Alphabeta Math
CounterexampleConstruction: AI-adaptedVerification: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-14
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A singular common level need not be a torus

Counterexample

For H(q,p)=(q2+p2)/2 on the symplectic plane, the zero-energy fibre is the single point (0,0), not a one-torus.

Facts & Assumptions

Given: The standard symplectic plane and the displayed Hamiltonian.

[F1]

In one degree of freedom, complete integrability requires one integral whose differential is independent on a dense regular locus. Completely integrable Hamiltonian system.

Verification

technique · direct
1.1

Here dH=q dq+p dp, so it is nonzero on R2∖{0}, an open dense set. Thus H supplies a completely integrable one-degree-of-freedom system in the sense of [F1].

F1givenalgebra
2.1

But H−1(0)={(0,0)} because it is a sum of squares, and dH(0,0)=0. The fibre is singular and zero-dimensional, hence cannot be diffeomorphic to S1. This shows why the regular-value hypothesis in the torus theorem is essential.

step 1.1algebra∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources