Understanding parallel transport (Gen Rel)

[Onamor]

Homework Statement

If I have a two curves $$\gamma_{1}, \gamma_{2}$$ with the same start and end points, lying on a smooth manifold $$M$$. For a vector $$v$$ at the "start" point, if I parallelly transport down both curves to the "end" point, will the two vectors at the "end" be different or the same?

Not a very formal question, but I'm very shakey on what parallell transport actually means.

Homework Equations

A vector $$v^{i}$$ is said to be parallelly transported from the point $$x^{k}$$ to a vector $$v^{i}+dv^{i}$$ at the point $$x^{i}+dx^{i}$$ if $$dv^{i}=\Gamma_{kj}^{i}v^{j}dx^{k}$$
where $$\Gamma_{kj}^{i}$$ are the Christoffel symbols/Connection.

(this is out of my notes - I don't see why the "starting" point should be $$x^{k}$$ and not $$x^{i}$$ though.)

The Attempt at a Solution

I think the answer is "yes" - it does matter which curve is chosen because, as I understand it, this difference is what gives rise to the Riemann Curvature tensor.

As I said what exactly it means to "parallelly transport" a vector is not clear to me - is it that the "final" vector will be parallel to the "starting" vector?

Also, the above description I give of parallel transport looks as if it is only "effective" (perhaps "holds" is a better word) over the infinitesimal distance $$dx^{i}$$, and not over an entire curve - is this correct?

Thank you kindly for any help, please let me know if i can be more specific.

 

[Dick]

Yes, it does depend on the curve. Try moving vectors around on great circles of the sphere. Move one from the pole to the equator, move around the equator by 90 degrees and then back to the pole. On great circles (geodesics) parallel transport probably means exactly what you might think. I.e. maintain a constant angle with respect to the curve tangent. Did it change when you came back to the pole?