Connection and tensor-issue with the proof

Homework Statement

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

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

Answers and Replies

WannabeNewton
Science Advisor
Your post is incomplete to say the least. What is ##\tau^{bc}## to start with?

That's a tensor.

a symmetric tensor.

George Jones
Staff Emeritus
Science Advisor
Gold Member
In a coordinate basis? A non-holonomic basis?

WannabeNewton
Science Advisor
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.

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

George Jones
Staff Emeritus
Science Advisor
Gold Member
Γabc are the christoffel symbols/connection and T^(bc) = (e^b,e^c)

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

yes.

George Jones
Staff Emeritus
Science Advisor
Gold Member
And $\left\{ e_a \right\}$ is an orthonormal basis?

Yes, it is,

George Jones
Staff Emeritus
Science Advisor
Gold Member
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?