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The group von Neumann algebra $L\Gamma$ is a factor if and only if the group $\Gamma$ is ICC (i.e. infinite conjugacy class property). Moreover if $\Gamma$ is nontrivial then $L\Gamma$ is a $\mathrm{II}_1$ factor.

Question 0.1 (Sakai, Problem 4.4.10 here, 1971): Is every $\mathrm{II}_1$ factor a group von Neumann algebra?
Answer (Connes here, 1975): No, because there is a $\mathrm{II}_1$ factor which is not anti-isomorphic to itself (whereas $L\Gamma$ is so).

Question 0.2: Is every $\mathrm{II}_1$ factor anti-isomorphic to itself a group von Neumann algebra?
Answer (Jones, here, 1980): No, because there is a $\mathrm{II}_1$ factor which is anti-isomorphic to itself but without involutory antiautomorphisms (whereas $L\Gamma$ has so).

The answers for the next two questions were pointed out by Jiang's comments.

Question 0.3 (Remark 5.7 in previous paper): Is a $\mathrm{II}_1$ factor with an involutory antiautomorphism a group von Neumann algebra?
Answer (Ioana here, 2010): No, because there are $\mathrm{II}_1$ factors $M$ whose amplifications $M^t$ (with $t \neq 1$) admit an involutory antiautomorphism but are not group von Neumann algebras (see Corollary F, Corollary 10.1 and Remark 10.3).

Question 0.4: Is a $\mathrm{II}_1$ factor with an involutory antiautomorphism stably isomormphic to a group von Neumann algebra (i.e. of the form $(L\Gamma)^t$)?
Answer (Boutonnet here, 2013): No, because there are crossed product von Neumann algebra $L^{\infty}(X,\mu) \rtimes_{\sigma} \Gamma$, with $σ$ a free ergodic pmp-action of an ICC group $Γ$ on a probability space $(X,μ)$, which are not stably isomorphic to a group von Neumann algebra.

Question: Is a $\mathrm{II}_1$ factor with an involutory antiautomorphism stably isomorphic to a crossed product $A \rtimes G$, with $A$ an abelian von Neumann algebra and $G$ a group?

Recall that if the action of the group $G$ on the abelian von Neumann algebra $A$ is free then $A \rtimes G$ is a factor iff the action is ergodic. Now $A \rtimes G$ can be a factor without the action being free, for example when $A = \mathbb{C}$, because then $A \rtimes G = LG$.

Remark: The existence of a Kac algebra generating a $\mathrm{II}_1$ factor not stably isomorphic to any $A⋊G$ (as above) would be very interesting.

The initial question and then Connes' paper were pointed out to me by Keshab Chandra Bakshi.

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    $\begingroup$ The answer is negative as shown in Ioana's paper ``W*-superrigidity for Bernoulli actions of property (T) groups", see the discussion after corollary F there. $\endgroup$
    – Jiang
    Commented Nov 12, 2019 at 15:12
  • $\begingroup$ @Jiang: Thanks! After Sakai, Connes, Jones and now Ioana, what should be the next updated question? Because you posted your answer as a comment and not as an answer, I understand that you implicitly agree that I update the question according to your comment. I just updated the question. Please let me know whether you appreciate and whether you also know the answer. $\endgroup$
    – Sebastien Palcoux
    Commented Nov 12, 2019 at 18:27
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    $\begingroup$ This was answered negatively again by theorem D in Remi Boutonnet's paper "W^*-superrigidity of mixing Gaussian actions of rigid groups". Indeed, the point is to find W^*-superrigid actions such that (Thm C in Ioana's paper holds) and the crossed product vn alg from this action is NOT a group vn alg. $\endgroup$
    – Jiang
    Commented Nov 12, 2019 at 21:31
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    $\begingroup$ I think that references to Ioana and Boutonnet would make a good answer, and that further questions could be asked separately (what I upvoted was the original question because of its simplicity, not its latest version). $\endgroup$
    – YCor
    Commented Nov 14, 2019 at 0:25
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    $\begingroup$ But I'm really not a specialist in vN algebras. That a question has been solved recently doesn't make it off-topic. I upvoted the initial question because it was telling to me, and I think would be worth an answer by @Jiang (and, fine, along with its first variant solved by Boutonnet), maybe with just a reference, or with some further hints. $\endgroup$
    – YCor
    Commented Nov 14, 2019 at 10:17

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