# Is the fundamental group of $II_{1}$ factors invariant under a relation?

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

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 $\mathcal F(N) = \{ 1 \}$.
• Thank you very much @JessePeterson ! I learn a lot by reading your answer. The only point I don't understand is : why $M \simeq M \star L(\mathbb{F}_{\infty}) \Rightarrow M \star M \hookrightarrow M$ ? – Sebastien Palcoux Jul 28 '13 at 8:42
• @SébastienPalcoux: This is because if $u \in L(\mathbb F_\infty) \subset M * L(\mathbb F_\infty)$ is a unitary with trace $0$ then $M$ is in free position from $u M u^*$, hence $M * M \cong W^*( M, u M u^* ) \subset M * L(\mathbb F_\infty) \cong M$. – Jesse Peterson Jul 28 '13 at 17:20