Variation of the Christoffel Symbols

[Breo]

So, it is defined that:

Γ^λ_μυ = Γ^λ_μυ + δΓ^λ_μυ

This makes obvious to see that the variation of the connection, which is defined as a difference of 2 connections, is indeed a tensor.

Therefore we can express it as a sum of covariant derivatives.

δΓ^λ_μυ = ½g^λν(-∇_λδg_μν + ∇_μδg_λν + ∇_νδg_λμ)

However, I do not know how to make this step. Help?

 

[Greg Bernhardt]

Thanks for the post! Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?

 

[jbunch]

So, it is defined that:

Γ^λ_μυ = Γ^λ_μυ + δΓ^λ_μυ

This makes obvious to see that the variation of the connection, which is defined as a difference of 2 connections, is indeed a tensor.

Therefore we can express it as a sum of covariant derivatives.

δΓ^λ_μυ = ½g^λν(-∇_λδg_μν + ∇_μδg_λν + ∇_νδg_λμ)

However, I do not know how to make this step. Help?

I think you might want to go over the equations you wrote down.

http://ned.ipac.caltech.edu/level5/March01/Carroll3/Carroll3.html

If you were looking for the anti-symmetrical properties of Christoffel Symbols (ie torsion) and the commutation relations of the covariant derivative...here is a link:

http://www.aias.us/documents/uft/a42ndpaper.pdf

 

[haushofer]

A simple way to do it is to apply the variation directly to the solution of the Christoffel symbol. You'll get partial derivatives of variations of the metric, but you know the result should be gct-covariant. So you can replace the partial derivatives by covariant ones. If you're not sure about this (think it through carefully!), you can check it explicitly, but that's a bit tedious.