# Bitensor covariant derivative commutation

1. Sep 23, 2012

### PLuz

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

where $R^{\alpha}_{\delta \rho \sigma}$ 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: $R^{\delta'}_{\beta' \rho \sigma}$ ...

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

Thank you

2. Sep 23, 2012

### Bill_K

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.