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

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The Killing form is nondegenerate on every reductive Lie algebra

Statement refuted

The Killing form of every finite-dimensional reductive Lie algebra is nondegenerate.

Facts & Assumptions

Given: A characteristic-zero field and the one-dimensional abelian algebra used below.

[L1]

A Lie algebra is reductive exactly when it is the direct sum of its center and a semisimple ideal (Equivalent characterizations of reductive Lie algebras).

[L2]

The Killing form is the adjoint trace form (Killing form).

Counterexample

technique · a central reductive algebra

1.1

Let $g = k$ be the one-dimensional abelian Lie algebra over a characteristic-zero field. It is reductive because $Z (g) = g$ and $[g, g] = 0$ , with the zero algebra semisimple. Thus $g$ is reductive by [L1].

L1given

2.1

Every adjoint endomorphism is zero, so [L2] gives $K_{g} = 0$ . On the nonzero one-dimensional space this form has radical all of $g$ and is degenerate. More generally, every nonzero central element lies in the Killing radical. Thus the displayed $g$ is a complete finite witness.

L2step 1.1algebra∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

9 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

Kirillov, An Introduction to Lie Groups and Lie Algebras, Killing form (standard reference, not scraped)