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

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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Adjoint exponential identity

Statement

Assume ${A C}_{ω}$ . Let $G$ be a finite-dimensional real Lie group with Lie algebra $g$ . For every $X \in g$ ,

${Ad}_{\exp_{G} X} = e^{{ad}_{X}} .$

Here, for $B \in End (g)$ , the linear-ODE exponential $e^{t B}$ denotes the unique solution $E_{B} (t)$ of

$E_{B}^{'} (t) = B \circ E_{B} (t), E_{B} (0) = I,$

and $e^{B} = E_{B} (1)$ . The countable-choice assumption is inherited exactly from the supplied exponential and $d Ad = ad$ results.

Facts & Assumptions

Given: ${A C}_{ω}$ , a finite-dimensional real Lie group $G$ with Lie algebra $g$ , and $X \in g$ .

[F1]

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

[F2]

The adjoint map is a smooth representation, so ${Ad}_{g h} = {Ad}_{g} {Ad}_{h}$ and ${Ad}_{e} = I$ . Adjoint is a smooth Lie-group representation.

[F3]

The unique one-parameter subgroup with initial velocity $X$ is the curve $t \mapsto \exp_{G} (t X)$ . One-parameter subgroups are exactly exponentials. Exponential scales one-parameter subgroups.

[F4]

Under $T_{I} GL (g) = End (g)$ , $d (Ad)_{e} X = {ad}_{X}$ . The differential of Ad is ad.

[F5]

A linear matrix initial-value problem on a compact interval has a unique solution on the whole interval. Linear matrix ODEs have unique global solutions on a fixed interval.

Proof

technique · direct

1.1F2F3F4

Put $A (t) = {Ad}_{\exp_{G} (t X)}$ . By [F2]--[F3], $A$ is smooth, $A (0) = I$ , and $A (t + s) = A (t) A (s) = A (s) A (t)$ . The chain rule and [F4] give $A^{'} (0) = d (Ad)_{e} X = {ad}_{X}$ .

2.1step 1.1algebra

Fix $t$ and differentiate $A (t + s) = A (s) A (t)$ with respect to $s$ at zero. Using step 1.1 gives $A^{'} (t) = {ad}_{X} \circ A (t)$ and $A (0) = I$ .

3.1F5step 2.1

In one fixed basis of $g$ , [F5] gives a unique solution of $E^{'} (t) = {ad}_{X} E (t)$ , $E (0) = I$ , on every compact interval containing zero. Solutions on overlapping intervals agree by uniqueness, so they define the global linear-ODE exponential $E (t) = e^{t {ad}_{X}}$ without any arbitrary selection. Step 2.1 and uniqueness give $A (t) = E (t)$ on every such interval. Evaluating at $t = 1$ yields ${Ad}_{\exp_{G} X} = e^{{ad}_{X}}$ .

4.1F1F2F3F4F5step 1.1step 2.1step 3.1∎

A Lie group is nonempty and boundaryless. If $\dim G = 0$ , both sides are the unique endomorphism of the zero space; in dimension one, ${ad}_{X} = 0$ and the ODE gives both sides equal to $I$ . Degenerate adjoint endomorphisms are allowed. The compact-interval solutions glue globally, so there is no endpoint issue, and no metric occurs. The only choice use is the stated ${A C}_{ω}$ inherited through [F3]--[F4]; one finite basis and uniquely determined ODE solutions add no choice. No biconditional is asserted.

Adjoint exponential identity

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources