# Example of a manifold with not “closed” geodesic

From my very limited experience and understanding, I have come to realize that some people study "closed" geodesics. I understand that to mean that for some a, b x(a)=x(b) for the geodesic curve x. In what situations would a manifold admit a geodesic that is not closed? Perhaps I am just not "seeing" it :-). Thank you for any help.

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How about $\mathbb R^n$? Then the geodesics are straight lines. – Jeff Nov 29 '11 at 20:58
Yes, thank you. – Aspar T. Ame Nov 29 '11 at 21:24

There is an additional condition that belongs here. A geodesic is called closed when your $x(a) = x(b)$ AND $x'(a) = x'(b),$ meaning that it makes a closed smooth curve, and keeps going over the same same set of points forever.

It is easy to find manifolds with self-intersecting geodesics that are not closed geodesics.

The traditional example, the surface of revolution $z = x^2 + y^2,$ is an exercise in do Carmo. If a geodesic goes through $(0,0,0)$ it just goes on forever. Otherwise, it intersects itself infinitely often, but is not closed. Here we go, pages 258-260 in Differential Geometry of Curves and Surfaces

I met do Carmo at a conference once. He sent me Celso Costa's dissertation. Which is why there are bound copies in the libraries of U.C. Berkeley and U.C. San Diego.

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Thank you; very good point about x'(a) = x'(b) – Aspar T. Ame Dec 12 '11 at 3:47

In the very simple case of the euclidean spaces, the geodesics are the straight lines and they are not closed (null curvature).

In the case of the sphere, geodesics are closed (strictly positive curvature).

I strongly suspect that whenever the curvature is negative, all geodesics cannot be closed (see the Poincaré half plane and its geodesics).

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Thanks; I guess I realized this but oh well :-). – Aspar T. Ame Nov 29 '11 at 21:24
@Glougloubarbaki: your latter comment is incorrect, or at least easily misinterpreted. There are always closed geodesics, for instance, in any hyperbolic 3-manifold whose fundamental group includes loxodromic elements. The axes of these elements will descend to closed geodesics of the manifold. It would be more precise and accurate to say that not every geodesic of a hyperbolic manifold is closed (as is also true of Euclidean manifolds). – Robert Haraway Nov 29 '11 at 21:53
I guess Glougloubarbaki meant that in presence of negative curvature, there are nonclosed geodesics. – PseudoNeo Nov 29 '11 at 22:37
@PseudoNeo : this is indeed what I meant. but like I said - I only think this is true, I have no proof in mind – Glougloubarbaki Nov 30 '11 at 15:43