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!


1 Answer 1


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.

  • 1
    $\begingroup$ 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"? $\endgroup$ Commented Mar 13, 2023 at 13:12
  • $\begingroup$ @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. $\endgroup$
    – J. De Ro
    Commented Mar 13, 2023 at 16:29
  • $\begingroup$ @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) $\endgroup$
    – Hanno
    Commented Mar 13, 2023 at 16:57

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