False statementConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-14

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Lie-algebra representations need not be completely reducible

Statement

Every representation of every Lie algebra is completely reducible.

Facts & Assumptions

Given: The asserted universal complete reducibility.

[L1]

Completely reducible means an algebraic direct sum of irreducible subrepresentations (Irreducible, completely reducible, and faithful representations).

Refutation

technique · direct counterexample

1.1

Let the one-dimensional abelian Lie algebra $k t$ act on $V = k e_{1} \oplus k e_{2}$ by $t e_{1} = 0$ and $t e_{2} = e_{1}$ . This is a representation because its sole action operator commutes with itself, and $k e_{1}$ is a proper nonzero stable line, so $V$ is not irreducible.

constructalgebra

2.1

Any stable line is spanned by an eigenvector of the nilpotent operator $t$ . Its eigenvalue must be zero, and $\ker t = k e_{1}$ , so $k e_{1}$ is the only stable line. Therefore $V$ cannot be a direct sum of two irreducible one-dimensional subrepresentations; since it is not itself irreducible, it has no decomposition of the form required by [L1].

step 1.1L1algebra

3.1

This two-dimensional representation is not completely reducible, refuting the universal claim over every field.

step 2.1∎

Depends on

Irreducible, completely reducible, and faithful representations

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources

Kirillov, An Introduction to Lie Groups and Lie Algebras, §4.3, printed pp. 52–53 (standard reference, not scraped)