If I consider the Einstein equations into the form $$ R_{\mu\nu}=\kappa \left(T_{\mu\nu}-\frac{1}{2}g_{\mu\nu}T\right)+\Lambda g_{\mu\nu} $$ and then linearize them, we should get by moving to harmonic coordinates $$ \Box g_{\mu\nu}=\Lambda g_{\mu\nu}+\text{terms}. $$ Therefore, the cosmological term appears to act like a genuine mass term (it could have the right sign depending on the sign of $\Lambda$). Thus, is it correct to say that, in a de Sitter metric, a graviton could acquire mass? Indeed, we would get a mass pole in the propagator.
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No, you don't get a mass term in the propagator, and there is no graviton mass for GR with a cosmological constant.
If you linearize about Minkowski space, $g_{\mu\nu}=\eta_{\mu\nu}+h_{\mu\nu}$, then you will find there is a tadpole term in the action. This is an instability indicating that Minkowski is not the correct background to expand around -- it is not a solution to the background equations of motion, and you can't trust perturbations about this solution to remain small.
The correct expansion is to linearize about de Sitter space, $g_{\mu\nu}=g^{dS}_{\mu\nu}+h_{\mu\nu}$, where $h_{\mu\nu}$ is the fluctuation and $g^{dS}_{\mu\nu}$ is the de Sitter metric. If you do this expansion properly, you will find the graviton does not have a mass. Instead, you get wave equation written in terms of the covariant derivative with respect to the de Sitter metric. There are two propagating degrees of freedom, consistent with a massless graviton (a massive graviton would have five).
Andrew
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