# Completely bounded norm for unital maps with completely positive sections

Consider a completely bounded unital map $\Phi: \mathbf M_h(\mathbb C) \to \mathbf M_k(\mathbb C)$. Suppose that $\Phi$ has right-inverse $\Psi$ 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 $\mathbf A = \mathbf M_h(\mathbb C)$?

• Is $\Psi$ also assumed to be unital? (And do you want trace-preserving?) – Yemon Choi Oct 25 '17 at 11:51
• @YemonChoi: (a) if $\Psi$ is a right-inverse of a unital map, is it not also ipso facto unital? (b) I am very sure that I do not want to restrict to trace-preserving unital maps --- I have the 'trace' tag only because I'm aware that trace-preservation is the dual property to unitality, so there is an equivalent version of this question involving trace-preserving maps which may not be unital, and their left inverses. – Niel de Beaudrap Oct 26 '17 at 7:36

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 cb-norm 2.
There is also a negative answer if you look for $\Phi$ being left-invertible 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 $$\|\Phi\| = 1 < 2 = \|\Phi\|_{\rm cb}.$$