CorollaryStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-14

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

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

1.1L1L2

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.

2.1step 1.1L3algebra

Now assume $k$ is algebraically closed and $V$ is finite-dimensional and irreducible. For $T \in {End}_{g} (V)$ , [L3] gives an eigenvalue $λ \in k$ . Then $T - λ I$ is still an intertwiner but is not invertible; the division-ring conclusion of step 1.1 forces $T - λ I = 0$ .

3.1step 1.1step 2.1∎

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

Schur’s lemma for irreducible Lie-algebra representations

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources