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

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A nonzero nilpotent Lie algebra has nonzero center

Statement

If $g$ is a nonzero nilpotent Lie algebra, then $Z (g) \neq 0$ .

Facts & Assumptions

Given: A nonzero nilpotent Lie algebra $g$ .

[L1]

Nilpotence means that the descending lower central series eventually vanishes, with $γ_{r + 1} = [g, γ_{r}]$ (Lower central series and nilpotent Lie algebras).

[L2]

The center consists of the elements $z$ satisfying $[g, z] = 0$ (Lie subalgebras, ideals, and center).

Proof

technique · direct

1.1

Because $γ_{1} (g) = g \neq 0$ and the series terminates by [L1], there is a largest index $c \geq 1$ with $γ_{c} (g) \neq 0$ . Its successor satisfies $[g, γ_{c}] = γ_{c + 1} = 0$ , so [L2] gives $0 \neq γ_{c} \subseteq Z (g)$ .

givenL1L2algebra

2.1

Hence the center contains the displayed nonzero subspace and is itself nonzero. The hypothesis $g \neq 0$ is essential: the zero algebra has zero center. No choice of a basis or of a central element is needed.

givenstep 1.1∎

Depends on

Used by

Codimension-one ideals in nilpotent Lie algebras Corollary
Malcev conjugacy of Levi subalgebras Theorem

Dependency tree · two levels

4 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(c) (standard reference, not scraped)