- #1
ianhoolihan
- 144
- 0
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/Parall...ng_the_connection_from_the_parallel_transport. I understand how to formulate the covariant derivative
<br /> \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}.<br />
(Actually not really the second equality -- is this letting V_{\gamma(0)}=\Gamma(\gamma)_0^0V_{\gamma(0)}?)
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 V 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.
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/Parall...ng_the_connection_from_the_parallel_transport. I understand how to formulate the covariant derivative
<br /> \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}.<br />
(Actually not really the second equality -- is this letting V_{\gamma(0)}=\Gamma(\gamma)_0^0V_{\gamma(0)}?)
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 V 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.
