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Given Hilbert spaces $\mathcal H_1,\mathcal H_2,\mathcal K_1,\mathcal K_2$ and bounded linear maps $S_i:\mathcal B^1(\mathcal H_i)\to\mathcal B^1(\mathcal K_i)$, $i=1,2$ between the respective trace classes one gets a unique linear map $$ S_1\odot S_2:\mathcal B^1(\mathcal H_1)\odot\mathcal B^1(\mathcal H_2)\to\mathcal B^1(\mathcal K_1)\odot\mathcal B^1(\mathcal K_2) $$ via the action on pure tensors, i.e. $(S_1\odot S_2)(A_1\otimes A_2)=S_1(A_1)\otimes S_2(A_2)$; here $\odot$ is the algebraic tensor product. Now $\mathcal B^1(\mathcal H_1)\odot\mathcal B^1(\mathcal H_2)$ naturally "sits inside" $\mathcal B^1(\mathcal H_1\otimes\mathcal H_2)$ so we can equip domain and co-domain of $S_1\odot S_2$ with the trace norm on $\mathcal H_1\otimes\mathcal H_2$. This leads to my question:

Under this identification does one have $\|S_1\odot S_2\|=\|S_1\|\|S_2\|$?

As $\geq$ is obvious only the converse $\leq$ remains to be shown.

What I tried so far:

Because $\|\cdot\|_1$ obviously is a cross norm my idea was to show that its action on $\mathcal B^1(\mathcal H_1)\odot\mathcal B^1(\mathcal H_2)$ coïncides with the action of one of the uniform normed space tensor norms for Banach spaces, such as the injective tensor norm, the projective tensor norm, or any other one of Grothendieck's natural norms. If so, my question would be a mere application of the known results for these natural norms---in particular the operator norm for operators between such spaces is known to be multiplicative.

To get the simple stuff out of the way: first I ruled out the injective tensor norm $\varepsilon$ with the simple counter-example $\mathcal H_i=\mathcal K_i=\mathbb C^2$ and $$ A=\begin{pmatrix}1&0&0&1\\0&0&0&0\\0&0&0&0\\0&0&0&0\end{pmatrix} $$ as $\|A\|_1=\sqrt2$ but $\varepsilon(A)=1$ (the problem here is that $A=|e_1\rangle\langle\psi|$ is of rank one where $e_1=e_1\otimes e_1$ is a pure tensor but $\psi=(1,0,0,1)^\top$ is entangled so no pure tensor $B\otimes C$ can create a maximal overlap between the two vectors.)

The more promising choice to me seemed to be the projective tensor norm $\pi$ because of a comment made in an older thread that the completion $\mathcal B^1(\mathcal H_1)\hat\otimes_\pi\mathcal B^1(\mathcal H_2)$ of the projective tensor product of $\mathcal B^1(\mathcal H_1)$ and $\mathcal B^1(\mathcal H_2)$ is equal to the full space $\mathcal B^1(\mathcal H_1\otimes\mathcal H_2)$, much like how $\mathcal B^1(\mathcal H_1)\odot\mathcal B^1(\mathcal H_2)$ is a dense subspace of $(\mathcal B^1(\mathcal H_1\otimes\mathcal H_2),\|\cdot\|_1)$. However, the thread this comment is from seems to imply that the canonical embedding $\iota:(\mathcal B^1(\mathcal H_1)\hat\otimes_\pi\mathcal B^1(\mathcal H_2),\pi)\to(\mathcal B^1(\mathcal H_1\otimes\mathcal H_2),\|\cdot\|_1)$ is not an isometric isomorphism (?) which I'm not sure how much of a problem that would be.

I also attempted to re-write the projective norm in this special case but only arrived at $$ \pi(A)=\sup_{\tilde S(\mathcal B(\mathcal B^1(\mathcal H_1),(\mathcal B^1(\mathcal H_2))^*),\|\tilde S\|=1}|\langle A,\tilde S\rangle|=\sup_{S(\mathcal B(\mathcal B^1(\mathcal H_1),\mathcal B(\mathcal H_2)),\|S\|=1}\Big|\sum_{i=1}^n\operatorname{tr}(Y_iS(X_i))\Big| $$ where $A=\sum_{i=1}^nX_i\otimes Y_i$ is an arbitrary representation. This does at least not resemble any formula for the trace norm that I am familiar with. The only connection I could think of is that maybe there is an isometric isomorphism $\Psi$ between $\mathcal B(\mathcal B^1(\mathcal H_1),\mathcal B(\mathcal H_2))$ and $\mathcal B(\mathcal H_1\otimes\mathcal H_2)$ via $\operatorname{tr}(A_2S(A_1))=\operatorname{tr}((A_1\otimes A_2)B)$---or $(\Psi(B))(A):=\operatorname{tr}_A(B)$ when using the partial trace with respect to a state, cf. here, Ch. 2.3.2. This would directly lead to the norm formula via the dual space $\|A\|_1=\sup_{B\in\mathcal B(\mathcal H_1\otimes\mathcal H_2),\|B\|=1}|\operatorname{tr}(AB)|$.

Maybe it is also possible to prove this directly, first for the special case $\mathcal H_2=\mathcal K_2$ and $S_2=\operatorname{id}$---similar to the proof for the norm formula for tensor products of bounded operators. If this works, however, then the techniques have to be substantially different as the proof I know heavily relies on the connection between the operator norm and the norm on the underlying Hilbert space (as well as the Pythagorean theorem, cf., e.g., here).

What's bugging me is that the (in)equality in question feels like it has to be true because it does hold for (almost?) all prominent cross norms as well as the analogous case of tensor products of bounded operators; this makes me think that the answer I am looking for is terribly obvious but I currently am too blind to see it. Either way thanks in advance for any answer or comment!

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    $\begingroup$ The norm induced on the tensor product of trace-class operators in the operator space projective tensor norm: this does not agree with any of the classical Banach space tensor norms. You need the $S_i$ to be completely bounded, not just bounded. For a counter-example, take $H_1=K_1$ and $S_1$ the identity, and $H_2=K_2$ with $S_2$ the adjoint map. The buzz-word to look for is "operator space". $\endgroup$ Jul 15, 2021 at 19:55
  • $\begingroup$ That's just was I was looking for, thank you! $\endgroup$ Jul 19, 2021 at 9:10

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