Let us consider two surfaces $S_1$ and $S_2$ such that both of them contain the same curve $\alpha: I \to S_1\cap S_2$, and their tangent planes in the points of this curve coincide. Let us denote by $Y$ the tangent vector along $\alpha$. I need to prove that the parallel transport of $Y$ along $\alpha$ on both surfaces is the same.
That is, in a point $t_0\in I$ we want $Y_0=Y(t_0)$ along $\alpha$ at the point $t_1$ to be the same in both surfaces.
Now, $\alpha: I \to S_1\cap S_2$ is a curve on the surface $S_1$ there exist a unique parallel vector field $Y$ along $\alpha$ with $Y(t_0)=Y_0$ then, by definition, $Y(t_1)$ is the parallel transport of $Y_0$ along $\alpha$ at the point $t_1$. The same holds for $S_2$. But then how to show that the only one vector field is the one parallel to $Y$ along $\alpha$ and it is unique?
Best Answer
Write $T=\alpha'(t)$. Assume $Y$ is any vector field along $\alpha$ that is tangent to the surface, then $\nabla_TY$ at $\alpha(t)$ is just the projection of $D_TY$ to the tangent plane at $\alpha(t)$; here $D_TY$ is the Euclidean directional derivative.
Now $Y$ is parallel along $\alpha$, iff $\nabla_TY=0$, iff $D_TY$ is everywhere perpendicular to the tangent plane. Now this last property holds for one surface iff it holds for the other.