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Fourier analysis in combinatorial number theory

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Russian Math . Surveys 65:3 513–567 c⃝ 2010 RAS(DoM) and LMS

Uspekhi Mat . Nauk 65:3 127–184 DOI 10.1070/RM2010v065n03ABEH004681

To Professor Valerii Vasil’evich Kozlov ,

Academician of the Russian Academy of Sciences,

on the occasion of his 60th birthday

Fourier analysis in combinatorial number theory

I. D. Shkredov

Abstract. In this survey applications of harmonic analysis to combinato-

rial number theory are considered. Discussion topics include classical prob-

lems of additive combinatorics, colouring problems, higher-order Fourier

analysis, theorems about sets of large trigonometric sums, results on esti-

mates for trigonometric sums over subgroups, and the connection between

combinatorial and analytic number theory.Bibliography: 162 titles.

Keywords: Fourier analysis, combinatorial number theory, additive com-

binatorics.

Contents

1. Introduction 513

2. Denitions 5163. Arithmetic progressions o length three 5194. Higher-order Fourier analysis 5275. Colouring problems 5326. Sets o large trigonometric sums 5397. Combinatorial congurations in sumsets 5458. Chebotarev’s theorem and the uncertainty principle 5529. Connection with analytic number theory 55410. Conclusion 559

Bibliography 560

1. Introduction

In the present survey we consider applications o Fourier analysis to problemso combinatorial number theory. Combinatorial number theory, or additive combi-natorics, as it is called nowadays, is a branch o mathematics in which combinato-rial questions connected with a group operation are studied (see the books [1], [2]and the surveys [3], [4]). Although the area under consideration is airly old

This research was supported by the Russian Foundation or Basic Research (grantno. 06-01-00383), a P. Deligne grant (the Balzan Foundation 2004), a grant o the Presidento the Russian Federation (grant no. MK-1959.2009.1), and the programme “Leading ScientifcSchools” (grant no. НШ-691.2008.1).

AMS 2010 Mathematics Subject Classifcation . Primary 11B25, 11B75, 42B99; Secondary05B10, 11B13, 11L07, 11N13, 11P70, 11T23.



514 I. D. Shkredov

(it is sucient to recall Cauchy’s result [5] in 1813 on addition o sets in the residuegroup Z/pZ, as well as the celebrated theorem o Schnirelmann [6] in 1933 on therepresentability o any positive integer as a sum o boundedly many primes), inter-est in this area o research has recently increased considerably, due rst o all to thesignicant success in combinatorial problems on arithmetic progressions [7]–[9], to

the emergence o new estimates or sums o products o sets in nite elds [ 10]–[23],to the progress in problems o combinatorial ergodic theory [24]–[35], and to the sub-stantial advances in the estimation o trigonometric sums over subgroups [14]–[23].

Not the smallest role here has been played by the method o Fourier analysis.Moreover, the strongest results in this area have been obtained precisely by usingharmonic analysis, or to say the least, these results have been stimulated by ques-tions o Fourier analysis.

The prominent English mathematician K. Roth was the rst to apply harmonicanalysis to problems o combinatorial number theory. In 1953 he proved the now

classical result on the density o subsets o a segment o the positive integers thatdo not contain arithmetic progressions o length three. The iteration approach usedby Roth in his proo ound its widest applications in combinatorial number theoryand was reworked rom the viewpoint o graph theory (see [36], [37]) and ergodictheory (see, or example, [38], [39]). At present the best results on Roth’s problemare due to Bourgain (see [40], [41]). The tools he created (rst o all, the methodo Bohr sets and the results on large trigonometric sums) are also systematicallyused in a whole series o problems in additive combinatorics.

A huge contribution to the development o Fourier analysis in its applications

to problems o combinatorial number theory was made by Gowers [7], [8]. Insolving the problem o density o a set without arithmetic progressions o lengthgreater than three and generalizing Roth’s method, Gowers created the so-calledhigher-order Fourier analysis. As is well known, classical Fourier analysis associateswith an arbitrary unction the set o its inner products with group characters (theFourier coecients). It turns out that in problems on arithmetic progressions o length greater than three, one needs to consider inner products o a unction anda character whose argument is some polynomial. The resulting analogues o Fouriercoecients play an extraordinarily important role in this whole area o research (seemore details in

§4). Gowers’ methods have been developed by various authors (see,

or example, [9], [14], [15] [42]–[46]). The most striking o the results obtained inthese papers is undoubtedly the theorem o Green and Tao asserting that the primenumbers contain arithmetic progressions o arbitrarily large length.

Finally, a connection was recently ound between certain problems o additivecombinatorics and problems o analytic number theory. For example, by usingestimates or sums o products in the group Z/pZ it was proved in [15] that everymultiplicative subgroup R ⊆ (Z/pZ)∗ is uniormly distributed i the size o R issuciently large, namely, |R| pε, where ε > 0 is any number. This signicantlystrengthens the analogous classical result (see, or example, the book [47]), where

the uniorm distribution property was known only or subgroups o size pc, c > 1/4(see [48]). The sum-product method made it possible to obtain non-trivial estimatesor trigonometric sums also in other problems (see [49], [50]). Conversely, as wasrecently pointed out in [51] and [52], the classical estimates or trigonometric sumsprove to be useul in problems related to sums o products. Applications o the



Fourier analysis in combinatorial number theory 515

methods o analytic number theory to combinatorial problems were also consideredin [53]–[56]. The author hopes that both o the research areas mentioned above willhave more mutual inuence on each other.

In the present survey we consider mainly problems in nite Abelian groups.Clearly, i some problem in additive combinatorics is posed in some concrete group,

then it can also be reormulated in any other group. In some groups G it is easy toanswer the question posed, but in others it is dicult, since, as Green said in [3],“not all Abelian groups were created equal.” Recently it became clear (see [40], [57],[58] and especially the excellent survey [3]) that it is extremely convenient to takethe groups (Z/pZ)n as the group G, where p is a small prime (or example, p = 2, 3,or 5). The act is that in such groups there is a natural vector space structure, andurthermore, the groups (Z/pZ)n themselves are not too complicated. Moreover,the general ideology o [3] consists in the ollowing: i some result o combinatorialnumber theory is proved or the groups (Z/pZ)n, then, as a rule, the corresponding

result can also be proved or an arbitrary nite Abelian group, and without involvingundamentally new ideas. All this tells us that it is airly reasonable to deal not withone concrete Abelian group (say, Z/N Z), but to discuss the entire set o Abeliangroups under consideration.

We now briey describe the structure o the present survey. In § 2 we recall somenotions o Fourier analysis and give the requisite denitions. § 3 is devoted to theproo o Roth’s theorem on arithmetic progressions o length three. We discuss indetail the case o the group (Z/3Z)n, as well as obtain the classical Roth theorem.In § 4 we touch upon a young and rapidly developing area o harmonic analysis, theso-called ‘higher-order Fourier analysis’. Here there are many open problems andunanswered questions which await investigation.

In solving linear equations Fourier analysis traditionally dealt with only ‘density’problems (see the denitions in § 5), such as Roth’s theorem. Recently the youngmathematician T. Sanders invented a method or using harmonic analysis in colour-ing problems, an area which traditionally belongs to pure combinatorics. Severalproblems and results here, such as Schur’s theorem [59] and its generalizations,are considered in § 5. Problems related to the study o sets o large trigonometricsums, which we discuss in § 6, is also a young and developing area. It is enough tosay that the rst general result here was obtained only in 2002. Nevertheless, any

progress in our understanding o the structure o sets o large trigonometric sumswill, in the author’s opinion, make advances possible in most o those problems inadditive combinatorics where Fourier analysis is used. The same idea was expressedby Gowers in the survey [60]. In any case, examples o such advances can be seen in[41], [61]–[66]. In what ollows (§ 7) we consider a new and interesting area o com-binatorial number theory involving the search or various congurations (rst andoremost, arithmetic progressions) in sums o several sets. The rst result in thisdirection is due to Bourgain [67]. It is interesting to point out that his method o proo does not t in the ramework o the iteration approach proposed by Roth. In

§ 8 we touch upon another aspect o harmonic analysis, namely, non-degeneracy (ina broad sense) o the Fourier transorm. We discuss the elegant paper o Tao devotedto the so-called ‘uncertainty principle’ in the group Z/pZ, where p is a prime. Itturns out that this result is a simple reormulation o the no less elegant theoremo our prominent mathematician Chebotarev on determinants o the minors o the



516 I. D. Shkredov

matrix o the Fourier transorm. In § 9 we consider connections between additivecombinatorics and analytic number theory. We discuss problems in estimating sumso products and their generalizations, as well as several combinatorial questions inwhose resolution purely analytic methods prove to be useul. Finally, in conclusionwe consider a number o open problems in combinatorial number theory or whose

solution Fourier analysis may have decisive importance.Unortunately, the choice o topics or the present survey was determined exclu-

sively by the author’s interests. Furthermore, o course, many branches o additivecombinatorics using Fourier analysis remained untouched, or example, the vastarea related to the Kakeya problem. The author can only suggest that the readerlearn about this area o research rom the excellent surveys [68], [69], as well asrom [70].

We mainly conne ourselves to a airly old area o classical harmonic analysis inthe light o its applications to combinatorial number theory. § 4 is the only section

devoted to the higher-order Fourier analysis developed in [7]–[9], [42], [71]–[76],and other sources. Nevertheless, we hope that ater reading our survey the readerwill be convinced that much which is new has lately happened also in this part o mathematics.

2. Defnitions

In this section we recall some notions o Fourier analysis. Let G = (G, +) bea locally compact Abelian group with additive group operation + . It is well knownthat on every such group there exists the Haar measure µG, that is, a non-negative

regular measure (all the denitions can be ound, or example, in [77], Chaps. 1, 2and Appendix E ) such that or any Borel set B and any x ∈ G

µG(B + x) = µG(B).

On the space L2(G, µG) = L2(G) an inner product o two complex unctions f andg is dened:

⟨f, g⟩ =

G

f (x)g(x) dµG(x).

Next, let

G be the dual group o G, that is, the locally compact Abelian group

o continuous homomorphisms ξ rom G into S1

= z ∈ C : |z| = 1, ξ : x →ξ · x = ξ(x), x ∈ G. The Fourier transorm (Φf )(ξ) = f (ξ) o an arbitrary complexunction f ∈ L1(G) is dened by the ormula

(Φf )(ξ) = f (ξ) =

G

f (x)ξ(x) dµG(x). (1)

Clearly, the ollowing inequality always holds:

∥f ∥∞

G

|f (x)| dµG(x) = ∥f ∥1.

Let Γ ⊆ G be some set o characters, and ε ∈ [0, 1] a real number. We dene theBohr set generated by the set Γ:

B(Γ, ε) =

x ∈ G : ∥ξ · x∥ ε

2πor all ξ ∈ Γ

, (2)




where ∥ · ∥ is the Euclidean metric on the circle. One can say that the quantity∥ξ · x∥ measures how much ξ difers rom the trivial character ξ0 ≡ 1. Simplearguments involving Dirichlet’s drawer principle enable one to prove the ollowinglemma (see, or example, [57]).

Lemma 1. I G is a compact Abelian group and |Γ| = d, then µG(B(Γ, ε)) εd.

In particular, the Bohr sets are always non-empty.

In what ollows we need the notion o convolution o two unctions f, g ∈ L1(G):

(f ∗ g)(x) =

G

f (y)g(x − y) dµG(x).

We have (see, or example, [77])

(f ∗ g)(ξ) = f (ξ)g(ξ).

The convolution operation is very important or additive problems o combinatorialnumber theory. Indeed, let A and B be arbitrary subsets o the group G. We denethe (Minkowski) sum o these sets:

A + B := c ∈ G : c = a + b, a ∈ A, b ∈ B.

Let f be the characteristic unction o the set A, and g the characteristic unctiono the set B. It is easy to see that (f ∗ g)(x) = 0 i and only i x ∈ A + B. Thesum o several sets is dened similarly. I in the group (G, +) there is anotherbinary operation, say, multiplication, then a product o sets can also be dened. Inwhat ollows we shall need the sum A B o distinct elements o sets A and B:

AB := c ∈ G : c = a + b, a ∈ A, b ∈ B, a = b.

We denote by A1 · · · Ad the set ormed by the sums o distinct elements o sets A1, . . . , Ad. The set consisting o the sums o d distinct elements o a set A is

denoted by dA.

We shall oten consider the situation where G is a nite Abelian group. It is wellknown that G = G in this case. We denote by N the cardinality o the group G.It is very convenient to dene the Fourier transorm o a unction f by the ormula

(Φf )(ξ) = f (ξ) =x∈G

f (x)ξ(x) . (3)



518 I. D. Shkredov

Thus, in contrast to (1), we do not normalize the sums with respect to G. We shallneed several well-known ormulae concerning the Fourier transorm:

∥f ∥22 :=x∈G

|f (x)|2 =1

N

ξ

∈ G

|f (ξ)|2 =1

N ∥f ∥22 (Parseval’s equality), (4)

⟨f, g⟩ :=x∈G

f (x)g(x) =1

N

ξ∈ G

f (ξ)g(ξ) =1

N ⟨f , g⟩ (Plancherel’s identity), (5)

x∈G

(f ∗ g)(x)2 =

x∈G

y∈G

f (y)g(x − y)

2 =1

N

ξ∈ G

|f (ξ)|2|g(ξ)|2, (6)

f (x) =1

N

ξ∈ G

f (ξ)ξ(x) (inversion ormula), (7)

f ∗ g = f g and ( fg)(x) = 1N

(f ∗ g)(x). (8)

For brevity we write

s instead o the sum

s∈G, and

ξ instead o

ξ∈ G.

We present two examples o groups which are used especially oten in our survey.

Example 1. Let N be a positive integer and let G = Z/N Z = ZN . Then all thecharacters are N th roots o unity, that is, e2πir/N , where r = 0, 1, . . . , N − 1. Wewrite eN (x) or even simply e(x) instead o e2πix/N . Then the Fourier transorm o a unction f is given by the ormula

(Φf )(ξ) = f (ξ) = x∈ZN

f (x)e(−ξx).

In the group ZN we can describe the structure o the Bohr sets (2). Simple argu-ments involving Dirichlet’s drawer principle enable us to prove the ollowing lemma.

Lemma 2. Let G = ZN and |Γ| = d. Then the Bohr set B(Γ, ε) contains an

arithmetic progression o length εN 1/d.

Thus, the Bohr set can be thought o as a union o arithmetic progressions.

Example 2. Let p be a prime and n a positive integer, and let G = (Z/pZ)n

. Thegroup G is a vector space with the inner product

x · y = x · y = ⟨x, y⟩ = x1y1 + · · · + xnyn (mod p).

The Fourier transorm o a unction f : Zn p → C is calculated by the ormula

f (r) =

x∈Znpf (x)e(−⟨r, x⟩),

where e(x) = e p(x) = e2πix/p

. The Fourier transorm in the group Zn2 has an

especially simple structure. Indeed, since e2(x) = (−1)x, or a real unction f the

numbers f (r) are real.Corresponding to the notion o a Bohr set or the groups Zn

p is the notion o an ane subspace. Let v1, . . . , vk be some linearly independent vectors, and let




ε1, . . . , εk be arbitrary elements o Z p. An ane subspace o codimension k isdened to be the set

P = P ε1,...,εk(v1, . . . , vk) =

x ∈ Zn p : ⟨x, v1⟩ = ε1, . . . , ⟨x, vk⟩ = εk

.

I all the εi are equal to zero, then we sometimes write P (v1, . . . , vk) instead o P 0,...,0(v1, . . . , vk). For a set W = (w1, . . . , w|W |), let P (W ) mean P (w1, . . . , w|W |).The Fourier transorm o the characteristic unction χP o the set P = P (W ) isextremely simple to compute. Let L be the linear space o dimension k spanned bythe vectors v1, . . . , vk, and let r ∈ Zn

p be an arbitrary vector. Clearly, P ⊥ = L andχP (r) = |P | or every r ∈ L. Next, it ollows rom Parseval’s equality

r∈Znp|χP (r)|2 = |P |N

that χP (r) = 0 or any r /∈ L. In other words,

χP (r) = χL(r)|P |, (9)

where χL is the characteristic unction o the set L. Thus, |χP (r)| either is zero, oris equal to |P |.

We make several more remarks on notation. I S ⊆ G is some set, then wedenote by S (x) its characteristic unction. In other words, S (x) = 1 i x ∈ S , and

zero otherwise. The numbers S are called the Fourier coecients o the set S . Thesymbol |S | is used or the cardinality o a set S . Here log means the logarithmto the base two, and ln means the natural logarithm. The signs ≪ and ≫ arethe usual Vinogradov symbols. I n is a positive integer, then we write [n] todenote the segment 1, . . . , n o positive integers. The symbol E(·) denotes themathematical expectation o a random quantity, and P(·) is the probability o someevent. Let D be the unit disc on the complex plane. Finally, Z∗N = ZN \ 0, and(m, n) is the largest common divisor o positive integers m and n.

3. Arithmetic progressions o length threeIn 1927 van der Waerden [78] proved his celebrated theorem on arithmetic pro-

gressions.

Theorem 1 (van der Waerden). Let h and k be positive integers. There exists

a number N (h, k) such that or any positive integer N N (h, k) and or an arbi-

trary partition o the set [N ] into h subsets, one o the subsets contains an arithmetic

progression o length k, that is, a sequence n, n + d, n + 2d , . . . , n + (k −1)d, where nand d are positive integers.

Let N be a positive integer. We set

ak(N ) =1

N max

|A| : A ⊆ [N ], A does not contain

arithmetic progressions o length k

.



520 I. D. Shkredov

In [79] Erdos and Turan stated the conjecture that in any set o positive densitythere exists an arithmetic progression o prescribed length. In other words, theyconjectured that or any k 3

ak(N )

→0 as N

→ ∞. (10)

Clearly, this conjecture implies van der Waerden’s theorem.

The conjecture (10) in the simplest case k = 3 was proved by Roth [80]. Usingthe Hardy–Littlewood method, Roth obtained the ollowing result in his paper.

Theorem 2 (Roth). For all positive integers N 3

a3(N ) ≪ 1

log log N .

Roth’s result was later improved by Szemeredi [81] and Heath-Brown [82]. Inde-pendently o each other, both these authors obtained the ollowing estimate ora3(N ).

Theorem 3 (Szemeredi, Heath-Brown). Let N be a positive integer , N 3. Then

a3(N ) ≪ 1

(log N )c,

where the constant c can be taken to be equal to 1/20.

At present the best result on upper estimates or the quantity a3(N ) is due toBourgain [41] (see also his earlier paper [40] and the paper [83] on Roth’s theoremin Rn).

Theorem 4 (Bourgain). Let N be a positive integer , N 3. Then

a3(N ) ≪ (log log N )3

(log N )2/3. (11)

The conjecture (10) was proved by Szemeredi or arbitrary k in 1975 [36], [84].In his proo Szemeredi uses dicult combinatorial arguments. His proo is basedon the so-called regularity lemma, which at present is a most important tool orstudying graphs (see, or example, [37], [85]). An alternative proo was proposedby Furstenberg [38]. His approach uses the methods o ergodic theory. Furstenbergshowed that Szemeredi’s theorem is equivalent to the assertion o multiple recur-rence or almost all points in an arbitrary dynamical system. We state his theoremin the ollowing orm.

Theorem 5 (Furstenberg). Let X be a metric space with metric d(·

,·

) and Borel

sigma-algebra o measurable sets Φ. Let T be a map o X into itsel preserving the

measure µ and let k 3. Then or almost all x ∈ X

lim in n→∞ max

d(T nx, x), d(T 2nx, x), . . . , d

T (k−1)nx, x

= 0.




We should point out that by using his method Furstenberg and his ollowersobtained a plethora o deep generalizations o Szemeredi’s theorem (see, or exam-ple, [24]–[35], [38], [39], [86]–[88]) which at the time could not be proved by com-binatorial methods. More details about Szemeredi’s theorem can be ound in thesurvey [4].

Behrend [89] obtained a lower estimate or a3(N ) (see also [90], [91]).

Theorem 6 (Behrend). Let ε > 0 be a real number . Then there exists an N ε ∈ Nsuch that or any positive integer N with N N ε

a3

N exp

−(1 + ε)C

log N

,

where C > 0 is some absolute constant .

In spite o its simplicity (the proo occupies only two pages), Behrend’s resultobtained in 1946 remains the best to the present day.

Rankin [92] generalized Behrend’s theorem to the case o all k 3.

Theorem 7 (Rankin). Let ε > 0 be a real number and let k 3 be a positive

integer . Then or all suciently large N

ak(N ) exp−(1 + ε)C k(log N )1/(k−1)

,

where C k is some positive efective constant depending only on k.

Unortunately, Szemeredi’s methods give very weak upper estimates or ak(N ).The ergodic approach does not give any estimates at all. Only in 2001 Gowers [8]obtained a quantitative result on the rate o convergence to zero o the quantityak(N ) or k 4. He proved the ollowing theorem.

Theorem 8 (Gowers). Let k 4. Then

ak(N ) ≪ 1

(log log N )ck,

where the constant ck > 0 depends only on k.

Gowers’ result is a signicant step on the road to proving another amous conjecture o Erdos and Turan about arithmetic progressions.

Conjecture 1 (Erdos–Turan). Let A = n1 < n2 < · · · be an innite sequence

o positive integers such that ∞

i=1

1

ni= ∞.

Then A contains an arithmetic progression o any length .

It is easy to show that Conjecture 1 is equivalent to the condition that or all

positive integers k 3 the series ∞l=1 ak(4l) converges (the proo o this equiva-lence can be ound, or example, in [4]). Consequently, an estimate o ak(N ) or allk 3 o the orm ak(N ) ≪ 1/(log N )1+ε with ε > 0 would imply Conjecture 1.

We discuss Theorem 8 in § 4, and we now give the proo o Theorem 2. Accordingto the general ideology o [3] (see also the Introduction) we consider rst an analogue



522 I. D. Shkredov

o Roth’s result in the groups Zn p , where p is a prime. Clearly, the notion o an

arithmetic progression o length three, or in other words, o three points x,y,zsatisying the equation x + y = 2z, is meaningless in the group Zn

2 . Thereore,the simplest o the groups Zn

p in which Roth’s problem can be considered is thegroup Zn

3 (other Abelian groups were studied in [93]–[95]).

Theorem 9 (Roth, the group Zn3 ). Let G = Zn

3 and let A be an arbitrary set

without arithmetic progressions o length three. Then

|A| ≪ N

log N . (12)

Proo . We need a rather general denition ([7], [8]).

Defnition 1. Let G be a nite Abelian group, let |G| = N , and let α ∈ (0, 1) bea real number. A unction f : G → C is said to be α-uniorm i

∥f ∥∞ αN.

A set A ⊆ G, |A| = δN , is said to be α-uniorm i the unction f (x) := A(x) − δ(the balance unction ) is α-uniorm.

We now discuss the main ideas o the proo o Theorem 9. Suppose that a setA ⊆ Zn

3 with |A| = δN does not contain arithmetic progressions o length three.The proo o Theorem 9 is an algorithm. At the rst step o this algorithm twosituations are possible: either the set A is α-uniorm with some α depending onlyon δ (in the proo, α is some power o δ), or it is not.

I A is an α-uniorm set, then the existence o arithmetic progressions in A canbe veried airly easily (see Lemma 3 below). Suppose now that the set A is notα-uniorm. One can show that the α-uniormity o A is equivalent to A beinguniormly distributed in the ane subspaces o codimension 1. More precisely, thecardinality o the intersection o an α-uniorm set A o cardinality δN with any suchsubspace H is approximately equal to δ|H |. Consequently, i A is not α-uniorm,then there exists a subspace H such that |A∩ H | = (δ + θ)|H |, where |θ| > 0. Moreprecise arguments enable one to prove that the subspace H can be chosen so thatthe quantity θ is positive and can be expressed explicitly in terms o the density δ(see Lemma 4).

Ater this, we consider the new set A′ = A ∩ H and apply to it our algorithm.Note that since A′ ⊆ A, the set A′ does not contain arithmetic progressions o lengththree. Furthermore, the density o A′ in H is at least δ + θ, where θ > 0. Conse-quently, at every step o the algorithm the density o the resulting sets increases bya positive quantity. On the other hand, the density never exceeds 1. This meansthat our algorithm terminates ater nitely many steps. Hence, at some step o the algorithm we obtain a subspace H o some dimension and an α-uniorm set Athat is contained in A ∩ H . As noted above, in this case A contains an arith-metic progression o length three. Thereore, the set A also contains an arithmetic

progression. We again obtain a contradiction to the assumption o no arithmeticprogressions in A.

We have presented the scheme o proo o Theorem 9. We now proceed to theproo itsel. An arithmetic progression x, x + d, x + 2d is said to be non-trivial i d = 0, and trivial otherwise.




Lemma 3 (the α-uniorm case, an arbitrary group without elements o order 2).Let G be a nite Abelian group that has no elements o order two, let |G| = N ,and let A,B,C ⊆ G be any sets, with |A| = δN , |B| = βN , |C | = γN . I the

set A is (δ(βγ )1/2/2)-uniorm and N > 2β −1γ −1, then there exists a solution o

the equation x + y = 2z with x

∈A, y

∈B, z

∈C such that x, y, z are distinct .

Indeed, the number o solutions o the equation x + y = 2z with x ∈ A, y ∈ B,z ∈ C is equal to

σ :=

x

(A ∗ B)(2x)C (x),

or in terms o the Fourier transorm,

σ =1

N

r

A(r) B(r) C (−2r) = δβγN 2 +1

N

r=0

A(r) B(r) C (−2r).

Let α = δ(βγ )1/2/2. From the α-uniormity o A, the Cauchy–Schwarz–Bunyakov-skii inequality, and Parseval’s equality we nd that

σ δβγN 2 − αN |B| |C |1/2 δβγN 2

2> |A|,

and all is proved, since the number o solutions o the equation x + y = 2z withx = y = z does not exceed |A|.

The meaning o Lemma 3 is extremely simple. Suppose, or simplicity, thatA = B = C and

|A

|= δN . Then by Lemma 3 any (δ2/2)-uniorm set A contains

a non-trivial arithmetic progression. We now consider the opposite (in a sense)situation.

Lemma 4 (the non-α-uniorm case, the group Zn3 ). Let A ⊆ Zn

3 be an arbitrary

set , let |A| = δN , and let α ∈ (0, 1] be a real number . I A is not α-uniorm , then

there exist a subspace H o codimension 1 and a vector x such that |A ∩ (H + x)| (δ + 2−1α)|H |.

Indeed, since A is not α-uniorm, there is an r = 0 such that |f (r)| αN , wheref (x) = A(x)

−δ. Let H 0 =

x

∈Zn3 :

⟨x, r

⟩= 0

, H 1 =

x

∈Zn3 :

⟨x, r

⟩= 1

,

H 2 = x ∈ Zn3 : ⟨x, r⟩ = −1, and aj = ⟨f, H j⟩, j ∈ [3]. Then j aj = 0 andj |aj | |f (r)| αN . By taking the sum o the last two relations we nd j0 such

that |aj0 | + aj0 αN/3. Hence aj0 > 0, and moreover aj0 αN/6. By settingH = H 0, or any x ∈ H j0 we get that aj0 = |A ∩ (H + x)| − δ|H | αN/6, and thelemma is proved.

We now veriy that Theorem 9 holds. For convenience we combine Lemmas 3, 4into one proposition.

Proposition 1. Let A ⊆ Zn3 with |A| = δN be an arbitrary set without arithmetic

progressions o length three. Suppose that

N > 2δ−2. (13)

Then there exist a subspace H o codimension 1 and a vector x such that

|A ∩ (H + x)| (δ + 2−1δ2)|H |.



524 I. D. Shkredov

We now show how Theorem 9 ollows rom Proposition 1. Suppose that a setA ⊆ Zn

3 does not contain arithmetic progressions o length three. As mentionedabove, the proo o Theorem 9 is an algorithm. We carry out the rst step o ouralgorithm. Let H 0 = Zn

3 , A0 = A, δ0 = δ. We have N > 2δ−2. By applyingProposition 1 we obtain a subspace H 1 o codimension 1 and a vector x1 such that

|A0 ∩ (H 1 + x1)| (δ0 + 2−1δ20)|H 1|. Then let A1 = (A0 − x1) ∩ H 1.Suppose that at the ith step o the algorithm, i 1, we have constructed a sub-

space H i o codimension i, a set Ai ⊆ H i that does not contain arithmetic progres-sions o length three, the number δi := |Ai|/|H i|, and a vector xi such that

|Ai−1 ∩ (H i + xi)| (δi−1 + 2−1δ2i−1)|H i|. (14)

We observe that at every step o the algorithm the density o the sets Ai in thesubspaces H i increases by the quantity 2−1δ2i−1 2−1δ2, since at the rst step o

the algorithm the density o A0 in H 0 is equal to δ. I

hi := |H i| > 2δ−2 2δ−2i , (15)

then we can carry out the (i + 1)th step o the algorithm. It is easy to see that themaximum number o steps K o our algorithm does not exceed 2 · 2δ−1 = 4δ−1,since the density o the sets Ai is bounded above by 1. Indeed, or the transitionrom density δ to density 2δ we need to use at most 2δ−1 steps; thereore, alto-gether we need at most 2δ−1 + 2(2δ)−1 + 2(4δ)−1 + · · · = 4δ−1 steps (these simplecalculations are presented more rigorously, or example, in [8] or [4]). This is not

yet the proo o Theorem 9, since we did not veriy that condition (15) holds atevery step o our algorithm. We have δ ≫ 1/log N . Hence

hK = 3−K N 3−4/δN > 2δ−2.

Since hi hK or all i K , the inequality (15) holds at every step o the algorithm.The theorem is proved.

In [95] Sanders improved the estimate N/log N in Theorem 9 or the group Zn4 .

Theorem 10 (Sanders). Let G = Zn4 and let A be an arbitrary set without triplesx, x + d, x + 2d with 2d = 0. Then

|A| ≪ N

log N · (log log N )1/6−ε, (16)

where ε > 0 is an arbitrary constant .

It would be very interesting to obtain some strengthening o the estimate (12),since this, apparently, would enable one to prove new upper estimates or the quan-

tity a3(N ) in any Abelian group. It would also be interesting to prove the Erd os–Turan conjecture in the group Zn

4 . As is easy to see, or that it suces to replacethe constant 1/6 in the estimate (16) by any quantity that is strictly greater than 1.

We now prove Theorem 2, using a simple modication o the arguments above.We shall need a technical lemma.




Lemma 5. Let s and N be positive integers, η ∈ (0, 1] a real number , and ϕ : ZN →ZN a linear unction , and suppose that N η−2·28. Then there exists an arithmetic

progression P ⊆ Z with 2−4ηs1/2 |P | 2−2ηs1/2 such that diam(ϕ(P )) s and

[N ] = l∈T

(P + l) ⊔Ω, (17)

where T is some set o translations and |Ω| ηN .

Proo . Let ϕ(x) = ax + b. I a = 0, then the lemma is obvious. Suppose thata = 0. Furthermore, suppose that t = ⌈ηN/(8s1/2)⌉ 2. Consider the t + 1residues 0, a, 2a , . . . , t a ∈ ZN . We divide the segment [1, N ] into t segments o length(N −1)/t. By Dirichlet’s drawer principle one o the segments contains two numbersin our sequence. Suppose that this segment contains the residues t1a and t2a, andlet t2 > t1. We set u = t2 − t1. Then 0 < u t and |ua| N/t. We partition [N ]

into the residue classes C j , j ∈ [u], modulo u. Obviously, |C j | = ⌊(N − j)/u⌋+1. Itis easy to see that the last number is either ⌊N/u⌋ or ⌈N/u⌉. Further, C 1 is a classo largest cardinality and C 1 = ⌈N/u⌉. Let C ′1 be C 1 without the right-most point.

Then [N ] =u−1

l=0 (C ′1 + l) ⊔ Ω∗, where |Ω∗| u.

Let w = ⌊st/N ⌋. We have w st/N N/(4u) ⌊N/u⌋/2. Consequently, everyarithmetic progression C ′1 + l can be partitioned into translates o some arithmeticprogression P with |P | = w and some arithmetic progression Ωl with diference uand length at most w. Clearly,

w =|P

|

st

2N 2−4ηs1/2.

Then

[N ] =

u−1l=0

l′

P l,l′

⊔ u−1

l=0

Ωl

⊔ Ω∗,

where the P l,l′ are some translates o the progression P . We set

Ω =

u−1l=0

Ωl

⊔ Ω∗.

Clearly, Ω is a segment o the orm [q, N ] and

|Ω| 4uw 8ust

N

8st2

N ηN.

Furthermore,

diam(ϕi(P )) |ur| · |P | N

t

st

N = s.

The lemma is proved.

We can now establish an analogue o Lemma 4.

Lemma 6 (the non-α-uniorm case, the group ZN ). Let A ⊆ ZN be an arbitrary

set with |A| = δN , let α ∈ (0, 1] be a real number , and suppose that N 212α−2.I A is not α-uniorm , then there exists a progression Q with |Q| 2−10α2N 1/2

such that |A ∩ Q| (δ + 2−3α)|Q|.



526 I. D. Shkredov

Proo . Indeed, since the set A is not α-uniorm, there exists an r = 0 such that

| A(r)| =

x∈A

e−2πixr/N

αN.

We apply Lemma 5 with the parameters η = α/4 and s = ⌊αN/(8π)⌋ 1 and withthe linear map ϕ(x) equal to rx. Then by the triangle inequality

|T |l=1

x∈P +l

f (x)e−2πixr/N

3αN

4,

where f (x) = A(x) − δ. Let xl be an arbitrary point in the progression P + l. Sincediam

ϕ(P + l)

s or any l ∈ [|T |], by applying once more the triangle inequality

we get that

|T |l=1

x∈P +l

f (x)

=

|T |l=1

x∈P +l

f (x)e−2πixlr/N

|T |l=1

x∈P +l

f (x)e−2πixr/N

−|T |l=1

α|P |4

αN

2.

Since

x f (x) = 0, we have

|T |l=1

x∈P +l

f (x) |Ω| αN

4.

Consequently,|T |l=1

x∈P +l

f (x)

+

x∈P +l

f (x)

αN

4,

and thereore or some l ∈ [|T |] we have

x∈P +l

f (x) = |A ∩ (P + l)| − δ|P | α|P |8

.

We set Q = P + l. Then Lemma 5 implies the estimate |Q| 2−4ηs1/2 2−10α2N 1/2, and Lemma 6 is completely proved.

We can now obtain Theorem 2 by using the same arguments as in the proo o Theorem 9. Let A ⊆ [N ] with |A| = δN be an arbitrary set without arithmeticprogressions o length three. We can assume that N is an odd number. Also, let

B = C = A∩⌊N/3⌋ + 1, ⌊N/3⌋ + 2, . . . , ⌊2N/3⌋. We identiy the set [N ] with thegroup ZN . Then the number o solutions o the equation x + y ≡ 2z (mod N ) withx ∈ A, y ∈ B, and z ∈ C coincides with the number o solutions o the equationx + y = 2z. As above, the proo o Theorem 2 is an algorithm, and at every stepo the algorithm we apply to the sets A, B, C Lemma 3, which is valid or any




Abelian group without elements o order two, in particular or ZN with N odd, andi this is impossible (that is, i A is not α-uniorm), then we use Lemma 6 (detailscan be ound in [8] or [4]). Then we obtain nested sets A0 = A ⊇ A1 ⊇ A2 ⊇ · · ·and arithmetic progressions Q0 = [N ] ⊇ Q1 ⊇ Q2 ⊇ · · · . As above, we can see thatthe algorithm works in at most K := Cδ−1 steps, where C > 0 is some absolute

constant. The main diference between Theorem 2 and Theorem 9 is the diferentnature o decrease o the lengths o the progressions Qi. It ollows rom Lemma 6that

|Qi| ≫ δ4|Qi−1|1/2 (18)

at step i. The last ormula involves the expression δ4, since, as is easy to see, thequantity α in Theorem 2, as in Theorem 9, is equal to α = cδ2, where c > 0 is someabsolute constant. We need to veriy the condition |QK | ≫ δ−4 at the last stepo the algorithm. By mathematical induction we can derive the inequality |Qi| ≫δ4iN 2

−i

rom the estimate (18). By hypothesis, δ

≫1/log log N . Consequently,

|QK | ≫ δ4K N 2−K

δ4C/δ N 2−C/δ ≫ δ−4,

and we obtain Theorem 2.In [43] Green somewhat strengthened Roth’s theorem. He proved that every

suciently dense subset o [N ] contains an arithmetic progression o length threewhose diference can be represented in the orm x2+y2, where x, y are some positiveintegers.

Theorem 11 (Green). Let N be a positive integer . There exists a positive efective

constant c > 0 such that every set A⊆

[N ] with

|A

| ≫N/(log log N )c contains an

arithmetic progression o length three whose diference can be represented in the

orm x2 + y2.

4. Higher-order Fourier analysis

In the preceding section we proved Roth’s theorem and mentioned Gowers’ the-orem on sets without arithmetic progressions. We state these results once more interms o the unctions ak.

Theorem 12 (Roth, Gowers). For all positive integers N 3 and k 3

ak(N ) ≪ 1

log log N .

As the reader remembers, the proo o Roth’s theorem (the case k = 3) wasa certain iteration procedure, and at every step o this procedure we consideredtwo variants: the rst, where the set A was α-uniorm with suciently small α,and the second, where there was no uniormity. We reormulate the main resultthat was used in the rst case (see the proo o Lemma 3).

Proposition 2 (the uniorm case). Let G be a nite Abelian group that has no

elements o order two, let |G| = N , and let A ⊆ G with |A| = δN be any α-uniorm set . Then the number o arithmetic progressions in A is equal to

δ3N 2 + θαδN 2, (19)

where |θ| 1.



528 I. D. Shkredov

Let f (x) = A(x) − δ be the balance unction o the set A. Also, let

Λ3(f ) =

x,d∈Gf (x)f (x + d)f (x + 2d),

where f is an arbitrary unction. For example, Λ3(A) is the number o arithmeticprogressions o length three in the set A. It is easy to see that Λ3(A) = δ3N 3+Λ3(f )and, consequently, Proposition 2 can be interpreted as ollows:

the norm ∥f ∥∞ is small =⇒ Λ3(f ) is small. (20)

Now let A be a set without arithmetic progressions o length k > 3. Gowers’ proo is a deep generalization o Roth’s method. Instead o the quantity Λ3 he considersits analogue Λk(f ) =

x,d∈G f (x)f (x + d) · · · f (x + (k − 1)d). The question arises:

does the implication (20) remain true i instead o Λ3 we take Λk? As Gowers

showed in [8], the answer here is negative (the ergodic aspect o this problem wasconsidered earlier in [87]). A counterexample is very simple. Let G = ZN , letx ∈ ZN , and let |x| mean the distance rom zero to x in the reduced system o residues. Then in the group ZN we can take, or example, the set A0 := x ∈ ZN :|x2| 2−1δN . By using the uniorm distribution o quadratic residues, it is easy

to see that |A0| ∼ δN , the quantity ∥f ∥∞ is small, but the value Λ4(f ) is large (seedetails in [8]). Thereore, or the problem with arithmetic progressions o lengthgreater than three we need to somehow modiy the implication (20).

In order to state a correct analogue o (20) we need the Gowers norms.

Let G be a nite Abelian group, let |G| = N , let d 0 be a positive integer,and let 0, 1d = ω = (ω1, . . . , ωd) : ωj ∈ 0, 1, j = 1, . . . , d be the usuald-dimensional cube. For ω ∈ 0, 1d let |ω| be equal to ω1 + · · · + ωd. I h =(h1, . . . , hd) ∈ Gd, then ω · h := ω1h1 + · · · + ωdhd. Also, let C mean the complexconjugation operator. I n is a positive integer, then C n means the application o the complex conjugation operator n times.

Defnition 2. Let f : G → C be an arbitrary unction. The Gowers uniorm norm

(or simply the Gowers norm) o this unction is given by the ormula

∥f ∥U d := 1N d+1

x∈G, h∈Gd

ω∈0,1d

C |ω|f ω(x + ω · h)1/2d . (21)

One can show that the expression (21) really denes a norm or all d 2 (see [8]or [4]). In particular, we have the triangle inequality

∥f + g∥U d ∥f ∥U d + ∥g∥U d . (22)

The Gowers norms, like the L p-norms, satisy the monotonicity inequality

∥f ∥U d−1 ∥f ∥U d (23)

or all d 2. We also point out one combinatorial property o the Gowers norms.Let h ∈ Gd. A set o 2d points o the orm ω · h = ω1h1 + · · · + ωdhd, ω ∈

0, 1d, is called a d-dimensional cube. Let A ⊆ G be some set. We say that A




contains a d-dimensional cube i A contains all the points o this cube. By using the

Cauchy–Schwarz–Bunyakovskii inequality it is easy to show that any set A ⊆ G o

cardinality δN always contains at least δ2d

N d+1 cubes o dimension d, and equalityis attained at ‘random subsets’ o G having density δ (see [8] or [4]). On the otherhand, the ollowing criterion holds (or simplicity we state it in the group ZN ).

Theorem 13 (the combinatorial meaning o the Gowers norms in ZN ). Suppose

that d 2, A ⊆ ZN , |A| = δN , and f is the balance unction o the set A. Then

∥f ∥U d = o(1) as N → ∞ i and only i A contains δ2d

N d+1 + o(N d+1) cubes.

Further properties o the Gowers norms can be ound in the recent paper [ 9] (seealso [75], [76]).

We consider the case d = 2 in a little more detail. By using the inversionormula (7) it is easy to see that the U 2-norm o a unction f can be expressed interms o the Fourier transorm:

∥f ∥U 2 = 1

N 4

r∈G

|f (r)|41/4

. (24)

Thus, the U 2-norm o some unction is simply the L4-norm o its Fourier transormmultiplied by N −1.

Lemma 7. Let f : G → D be an arbitrary unction . Then ∥f ∥∞ ∥f ∥4. Con-

versely , ∥f ∥U 2 = N −1∥f ∥4 ∥f ∥1/2∞ N 1/2.

The rst inequality o the lemma is obvious, and the second ollows rom Parse-val’s equality

∥f ∥44 =

r

|f (r)|4 ∥f ∥2∞∥f ∥22 = N ∥f ∥2∞∥f ∥22 ∥f ∥2∞N 2.

Lemma 7 tells us that or unctions f whose values are at most 1 in absolute valuethere is no big diference between ∥f ∥∞ and ∥f ∥U 2 . Thereore, the implication (20)can be reormulated as ollows:

∥f

∥U 2 is small =

⇒Λ3(f ) is small. (25)

It now became clear how the implication (20) can be generalized.

Assertion 1 (see [7], [8]). For all k 3

∥f ∥U k−1 is small =⇒ Λk(f ) is small . (26)

Thus, we have obtained an analogue o Proposition 2 or, which is the same, o Lemma 3 or the case o arithmetic progressions o length k > 3. It remains to nda correct version o Lemma 6. Using the inequalities or norms in Lemma 7, wereormulate Lemma 6 as ollows.

Lemma 8 (the non-uniorm case, the U 2-norm). Let A ⊆ ZN be an arbitrary set

with |A| = δN , let f be the balance unction o A, let α ∈ (0, 1] be a real number ,and suppose that N ≫ α−C . I ∥f ∥U 2 α, then there exists a progression Q with

|Q| ≫ αC N 1/2 such that |A ∩ Q| (δ + cαC )|Q|, where C, c > 0 are constants.



530 I. D. Shkredov

Above, the condition o non-α-uniormity o the set A was replaced by the esti-mate ∥f ∥U 2 α, which, o course, can be done in view o Lemma 7. We state ananalogue o Lemma 8 or the norms U k, k 3 (see [8]).

Lemma 9 (the non-uniorm case, the U k-norm). Let A ⊆ ZN be an arbitrary set

with |A

|= δN , let f be the balance unction o A, let α

∈(0, 1] be a real number ,

and suppose that N ≫ α−C . I ∥f ∥U k α, then there exists a progression Q with

|Q| ≫ αC N cαC such that

|A ∩ Q| (δ + cαC )|Q|, (27)

where c, C > 0 are constants.

Note that, in contrast to Lemma 8, in Lemma 9 the estimate or the length o theprogression Q is weaker. To some extent this is essential. At least, in the examplewith the set A0 (see p. 528) it is impossible to nd a progression Q in Z (or even

in ZN ) with length o order greater than N 1/2

or which the inequality (27) wouldhold.As we saw above (see Lemma 7), the smallness o the Fourier coecients o

a unction f is equivalent to the smallness o its U 2-norm. This is not at all so inthe case o the U k-norm, k 3. The same Gowers’ example o the set A0 showsthat there exist sets with large U 3-norm o their balance unction but with smallFourier coecients.

The act is that even in the simplest case k = 4 the uniorm Gowers U 3-normo a unction f can be expressed only in terms o the Fourier coecients o thediference unction ∆(f ; u)(x) := f (x)f (x + u):

∥f ∥8U 3 =1

N 6

u

r

|∆(f ; u)(r)|4.

It ollows rom this ormula that the U 3-norm o the balance unction o a set A issmall i and only i the set A ∩ (A − u) has small Fourier coecients or ‘almostall’ u (see [7], [8] or the survey [4]). Based on this expression or the U 3-normwe see that the proo o Lemma 9 presents signicant diculties even in the casek = 4, since the condition ∥f ∥U 3 α gives us inormation about the diference unc-

tions rather than about the unction f itsel. Nevertheless, in [8] Gowers solved thisinverse problem o additive number theory and obtained the ollowing inormationabout the original unction f .

Theorem 14 (Gowers). Let A ⊆ ZN be some set , f its balance unction , and η > 0a real number , and suppose that ∥f ∥U d η. Then there are absolute constants

C d, cd > 0 such that or all N exp(C dη−C d) there exist a polynomial pd−1(x) o

degree d − 1 and an arithmetic progression P with |P | ≫ cdηC dN cdηCd or which x∈P

f (x)e( pd−1(x))

cdηC d |P |. (28)

Using the quantitative version o Weil’s theorem on the uniorm distribution o the ractional parts o polynomials, Gowers relatively easily derived Lemma 9 romTheorem 14. One can even say that Theorem 14 is the core o his proo. It ispossible to generalize Theorem 14 a little.




Theorem 15 (Gowers). Let η > 0 be a real number , and f : ZN → D some unction

with

x f (x) = 0 and ∥f ∥U d η. Then there are absolute constants C d, cd > 0such that or all N exp(C dη−C d) there exist a polynomial pd−1(x) o degree d − 1

and an arithmetic progression P with |P | ≫ cdηC dN cdηCd such that

x∈P

f (x)e( pd−1(x)) cdηC d |P |. (29)

This assertion is called the weak inverse theorem or the Gowers norms. Herethe ‘weakness’ is in the act that, as we saw earlier, much more can be said in thecase d = 2:

∥f ∥U 2 η =⇒ there exists a polynomial p1(x) with deg p1 = 1

such that x f (x)e( p1(x))≫ ηC

N,

where C > 0 is some absolute constant. In other words, in the last ormula,unlike (29), there is no summation over the progression P . The weak inverse theo-rem or the Gowers norms in the situation where k = 3 was extended by Green andTao to the case o an arbitrary Abelian group G (see [9], as well as [46]). Roughlyspeaking, or a group G the arithmetic progression P in (29) must be replaced bya translate o a Bohr set. Inverse theorems including in their statement so-callednilsequences were obtained in [9]. About nilsequences, see [28], [31]–[34], [71], [74].

We mention one interesting conjecture o Green and Tao which is a ‘strong’ versiono Theorem 15, that is, a result that does not include in its statement a summationover the progression P . For simplicity let G = Zn

5 and k = 4.

Conjecture 2 (the polynomial conjecture or the Gowers norm). Let η > 0 be

a real number , and f : Zn5 → D a unction with

x f (x) = 0 and ∥f ∥U 3 η. Then

there are absolute constants C d, cd > 0 such that or all N exp(C dη−C d) there

exist a matrix M and a vector r such that

x∈Zn5 f (x)e(

⟨x,Mx

⟩+

⟨r, x

⟩+ b) cdηC dN. (30)

There is a connection between some inverse problems in additive combinatoricsand inverse problems in the theory o Gowers norms (see [96]). Conjecture 2 ollowsrom the ollowing conjecture o Katalin Marton, which is nowadays called theFreiman–Ruzsa conjecture.

Conjecture 3 (the polynomial Freiman–Ruzsa conjecture). Let A ⊆ Zn p , where p

is a prime, let K 1 be a real number , and suppose that |A + A| K |A|. Then

there exists a set A′

⊆A,

|A′

|

|A

|/C 1(K ), contained in an ane space o size at

most C 2(K )|A|, where the unctions C 1(K ), C 2(K ) depend on K polynomially .

The conjecture above holds i C 1(K ) and C 2(K ) depend on K exponentially(see, or example, [9]). Quite recently in [72] the equivalence o Conjectures 2 and 3was proved, as well as the equivalence o the corresponding conjectures in ZN .



532 I. D. Shkredov

The proo o Theorems 8 and 15 contains many new and beautiul ideas. Themethods in Gowers’ paper were developed by various authors (see, or example,[9], [14], [15], [42]–[46]). The most striking o the results obtained in these papersis undoubtedly the theorem o Green and Tao on progressions in the prime num-bers [42].

Theorem 16 (Green–Tao). For all positive integers k 3 the set o prime numbers

contains an arithmetic progression o length k.

In act, Green and Tao proved an even stronger result.Let A be an arbitrary subset o the set P o prime numbers, and let π(N ) be

the number o primes that do not exceed N . The upper density o A with respectto P is dened to be the quantity lim supN →∞ |A ∩ [N ]|/π(N ).

Theorem 17 (Green–Tao). Let A ⊆ P be an arbitrary set o positive upper density

with respect to P , and let k 3. Then A contains an arithmetic progression o

length k.

In their proo Green and Tao used a certain ‘strong’ version o Gowers’ The-orem 15, although their analogue o the inequality (29) involved a certain morecomplicated ‘dual’ unction instead o a polynomial [42].

For k = 3 Theorem 16 was proved in 1938 by Chudakov [97] (see also [98], [99]),and Theorem 17 by Green [100]. Green obtained an even stronger result. Letlog[1]N = log N , and or l 2 let log[l]N = log(log[l−1]N ). Thus, log[l]N is theresult o taking the logarithm o a number N successively l times.

Theorem 18 (Green). Let N be a suciently large positive integer , and A an arbitrary subset o P ∩ [N ] such that

|A| ≫N

log[5]N

log N ·

log[4]N .

Then A contains an arithmetic progression o length three.

We point out that Conjecture 1 implies both Theorems 16 and 17.

5. Colouring problems

Let k be a positive integer. We dene a k-colouring o the set o integers tobe an arbitrary map χ : Z → 1, 2, . . . , k. Here the segment o positive integers1, 2, . . . , k is associated with k diferent colours. In the case where the valueo k is unimportant, we simply speak o nite colourings. I f (x1, . . . , xn) = 0,

xi ∈ Z, is some equation, then a solution

x(0)1 , . . . , x

(0)n

o this equation is said to

be monochromatic i all the x(0)i have the same colour, in other words, there exists

an m∈

1, . . . , k

such that χx(0)

i = m or i = 1, . . . , n. An extensive literatureis devoted to the question o nding monochromatic solutions o various equations(see [101]–[106]). The rst result on colourings was proved by Schur [60] in 1916.

Theorem 19 (Schur). For any nite colouring o the set Z there exists a monochro-

matic solution o the equation x + y = z, where x, y, z are non-zero.




Another result on colourings is the celebrated theorem o van der Waerden [78](see § 3) proved in 1927.

Theorem 20 (van der Waerden). Let l and k be positive integers. For any colour-

ing o the integers with k colours there exists a monochromatic arithmetic progres-

sion o length l.Van der Waerden’s theorem has played a signicant role in the development o

two branches o mathematics: additive combinatorics and combinatorial ergodictheory (see, or example, the survey [4]). The theorem is itsel one o the unda-mental results o Ramsey theory [101]–[103], [107].

A classical result which completely answers the question o whether a givenlinear equation has a monochromatic solution or an arbitrary colouring is Rado’stheorem [105] (see also [108]). Let U = (uij) be an m × n matrix with integerelements. We consider the linear equation

nj=1

uijxj = 0, i = 1, . . . , m . (31)

Defnition 3. The system o equations (31) is said to be regular in Z i or anycolouring o Z with nitely many colours there exists a monochromatic solution o the system (31).

Note that the numbers x1, . . . , xn are not assumed to be distinct.

Theorem 21 (Rado). Let U = (uij) be an m×

n matrix with integer elements.The system o equations (31) is regular in Z i and only i there exist columns

C 1, . . . , C n and numbers ki with 1 k1 < · · · < kt = n such that the new columns

Ai =

kij=ki−1+1

C j

satisy the ollowing conditions:1) A1 is a zero column ;2) or i = 1, . . . , t the column Ai is a linear combination o C 1, . . . , C ki−1 with

rational coecients.

We call matrices satisying conditions 1) and 2) o Theorem 21 regular . In thecase m = 1 Rado’s theorem is written in the simplest orm.

Theorem 22 (Rado). Let n be a positive integer , and c1, . . . , cn non-zero integers.The system o equations

c1x1 + · · · + cnxn = 0 (32)

is regular in Z i and only i there exists a non-empty set I ⊆ [n] such that the sum

i∈I

ci is equal to zero.

For example, the equations x − 2y + z = 0 and x + y − z = 0 are regular, whilethe equation x + y − 5z = 0 is not. I x − 2y + z = 0, then the numbers x, y, z orman arithmetic progression. Note that or this equation Theorem 22 is trivial, sincewe can take x = y = z = 1. For the case k = 3 van der Waerden’s Theorem 20



534 I. D. Shkredov

allows one to assert that or every colouring o Z with nitely many colours thereexist distinct x, y, z o the same colour satisying x − 2y + z = 0.

Monochromatic solutions o linear equations, as well as o systems o linear equa-tions, have been studied not only in Z but also in other groups (see, or exam-ple, [106]). On the other hand, rather little is known about monochromatic solu-

tions o non-linear equations. For example, there is still no answer to the questiono whether there exists a monochromatic solution or the equation x2+ y2 = z2, aswell as to the question o whether or any nite colouring o Z there exist x + y, xyo the same colour (see [109], Problem 3). A solution o a more dicult questionis also unknown: is it true that or every colouring there exist x, y, x + y, xy thathave the same colour? Note that the existence o a non-zero monochromatic solu-tion or the equation xy = z, which uses solely multiplication, ollows easily romSchur’s theorem (it suces to consider the set 2nn∈N0). Concerning monochro-matic solutions o non-linear equations including multiplication, see [110]–[112].

Unortunately, these papers deal with equations that have either only multiplica-tion or only addition. Thereore, it would be interesting to have a solution o atleast one colouring problem that is not linear in one o the operations. However, inthis section we discuss exclusively linear equations.

Traditionally in colouring problems people mainly used combinatorial methods(see, or example, [107] or [102]), or an approach involving ergodic theory (see,or example, [38]), or topological methods, or example, the method o ultralters[113], [114]. Quite recently Sanders [115] applied harmonic analysis or the rst timein this circle o problems. The author o the present survey, A. Samorodnitsky, andA. Fish independently also proposed a method based on Fourier analysis or solvingcolouring problems (see the proo o Theorem 24).

Beore discussing the known results, we reormulate Theorem 19.

Theorem 23 (Schur). Let k be a positive integer . There exists a number S (k) such

that or any positive integer N S (k) and an arbitrary colouring o the set [N ]with k colours the equation x + y = z has a monochromatic solution .

Theorem 23, in contrast to Theorem 19, deals with nite sets (replacing the set o integers by the positive integers is quite unimportant). For this reason Theorem 23is called the nite version o Theorem 19. By using the general compactness princi-

ple (see, or example, [102] or [4]) it is easy to show that these results are equivalent.About the behaviour o the unction S (k) it is only known (see [102]) that

k ≪ log S (k) ≪ k log k. (33)

The upper estimate in (33) ollows rather easily rom inequalities or certain Ramseynumbers. The lower estimate is altogether elementary; we present here the relevantconstruction (an estimate or S (k) o the orm log S (k) Ck with the best-possibleconstant C > 1 can be ound in [102]). Indeed, we observe that there are nosolutions o Schur’s equation x + y = z among odd numbers. Thereore, taking an

arbitrary segment [N ] o the positive integers, we colour the set o odd numberswith one and the same colour, say, colour 1. The remaining set o even numbers isisomorphic to the segment [N/2]. Hence, we colour the set o ‘odd numbers’ inthis segment in another colour 2, and so on. In the end we obtain the estimateS (k) ≫ 2k.




When nite versions o the problems o van der Waerden and Rado are con-sidered, there appear analogues o the unction S (k), namely, the unctions W (k)and RU (k). Here U is the matrix o the linear equation (31). Rather little is knownabout the behaviour o these unctions (see [102]). Theorem 23 shows how we canstate the question o monochromatic solutions o the linear equation (31) in the

case o a nite Abelian group. For simplicity we discuss only the case m = 1.Let c1, . . . , cl be arbitrary elements o the group G. We consider the linear

equationc1x1 + · · · + clxl = 0. (34)

I c1 + · · · + cl = 0, then this equation is said to be ane, otherwise non-ane.For example, Schur’s equation is non-ane, while van der Waerden’s problem isane. A set A contains a solution o equation (34) i there exist x1, . . . , xl ∈ Asatisying (34).

Defnition 4.The Ramsey number o equation (34) is dened as the maximalpositive integer R = R(c) such that or any partition o the group G into R subsets

one o the subsets contains a solution o (34).

Let us illustrate the application proposed by Sanders o the method o Fourieranalysis to colouring problems. We consider the simplest situation, where G = Zn

2 .

Theorem 24 (Sanders). Let l be an odd number and let c = (1, . . . , 1), where 1occurs exactly l times in the vector c. Then

R(c) C ln1−1/(l−1), (35)

where C l > 0 is some constant depending only on l. I , however , l ≫ log n, then

R(c) ≫ n

log n. (36)

Remark 1. The upper estimate R(c) n that we have in the group Zn2 is obtained

quite trivially. Indeed, we colour any hyperplane o codimension 1 that does notcontain zero with colour 1, then we consider the remaining subspace H and observethat it is isomorphic to Zn−1

2 . Then we again take a hyperplane in H o codimen-sion 1 that does not contain zero and colour it with colour 2, and so on. Since the

number l is odd, in the end we colour Zn2 with n colours without monochromaticsolutions. In other words, R(c) n.

As in the case o Theorem 23, combinatorial methods enable one to prove thatR(c) ≫ n/log n (see [102]). Thus, or l ≫ log n this estimate and the inequality (36)o Theorem 24 coincide. For small l the combinatorial method works better thanthe analytic one.

Proo o Theorem 24. Let k = R(c) and suppose that the set Zn2 is partitioned

into k sets (colours) C 1, . . . , C k. We need to veriy that there exists a non-zero

monochromatic solution o the equationx1 + · · · + xl = 0. (37)

The proo o Theorem 24 is an algorithm. At an arbitrary step s o the algorithm,some subspace V s o codimension s is constructed. Next, with every C i in the set



536 I. D. Shkredov

o colours C 1, . . . , C k we associate a vector xsi . Furthermore, associated with the

set C i are the density (δ∗)si o C i in V s + xs

i and the number δsi = |C i ∩ V s|/|V s|.

The algorithm works so that the quantities (δ∗)si do not decrease, and the subspaces

V s are nested: V 0 ⊇ V 1 ⊇ · · · .At the zero step we put V 0 = Zn

2 and x01 =

· · ·= x0l = 0. Next, suppose that the

algorithm has completed s steps. We choose a colour C i with the maximal valueo δs

i . Clearly, δsi 1/k. We shall need the ollowing lemma.

Lemma 10. Let V ⊆ Zn2 be some subspace, and A ⊆ Zn

2 an arbitrary set without

non-zero solutions o equation (37). Let x0 ∈ Zn2 be any vector , let δ = |A∩V |/|V |,

and let δ∗ = |A ∩ (V + x0)|/|V |. Finally , suppose that δ∗ δ and |V | ≫ δ−l. Then

there exists a ξ /∈ V ⊥ such that A ∩ (V + x0)

(ξ) δ∗ + δ∗(δδ∗)1/(2(l−2))

|V |. (38)

The proo o the lemma is a modication o the arguments in Lemma 3. Insteado three sets in the latter lemma, we simply need to consider l sets A1, A2, . . . , Al andthen put A1 = A ∩ V and A2 = · · · = Al = A ∩ (V + x0). Since the characteristic o the group Zn

2 is equal to two, the number o solutions o equation (37) with xi ∈ Ai

is equal to zero (see a detailed proo in [115]).We now apply Lemma 10 to the set C i and then use the analogue o Lemma 4

in the group Zn2 . We nd a subspace H and a vector x such that

|C i ∩ (H + x)|

(δ∗)si + (δ∗)s

i

δs

i (δ∗)si

1/(2(l−2))|H |. (39)

Let V s+1 = H and xs+1i = x, and choose vectors xs+1j or j = i so as to maximizethe quantity |C j ∩ (V s+1 + z)|/|V s+1|. Then obviously δs+1

i δsi or i ∈ [k]. This

describes one step o the algorithm. We apply it also urther.Since always (δ∗)s

i δsi , it is easy to see that our algorithm terminates ater

at most Ol

k · k1/(l−2)

steps (see the similar arguments in § 3). Consequently,

i the inequality k1+1/(l−2) ≫ n holds, then we nd a monochromatic solution o equation (37). Recalling that k = R(c), we now obtain the estimate (35). Forproving the inequality (36) it is sucient to observe that in the case l ≫ log n theormula (40) can be rewritten in the orm

|C i ∩ (H + x)| (δ∗)si (1 + λ)|H |, (40)

where λ > 0 is some constant. Thereore, in this situation the algorithm makes atmost O(log k) steps. Hence the inequality (36) holds, and Theorem 24 is completelyproved.

We give another proo o (36) in the spirit o the arguments in § 3. The onlydiference o our approach rom Roth’s method is that at every step o the iterationprocedure it is required to nd a proper rather than an arbitrary ane subspacein which the density o the set A increases, where A does not contain solutionso (37). For that we have to trace the signs o the Fourier coecients. We shallneed several simple lemmas.

Lemma 11. Let A ⊆ Zn2 be an arbitrary set without solutions o equation (37).

Then there exists a ξ = 0 such that | A(ξ)| δ(δ/4)1/(l−2)N .




The proo o the lemma is a modication o the arguments in Lemma 3 orLemma 10.

Lemma 12. Let A ⊆ Zn2 be an arbitrary set without solutions o equation (37) and

suppose that

| A(ξ)

| δ(δ/4)1/(l−2)N or all ξ

= 0. Then the set

Rhuge := ξ : | A(ξ)| 2−1/(4l)|A|

has cardinality either 1 or 2.

Indeed, let R = Rhuge \ 0. Since A does not contain solutions o (37),

0 =

ξ

A l(ξ) = |A|l +

ξ: A(ξ)>0

A l(ξ) +

ξ: A(ξ)<0

A l(ξ) = |A|l + σ+ + σ−. (41)

In view o the last ormula,

2−1/4|A|l| R| ξ∈ R

| A(ξ)|l = σ+ + |σ−| = |A|l + 2σ+ 3

2|A|l.

Hence | R| < 2, and since 0 ∈ Rhuge always holds, we have |Rhuge| = 1 or 2.

It will be clear rom the arguments below that both variants |Rhuge| = 1 and

|Rhuge| = 2 can be realized.Lemma 13. Let l 5, let A ⊆ Zn

2 be an arbitrary set without solutions o equa-

tion (37), and let a vector ξ0 /∈ Rhuge be such that A(ξ0) 0 is minimal . Then

there exists a ξ = 0 such that

A(ξ) min

δ

δ

4

1/(l−4)N,

1

80| A(ξ0)|

. (42)

Proo . I there exists a ξ = 0 such that A(ξ) δ(δ/4)1/(l

−4)

N , then the lemma isproved. Thereore, we assume that the reverse inequality holds. Then by Lemma 12the cardinality o the set R := Rhuge \ 0 is equal to 0 or 1. By hypothesis theset A does not contain solutions o (37). Consequently,

0 = N

x1+···+xl=0

A(x1) · · · A(xl−1)A(xl)(−1)⟨x,ξ0⟩ =

ξ

A l−1(ξ) A(ξ + ξ0)

= |A|l−1

A(ξ0) + |A|

A l−1(ξ0) +

ξ

=0, ξ

=ξ0

A l−1(ξ)

A(ξ + ξ0) = σ′0 + σ′′0 + σ1. (43)

Since ξ0 /∈ Rhuge, we have | A(ξ0)| 2−1/(4l)|A|. Thus,

σ′0 + σ′′0 (1 − 2−1/12)|A|l−1| A(ξ0)| 1

20|A|l−1| A(ξ0)|.



538 I. D. Shkredov

In deriving the last ormula we used the inequality l 3. Returning to the iden-tity (43) and using the estimate | R| 1, we get that

− 1

20|A|l−1

A(ξ0)

ξ

=0, ξ

=ξ0

A l−1(ξ)

A(ξ + ξ0) (44)

= ξ=0, ξ=ξ0,ξ /∈ R

A l−1(ξ) A(ξ + ξ0) + ξ=0, ξ=ξ0, ξ∈ R

A l−1(ξ) A(ξ + ξ0) (45)

maxξ=0, ξ=ξ0

A(ξ + ξ0)| A(ξ0)|

ξ=0, ξ=ξ0

| A l−2(ξ)| + |A|l−1 maxξ=0, ξ=ξ0

A(ξ + ξ0).

(46)

Recalling that

A(ξ) δ(δ/4)1/(l−4)N and proceeding as in the proo o Lemma 12,

we have

ξ=0, ξ=ξ0

| A l−2(ξ)| 32 |A|l−2 2|A|l−2.

Combining the last ormula with (44) and (46), we obtain the estimate (42).Lemma 13 is proved.

Finally we prove the main lemma, which is interesting in its own right.

Lemma 14. Let l 5 and let A ⊆ Zn2 be an arbitrary set without solutions o

equation (37). Then the ollowing alternative holds:

1) either there exists a ξ = 0 such that A(ξ) 1

80 δδ

41/(l−4)

N ,

2) or there exist a subspace H with codim H = 1 and a vector x0 /∈ H such that

A ⊆ H + x0.

The example o a colouring above (see Remark 1, p. 535), which shows thatR(c) n in the group Zn

2 , demonstrates that the second variant o the alternativeis possible. It is clear how the inequality (36) ollows rom Lemma 14. Indeed,suppose we have a partition o the group Zn

2 into k sets (colours). We choosea colour o maximal cardinality C and apply to it Lemma 14. I part 1) holds, then

by the analogue o Lemma 4 inZn

2 we get that the subspace L := x : ⟨ξ, x⟩ = 0satises

|C ∩ L|

δ +1

80

δ

4

1/(l−4)|L|.

Then we apply Lemma 14 to the set C ∩ L, and so on. I at some stage the secondalternative holds, then the subspace H is coloured with k − 1 colours. Next, wechoose in H a colour o maximal cardinality and proceed as beore. By hypothesis,l ≫ log n; thereore, or every colour the rst alternative o the lemma can hold atmost O(log k) times (see ormula (40) in the proo o Theorem 24). Then in all our

algorithm completes at most O(k log k) steps, which is what gives the estimate (36).We now prove Lemma 14. Suppose that situation 1) does not hold. By Lemma 11

there exists a ξ = 0 such that | A(ξ)| δ(δ/4)1/(l−2)N δ(δ/4)1/(l−4)N . We have A(ξ) < 0. Let R− denote the set o such ξ. Suppose that there exists a ξ0 = 0such that ξ0 ∈ R− and ξ0 /∈ Rhuge. Then by Lemma 13 there exists a ξ such that




the estimate (42) holds, and thereore the rst alternative o Lemma 14 holds. Itremains to consider the case where R− = Rhuge. It ollows rom Lemmas 11 and12 that |Rhuge| = 2. Let Rhuge = 0 ⊔ ξ0. Then or all ξ = 0 with ξ = ξ0we have |

A(ξ)| δ(δ/4)1/(l−4)N . Let H = x : ⟨x, ξ0⟩ = 0 and H 1 = x :

⟨x, ξ0

⟩= 1

= H + x0, where x0 /

∈H . Also, let A0 = A

∩H and A1 = A

∩H 1.

We need to prove that the set A0 is empty. Let A′ = A1 − x0 ⊆ H . Since A(ξ0) =

|A0| − |A1| 2−1/(4l)|A| and l ≫ log n, we have | A′(ξ)| (δ/2)1/(l−4)|A1| := ε|A1|or all non-zero ξ ∈ H . Suppose that A0 is non-empty. Take z ∈ A0 and considerthe equation x1 + · · · + xl−1 + z = 0, xi ∈ A′. By hypothesis the set A does notcontain solutions o (37) and the number l is odd. Consequently,

0 =

ξ

A′(ξ)l−1(−1)⟨ξ,z⟩ |A1|l−1 − εl−3|A1|l−2|H | > 0.

This contradiction proves Lemma 14, and together with it the inequality (36).Another application o Fourier analysis to colouring problems can be ound

in [116].

6. Sets o large trigonometric sums

Let G be a nite Abelian group, and A ⊆ G some set with |A| = δ|G| = δN . Inthe two preceding sections we saw that or counting combinatorial congurationsin A (such as arithmetic progressions, solutions o linear equations, and so on) an

important role is played by large Fourier coecients o the set A. Thereore, it isnatural to consider the ollowing object:

R α = R α(A) = r ∈ G : | A(r)| αN , (47)

where α ∈ (0, δ] is a parameter. Such sets are called sets o large trigonometric

sums (or, in the terminology o the book [2], the spectrum o the set A).

For many problems o combinatorial number theory it is important to know thestructure o the set R α (see [60]). In other words, what non-trivial properties areenjoyed by sets o large trigonometric sums? Clearly, the question o the structureo R α belongs to inverse problems o additive number theory (see [1]).

We list the simplest properties o the set R α. It ollows rom the denition o R αthat 0 ∈ R α and R α = − R α in the sense that i r ∈ R α, then also −r ∈ R α. Fur-ther, Parseval’s equality (4) implies an estimate or the cardinality o R α, namely,| R α| δ/α2. Does the set R α possess any other non-trivial properties? The answerto this question turns out to be armative, although the rst general result on thestructure o the sets R α was obtained relatively recently by Chang [61] in 2002.

Theorem 25 (Chang). Let δ and α be real numbers with 0 < α δ 1, let A be

an arbitrary subset o G o cardinality δN , and let the set R α be dened by (47).Then there exists a set Λ = λ1, . . . , λ|Λ| ⊆ G with

|Λ| 2

δ

α

2log

1

δ(48)



540 I. D. Shkredov

such that every element r in R α can be represented in the orm

r =

|Λ|i=1

εiλi, εi ∈ −1, 0, 1. (49)

Thus, Theorem 25 asserts that the set R α, which a priori can have cardinalityδ/α2, is necessarily contained in the ‘linear hull’ o a relatively small set o cardi-nality 2(δ/α)2 log(1/δ). In other words, i we interpret the set Λ as a ‘basis’, thenwe can say that R α has a not too high dimension.

Note that or an arbitrary unction f the set o points where the Fourier trans-orm o this unction is large does not at all have to have any particular structure.The proo o Theorem 25 in [61] essentially used the act that the Fourier transormwas taken o the characteristic unction .

Developing the approach o [117], [118] (see also [119]), Chang applied her result

in a new proo o Freiman’s theorem [120] on sets with small sum. See otherapplications o Theorem 25 to problems o combinatorial number theory in [61]–[66].In our survey, applications o sets o large trigonometric sums and, in particular, o Chang’s theorem to problems o combinatorial number theory are discussed in §§ 7and 9.

Further results on sets o large trigonometric sums were obtained in [66] (seealso [121]). In particular, the ollowing theorem was proved in these papers.

Theorem 26. Let δ and α be real numbers with 0 < α δ, let A be an arbitrary

subset o G o cardinality δN , let k 2 be a positive integer , and let the set R α be

dened by (47). Also, let B ⊆ R α \ 0 be an arbitrary set . Then the quantity

T k(B) := |(r1, . . . , rk, r′1, . . . , r′k) ∈ B2k : r1 + · · · + rk = r′1 + · · · + r′k| (50)

is not less than

δα2k

δ2k|B|2k. (51)

Let us show that the assertion o Theorem 26 is non-trivial when the parameter δtends to zero as N tends to innity. Consider the simplest case k = 2. Suppose thatthe cardinality o the set R α is o order δ/α2. Then by Theorem 26 the number o

solutions o the equation

r1 + r2 = r3 + r4, where r1, r2, r3, r4 ∈ R α \ 0, (52)

is o order at least δ/α4. Among these solutions are three series o trivial solutions.The rst series: r1 = r3, r2 = r4; the second: r1 = r4, r2 = r3; and, nally,the third series: r1 = −r2, r3 = −r4. Consequently, equation (52) has at most3| R α|2 trivial solutions. Since the cardinality o the set R α does not exceed δ/α2,the quantity 3| R α|2 is bounded above by the number 3δ2/α4. We see that thisquantity is less than δ/α4 as δ tends to zero. Thus, Theorem 26 asserts that

equation (52) has non-trivial solutions. In this sense Theorem 26 shows that theset R α possesses some additive structure.

It is interesting to point out that i the parameter δ does not tend to zero asN → ∞, then the structure o the set R α can be airly arbitrary (in this regardsee [122], [123]). We present one o the known results here (see [122], [123], or [124]).




Theorem 27 (Nazarov). Let αr, r ∈ ZN , be arbitrary non-negative numbers such

that

r α2r N/1600. Then there exists a unction f : ZN → [0, 1] such that

∥f ∥1 = N/2 and

|f (r)| αr∥f ∥1

or all r ∈ ZN .

Thus, Theorem 27 tells us that i δ does not tend to zero (more precisely, i δ = 1/2), then the set o large trigonometric sums may contain an arbitrary set.

In the proo o Theorem 25 Chang used Rudin’s theorem [77] (also see [125]) ondissociative subsets o G. By denition a set D = d1, . . . , d|D | ⊆ G is dissociative

i an equation|D |

i=1εidi = 0 (mod N ) (53)

with εi ∈ −1, 0, 1 implies that all the εi are equal to zero.

Theorem 28 (Rudin). There exists an absolute constant C > 0 such that or an

arbitrary dissociative set D ⊆ G, arbitrary complex numbers an ∈ C, and all

positive integers p 2

1

N

x

∈G

n

∈D

ane(nx)

p

C √

p p

n

∈D

|an|2 p/2

. (54)

The proos o Theorem 28 and Chang’s theorem can also be ound in [124], [126].We now show how Rudin’s Theorem and Theorem 26 imply an analogue o Theo-rem 25 that difers rom Chang’s theorem only in a slightly weaker estimate or thecardinality o the set Λ. Indeed, consider an arbitrary dissociative subset o R α. ByTheorem 26 the set R α has an additive structure in the sense that any subset o ithas a large number o solutions o the equation in (50). But on the other hand, it iseasy to see that Rudin’s theorem implies that the number o such solutions must besmall. One can deduce rom this act that the cardinality o any dissociative subset

o R α is small. Let us take as a dissociative subset a maximal dissociative sub-set o R α (a ‘basis’ o it). Then it is easy to see that every element o R α can bewritten in the orm (49), and the number o elements in this expansion is small.We now give a rigorous proo.

Proposition 3. Let δ and α be real numbers with 0 < α δ 1, let A be an

arbitrary subset o G o cardinality δN , and let the set R α be dened by (47).Then there exists a set D = d1, . . . , d|D | ⊆ G with |D | 28C 2(δ/α)2 log(1/δ)such that every element r o the set R α can be represented in the orm

r =|D |i=1

εidi, (55)

where εi ∈ −1, 0, 1 and C is the absolute constant in Rudin’s inequality (54).



542 I. D. Shkredov

Proo . Let k = 2⌈log(1/δ)⌉, and let D ⊆ R α be a maximal dissociative subseto R α. Since D is a dissociative set, we have 0 /∈ D . By applying Theorem 26 weobtain the estimate

T k(D ) δα2k

24kδ2k|D |2k. (56)

On the other hand,

T k(D ) C 2k · 2kkk|D |k, (57)

where C is the absolute constant in Theorem 28. Indeed, let the numbers an in(54) be equal to D (n), and let p = 2k. Then the let-hand side in (54) is T k(D ),while the right-hand side is C 2k · 2kkk|D |k. We have k = 2⌈log(1/δ)⌉. Using(56) and (57), we get that |D | 28C 2(δ/α)2 log(1/δ). Since D is a maximaldissociative subset o R α, every element r o R α can be represented in the orm

r = |D |i=1 εidi (mod N ), where di

∈D and εi

∈ −1, 0, 1

. Note that the estimate

|D | 28C 2(δ/α)2 log(1/δ) difers rom the analogous estimate in Chang’s theoremonly by a constant actor. Proposition 3 is proved.

In exactly the same ashion (see [121]) one can derive rom Theorem 26 andgeneralizations o Rudin’s inequality a theorem o Bourgain [41] that he used in theproo o Theorem 4.

Theorem 29 (Bourgain). Let d be a positive integer , δ and α real numbers with

0 < α δ 1, and A an arbitrary subset o G o cardinality δN , and let the

set R α be dened by (47). Also, let Λ be a dissociative set . Then or all d 1

|dΛ ∩ R α| 8

δ

α

2logd

1

δ

. (58)

Theorem 29 clearly generalizes Theorem 25. The inequality (58) was slightlystrengthened in [121] or the group Zn

2 .

I multiplicative constants are disregarded, then as Green showed [127], Chang’stheorem is sharp. Developing his method in [128], the author proved a theoremthat is in a sense converse to Theorem 25. We state it in the simplest orm.

Theorem 30 (Green, Shkredov). Suppose that δ 1/8 and α are real numbers

such that 20N −1/2 < α δ/32 and Λ is an arbitrary dissociative set such that

|Λ| 2−11

δ

α

2log

1

δ

. (59)

Then there exists a set A ⊆ ZN with |A| = ⌊δN ⌋ such that R α(A) = 0⊔Λ⊔(−Λ).

Thus, Theorem 30 asserts that any dissociative set Λ o ‘admissible’ cardinality,

that is, satisying the inequality (59), is, together with (−Λ) and zero, a set o largetrigonometric sums o some set A. In particular, it ollows that the estimate (48)is sharp with respect to order. This cannot be said about the number o terms inthe representation (49). For example, the ollowing result was proved in [66].




Theorem 31. Let N be a positive integer such that (N, 2) = 1, let δ and α be real

numbers with 0 < α δ 1/16, let A be an arbitrary subset o ZN o cardinality

δN , and let the set R α be dened by (47). Then there is a set Λ∗ ⊆ ZN with

|Λ∗

| max212

δ

α2

log1

δ, 26 log21

δ (60)

such that or any residue r ∈ R α there exists a set λ∗1, . . . , λ∗M o at most 8 log(1/δ)elements o Λ∗ such that

r =M

i=1

εiλ∗i (mod N ), (61)

where εi ∈ −1, 0, 1.

In the same paper a similar result was obtained or subsets o the group Zn2 . It

is easy to see that the number o elements λ∗i in the representation (61) cannotbe decreased. Indeed, let α ≈ δ, and let A be a subset o Zn

2 with the propertythat | R α(A)| ≈ δ/α2 ≈ 1/δ. Such sets A do exist; or example, one can take asA any subspace o Zn

2 o cardinality δN . Then by Chang’s theorem there existsa set Λ∗ with |Λ∗| ≪ log(1/δ) such that the representation (61) holds or anyelement r ∈ R α(A). Since | R α(A)| ≈ 1/δ, it is easy to see that there exists a vectorr ∈ R α(A) such that k ≫ log(1/δ) vectors are required or representing it in the

orm (61). Indeed, since there are at most 2k

|Λ∗|

k

linear combinations in the span

o k arbitrary vectors in Λ∗, the inequality 2k|Λ∗|k ≫ 1/δ must hold, which impliesthe estimate k ≫ log(1/δ).

In [23], [18] other additive properties o the sets R α were used.

Proposition 4. Let δ and α be real numbers with 0 < α δ 1, and let A be an

arbitrary subset o G o cardinality δN . Then or any non-empty set B ⊆ R αz

(B ∗ Bc)(z) R α2/(2δ)(z) α2

2δ|B|2. (62)

Proo . Indeed, by the denition o the set

R α we get that

αN |B| r∈B

| A(r)| =

x

A(x)

r

B(r)e−i arg A(r)e(−rx).

Using the Cauchy–Schwarz–Bunyakovskii inequality and the triangle inequality, weget that

α2

δ|B|2N

x∈A

r

B(r)e−i arg A(r)e(−rx)

2

= r,r′

B(r)B(r′)e−i arg

A(r)+i arg

A(r′

) A(r − r′) z

(B ∗ Bc)(z)| A(z)|.

Summing only over z ∈ R α2/(2δ) in the last ormula, we obtain (62). Proposi-tion 4 is proved.



544 I. D. Shkredov

The papers [18] and [23] will be urther discussed in § 9. The question o thestructure o sets o large trigonometric sums in the case G = Z was studied byBourgain in [41]. See [65] or an application o these results to Freiman’s theoremon sets with small doubling.

It should be pointed out that the question o the structure o the set R α in the

case where the parameter α is close to δ had been studied in [129] much earlier thanChang’s paper; also see the survey [130]. The additive structure o sets o largetrigonometric sums in this case is also very clearly demonstrated by the ollowingtheorem (also see [131]).

Theorem 32 (Yudin). Let A ⊆ G be an arbitrary set with |A| = δ|G|, and let

ε, ε′ ∈ (0, 1). Then

R δ(1−ε)(A) + R δ(1−ε)(A) ⊆ R δ(1−2(ε+ε′))(A). (63)

Proo . Let r ∈ R δ(1−ε)(A) and r′ ∈ R δ(1−ε′)(A). Clearly, there exist angles ϕ andϕ′ such that Re

A(r)eiϕ (1 − ε)|A| and Re

A(r′)eiϕ′ (1 − ε′)|A|. Hence,

Rex∈A

2eiϕe(rx) + 2eiϕ′

e(rx′) − 3

= Re

2 A(r)eiϕ + 2 A(r′)eiϕ′ − 3

1 − 2(ε + ε′)|A|.

I we show that

Re2e

iϕ

e(rx) + 2e

iϕ′

e(r′x) − 3 Ree

iϕ+iϕ′

e(r + r′)xor every x, then it will be proved that r + r′ ∈ R δ(1−2(ε+ε′))(A). We x x ∈ G,

and let eiϕe(rx) = eiβ and eiϕ′

e(r′x) = eiβ′ , where β, β ′ ∈ (−π/2, π/2]. Then it issucient to veriy that

2cos β + 2 cos β ′ − 3 cos(β + β ′).

Using the upper convexity o the cosine on the interval (−π/2, π/2], we get that

2cos β + 2 cos β ′ − 3 4cosβ + β

′2 − 3

= 2 cos2β + β ′

2− 1 − 2

1 − cos

β + β ′

2

2 cos(β + β ′).

The theorem is proved.

We conclude this section by discussing the question o how small the Fouriercoecient o a characteristic unction can be. This problem was rst consideredin [132]. Our treatment ollows [131].

Theorem 33 (Konyagin–Lev). Let p be a prime number and let G = Z p. Then

1) or any set A with |A| = n and 3 n p − 1

minr∈G

| A(r)| > n−( p−3)/4; (64)




2) on the other hand , or every n o the orm n = 2k < p/20 with k a positive

integer there exists a set A ⊆ G such that

| A(r)| < n−(ln p)/(2ln2). (65)

We prove the rst part o Theorem 33. Part 2) is veried in a airly straightor-ward ashion — a certain dissociative set is taken as the set A.It is easy to see that the Fourier coecients o the characteristic unction o A

are non-zero. Indeed, it suces to observe that they are the values o the polyno-mial A(z) at roots o unity and that A(z) is not divisible by the minimal polynomial

1+z + · · ·+z p−1. The product m :=

r∈Z∗p A(r) is the norm o any root o A(z). In

particular,

r∈Z∗p A(r) ∈ Z\0. From the latter consideration, Parseval’s equality,

and the inequality between the arithmetic mean and the geometric mean, we getthat

1 r∈Z∗p

| A(r)|2 = | A(r0)|4 r∈Z∗p, r=±r0

| A(r)|2

| A(r0)|4

1

p − 3

r∈Z∗p, r=±r0

| A(r)|2 p−3

< | A(r0)|4

n( p − n)

p − 3

p−3 n p−3| A(r0)|4,

and all is proved in the case r0 = 0. I r0 = 0, then the inequality (64) is obvious.It would be airly interesting to reduce the gap between the upper and lowerestimates in Theorem 33.

7. Combinatorial confgurations in sumsets

In this section we consider problems about arithmetic progressions in sumsets,that is, in sets o the orm A1 + · · · + Ak. It turns out that such sets con-tain surprisingly long arithmetic progressions. This act was rst noted by Bour-gain [67]. Many papers on these problems have appeared recently (see, or example,

[62], [133]–[138]). We cannot even mention all the known results here but only touchupon some o them. In [67] Bourgain obtained the ollowing result.

Theorem 34 (Bourgain). Let A, B ⊆ [N ] be some sets, with |A| = γN and

|B| = δN . Then there exists an absolute constant c > 0 such that the set A + Bcontains an arithmetic progression o length at least

exp

c

(γδ log N )1/3 − log log N

.

On the other hand, Ruzsa [133] ound a lower estimate or the length o a maximalarithmetic progression in the set A + A.

Theorem 35 (Ruzsa). Let ε > 0 be arbitrary . Then there is a number p0(ε) such

that or all primes p with p > p0(ε) there exists a symmetric set A ⊆ Z p (that

is, A = −A) with |A| > (1/2 − ε) p such that A + A does not contain arithmetic

progressions o length greater than exp

(log p)2/3+ε

.




Defnition 5. Let S ⊆ ZN be some set and let ε ∈ (0, 1) be a parameter. The setS is said to be hereditarily non-ε-uniorm i

| S ′(r)| ε|S ′| ∀ r = 0

or every S ′ ⊆ S .The complements in ZN o sets that are hereditarily non-ε-uniorm have the

ollowing property.

Theorem 40 (Green). Let ε ∈ (0, 1) be a real number with

ε 4000(loglog N )/(log N )1/2,

and let S ⊆ ZN be a hereditarily non-ε-uniorm set . Then the complement o S contains an arithmetic progression o length eε

√logN/32.

Clearly, Theorem 37 ollows rom Theorem 40. Indeed, let S be the comple-ment in ZN o A + B. Also, let S ′ ⊆ S be an arbitrary set. Then, by denition,

x S ′(x)(A ∗ B)(x) = 0. We rewrite the last equation by using the Fourier trans-orm. We have

0 =1

N

r

S ′(r) A(r) B(r) =|S ′| |A| |B|

N +

1

N

r=0

S ′(r) A(r) B(r),

whence, by the Cauchy–Schwarz–Bunyakovskii inequality and Parseval’s equality,

we get that|A|1/2|B|1/2N max

r=0| S ′(r)| |S ′| |A| |B|,

and we see that the set S is hereditarily non-(αβ )1/2-uniorm. I (αβ )1/2 4000(loglog N )/(log N )1/2, then by Theorem 40 the complement o S , that is, theset A+B, contains an arithmetic progression o length exp

2−5(αβ log N )1/2

. But

i (αβ )1/2 4000(loglog N )/(log N )1/2, then Theorem 37 holds with the constantc = 4000/32 = 125.

Thus, we need to veriy Theorem 40. Let ω

∈(0, 1) be a parameter and let Ω

⊆S

be some set with |Ω| = ωN on which the minimum o the expression maxr=0 | Ω(r)|is attained. We denote this minimum by η|Ω|. Since S is not hereditarily ε-uniorm,we have η ε.

The main idea o the proo is that we should attempt to modiy the set Ω,that is, to turn it into a new set Ω′ with |Ω′| = |Ω|, so as to make the unctional

maxr=0 | Ω′(r)| as small as possible. But since it cannot become strictly less thanη|Ω|, this will mean that Ω (more precisely, its complement) possesses certain spe-cic structural properties. From this it is already not ar to the assertion that thecomplement o S has a special structure.

We shall modiy the set Ω as ollows: we remove a set X rom Ω, and then adda set D with |D| = |X | := t to Ω, where t is a parameter. O course, this mayresult in not a set but a multiset. We conront this diculty at the very end o theproo. Thus, Green’s method consists o two stages: construction o the set D ando the set X .



548 I. D. Shkredov

1. Construction o the set D. Here we use Theorem 25 in § 6 (it is also possibleto use Theorem 31). We consider the set o large trigonometric sums

R =

r : |

Ω(r)| η

|Ω|2

.

Lemma 15. The Bohr set B( R , η/64) contains an arithmetic progression P o

length

η3

27 log(1/ω)N η

2/(2 log(1/ω)).

Indeed, by Theorem 25 every element o the set R can be represented as a combi-nation o elements o the set Λ with coecients ±1. Hence, by the triangle inequal-ity we get that B

Λ, η/(64|Λ|) ⊆ B( R , η/64), and then we apply Lemma 2 to the

set BΛ, η/(64

|Λ

|). It is the transition rom the set R to the smaller set Λ that

constitutes the advantage o using Theorem 25.Let

C :=

x : |(P + x) ∩ Ω| 16ω

η|P |

.

We call these points o the group ZN good . Since

|P | |Ω| =

x

|(P + x) ∩ Ω| x /∈C

|(P + x) ∩ Ω| (N − |C |) 16ω

η|P |,

there are at least (1 − η/16)N good points. In other words, the set C is almost thewhole o ZN , and thus its Fourier coecients are small. Indeed, let C d = ZN \ C .Then or all r = 0

| C (r)| = | C d(r)| |C d| ηN

16 η

|C |8

,

and we can nally choose the set D. We need a variant o the probability lemmaon large deviations (see [62]).

Lemma 16. Let X j , j ∈ [n], be independent complex random variables, |X j | 1,with zero mathematical expectation and variance σ2j . Also, let t 0 be a real number

such that σ2 :=

j σ2j 6nt. Then

P

j

X j

nt

4e−n2t2/(8σ2).

We choose a set D ⊆ C at random: an element x belongs to D with probability p := t/|C |.Lemma 17. Let t ≫ η−2 log N . Then there exists a set D ⊆ C such that |D| = tand maxr=0 | D(r)| ηt/4.

The calculations are quite standard (see, or example, [141]). Let a set E ⊆ C be chosen at random with probability p. Then the rth Fourier coecient o the




set E is the sum o the independent random variables X (r)j (x) = E (x)e(−rx),

E E (r) = p C (r), and by Lemma 16,

P

|

E (r)| ηt

6

P

|

E (r) − E

E (r)| ηt

24

4e−η2t/5000.

Similarly,

P

|E | − t ηt

24

4e−η2t/5000.

By hypothesis, t ≫ η−2 log N , and consequently, with positive probability or allr = 0 we have

| E (r)| ηt/6 and |E | − t

ηt/24.

By removing or adding at most ηt/24 points in the set D we obtain the assertiono the lemma.

We now describe the second stage o the proo.2. Construction o the set X . Here we use quite simple arguments. We choosea set X ⊆ Ω at random. Then the proo o the ollowing assertion is similar to thato Lemma 17.

Lemma 18. Let βN t ≫ η−2 log N . Then there exists a set X ⊆ Ω with |X | = tsuch that or all r = 0

X (r) − t

|Ω|

Ω(r)

ηt

12.

We put Ω1 = (Ω \ X ) ∪ D. This object is a multiset. Nevertheless, it is possibleto estimate its Fourier coecients.

Lemma 19.

maxr∈R, r=0

| Ω1(r)| η|Ω1| − ηt

6, (66)

and or all r = 0

| Ω1(r)| η|Ω1|2

+ηt

3. (67)

Proo . From the construction o D and X we get that

Ω1(r) = Ω(r) − X (r) + D(r) =

1 − 1

|Ω| Ω(r) + Q,

where Q ηt/3. I r ∈ R and r = 0, then by recalling the denition o the quantity

η and taking into account the inequality | Ω(r)| η|Ω|/2 we veriy that (66) holds.

I, however, r /∈ R and r = 0, then | Ω(r)| η|Ω|/2 and the estimate (67) holds.Lemma 19 is proved.

Thus, we have constructed a multiset Ω1 all o whose Fourier coecients are less

than η|Ω1|. I, rst, Ω1 were a set and, second, the inclusion Ω1 ⊆ S held, then wewould obtain a contradiction to the extremal properties o Ω. We now try to shitthe set D so that the new multiset Ω′ has these two properties.

Let D = d1, . . . , dt and D′ = d1+ x1, . . . , dt + xt, where the xj are elementso the progression P , and let Ω′ = (Ω \ X ) ∪ D′. We prove the ollowing lemma.



550 I. D. Shkredov

Lemma 20. Let t ηωN/10. Then the inequality maxr=0 | Ω′(r)| < η|Ω′| holds

or the multiset Ω′.

Proo . Let r ∈ R . Then rom the inequality (67) and rom the act that the contri-bution o the set D′ to the Fourier coecients o Ω′ difers rom the corresponding

contribution o D to Ω1 by at most 2t we get that

| Ω′(r)| η|Ω′|2

+ 5t < η|Ω′|.

But i r ∈ R and r = 0, then

| Ω′(r) − Ω1(r)| t

j=1

e−r(dj + xj)− e(−rxj)

t maxj

|e(−rxj) − 1| ηt

8,

since P ⊆ B( R , η/64). Using (67), we obtain the assertion o the lemma.We now go directly to the proo o Theorem 40. Suppose that |(dj +P )∩(S \Ω)|

t or all j ∈ [t]. Then we take d1 and nd x1 ∈ P such that d1 + x1 ∈ S \ Ω. Next,we take d2 and nd x2 ∈ P such that d1 + x1 ∈ (S \ Ω) \ d1 + x1. And so on. Inthe end we nd translates o all elements dj that belong to S but not to Ω. Thisenables us to construct a set Ω′ ⊆ S with non-zero Fourier coecients that are lessthan η|Ω′|. But the last inequality is impossible because o the extremal propertieso Ω. Thereore, there exists a dj ∈ D such that |(dj + P ) ∩ (S \ Ω)| < t. Recallingthat the point dj is good, we get that

|S d ∩ (dj + P )| |P | − 16ω

η|P | − t

1 − 32ω

η

|P |, (68)

where the parameter t is chosen so that t 16ω|P |/η and S d = ZN \ S . Also,suppose that |P | η/(32β ). From the estimate (68) and elementary considerationso the average, it is easy to see that the set S d contains an arithmetic progres-sion o length η/(32β ). Recall that earlier we supposed that t ≫ η−2 log N andt ηωN/10. Furthermore, η ε always. By choosing the parameters

ω = e−η√logN /16, t ∼ η−2 log N

and perorming small calculations we obtain Theorem 40.We now prove a special case o Theorem 39 (see [140]). Let G = Zn

2 , k = 4, andA1 = A2 = A3 = A4. We consider this situation only or simplicity. In the generalcase airly similar arguments are used (c. Theorems 2 and 9 in § 3).

Theorem 41 (Sanders). Let G = Zn2 and let A ⊆ G with |A| = δN . Then 2A−2A

contains a subspace o codimension Cδ−1/2, where C > 0 is some absolute constant .

The reader can see that the estimates in Theorem 41 are better than in Theo-rem 39. As in Roth’s theorem, the proo uses an iteration lemma.

Lemma 21. Let A ⊆ G be some set with |A| = δN . Then the ollowing alternative

holds:1) either 2A − 2A = G;




2) or there exist a subspace V o codimension 1, a set A′ ⊆ V , and an x ∈ G

such that

(a) A′ + x ⊆ A,

(b) |A′| (δ + δ3/2)|V |.

Indeed, i 2A − 2A = G, then the set Ω := G \ (2A − 2A) is non-empty. Clearly,x Ω(x)(A ∗ A ∗ A ∗ A)(x) = 0. We rewrite the last equality in terms o the Fourier

coecients. We have

0 =

r

Ω(r) A4(r) = |Ω| |A|4 +r=0

Ω(r) A4(r).

Let αN := maxr=0 | A(r)|. By Parseval’s equality we nd that

|Ω

| |A

|4

|Ω

| |A

|N (αN )2,

whence α δ3/2. Thereore, | A(r)| = αN δ3/2N or some r = 0. Let H 0 =x ∈ G : ⟨x, r⟩ = 0 and H 1 = x ∈ G : ⟨x, r⟩ = 1. Then |H 0 ∩ A| + |H 1 ∩ A| =

|A| = δN and |H 0 ∩ A| − |H 1 ∩ A| = A(r). Since | A(r)| δ3/2N , either |H 0 ∩ A| (δ +δ3/2)N/2 or |H 1∩A| (δ +δ3/2)N/2. We obtain Lemma 21 by setting V = H 0,A′ = H 0 ∩ A, and x = 0 in the rst case, and V = H 0 and A′ = (H 1 ∩ A) − x orany x ∈ H 1 in the second case.

Proo o Theorem 21. We apply Lemma 21 to the set A. I the rst alternative o the

lemma holds, then there is nothing to prove. I, however, the second alternativeholds, then we apply the lemma to the set A′. And so on. Since the transitionrom the set A to the set A′ is accompanied by an increase o the density by thequantity δ3/2, it is easy to see that our iteration process cannot complete morethan Cδ−1/2 steps (see the proo o Theorem 9). Here C > 0 is some absoluteconstant. Thereore, at some point the iteration process will terminate at step 1).

In other words, some subspace W o codimension at most Cδ−1/2 is 2 A − 2 A orsome set

A. It is easy to see that there exists a translate o

A that is contained in A,

and hence 2 A −2 A ⊆

2A

−2A. Thereore, the set 2A

−2A contains a subspace W

o codimension at most Cδ−1/2, and the theorem is proved.

The methods o harmonic analysis considered in this section make it possible tond combinatorial congurations only in sums o suciently dense sets. Recentlythe paper [142] appeared, where, in particular, the ollowing result was proved.

Theorem 42 (Croot–Ruzsa–Schoen). Let k be an odd number . Also, let A, B ⊆[N ] be some sets with |A| |B| 6N 2−2/(k−1). Then A + B contains an arithmetic

progression o length k.

The method o proo o Theorem 42 is quite elementary and uses Dirichlet’sdrawer principle. Theorem 42 is non-trivial or suciently ‘lean’ sets, say, o car-dinality N 1−ε, where ε > 0 is some constant. Nevertheless, or dense sets themethods o Fourier analysis give better estimates, as is easy to see (or example,rom Theorem 39).



552 I. D. Shkredov

8. Chebotarev’s theorem and the uncertainty principle

Let G be a nite Abelian group, and f an arbitrary complex unction denedon the group. Let supp f be the support o f : supp f = x ∈ G : f (x) = 0. It iswell known that there is a connection between the cardinalities o the sets supp f and supp f (see, or example, [143] or [144]).

Assertion 2 (the uncertainty principle). Let f : G → C be an arbitrary unction

that is not identically equal to zero. Then

| supp f | · | supp f | |G|. (69)

Indeed, let maxξ∈ G |f (ξ)| be attained at a point ξ0. From the Cauchy–Schwarz–Bunyakovskii inequality and Parseval’s equality, we nd that

0 < |ˆ

f (ξ0)| x∈supp f |f (x)| | supp f |1/2x∈G |f (x)|

21/2

= | supp f |1/2

|G|−1

ξ∈supp f

|f (ξ)|21/2

| supp f |1/2|G|−1/2| supp f |1/2|f (ξ0)|,

whence we obtain the required inequality.Quite recently (see [144]) it was discovered that in the case G = Z/pZ with p

a prime the estimate (69) can be signicantly improved.

Theorem 43 (Tao, the uncertainty principle in the groupZ

/pZ

). Let p be a prime,and f : Z/pZ→ C an arbitrary unction that is not identically equal to zero. Then

| supp f | + | supp f | p + 1. (70)

As we shall now see, Tao’s theorem above is none other as a reormulation o thewell-known Chebotarev theorem on the minors o the Fourier matrix.

Theorem 44 (Chebotarev). Let p be a prime number . Then all the minors o the

matrix Φ = (e2πimn/p) pn,m=1 are non-zero.

Let us derive Theorem 43 rom Chebotarev’s theorem. Let A and B be twoarbitrary non-empty subsets o Z p o equal cardinality. Then Chebotarev’s theorem

asserts that the operator T A,B : f → f B, where f belongs to the space o unctionswith support in A, is invertible. Suppose that the inequality (70) is alse. In

other words, suppose that | supp f | + | supp f | p or some unction f that is not

identically equal to zero. We set A = supp f and choose B so that B ⊆ (Z p\supp f )and |B| = |A|. But then we obtain a contradiction to the invertibility o the operatorT A,B, and Theorem 43 is proved.

By using the invertibility o the operator T A,B it is easy to check (or exam-

ple, see [144]) that Theorem 43 is absolutely sharp in the sense that or any twonon-empty sets A and B with |A| + |B| p + 1 there exists a unction f with

supp f = A and supp f = B.Finally, we note that the inequality (70) does not hold in an arbitrary nite

Abelian group. For example, in the situation where G = Zn p : or the characteristic




unction f o a subspace P o dimension k n, we have | supp f | = |P | and

| supp f | = |G|/|P |, and (70) is alse.It is easy to see that Theorem 43 implies the well-known Cauchy inequality on

sums o sets in Z/pZ:

|A + B| min p, |A| + |B| − 1. (71)

Indeed, we point out rst that i |A|+|B| > p, then A+B = Z p and thereore theestimate (71) holds (the equality A + B = Z p ollows rom the act that or everyn ∈ Z p the sets n−A and B necessarily intersect). Suppose now that |A|+ |B| p.Then there exist two sets X and Y such that |X | = p + 1 − |A|, |Y | = p + 1 − |B|,and X ∪ Y = Z p. Hence, |X ∩ Y | = |X | + |Y | − p = p + 2 − |A| − |B|. Since theoperator T A,B is invertible, there exist unctions f and g that have the properties

supp f = A, supp f = X and supp g = B, supp g = Y . Consider the new unction

F = f ∗ g. Then supp F = X ∩ Y and supp F = A + B. By Theorem 43 we have

| supp F | + | supp F | = |A + B| + p + 2 − |A| − |B| p + 1,

and all is proved.We point out that recently Guo and Sun [145] rened the inequality (71) by

using Tao’s method.

Theorem 45 (Guo–Sun). Let p be a prime, and A, B, S non-empty subsets o

Z/pZ. Then

|a + b : a ∈ A, b ∈ B, a − b /∈ S | min p, |A| + |B| − 2|S | − 1.

It remains to prove Chebotarev’s Theorem 44. There exist quite a number o proos o this remarkable result (see, or example, [144], [146]). In our expositionwe ollow [146]. Let ω = e2πi/p and π = ω − 1. Then

ωrs = (1 + π)rs =rs

k=0

rs

k

πk (72)

or all positive integers r and s. Also, let M = |ωrisj

|i,j∈[l] be an arbitrary l×l minoro the matrix Φ. Using the multilinearity o the determinant and the identity (72),we get that

det M =

k10,...,kl0

Dk1,...,klπk1+···+kl , (73)

where Dk1,...,kl is the determinant o the matrix composed o the columnsris1k1

, . . . ,

risl

kl

i∈[l]

.

Consider the polynomial ps,k(x) =xs

k

o degree k. Then the columns o our deter-

minants are the values o the polynomials ps1,k1 , . . . , psl,kl at the points r1, . . . , rl.We x a positive integer d < l and suppose that in the sequence k1, . . . , kl thereexist d + 1 numbers kj that are less than d. Since the dimension o the space



554 I. D. Shkredov

o polynomials o degree t is equal to t + 1, the corresponding polynomials arelinearly dependent and the number Dk1,...,kl is equal to zero. It ollows rom ourconsiderations that all the coecients in the expansion (73) beore πa, where a = 0+1+2+· · ·+(l−1) = l(l−1)/2, are equal to zero. Let k1, . . . , kl = 0, 1, 2, . . . , l−1.We calculate the coecient o πa, that is, the determinant Dk1,...,kl . Denote this

integer by C . I we prove that C is coprime to p, then we will easily see that it isnot divisible by π in the ring Q[ω], and consequently det M = 0. Expanding eacho the polynomials ps,k(x) by the ormula ps,k(x) = sx(sx − 1) · · · (sx − k + 1)/k! =skxk/k! + gs,k(x), where gs,k(x) is a polynomial o degree k − 1, and using thearguments above, we get that

C =

σ

sgn(σ)s

σ(0)1 · · · s

σ(l−1)l

0!1! · · · (l − 1)!

i,j∈[l]

(rj − ri).

In deriving the last ormula we used the expression or the Vandermonde determi-nant. Using this expression once again, we get that

C =1

0!1! · · · (l − 1)!

i,j∈[l]

(rj − ri)

i,j∈[l]

sj − si

= 0 (mod p),

which is what was required.

9. Connection with analytic number theory

Let R ⊆ Z∗ p be an arbitrary multiplicative subgroup. The question o estimatesor trigonometric sums over subgroups is an old problem in analytic number theory(see, or example, [47] or [147]). In other words, we are interested in an upperestimate or the quantity

S (R) = maxa∈Z∗p

x∈R

e(ax)

.The classical estimate or S (R) is S (R)

√ p (see [147]). Clearly, this estimate

is trivial or subgroups o order less than √ p. In [48], [148], and [149] inequalitieswere obtained o the orm

S (R) |R| p−ε, (74)

where |R| pc, c > 0 is some constant, and ε = ε(c) > 0. For example, in [48]it was shown that the estimate (74) holds i c > 1/4 is arbitrary. Until recentlythis was the smallest size o a subgroup starting with which there was a non-trivialestimate (74). In [16] (see also [15]) the inequality (74) was proved or any positiveconstant c.

Theorem 46 (Bourgain–Konyagin). Let δ > 0 be a real number . Then there exist a positive integer p0(δ) and a number ε(δ) ∈ (0, 1) such that or all primes p with

p p0 and or any multiplicative subgroup R ⊆ Z∗ p with |R| pδ

S (R) |R| p−ε. (75)




The restriction |R| pδ in Theorem 46 was later weakened in [21].Since any subgroup R in Z∗ p has the orm R = xk : x ∈ Z p or some positive

integer k and since|R| = ( p − 1)/(k, p − 1),

Theorem 46 can be reormulated as ollows.Theorem 47 (Bourgain). Let δ be a real number , p a suciently large prime, and

k a positive integer such that k < p − 1 and (k, p − 1) < p1−δ . Then there exists an

ε = ε(δ) > 0 such that p−1x=1

e(axk)

< p1−ε.

In [17] (see also [15]) Bourgain obtained similar estimates or the sums

t1s=1

e(aθs) andt1

s,s′=1

e(aθs + bθss′), (76)

where a, θ = 0, θ is an element o order t, and t t1 > pδ.Theorem 47 was generalized in [18].

Theorem 48 (Bourgain). Let δ be a real number , let p be a suciently large prime,and let ki be positive integers or i ∈ [r] such that ki < p − 1,

(ki, p − 1) < p

1

−δ

, i ∈ [r], (77)and

(ki − kj , p − 1) < p1−δ, i, j ∈ [r], i = j. (78)

Let

f (x) =r

i=1

aixki , (ai, p) = 1. (79)

Then there exists an ε = ε(δ, r) > 0 such that

p−1x=1

e(f (x)) < p1−ε. (80)

The trigonometric sums in (80) are called Mordell’s sums. Earlier results onMordell’s sums can be ound in [150], [151]. Theorem 48 was rened in [152]. Wenote that condition (78) is necessary. Indeed (see [151]), or example, let r = 2,k1 = 1, k2 = ( p − 1)/2 + 1, and f (x) = x − x( p−1)/2+1. Then Weyl’s inequality(see, or example, [153]) easily implies that

p−1x=1

e(f (x)) =p − 1

2+

x(p−1)/2=−1

e(2x) + e(1) =p − 1

2+ O

√ p

,

since x( p−1)/2 is the Legendre symbol o a number x.



556 I. D. Shkredov

Polynomials o the orm (79) are called sparse polynomials. A denitive, ina sense, result describing all the sparse polynomials or which non-trivial estimateso trigonometric sums similar to (80) are possible was given in [154]. We shall needseveral denitions. Recall that the discrepancy o a sequence o points x1, . . . , xn ∈[0, 1) is dened to be the quantity

Dn = Dn(x1, . . . , xn) = sup0α<β1

|i : xi ∈ [α, β )|n

− (β − α).

Let f : Z p → Z p be an arbitrary unction. We denote the discrepancy o thesequence f (i)/p by D p(f ). Here Z p is identied with the sequence 0, 1, . . . , p−1.Finally, i pj is a sequence o prime numbers, then a sequence o polynomialsf j : Z pj → Z pj is said to be well distributed i or some ε > 0 we have D pj (f j) p−ε

j .It is well known (see, or example, [153]) that a non-trivial estimate o trigonometricsums over the points o a sequence implies that the sequence is well distributed. On

the other hand, or the polynomial f (x) = axk, a = 0, and an arbitrary prime pwe have D p(f ) k/p, since, as is easy to see, f (x) takes the same value deg f = ktimes. Thereore, we can reormulate Theorems 46 and 47 as ollows.

Proposition 5. A sequence o non-zero monomials f j = aj xkj is well distributed

i and only i there is an η > 0 such that (kj , pj − 1) < p1−ηj or any j.

By using Theorem 48 Konyagin obtained a similar criterion or arbitrary sparsepolynomials. We state only one result rom [154].

Theorem 49 (Konyagin). Let r be a positive integer and let f j∈Z pj [x] be a

sequence o polynomials such that

f j(x) =

kj∈Kj

a(kj )xkj ,

where K j ⊆ Z+ and |K j | r. The sequence f j is well distributed i and only i

there exists an η > 0 such that none o the sets K j can be represented as a disjoint

union o sets K j0 , . . . , K jsj having the ollowing properties:

(a) dj | k or all k ∈ K j0 ;

(b) dj | (k − k′) or every i = 1, . . . , sj and all k, k′ ∈ K j

i ;(c) there exists an xj

0 ∈ Z pj such that or every i = 1, . . . , sjk∈Kj

i

a(k)(xj0)

k = 0,

where dj > p1−ηj and dj | ( pj − 1).

See [22] or a generalization o Theorem 48 to a wider class o polynomials(roughly speaking, those representable as a sum o a sparse polynomial and a poly-

nomial o degree p1/2−δ, δ > 0).The undamental Theorem 46 ollows rom the result o Bourgain, Katz, and

Tao [10] on sums o products in the group Z p. See an independent proo o Theo-rem 46 in the remarkable paper [155]. See the earlier papers [156]–[159] about thetheory o sums o products in ordered groups such as R or Z.



558 I. D. Shkredov

See [160] or applications o the results o [49] to problems o theoretical com-puter mathematics. New inequalities or trigonometric sums over subgroups o thegroups Zq, where q is composite, were obtained in [19] and [23]. See [161] oran application o combinatorial methods or estimating trigonometric sums withmultiplicative characters.

Thus, we have seen that the methods o additive combinatorics make it possibleto prove new number-theoretic assertions. On the other hand, as became clearrecently (see [51], and also [162]), the classical estimates or trigonometric sumsover additive characters give new results on sums o products. We present onetheorem rom [51].

Theorem 52 (Garaev). Let p be a prime, and A ⊆ Z p an arbitrary set . Then

|A + A| |A · A| ≫ min

p|A|, |A|4

p

. (84)

Proo . We can assume that 0 /∈ A. Consider the equation

xy−1 + z = w, where x ∈ A · A, y ∈ A, z ∈ A, w ∈ A + A. (85)

Clearly, this has |A|3 solutions o the orm x = a1a2, y = a−12 , z = a3, w = a1+a3,where all the ai belong to A, and to diferent triples (a1, a2, a3) there corresponddiferent solutions x, y, z, w o equation (85). Let J be the number o solutions o equation (85). As we have just proved, J |A|3. On the other hand,

|A|3 J = 1 p

r∈Zp

x∈A·Ay∈A

z∈A

w∈A+A

er(xy−1 + z − w)

|A|2|A + A| |A · A| p

+1

p

r∈Z∗p

| A(−r)| |(A + A) (r)|

x∈A·A

y∈A−1

e(rxy)

.By applying the estimate or the trigonometric sum over products o sets (see, orexample, [153]) we get that

x∈A·A

y∈A−1

e(rxy) p|A · A| |A| , r = 0.

From this and the Cauchy–Schwarz–Bunyakovskii inequality we have

|A|3 J |A|2|A + A| |A · A|

p+

p|A · A| |A|2|A + A| ,

and the theorem is proved.

For example, i

|A

| ≫p2/3, then the estimate (84) implies the inequality

max|A + A|, |A · A| ≫ p1/2|A|1/2.Note that the last estimate cannot be improved (see, or example, [51]), and conse-quently Theorem 52 is sharp or sets A o cardinality ≫ p2/3. Here the lower bound




or the cardinality o A does not appear just by chance, since, as is well known, theanalytic methods work well only or suciently large sets.

In act, an asymptotic ormula or the number o solutions o (85) was obtainedin the proo o Theorem 52. Other equations with variables that belong to arbitrarysets in Z p were considered in [53]–[56]. We state only two results.

Theorem 53 [55]. Let A,B,C,D ⊆ Z p be arbitrary sets, and λ any element o Z∗ p.Then the number N 1(A,B,C,D) o solutions o the equation

ab + cd = λ, where a ∈ A, b ∈ B, c ∈ C, d ∈ D, (86)

satises the asymptotic ormula

N 1(A,B,C,D) =|A| |B| |C | |D|

p

−1

+ O

( p|A| |B| |C | |D|)1/2.

Theorem 54 [54]. For arbitrary sets A,B,C,D ⊆ Z p the number N 2(A,B,C,D)o solutions o the equation

a + b = cd, where a ∈ A, b ∈ B, c ∈ C, d ∈ D, (87)

satises the asymptotic ormula

N 2(A,B,C,D) =|A| |B| |C | |D|

p

−1

+ O

( p|A| |B| |C | |D|)1/2

.

The method o proo o Theorems 53 and 54 also amounts to using the classi-cal estimates or trigonometric sums. These theorems establish the solubility o certain equations in our unknowns running over arbitrary suciently dense setsA,B,C,D. For example, i |A|, |B|, |C |, |D| ≫ p1−ε0 , where ε0 > 0 is a sucientlysmall number, then equations (86) and (87) are soluble. Recently the author [116]presented an example o an equation in three variables o the orm f (x, y) = z,where f (x, y) is some polynomial, that is also soluble or any suciently dense sets.Clearly, there are no equations in two variables with a similar property.

10. Conclusion

We conclude the present survey by stating several unsolved problems o har-monic analysis related to problems o combinatorial number theory. Some o theseproblems were already stated in the preceding sections; or example, the polynomialFreiman–Ruzsa conjecture (see § 4).

The most dicult unsolved problem is still the Erdos–Turan Conjecture 1. Asshown in § 3, this conjecture is closely related to the problem o the behaviour o the unction ak(N ). We note that even in the simplest case k = 3 the precise order

o growth o a3(N ) is still unknown. Recently Green and Tao announced a result onan upper estimate or ak(N ) with k 4 o the orm ak(N ) ≪ 1/(log N )C k (whereC k > 0 is an absolute constant) and proved a similar result or the unction a4(N )in the groups Zn

p , where p is a prime, p = 2, 3 (see [46]). Concerning the behaviouro a3(N ) in the group Zn

3 , Green posed the ollowing conjecture (see [3]).



560 I. D. Shkredov

Conjecture 4 (Green). Let G = Zn3 . Then a3(N ) N −ε0 , where ε0 > 0 is some

absolute constant .

We state a conjecture (possibly too bold) whose validity or sets with cardinalityo order N/(log N )c, c > 1, would imply an armative solution o the Erdos–Turan

problem in the case k = 3, as well as Conjecture 4 (i the cardinality o the set isestimated rom below as N 1−c′ , c′ > 0).

Conjecture 5. Let A ⊆ ZN or A ⊆ Zn p , p 3, be an arbitrary set without

arithmetic progressions o length three, with some suciently large cardinality ( or

example, N/(log N )c, c > 1). Then there exists an r = 0 such that | A(r)| ρ|A|,where ρ > 0 is an absolute constant .

The validity o Conjecture 5 or sets o positive density ollows easily rom thearguments in § 3. We also note that Conjecture 5 is alse or sets without solutionso an arbitrary non-ane linear equation, say, x + y = z (see the example o a set inTheorem 2.11 o [128]). Finally, it is easy to see that Conjecture 5 is also not validor very small sets o cardinality, say, N 1/3. For example, one can take a randomset o cardinality N 1/3. The author is grateul to S. V. Konyagin or this remark.

In connection with the polynomial Freiman–Ruzsa Conjecture 3 (p. 531) theollowing question arises. Let p be a prime number, and S ⊆ Zn

p some set with|S | = δN . What is the codimension o the subspace contained in 2A − 2A? At themoment, the best result here is due to Sanders [140]; namely, this codimension c isO(δ−1/2) (see § 7). It is easy to veriy that the estimate c ≪ log(1/δ) implies theFreiman–Ruzsa conjecture.

The question o the structure o sets o large trigonometric sums is interestingin its own right. What are necessary and sucient conditions or some set to bea set o the orm R α? The answer is unknown even to the more special question:what can be said about the structure o the sets R α with cardinality close to theupper bound δ/α2 or the cardinality o a set o large trigonometric sums thatollows rom Parseval’s equality? Such results may help in solving problems o combinatorial number theory using Fourier analysis.

So ar the exact value o the Ramsey number has not been ound or any linearequations c1x1+· · ·+cnxn = 0 with n independent o the size o the group (see § 5).

It would be very interesting to have an example o such an equation or which weknow at least the order o the behaviour o R(c).In conclusion we point out that there remains open the problem (which is possibly

central in additive combinatorics) o nding necessary and sucient conditions ora set S to have the orm A + A. The author hopes that the methods o Fourieranalysis can be used in answering this question.

The author is grateul to S. V. Konyagin and K. S. Ryutin or a number o useuldiscussions and remarks.

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I. D. Shkredov

Moscow State University

E-mail : [email protected]

Received 07/OCT/09Translated by E. KHUKHRO





















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Fourier Analysis in rial Number Theory

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