Is $\mathbb{B}(H) \bigotimes \mathbb{B}(H)$ never equal to $\mathbb{B}(H\bigotimes H)$

Question

Given a Hilbert space $H$, if we use $\mathbb{B}(H)\bigodot\mathbb{B}(H)$ to denote the tensor product before taking the spatial norm closure, then we will have $\mathbb{B}(H)\bigodot\mathbb{B}(H)$ is dense in $\mathbb{B}(H\bigotimes H)$ in weak operator topology because $\mathbb{B}(H)\bigodot\mathbb{B}(H)$ contains all finite rank operators defined in $H\bigotimes H$. My question is:

1. If $H$ is infinite dimensional, is it true that $\mathbb{B}(H)\bigodot\mathbb{B}(H)$ is never dense in $\mathbb{B}(H\bigotimes H)$ in norm (or will $\mathbb{B}(H)\bigotimes\mathbb{B}(H)$ be always properly contained in $\mathbb{B}(H\bigotimes H))$?

2. If $H$ is finite dimensional, what will be the answer in question 1? In general, is it true that $M_n(\mathbb{C})\bigotimes M_m(\mathbb{C})$ is never equal to $M_{mn}(\mathbb{C})$?

Answer 1

The answer to question (1) is no. In case $H$ is a separable Hilbert space there are many significant differences between $B(H\otimes H)$ and the (spacial) tensor product $B(H)\otimes B(H)$. The former has a unique closed two-sided ideal, namely $K(H\otimes H)$ (compact operators), while the latter has at least three, namely $K(H)\otimes K(H)$, $K(H)\otimes B(H)$, and $B(H)\otimes K(H)$.

For non-separable Hilbert spaces, the situation is similar. Luft [1] has classified all closed ideals of $B(H)$, for an arbitrary Hilbert space $H$, and, in particular, he has shown that these ideals form a chain, meaning that for any two such ideals $I$ and $J$, either $I\subseteq J$ or $J\subseteq I$. On the other hand, in $B(H)\otimes B(H)$ the ideal $I=K(H)\otimes B(H)$ neither contains nor is contained in the ideal $J=B(H)\otimes K(H)$.

This actually shows that $B(H)\otimes B(H)$ not only differs from $B(H\otimes H)$, but these algebras are not even isomorphic.

In the finite dimensional context of (2) the situation is rather different: $M_n({\mathbb C})\otimes M_m({\mathbb C})$ is canonically isomorphic to $M_{nm}({\mathbb C})$ (rather than $M_{n+m}({\mathbb C})$, as observed by Peter) and this can be easily checked by defining a natural map and showing it to be an isomorphism by computing dimensions.

[1] Luft, E., The two-sided closed ideals of the algebra of bounded linear operators of a Hilbert space, Czech. Math. J. 18(93), 595-605 (1968). ZBL0197.11301.