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

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Cotangent lifts are symplectomorphisms

Statement

Assume ${A C}_{ω}$ . If $f : Q \to Q^{'}$ is a diffeomorphism, its cotangent lift

$\hat{f} : T^{*} Q \to T^{*} Q^{'}, \hat{f} (q, p) = (f (q), (d f_{q}^{- 1})^{*} p),$

is a symplectomorphism: ${\hat{f}}^{*} λ_{Q^{'}} = λ_{Q}$ and ${\hat{f}}^{*} ω_{Q^{'}} = ω_{Q}$ .

Facts & Assumptions

Given: ${A C}_{ω}$ and a diffeomorphism $f : Q \to Q^{'}$ .

[A1]

${A C}_{ω}$ is countable choice. The Axiom of Countable Choice ( ${A C}_{ω}$ ).

[F1]

The canonical cotangent form is $- d λ$ and is symplectic. The canonical cotangent two-form is symplectic.

[F2]

Exterior differentiation commutes with pullback. The exterior derivative commutes with pullback.

Proof

technique · direct

1.1

The formula for $\hat{f}$ is smooth with inverse $\hat{f^{- 1}}$ , and the projections satisfy $π_{Q^{'}} \circ \hat{f} = f \circ π_{Q}$ . For $ξ \in T_{(q, p)} T^{*} Q$ , the tautological definition gives $({\hat{f}}^{*} λ_{Q^{'}}) (ξ) = ((d f_{q}^{- 1})^{*} p) (d π_{Q^{'}} d \hat{f} ξ) = p (d π_{Q} ξ) = λ_{Q} (ξ)$ .

F1algebra

2.1

By [F1], [F2], and step 1.1, ${\hat{f}}^{*} ω_{Q^{'}} = - {\hat{f}}^{*} d λ_{Q^{'}} = - d ({\hat{f}}^{*} λ_{Q^{'}}) = - d λ_{Q} = ω_{Q}$ . Hence the diffeomorphism $\hat{f}$ is symplectic. The case $\dim Q = 0$ is included.

A1F1F2step 1.1∎

Depends on

Used by

The cotangent lift of a vector field is Hamiltonian Proposition

Dependency tree · two levels

12 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

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