Lemma: Smooth kernel of

Fix $f$, $g$, and a smooth $\varphi$ on the slab. There is a smooth symmetric contravariant tensor $\alpha =\alpha [f,g,\varphi ]$ on $M_T$ with $\mathrm{supp}\alpha \subset {\mathcal{D}}_f$ such that

Explicitly, with $b={{\mathsf{E}}_0}^{\mathrm{A}}(f\mathrm{d}\varphi )$, $b^{\prime }={{\mathsf{E}}_{\mathsf{m}}}^{\mathrm{A}}f$, and $(C^{\ast }b{)}^{\rho \sigma }=-{\partial }^{(\rho }{b}^{\sigma )}+\frac{1}{2}{\eta }^{\rho \sigma }{\partial }_{\mu }{b}^{\mu }$,

Proof. The coordinate term $-\int f{\mathsf{X}}_1[r]\cdot \partial \varphi$ is $-\text{Jc}(f\mathrm{d}\varphi )[r]$ in X-A's notation, whose kernel is ${\sigma }_{\epsilon }{C}^{\ast }({{\mathsf{E}}_0}^{\mathrm{A}}(f\mathrm{d}\varphi ))$ by X-A's kernel lemma; this gives the first term. For the scalar term, transposition gives $\int f{{\mathsf{E}}_{\mathsf{m}}}^{\mathrm{R}}{\mathsf{V}}_{1,g}[r]\varphi =\int b^{\prime }{\mathsf{V}}_{1,g}[r]\varphi$, and integrating by parts in the divergence (legitimate because $g$ has compact interior support),

Inserting ${a_1}^{\mu \nu }[r]=\frac{1}{2}{\eta }^{\mu \nu }\mathrm{tr}r-r^{\mu \nu }$, $b_1[r]=\frac{1}{2}\mathrm{tr}r$, and symmetrizing gives the second term of (eq:alphaexplicit). Supports: $b,b^{\prime }$ are supported in ${\mathcal{D}}_f$, and ${\mathcal{D}}_f$ is past-closed in $M_T$.