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Let $G$ be a countable discrete residually finite group.

Is there a way to characterise the actions of $G$ that are orbit-equivalent to profinite ones?

Ozawa and Popa introduced the concept of weakly compact actions. Weakly compact actions are stable under orbit equivalence and profinite actions are weakly compact.

Is it possible to find a weakly compact action that is not orbit equivalent to any profinite action?

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What does "orbit-equivalent" and "weakly compact" mean? –  Arno Kret May 20 '11 at 5:07
    
... and "profinite" action? –  Arno Kret May 20 '11 at 5:08
    
When you say "orbit equivalent to profinite ones" do you mean to profinite actions by the same group $G$ or do you allow the group to change? –  Jesse Peterson May 20 '11 at 11:55
    
Orbit equivalent means that the associated equivalence relations are isomorphic as measured equivalence relations. The profinite actions are projective limits of finite actions. The definition of weakly compact is somehow more technical and if you really want to know the definition, you should check the article of Ozawa and Popa. –  Alessandro Carderi May 20 '11 at 12:02
    
@Jesse I mean profinite actions of a fixed group $G$. But, if you know some other characterisation, please tell me. –  Alessandro Carderi May 20 '11 at 12:08
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1 Answer

up vote 8 down vote accepted

If a group has a free profinite action then in particular it must be residually finite. On the other hand, any measure preserving action of an amenable group is weakly compact. Thus, any free measure preserving action of infinite amenable group $G$ which is not residually finite cannot be orbit equivalent to a profinite action of $G$.

Of course by Ornstein and Weiss' theorem all such orbit equivalence relations are hyperfinte and so this doesn't work if you allow yourself to change the group $G$.

Update:

For residually finite groups an example will have to be more sophisticated since this will not work with amenable groups. However, there are enough orbit equivalence rigidity theorems around that this is still possible. The relevant theorem here is Ioana's rigidity theorem for profinite actions of property (T) groups (http://arxiv.org/abs/0805.2998):

Theorem[Adrian Ioana]: Let $\Gamma$, and $\Lambda$ be countable discrete groups and suppose that $\Gamma$ has property (T). Let $\Gamma \curvearrowright X$ be a free, ergodic, profinite, probability measure preserving action and suppose that $\Lambda \curvearrowright Y$ is a free probability measure preserving action which is orbit equivalent to the action $\Gamma \curvearrowright X$. Then there exists finite index subgroups $\Gamma_1 < \Gamma$ and $\Lambda_1 < \Lambda$ such that restricting to an $\Gamma_1$ (resp. $\Lambda_1$) ergodic component $X_1$ (resp. $Y_1$) we have that the actions $\Gamma_1 \curvearrowright X_1$ and $\Lambda_1 \curvearrowright Y_1$ are conjugate.

In particular, this shows that for orbit equivalent actions of property (T) groups if one action is profinite then so is the other. This is because a measure preserving action $\Gamma \curvearrowright X$ is profinite if and only if the weak closure $\overline \Gamma$ of $\Gamma$ in Aut$(X)$ is a profinite group, and $\overline \Gamma$ is invariant under conjugation. Thus one just needs to construct a weakly compact action of a residually finite property (T) group which is not profinite.

One such example is given by Dennis Sullivan (http://www.ams.org/mathscinet-getitem?mr=590825) who showed that for $n > 3$ there is a countable dense subgroup $\Gamma_n$ of $O(n + 1)$ which has property (T). $\Gamma_n$ will be residually finite by a classical result of Malcev since it is finitely generated and linear. Also, the Haar measure preserving action $\Gamma_n \curvearrowright O(n + 1)$ given by left multiplication will be free (since $\Gamma_n$ is a subgroup), ergodic (since $\Gamma_n$ is dense), and compact (since $O(n + 1)$ is compact), and hence also weakly compact. However, $\Gamma_n \curvearrowright O(n + 1)$ is not profinite since $O(n + 1) = \overline{\Gamma_n}$ is not profinite.

Note that this gives an example if you consider orbit equivalence with other groups $\Lambda$ as well.

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You're right, for non-residually finite groups the question is a kind of stupid. Do you know any counterexample in the residually finite world? I will add the assumption in the question. –  Alessandro Carderi May 20 '11 at 13:44
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