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Why isn't the Cotton tensor identical zero?

  1. Jan 29, 2010 #1
    Cotton tensor [tex]C_{\mu\varkappa\lambda}[/tex] is define as:
    where [tex]W^{\sigma}_{\phantom{M}\mu\varkappa\lambda}[/tex] is Weyl tensor and [tex] n [/tex] is dimension of space.

    Weyl tensor obey II. Bianchi identity (and all symetries of Rieamann tensor):
    [tex]\nabla_{\nu}W^{\sigma}_{\phantom{M}\mu\varkappa\lambda} + \nabla_{\varkappa}W^{\sigma}_{\phantom{M}\mu\lambda\nu}+\nabla_{\lambda}W^{\sigma}_{\phantom{M}\mu\nu\varkappa}=0[/tex]​
    and extra it is traceless:
    For a divergence of Weyl tensor can write:
    [tex]\nabla_{\sigma}W^{\sigma}_{\phantom{M}\mu\varkappa\lambda}=g^{\rho\sigma}\nabla_{\sigma}W_{\rho\mu\varkappa\lambda}=\left<\mbox{II. Bianchi identity}\right>=-g^{\rho\sigma}\left(\nabla_{\varkappa}W_{\rho\mu\lambda\sigma}+\nabla_{\lambda}W_{\rho\mu\sigma\varkappa}\right)=[/tex]
    Becose Weyl tensor is traceless therefor Cotton tensor must be identical zero!
    Is this correct?
  2. jcsd
  3. Jan 29, 2010 #2
    Let's do clean the equations a little bit and see what's wrong:

    From the second Bianchi Identity we have

    [tex]\nabla_{\nu}W^{\sigma}_{\mu \varkappa \lambda}+\nabla_{\varkappa }W^{\sigma}_{\mu \lambda \nu}+\nabla_{\lambda }W^{\sigma}_{\mu \nu\varkappa}=0[/tex]

    Contracting [tex]\nu[/tex] with [tex]\sigma[/tex] gives

    [tex]\nabla_{\sigma}W^{\sigma}_{\mu \varkappa \lambda}+\nabla_{\varkappa }W^{\sigma}_{\mu \lambda \sigma}+\nabla_{\lambda }W^{\sigma}_{\mu \sigma\varkappa}=0\Rightarrow[/tex]

    [tex]\nabla_{\sigma}W^{\sigma}_{\mu \varkappa \lambda}+\nabla_{\varkappa }W^{\sigma}_{\mu \lambda \sigma}-\nabla_{\lambda }W^{\sigma}_{\mu \varkappa\sigma}=0\Rightarrow[/tex]

    [tex]\nabla_{\sigma}W^{\sigma}_{\mu \varkappa \lambda}+g^{\rho \sigma}\nabla_{\varkappa }W_{\rho \mu \lambda \sigma}-g^{\rho \sigma}\nabla_{\lambda }W_{\rho \mu \varkappa\sigma}=0.[/tex] (1)

    Now I think by the property of tracelessness of Weyl tensor, or,

    [tex]g^{\rho \sigma}W_{\rho \mu \varkappa\sigma}=0,[/tex]

    you conclude from (1)

    [tex]\nabla_{\sigma}W^{\sigma}_{\mu \varkappa \lambda}=0.[/tex]

    So you put this into the equation

    [tex]\nabla_{\sigma}W^{\sigma}_{\phantom{M}\mu\varkappa \lambda}=\frac{n-3}{n-2}C_{\mu\varkappa\lambda}[/tex]

    and finally claim that the Cotton tensor vanishes. But unfortunately you did make a big mistake. I let you think about where this flaw arises in the above calculations! You can have a look at the fact that in a geodesic coordinates, for example, the second derivatives of metric tensors wrt coordinates do not vanish whereas their first derevatives do.

  4. Jan 29, 2010 #3
    If do you think case when [tex]n=3[/tex]. Yes, I did a big mistake. But I suppose [tex]n\geq4[/tex].
    Or Cotton tensor exist only for [tex]n=3[/tex] space (for [tex]n=2[/tex] it is also zero)?
  5. Jan 29, 2010 #4
    For n=3, the Cotton tensor does not vanish in general. But for [tex]n\geq4[/tex], if the weyl tensor vanishes, then the Cotton tensor always vanishes. For n less than 3 the Cotton tensor is not defined.

  6. May 19, 2010 #5
    This is correct, because I made another big mistake. Weyl tensor isn't governed of analog II. Bianchi identity. Therefore cotton tensor is non-zero generaly for [tex]n>2[/tex].
    Last edited: May 19, 2010
  7. May 20, 2010 #6
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