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

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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

1.1L1givenalgebra

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.

2.1step 1.1∎

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.

Orthogonal complements under invariant forms are ideals

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources