CounterexampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedjudge pass (gpt-6.1-sol)

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A closed orbit need not be stable: the trivial multiplicative-group action on a point

Statement refuted

Assume the Axiom of Choice inherited from the named suppliers. The assertion 'every point of a complex affine $G$ -variety whose orbit is closed is stable' is false. Counterexample: $G = G_{m}$ acts trivially on the one-point affine algebraic set $X = {0} = Spec C$ ; the unique orbit is closed, but the stabilizer $G_{0} = G$ is positive-dimensional, so the point is not stable (Stable points of an affine action, Reductive and linearly reductive complex algebraic groups).

Facts & Assumptions

Given: the multiplicative group $G = G_{m} = C^{\times}$ acting trivially on the one-point affine algebraic set $X = {0}$ , so that $t \cdot 0 = 0$ for every $t$ .

[F1]

Stable points. A point $x$ of an affine algebraic set with an algebraic $G$ -action is stable if its orbit $G x$ is closed in $X$ and its stabilizer $G_{x}$ is finite (Stable points of an affine action); by the same definition, $G_{x}$ is finite exactly when $\dim G_{x} = 0$ , so a positive-dimensional stabilizer is infinite.

[F2]

The trivial action is algebraic. An algebraic left action on an affine algebraic set is a morphism $G \times X \to X$ satisfying the identity and associativity laws; the constant map $G \times {0} \to {0}$ is a morphism and satisfies both (Classical complex affine algebraic actions and rational modules).

Verification

technique · direct

1.1F1F2

The trivial action of $G = G_{m}$ on $X = {0}$ is algebraic by [F2]; its unique orbit is ${0}$ , which is closed in $X$ because every subset of the one-point space is closed, and the stabilizer of $0$ is $G_{0} = G$ .

2.1F1step 1.1∎

The group $G_{m} = C^{\times}$ is infinite, since it contains $t$ for every nonzero complex number $t$ ; hence the stabilizer $G_{0}$ is not finite and the point $0$ is not stable by [F1], although its orbit is closed. This refutes the displayed assertion.

Remarks

This is Brion's Example 1.27(1) in its simplest form: the closedness of an orbit does not suffice for stability, and the finite-stabilizer hypothesis of the stable-locus theorem cannot be dropped.
The counterexample is the one prescribed by the AG-ACT-3 design; the companion B-page example ex-gm-quotient-of-affine-plane records the same phenomenon inside a positive-dimensional computation.

Depends on

Used by

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Dependency tree · two levels

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Sources

Michel Brion, Introduction to actions of algebraic groups, Les cours du CIRM 1 (2010), no. 1, 1-22 (standard reference, not scraped)