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This question is motivated by the following, given $M$ an amenable Von Neumann algebra and a Hilbert space H, is $M \overline{\otimes} \mathcal{B}(H)$ amenable?

I have two main questions:

1) Given any finite dimensional Hilbert space H, is $\mathcal{B}(H)$ amenable?

2) If the above is true, after proving that given two amenable VNA's M,N then $M \overline{\otimes} N$ is amenable, then can you conclude by approximation that $M \overline{\otimes} \mathcal{B}(H)$ is amenable?

ADA
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1 Answers1

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A working definition of "amenable" is AFD. This answers your questions trivially:

  • $B(H)$ is AFD if $H$ is separable (it is the wot limit of the "left upper corners").

  • When $H$ is not separable, $B(H)$ cannot be AFD, because it is not separable as a von Neumann algebra. In particular, $M\bar\otimes B(H)$ is not amenable when $H$ is not separable and $M$ is any von Neumann algebra.

  • any finite-dimensional von Neumann algebra is obviously approximately finite-dimensional.

  • If $M$ and $N$ are AFD, then $M\bar\otimes N$ is AFD, by tensoring the corresponding increasing sequences of finite-dimensional subalgebras.

For a proof that $B(H)$ is amenable when $H$ is separable, let $\{p_n\}$ be an increasing sequence of projections with $p_n\nearrow I$. So $p_n\to I$ sot. For any $x\in B(H)$, $\xi\in H$, $$ \langle p_nxp_n\xi,\xi\rangle\to\langle x\xi,\xi\rangle, $$ so $$ B(H)=\overline{\bigcup_n p_nB(H)p_n}, $$ and $p_nB(H)p_n\simeq M_{k_n}(\mathbb C)$ where $k_n=\text{Tr}\,(p_n)$.

Martin Argerami
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  • For separable Hilbert spaces AFD is equivalent to amenable. Yes, you second to last bullet is trivial, and the last you can also do by tensoring the conditional expectations for each amenable VNA.

    Thanks for the answer. Really the "crux" of my question is precisely your first bullet point, (why a seprable Hilbert space gives rise to $\mathcal{B}(H)$ being amenable).

     – ADA May 22 '17 at 20:34
  • Ok. I have added a short proof of that. – Martin Argerami May 22 '17 at 20:50
  • I have a question: If $N$ is a von Neumann subalgebra of an amenable von Neumann algebra M , can we have $N$ is amenable? – mathbeginner Mar 18 '24 at 07:49
  • There is no general answer. $B(H)$ for $H$ separable is amenable, so every von Neumann algebra with separable predual is a subalgebra of an amenable one. You also have $L(\mathbb F_2)\otimes B (H)$ which is non-amenable, ,so every von Neumann algebra with separable predual is a subalgebra of a non-amenable one. – Martin Argerami Mar 18 '24 at 13:23
  • If $K$ is the von Neumann subalgebra of $M$ and $K$ is isomorphic to the hyperfinite type II$_1$ factor, can we conclude that $K'\cap M$ is amenable? – mathbeginner Mar 26 '24 at 16:26
  • Not sure what you mean by "the" subalgebra. But no, there is no reason for the relative commutant to be amenable. Take $N$ any non-amenable algebra and put $M=K\otimes N$. Then $K'\cap M=1\otimes N$. – Martin Argerami Mar 26 '24 at 16:53