Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Disclaimer: The original question consisted of two parts. The first one has been answered negatively (see below the answers of Sam Lisi and Alejandro). It remains the second one.

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

Question
In Guillemin, Stenberg, Geometric Asymptotics, between pages 170-171, I have additionally found that, when all integral curves of $X_H$ are periodic, we can take $N=\text{per}$ in $(\ast),$ which should mean that in such a case $\text{per}$ is a smooth submanifold of $\mathbb R\times M.$

In order to justify this last point I was wondering myself:

  1. If $X$ is a non singular vector field on $M,$ all of whose integral curves are periodic, and $\tau(p)$ denotes the period of the orbit through $p,$ then $\tau:M\to\mathbb R$ is smooth?
  2. otherwise, how to prove that in such a case $\text{per}$ is a submanifold?

What I have tried about point 2
Probably I am missing something because my guess is that if there were a principal bundle structure $(M,p,X,\mathbb S^1)$ such that the $\mathbb S^1$-orbits are the trajectories of $X$ then the period $\tau:M\to\mathbb R$ should be smooth because of the relation $\zeta=\tau X_H,$ where $\zeta$ is the infinitesimal generator of the action.
But I don't know how to proceed without this additional hypothesis.

Edit1 (After Sebastian's answer about point 1): As illustration of my difficulties with point 1, I imagine that $M$ is the Moebius band $[0,1]\times\mathbb R/\sim$ and $X=\frac{\partial}{\partial x}$ then the period is $$\tau([(x,y)]_{\sim})=\begin{cases}1&\text{if }y=0\\\2&\text{if }y\neq 0\end{cases}$$ alt text

share|improve this question
    
A random thought: What about a harmonic oscillator? The periods are constant non-zero, except for the equilibria. But maybe you want to avoid this. –  Thomas Rot Aug 6 '12 at 8:00
    
@Thomas Rot thank you, so with the harmonic oscillator, we find a Hamiltonian system all of whose orbits are periodic for which $\tau$ is not smooth and $\text{per}_H$ is not a smooth manifold. But then what hypothesis are behind the statement in Guillemin and Sternberg ("when all orbits are periodic, we can take $N=\text{per{$ in Weinstein's formulation of the energy-period relation)? But probably I should restrict to periodic motions with positive period. –  Giuseppe Tortorella Aug 6 '12 at 8:25
add comment

3 Answers

You are confusing minimal period and period. The function $\tau(p)$ you computed on $M$ is the minimal period, which is a well-defined function, but is only lower semi-continuous. The period as discussed by Sebastian is only locally defined, and is actually multivalued if you think of it globally.

This multivalued period is what per is about. In your Moebius band example, every point has period 2 (also 4, 6, 8...), though the 0-section has minimal period 1 (I denote the 0 section by $\mathbf{0}$.

Then, per consists of $\bigcup_{k \ge 1} \{ 2k-1 \} \times \mathbf{0} \cup \bigcup_{k \ge 1} \{ 2k \} \times M \subset \mathbb{R} \times M $

share|improve this answer
add comment

Let $p\in M$ be a point such that $\Phi_t(p)=p.$ Let $U\subset M$ be a small neighborhood of $p$ and $N\subset U$ be a hypersurface such that $X$ is transversal to $N.$ Let $f\colon V\subset\mathbb R^{n-1}$ be a local parametrization of $N$ with $f(0)=p.$ Then $$F\colon\mathbb (t-\epsilon,t+\epsilon)\times V\to M;F(t,x)=\Phi_t(f(x))$$ is a local diffeomorphism to an open neighborhood of $p$ in $M.$ The preimage of $N$ by $F$ is a graph of a function from $\mathbb{R}^{n-1},$ the space where x libes in, to $\mathbb R,$ the space where t lives in. This function is exactly the "time period" function you look for.

share|improve this answer
    
What if $X(p)=0$? –  Liviu Nicolaescu Aug 6 '12 at 14:06
    
1) @Liviu Nicolaescu I have edited the question to point out that the stationary points aren't allowed, i.e. constant solutions aren't considered periodic 2) @Sebastian I have thought what you say, but as with Poincarè map, if the integral curve $\gamma$ starts at $q\in N,$ then its first return on $N$ could be at another point than $q.$ –  Giuseppe Tortorella Aug 6 '12 at 14:25
1  
For example, if $M$ is the Moebius' band $[0,1]\times\mathbb R/\sim$ and $X=\frac{\partial}{\partial x}$ then the period is$$\tau([x,y]_\sim)=\begin {cases}1&\text{if }y=0\\2&\text{if }y\neq 0\end{cases}.$$ –  Giuseppe Tortorella Aug 6 '12 at 14:30
add comment

Sam is completely right. In general, the period function $\tau\colon M\to\mathbb{R}$ is not even continuous.

A very nice reference for a (counter-)example to Giuseppe's question is the paper A counterexample to the periodic orbit conjecture, by Dennis Sullivan. In the paper, Sullivan constructs a singularity-free flow on a compact 5-manifold such that all its orbits are periodic and function $\tau$ is unbounded!

share|improve this answer
    
Initially I tried to justify the statement in Sternberg-Guillemin along the lines of Sebastian's answer. But it wasn't conclusive. Then I devised the very elementary counter-example in the Edit after which in general $\tau$ is not even continuous. So, under what hypothesis Guillemin and Sternberg state (between pages 170-171) that $\mathrm{per}_H$ is submanifold of $\mathbb R\times M$? I have thought to the presence on $M$ of a principal $S^1$-bundle structure whose fibers are the orbits of $X.$ But it seems to strong. –  Giuseppe Tortorella Aug 30 '12 at 6:43
    
Thanks for the answer, Alejandro. –  Giuseppe Tortorella Aug 30 '12 at 6:44
add comment

Your Answer

 
discard

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.