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](http://www.sciencedirect.com/science/article/pii/0022123672900043)) : 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](https://mathoverflow.net/questions/137520/whats-a-noncommutative-set)).


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

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](http://www.pnas.org/content/101/3/723.full)) : The fundamental group of a type $II\_{1}$ factor is the set of numbers $t > 0$ for which its
>  *amplification* by $t$ 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. 



Is the fundamental group of $II_{1}$ factors invariant under a relation?
Here is a counterexample. I don't see any easy way to augment the question to something more natural. 


Let $Q$ be a $w$-rigid II$\_1$ factor with trivial fundamental group, e.g., $Q = L( \mathbb Z^2 \rtimes SL\_2(\mathbb Z) )$. Let $\mathcal S \subset \mathbb R\_+^\*$ be a non-trival subgroup. Set $M = \*\_{s \in \mathcal S} Q^s$, and $N = Q \* M$. (We may take $M$ and $N$ separable if we take $\mathcal S$ countable, and $Q$ separable.) Clearly we have $M \hookrightarrow N$, and by a result of Dykema and Rădulescu (Theorem 1.5 from <http://www.ams.org/mathscinet-getitem?mr=1735079>) we have $M \cong M \* L(\mathbb F\_\infty)$ from which it follows easily that $N \hookrightarrow M \* M \hookrightarrow M$. 


(Note that by Umegaki's Theorem <http://www.ams.org/mathscinet-getitem?mr=68751> there always exist normal conditional expectations for von Neumann subalgebras of II$\_1$ factors.)


Corollary 6.5 in my paper with Ioana and Popa <http://www.ams.org/mathscinet-getitem?mr=2386109> shows that $\mathcal F(M) = \mathcal S$, while $\mathcal F(N) = \{ 1 \}$.


