This issue is in continuation of an answer I gave here about

noncommutative sets.

In order to define the equivalence relation, let's first recall the Tomita-Takesaki modular theory and conditional expectation for von Neumann algebras.

Let $H$ be a separable Hilbert space and $B(H)$ the algebra of bounded operators.

**Definition**: A von Neumann algebra is a *-subalgebra $M \subset B(H)$ stable under bicommutant: $M^{*} = M$ and $M'' = M$.

**Modular theory** : Let $M \subset B(H)$ be a von Neumann algebra. Let $\Omega \in H$ be a *cyclic* and *separating* vector (i.e., $M.\Omega$ and $M'.\Omega$ are dense in $H$). Let $S : H \to H$ be the closure of the anti-linear map $x\Omega \to x^{*}\Omega$. Then, $S$ admits a polar decomposition $S = J\Delta^{1/2}$, with $J$ anti-linear unitary and $\Delta$ positive. Then, $JMJ = M'$ and $\Delta^{it} M \Delta^{-it} = M$.

Let $\sigma_{\Omega}^{t}(x) = \Delta^{it} x \Delta^{-it}$ the modular action of $\mathbb{R}$ on $M$.

**Conditional expectation** (Takesaki 1972) : Let $N \subset M$ be an inclusion of von Neumann algebra, then there is a conditional expectation of $M$ onto $N$ with respect to $\Omega$ (cyclic and separating) if $N$ is invariant under the modular action (i.e., $\sigma_{\Omega}^{t}(N) = N)$.

**Notation** : if $\exists \Omega$ verifying the previous conditions, we note $N \subset_{e} M$.

**Remark** : The modular theory is trivial for $M = L(\Gamma) \subset B(H)$, with $\Gamma$ a discrete group and $H = l^{2}(\Gamma)$ (because $\Delta = I$). In particular, it's trivial for the abelian von Neumann algebras.

As a consequence, in this case: $N \subset M$ $\Leftrightarrow$ $N \subset_{e} M$.

**Notation** : Let $N$ and $M$ be two von Neumann algebras.

If $\exists P \simeq N$ such that $ P \subset_{e} M$, we note $N \hookrightarrow_{e} M$.

Equivalence relation: $M \sim N$ if $N \hookrightarrow_{e} M \hookrightarrow_{e} N$.

**Philosophy** : $M \sim N$ could significate they are isomorphic as *noncommutative sets* (see here).

**Examples** :

- Among $l^{\infty}(\{1,2,...,n \})$, $l^{\infty}(\mathbb{N})$ and $L^{\infty}([0,1])$ none is equivalent to another.
- $L^{\infty}([0,1])$, $L^{\infty}([0,1]\cup \{1,2,...,n \})$
and $L^{\infty}([0,1]\cup \mathbb{N})$ are pairwise equivalent,

because $L^{\infty}([0,1]) \subset L^{\infty}([0,1] \cup \{2,3,...,n\}) \subset L^{\infty}([0,1] \cup \mathbb{N}_{\geq 2}) \subset L^{\infty}(\mathbb{R})$

and $L^{\infty}([0,1]) \simeq L^{\infty}(\mathbb{R})$ - Obviously $L^{\infty}([0,1]) \not\sim B(H)$.
- Let $R \subset B(H)$ be the hyperfinite $II_{1}$ factor, $R_{\infty} = R \otimes B(H)$ the hyperfinite $II_{\infty}$ factor. $ B(H) \hookrightarrow_{e} R_{\infty} \hookrightarrow_{e} B(H \otimes H)$ and $B(H) \simeq B(H \otimes H)$. So, $R \not\sim B(H) \sim R_{\infty}$.
- Let $\Gamma$ be a non-amenable ICC discrete group. Then $L(\Gamma) \not\hookrightarrow_{e} B(H)$ and $L_{\infty}(\Gamma) = L(\Gamma) \otimes B(H) \not\hookrightarrow_{e} B(H \otimes H) $ so $L(\Gamma) \not\sim B(H) \not\sim L_{\infty}(\Gamma)$.
- Let $\mathbb{F}_{2} = \langle a,b \vert \ \rangle $ and $\mathbb{F}_{\infty} = \langle a_{1},a_{2},... \vert \ \rangle $.

Then $\mathbb{F}_{2} \hookrightarrow \mathbb{F}_{n} \hookrightarrow \mathbb{F}_{\infty} \hookrightarrow\mathbb{F}_{2} $ (the last injection is given by $a_{n} \to b^{-n}ab^{n}$).

Consequence : $L(\mathbb{F}_{2}) \sim L(\mathbb{F}_{n}) \sim L(\mathbb{F}_{\infty}) $

Fundamental group(see here) : The fundamental group of a type $II_{1}$ factor is the set of numbers $t > 0$ for which itsamplificationby $t$ is isomorphic to itself: $\mathcal{F}(M) = \{t>0 \ \vert \ M^{t}\simeq M \}$.

**Examples**:

- There is a semi-direct product $ \Gamma = \mathbb{Z}^{2} \rtimes SL(2,\mathbb{Z})$ such that $\mathcal{F}(L(\Gamma)) = \{1\}$
- It's countable for $II_{1}$ factors with property (T).
- $\mathcal{F}(R) = \mathcal{F}(L(\mathbb{F}_{\infty})) = \mathbb{R}_{+}^{*}$
**Open**: $\mathcal{F}(L(\mathbb{F}_{2})) = \{1\}$**or**$\mathbb{R}_{+}^{*}$, but we still do not know which it is.

This is a reformulation of the free group factor isomorphism problem: $L(\mathbb{F}_{2}) \simeq L(\mathbb{F}_{\infty}) $ ?

Question: Is the fundamental group $\mathcal{F}(M)$ of a $II_{1}$ factor $M$ invariant under $\sim$ ?

**Remark** : an affirmative answer would solve the free group factor isomorphism problem.

Because this problem is very difficult, if this question admits an affirmative answer, I do not expect that the proof will be given here without a colossal work, but I would be interested to know if (in your opinion) this way seems promising. If it admits a negative answer, then in addition to a possible counter-example, I would be interested to know if you see a manner to reformulate the question for becoming open.