Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 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.

share|improve this question

1 Answer 1

up vote 9 down vote accepted

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 $\mathcal F(N) = \{ 1 \}$.

share|improve this answer
    
Thank you very much @JessePeterson ! I learn a lot by reading your answer. The only point I don't understand is : why $M \simeq M \star L(\mathbb{F}_{\infty}) \Rightarrow M \star M \hookrightarrow M$ ? –  Sébastien Palcoux Jul 28 '13 at 8:42
3  
@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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.