Alphabeta Math
CorollaryStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-22
How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Semisimple Hamiltonian actions have a unique equivariant moment map when one exists

Statement

Assume ACω and let M be connected. Let g be a finite-dimensional real semisimple Lie algebra, and let a Hamiltonian action of a Lie group G with Lie algebra g on (M,ω) be given. Then there is at most one equivariant moment map for the action: if one equivariant moment map exists, it is the unique one.

Facts & Assumptions

Given: ACω, a connected symplectic G-manifold with g finite-dimensional real semisimple, and an equivariant moment map μ.

[A1]

ACω is countable choice; it is used only through [F1].

[F1]

Any two equivariant moment maps for the same action differ by a constant coadjoint-fixed covector δ∈(g∗)G. Moment maps for one action form an affine space over coadjoint-fixed covectors.

[F2]

If g is finite-dimensional semisimple over a characteristic-zero field then [g,g]=g. Semisimple Lie algebras are centerless and perfect, Simple, semisimple, and reductive Lie algebras.

[F3]

For α∈g∗ and ξ,ζ∈g, the coadjoint action satisfies ddt∣0⟨exp⁡G(tξ)⋅α,ζ⟩=−⟨α,[ξ,ζ]⟩. The coadjoint representation, action and orbits.

Proof

technique · direct
1.1F1given

Let μ1,μ2 be two equivariant moment maps. By [F1] there is δ∈(g∗)G with μ1−μ2=δ. We show δ=0.

2.1F3step 1.1

Since g⋅δ=δ for every g∈G, taking g=exp⁡G(tξ) and differentiating the constant function t↦⟨exp⁡G(tξ)⋅δ,ζ⟩ at t=0 gives 0=−⟨δ,[ξ,ζ]⟩ for all ξ,ζ∈g by [F3].

3.1step 2.1F2A1∎

Thus δ vanishes on the linear span of all brackets, that is on [g,g], which equals g by [F2]. Therefore δ=0 and μ1=μ2; an equivariant moment map, when it exists, is unique.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

26 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