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A symmetric formulation of the Croot-Lev-Pach-Ellenberg-Gijswijt capset bound

18 May, 2016 in expository, math.CO, math.RA | Tags: , , , , , , | by | 68 comments

A capset in the vector space {{\bf F}_3^n} over the finite field {{\bf F}_3} of three elements is a subset {A} of {{\bf F}_3^n} that does not contain any lines {\{ x,x+r,x+2r\}}, where {x,r \in {\bf F}_3^n} and {r \neq 0}. A basic problem in additive combinatorics (discussed in one of the very first posts on this blog) is to obtain good upper and lower bounds for the maximal size of a capset in {{\bf F}_3^n}.

Trivially, one has {|A| \leq 3^n}. Using Fourier methods (and the density increment argument of Roth), the bound of {|A| \leq O( 3^n / n )} was obtained by Meshulam, and improved only as late as 2012 to {O( 3^n /n^{1+c})} for some absolute constant {c>0} by Bateman and Katz. But in a very recent breakthrough, Ellenberg (and independently Gijswijt) obtained the exponentially superior bound {|A| \leq O( 2.756^n )}, using a version of the polynomial method recently introduced by Croot, Lev, and Pach. (In the converse direction, a construction of Edel gives capsets as large as {(2.2174)^n}.) Given the success of the polynomial method in superficially similar problems such as the finite field Kakeya problem (discussed in this previous post), it was natural to wonder that this method could be applicable to the cap set problem (see for instance this MathOverflow comment of mine on this from 2010), but it took a surprisingly long time before Croot, Lev, and Pach were able to identify the precise variant of the polynomial method that would actually work here.

The proof of the capset bound is very short (Ellenberg’s and Gijswijt’s preprints are both 3 pages long, and Croot-Lev-Pach is 6 pages), but I thought I would present a slight reformulation of the argument which treats the three points on a line in {{\bf F}_3} symmetrically (as opposed to treating the third point differently from the first two, as is done in the Ellenberg and Gijswijt papers; Croot-Lev-Pach also treat the middle point of a three-term arithmetic progression differently from the two endpoints, although this is a very natural thing to do in their context of {({\bf Z}/4{\bf Z})^n}). The basic starting point is this: if {A} is a capset, then one has the identity

\displaystyle \delta_{0^n}( x+y+z ) = \sum_{a \in A} \delta_a(x) \delta_a(y) \delta_a(z) \ \ \ \ \ (1)

 

for all {(x,y,z) \in A^3}, where {\delta_a(x) := 1_{a=x}} is the Kronecker delta function, which we view as taking values in {{\bf F}_3}. Indeed, (1) reflects the fact that the equation {x+y+z=0} has solutions precisely when {x,y,z} are either all equal, or form a line, and the latter is ruled out precisely when {A} is a capset.

To exploit (1), we will show that the left-hand side of (1) is “low rank” in some sense, while the right-hand side is “high rank”. Recall that a function {F: A \times A \rightarrow {\bf F}} taking values in a field {{\bf F}} is of rank one if it is non-zero and of the form {(x,y) \mapsto f(x) g(y)} for some {f,g: A \rightarrow {\bf F}}, and that the rank of a general function {F: A \times A \rightarrow {\bf F}} is the least number of rank one functions needed to express {F} as a linear combination. More generally, if {k \geq 2}, we define the rank of a function {F: A^k \rightarrow {\bf F}} to be the least number of “rank one” functions of the form

\displaystyle (x_1,\dots,x_k) \mapsto f(x_i) g(x_1,\dots,x_{i-1},x_{i+1},\dots,x_k)

for some {i=1,\dots,k} and some functions {f: A \rightarrow {\bf F}}, {g: A^{k-1} \rightarrow {\bf F}}, that are needed to generate {F} as a linear combination. For instance, when {k=3}, the rank one functions take the form {(x,y,z) \mapsto f(x) g(y,z)}, {(x,y,z) \mapsto f(y) g(x,z)}, {(x,y,z) \mapsto f(z) g(x,y)}, and linear combinations of {r} such rank one functions will give a function of rank at most {r}.

It is a standard fact in linear algebra that the rank of a diagonal matrix is equal to the number of non-zero entries. This phenomenon extends to higher dimensions:

Lemma 1 (Rank of diagonal hypermatrices) Let {k \geq 2}, let {A} be a finite set, let {{\bf F}} be a field, and for each {a \in A}, let {c_a \in {\bf F}} be a coefficient. Then the rank of the function

\displaystyle (x_1,\dots,x_k) \mapsto \sum_{a \in A} c_a \delta_a(x_1) \dots \delta_a(x_k) \ \ \ \ \ (2)

 

is equal to the number of non-zero coefficients {c_a}.

Proof: We induct on {k}. As mentioned above, the case {k=2} follows from standard linear algebra, so suppose now that {k>2} and the claim has already been proven for {k-1}.

It is clear that the function (2) has rank at most equal to the number of non-zero {c_a} (since the summands on the right-hand side are rank one functions), so it suffices to establish the lower bound. By deleting from {A} those elements {a \in A} with {c_a=0} (which cannot increase the rank), we may assume without loss of generality that all the {c_a} are non-zero. Now suppose for contradiction that (2) has rank at most {|A|-1}, then we obtain a representation

\displaystyle \sum_{a \in A} c_a \delta_a(x_1) \dots \delta_a(x_k)

\displaystyle = \sum_{i=1}^k \sum_{\alpha \in I_i} f_{i,\alpha}(x_i) g_{i,\alpha}( x_1,\dots,x_{i-1},x_{i+1},\dots,x_k) \ \ \ \ \ (3)

 

for some sets {I_1,\dots,I_k} of cardinalities adding up to at most {|A|-1}, and some functions {f_{i,\alpha}: A \rightarrow {\bf F}} and {g_{i,\alpha}: A^{k-1} \rightarrow {\bf R}}.

Consider the space of functions {h: A \rightarrow {\bf F}} that are orthogonal to all the {f_{k,\alpha}}, {\alpha \in I_k} in the sense that

\displaystyle \sum_{x \in A} f_{k,\alpha}(x) h(x) = 0

for all {\alpha \in I_k}. This space is a vector space whose dimension {d} is at least {|A| - |I_k|}. A basis of this space generates a {d \times |A|} coordinate matrix of full rank, which implies that there is at least one non-singular {d \times d} minor. This implies that there exists a function {h: A \rightarrow {\bf F}} in this space which is nowhere vanishing on some subset {A'} of {A} of cardinality at least {|A|-|I_k|}.

If we multiply (3) by {h(x_k)} and sum in {x_k}, we conclude that

\displaystyle \sum_{a \in A} c_a h(a) \delta_a(x_1) \dots \delta_a(x_{k-1})

\displaystyle = \sum_{i=1}^{k-1} \sum_{\alpha \in I_i} f_{i,\alpha}(x_i)\tilde g_{i,\alpha}( x_1,\dots,x_{i-1},x_{i+1},\dots,x_{k-1})

where

\displaystyle \tilde g_{i,\alpha}(x_1,\dots,x_{i-1},x_{i+1},\dots,x_{k-1})

\displaystyle := \sum_{x_k \in A} g_{i,\alpha}(x_1,\dots,x_{i-1},x_{i+1},\dots,x_k) h(x_k).

The right-hand side has rank at most {|A|-1-|I_k|}, since the summands are rank one functions. On the other hand, from induction hypothesis the left-hand side has rank at least {|A|-|I_k|}, giving the required contradiction. \Box

On the other hand, we have the following (symmetrised version of a) beautifully simple observation of Croot, Lev, and Pach:

Lemma 2 On {({\bf F}_3^n)^3}, the rank of the function {(x,y,z) \mapsto \delta_{0^n}(x+y+z)} is at most {3N}, where

\displaystyle N := \sum_{a,b,c \geq 0: a+b+c=n, b+2c \leq 2n/3} \frac{n!}{a!b!c!}.

Proof: Using the identity {\delta_0(x) = 1 - x^2} for {x \in {\bf F}_3}, we have

\displaystyle \delta_{0^n}(x+y+z) = \prod_{i=1}^n (1 - (x_i+y_i+z_i)^2).

The right-hand side is clearly a polynomial of degree {2n} in {x,y,z}, which is then a linear combination of monomials

\displaystyle x_1^{i_1} \dots x_n^{i_n} y_1^{j_1} \dots y_n^{j_n} z_1^{k_1} \dots z_n^{k_n}

with {i_1,\dots,i_n,j_1,\dots,j_n,k_1,\dots,k_n \in \{0,1,2\}} with

\displaystyle i_1 + \dots + i_n + j_1 + \dots + j_n + k_1 + \dots + k_n \leq 2n.

In particular, from the pigeonhole principle, at least one of {i_1 + \dots + i_n, j_1 + \dots + j_n, k_1 + \dots + k_n} is at most {2n/3}.

Consider the contribution of the monomials for which {i_1 + \dots + i_n \leq 2n/3}. We can regroup this contribution as

\displaystyle \sum_\alpha f_\alpha(x) g_\alpha(y,z)

where {\alpha} ranges over those {(i_1,\dots,i_n) \in \{0,1,2\}^n} with {i_1 + \dots + i_n \leq 2n/3}, {f_\alpha} is the monomial

\displaystyle f_\alpha(x_1,\dots,x_n) := x_1^{i_1} \dots x_n^{i_n}

and {g_\alpha: {\bf F}_3^n \times {\bf F}_3^n \rightarrow {\bf F}_3} is some explicitly computable function whose exact form will not be of relevance to our argument. The number of such {\alpha} is equal to {N}, so this contribution has rank at most {N}. The remaining contributions arising from the cases {j_1 + \dots + j_n \leq 2n/3} and {k_1 + \dots + k_n \leq 2n/3} similarly have rank at most {N} (grouping the monomials so that each monomial is only counted once), so the claim follows.

Upon restricting from {({\bf F}_3^n)^3} to {A^3}, the rank of {(x,y,z) \mapsto \delta_{0^n}(x+y+z)} is still at most {3N}. The two lemmas then combine to give the Ellenberg-Gijswijt bound

\displaystyle |A| \leq 3N.

All that remains is to compute the asymptotic behaviour of {N}. This can be done using the general tool of Cramer’s theorem, but can also be derived from Stirling’s formula (discussed in this previous post). Indeed, if {a = (\alpha+o(1)) n}, {b = (\beta+o(1)) n}, {c = (\gamma+o(1)) n} for some {\alpha,\beta,\gamma \geq 0} summing to {1}, Stirling’s formula gives

\displaystyle \frac{n!}{a!b!c!} = \exp( n (h(\alpha,\beta,\gamma) + o(1)) )

where {h} is the entropy function

\displaystyle h(\alpha,\beta,\gamma) = \alpha \log \frac{1}{\alpha} + \beta \log \frac{1}{\beta} + \gamma \log \frac{1}{\gamma}.

We then have

\displaystyle N = \exp( n (X + o(1))

where {X} is the maximum entropy {h(\alpha,\beta,\gamma)} subject to the constraints

\displaystyle \alpha,\beta,\gamma \geq 0; \alpha+\beta+\gamma=1; \beta+2\gamma \leq 2/3.

A routine Lagrange multiplier computation shows that the maximum occurs when

\displaystyle \alpha = \frac{32}{3(15 + \sqrt{33})}

\displaystyle \beta = \frac{4(\sqrt{33}-1)}{3(15+\sqrt{33})}

\displaystyle \gamma = \frac{(\sqrt{33}-1)^2}{6(15+\sqrt{33})}

and {h(\alpha,\beta,\gamma)} is approximately {1.013455}, giving rise to the claimed bound of {O( 2.756^n )}.

Remark 3 As noted in the Ellenberg and Gijswijt papers, the above argument extends readily to other fields than {{\bf F}_3} to control the maximal size of subset of {{\bf F}^n} that has no non-trivial solutions to the equation {ax+by+cz=0}, where {a,b,c \in {\bf F}} are non-zero constants that sum to zero. Of course one replaces the function {(x,y,z) \mapsto \delta_{0^n}(x+y+z)} in Lemma 2 by {(x,y,z) \mapsto \delta_{0^n}(ax+by+cz)} in this case.

Remark 4 This symmetrised formulation suggests that one possible way to improve slightly on the numerical quantity {2.756} by finding a more efficient way to decompose {\delta_{0^n}(x+y+z)} into rank one functions, however I was not able to do so (though such improvements are reminiscent of the Strassen type algorithms for fast matrix multiplication).

Remark 5 It is tempting to see if this method can get non-trivial upper bounds for sets {A} with no length {4} progressions, in (say) {{\bf F}_5^n}. One can run the above arguments, replacing the function

\displaystyle (x,y,z) \mapsto \delta_{0^n}(x+y+z)

with

\displaystyle (x,y,z,w) \mapsto \delta_{0^n}(x-2y+z) \delta_{0^n}(y-2z+w);

this leads to the bound {|A| \leq 4N} where

\displaystyle N := \sum_{a,b,c,d,e \geq 0: a+b+c+d+e=n, b+2c+3d+4e \leq 2n} \frac{n!}{a!b!c!d!e!}.

Unfortunately, {N} is asymptotic to {\frac{1}{2} 5^n} and so this bound is in fact slightly worse than the trivial bound {|A| \leq 5^n}! However, there is a slim chance that there is a more efficient way to decompose {\delta_{0^n}(x-2y+z) \delta_{0^n}(y-2z+w)} into rank one functions that would give a non-trivial bound on {A}. I experimented with a few possible such decompositions but unfortunately without success.

Remark 6 Return now to the capset problem. Since Lemma 1 is valid for any field {{\bf F}}, one could perhaps hope to get better bounds by viewing the Kronecker delta function {\delta} as taking values in another field than {{\bf F}_3}, such as the complex numbers {{\bf C}}. However, as soon as one works in a field of characteristic other than {3}, one can adjoin a cube root {\omega} of unity, and one now has the Fourier decomposition

\displaystyle \delta_{0^n}(x+y+z) = \frac{1}{3^n} \sum_{\xi \in {\bf F}_3^n} \omega^{\xi \cdot x} \omega^{\xi \cdot y} \omega^{\xi \cdot z}.

Moving to the Fourier basis, we conclude from Lemma 1 that the function {(x,y,z) \mapsto \delta_{0^n}(x+y+z)} on {{\bf F}_3^n} now has rank exactly {3^n}, and so one cannot improve upon the trivial bound of {|A| \leq 3^n} by this method using fields of characteristic other than three as the range field. So it seems one has to stick with {{\bf F}_3} (or the algebraic completion thereof).

Thanks to Jordan Ellenberg and Ben Green for helpful discussions.

Open question: best bounds for cap sets

23 February, 2007 in math.CO, question | Tags: , , , , | by | 31 comments

Earlier this month, in the previous incarnation of this page, I posed a question which I thought was unsolved, and obtained the answer (in fact, it was solved 25 years ago) within a week. Now that this new version of the page has better feedback capability, I am now tempted to try again, since I have a large number of such questions which I would like to publicise. (Actually, I even have a secret web page full of these somewhere near my home page, though it will take a non-trivial amount of effort to find it!)

Perhaps my favourite open question is the problem on the maximal size of a cap set – a subset of {\Bbb F}^n_3 ({\Bbb F}_3 being the finite field of three elements) which contains no lines, or equivalently no non-trivial arithmetic progressions of length three. As an upper bound, one can easily modify the proof of Roth’s theorem to show that cap sets must have size O(3^n/n) (see e.g. this paper of Meshulam). This of course is better than the trivial bound of 3^n once n is large. In the converse direction, the trivial example \{0,1\}^n shows that cap sets can be as large as 2^n; the current world record is (2.2174\ldots)^n, held by Edel. The gap between these two bounds is rather enormous; I would be very interested in either an improvement of the upper bound to o(3^n/n), or an improvement of the lower bound to (3-o(1))^n. (I believe both improvements are true, though a good friend of mine disagrees about the improvement to the lower bound.)

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