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

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Kernels, images, and the first isomorphism theorem for Lie algebras

Statement

For a Lie-algebra homomorphism $f : g \to h$ , the kernel is an ideal, the image is a Lie subalgebra, and

$g / \ker f ≅ im f$

as Lie algebras via $x + \ker f \mapsto f (x)$ .

Facts & Assumptions

Given: A homomorphism $f : g \to h$ of Lie algebras over the same field.

[L1]

Such an $f$ is linear and preserves brackets (Homomorphisms of possibly infinite-dimensional Lie algebras).

[L2]

The underlying linear map induces the vector-space isomorphism $g / \ker f \to im f$ , $x + \ker f \mapsto f (x)$ (First isomorphism theorem for modules: $M / \ker f ≅ im f$ ).

[L3]

Quotient brackets by ideals are those of Quotient Lie algebras.

Proof

technique · direct

1.1givenL1algebra

Linearity makes $\ker f$ and $im f$ linear subspaces. If $a \in \ker f$ and $x \in g$ , then $f ([x, a]) = [f (x), f (a)] = [f (x), 0] = 0$ , so $[x, a] \in \ker f$ ; hence the kernel is an ideal.

1.2L1algebra

If $u = f (x)$ and $v = f (y)$ are in the image, then $[u, v] = [f (x), f (y)] = f ([x, y])$ is again in the image. Thus the image is a Lie subalgebra.

1.3L1L2L3

Let $\overset{ˉ}{f}$ be the vector-space isomorphism of [L2]. By [L3], $\overset{ˉ}{f} ([x + \ker f, y + \ker f]) = f ([x, y]) = [f (x), f (y)] = [\overset{ˉ}{f} (x + \ker f), \overset{ˉ}{f} (y + \ker f)]$ , so it preserves brackets.

2.1step 1.3L2∎

A bijective bracket-preserving linear map has bracket-preserving inverse: for $u = f (x), v = f (y)$ , the inverse sends $[u, v] = f ([x, y])$ to $[x, y]$ . Hence $\overset{ˉ}{f}$ is a Lie-algebra isomorphism. The zero map gives $g / g ≅ 0$ , while an injective map gives $g / 0 ≅ im f$ ; these are included in the same proof.

Depends on

Dependency tree · 2 levels

Sources

Kirillov, An Introduction to Lie Groups and Lie Algebras, Lemma 5.17