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Antisymmetric connection (Torsion Tensor) 
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#1
Feb512, 05:28 AM

P: 19

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? 


#2
Feb612, 10:57 AM

P: 1,411

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?" 


#3
Feb612, 11:08 AM

P: 280

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?



#4
Feb612, 11:13 AM

P: 1,411

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.



#5
Feb612, 11:40 AM

P: 19

@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. 


#6
Feb612, 11:50 AM

P: 1,411

Check Eq. (3.6) in http://preposterousuniverse.com/grno...otesthree.pdf But do not read further than that!!!! 


#7
Feb712, 05:24 PM

P: 19

Solved.



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