Maximum number of closed simple geodesics in surface homeomorphic to sphere minus three points with $K < 0$

Let $$S$$ be a regular surface homeomorphic to a sphere minus three points with $$K < 0$$ everywhere. How many simple and closed geodesics can there be in $$S$$? Prove your result.

Since $$S$$ is homeomorphic to a sphere minus three points, it's homeomorphic to a plane minus two points. Now, I know the following result:

$$K < 0$$ everywhere $$\implies$$ $$\not\exists$$ a geodesic bounding a simply connected region

So there could be:

• a geodesic which bounds a region containing one point that's missing in the plane

• a geodesic which bounds a region containing the other point that's missing in the plane

• a geodesic which bounds a region containing both points missing in the plane

And there can't be a geodesic bounding a region wich doesn't contain both points (since then it wouldn't be missing any points and thus be simply connected), so $$S$$ can have at most three such geodesics. Is that it?

• It seems to me you're totally ignoring Gauss-Bonnet. – Ted Shifrin Jul 6 at 22:19
• @TedShifrin why? The result I used is a consequence of Gauss-Bonnet – Matheus Andrade Jul 6 at 22:36
• You know better than that. It's a very weak consequence. – Ted Shifrin Jul 6 at 23:06
• @TedShifrin well, I know $S$ is homeomorphic to a cylinder minus a point, so I could use a known consequence from Gauss Bonnet which tells me there's at most one simple closed geodesic on a surface homeomorphic to a cylinder with $K < 0$ everywhere. But here I'm missing a point, can I still use it? – Matheus Andrade Jul 6 at 23:24
• You can use the proof to deduce that there's at most one simple closed geodesic going around the "belt" of the cylinder (i.e., not nullhomotopic on the cylinder). But suppose you have closed geodesic(s) that are nullhomotopic on the cylinder but go around the missing point. How many can there be? – Ted Shifrin Jul 7 at 18:30