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

Proposition: Transformation of the mixed term

The mixed term, second-order boundary, and diagnostics › Action on the first dressing · lines 1353–1382 · proof included or linked

X-B-ST@1#prop:mixed-transformation, X-B-ST@1#eq:mixedchange

For $ξ \in g_{p}$ ,

M [h + K ξ] - M [h] = ϕ (- L_{1, g} [h] E_{A} D_{ξ} f) - ϕ (D_{ξ} c_{1} [h]) - ϕ (D_{ξ}^{2} f) .

This is a transformation law, not an invariance claim.

Proof. The two maps $h \mapsto L_{1, g} [h]$ and $h \mapsto c_{1} [h]$ are linear, while $c_{1} [K ξ] = D_{ξ} f$ . Expand

- (L_{1, g} [h] + L_{1, g} [K ξ]) E_{A} (c_{1} [h] + D_{ξ} f) .

On the plateau $L_{1, g} [K ξ] = Q_{K ξ}$ . For every compact test $a$ ,

- Q_{K ξ} E_{A} a = - D_{ξ} a + P_{m} (X_{ξ} E_{A} a) .

The second term vanishes in $A_{0}$ . Apply this with $a = c_{1} [h]$ and $a = D_{ξ} f$ .

Displayed formulas in this passage

X-B-ST@1#eq:mixedchange · line 1356
$M [h + K ξ] - M [h] = ϕ (- L_{1, g} [h] E_{A} D_{ξ} f) - ϕ (D_{ξ} c_{1} [h]) - ϕ (D_{ξ}^{2} f) .$
unlabeled · line 1370
$- (L_{1, g} [h] + L_{1, g} [K ξ]) E_{A} (c_{1} [h] + D_{ξ} f) .$
unlabeled · line 1376
$- Q_{K ξ} E_{A} a = - D_{ξ} a + P_{m} (X_{ξ} E_{A} a) .$