# A riemannian manifold with finitely many closed contractible geodesics

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.

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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

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.
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