Lemma: Plateau covariance of the density operator

For an admitted actual diffeomorphism $F$, and hence coefficientwise for (eq:formal-group),

The density pullback is $F_{\mathrm{dens}}^{\ast }r=(\det DF)(r\circ F)$.

Proof. The inverse-metric density ${\mathcal{A}}_{\gamma }$ is a contravariant tensor density, and ${\rho }_{\gamma }$ is a scalar density. Their coordinate transformation laws follow by matrix inversion and the determinant chain rule.

Because $F^{\ast }g=g$, the discrepancy between $F^{\ast }{\mathcal{A}}_g[\gamma ]$ and ${\mathcal{A}}_g[F^{\ast }\gamma ]$ is

Inside $U$, $1-g=0$. Outside $U$, $F$ and its derivative are the identity. The discrepancy therefore vanishes. The same argument applies to ${\rho }_g$.

For completeness, divergence of a vector density is natural: pair it against a compact scalar test, integrate by parts, and change variables by $F$. The transformed integral is exactly the integral of the pulled-back divergence. Applying this fact to ${\mathcal{A}}_g[\gamma {]}^{\mu \nu }{\partial }_{\nu }u$ and adding the mass density proves (eq:plateaucov). Formal expansion gives the coefficient identity.