TITLE: Do Burnside Group Factors have Gamma?
QUESTION [12 upvotes]: 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 
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
  $\Gamma$?

REPLY [5 votes]: The answer is no, the group von Neumann algebra of the Burnside group $B(2,665)$ does not have property Gamma. In fact, $B(2,665)$ is not inner amenable. 
To see this, first note that, as pointed out in the question, $B(2,665)$ is nonamenable, and the centralizer of every non-identity element of $B(2,665)$ is finite. However, every nonamenable inner amenable group $G$ contains a non-identity element whose centralizer is nonamenable. This is a consequence of applying the following well-known folklore lemma to the conjugation action of $G$ on $G-1$: 
Lemma: if an action of a nonamenable group $G$ on a set $X$ admits an invariant finitely additive probability measure $\mu$, then there is a point $x\in X$ whose stabilizer subgroup $G_x$ is nonamenable.
There are a variety of proofs of this Lemma. Here is a standard averaging argument: Suppose that every $G_x$ is amenable. Let $X_0\subset X$ contain one point from each $G$-orbit and for $x\in X_0$ let $\nu _x$ be a finitely additive probability measure on $G$ which is invariant under left translation by $G_x$. Then $x\mapsto \nu _x$ extends to a well-defined assignment on all of $X$ by taking $\nu _{g\cdot x} = g\cdot \nu _x$ for $x\in X_0$, $g\in G$. The barycenter $\nu = \int _X \nu _x \, d\mu (x)$ is then invariant under left translation by $G$, so $G$ is amenable, contrary to the hypothesis of the lemma.