A set of 2-recurrence whose perfect squares do not form a set of measurable recurrence

Fuente: arXiv
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Autore principale: Griesmer, John T.
Natura: Preprint
Pubblicazione: 2022
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author Griesmer, John T.
author_facet Griesmer, John T.
contents We say that $S\subset\mathbb Z$ is a set of $k$-recurrence if for every measure preserving transformation $T$ of a probability measure space $(X,μ)$ and every $A\subseteq X$ with $μ(A)>0$, there is an $n\in S$ such that $μ(A\cap T^{-n} A\cap T^{-2n}\cap \dots \cap T^{-kn}A)>0$. A set of $1$-recurrence is called a set of measurable recurrence. Answering a question of Frantzikinakis, Lesigne, and Wierdl, we construct a set of $2$-recurrence $S$ with the property that $\{n^2:n\in S\}$ is not a set of measurable recurrence.
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id arxiv_https___arxiv_org_abs_2207_11851
institution arXiv
publishDate 2022
record_format arxiv
spellingShingle A set of 2-recurrence whose perfect squares do not form a set of measurable recurrence
Griesmer, John T.
Dynamical Systems
Combinatorics
37A44, 11B30
We say that $S\subset\mathbb Z$ is a set of $k$-recurrence if for every measure preserving transformation $T$ of a probability measure space $(X,μ)$ and every $A\subseteq X$ with $μ(A)>0$, there is an $n\in S$ such that $μ(A\cap T^{-n} A\cap T^{-2n}\cap \dots \cap T^{-kn}A)>0$. A set of $1$-recurrence is called a set of measurable recurrence. Answering a question of Frantzikinakis, Lesigne, and Wierdl, we construct a set of $2$-recurrence $S$ with the property that $\{n^2:n\in S\}$ is not a set of measurable recurrence.
title A set of 2-recurrence whose perfect squares do not form a set of measurable recurrence
topic Dynamical Systems
Combinatorics
37A44, 11B30
url https://arxiv.org/abs/2207.11851