CounterexampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedverified 2026-10-15 (gpt-6-luna, gpt-6.1-sol, and delegated Codex review/repair agents; see completed run evidence)

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A finite-volume Hamiltonian map that is not ergodic

Statement refuted

Every nonzero-time map of a complete Hamiltonian flow preserving finite symplectic volume is ergodic with respect to the normalized volume measure.

Example

On the symplectic torus $T^{2} = (R / 2 π Z)^{2}$ with $ω = d q \land d p$ , let $H (q, p) = \cos q$ . The Hamiltonian flow is complete and preserves finite area, but every nonzero-time map is nonergodic: the invariant set $A = {(q, p) : \cos q > 0}$ has exactly half the total area. This gives a concrete reason volume preservation alone cannot identify time averages with whole-space ensemble averages.

Facts & Assumptions

Given: The symplectic torus, canonical Hamiltonian convention, and area measure.

[A1]

$q, p$ are angular coordinates modulo $2 π$ , $ω = d q \land d p$ , and $H (q, p) = \cos q$ .

[F1]

Symplectic form and symplectic manifold and Hamiltonian vector field and Hamiltonian function: a Hamiltonian vector field satisfies $ι_{X_{H}} ω = d H$ .

[F2]

Hamiltonian volume, recurrence, and ergodic scope: a complete invariant Hamiltonian flow map preserves symplectic area.

[F3]

Ergodicity relative to an invariant measure: ergodicity requires every invariant measurable set to have measure zero or full measure.

Counterexample

Given: [A1] and the canonical convention [F1].

1.1A1F1algebra

Write $X_{H} = a \partial_{q} + b \partial_{p}$ . Then $ι_{X_{H}} (d q \land d p) = a d p - b d q$ , while $d H = - \sin q d q$ . Equating coefficients gives $a = 0$ and $b = \sin q$ . The explicit solution is $ϕ_{t} (q, p) = (q, p + t \sin q (m o d 2 π)),$ which is defined for every real $t$ and is a diffeomorphism with inverse $ϕ_{- t}$ .

2.1A1F2F3step 1.1algebra∎

Since $H \circ ϕ_{t} = H$ , the set $A$ is invariant under every $ϕ_{t}$ . The torus has total area $(2 π)^{2} = 4 π^{2}$ . The allowed $q$ values in $A$ have total length $π$ , while $p$ ranges over length $2 π$ , so $m (A) = 2 π^{2}$ , strictly between zero and total area. The map preserves area by [F2], and the invariant set violates the criterion [F3]; hence it is not ergodic for any nonzero sampling time.

A finite-volume Hamiltonian map that is not ergodic

Statement refuted

Example

Facts & Assumptions

Counterexample

Depends on

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