PropositionStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-22

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The moment level is invariant under the coadjoint stabilizer

Statement

Assume ${A C}_{ω}$ . Let $(M, ω, G, μ)$ be a Hamiltonian $G$ -space, let $α \in g^{*}$ and let $G_{α} = {g \in G : g \cdot α = α}$ be the coadjoint stabilizer. Then $G_{α}$ preserves the level:

$g \cdot p \in μ^{- 1} (α) for all g \in G_{α}, p \in μ^{- 1} (α) .$

Facts & Assumptions

Given: ${A C}_{ω}$ , a Hamiltonian $G$ -space with equivariant moment map $μ$ , a covector $α$ , and $g \in G_{α}$ , $p \in μ^{- 1} (α)$ .

[A1]

${A C}_{ω}$ is countable choice; it is used only through the fundamental-field interface of the Hamiltonian action.

[F1]

$μ$ is coadjoint equivariant: $μ (g \cdot p) = g \cdot μ (p)$ . Moment map, component Hamiltonians and infinitesimal moment maps, Symplectic and Hamiltonian Lie-group actions.

[F2]

$G_{α} = {g \in G : g \cdot α = α}$ . The coadjoint representation, action and orbits.

Proof

technique · direct

1.1F1given

By [F1], $μ (g \cdot p) = g \cdot μ (p) = g \cdot α$ because $μ (p) = α$ .

1.2F2given

Since $g \in G_{α}$ , [F2] gives $g \cdot α = α$ .

2.1step 1.1step 1.2A1∎

Combining the two computations, $μ (g \cdot p) = α$ , that is $g \cdot p \in μ^{- 1} (α)$ . As $g$ and $p$ were arbitrary, $G_{α}$ preserves the level.

Depends on

Used by

Marsden--Weinstein--Meyer symplectic reduction Theorem

Dependency tree · two levels

18 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)
Eckhard Meinrenken, Symplectic Geometry (standard reference, not scraped)