LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedaudited 2026-09-22

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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

1.1

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

A1L1

2.1

$P (V)$ 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.

L3step 1.1

3.1

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

L1step 2.1

4.1

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].

L2L3L4step 2.1step 3.1

5.1

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 $λ$ .

step 4.1

6.1

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

step 5.1∎

Depends on

Used by

Dependency tree · two levels

27 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

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)