2 Fourier-analytic techniques

In this chapter we will assume that G is finite abelian.

G comes equipped with a group Ĝ of characters, i.e. homomorphisms γ:G. In fact, Ĝ is isomorphic to G.

See Representation Theory notes for more information about characters and proofs of this as well as some of the facts below.

Example 2.1.

Notation. Given BG nonempty, and any function g:B, let

𝔼xBg(x)=1|B|xBg(x).

Lemma 2.2. Assuming that:

  • γG^

Then
𝔼xGγ(x)={1if γ=10otherwise,

and for all xG,

γG^γ(x)={|G^|if x=00otherwise.

Proof. The first equality in eqch case is trivial. Suppose γ1. Then there exists yG with γ(y)1. Then

γ(y)𝔼zGγ(z)=𝔼zGγ(y+z)=𝔼zGγ(z)

So 𝔼zGγ(z)=0.

For the second part, note that given x0, there must by γG^ such that γ(x)1, for otherwise G^ would act trivially on x, hence would also be the dual group for Gx, a contradiction.

Definition 2.3 (Fourier transform). Given f:G, define its Fourier transform f^:G^ by

f^(γ)=𝔼xGf(x)γ(x)¯.

It is easy to verify the inversion formula: for all xG,

f(x)=γG^f^(γ)γ(x).

Indeed,

γG^f^(γ)γ(x)=γG^𝔼yGf(y)γ(y)¯γ(x)=𝔼yGf(y)γG^γ(xy)=|G| iff x=y=f(x)by Lemma 2.2

Given AG, the indicator or characteristic function of A, 𝟙A:G{0,1} is defined as usual.

Note that

𝟙A^(1)=𝔼xG𝟙A(x)1(x)=|A||G|.

The density of A in G (often denoted by α).

Definition (Characteristic measure). Given non-empty AG, the characteristic measure μA:G[0,|G|] is defined by μA(x)=α1𝟙A(x).

Note that 𝔼xGμA(x)=1=μA^(1).

Definition (Balanced function). The balanced function fA:G[1,1] is given by fA(x)=𝟙A(x)α. Note that 𝔼xGfA(x)=0=fA^(1).

Example 2.4. Let V𝔽pn be a subspace. Then for t𝔽pn^, we have

𝟙V^(t)=𝔼x𝔽pn𝟙V(x)e(xtp)=|V|pn𝟙V(t)

where V={t𝔽pn^:xt=0 xV} is the annihilator of V. In other words, 𝟙V^(t)=μV(t).

Example 2.5. Let RG be such that each xG lies in R independently with probability 12. Then with high probability

supγ1|𝟙R^(γ)|=O( log |G||G|).

This follows from Chernoff’s inequality: Given -valued independent random variables X1,X2,,Xn with mean 0, then for all 𝜃>0, we have

(|i=1nXi|𝜃i=1nXiL()2)4exp(𝜃24).

Example 2.6. Let Q={x𝔽pn:xx=0}𝔽pn with p>2. Then

|Q|pn=1p+O(pn2)

and supt0|𝟙Q^(t)|=O(pn2).

Given f,g:G, we write

f,g=𝔼xGf(x)g(x)¯andf^,g^=γG^f^(γ)g^(γ)¯.

Consequently,

fL2(G)2=𝔼xG|f(x)|2andf^l2(G^)2=γG^|f^(γ)|2.

Lemma 2.7. Assuming that:

  • f,g:G

Then
  • (i) fL2(G)2=f^l2(G^)2 (Parseval’s identity)
  • (ii) f,g=f^,g^ (Plancherel’s identity)

Proof. Exercise (hopefully easy).

Definition 2.8 (Spectrum). Let 1ρ>0 and f:G. Define the ρ-large spectrum of f to be

Specρ(f)={γG^:|f^(γ)|ρf1}.

Example 2.9. By Example 2.4, if f=𝟙V with V𝔽pn, then ρ>0,

Specρ(𝟙V)={t𝔽pn^:|𝟙V^(t)|ρ|V|pn}=V.

Lemma 2.10. Assuming that:

  • ρ>0

Then
|Specρ(f)|ρ2f22f12.

Proof. By Parseval’s identity,

f22=f^22=γG^|f^(γ)|2γSpecρ(f)|f^(γ)|2|Specρ(f)|(ρf1)2

In particular, if f=𝟙A for AG, then

f1=α=|A||G|=f22,

so |Specρ(𝟙A)|ρ2α1.

Definition 2.11 (Convolution). Given f,g:G, we define their convolution fg:G by

fg(x)=𝔼yGf(y)g(xy)xG.

Example 2.12. Given A,BG,

𝟙A𝟙B(x)=𝔼yG𝟙A(y)𝟙B(xy)=𝔼yG𝟙A(y)𝟙xB(y)=|A(xB)||G|=1|G|rA+B(x).

In particular, supp(𝟙A𝟙B)=A+B.

Lemma 2.13. Assuming that:

  • f,g:G

Then
fg^(γ)=f^(γ)g^(γ)γG^.

Proof.

fg^(γ)=𝔼xGfg(x)γ(x)¯=𝔼xG𝔼[y]Gf(y)g(xyu)γ(x)¯=𝔼uG𝔼[y]Gf(y)g(u)γ(u+y)¯=f^(γ)g^(γ)

Example 2.14.

𝔼x+y=z+wf(x)f(y)f(z)f(w)¯=f^l4(G^)4.

In particular,

𝟙A^l4(G^)4=E(A)|G|3

for any AG.

Theorem 2.15 (Bogolyubov’s lemma). Assuming that:

  • A𝔽pn be a set of density α

Then there exists V𝔽pn of codimension 2α2 such that VA+AAA.

Proof. Observe

2A2A=supp(𝟙A𝟙A𝟙A𝟙A=:g),

so wish to find V𝔽pn such that g(x)>0 for all xV. Let S=Specρ(𝟙A) with ρ=α2 and let V=S. By Lemma 2.10, codim(V)|S|ρ2α1. Fix xV.

g(x)=t𝔽pn^g^(t)e(xtp)=t𝔽pn^|𝟙A^(t)|4e(xtp)by Lemma 2.13=α4+t0|𝟙A^(t)|4e(xtp)=α4+tS{0}|𝟙A^(t)|4e(xtp)(1)+tS|𝟙A^(t)|4e(xtp)(2)

Note (1)(ρα)4 since xt=0 for all tS and

|(2)|suptS|𝟙A^(t)|2tS|𝟙A^|2suptS|𝟙A^(t)|2tS|𝟙A^|2(ρα)2𝟙A22by Parseval’s identity=ρ2α3

hence g(x)>0 (in fact, α42) for all xV and codim(V)2α2.

Example 2.16. The set A={x𝔽2n:|x|n2+n2} (where |x| counts the number of 1s in x) has density 18, but there is no coset C of any subspace of codimension n such that CA+A(=AA).

Lemma 2.17. Assuming that:

  • A𝔽pn of density α

  • ρ>0

  • supt0|𝟙A^(t)|ρα

Then there exists V𝔽pn of codimension 1 and x𝔽pn such that
|A(x+V)|α(1+ρ2)|V|.

PIC

Proof. Let t0 be such that |𝟙A^(t)|ρα, and let V=t. Write vj+V for j[p]={1,2,,p} for the p distinct cosets vj+V={x𝔽pn:xt=j} of V. Then

𝟙A^(t)=fA^(t)=𝔼x𝔽pn(𝟙A(x)α)e(xtp)=𝔼j[p]𝔼xvj+V(𝟙A(x)α)e(jp)=𝔼j[p](|A(vj+V)||vj+V|α=aj)e(jp)

By triangle inequality, 𝔼j[p]|aj|ρα. But note that 𝔼j[p]aj=0 so 𝔼j[p]aj+|aj|ρα, hence there exists j[p] such that aj+|aj|ρα. Then ajρα2.

Notation. Given f,g,h:G, write

T3(f,g,h)=𝔼x,dGf(x)g(x+d)h(x+2d).

Notation. Given AG, write

2A={2a:aA},

to be distinguished from 2A=A+A={a+a:a,aA}.

Lemma 2.18. Assuming that:

  • p3 prime

  • A𝔽pn of density α>0

  • supt0|𝟙A^(t)|𝜀

Then the number of 3-term arithmetic progressions in A differs from α3(pn)2 by at most 𝜀(pn)2.

Proof. The number of 3-term arithmetic progressions in A is (pn)2 times

T3(𝟙A,𝟙A,𝟙A)=𝔼x,d𝔽pn𝟙A(x)𝟙(x+d)𝟙A(x+2d)=𝔼x,y𝔽pn𝟙A(x)𝟙A(y)𝟙A(2yx)=𝔼yG𝟙A(y)𝔼xG𝟙A(x)𝟙A(2yx)=𝔼yG𝟙A(y)𝟙A𝟙A(2y)=𝟙2A,𝟙A𝟙A

By Plancherel’s identity and Lemma 2.13, we have

=𝟙2A^,𝟙A^2=t𝟙2A^(t)𝟙A^(t)2¯=α3+t0𝟙2A^(t)𝟙A^(t)2¯

but

|t0𝟙2A^(t)𝟙A^(t)2|supt0|𝟙A^(t)|t0|𝟙2A^(t)||𝟙A^(t)|CSsupt0|𝟙A^(t)|(t|𝟙2A^(t)|2t|𝟙A^(t)|2)12𝜀𝟙2A^2𝟙A^2=𝜀α

by Parseval’s identity.

Theorem 2.19 (Meshulam’s Theorem). Assuming that:

  • A𝔽pn a set containing no non-trivial 3 term arithmetic progressions

Then |A|=O(pnlogpn).

Proof. By assumption,

T3(𝟙A,𝟙A,𝟙A)=|A|(pn)2=αpn.

But as in (the proof of) Lemma 2.18,

|T3(𝟙A,𝟙A,𝟙A)α3|supt0|𝟙A^(t)|α,

so provided pn2α2, i.e. T3(𝟙A,𝟙A,𝟙A)α32 we have supt0|𝟙A^(t)|α22.

So by Lemma 2.17 with ρ=α2, there exists V𝔽pn of codimension 1 and x𝔽pn such that |A(x+V)|(α+α24)|V|.

We iterate this observation: let A0=A, V0=𝔽pn, α0=|A0||V0|. At the i-th step, we are given a set Ai1Vi1 of density αi1 with no non-trivial 3 term arithmetic progressions. Provided that pdim(Vi1)2αi12, there exists ViVi1 of codimension 1, xiVi1 such that

|(Axi)Vi|(αi1+(αi1)24)|Vi|.

Set Ai=(Axi)ViVi, has density αi1+(αi1)24, and is free of non-trivial 3 term arithmetic progressions.

Through this iteration, the density increases from α to 2α in at most α(α24)=4α1 steps.

2α to 4α in at most 2α((2α)24)=2α1 steps and so on.

So reaches 1 in at most

4α1(1+12+14+18+)8α1

steps. The argument must end with dim(Vi)n8α1, at which point we must have had pdim(Vi)<2αi122α2, or else we could have continued.

But we may assume that α2pn4 (or α2<2pn2) whence pn8α1pn2, or n22α1.

At the time of writing, the largest known subset of 𝔽3n containing no non-trivial 3 term arithmetic progressions has size (2.2202)n.

We will prove an upper bound of the form (2.756)n.

Theorem 2.20 (Roth’s theorem). Assuming that:

  • A[N]={1,,N}

  • A contains no non-trivial 3 term arithmetic progressions

Then |A|=O(Nlog log N).

Example 2.21 (Behrend’s example). There exists A[N] of size at least |A|exp(c log N)N containing no non-trivial 3 term arithmetic progressions.

Lemma 2.22. Assuming that:

  • A[N] of density α>0

  • N>50α2

  • A contains no non-trivial 3 term arithmetic progressions

  • p a prime in [N3,2N3]

  • let A=A[p]p

Then one of the following holds:
  • (i) supt0|𝟙A^(t)|α210 (where the Fourier coefficient is computed in p)
  • (ii) There exists an interval J[N] of length N3 such that |AJ|α(1+α400)|J|

Proof. We may assume that |A|=|A[p]|α(1α200)p since otherwise

|A[p+1,N]|αN(α(1α200)p)=α(Np)+α2200p(α+α2400)(Np)

so we would be in Case (ii) with J=[p+1,N]. Let A=A[p3,2p3]. Note that all 3 term arithmetic progressions of the form (x,x+d,x+2d)A×A×A are in fact arithmetic progressions in [N].

If |A[p3]| or |A[2p3,p]| were at least 25|A|, we would again be in case (ii). So we may assume that |A||A|5.

Now as in Lemma 2.18 and Theorem 2.19,

αp=|A|p2T3(𝟙A,𝟙A,𝟙A)=α(α)2+t𝟙A^(t)𝟙A ^(t)¯𝟙2A ^(t)

where α=|A|p and α=|A|p. So as before,

αα2supt0|𝟙A(t)|α ,

provided that αp12α(α)2, i.e. 2pαα. (Check this is satisfied).

Hence

supt0|𝟙A^(t)|αα 212(α(1α200))225α210.

Lemma 2.23. Assuming that:

  • m

  • φ:[m]p be given by xtx for some t0

  • 𝜀>0

Then there exists a partition of [m] into progressions Pi of length li[𝜀m2,𝜀m] such that
diam(φ(Pi))=maxx,yPi|φ(x)φ(y)|𝜀p

for all i.

Proof. Let u=m and consider 0,t,2t,,ut. By Pigeonhole, there exists 0v<wusuch that |wtvt|=|(wv)t|pu. Set s=wv, so |st|pu. Divide [m] into residue classes modulo s, each of which has size at least msm4. But each residue class can be divided into arithmetic progressions of the form a,a+s,,a+ds with 𝜀u2<d𝜀u. The diameter of the image of each progression under φ is |dst|dpu𝜀upu=𝜀p.

Lemma 2.24. Assuming that:

  • A[N] of density α>0

  • p a prime in [N3,2N3]

  • let A=A[p]p

  • |𝟙A^(t)|α220 for some t0

Then there exists a progression P[N] of length at least α2N500 such that |AP|α(1+α80)|P|.

Proof. Let 𝜀=α240π, and use Lemma 2.23 to partition [p] into progressions Pi of length

𝜀p2α240πN32α2N500

and diam(φ(Pi))𝜀p. Fix one xi from each of the Pi. Then

α220|fA^(t)|=|1pixPifA(x)e(xtp)|=1p|ixPifA(x)e(xitp)+ixPifA(x)(e(xtp)e(xitp))|1pi|xPifA(x)|+1pixPi|fA(x)||e(xtp)e(xitp)2π𝜀since |t(xxi)|𝜀p|

So

i|xPifA(x)|α240p.

Since fA has mean zero,

i(|xPifA(x)|+xPifA(x))α240p,

hence there exists i such that

|xPifA(x)|+xPifA(x)α280|Pi|

and so

xPifA(x)α2160|Pi|.

Definition 2.25 (Bohr set). Let ΓG^ and ρ>0. By the Bohr set B(Γ,ρ) we mean the set

B(Γ,ρ)={xG:|γ(x)1|<ρ γΓ}.

We call |Γ| the rank of B(Γ,ρ), and ρ its width or radius.

Example 2.26. When G=𝔽pn, then B(Γ,ρ)=Γ for all sufficiently small ρ.

Lemma 2.27. Assuming that:

  • ΓG^ of size d

  • ρ>0

Then
|B(Γ,ρ)|(ρ8)d|G|.

Proposition 2.28 (Bogolyubov in a general finite abelian group). Assuming that:

  • AG of density α>0

Then there exists ΓG^ of size at most 2α2 such that A+AAAB(Γ,ρ).

Proof. Recall 𝟙A𝟙A𝟙A𝟙A(x)=γG^|𝟙A^(γ)|4γ(x).

Let ΓSpecα2(𝟙A), and note that, for xB(Γ,12) and γΓ, Re(γ(x))>0. Hence, for xB(Γ,12),

ReγG^|𝟙A^(γ)|4γ(x)=ReγΓ|𝟙A^(γ)|4γ(x)α4+ReγΓ|𝟙A^(γ)|4γ(x)

and

|ReγΓ|𝟙A^(γ)|4γ(x)|supγΓ|𝟙A^(γ)|2γΓ|𝟙A^(γ)|2(α2α)2α=α42.