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Exercise [14.19]
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Joined: 26 Mar 2010, 04:39
Posts: 109
Exercise [14.19]
First notice that

In other words

Hence

In particular, if the torsion vanishes we have:

It's worth comparing this with the first expression for . Notice that in the torsion-free case we can just swap partial derivatives for covariant derivatives (or vice versa). In other words, the Lie derviative is independent of the torsion-free connection.

06 May 2010, 08:55

Joined: 12 Jul 2010, 07:44
Posts: 154
Re: Exercise [14.19]
Sorry Robin, I don't quite understand how you justify your first line:

This is clearly true if you apply to a scalar, but it's not obviously true in general; in fact it's a statement of the result we're trying to show, for the particular choice of the coordinate connection ().

I did it this way:

Noting that given any particular connection , the action of on an arbitrary tensor is , we get:

But we have assumed our (given) is torsion-free, so the third term in the brackets vanishes, and therefore:

for all torsion-free connections .

Last edited by deant on 10 Jan 2012, 14:47, edited 2 times in total.

24 Oct 2010, 07:23
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Joined: 07 Jun 2008, 08:21
Posts: 235
Re: Exercise [14.19]
This is deant's solution attached to overcome the Latex problem:
Attachment:
RRpost.pdf [52.27 KiB]