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

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Nilpotent adjoint action yields a central series

Statement

If every ${ad}_{x}$ is nilpotent on a finite-dimensional Lie algebra $g$ , then there is a finite filtration

$g = A_{0} \supset A_{1} \supset \dots \supset A_{m} = 0$

such that $[g, A_{i}] \subseteq A_{i + 1}$ for every $i < m$ . Consequently the upper central series reaches $g$ .

Facts & Assumptions

Given: A finite-dimensional Lie algebra $g$ for which every ${ad}_{x}$ is nilpotent.

[L1]

Engel triangularization gives a flag lowered by a nil representation (Engel triangularization theorem).

[L2]

Engel's theorem identifies the hypothesis with nilpotence of $g$ (Engel's theorem).

[L3]

The upper central series has $Z_{0} = 0$ , and $[g, B] \subseteq Z_{r}$ implies $B \subseteq Z_{r + 1}$ (Upper central series of a Lie algebra).

Proof

technique · direct

1.1

Apply [L1] to the adjoint representation. If $0 = V_{0} \subset \dots \subset V_{m} = g$ is its lowered flag, put $A_{i} = V_{m - i}$ . Then $A_{0} = g$ , $A_{m} = 0$ , and $[g, A_{i}] \subseteq A_{i + 1}$ . In particular [L2] also confirms that $g$ is nilpotent.

givenL1L2algebra

2.1

We prove $A_{m - r} \subseteq Z_{r}$ for $0 \leq r \leq m$ . At $r = 0$ this is $A_{m} = 0 = Z_{0}$ . If it holds at $r$ , then $[g, A_{m - r - 1}] \subseteq A_{m - r} \subseteq Z_{r}$ , so [L3] gives $A_{m - r - 1} \subseteq Z_{r + 1}$ . At $r = m$ this yields $g = A_{0} \subseteq Z_{m}$ , hence equality. For $g = 0$ , take $m = 0$ .

L3step 1.1algebra∎

Depends on

Used by

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Dependency tree · two levels

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Sources

Milne, Lie Algebras, Theorem 2.8 and Corollary 2.11 (standard reference, not scraped)