Angle between two vectors on manifold I'm parallel transporting a vector along a curve and trying to calculate how much this vector rotates relative to the curve's tangent vector.  So if the path is a geodesic then I will get an answer of zero.  
I have looked at paths that are not geodesics and calculated the end form of the vector.  I now want to compare it to its original form.
My textbooks says that if two vectors ($\Bbb{X},\Bbb{Y}$) are at the same point then the angle between them is:
$\cos(\theta)=\frac{<\Bbb{X},\Bbb{Y}>}{|\Bbb{X}|\cdot |\Bbb{Y}|}$
I was wondering if I'm allowed to do this if the vectors are not at the same point i.e. I want to compare the starting vector to its end form even though they are at different places.
My thoughts: intuitively it makes sense that this is possible.  Between the vector's start and end points there is a geodesic.  I could parallel transport the vector to the end point via this geodesic.  The angle of my vector to the geodesic's tangent will be preserved.  Now I could compare this vector to the vector obtained via the non-geodesic path.  Would this give me the same answer as if I just used the above formula?  Thanks!
 A: The problem arises when there is not a unique geodesic between the two points, and these geodesics may each give you different parallel transport maps from the first point to the second. For instance, if your manifold is the usual round $S^2$, then between the north and south poles there are infinitely many geodesics, namely the lines of longitude, and these do indeed give different parallel transport maps from the tangent space at the south pole to the tangent space at the north pole.
So in summary, no, you are not allowed to do that, because what you get would not be well-defined.
A: Suppose that $I$ is an interval in $\Bbb{R}$ containing $t_0$, and let $\sigma : I \to M$ is a curve in the manifold $M$. Suppose $\dot{\sigma}(t_0) = X_{t_0}$ and let $//_{t_0}^{t}(\sigma)$ denote parallel transport along $\sigma$ starting at time $t_0$ and ending at time $t\in I$. If I understand your question correctly, what you are asking is the following: Let $X_t = \dot{\sigma}(t)$ and $Y_t = //_{t_0}^t(\sigma) X_{t_0}$; how do $X_t$ and $Y_t$ compare? In this case, you can use the metric to evaluate $\langle X_t, Y_t\rangle$ since they are both in $T_{\sigma(t)}M$. Alternatively, you could compare the "ending vector" with the "starting vector" by letting $\tilde{X}_{t_0} = //_t^{t_0}(\sigma)X_t$ and using the metric to look at $\langle \tilde{X}_{t_0}, X_{t_0}\rangle$.
