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The Casimir comparison on a weight space

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let g be a finite-dimensional complex semisimple Lie algebra with Cartan subalgebra h and root system Φ with a chosen positive system Φ+, and let C∈U(g) be the quadratic Casimir element of The quadratic Casimir element. Let h1,…,hr and h1,…,hr be bases of h dual with respect to the Killing form B of The Killing form of a semisimple Lie algebra, so that B(hj,hk)=δjk, and for every α∈Φ+ choose eα∈gα and fα∈g−α with B(eα,fα)=1; then [eα,fα]=Hα is the Killing-dual vector of α (Opposite root spaces bracket to the Killing-dual line). Then C=∑j=1rhjhj+∑α∈Φ+(eαfα+fαeα) in U(g), and for every dominant integral λ∈Λ+ and every μ∈h∗, writing mλ(μ)=dim⁡L(λ)μ for the multiplicity of μ as a weight of the finite-dimensional simple module L(λ) (Highest-weight classification), tr⁡L(λ)μ(C)=mλ(μ)(μ,μ)+∑α∈Φ+tr⁡L(λ)μ(eαfα+fαeα). Since C acts on L(λ) by the scalar (λ,λ+2ρ) (The quadratic Casimir eigenvalue on a highest-weight module is (λ,λ+2ρ)), where ρ is the Weyl vector (The Weyl vector rho for a chosen positive system) and the pairing on h∗ is the one induced by B, also tr⁡L(λ)μ(C)=mλ(μ)(λ,λ+2ρ), hence ((λ,λ+2ρ)−(μ,μ))mλ(μ)=∑α∈Φ+tr⁡L(λ)μ(eαfα+fαeα).

Facts & Assumptions

Given: The Axiom of Choice, such g and h with chosen positive system Φ+, the Casimir element C, the two dual bases hj,hj of h, and vectors eα,fα with B(eα,fα)=1 for each α∈Φ+; also λ∈Λ+ and μ∈h∗.

[A1]

The Axiom of Choice is assumed; it enters through the root-space theory supplying [F1] and [F3] and through the classification of [F6] (The Axiom of Choice).

[F1]

Under Choice, the supplied Cartan subalgebra is maximal toral by Cartan subalgebras are exactly maximal toral subalgebras. Hence g has the finite root-space decomposition g=h⊕⨁α∈Φgα, every root space is one-dimensional, B restricts nondegenerately to h, pairs opposite root spaces perfectly, and pairs no other two weight spaces, and the Killing-dual vector Hα satisfies B(Hα,h)=α(h) for h∈h (Finite semisimple Cartan, root and string structure).

[F2]

For any B-dual bases x1,…,xn and x1,…,xn of g one has C=∑ixixi, independent of the choice of dual bases (The quadratic Casimir element).

[F3]

If x∈gα and y∈g−α, then [x,y]=B(x,y)Hα, and B(x,y)=0 whenever x∈gα, y∈gβ with α+β≠0 (Opposite root spaces bracket to the Killing-dual line, Under Choice, the Killing form pairs only opposite root spaces).

[F4]

Vμ={v∈V:H⋅v=μ(H)v for all H∈h} for every representation V of g (Weight and weight space); if x∈gα then x⋅v∈Vμ+α for v∈Vμ (Root vectors shift weights); and the action of g extends to a unital action of U(g) under which a product xy acts by composition of the operators (Lie representations are U(g)-modules).

[F5]

The Killing form induces a pairing ( , ) on h∗ and μ(Hα)=(μ,α) for every μ∈h∗ and every root α (The Killing form of a semisimple Lie algebra, The quadratic Casimir eigenvalue on a highest-weight module is (λ,λ+2ρ)).

[F6]

For λ∈Λ+ the module L(λ) is a cyclic highest-weight module of highest weight λ, and C acts on it by the scalar (λ,λ+2ρ); it is finite-dimensional, so each L(λ)μ is finite-dimensional (Highest-weight classification, The quadratic Casimir eigenvalue on a highest-weight module is (λ,λ+2ρ), The quadratic Casimir element is central).

Proof

technique · direct
1.1F1F2F3algebraA1

The union {h1,…,hr}∪{eα}α∈Φ+∪{fα}α∈Φ+ is a basis of g, because [F1] decomposes g into h and the one-dimensional root spaces, and its B-dual basis is {h1,…,hr}∪{fα}∪{eα}, because B(hj,hk)=δjk by hypothesis, B(eα,fβ)=δαβ by [F3] and B(eα,fα)=1, and every other pairing of these vectors vanishes by [F3]; with [F2] this gives the displayed expansion of C.

2.1F1F4F5step 1.1algebra

The element hjhj acts on L(λ)μ by the scalar μ(hj)μ(hj) by [F4], so the trace of the Cartan part is dim⁡L(λ)μ∑jμ(hj)μ(hj); to identify the sum, let Hμ∈h be the vector with B(Hμ,h)=μ(h), which exists and is unique because B is nondegenerate on h, and note that B(Hμ,hj)=μ(hj), so the vector ∑jB(Hμ,hj)hj pairs with each hk as B(Hμ,hk) and therefore equals Hμ; applying B(Hμ,⋅) gives ∑jμ(hj)μ(hj)=B(Hμ,Hμ)=(μ,μ) by symmetry of B and the definition of the induced pairing, while μ(Hα)=B(Hμ,Hα)=(μ,α) by the same definition.

2.2step 1.1algebra

Since the trace is linear and C is the sum of the Cartan part and the root part by step 1.1, tr⁡L(λ)μ(C)=tr⁡L(λ)μ(∑jhjhj)+∑α∈Φ+tr⁡L(λ)μ(eαfα+fαeα).

3.1step 2.1step 2.2F6∎

Combining steps 2.1 and 2.2, tr⁡L(λ)μ(C)=mλ(μ)(μ,μ)+∑α∈Φ+tr⁡L(λ)μ(eαfα+fαeα), and by [F6] the same trace equals mλ(μ)(λ,λ+2ρ); subtracting the Cartan term from both expressions for the trace gives the stated comparison identity.

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