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.

  • $\begingroup$ 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, 2016 at 18:25

2 Answers 2


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.

  • 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, 2016 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, 2016 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, 2016 at 16:36
  • $\begingroup$ This is Proposition 8.1.10 in Effros-Ruan. $\endgroup$ May 9, 2016 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, 2016 at 20:04

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.)

  • $\begingroup$ 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, 2016 at 15:43

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