# Metric determinant and its partial and covariant derivative

question : $$\nabla_a \nabla_b \sqrt{g} \phi =\partial_a \sqrt{g} \partial_b \phi$$ is true?

because $$\nabla_a \sqrt{g}=0$$ so we can write $$\sqrt{g} \nabla_a \nabla_b \phi$$ , but because the determinant of the metric does not transform like a scalar, we can not write partial derivative instead of covariant derivative.

• $\sqrt{g}$ is not a tensor and there we cannot define a covariant derivative action on it, so I don't know what you mean by $\nabla_a \sqrt{g}$. Can you clarify that? Apr 21, 2015 at 18:44
• @Prahar It is possible to define a covariant derivative on a scalar density $\rho(x)$ as $\nabla_i\rho=(\partial_i-\Gamma^l{}_{il})\rho$, cf. N. Straumann General Relativity (2013), pp. 663. It may be easily verified that this satisfies $\nabla_i\sqrt{g}=0$. Apr 22, 2015 at 2:11
• @0celo7 - cool. Did not know that. Sep 19, 2015 at 0:51

$$g=\frac{1}{4!}\varepsilon^{abcd}\varepsilon^{efgh}g_{ae}g_{bf}g_{dg}g_{dh}\\ \therefore \quad \nabla_m g = \frac{1}{3!} \varepsilon^{abcd}\varepsilon^{efgh}g_{ae}g_{bf}g_{dg}\nabla_m g_{dh}\\=0\;.$$ Note that $\varepsilon^{abcd}$ is Levi-Civita symbol, it is constant. It is tensor density of weight 0.5 and make $g$ be a tensor density of weight 1.
Yes it's true, since you can use the Leibniz rule like you do and since the covariant derivative of $$\sqrt g$$ is zero, like you say.