From Liouville's integrability theorem we know that:

If a system with $n$ degrees of freedom exhibits at least $n$ globally defined integrals of motion (i.e. first integrals), where all such conserved variables are in Poisson involution with one another, then the Hamiltonian system is Liouville integrable.

More formally: (from here)

in the case when the phase space M is compact, almost all orbits are n-dimensional tori $T^n$ (Liouville tori), and classical Liouville theorem says that the Hamiltonian action takes a very simple standard form in some symplectic coordinates (existing in a neighborhood of each Liouville torus). So, the topological structure of an integrable Hamiltonian system is quite clear in some neighborhood of a Liouville torus.

I understand very little from the above, which brings me to my question:


  • Is there an intuitive way of seeing why the topological structure of the phase space of our system is torus-like when the Hamiltonian system is integrable?
  • If there's no intuition behind, then can someone maybe simplify the above quote?
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    $\begingroup$ Comment to the question (v3): See V.I. Arnold, Mathematical methods of Classical Mechanics, $\S49$. $\endgroup$ – Qmechanic Oct 22 '14 at 19:45

i will try this one.

A Hamiltonian system is (fully) integrable, which means there are $n$ ($n=$ number of dimensions) independent integrals of motion (note that completely integrable hamiltonian systems are very rare, almost all hamiltonian systems are not completely integrable).

What this states in essence (and intuitively) is that the hamiltonian system of dimension $n$ can be decomposed into a cartesian product of a set of $n$ independent sub-systems (e.g in action-angle representation) which are minimally coupled to each other.

This de-composition into a cartesian product of $n$ independent systems (each of which has bounded energy as the whole system has bounded energy), means topologically is the $n$-dimensional torus $S^1 \times S^1 \times ... \times S^1$ ($n$ factors) which is compact (bounded system is topologically compact).

note $S^1$, literaly means topological circle or topological $1$-dimensional sphere. What it means, is that it represents (since this is topology and not geometry) a compact, bounded 1-dimensional space (1-parameter space). So a hamiltonian system with $n$ independent parameters (integrable) is (should be, locally) topologicaly the cartesian product of $n$ (abstract) $S^1$ spaces ($1$ for each parameter/dimension)

Each $S^1$ space represents a simple harmonic oscilator (a simple periodic system, or in other words a system moving on a circle, see the connection with $S^1$ spaces).

When a (completely) integrable hamiltonian system is de-composed into $n$ indepenent sub-systems, in essence this means that (locally, at every neighborhood of a point of the system phase-space) it can be linearised and represented as a stack of (independent) harmonic oscilators (stacks of $S^1$ spaces). This is the basic theorem of Liouville-Arnold on hamiltonian dynamics

For a simple example of a 3-dimensional (actually 2-dimensional, since the configuration space is the surface of a sphere) hamiltonian system which is completely integrable, see the spherical pendulum and analysis thereof

spherical pendulum

The spherical pendulum is 2-dimensional system (thus the phase-space is 4-dimesnional) and has a second integral of motion the moment about the vertical axis.

(a link on a more advanced analysis on the dynamics of pendula).

In other words the whole is just the sum of its parts.

What would be the hamiltonian space of a (for example $2$-dimensional) system which the dimensions are not independent (not-integrable).

This means the dimensions are correlated and cannot be de-composed into independent sub-systems (i.e a $2$-dimensional torus $S^1 \times S^1$), so topologically it is a $2$-dimensional sphere ($S^2$).

2-d torus

2-d sphere

In a $2$-dimensional sphere the $2$ dimensions are correlated and cannot be made flat (i.e cannot be linearised and mapped into a flat space of same dimension, unlike a $2$-dimensional torus, in other words has what is refered as intrinsic curvature).

Elaborating a little on this.

Of course, if one sees the 2-dimensional torus as a 3D object (in effect this means embedded in a flat 3D euclidean space), it has curvature. This is refered as "external" curvature stemming from the embedding into a 3D space. But if one sees the 2-dim torus as a 2-dimensional surface on its own, it has no (zero) curvature. This is refered to as (intrinsic) curvature (in the riemannian sense).

If one takes the 2-dim torus and cuts it and unfold it, one gets the 2-dimensional cylinder . If further one cuts the 2-dim cylinder and unfold it, one gets a 2-dimensional flat surface. This means the (intrinsic) curvature of the 2-dim torus is zero and can be mapped into a flat space of the same dimension.

For the 2-dim sphere, this is not possible. There is no way it can be cut and mapped into a flat surface of the same dimension. It has (intrinsic) curvature non-zero and this is also a measure away from flatness (and also a measure of dimension correlation). One example is maps of earth (2-dimensional spherical surface) on a flat paper, one can see that the map contains distortions, since there is no mapping of a sphere into a flat surface.

On the other hand if one takes a flat 2-dim surface and makes one boundary periodic, one gets a 2-dim cylinder, if further makes the other boundary also periodic, one gets the 2-dim torus.

In general the conditions under which any given hamiltonian system is (completely) integrable is a very difficult problem.

Still another way to see this is an analogy with probability spaces. Consider 2 event spaces of 2 physical systems consisting of 2 parameters (lets say 2 coins) $\Omega_{12}$ and $\Omega_{AB}$.

When the system is integrable (i.e the parameters are independent, meaning $P(1|2) = P(1)$) then the event space $\Omega_{12}$ is the cartesian product of each sub-space $\Omega_1 \times \Omega_2$. And each outcome of the total system is just the product of the probabilities of each sub-system.

Now consider a second system where the coins are correlated, meaning $P(A|B) \ne P(A)$.

This space $\Omega_{AB}$ cannot be de-composed into 2 independent sub-spaces $\Omega_A$, $\Omega_B$ as their cartesian product since the sub-spaces are not independent. This corresponds to a non-integrable Hamiltonian system (and a topological $2$-d sphere).

The analog of statistical independence in probability event spaces for hamiltonian systems is exactly the existence and functional (more correctly poisson) independence of the appropriate number of integrals of motion (complete integrability).

In other words for a non-integrable system the whole is more than the sum of its parts.

Hope this is useful to you

PS. You might also want to check: Holonomic System, Non-holonomic System, Integrable System

  • 1
    $\begingroup$ @Phonon, ah thanks, the main difference (which relates to non-integrability) is mentioned, a sphere cannot be made flast (has intrinsic curvature). Curvature of cousre is also a measure of correlation between dimensions. If you have sth else in mind, i can update $\endgroup$ – Nikos M. Oct 23 '14 at 10:12
  • $\begingroup$ @Phonon, sure i'll update give me a couple of minutes to find figures and references for a pendulum system as an example $\endgroup$ – Nikos M. Oct 23 '14 at 10:52
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    $\begingroup$ @Phono, updated the answer, hope this reflects your comments, although i disagree with the "generalisation hint". Plus i would not like to add hamiltonian flows and symplectic manifodls etc.. (they are nice no argue about that). i think sometimes they can obfuscate what should be an intuitive picture behind terminology and convention, but we can discuss this some other time $\endgroup$ – Nikos M. Oct 23 '14 at 12:14
  • $\begingroup$ @user929304, this is a different (very difficult) problem under what conditions a hamiltonian system is integrable $\endgroup$ – Nikos M. Oct 23 '14 at 12:51
  • $\begingroup$ @user929304, hmm,ok still effectively this involves, the conditions under which one can prove a given system integrable or not, i'll be back and update on your comment in a while, thankx $\endgroup$ – Nikos M. Oct 23 '14 at 13:01

Liouville Arnol'd theorem state that given an integrable Hamiltonian system and denoting with $$M_a'=\{(q,p)\in\Gamma:f_i(q,p,t)=a_i\}$$ the connected component of the level sets of all of the first integral, then the restriction of the foundamental form on $M_a'$ $$p\cdot dq\Big|_{M_a'}$$ equal $dS(q,a)$ where $S$ is the generating functional of the canonical transformation$^\dagger$ $\mathscr{C}:(q,p)\mapsto(b=\partial_aS,a)$, i.e. that transformation such that the transformed hamiltonian depend on new momenta only $$H\circ\mathscr{C}^{-1}(b,a)=H(q,\partial_qS)=K(a).$$ Thus $\mathscr{C}$ maps your initial system into one for which the dynamic is really easy in fact from the new hamiltonian one get $$\begin{array}{ccc}\dot{b}&=&\partial_aK\\\dot{a}&=&-\partial_bK=0.\end{array}$$ If $M_a'$ is compact then our dynamical variables $(b,a)$ can be thought as an angle $\varphi$ on a torus labelled by a constant action $I$. In this case the solutions of the above equations are $$\begin{array}{cccc}\varphi(t)&=&\varphi_0+\partial_IKt&=\varphi_0+\omega_0t\\I(t)&=&I_0&=const.\end{array}$$ $\dagger$ One way to define a canonical transformation from the ''old'' system $(q,p,H,t)$ to the ''new one'' $(Q,P,K,T)$, is to ask that $\mathscr{C}$ preserves the Poicare-Cartan 1-form $$dA=p\cdot dq-Hdt$$ in the sense that $A_{old}=cA_{new}+\Delta F(q,Q,t)$. From this easily follow that $\partial_qF=p$, $\partial_QF=-P$ and $\partial_tF=K-H$, then performing the Legendre transformation of $F$ one obtain $S=F+P\cdot Q$ and the last set of equations became $\partial_qS=p$, $\partial_PS=Q$ and $\partial_tS=K-H$. The last equation is the so called Hamilton-Jacobi equation.


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