Manuscript III@2 · III.tex · proposition · 82dad3e5

Proposition: Difference of compensators

Compensator choice and classical kernel comparison › Different right inverses need not give the same observable · lines 1482–1503 · proof included or linked

III@2#prop:choices

Back to manuscript

Let $C, C^{'}$ be linear maps on a common metric domain, valued in smooth vector fields, with

C K = C^{'} K = id

on a common gauge class. For $A = C^{'} - C$ , one has $A K = 0$ , and

F_{f}^{C^{'}} - F_{f}^{C} = - κ \int_{M} f X_{A h} ϕ_{0} d^{4} x .

The difference is invariant under the common coefficient gauge action, but need not vanish.

Proof. Subtract the right-inverse identities. Under $h \mapsto h + K ξ$ , $A h$ is unchanged. Subtract the complete coefficient functionals.

Displayed formulas in this passage

unlabeled · line 1486
$C K = C^{'} K = id$
unlabeled · line 1491
$F_{f}^{C^{'}} - F_{f}^{C} = - κ \int_{M} f X_{A h} ϕ_{0} d^{4} x .$

Nearby in this section

section: Different right inverses need not give the same observable