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CorollaryStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-22
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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 αg and ξ,ζg, the coadjoint action satisfies ddt0expG(tξ)α,ζ=α,[ξ,ζ]. The coadjoint representation, action and orbits.

Proof

technique · direct
1.1

Let μ1,μ2 be two equivariant moment maps. By [F1] there is δ(g)G with μ1μ2=δ. We show δ=0.

F1given
2.1

Since gδ=δ for every gG, taking g=expG(tξ) and differentiating the constant function texpG(tξ)δ,ζ at t=0 gives 0=δ,[ξ,ζ] for all ξ,ζg by [F3].

F3step 1.1
3.1

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.

step 2.1F2A1

Depends on

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