# Why is the covariant derivative of the determinant of the metric zero?

Comments to the post (v2):

Note that $\sqrt{|g|}$ transforms as a density rather than a scalar under general coordinate transformations. In particular, the covariant derivative of $\sqrt{|g|}$ does not necessarily coincide with the partial derivative of $\sqrt{|g|}$.

Here is a heuristic explanation using local coordinates. The Levi-Civita connection is compatible with the metric $g_{\mu\nu}\mathrm{d}x^{\mu}\odot \mathrm{d}x^{\nu}$. That a connection $\nabla$ is compatible with a metric means that $\nabla_{\lambda}g_{\mu\nu}=0$. Using linearity and Leibniz rule, the covariant derivative $\nabla_{\lambda}$ then annihilates any sufficiently nice function $f(g_{00},g_{01}, \ldots)$ of the metric. In particular, the square root of the determinant $\sqrt{|g|}$, so $\nabla_{\lambda}\sqrt{|g|}=0$.

OK. Let us take ordinary derivative of determinant of some covariant 2-tensor $A_{\mu\nu}$. Let call it $A$. But it is more convenient to allow us to think about $A_{\mu\nu}$ like a matrix with covariant indices. So $$\det{A_{\mu\nu}} = A$$ Next, let's do the following calculations: $$\delta\ln{\det{A_{\mu\nu}}} = \ln{\det{(A_{\mu\nu}}+\delta A_{\mu\nu})}-\ln{\det{A_{\mu\nu}}} = \ln{\det({A^{\mu\sigma}(A_{\sigma\nu}+\delta A_{\sigma\nu}))}},$$ where $\delta$ is like differential and $A^{\mu\sigma}$ denotes contravriant 2-tensor with the following property: $A^{\mu\sigma}A_{\sigma\nu} = \delta^{\mu}_{\,\nu}$, in other words, "inverse" tensor.

Let's continue $$\ln{\det({A^{\mu\sigma}(A_{\sigma\nu}+\delta A_{\sigma\nu}))}} = \ln{\det{(I+A^{\mu\sigma}\delta A_{\sigma\nu})}} = \ln{(1 + \mathrm{Tr}\,{A^{\mu\sigma}\delta A_{\sigma\nu}})} = \mathrm{Tr}\,{A^{\mu\sigma}\delta A_{\sigma\nu}}.$$ But $$\mathrm{Tr}\,{A^{\mu\sigma}\delta A_{\sigma\nu}} = A^{\mu\sigma}\delta A_{\sigma\mu}$$

Divided by $dx^{\lambda}$ it gives $$\partial_{\lambda}\ln{\det{A_{\mu\nu}}} = A^{\mu\sigma}\partial_{\lambda} A_{\sigma\mu}.$$

Therefore, $$\frac{\partial_{\lambda}A}{A} = A^{\mu\sigma}\partial_{\lambda} A_{\sigma\mu}$$

Or $$\partial_{\lambda}g = g g^{\mu\sigma}\partial_{\lambda} g_{\sigma\mu}$$ The following step is pretty fun. Let's replace all ordinary partials by absolute (covariant). So we have $$\nabla_{\lambda}g = g g^{\mu\sigma}\nabla_{\lambda} g_{\sigma\mu}.$$ But $$\nabla_{\lambda} g_{\sigma\mu} = 0.$$ QED. The last is not hard exercise. Indeed, in the geodesic coordinates it is always true because in these coordinates $\nabla_{\nu} = \partial_{\nu}$. But if some tensor is equal to zero in one reference frame then it is zero in every reference frame.

But I am not sure about the same trick with arbitrary matrix (although it may turn out the same). It would be better to use the following. Since $$\det{g^{\mu\nu}A_{\nu\sigma}}$$ is a scalar, we can use ordinary derivative for this. But on the other hand, we could use covariant derivative for it. For scalar it is the same. So $$\nabla_{\nu}(\det{g^{\mu\nu}A_{\mu\nu}}) = g^{-1}\nabla_{\nu}A + A\nabla_{\nu}g^{-1} = g^{-1}\partial_\nu A + A\partial_\nu g^{-1}$$ Let us continue calculations $$\nabla_{\nu}A = \partial_{\nu} A - A\frac{\partial_\nu g}{g}$$ Where we used $\nabla_\nu g = 0$. Partial derivatives we can find from the previous equations.

Here's a heuristic calculation: Let $\{E_i\}$ be an orthonormal frame ($g(E_i,E_j)=\epsilon_i\delta_{ij}, \epsilon_i=\pm 1$). Then $\mu$ is the canonical volume form $\sqrt{g}\,\mathrm{d}x^1\wedge\cdots\wedge\mathrm{d}x^n$ iff $\mu(E_1,\dotsc, E_n)=1$. Then $$(\nabla_X\mu)(E_1,\dotsc,E_n)=\nabla_X(\mu(E_1,\dotsc,E_n))-\sum \mu(E_1,\dotsc,\nabla_X E_i,\dotsc,E_n)=-\sum \epsilon_ig(E_i,\nabla_X E_i)\mu(E_1,\dotsc,E_i,\dotsc,E_n)=-\sum \epsilon_ig(E_i,\nabla_X E_i)=-\frac{1}{2}\sum \epsilon_i\nabla_X g(E_i,E_i)=0$$ for all vector fields $X$. Then using the derivation property of the connection, and $\nabla_X(\mathrm{d}x^1\wedge\cdots\wedge\mathrm{d}x^n)=0$ for all $X$, one has $$\nabla_X\mu=(\nabla_X\sqrt{g})\,\mathrm{d}x^1\wedge\cdots\wedge\mathrm{d}x^n=0$$ whence $\nabla_X\sqrt{g}=0$ for all $X$ and in some chart.

To be absolutely pedantic, one should adapt the defintion of the connection in terms of parallel transport to tensor densities. This is done in e.g. Straumann, *General Relativity* (2013). For a scalar density $\rho$ one finds in local coordinates $\nabla_i\rho=(\partial_i-\Gamma^l{}_{il})\rho$. From the standard expression for $\Gamma^l{}_{li}$ it is easy to verify that $\nabla_i\sqrt{g}=0$.