Vanishing action integral for Gravitation Field

Question

For a Gravitation Field Action Integral looks like:

\begin{equation}\label{1} S_{gravity} = \frac{c^3}{16\pi G}\int R\sqrt{-g} d^4x. \end{equation}

А Least Action Principle says the $\delta S_{gravity} + \delta S_{matter} = 0$.

But we know the $\delta S_{gravity}$ itself is zero: \begin{equation} \delta S_{gravity} = \frac{c^3}{16\pi G} \int G_{\mu\nu} \sqrt{-g} \delta g^{\mu\nu}d^4x = \frac{c^3}{16\pi G} \int \sqrt{-g} dx^4 G_{\mu\nu}^{\,\, ;\nu}\xi^{\mu} = 0. \end{equation} Where $G_{\mu\nu}$ --- Einstein tensor, $\xi^{\mu}$ --- Killing vector.

Also, for matter

\begin{equation}\label{2} \delta S_{matter} = \int T_{\mu\nu}\sqrt{-g} \delta g^{\mu\nu}d^4x = \int \sqrt{-g}T^{\mu\nu}_{\,\, ;\nu}\xi_{\mu}d^4x = 0. \end{equation}

So, the equation $\delta S_{gravity} + \delta S_{matter} = 0$ looks like summation of zeroth $0 + 0 =0$ in contrast, for example, with Maxwell's theory where $\delta S_{EMF} + \delta S_{matter} = 0$ each therm are non zero.

So, why so for gravity? I really thought that only the sum of two $\delta S $ is zero, but not by themselves.

Source http://www.roma1.infn.it/teongrav/VALERIA/TEACHING/ONDE_GRAV_STELLE_BUCHINERI/AA2006_2007/DISPENSE/variational_principle.pdf

Answer 1

1. It seems that OP is only considering infinitesimal gauge transformations $$\delta g_{\mu\nu}~=~\nabla_{\{\mu}\xi_{\nu\}}.$$

2. The stationary action principle for $S_{\rm gravity}+S_{\rm matter}$ holds for arbitrary infinitesimal variations $\delta g_{\mu\nu}$ (that satisfy appropriate boundary conditions).

   In the latter case, $\delta S_{\rm gravity}$ is only zero in vacuum without matter.