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Covariant Derivative Commutation

  1. Sep 22, 2012 #1

    Can anyone tell me the general formula for commuting covariant derivatives, I mean, given a (r,s)-tensor field what is the formula to commute covariant derivatives?

    I found a formula here page 25, Eq.6.18 but it doesn't seem right, since for a vector field one would write:

    [itex](\nabla_{\alpha} \nabla_{\beta}- \nabla_{\beta}\nabla_{\alpha})U^{\gamma}=R^{\gamma}\hspace{.5 mm}_{\delta \alpha \beta}U^{\delta}[/itex]

    And according to the formula in the link it would be, for a vector field

    [itex](\nabla_{\alpha} \nabla_{\beta}- \nabla_{\beta}\nabla_{\alpha})U^{\gamma}=-R^{\gamma}\hspace{.5 mm}_{\delta \alpha \beta}U^{\delta}[/itex]

    Thank you
  2. jcsd
  3. Sep 23, 2012 #2


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    I see only a difference in minus signs between the two expressions which can be accounted for in the convention used to define the Riemann tensor. I believe this formula is correct (at least for a coordinate basis, I cannot be sure if there are more terms for a non-coordinate basis).
  4. Sep 25, 2012 #3


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    For non-coordinate (aka anholonomic) basis, compute the Riemann tensor (it doesn't matter which sign convention you use) with the connection coefficients as provided in MTW page 210, formula 8.24b.
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