# Variation of the Einstein-Hilbert action to derive the metricity condition

Consider the Einstein-Hilbert action: $$S=\int d^{4} x \sqrt{-g} g^{\mu \nu} R_{\mu \nu}$$ If we vary it with respect to the connection, assuming no prior relation between the metric and the connection, we have $$\delta S=\int d^{4} x \sqrt{-g} g^{\mu \nu} \delta R_{\mu \nu}$$

$$= \int d^{4} x \sqrt{-g} g^{\mu \nu} ( \nabla_{\rho} \delta \Gamma^{\rho}_{\mu \nu} - \nabla_{\nu} \delta \Gamma^{\rho}_{\mu \rho} )$$

$$= \int d^{4} x \sqrt{-g}(\nabla_{\rho} g^{\mu \nu}-\delta^{\nu}_{\rho} \nabla_{\alpha} g^{\mu \alpha}) \delta \Gamma^{\rho}_{\mu \nu}$$

Where the second equality has been written cutting a long story short. Let us also assume that the connection is symmetric in the lower indices, then, for the action to be stationary, $$\nabla_{\rho} g^{\mu \nu} - \frac{1}{2} \delta^{\nu}_{\rho} \nabla_{\alpha} g^{\mu \alpha} - \frac{1}{2} \delta^{\mu}_{\rho} \nabla_{\alpha} g^{\nu \alpha} =0$$

The question is how the metricity condition $$\nabla_{\alpha} g^{\mu \nu}=0$$ can be deduced from this?

## 1 Answer

To see that this implies the covariant conservation of the metric, we simply look at some of the traces and contractions of the equations presented here. For instance, tracing over $$\mu$$ and $$\rho$$ gives $$\nabla_{\rho} g^{\rho \nu} - \frac{1}{2} \delta^{\nu}_{\rho} \nabla_{\alpha} g^{\rho \alpha} - \frac{1}{2} \delta^{\rho}_{\rho} \nabla_{\alpha} g^{\nu \alpha} =0 \\ \nabla_{\rho} g^{\rho \nu} - \frac{1}{2}\nabla_{\alpha} g^{\nu \alpha} - 2 \nabla_{\alpha} g^{\nu \alpha} =0 \\ -\frac{3}{2}\nabla_{\rho} g^{\rho \nu} =0$$

This shows that the traced non-metricity terms must vanishes. Then simply substituting this back into the original equation yields the desired result $$\nabla_{\rho} g^{\mu \nu} =0$$