PropositionStatement: AI-adaptedProof: AI-adaptedPipeline-generatedprecheck passaudited 2026-09-14

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Commuting Lie-algebra elements have multiplicative exponentials

Statement

Assume ${A C}_{ω}$ . Let $G$ be a finite-dimensional real Lie group with Lie algebra $g$ . If $X, Y \in g$ satisfy $[X, Y]_{G} = 0$ , then

$\exp_{G} (X + Y) = \exp_{G} (X) \exp_{G} (Y) = \exp_{G} (Y) \exp_{G} (X) .$

The countable-choice assumption is used exactly through the supplied tangent bracket, invariant-field, and exponential results.

Facts & Assumptions

Given: ${A C}_{ω}$ , a finite-dimensional real Lie group $G$ with identity $e$ and Lie algebra $g$ , and $X, Y \in g$ with $[X, Y]_{G} = 0$ .

[F1]

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

[F2]

Lie-group multiplication $m : G \times G \to G$ is smooth. Lie group.

[F3]

The tangent bracket satisfies $[X^{L}, Y^{L}] = [X, Y]_{G}^{L}$ . Lie bracket on the tangent space of a Lie group.

[F4]

Two smooth vector fields have commuting local flows if and only if their bracket vanishes. Two vector fields commute if and only if their local flows commute.

[F5]

The one-parameter subgroup $γ_{X}$ is the global identity integral curve of $X^{L}$ . One-parameter subgroups are integral curves of left-invariant fields.

[F6]

The scaling and additive-parameter identities are $γ_{X} (t) = \exp_{G} (t X)$ and $\exp_{G} ((s + t) X) = \exp_{G} (s X) \exp_{G} (t X)$ . Exponential scales one-parameter subgroups.

[F7]

A one-parameter subgroup is a smooth homomorphism $(R, +) \to G$ . One-parameter subgroup of a Lie group.

[F8]

A one-parameter subgroup with initial velocity $Z$ is exactly the curve $t \mapsto \exp_{G} (t Z)$ . One-parameter subgroups are exactly exponentials.

[F9]

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

Proof

technique · direct

1.1

By [F3], $[X^{L}, Y^{L}] = [X, Y]_{G}^{L} = 0$ . The forward implication of [F4] therefore says that the global flows of $X^{L}$ and $Y^{L}$ commute.

F3F4

2.1

By [F3], $X^{L}$ and $Y^{L}$ are left invariant. For fixed $g$ , the chain rule and left invariance show that the left translates of the identity integral curves in [F5] are integral curves through $g$ . Hence [F5] and [F6] give the global flows $Φ_{t}^{X} (g) = g \exp_{G} (t X)$ and $Φ_{s}^{Y} (g) = g \exp_{G} (s Y)$ . Evaluating the commutation identity from step 1.1 at $e$ gives $\exp_{G} (s Y) \exp_{G} (t X) = \exp_{G} (t X) \exp_{G} (s Y)$ for every $s, t \in R$ .

F3F4F5F6F9step 1.1

3.1

Define $c (t) = \exp_{G} (t X) \exp_{G} (t Y)$ . It is smooth by [F2], [F5], and [F6]. For $s, t \in R$ , [F6] and step 2.1 give $\begin{aligned} c (s + t) & = \exp_{G} (s X) \exp_{G} (t X) \exp_{G} (s Y) \exp_{G} (t Y) \\ = \exp_{G} (s X) \exp_{G} (s Y) \exp_{G} (t X) \exp_{G} (t Y) = c (s) c (t) . \end{aligned}$ Also $c (0) = e$ , so $c$ is a one-parameter subgroup by [F7].

F2F6F7step 2.1algebra

4.1

Let $m : G \times G \to G$ be multiplication and put $α (t) = \exp_{G} (t X)$ and $β (t) = \exp_{G} (t Y)$ . The identities $m (g, e) = g$ and $m (e, h) = h$ show that $d m_{(e, e)} (X, 0) = X$ and $d m_{(e, e)} (0, Y) = Y$ . Since a differential is linear, $d m_{(e, e)} (X, Y) = X + Y .$ Hence [F5], [F6], and [F9] give $c^{'} (0) = X + Y$ .

F2F5F6F9step 3.1algebra

5.1

By [F8], the one-parameter subgroup $c$ with initial velocity $X + Y$ is $c (t) = \exp_{G} (t (X + Y))$ . At $t = 1$ , $\exp_{G} (X + Y) = \exp_{G} (X) \exp_{G} (Y) .$ Step 2.1 with $s = t = 1$ also gives $\exp_{G} (X) \exp_{G} (Y) = \exp_{G} (Y) \exp_{G} (X)$ .

F7F8step 2.1step 3.1step 4.1

6.1

Lie groups are nonempty and boundaryless. If $\dim G = 0$ , then $X = Y = 0$ and all terms equal $e$ ; in dimension one the proof is unchanged. All flows and exponential curves are global, so no endpoint issue occurs. No metric or nondegeneracy condition occurs. The only choice use is the stated ${A C}_{ω}$ , inherited through [F3], [F5], [F6], and [F8]; fixing two fields and finitely many parameters adds no choice. No biconditional is asserted.

F1F2F3F4F5F6F7F8F9step 1.1step 2.1step 3.1step 4.1step 5.1∎

Depends on

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Sources

Alexander Kirillov Jr., An Introduction to Lie Groups and Lie Algebras (standard reference, not scraped)