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

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Adjoint intertwines the exponential map

Statement

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

$g \exp_{G} (X) g^{- 1} = \exp_{G} ({Ad}_{g} X) .$

The countable-choice assumption is inherited exactly from exponential naturality.

Facts & Assumptions

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

[F1]

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

[F2]

Conjugation $C_{g} (h) = g h g^{- 1}$ is a Lie-group automorphism and $d (C_{g})_{e} = {Ad}_{g}$ . Conjugation and the adjoint representation of a Lie group.

[F3]

Every Lie-group homomorphism $F$ satisfies $F (\exp X) = \exp (d F_{e} X)$ , assuming ${A C}_{ω}$ . Exponential map is natural for Lie-group homomorphisms.

Proof

technique · direct

1.1

Apply exponential naturality [F3] to the conjugation automorphism $C_{g}$ from [F2]. Since $d (C_{g})_{e} = {Ad}_{g}$ , it gives $C_{g} (\exp_{G} X) = \exp_{G} ({Ad}_{g} X)$ . Expanding the definition of $C_{g}$ is the claimed identity.

F2F3

2.1

A Lie group is nonempty and boundaryless. If $\dim G = 0$ , both sides equal the conjugate of the identity, and dimension one requires no change; $X = 0$ gives $e$ on both sides. No metric, degeneracy, interval, endpoint, or biconditional occurs. The only choice use is the stated ${A C}_{ω}$ inherited through [F3]; fixing one $g$ and one $X$ adds no choice.

F1F2F3step 1.1∎

Depends on

Used by

Dependency tree · two levels

21 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

Anthony W. Knapp, Lie Groups Beyond an Introduction, 2nd ed. (standard reference, not scraped)
Alexander Kirillov Jr., An Introduction to Lie Groups and Lie Algebras (standard reference, not scraped)