# Fundamental group and group measure space construction

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, 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 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) 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.

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.