Lemma: Classical coupled identity

For compact $\xi$,

Therefore ${\mathcal{V}}_{\xi }{O}_f=0(\mathrm{mod}{s}^3)$.

Proof. The gauge fields satisfy

which give $\ell (K\xi )=0$. The time displacement cancels the background spatial scale variation, giving

Its divergence is $-\Delta {\xi }^a$. The decaying Newton inverse gives $\mathcal{R}(K\xi {)}^a={\xi }^a$.

Pullback maps clock slices, their induced metrics, and their harmonic equations to the corresponding new data. Uniqueness of the decaying rod coefficients identifies the transformed rods with the old rods pulled back. The raw lapse is a scalar. Hence its composition with the inverse clock-and-rod map is unchanged. Expanding this equality through second order gives the last identity.

Explicitly, the polarized rod contribution is

It is part of the identity, not a separately invariant piece. Substitution into (eq:O) proves the final statement.