- #1
PLuz
- 60
- 0
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
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