Can one raise indices on covariant derivative and products thereof? Can the following be true?


*

*$g^{\sigma\rho}\nabla_{\rho}\nabla_{\mu} = \nabla^{\sigma}\nabla_{\mu}$

*$g^{\sigma\rho}\nabla_{\nu}\nabla_{\sigma} = \nabla_{\nu}\nabla^{\rho}$

*$g^{\sigma\rho}\nabla_{\nu}\nabla_{\mu}T_{\sigma\rho} = \nabla_{\nu}\nabla_{\mu}T$
 A: A little subtlety on this. If this is general relativity in its usual formulation, this is all true. The covariant derivative then involves a connection which is usually known as the Levi-Civita or Christoffel connection, which has a simple construction based on the metric. This covariant derivative does commute with the metric - in the jargon, it is "metric compatible". However, it is possible to define connections and associated covariant derivatives which are not metric compatible. But it's very unlikely you're dealing with those - if you're looking at GR, the comments above are perfectly correct.
A: *

*This is true - in fact you could define $\nabla^\sigma = g^{\sigma\rho} \nabla_\rho$.

*I assume this meant to say
$$ g^{\sigma\rho} \nabla_\nu \nabla_\sigma = \nabla_\nu \nabla^\rho. $$
Again, this is true, but for a slightly less trivial reason than (1). To employ (1) to prove this, you need to be able to switch $g^{\sigma\rho}$ with $\nabla_\nu$, which you are able to do because one of the axioms we start with when defining the covariant derivative is that it commutes with the metric (i.e., the metric has vanishing covariant derivative, so that other term in the product rule drops out).

*This also holds, following the same reasoning as in (2).
