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Connection and tensor-issue with the proof

  1. Oct 7, 2013 #1
    1. The problem statement, all variables and given/known data

    I am tying to prove the following:
    [itex]\Gamma^{a}_{bc}[/itex] T[itex]^{bc}[/itex] =0

    2. Relevant equations



    3. The attempt at a solution
    I approached this problem as follows:
    [itex]dx_{b}/dx^{c} * e^{a} (e^{b} . e^{c})[/itex] but it did not yield anything.
    Then I expanded the christoeffel symbols into g s and again I am not sure what to do next.So any hints please
     
  2. jcsd
  3. Oct 7, 2013 #2

    WannabeNewton

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    Your post is incomplete to say the least. What is ##\tau^{bc}## to start with?
     
  4. Oct 7, 2013 #3
    That's a tensor.
     
  5. Oct 7, 2013 #4
    a symmetric tensor.
     
  6. Oct 7, 2013 #5

    George Jones

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    In a coordinate basis? A non-holonomic basis?
     
  7. Oct 7, 2013 #6

    WannabeNewton

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    Again you are not being specific enough. ##\Gamma^{a}_{bc}\tau^{bc} = 0## is certainly not true in general for any symmetric tensor ##\tau^{bc}## if ##\Gamma^{a}_{bc}## are the coefficients of the Levi-Civita connection. It is only true in general if ##\tau^{bc}## is antisymmetric so you must specify what exactly the tensor ##\tau^{bc}## is.
     
  8. Oct 7, 2013 #7
    Γabc are the christoffel symbols/connection and T^(bc) = (e^b,e^c)
     
  9. Oct 7, 2013 #8

    George Jones

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    Do you mean [itex]T^{bc} = T \left( e^b , e^c \right)[/itex]?
     
  10. Oct 7, 2013 #9
    yes.
     
  11. Oct 7, 2013 #10

    George Jones

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    And [itex]\left\{ e_a \right\}[/itex] is an orthonormal basis?
     
  12. Oct 7, 2013 #11
    Yes, it is,
     
  13. Oct 7, 2013 #12

    George Jones

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    First, In don't think that you should write [itex]\Gamma^a{}_{bc}[/itex] instead of [itex]\Gamma^a_{bc}[/itex].

    Second, what anti-symmetry property does the Levi-Civita connection have with respect to an orthonormal basis?
     
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