Proposition: Perfect-copy screening and doubling

For a perfect-record tower:

1. If some $j\in D(r,r^{\prime })$ has a member outside $S$, then ${{\Gamma }^S}_{r{r}^{\prime }}=0$.

2. If every member of every differing class lies in $S$, then the left and right supports of ${{\Gamma }^S}_{r{r}^{\prime }}$ are orthogonal, and

   $$\Vert {{\Gamma }^S}_{r{r}^{\prime }}+{{\Gamma }^{S†}}_{r{r}^{\prime }}{\Vert }_1=2v_{r{r}^{\prime }}(S).$$

Proof. An inaccessible differing member contributes $\mathrm{tr}|r_j⟩⟨{r^{\prime }}_j|=0$, proving the first assertion. Otherwise, the reduced cross operator retains the differing perfect record factors on $S$. They place its left and right supports in orthogonal computational subspaces. For an operator $P\Gamma Q$ with $PQ=0$, the Hermitian sum has block form

Its singular values are those of $B$, each occurring twice.