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

Let $\mathcal{R}$ be the hyperfinite factor of type $\rm{II}_1$ and let $\mathbb{F}_n$ be a free group with $n$ generators. Let $\alpha$ be an action of $\mathbb{F}_n$ on $\mathcal{R}$.

Does the von Neumann crossed product $\mathcal{R}\rtimes_{\alpha}\mathbb{F}_n$ have the QWEP?

Remarks: Since $\mathbb{F}_n$ is a residually finite group, the group von Neumann algebra $\rm{VN}(\mathbb{F}_n)$ has the QWEP. Moreover $\mathcal{R}$ has the QWEP.

share|improve this question
1  
By the way, it's good practice to define your notation (QWEP?). You wouldn't want to make the mistake of Serre, "Let X be an E.C. with. C.M." where "with." stands, of course, for "without". =) –  Harry Gindi Jul 10 '10 at 7:52
    
@Harry: normally I agree, but I think almost everyone in the intended audience for this question would have heard of, or at least know how to interpret Google results for, "QWEP". Not that I have any clue as to the answer, mind you. –  Yemon Choi Jul 10 '10 at 9:27
3  
QWEP = Quotient Weak Expectation Property –  Jesse Peterson Jul 10 '10 at 9:32

1 Answer 1

up vote 8 down vote accepted

Yes. If $a$ and $b$ are generators of $\mathbb F_2$ then $\mathcal R \rtimes_\alpha \mathbb F_2$ decomposes as an amalgamated free product of $(\mathcal R \rtimes_\alpha \langle a \rangle)$ and $(\mathcal R \rtimes_\alpha \langle b \rangle)$ over $\mathcal R$, where each of these are hyperfinite. Brown, Dykema, and Jung showed in http://arxiv.org/abs/math/0609080 that for separable finite von Neumann algebras being embeddable into $\mathcal R^\omega$ is stable under amalgamated free products over a hyperfinite von Neumann algebra. Thus $\mathcal R \rtimes_\alpha \mathbb F_2$ is embeddable into $\mathcal R^\omega$, which is equivalent to QWEP. Induction then gives the case when $2 \leq n < \infty$, and the case $n = \infty$ then follows since QWEP is preserved under (the weak-closure of) increasing unions.

Related to this, Collins and Dykema in http://arxiv.org/abs/1003.1675 have recently shown that the class of Sophic groups is stable under taking amalgamated free products over amenable groups.

I believe this is an open problem however if we consider arbitrary residually finite groups instead of only $\mathbb F_n$.

share|improve this answer
    
I am very grateful to you for showing me all this! A last question: do you know a reference for the following assertion $\mathcal{R}\rtimes_{\alpha} \mathbb{F}_2$ decomposes as an amalgamated free product of $\mathcal{R} \rtimes_{\alpha}<a>$ and $\mathcal{R}\rtimes_{\alpha}<b>$ over $\mathcal{R}$. –  BigBill Jul 10 '10 at 13:24
1  
This follows, more or less, directly from the definition of amalgamated free products for tracial von Neumann algebras. You just need to show that $\mathcal R \rtimes_\alpha \langle a \rangle$ and $\mathcal R \rtimes_\alpha \langle b \rangle$ are freely independent relative to $\mathcal R$. See for instance springerlink.com/content/k0u6213wg2227v00. –  Jesse Peterson Jul 10 '10 at 18:22
    
Thank you very much. –  BigBill Jul 13 '10 at 14:54

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.