7
$\begingroup$

The Lyusternik-Fet theorem states that every compact Riemannian manifold has at least one closed geodesic.

Are there any easy-to-construct1 examples of compact Riemannian manifolds for which it is easy to see they only have one closed geodesic?2

If there aren't any such examples, are there any easy-to-construct examples that only have one closed geodesic but where proving this might be difficult?

And if there aren't any examples of this, are there any examples at all of compact manifolds with only one closed geodesic?


1 Of course, the $1$-sphere $S^1$ contains just one closed geodesic, but I'm interested in examples besides this one.

2 By the theorem of the three geodesics, this example cannot be a topological sphere.

$\endgroup$
0
14
$\begingroup$

First of all, you have to exclude constant maps $S^1\to M$ from consideration: They are all closed geodesics. Secondly, you have to talk about geometrically distinct closed geodesics: Geodesics which have the same image are regarded as "the same". Then, it is a notorious conjecture/open problem:

Conjecture. Every compact Riemannian manifold of dimension $n >1$ contains infinitely many geometrically distinct nonconstant geodesics.

See for instance this survey article by Burns and Matveev.

This is known for surfaces (with the only hard case when the surface is diffeomorphic to $S^2$ in which case the result is due to Bangert and Franks) and for many higher-dimensional manifolds. However, the problem is open already when $M$ is diffeomorphic to the sphere $S^n$, $n\ge 3$.

Edit. I recently found a preprint which claims to prove the conjecture on closed geodesics:

S. Charles, The Existence of Infinitely Many Geometrically Distinct Non-constant Prime Closed Geodesics on Riemannian Manifolds, 2018.

The paper is still unpublished and I do not know if the proof is correct.

$\endgroup$
7
$\begingroup$

If you analyze the geodesics using Clairaut's relation, you'll find that the only closed geodesic on a hyperboloid of one sheet is the central circle. Indeed, the same holds for a concave surface of revolution of the same "shape" as the hyperboloid of one sheet.

EDIT: Apologies for missing the crucial compactness hypothesis.

$\endgroup$
3
  • $\begingroup$ Lovely example, but a hyperboloid isn't compact, right? $\endgroup$ Apr 12 '19 at 23:09
  • $\begingroup$ Oops. Sloppy reading. I'll delete. $\endgroup$ Apr 12 '19 at 23:10
  • 1
    $\begingroup$ It's a nice example; you should leave it. $\endgroup$ Apr 12 '19 at 23:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.