Popular Difference Sets
نویسنده
چکیده
We provide further explanation of the significance of an example in a recent paper of Wolf in the context of the problem of finding large subspaces in sumsets. Throughout these notes G will denote the group F2 , that is the n-dimensional vector space over F2, and PG will denote the normalized counting measure on G. The convolution of two functions f, g ∈ L(PG) is defined to be f ∗ g(x) := ∫ f(y)g(x− y)dPG(y), and is of particular importance in additive combinatorics. To see why recall that if A,B ⊂ G then we write A+B for the sumset {a+ b : a ∈ A, b ∈ B}. Convolution then yields the following important identity A+B = supp 1A ∗ 1B . The sumset A+B can thus be studied through the convolution 1A ∗ 1B ; a typical example of this is in the following problem. Problem 1.1 (Bourgain-Green, [Bou90, Gre02, Gre05]). Suppose that A ⊂ G := F2 has density Ω(1). How large a subspace V with V ⊂ A + A can we guarantee may be found? All known approaches to this problem proceed by harmonic analysis of 1A ∗ 1A. Since the arguments are analytic the methods are not good at distinguishing between when 1A ∗ 1A is small and when it is very small. To be more precise we introduce a definition. Given a set A ⊂ G of density α > 0 and a parameter c ∈ [0, 1] we define the popular difference set with parameter c to be Dc(A) := {x ∈ G : 1A ∗ 1A(x) > cα}. Our definition employs a slightly different normalization to those presented elsewhere so that c naturally lies in the range [0, 1]. Indeed, it is easy to see that if c is greater than 1 then even for large sets A we may have Dc(A) = {0G}. At the other end of the spectrum we have D0(A) = A+A. Now, the arguments for tackling Problem 1.1 all fail to meaningfully distinguishing between the set A + A = D0(A) and Dc(A) when c is small. In particular, for example, Green effectively proves the following result in [Gre05]. 1In particular this occurs generically: if c > 1, |G| is large enough in terms of c, and A is chosen uniformly at random from sets of size |G|/2 then (with high probability) Dc(A) = {0G}.
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