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!

 

[blue_raver22]

Hello,

Thank you for reaching out. The third term in the derivation is obtained by applying the Leibniz rule for covariant derivatives. This rule states that for any two vector fields, X and Y, and a third vector field Z, the covariant derivative of their commutator [X,Y] with respect to Z is given by:

$$ \nabla_Z [X,Y] = [\nabla_Z X, Y] + [X, \nabla_Z Y] $$

Using this rule, the third term in the derivation can be expanded as follows:

$$ \nabla_Z [X,Y] = [\nabla_Z \nabla_X Y - \nabla_Z \nabla_Y X, Z] + [\nabla_X Y, \nabla_Z Z] + [\nabla_Z X, \nabla_Z Y] $$

Since the covariant derivative is linear, we can rearrange these terms to get:

$$ \nabla_Z [X,Y] = \nabla_Z \nabla_X Y - \nabla_Z \nabla_Y X + [\nabla_Z X, \nabla_Z Y] + [Z, [\nabla_X Y, Z]] $$

Finally, since we are in a torsion-free space, the torsion tensor T(X,Y) = 0, so the last term can be simplified to [Z, [X,Y]]. This gives us the third term in the original derivation:

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

I hope this explanation helps clarify the derivation. Please let me know if you have any further questions.

Best,