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

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Every finite-dimensional representation of a reductive Lie algebra is completely reducible

Statement refuted

Every finite-dimensional representation of a reductive Lie algebra in characteristic zero is completely reducible.

Facts & Assumptions

Given: A characteristic-zero field and the module displayed below.

[L1]

The reductive characterization allows a nonzero center (Equivalent characterizations of reductive Lie algebras).

[L2]

Weyl complete reducibility applies to a semisimple acting algebra (Weyl's complete reducibility theorem).

Counterexample

technique · a nilpotent action of the center

1.1

Let the one-dimensional abelian—and hence reductive—algebra $k t$ act on $V = k e_{1} \oplus k e_{2}$ by $t e_{1} = 0$ and $t e_{2} = e_{1}$ . The line $k e_{1}$ is invariant, and $k t$ is reductive by [L1].

L1givenalgebra

2.1

Any complementary line has a generator $e_{2} + a e_{1}$ . Its image under $t$ is $e_{1}$ , which does not belong to that line. Thus $k e_{1}$ has no invariant complement and the representation is not completely reducible. This does not contradict [L2], whose acting algebra must be semisimple: here the nonzero center acts by a nilpotent Jordan block.

L2step 1.1algebra∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

13 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

Milne, Lie Algebras, Theorem 6.14 and surrounding discussion (standard reference, not scraped)