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

This question, metric determinant and its partial and covariant derivative, seems to indicate $$\nabla_a \sqrt{g}=0.$$ Why is this the case? I've always learned that $$\nabla_a f= \partial_a f,$$ hence surely $$\nabla_a \sqrt{g}= \partial_a\sqrt{g} \neq 0.$$

Where's the hole in my logic?

• Notice that in their own definition $\nabla_{\mu}f\neq \partial_{\mu}f$. – Giorgio Comitini Apr 16 '16 at 12:50
• Why would we have the term "covariant derivative" if it was just the same as the normal partial derivative? – Danu Apr 16 '16 at 13:09
• @Danu. I was under the impression that this is always true, see p39 here: damtp.cam.ac.uk/user/hsr1000/lecturenotes_2012.pdf . How is its action normally defined on a function rather than a tensor? – user138901 Apr 16 '16 at 21:25
• @user138901 Ah, that comment makes it clear what you're confused about: $\sqrt{\lvert g\rvert}$ is not a scalar (as pointed out in the answer by Qmechanic)! I initially misinterpreted your question. – Danu Apr 16 '16 at 21:36

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.

• By the way, I am not sure about the same trick with arbitrary matrix. See the following. – LRDPRDX Oct 21 '16 at 12:37
• @7919, I have chosen $\nabla_{\nu}$ for the covariant derivative. – LRDPRDX Feb 21 '18 at 5:18

1. 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 $\sqrt{|g|}$.

2. 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 $\nabla_{\lambda}\sqrt{|g|}=0$.

• Qmechanic I do not really understand your heuristic explanation. Following your logic it would seem that any simple product of metric components should be covariantly constant. But it's not – magma Apr 28 '16 at 7:42
• @magma : Which simple counterexample do you have in mind? – Qmechanic Apr 28 '16 at 9:04
• We probably misunderstand each other, but -from what you write - it would seem that any polynomial function of metric components , such as $f=g_{00}^2$, is covariantly constant which is not the case. I do not think at the moment that there is a simple heuristic/intuitive justification for the constancy of the determinant. You just have to take the definition and make the calculations – magma Apr 28 '16 at 9:43
• In the example $f=g_{00}g_{00}$, one could apply the covariant derivative $\nabla_{\lambda}$ to the $(0,4)$ tensor $T=(g_{\mu\nu}\mathrm{d}x^{\mu} \odot \mathrm{d}x^{\nu})\odot (g_{\rho\sigma}\mathrm{d}x^{\rho} \odot \mathrm{d}x^{\sigma})$, and compare the $0000$-component before and after. – Qmechanic Apr 28 '16 at 10:48

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 $\nabla_i\sqrt{g}=0$.

• My diff. geometry is very rusty, but I was wondering whether it would suffice to show that $\nabla_i\sqrt{g}=0$ in normal coordinates (where it is trivial)? Wouldn't the general case automatically follow? – AccidentalFourierTransform Apr 19 '16 at 18:28
• @AccidentalFourierTransform That's what I do in the "to be absolutely pedantic" section. (Not in normal coordinates, but in normal coordinates one has $\Gamma\equiv 0$ so it is indeed trivial.) The issue is that a density always refers to a coordinate system, that's what makes it a density. So I'm not sure if doing the calculation in one coordinate system is sufficient -- this object $\nabla_i\sqrt{g}$ is not a tensor. – Ryan Unger Apr 19 '16 at 19:40
• IIRC, local coordinates$\neq$normal coordinates. What I mean is, in normal coordinates at $p$ we have $g_{ij}(p)=\delta_{ij}$ and $\Gamma^i_{jk}(p)=0$. Therefore, $g=1$ and $\nabla_i g=\partial_i g=0$. As $\nabla_i g$ is a (pseudo)tensor, if its zero in these coordinates, it must be zero in any coordinate system. [The terminology is probably very wrong because I studied these issues a couple of years ago]. I would phrase my reasoning as "we found a system of reference where $\nabla_i g=0$, and therefore, by covariance, it must be zero in any system of reference".Does this make sense to you? – AccidentalFourierTransform Apr 19 '16 at 19:47
• @AccidentalFourierTransform Normal coordinates are a special kind of local coordinates. (All coordinates are local coordinates.) The issue is that a priori, it is not clear what $\nabla_ig$ even means. – Ryan Unger Apr 19 '16 at 20:53
• @AccidentalFourierTransform The axioms of the connection allow us only to take derivatives of functions and vector fields. We can extend this to one-forms (and simple tensors) by using the Leibnitz rule and to arbitrary tensors by factoring into simple tensors. It turns out one has to define an extra structure to make sense of $\nabla_i\rho$, where $\rho$ is a tensor density. – Ryan Unger Apr 19 '16 at 21:23