Is $\mathcal{B}(H)$ amenable for any Hilbert space H?

183 Views Asked by At

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?

1

There are 1 best solutions below

6
On BEST ANSWER

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 $k_n=\text{Tr}\,(p_n)$.