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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?

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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.

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  • $\begingroup$ thanks, so we compute using the formula given here, (mathworld.wolfram.com/CovariantDerivative.html) right? $\endgroup$
    – Student
    Feb 18 at 7:54
  • $\begingroup$ I am asking because I noticed that there the index $\kappa$ is at the bottom so I was just confused. $\endgroup$
    – Student
    Feb 18 at 7:54
  • $\begingroup$ Yes, but with an additional term, because $g$ has rank 2. I will elaborate on my answer in a moment. $\endgroup$
    – scaphys
    Feb 18 at 7:56

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