2
$\begingroup$

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)

$\,$

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.

  • 1
    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.2017-01-26
  • 1
    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.2017-01-26
  • 1
    that explains why it was so hard to solve... thanks!2017-01-26
  • 2
    You also might want to run away form the person who suggested these problems.2017-01-26

1 Answers 1

2

Part ii) might be false, do Carmo writes:

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 $K>0$.