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LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-14
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The semidirect-product bracket satisfies Jacobi

Statement

The bracket defining g⋉ρh is bilinear, alternating, and satisfies the Jacobi identity.

Facts & Assumptions

Given: Lie algebras g,h and a Lie homomorphism ρ:g→Der⁡(h), with the bracket of Semidirect products of Lie algebras.

[L1]

Each ρ(x) is a derivation: ρ(x)[u,v]=[ρ(x)u,v]+[u,ρ(x)v].

[L2]

Bracket preservation says [ρ(x),ρ(y)]=ρ([x,y]).

Proof

technique · direct
1.1givenalgebra

Every term in the displayed bracket is bilinear. Substituting the same pair twice gives ([x,x],ρ(x)u−ρ(x)u+[u,u])=(0,0), so the bracket is alternating.

1.2given

The first component of the cyclic Jacobi sum for (x,u),(y,v),(z,w) is [x,[y,z]]+[y,[z,x]]+[z,[x,y]]=0 by Jacobi in g.

1.3L1algebra

In the second component, the terms containing two elements of h and one action are, cyclically, ρ(x)[v,w]−[ρ(x)v,w]−[v,ρ(x)w]; each vanishes by [L1]. The terms with three elements of h form their Jacobi sum and vanish.

1.4L2algebra

The remaining terms are ([ρ(x),ρ(y)]−ρ([x,y]))w and its two cyclic analogues. They vanish by [L2]. Thus the entire second component is zero.

2.1step 1.2step 1.3step 1.4∎

Both components of the Jacobi sum vanish, proving the claim. If one summand is zero, or if ρ=0, the calculation reduces to the componentwise Jacobi identity, so all boundary cases are already included and no choices are made.

Depends on

Used by

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Sources