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.



"Identity tensor transpose" as a map $M_n \hat{\otimes} M_n \to M_n \overline{\otimes} M_n$
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.


