Problem 9-1 in Lee's "Introduction to Riemannian Manifolds", the author asks the following:
Let $(M,g)$ be an oriented Riemannian 2-manifold with nonpositive Gaussian curvature everywhere. Prove that there are no geodesic polygons with exactly 0,1, or 2 ordinary vertices. Give examples of all three if the curvature hypothesis is not satisfied.
I was able to prove the first part of the problem by a simple application of the Gauss-Bonnet formula. I was also able to come up with counter examples for the second part in the case of 0 or 2 ordinary vertices by considering the round 2-sphere (a great circle for the 0 case and two half great circles for the 2 case). I am stuck however coming up with an example for the 1 vertex case. Does anyone have any suggestions?
Best Answer
Take a Euclidean square $\square PQRS$ and bend it into a cone so as to identify the sides $\overline{QP}$ and $\overline{QR}$. The two right angles $\angle QPS$ and $\angle QRS$ join to form a straight angle, and the two sides $\overline{PS}$ and $\overline{RS}$ join to form a closed geodesic, but the angle at its base point $S$ is not a straight angle. Try it with a square piece of paper and a strip of tape!
Roughly speaking this is a 1-sided geodesic polygon, except that the vertex $Q$ is not a smooth point. So smooth it! Cut out a cone neighborhood of $Q$ and smoothly reattach a nice smooth disc neighborhood.
If you like, you can embed this into a Riemannian manifold diffeomorphic to the 2-sphere.