CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-14

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A proper nonfree action is not a principal bundle

Statement refuted

False claim: every smooth proper Lie-group action makes its orbit projection a principal bundle for that action.

Facts & Assumptions

Given: The standard rotation action of $S O (2)$ on $R^{2}$ .

[F1]

Every continuous action of a compact Lie group on a manifold is proper. Compact Lie-group actions are proper.

[F2]

Freeness means that every stabilizer is trivial. Free and proper Lie-group actions.

[F3]

The group action in a principal bundle is free and transitive on every fibre. Principal g bundle and associated fiber bundle.

Counterexample

technique · constructive

1.1

Let $S O (2)$ act on $R^{2}$ by matrix multiplication. This is a smooth action, and $S O (2)$ is compact, so [F1] makes the action proper.

givenF1construct

2.1

Every rotation fixes the origin. Thus the stabilizer of $0$ is all of $S O (2)$ rather than the trivial group, and the action is not free by [F2].

F2step 1.1

3.1

If the orbit projection $R^{2} \to R^{2} / S O (2)$ were a principal $S O (2)$ -bundle for this action (or for the equivalent right action $x \cdot g = g^{- 1} x$ ), [F3] would make the action free on the fibre over the orbit of $0$ . Step 2.1 contradicts this. Hence properness without freeness does not yield a principal bundle. No choice principle is used.

F3step 1.1step 2.1discharge-construct: counterexample complete∎

Depends on

Used by

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Dependency tree · two levels

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Sources

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