[Physics] Covariant derivative of a covariant tensor wrt superscript

differential-geometrydifferentiationgeneral-relativitytensor-calculus

Is it true that when you take the covariant derivative of a covariant tensor, do you always have to do with a subscript? What if you do it wrt a superscript?Does the first term (with the partial derivative) take a minus sign? More specifically, is this true?

$$\nabla^{\mu}R_{\mu\nu} = -\frac{\partial R_{\mu\nu}}{{\partial x_{\mu}}} + \text{(Christoffels)}$$

Where does the minus sign come from? Is there a proof for this, or is it just a definition?
Also, is there a change in the signs for Christoffel symbols(not the change if the the tensor's indices change position, but the change when the index of the covariant differential changes)?

I want to know the PROOF/REASON behind the minus sign.

Best Answer

No. The subscript is the defined thing. If you have the superscript, you just assume raising with the metric tensor, so:

$$\nabla^{\mu}R_{\mu\nu} \equiv g^{\mu\alpha}\nabla_{\alpha}R_{\mu\nu}$$

which you expand normally with partial derivatives and Christoffels. Of course, since we know that $\nabla^{a}\left(R_{ab} - \frac{1}{2}Rg_{ab} \right)= 0$, we know right away that we can simplify $\nabla^{\mu}R_{\mu\nu}$ to $\frac{1}{2}\nabla_{\nu}R$

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