# Covariant derivative of a metric determinant

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?

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 $$\nabla_\sigma\sqrt{\text{-det g}}= \frac{1}{2}\sqrt{\text{-det g}}\;g^{\alpha\beta}\nabla_\sigma g_{\alpha\beta}=\;0.$$

• Thoroughly explained. Commented Apr 19, 2020 at 3:08
• What is the reason to involve the square root? Commented Apr 19, 2020 at 8:51
• Why did not you start from here?$$g_{;\sigma}=g g^{\alpha\beta}g_{\alpha\beta;\sigma}$$ Commented Apr 19, 2020 at 9:09