When do the compact operators form a Von Neumann algebra?

Question

Let $H$ be a Hilbert space. If it is finite-dimensional, the identity is compact, so as $K(H) \trianglelefteq B(H)$, it follows that $K(H) = B(H)$, which is a Von Neumann algebra.

I suspect that it is not so if $H$ is infinite-dimensional. Indeed, if then $K(H)$ were one, it would be unital. This would mean that its unit would not be id$_H$. I can’t get any further though…

Answer 1

The finite-rank operators are sot/wot dense in $B(H)$; equivalently, $K(H)'=\mathbb C\,I$. So $K(H)''=B(H)$. In other words, $K(H)$ is a von Neumann algebra if and only if $\dim H<\infty$.