Consider an integrable $n$ degree of freedom Hamiltonian $H$ with $n$ integrals $I_{1},...,I_{n}$ in involution. Take a level set $M$ of the integrals. According to Lioville-Arnold Theorem:

  1. If on the level set $M$ the functions $dI_{i}$ are independent, then that level set is a smooth invariant manifold for $H$

  2. If $M$ is compact and connected, it is diffeomorphic to an $n$-torus.

  3. Action-angle variables may be introduced.

I don't understand what happens on "separatrix" level sets, and in particular which of the above conditions fail and why action-angle variables can't be used.

For example take a pendulum $H(q,p) = \frac{p^{2}}{2} + \cos q.$ Phase space is $\mathbb{T} \times \mathbb{R}$. Then the separatix level set is $H=1$, connecting the hyperbolic fixed point $(0,0)$ to itself (since we take $q$ modulo $2\pi$). Since we only have one integral $H = I$, what does this mean about "independence" requirement? And isn't $H=1$ a connected compact set?

Edit. I assume that condition 1) above is what fails in my pendulum example (since at the hyperbolic fixed point we $dH=(0,0)$ i.e. $dH$ vanishes; this point is connected by the separatrix hence by continuity on the separatrix level $dH$ does not satisfy condition 1).

Would be grateful if someone could shed more light on this.


In this case Liouville-Arnold doesn't tell much for the separitrix. Indeed as you've noticed $dH(0,0)=0$ and this means that $1$ is a critical value of $H$. Now, it is known that $H^{-1}(1)\setminus\{(0,0)\}$ is a submanifold of your phase space, but this is not compact, thus you can't apply the theorem. It is true that the preimage of a critical value can be a submanifold (and thus may be possible to apply the theorem) but in this case it doesn't happen.

| cite | improve this answer | |
  • $\begingroup$ Thanks. Could you please clarify why $H^{-1}(1) \ \{(0,0)\}$, i.e. the separatrix, is not compact? @Uskebasi $\endgroup$ – Alex Apr 13 '18 at 15:35
  • $\begingroup$ $H^{-1}(1)=\{(q,p)\in T\times\mathbb{R},\ s.t. p=\pm\sqrt{2-2\cos(q)}\}$, which is an $x$-shaped "curve" on the cylinder. If you take off $(0,0)$ from this set you get a set which is open, thus it cannot be compact $\endgroup$ – Uskebasi Apr 13 '18 at 22:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.