Lemma: Gauge-fixed operator

${\mathcal{P}}^2=1$, and for every smooth symmetric $h$,

Thus $P_{\mathrm{gr}}=-\frac{1}{4}\mathcal{P}\square =-\frac{1}{4}\square \mathcal{P}$, a diagonal normally hyperbolic operator on symmetric tensors. Its retarded and advanced inverses are

and $\mathrm{supp}{{\Delta }_{\mathrm{R}}}^{\mathrm{gr}}S\subset J^{+}(\mathrm{supp}S)$ for past-compact $S$.

Proof. ${\mathrm{tr}}_{\eta }\bar{h}=-H$, so $\mathcal{P}\bar{h}=\bar{h}+\frac{1}{2}\eta H=h$. Substitute $h=\bar{h}-\frac{1}{2}\eta \mathrm{tr}\bar{h}$ into (eq:G2): the terms ${\partial }_{\mu }{\partial }^{\rho }{h}_{\rho \nu }+{\partial }_{\nu }{\partial }^{\rho }{h}_{\rho \mu }-{\partial }_{\mu }{\partial }_{\nu }H$ become ${\partial }_{\mu }{C}_{\nu }+{\partial }_{\nu }{C}_{\mu }$, and $-{\eta }_{\mu \nu }{\partial }^{\rho }{\partial }^{\sigma }{h}_{\rho \sigma }+{\eta }_{\mu \nu }\square H-\square h_{\mu \nu }$ becomes $-\square {\bar{h}}_{\mu \nu }-{\eta }_{\mu \nu }\partial \cdot C$; halve. For the gauge term, $\delta C^{\rho }={\partial }_{\sigma }\delta h^{\sigma \rho }-\frac{1}{2}{\partial }^{\rho }{\eta }_{\alpha \beta }\delta h^{\alpha \beta }$ gives $\delta S_{\mathrm{gf}}=\int \delta h^{\mu \nu }(-\frac{1}{2}{\partial }_{(\mu }{C}_{\nu )}+\frac{1}{4}{\eta }_{\mu \nu }\partial \cdot C)$, which cancels the two $C$-terms of $\frac{1}{2}{G}_{\mathrm{lin}}$. Since $\mathcal{P}$ is constant and invertible, inverses of $P_{\mathrm{gr}}$ are $-4\mathcal{P}$ times those of $\square$.