Theorem: Retained-pair obstruction

Suppose $B_f\otimes W_g\in {\mathcal{D}}_2$, both actual support orderings hold, and ${\mathcal{C}}_{f,g}\ne 0$. Then no symmetric causally factorizing map ${\mathsf{T}}_2$ exists on ${\mathcal{D}}_2$.

Proof. Factorization and symmetry would require

This contradicts (eq:commutator). The nonzero order-$\hslash$ coefficient remains nonzero modulo ${\hslash }^4$. No higher coefficient or contact can make the two prescribed free values equal.