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Subtag of the [oa.operator-algebras] tag for questions about von Neumann algebras, that is, weak operator topology closed, unital, *-subalgebras of bounded operators on a Hilbert space.

6
votes
Let me first point out that $X_1$ and $Y=h X_2 (1-h)$ are not freely independent. This is most easily seen if $h$ has trace 1/2, in which case $Y$ has range and support projections $h$ and $(1-h)$, r …
answered Apr 7 '11 by Dima Shlyakhtenko
4
votes
Yes, one can get a similar characterization, only you will need to talk about $\ast$-cumulants instead of cumulants (in other words, you will need to allow both $x_{ij}$'s and $x_{ji}^\ast$'s in the f …
answered Apr 18 '11 by Dima Shlyakhtenko
5
votes
Perhaps I am missing some hypothesis, but I think the proof is just about the same whether or not $B$ is a factor. Here is the proof from Jones' original paper, and I believe it does not use factoria …
answered Apr 2 '11 by Dima Shlyakhtenko
12
votes
I am guessing that the answer is "yes" if you interpret the question in the following way. Let $A_i$ be some subalgebras of a von Neumann algebra $(M,\tau)$ and assume that there are mutually orthogo …
answered Feb 1 '11 by Dima Shlyakhtenko
1
vote
I am not sure of what you want, but let me offer a few ideas. First, let me give you a silly answer: by the bicommutant theorem, if $M\subset B(H)$ is a von Neumann algebra, then $M = (M') '$. Now p …
answered Apr 7 '11 by Dima Shlyakhtenko