# A lower positive bound on the number of closed orbits with given energy for a mechanical system

Let be given a mechanical system with configuration manifold $M,$ potential energy $V$ and kinetic energy $K$ corresponding to a riemannian metric on $M.$ Its dynamics is determined by the Euler-Lagrange vector field on $TM$ corresponding to $L=K-V.$

For a given value $e$ of the total energy $K+V,$ let us denote by $n(e)$ the number of closed trajectories of motion with energy, (here I am identifying motions differing just by a time translation).

Question.1 Under what hypothesis on the whole mechanical system, or only on the value $e$, is it possible to give a positive lower bound on $n(e)$?

I was motivated even for having been puzzled, probably for lacking preparation, by a statement in §4.2.1 in Arnol'd, Kozlov, Neishtadt Mathematical Aspects of Classical and Celestial Mechanics, (3rd Edn).
There I found that, invoking Hadamard(1898) (in any homotopy free class of a not simply connected riemannian manifold there exists a closed geodesic) and Maupertuis' principle, when $M$ is not simply connected and $\sup_M V<e,$ it is possile to extimate from below $n(e).$

Question.2 What kind of lower bound they are referring to? Probably it is elementary, and I am missing something, but what?

Excuse me if the question is not well-posed, any suggestion in order to improve the terms of the problem are welcome, just as the answers of course.

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Haven't worked out the details, but most likely the idea is to directly generalise Hadamard's theorem (which is one about calculus of variations) into minimising the "length function" $\int_\gamma ds \sqrt{e - V\circ \gamma}$. This maybe even cast as a Finsler geometry (I'm not sure) in which case Hadamard's theorem can be applied directl. The constraint $\sup V < e$ is just so that the for constant energy curves, you cannot visit the points of potential energy higher than the total energy, and thus the sublevel set of $V < e$ may not be the whole manifold and have different topology. –  Willie Wong Feb 10 '12 at 8:02
@Willie Wong, My confusion is: ok, in any homotopy free class there exists a geodesic (for the Jacobi metric $(e-V)g,$ and so it is a closed trajectory with energy $e$ for the mechanical system), but I cannot conclude $n(e)\geq|\pi_\text{free}(M)|$. For example, if $\gamma$ is a closed geodesic, then the last term distinguish $[\gamma]^n$ for different $n,$( or not?) –  Giuseppe Tortorella Feb 10 '12 at 10:34
Ah, no, you cannot get $|\pi_{\texrm{free}}(M)|$. A very rough bound you can get is a lower bound by the number of generators of the free homotopy group. Simples example, let $V = 0$ and $M = \mathbb{S}^1$. The only closed geodesic for a fixed energy up to reparametrisation is the circle $\gamma$. But the fundamental group is $\mathbb{Z}$. But this can vastly underestimate: For example, think again about $\mathbb{T}^2$. There's probably some facts about group theory which may help, but I am not familiar enough with them to say. Sorry. –  Willie Wong Feb 10 '12 at 13:11
Dear Willie Wong, thank you very much. –  Giuseppe Tortorella Feb 10 '12 at 13:19
You may be interested in this paper, the references therein, and so forth. Leave this question here for a while. If you don't get better answers, you should consider asking on MathOverflow. –  Willie Wong Feb 10 '12 at 13:47