# Variation of the Christoffel Symbols

1. Oct 16, 2014

### Breo

So, it is defined that:

Γλμυ = Γλμυ + δΓλμυ

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.

δΓλμυ = ½gλν(-∇λδgμν + ∇μδgλν + ∇νδgλμ)

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

Last edited: Oct 16, 2014
2. Oct 21, 2014

### 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?

3. Oct 22, 2014

### jbunch

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

Last edited: Oct 22, 2014
4. Oct 22, 2014

### 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.