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?
If you have a $(0,2)$ tensor $\mathbf{T}$, then $T_{\mu\nu~;\kappa}$ is often used as an abbreviation for $\nabla_{\kappa}T_{\mu\nu}$ which is itself an abbreviation for $(\nabla T)_{\mu\nu\kappa}$. (See semicolon derivative.) Because this covariant derivative of $\mathbf{T}$ transforms as a tensor, it is sensible to raise and lower indices: $$T_{\mu\nu ~;}{}^\kappa=g^{\lambda\kappa}T_{\mu\nu~;\lambda}$$ And we can of course contract as well: $$T_{\mu\nu~;}{}^\kappa=g^{\lambda\kappa}T_{\mu\kappa~;\lambda}$$ I suppose in your case $\mathbf{T}=-\dot{\mathbf g}$.