Manuscript X-B-ST@1 · X-B-ST.tex · proposition · d03c150c

Proposition: Failure of input descent to free classes

The normalized scalar retarded map › The insertion domain is genuinely off shell · lines 1022–1051 · proof included or linked

X-B-ST@1#prop:input, X-B-ST@1#eq:inputwarning

Although its output is in $A_{0}$ , the map need not factor through the free class of its input:

R_{1} (V_{1, g} [h]; F_{P_{m} a}) = ϕ (- L_{1, g} [h] a)

need not vanish.

Proof. Use $E_{A} P_{m} a = a$ . For a concrete example let $m = 0$ , choose a real bump $r$ supported where $g = 1$ , and take $h = r η$ . Then

L_{1, g} [h] u = \partial_{μ} (r \partial^{μ} u) .

Choose $a = \partial_{t} r$ , with $\partial_{t} r \neq = 0$ . Against the smooth free solution $u = t$ ,

⟨ - L_{1, g} [h] a, t ⟩ = - ⟨ a, L_{1, g} [h] t ⟩ = - \int (\partial_{t} r)^{2} d^{4} x \neq = 0.

Every element of the equation ideal vanishes on smooth free solutions. Hence the output class is nonzero, although $[F_{P_{0} a}] = 0$ .

Displayed formulas in this passage

X-B-ST@1#eq:inputwarning · line 1026
$R_{1} (V_{1, g} [h]; F_{P_{m} a}) = ϕ (- L_{1, g} [h] a)$
unlabeled · line 1038
$L_{1, g} [h] u = \partial_{μ} (r \partial^{μ} u) .$
unlabeled · line 1043
$⟨ - L_{1, g} [h] a, t ⟩ = - ⟨ a, L_{1, g} [h] t ⟩ = - \int (\partial_{t} r)^{2} d^{4} x \neq = 0.$