The ledger on the witness

Take (eq:witness). Then $A\gamma =0$ everywhere, so Corollary cor:plateau applies, and $g=1$ on the whole observation dependence cylinder, so ${\mathcal{D}}_f\cap J^{+}(\mathrm{supp}\xi )\subset P_g$ and the sum is zero. The sheet asks that the contributions be printed before they are added. In the cylinder, with $r={\mathcal{L}}_{\xi }\gamma$, $s_2=\frac{1}{2}{\mathcal{L}}_{\xi }K\xi$, and $p:=a{a}^{\prime \prime }+2a^{\prime 2}$,

{L1, held-fixed cross term, linear in $\xi$.} XI-E recomputed ${{\mathsf{X}}_1}^0[r]=ta-2{{\int }_0}^{t}a$, ${\mathsf{V}}_{1,g}[r]\varphi =-4t{a}^{\prime }-2t^2{a}^{\prime \prime }$, $-{{\mathsf{E}}_0}^{\mathrm{R}}{\mathsf{V}}_{1,g}[r]\varphi =2t^{2}a-4{{\int }_0}^{t}sa$, whence the density $4{{\int }_0}^t(t-s)a(s)\mathrm{d}s$ and

{L2, held-fixed cross term, quadratic in $\xi$.} ${a_1}^{00}[s_2]=-\frac{1}{2}p$, ${a_1}^{ij}[s_2]=-\frac{1}{2}p{\delta }^{ij}$, ${{\mathsf{X}}_1}^0[s_2]={{\mathsf{E}}_0}^{\mathrm{R}}(\frac{1}{2}{p}^{\prime })=\frac{1}{2}P(t)$ with $P(t)={{\int }_0}^{t}p$, ${\mathsf{V}}_{1,g}[s_2]\varphi ={\partial }_0(-pt)+{a_1}^{ij}{\partial }_i{\partial }_j\varphi =-2p-t{p}^{\prime }$, and $-{{\mathsf{E}}_0}^{\mathrm{R}}{\mathsf{V}}_{1,g}[s_2]\varphi ={{\int }_0}^t(t-s)(2p+s{p}^{\prime })\mathrm{d}s=tP(t)$. The density is $tP-\frac{1}{2}P\cdot 2t=0$:

3, covariant part of the cubic response, linear.

4, covariant part of the cubic response, quadratic.

5, gauge-fixing response. ${\zeta }_{\mathrm{gf}}[s_{\xi }]=z(t){\partial }_t$ with $z=-{{\mathsf{E}}_0}^{\mathrm{R}}(t{a}^{\prime \prime }+\frac{1}{2}{p}^{\prime })=-(ta-2{{\int }_0}^{t}a+\frac{1}{2}P)$, $z(0)=z^{\prime }(0)=0$. For any such $\zeta =z{\partial }_t$ in the $g=1$ cylinder, $(K\zeta {)}_{00}=2z^{\prime }$, ${a_1}^{00}[K\zeta ]=-z^{\prime }$, ${a_1}^{ij}[K\zeta ]=-z^{\prime }{\delta }^{ij}$, ${\mathsf{V}}_{1,g}[K\zeta ]\varphi =-4z^{\prime }-2t{z}^{\prime \prime }$, $-{{\mathsf{E}}_0}^{\mathrm{R}}{\mathsf{V}}_{1,g}[K\zeta ]\varphi =2tz$, ${\mathsf{X}}_1[K\zeta ]=\zeta$, and the density is $2tz-z{\partial }_0\varphi =2tz-2tz=0$. This is Lemma lem:switch verified by hand:

(For $t$ beyond $\mathrm{supp}a$, $z$ is the constant $2\int a-\frac{1}{2}\int a^{\prime 2}$: the de Donder parameter ends as a rigid time translation. This is a remark, not an input.)

6, Einstein-shell term.

7, tree. $\mathcal{T}$ has no metric argument: $\text{L7}=0$.

This is Corollary cor:plateau evaluated, and the positive witness pairing (eq:I) verifies the sign: the held-fixed $-I$ is removed by the covariant part of the response generated by the cubic vertex, not by the gauge-fixing response, which pairs to zero.