# Ricci Identity with Torsion Proof

In the notes I'm using for General Relativity, the author begins their proof of the Ricci identity with torsion by writing

$$\nabla_{[\mu}\nabla_{\nu]}Z^{\sigma}=\partial_{[\mu}(\nabla_{\nu]}Z^{\sigma})+\Gamma^{\sigma}_{[\mu|\lambda|}\nabla_{\nu]}Z^{\lambda}-\Gamma^{\rho}_{[\mu\nu]}\nabla_{\rho}Z^{\sigma}$$

I can't for the life of me see how to get the last term, so if someone could help me out here that would be great!

Hint: You can, in a first step, expand the outer derivative (write $$D_\nu Z^\sigma=A_\mu^\sigma$$ if you wish). You will get a partial derivative acting on $$A_\mu^\sigma$$ and two terms with Christoffel symbols.

Can you take it from here?

• Ah I see, it's just the covariant derivative of a (1,1) tensor. That did it, thanks! Commented Apr 26, 2022 at 11:04

Remember that $$\nabla_{\nu} Z^{\sigma}$$ is a type $$(1,1)$$ tensor, so $$\nabla_{[\mu} \nabla_{\nu]} Z^{\sigma}$$ will spit out terms like $$\Gamma_{[\mu \nu]}^{\lambda} \nabla_{\lambda} Z^{\sigma}$$.

We have

$$\nabla_\mu \nabla_\nu Z^\sigma=\partial_\mu(\nabla_\nu Z^\sigma)+\Gamma_{\lambda\mu}^\sigma \nabla_\nu Z^\lambda - \Gamma_{\nu\mu}^\rho \nabla_\rho Z^\sigma$$

as $$\nabla_\mu$$ is acting on the $$(1,1)$$ tensor $$\nabla_\nu Z^\sigma$$.
Using the equivalent expression for $$\nabla_\nu\nabla_\mu Z^\sigma$$ and the fact that $$\nabla_{[\mu}\nabla_{\nu]}Z^\sigma = \frac 12 (\nabla_\mu \nabla_\nu Z^\sigma-\nabla_\nu\nabla_\mu Z^\sigma)$$, we immediately can find the right hand side expression.