Theorem: Normally hyperbolic coupled realization

The operator

has differential principal part ${\square }_{\bar{g}}{\mathrm{id}}_{\mathcal{V}}$. All its lower-order coefficients are smooth on $\mathcal{M}$. It is formally self-adjoint for the nondegenerate pairing

Proof. The second-derivative part of the linearized Einstein tensor is

Therefore the metric block of $\mathcal{H}$ has second-derivative part

The second-derivative part of $r{\mathcal{F}}^{†}\mathcal{F}$ is $r{D}^{†}D$. Equation (eq:beta) cancels it exactly.

The scalar block has principal part ${\square }_{\bar{g}}$. Mixed metric–scalar entries have at most one derivative: the background scalar gradient and its derivatives are fixed coefficients. The scalar term in $\mathcal{F}$ likewise creates only first- and zeroth-order mixed entries.

Thus $\mathcal{B}=\mathcal{A}{\square }_{\bar{g}}+\text{lower differential order}$. All coefficients are smooth because $t$ stays away from zero. Finally, $\mathcal{B}$ is self-adjoint for $(\cdot ,\cdot {)}_0$; using $\mathcal{A}\mathcal{L}=\mathcal{B}$ proves the asserted weighted self-adjointness.