Parallel transport curve with corners 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?
 A: 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).  
A: 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.
