Alphabeta Math
PropositionStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedaudited 2026-09-22
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An irreducible module is generated by its 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, let V be a finite-dimensional irreducible representation of g (Irreducible, completely reducible, and faithful representations), and let v∈V be a highest weight vector (Highest-weight vectors and modules). Then the subrepresentation generated by v is V; equivalently U(g)v=V.

Facts & Assumptions

Given: The Axiom of Choice, such g,h, a finite-dimensional irreducible module V, and a highest weight vector v∈V.

[A1]

The Axiom of Choice is assumed; it enters through the definition of a highest weight vector, whose contract carries the assumption (The Axiom of Choice).

[L1]

The action of g extends uniquely to a unital action of U(g) on V, and the subrepresentation generated by v is U(g)v (Lie representations are U(g)-modules, Highest-weight vectors and modules).

[L2]

A nonzero subrepresentation of an irreducible representation is the whole representation; the zero subspace and V are the only subrepresentations of an irreducible V≠0 (Irreducible, completely reducible, and faithful representations, Subrepresentations, quotient representations, and intertwiners).

Proof

technique · direct
1.1L1A1

The vector v is nonzero and lies in U(g)v (as the image of v under the unit of U(g)), so U(g)v is a nonzero subrepresentation of V.

2.1L1L2step 1.1

Since V is irreducible, [L2] applied to the nonzero subrepresentation U(g)v gives U(g)v=V; by [L1] this is the subrepresentation generated by v.

3.1step 2.1∎

The subrepresentation generated by any highest weight vector is therefore all of V.

Depends on

Used by

Dependency tree · two levels

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Sources