Manuscript IX-E-3@1 · IX-E-3.tex · proposition · 8c41417a

Proposition: Retained-pair obstruction criterion

Conjugate transfer and fixed-domain obstruction · lines 1027–1043 · proof included or linked

If a retained pair has both actual orderings and $C_{f, g} \neq = 0$ , no symmetric causally factorizing map $T_{2} : D_{2} \to V^{[3]}$ exists.

Proof. Factorization and symmetry would require

B_{f} ⋆_{c} W_{g} = T_{2} (B_{f}, W_{g}) = T_{2} (W_{g}, B_{f}) = W_{g} ⋆_{c} B_{f} .

This contradicts (eq:commutator). A higher coefficient or contact cannot equalize unequal prescribed free values.

Displayed formulas in this passage

unlabeled · line 1035
$B_{f} ⋆_{c} W_{g} = T_{2} (B_{f}, W_{g}) = T_{2} (W_{g}, B_{f}) = W_{g} ⋆_{c} B_{f} .$