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A Difficulty in understanding contracted Bianchi identities

  1. Jul 7, 2016 #1
    I am confused about the contraction in the proof of the contracted Bianchi identities in

    https://en.wikipedia.org/wiki/Proofs_involving_covariant_derivatives

    from the step
    [tex] {g^{bn}}(R_{bmn;l}^m - R_{bml;n}^m + R_{bnl;m}^m) = 0[/tex]
    it seems that the following two quantities are equal
    [tex]{g^{bn}}R_{bml;n}^m = R_{l;n}^n[/tex]
    [tex]- {g^{bn}}R_{bnl;m}^m = R_{l;m}^m[/tex]
    but I don't understand how is this done if I write them explicitly
    [tex]{g^{bn}}({\nabla _n}R)_{bml}^m[/tex]
    [tex]- {g^{bn}}({\nabla _m}R)_{bnl}^m[/tex]
    Can anybody help me? I am new to this field and I feel there is something missing. Please help to point out.
     
    Last edited: Jul 7, 2016
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  3. Jul 7, 2016 #2

    Orodruin

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    The only difference between the terms is the name of the summation index. Sums do not depend on what you call the summation indices.
     
  4. Jul 7, 2016 #3
    Thanks for replying. you are right it doesn't matter what dummy indices are used for summation. But I feel that here is a game of cancling indecis, which does not make sense. does the proof imply
    [tex]{g^{bn}}R_{bml}^m = R_l^n[/tex]
    and
    [tex]{g^{bn}}R_{bnl}^m = -R_l^m[/tex]
    in this case
    [tex]- {g^{bn}}R_{bnl}^m = R_l^m = {g^{bm}}R_{bkl}^k[/tex]
    I just don't see how this is true
     
  5. Jul 7, 2016 #4

    Orodruin

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    There is no cancelling of indices, just the definition of the Ricci tensor and use of the symmetries of the Riemann tensor.
     
  6. Jul 8, 2016 #5
    Can you be more specific? is this relationship right?

    [tex] - {g^{bn}}R_{bnl}^m = R_l^m = {g^{bm}}R_{bkl}^k[/tex]

    sorry for haunting you. I have been stuck here for 2 days.
     
  7. Jul 8, 2016 #6

    Orodruin

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    I suggest using the symmetries and antisymmetries of the Riemann tensor with all indices lowered and you should be able to figure it out.
     
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