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.


1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy