Separating Bohr denseness from measurable recurrence

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.