# 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?

• 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? – user125763 Mar 18 '14 at 11:47
• 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. – Emil Mar 18 '14 at 11:51
• @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. – Paul Siegel Mar 18 '14 at 12:18

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.
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.