Consider a completely bounded unital map $\Phi: \mathbf M_h(\mathbb C) \to \mathbf M_k(\mathbb C)$. Suppose that $\Phi$ has rightinverse $\Psi$ which is completely positive. Is the operator norm of $\Phi$ stable under tensor products with other spaces — i.e., is $$\lVert \Phi \rVert \stackrel?= \lVert \Phi \rVert_{\mathrm{cb}} = \bigl\lVert \Phi \otimes \mathrm{id}_{\mathbf A} \bigr\rVert$$ for $\mathbf A = \mathbf M_h(\mathbb C)$?

$\begingroup$ Is $\Psi$ also assumed to be unital? (And do you want tracepreserving?) $\endgroup$– Yemon ChoiOct 25, 2017 at 11:51

$\begingroup$ @YemonChoi: (a) if $\Psi$ is a rightinverse of a unital map, is it not also ipso facto unital? (b) I am very sure that I do not want to restrict to tracepreserving unital maps  I have the 'trace' tag only because I'm aware that tracepreservation is the dual property to unitality, so there is an equivalent version of this question involving tracepreserving maps which may not be unital, and their left inverses. $\endgroup$– Niel de BeaudrapOct 26, 2017 at 7:36
1 Answer
Unfortunately, the answer is no.
Suppose $\Phi : M_4 \rightarrow M_2$ and $\Psi : M_2 \rightarrow M_4$ are given by $$\Phi\left(\left[\begin{array}{cc} A & B\\ C& D\end{array}\right]\right) = A + 10B^T \ \ \textrm{and} \ \ \Psi(A) = \left[\begin{array}{cc} A&0\\ 0&A\end{array}\right] $$ where $A,B,C,D\in M_2$. Then $\Phi$ is unital, completely bounded, $\Psi$ is unital, completely positive and $\Phi\circ\Psi = I_{M_2}$ but $$\\Phi\ \leq 11 \ \ \textrm{and} \ \ \\Phi\_{\rm cb} \geq 20$$ since the transpose map is cbnorm 2.
There is also a negative answer if you look for $\Phi$ being leftinvertible by a ucp map:
In this case a counterexample is given by $\Phi : M_2 \rightarrow M_4$ and $\Psi : M_4 \rightarrow M_2$ defined by $$ \Phi(A) = \left[\begin{array}{cc}A & 0 \\ 0 & A^T\end{array}\right]\ \ \textrm{and} \ \ \Psi\left(\left[\begin{array}{cc} A & B \\ C & D\end{array}\right]\right) = A $$ where $A,B,C,D \in M_2$. Then $\Phi$ is ucb, $\Psi$ is ucp and $\Psi\circ\Phi = I_{M_2}$ but $$ \\Phi\ = 1 < 2 = \\Phi\_{\rm cb}. $$

$\begingroup$ This looks like a pretty straightforward example, thanks! It appears that I still am not asking quite the right question, but your answer is helpful in getting me to the right one. $\endgroup$ Oct 28, 2017 at 10:32