Proposition: Classical rod images are spatially nonlocal

There are compact perturbations for which $Y_1$ is not compactly supported. Consequently, the rod part of the Hessian of a compactly label-smeared lapse is not automatically a compact joint field kernel.

Proof. Take $\phi =h_{00}=h_{0i}=0$ and $h_{ij}=2a^{2}b{\delta }_{ij}$, where $b$ is compact and has nonzero spatial integral at some time. Then $C=2bI$, $B_1=bI$, and ${Y_1}^i=-\mathsf{G}{\partial }_{i}b$. The displayed asymptotic shows its noncompact tail.

Polarize the rod term $-2{Y_1}^i{\partial }_i{n}_1$ against a second perturbation having $h_{00}\ne 0$ near the label-smearing support. Its dependence on the first perturbation is through the Newton kernel even at distant spatial points. No compact support in those first-input variables follows from compact support of the output test.