# Covariant derivative of a commutator (deriving Bianchi identity)

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##.

Last edited:
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