[Math] Parallel Transport Isometry on the 2-Sphere

differential-geometry

I am having trouble grasping what this question is even asking… I was hoping that I could get some help here (my first post).

The problem is 4.4.22 out of do Carmo, it reads

Let $S^2 = [(x,y,z) \in R^3; x^2 + y^2 + z^2 = 1]$ and let $p \in S^2$. For each piece-wise parameterized curve $\alpha:[0,l] \rightarrow S^2$ with $\alpha(0)=\alpha(l) = p$, let $P_\alpha:T_p(S^2) \rightarrow T_P(S^2)$ be the map which assigns to each $v \in T_p(S^2)$ its parallel transport along $\alpha$ back to p. By prop.1, $P_\alpha$ is an isometry. Prove that for every such rotation R of $T_p(S)$ there exists an $\alpha$ such that $R=P_\alpha$.

I am stuck.

I understand that for any given curve $\alpha$ and starting vector $v$ there is a unique parallel transport along $\alpha$ so that $v(t)$ is a parallel vector field. However, I am hung up on what the significance of the rotation of the tangent plane is. First I do not know how I would ever incorporate this rotation into anything mathematically (like what qualifies as a rotation?). Second I'm not convinced that rotating the tangent plane does anything, shouldn't this effectively just be a change of basis for the vector $v$? Any help would be appreciated.

Best Answer

A rotation is really what you think of in $\mathbb R^2$: Pick any orthonormal basis $e_1, e_2$ of $T_pM$, a rotation $R$ is given by

$$R(e_1) = \cos\theta e_1 + \sin\theta e_2, \ \ R(e_2)= -\sin\theta e_1 + \cos\theta e_2$$

First of all, every parallel transport preserves the metric, we have $\langle v, w\rangle = \langle P_\alpha v, P_\alpha w\rangle$ for all $v, w\in T_pM$. Also, we must have $\det P_\alpha = 1$, as $\det$ is a continuous function and the properties on metric imply that the determinant has to be $\pm 1$.

These two conditions imply that $P_\alpha$ has to be a rotational metric. But the question is., whether every rotations can be realized as a parallel transport. For example, if $M$ is the plane, it turns out that only the identity matrix can be realized as a parallel transport. This is due to the fact that the plane has Euclidean curvature.

Thus parallel transport is a measurement of the curvature of your surfaces. In higher dimensional case, the holomony group (groups of all parallel transport) has been classified, and different holomony group gives us quite different geometry.