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CorollaryStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-14
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Schur’s lemma for irreducible Lie-algebra representations

Statement

A nonzero intertwiner between irreducible g-representations is an isomorphism, and the endomorphism ring of an irreducible representation is a division ring. If k is algebraically closed and the representation is finite dimensional, every intertwining endomorphism is scalar.

Facts & Assumptions

Given: Irreducible representations of one Lie algebra g over k.

[L1]

Lie representations and their intertwiners are respectively unital U(g)-modules and module maps (Lie representations are U(g)-modules).

[L2]

Schur's lemma for modules makes a nonzero map between simple modules an isomorphism and the endomorphism ring of a simple module a division ring (Schur's lemma for simple modules).

[L3]

A linear operator on a positive finite-dimensional vector space over an algebraically closed field has an eigenvalue (Every endomorphism of a nonzero finite-dimensional vector space over an algebraically closed field has an eigenvalue).

Proof

technique · direct
1.1

A stable subspace is exactly a U(g)-submodule under [L1], so an irreducible Lie representation is a simple nonzero U(g)-module. Applying [L2] proves the first two assertions.

L1L2
2.1

Now assume k is algebraically closed and V is finite-dimensional and irreducible. For TEndg(V), [L3] gives an eigenvalue λk. Then TλI is still an intertwiner but is not invertible; the division-ring conclusion of step 1.1 forces TλI=0.

step 1.1L3algebra
3.1

Thus every such T equals λI. The finite-dimensional and algebraic-closure hypotheses are used only in step 2.1 and are not claimed in the division-ring statement.

step 1.1step 2.1

Depends on

Used by

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