Disclaimer: I don't know if this question is well suited for this site, but I have posted this question on Math.StackExchange with no answer, so I have thought to post it even here.
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.)
In Guillemin, Stenberg, Geometric Asymptotics, on page 170, 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 understand this last point I was wondering myself:
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? otherwise how to prove that in such a case $\text{per}$ is a submanifold?
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.
Edit (After Sebastian's answer): 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 $$\tau([(x,y)]_{\sim})=\begin{cases}1&\text{if }y=0\\\2&\text{if }y\neq 0\end{cases}$$ alt text http://upload.wikimedia.org/wikipedia/commons/thumb/1/19/Moebius_strip.svg/267px-Moebius_strip.svg.png