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

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A central extension of a nilpotent Lie algebra is nilpotent

Statement

Let $i \subseteq Z (g)$ be an ideal. If $g / i$ is nilpotent of class $c$ , then $g$ is nilpotent of class at most $c + 1$ .

Facts & Assumptions

Given: A Lie algebra $g$ and a central ideal $i \subseteq Z (g)$ whose quotient has class $c$ .

[L1]

The lower central series has $γ_{r + 1} = [g, γ_{r}]$ (Lower central series and nilpotent Lie algebras).

[L2]

The quotient map $π : g \to g / i$ is a surjective Lie homomorphism (Quotient Lie algebras).

[L3]

Centrality means $[g, i] = 0$ (Lie subalgebras, ideals, and center).

Proof

technique · direct

1.1

Surjectivity and bracket preservation in [L2] give $π (γ_{r} (g)) = γ_{r} (g / i)$ by induction on $r$ . Since the quotient has class $c$ , its $(c + 1)$ st term vanishes, and therefore $γ_{c + 1} (g) \subseteq \ker π = i$ .

givenL1L2algebra

2.1

Now [L1], step 1.1, and centrality [L3] give $γ_{c + 2} (g) = [g, γ_{c + 1} (g)] \subseteq [g, i] = 0$ . Thus $g$ has class at most $c + 1$ . When the quotient is zero ( $c = 0$ ), this says precisely that the central algebra $g = i$ is abelian.

L1L3step 1.1∎

Depends on

Used by

The derived algebra of a solvable Lie algebra is nilpotent in characteristic zero Corollary

Dependency tree · two levels

6 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, Proposition 2.5(b) (standard reference, not scraped)