# is any hamiltonian system with just one degree of freedom completely integrable?

An hamiltonian system with $n$ degree of freedom is said to be completely integrable when there exists an system $f_1,\ldots,f_n$ of first integrals mutually Poisson-commuting, such that $df_1(x),\ldots,df_n(x)$ are linearly independent for any $x$ in a dense subset.

Consequently any hamiltonian system with one degree of freedom, whose non fixed points constitute a dense subset, is completely integrable, (infact in this case just its hamilton function gives a maximal set of Poisson-commuting first integrals which are independent on a dense subset).

Now in a paper I read that any hamiltonian system with one degree of freedom is completely integrable, because there should exist always a smooth function which is Poisson-commuting with the Hamilton function and whose regular points are every-where dense.

I have some difficulty in understanding this statetement, so is there someone who could give me an hint?
My guess is that, given an Hamilton function $H$ on $(M,\omega)$ whose regular points constitute an open subset $U$, if it is possible to prove that there exists a smooth function $\delta$ vanishing on $U$ and whose regular points are dense in $M\setminus\overline{U}$, then we could take $H+\delta$.

• Isn't this function the Hamiltonian itself? – Fabian May 2 '11 at 17:41
• sure, but only when the non-fixed points of the hamiltonian system constitute a dense subset. – agtortorella May 2 '11 at 19:28
• My question is about the remaining case. – agtortorella May 2 '11 at 19:28

I know this is very late, but what you are concerned about is if the set of points where $dH = 0$ contains an open set. If you consider the interior of this set $dH=0$ and take any function $f$ whose support is contained in here, then $H + f$ commutes with $H$. By choosing $f$ generically, $H+f$ shouldn't have any critical points in this set, and I believe you are done.