# Can one raise indices on covariant derivative and products thereof?

Can the following be true?

1. $g^{\sigma\rho}\nabla_{\rho}\nabla_{\mu} = \nabla^{\sigma}\nabla_{\mu}$
2. $g^{\sigma\rho}\nabla_{\nu}\nabla_{\sigma} = \nabla_{\nu}\nabla^{\rho}$
3. $g^{\sigma\rho}\nabla_{\nu}\nabla_{\mu}T_{\sigma\rho} = \nabla_{\nu}\nabla_{\mu}T$
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yes, it is the inherent property (definition) of covariant derivative construct. –  Grisha Kirilin Jan 22 '13 at 21:28

1. This is true - in fact you could define $\nabla^\sigma = g^{\sigma\rho} \nabla_\rho$.
2. 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).