How to compute divergence of a metric tensor? I am reading a paper where the author defines the divergence to be
$$\left(\delta_{g} \dot{g}\right)_{\mu}:=-\dot{g}_{\mu \kappa;}{}^{\kappa}$$
where $g$ looks like the De Sitter metric,
$$g=(3 / \Lambda) \frac{-d \tau^{2}+h(x, d x)}{\tau^{2}}.$$
I am not sure if I understand what $;$ means in the subscript of $\dot{g}_{\mu \kappa;}{}^{\kappa}.$ Could someone please explain?
 A: Often, the semicolon is used to denote covariant derivatives (just like a comma denotes a partial derivative), so:
\begin{align}
\dot g_{\mu\kappa;}{}^\kappa &\equiv \nabla^\kappa \dot g_{\mu\kappa}\\
&= g^{\alpha\kappa}\nabla_\alpha \dot g_{\mu\kappa} \\
&= \nabla_\alpha g^{\alpha\kappa} \dot g_{\mu\kappa} \\
&= \nabla_\alpha\dot g_{\mu}{}^\alpha
\end{align}
We have used metric compatibility of the metric to pull the metric past the covariant derivative (as an answer to your comment). In this post, I assume the Levi-Civita connection, as usual in general relativity.
For a tensor field $t^\mu{}_\nu$, the covariant derivative is given by
$$
\nabla_\alpha t^\mu{}_\nu = \partial_\alpha t^\mu{}_\nu + \Gamma^\mu{}_{\alpha\beta}t^\beta{}_\nu - \Gamma^\beta{}_{\alpha\nu} t^\mu{}_\beta,
$$
with the Cristoffel symbols $\Gamma^\mu{}_{\alpha\beta}$.
This generalizes straightforwardly to tensor fields of higher rank.
The covariant divergence is thus
$$
\nabla_\mu t^\mu{}_\nu = \partial_\mu t^\mu{}_\nu + \Gamma^\mu{}_{\mu\beta}t^\beta{}_\nu - \Gamma^\beta{}_{\mu\nu} t^\mu{}_\beta.
$$
It can, however, be shown (using Cramer's rule) that
$$
\Gamma^\mu{}_{\mu\beta} = \frac{1}{\sqrt{-g}}\partial_\beta\sqrt{-g},
$$
where $g$ denotes the determinant of the metric. Inserting into the above expression, this yields
\begin{align}
\nabla_\mu t^\mu{}_\nu &= \partial_\mu t^\mu{}_\nu + \frac{1}{\sqrt{-g}}t^\beta{}_\nu\partial_\beta\sqrt{-g} - \Gamma^\beta{}_{\mu\nu} t^\mu{}_\beta\\
&= \frac{1}{\sqrt{-g}}\partial_\mu(\sqrt{-g}\ t^\mu{}_\nu) - \Gamma^\beta{}_{\mu\nu} t^\mu{}_\beta,
\end{align}
where we have used the product rule in the last step. This can now directly applied to your case.
