Connection coefficients as derivatives of parallel propagator


by ianhoolihan
Tags: coefficients, connection, derivatives, parallel, propagator
ianhoolihan
ianhoolihan is offline
#1
Mar25-12, 07:33 PM
P: 145
Hi all,

I've been fiddling around with this problem for a while. I intuitively understand that the parallel propagator is the path integral of the connection. I would like to be able to show the converse (connection is derivative of parallel propagator) mathematically, and I am having a little trouble.

I've been thinking of the parallel propagator as in http://en.wikipedia.org/wiki/Paralle...llel_transport. I understand how to formulate the covariant derivative
[tex]
\nabla_X V = \lim_{h\to 0}\frac{\Gamma(\gamma)_h^0V_{\gamma(h)} - V_{\gamma(0)}}{h} = \frac{d}{dt}\Gamma(\gamma)_t^0V_{\gamma(t)}\Big|_{t=0}.
[/tex]
(Actually not really the second equality -- is this letting [itex]V_{\gamma(0)}=\Gamma(\gamma)_0^0V_{\gamma(0)}[/itex]?)

However, the above link doesn't really show what the connection is. Yet, if you evaluate the last term of the above equation, then using the product rule you've got a term with the derivative of [itex]V[/itex] and a term with the derivative of the parallel propagator. This is what you'd expect for the covariant derivative of a vector, where the connecton coefficients are the derivatives of the parallel propagator.

Nonetheless, I'm unable to make this all work out mathematically, so I was wondering if anyone could give me a hint? That is, on how the connection coefficients are the derivatives of the parallel propagator.

Cheers.
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ianhoolihan
ianhoolihan is offline
#2
Mar26-12, 03:39 PM
P: 145
Well it looks like lots of people have viewed this thread, so I guess that means some interest. Can anyone shed some light, or offer a suggestion?

Cheers


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