The free group II$_{\infty}$ factor isomorphism problem

Question

Let $\Gamma$ be an infinite discrete group, and $H = l^{2}(\Gamma)$ the separable infinite dimensional Hilbert space.
Let $\rho$ be the left regular representation of $\Gamma$ on $H$.

Definition : Let $L(\Gamma) := \rho (\Gamma)''$ the von Neumann algebra generated by $\Gamma$.

Proposition : $L(\Gamma)$ is a $II_{1}$ factor iff $\Gamma$ is ICC (infinite conjugacy class).

Let $\Gamma = \mathbb{F}_{n} = \langle a_{1}, a_{2},..., a_{n} \vert \rangle$ the free group with $n$ generator. It's ICC iff $n \geq 2$.

Free group factor isomorphism problem : $L(\mathbb{F}_{n}) ≃ L(\mathbb{F}_{m})$, $∀n,m≥2$ ?

Voiculescu's formula (see here p 3) : $L(\mathbb{F}_{n}) \simeq L(\mathbb{F}_{(n-1)k^{2}+1}) \otimes M_{k}(\mathbb{C})$

Example: $L(\mathbb{F}_{2}) \simeq L(\mathbb{F}_{5}) \otimes M_{2}(\mathbb{C})$

Corollary: $\bigotimes_{i} L(\mathbb{F}_{n_{i}}) \simeq \bigotimes_{i} L(\mathbb{F}_{m_{i}})$ if $\prod_{i}(n_{i}-1) = \prod_{i}(m_{i}-1)$

Observation : $L(\mathbb{F}_{n}) \otimes B(H) \simeq L(\mathbb{F}_{(n-1)k^{2}+1}) \otimes M_{k}(\mathbb{C}) \otimes B(H) \simeq L(\mathbb{F}_{(n-1)k^{2}+1}) \otimes B(H)$

However, "$\exists k_{1}, k_{2} $ : $(n-1)k_{1}^{2}+1 = (m-1)k_{2}^{2}+1$" $\Leftrightarrow$ "$(n-1)(m-1)$ is a square".

Corollary: $L(\mathbb{F}_{n}) \otimes B(H) \simeq L(\mathbb{F}_{m}) \otimes B(H)$ if $(n-1)(m-1)$ is a square.

Question : Is it known to be true in general : $L(\mathbb{F}_{n}) \otimes B(H) \simeq L(\mathbb{F}_{m}) \otimes B(H)$ $\forall n, m \ge 2$ ?
(if "no", is it equivalent to the free group factor isomorphism problem ?)

In general: $L(\mathbb{F}_{n}) \simeq L(\mathbb{F}_{m})^{\alpha} $ with $\alpha = (\frac{m-1}{n-1})^{1/2}$.

Definition: Let $\mathcal{M}$ be a $II_{1}$ factor, then $\mathcal{M}^{t} \simeq p(M_{k}(\mathbb{C}) \otimes \mathcal{M})p \simeq q(B(H) \otimes \mathcal{M})q$,
with $q \in B(H) \otimes \mathcal{M}$ and $p \in M_{k}(\mathbb{C}) \otimes \mathcal{M}$ projections, $Tr(q) = t$, $k=\lceil t \rceil$ and $\tau(p) = \frac{t}{k}$.

Remark : Let $n, m \ge 2$ such that $(n-1)(m-1)$ is a square.
Let the $II_{\infty}$ factor $\mathcal{N} \simeq L(\mathbb{F}_{n}) \otimes B(H) \simeq L(\mathbb{F}_{m}) \otimes B(H)$.
Let $Tr$ (resp. $Tr'$) be the trace on $\mathcal{N}_{+}$ coming from the unique trace of $L(\mathbb{F}_{n})$ (resp. $L(\mathbb{F}_{m})$).
Let the projection $p, q \in \mathcal{N}$ such that $Tr(p) = Tr'(q) = 1$.
Then $p \mathcal{N} p \simeq L(\mathbb{F}_{n})$ and $q \mathcal{N} q \simeq L(\mathbb{F}_{m})$. So, $Tr' = \alpha Tr$ with $\alpha = (\frac{n-1}{m-1})^{1/2}$.

Answer 1

Yes, this is true.

Radulescu introduced the interpolated free group factors $L(\mathbb{F}_r)$ for all $1<r<\infty$. They obey the rule that if $r\in\mathbb{N}$ then we get the proper free group factor and they also obey,

$L(\mathbb{F}_r)^s=L(\mathbb{F}_{r'})$ where $r'={1+\frac{r-1}{s^2}}$. This gives that all free group factors are amplifications of each other. Which implies that they all generate the same $II_\infty$ factors. (ie $L(\mathbb{F}_n)\overline\otimes\mathcal{B(H)}\simeq L(\mathbb{F}_m)\overline\otimes\mathcal{B(H)} $.

Further, this reduces the free group factor isomorphism problem to the calculation of the fundamental groups one of the finite free group factors.

It is known that it is either

I.) All of $\mathbb{R}^+$ in which case they are all isomorphic.

II.) Or $\{1\}$, in which case they are all non-isomorphic.

Here is part of Dykema's thesis where this is proved.

http://arxiv.org/pdf/funct-an/9211012v1.pdf