Connection and tensor-issue with the proof

  • Thread starter rsaad
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  • #1
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Homework Statement



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

Homework Equations





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
 

Answers and Replies

  • #2
WannabeNewton
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Your post is incomplete to say the least. What is ##\tau^{bc}## to start with?
 
  • #3
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That's a tensor.
 
  • #4
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a symmetric tensor.
 
  • #5
George Jones
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In a coordinate basis? A non-holonomic basis?
 
  • #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.
 
  • #7
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Γabc are the christoffel symbols/connection and T^(bc) = (e^b,e^c)
 
  • #8
George Jones
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Γabc are the christoffel symbols/connection and T^(bc) = (e^b,e^c)

Do you mean [itex]T^{bc} = T \left( e^b , e^c \right)[/itex]?
 
  • #10
George Jones
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And [itex]\left\{ e_a \right\}[/itex] is an orthonormal basis?
 
  • #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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