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Freudenthal's weight multiplicity recursion

Statement

Assume the Axiom of Choice (The Axiom of Choice). For every dominant integral weight λ∈Λ+ and every μ∈h∗, ((λ+ρ,λ+ρ)−(μ+ρ,μ+ρ))mλ(μ)=2∑α∈Φ+∑j≥1(μ+jα,α) mλ(μ+jα), with mλ(ν)=0 for every ν that is not a weight of L(λ) and with ρ the Weyl vector (The Weyl vector rho for a chosen positive system); the inner sum is finite by Positive root strings sum the Freudenthal correction.

Facts & Assumptions

Given: The Axiom of Choice, a dominant integral weight λ∈Λ+, an element μ∈h∗, the positive system Φ+, the Weyl vector ρ, and the multiplicities mλ(ν)=dim⁡L(λ)ν of the finite-dimensional simple module of highest weight λ.

[A1]

The Axiom of Choice is assumed; it is inherited from the published Casimir and classification suppliers used in [F1] and [F2] (The Axiom of Choice).

[F1]

The Casimir comparison on the weight space L(λ)μ reads ((λ,λ+2ρ)−(μ,μ))mλ(μ)=∑α∈Φ+tr⁡L(λ)μ(eαfα+fαeα) (The Casimir comparison on a weight space).

[F2]

Each positive root contributes its string trace tr⁡L(λ)μ(eαfα+fαeα)=mλ(μ)(μ,α)+2∑j≥1mλ(μ+jα)(μ+jα,α), the sum being finite and the coefficients vanishing off the weights of L(λ) (Positive root strings sum the Freudenthal correction).

[F3]

ρ=12∑α∈Φ+α, so the bilinear form gives ∑α∈Φ+(μ,α)=(μ,2ρ)=2(μ,ρ), and mλ(ν)=dim⁡L(λ)ν=0 for every ν that is not a weight (The Weyl vector rho for a chosen positive system, The formal character of a finite-dimensional weight module).

[F4]

Expanding the shifted squares with bilinearity and symmetry of ( , ) gives (λ+ρ,λ+ρ)−(μ+ρ,μ+ρ)=(λ,λ+2ρ)−(μ,μ)−2(μ,ρ).

Proof

technique · direct
1.1F3F4algebraA1

By [F3] the sum of the pairings over the positive roots is ∑α∈Φ+(μ,α)=2(μ,ρ), and by [F4] the Casimir coefficient and the shifted-norm difference are related by (λ+ρ,λ+ρ)−(μ+ρ,μ+ρ)=(λ,λ+2ρ)−(μ,μ)−2(μ,ρ).

2.1F1F2step 1.1algebra

Substituting [F2] into the right side of [F1] gives ((λ,λ+2ρ)−(μ,μ))mλ(μ)=mλ(μ)∑α∈Φ+(μ,α)+2∑α∈Φ+∑j≥1mλ(μ+jα)(μ+jα,α), and step 1.1 turns the first term into 2(μ,ρ)mλ(μ).

3.1F2F3step 1.1step 2.1algebra∎

Subtracting 2(μ,ρ)mλ(μ) from both sides of step 2.1 and using the coefficient identity of step 1.1 gives ((λ+ρ,λ+ρ)−(μ+ρ,μ+ρ))mλ(μ)=2∑α∈Φ+∑j≥1(μ+jα,α)mλ(μ+jα), which is the asserted recursion; the inner sums are finite and the coefficients vanish off the weights of L(λ) by [F2] and [F3].

Depends on

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