About the geometry of completely integrable systems - MathOverflow most recent 30 from http://mathoverflow.net2013-05-21T19:57:52Zhttp://mathoverflow.net/feeds/question/63836http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/63836/about-the-geometry-of-completely-integrable-systemsAbout the geometry of completely integrable systemsGiuseppe2011-05-03T18:34:01Z2011-05-04T13:40:53Z
<p>During a conversation I heard an assertion that I found at least dubious for the lack of adeguate hypothesis, but I am not able to imagine a counterexample, even if it is probably obvious to some of you.<br>
My question: is there someone who can point out to me a counterexample for the following implication?</p>
<p>This is the setting:<br>
Let $(M,\omega)$ be a $2n$-dimensional symplectic manifold, and $f_1,\ldots,f_n$ independent functions on $M$ mutually Poisson-commuting, and such that the hamiltonian vector fields $X_{f_1},\ldots,X_{f_n}$ are complete.<br>
Let $\mathcal{F}$ denote the lagrangian foliation of $M$ determined by the integrable distribution $D$ generated by $X_{f_1},\ldots,X_{f_n}$.</p>
<p>This is the conclusion that I find not well justified:<br>
For any $x$ in $M$ there exists a local manifold $\Sigma_x$ which is lagrangian, transversal to $D$, and doesn't intersect any leaf of $M$ at two distinct points.<br>
(I know that this condition is necessary and sufficient for the existence of the manifold of the leaves of $\mathcal{F}$)</p>
http://mathoverflow.net/questions/63836/about-the-geometry-of-completely-integrable-systems/63838#63838Answer by DamienC for About the geometry of completely integrable systemsDamienC2011-05-03T19:10:57Z2011-05-04T13:40:53Z<h3>preliminary remark</h3>
<p>I assume that being independant for functions here means that their differentials at <strong>any</strong> point are linearly indenpendant (and not at <strong>almost</strong> any point, like it is sometimes assumed... in which case a counter-example is easy to find for $n=2$: take $f_1$ to be the square of the norm on $\mathbb{R}^2$, and look at the origin). </p>
<h3>the statement</h3>
<p>I claim that one can prove the following weak version of the action-angle coordinate Theorem, under the hypothesis of the question: </p>
<blockquote>
<p>For any $x\in M$ there exists Darboux coordinates $(p_1,\dots,p_n,q_1,\dots, q_n)$<br>
around $x$ such that the leaves of the Lagrangian foliation are $(q_1,\dots,q_n)=cst$. </p>
</blockquote>
<p>Then the local manifold <code>$\{p_1=\dots=p_n=0\}$</code> satisfies your requirement. </p>
<p>To prove the statement, consider $q_i=f_i-f_i(x)$ and extend them to a Darboux chart $(q_1,\dots,q_n,p_1,\dots,p_n)$ around $x$. </p>
<h3>source of confusion</h3>
<ol>
<li><p>foliations may be very wilde... but here we actually have a submersion. </p></li>
<li><p>in usual action-angle coordinates Theorem on needs some properness assumption. But the usual Theorem tell us about properties of semi-global coordinates. Here we were dealing with a purely local statement and we don't need any properness assumption. </p></li>
</ol>