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 1


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 $$\nabla_{\rho} g^{\mu \nu} =0$$


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