LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-22

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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.

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The differential of the moment map and the orbit-orthogonal identity

Statement

Assume ${A C}_{ω}$ . Let a Hamiltonian action of $G$ on $(M, ω)$ have moment map $μ$ , and let $p \in M$ . Then

$\ker d μ_{p} = (T_{p} (G \cdot p))^{ω}, im d μ_{p} = ann (g_{p}),$

where $(T_{p} (G \cdot p))^{ω}$ is the symplectic orthogonal of the tangent space of the orbit of $p$ and $g_{p}$ is the infinitesimal stabilizer.

Facts & Assumptions

Given: ${A C}_{ω}$ , a Hamiltonian $G$ -space with moment map $μ$ , and a point $p \in M$ .

[A1]

${A C}_{ω}$ is countable choice; it is used only through the orbit and fundamental-field interface of the two suppliers cited in [F3], both of which carry the same assumption.

[F1]

$d μ^{ξ} = - ι_{ξ_{M}} ω$ for every $ξ \in g$ . Moment map, component Hamiltonians and infinitesimal moment maps.

[F3]

The infinitesimal orbit map $ξ \mapsto ξ_{M} (p)$ has image $T_{p} (G \cdot p)$ , the tangent space of the orbit with its canonical structure, and kernel $g_{p}$ . Kernel of the infinitesimal orbit map, Every orbit is an injectively immersed homogeneous space.

[F4]

For a subspace $W$ of a finite-dimensional symplectic vector space, $\dim W + \dim W^{ω} = \dim V$ and $(W^{ω})^{ω} = W$ . Symplectic double-orthogonal and dimension identities, Symplectic orthogonal complement.

Proof

technique · direct

1.1F1F3

For $v \in T_{p} M$ and $ξ \in g$ , [F1] gives $⟨ d μ_{p} (v), ξ ⟩ = d μ_{p}^{ξ} (v) = - ω_{p} (ξ_{M} (p), v)$ . Hence $d μ_{p} (v) = 0$ if and only if $ω_{p} (ξ_{M} (p), v) = 0$ for every $ξ$ , that is, if and only if $v$ is symplectically orthogonal to the span of the values $ξ_{M} (p)$ ; by [F3] that span is $T_{p} (G \cdot p)$ . Therefore $\ker d μ_{p} = (T_{p} (G \cdot p))^{ω}$ .

2.1step 1.1F3

The image is contained in the annihilator: if $ξ \in g_{p}$ , then $ξ_{M} (p) = 0$ by [F3], so for every $v \in T_{p} M$ the same identity gives $⟨ d μ_{p} (v), ξ ⟩ = - ω_{p} (0, v) = 0$ , so $im d μ_{p} \subseteq ann (g_{p})$ .

3.1step 1.1step 2.1F3F4A1∎

Dimension count: by step 1.1 and [F4], $\dim \ker d μ_{p} = \dim M - \dim T_{p} (G \cdot p)$ , and by [F3] $\dim T_{p} (G \cdot p) = \dim g - \dim g_{p}$ . Hence $\dim im d μ_{p} = \dim g - \dim g_{p} = \dim ann (g_{p})$ . Since step 2.1 gives containment between spaces of equal dimension, $im d μ_{p} = ann (g_{p})$ .

Depends on

Used by

Dependency tree · two levels

25 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

Eckhard Meinrenken, Symplectic Geometry (standard reference, not scraped)
Ana Cannas da Silva, Lectures on Symplectic Geometry (standard reference, not scraped)