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How to Prove contracted Bianchi Identity

  1. May 18, 2007 #1
    How to prove
    [itex]g^{im}
    \nabla_{\partial_j}R_{ilkm}=\nabla_{\partial_j}R_{lk}[/itex].

    of cause [itex]g^{im}R_{ilkm}=R_{lk}[/itex], but I don't know how the contraction can pass through the covariant derivative?
     
    Last edited: May 18, 2007
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  3. May 18, 2007 #2

    Hurkyl

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    Because derivatives and (finite) sums commute. e.g.

    [tex]
    \frac{d}{dx} \sum_{i = a}^b f_i(x) = \sum_{i = a}^b \frac{df_i}{dx}(x)
    [/tex]
     
  4. May 18, 2007 #3

    robphy

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    The derivative operator is usually taken to be "metric compatible", i.e.,
    [tex]\nabla_a g_{bc}=0[/tex].
     
  5. May 18, 2007 #4
    Yes, I know that. But [tex]\nabla_{\partial_j}R_{ilkm}=(\nabla_{\partial_j}R)(\partial_i,\partial_l,\partial_k,\partial_m)[/tex] How can the contraction really happen?(since [tex]\nabla_a g_{bc}=0[/tex] means [tex](\nabla_a g)(\partial_b,\partial_c)=0[/tex])
    Would you like to give me more detail? Thank you!
     
    Last edited: May 18, 2007
  6. May 18, 2007 #5

    Chris Hillman

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    This is OT, but George Jones just pointed out that today's issue of the daily paper in Toronto featured a picture of the uncontracted Bianchi identities. For some reason a politician is in the foreground.

    Industrious students can look for papers pointing out that Bianchi himself credited these identities to someone else.
     
  7. Jun 8, 2007 #6
    [itex]g^{im}\nabla_{\partial_j}R_{ilkm}=\nabla_{\partial_j}(g^{im}R_{ilkm}) - R_{ilkm}\nabla_{\partial_j}g^{im}=\nabla_{\partial_j}R_{ lk}[/itex]

    because [tex]\nabla_a g_{bc}=0[/tex]

    it respects the Leibniz derivation
     
    Last edited: Jun 8, 2007
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