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Suppose $M$ is a connected, compact orientable 2-dimensional Riemannian manifold, with positive Gaussian curvature. I'd like to show that two non-self-intersecting closed geodesics must intersect each other.

I tried to use Gauss-Bonnet Theorem to prove this, but I wasn't successful. Can somebody help me ? Thank you.

thetruth
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  • $M$ is connected? – Emanuele Paolini Dec 15 '13 at 21:01
  • Yes. I will edit it. Thank you. – thetruth Dec 15 '13 at 21:01
  • Compactness says there is a positive lower bound on curvature. Toponogov. Part of it is that your manifold is a sphere. – Will Jagy Dec 15 '13 at 21:14
  • Actually, Gauss-Bonnet and some topology works. The fact that you have a sphere is attributed to Synge. Where did you get this problem? It is a jump in difficulty from your previous one on geodesics. – Will Jagy Dec 15 '13 at 21:27
  • http://en.wikipedia.org/wiki/Synge%27s_theorem – Will Jagy Dec 15 '13 at 21:37
  • Hello Will. I'm learning Riemannian Geometry for the first time and this was a homework question from the notes we are following in class. I have to say that the most sophisticated tool i have is Gauss-Bonnet thrm and what is likely to be teached in a first course of Riem. Geometry. Can you please be a little bit more comprehensive in your comment/answer? Thank you – thetruth Dec 15 '13 at 21:38
  • ah... i will check the link now. – thetruth Dec 15 '13 at 21:38
  • In parallel to your question the following statement is true:

    Assume that we have a Riemannian metric on the plane with at least two closed geodesics $\gamma {1}$ and $\gamma{2}$ such that $\gamma_{1} $ lies in the interior of $\gamma_{2}$. then there is a point in the annular region between two closed geodesics with zero curvature:

    Otherwise we glue two copies of the annular region along the boundaries so we obtain a torus with non zero curvature, contradiction to the Gauss Bonnet theorem.

     – Ali Taghavi Jun 15 '16 at 09:27
  • I had already used this fact to construct the following question: http://mathoverflow.net/questions/160945/limit-cycles-as-closed-geodesicsin-negatively-curved-space – Ali Taghavi Jun 15 '16 at 09:28

1 Answers1

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Fine. It's a sphere, by Synge. The (smooth) Jordan Curve Theorem says that a closed curve bounds two hemispheres, not necessarily of equal volume but of equal total curvature. If your two geodesics are not identical, there is an overlap of each "hemisphere" for one geodesic with a hemisphere for the other, making for positive total curvature on the intersections. So these are not disjoint and the boundary curves intersect each other.

Put another way, if the geodesics do not intersect the sphere is divided into two disjoint disks and an intermediate annulus, like the Earth above 45 degrees North Latitude, below 45 degrees South Latitude, and a ring in between that includes the Equator. Gauss Bonnet says that the two disks, together, have total curvature equal to the entirety of the sphere. So the supposed annulus has zero total curvature and is one-dimensional, that is the two geodesics are identical.

Will Jagy
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  • Thank you very much Will. Your answer is clear and quite elegant, though I'm not supposed to know Synge's Theorem - even if it says on wikipedia that is a classic result. Is there another (not high-tech) way to prove that $M$ must be a sphere ? Thank you so much again :) – thetruth Dec 15 '13 at 21:55
  • I mean, it's compact and has genus zero, by Gauss Bonnet. So topologically is homeomorphic to a sphere. Is it enough to our case ? Or do we need an isometry or something else... – thetruth Dec 15 '13 at 21:58
  • In this dimension, yes. Dimension above six, settled only quite recently (quarter-pinched sectional curvatures), methods due to Hamilton and Perelman, finished by Schoen and Brendle. – Will Jagy Dec 15 '13 at 22:01