MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

By a closed geodesic, I mean a smooth periodic geodesic $\mathbb{R} \rightarrow (M,g)$. I will consider them up to geometric distinction. This means that any two closed geodesics are equivalent if they have the same image in $(M,g)$.

Manifolds with constant curvature $\leq 0$, by Cartan's theorem, cannot have any closed contractible geodesics, and every riemannian metric on $S^2$ has infinitely many closed geodesics (for $n\geq 3$, the analogous theorem for $S^n$ is not known). Moreover, if the sequence of Betti numbers of the loops space $\Omega(M)$ is unbounded and $M$ is simply-connected, then $(M,g)$ contains infinitely many (contractible) closed geodesics.

Are there any known examples of riemannian manifolds with finitely and positively many closed contractible geodesics (or even just closed geodesics)?

There is a theorem associated with Gromov asserting that the word problem of $\pi_1 M$ is solvable if there is a metric $g$ on $M$ with only finitely many contractible closed geodesics. I was wondering if there are any non-trivial examples for this theorem.

share|cite|improve this question
What does the question mean, exactly? If $\gamma$ is a geodesic, then so is $\gamma^k,$ for any $k$ so what additional property are your geodesics supposed to have? – Igor Rivin Oct 26 '12 at 1:11
They are counted up to geometric distinction, i.e. any two closed geodesics are equivalent if they have the same image in $(M,g)$. – Malte Oct 26 '12 at 9:17
up vote 11 down vote accepted

I think if you take the metric on $\mathbb{R}^2$ obtained by rotating a curve which is $\sqrt{1-x^2}$ for $-1\leq x\leq 0$, and $x^2+1$ for $x\geq 0$ around the $x$-axis, then I think there will be a single closed contractible geodesic obtained by rotating the point $(0,1)$ around the $x$-axis.

share|cite|improve this answer

Ellipsoids with almost but not quite equal axes have exactly three simple closed geodesics.

share|cite|improve this answer
This looks/sounds convincing. However, if this were true, it would contradict the theorem I mentioned above. The theorem is due to Franks-Bangert and asserts that every riemannian metric on $S^2$ has infinitely many prime closed geodesics. The "prime" relation is the same as the geometric distinction: $c_0$ is prime if it is not a multiple of another closed geodesic. – Malte Oct 26 '12 at 20:51
I figured out the problem here. The result in Igor Rivin's post is (almost) the Ljusternik-Schnirelmann theorem, which asserts that an ellipsoid with axes very close to the round sphere has exactly three simple closed gedesics (namely, the intersections of the ellipsoid and the coordinate planes). Moreover, it seems that Calabi once conjectured that there are no compact mf. with finitely many closed geodesics. (This does not mean, however, that there are no non-simply-connected mf. with only finitely many contractible closed geodesics). I guess the question has no answer (yet). – Malte Oct 27 '12 at 14:18
Yes, correct (re simple vs not)... – Igor Rivin Oct 27 '12 at 18:35

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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