Is the restriction of a strictly contractive channel (SCC) to a subspace necessarily still SCC? (impossibility of perfect QEC for SCCs)

This paper shows the impossibility of perfect error correction for strictly contractive quantum channels, i.e., for channels such that $$||\mathcal{E}(\rho)-\mathcal{E}(\sigma) ||\leq k ||\rho-\sigma||$$, for $$0\leq k <1$$.

The requirement for perfect error correction of a subspace $$K$$ is that there exists a channel $$S$$ such that $$S$$ is the inverse of the restriction of $$\mathcal{E}$$ to the subspace $$K$$.

The proof of impossibility uses the fact that this would require $$||S\mathcal{E}(|u\rangle\langle u|)-S\mathcal{E}(|v\rangle\langle v|)|| = |||u\rangle\langle u|-|v\rangle\langle v|||$$, for some basis vectors $$u,v$$, which would contradict strict contractivity.

My confusion is concerning how this contradiction argument doesn't seem take into consideration the fact that we should restrict to the subspace $$K$$. In other words, if $$P$$ is the projector onto the subspace $$K$$, is it generally true that if $$\mathcal{E}$$ is strictly contractive, then $$||P(\mathcal{E}(\rho))-P(\mathcal{E}(\sigma)) ||<||P(\rho)-P(\sigma)||$$?

• No, that is negated by noting that the projector itself is a CP-map: consider $P(\rho) = \sum\limits_{j} \Pi_{j} \rho \Pi_{j}$ and notice that it is already in a Kraus form (and hence automatically CP); although it is not TP (unless it is simply a dephasing operator). Therefore, $\left\Vert P \circ \mathcal{E} (\rho - \sigma) \right\Vert_{1} \leq \left\Vert \mathcal{E} (\rho - \sigma) \right\Vert_{1} \leq \left\Vert \rho - \sigma \right\Vert_{1}$. So, no, the projector will only contract'' the $1$-norm distance even more. Jul 29 '20 at 7:46
• I understand that the projector will cause the trace distance to contract compared to the original space, but what happens when we compare $||P\circ \mathcal{E}(\rho-\sigma)||$ to $||P(\rho -\sigma)||$ not $||\rho -\sigma||$? Jul 29 '20 at 7:51
I am no longer confused about this, since now I see in this equation we are already restricting to a subspace $$||S\mathcal{E}(|u\rangle\langle u|)-S\mathcal{E}(|v\rangle\langle v|)|| = |||u\rangle\langle u|-|v\rangle\langle v|||$$, and the contracting map has to contract every subspace.