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Every finite-dimensional irreducible module has a highest-weight vector

Statement

Assume the Axiom of Choice. Let $g$ be a finite-dimensional complex semisimple Lie algebra with Cartan subalgebra $h$ and a chosen positive system, and let $V \neq 0$ be a finite-dimensional irreducible representation of $g$ (Irreducible, completely reducible, and faithful representations). Then $V$ contains a highest weight vector for the chosen positive roots (Highest-weight vectors and modules); consequently $V$ is a highest weight module for some highest weight $λ$ .

Facts & Assumptions

Given: The Axiom of Choice, such $g, h$ , a chosen positive system, and a nonzero finite-dimensional irreducible module $V$ .

[A1]

The Axiom of Choice is assumed; among the facts used below, it enters through the weight decomposition [L1] (The Axiom of Choice). The algebraic shift property [L2] has no choice hypothesis.

[L1]

$V = ⨁_{μ} V_{μ}$ is a direct sum over its finitely many weights, and each weight space is finite dimensional (Finite-dimensional modules decompose into weight spaces).

[L2]

If $x \in g_{α}$ and $w \in V_{μ}$ , then $x \cdot w \in V_{μ + α}$ , with $0$ allowed (Root vectors shift weights).

[L3]

The root order $\leq$ is a partial order on the weights, and a positive root $α$ satisfies $α \in Q_{+} ∖ {0}$ , so $λ + α > λ$ for every weight $λ$ (Root order on weights, Simple roots form a signed integral basis).

[L4]

$n^{+}$ is the sum of the root spaces $g_{α}$ with $α \in Φ^{+}$ (Positive and negative nilpotent subalgebras and the Borel).

Proof

technique · direct

1.1A1L1

By [L1] the set $P (V)$ of weights of $V$ is finite and nonempty, because $V \neq 0$ has a nonzero weight space.

2.1L3step 1.1

$P (V)$ has a maximal element: enumerating $P (V) = {λ_{1}, \dots, λ_{N}}$ and starting from $λ_{1}$ , replace the current element by a strictly larger element of $P (V)$ whenever one exists; the resulting chain is strictly increasing in the partial order [L3] and therefore has at most $N$ terms, so the procedure stops at a weight $λ$ above which no weight of $V$ lies.

3.1L1step 2.1

Choose $0 \neq v \in V_{λ}$ , which is possible because $λ$ is a weight by step 2.1.

4.1L2L3L4step 2.1step 3.1

Let $α \in Φ^{+}$ and $x \in g_{α}$ . If $x \cdot v \neq 0$ , then $x \cdot v \in V_{λ + α}$ by [L2], so $λ + α$ would be a weight of $V$ strictly above $λ$ by [L3], contradicting the maximality of $λ$ from step 2.1. Hence $x \cdot v = 0$ for every $x$ in every positive root space, and therefore $n^{+} \cdot v = 0$ by [L4].

5.1step 4.1

By step 4.1 the vector $v$ is a highest weight vector of weight $λ$ (Highest-weight vectors and modules); since $V$ is irreducible and nonzero, the subrepresentation generated by $v$ is all of $V$ , so $V$ is a highest weight module of highest weight $λ$ .

6.1step 5.1∎

The existence of a highest weight vector in $V$ is proved.

Depends on

Used by

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Sources

Alexander Kirillov Jr., An Introduction to Lie Groups and Lie Algebras (standard reference, not scraped)
Anthony W. Knapp, Lie Groups Beyond an Introduction, 2nd ed. (standard reference, not scraped)