6
$\begingroup$

Equipping $M_n$ with its usual operator space structure, $\newcommand{\ptp}{\widehat{\otimes}}$ we can form the projective tensor product of operator spaces $M_n\ptp M_n$. In particular this puts a Banach space norm on the algebraic tensor product $M_n\otimes M_n$.

Now consider the transpose map $T:M_n \to M_n$. It is a standard calculation to show that $T$ is not completely bounded, $\newcommand{\stp}{\overline{\otimes}}$ and in particular one can show that the map $\iota \otimes T : M_n \stp M_N \to M_n\stp M_n$ has norm $n$, where $\stp$ denotes the spatial tensor product (in this setting the same as the injective tensor product of operator spaces).

Question 1. What is the asymptotic behaviour (as $n\to \infty$) of $\Vert \iota \otimes T : M_n \ptp M_N \to M_n\stp M_n\Vert$?

Note that I'm asking merely about the norm as a map between two Banach spaces, not about the cb norm. If precise asymptotics are tricky, how about the following sub-question:

Question 2. In particular, does the norm of this map tend to infinity as $n\to\infty$?

This feels like something that should follow by tweaking a standard example or exercise in one of the usual books on Operator Spaces, but I couldn't succeed in converting the usual examples to get something that answers the question above.

Remark: the usual way to get a lower bound on $\Vert\iota \otimes T : M_n \stp M_N \to M_n\stp M_n \Vert$ is to consider what this map does to the tensor $x = \sum_{i,j=1}^n E_{ij} \otimes E_{ji}$, the point being that $x$ has norm $1$ when viewed as an element of $M_{n^2}$ while $(\iota\otimes T)(x)$ has norm $n$ as an element of $M_{n^2}$. However, since matrix multiplication gives a complete contraction $M_n \ptp M_n \to M_n$, I think it can be shown that $x$ has norm $n$ as an element of $M_n\ptp M_n$.


Update 2016-05-04: I think I've now found a proof that this map is contractive for all $n$, which moreover works if you replace proj tp with Haagerup tp. Previously I thought that this stronger claim (with the Haagerup tp) was false by adapting the usual argument to show the claim fails for min tp; however, this was based on a stupid miscalculation. based on an interpolation argument. If the details work then I'll leave them as an answer.

$\endgroup$
1
  • $\begingroup$ Not that I mind too much, but would whoever dowvoted like to explain why? I'm particularly interested if you know of an easier proof that Question 2 has a negative answer; as shown by the solution I gave below it appears to be somewhat tricky, but perhaps you've seen something I missed $\endgroup$
    – Yemon Choi
    Oct 2 '16 at 18:25
3
$\begingroup$

Well I guess I should write something quickly, even if it doesn't have all the details, otherwise I'll keep putting it off. And maybe someone will spot a mistake...

Fix Hilbert spaces $V$ and $W$, which we think of as having column OSS. Equip $B(V)$ and $B(W)$ with their usual OS structures. Then the linear map $$ \iota\otimes \top : B(V) \otimes B(W) \to B(V \otimes_2 W)$$ extends to a contractive linear map $B(V) \ptp B(W) \to B(V\otimes_2 W)$.

The proof goes in stages.

Step 1. If $x = \sum_i a_i \otimes b_i \in B(V)\otimes B(W)$, then $$ \Vert \sum\nolimits_i a_i \otimes b_i^\top \Vert_{B(V\otimes_2 W)} = \sup \left\{ \Vert \sum\nolimits_i a_i c b_i \Vert_{HS(W,V)} \;\colon\; c \in HS(W,V), \Vert c\Vert_{HS(W,V)} \leq 1 \right\} $$

(This is not hard to hack out by hand, but with hindsight can also be found in various places, for instance I think it is in Pisier's book somewhere early on.)

Step 2. Note that there are have natural completely contractive maps$\newcommand{\ptp}{\widehat{\otimes}}$ $$ B(V) \ptp V \to V\quad,\quad W^* \ptp B(W) \to W^* $$ where the first is the usual action $a\otimes v \mapsto av$ and the second is the transposed action $\psi\otimes b \mapsto \psi\circ L_b$, $L_b$ being the action $w\mapsto bw$. Therefore by general stuff on operator space tensor products, we have complete contractions$\newcommand{\itp}{\otimes_{\rm min}}$ $$ B(V) \ptp( V\ptp W^*) \ptp B(W) \to V\ptp W^* $$ $$ B(V) \ptp( V\itp W^*) \ptp B(W) \to V\itp W^* $$

where $\itp$ denotes injective tensor product of operator spaces. Note that if we identify $V\otimes W^*$ with the space of finite rank operators $W\to V$, then the two maps above just correspond to $a\otimes c \otimes b \mapsto acb$.

Step 3. Under the natural identification of $V\otimes W^*$ with the finite-rank operators $W\to V$, we have isometric isomorphisms $V\ptp W^*\cong S_1(W,V)$ and $V\itp W^* \cong S_\infty(W,V)$, where $S_1$ denotes trace-class operators and $S_\infty$ the compact operators.

Step 4. Let $x$ be as in Step 1 and WLOG assume its norm in $B(V)\ptp B(W)$ is $1$. Let $E_x$ denote the elementary operator on $B(W,V)$ defined by $c\mapsto \sum_i a_icb_i$. We wish to show that the norm of $E_x$ as a map $HS(W,V)\to HS(W,V)$ is $\leq 1$. But now by Steps 2 and 3, we know that $E_x$ is (completely) contractive from $S_1(W,V)$ to $S_1(W,V)$, and (completely) contractive from $S_\infty(W,V)$ to $S_\infty(W,V)$. By classical complex interpolation results, $E_x$ is therefore contractive on all the intermediate Schatten classes, in particular on the Hilbert-Schmidt operators, and we are done.

Remark. Of course we can interpolate in the category of operator spaces. The argument above, if correct, seems to actually show that $E_x$ is completely contractive on $HS(W,V)$ when we equip this space with the "self-dual" OSS. If I denote this operator space by OH temporarily, then we can rephrase this as: $\iota\otimes \top$ is contractive from $B(V)\ptp B(W)$ to $CB(OH)$. It seems plausible that we actually get a complete contraction, but I haven't yet done the book-keeping required to check this.

$\endgroup$
5
  • 1
    $\begingroup$ It seems to me you implicitly re-bracket $B(V) \ptp( V\itp W^*) \ptp B(W)$ to $(B(V) \ptp V)\itp (W^* \ptp B(W))$ in order to then apply your two maps. Why can you do that? $\endgroup$ May 7 '16 at 12:46
  • $\begingroup$ Good point - this is the part I was sweeping under the carpet. I guess I could try to duck the issue by saying that the LHS certainly embeds completely contractively into "B(V) Haagerup midlle thing Haagerup B(W)", then the" middle" is compact operators W to V, and we know that multiplication of op algebras is bounded wrt Haagerup tensor product. (This was my original line of thought) $\endgroup$
    – Yemon Choi
    May 7 '16 at 15:53
  • $\begingroup$ Alternatively, I think I remember reading in Effros-Ruan that there is a tensor interchange map for $\hat{\otimes}$ and $\otimes_{\min}$, which should go $E \hat{\otimes} (F \otimes_{\min} G) \to (E\hat{\otimes} F) \otimes_{\rm min} G$. Repeated use of this seems to then yield a complete contraction from the first thing in your comment to the second one. $\endgroup$
    – Yemon Choi
    May 7 '16 at 16:36
  • $\begingroup$ This is Proposition 8.1.10 in Effros-Ruan. $\endgroup$ May 9 '16 at 19:54
  • $\begingroup$ @MatthewDaws Thanks - I thought I remembered seeing it when trying to prove something else. I can prove the BSp version by hand (checked this after your comment) which made be 95% sure the OS version should work... $\endgroup$
    – Yemon Choi
    May 9 '16 at 20:04
0
$\begingroup$

My gut feeling is that these maps should have norm 1 for all $n$, here is an attempted argument:

For $n$ fixed and $k\geq 1$ we identify the algebraic tensor product $M_k\otimes M_n$ with $M_k(M_n)$ in the usual way. If we fix an operator space structure on $M_n$, then for each $k$ the norm on $M_k(M_n)$ induces a norm on $M_k\otimes M_n$. If I'm not mistaken, it's a result of Blecher and Paulsen [Tensor Products of Operator Spaces, JFA 1991] that the map $$ M_n\widehat{\otimes}M_n\to M_n\otimes_{max} M_n $$ is contractive (as a map between Banach spaces), where $max$ is the norm on $M_n\otimes M_n\cong M_n(M_n)$ induced by the maximal operator space structure on $M_n$. On the other hand, if we give $M_n$ its maximal operator space structure, then every contractive map $T:M_n\to B(H)$ is completely contractive, so the map in question factors as $$ M_n\widehat{\otimes}M_n\to M_n\otimes_{max} M_n \to M_n\overline{\otimes}M_n $$ where the first map is the identity, and the second is $id\otimes T$. But $M_n\overline{\otimes}M_n$ is the norm on $M_n(M_n)$ induced by the minimal operator space norm, and hence $id\otimes T$ is contractive (since $T$ is completely contractive out of the $max$ norm.)

EDIT: The above argument is faulty: in particular, it's not clear to me that the 1st level norm on the operator space tensor product $M_n\widehat{\otimes}M_n$ coincides with the projective tensor product of $M_n$ with $M_n$ in the category of Banach spaces. (There's a remark to this effect for general $X\widehat{\otimes}Y$ in the Blecher-Paulsen paper. However I'm not sure whether or not its true in the particular case of $X=Y=M_n$ with the usual operator space structure; I'll leave the above argument in place for now in case someone else knows how to fix it.)

$\endgroup$
1
  • $\begingroup$ Thanks for having a go at the question! I should note, writing $\otimes_\gamma$ for proj t.p. in category of Banach spaces, that "identity tensor transpose" is an isometry on from $M_n \otimes_\gamma M_n \to M_n \otimes_\gamma M_n$, because transpose is an isometric involution. Therefore, if $M_n \otimes_\gamma M_n$ coincides (up to equivalence of norms) with $M_n \widehat{\otimes} M_n$ then my problem is immediately solved... $\endgroup$
    – Yemon Choi
    Apr 28 '16 at 15:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.