I'm trying to solve this problem:
Prove that if the parallel transport is path independent, i.e., given two points $p,q \in S$ the parallel transport from $p$ to $q$ is the same, no matter the curve chosen, then the Gaussian curvature of the surface must be zero.
What I have is:
Given the points $p, q \in S$, Let $\alpha : I \rightarrow S$ and $\beta : I \rightarrow S$ be two curves, which $\alpha(0) = p = \beta(0)$ and $\alpha(t_1) = q = \beta(t_2)$. And $w_{\alpha}(t)$ a parallel vector field along $\alpha$ and $w_{\beta}(t)$ a parallel vector field along $\beta$ such that $w_0 = w_\alpha(0) = w_\beta(0)$.
Since parallel transport is the same at point $q$ then, must be $w_\alpha(t_1) = w_\beta(t_2)$. And also we have that $D_{\alpha'(t)}w_\alpha = D_{\beta'(t)}w_\beta = 0$
My idea is to prove that the Christoffel symbols are all zero, and consequently K = 0.
Is that the right approach here?
I also ask to not use the concepts of Riemannian Manifolds, tensor calculus, if possible.
Thank you
Best Answer
Here's a good place to use Gauss-Bonnet. If you parallel transport a vector around a small smooth closed curve $C$, the angle through which it turns is $2\pi - \int_C \kappa_g(s)\,ds$. It follows that $\iint_R K\,dA = 0$ for all small regions $R$. By continuity, $K=0$ everywhere.