Theorem: Exact complete-complement criterion

The following are equivalent:

1. there is a CPTP decoder $D$ with $D\mathcal{N}={\mathrm{id}}_C$;

2. there is a state ${\tau }_B$ with

   $$\mathcal{C}(X)=\mathrm{Tr}(X){\tau }_B(X\in \mathcal{B}(C));$$

3. for all code matrix units, $\mathcal{C}(E_{ij})={\delta }_{ij}{\tau }_B$.

When these conditions hold and $r=\mathrm{rank}{\tau }_B$,

Recovery preserves every independent external reference.

Proof. Dilate an exact decoder by an isometry $U:R\to C\otimes G$. Then $Z=(U\otimes I_B)W$ dilates identity. Purity of the decoded output on basis input $|i⟩$ gives

Exact recovery of off-diagonal matrix units gives $⟨{\chi }_j,{\chi }_i⟩=1$. All normalized vectors ${\chi }_i$ coincide. Tracing $CG$ proves constancy of the complete complement.

Conversely, diagonalize ${\tau }_B={{\sum }_{a=1}}^r{\lambda }_a|a⟩⟨a|$, with ${\lambda }_a>0$. There is no component of $W|i⟩$ in the kernel of ${\tau }_B$, so write

Constancy on matrix units gives

The $dr$ vectors are orthonormal, proving the dimension bound.

On their span define the isometry $A^{†}{u}_{ia}=|i⟩|a⟩$, equivalently $A:C\otimes {\mathbb{C}}^r\to R$. The channel

is CPTP for any fixed state ${\omega }_C$ and recovers every matrix unit. Linearity proves equivalence with condition three. Channel equality extends after tensoring with every reference.