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arxiv: 1204.0829 · v2 · pith:H4Q6UP7Tnew · submitted 2012-04-03 · 🧮 math.LO · math.DS

Finite generators for countable group actions in the Borel and Baire category settings

classification 🧮 math.LO math.DS
keywords borelactioncountablegeneratorpolishansweraperiodicfinite
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For a continuous action of a countable discrete group $G$ on a Polish space $X$, a countable Borel partition $P$ of $X$ is called a generator if $G \cdot P := \{ gC : g \in G, C \in P \}$ generates the Borel $\sigma$-algebra of $X$. For $G = Z$, the Kolmogorov--Sinai theorem gives a measure-theoretic obstruction to the existence of finite generators: they do not exist in the presence of an invariant probability measure with infinite entropy. It was asked by Benjamin Weiss in the late 80s whether the nonexistence of any invariant probability measure guarantees the existence of a finite generator. We show that the answer is positive (in fact, there is a 32-generator) for an arbitrary countable group $G$ and $\sigma$-compact $X$ (in particular, for locally compact $X$). We also show that any continuous aperiodic action of $G$ on an arbitrary Polish space admits a 4-generator on a comeager set, thus giving a positive answer to a question of Alexander Kechris asked in the mid-90s. Furthermore, assuming a positive answer to Weiss's question for arbitrary Polish spaces and $G = Z$, we prove the following dichotomy: every aperiodic Borel action of $Z$ on a Polish space $X$ admits either an invariant probability measure of infinite entropy or a finite generator. As an auxiliary lemma, we prove the following statement, which may be of independent interest: every aperiodic Borel action of a countable group $G$ on a Polish space $X$ admits a $G$-equivariant Borel map to the aperiodic part of the shift action of $G$ on $2^G$. We also obtain a number of other related results, among which is a criterion for the nonexistence of non-meager weakly wandering sets for continuous actions of $Z$. A consequence of this is a negative answer to a question asked by Eigen--Hajian--Nadkarni, which was also independently answered by Benjamin Miller.

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