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I know that $\nabla_\sigma g_{\mu\nu}=0$, being $\nabla_\sigma$ the covariant derivative, due to metric compatibility. I've also seen somewhere else that $\nabla_\sigma g=0$ as well, being $g$ the determinant of the metric tensor $g_{\mu\nu}$. Nonetheless, I don't know if the property:

$\nabla\left(S\otimes T\right)=\left(\nabla S\right)\otimes T+S\otimes\left(\nabla T\right)$

Holds in the case of the determinant, and therefore if $\nabla_\sigma\sqrt{g}g_{\mu\nu}=0$ for all values of $g_{\mu\nu}$.

Antoniou
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  • The determinant of the metric, regarded as an $\Bbb R$-valued function, depends on the choice of the coordinates. You can see that fact by inspecting the change-of-basis formula for the matrix representation of a bilinear form. – Travis Willse Jul 18 '23 at 23:23
  • The corresponding covariantly constant object on an oriented Riemannian manifold $(M, g)$ is a volume form, that is, a parallel section $\operatorname{vol}:=\bigwedge^n T^*M$ , where $(M, g)$ is the Riemannian manifold, satisfying $\operatorname{vol}(E_1, \ldots, E_n) = 1$, where $(E_1, \ldots, E_n)$ is a (local) orthonormal frame. One can construct $\operatorname{vol}$ from $g$ purely algebraically. – Travis Willse Jul 18 '23 at 23:23
  • Hey Travis, thanks for being around. I don't see how that responds to the question. – Antoniou Jul 19 '23 at 07:12
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    I think the underlying question here may amount to a confusion about notation. The determinant $|g| := \det (g_{ij})$ of the metric tensor (with respect to a frame) is not an invariant of the Riemannian metric $g$, so we shouldn't expect it to be invariant, which for a scalar-valued function like $|g|$ just means constant (on connected components). The metric compatibility condition is often written in index-free notation , as $\nabla g = 0$, and I suspect that's the formula you're referencing. – Travis Willse Jul 19 '23 at 07:26
  • Given oriented coordinates $(x^i)$, the volume form can be expressed in the coordinate patch as $\operatorname{vol} = \sqrt{|g|} , dx^1 \wedge \cdots \wedge dx^n$, and again this quantity is invariant, that is, satisfies $\nabla \operatorname{vol} = 0$. – Travis Willse Jul 19 '23 at 07:31
  • You might find this answer on physics.se useful: https://physics.stackexchange.com/a/589097 – Travis Willse Jul 19 '23 at 07:35
  • What do you mean by $g$? The determinant of the matrix $[g_{\mu\nu}]$ depends on the coordinates and therefore is not a well defined function on the manifold. Or do you mean the $n$-form we call the volume form? – Deane Jul 19 '23 at 10:54
  • I was referring to the determinant, since the volume form (the integral of $\sqrt{g}$ over a region $\Omega$ with volume differential $dV$) remains the same for any given metric independent of the coordinate system. A sphere will have the same volume in Cartesian and Spherical coordinates given any metric $g_{\mu\nu}$. – Antoniou Jul 19 '23 at 12:25

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Since you know that metric compatibility of the connection implies $$ \nabla_\sigma\,g_{\mu\nu}=0 $$ we only have to examine $\nabla_\sigma\sqrt{g}$ which is a partial derivative since $\sqrt{g}$ is a scalar. By the product rule, \begin{align}\require{cancel} \nabla_\sigma(\sqrt{g}\,g_{\mu\nu})=(\partial_\sigma\sqrt{g})g_{\mu\nu}+\cancel{\sqrt{g}\,\nabla_\sigma\,g_{\mu\nu}}\,. \end{align} This is zero, if either the component $g_{\mu\nu}$ is zero, or if $\partial_\sigma\sqrt{g}$ is zero. To figure out when the later holds we use one of the very useful formulas for the contracted Christoffel symbol $$ {\Gamma^\color{red}{\rho}}_{\color{red}{\rho}\,\sigma}=\partial_\sigma\log\sqrt{g}=\frac{\partial_\sigma\sqrt{q}}{\sqrt{q}}\,. $$

Kurt G.
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