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 $$Div \cdot V = \nabla_{\partial_i}V^i = \partial_iV^i+\Gamma_{ij}^iV^i$$
Best Answer
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.