It is a well-known fact that the covariant derivative of a metric is zero. In a textbook, I found that the covariant derivative of a metric determinant is also zero.
I know
$$
g_{\alpha \beta;\sigma}=0
$$
So, $g=\det g_{\alpha\beta}$ is a metric determinant. $g_{;\sigma}$ is a covariant derivative of a metric determinant which is equal to an ordinary derivative of $g$.
$$
g_{;\sigma}=g_{,\sigma}=g g^{\alpha\beta}g_{\alpha\beta,\sigma}
$$
My question is why this must be zero?
Best Answer
It's because the determinate of the metric isn't a function - it's a tensor density.
Using the identity $$\text{ln(det g)=Tr(ln g)}$$
and taking the partial derivative, one can show
$$\frac{1}{\text {det g}}\partial_\sigma\text{det g}=g^{\alpha\beta}\partial_\sigma g_{\alpha\beta}$$
$$\partial_\sigma\sqrt{\text{-det g}}=\frac{1}{2}\sqrt{\text{-det g}}\;g^{\alpha\beta}\partial_{\sigma}g_{\alpha\beta}.\;\;(1)$$
Writing the partial derviatives of the metric tensor in terms of the Christoffel symbol
$$\Gamma^{\rho}_{\alpha\beta}=\frac{1}{2}g^{\rho\sigma}(\partial_\alpha g_{\beta\sigma}+\partial_\beta g_{\sigma\alpha}-\partial_\sigma g_{\alpha\beta})$$
and contracting the indices yields
$$\Gamma^{\alpha}_{\sigma\alpha}=\frac{1}{2}g^{\alpha\beta}\partial_\sigma g_{\alpha\beta}.\;\;(2)$$
Substituting equation $(2)$ into equation $(1)$
$$\partial_\sigma\sqrt{\text{-det g}} = \sqrt{\text{-det g}}\; \Gamma^{\alpha}_{\sigma\alpha} $$
and then combining the partial derivative with the Christoffel symbol implies $$ \nabla_\sigma\sqrt{\text{-det g}}= \frac{1}{2}\sqrt{\text{-det g}}\;g^{\alpha\beta}\nabla_\sigma g_{\alpha\beta}=\;0. $$