I've been asked to show the following: For a vector field $V$ on a semi-Riemannian manifold with metric $g$ that $$Div \cdot V = \frac{1}{\sqrt{\det(g)}}\partial_i\left(\sqrt{\det(g)}V^i\right)$$ I know we're supposed to use Christoffel symbols as well as a few matrix formulas, but I'm not sure how to proceed. In particular, we were given that for a (invertible) matrix $M$ with some parameter $s$, that $$\frac{d}{ds}\det M(s)=\det M(s) \cdot tr\left(M(s)^{-1}\frac{d}{ds}M(s)\right)$$ and $$\frac{d}{ds}(M(s)^{-1})=-M(s)^{-1}M'(s)M(s)^{-1}$$ Any help would be greatly appreciated.

The definition of divergence that we were given was $$Div \cdot V = \nabla_{\partial_i}V^i = \partial_iV^i+\Gamma_{ij}^iV^i$$


Let's work backwards.

The product rule gives $$\begin{align*} \frac{1}{\sqrt{\det g}}\partial_i\left(\sqrt{\det g}\,V^i \right) & = \frac{1}{\sqrt{\det g}}\left[ \sqrt{\det g}\ \partial_iV^i + \partial_i\left(\sqrt{\det g}\right)V^i \right] \\ & = \partial_iV^i + \frac{1}{\sqrt{\det g}}\partial_i\left(\sqrt{\det g}\right)V^i \\ & = \partial_iV^i + \frac{1}{\sqrt{\det g}} \frac{1}{2\sqrt{\det g}}\partial_i\left(\det g\right)V^i \end{align*}$$

We can now apply the formula for the derivative of a determinant to get: $$\begin{align*} \frac{1}{\sqrt{\det g}}\partial_i\left(\sqrt{\det g}\,V^i \right) & = \partial_iV^i + \frac{1}{2} \text{tr}(g^{-1}\partial_ig)\,V^i \\ & = \partial_iV^i + \frac{1}{2} \text{tr}(g^{jk}\partial_ig_{kl})\,V^i \\ & = \partial_iV^i + \frac{1}{2} g^{jk} \partial_i g_{kj}\,V^i \end{align*}$$

Now, it is a general fact that for the Levi-Civita connection, the Christoffel symbols satisfy $$\partial_ig_{kj} = \Gamma^l_{ik}g_{lj} + \Gamma^l_{ij}g_{kl}.$$ Therefore, $$\begin{align*} \frac{1}{\sqrt{\det g}}\partial_i\left(\sqrt{\det g}\,V^i \right) & = \partial_iV^i + \frac{1}{2} g^{jk} \left(\Gamma^l_{ik}g_{lj} + \Gamma^l_{ij}g_{kl} \right)\,V^i \\ & = \partial_iV^i + \frac{1}{2}\left(\Gamma^l_{ik}\delta^k_l + \Gamma^l_{ij}\delta^j_l \right)\,V^i \\ & = \partial_iV^i + \Gamma^l_{il}\,V^i \\ & = Div \cdot V \end{align*}$$ as was to be shown.

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  • $\begingroup$ Actually, I think the formula $\partial_i g_{kj} = \Gamma^l_{ik}g_{lj} + \Gamma^l_{ij}g_{kl}$ is true more generally for any connection compatible with the metric, but don't quote me on that. $\endgroup$ – Jesse Madnick Apr 27 '12 at 9:15
  • $\begingroup$ This (the formula for the derivative of the metric tensor) seems to be a direct consequence of corrollary 7 in chapter 6 of the second volume of Spivak's comprehensive introduction to differential geometry. $\endgroup$ – user20266 Apr 27 '12 at 10:28
  • $\begingroup$ I just know it as a consequence of the Koszul formula, a.k.a. one of the Christoffel identities. (I don't own any of Spivak's diff geo books besides Calculus on Manifolds.) $\endgroup$ – Jesse Madnick Apr 27 '12 at 15:19

$\def\div[0]{\operatorname{div}}$ My favourite reason this formula is true: let $\phi$ be an arbitrary smooth function with compact support contained in a single chart $(U,x)$. Then integrating by parts we get

$$ \int_U \phi \div V d\mu_g= -\int_U V^i \partial_i \phi\, d\mu_g = -\int_U V^i \partial_i \phi \sqrt {\det g}\, dx.$$ Now integrate by parts again, but this time in the chart geometry (i.e. use the product rule for partial derivatives instead of the one for Riemannian divergence, and the measure $dx$ instead of $d \mu_g$):

$$ -\int_U \partial_i \phi\, V^i \sqrt {\det g}\, dx= \int_U \phi\, \partial_i (\sqrt{\det g}\, V^i) dx = \int_U \phi\, \frac 1{\sqrt{\det g}} \partial_i(\sqrt{\det g}\, V^i)\ d\mu_g.$$

Since this is true for every $\phi$ we get the desired formula for $\div V$.

In some sense this is just a weak version of Giuseppe's answer, but seeing it was the first time this formula really clicked for me, so I figured I'd write it down.

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  • $\begingroup$ This is nice, but I have two questions: 1) Isn't it true that the "integration by parts formula" for the divergence is proved via its coordinate formula (you need to know in advance that $\operatorname{div}(fX)=f\operatorname{div}(X)+\langle \operatorname{grad f}, X\rangle$). The only way I know how to do this is via coordinates, which of course would make your argument circular (but I guess it is just my ignorance). 2) I am not sure what happend in your last equality. $\endgroup$ – Asaf Shachar Jun 27 '16 at 14:18
  • $\begingroup$ 1) I did my whole first course in Riemannian geometry without even seeing this formula for the divergence - you can do everything with its Christoffel-style coordinate form $\rm{div} X = \partial_i X^i + \Gamma_{ij}^i X^j$ if you need a coordinate expression for some reason. 2) Thanks, fixed a typo there. $\endgroup$ – Anthony Carapetis Jun 27 '16 at 14:22

The metric volume form is $\omega=\sqrt{|\det(g)|}dx^1\wedge\ldots\wedge dx^n.$
The divergence of $V=V^i\partial_i$ is determined by $(\text{div}V)\omega=d(V\lrcorner\omega)\equiv V(\omega),$ hence we get: $$(\text{div}V)\omega=\left[V^i\partial_i(\sqrt{|\det(g)|})+\sqrt{|\det(g)|}\partial_iV^i\right]dx^1\wedge\ldots\wedge dx^n,$$ Where we used the obvious formula $V(dx^1\wedge\ldots\wedge dx^n)=(\partial_iV^i)dx^1\wedge\ldots\wedge dx^n.$
Therefore $$\text{div}V=\frac{1}{\sqrt{|\det(g)|}}\frac{\partial}{\partial x_i}\left[\sqrt{|\det(g)|}V^i\right]$$

Edit added in reply to Asaf Shachar's comment

Probably, in the original version of my answer, I should have spent a few words justifying the following formula $$V(dx^1\wedge\ldots\wedge dx^n)=(\partial_iV^i)dx^1\wedge\ldots\wedge dx^n.\tag{$\star$}$$ Here, on the left-hand side, as elsewhere in my answer, I am using the expression $V(\eta)$ to denote the Lie derivative, along the vector field $V$, of a differential form $\eta$.

Recall that the vector field $V$ is locally given by $V=V^i\partial_i$ in terms of the holonomic frame $\partial_1,\ldots,\partial_n$ corresponding to the local coordinate system $x^1,\ldots,x^n$, so that $$V(x^i)=V^i\tag{1}.$$ Moreover, since the Lie derivative $V(-)$ is a (degree $0$) derivation of the differential graded algebra $(\Omega^\bullet(X),\wedge,d_{dR})$, we get immediately that $$V(dx^1\wedge\ldots\wedge dx^n)=\sum_{i=1}^n dx^1\wedge\ldots\wedge d(V(x^i))\wedge\ldots\wedge dx^n.\tag{2}$$ Now, from $(1)$ and $(2)$, it follows $(\star)$, as we needed.

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    $\begingroup$ Can you please explain the "obvious formula"? $\endgroup$ – Asaf Shachar Jun 26 '16 at 20:00
  • $\begingroup$ Dear @AsafShachar I have edited my answer according with your request. Let me know it is now ok for you. $\endgroup$ – agtortorella Jun 26 '16 at 20:52
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    $\begingroup$ Thanks! this really helped. I now see your derivation uses indeed the fact that the Lie derivative is a derivation, together with its commutation with the exterior derivative (The latter is required in order to prove $\mathcal{L}_V(dx^i)=dV^i$). $\endgroup$ – Asaf Shachar Jun 27 '16 at 8:44
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    $\begingroup$ By the way, this answer here (math.stackexchange.com/a/1758100/104576) contains alternative proof which is based on evaluations of the relevant differential forms on $n$-vectors. Interestingly, it does not use the commutation of the Lie derivative with the exterior derivative. $\endgroup$ – Asaf Shachar Jun 27 '16 at 9:25

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