When looking at the linearised graviton propagator around a background, one can use several decompositions, e.g. decomposing the perturbation to the background metric $g_{\mu\nu}$ into
$h_{\mu\nu} = h^\perp_{\mu\nu} + \nabla_\mu A_\nu +\nabla_\nu A_\mu + \left(\nabla_\mu \nabla_\nu -\frac{1}{4}\nabla^2 g_{\mu\nu} \right) B + \frac{1}{4}g_{\mu\nu} h$
or rewriting the linearised equations of motion in terms of
$\theta_{\mu\nu}=g_{\mu\nu} - \frac{1}{\nabla^2}\nabla_\mu \nabla_\nu $ and $\omega_{\mu\nu} = \frac{1}{\nabla^2}\nabla_\mu \nabla_\nu$.
and then using $T_{\mu\nu} = \Pi^{-1}{}_{\mu\nu}{}^{\rho\sigma} h_{\rho\sigma}$ to find the inverse propagator $\Pi^{-1}$. If negative residues are produced in the propagator using these decompositions, then we know there must be a ghost. However, will negative residues coming directly from the equations of motion still produce a ghost?
e.g. if we have the equation of motion $T_{\mu\nu} = (k^2+m^2)(k^2-m^2)g_{\rho\sigma}\nabla_\mu \nabla_\nu h^{\rho\sigma}+\cdots$
which clearly produces a negative residue when inverted, can we say this will produce a ghost without needing to do one of the above decompositions?
