particular we have

∑_{x∈A·A} ∑_{a∈A}1_{a·A}(x) = |A|^{2}.

Since |A·A| ≤ K|A|, we may apply Cauchy-Schwarz and conclude

∑_{x∈A·A}(∑_{a∈A}1_{a·A}(x))^{2} ≥ |A|^{3}/K.

We rearrange this as

∑_{a,b∈A}|(a·A)∩(b·A)| ≥ |A|^{3}/K.

By the pigeonhole principle we can thus find a b ∈ A such that

∑_{a∈A}|(a·A)∩(b·A)| ≥ |A|^{2}/K.

Fix this b. Setting A′ to be the set of all a ∈ A′ such that

|(a·A)∩(b·A)| ≥ |A|/2K

we conclude that

∑_{a∈A′} |(a·A)∩(b·A)| ≥ |A|^{2}/2K.

and hence |A′| ≤ |A|/2K. By shrinking A′ by one if necessary we may assume b ∉ A′. Now recall the Ruzsa distance d(A,B) := log \frac{|A−B|}{|A|^{1/2}|B|^{1/2}} , and observe that d(a·A,a·B) = d(A,B) whenever a is not a zero divisor. Then d(A,A) ≤ 2d(A,−A) = 2logK, and hence

d(a·A,a·A) = d(b·A,b·A) = d(A,A) ≤ 2logK for all a ∈ A′.

Since (a·A)∩(b·A) is a large subset of a·A and b·A, one can the compute

d(a·A,a·A∩b·A),d(b·A,a·A∩b·A) = O(1+logK)

and hence by the Ruzsa triangle inequality

d(a·A,b·A) = O(1+logK) for all a ∈ A′.  (2.39)

Dilating this, we obtain

d(a_{1}a_{2} ·A,ba_{2} ·A),d(ba_{2} ·A,b^{2} ·A) = O(1+logK) for all a_{1},a_{2} ∈ A′ and hence by the Ruzsa triangle inequality

d(a_{1}a_{2} ·A,b^{2} ·A) = O(1+logK) for all a_{1},a_{2} ∈ A′.  (2.40)

To proceed further we need to “invert” elements in A. For any a ∈ A let â := ∏_{a′∈A\{a}} a′ ∈ Z^{×}. By dilating (2.39) (with a replaced by a_{3}) by a_{1}a_{2} ∏_{a′∈A\{a_{3},b}} a′ for a_{1},a_{2},a_{3} ∈ A′, we obtain

d(a_{1}a_{2}b̂·A,a_{1}a_{2}â_{3} ·A) = O(1+logK) for all a_{1},a_{2},a_{3} ∈ A′.

Meanwhile, from dilating (2.40) we have

d(a_{1}a_{2}b̂·A,b^{2}b̂·A) = O(1+logK) for all a_{1},a_{2},a_{3} ∈ A′.

Applying the Ruzsa triangle inequality, we thus have

d(a_{1}a_{2}â_{3} ·A,a′_{1}a′_{2}â′_{3} ·A) = O(1+logK) for all a_{1},a_{2},a_{3},a′_{1},a′_{2},a′_{3} ∈ A′

and thus

|a_{1}a_{2}â_{3} ·A−a′_{1}a′_{2}â′_{3} ·A| = O(K^{O(1)})|A|.

In particular we have

∑_{x,y∈A′·A′·Â′} |x·A−y ·A| = O(K^{O(1)})|A||A′ ·A′ ·Â′|^{2}

where Â′ := {â : a ∈ A′}. But since |A·A| ≤ K|A| and |A′| ≥ |A|/2K − 1, we see from the multiplicative version of sum set estimates (working in the formal multiplicative group generated by the cancellative monoid Z^{∗}) that |A·A·Â| = O(K^{O(1)}|A|). We thus have

∑_{x,y∈A′·A′·Â′} |x·A−y ·A| ≤ O(K^{O(1)}|A′|^{3}).

We rewrite the left-hand side as

∑_{z∈Z} |{(x,y,a,b) : x,y ∈ A′ ·A′ ·Â′;a,b ∈ A′;z = xa−yb}|.

Write ω := ∏_{a∈A} a, and observe that whenever a_{1},a_{2},a_{3},a_{4} ∈ A′, the number ω(a_{1}a_{2} − a_{3}a_{4}) has at least |A′|^{2} representations of the form xa−yb with x,y ∈ A′ ·A′ ·Â′ and a,b ∈ A′, thanks to the identity

ω(a_{1}a_{2} −a_{3}a_{4}) = (a_{1}a_{2}â)a−(a_{3}a_{4}b̂)b.

Thus

|ω ·(A′ ·A′ −A′ ·A′)| = O(K^{O(1)}|A′|)

and the claim follows since ω ∈ Z^{∗}.

A modification of the above argument also gives the following statement, which can be viewed as a variant of Corollary 2.22 for the sumproduct setting; we leave the proof to Exercise 1.

Lemma 2.50 [42] Let Z be a commutative ring, and let A ⊆ Z^{×} be a finite non-empty set such that |A · A − A · A| ≤ K|A|. Then we have |A^{k} −A^{k}| ≤ K^{O(k)}|A| for all k ≥ 1, where A^{k} = A·...·A is the k-fold product set of A.

We can now classify those finite subsets of fields with small additive doubling and multiplicative doubling constant, up to polynomial losses:

Theorem 2.51 (Freiman theorem for sum-products) Let A be a finite non-empty subset of a field F, and let K ≥ 1. Then the following statements are equivalent up to constants, in the sense that if the j^{th} property holds for some absolute constant C_{j}, then the k^{th} property will also hold for some absolute constant C_{k} depending on C_{j}:

(i) |A+A| ≤ C_{1}K^{C_{1}}|A| and |A·A| ≤ C_{1}K^{C_{1}}|A|.

(ii) Either |A| ≤ C_{2}K^{C_{2}}, or else there exists a subfield G of F, an non-zero element x ∈ F, and a set X in F such that |G| ≤ C_{2}K^{C_{2}}|A|, |X| ≤ C_{2}K^{C_{2}}, and A ⊆ x·G∪X.

This is a slight strengthening of a result in [42], [43].

Proof We shall only show the forward implication, leaving the easy backward implication to Exercise 2. By relabeling C_{1}K^{C_{1}} as K, we may thus assume that |A + A| ≤ K|A| and |A · A| ≤ K|A|. We may assume that |A| ≥ C_{0}K^{C_{0}} for some large absolute constant C_{0}, since the claim is trivial otherwise. We may also remove 0 from A without any difficulty, thus we may assume A ⊆ F^{∗}. Applying Lemma 2.49 and Lemma 2.50, we may find a subset A′ of A with |A′| = Ω(K^{−O(1)}|A|) and |(A′)^{k} −(A′)^{k}| = O(K)^{O(k)}|A′| for all k ≥ 1. By Corollary 2.22 this implies that

|n(A′)^{k} −m(A′)^{k}| ≤ O(K)^{O_{k,n,m}(1)}|A′| for all n,k,m ≥ 1. (2.41)

By dilating A by a non-zero factor if necessary we may assume 1 ∈ A′ (noting that the hypothesis and conclusion of the theorem are invariant under such dilations).

Now we apply Corollary 2.48. Let D := (A′ − A′)\{0} and G := Q[A′] = (A′ −A′)/D. Using lowest common denominators, we observe that

A′+G·G·A′ ⊆ (A′·D·D−(A′−A′)·(A′−A′)·A′)/D^{2} ⊆ (4(A′)^{3}−4(A′)^{3})/D^{2}.

on the other hand, from (2.41) we have

|(4(A′)^{3} −4(A′)^{3})·D^{2}| = O(K^{O(1)}|A′|),

so by the multiplicative version of Proposition 2.11 we see that

|A′ +G·G·A′| = O(K^{O(1)}|A′|) < |A′|^{2}

if C_{0} is sufficiently large. A similar argument gives |A′ +(G+G)·A′| = O(K^{O(1)}|A′|) < |A′|^{2}. Applying Corollary 2.48 we see that G is in fact a field.

Now let x be a non-zero element of A′, and let y be an element of A′. Then (a−y)/x ∈ Q[A′] = G for all a ∈ A′, thus

A′ ⊆ x·G+y.

Thus

x·G+y = A′ +x·G ⊆ A′ +A′ ·Q[A′]

and hence

(x·G+y)^{2} ⊆ (A′ +A′ ·Q[A′])^{2}.

But an argument using (2.41) and Proposition 2.11 as before gives |(A′+ A′ ·Q[A′])^{2}| = O(K^{O(1)}|A′|) ≤ O(K^{O(1)}|G|). Direct computation shows that |(x·G+y)^{2}| ≥ |G|^{2} unless y ∈ x·G. Thus (if C_{0} is sufficiently large) we can take y ∈ x·G. Since A′ contains 1, we thus have A′ ⊆ G.

Since |A+A′| ≤ K|A| = O(K^{O(1)}|A′|), we may apply Ruzsa’s covering lemma (Lemma 2.13) and cover A by O(K^{O(1)}) translates of A′ − A′, and hence by O(K^{O(1)}) translates of G. A similar argument using the multiplicative version of this lemma (and temporarily removing the noninvertible 0 element from A if necessary) covers A by O(K^{C}) dilates of G. On the other hand, we have |(G·x)∩(G+y)| ≤ 1 whenever x ≠ 1. Thus we have |A\G| = O(K^{O(1)}), and the claim follows.

This theorem implies that at least one of A+A or A·A is large if A does not intersect with a subfield of F:

Corollary 2.52 (Sum-product estimate) [42],[43] Let A be a finite non-empty subset of a field F, and suppose that K ≥ 1 is such that there is no finite subfield G of F of cardinality |G| ≤ K|A| and no x ∈ F such that |A\(x · G)| ≤ K. Then we have either |A| = O(K^{O(1)}) or |A+A|+|A·A| = Ω(K^{c}|A|) for some absolute constant c > 0.

Remark 2.53 In the particular case when F has no finite subfields we thus obtain |A+A|+|A·A| = Ω(|A|^{1+ε}) for some absolute constant ε > 0; this result was first obtained (when F = R) by Erdös and Szemerédi [88]. In the setting of the real line it is was in fact conjectured in [88] that one can take ε arbitrarily close to 1 in the above estimate. For the most recent value of ε, see Theorem 8.15.

In the particular case of the field F = F_{p} of prime order, which has no subfields other than {1} and F_{p}, one obtains

Corollary 2.54 (Sum-product estimate for F_{p}) [42],[43] Let A be a non-empty subset of F_{p}. Then

|A+A|+|A·A| = Ω(min(|A|,|F_{p}|/|A|)^{c}|A|)

for some absolute constant c > 0.

We give a simple application of this estimate below; a somewhat stronger Fourier-analytic version of this estimate will be proven in Theorem 4.40.

Corollary 2.55 Let H be a multiplicative subgroup of F_{p} such that |H| ≥ p^{δ} for some 0 < δ ≤ 1. Then there exists an integer k = k(δ) ≥ 1 depending only on δ such that kH = F_{p}.

Proof Observe that since H is a multiplicative subgroup, kH + kH = 2kH and kH ·kH ⊆ k^{2}H. Thus for k ≥ 2 we may apply Corollary 2.54 and conclude that

|k^{2}H| ≥ cp^{cδ}|kH| whenever |kH| ≤ p^{3/4}

(for instance). Iterating this starting with the observation |2H| ≥ |H| ≥ p^{δ}, this shows that for k sufficiently large depending only on δ, that |kH| > p^{3/4}. To conclude one can use Lemma 4.10 from Chapter 4.

By combining Corollary 2.55 with the asymmetric Balog-Szemerédi theorem, we can show that multiplicative subgroups of F_{p} cannot have high additive energy:

Corollary 2.56 Let H be a multiplicative subgroup of F_{p} such that |H| ≥ p^{δ} for some 0 < δ ≤ 1. Then there exists an ε = ε(δ) > 0 depending only on δ, such that E(A,H) ≤ p^{−ε}|A||H|^{2} for all A ⊆ F_{p} with 1 ≤ |A| ≤ p^{1−δ}, if p is sufficiently large depending on δ. In particular, from (2.9) we have the expanding inequality |A+H| ≥ p^{ε}|A|.

Proof Let ε′ = ε′(δ) > 0 be a small number to be chosen later, and let ε = ε(ε′,δ′) > 0 be an even smaller number to be chosen later. Suppose for contradiction that there existed a set A such that E(A,H) ≥ p^{−ε}|A||H|^{2}. Applying Corollary 2.35 (with L := p and ε replaced by ε′) we obtain that

p = |kH| ≤ |A+kH| = O_{k,ε,ε′}(p^{O_{k,ε′} (ε)+O_{k}(ε′)}|A|) ≤ O_{k,ε,ε′}p^{O_{k,ε′} (ε)+O_{k}(ε′)−δ}p,

where k = k(δ) is the quantity in Corollary 2.55. But this yields a contradiction if ε′ is chosen sufficiently small depending on k, δ, and then ε is chosen sufficiently small depending on k, δ, ε′, and p is assumed sufficiently large depending on k,ε,ε′,δ.

We shall apply this to exponential sums over multiplicative subgroups; see Theorem 4.40. For a variant of this estimate, see Lemma 9.41.

It seems of interest to obtain estimates of this type for more general commutative rings, and possibly even to non-commutative rings by combining these arguments with those in the preceding section. In this direction, Bourgain has established

Theorem 2.57 [40] Let p be a large prime, and let A be a subset of the commutative ring F_{p} ×F_{p} (endowed with the product structure (a,b)· (c,d) = (ac,bd)) be such that |A| ≥ p^{δ} and |A + A|,|A · A| ≤ p^{ε}|A| for some δ,ε > 0. Then there exists a set G of F_{p} × F_{p} such that |G| ≤ p^{O_{δ}(ε)}|A| and |A ∩ G| ≥ p^{−O_{δ}(ε)}|A|, where G is one of the following objects:

• The whole space G = F_{p} ×F_{p};

• A horizontal line G = F_{p} ×{a} for some a ∈ F_{p};

• A vertical line G = {a}×F_{p} for some a ∈ F_{p};

• A line G = {(x,ax) : x ∈ F_{p}} for some a ∈ F_{p}^{×}.

We sketch a proof of this proposition in the exercises. This is not as complete a characterization of sets with small sum-product as Theorem 2.51 - in particular, it does not address the case of very small A - but is already sufficient to control a number of exponential sums of importance in number theory and cryptography. See [40], [39].

The problem of obtaining good sum-product estimates when the ambient commutative ring is the integers Z = Z has attracted a lot of interest. In this case it has been conjectured by Erdös and Szemerédi [88] that

|kA|+|A^{k}| = Ω_{k,ε}(|A|^{k−ε})  (2.42)

for all ε > 0, all k ≥ 2 and all additive sets A ⊂ Z. Even the k = 2 case is open (and considered very difficult); this k = 2 case has currently been verified for all ε > \frac{8}{11}, see Theorem 8.15. In another direction towards (2.42), a recent result of Bourgain and Chang [41] has shown that for every m > 1 there exists an integer k = k(m) ≥ 1 such that

|kA|+|A^{k}| = Ω_{m}(|A|^{m})  (2.43)

for all additive sets A ⊂ Z. This last result is rather deep, in particular using an intricate “induction on scales” argument, coupled with some quantitative Freiman-type theorems.

2.8.1 [40] Modify the proof of Lemma 2.49 to prove Lemma 2.50. (Hint: first use multiple applications of the triangle inequality to obtain control on |x·A−y ·A| for all x,y ∈ A^{k} ·Â).

2.8.2 Prove the remaining implication in Theorem 2.51. 2.8.3 Deduce Corollary 2.52 and Corollary 2.54 from Theorem 2.51. 2.8.4 [43], [42] Let A,A′,B be non-empty subsets of a field F such

that 0 ∉ B. Using the first moment method, show that there exists ξ ∈ B such that

E(A,ξ ·A′) ≤ \frac{|A|^{2}|A′|^{2}}{|B|} +|A||A′|

and conclude from (2.9) that

|A+ξ ·A′| ≥ \frac{|A||A′||B|}{|A||A′|+|B|}.

2.8.5 [43] Let A be a subset of a finite field F such that |A| > |F |^{1/2}. Show that |(A − A) · A + (A − A) · A| ≥ sup_{x∈F} |A + x · A| ≥ \frac{|F |}{2} and then conclude that

F = (A−A)·A+(A−A)·A+(A−A)·A+(A−A)·A.

Hints: the first inequality follows easily from Corollary 2.47. For the second inequality, use Exercise 4.

2.8.6 (Croot, personal communication) Let A be a subset of a finite field F such that |A| > |F |^{1/k} for some integer k ≥ 2. Show that |Q[A]| ≥ |F |^{1/(k−1)}; this clearly generalizes Corollary 2.47. (Hint: exploit the fact that the maps (a_{1},...,a_{k}) ↦ x_{1}a_{1} +...+ x_{k}a_{k} fail to be injective for arbitrary x_{1},...,x_{k} ∈ F).

2.8.7 [42] Let A be a subset of a field F such that |A| ≥ |F |^{ε} for some ε > 0. Show that there exists an integer k = k(ε) > 1 depending only on ε such that k(A^{k})−k(A^{k}) = G for some subfield G of F. (Use Exercise 5 or Lemma 4.10).

2.8.8 [40] Let F_{p} be a field of prime order p and Z = F_{p}×F_{p}. Let A ⊆ Z be such that |A∩({a}×F_{p})| ≥ p^{δ} and |A∩({b}×F_{p})| ≥ p^{δ} for some 0 < δ < 1 and a,b ∈ F_{p}. Show that for some k = k(δ) > 0 we have k(A^{k})−k(A^{k}) = Z. (Hint: use Exercise 7).

2.8.9 [40] Let F_{p}, Z, be as in Exercise 8, and let π_{1} : Z → F_{p}, π_{2} : Z → F_{p} be the co-ordinate projections. Suppose that A ⊆ Z is such that |π_{1}(A)|,|π_{2}(A)| ≥ p^{δ} for some 0 < δ < 1 and such that at least one of π_{1},π_{2} is not injective. Show that for some k = k(δ) > 0 we have k(A^{k})−k(A^{k}) = Z. (Hint: by Exercise 8 it suffices to find some k′ such that k′(A^{k′})−k′(A^{k′}) contains a large intersection with either a horizontal line or a vertical line).

2.8.10 [40] Let F_{p}, Z, π_{1}, π_{2} be as in Exercises 8, 9. Suppose that that A ⊆ Z is such that |π_{1}(A)|,|π_{2}(A)| ≥ p^{δ} for some 0 < δ < 1. Show that either A is contained in a line {(x,ax) : x ∈ F_{p}} for some a ∈ F_{p}^{×}, or else k(A^{k})−k(A^{k}) = Z for some k = k(δ) > 0. (Hint: by Exercise 7 one can reduce to the case where π_{p}(A) = π_{2}(A) = F_{p}. Now divide into two cases depending on whether π_{1} or π_{2} is injective on 2A−2A or not).

2.8.11 [40] Use Exercise 10 and Lemmas 2.49, 2.50 to deduce Theorem 2.57. (You will have to take a small amount of care concerning the zero-divisors {0}×F_{p} ∪F_{p} ×{0}).

2.8.12 Z be a commutative ring, and let A_{1},A_{2},A_{3},A_{4} be subsets of Z^{×} such that |A_{1}| = |A_{2}| = |A_{3}| = |A_{4}| = N and |A_{1}·A_{2}−A_{3}·A_{4}| ≤ KN. Show that |A_{j} ·A_{j} −A_{j} ·A_{j}| ≤ K^{O(1)}N for all j = 1,2,3,4 and some absolute constant C. This lemma allows one to extend several of the above results to the setting where the single set A is replaced by a number of sets of comparable cardinality.

2.8.13 [39] Prove the following variant of Corollary 2.55: for any δ > 0 there exists a ε > 0 and k ≥ 1 such that whenever H is a subset of F_{p} with |H| ≥ p^{δ} and p sufficiently large depending on δ, then either |kH| ≥ p^{1−δ} or |H ·H| ≥ p^{ε}|H|. (Hint: assume that |H ·H| < p^{ε}|H| and use the multiplicative version of Proposition 2.25 to replace H by an approximate multiplicative group. One can in fact ensure that kH = F_{p} for k large enough, but this requires some additional effort; see [39]).

2.8.14 [39] Prove the following variant of Corollary 2.56: for any δ > 0 there exists a ε > 0 such that whenever H,A are subsets of F_{p} with |H| ≥ p^{δ}, |H · H| ≤ p^{ε}|H|, and 1 < |A| < p^{1−δ}, then E(A,H) = O_{δ}(p^{−ε}|A||H|^{2}). In particular we have |A + H| = Ω_{δ}(p^{ε}|H|).

2.8.15 [18] Let A be an additive set in F_{p} such that |A| < p^{1−δ} for some δ > 0. Show that there exists an ε > 0 depending on δ such that |{(a,b,c,d,e,f) ∈ A^{6} : ab + c = de + f}| = O_{ε,δ}(|A|^{5−ε}). (Hint: use the Balog-Szemerédi-Gowers theorem in both the additive and multiplicative forms, together with Corollary 2.54). This estimate is used in [18] to show that iterations of the map X ↦ X_{1} ·X_{2} +X_{3} on random variables in F_{p} (where X_{1},X_{2},X_{3} are independent trials of X) converge in a certain sense to the uniform distribution, which has applications to random number generation.

3

Additive geometry

In Chapter 2 we studied the elementary theory of sum sets A+B for general subsets A, B of an arbitrary additive group Z. In order to progress further with this theory, it is important to first understand an important subclass of such sets, namely those sets with a strong geometric and additive structure. Examples include (generalized) arithmetic progressions, convex sets, lattices, and finite subgroups. We will term the study of such sets (for want of a better name) additive geometry; this includes in particular the classical convex geometry of Minkowski (also known as geometry of numbers). Our aim here is to classify these sets and to understand the relationship between their geometrical structure, their dimension (or rank), their size (or volume, or measure), and their behavior under addition or subtraction. Despite looking rather different at first glance, it will transpire that progressions, lattices, groups, and convex bodies are all related to each other, both in a rigorous sense and also on the level of heuristic analogy. For instance, progressions and lattices play a similar role in arithmetic combinatorics that balls and subspaces play in the theory of normed vector spaces. In later sections, by combining methods of additive geometry, sum set estimates, Fourier analysis, and Freiman homomorphisms, we will be able to prove Freiman’s theorem, which shows that all sets with small doubling constant can be efficiently approximated by progressions and similarly structured sets.

Closely related to all of these additive geometric sets are the Bohr sets, which are in many ways the dual object to progressions, but we shall postpone the discussion of these sets (and their relationship with progressions) in Section 4.4, once we have introduced the Fourier transform.

3.1 Additive groups

We first review the theory of additive groups, which we introduced in Definition 0.1, obtaining in particular the classification theorem for finitely generated additive groups (Corollary 3.9). This is a fundamental result in additive group theory, but it will also motivate similar results concerning other additively structured sets such as progressions, Bohr sets, and the intersection of convex sets and lattices.

Typical examples of additive groups include the integers Z, the reals R, the lattices Z^{d}, the Euclidean spaces R^{d}, the torus groups R^{d}/Z^{d}, and the cyclic groups Z_{N} := Z/N ·Z. Note that the direct sum Z ⊕Z′ of two additive groups is again an additive group. We now make an important distinction between torsion groups and torsion-free groups.

Definition 3.1 (Torsion) If Z is an additive group and x ∈ Z, we let ord(x) be the least integer n ≥ 1 such that n·x = 0, or ord(x) = +∞ if no such integer exists. We say that Z is a torsion group if ord(x) is finite for all x ∈ Z, and we say that it is an r-torsion group for some r ≥ 1 if ord(x) divides r for all x ∈ Z. We say that Z is torsion-free if ord(x) = +∞ for all x ∈ Z.

Examples 3.2 The groups Z, R, Z^{d}, R^{d} are torsion-free, whereas any finite group such as Z_{N} is a torsion group.

A group homomorphism ϕ : Z → Z′ between two additive groups Z,Z′ is any map which preserves addition, negation, and zero (thus ϕ(x+y) = ϕ(x)+ϕ(y), ϕ(−x) = −ϕ(x), and ϕ(0) = 0 for all x,y ∈ Z). If ϕ is also invertible, then the inverse ϕ^{−1} is automatically a group homomorphism, and we say that ϕ is an group isomorphism, and Z and Z′ are group isomorphic. Since all of our notions here shall be defined in terms of the addition, negation, and zero operations, they will all be preserved by group isomorphism, and so we will treat group isomorphic groups to be essentially equivalent. Later on we shall develop a weaker notion of Freiman homomorphism and Freiman isomorphism which is more suitable for the study of “approximate groups” (sets that are “almost” closed under addition); see Section 5.3.

If G is a subgroup of an additive group Z, then we can form the quotient group

Z/G := {x+G : x ∈ Z}

formed by taking all the cosets of G; this is easily verified to be a group (though it is no longer a subgroup of Z). For instance, the cyclic group Z/N ·Z is the quotient of the integers Z by the subgroup N ·Z. Observe that the map π : Z → Z/G defined by π(x) := x + G is a surjective homomorphism.

The sumset G+H and intersection G∩H of two subgroups are still subgroups. Indeed, the arbitrary intersection of a family of subgroups is still a subgroup. Hence, given any subset X of Z, we can define the span ⟨X⟩ of Z to be the smallest subgroup of Z which contains X; equivalently, ⟨X⟩ is the space of all finite Z-linear combinations of elements of X. Thus for instance if x ∈ Z, then ⟨x⟩ is a group with cardinality ord(x). We say that a additive group Z is finitely generated if it can be written as the span Z = ⟨X⟩ of some finite set X. Clearly, every additive set X is contained in at least one finitely generated group, namely ⟨X⟩. Thus in the theory of additive sets one can usually reduce to the case when the ambient group Z is finitely generated (though it is sometimes convenient to work in some selected non-finitely generated additive groups, such as Q, R, or R^{d}). In Corollary 3.9, we shall completely classify all finitely generated additive groups up to isomorphism.

Let v = (v_{1},...,v_{d}) ∈ Z^{d} denote a d-tuple of elements in Z. We can rewrite the span ⟨v⟩ := ⟨{v_{1},...,v_{d}}⟩ of this d-tuple in the following manner. For any element n = (n_{1},...,n_{d}) ∈ Z^{d} we define the dot product n·v in the usual manner as

n·v := n_{1}v_{1} +...+n_{d}v_{d}.

The map n ↦ n·v is then a homomorphism from Z^{d} to Z, and its image Z^{d} ·v is precisely the span of v:

⟨v⟩ = Z^{d} ·v.

The notion of a progression, introduced in Definition 0.2, is a truncated version of the concept of a span, in which the infinite lattice Z^{d} is replaced instead by a box. Alternatively, one can think of lattices as infinite progressions.

3.1.1 Lattices We now study a special type of additive group, namely the lattices in Euclidean space.

Definition 3.3 (Lattices) A lattice Γ in R^{d} is any additive subgroup of the Euclidean space R^{d} which is discrete (i.e. every point in Γ is isolated). We define the rank k of Γ to be the dimension of the linear space spanned by the elements of Γ, thus 0 ≤ k ≤ d. If k = d, we say that Γ has full rank. If Γ′ is another lattice in R^{d} which is contained in Γ, we say that Γ′ is a sub-lattice of Γ.

Thus for instance Z^{d} is a lattice of full rank in R^{d}. More generally, a typical example of a lattice of rank k is the set Z^{k} · v, where v = (v_{1},...,v_{k}) is a collection of linearly independent vectors in R^{d} for some 0 ≤ k ≤ d. In fact, this is the only possible type of lattice, as we shall see in Lemma 3.4. We observe that if T : R^{d} → R^{d} is an invertible linear transformation on R^{d}, and Γ is a lattice, then T (Γ) is also a lattice with the same rank as Γ.

If Γ is a lattice, then the quotient space R^{d}/Γ is a smooth manifold with a natural Lebesgue (or Haar) measure induced from R^{d}. If Γ has full rank, it is easy to see that R^{d}/Γ is also compact, and thus has a volume mes(R^{d}/Γ), which we refer to as the covolume of Γ.

Next, we classify all lattices in R^{d}. Call a vector v in Γ irreducible if v/n ∉ Γ for any integer n ≥ 0.

Lemma 3.4 (Fundamental theorem of lattices) If Γ is a lattice in R^{d} of rank k, then there exist linearly independent vectors v_{1},...,v_{k} in R^{d} such that Γ = Z^{k} ·v. In particular every lattice of rank k is finitely generated and is isomorphic (via an invertible linear transformation from the linear span of Γ to R^{d}) to the standard lattice Z^{k}. Furthermore, if w is an irreducible vector in Γ, we may choose the above representation Γ = Z^{k} ·v so that v_{1} = w.

Proof We first observe that we may assume that the vectors in Γ span R^{d}, else we could pass from R^{d} to a smaller vector space and continue the argument. In other words, we may assume that the rank k of Γ is equal to d. We may also clearly assume that d ≥ 1, since the d = 0 case is vacuously true.

Observe that Γ contains at least one irreducible vector w, since one can start with any non-zero vector v in Γ and take w to be the smallest vector of the form v/n (such a vector must exist since Γ is discrete). Now let w be an irreducible vector. By the full rank assumption, we can find d linearly independent vectors v_{1},...,v_{d} in Γ with v_{1} = w, so in particular the volume |v_{1} ∧...∧v_{d}| of the parallelopiped spanned by v_{1},...,v_{d} is strictly positive. Since Γ contains Z^{d} ·(v_{1},...,v_{d}), we obtain an upper bound for the covolume:

|v_{1} ∧...∧v_{d}| ≥ mes(R^{d}/Γ).

We now use the method of descent. If Z^{d} · (v_{1},...,v_{d}) is equal to Γ then we are done. Otherwise, the half-open paralleopiped {∑_{i=1}^{d} t_{i}v_{i} : 0 ≤ t_{i} < 1} generated by the vectors v_{1},...,v_{d}, being a fundamental domain of Z^{d} ·(v_{1},...,v_{d}), must contain a non-zero lattice point x in Γ. Write x = ∑_{i=1}^{d} t_{i}v_{i}; note that at least one of t_{2},...,t_{d} must be non-zero otherwise we would have tw ∈ Γ for some 0 < t < 1, which (by the Euclidean algorithm) contradicts the irreducibility of w. By permuting the indices 1,...,d if necessary we may assume that t_{d} > 0. We may also assume that t_{d} ≤ 1/2 since we could replace w by v_{1} +...+v_{d} −x otherwise. Then the volume |v_{1} ∧ ... ∧ v_{d−1} ∧ x| is at most half that of |v_{1}∧...∧v_{d}|, but is still non-zero. We thus replace v_{d} by x and repeat the above argument. Because of our absolute lower bound on the volume of paralleopipeds, this argument must eventually terminate, at which point we have found the desired presentation for Γ. Note that this procedure will never alter v_{1} and hence v_{1} is equal to w as desired.

Corollary 3.5 (Splitting lemma) Let Γ be a lattice of rank k, and let v be an irreducible vector in Γ. Then there exists a sublattice Γ′ of Γ of rank k−1 such that Γ is the direct sum of Z·v and Γ′, i.e. Γ = Z·v+Γ′ and Z·v ∩Γ′ = {0}.

Proof Apply Lemma 3.4 with v_{1} := v, and set Γ′ := Z^{k−1} ·(v_{2},...,v_{k}); the claim then follows from the linear independence of v_{1},...,v_{k}.

Corollary 3.6 (Fundamental theorem of finitely generated torsionfree additive groups) Let Z be a finitely generated torsion-free additive group. Then Z is isomorphic to Z^{d} for some d ≥ 0.

Proof We shall use the homomorphism theorems (Exercise 1). Since Z is finitely generated, we may find elements v_{1},...,v_{n} in Z such that Z^{n}·(v_{1},...,v_{n}) = Z. Now let Γ be the set Γ := {n ∈ Z^{n} : n·(v_{1},...,v_{n}) = 0}, then Γ is a sub-lattice of Z^{n} and Z is isomorphic to Z^{n}/Γ. In particular, Z^{n}/Γ is torsion free. We shall show that this implies that Z^{n}/Γ is isomorphic to some Z^{d}, as desired. We induct on n, the case n = 0 being trivial. If Γ = {0} we are done, so suppose Γ contains a nonzero vector v ∈ Γ, which we may assume without loss of generality to be irreducible in Γ. It is also irreducible in Z^{n}, for if v = m·w for some w ∈ Z^{d} and m > 1, then w+Γ would be a non-zero element of Z^{n}/Γ such that m·(w +Γ) = 0+Γ, contradicting the torsion-free assumption. By Lemma 3.4 or Corollary 3.5, this implies that Z^{n}/(Z·w) is isomorphic to Z^{n−1}. Since Z^{n}/Γ is isomorphic to (Z^{n}/(Z·w))/(Γ/(Z·w)), the claim then follows from the induction hypothesis.

3.1.2 Quotients of lattices Let G be a finitely generated additive group generated by d elements v_{1},...,v_{d} ∈ G. If we write v := (v_{1},...,v_{d}), and let Γ ⊂ Z^{d} be the lattice Γ := {n ∈ Z^{d} : n·v = 0}, it is easy to see that G is isomorphic to the quotient Z^{d}/Γ. Thus it is of interest to understand the quotient of two lattices. A basic tool for doing so is

Theorem 3.7 (Smith normal form) Let Γ and Γ′ be two lattices of full rank in R^{d} such that Γ′ is a sublattice of Γ. Then there exist linearly independent vectors v_{1},...,v_{d} in Γ such that

Γ = Z^{d} ·(v_{1},...,v_{d})

and

Γ′ = Z^{d} ·(N_{1}v_{1},...,N_{d}v_{d})

where 1 ≤ N_{1} ≤ ... ≤ N_{d} are positive integers such that N_{j} divides N_{j+1} for all j = 1,...,d−1.

Note that by applying an invertible linear transformation one can set (v_{1},...,v_{d}) equal to the standard basis (e_{1},...,e_{d}), so that Γ becomes just the standard lattice Z^{d}, while Γ′ is the sublattice of Z^{d} of vectors whose j^{th} co-ordinate is a multiple of N_{j} for j = 1,...,d.

Proof We induct on d. For d = 0 the statement is vacuously true, so suppose d ≥ 1 and the claim has already been proven for d−1. Given any non-zero vector v ∈ Γ, define the index of v to be the largest positive integer n such that v/n ∈ Γ; note that the index is finite since Γ is discrete. Note that the index of v is n if and only if v = nw for some irreducible vector w in Γ.

Since Γ′ has full rank, it contains non-zero vectors, each of which has an index. Let N_{1} denote the minimum index of all such vectors. By the well-ordering principle, this index is attained, and thus there exists an irreducible vector v_{1} ∈ Γ such that N_{1}v_{1} ∈ Γ′.

Using Lemma 3.4, we may apply an invertible linear transformation to map Γ to Z^{d}, in such a way that v_{1} is now equal to the standard basis vector e_{1}. Now let (n_{1},...,n_{d}) be any vector in Γ′. Observe that n_{1},...,n_{d} are integers; furthermore, n_{1} must be a multiple of N_{1}, otherwise by subtracting off a multiple of N_{1}e_{1} we could ensure that |n_{1}| < N_{1}, which contradicts the definition of N_{1} as the minimal index of Γ′. Thus we may factorize Γ′ = N_{1}Z·e_{1} +Γ′′, where Γ′′ is some sub-lattice of Z^{d−1} (which we think of as the span of e_{2},...,e_{d}). Note that if x ∈ Γ′′, then (N_{1},x) ∈ Γ′, and hence (since (N_{1},x) must have index at least N_{1}), x must be a multiple of N_{1}. Thus Γ′′ actually lies in N_{1} · Z^{d−1}, and we may therefore write Γ′ = N_{1}(Z·e_{1} +Γ′′′) for some sub-lattice Γ′′′ of Z^{d−1}. Note that Γ′′′ must have rank d−1 since Γ′ has rank d.

We now apply the inductive hypothesis, and by applying an invertible linear transformation to Z^{d−1} if necessary we may assume that

Γ′′′ = {(n_{2}M_{2},...,n_{d}M_{d}) : n_{2},...,n_{d} ∈ Z}

for some 1 ≤ M_{2} ≤ ... ≤ M_{d} such that M_{j} divides M_{j+1} for all j = 2,...,d−1. The claim follows by setting N_{j} := N_{1}M_{j} for j = 2,...,d.

We can now obtain the well known classification of finite and finitely generated additive groups:

Corollary 3.8 (Fundamental theorem of finite additive groups) Every finite additive group G is isomorphic to the direct sum of a finite number of cyclic groups Z/(N ·Z).

Proof Let g_{1},...,g_{d} be a finite set of generators for G. Then the map ϕ : Z^{d} ↦ G defined by ϕ(n) := n·(g_{1},...,g_{d}) is a surjection, and thus G is isomorphic to Z^{d}/ϕ^{−1}(0), which is a subgroup of R^{d}/ϕ^{−1}(0). The kernel ϕ^{−1}(0) is clearly a lattice of some rank 0 ≤ k ≤ d, and hence by Lemma 3.4 is generated by k linearly independent vectors v_{1},...,v_{k} in Z^{d}. Observe that we must have full rank k = d, otherwise Z^{d}/ϕ^{−1}(0) (and hence G) will be infinite. Using the Smith normal form, we can after an isomorphism write ϕ^{−1}(0) as the lattice generated by N_{1}e_{1},...,N_{d}e_{d} for some integers N_{1},...,N_{d} ≥ 1; this makes G isomorphic to G ≡ Z/N_{1}Z⊕...⊕Z/N_{d}Z, as desired (indeed we even obtain a normal form in which N_{j} divides N_{j+1} for j = 1,...,d−1).

Corollary 3.9 (Fundamental theorem of finitely generated additive groups) Every finitely generated additive group G is isomorphic to the direct sum of a finite number of cyclic groups Z/(N ·Z), and some Z^{d} for some d ≥ 0.

Proof Let G̃ := {x ∈ G : nx = 0 for some n > 0} be the torsion group of G, then by Corollary 3.8 G̃ is the direct sum of cyclic groups. The quotient group G/G̃ is torsion free and is thus isomorphic to Z^{d} for some d ≥ 0 by Corollary 3.6. If we let ẽ_{1},...,ẽ_{d} be arbitrary representatives in G of the standard basis e_{1},...,e_{d} of Z^{d}, we thus see that G is the direct sum of G̃ and Z·ẽ_{1},...,Z·ẽ_{d}, and the claim follows.

3.1.1 (Homomorphism theorems) If ϕ : Z → Z′ is a homomorphism between groups, show that the range ϕ(Z′) is a group which is isomorphic to the quotient group Z/ϕ^{−1}(0). If G,H are subgroups of Z, show that (G+H)/G is isomorphic to H/(G∩H). If furthermore G ⊂ H, show that H/G is a subgroup of Z/G and that (Z/G)/(H/G) is isomorphic to Z/H. If G′ is a subgroup of Z′, show that (Z ⊕ Z′)/(G ⊕ G′) is isomorphic to (Z/G) ⊕ (Z′/G′).

3.1.2 (Cauchy’s theorem) Show that if G is a subgroup of a finite additive group Z, then |Z/G| = |Z|/|G| (and in particular |G| must divide |Z|). By considering the groups ⟨x⟩ for various x ∈ Z, conclude that every finite additive group Z is an |Z|- torsion group; in particular, ord(x) divides |Z| for all x ∈ Z.

3.1.3 Show that if x is any element of a additive group Z, then the group ⟨x⟩ = Z · x has cardinality ord(x). More generally, if v = (v_{1},...,v_{d}) ∈ Z^{d} show that the group Z^{d} ·v has cardinality at most ord(v_{1})...ord(v_{d}), but at least as large as the least common multiple of the ord(v_{j}).

3.1.4 Let Z be an additive group. Show that Z is an N-torsion group if and only if for every x ∈ Z, the torsion of x is a divisor of N. Show that Z is torsion-free if and only if Z contains no finite subgroups other than the trivial subgroup {0}.

3.1.5 Let Z = Z_{1} ⊕Z_{2} be a direct sum of additive groups and r ≥ 1. Show that Z is torsion-free (resp. r-torsion) if and only if Z_{1} and Z_{2} are torsion-free (resp. r-torsion).

3.1.6 Prove that Q and R are not finitely generated. 3.1.7 If x,y are elements of an additive group Z with finite order,

show that x+y also has finite order, and that ord(x+y) divides the least common multiple of ord(x) and ord(y). Conclude that the set tor(Z) := {x ∈ Z : ord(x) < ∞} is a torsion group; we refer to it as the torsion subgroup of Z. It is clearly the largest subgroup of Z which is a torsion group. Show that the quotient group Z/tor(Z) is torsion-free, and is in fact the largest quotient which is torsion-free (in the sense that all other torsionfree quotients are quotients of Z/tor(Z)).

3.1.8 Show that Corollary 3.5 fails whenever v is not irreducible.

3.2 Progressions

We now study a basic example of an additive set, namely that of a generalized arithmetic progression (or progression for short), as defined in Definition 0.2. These will be model examples of additive sets with large amounts of additive structure; they can be viewed as a hybrid between a lattice and a convex set. (For a more quantitative realization of this heuristic, see Lemma 3.36 below.)

Note that progressions with the same set of basis vectors add very easily

(a+[0,N]·v)+(a′ +[0,N′]·v) = (a+a′)+[0,N +N′]·v  (3.1)

(so in particular the rank and basis vectors do not change), whereas progressions with different basis vectors add via the formula

(a+[0,N]·v)+(a′ +[0,N′]·v′) = (a+a′)+[0,N ⊕N′]·(v ⊕v′). (3.2)

Note the progression on the right-hand side of (3.2) is likely to be highly improper if v and v′ share some basis vectors in common. Also one can replace the box [0,N] by another box and also obtain a progression:

a+[N,M]·v = (a+N ·v)+[0,M −N]·v.

Similarly if one uses boxes such as [N,M), etc. In particular, the negation of a progression is also a progression:

−(a+[0,N]·v) = (−a)+[0,N]·(−v) = (−a−N ·v)+[0,N]·v. (3.3)

From this and (3.2) we see that the sum or difference of two progressions is again a progression. Finally, we make the easy observation that the Cartesian product of two progressions is again a progression.

We now show that, up to errors of O(1)^{d}, that progressions of rank d are essentially closed under addition.

Lemma 3.10 Let P = a + [0,N] · v be a progression of rank d in an additive group Z; we do not require that P is proper. Then for any integers n < m and any b ∈ Z, we can cover b+[nN,mN]·v by (m−n)^{d} translates of P. In particular for any n,m ≥ 0 with (n,m) ≠ (0,0), we can cover nP −mP by (n+m)^{d} translates of P, and in particular

|nP −mP | ≤ (n+m)^{d}|P |.

Furthermore, nP − mP is also a progression of rank d and volume at most vol(nP −mP) ≤ (n+m)^{d}vol(P).

Proof The first claim is clear since

[n·N,m·N]·v = [0,N]·v +[(n,...,n),(m,...,m)]·(N_{1}v_{1},...,N_{d}v_{d}).

From (3.1) we have

nP −mP = (na−ma−mN ·v)+[0,(n+m)N]·v

from which the remaining claims follow.

From this lemma we see in particular that if P is a symmetric progression of rank d and contains the origin (e.g. if P = [−N,N]·v), then P is a 2^{d}-approximate group in the sense of Definition 2.24. Indeed one can think of (symmetric) progressions of small rank as substitutes for subgroups in torsion-free settings (since torsion-free groups cannot contain finite subgroups). They also are the arithmetic analogue of boxes (or more generally, parallelopipeds) in Euclidean space, and in fact many of the results from real-variable harmonic analysis regarding covering by boxes (in physical space, Fourier space, or both) will have analogues for progressions.

In the special case when the rank d is equal to 1, a generalized arithmetic progression is the same as an ordinary arithmetic progression (or arithmetic progression for short)

P = a+[0,N]·v = {a+nv : 0 ≤ n ≤ N}

with base point a ∈ Z, basis vector or step v ∈ Z, and length N + 1. Note again that the cardinality of P may be less than N +1 if P is not proper, though in a torsion-free group this is only possible if the step v is zero.

We record a trivial lemma, that asserts that the sumset of a progression and a small set can be contained (somewhat inefficiently) in another progression.

Lemma 3.11 If P is a progression of rank d, and P + w_{1},...,P + w_{K} are translates of P, then all the translates P + w_{1},...,P + w_{K} can be contained inside a single progression of rank d + K − 1 and volume 2^{K−1}vol(P).

Proof Write P = a + [0,N] · v. By translation invariance we may set w_{K} = 0. Then the claim follows by using the progression a+[0,N]·v + [0,1]^{K−1} ·(w_{1},...,w_{K−1}).

Thus if one adds a small number of elements to an progression, one can still place the combined set inside a progression of slightly larger rank and volume, although the volume can grow exponentially in |A|. This is unavoidable, see Exercise 2. Because of this exponential loss, it is sometimes better not to invoke this lemma, and deal with multiple shifts of a single progression rather than trying to contain everything inside a single progression. Note that we have not guaranteed that the progressions in Lemma 3.11 are proper; we will return to this point in section 3.6.

3.2.1 Let N = (N_{1},...,N_{d}) be a collection of non-negative integers. Show that every proper ordinary arithmetic progression of length (N_{1} + 1)...(N_{d} + 1) is equal (as a set) to a proper generalized arithmetic progression of dimension N. (This example shows that the rank of a progression cannot be uniquely determined from the set of its elements, even if we restrict the progression to be proper).

3.2.2 Let K ≥ 1 and d ≥ 0 be integers, and P = a + [0,N] · v be a rank d progression in an additive group Z for some basis vectors v = (v_{1},...,v_{d}), and let X = {e_{1},...,e_{K}} be a set of K elements in Z. Suppose that the elements v_{1},...,v_{d},e_{1},...,e_{K} are linearly independent over Z. Show that any progression which contains P +X must necessarily have rank at least d+K −1 and volume at least 2^{K−1}vol(P), which shows that Lemma 3.11 is sharp.

3.2.3 Show that in a torsion-free additive group, the intersection of two ordinary arithmetic progressions is again an ordinary arithmetic progression. What happens if the torsion-free hypothesis is removed? What happens if one or both of the progressions is allowed to have rank greater than one?

3.2.4 Show that every finite additive group is also a proper progression.

3.2.5 Let P be a progression of rank d. Show that P contains an arithmetic progression Q with |Q| ≥ |P |^{1/d}, and furthermore that Q is proper if P is, and Q can be chosen to be symmetric around the origin if P is.

3.2.6 Let P be a proper progression of rank d, and let A be a subset of P such that |A| ≤ ε|P | for some 0 < ε < 1. Show that P \A contains a proper progression Q of rank d with |Q| ≥ C^{−d}/ε for some absolute constant C.

3.2.7 Let A be an additive set in an ambient group Z, and let v ∈ Z. Show that |(A+v)\A| ≤ 1 if and only if A is equal to a proper arithmetic progression of step v, union a finite (possibly zero) number of translates of the group ⟨v⟩. In particular, if |A| < ord(v), then |(A+v)\A| > 0, and |(A+v)\A| = 1 if and only if A is a proper arithmetic progression of step v.

3.3 Convex bodies

We now review some of the theory of convex bodies in R^{d}, which are in some sense the continuous analogue of generalized arithmetic progressions. This is of course a vast field, and we shall restrict ourselves with just a small sample of results, relating to the additive theory of such sets, to covering lemmas, and the relationship between addition and volume.

We shall use mes(A) to denote the volume of a set A in R^{d}; to avoid issues with measurability we shall mostly concern ourselves with bounded open sets A. If A ∈ R^{d} and λ ∈ R, we use λ·A to denote the dilation λ·A := {λx : x ∈ A}. Observe that mes(λA) = |λ|^{d}mes(A).

Recall that a set A in R^{d} is convex if we have (1 − θ)x + θy ∈ A whenever x,y ∈ A and 0 ≤ θ ≤ 1; equivalently, a set is convex if and only if

a·A+b·A = (a+b)·A

for all real a,b ≥ 0 (Exercise 3). In particular we have nA = |n|·A for any integer A. We call A a convex body if it is convex, open, non-empty, and bounded. In particular we see that if A is a convex body, then

mes(A+A) = mes(2·A) = 2^{d}mes(A),  (3.4)

so convex bodies have small doubling constant. As for A − A, we can use

Lemma 3.12 [284] For any bounded open subsets A,B,C of R^{d} (not necessarily convex), we have

mes(A−C)mes(B) ≤ mes(A−B)mes(B −C).

This is proven by modifying the proof of Lemma 2.6 appropriately and is left as an exercise. From this Lemma (with A = C and B = −A) and (3.4) we obtain

mes(A−A) ≤ 4^{d}mes(A);  (3.5)

compare these bounds with Lemma 3.10. In the converse direction, the Brunn-Minkowski inequality (Theorem 3.16 below) will give mes(A − A) ≥ 2^{d}mes(A).

Call a convex body A symmetric if A = −A; thus for us symmetry will always be with respect to the origin. The following theorem of John essentially classifies all convex bodies (symmetric and non-symmetric) up to a (dimension-dependent) constant factor.

Theorem 3.13 (John’s theorem) [185] Let A be a convex body in R^{d}. Then there exists an invertible linear transformation T : R^{d} → R^{d} on R^{d} and a point x_{0} ∈ A such that

B_{d} ⊆ T (A−x_{0}) ⊆ d·B_{d},

where B_{d} is the unit ball {(x_{1},...,x_{d}) ∈ R^{d} : x_{1}^{2} +...+x_{d}^{2} < 1}. If A is symmetric, then we can improve these inclusions to

B_{d} ⊆ T (A) ⊆ \sqrt{d}·B_{d}.

The constants d and \sqrt{d} are sharp; see exercises.

Proof We will use a variational argument. Define an ellipsoid to be any set E of the form E = L(B_{d})+x_{0}, where B_{d} is the unit ball, x_{0} ∈ R^{d}, and L is a (possibly degenerate) linear transformation in R^{d}; we allow the ellipsoid to degenerate for compactness reasons. Since A is open and bounded, it is easy to see that the set of all ellipsoids E contained in A is a compact set (with respect to the usual topology on L and x_{0}). Also the volume of the ellipsoid E is mes(E) = |det(L)|, whch is clearly a continuous function of E. Thus there exists an ellipsoid E = L(B_{d}) + x_{0} in A which maximizes the volume mes(E); since A is open, this volume is non-zero, and hence L is invertible. By applying L^{−1} if necessary (observing that the conclusion of the lemma is invariant under invertible linear transformations) we may thus assume that E is a translate E = B_{d} +y_{0} of the unit ball, where y_{0} = L^{−1}(x_{0}).

Let us now restrict to the case where A is symmetric. Observe that if A contains B_{d} +y_{0} then it also contains B_{d} −y_{0} by symmetry, and hence contains B_{d}, which is in the convex hull of B_{d} + y_{0} and B_{d} − y_{0}. To conclude the proof of the lemma in this case we need to show that A is contained in \sqrt{d}·B_{d}. Suppose for contradiction that A was not contained in \sqrt{d}·B_{d}; without loss of generality (and using the hypothesis that A is open) we may then suppose that re_{1} ∈ A for some r > \sqrt{d}, where e_{1} is the first basis vector. Observe now from elementary geometry that if ω is any point on the boundary of B_{d} making an angle ∠(ω,e_{1}) < arctan(r^{2} −1), then the line segment connecting ω to re_{1} is disjoint from (and not tangent to) B_{d}, and since B_{d} and re_{1} both lie in the convex set A we thus see that ω also lies in the open set A. By symmetry, the same is true if ∠(ω,−e_{1}) < arctan(\sqrt{r^{2} −1}).

We now perturb the ball B_{d} by an epsilon. Now let δ > 0 be a small number, let ε > 0 be an even smaller number, and consider the ellipsoid L_{ε,δ}(B_{d}), where

L_{ε,δ}(x_{1},...,x_{d}) := ((1+(d−1+δ)ε)x_{1},(1−ε)x_{2},...,(1−ε)x_{d})).

When ε = 0, L_{ε,δ}(B_{d}) is just B_{d}. Now consider how L_{ε,δ}(B_{d}) evolves in ε. The determinant of this transformation is (1+(d−1+δ)ε)(1−ε)^{d−1}, which has a positive ε-derivative at ε = 0. Thus L_{ε,δ}(B_{d}) has larger volume than B for sufficiently small ε (depending on δ). Now we check which points on the surface of L_{ε,δ}(B_{d}) expand away from the origin, and which ones contract. A simple computation shows that for any ω = (ω_{1},...,ω_{d}) on the boundary of B_{d}, the derivative

\frac{d}{dε}∥L_{ε,δ}(ω)∥^{2}|_{ε=0},

where ∥(y_{1},...,y_{d})∥^{2} := y_{1}^{2} +...+y_{d}^{2}, is negative unless

(d−1+δ)ω_{1}^{2} −ω_{2}^{2} −...−ω_{d}^{2} ≥ 0,

or in other words that

∠(ω,±e_{1}) ≤ arctan(\sqrt{d−1+δ}).

But if δ is small enough depending on r, this region is contained entirely within the interior of A by the previous discussion. Thus for ε small enough L_{ε,δ}(B_{d}) is completely contained inside A but has larger volume, contradicting the maximality of B_{d}, and we are done.

Now suppose that A is not symmetric. In this case we may translate so that y_{0} = 0. Thus again we have B_{d} ⊆ A, and the task is to show that A ⊆ n·B_{d}. Suppose again for contradiction that re_{1} ∈ A for some r > d; again this means that every point ω in the boundary of B_{d} with ∠(ω,e_{1}) < arctan(r^{2} −1) will lie in the interior of A. Now let δ,ε > 0 and consider the ellipsoid

L_{ε,δ}(x_{1},...,x_{d})+(d−1+δ)εe_{1};

again, this ellipsoid has larger volume than B_{d} if ε is sufficiently small. Also, we see that

\frac{d}{dε}∥L_{ε,δ}(ω)+(d−1+δ)εe_{1}∥^{2}|_{ε=0}

is negative unless

(d−1+δ)ω_{1}^{2} +(d−1+δ)ω_{1} −ω_{2}^{2} −...−ω_{d}^{2} ≥ 0,

which can be rewritten (using ∥ω∥ = 1) as

((d+δ)ω_{1} −1)(ω_{1} +1) ≥ 0,

or equivalently

∠(ω,e_{1}) ≤ arctan(\sqrt{(d+δ)^{2} −1}).

We now argue as in the symmetric case to again obtain the desired contradiction, if δ is chosen so that d+δ < r.

As a corollary of Theorem 3.13 we see that if A is a convex body, we can cover A+A or A−A by a relatively small number of copies of A:

A±A can be covered by O(d)^{d} translates of A.  (3.6)

This follows immediately from the geometric observation that d·B_{d} +d· B_{d} = 2d·B_{d} can be covered by O(d)^{d} translates of B_{d}. If A is symmetric, we can improve this somewhat. In the special case when A is a cube or a box, it is clear that A ± A can be covered by 2^{d} translates of A (cf. Lemma 3.10), but one cannot hope for this in general; for instance if A is a disk in R^{2} then one needs six copies of A to cover A ± A. In the general case, we will need the following continuous version of

Lemma 3.14 (Ruzsa’s covering lemma) [287], [239] For any bounded subsets A,B of R^{d} with positive measure (not necessarily convex), we can cover B by at most min(\frac{mes(A+B)}{mes(A)}, \frac{mes(A−B)}{mes(A)}) translates of A−A.

The proof of this lemma is nearly identical to that of Lemma 2.13 and is left as an exercise. As a consequence we can improve (3.6) for symmetric convex bodies:

Corollary 3.15 Let A ⊂ R^{d} be a convex body, and let λ,μ > 0 be real. Then λ · A can be covered by at most (λ + 1)^{d} translates of A − A, and λ·A−μ·A can be covered by (2max(λ,μ)+1)^{d} translates of A−A. If A is symmetric, then λ·A can be also covered by (2λ+1)^{d} translates of A.

Proof The first claim follows from Lemma 3.14 since mes(λ·A+A) = (λ + 1)^{d}mes(A). To prove the second claim, we may take λ ≥ μ. The first claim implies that 2λ·A can be covered by (2λ+1)^{d} translates of A−A = 2·A, and the third claim follows by rescaling by 1/2. Finally, the second claim follows by applying the third claim to A−A.

Observe that all the bounds obtained here tend to be exponential in d or worse. Thus when using the theory of convex bodies to obtain explicit estimates, it is often important to keep the dimension d as low as possible, even at the cost of making some other parameters larger than what would otherwise be necessary. See [239] for further discussion of sum set and covering estimates for convex bodies.

We have not yet seen what happens to the sum or difference of two unrelated convex bodies A and B. The relationship here is given by the Brunn-Minkowski inequality, which we turn to next.

3.3.1 Prove Lemma 3.12. 3.3.2 Prove Lemma 3.14. 3.3.3 Verify that the two definitions of convexity given are indeed

equivalent. 3.3.4 Let A be an open bounded subset of R^{d}. Show that A is convex

if and only if 2A = 2·A, and that A is convex and symmetric if and only if 2A = −2·A.

3.3.5 For any s > 0 let Γ(s) := ∫_{0}^{∞} e^{−x}x^{s−1} dx denote the Gamma function. Show that Γ(s+1) = sΓ(s) for all s > 0, that Γ(d) = (d−1)! for all d ≥ 1, that Γ(1/2) = \sqrt{π}, and we have the Stirling formula

logΓ(s) = slogs−s+O(logs)  (3.7)

for all large s. (Hint: use (1.52) and the monotonicity of the Γ function.)

3.3.6 Let B_{d} be the unit ball in R^{d}. By evaluating the integral ∫_{R^{d}} e^{−π|x|^{2}} dx in both Cartesian and polar co-ordinates, and using the preceding exercise, establish the volume formula

mes(B_{d}) = \frac{Γ(3/2)^{d}2^{d}}{Γ(d/2+1)} = (2πe+o(1))^{d/2}d^{−d/2}.  (3.8)

3.3.7 Let O_{d} be the octahedron given by the convex hull of ±e_{1},...,±e_{d} in R^{d}. Show that mes(O_{d}) = 2^{d}/d! = (2e+o(1))^{d}d^{−d}. Thus in large dimension the octahedron becomes considerably smaller than the circumscribing ball B_{d} which contains it, which in turn is considerably smaller than the circumscribing cube.

3.3.8 Show that the constants d and \sqrt{d} in Theorem 3.13 cannot be improved. (For the non-symmetric case, take A to be a d-simplex (the convex hull of d points in R^{d}); for the symmetric case, take A to be a cube.)

3.3.9 If A and A′ are two symmetric convex bodies in R^{d}, show that there exists an invertible linear transformation T : R^{d} → R^{d} such that

A ⊆ T (A′) ⊂ d·A.

State and prove a similar result in the case when A and A′ are not necessarily symmetric.

3.3.10 Let A,B be open bounded sets. Show that

mes((A−A)∩(B −B)) ≥ \frac{mes(A)mes(B)}{mes(A±B)}

for either choice of sign ±, by developing a continuous analogue of the arguments used to prove (2.9). (Alternatively, one can try to discretize A and B to replace them with finite sets, and then use (2.9) directly).

3.4 The Brunn-Minkowski inequality

The purpose of this section is to prove the following lower bound for the volume mes(A+B) of a sum set.

Theorem 3.16 (Brunn-Minkowski inequality) If A and B are nonempty bounded open subsets of R^{d}, then

mes(A+B)^{1/d} ≥ mes(A)^{1/d} +mes(B)^{1/d}.

This inequality is sharp (Exercise 2). The theorem also applies if A and B are merely measurable (as opposed to being bounded and open), though one must then also assume that A + B is measurable; we will not prove this here. In general, there is no upper bound for mes(A+B); consider for instance the case when A is the x-axis and B is the y-axis in R^{2}, then A, B both have measure zero but A+B is all of R^{2}. One can easily modify this example to show that there is no upper bound for mes(A+B) in terms of mes(A) and mes(B) when A,B are bounded open sets. See [122] for a thorough survey of the Brunn-Minkowski inequality and related topics.

To prove this theorem, it suffices to prove the following dimensionindependent version:

Theorem 3.17 If A and B are non-empty bounded open subsets of R^{d}, and 0 < θ < 1, then

mes((1−θ)·A+θ ·B) ≥ mes(A)^{1−θ}mes(B)^{θ}.

To see why Theorem 3.17 implies the Brunn-Minkowski inequality, apply Theorem 3.17 with A and B replaced by mes(A)^{−1/d} · A and mes(B)^{−1/d} ·B to obtain

mes(\frac{1−θ}{mes(A)^{1/d}} ·A+ \frac{θ}{mes(B)^{1/d}} ·B) ≥ 1

for any 0 < θ < 1. Setting θ := \frac{mes(B)^{1/d}}{mes(A)^{1/d}+mes(B)^{1/d}} we obtain the result. Conversely, one can easily deduce Theorem 3.17 from the Brunn-Minkowski inequality (Exercise 1).

It remains to prove Theorem 3.17. We begin by first proving

Lemma 3.18 (One-dimensional Brunn-Minkowski inequality) If A and B are non-empty bounded open subsets of R, then mes(A+B) ≥ mes(A)+mes(B).

Proof The hypotheses and conclusion of this lemma are invariant under independent translations of A and B, so we can assume that sup(A) = 0 and inf(B) = 0, so in particular A and B are disjoint. But then we see that A+B contains both A and B separately, and we are done.

Using this Lemma, we deduce

Proposition 3.19 (One-dimensional Prékopa-Leindler inequality) Let 0 < θ < 1, and let f,g,h : R → [0,∞) be lower semi-continuous, compactly supported non-negative functions on R such that

h((1−θ)x+θy) ≥ f(x)^{1−θ}g(y)^{θ}

for all x,y ∈ R. Then we have

∫_{R} h ≥ (∫_{R} f)^{1−θ}(∫_{R} g)^{θ}.

Proof By multiplying f,g,h by appropriate positive constants we may normalize ∫_{R} f = ∫_{R} g = 1; our task is then to show that ∫_{R} h ≥ 1.

Let λ > 0 be arbitrary. Observe that if f(x) > λ and g(y) > λ, then by hypothesis h((1−θ)x+θy) > λ. Thus we have

{z ∈ R : h(z) > λ} ⊆ (1−θ)·{x ∈ R : f(x) > λ}+θ·{y ∈ R : g(x) > λ}.

Since f,g,h are lower semi-continuous and compactly supported, all the sets above are open and bounded, hence by Lemma 3.18

mes({z ∈ R : h(z) > λ}) ≥ (1−θ)mes({x ∈ R : f(x) > λ})+θmes({y ∈ R : g(x) > λ}).

Integrating this for λ ∈ [0,∞) and using Fubini’s theorem (cf. (1.6)), the claim follows.

Now we iterate this to higher dimensions.

Proposition 3.20 (Higher-dimensional Prékopa-Leindler inequality) Let 0 < θ < 1, d ≥ 1, and let f,g,h : R^{d} → [0,∞) be lower semicontinuous, compactly supported non-negative functions on R^{d} such that

h((1−θ)x+θy) ≥ f(x)^{1−θ}g(y)^{θ}

for all x,y ∈ R^{d}. Then we have

∫_{R} h ≥ (∫_{R} f)^{1−θ}(∫_{R} g)^{θ}.

Proof We induct on d. When d = 1 this is just Proposition 3.19. Now assume inductively that d > 1 and the claim has already been proven for all smaller dimensions d. Define the one-dimensional function h_{d} : R → [0,∞) by

h_{d}(x_{d}) := ∫_{R^{d−1}} h(x_{1},...,x_{d}) dx_{1} ...dx_{d−1},

and similarly define f_{d}, g_{d}. One can easily check that (using Fatou’s lemma) that these functions are lower semi-continuous and compactly supported. Also, applying the inductive hypothesis at dimension d−1 we see that

h_{d}((1−θ)x_{d} +θy_{d}) ≥ f_{d}(y_{d})^{1−θ}g_{d}(y_{d})^{θ}

for all x_{d},y_{d} ∈ R. If we then apply the one-dimensional Prékopa-Leindler inequality, we obtain the desired result.

If we apply Proposition 3.20 with f := 1_{A}, g := 1_{B}, and h := 1_{(1−θ)A+θB} we obtain Theorem 3.17, and the Brunn-Minkowski inequality follows.

3.4.1 Show that Theorem 3.16 implies Theorem 3.17. 3.4.2 Show that equality in Theorem 3.17 can occur when A is convex,

and B = λ · A + x_{0} for some λ,x_{0} ∈ R^{n}. Conversely, if A and B are non-empty bounded open subsets of R^{d}, show that the preceding situation is in fact the only case in which equality can be attained. (The case when A and B are merely measurable is a bit trickier, and is of course only true up to sets of measure zero; see [122] for further discussion).

3.4.3 Let A be a convex body in R^{d}. Using Theorem 3.17, show that the cross-sectional areas f(x_{d}) := mes({x′ ∈ R^{d−1} : (x′,x_{d}) ∈ A}) are a log-concave function of x_{d}, i.e. f((1−λ)x_{d} +λy_{d}) ≥ f(x_{d})^{1−λ}f(y_{d})^{λ} for all 0 ≤ λ ≤ 1 and x_{d},y_{d} ∈ R; this is known as Brunn’s inequality.

3.4.4 Let A be a bounded open set with smooth boundary ∂A, and let B be a ball with the same volume as A. Prove the isoperimetric inequality mes(∂A) ≥ mes(∂B). (Hint: Use the Brunn-Minkowski inequality to estimate \frac{mes(A+ε·B)−mes(A)}{ε} for ε > 0 small, and then let ε → 0.

3.4.5 Let A,B be symmetric convex bodies in R^{d}. Show by examples that there is no upper bound for mes(A+B) in terms of mes(A), mes(B), amd d alone, except in the d = 1 case. However, by using Lemma 3.12, show that mes(A+B) ≤ 4^{d} \frac{mes(A)mes(B)}{mes(A∩B)}.

3.4.6 Let A be a convex body, and let a,b > 0 be real numbers. For fixed values of mes(A), a, b, obtain upper and lower bounds for the quantity mes(a·A−b·A) that are as sharp as you can manage; try some examples (e.g. A being a ball, cube, or simplex) to see whether your bounds are sharp. Note that this task is rather trivial if A is symmetric around the origin.

3.5 Intersecting a convex set with a lattice

In previous sections we have studied lattices, which are discrete but unbounded, and convex sets, which are bounded but continuous. We now study the intersection B ∩ Γ of a convex set B and a lattice Γ in a Euclidean space R^{d}, which is then necessarily a finite set. A model example of such set is the discrete box [0,N) for some N = (N_{1},...,N_{d}), which is the intersection of the convex body {(x_{1},...,x_{d}) : −1 < x_{i} < N_{i} for all 1 ≤ i ≤ d} with the Euclidean lattice Z^{d}. One of the main objectives of this section shall to show a “discrete John’s lemma” which shows that all intersections B ∩Γ can be approximated in a certain sense by a discrete box.

We begin with some elementary estimates.

Lemma 3.21 Let Γ be a lattice in R^{d}. If A ⊂ R^{d} is an arbitrary bounded set and P ⊂ R^{d} is a finite non-empty set, then

|A∩(Γ+P)| ≤ |(A−A)∩(Γ+P −P)|.  (3.9)

If B is a symmetric convex body, then

(k ·B)∩Γ can be covered by (4k +1)^{d} translates of B ∩Γ  (3.10)

for all k ≥ 1. If furthermore Γ′ is a sub-lattice of Γ of finite index |Γ/Γ′|, then we have

|B ∩Γ′| ≤ |B ∩Γ| ≤ 9^{d}|Γ/Γ′||B ∩Γ′|.  (3.11)

Proof We first prove (3.9). We may of course assume that A∩(Γ+P) contains at least one element a. But then A ∩ (Γ + P) ⊆ ((A − A) ∩ (Γ + P − P)) + a, and the claim follows. Now we prove (3.10). The lower bound is trivial, so it suffices to prove the upper bound. By the preceding argument we can cover |(\frac{1}{2} ·B +x)·Γ| by a translate of B ∩Γ for any x ∈ R^{d}. But by Corollary 3.15 we can cover k ·B by (4k +1)^{d} translates of \frac{1}{2} ·B, and the claim (3.10) follows.

Finally, we prove (3.11). The lower bound is trivial. For the upper bound, observe that Γ is the union of |Γ/Γ′| translates of Γ, so it suffices to show that |B ∩(Γ′ +x)| ≤ 9^{d}|B ∩Γ′| for all x ∈ R^{d}. But by (3.9) and (3.10) we have

|B ∩(Γ′ +x)| ≤ |(2·B)∩Γ′| ≤ 9^{d}|B ∩Γ′|

as desired.

Next, we recall a result of Gauss concerning the intersection of a very large convex body with a lattice of full rank.

Lemma 3.22 Let Γ ⊂ R^{d} be a lattice of full rank, let v_{1},...,v_{d} ∈ Γ be a set of generators for Γ, and let B be a convex body in R^{d}. Then for large R > 0, we have

|(R·B)∩Γ| = (R^{d} +O_{Γ,B,d}(R^{d−1}))\frac{mes(B)}{|v_{1} ∧...∧v_{d}|}.

Here |v_{1} ∧ ... ∧ v_{d}| denotes the volume of the parallelopiped with edges v_{1},...,v_{d}.

Proof We use a “volume packing argument”. Since Γ has full rank, v_{1},...,v_{d} are linearly independent. By applying an invertible linear transformation we may assume that v_{1},...,v_{d} is just the standard basis e_{1},...,e_{d}, so that Γ = Z^{d}. Now let Q be the unit cube centered at the origin. Observe that the sets {x+Q : x ∈ (R·B)∩Z^{d}} are disjoint up to sets of measure zero, and their union differs from R·B only in the \sqrt{d}- neighbourhood of the surface of R·B, which has volume O_{Γ,B,d}(R^{d−1}). The claim follows.

Remark 3.23 The task of improving the error term O_{Γ,B,d}(R^{d−1}) for various lattices and convex bodies is a deep and important problem in number theory and harmonic analysis, but we will not discuss this issue in this book; our only concern is that the error term is strictly lower order than the main term.

If Γ is a lattice, we define a fundamental parallelopiped for Γ to be any parallelopiped whose edges v_{1},...,v_{d} generate Γ. From the above lemma we conclude that all fundamental parallelopipeds have the same volume; indeed this volume is nothing more than the covolume mes(R^{d}/Γ) of Γ. Thus for instance mes(R^{d}/Z^{d}) = 1.

By another volume-packing argument we can establish

mes(R^{d}/Γ)|Γ/Γ′| = mes(R^{d}/Γ′)  (3.12)

whenever Γ′ ⊆ Γ ⊂ R^{d} are two lattices of full rank; see exercises. In particular we see that the quotient group |Γ/Γ′| is finite.

Yet another volume packing argument gives the following continuous and periodic analogue of (2.9).

Lemma 3.24 (Volume packing lemma) Let Γ ⊂ R^{d} be a lattice of full rank, let V be a bounded open subset of R^{d}, and let P be a finite non-empty set in R^{d}. Then

|(V −V)∩(Γ+P −P)| ≥ \frac{mes(V)|P |}{mes(R^{d}/Γ)}.

In particular, we have

|(V −V)∩Γ| ≥ \frac{mes(V)}{mes(R^{d}/Γ)}.

Proof Let B be the unit ball on R^{d}, and let R > 0 be a large number. Consider the integral of the function

f(x) := ∑_{y∈Γ∩(R·B)} ∑_{p∈P} 1_{V +y+p}(x).

On the one hand we can compute this integral using Lemma 3.22 as

∫_{R^{d}} f(x) dx = ∑_{y∈Γ∩(R·B)} ∑_{p∈P} mes(V +y)

= |Γ∩(R·B)||P ||mes(V)|

= (R^{d} +O_{Γ,B,d}(R^{d−1}))|P |\frac{mes(B)mes(V)}{mes(R^{d}/Γ)}

On the other hand, from (3.9) we have

f(x) ≤ |(x−V)∩(Γ+P −P)| ≤ |(V −V)∩(Γ+P −P)|.

Furthermore f(x) is only non-zero when x lies in R·B +V +P ⊂ (R+ O_{V,P} (1))·B, which has volume R^{d} +O_{V,P,d}(R^{d−1}). Thus

∫_{R^{d}} f(x) dx ≤ |(V −V)∩(Γ+P −P)|R^{d} +O_{V,P,d}(R^{d−1}).

Combining these inequalities, dividing by R^{d}, and taking limits as R → ∞, we obtain the result.

To see the utility of this lemma, let us pause to establish the following classical result in number theory, which we will need later in this book. Let ∥x∥_{R/Z} denote the distance from x to the nearest integer.

Corollary 3.25 (Kronecker approximation theorem) Let α_{1},...,α_{d} be real numbers, and let 0 < θ_{1},...,θ_{d} ≤ 1/2. Then for any N > 0, we have

|{n ∈ (−N,N) : ∥nα_{j}∥_{R/Z} < θ_{j} for all j = 1,...,d}| ≥ Nθ_{1} ...θ_{d}.

In particular, if Nθ_{1} ...θ_{d} ≤ 1, then there exists an integer 0 < n < N such that ∥nα_{j}∥_{R/Z} ≤ θ_{j} for all j = 1,...,d.

Proof Apply Lemma 3.24 with Γ := Z^{d},

V := {(t_{1},...,t_{d})+Z^{d} : 0 < t_{j} < θ_{j} for all 1 ≤ j ≤ d},

and P equal to the arithmetic progression P = [0,N)·(α_{1},...,α_{d}) in R^{d}.

Now let us apply Lemma 3.24 to the case V = \frac{1}{2} · B and P = \{0\}, where B is a symmetric convex body, we obtain the lower bound

|B ∩Γ| ≥ \frac{mes(B)}{2^{d}mes(R^{d}/Γ)}.  (3.13)

The assumption of symmetry is essential; consider for instance Γ := Z^{2} and a convex set of the form B := {(x,y) : 1/3 < x < 2/3;−N < y < N} for arbitrarily large N. Even when B is symmetric, it is possible for |B ∩Γ| to be extremely large compared to \frac{mes(B)}{2^{d}mes(R^{d}/Γ)}; consider for instance Γ := Z^{2} and B := {(x,y) : −1/N^{2} < x < 1/N^{2};−N < y < N}. However, if B∩Γ has full rank, then we can complement the lower bound (3.13) with an upper bound:

Lemma 3.26 Let Γ be a lattice of full rank in R^{d}, and let B be a symmetric convex body in R^{d} such that the vectors in B ∩ Γ linearly span R^{d}. Then

|B ∩Γ| ≤ \frac{3^{d}d!mes(B)}{2^{d}mes(R^{d}/Γ)}.  (3.14)

This bound is with a factor of 3^{d}/(2d+1) of being sharp, as can be seen by the example where Γ = Z^{d} and B is (a slight enlargement of) the octahedron with vertices ±e_{1},...±e_{d}. Indeed this example motivates the volume packing argument used in the proof.

Proof By hypothesis, B ∩ Γ contains a d-tuple (v_{1},...,v_{d}) of linearly independent vectors. Since B ∩ Γ is finite, we can choose v_{1},...,v_{d} in order to minimize the volume mes(O) = \frac{2^{d}}{d!} |v_{1}∧...∧v_{d}| of the octahedron with vertices ±v_{1},...,±v_{d}. Since B is symmetric and convex, we see that O ⊆ B. Also O does not contain any elements of Γ other than v_{1},...,v_{d}, since otherwise one could replace one of v_{1},...,v_{d} with this element and reduce the volume of O, a contradiction. Thus we see that the sets \{x + \frac{1}{2} · O : x ∈ B ∩ Γ\} are all disjoint and are contained in B + \frac{1}{2} ·O ⊆ \frac{3}{2} ·B. Thus

|B ∩Γ| ≤ \frac{mes(\frac{3}{2} ·B)}{mes(\frac{1}{2} ·O)} = \frac{3^{d}d!}{2^{d}|v_{1} ∧...∧v_{d}|}mes(B).

Since |v_{1} ∧...∧v_{d}| ≥ mes(R^{d}/Γ), the claim follows.

As an immediate consequence of (3.13) we also obtain

Lemma 3.27 (Blichtfeld’s lemma) Let Γ ⊂ R^{d} be a lattice of full rank, and let V be an open set in R^{d} such that mes(V) > mes(R^{d}/Γ). Then there exists distinct x,y ∈ V such that x−y ∈ Γ.

Now let B be a symmetric convex body with mes(B) ≥ 2^{d}mes(R^{d}/Γ). Applying the above lemma with V = \frac{1+ε}{2} · B and letting ε → 0, we conclude

Theorem 3.28 (Minkowski’s first theorem) Let Γ be a lattice of full rank, and let B be a symmetric convex body such that mes(B) ≥ 2^{d}mes(R^{d}/Γ). Then the closure of B must contain at least one non-zero element of Γ (in fact it contains at least two, by symmetry). If we have strict inequality mes(B) ≥ 2^{d}mes(R^{d}/Γ), then we can replace the closure of B with the interior of B in the above statement.

The constant in Minkowski’s first theorem is sharp; we may apply an invertible linear transformation to set Γ := Z^{d}, and then the example of the cube A := {(t_{1},...,t_{d}) : −1 < t_{j} < 1 for all j = 1,...,d} shows that the constant 2^{d} cannot be improved. Nevertheless, it is possible to improve Minkowski’s first theorem by generalizing it to a “multiparameter” version as follows.

Definition 3.29 (Successive minima) Let Γ be a lattice in R^{d} of rank k, and let B be a convex body in R^{d}. We define the successive minima λ_{j} = λ_{j}(B,Γ) for 1 ≤ j ≤ k of B with respect to Γ as

λ_{j} := inf{λ > 0 : λ·B contains k linearly independent elements of Γ}.

Note that 0 < λ_{1} ≤ ... ≤ λ_{k} < ∞.

Thus, for instance, if Γ = Z^{d} and B is the box

B := {(t_{1},...,t_{d}) : |t_{j}| < a_{j} for all j = 1,...,d}

for some a_{1} ≥ a_{2} ≥ ... ≥ a_{d} > 0, then λ_{j} = 1/a_{j} for j = 1,...,d. Note that the assumption that Γ has rank k ensures that the λ_{j} are both finite and non-zero.

Theorem 3.30 (Minkowski’s second theorem) Let Γ be a lattice of full rank in R^{d}, and let B be an symmetric convex body in R^{d}, with successive minima 0 < λ_{1} ≤ ... ≤ λ_{d}. Then there exists d linearly independent vectors v_{1},...,v_{d} ∈ Γ with the following properties:

• For each 1 ≤ j ≤ d, v_{j} lies in the boundary of λ_{j} ·B, but λ_{j} ·B itself does not contain any vectors in Γ outside of the span of v_{1},...,v_{j−1}.

• The octahedron with vertices ±v_{j} contains no elements of Γ in its interior, other than the origin.

• We have

\frac{2^{d}|Γ/(Z^{d} ·(v_{1},...,v_{d}))|}{d!} ≤ \frac{λ_{1} ...λ_{d}mes(B)}{mes(R^{d}/Γ)} ≤ 2^{d}.  (3.15)

In particular, the sub-lattice Z^{d} ·(v_{1},...,v_{d}) of Γ has bounded index:

|Γ/(Z^{d} ·(v_{1},...,v_{d}))| ≤ d!.  (3.16)

One can state (3.15) rather crudely as

λ_{1} ...λ_{d}mes(B) = d^{O(d)}mes(R^{d}/Γ)

thus relating the successive minima to the volume of the body B and the covolume of the lattice Γ.

Note that if B contains no non-zero elements of Γ then λ_{j} ≥ 1 for all j, so Minkowski’s second theorem implies Minkowski’s first theorem. Conversely, we shall see from the proof that Minkowski’s second theorem can be obtained from Minkowski’s first theorem by a non-isotropic dilation. The basis v_{1},...,v_{d} is sometimes referred to as a directional basis for A with respect to Γ, although one should caution that this basis does not quite generate Γ (the index in (3.16) is bounded but not necessarily equal to 1).

Proof By definition of λ_{1}, we may find a vector v_{1} ∈ Γ such that v_{1} lies in the closure of λ_{1} ·B, but that λ·B contains no non-zero elements of Γ for any λ ≤ λ_{1}. By definition of λ_{2}, we can then find a vector v_{2} ∈ Γ, linearly independent from v_{1}, such that v_{2} lies in the closure of λ_{2}B, but that λ · B contains no elements of Γ outside of the span of v_{1} for any λ ≤ λ_{2}. Continuing inductively we can eventually find a linearly independent set v_{1},...,v_{d} in Γ such that v_{j} lies in the boundary of λ_{j} ·B, but λ_{j} ·A itself does not contain any vectors in Γ outside of the span of v_{1},...,v_{j−1}, for all 1 ≤ j ≤ n.

The set v_{1},...,v_{d} is a basis of R^{d}; by applying an invertible linear transformation we may assume it is the standard basis e_{1},...,e_{d} (this changes both B and Γ, but one may easily verify that the conclusion of the theorem remains unchanged). In particular this forces Γ to contain Z^{d}, hence by (3.12)

mes(R^{d}/Γ) = mes(R^{d}/Z^{d})/|Γ/Z^{d}| = 1/|Γ/Z^{d}| ≤ 1.  (3.17)

Let O^{d} be the open octahedron whose vertices are ±e_{1},...,±e_{d}. We need to verify that O^{d} contains no lattice points from Γ other than the origin. Suppose for contradiction that O^{d} ∩Γ contained w = t_{1}e_{1} +...+ t_{j}e_{j} where 1 ≤ j ≤ d and t_{j} ≠ 0. Then (1 + ε)w would be a linear combination of ±e_{1},...,±e_{j} for some ε > 0. All of these points lie in λ_{j} ·B, hence w lies in the interior of λ_{j} ·B, but does not lie in the span of e_{1},...,e_{j−1}. But this contradicts the construction of v_{j} = e_{j}. Hence O^{d} ∩Γ = {0}.

Next, observe that for each 1 ≤ j ≤ d, ±v_{j} = ±e_{j} lies on the boundary of λ_{j} ·B. Thus B contains the open octahedron whose vertices are ±e_{1}/λ_{1},...,±e_{d}/λ_{d}. This octahedron is easily verified to have volume \frac{2^{d}}{d!λ_{1}...λ_{d}}; indeed one can rescale to the case when all the λ_{j} are equal to 1, and then one can decompose the octahedron into 2^{d} simplices, each of which has volume 1/d!. This, together with (3.17), establishes the lower bound in (3.15).

Now we establish the upper bound in (3.15). We need the following lemma.

Lemma 3.31 (Squeezing lemma) Let B be a symmetric convex body in R^{d}, let A be an open subset of B, let V be a k-dimensional subspace of R^{d}, and let 0 < θ ≤ 1. Then there exists an open subset A′ of B such that mes(A′) = θ^{k}mes(A) and (A′ −A′)∩V ⊆ θ ·(A−A)∩V.

Note that we do not assume any convexity on A or A′. Indeed the squeezing operation we define in the proof below does not preserve the convexity of A.

Proof Without loss of generality we may take V = R^{k}, and write R^{d} = R^{k} × R^{d−k}. Let π : R^{d} → R^{d−k} be the orthogonal projection map, which restricts to a map π : B → π(B). Let f : π(B) → B be any continuous right-inverse of π; thus for instance f(y) could be the center of mass of π^{−1}(y). We define

A′ := {(1−θ)f(π(x))+θx : x ∈ A}.

From the convexity of B we see that A′ is an open subset of B, and from Cavalieri’s principle (or Fubini’s theorem) we see that mes(A′) = θ^{k}mes(A). Also if z := (1−θ)f(π(x))+θx and z′ := (1−θ)f(π(x′))+θx′ are such that z −z′ ∈ V, then we have π(z) = π(z′), hence π(x) = π(x′), hence z −z′ = θ(x−x′). This establishes the inclusion (A′ −A′)∩V ⊂ θ ·(A−A)∩V.

We apply the squeezing lemma iteratively, starting with A_{0} := \frac{λ_{d}}{2} ·B, to create open sets A_{1},...,A_{d−1} ⊆ \frac{λ_{d}}{2} ·B such that

mes(A_{j}) ≤ ∏_{i=1}^{j}(\frac{λ_{i}}{λ_{i+1}} )^{i}2^{−d}λ_{d}^{d}mes(B)

and

(A_{j} −A_{j})∩R^{j} ⊆ \frac{λ_{j}}{λ_{j+1}} ·(A_{j−1} −A_{j−1})∩R^{j}

for all 1 ≤ j ≤ d−1, where R^{j} is the span of e_{1},...,e_{j}. In particular we have

mes(A_{d−1}) ≤ λ_{1} ...λ_{d}2^{−d}mes(B)  (3.18)

and

(A_{d−1} −A_{d−1})∩R^{j} ⊆ \frac{λ_{j}}{λ_{d}} (\frac{λ_{d}}{2} ·B − \frac{λ_{d}}{2} ·B)∩R^{j} = λ_{j} ·B ∩R^{j}

for all 1 ≤ j ≤ d. Now recall from the normalization v_{j} = e_{j} that λ_{j} ·B ∩R^{j} does not contain any lattice point in Γ, except for those in R^{j−1}. This implies that A_{d−1} −A_{d−1} does not contain any point in Γ other than the origin. Applying Blichtfeld’s lemma, we conclude that

mes(A_{d−1}) ≤ mes(R^{d}/Γ),

which when combined with (3.18) gives the upper bound in (3.15).

We now give several applications of this theorem. First we “factorize” a convex body B as the finitely overlapping sum of a subset of Γ and and a dilate of a small convex body B′, up to some powers of scaling factors of O(d):

Lemma 3.32 Let B be a symmetric convex body in R^{d}, and let Γ be a lattice in R^{d}. Then there exists a symmetric convex body B′ ⊆ B such that B′ contains no non-zero elements of Γ, and such that B ⊆ O(d^{3/2}) · B′ + ((O(d^{3/2}) · B) ∩ Γ. In particular, the projection of B in R^{d}/Γ is contained in the projection of O(d^{3/2}) · B′. Furthermore, we have the bounds

\frac{mes(B)}{O(d)^{5d/2}|B ∩Γ|} ≤ mes(B′) ≤ O(1)^{d} \frac{mes(B)}{|B ∩Γ|}.  (3.19)

Proof By using John’s theorem and an invertible linear transformation we may assume that B_{d} ⊆ B ⊆ \sqrt{d}·B_{d}, where B_{d} is the unit ball. We may assume that the vectors in B ∩ Γ generate Γ, since otherwise we could replace Γ by the lattice generated by B ∩Γ.

Let us temporarily assume that Γ has full rank, and thus that the linear span of B ∩Γ is R^{d}. Thus if we let λ_{1} ≤ ... ≤ λ_{d} be the successive minima of B, then we have λ_{j} ≤ 1 for all j.

Now we take a directional basis v_{1},...,v_{d} of Γ, and let B′ be the open octahedron with vertices ±v_{j}; this octahedron then contains no non-zero elements of Γ, and is also contained in B (since ±v_{j}/λ_{j} already lies on the boundary of B). Observe that d·B′ contains a parallelopiped with edges v_{1},...,v_{d}, and hence d·B′ +Γ = R^{d}. Thus

B ⊆ d·B′ +((B −d·B′)∩Γ) ⊆ d·B′ +(((d+1)·B)∩Γ)

as desired (with about d^{1/2} room to spare). In particular we have

mes(B) ≤ mes(d·B′)|(d+1)·B ∩Γ| ≤ (d(4d+5))^{d}mes(B′)|B ∩Γ|

thanks to (3.10); this proves the lower bound in (3.19) (with a factor of d^{d/2} to spare). Conversely, the sets \{x+ \frac{1}{2} ·B′ : x ∈ B ∩Γ\} are disjoint (since B′ contains no non-zero elements of Γ) and contained in 2 · B, hence

|B ∩Γ|mes(\frac{1}{2} ·B′) ≤ mes(2·B)

which gives the upper bound in (3.19). This concludes the proof when Γ has full rank.

Now suppose that Γ has rank r < d, then after a rotation we may assume that Γ is contained in R^{r} ×{0} ⊂ R^{r} ×R^{d−r}. The point is that the behavior in the d−r dimensions orthogonal to R^{r} is rather trivial and can be easily dealt with as follows. Let B̃ ⊂ R^{r} be the intersection of B with R^{r} ×{0}, identifying R^{r} ×{0} with R^{r} in the usual manner. Then by John’s theorem we have the inclusions

\frac{1}{2} ·(B̃ ×B_{d−r}) ⊆ B ⊆ \sqrt{d}·(B̃ ×B_{d−r}).

Applying the previous arguments to B̃ to obtain a set B̃′ ⊆ B̃, and then defining B′ := \frac{1}{2} ·(B̃′×B_{d−r}), we can verify the claim in this case (losing some additional factors of d^{1/2} and d^{d/2}); we omit the details.

As another consequence of Minkowski’s second theorem, we show how to find large proper progressions inside sets of the form B ∩Γ.

Lemma 3.33 Let B be a convex symmetric body in R^{d}, and let Γ be a lattice in R^{d}. Then there exists a proper progression P in B ∩Γ of rank at most d such that |P | ≥ O(d)^{−3d}|B ∩Γ|.

Proof Applying John’s theorem (Theorem 3.13) followed by a linear transformation, we may reduce to the case where B is the unit ball B = B_{d} in R^{d}. We may assume that B ∩Γ spans R^{d}, since otherwise we may restrict B to the linear span of B ∩Γ, which is then isomorphic to a Euclidean space of some lower dimension. In particular this means Γ has full rank, and that the successive minima 0 < λ_{1} ≤ ... ≤ λ_{d} of B with respect to Γ cannot exceed 1. Let v_{1},...,v_{d} ∈ Γ ∩ B be the corresponding directional basis. Let Q denote the parallelopiped

Q := {t_{1}v_{1} +...t_{d}v_{d} : 0 ≤ t_{j} < 1/2 for all j ∈ [1,d]}.

By (3.16), each translate of Q − Q, being a fundamental domain for Z^{d} ·(v_{1},...,v_{d}), contains at most d! elements of Γ. By Lemma 2.13, we can cover B by at most \frac{mes(B+Q)}{mes(Q)} translates of Q−Q, and thus

|B| ≤ d!\frac{mes(B +Q)}{mes(Q)}.

Since the v_{1},...,v_{d} lie in the unit ball B, we see that Q ⊆ \frac{d}{2} ·B and hence B + Q ⊆ (\frac{d}{2} + 1) · B. Crudely bounding d! = O(d^{d}), we thus conclude that

|B_{∩}Γ| ≤ O(d)^{2d}/mes(Q).

From (3.15) we have

λ_{1} ...λ_{d} ≤ O(1)^{d}mes(Z^{d}/Γ) ≤ O(1)^{d}mes(Q)

and thus

|B ∩Γ| ≤ O(d)^{2d}/λ_{1} ...λ_{d}.

The claim now follows by setting P := [−N,N]·(v_{1},...,v_{d}), where N_{j} := v_{j}/2d for j ∈ [1,d]; note that one can easily verify that P is contained in B ∩Γ.

We now give an alternate approach that gives results similar to Lemma 3.33. We first need a lemma to modify the directional basis given by Minkowski’s second theorem (which only spans a sub-lattice of Γ, see (3.16)) into a genuine basis.

Theorem 3.34 (Mahler’s theorem) Let Γ be a lattice of full rank in R^{d}, and let B be an symmetric convex body in R^{d}, with successive minima 0 < λ_{1} ≤ ... ≤ λ_{d}. Let v_{1},...,v_{d} be a directional basis for Γ. Then there exists a basis w_{1},...,w_{d} of Γ such that w_{1} lies in the closure of λ_{1} ·B, and w_{i} lies in the closure of \frac{iλ_{i}}{2} ·B for all 2 ≤ i ≤ d. Furthermore, if V_{i} is the linear span of v_{1},...,v_{i}, then w_{1},...,w_{i} forms a basis for Γ∩V_{i}.

The basis w_{1},...,w_{d} is sometimes known as a Mahler basis for Γ.

Proof We choose w_{1} := v_{1}; clearly w_{1} forms a basis for Γ ∩ V_{1}. Now suppose inductively that 2 ≤ i ≤ d and w_{1},...,w_{i−1} have already been chosen with the desired properties. The lattice Γ ∩ V_{i} has one higher rank than Γ∩V_{i−1} and hence there exists a vector w_{i} in Γ∩(V_{i}\V_{i−1}) which, together with Γ∩V_{i−1}, generates Γ∩V_{i}; in particular, w_{1},...,w_{i} will generate Γ∩V_{i}. Since v_{1},...,v_{i} linearly span V_{i} we may write w_{i} = t_{1}v_{1} + ... + t_{i−1}v_{i−1} + t_{i}v_{i} for some real numbers t_{1},...,t_{i} with t_{i} ≠ 0. Since v_{i} lies in Γ ∩ V_{i}, we must have t_{i} = ±1/n for some integer n. If |t_{i}| = 1, then Γ ∩ V_{i} is generated by Γ ∩ V_{i−1} and v_{i}, and we can take w_{i} := v_{i}. Thus we may assume |t_{i}| ≤ 1/2. Also, by subtracting integer multiples of v_{1},...,v_{i−1} from w_{∗} if necessary (which will not affect the fact that Γ ∩ V_{i} is generated by Γ ∩ V_{i−1} and w_{i}) we may assume that |t_{j}| ≤ 1/2 for all 1 ≤ j < i. But since each v_{j} lies in the closure of λ_{j} ·B and hence λ_{i} ·B, we conclude that w_{i} lies in the closure of \frac{iλ_{i}}{2} ·B, and so we can continue the iterative construction. Setting i = d we obtain the remaining claims in the theorem.

As an application we give

Corollary 3.35 Let Γ be a lattice of full rank in R^{d}. Then there exists linearly independent vectors w_{1},...,w_{d} ∈ Γ which generate Γ, and such that

mes(R^{d}/Γ) = |w_{1} ∧...∧w_{d}| ≥ O(d)^{−3d/2}|w_{1}|...|w_{d}|.  (3.20)

Proof Let w_{1},...,w_{d} be a Mahler basis for Γ with respect to the unit ball B, and let λ_{1},...,λ_{d} be the successive minima. Then by Theorem 3.34 we have

|w_{1}|...|w_{d}| ≤ λ_{1} ∏_{i=2}^{d} \frac{iλ_{i}}{2} .

Applying (3.15) we obtain

|w_{1}|...|w_{d}| ≤ \frac{2d!}{mes(B)}mes(R^{d}/Γ).

On the other hand, from (3.8) we have

mes(B) = \frac{Γ(3/2)^{d}2^{d}}{Γ(d/2+1)} = (2πe+o(1))^{d/2}d^{−d/2}.

Crudely bounding d! = O(d^{d}), the claim follows.

As a consequence, we can give a “discrete John’s theorem” to characterize the intersection of a convex symmetric body with a lattice.

Lemma 3.36 (Discrete John’s theorem) Let B be a convex symmetric body in R^{d}, and let Γ be a lattice in R^{d} of rank r. Then there exists a r-tuple w = (w_{1},...,w_{r}) ∈ Γ^{r} of linearly independent vectors in Γ and and a r-tuple N = (N_{1},...,N_{r}) of positive integers such that

(r^{−2r} ·B)∩Γ ⊆ (−N,N)·w ⊆ B ∩Γ ⊆ (−r^{2r}N,r^{2r}N)·w.

Proof We first observe using John’s theorem and an invertible linear transformation that we may assume without loss of generality that B_{d} ⊆ B ⊆ d·B_{d}, where B_{d} is the unit ball in R^{d}. We may assume that Γ has full rank r = d, for if r < d then we may simply restrict B to the linear span of Γ, which can then be identified with R^{r}. We may assume d ≥ 2 since the claim is easy otherwise. We may also assume d ≥ 2, since the claim is easy otherwise.

Now let w = (w_{1},...,w_{d}) be as in Lemma 3.35. For each j, let L_{j} be the least integer greater than 1/d|w_{j}|. Then from the triangle inequality we see that |l_{1}w_{1}+...+l_{d}w_{d}| < 1 whenever |l_{j}| < L_{j}, and hence (−L,L)· w is contained in B_{d} and hence in B.

Now let x ∈ B ∩Γ. Since w generates Γ, we have x = l_{1}w_{1} +...+l_{d}w_{d} for some integers l_{1},...,l_{d}; since B ⊆ d·B_{d}, we have |x| ≤ d. Applying Cramer’s rule to solve for l_{1},...,l_{d} and (3.20), we have

|l_{j}| = \frac{|x∧w_{1} ...w_{j−1} ∧w_{j+1} ∧w_{d}|}{|w_{1} ∧...∧w_{d}|} ≤ \frac{|x||w_{1}|...|w_{d}|}{|w_{j}||w_{1} ∧...∧w_{d}|} = \frac{|x|mes(R^{d}/Γ)}{|w_{j}|} ≤ \frac{2d·d!}{|w_{j}|},

which is certainly at most d^{2d}L_{j}. It follows that x ∈ (−d^{2d}L,d^{2d}L)·w, which is what we wanted to prove. A more-or-less identical argument gives the inclusion (d^{−2d} ·B)∩Γ ⊆ (−L,L)·w.

It would be of interest to see if the constant r^{2r} could be significantly improved here, for instance to e^{O(r)} or even r^{O(1)}. Progress on this issue may well have applications to improvements for Freiman’s theorem (see Chapter 5), which can be viewed as a variant of the above theorem in which the set B ∩Γ is replaced by a more general set of small doubling.

3.5.1 Prove (3.12). 3.5.2 Let α be an irrational number, and let I be any open interval

in R. Show that Z · α and I + Z have non-empty intersection. (In other words, the integer multiples of α are dense in R/Z).

3.5.3 Let Γ be a lattice in R^{d}, and let A be a convex body (possibly

asymmetric). Show that Dbl[A∩Γ] ≤ O(1)^{d}. 3.5.4 Let v_{1},...,v_{d} be any vectors in a lattice Γ ⊂ R^{d} of full rank.

Show that |v_{1} ∧ ... ∧ v_{d}| is an integer multiple of the covolume mes(R^{d}/Γ).

3.5.5 Let Γ be a lattice of full rank in R^{d}, let B be a symmetric convex body, and let v_{1},...,v_{d} be a directional basis with successive minima λ_{1} ≤ ... ≤ λ_{d}. Let O be the open octahedron with vertices ±v_{j}/λ_{j}. Show that O ⊆ B ⊆ O(d)^{d} ·O. Thus Minkowski’s second theorem can be used to give a rather weak version of John’s theorem.

3.5.6 Let Γ be a lattice of full rank in R^{d}, let B be a symmetric convex body, and let λ_{1} ≤ ... ≤ λ_{d} be the successive minima of B. Establish the bounds

O(d)^{−O(d)} ∏_{1≤i≤d}max(1, \frac{1}{λ_{i}}) ≤ |B∩Γ| ≤ O(d)^{O(d)} ∏_{1≤i≤d}max(1, \frac{1}{λ_{i}}).

(3.21)

3.5.7 Generalize Lemma 3.32 and Lemma 3.36 to the case when B is

an asymmetric convex body. 3.5.8 Let A be a bounded open subset of R^{d}, and let B,C be open

subsets of A. Prove that

mes((B −B)∩(C −C)) ≥ \frac{mes(B)mes(C)mes(A)}{mes(A−B)mes(A−C)}.

(Hint: use the volume-packing argument to locate a large set of the form (x+B)∩(y +C) where x ∈ A−B and y ∈ A−C.)

3.6 Progressions and proper progressions

In this section we work in a fixed additive group Z, which may or may not be torsion free.

Recall that a progression P = a+[0,N]·v is proper if the map n ↦ n · v is injective on [0,N]. Not all progressions are proper; however it turns out that, just as John’s theorem (Theorem 3.13) shows that all convex sets are in some sense comparable to ellipsoids, all progressions are comparable to proper progressions. This is most obvious in the rank 1 case, in which every arithmetic progression is equal (as a set) to a proper arithmetic progression:

Lemma 3.37 Let a+[0,N]·v be an arithmetic progression in an additive group Z. Then there exists an n > 0 such that a+[0,n)·v is a proper arithmetic progression and a+[0,n)·v = a′ +[0,N′]·v′.

Proof If a + [0,N] · v is already proper, then we are done. Otherwise, there exist distinct n_{1},n_{2} ∈ [0,N] such that a + n_{1} · v = a + n_{2} · v. In particular, there exists n ∈ [1,N] such that n · v = 0. Let n be the least integer in [1,N] with this property. Then a+[0,n)·v is necessarily proper, and by the Euclidean algorithm it is clear that a + [0,n) · v = a+[0,N]·v.

We now consider the higher rank case; as with John’s theorem, the constants will deteriorate worse than exponentially in d. We first show the easier of the two containments, namely that every progression contains a large proper progression of equal or lesser rank.

Theorem 3.38 Let P be a progression of rank d in a additive group Z. Then P contains a proper progression of rank at most d and volume at least O(d)^{−9d/2}|P |.

Remark 3.39 For a result of similar flavor (but proven by completely different methods), see Theorem 4.41 below. Note that the d = 1 case already follows from Lemma 3.37 (with a constant of 1 instead of O(d)^{−9d/2}.

Proof The idea is to pass to a convex body, apply Lemma 3.32 to obtain a “proper” subset of this body, and then use Lemma 3.33 to pass back to a progression.

By translating and enlarging P slightly we may assume P = [−N,N]· v. We may assume that none of the components N_{j} of N are equal to 0 or 1, since otherwise we could refine P by at worst a factor of 3^{d} to eliminiate those dimensions. Now consider the set Γ := {n ∈ Z^{d} : n·v = 0}, which is clearly a sub-lattice of Z^{d}, and let A be the symmetric convex box

A := {(x_{1},...,x_{d}) ∈ R^{d} : −N_{j} ≤ x_{j} ≤ N_{j} for all 1 ≤ j ≤ d}.

By Lemma 3.32, we may find a symmetric convex subset A′ of A such that A′ −A′ is disjoint from Γ−{0}, and such that A ⊂ O(d)^{3/2} ·A′ +Γ for some x ∈ R^{d}. From Corollary 3.15, we thus see that A can be covered by O(d)^{3d/2} translates of \frac{1}{2} · A′ + Γ. Since [−N,N] = A ∩ Z^{d} and Γ ⊆ Z^{d}, we conclude that [−N,N] can be covered by O(d)^{3d/2} sets of the form [(\frac{1}{2} · A′ + x) ∩ Z^{d}] + Γ. Taking inner products with v, we conclude that P = [−N,N] · v can be covered by O(d)^{3d/2} sets of the form [(\frac{1}{2} ·A′ +x)∩Z^{d}]·v. By the pigeonhole principle, there must thus exist an x such that

|(\frac{1}{2} ·A′ +x)∩Z^{d}| ≥ O(d)^{−3d/2}|P |

and hence by (3.9)

|A′ ∩Z^{d}| ≥ O(d)^{−3d/2}|P |.

We now apply Lemma 3.33 to find a proper progression P̃ ⊆ A′ ∩Z^{d} ⊆ [0,N] of rank at most d such that

|P̃| ≥ O(d)^{−3d}|A′ ∩Z^{d}| ≥ O(d)^{−9d/2}|P |.

The set P̃ ·v is then clearly a progression of rank at most d contained in P; it is proper since A′ −A′ is disjoint from Γ−{0}, so in particular |P̃ ·v| = |P̃|). The claim follows.

Now we show the more difficult containment, that every progression can be contained inside a proper progression of equal or lesser rank, but somewhat larger volume.

Theorem 3.40 [26] Let P be a progression of rank d in an additive group Z. Then P is contained in a proper progression Q of rank at most d and volume at most d^{C_{0}d^{3}}|P | for some absolute constant C_{0} > 0. Also, Q is contained in a translate of d^{C_{0}d^{2}}P. If d ≥ 2 and P is not proper, then Q can be chosen to have rank at most d−1. Finally, if Z is torsion-free and P is symmetric, then one can ensure that Q is symmetric also.

Remark 3.41 This result was discovered independently by Ruzsa (private communication) and Gowers and Walters (private communication). Comparison with Theorem 3.38 suggests that the factor d^{C_{0}d^{3}} is probably not best possible, but we do not know what the correct constant here should be. This theorem can be thought of as the analogue of Corollary 3.8 or Corollary 3.9, but for progressions rather than finitely generated additive groups.

Proof This claim is analogous to the basic linear algebra statement that every linear space spanned by d vectors is equal to a linear space with a basis of at most d vectors. Recall that the proof of that linear algebra fact proceeds by a descent argument, showing that if the d spanning vectors were linearly dependent, then one could exploit that dependence to “drop rank” and span the same linear space with d−1 vectors. Our proof of Theorem 3.40 shall be based on a similar strategy.

We shall work only in the case when Z is torsion-free; the general case is proven similarly but contains a few additional technicalities, and we leave it as an exercise (Exercise 3).

We induct on d. When d = 1 the claim follows from Lemma 3.37. Now suppose inductively that d ≥ 2, and the claim has already been proven for d − 1 (for arbitrary groups Z and arbitrary progressions P). Let P = a+[0,N]·v be a progression in Z of rank d, where N = (N_{1},...,N_{d}) and v = (v_{1},...,v_{d}) we may translate so that the base point a equals 0. If P is proper, then we are done. Similarly, if one of the N_{j} is equal to zero, then we done by induction hypothesis. Suppose instead that P is not proper and all the N_{j} are at least one, then there exist distinct n,n′ ∈ [0,N] such that n·v = n′ ·v. If we then let Γ_{0} ⊆ Z^{d} denote the lattice {m ∈ Z^{d} : m·v = 0}, then we see that Γ_{0} ∩[−N,N] contains at least one non-zero element, namely n′ −n.

Let m = (m_{1},...,m_{d}) be a non-zero element of Γ_{0} ∩[−N,N], thus

m_{1} ·v_{1} +...+m_{d} ·v_{d} = 0.  (3.22)

We may assume without loss of generality that m is irreducible in Γ_{0}. Since Z is torsion-free, this also implies that m is irreducible in Z^{d} (i.e. that m_{1},...,m_{d} have no common divisor) unless Z is torsion-free. The strategy shall be to contain P inside a progression Q of rank d−1 and size

|Q| ≤ d^{O(d^{2})}|P |,  (3.23)

such that Q is contained in a translate of d^{O(d)}P. If we can achieve this, then by the induction hypothesis we can contain Q inside a proper progression R of rank at most d−1 and cardinality

|R| ≤ (d−1)^{C_{0}(d−1)^{3}}(O(d))^{O(d^{2})}|P |

and which is contained in a translate of d^{C_{0}(d−1)^{2}}d^{O(d)}P. If C_{0} is sufficiently large, we have closed the induction hypothesis.

It remains to cover P by a progression of rank at most d−1 with the bound (3.23) and contained in a translate of d^{O(d)}P. Observe that m lies in [−N,N], so the rational numbers m_{1}/N_{1},...,m_{d}/N_{d} lie between −1 and 1. Without loss of generality we may assume that m_{d}/N_{d} has the largest magnitude, thus

|m_{d}|/N_{d} ≥ |m_{j}|/N_{j}  (3.24)

for all 1 ≤ j ≤ d. By replacing v_{d} with −v_{d} if necessary, we may also assume that m_{d} is positive.

To exploit the cancellation in (3.22) we introduce the rational vector q ∈ \frac{1}{m_{d}} ·Z^{d−1} by the formula

q := (−\frac{m_{1}}{m_{d}} ,...,−\frac{m_{d−1}}{m_{d}}).

Since m is irreducible in Z^{d}, we see that for any integer n, that n · q lies in Z^{d−1} if and only if n is a multiple of m_{d}, because (m_{1},...,m_{d}) is irreducible in Z^{d}.

Next, let Γ ⊂ R^{d−1} denote the lattice Γ := Z^{d−1} + Z · q. Since q is rational, this is indeed a lattice; since it contains Z^{d−1}, it is certainly full rank. We define the homomorphism f : Γ → Z by the formula

f((n_{1},...,n_{d−1})+n_{d}q) := (n_{1},...,n_{d})·v.

the condition (3.22) ensures that this homomorphism is indeed well defined, in the sense that different representations v = (n_{1},...,n_{d−1})+n_{d}q of the same vector v ∈ Γ give the same value of f(v). We also let B ⊆ R^{d−1} denote the convex symmetric body

B := {(t_{1},...,t_{d−1}) ∈ R^{d−1} : −3N_{j} < t_{j} < 3N_{j} for all 1 ≤ j ≤ d−1}.

We now claim the inclusions

P ⊆ f(B ∩Γ) ⊆ 5P −5P.

To see the first inclusion, let n·v ∈ P for some n ∈ [0,N], then we have n·v = f((n_{1},...,n_{d−1})+n_{d}q); from (3.24) we see that the j^{th} co-efficient of (n_{1},...,n_{d−1})+n_{d}q has magnitude at most 3N_{j}, and thus n·v lies in f(B ∩Γ) as claimed. To see the second inclusion, let (n_{1},...,n_{d−1})+n_{d}q be an element of B ∩Γ. By subtracting an integer multiple of m_{d} from n_{d} if necessary (and thus adding integer multiples of m_{1},...,m_{d−1} to n_{1},...,n_{d−1}) we may assume that |n_{d}| ≤ |m_{d}|/2. By (3.24) and the definition of B, this forces |n_{j}| ≤ 5N_{j} for all 1 ≤ j ≤ d, and hence

f((n_{1},...,n_{d−1})+n_{d}q) = (n_{1},...,n_{d})·v ⊆ [−5N,5N]·v = 5P −5P.

Next, we apply Theorem 3.36 to find vectors w_{1},...,w_{d−1} ∈ Γ and M_{1},...,M_{d−1} such that

(−M,M)·w ⊆ B ∩Γ ⊆ (−d^{O(d)}M,d^{O(d)}M)·w.

Applying the homomorphism f, we obtain

(−M,M)·f(w) ⊆ f(B ∩Γ) ⊆ (−d^{O(d)}M,d^{O(d)}M)·f(w)

where f(w) := (f(w_{1}),...,f(w_{d−1}). Observe that (−d^{O(d)}M,d^{O(d)}M)· f(w) is a progression of rank d−1 which contains f(B ∩Γ) and hence contains P. Furthermore, by two applications of Lemma 3.10 we have

|(−d^{O(d)}M,d^{O(d)}M)·f(w)| ≤ (O(d))^{O(d^{2})}|f(B ∩Γ)|
≤ (O(d))^{O(d^{2})}|5P −5P |
≤ (O(d))^{O(d^{2}))}O(1)^{d}|P |

which proves (3.23). Also, since (−M,M)·f(w) is contained in f(B∩Γ), which is contained in 5P −5P, which is a translate of 10P, we see that (−d^{O(d)}M,d^{O(d)}M) · f(w) is contained in a translate of d^{O(d)}P. This completes the induction and proves the theorem. When P is symmetric, one can easily modify the above argument to ensure that all progressions in the above construction are also symmetric; we leave this modification to the interested reader.

3.6.1 Let P = a+[0,N]·v be a progression of rank d in some additive group Z, and let Γ := {n ∈ Z^{d} : n · v = 0} be the associated sublattice of Z^{d}. Prove the inequalities

\frac{|[0,N]|}{|P |} ≤ |[−N,N]∩Γ| ≤ 3^{d} \frac{|[0,N]}{|P |}.

Thus the ratio between the volume and cardinality of a progression P is essentially controlled by the quantity |[−N,N] ∩ Γ|. Hints: for the lower bound, first use Cauchy-Schwarz to obtain a lower bound for {(n,n′) ∈ [0,N] : n·v = n′ ·v}. For the upper bound, consider the multiplicity of the map f : [−N,2N] → Z defined by f(n) := n·v.

3.6.2 Let [0,N] be a box in Z^{d}, and let Γ be a sublattice of Z^{d}. Show

that |[−kN,kN]∩Γ| ≤ (2k)^{d}|[−N,N]∩Γ| for all integers k ≥ 1. 3.6.3 Prove Theorem 3.40 in the case when Z is not necessarily torsion-

free. (The main new difficulty is that the vector m is not always irreducible in Z^{d}; in such a case one will have to “quotient out” a finite cyclic group from P before proceeding with the rest of the argument. However, this will only introduce additional factors of C^{d} into the inductive bound (3.23), which is acceptable). Note that the second part of the Theorem does not extend to the torsion case, as can already be seen by considering P = Z = Z_{2}.

3.6.4 Prove an extension of Theorem 3.40 in the torsion-free case in which one requires that kQ is also proper for some fixed constant k ≥ 1 (of course, the bounds on Q will depend on k). Note that the torsion-free hypothesis is now essential, as can be seen by considering the case when P = [1,N]·1 in Z_{N}.

3.6.5 [334] Let N_{1},N_{2},a_{1},a_{2} be positive integers such that 0 < a_{2} < N_{1}/5 and 0 < a_{1} < N_{2}/5, and a_{1},a_{2} are coprime. Use the Chinese remainder theorem to show the inclusion

[\frac{1}{5}(a_{1}N_{1} +a_{2}N_{2}), \frac{4}{5}(a_{1}N_{1} +a_{2}N_{2})] ⊆ [0,(N_{1},N_{2})]·(a_{1},a_{2}).

Conclude that if P is any progression of rank 2 in the integers of dimensions N_{1},N_{2} and steps v_{1},v_{2} with 0 < v_{2} < N_{1}/5 and 0 < v_{1} < N_{2}/5, then P contains a proper arithmetic progression of length \frac{3(N_{1}v_{1}+N_{2}v_{2})}{5gcd(v_{1},v_{2})} and spacing gcd(v_{1},v_{2}).

3.6.6 [334] Let A be an additive set in an ambient group Z. Show that there exists d = O(log|A|) and distinct elements v_{1},...,v_{d} ∈ A such that the cube [0,1]^{d} · (v_{1},...,v_{d}) has cardinality at least \frac{1}{4}|A|. (Hint: Using (2.22), show that if S is any additive set in Z such that |S| < \frac{1}{4}|A| then there exists a ∈ A such that |S ∪(S +a)| ≥ \frac{3}{2}|S|. Then use the greedy algorithm.)

534 12 Bibliography

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