Weakly solid factors? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T03:06:53Z http://mathoverflow.net/feeds/question/48572 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/48572/weakly-solid-factors Weakly solid factors? Jon Bannon 2010-12-07T16:07:20Z 2011-01-12T22:40:40Z <p>A type $II_{1}$ factor $\mathcal{M}$ is <em>solid</em> if for every diffuse von Neumann subalgebra $\mathcal{A}$ of $\mathcal{M}$, the relative commutant $\mathcal{A}'\cap \mathcal{M}$ is injective. (See Ozawa's excellent paper: <a href="http://arxiv.org/abs/math/0302082" rel="nofollow">http://arxiv.org/abs/math/0302082</a>)</p> <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>We say that a type $II_{1}$ factor $M$ with trace $\tau$ is $\Gamma$-<em>solid</em> if for every diffuse von Neumann subalgebra $\mathcal{A}$ of $\mathcal{M}$, the relative commutant $\mathcal{A}'\cap \mathcal{M}$ has property $\Gamma$.</p> <p>Since every injective von Neumann algebra has property $\Gamma$, the notion of $\Gamma$-solidity is weaker than that of solidity. </p> <blockquote> <p>Is every $\Gamma$-solid factor solid?</p> </blockquote> http://mathoverflow.net/questions/48572/weakly-solid-factors/51850#51850 Answer by Sven Raum for Weakly solid factors? Sven Raum 2011-01-12T15:38:12Z 2011-01-12T18:18:02Z <p>Sorry, this was a wrong example. I should have thought about it better.</p>