# Coordinate and tensor divergences

Using our previous results regarding the divergence and the relation $${\overline{\Gamma}^{\mu}{}_{\lambda\mu}=\Gamma^{\mu}{}_{\lambda\mu}-T^{\mu}{}_{\mu\lambda}}$$, we can derive many useful coordinate dependent relations. Adopting the common abbreviation

$$\displaystyle \sqrt{g}\equiv\sqrt{\left|\det\left(g_{\mu\nu}\right)\right|}$$

and including torsion for completeness, we expand both sides of the coordinate divergence expression to get

\begin{aligned}\partial_{\lambda}\sqrt{g} & =\sqrt{g}\left(\Gamma^{\mu}{}_{\lambda\mu}-T^{\mu}{}_{\mu\lambda}\right)\\
& =\sqrt{g}\ \overline{\Gamma}^{\mu}{}_{\lambda\mu}, \end{aligned}

which along with our previous expressions for the metric derivative yields

\begin{aligned}\partial_{\lambda}\left(\sqrt{g}g^{\mu\nu}\right) & =\sqrt{g}\left(g^{\mu\nu}\Gamma^{\sigma}{}_{\lambda\sigma}-g^{\mu\nu}T^{\sigma}{}_{\sigma\lambda}-\Gamma^{\mu\nu}{}_{\lambda}-\Gamma^{\nu\mu}{}_{\lambda}\right)\\
\Rightarrow\partial_{\nu}\left(\sqrt{g}g^{\mu\nu}\right) & =-\sqrt{g}\left(\Gamma^{\mu\nu}{}_{\nu}-T^{\nu\mu}{}_{\nu}\right). \end{aligned}

From $${\det\left(\exp\left(g\right)\right)=\exp\left(\mathrm{tr}\left(g\right)\right)\Rightarrow\ln\left(\det\left(g\right)\right)=\mathrm{tr}\left(\ln\left(g\right)\right)}$$, we can take the derivative of the components upon which it turns out that

\begin{aligned}\frac{1}{\det\left(g\right)}\partial_{\lambda}\left(\det\left(g\right)\right) & =\mathrm{tr}\left(g^{-1}\partial_{\lambda}g\right)\\
& =g^{\mu\nu}\partial_{\lambda}g_{\mu\nu}\\
\Rightarrow\partial_{\lambda}\sqrt{g} & =\frac{1}{2}\sqrt{g}g^{\mu\nu}\partial_{\lambda}g_{\mu\nu}\\
\Rightarrow\frac{1}{2}g^{\mu\nu}\partial_{\lambda}g_{\mu\nu} & =\Gamma^{\mu}{}_{\lambda\mu}-T^{\mu}{}_{\mu\lambda}\\
& =\overline{\Gamma}^{\mu}{}_{\lambda\mu}.\end{aligned}

By considering the inverse matrix, we see that these expressions are also valid with $${g^{\mu\nu}\partial_{\lambda}g_{\mu\nu}\rightarrow-g_{\mu\nu}\partial_{\lambda}g^{\mu\nu}}$$. The first line above may also be applied to the tensor Lie derivative in terms of coordinates, yielding

\begin{aligned}L_{u}\sqrt{g} & =\sqrt{g}\ \mathrm{div}\left(u\right)\\
\Rightarrow\mathrm{div}\left(u\right) & =\frac{1}{2}g^{\mu\nu}L_{u}g_{\mu\nu}.
\end{aligned}

If we consider an anti-symmetric tensor $${F^{\mu\nu}}$$ and a symmetric tensor $${G^{\mu\nu}}$$, it is not hard to see that

\begin{aligned}\nabla_{\nu}F^{\mu\nu}-T^{\lambda}{}_{\lambda\nu}F^{\mu\nu} & =\frac{1}{\sqrt{g}}\partial_{\nu}\left(\sqrt{g}F^{\mu\nu}\right)-\frac{1}{2}T^{\mu}{}_{\lambda\nu}F^{\lambda\nu},\\
\nabla_{\nu}G^{\mu\nu}-T^{\lambda}{}_{\lambda\nu}G^{\mu\nu} & =\frac{1}{\sqrt{g}}\partial_{\nu}\left(\sqrt{g}G^{\mu\nu}\right)+\Gamma^{\mu}{}_{\lambda\nu}G^{\lambda\nu},\\
\nabla_{\nu}G_{\mu}{}^{\nu}-T^{\lambda}{}_{\lambda\nu}G_{\mu}{}^{\nu} & =\frac{1}{\sqrt{g}}\partial_{\nu}\left(\sqrt{g}G_{\mu}{}^{\nu}\right)-\Gamma^{\lambda}{}_{\mu\nu}G_{\lambda}{}^{\nu}\\
& =\frac{1}{\sqrt{g}}\partial_{\nu}\left(\sqrt{g}G_{\mu}{}^{\nu}\right)-\frac{1}{2}\partial_{\mu}g_{\lambda\nu}G^{\lambda\nu}+T^{\lambda}{}_{\mu\nu}G_{\lambda}{}^{\nu}.
\end{aligned}

The above expressions are more commonly presented with zero torsion, with $${\overline{\nabla}_{\nu}}$$ defining the “divergence” of the tensor. It can also be shown ( p. 365) that the “divergence” of an exterior $${k}$$-form expressed as an anti-symmetric tensor can be written in terms of the hodge star as

\begin{aligned}\overline{\nabla}^{\nu}F_{\nu\mu_{2}\cdots\mu_{k}}&\equiv g^{\nu\mu_{1}}\overline{\nabla}_{\nu}F_{\mu_{1}\cdots\mu_{k}}\\&=-\left(\delta F\right)_{\mu_{2}\cdots\mu_{k}}\\&=\left(-1\right){}^{n(k+1)+s}\left(*\mathrm{d}\left(*F\right)\right)_{\mu_{2}\cdots\mu_{k}}.
\end{aligned}