Separating Bohr denseness from measurable recurrence

- Applied Mathematics and Statistics, Colorado School of Mines

*Discrete Analysis*, September. https://doi.org/10.19086/da.26859.

### Editorial introduction

Separating Bohr denseness from measurable recurrence, Discrete Analysis 2021:9, 20 pp.

This paper is about the structure of difference sets of dense sets of integers. More precisely, a set \(A\subset\mathbb Z\) is said to have *positive upper Banach density* if there exists \(\delta>0\) such that for every positive integer \(N\) there is an interval \(I\) of length \(n\geq N\) such that \(|A\cap I|\geq\delta n\), and the *difference set* of a set \(A\) is the set \(A-A=\{a-b:a,b\in A\}\).

A common theme in additive combinatorics is that in general the difference set of a set \(A\) has much more additive structure than the set \(A\) itself, particularly when \(A\) is dense. And a more precise version of the theme is that typically a difference set of a dense set will contain almost all of some large “structured set”, and that the difference set of a difference set will contain *all* of a structured set of the same type.

There are various possible notions of structured set, but probably the most important one is that of a Bohr set: given an Abelian group \(G\), a Bohr set in \(G\) is an intersection of boundedly many sets of the form \(\{x:|\chi(x)-1|\leq\delta\}\), where \(\delta>0\) and \(\chi\) is a character. This paper uses a very similar definition: a subset of \(\mathbb Z\) is a *Bohr neighbourhood* if it is of the form \(\{n\in\mathbb Z:n\mathbf{\alpha}\in U\}\), where \(\alpha\in\mathbb T^k\) for some bounded \(k\), and \(U\) is a non-empty open subset of \(\mathbb T^k\). (To see that it is similar, consider the case where \(U\) is a product of open intervals.)

Several authors have posed the problem of whether the difference set of a subset of \(\mathbb Z\) of positive upper Banach density must contain a non-empty Bohr neighbourhood. Given the heuristic mentioned earlier, one would expect the answer to be no (but that it would contain almost all of a Bohr neighbourhood, in some suitable sense), but that turned out not to be straightforward to prove: in fact, it is not completely easy to prove even for finite groups, and there is then not an obvious way to generalize the constructions for finite groups to \(\mathbb Z\).

As an aside, one example of such a construction that works in the group \(\mathbb F_2^n\) is to take \(A\) to be the set of all sequences with at most \(n/2-\sqrt n\) 1s. The difference set \(A-A\) then consists of all sequences with at most \(n-2\sqrt n\) 1s. In this context, a Bohr neighbourhood is an affine subspace of bounded codimension. Since any such subspace contains a sequence with more than \(n-2\sqrt n\) 1s, it is not contained in \(A-A\). A similar but more complicated construction of Ruzsa, known as a *niveau set*, gives an example in \(\mathbb Z_N\).

This paper solves the problem for \(\mathbb Z\) with a delicate construction that shows that the answer is indeed negative. The result can also be formulated in dynamical terms. A set of integers \(S\) is said to be a *set of measurable recurrence* if for every measure-preserving dynamical system \((X,\mu,T)\) and every subset \(A\subset X\) with \(\mu(A)>0\) there exists some \(n\in S\) with \(\mu(A\cap T^{-n}A)>0\). The main result of this paper is equivalent to the statement that there is a set that is dense in the Bohr topology (that is, it intersects all non-empty Bohr neighbourhoods) but is not a set of measurable recurrence.