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Riemann tensor cyclic identity (first Bianchi) and noncoordinate basis

  1. Jul 12, 2012 #1
    I got trouble to understand the cyclic sum identity (the first Bianchi identity) of the Riemann curvature tensor:

    [itex]{R^\alpha}_{[ \beta \gamma \delta ]}=0[/itex]

    or equivalently,

    [itex]{R^\alpha}_{\beta \gamma \delta}+{R^\alpha}_{\gamma \delta \beta}+{R^\alpha}_{\delta \beta \gamma}=0[/itex].

    I can understand the proofs from the most general textbooks, for example:
    Wald P.39 (3.2.18), Padmanabhan P.200 (5.41), Weinberg P.141 (6.6.5) and so on.

    But I found that all of them do the proof under a coordinate basis, so commutation coefficients are zero, [itex]{c_{\mu\nu}}^\alpha=0[/itex]. In this case, the Riemann tensors are
    [itex]{R^\alpha}_{\beta \gamma \delta}
    =
    {\Gamma^\alpha}_{\beta \gamma , \delta}
    -{\Gamma^\alpha}_{\beta \gamma , \delta}
    +{\Gamma^\alpha}_{\mu \delta}{\Gamma^\alpha}_{\beta \gamma}
    -{\Gamma^\alpha}_{\mu \gamma}{\Gamma^\alpha}_{\beta \delta}
    [/itex]
    and
    [itex][\nabla_\gamma, \nabla_\delta]\,v^\alpha
    ={R^\alpha}_{\beta \gamma \delta}\,v^\beta[/itex]
    for a vector [itex]v=v^\alpha \,e_\alpha[/itex]. I've checked that we can easily prove the above cyclic sum identity by starting from either of these.

    But in a noncoordinate basis case, the Riemann tensors are
    [itex]{R^\alpha}_{\beta \gamma \delta}
    =
    {\Gamma^\alpha}_{\beta \gamma , \delta}
    -{\Gamma^\alpha}_{\beta \gamma , \delta}
    +{\Gamma^\alpha}_{\mu \delta}{\Gamma^\alpha}_{\beta \gamma}
    -{\Gamma^\alpha}_{\mu \gamma}{\Gamma^\alpha}_{\beta \delta}
    -{\Gamma^\alpha}_{\beta \mu} \,{c_{\delta \gamma}}^\mu
    [/itex]
    and
    [itex]([\nabla_\gamma, \nabla_\delta]-{c_{\gamma \delta}}^\mu\nabla_\mu)\,v^\alpha
    ={R^\alpha}_{\beta \gamma \delta}\,v^\beta[/itex].

    I've tried the proof again for a noncoordinate basis and stuck:
    [itex]3{R^\alpha}_{[ \beta \gamma \delta ]}
    =
    {c_{\delta \beta}}^\alpha{}_{,\gamma}
    +{c_{\beta \gamma}}^\alpha{}_{,\delta}
    +{c_{\gamma \delta}}^\alpha{}_{,\beta}
    +{c_{\gamma \mu}}^\alpha{c_{\delta \beta}}^\mu
    +{c_{\delta \mu}}^\alpha{c_{\beta \gamma}}^\mu
    +{c_{\beta \mu}}^\alpha{c_{\gamma \delta}}^\mu
    [/itex]

    I think the identity will be true in a noncoordinate basis too. But how can I proove or understand it in a noncoordinate basis?
     
    Last edited: Jul 12, 2012
  2. jcsd
  3. Jul 13, 2012 #2

    pervect

    User Avatar
    Staff Emeritus
    Science Advisor

    I think all you need is to note that [itex]{R^\alpha}_{[ \beta \gamma \delta ]}[/itex] is a tensor because it's the sum of three tensors, and that if a tensor is zero in one basis it's zero in all.
     
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