Bitensor covariant derivative commutation

  • Thread starter PLuz
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  • #1
PLuz
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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
 

Answers and Replies

  • #2
Bill_K
Science Advisor
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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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