# Integration over the hypersurface in Einstein-Hilbert action

In the variation of Einstein-Hilbert action the integral of the term $$d^4x\sqrt{-g}g^{\mu\nu}\delta R_{\mu\nu}$$ over the spacetime manifold is equal to zero. $$I=\underset{\mathcal M}\int d^4x\sqrt{-g}g^{\mu\nu}\delta R_{\mu\nu}=\underset{\mathcal M}\int d^4x\sqrt{-g}D_{\mu}(g^{\alpha\beta}\delta\Gamma^{\mu}_{\alpha\beta}-g^{\alpha\mu}\delta\Gamma^{\beta}_{\alpha\beta})$$

Denoting $$g^{\alpha\beta}\delta\Gamma^{\mu}_{\alpha\beta}-g^{\alpha\mu}\delta\Gamma^{\beta}_{\alpha\beta}=\delta\mathcal V^{\mu}:$$ $$\underset{\mathcal M}\int d^4x\sqrt{-g}D_{\mu}\mathcal \delta V^{\mu}=\underset{\partial \mathcal M}\oint d\Sigma_{\mu}\delta\mathcal V^{\mu}=0$$ where $$d\Sigma_{\mu}=n_{\mu}\sqrt{|\gamma|}d^3\xi$$, $$\gamma$$ is $$3$$-dimensional induced metric. What is the reason of being zero of the integral?

• Usually when one is performing the variation of the action to get the equations of motion of the theory and suddenly throws boundary terms away, the non-stated assumption is that the variations of the fields are supposed to be of compact support, meaning that they vanish outside of a compact set. In that case these boundary integrals end up being set to zero. But if the variations are not compactly supported one cannot throw these terms away. In particular they are relevant in defining the sympletic form following Wald's method. Please see e.g. aip.scitation.org/doi/10.1063/1.528801.
– Gold
Jan 29 '20 at 18:37