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

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Orthogonal complements under invariant forms are ideals

Statement

Let $B$ be a symmetric invariant bilinear form on a Lie algebra $g$ , and let $i$ be an ideal. Then

$i^{⊥} = {u \in g : B (u, v) = 0 for every v \in i}$

is an ideal. In particular, the radical $g^{⊥}$ of $B$ is an ideal.

Facts & Assumptions

Given: A Lie algebra $g$ , an ideal $i$ , and a symmetric bilinear form satisfying $B ([z, u], v) + B (u, [z, v]) = 0$ .

[L1]

An ideal is a linear subspace $i$ with $[g, i] \subseteq i$ (Lie subalgebras, ideals, and center).

Proof

technique · direct

1.1

The equations defining $i^{⊥}$ are linear in $u$ , so it is a linear subspace. If $u \in i^{⊥}$ , $z \in g$ , and $v \in i$ , invariance gives $B ([z, u], v) = - B (u, [z, v])$ . The second argument on the right belongs to $i$ by [L1], so the right side is zero. Hence $[z, u] \in i^{⊥}$ , proving ideality.

L1givenalgebra

2.1

Taking $i = g$ in step 1.1 gives that $g^{⊥}$ is an ideal. This includes the zero algebra, the zero form, and the nondegenerate case $g^{⊥} = 0$ without any separate choice.

step 1.1∎

Depends on

Lie subalgebras, ideals, and center

Used by

Cartan's semisimplicity criterion Theorem
Derivations of semisimple Lie algebras are inner Theorem
Second Whitehead lemma Theorem
Semisimple Lie algebras decompose into simple ideals Theorem
Weyl's complete reducibility theorem Theorem

Dependency tree · two levels

3 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, Lemma 4.6 (standard reference, not scraped)