Theorem: Complete coupled classical Ward calculation

On the smooth spatially compact domain, and on the spatial Schwartz extension of XX-D,

Thus

Holding $\zeta$ fixed would instead leave $-s^2{L}_f(A_{\xi }z)$, which is not generally zero.

Proof. The first identity is Lemma lem:rodgauge.

For the second, the raw lapse is a scalar, the clock slices transform by pullback, and the slice-harmonic equation is geometrically covariant. The transformed original rods have the same asymptotic labels. Uniqueness of the decaying coefficientwise Newton solution identifies them with the pullbacks of the old rods.

If $\mathcal{C}$ is the full clock-and-rod map, this gives

Therefore $N\circ \alpha$ is invariant. Expanding this equality under the flow generated by $s\xi$, with (eq:jetgauge), gives

Both coefficients vanish.

This argument includes the original rod term. Indeed, its polarized gauge contribution is

using (eq:rodgauge). It cannot be omitted from the coefficient identity.

If the second jet is held fixed, the $s^2{L}_f(A_{\xi }z)$ contribution is lost. The surviving variation is the stated residual.

It is not identically zero: choose a scalar $\phi$ locally constant and nonzero and a compact time component ${\xi }^0$ with nonzero time derivative in that region. The metric part of $A_{\xi }z$ can be taken zero. A compact output test detects the resulting nonzero $\ell (A_{\xi }z)$.