TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-14

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Differential of a Lie-group homomorphism is a Lie-algebra homomorphism

Statement

Assume ${A C}_{ω}$ . Let $F : G \to H$ be a homomorphism of finite-dimensional real Lie groups, with Lie algebras $g = T_{e} G$ and $h = T_{e_{H}} H$ . Then

$d F_{e} : g ⟶ h$

is a Lie-algebra homomorphism. The countable-choice assumption is used exactly through the supplied smooth invariant-field and tangent-bracket results.

Facts & Assumptions

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

[F1]

${A C}_{ω}$ is countable choice. The Axiom of Countable Choice ( ${A C}_{ω}$ ).

[F2]

A Lie-algebra homomorphism is a linear map preserving brackets. Lie-algebra homomorphism.

[F3]

The map $F$ is smooth, preserves identities, and satisfies $F (g h) = F (g) F (h)$ . Lie-group homomorphism, isomorphism, and automorphism.

[F4]

Pairs of related smooth vector fields have related brackets. Related vector fields have related Lie brackets.

[F5]

Assuming ${A C}_{ω}$ , each tangent vector has a unique left-invariant smooth extension $X_{g}^{L} = d (L_{g})_{e} X$ . Left-invariant vector fields evaluate isomorphically at the identity.

[F6]

The tangent bracket is characterized by $[X^{L}, Y^{L}] = [X, Y]_{g}^{L}$ , and similarly for $h$ . Lie bracket on the tangent space of a Lie group.

[F7]

The differential of a smooth map at a point is linear. The differential sends derivations to derivations and is linear.

[F8]

Differentials of smooth maps obey the chain rule. The chain rule for differentials of smooth maps.

Proof

technique · direct

1.1F3F5F8algebra

Fix $X \in g$ . By [F5], $X$ and $d F_{e} X$ have left-invariant extensions $X^{L}$ on $G$ and $(d F_{e} X)^{L}$ on $H$ . For every $g \in G$ , the homomorphism law [F3] gives $F \circ L_{g} = L_{F (g)} \circ F$ . Therefore [F8] gives $d F_{g} (X_{g}^{L}) = d F_{g} d (L_{g})_{e} X = d (L_{F (g)})_{e_{H}} d F_{e} X = (d F_{e} X)_{F (g)}^{L} .$ Thus $X^{L}$ and $(d F_{e} X)^{L}$ are $F$ -related.

2.1F4F5F6step 1.1

Apply [F4] to the related pairs from step 1.1 for $X$ and $Y$ . Then $[X^{L}, Y^{L}]$ is $F$ -related to $[(d F_{e} X)^{L}, (d F_{e} Y)^{L}]$ . Evaluating relatedness at $e$ and using [F6] on both groups yields $d F_{e} ([X, Y]_{g}) = [d F_{e} X, d F_{e} Y]_{h} .$

3.1F2F6F7step 2.1

By [F7], $d F_{e}$ is linear, and step 2.1 proves bracket preservation. Hence $d F_{e}$ is a Lie-algebra homomorphism by [F2].

4.1F1F2F3F4F5F6F7F8step 1.1step 2.1step 3.1∎

Lie groups are nonempty and boundaryless. If either Lie algebra is zero-dimensional, the same related-field calculation applies and all relevant source vectors or target values are zero; in dimension one the proof is unchanged. No metric, nondegeneracy, interval, or endpoint occurs. The only choice use is the stated ${A C}_{ω}$ , inherited through [F5] and [F6]; fixing two tangent vectors adds no choice. No biconditional is asserted.

Depends on

Dependency tree · 2 levels

Sources

Alexander Kirillov Jr., An Introduction to Lie Groups and Lie Algebras