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Tensor product with a minuscule representation

Statement

Assume the Axiom of Choice. Let $ω \in Λ^{+}$ be a minuscule weight (Minuscule weights) of a finite-dimensional complex simple Lie algebra $g$ , and let $λ \in Λ^{+}$ . Then $L (ω) \otimes L (λ) ≅ ⨁_{γ \in W ω} L (λ + γ),$ where $L (λ + γ)$ is read as $0$ when $λ + γ \notin Λ^{+}$ ; equivalently the sum runs over those $γ \in W ω$ with $λ + γ$ dominant integral and each such summand occurs once. In characters, $ch (L (ω) \otimes L (λ)) = \sum_{γ : λ + γ \in Λ^{+}} ch L (λ + γ) .$

Facts & Assumptions

Given: AC, a minuscule weight $ω \in Λ^{+}$ , a dominant integral weight $λ$ , the Weyl orbit $W ω$ , and the alternation operator $A$ with $A (η) = \sum_{w \in W} (- 1)^{ℓ (w)} e^{w η}$ in the completed character ring $R$ (The Weyl alternation operator, The completed formal character ring).

[F1]

Orbit-sum character: $ch L (ω) = \sum_{γ \in W ω} e^{γ}$ (Minuscule weights have exactly the Weyl orbit as their weights, Minuscule weights).

[F2]

Weyl character formula: $A (ρ) ch L (λ) = A (λ + ρ)$ and, for every $η \in Λ^{+}$ , $A (ρ) ch L (η) = A (η + ρ)$ ; formal characters are multiplicative on tensor products, $A (ρ)$ is invertible in $R$ , and in any finite-dimensional module the coefficients of the simple characters are their multiplicities. Every finite-dimensional $g$ -module is completely reducible (The Weyl character formula, Formal characters are additive and multiplicative, Geometric series are invertible in the completed character ring, Tensor-product multiplicities are character structure constants, Weyl's complete reducibility theorem).

[F3]

Alternant vanishing on walls: if a reflection $s \in W$ fixes $η$ , then $A (η) = 0$ ; equivalently $A$ is skew-invariant, $A (s η) = - A (η)$ for $s$ a reflection, so $A (η) = 0$ whenever $η$ lies on a wall (Weyl alternants are skew-invariant, The Weyl alternation operator).

[F4]

For every $γ \in W ω$ one has $⟨ γ, α^{\lor} ⟩ \geq - 1$ for every positive root $α$ : by Minuscule weights the pairing of $ω$ with every coroot lies in ${- 1, 0, 1}$ , and $γ = w ω$ with $⟨ w ω, α^{\lor} ⟩ = ⟨ ω, w^{- 1} α^{\lor} ⟩$ , a pairing of $ω$ with a coroot. If a weight $η$ has $⟨ η, α_{i}^{\lor} ⟩ = - 1$ for a simple coroot, then $⟨ η + ρ, α_{i}^{\lor} ⟩ = 0$ : the positive-root half-sum definition of $ρ$ and the fact that $s_{i}$ permutes the positive roots other than $α_{i}$ give $s_{i} ρ = ρ - α_{i}$ and therefore $⟨ ρ, α_{i}^{\lor} ⟩ = 1$ (The Weyl vector rho for a chosen positive system, Finite Weyl positive roots and simple reflections). Hence $η + ρ$ is fixed by $s_{i}$ and lies on its wall (Finite Weyl root system, lattice and chamber conventions, Integral, dominant, and strictly dominant weights, Finite Weyl closed chambers and stabilizers).

Proof

1.1F1F2givenalgebra

By [F1], [F2] and multiplicativity, $A (ρ) ch (L (ω) \otimes L (λ)) = (\sum_{γ \in W ω} e^{γ}) A (λ + ρ) = \sum_{w \in W} \sum_{γ \in W ω} (- 1)^{ℓ (w)} e^{γ + w (λ + ρ)} .$ For each fixed $w$ , the map $γ \mapsto w γ$ permutes the orbit $W ω$ , so the inner sum is unchanged when $e^{γ}$ is replaced by $e^{w γ}$ . Reindexing the finite double sum therefore gives $\sum_{w \in W} \sum_{γ \in W ω} (- 1)^{ℓ (w)} e^{w (λ + ρ + γ)} = \sum_{γ \in W ω} A (λ + γ + ρ) .$

1.2F2F3F4givenalgebra

Non-dominant translates vanish. Let $γ \in W ω$ with $λ + γ \notin Λ^{+}$ . Since $λ \in Λ^{+}$ and by [F4] $⟨ γ, α^{\lor} ⟩ \geq - 1$ for every positive root, there is a simple coroot $α_{i}^{\lor}$ with $⟨ λ + γ, α_{i}^{\lor} ⟩ < 0$ ; then $⟨ λ + γ, α_{i}^{\lor} ⟩ = - 1$ , because $⟨ λ, α_{i}^{\lor} ⟩ \geq 0$ and $⟨ γ, α_{i}^{\lor} ⟩ \geq - 1$ , and hence $⟨ λ + γ + ρ, α_{i}^{\lor} ⟩ = 0$ . Thus $λ + γ + ρ$ is fixed by the reflection $s_{i}$ , and $A (λ + γ + ρ) = 0$ by [F3].

2.1F1F2step 1.1step 1.2algebra

For a translate with $λ + γ \in Λ^{+}$ , $A (λ + γ + ρ) = A (ρ) ch L (λ + γ)$ by the character formula [F2]. Substituting these dominant terms and the vanishing terms of step 1.2 into step 1.1 gives $A (ρ) ch (L (ω) \otimes L (λ)) = A (ρ) \sum_{γ : λ + γ \in Λ^{+}} ch L (λ + γ) .$ Cancelling the invertible element $A (ρ)$ in the ring $R$ [F2] gives the asserted character identity $ch (L (ω) \otimes L (λ)) = \sum_{γ : λ + γ \in Λ^{+}} ch L (λ + γ)$ ; the sum is finite because $W ω$ is finite.

3.1F2step 2.1algebra∎

Decomposition. By Weyl complete reducibility, both finite-dimensional modules in the character identity of step 2.1 decompose as finite direct sums of the pairwise non-isomorphic simples $L (ν)$ . The right-hand side is finite because $W ω$ is finite. Equality of their characters, together with the multiplicity-uniqueness clause of [F2], forces the multiplicities of each $L (ν)$ to agree. This gives the asserted module isomorphism, with every surviving summand occurring once.

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Three times three for sl3 Example

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Sources

P. Etingof, Lie Groups and Lie Algebras II (MIT 18.755, Spring 2024), complete lecture notes (standard reference, not scraped)
R. Goodman and N. R. Wallach, Symmetry, Representations, and Invariants, Graduate Texts in Mathematics 255, Springer 2009 (standard reference, not scraped)