Gauss-Bonnet Like Statement Connecting Parallel Transport and Curvature

Solution 1:

You can prove your result in the same way one proves the (local) Gauss-Bonnet theorem. Assume for simplicity that $\gamma$ is smooth and parametrized by unit length. Performing Gram-Schmidt on the local oriented coordinate frame associated with the chart $\varphi$, you can get a local oriented orthonormal frame $E_1, E_2$ defined on $U$. Now write

$$ P_{\gamma, 0, t}(\dot{\gamma(0)}) = a(t) \cdot E_1(\gamma(t)) + b(t) \cdot E_2(\gamma(t)) := E(t) $$

where $P_{\gamma,0,t}$ is the parallel transport along $\gamma$ from $T_{\gamma(0}M$ to $T_{\gamma(t)} M$. Since the parallel transport is an isometry, we must have $a(t)^2 + b(t)^2 \equiv 1$ and hence we can find a smooth function $\theta \colon [0,1] \rightarrow \mathbb{R}$ such that $(a(t),b(t)) = (\cos \theta(t), \sin \theta(t))$ so

$$ P_{\gamma,0,t}(\dot{\gamma}(0)) = \cos \theta(t) \cdot E_1(\gamma(t)) + \sin \theta(t) \cdot E_2(\gamma(t)). $$

The parallel transport map $P_{\gamma,0,1}$ acts on $\dot{\gamma}(0)$ (and hence on any other vector) by rotating it by $\theta(1) - \theta(0)$ counterclockwise (with respect to the orientation determined by the frame $E_1,E_2$) and so what you want to show is that

$$ \theta(1) - \theta(0) = \int_0^1 \dot{\theta}(t) \, dt = \int_{\Omega} K \, dA. $$

Now proceed as in the proof of the Gauss-Bonnet theorem. Define a one-form $\omega$ on $U$ by $$\omega(X) = \left< E_1, \nabla_X E_2 \right> = -\left< \nabla_X E_1, E_2 \right>. $$ A direct caculation shows that $d\omega = K \, dA$ while

$$ 0 = \frac{DE}{dt} = -(\dot{\theta} \sin \theta) \cdot E_1 + \cos \theta \cdot \nabla_{\dot{\gamma}} (E_1) + (\dot{\theta} \cos \theta) \cdot E_2 + \sin \theta \cdot \nabla_{\dot{\gamma}} (E_2) = \\ (\omega(\dot{\gamma}) - \dot{\theta})( \sin \theta \cdot \, E_1 -\cos \theta \cdot E_2) $$

where we used in the calculation that $\left< E_1, \nabla_X E_1 \right> = \left< E_2, \nabla_X E_2 \right> = 0$ (a consequence of $\| E_1 \| = \| E_2 \| \equiv 1$).

Hence, we get $\omega(\dot{\gamma}(t)) - \dot{\theta}(t) \equiv 0$ and the result follows from Stokes' theorem:

$$ \theta(1) - \theta(0) = \int_0^1 \dot{\theta}(t) \, dt = \int_0^1 \omega(\dot{\gamma}(t))\, dt = \int_{\gamma} \omega = \int_{\Omega} d\omega = \int_{\Omega} K \, dA. $$