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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!

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3 Answers 3

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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?

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  • $\begingroup$ Ah I see, it's just the covariant derivative of a (1,1) tensor. That did it, thanks! $\endgroup$ Commented Apr 26, 2022 at 11:04
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Remember that $\nabla_{\nu} Z^{\sigma}$ is a type $(1,1)$ tensor, so $\nabla_{[\mu} \nabla_{\nu]} Z^{\sigma}$ will spit out terms like $\Gamma_{[\mu \nu]}^{\lambda} \nabla_{\lambda} Z^{\sigma}$.

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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.

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