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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 $\Phi(pL(\Gamma)p)$ = $qXq$.

Can we choose $\Phi$ such that $X = L(\Gamma)$?

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

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.

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).

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 $\Phi(pL(\Gamma)p)$ = $qXq$.

Can we choose $\Phi$ such that $X = L(\Gamma)$?

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

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 $\mathcal{F}(L(\Gamma))$?

A reference related to such question is the paper On the fundamental group of ${\rm II}_1$ factors and equivalence relations arising from group actions, of Sorin Popa and Stefaan Vaes.

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

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).