Do Baumslag-Solitar Group von Neumann algebras have Property $\Gamma$? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T11:37:44Z http://mathoverflow.net/feeds/question/96586 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/96586/do-baumslag-solitar-group-von-neumann-algebras-have-property-gamma Do Baumslag-Solitar Group von Neumann algebras have Property $\Gamma$? Jon Bannon 2012-05-10T16:55:12Z 2012-05-13T23:06:13Z <p>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 $||ux_{j}-x_{j}u||_{2}&lt;\epsilon$ for all $1 \leq j \leq n$. (Here $||T||_2=(\tau(T^{*}T))^{1/2}$ for $T\in M$.) </p> <p>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 $g \in G.$</p> <p>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. <a href="http://arxiv.org/PS_cache/arxiv/pdf/0909/0909.1485v1.pdf" rel="nofollow">inner amenable groups whose group von Neumann algebras don't have $\Gamma$</a>, as recently shown by Stefaan Vaes.</p> <blockquote> <p>Given a non-residually finite <a href="http://projecteuclid.org/DPubS?service=UI&amp;version=1.0&amp;verb=Display&amp;handle=euclid.bams/1183524561" rel="nofollow">Baumslag-Solitar group</a> $$BS(m,n) = \langle b,s\mid s^{-1}b^ms = b^n\rangle$$ does its group von Neumann algebra have property $\Gamma$?</p> </blockquote> <p>It is known that all such groups are inner amenable, and it <a href="http://www.math.jussieu.fr/~pfima/Documents/Baumslag-Solitar-Groups.pdf" rel="nofollow">recently has been shown that</a> the associated group factors have no Cartan subalgebra, are prime and yet are not solid.</p> http://mathoverflow.net/questions/96586/do-baumslag-solitar-group-von-neumann-algebras-have-property-gamma/96857#96857 Answer by Narutaka OZAWA for Do Baumslag-Solitar Group von Neumann algebras have Property $\Gamma$? Narutaka OZAWA 2012-05-13T23:06:13Z 2012-05-13T23:06:13Z <p>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., $| T_l^{-1}(E) \bigtriangleup E | &lt; \epsilon |E|$ for $l=m,n$. Then, Abert--Nikolov's argument (see Chifan--Ioana's paper arXiv:0802.2353 Lemma 10) allows one to widen $E$. </p>