Gauge invariance of linearized gravity with an arbitrary background spacetime

Question

Consider here a background metric $g_{\mu\nu}$, we impose a perturbation $g_{\mu\nu}+\epsilon h_{\mu\nu}$ with $\epsilon\ll1$. Then we can write down the modified Einstein-Hilbert action with zero cosmological constant: \begin{align} S&=\int\mathrm{d}^{D}x\sqrt{-g}\left(1-\epsilon\frac{1}{2}h_{\mu\nu}g^{\mu\nu}\right)\left(R^{0}+\epsilon R^{1}\right)\\ &=S^{0}+\epsilon\int\mathrm{d}^{D}x\sqrt{-g}\left(R^{1}-\frac{1}{2}R^{0}g_{\mu\nu}h^{\mu\nu}\right)+\mathcal{O}(\epsilon^2)\, \end{align} where $R^{0}$ is the Ricci scalar w.r.t $g_{\mu\nu}$, and $R^{1}$ is the correction of $R$ at the leading order which can be found in this answer.

To my knowledge, if for any: $$\delta_{\zeta}h_{\mu\nu}=\nabla_{\mu}\zeta_{\nu}+\nabla_{\nu}\zeta_{\mu}=\mathcal{L}_{\zeta}g_{\mu\nu} \, \ \ \ \delta_{\zeta}g_{\mu\nu}=0 \,$$ the action is invariant at each order, this theory is gauge invariant, and the perturbation field $h_{\mu\nu}$ is a spin-2 gauge field. For example, when the background is flat $g_{\mu\nu}=\eta_{\mu\nu}$, $R^{0}=0$, one can check this easily, for an arbitrary $\zeta \ $: \begin{align} \delta_{\zeta}S&=\delta_{\zeta}S^0+\epsilon\int\mathrm{d}^{D}x\sqrt{-g}\delta_{\zeta}R^{1} +\mathcal{O}(\epsilon^2)\\ &=0. \end{align} But for a curved background $R^{0}\ne 0$, in principle one should show whether: $$\int\mathrm{d}^Dx\sqrt{-g}\left(\delta_{\zeta}R^{1}-\frac{1}{2}R^{0}g_{\mu\nu}\mathcal{L}_{\zeta}g^{\mu\nu}\right)=0 \ .$$ According to the P4 eq(2.4) in this paper 2110.13512, one find \begin{equation} \delta_{\zeta}R^{1}=\mathcal{L}_{\zeta}R^{0}. \end{equation} Thus, for any $g_{\mu\nu}$ \begin{align} \mathcal{L}_{\zeta}R-\frac{1}{2}Rg_{\mu\nu}\mathcal{L}_{\zeta}g^{\mu\nu}&=\mathcal{L}_{\zeta}(g^{\mu\nu}R_{\mu\nu})-\frac{1}{2}Rg_{\mu\nu}\mathcal{L}_{\zeta}g^{\mu\nu}\\ &=g^{\mu\nu}\mathcal{L}_{\zeta}R_{\mu\nu}+\left[R_{\mu\nu}-\frac{1}{2}Rg_{\mu\nu}\right]\mathcal{L}_{\zeta}g^{\mu\nu}\\ &=g^{\mu\nu}\mathcal{L}_{\zeta}R_{\mu\nu}+\mathrm{EoM}\mathcal{L}_{\zeta}g^{\mu\nu} \end{align} if $g_{\mu\nu}$ is a solution of field equation, the gauge invariance of a linearized theory if and only if $\delta_{\zeta}R^1_{\mu\nu}=\mathcal{L}_{\zeta}R^0_{\mu\nu}=0$, as expected. But it is not easy to calculate. So on the other hand, we express it as \begin{align} \mathcal{L}_{\zeta}R-\frac{1}{2}Rg_{\mu\nu}\mathcal{L}_{\zeta}g^{\mu\nu}&=\mathcal{L}_{\zeta}R+\frac{1}{2}Rg^{\mu\nu}\mathcal{L}_{\zeta}g_{\mu\nu}\\ &=\zeta^\mu\nabla_\mu R+R\nabla_\mu\zeta^\mu\\ &=\nabla_\mu(\zeta^\mu R) \end{align} This equation states that a theory is gauge invariant if and only if $\zeta^\mu R^{0}\big|_{\partial\mathcal{M}}=0$. This result is suspicious to me, is something wrong here?

Especially, for $dS$ spacetime, this boundary term is non-zero. So is there a gauge-invariant linearized theory in $dS$?

Answer 1

Gauge symmetry for a linearized theory

Consider a gravity theory allowing a cosmological constant described by the Hilbert-Einstein action $$S[g]=\int_\mathcal{M}\mathrm{d}^nx\sqrt{-g}\left(R-2\Lambda\right)\, ,$$ which is invariant under an arbitrary diffeomorphism transformation $$x^\mu\rightarrow x^\mu+\xi^\mu(x)\, ,$$ but usually, a distinct viewpoint attributing all transformations on the field is more useful $$g_{\mu\nu}\rightarrow g_{\mu\nu}+(\nabla_{\mu}\xi_{\nu}+\nabla_{\nu}\xi_{\mu})\, .$$ Now we divide the whole spacetime metric into two parts $$g_{\mu\nu}+\bar{g}_{\mu\nu}+ h_{\mu\nu}\, ,$$ where we set $\bar{g}_{\mu\nu}$ as a background field, which is fixed thus with no dynamics - a given solution of field equation, $$S^\prime[h]=S[h+\bar{g}]\, .$$ Thus the theory of $h_{\mu\nu}$ inherits the diffeomorphism symmetry from the whole theory of $g_{\mu\nu}$, i.e. under \begin{align} \delta_\xi \bar{g}_{\mu\nu}&=0\\ \delta_\xi h_{\mu\nu}&=\nabla_{\mu}\xi_{\nu}+\nabla_{\nu}\xi_{\mu}\, \\ &=\bar{\nabla}_{\mu}\xi_{\nu}+\bar{\nabla}_{\nu}\xi_{\mu}-2\delta\Gamma^{\rho}_{\ \mu\nu}\xi_\rho\, ,\\ &=\bar{\nabla}_{\mu}\xi_{\nu}+\bar{\nabla}_{\nu}\xi_{\mu}+\mathcal{O}(h\xi)\\ \delta\Gamma^{\rho}_{\ \mu\nu}&=\frac{1}{2}[\bar{\nabla}_\mu h_{\nu}^{\ \rho}+\bar{\nabla}_\nu h_{\mu}^{\ \rho}-\bar{\nabla}^\rho h_{\mu\nu}]\propto \mathcal{O}(h)\, . \end{align} the $S^\prime$ is invariant.