## 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?

## 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 $S'+T \cup S+T' = S+T.$

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 $S'+T \cup S+T' = S+T$, 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.

## An incidence conjecture of Bourgain over fields of positive characteristic (with Hablicsek)

Marci Hablicsek (a finishing Ph.D. student at UW) and I recently posted a new preprint, “An incidence conjecture of Bourgain over fields of finite characteristic.”

The theme of the paper is a beautiful theorem of Larry Guth and Nets Katz, one of the early successes of Dvir’s “polynomial method.”  They proved a conjecture of Bourgain:

Given a set S of points in R^3, and a set of N^2 lines such that

• No more than N lines are contained in any plane;
• Each line contains at least N points of S;

then S has at least cN^3 points.

In other words, the only way for a big family of lines to have lots of multiple intersections is for all those lines to be contained in a plane.  (In the worst case where all the lines are in a plane, the incidences between points and lines are governed by the Szemeredi-Trotter theorem.)

I saw Nets speak about this in Wisconsin, and I was puzzled by the fact that the theorem only applied to fields of characteristic 0, when the proof was entirely algebraic.  But you know the proof must fail somehow in characteristic p, because the statement isn’t true in characteristic p.  For example, over the field k with p^2 elements, one can check that the Heisenberg surface

$X: x - x^p + yz^p - zy^p = 0$

has a set of p^4 lines, no more than p lying on any plane, and such that each line contains at least p^2 elements of X(k).  If the Guth-Katz theorem were true over k, we could take N = p^2 and conclude that |X(k)| is at least p^6.  But in fact, it’s around p^5.

It turns out that there is one little nugget in the proof of Guth-Katz which is not purely algebraic.  Namely:  they show that a lot of the lines are contained in some surface S with the following property;  at every smooth point s of S, the tangent plane to S at s intersects S with multiplicity greater than 2.  They express this in the form of an assertion that a certain curvature form vanishes everywhere.  In characteristic 0, this implies that S is a plane.  But not so in characteristic p!  (As always, the fundamental issue is that a function in characteristic p can have zero derivative without being constant — viz., x^p.)  All of us who did the problems in Hartshorne know about the smooth plane curve over F_3 with every point an inflection point.  Well, there are surfaces like that too (the Heisenberg surface is one such) and the point of the new paper is to deal with them.  In fact, we show that the Guth-Katz theorem is true word for word as long as you prevent lines not only from piling up in planes but also from piling up in these “flexy” surfaces.

It turns out that any such surface must have degree at least p, and this enables us to show that the Guth-Katz theorem is actually true, word for word, over the prime field F_p.

If you like, you can think of this as a strengthening of Dvir’s theorem for the case of F_p^3.  Dvir proves that a set of p^2 lines with no two lines in the same direction fills up a positive-density subset of the whole space.  What we prove is that the p^2 lines don’t have to point in distinct directions; it is enough to impose the weaker condition that no more than p of them lie in any plane; this already implies that the union of the lines has positive density.  Again, this strengthening doesn’t hold for larger finite fields, thanks to the Heisenberg surface and its variants.

This is rather satisfying, in that there are other situations in this area (e.g. sum-product problems) where there are qualitatively different bounds depending on whether the field k in question has nontrivial subfields or not.  But it is hard to see how a purely algebraic argument can “see the difference” between F_p and F_{p^2}.  The argument in this paper shows there’s at least one way this can happen.

Satisfying, also, because it represents an unexpected application for some funky characteristic-p algebraic geometry!  I have certainly never needed to remember that particular Hartshorne problem in my life up to now.

## “Kakeya sets over non-archimedean local rings,” by Dummit and Hablicsek

A new paper posted this week on the arXiv this week by UW grad students Evan Dummit and Márton Hablicsek answers a question left open in a paper of mine with Richard Oberlin and Terry Tao.  Let me explain why I was interested in this question and why I like Evan and Marci’s answer so much!

Recall:  a Kakeya set in an n-dimensional vector space over a field k is a set containing a line (or, in the case k = R, a unit line segment) in every direction.  The “Kakeya problem,” phrased loosely, is to prove that Kakeya sets cannot be too small.

But what does “small” mean?  You might want it to mean “measure 0” but for the small but important fact that in this interpretation the problem has a negative answer:  as Besicovitch discovered in 1919, there are Kakeya sets in R^2 with measure 0!  So Kakeya’s conjecture concerns a stronger notion of “small”  — he conjectures that a Kakeya set in R^n cannot have Hausdorff or Minkowski dimension strictly smaller than n.

(At this point, if you haven’t thought about the Kakeya conjecture before, you might want to read Terry’s long expository post about the Kakeya conjecture and Dvir’s theorem; I cannot do it any better here.)

The big recent news in this area, of course, is Dvir’s theorem that that the Kakeya conjecture is true when k is a finite field.

Of course one hopes that Dvir’s argument will give some ideas for an attack on the original problem in R^n.  And that hasn’t happened yet; though the “polynomial method,” as the main idea of Dvir’s theorem is now called, has found lots of applications to other problems in real combinatorial geometry (e.g. Guth and Katz’s proof of the joints conjecture.)

Why not Kakeya?  Well, here’s one clue.  Dvir actually proves more than the Kakeya conjecture!  He proves that a Kakeya set in F_q^n has positive measure.

(Note:  F_q^n is a finite set, so of course any nonempty subset has positive measure; so “positive measure” here is shorthand for “there’s a lower bound for the measure which is bounded away from 0 as q grows with n fixed.”)

What this tells you is that R really is different from F_q with respect to this problem; if Dvir’s proof “worked” over R, it would prove that a Kakeya set in R^n had positive measure, which is false.

So what’s the difference between R and F_q?  In my view, it’s that R has multiple scales, while F_q only has one.  Two elements in F_q are either the same or distinct, but there is nothing else going on metrically, while distinct real lines can be very close together or very far apart.  The interaction between distances at different scales is your constant companion when working on these problems in the real setting; so maybe it’s not so shocking that a one-scale field like F_q is not a perfect model for the phenomena we’re trying to study.

Which leads us to the ring F_q[[t]] — the “non-archimedean local ring” which Dummit and Hablicsek write about.  This ring is somehow “in between” finite fields and real numbers.  On the one hand, it is “profinite,” which is to say it is approximated by a sequence of larger and larger finite rings F_q[[t]]/t^k.  On the other hand, it has infinitely many scales, like R.  From the point of view of Kakeya sets, is it more like a finite field, or more like the real numbers?  In particular, does it have Kakeya sets of measure 0, making it potentially a good model for the real Kakeya problem?

This is the question Richard, Terry, and I asked, and Evan and Marci show that the answer is yes; they construct explicitly a Kakeya set in F_q[[t]]^2 with measure 0.

Now when we asked this question in our paper, I thought maybe you could do this by imitating Besicovitch’s argument in a straightforward way.  I did not succeed in doing this.  Evan and Marci tried too, and they told me that this just plain doesn’t work.  The construction they came up with is (at least as far as I can see) completely different from anything that makes sense over R.  And the way they prove measure 0 is extremely charming; they define a Markov process such for which the complement of their Kakeya set is the set of points that eventually hit 0, and then show by standard methods that their Markov process goes to 0 with probability 1!

Of course you ask:  does their Kakeya set have Minkowski dimension 2?  Yep — and indeed, they prove that any Kakeya set in F_q[[t]]^2 has Minkowski dimension 2, thus proving the Kakeya conjecture in this setting, up to the distinction between Hausdorff and Minkowski dimension.  (Experts should feel free to weigh in an tell me how much we should worry about this distinction.)  Note that dimension 2 is special:  the Kakeya conjecture in R^2 is known as well.  For every n > 2 we’re in the dark, over F_q[[t]] as well as over R.

To sum up:  what Dummit and Hablicsek prove makes me feel like the Kakeya problem over  F_q[[t]] is, at least potentially, a pretty good model for the Kakeya problem over R!  Not that we know how to solve the Kakeya problem over F_q[[t]]…..