# A question about tensor product of Hilbert spaces and operators

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!

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:

If $$H$$ is a Hilbert space, and $$\theta_{x,y}(z) = \langle z,y\rangle x$$, then the closed span of the operators $$\{\theta_{x,y}: x,y \in H\}$$ is equal to $$\mathcal{K}(H)$$.

Then use the fact that

$$\theta_{x,y}\otimes \theta_{s,t}= \theta_{x\otimes s,y \otimes t}$$ as operators in $$B(H_1\otimes H_2)$$

to conclude that

$$\mathcal{K}(H_1\otimes H_2)= \mathcal{K}(H_1)\otimes \mathcal{K}(H_2)$$.

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.

• Beautiful answer; one small comment: in the last sentence, shouldn't it be "if and only if $H_1$ or $H_2$ is finite dimensional"? Commented Mar 13, 2023 at 13:12
• @alepopoulo110 Thanks for the comment. You are of course right. Here is a way to show the equivalence. If $H_1$ is finite-dimensional, say $H_1 \cong \mathbb{C}^n$, we have a chain of isomorphisms $B(H_1)\otimes B(H_2) \cong M_n(\mathbb{C})\otimes B(H_2) \cong M_n(B(H_2)) \cong B(H_2^{\oplus n}) \cong B(\mathbb{C}^n\otimes H_2) \cong B(H_1 \otimes H_2)$. Calculating this composition gives the canonical map $B(H_1)\otimes B(H_2)\to B(H_1\otimes H_2)$. This shows one of the implications in the claim in my answer. The converse is the same argument as in the $\ell^2$-case. Commented Mar 13, 2023 at 16:29
• @alepopoulo110 All this is related to $\mathcal K(H)$ being a (completed) tensor product, but $\mathcal B(H)$ is not, if $H$ is infinite-dimensional. Cf [ Does B(H)=H⊗H∗ in infinite dimensions?](math.stackexchange.com/q/2712906/316749) Commented Mar 13, 2023 at 16:57