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Reductive Affine Invariant Theory and Geometric Quotients — Examples

1 · Prerequisites

2 · Summary

The hyperbolic example computes an invariant ring in full. For Gm=C× acting on the plane by t⋅(x,y)=(tx,t−1y) the monomial xayb is scaled by tb−a, so a single test value already forces every off-diagonal coefficient of an invariant polynomial to vanish; the invariant ring is C[xy], a polynomial ring in one variable, and the categorical quotient of the plane is π(x,y)=xy with target the affine line.

The worked quotient example then reads off the orbit picture: the non-zero fibres of π are the hyperbolae xy=c, each a single closed orbit; the two punctured coordinate axes are orbits whose closures contain the origin; and the origin is fixed, hence closed, with stabilizer all of Gm. The stable locus is therefore {xy≠0}, and there the explicit product decomposition (x,y)↦(x,xy), with inverse (t,c)↦(t,c/t), exhibits the quotient as the projection of a trivial Gm-bundle over C× with the structure group acting on the first factor. The origin is unstable although its orbit is closed.

The counterexample isolates that last phenomenon in its simplest form: the trivial action of Gm on a one-point set has a closed orbit but an infinite stabilizer, so closedness of the orbit alone does not imply stability. The final remark compares hypotheses rather than proving a theorem: Noether's finiteness theorem for finite groups requires finiteness, which fails for Gm, so it is not a substitute for the Reynolds-operator proof of finite generation for positive-dimensional reductive groups; no reduction of a reductive action to a finite-group action is asserted. The Axiom of Choice is inherited from the quotient and stable-locus suppliers used by the worked example.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6.1-sol)Open item page →

Invariants of the hyperbolic action of the multiplicative group on the plane

Statement

Let Gm=C× act on C2 by t⋅(x,y)=(tx,t−1y) (Classical complex affine algebraic actions and rational modules). Then the invariant subring of C[x,y] (The coordinate ring of a classical affine algebraic set) is C[x,y]Gm=C[xy], a polynomial ring in one variable; explicitly, a polynomial f=∑a,bcabxayb is invariant if and only if cab=0 for every pair with a≠b (Monomials, coefficients, degree in each variable and total degree in F[x1,…,xn]).

Facts & Assumptions

Given: the multiplicative group Gm=C× acting on C2 by t⋅(x,y)=(tx,t−1y), and a polynomial f=∑a,bcabxayb∈C[x,y] written in its unique finite monomial expansion.

[F1]

The action and the action on functions. The map t⋅(x,y)=(tx,t−1y) is an algebraic action of Gm on the affine plane, and the induced left action on functions is (t⋅f)(p)=f(t−1⋅p), so that (t⋅f)(x,y)=f(t−1x,ty) (Classical complex affine algebraic actions and rational modules).

[F2]

Unique monomial expansion. Every polynomial in F[x1,…,xn] has a unique finite expansion f=∑tctxt over finitely many multi-indices; in particular the monomials are linearly independent over F and a polynomial is zero exactly when all its coefficients vanish (Monomials, coefficients, degree in each variable and total degree in F[x1,…,xn]).

[F3]

The coordinate ring of the plane. For an affine algebraic set X⊆kn the coordinate ring is k[X]=k[x1,…,xn]/I(X); for X=C2 this is C[x,y], with x and y the coordinate functions (The coordinate ring of a classical affine algebraic set).

Proof

technique · direct
1.1F1

For t∈Gm and a monomial xayb the function action of [F1] gives (t⋅(xayb))(x,y)=(t−1x)a(ty)b=tb−axayb, so the multiplicative group acts on each monomial by the scalar tb−a: monomials on the diagonal a=b are fixed, while off-diagonal monomials are scaled by a nonconstant character.

2.1F1F2step 1.1

An element f=∑a,bcabxayb∈C[x,y] is invariant under Gm if and only if ∑a,bcab(tb−a−1)xayb=0 for every t∈Gm; by step 1.1 this is the difference between t⋅f and f, and by uniqueness of the monomial expansion it holds if and only if cabtb−a=cab for all pairs (a,b) and all t∈Gm.

3.1F2step 2.1

The coefficient conditions hold if and only if cab=0 for every pair with a≠b. If f is invariant, fix a pair with a≠b; evaluating the condition of step 2.1 at the nonzero complex number t=2 gives cab(2b−a−1)=0, and 2b−a≠1 because 2b−a>1 for b>a and 0<2b−a<1 for b<a; hence cab=0. Conversely, if cab=0 whenever a≠b, then f=∑acaxaya is a sum of diagonal monomials, and by step 1.1 each such monomial satisfies t⋅(xaya)=ta−axaya=xaya, so t⋅f=f for every t and f is invariant.

4.1F2F3step 3.1∎

Consequently a polynomial f∈C[x,y] is Gm-invariant if and only if its expansion has no off-diagonal terms, that is, if and only if f=∑acaxaya lies in the subalgebra C[xy] generated by the invariant function xy. Hence C[x,y]Gm=C[xy]. The substitution φ:C[z]→C[x,y], z↦xy, is injective: if φ(g)=0 for g=∑ndnzn, then ∑ndnxnyn=0, and uniqueness of the monomial expansion forces every dn=0, so g=0. Therefore C[xy]≅C[z] is a polynomial ring in one variable, and the plane's invariant ring is C[xy].

Remarks

  • This is the invariant computation behind Brion's Example 1.27(2): the quotient of C2 by the hyperbolic Gm-action is the affine line with coordinate xy. The companion example uses exactly this lemma to identify the invariant ring with C[xy] and to describe the quotient map (x,y)↦xy; that the raw level set {xy=0} contains both the origin and the two coordinate axes is the phenomenon behind the stable-locus analysis there.
  • No choice is used: a single test value t=2 already separates every off-diagonal monomial from the invariant ones, so no infinite family of excluded values has to be avoided.
CounterexampleConstruction: AI-generatedVerification: AI-generatedjudge pass (gpt-6.1-sol)Open item page →

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=Gm acts trivially on the one-point affine algebraic set X={0}=Spec⁡C; the unique orbit is closed, but the stabilizer G0=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=Gm=C× acting trivially on the one-point affine algebraic set X={0}, so that t⋅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 Gx is closed in X and its stabilizer Gx is finite (Stable points of an affine action); by the same definition, Gx is finite exactly when dim⁡Gx=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×X→X satisfying the identity and associativity laws; the constant map G×{0}→{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=Gm 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 G0=G.

2.1F1step 1.1∎

The group Gm=C× is infinite, since it contains t for every nonzero complex number t; hence the stabilizer G0 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.
RemarkRemark: AI-adaptedProof: Not applicablejudge pass (gpt-6.1-sol)Open item page →

The finite-group Noether theorem does not supply invariant finite generation for positive-dimensional groups

Remark

Noether's finiteness theorem for invariants of a finite group (Noether's finiteness theorem: the invariants of a finite group acting on a finite-type algebra over a Noetherian ring form an algebra of finite type) requires the acting group to be finite. It therefore does not supply the finite generation of invariant rings of positive-dimensional groups: applied to the action of Gm=C× on C2 of Invariants of the hyperbolic action of the multiplicative group on the plane the hypothesis fails, since Gm is not finite (Reductive and linearly reductive complex algebraic groups), even though the invariant ring there is C[xy]; and no argument on this page reduces a reductive group action to a finite-group action. The finite-group theorem is thus not a substitute for the Reynolds-operator proof of finite generation for complex reductive groups given by the bridge theorem of this page.

The remark is a hypothesis comparison, not a proof step: it records the design caveat that the finite-group Noether bound is not invoked anywhere in the finite-generation argument for reductive G, where the Reynolds operator and the graded ideal argument do all the work. No reduction of a Gm- or reductive action to a finite-group action is asserted or used.

ExampleConstruction: Literature-sourcedVerification: AI-adaptedOpen item page →

The quotient of the plane by the hyperbolic multiplicative-group action

Example

Assume the Axiom of Choice inherited from the named suppliers. In the setting of Invariants of the hyperbolic action of the multiplicative group on the plane, let G=Gm act on X=C2 by t⋅(x,y)=(tx,t−1y). Then: (i) the invariant ring is C[X]G=C[xy], so the categorical quotient is π:X→X/ ⁣/G=C, π(x,y)=xy (Finite generation of invariants and the affine categorical quotient); (ii) the orbits are the hyperbolae xy=c for c≠0, the two punctured coordinate axes {x=0,y≠0}, {y=0,x≠0}, and the origin; the two punctured-axis orbits are not closed and their closures are the corresponding axes through the origin; the origin is closed and is the unique closed orbit in the fibre π−1(0); (iii) the stable locus is Xs={xy≠0}, π(Xs)=C×, and πs:Xs→C× is a geometric quotient, a principal Gm-bundle (The stable locus has a geometric quotient, Stable points of an affine action); the origin is unstable although its orbit is closed, its stabilizer being all of Gm.

Facts & Assumptions

Given: the group G=Gm=C× acting on X=C2 by t⋅(x,y)=(tx,t−1y), its invariant ring C[X]G=C[xy], and the quotient π:X→X/ ⁣/G, π(x,y)=xy.

[F1]

The invariant ring. C[x,y]Gm=C[xy], a polynomial ring in one variable (Invariants of the hyperbolic action of the multiplicative group on the plane).

[F2]

The affine quotient. For a complex reductive group with an algebraic action on an affine algebraic set, the invariant ring is finitely generated, the quotient π is a categorical quotient, and every fibre of π contains exactly one closed orbit (Finite generation of invariants and the affine categorical quotient).

[F3]

The stable locus is a geometric quotient. The stable locus Xs is open and saturated, and πs:Xs→π(Xs) is a geometric quotient with fibres exactly the orbits, each orbit in Xs closed in X (The stable locus has a geometric quotient).

[F4]

Stable points. A point is stable exactly when its orbit is closed in X and its stabilizer is finite, equivalently of dimension zero (Stable points of an affine action).

[F5]

Geometric quotient. A G-invariant morphism is a geometric quotient when it is surjective with fibres exactly the orbits, defines the quotient topology on its image, and pulls regular functions back isomorphically onto the invariant regular functions (Categorical and geometric quotients of classical varieties).

[F6]

Reductivity of Gm. A complex affine group is reductive when it has no non-trivial closed normal subgroup for which every non-zero rational module has a non-zero fixed vector (Reductive and linearly reductive complex algebraic groups).

Verification

technique · direct
1.1F1F2F6

First Gm is reductive: every non-trivial subgroup U has an element u≠1, and its rational scalar action on the one-dimensional space C has no non-zero fixed vector, since (u−1)v=0 forces v=0. Thus no such closed subgroup is unipotent in the fixed-vector sense of [F6]. Part (i): by [F1] the invariant ring is C[xy], a polynomial ring in the single variable xy, and by [F2] the quotient is π:X→X/ ⁣/G=C with π(x,y)=xy, a categorical quotient.

1.2F1F2

Part (ii): for c≠0 the fibre π−1(c)={(x,y):xy=c} is a single orbit, because for (x,y),(x′,y′) with xy=x′y′=c the element t=x′/x is nonzero and t⋅(x,y)=(x′,t−1y)=(x′,(x/x′)y)=(x′,c/x′)=(x′,y′). The hyperbola xy=c is closed in the plane, being the zero set of the polynomial xy−c. On {x=0,y≠0} the action is transitive by the second coordinate and the orbit is not closed: its closure contains the origin, and a polynomial vanishing on {x=0,y≠0} vanishes on the whole axis {x=0} because a one-variable polynomial with infinitely many roots is zero; similarly for {y=0,x≠0}. The origin is fixed by the action, so it is an orbit, it is closed, and it is the unique closed orbit in the zero fibre by [F2]; the zero fibre consists of exactly the two punctured axes and the origin.

2.1F4step 1.2

Stabilizers and stability: at a point with x≠0 the condition tx=x gives t=1, and at a point with y≠0 the condition t−1y=y gives t=1; hence every point of X∖{0} has trivial stabilizer, while the origin has stabilizer Gm=C×, which is infinite. By [F4] and step 1.2 the stable points are exactly those with xy≠0: there the orbit is a closed hyperbola and the stabilizer is trivial, while the punctured-axis points and the origin are unstable.

3.1F3step 2.1

Part (iii): by step 2.1 the stable locus is Xs={xy≠0} and π(Xs)=C×, and by [F3] the restriction πs:Xs→C× is a geometric quotient whose fibres are exactly the orbits and whose orbits are closed in X.

4.1F5step 3.1

The bundle statement: the map Φ:Gm×Gm→Xs, Φ(t,c)=(t,c/t), is an isomorphism with inverse (x,y)↦(x,xy), and it is G-equivariant for the action on the first factor because Φ(t′t,c)=(t′t,c/(t′t))=t′⋅(t,c/t)=t′⋅Φ(t,c). Under Φ the quotient πs corresponds to the projection Gm×Gm→Gm, (t,c)↦c, which is a trivial principal Gm-bundle with structure group acting on the first factor and a geometric quotient; so πs is a principal Gm-bundle in this explicit trivialized sense, and no identification with any other principal-bundle theory is used.

5.1F2F3step 1.1step 1.2step 3.1step 4.1∎

Assembly: (i) is step 1.1, (ii) is step 1.2 together with the closed-orbit count of [F2], (iii) is steps 2.1, 3.1 and 4.1; the origin is unstable although its orbit is closed because its stabilizer is all of Gm. All Axiom of Choice content is inherited from the affine quotient and stable-locus suppliers.

Remarks

  • This is Brion's Example 1.27(2) (printed p. 10). The computation exhibits directly why the stable locus must exclude the punctured axes as well as the origin: the axes have non-closed orbits and the origin has an infinite stabilizer.
  • The phrase "principal bundle". Only the explicit product decomposition Φ:Gm×Gm→Xs of step 4.1 is asserted, with the structure group acting on the first factor; this is a trivial bundle in the elementary sense. The topological definition of a principal bundle is a different register and is not invoked, and no algebraic principal-bundle theory is assumed anywhere in this pair.

Sources