

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer

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. $g^{\sigma\rho}\nabla_{\nu}\nabla_{\mu}T_{\sigma\rho} = \nabla_{\nu}\nabla_{\mu}T$

share|cite|improve this question
yes, it is the inherent property (definition) of covariant derivative construct. – Grisha Kirilin Jan 22 '13 at 21:28
up vote 9 down vote accepted
  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).

share|cite|improve this answer
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. – joshphysics Jan 22 '13 at 21:57

Your Answer


