# On classifying von Neumann algebras with respect to some property

For which class of von Neumann algebras have the property$:T \in B(\mathcal{H})\text{ such that } \text{Co}_{M}(T)^{-}\cap M'\text{ nonempty }$. Where $\text{Co}_{M}(T)^{-}$ is weak operator closure of convex hull of $\{uTu^{*}:u \in \mathcal{u}(M)\}$, where $M$ is vN algebra in $B(\mathcal{H})$.

• What does the $'$ mean in $M'$? – rschwieb Sep 7 '18 at 16:07
• @rschwieb commutant of $M$$=$$M'$ – mathlover Sep 7 '18 at 16:12
• @rschwieb: that notation is completely standard in von Neumann algebras. I don't think I have never seen any other notation. – Martin Argerami Sep 8 '18 at 2:34
• @MartinArgerami Dear Martin: I suppose that would’ve something natural to say if I were disputing something about it, but if you reread it as someone who is asking a naive question about the most overloaded notation in mathematics, I think you might find it somewhat off-putting. Sometimes people who don’t look like newbies can be newbies in you favorite field, but they don’t want to be bitten either. Regards – rschwieb Sep 8 '18 at 20:34
• @rschwieb: I wasn't trying to be aggressive in any way, my apologies if it came out like that. I thought it was important to mention because it is such a basic thing when dealing with von Neumann algebras. – Martin Argerami Sep 8 '18 at 21:51

## 1 Answer

It is always nonempty. Even if you take norm closure instead of wot closure. That's exactly what the Dixmier Approximation Theorem says: the set $\overline{\operatorname{Co}_M}(T),$ where the closure is in norm, always intersects the centre of $M$. So there is always $X\in\overline{\operatorname{Co}_M}(T)$ such that $X\in M\cap M'$.