Resolution of the Cap Set Problem: ProgressionFree Subsets of Fn

q are exponentially small

Jordan S. Ellenberg, Dion GijswijtPresented by: Omri Ben-Eliezer

Tel-Aviv University

January 25, 2017

The Cap Set Problem

A ⊆ Fnq is a cap set if it contains no arithmetic progressions, i.e.

no distinct a, b, c ∈ A with b = (a + c)/2.In other words: no nontrivial solutions for a− 2b + c = 0 in A.

Equivalent definitions in Fn3:

I A is a cap set if it contains no lines.

I A is a cap set if a + b + c 6= 0 for any three distinct elementsa, b, c ∈ A.

What is the maximal size of a cap set A ⊆ Fnq?

Denote this size by r(Fnq).

Background

In Zn, maximal size of a set with no arithmetic progression isn1−o(1) [Behrend 1946, Roth 1953]. What about Fn

3, and Fnq in

general?

I Brown, Buhler [1982]: r(Fn3) = o(3n) (using regularity lemma)

I Meshulam [1995]: r(Fn3) = O(3n/n) (using Fourier analysis)

I Tao [2007]: ”Perhaps my favorite open question”.

I Bateman, Katz [2012]: r(Fn3) = o(3n/n1+ε) (complicated

Fourier arguments)

I Croot, Lev, Pach [2016]: r(Fn4) < 3.62n

I Ellenberg, Gijswijt [2016]: r(Fnq) < cn with c < q, for any

finite field Fq (polynomial method)

Background

In Zn, maximal size of a set with no arithmetic progression isn1−o(1) [Behrend 1946, Roth 1953]. What about Fn

3, and Fnq in

general?

I Ellenberg, Gijswijt [2016]: r(Fnq) < cn with c < q, for any

prime q (polynomial method)

Example: q = 3

Upper bound (Ellenberg, Gijswijt 2016) r(Fn3) = O(2.756n)

Trivial lower bound r(Fn3) ≥ 2n

Best lower bound (Edel 2004) r(Fn3) = Ω(2.217n)

Definitions

Mn Set of monomials in x1, . . . , xn with degree ≤ q − 1in each variable. There are qn monomials.

Sn Span of monomials from Mn over Fq. Linear spaceof dimension qn over Fq.

e : Sn → FFnq

q Evaluation map: each polynomial is mapped to theset of values it has for any assignment of n elementsin Fq to its variables.This is an isomorphism: the polynomial∏n

i=1(1− (xi − ai )q−1) is mapped to the indicator of

a = (a1, . . . , an) ∈ Fnq.

Definitions

Mdn Set of monomials from Mn with total degree ≤ d in

each variable.

Sdn Subspace of Sn spanned by Md

n .

md Dimension of Sdn over Fq. Equivalently, md = |Md

n |.

Structural lemma on low-degree polynomials

LemmaLet Fq be a finite field and let A ⊆ Fn

q. Let α, β, γ ∈ Fq with

α + β + γ = 0. Suppose P ∈ Sdn satisfies P(αa + βb) = 0 for

every a 6= b ∈ A. Then the number of a ∈ A with P(−γa) 6= 0 isat most 2md/2.

RemarkCroot, Lev and Pach proved a special case of this lemma: if(α, β, γ) = (1,−1, 0) and |A| > 2md/2, then P(0) = 0. Thestronger result above is needed for our proof.

Structural lemma on low-degree polynomials

LemmaLet Fq be a finite field and let A ⊆ Fn

q. Let α, β, γ ∈ Fq with

α + β + γ = 0. Suppose P ∈ Sdn satisfies P(αa + βb) = 0 for

every a 6= b ∈ A. Then the number of a ∈ A with P(−γa) 6= 0 isat most 2md/2.

ProofLet B be the A× A matrix whose a, b entry is P(αa + βb). ThenB is diagonal.Since P(−γa) = P(αa + βa) is the a, a entry of B, we need tobound the number of non-zero entries on the diagonal. This is therank of B.

Structural lemma on low-degree polynomials

LemmaLet Fq be a finite field and let A ⊆ Fn

q. Let α, β, γ ∈ Fq with

α + β + γ = 0. Suppose P ∈ Sdn satisfies P(αa + βb) = 0 for

every a 6= b ∈ A. Then the number of a ∈ A with P(−γa) 6= 0 isat most 2md/2.

ProofSince P ∈ Sd

n we can write

P(αx + βy) =∑

m,m′∈Mdn

deg(mm′)≤d

cm,m′m(x)m′(y)

In each summand one of m, m′ has degree at most d/2.

Structural lemma on low-degree polynomials

LemmaLet Fq be a finite field and let A ⊆ Fn

q. Let α, β, γ ∈ Fq with

α + β + γ = 0. Suppose P ∈ Sdn satisfies P(αa + βb) = 0 for

every a 6= b ∈ A. Then the number of a ∈ A with P(−γa) 6= 0 isat most 2md/2.

ProofWe can write (maybe not uniquely)

P(αx + βy) =∑

m∈Md/2n

m(x)Fm(y) +∑

m′∈Md/2n

m′(y)Gm′(x)

Put x = a and y = b. Then we get

Bab = P(αa + βb) =∑

m∈Md/2n

m(a)Fm(b) +∑

m′∈Md/2n

m′(b)Gm′(a)

Structural lemma on low-degree polynomials

LemmaLet Fq be a finite field and let A ⊆ Fn

q. Let α, β, γ ∈ Fq with

α + β + γ = 0. Suppose P ∈ Sdn satisfies P(αa + βb) = 0 for

every a 6= b ∈ A. Then the number of a ∈ A with P(−γa) 6= 0 isat most 2md/2.

ProofFor any m ∈ M

d/2n Define A× A matrices Cm and Dm as follows:

Cmab = m(a)Fm(b) Dm

ab = m(b)Gm(a)

Each Cm (and each Dm) is a matrix of rank 1: all of its columnsare multiples of the vector (m(a))a∈A.

Structural lemma on low-degree polynomials

LemmaLet Fq be a finite field and let A ⊆ Fn

q. Let α, β, γ ∈ Fq with

α + β + γ = 0. Suppose P ∈ Sdn satisfies P(αa + βb) = 0 for

every a 6= b ∈ A. Then the number of a ∈ A with P(−γa) 6= 0 isat most 2md/2.

ProofFor (a, b) ∈ A× A we have

Bab =∑

m∈Md/2n

m(a)Fm(b) +∑

m′∈Md/2n

m′(b)Gm′(a)

=∑

m∈Md/2n

Cmab +

∑m′∈Md/2

n

Dm′ab

Structural lemma on low-degree polynomials

LemmaLet Fq be a finite field and let A ⊆ Fn

q. Let α, β, γ ∈ Fq with

α + β + γ = 0. Suppose P ∈ Sdn satisfies P(αa + βb) = 0 for

every a 6= b ∈ A. Then the number of a ∈ A with P(−γa) 6= 0 isat most 2md/2.

Proof

B =∑

m∈Md/2n

Cm +∑

m′∈Md/2n

Dm′

So B is the sum of 2md/2 matrices of rank 1.Thus, B has rank at most 2md/2, and we are finished.

Main Theorem

TheoremLet α, β, γ ∈ Fq with α + β + γ = 0 and γ 6= 0. Let A ⊆ Fn

q suchthat the equation

αa1 + βa2 + γa3 = 0

has no solutions (a1, a2, a3) ∈ A3 other than those witha1 = a2 = a3. Then |A| ≤ 3m(q−1)n/3.

Remark: A cap set A satisfies the above requirements withα = γ = 1, β = −2.

Main Theorem

TheoremLet α, β, γ ∈ Fq with α + β + γ = 0 and γ 6= 0. Let A ⊆ Fn

q suchthat the equation

αa1 + βa2 + γa3 = 0

has no solutions (a1, a2, a3) ∈ A3 other than those witha1 = a2 = a3. Then |A| ≤ 3m(q−1)n/3.

ProofLet d be an integer in [0, (q − 1)n] and let V be the subspace ofpolynomials in Sd

n vanishing on the complement of −γA:

∀P ∈ V ∀x /∈ (−γA) : P(x) = 0

Proof strategy: we will bound dimV from above and below, with|A| appearing in the lower bound. This gives upper bound for |A|.

Main Theorem

TheoremLet α, β, γ ∈ Fq with α + β + γ = 0 and γ 6= 0. Let A ⊆ Fn

q suchthat the equation

αa1 + βa2 + γa3 = 0

has no solutions (a1, a2, a3) ∈ A3 other than those witha1 = a2 = a3. Then |A| ≤ 3m(q−1)n/3.

Proof

I The number of monomials of degree > d is the same as thenumber of monomials of degree < n(q − 1)− d : mn(q−1)−d .

I The dimension of the space of polynomials vanishing outside−γA is |A|: it is spanned by the indicator polynomials of allelements in −γA.

I The dimension of V is ≥ |A| −mn(q−1)−d .

Main Theorem

TheoremLet α, β, γ ∈ Fq with α + β + γ = 0 and γ 6= 0. Let A ⊆ Fn

q suchthat the equation

αa1 + βa2 + γa3 = 0

has no solutions (a1, a2, a3) ∈ A3 other than those witha1 = a2 = a3. Then |A| ≤ 3m(q−1)n/3.

Proof

I Let S(A) ⊆ Fq be the set of elements αa1 + βa2 witha1 6= a2 ∈ A. S(A) is disjoint from −γA.

I Thus, any P ∈ V vanishes on S(A).

I By Lemma, P(−γa) 6= 0 for at most 2md/2 points a ∈ A.

I P is in V so it vanishes outside −γA.

Main Theorem

TheoremLet α, β, γ ∈ Fq with α + β + γ = 0 and γ 6= 0. Let A ⊆ Fn

q suchthat the equation

αa1 + βa2 + γa3 = 0

has no solutions (a1, a2, a3) ∈ A3 other than those witha1 = a2 = a3. Then |A| ≤ 3m(q−1)n/3.

Proof

I Define support of a function f as supp(f ) = x : f (x) 6= 0.I Let P ∈ V with support Σ of maximal size. Then|Σ| ≥ dimV : the subspace of polynomials in V vanishing onΣ has dimension ≥ dimV − |Σ|; but if Q ∈ V is a nonzeropolynomial in this subspace, then P + Q ∈ V has largersupport, a contradiction.

Main Theorem

TheoremLet α, β, γ ∈ Fq with α + β + γ = 0 and γ 6= 0. Let A ⊆ Fn

q suchthat the equation

αa1 + βa2 + γa3 = 0

has no solutions (a1, a2, a3) ∈ A3 other than those witha1 = a2 = a3. Then |A| ≤ 3m(q−1)n/3.

Proof

I Thus |A| −m(q−1)n−d ≤ dimV ≤ |Σ| ≤ 2md/2, orequivalently |A| ≤ 2md/2 + m(q−1)n−d .

I Picking d = 2(q−1)n3 , we get |A| ≤ 3m(q−1)n/3 as desired.

Bounding m(q−1)n/3

m(q−1)n/3 is exponentially small with respect to qn with q fixedand n large. Most monomials have degree around (q-1)n/2.

Indeed, let Xi for i = 1, . . . , n be i.i.d. random variables takingvalues from 0, 1, . . . , q − 1 uniformly at random. Therefore

m(q−1)n/3

qn= P

(1

n

n∑i=1

Xi ≤ (q − 1)/3

)

This is a large deviation problem. Original proof uses a generalmethod of Cramer for such problems.However, we will see a more elementary method, proposed by Taowho gave an entirely different proof for the theorem in his blog(recommended!). For example, take q = 3.

Bounding m2n/3 over F3

We can count monomials of total degree ≤ d and degree ≤ 2 ineach variable as follows: choose a, b, c ≥ 0 such thata + b + c = n. Pick a, b, c variables to be of degree 0, 1, 2respectively, where the degree of the monomial is b+ 2c ≤ d . Thus

m2n/3 =∑

a,b,c≥0: a+b+c=n,b+2c≤2n/3

n!

a!b!c!

Write a = αn, b = βn, c = γn. Stirling formula gives

n!

a!b!c!∼ nn

aabbcc=

(1

ααββγγ

)n

= exp (n · h(α, β, γ))

Where h is the entropy function

h(α, β, γ) = α log1

α+ β log

1

β+ γ log

1

γ

Bounding m2n/3 over F3

Up to polynomial factor, m2n/3 is equal to

exp (n ·max h(α, β, γ))

where the maximum is taken over all

α, β, γ ≥ 0; α + β + γ = 1; β + 2γ ≤ 2/3

A standard Lagrange multiplier computation implies that themaximal h(α, β, γ) under the above constraints is ∼ 1.013455,implying that |A| ≤ 3m2n/3 = O(2.756n).