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Products and opposites of symplectic moment maps

Statement

Assume ${A C}_{ω}$ . Let $(M, ω_{M})$ and $(N, ω_{N})$ be Hamiltonian $G$ -spaces with equivariant moment maps $μ_{M}$ and $μ_{N}$ .

On the product $M \times N$ with the diagonal action $g \cdot (p, q) = (g \cdot p, g \cdot q)$ and the product form $Ω = {pr}_{M}^{*} ω_{M} + {pr}_{N}^{*} ω_{N}$ (Products and opposites of symplectic manifolds), the map $μ (p, q) : = μ_{M} (p) + μ_{N} (q)$ is an equivariant moment map, with components $μ^{ξ} (p, q) = μ_{M}^{ξ} (p) + μ_{N}^{ξ} (q)$ .
On $(M, - ω_{M})$ with the same action, $- μ_{M}$ is an equivariant moment map: component equations and equivariance are those of $μ_{M}$ with the signs of the form and the map reversed.

Facts & Assumptions

Given: ${A C}_{ω}$ , Hamiltonian $G$ -spaces $(M, ω_{M}, μ_{M})$ and $(N, ω_{N}, μ_{N})$ with equivariant moment maps.

[A1]

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

[F1]

$Ω = {pr}_{M}^{*} ω_{M} + {pr}_{N}^{*} ω_{N}$ is symplectic on $M \times N$ , and $(M, - ω_{M})$ is symplectic. Products and opposites of symplectic manifolds.

[F2]

The fundamental field of a product action is the pair of fundamental fields: $ξ_{M \times N} (p, q) = (ξ_{M} (p), ξ_{N} (q))$ , and on $(M, - ω_{M})$ the fundamental field is unchanged, equal to $ξ_{M}$ . Fundamental vector fields for a left action, Symplectic and Hamiltonian Lie-group actions.

[F3]

$μ_{M}, μ_{N}$ are equivariant moment maps: $d μ_{M}^{ξ} = - ι_{ξ_{M}} ω_{M}$ , $d μ_{N}^{ξ} = - ι_{ξ_{N}} ω_{N}$ , $μ_{M} (g \cdot p) = g \cdot μ_{M} (p)$ and $μ_{N} (g \cdot q) = g \cdot μ_{N} (q)$ . Moment map, component Hamiltonians and infinitesimal moment maps.

Proof

technique · direct

1.1

On the product, the contraction of the product form with the fundamental field splits: by [F1] and [F2], $ι_{ξ_{M \times N}} Ω = {pr}_{M}^{*} (ι_{ξ_{M}} ω_{M}) + {pr}_{N}^{*} (ι_{ξ_{N}} ω_{N}),$ because each summand of $Ω$ is pulled back from one factor and the fundamental field has the corresponding component there.

F1F2

1.2

Equivariance of $μ$ : $μ (g \cdot (p, q)) = μ_{M} (g \cdot p) + μ_{N} (g \cdot q) = g \cdot μ_{M} (p) + g \cdot μ_{N} (q) = g \cdot (μ_{M} (p) + μ_{N} (q)) = g \cdot μ (p, q)$ by linearity of the coadjoint action.

2.1

Hence, using [F3], $d μ^{ξ} = {pr}_{M}^{*} d μ_{M}^{ξ} + {pr}_{N}^{*} d μ_{N}^{ξ} = - {pr}_{M}^{*} (ι_{ξ_{M}} ω_{M}) - {pr}_{N}^{*} (ι_{ξ_{N}} ω_{N}) = - ι_{ξ_{M \times N}} Ω,$ so the components of $μ = μ_{M} + μ_{N}$ satisfy the component moment equations for the diagonal action.

step 1.1F3

2.2

For the opposite form, [F3] and [F2] give, with $ν : = - μ_{M}$ , $d ν^{ξ} = - d μ_{M}^{ξ} = ι_{ξ_{M}} ω_{M} = - ι_{ξ_{M}} (- ω_{M}),$ so $ν$ satisfies the component equations on $(M, - ω_{M})$ ; and $ν (g \cdot p) = - μ_{M} (g \cdot p) = - g \cdot μ_{M} (p) = g \cdot ν (p)$ by linearity of the coadjoint action, so $ν$ is equivariant.

step 1.2F2F3

3.1

Steps 2.1 and 1.2 show that $μ_{M} + μ_{N}$ is an equivariant moment map on the product, and step 2.2 that $- μ_{M}$ is an equivariant moment map on $(M, - ω_{M})$ .

step 2.1step 1.2step 2.2A1∎

Depends on

Used by

Diagonal action and addition of angular momenta Example
The shifting trick identifies reduction at a value with a zero reduction Proposition

Dependency tree · two levels

18 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)