Connection and tensor-issue with the proof

[rsaad]

Homework Statement

I am tying to prove the following:
$\Gamma^{a}_{bc}$ T$^{bc}$ =0

Homework Equations

The Attempt at a Solution

I approached this problem as follows:
$dx_{b}/dx^{c} * e^{a} (e^{b} . e^{c})$ 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

 

[WannabeNewton]

Your post is incomplete to say the least. What is $\tau^{bc}$ to start with?

 

[rsaad]

That's a tensor.

 

[rsaad]

a symmetric tensor.

 

[George Jones]

In a coordinate basis? A non-holonomic basis?

 

[WannabeNewton]

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.

 

[rsaad]

Γabc are the christoffel symbols/connection and T^(bc) = (e^b,e^c)

 

[George Jones]

Γabc are the christoffel symbols/connection and T^(bc) = (e^b,e^c)

Do you mean $T^{bc} = T \left( e^b , e^c \right)$?

 

[rsaad]

yes.

 

[George Jones]

And $\left\{ e_a \right\}$ is an orthonormal basis?

 

[rsaad]

Yes, it is,

 

[George Jones]

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

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