Given a vector field all of whose integral curves are closed, is the period a smooth function? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T14:17:05Z http://mathoverflow.net/feeds/question/104080 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/104080/given-a-vector-field-all-of-whose-integral-curves-are-closed-is-the-period-a-smo Given a vector field all of whose integral curves are closed, is the period a smooth function? Giuseppe 2012-08-06T07:19:29Z 2012-08-30T07:04:02Z <blockquote> <p><strong>Disclaimer</strong>: 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.</p> </blockquote> <p><strong>Background</strong><br> I am reading about the energy-period relation for Hamiltonian Systems.<br> In Weinstein's formulation (cf. Abraham, Marsden, Foundations of Mechanics 2nd Ed, page 198) this relation amounts to:</p> <blockquote> <p>$(\ast)$ Given an Hamiltonian system $(M,\omega, X_H)$, let be $\Phi$ the flow of $X_H$ and $\text{per}:=\{(t,x)\in\mathbb R\times M\mid\Phi(t,x)=x\}.$<br> 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.)</p> </blockquote> <p><strong>Question</strong><br> In Guillemin, Stenberg, Geometric Asymptotics, between pages <a href="http://books.google.it/books?id=58PgdwJzirUC&amp;lpg=PP1&amp;ots=GDIGlP2eaz&amp;dq=guillemin%20sternberg%20Geometric%20Asymptotics&amp;pg=PA170#v=onepage&amp;q&amp;f=false" rel="nofollow">170-171</a>, 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 $\mathbb R\times M.$ </p> <p>In order to justify this last point I was wondering myself:</p> <blockquote> <ol> <li>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? </li> <li>otherwise, how to prove that in such a case $\text{per}$ is a submanifold?</li> </ol> </blockquote> <p><strong>What I have tried about point 2</strong><br> 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.<br> But I don't know how to proceed without this additional hypothesis. </p> <p><strong>Edit1 (After Sebastian's answer about point 1):</strong> 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 $$\tau([(x,y)]_{\sim})=\begin{cases}1&amp;\text{if }y=0\\2&amp;\text{if }y\neq 0\end{cases}$$ <img src="http://upload.wikimedia.org/wikipedia/commons/thumb/1/19/Moebius_strip.svg/267px-Moebius_strip.svg.png" alt="alt text"></p> http://mathoverflow.net/questions/104080/given-a-vector-field-all-of-whose-integral-curves-are-closed-is-the-period-a-smo/104094#104094 Answer by Sebastian for Given a vector field all of whose integral curves are closed, is the period a smooth function? Sebastian 2012-08-06T09:59:49Z 2012-08-06T09:59:49Z <p>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.</p> http://mathoverflow.net/questions/104080/given-a-vector-field-all-of-whose-integral-curves-are-closed-is-the-period-a-smo/104135#104135 Answer by Sam Lisi for Given a vector field all of whose integral curves are closed, is the period a smooth function? Sam Lisi 2012-08-06T18:40:54Z 2012-08-06T18:40:54Z <p>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. </p> <p>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 $\mathbf{0}$. </p> <p>Then, per consists of $\bigcup_{k \ge 1} \{ 2k-1 \} \times \mathbf{0} \cup \bigcup_{k \ge 1} \{ 2k \} \times M \subset \mathbb{R} \times M$</p> http://mathoverflow.net/questions/104080/given-a-vector-field-all-of-whose-integral-curves-are-closed-is-the-period-a-smo/105873#105873 Answer by Alejandro for Given a vector field all of whose integral curves are closed, is the period a smooth function? Alejandro 2012-08-29T19:48:33Z 2012-08-29T19:48:33Z <p>Sam is completely right. In general, the period function $\tau\colon M\to\mathbb{R}$ is not even continuous. </p> <p>A very nice reference for a (counter-)example to Giuseppe's question is the paper <a href="http://www.numdam.org/item?id=PMIHES_1976__46__5_0" rel="nofollow">A counterexample to the periodic orbit conjecture</a>, 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!</p>