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Covariant derivative of a commutator (deriving Bianchi identity)

  1. Sep 5, 2013 #1
    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?
     
  2. jcsd
  3. Sep 5, 2013 #2

    WannabeNewton

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    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: Sep 5, 2013
  4. Sep 9, 2013 #3
    Thanks WbN! Was the point of choosing normal coordinates that you could immediately set ##\partial_\nu \to \nabla_\nu## within the neighbourhood?
     
  5. Sep 9, 2013 #4

    WannabeNewton

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