Normalization: the classical retarded response

The relational dressing of the backreaction is the double retarded integral of the de Donder violation sourced by $\partial g$. It vanishes if and only if the switched source is conserved, that is, if and only if $({\partial }_{\nu }g)T^{\nu \mu }[{\varphi }_0]=0$: the same tensor as in Proposition prop:bianchi. The tree therefore measures, at the level of the observable, exactly the quantity whose vanishing the coupled jet would require. The ${\rho }_1$ term of (eq:classical-tree) is X-A's scalar correction at the background $k_1$; it does not vanish for conserved sources and is the part of the tree that survives when the dressing does not.

(i) If ${\partial }_{\nu }{S}^{\nu \mu }=0$ for a smooth source supported in $t>2\delta$, then ${\mathsf{X}}_1[{{\Delta }_{\mathrm{R}}}^{\mathrm{gr}}S]={{\mathsf{E}}_0}^{\mathrm{R}}(C[{{\Delta }_{\mathrm{R}}}^{\mathrm{gr}}S])=0$ by Lemma lem:violation: a conserved source is not dressed. (ii) If $h_{\mathrm{cl}}=K\xi$, $k_{\mathrm{cl}}=\frac{1}{2}{\mathcal{L}}_{\xi }K\xi$, $\xi \in {\mathfrak{g}}_U$, then $G_{\mathrm{lin}}[K\xi ]=0$, so the pure-gauge background is not a source of the metric equation at first order, and the classical completed observable is unchanged by X-D's Corollary; ${\mathsf{T}}_{\mathrm{tree}}$ does not depend on $(h_{\mathrm{cl}},k_{\mathrm{cl}})$ at all. Both checks are consistent with Proposition prop:normalization; neither replaces it.