ExampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedjudge 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 and ad for a matrix Lie group

Example

Assume ${A C}_{ω}$ . If $G \subseteq {GL}_{n} (R)$ is a matrix Lie group with Lie algebra $g \subseteq M_{n} (R)$ , then

${Ad}_{g} X = g X g^{- 1}, {ad}_{X} Y = X Y - Y X .$

Facts & Assumptions

Given: $g \in G$ and $X, Y \in g$ .

[F1]

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

[F2]

The group differential satisfies $d (Ad)_{I} (X) = {ad}_{X}$ . The differential of Ad is ad.

[F3]

For a matrix Lie group, $\exp_{G} (t X) = e^{t X}$ . Matrix exponential as the Lie-group exponential.

[F4]

The choice assumption used by [F2] and [F3] is countable choice. The Axiom of Countable Choice ( ${A C}_{ω}$ ).

Verification

technique · differentiate the displayed matrix curves

1.1

The tangent curve $c (t) = I + t X + o (t)$ gives $C_{g} (c (t)) = I + t (g X g^{- 1}) + o (t)$ . By [F1], differentiating at zero proves ${Ad}_{g} X = g X g^{- 1}$ .

F1algebra

2.1

By [F3], a curve through the identity with velocity $X$ is $e^{t X} = I + t X + o (t)$ , whose inverse is $e^{- t X} = I - t X + o (t)$ . Step 1.1 therefore gives ${Ad}_{e^{t X}} Y = e^{t X} Y e^{- t X} = Y + t (X Y - Y X) + o (t)$ .

F3step 1.1algebra

3.1

Differentiating step 2.1 at zero yields $d (Ad)_{I} (X) (Y) = X Y - Y X$ ; [F2] identifies the left side with ${ad}_{X} (Y)$ and proves the second formula. For $n = 0$ all matrices and maps are uniquely zero; for $X = 0$ or $Y = 0$ the commutator vanishes as the formula says. No invertibility is required of $X$ or $Y$ , there is no metric or endpoint condition, and no iff is asserted. ${A C}_{ω}$ is used exactly through [F2] and [F3]; differentiating the fixed curves adds no choice.

F2F3F4step 1.1step 2.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

35 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)