A type $II_{1}$ factor $\mathcal{M}$ is solid 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: http://arxiv.org/abs/math/0302082)

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 $T\in M$.)

We say that a type $II_{1}$ factor $M$ with trace $\tau$ is $\Gamma$-solid if for every diffuse von Neumann subalgebra $\mathcal{A}$ of $\mathcal{M}$, the relative commutant $\mathcal{A}'\cap \mathcal{M}$ has property $\Gamma$.

Since every injective von Neumann algebra has property $\Gamma$, the notion of $\Gamma$-solidity is weaker than that of solidity.

Is every $\Gamma$-solid factor solid?

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    $\begingroup$ Hey Jon, There are weaker forms, such as semi-solidity, or even say prime. Of course these are weakenings of a different form. For instance, semi-solid you restrict which algebras you are taking relative commutants of, while $\Gamma$-solidity eases the requirement of the relative commutant itself. Also I was just wondering, is this notion of $\Gamma$-solid in the literature, i haven't heard of it before. $\endgroup$ Jan 12, 2011 at 16:23
  • $\begingroup$ Thanks, Owen! You are right, there are other weakenings of solidity. As far as I know, this notion of Γ solidity hasn't appeared anywhere. I started thinking about it, wondering how "rigid" the notion of solidity should be in various regards. If this is completely inane, Sorin would know. $\endgroup$
    – Jon Bannon
    Jan 12, 2011 at 22:49

1 Answer 1


Sorry, this was a wrong example. I should have thought about it better.

  • $\begingroup$ You haven't actually shown that $L(G)$ is $\Gamma$-solid. $\endgroup$ Jan 12, 2011 at 16:18

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