Question

The Free Burnside group $G=B(2,665)=\langle a,b|g^{665} \rangle$ is infinite, by the work of Adyan and Novikov. Furthermore, the centralizer of any nonidentity element in $G$ is finite cyclic, and so the group is an i.c.c. group and the associated left group von Neumann algebra $LG$ is a type $II_{1}$ factor. It is a fact, due to of Adyan, that this group is not amenable, so the group von Neumann algebra is not injective.

A type $II_{1}$ factor $M$ with trace $\tau$ has Property $\Gamma$ if for every finite subset ${ x_{1}, x_{2},..., x_{n} } \subseteq M$ and each $\epsilon >0$, there is a unitary element $u$ in $M$ with $\tau (u)=0$ and $||ux_{j}-x_{j}u||_{2}<\epsilon$ for all $1 \leq j \leq n$. (Here $||T||_2=(\tau(T^{*}T))^{1/2}$ for $T\in M$.)

I should mention that if a group is not inner amenable in the sense described in

http://mathoverflow.net/questions/27233/is-there-an-i-c-c-nonamenable-simple-group-that-is-inner-amenable

then its left group von Neumann algebra does not have property $\Gamma$. (There exist i.c.c. inner amenable groups whose group von Neumann algebras don't have $\Gamma$, as recently shown by Stefaan Vaes: http://arxiv.org/PS_cache/arxiv/pdf/0909/0909.1485v1.pdf.)

My question is:

Does the group von Neumann algebra $LG$ have Property $\Gamma$?

Answer 1

If $G$ has a homomorphism onto a finite cyclic group, then it is inner amenable? (For every subset count the number of elements in the factor.) If so then the free Burnside group is certainly inner amenable. On the other hand, by Zelmanov's theorem, there exists a finite index subgroup of $B(2,665)$ which does not have finite factors. That group also is i.c.c., etc. So you may want to ask your question about that group. What is the reason for this question? There is a "similar" and quite popular question of whether $B(2,n)$, $n\gg 1$, has property (T). Y. Shalom conjectured in his ICM talk that it has. Gromov (unpublished) conjectured that it has not.

Edit: I was wrong in the first statement. The counting does not produce a measure because two disjoint subsets can map to the same set in the factor-group. So the question about inner amenability of $B(2,665)$ is open.