4 Further Topics

In 𝔽pn, we can do much better.

Theorem 4.1 (Ellenberg-Gijswijt, following Croot-Lev-Pach). Assuming that:

  • A𝔽3n contains no non-trivial 3 term arithmetic progressions

Then |A|=o(2.756)n.

Notation. Let Mn be the set of monomials in x1,,x2 whose degree in each variable is at most 2. Let Vn be the vector space over 𝔽3 whose basis is Mn. For any d[0,2n], write Mnd for the set of monomials in Mn of (total) degree at most d, and Vnd for the corresponding vector space. Set md=dim(Vnd)=|Mnd|.

Lemma 4.2. Assuming that:

  • A𝔽3n

  • PVnd is a polynomial

  • P(a+a)=0 for all aaA

Then
|{aA:P(2a)0}|2md2.

Proof. Every PVnd can be written as a linear combination of monomials in Mnd, so

P(x+y)=m,mMnddeg(mm)dcm,mm(x)m(y)

for some coefficients cm,m. Clearly at least one of m,m must have degree d2, whence

P(x+y)=mMnd2m(x)Fm(y)+mMnd2m(y)Gm(x),

for some families of polynomials (Fm)mMnd2, (Gm)mMnd2.

Viewing (P(x+y))x,yA as a |A|×|A|-matrix C, we see that C can be written as the sum of at most 2md2 matrices, each of which has rank 1. Thus rank(C)2md2. But by assumption, C is a diagonal matrix whose rank equals |{aA:P(a+a)0}|.

Proposition 4.3. Assuming that:

  • A𝔽3n a set containing no non-trivial 3 term arithmetic progressions

Then |A|3m2n3.

Proof. Let d[0,2n] be an integer to be determined later. Let W be the space of polynomials in Vnd that vanish on (2A)c. We have

dim(W)dim(Vnd)|(2A)c|=md(3n|A|).

We claim that there exists PW such that |supp(P)|dim(W). Indeed, pick PW with maximal support. If |supp(P)|<dim(W), then there would be a non-zero polynomial QW vanishing on supp(P), in which case supp(P+Q)supp(P), contradicting the choice of P.

PIC

Now by assumption,

{a+a:aaA}2A=.

So any polynomial that vanishes on (2A)c vanishes on {a+a:aaA}. By Lemma 4.2 we now have that,

|A|(3nmd)=md(3n|A|)dim(W)|supp(P)|=|{x𝔽3n:P(x)0}|=|{aA:P(2a)0}|2md2

Hence |A|3nmd+2md2. But the monomials in MnMnd are in bijection with the ones in M2nd via x1α1xnαnx12α1xn2αn, whence 3nmd=m2nd. Thus setting d=4n3, we have |A|m2n3+2m2n3=3m2n3.

You will prove Theorem 4.1 on Example Sheet 3.

We do not have at present a comparable bound for 4 term arithmetic progressions. Fourier techniques also fail.

Example 4.4. Recall from Lemma 2.18 that given AG,

|T3(𝟙A,𝟙A,𝟙A)α3|supγ1|𝟙A^(γ)|.

But it is impossible to bound

T4(𝟙A,𝟙A,𝟙A,𝟙A)α4=𝔼xd𝟙A(x)𝟙A(x+d)𝟙A(x+2d)𝟙A(x+3d)α4

by supγ1|𝟙A^(γ)|. Indeed, consider Q={x𝔽pn:xx=0}. By Problem 11(ii) on Sheet 1,

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

and

supt0|𝟙Q^(t)|=O(pn2).

But given a 3 term arithmetic progression x,x+d,x+2dQ, by the identity

x23(x+d)2+3(x+2d)2(x+3d)2=0x,d,

x+3d automatically lies in Q, so

T4(𝟙A,𝟙A,𝟙A,𝟙A)=T3(𝟙A,𝟙A,𝟙A)=(1p)3+O(pn2)

which is not close to (1p)4.

Definition 4.5 (U2-norm). Given f:G, define its U2-norm by the formula

fU2(G)4=𝔼x,a,bGf(x)f(x+a)f(x+b)¯f(x+a+b).

Problem 1(i) on Sheet 2 showed that fU2(G)=f^l4(G^), so this is indeed a norm.

Problem 1(ii) asserted the following:

Lemma 4.6. Assuming that:

  • f1,f2,f3:G

Then
|T3(f1,f2,f3)|mini[3]fiU2(G)jifjL(G).

Note that

supγG^|f^(γ)|4γG^|f^(γ)|4supγG^|f^(γ)|2γG^|f^(γ)|2

and thus by Parseval’s identity,

fU2(G)4=f^l(G^)4f^l(G^)2fL2(G)2.

Hence

f^l(G^)f^l4(G^)=fU2(G)f^l(G^)12fL2(G)12.

Moreover, if f=fAA=𝟙Aα, then

T3(f,f,f)=T3(𝟙Aα,𝟙Aα,𝟙Aα)=T3(𝟙A,𝟙A,𝟙A)α3.

We may therefore reformulate the first step in the proof of Meshulam’s Theorem as follows: if pn2α2, then by Lemma 4.6,

α32|αpnα3|=|T3(fAA,fAA,fAA)|fAAU2(𝔽pn).

It remains to show that if fAAU2(𝔽pn) is non-trivial, then there exists a subspace V𝔽pn of bounded codimension on which A has increased density.

Theorem 4.7 (U2 Inverse Theorem). Assuming that:

  • f:𝔽pn

  • fL(𝔽pn)1

  • δ>1

  • fU2(𝔽pn)δ

Then there exists b𝔽pn such that
|𝔼x𝔽pnf(x)e(xbp)|δ2.

In other words, |f,ϕ|δ2 for ϕ(x)=e(xbp) and we say “f correlates with a linear phase function”.

Proof. We have seen that

fU2(𝔽pn)2f^l(𝔽pn^)fL2(𝔽pn)f^l(𝔽pn^),

so

δ2f^l(𝔽pn^)=supt𝔽pn^|𝔼xf(x)e(xtp)|.

PIC

Definition 4.8 (U3 norm). Given f:G, define its U3 norm by

fU3(G)8:=𝔼x,a,b,cf(x)f(x+a)f(x+b)f(x+c)¯f(x+a+b)f(x+b+c)f(x+a+c)f(x+a+b+c)¯=𝔼x,h1,h2,h3G𝜀{0,1}3C|𝜀|f(x+𝜀h)

where Cg(x)=g(x)¯ and |𝜀| denotes the number of ones in 𝜀.

It is easy to verify that 𝔼cGΔcfU2(G)4 where Δcg(x)=g(x)g(x+c)¯.

Definition 4.9 (U3 inner product). Given functions f𝜀:G for 𝜀{0,1}3, define their U3 inner product by

(f𝜀)𝜀{0,1}3U3(G)=𝔼x,h1,h2,h3G𝜀{0,1}3C|𝜀|f𝜀(x+𝜀h).

Observe that f,f,f,f,f,f,f,fU3(G)=fU3(G)8.

Lemma 4.10 (Gowers–Cauchy–Schwarz Inequality). Assuming that:

  • f𝜀:G, 𝜀{0,1}3

Then
|(f𝜀)𝜀{0,1}3U3(G)𝜀{0,1}3f𝜀U3(G).

Setting f𝜀=f for 𝜀{0,1}2×{0} and f𝜀=1 otherwise, it follows that fU2(G)4fU3(G)4 hence fU2(G)fU3(G).

Proposition 4.11. Assuming that:

  • f1,f2,f3,f4:𝔽5n

Then
T4(f1,f2,f3,f4)mini[4]fiU3(G)jifjL(𝔽5n).

Proof. We additionally assume f=f1=f2=f3=f4 to make the proof easier to follow, but the same ideas are used for the general case. We additionally assume fL(𝔽5n)1, by rescaling, since the inequality is homogeneous.

Reparametrising, we have

T4(f,f,f,f)=𝔼a,b,c,d𝔽5nf(3a+2b+c)f(2a+bd)f(ac2d)f(b2c3d)|T4(f,f,f,f)|8(𝔼a,b,c|𝔼df(2a+bd)f(ac2d)f(b2c3d)|2)4=(𝔼d,d𝔼a,bf(2a+b+d)f(2a+bd)¯𝔼cf(ac2d)f(ac2d)¯f(b2c3d)f(b2c3d)¯)4(𝔼d,d𝔼a,b|𝔼cf(ac2d)f(ac2d)¯f(b2c3d)f(b2c3d)¯|2)2=(𝔼c,c,d,d𝔼af(ac2d)f(ac2d)¯f(ac2d)¯f(ac2d)𝔼bf(b2c3d)f(b2c3d)¯f(b2c3d)¯f(b2c3d))2𝔼c,c,d,d,a|𝔼bf(b2c3d)f(b2c3d)¯f(b2c3d)¯f(b2c3d)|2=𝔼b,b,c,c,d,df(b2c3d)f(b2c3d)¯f(b2c3d)¯f(b2c3d)f(b2c3d)¯f(b2c3d)f(b2c3d)f(b2c3d)¯

Theorem 4.12 (Szemeredi’s Theorem for 4-APs). Assuming that:

  • A𝔽5n a set containing no non-trivial 4 term arithmetic progressions

Then |A|=o(5n).

Idea: By Proposition 4.11 with f=fA=𝟙Aα,

T4(𝟙AfA+α,𝟙AfA+α,𝟙AfA+α,𝟙AfA+α)α4=T4(fA,fA,fA,fA)+

where consists of 14 other terms in which between one and three of the inputs are equal to fA.

These are controlled by

fAU2(𝔽5n)fAU3(G),

whence

|T4(𝟙A,𝟙A,𝟙A,𝟙A)α4|15fAU3(G).

So if A contains no non-trivial 4 term arithmetic progressions and 5n>2α3, then fAU3(G)α430.

What can we say about functions with large U3 norm?

Example 4.13. Let M be an n×n symmetric matrix with entries in 𝔽5. Then f(x)=e(xMx5) satisfies fU3(G)=1.

Theorem 4.14 (U3 inverse theorem). Assuming that:

  • f:𝔽5n

  • fL(𝔽5n)1

  • fU3(G)δ for some δ>0

Then there exists a symmetric n×n matrix M with entries in 𝔽5 and b𝔽5n such that
|𝔼xf(x)e((xMx+bx)p)|c(δ)

where c(δ) is a polynomial in δ. In other words, |f,ϕ|c(δ) for ϕ(x)=e((xMx+bx)p) and we say “f correlates with a quadratic phase function”.

Proof (sketch). Let Δhf(x) denote f(x)f(x+h)¯.

fU3(G)=(𝔼hΔhfU24)18.

  • STEP 1: Weak linearity. See reference.
  • STEP 2: Strong linearity. We will spend the rest of the lecture discussing this in detail.
  • STEP 3: Symmetry argument. Problem 8 on Sheet 3.
  • STEP 4: Integration step. Problem 9 on Sheet 3.

STEP 1: If fU3(G)8=𝔼hΔhU24δ8, then for at least a δ82-proportion of h𝔽5n, δ82ΔhfU24Δhf^l2. So for each such h𝔽5n, there exists th such that |Δhf^(th)|2δ82.

Proposition 4.15. Assuming that:

  • f:𝔽5n

  • f1

  • fU3(G)δ

  • |𝔽5n|=Ωδ(1)

Then there exists S𝔽5n with |S|=Ωδ(|𝔽5n|) and a function ϕ:S𝔽5n^ such that
  • (i) |Δhf^(ϕ(h))|=Ωδ(1);
  • (ii) There are at least Ωδ(|𝔽5n|3) quadruples (s1,s2,s3,s4)S4 such that s1+s2=s3+s4 and ϕ(s1)+ϕ(s2)+ϕ(s4).

STEP 2: If S and ϕ are as above, then there is a linear function ψ:𝔽5n𝔽5n^ which coincides with ϕ for many elements of S.

Proposition 4.16. Assuming that:

Then there exists n×n matrix M with entries in 𝔽5 and b𝔽5n such that ψ(x)=Mx+b (ψ:𝔽5n𝔽5n^) satisfies ψ(x)=ϕ(x) for Ωδ(|𝔽5n|) elements xS.

Proof. Consider the graph of ϕ, Γ={(h,ϕ(h)):hS}𝔽5n×𝔽5n^. By Proposition 4.15, Γ has Ωδ(|𝔽5n|3) additive quadruples.

By Balog–Szemeredi–Gowers, Schoen, there exists ΓΓ with |Γ|=Ωδ(|Γ|)=Ωδ(|𝔽5n|) and |Γ+Γ|=Oδ(|Γ|). udefine SS by Γ={(h,ϕ(h)):hS} and note |S|=Ωδ(|𝔽5n|).

By Freiman-Ruzsa applied to Γ𝔽5n×𝔽5n^, there exists a subspace H𝔽5n×𝔽5n^ with |H|=Oδ(|Γ|)=Oδ(|𝔽5n|) such that ΓH.

Denote by π:𝔽5n×𝔽5n^𝔽5n the projection onto the first n coordinates. By construction, π(H)S. Moreover, since |S|=Ωδ(|𝔽5n|),

|ker(π|H)|=|H||Im(π|H)|=Oδ(|𝔽5n|)|S|=Oδ(1).

We may thus partition H into Oδ(1) cosets of some subspace H such that π|H is injective on each coset. By averaging, there exists a coset x+H such that

|Γ(x+H)|=Ωδ(|Γ|)=Ωδ(|𝔽5n|).

Set Γ=Γ(x+H), and define S accordingly.

Now π|x+ is injective and surjective onto V:=Im(π|x+H). This means there is an affine linear map ψ:V𝔽5n^ such that (h,ψ(h))Γ for all hS.

Then do steps 3 and 4.

What to do if you have lots of additive quadruples?

Balog-Szemeredi-Gowers

What to do if you have small doubling constant?

Freiman-Ruzsa (or Polynomial-Freiman-Ruzsa)