Manuscript IX-D-ST@1 · IX-D-ST.tex · lemma · b1dc631a

Lemma: Free coherence does not imply causal factorization

Scalar non-lift and an explicit causal separation · lines 855–881 · proof included or linked

IX-D-ST@1#lem:separation

The inherited free coherence is compatible with the symmetric, admitted auxiliary map $M_{2} (F, G) = F G$ , although that map fails causal factorization on a retained ordered first-stage pair.

Proof. Choose nonzero tests with

supp f \subset {0 < t < a}, supp g \subset {t > b}, b > a .

The support bounds imply $W_{g} ≳ B_{f}$ . Both free orderings belong to the first-stage domain.

Classical closure makes $M_{2}$ admitted and symmetric. It changes none of the inherited free data. But

S (W_{g} \otimes B_{f}) = W_{g} B_{f} + ℏ Q^{R}_{g, f}, Q^{R}_{g, f} \neq = 0

by the inherited injective presentation map. Thus $M_{2} (W_{g}, B_{f}) \neq = S (W_{g} \otimes B_{f})$ .

Displayed formulas in this passage

unlabeled · line 865
$supp f \subset {0 < t < a}, supp g \subset {t > b}, b > a .$
unlabeled · line 875
$S (W_{g} \otimes B_{f}) = W_{g} B_{f} + ℏ Q^{R}_{g, f}, Q^{R}_{g, f} \neq = 0$

Lemma: Free coherence does not imply causal factorization

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