CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedaudited 2026-09-22

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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The zero angular-momentum level has nonfree points and no regular reduction

Statement refuted

On $T^{*} R^{3}$ with the rotation action of $S O (3)$ , the zero level of the angular-momentum moment map is a free $S O (3)$ -space, $0$ is a regular value, and the quotient is the smooth symplectic manifold produced by regular reduction. This is false: the level contains the fixed origin and points with circle stabilizers, and $0$ is a critical value, so those hypotheses fail.

Facts & Assumptions

Given: ${A C}_{ω}$ , $M = T^{*} R^{3}$ with the cotangent-lift rotation action of $S O (3)$ , moment map $μ (q, p) = q \times p$ , and the value $0$ .

[A1]

${A C}_{ω}$ is countable choice; it is used only through the fundamental-field and reduction suppliers.

[F1]

The moment map of the rotation action is $μ (q, p) = q \times p$ under the identification $s o (3)^{*} ≃ R^{3}$ , and it is an equivariant moment map. Angular momentum as the moment map for rotations of a cotangent bundle, SU(2) and SO(3): same local Lie theory, different groups.

[F2]

A value of a moment map is regular exactly when the infinitesimal stabilizers of the points of its level vanish; the reduction theorem requires a regular value and a free proper stabilizer action on the level. Regularity of a moment map is equivalent to local freeness, Marsden--Weinstein--Meyer symplectic reduction.

[F3]

On a regular level the characteristic kernel is exactly the tangent space of the stabilizer orbit. The characteristic kernel on a regular moment level.

Counterexample

technique · direct

1.1

The origin $(q, p) = (0, 0)$ lies on the zero level because $0 \times 0 = 0$ , and it is fixed by every rotation; in particular its stabilizer is all of $S O (3)$ and the infinitesimal stabilizer is the full Lie algebra $s o (3) \neq 0$ .

F1given

2.1

The point $(q, p) = (e_{1}, e_{1})$ also lies on the zero level, because $e_{1} \times e_{1} = 0$ . Its stabilizer is the circle of rotations about the $e_{1}$ -axis: indeed $g \cdot (e_{1}, e_{1}) = (g e_{1}, g e_{1}) = (e_{1}, e_{1})$ exactly when $g e_{1} = e_{1}$ . The orbit of this point therefore has dimension $2$ , while the orbit of the origin has dimension $0$ .

step 1.1F1

2.2

By [F2] the value $0$ is not regular, since its level contains a point, the origin, with nonzero infinitesimal stabilizer; and the action on the level is not free because the origin is fixed. Hence the hypotheses of the reduction theorem both fail at this value.

step 1.1F2

3.1

Consequently no smooth reduced symplectic manifold is obtained for the value $0$ by the theorem: the level is a stratified space whose strata carry orbits of dimensions $0$ and $2$ , and the characteristic-kernel description of regular levels does not apply.

step 2.2F2F3

4.1

The zero angular-momentum level therefore has nonfree points and cannot be reduced by the free proper regular theorem; the false statement is refuted.

step 2.1step 3.1A1∎

Depends on

Used by

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Sources

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