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

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A graph of a one-form is Lagrangian exactly when the form is closed

Statement

Assume ${A C}_{ω}$ . For $α \in Ω^{1} (Q)$ , its graph $s_{α} (Q) \subset (T^{*} Q, - d λ)$ is Lagrangian if and only if $d α = 0$ .

Facts & Assumptions

Given: ${A C}_{ω}$ , a smooth $n$ -manifold $Q$ , and $α \in Ω^{1} (Q)$ .

[A1]

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

[F1]

The canonical form on $T^{*} Q$ is $ω_{c a n} = - d λ$ . The canonical cotangent two-form is symplectic.

[F2]

A submanifold is Lagrangian when its tangent spaces are Lagrangian. Isotropic, coisotropic, symplectic, and Lagrangian submanifolds.

[F3]

An isotropic half-dimensional subspace is Lagrangian. Equivalent characterizations of Lagrangian subspaces.

[F4]

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

Proof

technique · direct

1.1F1F4algebra

Since $π \circ s_{α} = {id}_{Q}$ , the tautological formula gives $s_{α}^{*} λ = α$ . Therefore [F1] and [F4] give $s_{α}^{*} ω_{c a n} = - d α$ .

2.1A1F2F3step 1.1∎

The graph section is an embedding and its image has dimension $n$ , half of $\dim T^{*} Q = 2 n$ . By [F2]--[F3], it is Lagrangian exactly when the pulled-back symplectic form vanishes. Step 1.1 says this occurs exactly when $d α = 0$ , proving both directions, including $n = 0$ .

Depends on

Used by

Graphs of exact and closed one-forms as Lagrangians Example

Dependency tree · two levels

17 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

Eckhard Meinrenken, Symplectic Geometry (standard reference, not scraped)