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I have encountered expression $$\frac{1}{2}\left(2 \dot{g}_{\mu}{}^{\lambda ; \mu}-\dot{g}_{\mu}{}^{\mu ; \lambda}\right)$$ in a GR paper.

Here we assume to be working with the de Sitter metric $g$ and $\dot{g}$ is some two tensor. I know that in general $$F_{\mu\nu;\kappa}=\partial_{\kappa} F_{\mu \nu}-\Gamma(g)_{\mu \kappa}^{\lambda} F_{\lambda \nu}-\Gamma(g)_{\nu \kappa}^{\lambda} F_{\mu \lambda},$$ but I am not sure how I can apply this to two terms where one index is at the bottom and the other one is at the top. I tried to lower everything as follows.

Thus for instance for the first term, \begin{align} \dot{g}_{\mu}{}^{\lambda ; \mu} =\nabla^\mu \dot{g}_{\mu}{}^{\lambda}=g^{\mu \alpha}\nabla_{\alpha}(g^{\lambda \gamma}\dot{g}_{\mu \gamma}). \end{align} However, now I have to take the derivative of the product of two tensors which is not very nice. Is there a way to write a direct formula just like the one for $F$?

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  • $\begingroup$ A link to the paper would also be useful $\endgroup$
    – Eletie
    Mar 8 at 0:28

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In General Relativity, the covariant derivative is always taken to be compatible with the metric. In other words, $\nabla_{\mu} g_{\nu\tau} = 0$ and $\nabla_{\mu} g^{\nu\tau} = 0$. This implies that $\nabla_{\alpha}(g^{\lambda \gamma}\dot{g}_{\mu \gamma}) = g^{\lambda \gamma} \nabla_{\alpha}\dot{g}_{\mu \gamma}$.

As for the remaining steps of the calculation, I would do it in the same manner: bring all the derivative indices down and proceed as usual.

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