Here is the idea of Roth’s proof of the k=3 case of Szemeredi’s Theorem. (See yesterday’s post if you have no idea what I am talking about.) We have a subset A of ZN of size dN and we want to find a length-3 arithmetic progression in A. A first observation is that if d > 2/3 then we are done, because if we pick a,b at random there is positive probability that a,a+b,a+b+b are all in A. (Use the union bound.) The proof will work by “induction” on d, showing that the truth of the theorem for a smaller value of d can be derived from the truth of the theorem for a larger value d. Of course d is a continous parameter, so it will not really be induction, but that’s how we should think of it.

The main result is an “algorithm” that given A subset of ZN of size dN does one of the following two things:

  1. It immediately finds a length-3 progression in A; or
  2. it constructs a subset A’ of ZN’ such that (i) if A’ has a length-3 progression then so does A; (ii) A’ has size at least about (d+d2)*N’; and (iii) N’ is about sqrt(N).

So, given A, we run the algorithm, and we either immediately find the progression, or we get A’. In the latter case, we run the algorithm on A’, and we either get a progression in A’ (implying a progression in A) or we get a new set A”, and so on. But how many times do we have to keep doing this? At each step, the density of the set increases, and if it ever goes above 2/3 then we are also done. So, how many times can you do the d -> d + d2 transformation until you get 2/3? Certainly no more than O(1/d2) times, but actually O(1/d) is enough. This means that we will always find a length-3 progression in A, provided that N is large enough that we can take its square root O(1/d) times. So if N is doubly exponential in 1/d we have our proof.

Gowers’s proof for the general case has the same outline. Now we want to find a length-k progression in A. If the density of A is more than (k-1)/k, we are done. Gowers describes an algorithm that, given A, either finds a length-k progression or constructs a set A’ in ZN’ such that if A’ has a length-k progression that so does A’. Furthermore, A’ has density d+poly(d) and N’ is a constant root of N.

How do these arguments work? Let us look at Roth’s proof in the case in which A is a subset of Fnp with prime p; for concreteness, take p=3. Given A, consider its characteristic function (that we denote again A) A: Fn3 -> {0,1} and compute its Fourier transform.

Consider now the probability that, when picking a,b at random, the three points a,a+b,a+b+b are all in A. This is the same as

(*) Ea,b A(a)*A(a+b)*A(a+b+b)

which can be expressed really cleanly in terms of the Fourier coefficients of A. In particular, the expression (*) is d3, which is the number you would get if you were looking at 3 independent points, plus or minus an error term that depends on the largest Fourier coefficient of A. If all Fourier coefficients are less than, say, d2/2, then epression (*) is at least d3/2, and we have plenty of length-3 progressions in A; this is case (1) of the proof. In this case, we should think of A as a “pseudorandom set against adversaries that look for random length-3 progressions.”

What happens if A has a large Fourier coefficient? This means that A is correlated with a linear function, and so there is an affine subspace V in Fn3 of dimension n-1 such that A is denser in V than in the rest of Fn3. Now, map V to Fn-13 via an invertible affine map, and let A’ be the image of A under this map. If there is a length-3 progression in A’, that’s just 3 points on a line; but then it means that also in A we had 3 points on a line. The set A’ has density about d+d2 in Fn-13, and so we have part (2) of the argument.

As a bonus, we only lost a constant fraction of our group size at each step, so d can be as large as about 1/logN, where N=3n is the size of the initial group containing A.

Note the “win-win” structure of the argument. If A has small Fourier coefficients, then it is “pseudorandom”, and we get what we want from the pseudorandomness. Otherwise, A has some “linear structure,” and we can take advantage of it to reduce to a simpler instance of our original problem.

This “either we get what we want or we reduce to a simpler case” proof strategy has been used in some constructions of extractors, but it seems such a good idea that it may apply more widely, and one should keep it mind.

What about lower bounds? Excellent question. There is a very simple construction due to Behrend of a subset of size N/exp(sqrt(log N)) of {1,…,N} that has no length-3 progression, so N must be super-polynomial in 1/d when we prove the theorem over the integers (or over ZN with prime N, the two cases are essentially equivalent). Note that if you try to find a large set with no length-3 progression using the probabilistic method, you will not go very far. This is a better than random explicit construction, a rarity in extremal combinatorics. By the way, Behrend’s construction is very simple (here is a half-page description), it is 60 year old, and nobody has been able to improve it ever since.

What about lower bounds for Fn3? There is no known construction of a large subset avoiding length-3 progressions. The best lower bounds are of the form cn, with c<3. Is there a (3-o(1))n lower bound? This is a major open question. If there were a (2.999)n upper bound, then it would be a stunning result, because it would show that the business of translating proofs from finite fields to the integers using Bohr sets cannot be applied as a black box, but it works on some proofs and does not work on other proofs.

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