# Covariant derivative with upper index

I just need clarification, that is, to see that I'm doing the right thing.

When calculating central charge for certain metric, I need to solve an integral that contains Lie brackets etc. And I have some expression, that contains terms like:

$$\xi_\sigma D^\nu h^{\mu\sigma}$$

Where $D_\nu$ should be covariant derivative. Now, can I just say:

$$D^\nu=g^{\nu\mu}D_\mu$$

And similarly for the $h^{\mu\sigma}$, and use the standard definition of covariant derivatives?

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Just a small point : One can define $D^{\nu}$ either as $D^{\nu}F=g^{\nu\mu}D_{\mu}F$ or as $D^\nu F=D_{\mu} (g^{\nu\mu}F)$ because covariant derivative of metric itself is zero. – user10001 Jan 22 '13 at 20:34
Hmm, but what if I have $D_\mu h^{\sigma\nu}$, and I need to lower the indices of the $h^{\sigma\nu}$, does that mean, that I can put the g's that will lower the indices of h outside also? – dingo_d Jan 23 '13 at 7:35
Yes, you are right. You can put your $g$'s anywhere inside or outside of covariant derivative. – user10001 Jan 23 '13 at 10:29
Hmmm, thanks, that could help, I hope :D – dingo_d Jan 23 '13 at 10:36

Yes, that is generally what it means. Typically, it also implies that any derivatives appearing are with respect to covariant coordinates instead of contravariant.

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"Typically, it also implies that any derivatives appearing are with respect to covariant components instead of contravariant." What if I have mixed terms inside? I mean, in the expression I also have scalars, covariant vectors, contravariant vectors, covariant and contravariant tensors... – dingo_d Jan 22 '13 at 16:23
Sorry, meant to say covariant coordinates. The ranks and makeup of the objects you're acting on with the covariant derivative don't really matter; it's just their position dependence that needs to be accounted for. In practice, having $D^a$ is just going to add an index upstairs to everything it acts on, and otherwise, you don't worry about it. – Muphrid Jan 22 '13 at 16:26
Thanks for the clarification, I'll try to make sth out of this expression :D – dingo_d Jan 22 '13 at 16:30