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 $g ⋉_{ρ} h$ is bilinear, alternating, and satisfies the Jacobi identity.

Facts & Assumptions

Given: Lie algebras $g, h$ and a Lie homomorphism $ρ : g \to Der (h)$ , 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

1.1

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.

givenalgebra

1.2

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 $g$ .

given

1.3

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.

L1algebra

1.4

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.

L2algebra

2.1

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.

step 1.2step 1.3step 1.4∎

Depends on

Semidirect products of Lie algebras

Used by

A semidirect-product Lie algebra from a linear action Example

Dependency tree · two levels

5 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

Kirillov, An Introduction to Lie Groups and Lie Algebras, semidirect products in §3.3 (standard reference, not scraped)