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.


1 Answer 1


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 $E$.

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    $\begingroup$ 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. $\endgroup$ May 15, 2012 at 0:08

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