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

http://en.wikipedia.org/wiki/Parallel_transport

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?

## Best Answer

When you do the infinitesimal parallel transport near N, the vector does

notend 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.