Understanding parallel transport (Gen Rel)

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SUMMARY

The discussion centers on the concept of parallel transport in the context of differential geometry on smooth manifolds. It is established that parallel transport of a vector along two different curves with the same endpoints results in different vectors at the endpoint due to the influence of the Riemann Curvature tensor. The mathematical representation of parallel transport is given by the equation dv^{i} = \Gamma_{kj}^{i}v^{j}dx^{k}, where \Gamma_{kj}^{i} are the Christoffel symbols. The discussion emphasizes that parallel transport is effective over infinitesimal distances and is dependent on the chosen curve.

PREREQUISITES
  • Understanding of smooth manifolds and their properties
  • Familiarity with differential geometry concepts, particularly the Riemann Curvature tensor
  • Knowledge of Christoffel symbols and their role in parallel transport
  • Basic grasp of vector calculus in curved spaces
NEXT STEPS
  • Study the properties of the Riemann Curvature tensor in detail
  • Learn about geodesics and their significance in parallel transport
  • Explore the implications of parallel transport on vector fields in curved spaces
  • Investigate the relationship between parallel transport and the geometry of spheres
USEFUL FOR

This discussion is beneficial for students and researchers in mathematics and physics, particularly those focusing on general relativity, differential geometry, and the study of curved manifolds.

Onamor
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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.
 
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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?
 

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