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.



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!

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