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The normalizer of the torus permutes weight spaces

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) and let n∈NG(T)(R) for a k-algebra R. If v∈Vλ, then n⋅(v⊗1)∈(V⊗kR)nλ, where nλ:TR→Gm,R is the character t↦λ(n−1tn), tested over every R-algebra. If R is disconnected, this character can vary between components; the target denotes its eigensubmodule over R. Consequently, for every w∈W(G,T) the weight spaces Vλ and Vwλ have the same dimension, and the set of weights of (V,r) is stable under W (The Weyl group, Borel subgroups and chambers, The root datum of a split reductive group).

Facts & Assumptions

Given: A rational representation (V,r) of the split reductive group (G,T), a weight vector v∈Vλ and a point n∈NG(T)(R) for a k-algebra R.

[F1]

The character nλ. For n∈NG(T)(R) the assignment t↦λ(n−1tn) is a character nλ∈X(T)R of TR: it is a morphism in t and multiplicative because conjugation is a group homomorphism (The root datum of a split reductive group, Weights, dominant weights and the highest-weight order of a rational representation).

[F2]

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

[F3]

Weyl group. W(G,T)=NG(T)(k)/T(k), and every w∈W has representatives n∈NG(T)(k) and n−1 for w−1 (The Weyl group, Borel subgroups and chambers, The root datum of a split reductive group).

Proof

technique · direct
1.1F1F2given

For t∈T(R) one computes t⋅(n⋅v)=n⋅((n−1tn)⋅v)=n⋅(λ(n−1tn)v)=λ(n−1tn) (n⋅v)=(nλ)(t) (n⋅v) in VR: the first equality uses that the action is a left action and n−1tn∈T(R), and the second uses [F2] and R-linearity of the action of n. Hence n⋅(v⊗1) belongs to the stated eigensubmodule of V⊗kR, with the calculation valid over every R-algebra.

2.1F3step 1.1

Applying step 1.1 to n∈NG(T)(R) gives a bijection Vλ⊗kR→(V⊗kR)nλ, v↦n⋅v, with inverse given by n−1; taking R=k and representatives of w shows dim⁡kVλ=dim⁡kVwλ for every w∈W.

3.1step 2.1

Since Vχ≠0 exactly for the weights of V, step 2.1 shows that the set of weights is stable under the action w ⁣:λ↦wλ of W.

4.1step 1.1step 2.1step 3.1∎

Steps 1.1, 2.1 and 3.1 establish the three assertions.

Depends on

Used by

Dependency tree · two levels

23 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources