# Bitensor covariant derivative commutation

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

## Answers and Replies

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.