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. 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: **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.