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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3$\begingroup$ Since gravity has always been observed to travel at $c$, even in an expanding universe with $\Lambda\approx1.1056\times10^{-52}$, so it's mass has to be zero. $\endgroup$– controlgroupCommented Jul 8 at 20:50
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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).
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2$\begingroup$ @ Andrew Please check this paper arxiv.org/pdf/1510.07856, equation 9. They have linearised it around de Sitter space and obtained this equation. Using Fourier modes for the field $h_{\mu\nu}$, we get a dispersion relation like $p^2 = \Lambda/3$. Despite the curvature of de-Sitter giving rise to a "mass" like term, I think we cannot interpret this as a mass of the graviton because the representation theory of de Sitter is different than flat space and naive comparison wont work. I want your opinion on it. $\endgroup$ Commented Jul 8 at 16:30
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2$\begingroup$ @PratikChatterjee Correct, the graviton does not have a mass on de Sitter space. The relevant group of isometries for which we want to find unitary transformations would be the de Sitter group instead of the Poincaire group. A massless graviton on de Sitter has two degrees of freedom, like a massless graviton on Minkowski, and that is what you find if you perturb Einstein's equations about a de Sitter background and count the degrees of freedom. $\endgroup$– AndrewCommented Jul 8 at 16:38
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2$\begingroup$ @andrew, is it true that a graviton mass would imply noninfinite range for gravitational effects? $\endgroup$ Commented Jul 8 at 17:56
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3$\begingroup$ @nielsnielsen Pertubartively, a massive graviton behaves like any other massive particle, with a Yukawa potential, and a modified dispersion relationship for gravitational waves. This leads to fairly stringent constraints on the graviton mass. Non-perturbatively, it's a challenge to define a non-linear theory of massive gravity, because of various pathologies (ghosts, superluminality...). There's a ghost-free theory of massive gravity, and the hope was to use it in cosmology to explain why the cosmological constant is degravitated, but no stable cosmological solution has been shown to exist. $\endgroup$– AndrewCommented Jul 8 at 19:26
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3$\begingroup$ Some references: arxiv.org/abs/1606.08462, arxiv.org/abs/1108.5231, arxiv.org/abs/hep-th/0703027, arxiv.org/abs/1401.4173, arxiv.org/abs/1105.3735 $\endgroup$– AndrewCommented Jul 8 at 19:27