# Do Baumslag-Solitar Group von Neumann algebras have Property $\Gamma$?

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 $T\in M$.)

A countable discrete group $G$ is inner amenable if there is a finitely additive measure $m$ on the subsets of $G \backslash${$e$} with total mass 1 and satisfying $m(gXg^{-1})=mX$ for all $X\subseteq G \backslash${$e$} and all $g \in G.$

I should mention that if the left group von Neumann algebra of an i.c.c. group has property $\Gamma$ then the group is inner amenable, however there exist i.c.c. inner amenable groups whose group von Neumann algebras don't have $\Gamma$, as recently shown by Stefaan Vaes.

Given a non-residually finite Baumslag-Solitar group $$BS(m,n) = \langle b,s\mid s^{-1}b^ms = b^n\rangle$$ does its group von Neumann algebra have property $\Gamma$?

It is known that all such groups are inner amenable, and it recently has been shown that the associated group factors have no Cartan subalgebra, are prime and yet are not solid.

Yes, it has property $(\Gamma)$. This follows from Stalder's proof of inner amenability plus a fact that the semigroup $\langle T_m, T_n \rangle$ admits an approximately invariant subsets having proportional measures. Here, $T_m$ is the $m$-times map on $[0,1)$, $T_m x = mx \mod 1$. As Stalder proves, $\sum_{1\le i,j \le n} z^{m^i n^j}$ is approximately invariant under $T_m$ and $T_n$. By a standard procedure, it gives rise a subset $E \subset [0,1)$ which is approximately invariant under $T_m$ and $T_n$, i.e., $| T_l^{-1}(E) \bigtriangleup E | < \epsilon |E|$ for $l=m,n$. Then, Abert--Nikolov's argument (see Chifan--Ioana's paper arXiv:0802.2353 Lemma 10) allows one to widen $E$.
• Yes, but I was not thinking of a specific statement. Since I didn't explain so well, let me do it now. My $z$ above is $b$ for $BS(m,n) = \langle b, s\rangle$, and is considered as $\exp(2\pi i t)$ on $[0,1)$. The function $K^{-1}\sum_{1\le i,j \le K} z^{m^i n^j}$ in $L^2$ is approximately invariant under $T_m$ and $T_n$, and Abert--Nikolov argument (adapted to semigroups) provides an approximately invariant subset $F$ of measure $1/2$. Now $p:=\chi_{T_m^{-1}(F)}$ is in $W^*(b^m)\subset W^*(b) \cong L^\infty[0,1)$ and $sps^{-1}=\chi_{T_n^{-1}(F)}\approx p$. So, $p$ is approximately central. May 15, 2012 at 0:08