For a Hilbert space $H$, let $\mathcal B(H)$ and $\mathcal K(H)$ denote the spaces of bounded linear operators and compact operators on $H$ respectively.
If $H_1$ and $H_2$ are two Hilbert spaces (infinite-dimensional), let $H=H_1\otimes H_2$ be the tensor product of $H_1$ and $H_2$. How could I show that the inclusion $\mathcal B(H_1)\otimes \mathcal B(H_2)\subset \mathcal B(H)$ is proper?
Similar question exists for compact operators. Namely, is the inclusion $\mathcal K(H_1)\otimes \mathcal K(H_2)\subset \mathcal K(H)$ proper?
The tensor product of $C^{\ast}$-algebra is too abstract for me, and I don't have any idea on above questions. Can you help me to solve these problems? Thank you very much!
Best Answer
The inclusion $\mathcal{K}(H_1) \otimes \mathcal{K}(H_2) \subseteq \mathcal{K}(H)$ is an equality. To see this, note that rank-one operators on $H$ split as a tensor product of rank-one operators on $H_1$ and rank-one operators on $H_2$.
More formally, use the following:
Then use the fact that
to conclude that
The inclusion $\mathcal{B}(H_1)\otimes \mathcal{B}(H_2)\subseteq \mathcal{B}(H)$ is almost never an equality.
Here is a concrete example that illustrates this:
If $H_1 = H_2 = \ell^2$, then $\mathcal{B}(\ell^2) \otimes \mathcal{B}(\ell^2)$ has more than one non-zero ideal, while $\mathcal{B}(H)= \mathcal{B}(\ell^2\otimes \ell^2)$ only has one non-zero ideal, so $\mathcal{B}(\ell^2)\otimes \mathcal{B}(\ell^2)\not\cong \mathcal{B}(\ell^2 \otimes \ell^2) $. In particular, the canonical inclusion is not surjective.
More generally, one can show that this inclusion is an equality if and only if $H_1$ or $H_2$ is finite-dimensional.