Proposition: Exact erasure benchmark

For this channel,

For every $0<p<1$,

so the exact rank-capacity condition fails, although optimal error tends to zero as $p\downarrow 0$.

Proof. Return the transmitted code state on success and prepare $I_C/d$ on erasure. The complete decoded channel is $(1-p)\mathrm{id}+p{\mathcal{T}}_{I_C/d}$, with error $p(1-d^{-2})$.

For any decoder, the success contribution to Bell overlap is at most $1-p$. The erasure contribution has a product Bell output and overlap exactly $p/d^2$. The Bell-projector effect therefore gives the matching lower bound. The rank calculation is immediate from the displayed complement.