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

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Levi subalgebras are literally unique

Statement refuted

A finite-dimensional characteristic-zero Lie algebra has at most one Levi subalgebra.

Facts & Assumptions

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

[L1]

Malcev's theorem asserts conjugacy, rather than equality, of Levi subalgebras (Malcev conjugacy of Levi subalgebras).

Counterexample

technique · move a complement in a semidirect product

1.1givenconstruct

Let $V = k^{2}$ be the standard nontrivial ${s l}_{2}$ -module and $g = V ⋊ {s l}_{2}$ , with $V$ abelian. Its radical is $V$ and the standard copy $s = 0 \oplus {s l}_{2}$ is a Levi factor. Choose $v \in V$ and $x \in {s l}_{2}$ with $x v \neq 0$ .

2.1L1step 1.1algebra∎

Since $V$ is abelian, $(ad v)^{2} = 0$ , so $T = \exp (ad v) = 1 + ad v$ is an automorphism. It maps $(0, x)$ to $(- x v, x)$ , so $T (s)$ is a Levi factor distinct from $s$ . They are conjugate exactly as [L1] predicts, but are not equal. This finite witness refutes literal uniqueness.

Depends on

Malcev conjugacy of Levi subalgebras

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

7 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, proof of Theorem 6.25 (standard reference, not scraped)