Take the 2-minute tour ×

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 $\Gamma$?

share|improve this question

2 Answers 2

up vote 5 down vote accepted

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.

share|improve this answer

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.

share|improve this answer
Also, thanks for the question regarding Property (T)! I wasn't aware of this. –  Jon Bannon Sep 22 '10 at 18:30
It is an interesting question. The non-amenability of $B(2,n)$ and Tarski monsters was proved using Kesten-Grigorchuk combinatorial criterion of amenability. Is there a similar "co-growth" criterium for inner amenability? If one can formulate the property in terms of words (= walks on the Cayley graph), one would be able to answer your question. –  Mark Sapir Sep 22 '10 at 18:33
That is precisely the content of my question: mathoverflow.net/questions/27233/… I would absolutely love to see if such a "co-growth" condition for inner amenability were feasible. –  Jon Bannon Sep 22 '10 at 19:17
By the way, unlike free Burnside group $B(2,n)$ which is unique, there are many Tarski monsters. Some better than others. There is a class of groups called ``lacunary hyperbolic groups" which contain many Tarski monsters. See front.math.ucdavis.edu/0701.5365 . –  Mark Sapir Sep 22 '10 at 19:42
Thanks for this. I'll have a look. –  Jon Bannon Sep 22 '10 at 20:23

Your Answer


