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 $B(\mathcal{H})$.
$\begingroup$
$\endgroup$
5
-
$\begingroup$ What does the $'$ mean in $M'$? $\endgroup$– rschwiebCommented Sep 7, 2018 at 16:07
-
1$\begingroup$ @rschwieb commutant of $M$$=$$M'$ $\endgroup$– mathloverCommented Sep 7, 2018 at 16:12
-
$\begingroup$ @rschwieb: that notation is completely standard in von Neumann algebras. I don't think I have never seen any other notation. $\endgroup$– Martin ArgeramiCommented Sep 8, 2018 at 2:34
-
1$\begingroup$ @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 $\endgroup$– rschwiebCommented Sep 8, 2018 at 20:34
-
1$\begingroup$ @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. $\endgroup$– Martin ArgeramiCommented Sep 8, 2018 at 21:51
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
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 $X\in M\cap M'$.
answered Sep 8, 2018 at 2:33