Luke Pebody on sharp bounds for tri-colored sum-free sets

Quick update on this post, where I listed three variations on the problem of finding large subsets of an abelian group A with no three terms in arithmetic progression.  The quantities I called G_2(F_q^n) and G_3(F_q^n) in that post are both bounded above by M(F_q^n), the number of monomials of total degree at most (q-1)n/3 and degree at most q-1 in each variable.  There’s already been a lot of motion in the few weeks since I wrote that post!  A result of Kleinberg, Sawin, and Speyer shows that G_2(F_q^n) is bounded between c_q^n and c_q^{n-epsilon} for some constant c_q, which they explicitly describe but which is kind of hard to compute.  But it’s kind of a win-win.  Either c_q^n is less than M(F_q^n), in which case, great, improvement over the results of CLP/EG, or not, in which case, great, the bounds on tri-colored sum-free sets in CLP/EG are tight up to subexponential factors!  And now Luke Pebody has posted a preprint showing that the latter is the case.

To sum up:  the quantities G_2(F_q^n) and G_3(F_q^n) which I alluded to in the original post are now bounded above by M(F_q^n) and below by M(F_q^n)^{1-epsilon}.  Wonderful!

This only heightens the interest in the original problem of estimating G_1(F_q^n), the size of the largest subset of F_q^n containing no three-term arithmetic progession.  Is the bound M(F_q^n) essentially sharp?  Or is G_1(F_q^n) much smaller?

Arithmetic progression joke

Why was 7.333… disgusted by 7.666….?

Variations on three-term arithmetic progressions

Here are three functions.  Let N be an integer, and consider:

•  G_1(N), the size of the largest subset S of 1..N containing no 3-term arithmetic progression;
•  G_2(N), the largest M such that there exist subsets S,T of 1..N with |S| = |T| = M such that the equation s_i + t_i = s_j + t_k has no solutions with (j,k) not equal to (i,i).  (This is what’s called  a tri-colored sum-free set.)
• G_3(N), the largest M such that the following is true: given subsets S,T of 1..N, there always exist subsets S’ of S and T’ of T with |S’| + |T’| = M and

You can see that G_1(N) <= G_2(N) <= G_3(N).  Why?  Because if S has no 3-term arithmetic progression, we can take S = T and s_i = t_i, and get a tri-colored sum-free set.  Now suppose you have a tri-colored sum-free set (S,T) of size M; if S’ and T’ are subsets of S and T respectively, and , then for every pair (s_i,t_i), you must have either s_i in S’ or t_i in T’; thus |S’| + |T’| is at least M.

When the interval 1..N is replaced by the group F_q^n, the Croot-Lev-Pach-Ellenberg-Gijswijt argument shows that G_1(F_q^n) is bounded above by the number of monomials of degree at most (q-1)n/3; call this quantity M(F_q^n).  In fact, G_3(F_q^n) is bounded above by M(F_q^n), too (see the note linked from this post) and the argument is only a  modest extension of the proof for G_1.  For all we know, G_1(F_q^n) might be much smaller, but Kleinberg has recently shown that G_2(F_2^n) (whence also G_3(F_2^n)) is equal to M(F_2^n) up to subexponential factors, and work in progress by Kleinberg and Speyer has shown this for several more q and seems likely to show that the bound is tight in general.  On the other hand, I have no idea whether to think G_1(F_q^n) is actually equal to M(F_q^n); i.e. is the bound proven by me and Dion sharp?

The behavior of G_1(N) is, of course, very much studied; we know by Behrend (recently sharpened by Elkin) that G_1(N) is at least N/exp(c sqrt(log N)).  Roth proved that G_1(N) = o(N), and the best bounds, due to Tom Sanders, show that G_1(N) is O(N(log log N)^5 / log N).  (Update:  Oops, no!  Thomas Bloom has an upper bound even a little better than Sanders, change that 5 to a 4.)

What about G_2(N) and G_3(N)?  I’m not sure how much people have thought about these problems.  But if, for instance, you could show (for example, by explicit constructions) that G_3(N) was closer to Sanders than to Behrend/Elkin, it would close off certain strategies for pushing the bound on G_1(N) downward. (Update:  Jacob Fox tells me that you can get an upper bound for G_2(N) of order N/2^{clog* N} from his graph removal paper, applied to the multicolored case.)

Do we think that G_2(N) and G_3(N) are basically equal, as is now known to be the case for F_q^n?

Sumsets and sumsets of subsets

Say that ten times fast!
Now that you’re done, here’s an interesting fact.  I have been turning over this argument of Croot-Lev-Pach and mine and Gijswijt’s for a couple of weeks now, trying to understand what it’s really doing that leads to control of subsets of F_q^n without arithmetic progressions.

It turns out that there’s a nice refinement of what we prove, which somehow feels like it’s using more of the full strength of the Croot-Lev-Pach lemma.  The critical input is an old result of Roy Meshulam on linear spaces of low-rank matrices.

So here’s a statement.  Write M(q,n) for the CLP/EG upper bound on subsets of F_q^n with no three-term AP.

Then Theorem:  every subset S of F_q^n contains a subset S’ of size at most M(q,n) such that S’+S = S+S.

(Exercise:   Show that this immediately implies the bound on subsets with no three-term AP.)

I find this result very funny, so much so that I didn’t believe it at first, but I think the proof is right..!  Well, see for yourself, here it is.

Two natural questions:  is the bound on S’ sharp?  And is there any analogue of this phenomenon for the integers?

Update:  Of course right after I post this I realize that maybe this can be said more simply, without the invocation of Meshulam’s result (though I really like that result!)  Namely:  it’s equivalent to say that if |S| > M(q,n), you can remove ONE element from S and get an S’ with S’+S = S+S.  Why is this so?  Well, suppose not.  Choose some s_1.  We know it can’t be removed, so there must be some s_1 + s’_1 which is not expressible as a sum in S+T any other way.  The same applies to s_2, s_3, and so on.  So you end up with a set U of “unique sums” s_i + s’_i.  Now you can apply the CLP/EG argument directly to this situation; let P be a polyomial vanishing off U, this makes the matrix P(s+t) on S have a single 1 in each row and each column, and this is just as good as diagonal from the point of view of the argument in EG, so you can conclude just as there that |S| <= M(q,n).  Does that make sense?  This is the same spirit in which the polynomial method is used by Blasiak-Church-Cohn-Grochow-Umans to control multicolored sum-free sets, and the multicolored sum-free set of size (2^(4/3))^n constructed by Alon, Shpilka, and Umans also gives a lower bound for the problem under discussion here.

I still like the one-step argument in the linked .pdf better!  But I have to concede that you can prove this fact without doing any fancy linear algebra.

Update to Update (Jun 9):  Actually, I’m not so sure this argument above actually proves the theorem in the linked note.  So maybe you do need to (get to!) use this Meshulam paper after all!  What do you guys think?

Update:  The bound is sharp, at least over F_2!  I just saw this paper of Robert Kleinberg, which constructs a multicolored sum-free set in F_2^n of size just under M(2,n)!  That is, he gives you subsets S and T, both of size just under M(2,n), such that S’+T union S+T’ can’t be all of S+T if S’ and T’ are smaller than (1/2)S and (1/2)T, if I worked this out right.

The construction, which is actually based on one from 2014 by Fu and Kleinberg, actually uses a large subset of a cyclic group Z/MZ, where M is about M(2,n), and turns this into a multicolored sum-free set in (F_2)^n of (about) the same size.  So the difference between the upper bound and the lower bound in the (F_2)^n case is now roughly the same as the difference between the (trivial) upper bound and the lower bound in the case of no-three-term-AP sets in the interval.  Naturally you start to wonder:  a) Does the Fu-Kleinberg construction really have to do with characteristic 2 or is it general?  (I haven’t read it yet.)  b) Can similar ideas be used to construct large 3-AP-free subsets of (F_q)^n?  (Surely this has already been tried?) c) Is there a way to marry Meshulam’s Fourier-analytic argument with the polynomial method to get upper bounds on order (1/n)M(q,n)?  I wouldn’t have thought this worthwhile until I saw this Kleinberg paper, which makes me think maybe it’s not impossible to imagine we’re getting closer to the actual truth.

 

Bounds for cap sets

Briefly:  it seems to me that the idea of the Croot-Lev-Pach paper I posted about yesterday can indeed be used to give a new bound on the size of subsets of F_3^n with no three-term arithmetic progression! Such a set has size at most (2.756)^n. (There’s actually a closed form for the constant, I think, but I haven’t written it down yet.)

Here’s the preprint. It’s very short. I’ll post this to the arXiv in a day or two, assuming I (or you) don’t find anything wrong with it, so comment if you have comments! Note: I’ve removed the link, since the official version of this result is now the joint paper by me and Gijswijt, and the old version shouldn’t be cited.

Update:  Busy few days of administrative stuff and travel, sorry for not having updated the preprint yet, will try to finish it today.  One note, already observed below in the comments:  you get a similar bound for subsets of (F_q)^n free of solutions to (ax+by+cz=0) for any (a,b,c) with a+b+c=0; the cap set case is q=3, (a,b,c) = (1,1,1).

Update 2:  Dion Gijswijt and I will be submitting this result as a joint paper, which will amalgamate the presentations of our essentially identical arguments.  Dion carried out his work independently of mine at around the same time, and the idea should be credited to both of us.  Our joint paper is available on the arXiv.

 

Croot-Lev-Pach on AP-free sets in (Z/4Z)^n

As you know I love the affine cap problem:  how big can a subset of (Z/3Z)^n be that contains no three elements summing to 0 — or, in other words, that contains no 3-term arithmetic progression?  The best upper bounds, due to Bateman and Katz, are on order 3^n / n^(1+epsilon).  And I think it’s fair to say that all progress on this problem, since Meshulam’s initial results, have come from Fourier-analytic arguments.

So I’m charmed by this paper of Ernie Croot, Vsevolod Lev, and Peter Pach which proves a much stronger result for A = (Z/4Z)^n:  a subset with no 3-term arithmetic progression has size at most c^n for c strictly less than 4.  Better still (for an algebraic geometer) the argument has no harmonic analysis at all, but proceeds via the polynomial method!

This is surprising for two reasons.  First, it’s hard to make the polynomial method work well for rings, like Z/4Z, that aren’t fields; extending our knowledge about additive combinatorics to such settings is a long-standing interest of mine.  Second, the polynomial method over finite fields usually works in the “fixed dimension large field” regime; problems like affine cap, where the base ring is fixed and the dimension are growing, have so far been mostly untouched.

As for the first issue, here’s the deal.  This looks like a problem over Z/4Z but is really a problem over F_2, because the condition for being a 3-term AP

a – 2b + c = 0

has a 2 in it.  In other words:  the two outer terms have to lie in the same coset of 2A, and the middle term is only determined up to 2A.

 So CLP recast the problem as follows.  Let S be a large subset of A with no 3-term AP.   Let V be 2A, which is an n-dimensional vector space over F_2.  For each v in V, there’s a coset of V consisting of the solutions to 2a = v, and we can let S_v be the intersection of S with this coset.

We want to make this a problem about V, not about A.  So write T_v for a translate of S_v by some element of the coset, so T_v now sits in V.  Which element?  Doesn’t matter!

We can now write the “no 3-term AP” condition strictly in terms of these subsets of V.  Write (T_v – T_v)^* for the set of differences between distinct elements of T_v.  Write U for the set of v in V such that T_v is nonempty.  Then the union over all v in U of

(T_v – T_v)^* + v

is disjoint from U.

I leave it as an exercise to check the equivalence.

Now we have a combinatorial question about vector spaces over F_2; we want to show that, under the condition above, the sum of |T_v| over all v in U can’t be too large.

This is where the polynomial method comes in!  CLP show that (over any field, not just F_2), a polynomial of low degree vanishing on (T_v – T_v)^* has to vanish at 0 as well; this is Lemma 1 in their paper.  So write down a polynomial P vanishing on V – U; by dimension considerations we can choose one which doesn’t vanish on all of V.  (This uses the fact that the squarefree monomials of degree up to d are linearly independent functions on F_2^n.)  If U is big, we can choose P to have lowish degree.

Since P vanishes on V-U, P has to vanish on (T_v – T_v)^* + v for all v.  Since P has low degree, it has to vanish on v too, for all v.  But then P vanishes everywhere, contrary to our assumption.

The magic of the paper is in Lemma 1, in my view, which is where you really see the polynomial method applied in this unusual fixed-field-large-dimension regime.  Let me say a vague word about how it works.  (The actual proof is less than a page, by the way, so I’m not hiding much!)  Let P be your polynomial and d its degee.  You send your vector space into a subvariety of a much larger vector space W via degree-d Veronese embedding F_d. In fact you do this twice, writing

V x V -> W x W.

Now if P is your polynomial of degree-d, you can think of P(v_1 – v_2) as a bilinear form <,> on W x W.  Suppose S is a subset of V such that P(s_1 – s_2) vanishes for all distinct s_1, s_2 in S.   That means

<F_d(s_1), F_d(s_2)> = 0

for all distinct s_1, s_2 in S.  On the other hand,

<F_d(s_1), F_d(s_1)>

doesn’t depend on s_1; it just takes the value P(0).  So if P(0) is not equal to 0, you have |S| vectors of nonzero norm which are mutually orthogonal under this bilinear form, and so there can be at most dim W of these, and that’s the bound on |S| you need.

This is very slick and I hope the idea is more generally applicable!