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Weyl alternants are skew-invariant

Statement

Let W act on the finite-support elements of the completed character ring R by w⋅eμ=ewμ as in The Weyl alternation operator, and let A(ν) be the alternant of The Weyl alternation operator. Then for all w∈W and ν∈h∗, w⋅A(ν)=(−1)ℓ(w)A(ν), and if a simple reflection si fixes ν, that is siν=ν, then A(ν)=0.

Facts & Assumptions

Given: The root system with Weyl group W and length function ℓ, the completed character ring R with its action of W on finite-support elements, the alternants A(ν), and elements w∈W, ν∈h∗.

[F1]

For finite-support g=∑μcμeμ one has w⋅g=∑μcμewμ, and A(ν)=∑x∈W(−1)ℓ(x)exν is a finite-support element of R (The Weyl alternation operator, The completed formal character ring).

[F2]

The sign (−1)ℓ is a homomorphism: for all u,v∈W, (−1)ℓ(uv)=(−1)ℓ(u)(−1)ℓ(v), and (−1)ℓ(w−1)=(−1)ℓ(w) (The sign of the Weyl length is multiplicative).

[F3]

The action of W on h∗ is a group action by the root reflections sα(λ)=λ−⟨λ,α∨⟩α; the simple reflection si satisfies siαi=−αi≠αi, so si is not the identity, and ℓ is the least number of simple reflections in an expression for an element, so ℓ(si)=1 (Root reflections and the Weyl group action, The Weyl group is finite and faithful, Finite Weyl root system, lattice and chamber conventions).

Proof

technique · direct
1.1F1F2algebra

Since A(ν) is a finite sum, [F1] gives w⋅A(ν)=∑x∈W(−1)ℓ(x)ewxν; reindexing by y=wx and using [F2] yields w⋅A(ν)=∑y∈W(−1)ℓ(w−1y)eyν=(−1)ℓ(w)∑y∈W(−1)ℓ(y)eyν=(−1)ℓ(w)A(ν).

2.1F1F2F3algebra∎

If siν=ν, then A(ν)=A(siν)=∑x∈W(−1)ℓ(x)exsiν; reindexing by y=xsi gives A(siν)=∑y∈W(−1)ℓ(ysi)eyν=(−1)ℓ(si)A(ν)=−A(ν) by [F2] and [F3], since ℓ(si)=1, so 2A(ν)=0 and A(ν)=0 because its coefficients are integers.

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