Antisymmetric connection (Torsion Tensor)
How to show:
T^{a}_{bc} = [itex]\Gamma[/itex]^{a}_{bc}  [itex]\Gamma[/itex]^{a}_{cb} is a Tensor of rank (1,2) Attempted solution: 1. Using definition of Covariant Derivative: D_{b}T^{a}= ∂_{a}T^{a}+[itex]\Gamma[/itex]^{a}_{bc}T^{c} (1) D_{c}T^{a}= ∂_{c}T^{a}+[itex]\Gamma[/itex]^{a}_{cb}T^{b} (2) I subtracted (2) from (1) but I couldn't really get a Tensor out of it. I just got lost in the mess. Is this is the right way to start it? 
Re: Antisymmetric connection (Torsion Tensor)
Do you have to use covariant derivatives in your problem? Is it a hint in your problem? There are several ways to show your property.
And why do you say "antisymmetric connection?" 
Re: Antisymmetric connection (Torsion Tensor)
I am also in the process of learning tensor calculus, so I may not be right, but wouldn't it work if you raised the indices and made every tensor abcontravariant?

Re: Antisymmetric connection (Torsion Tensor)
Which text are you using? There are different ways of showing your property, but the method should be adapted to what you already know.

Re: Antisymmetric connection (Torsion Tensor)
@arkajad: Covariant derivative is not a hint in the problem. I am just trying to solve that way. I am following various kind of textbooks. So, any way would work for me.
@meldraft: I am sure if that will work. Since the purpose of this exercise is to show how the difference between two Christoffel symbols that are asymmetric gives rise to torsion tensor. 
Re: Antisymmetric connection (Torsion Tensor)
Quote:
Check Eq. (3.6) in http://preposterousuniverse.com/grno...otesthree.pdf But do not read further than that!!!! 
Re: Antisymmetric connection (Torsion Tensor)
Solved.

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