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An invariant horizontal form on a free proper quotient descends uniquely

Statement

Assume ${A C}_{ω}$ . Let a Lie group $G$ act smoothly, freely and properly on a smooth manifold $M$ with quotient map $π : M \to M / G$ . A smooth $k$ -form $σ$ on $M$ is the pullback $σ = π^{*} \overset{ˉ}{σ}$ of a smooth $k$ -form on $M / G$ if and only if

$σ$ is $G$ -invariant: $a_{g}^{*} σ = σ$ for all $g \in G$ , where $a_{g} (x) = g \cdot x$ ; and
$σ$ is horizontal: $σ_{x} (v_{1}, \dots, v_{k}) = 0$ whenever some $v_{i}$ is tangent to the orbit $G \cdot x$ ; equivalently $ι_{ξ_{M}} σ = 0$ for every fundamental field $ξ_{M}$ .

The form $\overset{ˉ}{σ}$ is then unique.

Facts & Assumptions

Given: ${A C}_{ω}$ , a smooth free proper action of $G$ on $M$ , the quotient map $π$ , and a smooth $k$ -form $σ$ on $M$ .

[A1]

${A C}_{ω}$ is countable choice; it is used only through the quotient and slice suppliers cited below.

[F1]

$M / G$ is a smooth manifold and $π$ is a smooth surjective submersion. Free proper action quotient manifold.

[F2]

$\ker d π_{x} = T_{x} (G \cdot x)$ , and $π \circ a_{g} = π$ for all $g$ . Tangent space of a free proper quotient, Free proper action quotient manifold.

[F3]

Every $x$ has a slice $S$ such that $G \times S \to G \cdot S$ is a diffeomorphism onto an open saturated neighbourhood and $q ∣_{S}$ is a diffeomorphism onto its image. Local slice for a free proper action, Free proper action quotient manifold.

[F4]

A submersion admits smooth local sections, and a surjective submersion is a quotient map; a form on the base that pulls back to zero is zero because $d π$ is pointwise onto. The constant-rank theorem for manifolds, For a quotient map $q : X \to Y$ , a map out of $Y$ is continuous iff its composite with $q$ is; a continuous map on $X$ constant on the fibres of $q$ factors uniquely through $q$ ; and a composite of quotient maps is a quotient map.

Proof

technique · direct

1.1F1F2

The conditions are necessary. If $σ = π^{*} \overset{ˉ}{σ}$ , then $a_{g}^{*} σ = a_{g}^{*} π^{*} \overset{ˉ}{σ} = (π \circ a_{g})^{*} \overset{ˉ}{σ} = π^{*} \overset{ˉ}{σ} = σ$ by [F2], so $σ$ is $G$ -invariant. If $v = ξ_{M} (x)$ is vertical, then $d π_{x} v = 0$ by [F2] and therefore $σ_{x} (v, w_{2}, \dots, w_{k}) = {\overset{ˉ}{σ}}_{π (x)} (0, d π w_{2}, \dots, d π w_{k}) = 0$ , so $σ$ is horizontal.

1.2F1F3given

For the converse, fix $x_{0} \in M$ and let $V : = π (G \cdot S)$ for a slice $S$ through $x_{0}$ from [F3]; $V$ is open in $M / G$ . For $π (x) \in V$ and $v_{1}, \dots, v_{k} \in T_{π (x)} (M / G)$ , choose lifts ${\tilde{v}}_{i} \in T_{x} M$ with $d π_{x} {\tilde{v}}_{i} = v_{i}$ and set ${\overset{ˉ}{σ}}_{π (x)} (v_{1}, \dots, v_{k}) : = σ_{x} ({\tilde{v}}_{1}, \dots, {\tilde{v}}_{k}) .$

2.1step 1.2F2

The prescription of step 1.2 does not depend on the lifts: two lifts of the same $v_{i}$ differ by an element of $\ker d π_{x} = T_{x} (G \cdot x)$ by [F2], and expanding multilinearly every resulting difference term contains a vertical entry, hence vanishes by horizontality.

2.2step 1.2F2

It does not depend on the chosen point in the fibre: if $x^{'} = h \cdot x$ and ${\tilde{v}}_{i}$ are lifts at $x$ , then $d (a_{h}) {\tilde{v}}_{i}$ are lifts of the same $v_{i}$ at $x^{'}$ by [F2], and $G$ -invariance of $σ$ gives $σ_{x^{'}} (d (a_{h}) {\tilde{v}}_{1}, \dots) = σ_{x} ({\tilde{v}}_{1}, \dots)$ .

3.1step 2.1step 2.2F4

Hence $\overset{ˉ}{σ}$ is well defined on all of $M / G$ . It is smooth: near any point of $M / G$ the submersion $π$ admits a smooth local section $s$ by [F4], and there $\overset{ˉ}{σ} = s^{*} σ$ because $d s$ provides the lifts; the local definitions agree on overlaps since both pull back to $σ$ , and forms on the base with equal pullback are equal by [F4].

4.1step 3.1F4A1∎

By construction $π^{*} \overset{ˉ}{σ} = σ$ . If ${\overset{ˉ}{σ}}^{'}$ is another such form, then $π^{*} (\overset{ˉ}{σ} - {\overset{ˉ}{σ}}^{'}) = 0$ , so $\overset{ˉ}{σ} - {\overset{ˉ}{σ}}^{'} = 0$ by [F4]; the descended form is therefore unique.

Depends on

Used by

Marsden--Weinstein--Meyer symplectic reduction Theorem

Dependency tree · two levels

34 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

John M. Lee, Introduction to Smooth Manifolds, 2nd ed. (standard reference, not scraped)
Eckhard Meinrenken, Symplectic Geometry (standard reference, not scraped)