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Let's consider the issue of parallel transport in relation to the figure on the following Wikipedia link:

With reference to the Figure on the link:

Instead of parallel transporting the vector from A to N lets (parallel) transport it from A to N' along the meridian where N' is a point just below N [say latitude=89.9999 degrees]. Now we move the vector parallel to itself along the line of latitude passing through N'to reach the corresponding point on the meridian NB.The vector is now almost parallel to the meridian NB[Since the concerned line of latitude is not a geodesic I have used the word almost]. The vector is moved down and then moved back to A along the equator. It turns by a very small amount.The exclusion of a tiny[you could make it microscopic] spherical triangle is causing so much of difference.Why?

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As Ron tells you, you are doing parallel transport incorrectly. It's good to have a particular example and a general enough answer. Consider the surface of a sphere and do parallel transport around a curve C. Then the transformation induced on the tangent space is nothing else than the rotation by X radians where X is the solid angle of the encircle area in steradians. For example, a parallel transport on the equator is the rotation by $2\pi$ because hemisphere is $2\pi$ steradians. If you add a tiny triangle, tiny solid angle, the change in the parallel transport will be tiny. – Luboš Motl Dec 2 '12 at 8:14
I am referring to the change between the initial and the final positions of the vector when it goes round a loop on a curved surface(by parallel transport). You may connect the point N'(in the original posting) with some point on the meridian NB by a $small{\;\;}$ curve so that the transported vector on landing on the meridian becomes tangential parallel or nearly tangential to the meridian NB. – Anamitra Palit Dec 3 '12 at 4:41
Does "parallel transport" move a vector parallel to itself on a curved surface even in the infinitesimal sense? You may think of two adjacent tangent planes on a curved surface.Is it always possible to have parallel vectors at the points of contact(one vector being preassigned) even if the planes are awkwardly inclined? – Anamitra Palit Dec 4 '12 at 13:23

When you do the infinitesimal parallel transport near N, the vector does not end up nearly parallel to the meridian through B. It ends up in nearly exactly the same direction as it was when you start the parallel transport. You must remember that the space near the north pole is nearly flat (as it is everywhere on a sphere, the sphere is a manifold), and parallel transport on a flat region does nothing. So your argument is just incorrect. Removing the infinitesimal triangle does nothing.

In general, in two dimensions, the angle defect you get when you parallel transport over a loop is the integral of the curvature over the interior of the loop. When the loop is small, the area vanishes, and there is no change in angle.

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Only if you consider a geodesic the an infinitesimally think space round it and parallel to it is nearly flat--the tangent vector propagates parallel to itself – Anamitra Palit Dec 2 '12 at 3:10
Quoting Ron Maimon:"It ends up in nearly exactly the same direction as it was when you start the parallel transport"---in such a situation if you move back to A the vector does not turn by alpha. It turns by a much smaller amount! On removing the triangle the vector in the initial and the final situation (at A) make a very small angle. If the triangle is there it turns by 90 degrees after looping round – Anamitra Palit Dec 2 '12 at 3:33
@AnamitraPalit: I think you misunderstand parallel transport--- if you parallel transport a vector in flat space around a circle, it never changes direction. You are thinking that the changing direction of the curve leads to a changing direction for the transported vector, this is not so--- the vector tries to stay parallel to itself as best as possible while staying in the tangent plane to the sphere. – Ron Maimon Dec 2 '12 at 6:03
@AnamitraPalit: The change in orientation is infinitesimal in dx, and when you traverse a loop, the change in angle is the area of the loop times the curvature. So when you go along the meridian, basically nothing happens. Only as much change in angle as the teeny triangle has area. – Ron Maimon Dec 3 '12 at 7:33
@AnamitraPalit: The Christoffel symbols are differential they give you the rate of change of the vector, you need to mulitply by the (infinitesimal) displacement to get the actual change. Your example is bogus, it's a misunderstanding, try drawing it out on an actual sphere, you will see that moving the vector on the little segment of lattitude near the north pole does next to nothing. It only gives a tiny differential change. – Ron Maimon Dec 3 '12 at 11:33

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