# Covariant derivative contracted with a metric

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

• Are you supposing metric compatibility (that the metric is covariantly constant)? If so you can move the derivative past the gs straight away
– lux
Nov 5 '19 at 15:22
• I'm supposing that the background geometry satisfies Einsteins field equations, which I think would make what you said valid? Nov 5 '19 at 15:34
• It would, although beware that the two are implicitly (rather than explicitly) connected. Look up metric compatibility to be sure you understand this
– lux
Nov 5 '19 at 17:49

For the covariant derivative compatible with the metric, which is probably what you meant, $$\nabla_\mu g^{\alpha \beta} = 0$$, so you get
$$\nabla_\mu \left( g^{\mu \alpha} g^{\nu \beta} \nabla_\alpha \kappa_\beta \right) = \nabla_\mu \nabla^\mu \kappa^\nu .$$
• Yeah, that makes sense. If $\kappa$ was a killing vector what would that say about $\nabla_\mu\nabla^\mu\kappa^\nu - \nabla_\mu\nabla^\nu\kappa^\mu$? Nov 5 '19 at 15:47