## Difference sets missing a Hamming sphere

I tend to think of the Croot-Lev-Pach method (as used, for instance, in the cap set problem) as having to do with n-tensors, where n is bigger than 2.  But you can actually also use it in the case of 2-tensors, i.e. matrices, to say things that (as far as I can see) are not totally trivial.

Write m_d for the number of squarefree monomials in x_1, .. x_n of degree at most d; that is,

$m_d = 1 + {n \choose 1} + {n \choose 2} + \ldots + {n \choose d}.$

Claim:  Let P be a polynomial of degree d in F_2[x_1, .. x_n] such that P(0) = 1.  Write S for the set of nonzero vectors x such that P(x) = 1.  Let A be a subset of F_2^n such that no two elements of A have difference lying in S.  Then |A| < 2m_{d/2}.

Proof:  Write M for the A x A matrix whose (a,b) entry is P(a-b).  By the Croot-Lev-Pach lemma, this matrix has rank at most 2m_{d/2}.  By hypothesis on A, M is the identity matrix, so its rank is |A|.

Remark: I could have said “sum” instead of “difference” since we’re in F_2 but for larger finite fields you really want difference.

The most standard context in which you look for large subsets of F_2^n with restricted difference sets is that of error correcting codes, where you ask that no two distinct elements of A have difference with Hamming weight (that is, number of 1 entries) at most k.

It would be cool if the Croot-Lev-Pach lemma gave great new bounds on error-correcting codes, but I don’t think it’s to be.  You would need to find a polynomial P which vanishes on all nonzero vectors of weight larger than k, but which doesn’t vanish at 0. Moreover, you already know that the balls of size k/2 around the points of A are disjoint, which gives you the “volume bound”

|A| < 2^n / m_{k/2}.

I think that’ll be hard to beat.

If you just take a random polynomial P, the support of P will take up about half of F_2^n; so it’s not very surprising that a set whose difference misses that support has to be small!

Here’s something fun you can do, though.  Let s_i be the i-th symmetric function on x_1, … x_n.  Then

$s_i(x) = {wt(x) \choose i}$

where wt(x) denotes Hamming weight.  Recall also that the binomial coefficient

${k \choose 2^a}$

is odd precisely when the a’th binary digit of k is 1.

Thus,

$(1-s_1(x))(1-s_2(x))(1-s_4(x))\ldots(1-s_{2^{b-1}}(x))$

is a polynomial of degree 2^b-1 which vanishes on x unless the last b digits of wt(x) are 0; that is, it vanishes unless wt(x) is a multiple of 2^b.  Thus we get:

Fact:  Let A be a subset of F_2^n such that the difference of two nonzero elements in A never has weight a multiple of 2^b.  Then

$|A| \leq 2m_{2^{b-1} - 1}$.

Note that this is pretty close to sharp!  Because if we take A to be the set of vectors of  weight at most 2^{b-1} – 1, then A clearly has the desired property, and already that’s half as big as the upper bound above.  (What’s more, you can throw in all the vectors of weight 2^{b-1} whose first coordinate is 1; no two of these sum to something of weight 2^b.  The Erdös-Ko-Rado theorem says you can do no better with those weight 2^{b-1} vectors.)

Is there an easier way to prove this?

When b=1, this just says that a set with no differences of even Hamming weight has size at most 2; that’s clear, because two vectors whose Hamming weight has the same parity differ by a vector of even weight.  Even for b=2 this isn’t totally obvious to me.  The result says that a subset of F_2^n with no differences of weight divisible by 4 has size at most 2+2n.  On the other hand, you can get 1+2n by taking 0, all weight-1 vectors, and all weight-2 vectors with first coordinate 1.  So what’s the real answer, is it 1+2n or 2+2n?

Write H(n,k) for the size of the largest subset of F_2^n having no two vectors differing by a vector of Hamming weight exactly k.  Then if 2^b is the largest power of 2 less than n, we have shown above that

$m_{2^{b-1} - 1 } \leq H(n,2^b) \leq 2m_{2^{b-1} - 1}$.

On the other hand, if k is odd, then H(n,k) = 2^{n-1}; we can just take A to be the set of all even-weight vectors!  So perhaps H(n,k) actually depends on k in some modestly interesting 2-adic way.

The sharpness argument above can be used to show that H(4m,2m) is as least

$2(1 + 4m + {4m \choose 2} + \ldots + {4m \choose m-1} + {4m-1 \choose m-1}). (*)$

I was talking to Nigel Boston about this — he did some computations which make it looks like H(4m,2m) is exactly equal to (*) for m=1,2,3.  Could that be true for general m?

(You could also ask about sets with no difference of weight a multiple of k; not sure which is the more interesting question…)

Update:  Gil Kalai points out to me that much of this is very close to and indeed in some parts a special case of the Frankl-Wilson theorem…  I will investigate further and report back!

## 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.