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LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6.1-sol)
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Expansion of a root-group translate of a weight vector

Statement

Assume the Axiom of Choice inherited from the named suppliers. Let (V,r) be a rational representation of a split reductive group (G,T), let v∈Vλ be a weight vector, and let α∈Φ with root-group isomorphism uα:Ga→Uα (Root subgroups of a split reductive group). These coordinates satisfy t uα(c) t−1=uα(α(t)c) over every k-algebra. Then there are vectors vi∈Vλ+iα (i≥1), only finitely many nonzero, such that uα(c)⋅v=v+∑i≥1civi for every c (and every k-algebra after base change). In particular the orbit map c↦uα(c)⋅v is polynomial with constant term v and higher coefficients in the stated weight spaces.

Facts & Assumptions

Given: A split reductive group (G,T) with root α∈Φ, the root-group isomorphism uα:Ga→Uα, a rational representation (V,r) and a weight vector v∈Vλ (Weights, dominant weights and the highest-weight order of a rational representation).

[F1]

Finite-dimensional orbit. The vector v lies in a finite-dimensional subcomodule W⊆V; the action of Uα preserves W, so uα(c)⋅v∈W for all c (Every element of a comodule lies in a finite-dimensional subcomodule, Rational representations and comodules of an affine group scheme).

[F2]

Rational action is polynomial. For a finite-dimensional rational representation W of Ga the matrix coefficients of c↦r(uα(c)) are polynomial functions of c; hence c↦uα(c)⋅v is given by a polynomial in c with values in W, and a basis of W exhibits it as uα(c)⋅v=∑i≥0civi with vi∈W, only finitely many nonzero (Rational representations and comodules of an affine group scheme).

[F3]

Conjugation formula. For t∈T(R) and c∈R one has t uα(c) t−1=uα(α(t)c) (Root subgroups of a split reductive group, Roots and root groups of a split reductive group). To justify the formula from the supplied T-stability and Lie weight, work with the universal torus point over A=k[x1±1,…,xr±1], a domain. Conjugation induces a polynomial automorphism of A[c] fixing c=0; its polynomial inverse and the degree-of-composition identity over a domain force it to be c↦a(t)c, with a(t) a unit. Its differential at c=0 is the adjoint character α(t), hence a(t)=α(t). This universal identity specializes to every R, including nonreduced base algebras.

[F4]

Weight vectors. t⋅v=λ(t)v for all t∈T(R), and the weight spaces Vχ are the eigenspaces of the T-action (Weights, dominant weights and the highest-weight order of a rational representation).

Proof

technique · direct
1.1F1F2given

By [F1] and [F2] there is a finite expansion uα(c)⋅v=∑i≥0civi with vi∈W and only finitely many nonzero. Setting c=0 gives v0=v.

2.1F3F4step 1.1

For t∈T(R) one has t⋅(uα(c)⋅v)=(t uα(c) t−1)⋅(t⋅v)=uα(α(t)c)⋅(λ(t)v)=λ(t)∑i≥0α(t)icivi using [F3] and [F4].

3.1F4step 1.1step 2.1

On the other hand t⋅(uα(c)⋅v)=∑i≥0ci(t⋅vi) by linearity of the action over R[c]. Comparing coefficients of ci in the two polynomial expressions for all R-points t shows t⋅vi=λ(t)α(t)ivi for every i, that is, vi∈Vλ+iα: the character χi with t⋅vi=χi(t)vi satisfies χi(t)=λ(t)α(t)i for all t and all R. Together with step 1.1 this gives the asserted expansion.

4.1step 1.1step 3.1∎

Since v0=v and vi∈Vλ+iα for i≥1 with only finitely many nonzero, the orbit map c↦uα(c)⋅v is polynomial with constant term v and higher coefficients in the stated weight spaces.

Remarks

  • The base-change clause of the statement is included because the computation is carried out for an arbitrary k-algebra R of values of c and points t∈T(R); the expansion is the same polynomial for every R.
  • If vi=0 for all i≥1 the vector v is fixed by Uα; the vanishing of all higher coefficients is what makes a primitive vector fixed by every positive root group.

Depends on

Used by

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Sources