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PropositionStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedaudited 2026-09-22
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Invariant Hamiltonians descend to reduced Hamiltonians

Statement

Assume ACω. Let (M,ω,G,μ) be a Hamiltonian G-space with a G-invariant Hamiltonian HC(M)G, let α be a regular value of μ, and suppose Gα acts freely and properly on the level μ1(α), with reduction (Mα,ωα) and quotient map π. Then:

  1. XH is tangent to the level μ1(α) and is Gα-invariant, so it pushes forward to a smooth vector field Y=dπ(XHμ1(α)) on Mα;
  2. Hμ1(α) is Gα-invariant and descends to a unique smooth hC(Mα) with πh=ιH;
  3. Y=Xh; consequently every integral curve of XH that lies in the level projects under π to an integral curve of the flow of h on Mα.

Facts & Assumptions

Given: ACω, a Hamiltonian G-space with invariant Hamiltonian H, a regular value α, and a free proper Gα-action on the level.

[A1]

ACω is countable choice; it is used only through the reduction, fundamental-field and descent suppliers cited below.

[F1]

If H is G-invariant then {μξ,H}=0 for all ξ, and μ is constant along the flow of XH. Noether's conservation law for Hamiltonian actions.

[F2]

XH is the unique field with ιXHω=dH, and XG(F)={F,G} for the Poisson bracket. Hamiltonian vector fields exist uniquely for smooth functions, Poisson bracket on a symplectic manifold.

[F3]

Reduction gives the smooth quotient and πωα=ιω (Marsden--Weinstein--Meyer symplectic reduction). The free proper action quotient theorem makes π a smooth surjective submersion for this quotient structure (Free proper action quotient manifold).

[F5]

Integral curves of a smooth vector field through a given initial point are unique. Through each point there is a unique maximal integral curve.

[F6]

At a regular level, Tpμ1(α)=kerdμp (The tangent space of a regular level set is the kernel).

Proof

technique · direct
1.1

XH is tangent to the level: for every ξ, dμξ(XH)=XH(μξ)={μξ,H}=0 by [F1] and [F2], so XH lies in kerdμ and hence in Tμ1(α) by [F6].

F1F2F6
1.2

XH is invariant under the action: since H is G-invariant, for gG and pM, ωgp(d(ag)pXH(p),d(ag)pv)=ωp(XH(p),v)=dHp(v)=dHgp(d(ag)pv) for all v, so nondegeneracy gives d(ag)pXH(p)=XH(gp).

F1F2given
2.1

By steps 1.1 and 1.2 the field XH is Gα-invariant and tangent to the level, so Y[p]:=dπp(XH(p)) is well defined: for p=hp with hGα one has dπpXH(p)=dπpd(ah)XH(p)=d(πah)pXH(p)=dπpXH(p) because πah=π. It is smooth: by [F4], submersion coordinates for π have the form (u,v)u; fixing v=v0 gives a smooth local section s. There Y=dπXHs, interpreting XH as its tangent restriction to the level, so this local expression is smooth.

step 1.1step 1.2F3F4
2.2

By invariance, Hμ1(α) is constant on the fibres of π, so [F4] gives a unique continuous h:MαR with πh=ιH. Near every point of Mα, the submersion π has a smooth local section s by [F3] and [F4], by fixing the fibre coordinates as above, and there h=Hιs; hence h is smooth.

step 1.2F3F4
3.1

The projected field is the Hamiltonian field of h: for vTpμ1(α), dh[p](dπpv)=d(πh)p(v)=d(ιH)p(v)=dHp(v)=ωp(XH(p),v)=(πωα)p(XH(p),v)=ωα(Y[p],dπpv), using [F3]; since dπp is onto, ιYωα=dh, and uniqueness of Hamiltonian fields [F2] gives Y=Xh.

step 2.1step 2.2F2F3
4.1

If γ is an integral curve of XH lying in the level, then πγ is an integral curve of Y=Xh by the chain rule, and by uniqueness of integral curves [F5] it agrees on its interval of definition with the reduced integral curve through the projected initial point. Thus the restricted flow projects wherever the original curve is defined; no completeness or equality of maximal time intervals is asserted. If the level is empty, there is a unique empty descended function and vector field and every assertion is vacuous.

step 3.1F5A1

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