CounterexampleConstruction: AI-adaptedVerification: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-14

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Poisson-commuting functions with dependent differentials do not give Liouville–Arnold coordinates

Counterexample

On standard $R^{4}$ , take $F_{1} = H = q^{1}$ and $F_{2} = (q^{1})^{2}$ . They Poisson commute everywhere, but their differentials are dependent everywhere, so they do not yield Liouville–Arnold coordinates.

Facts & Assumptions

Given: The standard form $d q^{1} \land d p_{1} + d q^{2} \land d p_{2}$ .

[F1]

Complete integrability requires both involution and independence of $d F_{1}, \dots, d F_{n}$ on a dense open subset; that subset is the regular locus. Completely integrable Hamiltonian system.

Verification

technique · direct

1.1

Both functions depend only on $q^{1}$ . Their Hamiltonian fields are scalar multiples of $\partial_{p_{1}}$ , so their symplectic pairing and hence ${F_{1}, F_{2}}$ vanish.

givenalgebra

2.1

Nevertheless $d F_{2} = 2 q^{1} d F_{1}$ at every point, so the integral map $(F_{1}, F_{2})$ has rank at most one instead of two and has no regular locus of the required rank. By [F1], it is not a completely integrable system, and Liouville–Arnold cannot furnish action–angle coordinates for it.

F1step 1.1algebra∎

Depends on

Completely integrable Hamiltonian system

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources

Ana Cannas da Silva, Lectures on Symplectic Geometry (standard reference, not scraped)