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Let $\mathcal{H}$ be a finite-dimensional Hilbert space and $\mathcal{A}\subseteq\mathcal{B}(\mathcal{H})$ be an operator system. Is it possible to extend the projective norm (the greatest cross norm) of $\mathcal{A}\otimes M_n$ to a cross norm on $\mathcal{B}(\mathcal{H})\otimes M_n$?

Any comment is highly appreciated. Please suggest to me some reference in case it exists in the literature.

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    $\begingroup$ Put the projective norm on $\mathcal B(\mathcal H)\otimes M_n$. We know that the inclusion $\mathcal A\otimes M_n \rightarrow \mathcal B(\mathcal H)\otimes M_n$ is norm-decreasing. But the projective norm is the greatest crossnorm. So if your question has a positive answer, the projective norm is the only choice, and we need to prove that the inclusion is an isometry... $\endgroup$
    – Matthew Daws
    Commented Jun 29, 2022 at 11:00
  • $\begingroup$ @MatthewDaws I want to mention that the author of the msp.org/pjm/1967/22-1/pjm-v22-n1-p09-p.pdf (page 2, paragraph 2) wrote that Schatten has shown that for S, T, being subspace of U, V, the greatest cross norm topology on $U\otimes V$ is not an extension of the greatest cross norm topology on $S\otimes T$ in general. $\endgroup$
    – Piku
    Commented Jun 29, 2022 at 11:21
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    $\begingroup$ Yes, exactly... So this will not be automatic from general theory, but of course you are in a special case when it might (sometimes) be true. $\endgroup$
    – Matthew Daws
    Commented Jun 29, 2022 at 11:23
  • $\begingroup$ @MatthewDaws I understand that if my question has a positive answer, say, $\Vert .\Vert$. Then for $a\in\mathcal{A}\otimes M_n$ we have $\Vert a\Vert=\Vert a\Vert_{\mathcal{A}\otimes M_n}=\Vert a\Vert_{\mathcal{B(\mathcal{H})}\otimes M_n}$. I do not understand why you are saying $\Vert .\Vert$ is necessarily the projective norm on $\mathcal{B(\mathcal{H})}\otimes M_n$. Could you please ellaborate? $\endgroup$
    – Piku
    Commented Jun 29, 2022 at 16:25
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    $\begingroup$ You should probably ask this as a new question. But, immediately, I have no ideas in this direction. $\endgroup$
    – Matthew Daws
    Commented Jul 1, 2022 at 9:23

1 Answer 1

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By request, some expansion of my easy comment. Given normed spaces $E,F$ and $\newcommand{\mc}{\mathcal}u\in E\otimes F$ write $\pi(u; E\otimes F)$ for the projective norm of $u$ in $E\otimes F$.

The question asks if there is a crossnorm $\|\cdot\|$ on $\mc B(\mc H)\otimes M_n$ with $$ \|u\| = \pi(u; \mc A\otimes M_n) \qquad (u\in\mc A\otimes M_n). $$ However, we know two things:

  • $\|u \| \leq \pi(u; \mc B(\mc H)\otimes M_n)$ for any $u\in\mc B(\mc H)\otimes M_n$ because $\pi$ is the greatest crossnorm;
  • $\pi(u; \mc B(\mc H)\otimes M_n) \leq \pi(u; \mc A\otimes M_n)$ by definition of the projective norm.

Putting these together gives $$ \pi(u; \mc A\otimes M_n) = \|u\| \leq \pi(u; \mc B(\mc H)\otimes M_n) \leq \pi(u; \mc A\otimes M_n) $$ for each $u\in\mc A\otimes M_n$. So we have equality throughout, meaning:

  • $\|\cdot\| = \pi(\cdot; \mc B(\mc H)\otimes M_n)$ on $\mc A\otimes M_n$;
  • So the original question has a positive answer exactly when $\pi(u; \mc B(\mc H)\otimes M_n) = \pi(u; \mc A\otimes M_n)$ for all $u$.

That is, maybe there is some choice in $\|\cdot\|$ but that choice is unimportant, because if any crossnorm will work, the projective norm will work.

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  • $\begingroup$ This was fine. But you said that if my question has a positive answer then $\Vert . \Vert$ has to be the projective norm on $\mathcal{B}\mathcal{(H)}\otimes M_n$. For that one also need to show $\pi(u; \mathcal{B}\mathcal{(H)}\otimes M_n)=\Vert u\Vert$ for all $u\in (\mathcal{A}\otimes M_n) ^c$, which is not clear to me. That is what I requested to ellaborate. $\endgroup$
    – Piku
    Commented Jun 29, 2022 at 20:49
  • $\begingroup$ You are right, I was rather unclear here. I'd adjusted my answer to hopefully make it (a) correct; and (b) less ambiguous. $\endgroup$
    – Matthew Daws
    Commented Jun 30, 2022 at 8:49

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