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What is the covariant derivative of a metric tensor, this particular one to be specific $\nabla_{\mu} g^{\mu\nu}$? Notice we've got repetitive indices here. Is it zero and has it got to do anything with $$\nabla_{\alpha} g^{\mu\nu}=0~?$$

Here $\nabla_{\mu}$ is the covariant derivative and the connection is given by the Christoffel symbol $\Gamma^{\mu}_{\alpha \beta}$.

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Yes, it is zero. $\nabla_{\alpha} g^{\mu\nu}$ is a three-index tensor. If any tensor is zero, all of its contractions are zero.

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  • $\begingroup$ Could you elaborate that mathematically? $\endgroup$
    – Geeth
    May 27, 2019 at 17:00
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    $\begingroup$ A contraction is a particular sum of components in a basis. But if all components are 0, then this particular sum is 0. $\endgroup$
    – DanielC
    May 27, 2019 at 17:01
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To add a bit on the subject, when you don't assume that the connection needs to be metric-compatible, then you may have what is called non-metricity, defined as follows: $$Q_{\alpha \beta \gamma}\equiv \nabla_{\alpha}g_{\beta \gamma}$$.

And then the object you are mentioning is no longer zero but rather a non-vanishing vector $\hat{Q}^\alpha$, where $$ \hat{Q}^\alpha \equiv g_{\mu \nu}Q^{\mu \nu \alpha} \propto \nabla_\mu g^{\mu \nu}$$

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