Euler sign

With signature $(+,-,-,-)$ and the density (eq:Lg),

The leading part of ${\mathsf{E}}_g[q]$ is $\square +m^2={{\partial }_t}^2+{{\Omega }_m}^2$, and the first-order equation is $(\square +m^2){\rho }_{1,g}=-{\mathsf{V}}_{1,g}{\varphi }_0$, hence ${\rho }_{1,g}=-{\mathsf{E}}_m{\mathsf{V}}_{1,g}{\varphi }_0$ with the positive Duhamel kernel (eq:Duhamel). The retarded operator for sources supported in $\{t\ge 0\}$ agrees with ${\mathsf{E}}_m$. A convention with the opposite sign for the perturbation operator must change the response sign as well; the two cannot be mixed.