# 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?

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.
• I am asking because I noticed that there the index $\kappa$ is at the bottom so I was just confused. Feb 18 at 7:54
• Yes, but with an additional term, because $g$ has rank 2. I will elaborate on my answer in a moment. Feb 18 at 7:56