I am trying to understand the parallel transport along the triangle looking path on the sphere, as in http://en.m.wikipedia.org/wiki/File:Parallel_transport.png. The thing that confuses me is that the curve looks only piecewise continuous, so the only reason the vector points in a new direction is that the parallel transport ignores the rotation that occurs in the corners? I thought curves had to be smooth? What am I missing?
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$\begingroup$ can you not smooth it out using a bump function in a small neighbourhood of the corner? if you work in a ball all paths are homotopic so parallel transport shouldn't change. right? $\endgroup$– user125763Mar 18, 2014 at 11:47
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$\begingroup$ I am just using my intuition. It feels like walking around with a spear held fixed along that curve requires me to either rotate or start walking sideways in the corners, parallell transport equaling walking sideways... Don't know about homotopic paths in a ball and parallell transport. $\endgroup$– EmilMar 18, 2014 at 11:51
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$\begingroup$ @user125763 Homotopic paths do not induce the same parallel transport maps. Your construction would work if you confined the distortion to a tiny neighborhood of the corner, but ultimately it's unnecessary. $\endgroup$– Paul SiegelMar 18, 2014 at 12:18
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2 Answers
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Let $\gamma$ be a smooth curve starting at $p_1$ and ending at $p_2$. "Parallel transport" is a linear map from the tangent space at $p_1$ to the tangent space at $p_2$ determined by $\gamma$ (as well as the chosen connection on the underlying manifold). So if you have two curves - the first from $p_1$ to $p_2$ and the second from $p_2$ to $p_3$ - then parallel transport along their concatenation is just the composition $T_{p_1} \to T_{p_2} \to T_{p_3}$ of the parallel transport maps associated to the two curves.
In particular, there is no "rotation around the corners". That sort of consideration would be relevant if you wanted to explore second-order phenomena such as the curvature of a curve, but not in parallel transport. Perhaps the following example will help your intuition. Take a vector on the equator of a sphere which points towards the north pole. If you parallel transport the vector along the equator it will still point north (not true along any curve, but true along the equator). But if you take two northward pointing vectors on the equator and parallel transport them both up to the north pole along geodesics then they will not point in the same direction (their angle will depend on how far apart they were at the equator).
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$\begingroup$ Okay I should think of parallell transport as statue pushing I guess! $\endgroup$– EmilMar 18, 2014 at 12:21
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Do parallel transport of the vector ${\bf v}:=(1,0)$ along the unit circle beginning and ending at $(1,0)$. Then the vector ${\bf v}$ will be horizontal pointing to the right at all times, even though the circle changes its forward direction continuously by a total amount of $2\pi$.
This is telling you that you should not turn the transported vector ${\bf v}$ at points where the curve $\gamma$ along which you transport has corners.
answered Mar 18, 2014 at 12:17