2
$\begingroup$

Let $M$ be a $2$-dimensional orientable Riemannian manifold and $p$ be a point in $M$. Let $(U, \varphi)$ be a coordinate chart about $p$ such that $\text{Im}(\varphi)$ is a ball in $\mathbf R^2$. Let $\gamma$ be a piecewise smooth simple loop based at a point $p$ in $M$ such that $\text{Im}(\gamma)$ is contained in $U$.

I want to show that $\int_\Omega K\ dA \equiv \text{rot}_\gamma \pmod{2\pi}$

Here $K$ denotes the sectional curvature, $\Omega$ is the interior of $\gamma$, $dA$ is the Riemannian volume form, and $\text{rot}_\gamma$ is the rotation of $T_pM$ caused by parallel transport around the loop $\gamma$.

The statement above is clearly true if $\gamma$ formed a geodesic triangles, for we can just apply the Gauss-Bonnet formula which states that the angle defect of the triangle is same as the integral $\int_\Omega K\ dA$. Similarly, one can prove this result if $\gamma$ was a geodesic polygon. From there perhaps one could try to argue by approximating an arbitrary $\gamma$ by geodesic polygons. But that seems inelegant to me.

$\endgroup$
  • $\begingroup$ You might check out pp. 80-83 or so of my differential geometry text. $\endgroup$ – Ted Shifrin Dec 15 '17 at 22:53
  • $\begingroup$ At the bottom of pg. 79, you define $e_1$ and $e_2$. The terms $x_u, x_v, E$ and $G$ appear there. Can you please explain the meaning of these? Also, here is a similar question math.stackexchange.com/questions/2568361/… $\endgroup$ – caffeinemachine Dec 16 '17 at 6:09
  • $\begingroup$ Obviously, you will need to go back to the beginning of chapter 2 for notation. You might need to do a bit of reading. $\endgroup$ – Ted Shifrin Dec 16 '17 at 6:21
2
+50
$\begingroup$

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 $\omega(\dot{\gamma}) - \dot(\theta) \equiv 0$ and the result follows from Stokes' theorem:

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

$\endgroup$
  • $\begingroup$ Somehow I'm unable to award bounty right away. Will try tomorrow. Thanks for the great answer. $\endgroup$ – caffeinemachine Dec 18 '17 at 16:41
  • $\begingroup$ I think we need to assume that the orientation on $\text{Im}(\gamma)$ coming from the map $\gamma:I\to M$ is same as the Stokes' orientation when $\text{Im}(\gamma)$ is thought of as the boundary of $\Omega$. This needed in the last step where we use Stokes' theorem. Am I right? $\endgroup$ – caffeinemachine Dec 28 '17 at 19:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.