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I would like to calculate $\nabla_\mu(g^{\mu\alpha}g^{\nu\beta}\nabla_\alpha \kappa_\beta)$. How would this expand?

Where $\nabla$ is the covariant derivative, g the metric and $\kappa_\beta$ a 1-form

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    $\begingroup$ Are you supposing metric compatibility (that the metric is covariantly constant)? If so you can move the derivative past the gs straight away $\endgroup$
    – lux
    Nov 5 '19 at 15:22
  • $\begingroup$ I'm supposing that the background geometry satisfies Einsteins field equations, which I think would make what you said valid? $\endgroup$ Nov 5 '19 at 15:34
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    $\begingroup$ It would, although beware that the two are implicitly (rather than explicitly) connected. Look up metric compatibility to be sure you understand this $\endgroup$
    – lux
    Nov 5 '19 at 17:49
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For the covariant derivative compatible with the metric, which is probably what you meant, $\nabla_\mu g^{\alpha \beta} = 0$, so you get

$$\nabla_\mu \left( g^{\mu \alpha} g^{\nu \beta} \nabla_\alpha \kappa_\beta \right) = \nabla_\mu \nabla^\mu \kappa^\nu .$$

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  • $\begingroup$ Yeah, that makes sense. If $\kappa$ was a killing vector what would that say about $\nabla_\mu\nabla^\mu\kappa^\nu - \nabla_\mu\nabla^\nu\kappa^\mu$? $\endgroup$ Nov 5 '19 at 15:47

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