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Semisimple Hamiltonian actions have a unique equivariant moment map when one exists
Statement
Assume and let be connected. Let be a finite-dimensional real semisimple Lie algebra, and let a Hamiltonian action of a Lie group with Lie algebra on 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: , a connected symplectic -manifold with finite-dimensional real semisimple, and an equivariant moment map .
is countable choice; it is used only through [F1].
Any two equivariant moment maps for the same action differ by a constant coadjoint-fixed covector . Moment maps for one action form an affine space over coadjoint-fixed covectors.
If is finite-dimensional semisimple over a characteristic-zero field then . Semisimple Lie algebras are centerless and perfect, Simple, semisimple, and reductive Lie algebras.
For and , the coadjoint action satisfies . The coadjoint representation, action and orbits.
Proof
Let be two equivariant moment maps. By [F1] there is with . We show .
Since for every , taking and differentiating the constant function at gives for all by [F3].
Thus vanishes on the linear span of all brackets, that is on , which equals by [F2]. Therefore and ; an equivariant moment map, when it exists, is unique.
Depends on
Used by
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Dependency tree · two levels
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Sources
- Ana Cannas da Silva, Lectures on Symplectic Geometry (standard reference, not scraped)
- Eckhard Meinrenken, Symplectic Geometry (standard reference, not scraped)