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

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The Lie-group exponential map is smooth with identity differential at zero

Statement

Assume ${A C}_{ω}$ . For a finite-dimensional real Lie group $G$ , the exponential map

$\exp_{G} : g = T_{e} G ⟶ G$

is smooth, satisfies $\exp_{G} (0) = e$ , and, under the canonical identification $T_{0} g ≃ g$ , has differential

$d (\exp_{G})_{0} = {id}_{g} .$

The countable-choice assumption is used exactly through the supplied smooth tangent-bundle trivialization and one-parameter-subgroup results.

Facts & Assumptions

Given: ${A C}_{ω}$ and a finite-dimensional real Lie group $G$ with identity $e$ and Lie algebra $g = T_{e} G$ .

[F1]

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

[F2]

The exponential map is the total map $\exp_{G} (Y) = γ_{Y} (1)$ . Exponential map of a Lie group.

[F3]

Assuming ${A C}_{ω}$ , the map $(g, Y) \mapsto d (L_{g})_{e} Y$ is a smooth vector-bundle trivialization $G \times g \to T G$ . Translations are diffeomorphisms and their differentials trivialize the tangent bundle.

[F4]

Assuming ${A C}_{ω}$ , every $Y \in g$ determines a unique global one-parameter subgroup $γ_{Y}$ , and $γ_{Y}^{'} (t) = d (L_{γ_{Y} (t)})_{e} Y$ . One-parameter subgroups are integral curves of left-invariant fields.

[F5]

The maximal flow of a smooth vector field has open domain, is smooth on that domain, and is uniquely determined by its maximal integral curves. The fundamental theorem on flows.

[F6]

The scaling identity is $γ_{Y} (a) = \exp_{G} (a Y)$ for all $a \in R$ . Exponential scales one-parameter subgroups.

[F7]

Every one-parameter subgroup satisfies $γ (0) = e$ . One-parameter subgroup of a Lie group.

[F8]

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

Proof

technique · direct

1.1F3construct

On the product manifold $M = G \times g$ , define $X_{(g, Y)} : = (d (L_{g})_{e} Y, 0) \in T_{g} G \times T_{Y} g = T_{(g, Y)} M .$ Its first component is the smooth map supplied by [F3], and its second component is the zero field on the vector space $g$ . Hence $X$ is a smooth vector field on $M$ .

2.1F3F4F5F8step 1.1algebra

For $(g, Y) \in M$ , define $c_{g, Y} : R \to M$ by $c_{g, Y} (t) = (g γ_{Y} (t), Y)$ . By [F4], [F8], and $L_{g} \circ L_{γ_{Y} (t)} = L_{g γ_{Y} (t)}$ , $c_{g, Y}^{'} (t) = (d (L_{g})_{γ_{Y} (t)} d (L_{γ_{Y} (t)})_{e} Y, 0) = (d (L_{g γ_{Y} (t)})_{e} Y, 0) = X_{c_{g, Y} (t)} .$ Thus $c_{g, Y}$ is an integral curve of $X$ through $(g, Y)$ and is defined on all of $R$ . Every maximal integral curve of $X$ is therefore global. By [F5], the global flow is smooth and is $Ψ (t, (g, Y)) = (g γ_{Y} (t), Y) .$

3.1F2F5step 2.1

Restricting this smooth flow to $(t, (e, Y))$ and projecting to $G$ shows that $(t, Y) \mapsto γ_{Y} (t)$ is smooth. Restricting further to $t = 1$ gives the map $Y \mapsto γ_{Y} (1) = \exp_{G} (Y)$ , which is smooth by [F2].

4.1F4F6F7F8step 3.1algebra

By [F6] with $a = 0$ and [F7], $\exp_{G} (0) = γ_{X} (0) = e$ . Fix $X \in g$ and consider $q_{X} (a) = a X$ . Applying [F8] to $\exp_{G} \circ q_{X}$ and then [F6] gives $d (\exp_{G})_{0} (X) = {\frac{d}{d a} ∣}_{a = 0} \exp_{G} (a X) = {\frac{d}{d a} ∣}_{a = 0} γ_{X} (a) = γ_{X}^{'} (0) = X .$ Hence $d (\exp_{G})_{0}$ is the identity under $T_{0} g ≃ g$ .

5.1F1F2F3F4F5F6F7F8step 1.1step 2.1step 3.1step 4.1∎

Lie groups are nonempty and boundaryless. If $\dim G = 0$ , then $g = 0$ , the exponential maps the unique vector to $e$ , and its differential is the identity of the zero space; in dimension one the proof is unchanged. The global curves in step 2.1 remove finite-time endpoint issues. No metric or nondegeneracy condition occurs. The only choice use is the stated ${A C}_{ω}$ , inherited through [F2]--[F4] and especially the smooth tangent-bundle trivialization [F3]; forming one product field and restricting its flow adds no choice. No biconditional is asserted.

The Lie-group exponential map is smooth with identity differential at zero

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources