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

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$n$ independent first integrals automatically form a completely integrable system

Statement refuted

On a $2 n$ -dimensional phase space, any $n$ independent first integrals automatically form a completely integrable system.

Facts & Assumptions

Given: The proposed sufficiency claim.

[F1]

Complete integrability also requires pairwise zero Poisson brackets. Completely integrable Hamiltonian system.

[F2]

Hamiltonian fields satisfy $ι_{X_{F}} ω = d F$ , and ${F, G} = ω (X_{F}, X_{G})$ . Hamiltonian vector field and Hamiltonian function, Poisson bracket on a symplectic manifold.

Refutation

technique · direct

1.1

On standard $R^{4}$ take the Hamiltonian $H = 0$ and the two functions $F_{1} = q^{1}$ , $F_{2} = p_{1}$ . Every function is a first integral of the zero flow, and $d F_{1}, d F_{2}$ are independent everywhere.

given

2.1

With $ω = d q^{1} \land d p_{1} + d q^{2} \land d p_{2}$ , [F2] gives $X_{q^{1}} = - \partial_{p_{1}}$ and $X_{p_{1}} = \partial_{q^{1}}$ , hence ${F_{1}, F_{2}} = ω (- \partial_{p_{1}}, \partial_{q^{1}}) = 1$ . Thus the functions are not in involution and fail the separate requirement in [F1], refuting the claim for $n = 2$ .

F1F2step 1.1∎

Depends on

Used by

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Dependency tree · two levels

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Sources

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

nn independent first integrals automatically form a completely integrable system