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Bitensor covariant derivative commutation

  1. Sep 23, 2012 #1
    Hi everyone,

    I'm trying to prove a relation in which I need do commute covariant derivatives of a bitensor. The equation is quite long but I need to write something like this:

    Given a bitensor [itex]G^{\alpha}_{\beta'}(x,x')[/itex], where the unprimed indexes ([itex]\alpha[/itex],[itex]\beta[/itex], etc) are assigned to the point [itex]x[/itex] and the primed indexes to the point [itex]x'[/itex]. If I use the usual law for commutating covariant derivatives of a (r,s)-tensor one can write:[tex]\nabla_{\rho} \nabla_{\sigma} G^{\alpha}_{\beta'}=\nabla_{\sigma}\nabla_{\rho} G^{\alpha} _{\beta'}+R^{\alpha}_{\delta \rho \sigma}G^{\delta}_{\beta'}-R^{\delta'}_{\beta' \rho \sigma}G^{\alpha}_{\delta'} ,[/tex]

    where [itex]R^{\alpha}_{\delta \rho \sigma}[/itex] is the Riemann tensor. But this formula can't be right, at least is not well defined since I don't know what is this object: [itex]R^{\delta'}_{\beta' \rho \sigma}[/itex] ...

    Does anyone know how to commute covariant derivatives of a bitensor?

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

    Bill_K

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    Science Advisor

    You're only differentiating wrt the unprimed point x, so the primed indices are effectively out of the picture. Keep the first Riemann tensor term and dscard the second.
     
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