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Suppose $G$ is a discrete group acting by homeomorphisms on a compact Hausdorff space $X$, such that the action is minimal. Fix an invariant Radon measure $\nu$ on $X$. Is topologically free (the fixed point set $\operatorname{Fix}(g)$ has empty interior for $g \neq e$) equivalent to essentially free ($\operatorname{Fix}(g)$ has $\nu$-measure zero for $g \neq e$)?

"Essentially free $\implies$ topologically free" is easy - this follows from $\nu$ having full support. The converse feels like it shouldn't be true, but I can't seem to come up with any counterexample. A positive answer even in special cases would be interesting, for example amenable $G$ or metrizable $X$.

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  • $\begingroup$ Interesting question! I think I'm missing something. Why for a minimal shift does having an essentially free invariant measure with full support imply topological freeness? (Without the minimality assumption the Full/Bernoulli shift, and measure of maximal entropy, is a counterexample.) $\endgroup$
    – Josh F
    Feb 7, 2021 at 8:00
  • $\begingroup$ @JoshF Because if you had a fixed point set with nonempty interior, then it would have to have positive measure by $\nu$ having full support (every open set has positive measure). In fact, this is why I require minimality - any invariant measure necessarily has full support. $\endgroup$
    – Dan Ursu
    Feb 7, 2021 at 16:31
  • $\begingroup$ Right, but the set of points fixed by a given g is a closed set and can have empty interior even for minimal dynamical systems. For example, the free group acting on its (Gromov) boundary only has countably many points with non-trivial stabilizer. $\endgroup$
    – Josh F
    Feb 7, 2021 at 19:20
  • $\begingroup$ @JoshF I know - I'm asking if the fixed point sets all having empty interior will imply that the measure of the fixed point sets are necessarily zero with respect to a given invariant measure. (Note that the Gromov boundary of $F_n$ has no invariant measures). $\endgroup$
    – Dan Ursu
    Feb 8, 2021 at 1:57

2 Answers 2

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No. It sounds exotic, but it is actually easy to construct examples of actions on rooted trees which are topologically free, but the fixed point set has positive measure. You just have to know what you are looking for. Check out this paper by Groger & Lukina for a discussion, and how to construct examples - https://arxiv.org/abs/1911.00680

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  • $\begingroup$ Would you be more specific about the group $G$? For instance, is it true that every infinite countable group has a minimal action on the Cantor set, that is topologically free, and preserves a Radon probability measure for which the action is not essentially free? $\endgroup$
    – YCor
    Dec 15, 2022 at 11:37
  • $\begingroup$ This is not true for every infinite countable infinite group. For example, $\mathbb{Z}$ would be a counterexample, since every minimal $\mathbb{Z}$ action on the Cantor set is free. $\endgroup$
    – Owen Tanner
    May 15, 2023 at 9:20
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Sorry for the lateness to this thread, but in case you are still interested, what you are talking about is the notion of an allosteric action. It's an active research question to determine which groups admit allosteric actions, and is very relevant to the study of which group actions give rise to classifiable crossed product C*-algebras since it is known that almost finiteness implies essentially free. For this reason it is known that every topologically free action of an elementary amenable group is almost finite and is therefore essentially free. But even basic examples of amenable-but-not-elementary amenable groups admit allosteric actions.

See for example work by Matthieu Joseph for examples of allosteric actions of amenable groups on the Cantor space:

https://arxiv.org/abs/2301.07616

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