This question is related to a previous one; now I better understand the problem and I can more clearly state what is the question.


I refer to the following concepts:

Liouville integrability: a Hamiltonian with $n$ degrees of freedom has $n$ independent integrals of motion in involution; we know that the Hamiltonian can be brought to the form $H(p_1, \dots, p_n)$ (i.e. independent on the $q$s) by a canonical transformation.

Poincare' reduction: the Hamiltonian has one integral of motion (just to simplify!); we can bring the Hamiltonian to the form $H(p_1, q_1, \dots, p_{n-1}, q_{n-1}, p_n)$ (i.e. we remove the dependence on one of the $q$s, $q_n$) by means of a canonical transformation.

Global versus local

There are examples of both operations (Liouville integration or Poincare' reduction) that are performed globally.

We also know that the Liouville integration is guaranteed to be feasible globally under a non-degeneracy condition (Arnold-Liouville theorem). Else, if the non-degenerasy condition is not met, the integrability is guaranteed only locally.

On the other hand, the feasibility of Poincare' reduction is discussed only locally (for what I've seen, e.g. Arnold, "Mathematical Aspects of Classical and Celestial Mechanics", proposition 3.2).

What I'm asking

I would like to know if there is any result which guarantees the global feasibility of the Poincare' reduction. This could possibly include a non-degeneracy condition (as in the case of Liouville's integration) or a wider range of forms of the reduced Hamiltonian.

  • $\begingroup$ In the case of a single integral (which i understand is what interests you) isn't the degeneracy condition of prop. 3.2, p.116, $dF(z)\neq 0$ for all points $z$ of the phase space manifold, what you are asking for? $\endgroup$ Commented Jun 6, 2019 at 16:16
  • $\begingroup$ Also: in the second paragraph of the background section you say "Poisson reduction". I guess you mean "Poincare reduction" ? $\endgroup$ Commented Jun 6, 2019 at 17:52
  • $\begingroup$ I fixed the errors, thank you for the help in both questions. For the suggestion $dF(z)\ne 0$ for all points $z$, on one side, I do not think that it works. I have a counter-example, in which $dF(z)\ne 0$ everywhere, but the reduction does not work. If you are interested I can show a document. Moreover, if you inspect the proof (of the rectification of the trajectory, as suggested by Arnold), you see that it is not enough to cover the space with the neighborhoods to make the property global. $\endgroup$ Commented Jun 6, 2019 at 21:11
  • $\begingroup$ ok! thanks for the feedback. unfortunately, i do not know more to say. If you can share the link you are mentioning (the counter-example) that would be great! $\endgroup$ Commented Jun 6, 2019 at 21:25
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    $\begingroup$ I have an updated version, but, for further discussion, please join this chat: chat.stackexchange.com/rooms/94648/global-poincare-reduction $\endgroup$ Commented Jun 7, 2019 at 16:58

1 Answer 1


It's probably easier to think about this geometrically: suppose we have a function $F:M\to\mathbb{R}$ (where $M$ is your phase space) such that $\lbrace H, F\rbrace = 0$, and also the level sets $F^{-1}(c)$ are all (embedded) submanifolds of $M$. Then we can take $P_n=F$, and from that construct canonical coordinates $Q_i, P_i$. The fact that $\lbrace H, F\rbrace = 0$ means that $H$ is invariant under the flow produced by taking $F=P_n$ as the Hamiltonian. This flow maps $Q_n$ to $Q_n+t$, and preserves all other coordinates. Hence $H$ is independent of $Q_n$.

The requirement $dF \ne 0$ everywhere ensures that $F^{-1}(c)$ is actually an embedded submanifold for each $c$ (this is often referred to as the Submersion Theorem). I believe the reason your counterexample (linked in the chatroom) fails is that it does not satisfy this condition: the Henon-Heiles Hamiltonian $H$ has several points where $dH = 0$, even if you exclude the point $(q_1,q_2,p_1,p_2)=(0,0,0,0)$ (incidentally, these are the equilibrium points of the system).

Edit: I realise now I'm not really addressing your actual question, but more the counterexample (linked to in the chatroom). One issue with the discussion above is that the new canonical coordinates $Q_i, P_i$ are only defined locally in general, and you're correct in asserting that what you call Poincare reduction is really only a local construction. Marsden-Weinstein reduction (which is a global construction) would be applicable in the case that the space of trajectories $F^{-1}(c)/\mathbb{R}$ of the Hamiltonian $F$ in $F^{-1}(c)$ can be given a smooth structure such that $\pi:F^{-1}(c)\to F^{-1}(c)/\mathbb{R}$ is a submersion. One criterion for this is that the $\mathbb{R}$-action on $M$ determined by $F$ is proper, but I'm not sure how useful this would be for you (sorry).

  • $\begingroup$ Thank you for your contribution! If I correctly understand, the condition that the gradient of $F$ does not vanish in $M$ implies that it can be taken equal to $P_n$. This is a quite hevy consequence, since it means that it has many functions in involution, such as all the $P$s and the $Q$s except $Q_n$. Is this correct? However, the idea of removing the "bad" points from $M$ leads me to think that there must be the condition that $M$ does not have "holes", such as the complementary space of $M$ is simply connected. Do you confirm the presence of this condition? $\endgroup$ Commented Jun 12, 2019 at 16:03
  • $\begingroup$ It's correct that this implies the other $P$s and $Q$s are in involution, but as I mentioned in the edit these are only locally defined functions in general. For example, in the extremely simple case of a 1D harmonic oscillator (with the origin removed), and taking $F=H=P_1$, the complementary coordinate $Q_1$ would be an angular variable, which is not defined globally. I'm not quite sure what you're asking in your second question, I'm afraid - perhaps you could clarify? $\endgroup$
    – user17945
    Commented Jun 12, 2019 at 16:24
  • $\begingroup$ If you agree that the Poincare' method works only locally, wether $dF\ne 0$ on $M$ or not, then the question on the topology of $M$ does not matter. On the other hand, I would like to better understand the Marsden-Weinstein reduction. What does it say on the case of my counter-example? Can we apply it? And what is the form of the reduced system? $\endgroup$ Commented Jun 12, 2019 at 20:05
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    $\begingroup$ I'm not too knowledgeable about chaotic systems, but no, Marsden-Weinstein reduction isn't applicable in this case. The reason is that for Henon-Heiles, the space of trajectories $H^{-1}(E)/\mathbb{R}$ is not even a smooth manifold (some trajectories are quasiperiodic, some are homoclinic, some are dense in an open subset of the energy surface $H^{-1}(E)$). Your counter-example is a bit artificial, since you're essentially adding together two (non-coupled) Hamiltonian systems, and taking one system's Hamiltonian as your first integral. But the issue exists in the first system on its own. $\endgroup$
    – user17945
    Commented Jun 12, 2019 at 20:59
  • $\begingroup$ Correction: I meant periodic, not quasiperiodic. $\endgroup$
    – user17945
    Commented Jun 12, 2019 at 21:10

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