About the geometry of completely integrable systems - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T19:57:52Z http://mathoverflow.net/feeds/question/63836 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/63836/about-the-geometry-of-completely-integrable-systems About the geometry of completely integrable systems Giuseppe 2011-05-03T18:34:01Z 2011-05-04T13:40:53Z <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#63838 Answer by DamienC for About the geometry of completely integrable systems DamienC 2011-05-03T19:10:57Z 2011-05-04T13:40:53Z <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>