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    i'm trying to show that R_{[abc]}{}^d=0
    the proof says:
    Note for an arbitrary dual vector field \omega_a and any derivative operator \nabla_a we have \nabla_{[a} \nabla_b \omega_{c]}=0
    this equation can be proven from
    \nabla_a T^{b_1 . . . b_k}{}_{c_1 . . . c_l}= \tilde{\nabla_a} T^{b_1 . . . b_k}{}{c_1 . . . c_l} + \displaystyle\sum_i C^{b_i}{}_{ad} T^{b_1 . . d . . b_k}{}_{c_1 . . . c_l} - \displaystyle\sum_j  C^{d}_{a c_j} T^{b_1 . . . b_k}{}_{c_1 . . d . . c_l}
    where the ordinary derivative \partial_a has been substituted for \tilde{\nabla_a} and we have made use of the commutativity of ordinary derivatives and the symmetry of C^{c}{}_{ab}=\Gamma^{c}{}_{ab}. In differential forms notation this statement is d^2 \omega=0. Thus,
    0=2 \nabla_{[a} \nabla_{b} \omega_{c]} = \nabla_{[a} \nabla_{b} \omega_{c]} - \nabla_{[b} \nabla_a \omega_{c]} = R_{[abc]}{}^d \omega_d \forall \omega_d
    \Rightarrow R_{[abc]}{}^d=0

    now i have quite a lot of problems following this:
    (i) terminology: it talks about \omega_d as a dual vector field. now apparently there are two types of dual space - algebraic dual space and continuous dual space. if its the algebraic one then we can refer to it as a 1 form. my question here is basically which is it? am i entitled to call it a 1 form?

    (ii)ok so i have no idea how to get to \nabla_{[a} \nabla_b \omega_{c]}=0
    using the method they descirbe.

    (iii) i assume that since \nabla_{[a} \nabla_b \omega_{c]}=0, they just double both sides to get 2 \nabla_{[a} \nabla_b \omega_{c]}=0 but then why is
    2 \nabla_{[a} \nabla_b \omega_{c]}= \nabla_{[a} \nabla_b \omega_{c]} - \nabla_{[b} \nabla_a \omega_{c]}?

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Updated: October 13, 2009

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