Theorem: Fixed-domain obstruction

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

Proof. Factorization and symmetry would give

This contradicts Proposition prop:commutator. The nonzero coefficient at order $\hslash$ remains nonzero in $\mathbb{C}[\hslash ]/({\hslash }^4)$. The argument is independent of every contact choice and every higher coefficient.