Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. 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

I am reading about the energy-period relation for Hamiltonian Systems.
In Weinstein's formulation (cf. Abraham, Marsden, Foundations of Mechanics 2nd Ed, page 198) this relation amounts to:

$(\ast)$ Given an Hamiltonian system $(M,\omega, H)$, let be $\Phi$ the flow of $X_H$ and $\text{per}_H:=\{(t,x)\mid\Phi(t,x)=x\}.$
If $N$ is a smooth submanifold contained in $\text{per}_H,$ then $\left.dt\wedge dH\right|_N=0,$ i.e. $t=t(H)$ on $N,$ (the period depends only on the energy.)

In Guillemin, Stenberg, Geometric Asymptotics, on page 170, I have additionally found that, when all integral curves of $X_H$ are closed, we can take $N=\text{per}_H$ in $(\ast),$ which should mean that in such a case $\text{per}_H$ is a smooth submanifold of $\mathbb R\times M.$

Starting from this I was wondering myself:

If all integral curves of $X_H$ are closed, and $\tau(p)$ denotes the period of the orbit through $p,$ then $\tau:M\to\mathbb R$ is smooth? otherwise how to prove that in such a case $\text{per}_H$ is a submanifold?

My guess is that if there were a principal bundle structure $(M,\pi,X,\mathbb S^1)$ such that the $\mathbb S^1$ orbits are the trajectories of $X$ then $\tau:M\to\mathbb R$ should be smooth because of the relation $\zeta=\tau X_H,$ where $\zeta$ is the infinitesimal generator of the $\mathbb S^1$-action. But I don't know how to proceed without this further assumption.

Probably my doubts are silly, but as illustration of my difficulties I imagine that $M$ is the Moebius band $[0,1]\times\mathbb R/\sim$ and $X=\frac{\partial}{\partial x}$ then the period is $$\tau([(x,y)]_{\sim})=\begin{cases}1&\text{if }x=0,\\2&\text{if }x\neq 0.\end{cases}$$ Consequently $\tau:M\to\mathbb R$ isn't smooth and $\text{per}$ isn't a smooth submanifold of $M\times \mathbb R.$ alt text

share|cite|improve this question
Warning I have posted the question even on… – Giuseppe Aug 6 '12 at 7:20

Your Answer


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

Browse other questions tagged or ask your own question.