Diagnostics

Let $m=0$, ${\varphi }_0=t$, and near a causal cylinder containing the dependence region of $K$ let $g=g_0(t)$, with the usual spatial cutoffs outside it. Then $T^{\mu \nu }[t]=n^{\mu }{n}^{\nu }-\frac{1}{2}{\eta }^{\mu \nu }$, $\mathcal{P}T=n\otimes n$, and (eq:k1) gives

Hence $a^{00}[k_1]=G_2$, $({\partial }_{\nu }g)T^{\nu \mu }=\frac{1}{2}{g}_0^{\prime }{n}^{\mu }$, $C^{\mu }[k_1]=-G_1{n}^{\mu }$, and ${{\mathsf{X}}_1}^0[k_1]=-G_3$, in agreement with (eq:dressing). The scalar part is ${\rho }_1=-{{\mathsf{E}}_0}^{\mathrm{R}}{\partial }_t(g_0{G}_2)=-{{\int }_0}^t{g}_0{G}_2$, so

This agrees with (eq:classical-tree) and is nonnegative for $f\ge 0$. It vanishes if $g_0\equiv 1$ on $[0,t]$, consistent with the fact that for ${\varphi }_0=t$ the first-order relational scalar is identically zero on plateau-supported backgrounds: $t$ is itself a harmonic coordinate. The witness of Proposition prop:witness therefore uses ${\varphi }_0=t{x}^1$, not ${\varphi }_0=t$.

Let $\xi \in {\mathfrak{g}}_U$, $\zeta =0$, $h_{\mathrm{cl}}=K\xi$, $k_{\mathrm{cl}}=\frac{1}{2}{\mathcal{L}}_{\xi }K\xi$. The classical completed observable equals its value at $(0,0)$ modulo $\mathfrak{J}$ by X-D's Corollary. With $\hat{h}$ present and fixed, Theorem thm:ward gives the additional term $-{\kappa }^2{\hat{A}}_1[{\mathcal{L}}_{\xi }\hat{h}]$, which by Lemma lem:gamma is $2{\kappa }^2{\hat{A}}_1[{\xi }^{\lambda }{{\Gamma }^{(1)}}_{\lambda \cdot \cdot }[\hat{h}]]$ modulo $\mathfrak{J}$, and by Proposition prop:witness is nonzero for suitable inputs. The two-jet check of X-D passes; the quantum-metric check at ${\kappa }^2$ does not. That is the order boundary.