A riemannian manifold with finitely many closed contractible geodesics - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T14:13:57Z http://mathoverflow.net/feeds/question/110709 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/110709/a-riemannian-manifold-with-finitely-many-closed-contractible-geodesics A riemannian manifold with finitely many closed contractible geodesics Malte 2012-10-25T22:20:21Z 2013-05-12T08:17:31Z <p>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 $(M,g)$. </p> <p>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.</p> <p><strong>Are there any known examples of riemannian manifolds with finitely and positively many closed contractible geodesics (or even just closed geodesics)?</strong></p> <p>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.</p> http://mathoverflow.net/questions/110709/a-riemannian-manifold-with-finitely-many-closed-contractible-geodesics/110764#110764 Answer by Agol for A riemannian manifold with finitely many closed contractible geodesics Agol 2012-10-26T15:53:53Z 2012-10-26T15:53:53Z <p>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.</p> http://mathoverflow.net/questions/110709/a-riemannian-manifold-with-finitely-many-closed-contractible-geodesics/110768#110768 Answer by Igor Rivin for A riemannian manifold with finitely many closed contractible geodesics Igor Rivin 2012-10-26T16:38:11Z 2013-05-12T08:17:31Z <p>Ellipsoids with almost but not quite equal axes have exactly three simple closed geodesics.</p>