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 $\mathbf A = \mathbf M_h(\mathbb C)$?

  • $\begingroup$ Is $\Psi$ also assumed to be unital? (And do you want trace-preserving?) $\endgroup$
    – Yemon Choi
    Oct 25, 2017 at 11:51
  • $\begingroup$ @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. $\endgroup$ Oct 26, 2017 at 7:36

1 Answer 1


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 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 $$ \|\Phi\| = 1 < 2 = \|\Phi\|_{\rm cb}. $$

  • $\begingroup$ 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

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