Covariant derivative of a commutator (deriving Bianchi identity)

[center o bass]

Hi. I'm trying to understand a derivation of the Bianchi idenity which starts from the torsion tensor in a torsion free space;

$$ 0 = T(X,Y) = \nabla_X Y - \nabla_Y X - [X,Y]$$

according to the author, covariant differentiation of this identity with respect to a vector Z yields

$$$ 0 = \nabla_Z \{\nabla_X Y - \nabla_Y X - [X,Y]\} = \nabla_Z\nabla_X Y - \nabla_Z \nabla_Y X - \{ \nabla_{[X,Y]}Z + [Z,[X,Y]] \}$$.

The two first terms are obvious, but how does he arrive at the third term?

 

[WannabeNewton]

Let $p \in M$ and choose normal coordinates on a neighborhood of $p$.

Then $[Z,[X,Y]]^{\mu}|_p\\ = Z^{\nu}\partial_{\nu}[X,Y]^{\mu}|_p - [X,Y]^{\nu}\partial_{\nu} Z^{\mu}|_p\\ = Z^{\nu}\nabla_{\nu}[X,Y]^{\mu}|_p - [X,Y]^{\nu}\nabla_{\nu}Z^{\mu}|_p$

so $\nabla_{Z}[X,Y]|_p = \nabla_{[X,Y]}Z|_p + [Z,[X,Y]]|_p$.

 

[center o bass]

Let $p \in M$ and choose normal coordinates on a neighborhood of $p$.

Then $[Z,[X,Y]]^{\mu}|_p\\ = Z^{\nu}\partial_{\nu}[X,Y]^{\mu}|_p - [X,Y]^{\nu}\partial_{\nu} Z^{\mu}|_p\\ = Z^{\nu}\nabla_{\nu}[X,Y]^{\mu}|_p - [X,Y]^{\nu}\nabla_{\nu}Z^{\mu}|_p$

so $\nabla_{Z}[X,Y]|_p = \nabla_{[X,Y]}Z|_p + [Z,[X,Y]]|_p$.

Thanks WbN! Was the point of choosing normal coordinates that you could immediately set $\partial_\nu \to \nabla_\nu$ within the neighbourhood?

 

[WannabeNewton]

Yep!