Let $N$ be a type ${\rm II}$ factor, with trace $\tau$. Consider its fundamental group$$ \mathcal{F}(N)= \{ \tau(p)/\tau(q) \ | \ p,q \text{ non-zero finite projections in } N \text{ and } pNp \simeq qNq \}. $$


Let $\alpha$ be a free ergodic measure preserving action of a countable ICC group $\Gamma$ on a $\sigma$-finite standard Borel measure space $(X,\mu)$. Then $L^{\infty}(X,\mu) \rtimes\_{\alpha} \Gamma$ and $L(\Gamma)$ are type ${\rm II}$ factors. 



> 
> **Question**: Is $\mathcal{F}(L^{\infty}(X,\mu) \rtimes\_{\alpha} \Gamma)$ a subgroup of $\mathcal{F}(L(\Gamma))$?
> 
> 
> 


I did not find a counterexample in the following reference: [On the fundamental group of ${\rm II}\_1$ factors and equivalence relations arising from group actions](https://lirias.kuleuven.be/bitstream/123456789/290905/2/fundSoStC.pdf), by Sorin Popa and Stefaan Vaes. 


*Application*: A positive answer would solve the free group factor isomorphism problem.  

*Proof*: The group measure space construction $\mathcal{M} = L^{\infty}(\mathbb{S}^{1}, Leb) \rtimes\_{\alpha} \mathbb{F}\_{2}$ in [this post](https://mathoverflow.net/q/141577/34538) is a ${\rm III}\_1$ factor, so that its core $\widetilde{\mathcal{M}} = \mathcal{M} \rtimes\_{\sigma} \mathbb{R} = L^{\infty}(\mathbb{S}^1 \times \mathbb{R}\_{+}^\*, Leb) \rtimes\_{\widetilde{\alpha}} \mathbb{F}\_2$ (see [this answer](https://mathoverflow.net/a/309471/34538)) is a ${\rm II}\_{\infty}$ factor of fundamental group $\mathbb{R}\_{+}^\*$ (moreover $\widetilde{\alpha}$ is free and ergodic). But by assumption $\mathcal{F}(\widetilde{\mathcal{M}})$ would be a subgroup of $\mathcal{F}(L(\mathbb{F}\_{2}))$, so that $\mathcal{F}(L(\mathbb{F}\_{2})) = \mathbb{R}\_{+}^\*$ also, implying that for all $n \ge 2$, $L(\mathbb{F}\_{n}) \simeq L(\mathbb{F}\_{2})$, by the works of Voiculescu and Radulescu.


*Bonus question*: Can every subgroup of $\mathcal{F}(L(\Gamma))$ be realized as $\mathcal{F}(L^{\infty}(X,\mu) \rtimes\_{\alpha} \Gamma)$?


For this bonus question, I expect at most a counter-example (because a proof could be very hard). 




---


**Naive approach** for a positive answer to the main question: 


Let $N$ be $L^{\infty}(X,\mu) \rtimes\_{\alpha} \Gamma$, as specified above. First of all, $L(\Gamma)$ can be taken as a subfactor of $N$. Take $t \in \mathcal{F}(N)$, then there are projections $p,q \in L(\Gamma) \subset N$ such that $\tau(p)/\tau(q) = t$. Then, by definition of the fundamental group, $pNp$ is isomorphic to $qNq$ (because the isom. class of such compression depends only on the trace of the projection). Let $\Phi: pNp \to qNq$ be an isomorphism. Then $\Phi(pL(\Gamma)p)$ = $qKq$, for some $K \subset N$.


Can we choose $\Phi$ such that we can take $K = L(\Gamma)$?


If so, $t$ is in $\mathcal{F}(L(\Gamma))$, and the result follows.



Fundamental group and group measure space construction
I asked Stefaan Vaes by email, below is his answer (reproduced with his authorization): 



> 
> For instance, take $G$ to be the semidirect product of $\mathbb{Z}^2$
>  and ${\rm SL}(2,\mathbb{Z})$. The group von Neumann algebra $L(G)$ has
>  trivial fundamental group and this was even the very first ${\rm
>  II}\_1$ factor that was known to have trivial fundamental group (paper
>  of Popa in Annals of Math). Now fix a prime number $p$ and take the
>  essentially free, ergodic, pmp action of $G$ on $\mathbb{Z}\_p^2$
>  (where $\mathbb{Z}\_p$ are the $p$-adic integers). The action of $G$ on
>  $X=\mathbb{Z}\_p^2$ is given as follows: $\mathbb{Z}^2$ acts by
>  translation and ${\rm SL}(2,\mathbb{Z})$ acts in the "obvious" way.
> 
> 
> Then consider the subset $Y$ of $X$ given by $Y = (p \mathbb{Z}\_p)^2$.
>  The restriction of the orbit equivalence relation to $Y$ is precisely
>  the orbit equivalence relation of the semidirect product of $(p
>  \mathbb{Z})^2$ and ${\rm SL}(2,\mathbb{Z})$. So, multiplication by $p$
>  will make sure that $p^2$ belongs to the fundamental group of the
>  orbit equivalence relation, and thus to the fundamental group of the
>  crossed product ${\rm II}\_1$ factor.
> 
> 
> Conclusion: the crossed product has a nontrivial fundamental group,
>  but $L(G)$ has trivial fundamental group.
> 
> 
> 


Now, the topology of $\mathbb{Z}\_p$ is that of a Cantor set. Then: 


Can we still hope that the question has a positive answer if we restrict to a more "regular" case like a continuous action on a topological manifold (with or without boundary), or even (if necessary) a smooth action on a smooth manifold (with or without boundary)? 


Here is an other answer of Stefaan: 



> 
> When it comes to von Neumann algebras or ergodic equivalence
>  relations, there is no way to "see" such topological or geometric
>  properties of group actions. So I am sure that there are also
>  counterexamples that have natural, nice topological or geometric
>  models. It is just a bit harder to find such counterexamples.
> 
> 
> 


