Resolution of the Cap Set Problem: Progression Free Subsets of F n q are exponentially small

Transcription

1 Resolution of the Cap Set Problem: Progression Free Subsets of F n q are exponentially small Jordan S. Ellenberg, Dion Gijswijt Presented by: Omri Ben-Eliezer Tel-Aviv University January 25, 2017

2 The Cap Set Problem A F n q 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 F n 3 : A is a cap set if it contains no lines. A is a cap set if a + b + c 0 for any three distinct elements a, b, c A. What is the maximal size of a cap set A F n q? Denote this size by r(f n q).

3 Background In Z n, maximal size of a set with no arithmetic progression is n 1 o(1) [Behrend 1946, Roth 1953]. What about F n 3, and Fn q in general? Brown, Buhler [1982]: r(f n 3 ) = o(3n ) (using regularity lemma) Meshulam [1995]: r(f n 3 ) = O(3n /n) (using Fourier analysis) Tao [2007]: Perhaps my favorite open question. Bateman, Katz [2012]: r(f n 3 ) = o(3n /n 1+ɛ ) (complicated Fourier arguments) Croot, Lev, Pach [2016]: r(f n 4 ) < 3.62n Ellenberg, Gijswijt [2016]: r(f n q) < c n with c < q, for any finite field F q (polynomial method)

4 Background In Z n, maximal size of a set with no arithmetic progression is n 1 o(1) [Behrend 1946, Roth 1953]. What about F n 3, and Fn q in general? Ellenberg, Gijswijt [2016]: r(f n q) < c n with c < q, for any prime q (polynomial method) Example: q = 3 Upper bound (Ellenberg, Gijswijt 2016) r(f n 3 ) = O(2.756n ) Trivial lower bound r(f n 3 ) 2n Best lower bound (Edel 2004) r(f n 3 ) = Ω(2.217n )

5 Definitions M n Set of monomials in x 1,..., x n with degree q 1 in each variable. There are q n monomials. S n e : S n F Fn q q Span of monomials from M n over F q. Linear space of dimension q n over F q. Evaluation map: each polynomial is mapped to the set of values it has for any assignment of n elements in F q to its variables. This is an isomorphism: the polynomial n i=1 (1 (x i a i ) q 1 ) is mapped to the indicator of a = (a 1,..., a n ) F n q.

6 Definitions M d n Set of monomials from M n with total degree d in each variable. S d n Subspace of S n spanned by M d n. m d Dimension of S d n over F q. Equivalently, m d = M d n.

7 Structural lemma on low-degree polynomials Lemma Let F q be a finite field and let A F n q. Let α, β, γ F q with α + β + γ = 0. Suppose P S d n satisfies P(αa + βb) = 0 for every a b A. Then the number of a A with P( γa) 0 is at most 2m d/2. Remark Croot, Lev and Pach proved a special case of this lemma: if (α, β, γ) = (1, 1, 0) and A > 2m d/2, then P(0) = 0. The stronger result above is needed for our proof.

8 Structural lemma on low-degree polynomials Lemma Let F q be a finite field and let A F n q. Let α, β, γ F q with α + β + γ = 0. Suppose P S d n satisfies P(αa + βb) = 0 for every a b A. Then the number of a A with P( γa) 0 is at most 2m d/2. Proof Let B be the A A matrix whose a, b entry is P(αa + βb). Then B is diagonal. Since P( γa) = P(αa + βa) is the a, a entry of B, we need to bound the number of non-zero entries on the diagonal. This is the rank of B.

9 Structural lemma on low-degree polynomials Lemma Let F q be a finite field and let A F n q. Let α, β, γ F q with α + β + γ = 0. Suppose P S d n satisfies P(αa + βb) = 0 for every a b A. Then the number of a A with P( γa) 0 is at most 2m d/2. Proof Since P S d n we can write P(αx + βy) = m,m M d n deg(mm ) d c m,m m(x)m (y) In each summand one of m, m has degree at most d/2.

10 Structural lemma on low-degree polynomials Lemma Let F q be a finite field and let A F n q. Let α, β, γ F q with α + β + γ = 0. Suppose P S d n satisfies P(αa + βb) = 0 for every a b A. Then the number of a A with P( γa) 0 is at most 2m d/2. Proof We can write (maybe not uniquely) P(αx + βy) = m M d/2 n m(x)f m (y) + Put x = a and y = b. Then we get B ab = P(αa + βb) = m(a)f m (b) + m M d/2 n m M d/2 n m M d/2 n m (y)g m (x) m (b)g m (a)

11 Structural lemma on low-degree polynomials Lemma Let F q be a finite field and let A F n q. Let α, β, γ F q with α + β + γ = 0. Suppose P S d n satisfies P(αa + βb) = 0 for every a b A. Then the number of a A with P( γa) 0 is at most 2m d/2. Proof For any m Mn d/2 Define A A matrices C m and D m as follows: C m ab = m(a)f m(b) D m ab = m(b)g m(a) Each C m (and each D m ) is a matrix of rank 1: all of its columns are multiples of the vector (m(a)) a A.

12 Structural lemma on low-degree polynomials Lemma Let F q be a finite field and let A F n q. Let α, β, γ F q with α + β + γ = 0. Suppose P S d n satisfies P(αa + βb) = 0 for every a b A. Then the number of a A with P( γa) 0 is at most 2m d/2. Proof For (a, b) A A we have B ab = m M d/2 n = m M d/2 n m(a)f m (b) + C m ab + m M d/2 n m M d/2 n D m ab m (b)g m (a)

13 Structural lemma on low-degree polynomials Lemma Let F q be a finite field and let A F n q. Let α, β, γ F q with α + β + γ = 0. Suppose P S d n satisfies P(αa + βb) = 0 for every a b A. Then the number of a A with P( γa) 0 is at most 2m d/2. Proof B = m M d/2 n C m + m M d/2 n D m So B is the sum of 2m d/2 matrices of rank 1. Thus, B has rank at most 2m d/2, and we are finished.

14 Main Theorem Theorem Let α, β, γ F q with α + β + γ = 0 and γ 0. Let A F n q such that the equation αa 1 + βa 2 + γa 3 = 0 has no solutions (a 1, a 2, a 3 ) A 3 other than those with a 1 = a 2 = a 3. Then A 3m (q 1)n/3. Remark: A cap set A satisfies the above requirements with α = γ = 1, β = 2.

15 Main Theorem Theorem Let α, β, γ F q with α + β + γ = 0 and γ 0. Let A F n q such that the equation αa 1 + βa 2 + γa 3 = 0 has no solutions (a 1, a 2, a 3 ) A 3 other than those with a 1 = a 2 = a 3. Then A 3m (q 1)n/3. Proof Let d be an integer in [0, (q 1)n] and let V be the subspace of polynomials in S d n vanishing on the complement of γa: P V x / ( γa) : P(x) = 0 Proof strategy: we will bound dim V from above and below, with A appearing in the lower bound. This gives upper bound for A.

16 Main Theorem Theorem Let α, β, γ F q with α + β + γ = 0 and γ 0. Let A F n q such that the equation αa 1 + βa 2 + γa 3 = 0 has no solutions (a 1, a 2, a 3 ) A 3 other than those with a 1 = a 2 = a 3. Then A 3m (q 1)n/3. Proof The number of monomials of degree > d is the same as the number of monomials of degree < n(q 1) d: m n(q 1) d. The dimension of the space of polynomials vanishing outside γa is A : it is spanned by the indicator polynomials of all elements in γa. The dimension of V is A m n(q 1) d.

17 Main Theorem Theorem Let α, β, γ F q with α + β + γ = 0 and γ 0. Let A F n q such that the equation αa 1 + βa 2 + γa 3 = 0 has no solutions (a 1, a 2, a 3 ) A 3 other than those with a 1 = a 2 = a 3. Then A 3m (q 1)n/3. Proof Let S(A) F q be the set of elements αa 1 + βa 2 with a 1 a 2 A. S(A) is disjoint from γa. Thus, any P V vanishes on S(A). By Lemma, P( γa) 0 for at most 2m d/2 points a A. P is in V so it vanishes outside γa.

18 Main Theorem Theorem Let α, β, γ F q with α + β + γ = 0 and γ 0. Let A F n q such that the equation αa 1 + βa 2 + γa 3 = 0 has no solutions (a 1, a 2, a 3 ) A 3 other than those with a 1 = a 2 = a 3. Then A 3m (q 1)n/3. Proof Define support of a function f as supp(f ) = {x : f (x) 0}. Let P V with support Σ of maximal size. Then Σ dim V : the subspace of polynomials in V vanishing on Σ has dimension dim V Σ ; but if Q V is a nonzero polynomial in this subspace, then P + Q V has larger support, a contradiction.

19 Main Theorem Theorem Let α, β, γ F q with α + β + γ = 0 and γ 0. Let A F n q such that the equation αa 1 + βa 2 + γa 3 = 0 has no solutions (a 1, a 2, a 3 ) A 3 other than those with a 1 = a 2 = a 3. Then A 3m (q 1)n/3. Proof Thus A m (q 1)n d dim V Σ 2m d/2, or equivalently A 2m d/2 + m (q 1)n d. Picking d = 2(q 1)n 3, we get A 3m (q 1)n/3 as desired.

20 Bounding m (q 1)n/3 m (q 1)n/3 is exponentially small with respect to q n with q fixed and n large. Most monomials have degree around (q-1)n/2. Indeed, let X i for i = 1,..., n be i.i.d. random variables taking values from 0, 1,..., q 1 uniformly at random. Therefore ( ) m (q 1)n/3 1 n q n = P X i (q 1)/3 n i=1 This is a large deviation problem. Original proof uses a general method of Cramér for such problems. However, we will see a more elementary method, proposed by Tao who gave an entirely different proof for the theorem in his blog (recommended!). For example, take q = 3.

21 Bounding m 2n/3 over F 3 We can count monomials of total degree d and degree 2 in each variable as follows: choose a, b, c 0 such that a + b + c = n. Pick a, b, c variables to be of degree 0, 1, 2 respectively, where the degree of the monomial is b + 2c d. Thus m 2n/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 a a b b c c = 1 n α α β β γ γ = exp (n h(α, β, γ)) Where h is the entropy function h(α, β, γ) = α log 1 α + β log 1 β + γ log 1 γ

22 Bounding m 2n/3 over F 3 Up to polynomial factor, m 2n/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 the maximal h(α, β, γ) under the above constraints is , implying that A 3m 2n/3 = O(2.756 n ).