$\sigma$-weak closure of $B(H) \otimes B(K)$.

Question

Let $H$ and $K$ be Hilbert spaces. I'm trying to prove that $B(H) \otimes B(K)$ (= minimal tensor product of $C^*$-algebras) is $\sigma$-weakly dense (= weak$^*$-dense) in $B(H \otimes K)$.

My progress so far:

I managed to show:

• $B(H) \otimes B(K)$ is strongly dense in $B(H \otimes K)$
• $B(H) \otimes B(K)$ contains the compact operators on $H \otimes K$, hence it suffices to show that the compact operators are $\sigma$-weakly dense in $B(H \otimes K)$.

Any ideas how I can proceed?

Answer 1

Yes, the compact operators on any Hilbert space $H$ form a $\sigma$-weakly dense subset. This is because otherwise there would be a nonzero trace class operator $S$ such that $\text{tr}(TS)=0$, for all compact operator $T$, and it is not hard to see that this is impossible.

Answer 2

Here is a more or less alternative argument.

By looking at what the rank-one operators look like, it is easy to see that $\mathcal K(H\otimes K)=\mathcal K(H)\otimes \mathcal K(K)$, this last product as C$^*$-algebras (no ambiguity due to nuclearity).

Then $\mathcal K(H)\otimes \mathcal K(K)$ is $\sigma$-weakly dense in $B(H\otimes K)$ because the locally compact space $\mathcal K(H\otimes K)$ is weak$^*$-dense in its double dual $B(H\otimes K)$.