Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The hyperfinite $II_1$ factor arises as the group von Neumann algebra of any infinite amenable group such that every conjugacy class but that of the identity has infinite cardinality. The unitary group of this von Neumann algebra contains every countable discrete amenable group as a subgroup.

I'd like to collect here some exotic (countable discrete) subgroups of the unitary group of the hyperfinite $II_{1}$ factor that aren't amenable. As subjective as it sounds, here's my question:

What is the most interesting discrete subgroup appearing as a subgroup of the unitary group of the hyperfinite $II_1$ factor?

(An invariant mean on a group G gives rise to a hypertrace for LG, but if I have a subgroup of the unitary group that gives rise to a representation not the regular representation, this no longer works.)

share|improve this question
    
I added the examples-counterexamples tag to your question. –  Alex Becker Jan 6 '12 at 20:26
    
Thank you Alex! –  Jon Bannon Jan 6 '12 at 20:28

1 Answer 1

Any residually amenable group embeds into the unitary group of the hyperfinite II$_1$ factor $R$. Indeed, if $\Gamma_n \lhd \Gamma$ is a decreasing family of normal subgroups such that $\Gamma/\Gamma_n$ is amenable then the representation $\pi = \oplus_n \lambda_{\Gamma/\Gamma_n}$ will be a faithful representation which embeds $\Gamma$ into the unitary group of $\oplus_n L(\Gamma/\Gamma_n)$ which is an amenable finite von Neumann algebra and hence appears as a subalgebra of $R$.

I'm not aware of any countable subgroup of $\mathcal U(R)$ which is known to not be residually amenable.

share|improve this answer
    
This is very interesting, Jesse. Is it conjectured that all discrete subgroups of the unitary group of R are residually amenable? –  Jon Bannon Mar 18 at 23:54
    
@Jon: This is true for property (T) groups by a result of Kirschberg, but I think it's unlikely to be true for all groups. –  Jesse Peterson May 28 at 22:30

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.