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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?

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  • $\begingroup$ @Kajelad what is the issue if $M$ is not compact? I didn’t see a problem in my proof $\endgroup$ Jun 3 at 19:21
  • $\begingroup$ @AdamMartens Gauss-bonnet does not apply when the interior of the region is not compact. $\endgroup$
    – Kajelad
    Jun 3 at 19:37
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    $\begingroup$ 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. $\endgroup$
    – Jack Lee
    Jun 3 at 20:19
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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.

If you like, you can embed this into a Riemannian manifold diffeomorphic to the 2-sphere.

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  • $\begingroup$ Clever thanks! This example has the added bonus that the curve you describe is a geodesic by default since “bending” the piece of (flat) paper is a local isometry and the curve is the result of a straight line. Cool $\endgroup$ Jun 3 at 23:47

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