# Example of a Riemannian 2-manifold that has a geodesic polygon with only one vertex

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?

• @Kajelad what is the issue if $M$ is not compact? I didn’t see a problem in my proof Jun 3 at 19:21
• @AdamMartens Gauss-bonnet does not apply when the interior of the region is not compact. Jun 3 at 19:37
• I should point out that the definition of "geodesic polygon" that I use in my book (p. 271) includes the assumption that the polygon is the boundary of a precompact open set whose closure is contained in a single coordinate domain. With this definition, it's not necessary to assume $M$ is compact. Jun 3 at 20:19

Take a Euclidean triangle $$\triangle PQR$$ and bend it into a cone so as to identify the sides $$\overline{PQ}$$ and $$\overline{QR}$$. The third side $$\overline{PR}$$ is then a closed geodesic and the angle at its base point $$P=R$$ is not a straight angle. Try it with a triangular 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.