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

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The canonical cotangent two-form is symplectic

Statement

Assume ${A C}_{ω}$ . On $T^{*} Q$ the canonical form $ω_{c a n} = - d λ$ is symplectic and, in cotangent coordinates,

$ω_{c a n} = \sum_{i = 1}^{n} d q^{i} \land d p_{i} .$

Facts & Assumptions

Given: ${A C}_{ω}$ , a smooth $n$ -manifold $Q$ , and the tautological form on $T^{*} Q$ .

[A1]

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

[F1]

In cotangent coordinates, $λ = \sum_{i} p_{i} d q^{i}$ . The tautological one-form is intrinsic and smooth.

Proof

technique · direct

1.1

From [F1], $- d λ = - \sum_{i} d p_{i} \land d q^{i} = \sum_{i} d q^{i} \land d p_{i}$ . Also $d ω_{c a n} = - d^{2} λ = 0$ .

F1algebra

2.1

For $X = \sum_{i} (a^{i} \partial_{q^{i}} + b_{i} \partial_{p_{i}})$ , step 1.1 gives $ι_{X} ω_{c a n} = \sum_{i} (a^{i} d p_{i} - b_{i} d q^{i})$ . This vanishes only when all $a^{i}, b_{i}$ vanish, so the form is nondegenerate. For $n = 0$ the same assertion is vacuous. Thus it is symplectic.

A1F1step 1.1∎

Depends on

Used by

Canonical symplectic model near the zero section of T^*L Definition
The cotangent bundle of a circle as a symplectic cylinder Example
The zero section and cotangent fibres as Lagrangians Example
A graph of a one-form is Lagrangian exactly when the form is closed Proposition
Cotangent lifts are symplectomorphisms Proposition
Hamilton equations in canonical cotangent coordinates Theorem

Dependency tree · two levels

8 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)