User - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T18:32:54Z http://mathoverflow.net/feeds/user/12998 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/57647/injective-flow-lines-of-a-vector-field-near-a-closed-orbit Injective flow lines of a vector field near a closed orbit silentchild 2011-03-07T09:23:17Z 2011-03-07T19:40:50Z <p>Suppose $v$ is a vector field on a manifold $X$ with flow $\phi^t$. Suppose $v$ carries a first integral $f:X \rightarrow \mathbb{R}$ (i.e. $f$ is constant on the orbits of $v$). Suppose $\gamma:[0,T]\rightarrow X$ is a closed orbit of $v$, that is, $\gamma(x)=\phi^t(x)$ with $\phi^T(x)=x$. Suppose $f(\gamma)\equiv c$. Suppose $U \subset f^{-1}(c)$ is a compact neighborhood of $\mathrm{Im}(\gamma)$ in $f^{-1}(c)$ such that $U$ contains no other closed orbits of $v$ other than $\gamma$ (or it's translates).</p> <p>What conditions are sufficient to impose on $v$ so that there exists some $t_* \ge 0$ such that no flow line of $v$ apart from $\gamma$ can live in $U$ for time longer than $t_*$. More precisely, if $\gamma_1$ is another flow line of $v$ such that $\mathrm{Im}(\gamma_1) \cap U \ne \emptyset$, and $[a,b] \subset \mathbb{R}$ is any interval such that $\gamma_1([a,b]) \subset U$ then $|b-a| \le t_*$.</p> <p><strong>Edit</strong></p> <p>(Thanks to Willie Wong)</p> <p>Assume $X$ has dimension $\ge 3$. The question is perhaps better phrased as "are there well known conditions that always guarantee such a bound?".</p> <p><strong>Edit</strong></p> <p>(Thanks to Pietro Majer)</p> <p>Sorry the question is badly phrased. I've tried to improve it. I'm most interested in the case where $v$ carries a first integral $f:X \rightarrow \mathbb{R}$ (i.e. $f$ is constant on the orbits of $v$), and we only care about orbits $\gamma_1$ such that $f(\gamma_1)=f(\gamma)$. I have thus rewritten the question. But the alternative question where the orbit $\gamma_1$ is required to have the same period as $\gamma$ is also interesting to me.</p> http://mathoverflow.net/questions/55913/contact-manifolds-that-are-not-cooriented Contact manifolds that are not cooriented silentchild 2011-02-18T20:52:33Z 2011-02-18T22:11:43Z <p>One always sees the definition of a contact manifold $(X,\xi)$ as an odd dimensional manifold with a hyperplane distribution $\xi$ which can locally be expressed as $\xi = \ker \alpha$ for a 1-form $\alpha$. But in fact, it seems that in every example I know of, one always assumes that $\xi$ is <strong>cooriented</strong>, and hence we can write $\xi = \ker \alpha$ <em>globally</em>.</p> <p>Is there a reason (other than historical) as to why coorientation wasn't built in automatically in the definition of a contact manifold? It seems strange that this isn't required in the definition.</p> http://mathoverflow.net/questions/55547/a-c2-small-autonomous-hamiltonian-has-only-constant-1-periodic-orbits A $C^2$ small autonomous Hamiltonian has only constant 1-periodic orbits silentchild 2011-02-15T19:25:30Z 2011-02-15T19:28:14Z <p>Consider a autonomous Hamiltonian $h:W\rightarrow \mathbb{R}$, where $W$ is a symplectic manifold. Let $\mathrm{sgrad}(h)$ denote the vector field on $W$ that is dual to the differential $Dh$ using the symplectic form. </p> <p>Then it seems to be often stated that if $h$ is $C^2$ small then the only 1-periodic orbits of $\mathrm{sgrad}(h)$ are constants. Why is this?</p> http://mathoverflow.net/questions/55547/a-c2-small-autonomous-hamiltonian-has-only-constant-1-periodic-orbits/55548#55548 Comment by 2011-02-15T19:34:19Z 2011-02-15T19:34:19Z thanks! that was very quick!