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The tautological cotangent moment map is equivariant

Statement

Assume ${A C}_{ω}$ . For the cotangent-lifted action of $G$ on $T^{*} Q$ and the tautological moment map $⟨ μ (q, p), ξ ⟩ = - p (ξ_{Q} (q))$ , coadjoint equivariance holds:

$μ (\hat{g} (q, p)) = g \cdot μ (q, p) (g \in G, (q, p) \in T^{*} Q) .$

Together with the component equations of the companion proposition this makes $μ$ an equivariant moment map.

Facts & Assumptions

Given: ${A C}_{ω}$ , a smooth left action of $G$ on $Q$ , the lifted action on $T^{*} Q$ , and the tautological moment map.

[A1]

${A C}_{ω}$ is countable choice; it is used only through the fundamental-field interface cited in [F2].

[F1]

The lifted action is $\hat{g} (q, p) = (g \cdot q, (d (a_{g})_{q}^{- 1})^{*} p)$ , the tautological moment map is $⟨ μ (q, p), ξ ⟩ = - p (ξ_{Q} (q))$ , and the lifted action is a smooth left action preserving $ω_{c a n}$ . The cotangent lift of an action is Hamiltonian with the tautological moment map, Cotangent lifts are symplectomorphisms.

[F2]

For every $g \in G$ and $ξ \in g$ the fundamental fields are intertwined by the action: $({Ad}_{g} ξ)_{Q} (g \cdot q) = d (a_{g})_{q} ξ_{Q} (q)$ , equivalently $({Ad}_{g^{- 1}} ξ)_{Q} (q) = (d (a_{g})_{q}^{- 1}) ξ_{Q} (g \cdot q)$ . Adjoint intertwines the exponential map, Fundamental vector fields for a left action, The cotangent lift of an action is Hamiltonian with the tautological moment map.

Proof

technique · direct

1.1F1given

Fix $g \in G$ , $ξ \in g$ and $(q, p) \in T^{*} Q$ . The lifted action acts on the fibre over $g \cdot q$ by the inverse transpose of $d (a_{g})_{q}$ , so $⟨ μ (\hat{g} (q, p)), ξ ⟩ = - ((d (a_{g})_{q}^{- 1})^{*} p) (ξ_{Q} (g \cdot q)) = - p ((d (a_{g})_{q}^{- 1}) ξ_{Q} (g \cdot q)) .$

2.1step 1.1F2

By [F2] the argument of $p$ in step 1.1 is $({Ad}_{g^{- 1}} ξ)_{Q} (q)$ , so $⟨ μ (\hat{g} (q, p)), ξ ⟩ = - p (({Ad}_{g^{- 1}} ξ)_{Q} (q)) = ⟨ μ (q, p), {Ad}_{g^{- 1}} ξ ⟩$ .

3.1step 2.1F1A1∎

By the definition of the coadjoint action, $⟨ μ (q, p), {Ad}_{g^{- 1}} ξ ⟩ = ⟨ g \cdot μ (q, p), ξ ⟩$ ; since $ξ$ was arbitrary and $g, (q, p)$ were arbitrary, $μ (\hat{g} (q, p)) = g \cdot μ (q, p)$ for all $g$ and $(q, p)$ . Hence $μ$ is coadjoint equivariant.

Depends on

Used by

Cotangent reduction for a principal bundle at zero Example
Diagonal action and addition of angular momenta Example
Moment maps are unique without normalization False statement

Dependency tree · two levels

24 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)