# simple application: gauss bonnet and geodesics

I have two (supposedly simple) problems that are solvable by the local Gauss-Bonnet Theorem. Any help is extremely appreciated!

Let $D$ be a compact surface in $\mathbb R^3$ homeomorphic to a disk, with a piecewise smooth boundary. Then Gaussian curvature satisfies $$\int_D K \, dA + \int_{\partial D} \kappa_g \, ds + \sum_{i=1}^n \alpha_i = 2 \pi,$$ where $\kappa_g$ is the geodesic curvature wrt. to the interior of $D$, $\alpha_i$ are the outer angles at the corners in $\partial D$ and $n$ is the number of these corners.

i) Let $F$ be surface in $\mathbb R^3$ that is homeomorphic to a disk, and let $\int_F K \, dA \leq \pi$. Show that geodesics don't intersect. (Hint: use the above stated theorem)

ii) Let $M$ be a compact oriented surface in $\mathbb R^3$, and let $K > 0$. Show that simple closed geodesics don't intersect. (Hint: use the above stated theorem)

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edit. i know now, what the above questions should look like:

i) Let $F$ be surface in $\mathbb R^3$ that is homeomorphic to a disk, and let $\int_F K \, dA \leq \pi$. Show that geodesics don't intersect twice.

ii) Let $M$ be a compact oriented surface in $\mathbb R^3$, and let $K > 0$. Show that simple closed geodesics do intersect.

• Part ii seems false: the unit sphere is compact and oriented with $K = 1$, and any great circle is a simple closed geodesic, but any two great circles intersect. Perhaps I'm missing something, though. – John Hughes Jan 26 '17 at 17:04
• Part i seems false, too: take the unit disk in the plane. Its total curvature is 0. But any two diameters intersect, and all straight lines in the disk are geodesic. – John Hughes Jan 26 '17 at 17:05
• that explains why it was so hard to solve... thanks! – augustin souchy Jan 26 '17 at 17:13
• You also might want to run away form the person who suggested these problems. – John Hughes Jan 26 '17 at 17:18

If the two geodesics didn't meet, then they would be the boundary of a region $R$ with Euler characteristics $\chi(R)=0$. Hence by Gauss Bonnet we have $\int K dA = 0$ and that's a contradiction since $K>0$.