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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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    $\begingroup$ yes, it is the inherent property (definition) of covariant derivative construct. $\endgroup$
    – Grisha Kirilin
    Commented Jan 22, 2013 at 21:28
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    $\begingroup$ @GrishaKirilin That's not true - it only works if the covariant derivative is with respect to a metric-compatible connection. $\endgroup$
    – tparker
    Commented Jul 28, 2017 at 21:52

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  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).

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

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    $\begingroup$ To add to what Chris said, raising and lowering indices with the metric can be done on the indices of any tensor. Since the covariant derivative of a tensor is a new tensor with an additional index, raising its indices is a special case of this fact. $\endgroup$
    – joshphysics
    Commented Jan 22, 2013 at 21:57
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    $\begingroup$ @joshphysics That's true, but I think a bit misleading. For a metric-incompatible connection, there are multiple inequivalent valid ways to use the metric to raise and lower indices, and the OP's notation is ambiguous. For example, if $f_\nu$ is a one-form, then the notation $\nabla_\mu f^\mu$ could be interpreted to mean either $g^{\mu \nu} \nabla_\mu f_\nu$ or $\nabla_\mu (g^{\mu \nu} f_\nu) = f_\nu \nabla_\mu g^{\mu \nu} + g^{\mu \nu} \nabla_\mu f_\nu$, which are both valid (but inequivalent) tensor expressions. $\endgroup$
    – tparker
    Commented Jul 28, 2017 at 22:09
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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.

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