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

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The Lie algebra of a Lie subgroup is a Lie subalgebra

Statement

Assume ${A C}_{ω}$ . If $i : H \to G$ is a Lie-subgroup inclusion, then

$d i_{e} : Lie (H) = T_{e} H ⟶ Lie (G) = T_{e} G$

is injective and identifies $Lie (H)$ with the Lie subalgebra $d i_{e} (T_{e} H)$ of $Lie (G)$ .

Facts & Assumptions

Given: ${A C}_{ω}$ , a finite-dimensional real Lie group $G$ , and a Lie subgroup $i : H \to G$ .

[F1]

A Lie-subgroup inclusion is an injective immersion and a smooth Lie-group homomorphism. Immersed, embedded, and closed Lie subgroups.

[F2]

The differential of an immersion is injective at every point. Immersed, embedded, and closed Lie subgroups.

[F3]

Under ${A C}_{ω}$ , the identity differential of a smooth Lie-group homomorphism is linear and bracket preserving. The Axiom of Countable Choice ( ${A C}_{ω}$ ), Differential of a Lie-group homomorphism is a Lie-algebra homomorphism.

[F4]

A Lie subalgebra is a linear subspace closed under the ambient bracket. Lie subalgebras and ideals.

Proof

technique · identify the tangent algebra with the image of the identity differential

1.1

Since $i$ is an immersion by [F1], its differential $d i_{e} : T_{e} H \to T_{e} G$ is injective by [F2]. It is therefore a linear isomorphism from $T_{e} H$ onto the linear subspace $h : = d i_{e} (T_{e} H) \subseteq T_{e} G$ .

F1F2algebra

2.1

Because $i$ is also a smooth Lie-group homomorphism, [F3] gives $d i_{e} ([X, Y]_{H}) = [d i_{e} X, d i_{e} Y]_{G}$ for all $X, Y \in T_{e} H$ . Hence the bracket of any two vectors in $h$ again lies in $h$ .

F1F3step 1.1

3.1

Thus $h$ is a bracket-closed linear subspace of $T_{e} G$ and so is a Lie subalgebra by [F4]. Step 1.1 identifies $Lie (H)$ with it, and step 2.1 shows that this identification respects Lie brackets. This includes the zero-dimensional and full-dimensional cases. The only choice used is the stated ${A C}_{ω}$ inherited by [F3].

F3F4step 1.1step 2.1∎

Depends on

Used by

Cartan closed subgroup theorem Theorem

Dependency tree · two levels

22 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

John M. Lee, Introduction to Smooth Manifolds, 2nd ed. (standard reference, not scraped)
Alexander Kirillov Jr., An Introduction to Lie Groups and Lie Algebras (standard reference, not scraped)