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Whitehead's second lemma removes the infinitesimal equivariance obstruction for semisimple actions

Statement

Assume ACω, connected G, connected M, and suppose an infinitesimal moment map μ:Mg is supplied for the action: that is, each closed one-form ιξMω has a chosen Hamiltonian function μξ, depending linearly on ξ. If g is finite-dimensional real semisimple, then there is a covector bg such that

μb: Mg,(μb)ξ=μξb(ξ),

is a coadjoint-equivariant moment map. Thus constants can be added to a supplied infinitesimal moment map to make it equivariant. The argument assumes the linear choice of Hamiltonians and does not prove that the component one-forms ιξMω are exact; it removes only the obstruction to equivariance.

Facts & Assumptions

Given: ACω, connected G and M, an infinitesimal moment map μ for the action, and a finite-dimensional real semisimple g.

[A1]

ACω is countable choice; it is used through the moment-map, defect, and equivariance interfaces cited in [F1], [F2], and [F6].

[F1]

The components satisfy dμξ=ιξMω for all ξ. Moment map, component Hamiltonians and infinitesimal moment maps.

[F2]

The nonequivariance defect c(ξ,η)={μξ,μη}μ[ξ,η] is a constant on M and is a Chevalley--Eilenberg two-cocycle with trivial coefficients. The nonequivariance defect of an infinitesimal moment map is a constant Lie-algebra two-cocycle.

[F3]

For a finite-dimensional semisimple g over a characteristic-zero field, H2(g,M)=0 for every finite-dimensional module M, in particular H2(g,R)=0 for the trivial module. Second Whitehead lemma, Lie algebra cohomology, Simple, semisimple, and reductive Lie algebras.

[F4]

With the zero-based convention, the Chevalley--Eilenberg differential of a one-cochain b is (db)(ξ,η)=b([ξ,η]); hence c=db means c(ξ,η)=b([ξ,η]). Chevalley–Eilenberg differential.

[F5]

Constants are Poisson-central: adding a constant to a function changes no Hamiltonian vector field and the Poisson bracket of a constant with any function vanishes. Poisson bracket on a symplectic manifold.

[F6]

The defect vanishes identically on the connected manifold M if and only if μ is coadjoint equivariant, and for connected G the bracket identity is equivalent to equivariance. The nonequivariance defect of an infinitesimal moment map is a constant Lie-algebra two-cocycle, For connected groups, equivariance is equivalent to the moment-map Poisson bracket identity.

Proof

technique · direct
1.1

By [F1] the map μ is an infinitesimal moment map, so its defect c from [F2] is a constant two-cocycle with trivial coefficients; identifying the trivial module with R, [F3] gives H2(g,R)=0.

F1F2F3given
2.1

Since c represents the zero class and c is a two-cocycle, it is a coboundary c=db for some one-cochain bC1=g, so c(ξ,η)=b([ξ,η]) for all ξ,η by [F4].

step 1.1F4
3.1

Define μ:=μb, that is μξ=μξb(ξ). Its components differ from those of μ by constants, so by [F5] the Poisson bracket is unchanged and the defect of μ is c(ξ,η)={μξ,μη}(μ[ξ,η]b([ξ,η]))=c(ξ,η)+b([ξ,η])=c(ξ,η)(db)(ξ,η)=0.

step 2.1F4F5
4.1

The component moment equations hold for μ as well, because the components differ from those of μ by constants, which have zero differential.

step 3.1F1F5
5.1

Since c0 on the connected manifold M and G is connected, [F6] shows that μ is coadjoint equivariant; combined with step 4.1, μ is an equivariant moment map.

step 3.1step 4.1F6A1

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