Let $H_1$ and $H_2$ be Hilbert spaces.
Let $A\subset B(H_1)$ be a factor and $A'$ its commutant.

If a von Neumann algebra $M\subset B(H_1\otimes H_2)$ contains $A\otimes 1$ and commutes with $A'\otimes 1$, is it then necessarily of the form $M=A\otimes B$ for some von Neumann algebra $B\subset B(H_2)$?

I think that I know how to prove this if $A$ is hyperfinite, and I wonder if it's true in general.


What about $A = A_1 \oplus A_2$ and $M = (A_1\otimes 1) \oplus (A_2 \otimes B(H_2))$?

It seems like your condition just says that $A\otimes 1 \subseteq M \subseteq A \otimes B(H_2)$, did you leave something out?

Edit: in case $A$ is a factor, the answer is yes. Ge and Kadison, On tensor products of von Neumann algebras, Invent. Math. 123 (1996), 453-466.

  • 1
    $\begingroup$ You're right. I forgot to say that A should be a factor. I'll edit the question. $\endgroup$ – André Henriques Sep 20 '13 at 17:58
  • $\begingroup$ Wonderful! I don't feel too bad that I wasn't able to do this "exercise" ;-) $\endgroup$ – André Henriques Sep 20 '13 at 19:07
  • $\begingroup$ Yeah ... not trivial. This was one of Liming's earliest papers, when he was just starting to get famous. $\endgroup$ – Nik Weaver Sep 20 '13 at 20:21

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