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Is the set of all ICC amenable groups countable?

If "yes", then in general, the classes of all countable ICC groups that give rise to the same von Neumann algebra (factor) -- are these classes always countable, do we know? Is it useful to consider this as an equivalence relation?

How about the same question with the crossed product construction of factors, and equivalence classes of (measure space with a group action) setups?

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    $\begingroup$ No, there are continuum many non-isomorphic simple locally finite groups (there are continuum many locally finite fields, take their $PSL_2$). $\endgroup$
    – YCor
    Commented Aug 28, 2020 at 6:37
  • $\begingroup$ Thank you. Shouldn't you post it as an answer so that the question doesn't stay near the top? $\endgroup$
    – Chilperic
    Commented Aug 28, 2020 at 8:17
  • $\begingroup$ Done. Just to help the reader, recall that "icc" means the FC-center, defined as the union of all finite conjugacy classes, is reduced to $\{1\}$. That is, all conjugacy classes other than $\{1\}$ are infinite. $\endgroup$
    – YCor
    Commented Aug 28, 2020 at 8:34

2 Answers 2

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The set of countable ICC groups doesn't exist, so I think you're asking about isomorphism classes.

There are continuum many non-isomorphic locally finite fields (e.g., take, for $S$ any set of primes, the invariants by $\prod_{p\in S}\mathbf{Z}_p$, which has Galois group $\prod_{p\notin S}\mathbf{Z}_p)$ in the algebraic closure of $\mathbf{F}_p$.

It's known that non-isomorphic fields give rise to non-isomorphic groups $\mathrm{PSL}_2$. Hence, taking $\mathrm{PSL}_2$ of such fields yield continuum many non-isomorphic locally finite (hence amenable) groups, which are infinite simple, hence icc.

(Actually there are continuum many non-isomorphic solvable icc finitely generated groups as well, by another argument.)

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As for the second question, the equivalence relation on the class of countable ICC groups given by $G \sim \Gamma$ if and only if $L(G) \cong L(\Gamma)$ is very interesting and usually called $W^*$-equivalence of $G$ and $\Gamma$.

On page 45 of [Con82], Connes conjectures that the $W^*$-equivalence class of an ICC property (T) group is a singleton (up to isomorphism). Such groups are now called $W^*$-superrigid. This conjecture is wide open. At the moment, there is no counterexample, but also no example: there is no known $W^*$-superrigid property (T) group.

In Section 4 of [Pop07] it is proven that an ICC property (T) group is $W^*$-equivalent with at most countably many nonisomorphic groups.

In [IPV10] we introduced the first family $W^*$-superrigid ICC groups. These examples are given by a generalized wreath product construction and do not have property (T).

For all countably infinite abelian groups $\Gamma$, the group von Neumann algebra $L(\Gamma)$ is the unique diffuse abelian von Neumann algebra. One can deduce that for all countably infinite abelian groups $\Gamma_1,\Gamma_2$, the ICC group $\Gamma_1 * \Gamma_2$ is $W^*$-equivalent to the free group $\mathbb{F}_2$. More generally, by Theorem 4.6 in [Dyk92], whenever $\Gamma_1$ and $\Gamma_2$ are infinite amenable groups, we have that their free product $\Gamma_1 * \Gamma_2$ is $W^*$-equivalent to $\mathbb{F}_2$.

[Con82] A. Connes, Classification des facteurs. In Operator algebras and applications, Part 2 (Kingston, 1980), Proc. Sympos. Pure Math. 38, Amer. Math. Soc., Providence, 1982, pp. 43–109.

[Pop07] S. Popa, Deformation and rigidity for group actions and von Neumann algebras. In International Congress of Mathematicians (Madrid 2006), Eur. Math. Soc., Zürich, 2007, pp. 445-477.

[IPV10] A. Ioana, S. Popa and S. Vaes, A class of superrigid group von Neumann algebras. Ann. of Math. 178 (2013), 231-286.

[Dyk92] K. Dykema, Free products of hyperfinite von Neumann algebras and free dimension. Duke Math. J. 69 (1993), 97–119.

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  • $\begingroup$ Thanks very much, Stefaan and YCor. I was aware of some of this work on cardinality of these classes. I was wondering more about thinking of them along the results we have for countable Borel equivalence relations. Does Silver's dichotomy hold -- either countable number of classes or an uncountable number of singletons? Is it possible to have a group action so that the classes are orbits? That sort of thing. $\endgroup$
    – Chilperic
    Commented Aug 29, 2020 at 6:05
  • $\begingroup$ I think that in your question, your are only mentioning countability of equivalence classes. Maybe you can post a new and more precise question about these other points that you have in mind. $\endgroup$ Commented Aug 30, 2020 at 8:38

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