Separating measurable recurrence from strong recurrence via rigidity sequences

Abstract

If $G$ is an abelian group, we say $S\subset G$ is a set of recurrence if for every probability measure preserving $G$-system $(X,\hat{I}¼,T)$ and every $D\subset X$ having $\hat{I}¼(D)>0$, there is a $g\in S$ such that $\hat{I}¼(D\cap T^{g}D)>0$. We say $S$ is a set of strong recurrence if for every set $D$ having $\hat{I}¼(D)>0$ there is a $c>0$ such that $\hat{I}¼(D\cap T^{g}D)>c$ for infinitely many $g\in S$. We call $S$ measure expanding if for all $g\in G$, the translate $S+g$ is a set of recurrence. A rigidity sequence for $(X,\hat{I}¼,T)$ is a sequence of elements $s_n\in G$ satisfying ${\lim }_{n\to \infty }\hat{I}¼(D△T^{s_n}D)=0$ for all measurable $D\subset X$. For all but countably many countable abelian groups $G$, we prove that if $S$ is measure expanding, there is a sequence of elements $s_n\in S$ such that $\{s_n:n\in \mathbb{N}\}$ is also measure expanding and every translate of $(s_n)$ is a rigidity sequence for some free weak mixing measure preserving $G$-system. The special case where $S=G$ proves a conjecture of Ackelsberg. As a consequence, we prove that for every countably infinite abelian group $G$ and every measure expanding set $S\subset G$ there is a subset $S^{\prime }\subset S$ such that $S^{\prime }$ is measure expanding and no translate of $S^{\prime }$ is a set of strong recurrence.

30 pages. v4 incorporates referee comments