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The classical definition of completely integrable (Hamiltonian) systems goes as follows.

  • One considers a symplectic manifold $(M,\omega)$ and $n$ smooth functions $h_i\in C^\infty(M)$, where $\dim(M)=2n$.
  • The 'Hamiltonians' are in involution with respect to the Poisson bracket: $\{h_i,h_j\}=0$ for all $i,j\leq n$.
  • The differentials $\mathrm{d}h_i$ are linearly independent.

Often, condition 3 is not required on all of $M$:

  1. In some references, it is only required on a dense subset of $M$ (e.g. [1, 2]).
  2. In some references, it is only required to hold almost everywhere with respect to the Liouville measure on $M$ (e.g. [3, 4]).

This is where my question arises. How are these 2 points reconciled? In general, (Borel) sets of full measure are not necessarily dense, nor does the reverse inclusion hold. Is there a particular property of the Liouville measure that makes these two conditions coincide?

Is perhaps the inclusion "full measure $\subset$ dense" (obtained by combining the answer in [5] together with the strict positivity of the Liouville measure, which follows from nondegeneracy of the symplectic form) the relation I am looking for?

References:

[1] Jovanović, B. (2011). What are completely integrable Hamilton systems. The Teaching of Mathematics, (26), 1-14.

[2] https://mathoverflow.net/questions/6379/what-is-an-integrable-system

[3] Bolsinov, A. V., & Fomenko, A. T. (2004). Integrable Hamiltonian systems: geometry, topology, classification. CRC press.

[4] Hohloch, S. (2021). Integrable Hamiltonian Systems. Lecture notes from University of Antwerp, Belgium.

[5] Is a set of full measure also dense?

NDewolf
  • 1,979
  • Yes, the symplectic measure (or Liouville measure) is positive for every non-empty open set (and yes, this boils down to non-degeneracy), so every full-measure set is dense. Similarly for semi-Riemannian measures. – peek-a-boo Oct 15 '24 at 20:18

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