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Freudenthal recursion for the sl3 adjoint zero weight

Example

Assume the Axiom of Choice (The Axiom of Choice). Keep the setting of Kostant multiplicity in the sl3 adjoint module: g=sl3, λ=θ=α1+α2=ρ, μ=0. Then (λ+ρ,λ+ρ)−(μ+ρ,μ+ρ)=(2θ,2θ)−(θ,θ)=3(θ,θ), and in the right side of Freudenthal's weight multiplicity recursion only j=1 contributes, since 2α,3α,… are not weights of the adjoint module L(θ); hence the recursion reads 3(θ,θ)mλ(0)=2∑α∈Φ+(α,α)mλ(α). In the realization α1=e1−e2, α2=e2−e3, θ=e1−e3 of Root systems of the classical complex Lie algebras the three positive roots have the same length, and mλ(α1)=mλ(α2)=mλ(θ)=1, so 3(θ,θ)mλ(0)=2⋅3(θ,θ) and mλ(0)=2, recovering Kostant multiplicity in the sl3 adjoint module. The extremal weights ±α1,±α2,±θ=Wλ are the Weyl orbit of the top weight and each has multiplicity one (Extremal Weyl-orbit weights); among them only the top weight θ=λ has the vanishing recursion coefficient of the indeterminate case 0=0 in Freudenthal recursion terminates from the highest weight, whose base value mλ(λ)=1 is stated there.

Facts & Assumptions

Given: The Axiom of Choice, the realization of sl3 with positive roots α1,α2,θ=α1+α2 of equal length, the Weyl vector ρ=θ, the top weight λ=θ, the weight μ=0, the adjoint module L(θ), and the multiplicities mθ(ν).

[A1]

The Axiom of Choice is assumed; it enters through the Freudenthal recursion and the highest-weight suppliers below (The Axiom of Choice).

[F1]

In the realization α1=e1−e2, α2=e2−e3 the positive roots are α1,α2,θ with (α1,α1)=(α2,α2)=(θ,θ), and ρ=α1+α2=θ (Root systems of the classical complex Lie algebras, Classical complex matrix Lie algebras, The Weyl vector rho for a chosen positive system).

[F2]

The adjoint module is L(θ) and its weights are 0 with multiplicity 2, and ±α1,±α2,±θ each with multiplicity 1 (The adjoint highest weight is the highest root, Kostant multiplicity in the sl3 adjoint module).

[F3]

Freudenthal's recursion reads ((λ+ρ,λ+ρ)−(μ+ρ,μ+ρ))mλ(μ)=2∑α∈Φ+∑j≥1(μ+jα,α)mλ(μ+jα) (Freudenthal's weight multiplicity recursion).

[F4]

The extremal weights Wλ each have multiplicity one, and the recursion has the base value mλ(λ)=1 with the indeterminate case 0=0 occurring, among actual weights of L(λ), only at μ=λ (Extremal Weyl-orbit weights, Freudenthal recursion terminates from the highest weight).

Verification

1.1F2F3A1

With λ=θ=ρ and μ=0 the recursion coefficient of [F3] is (2θ,2θ)−(θ,θ)=3(θ,θ), and the weights 0+jα=jα of L(θ) are nonzero only for j=1 by the weight list [F2], so each inner sum of 2∑α∈Φ+∑j≥1(αj,α)m(jα) reduces to its j=1 term (α,α)m(α).

1.2F1F2

By [F1] all three positive roots have the same squared length, and by [F2] m(α1)=m(α2)=m(θ)=1, so the right side of the recursion is 2((α1,α1)+(α2,α2)+(θ,θ))=6(θ,θ).

2.1F2step 1.1step 1.2

Combining steps 1.1 and 1.2, the recursion reads 3(θ,θ)mθ(0)=6(θ,θ), and since (θ,θ)≠0 this gives mθ(0)=2, recovering the value computed by Kostant's formula in [F2].

3.1F4step 1.1∎

The extremal weights are the six Weyl translates of λ=θ, each of multiplicity one by [F4]; the recursion coefficient 4(θ,θ)−(μ+ρ,μ+ρ) vanishes, among actual weights of L(θ), only at μ=λ by [F4], so among the extremal weights only the top weight is the indeterminate case of the recursion, whose value mθ(θ)=1 is the stated base case; this is consistent with the recursion fixing every other multiplicity from that base.

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