False statementConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedprecheck passaudited 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.

The plus exponential convention is not a homomorphism for left actions

False statement

Assume ${A C}_{ω}$ . For a smooth left action, the plus-sign assignment

$X ⟼ {\hat{X}}_{M}, {\hat{X}}_{M} (p) = {\frac{d}{d t} ∣}_{0} \exp (t X) \cdot p$

is a Lie-algebra homomorphism.

Facts & Assumptions

Given: ${A C}_{ω}$ and a smooth left action of a Lie group $G$ on $M$ . Write $X_{M}$ for the library's minus-sign fundamental field and ${\hat{X}}_{M}$ for the plus-sign field in the false claim.

[A1]

The standing definition is $X_{M} (p) = {\frac{d}{d t} ∣}_{0} \exp (- t X) \cdot p$ , and its field assignment is a Lie-algebra homomorphism. The Axiom of Countable Choice ( ${A C}_{ω}$ ), Fundamental vector fields for a left action, Fundamental vector fields form a Lie-algebra homomorphism.

[F1]

A Lie group has smooth multiplication and inversion, and $E_{i j}$ denotes the matrix with its single nonzero entry $1$ in position $(i, j)$ . Lie group, Matrix units $E_{i j}$ and the Kronecker delta. The determinant is the usual finite polynomial. For $n \geq 1$ , the determinant over a commutative ring by the Leibniz formula, and $∣ \det A ∣$ for a real matrix.

Refutation

technique · compute the sign and evaluate it on a nonabelian action

1.1

Replacing $t$ by $- t$ in [A1] gives ${\hat{X}}_{M} = - X_{M}$ . Therefore bilinearity and the theorem in [A1] give $[{\hat{X}}_{M}, {\hat{Y}}_{M}] = [X_{M}, Y_{M}] = [X, Y]_{M} = - {\hat{[X, Y]}}_{M}$ . Thus the plus-sign assignment is an antihomomorphism.

A1algebra

2.1

Let $G = {GL}_{2} (R)$ act on itself by left multiplication. The determinant-nonzero locus is open in $M_{2} (R)$ ; multiplication is polynomial and the formula $A^{- 1} = (\det A)^{- 1} adj (A)$ makes inversion smooth there, so [F1] makes $G$ a Lie group and its left action smooth. Take $X = E_{01}$ and $Y = E_{10}$ . Direct matrix multiplication gives $[X, Y] = E_{00} - E_{11} \neq 0$ . At the identity, the plus fundamental field of this bracket has value ${\frac{d}{d t} ∣}_{0} \exp (t (E_{00} - E_{11})) = E_{00} - E_{11} \neq 0$ . Hence step 1.1 yields $[{\hat{X}}_{M}, {\hat{Y}}_{M}] = - {\hat{[X, Y]}}_{M} \neq {\hat{[X, Y]}}_{M}$ , so the claimed homomorphism identity fails.

F1step 1.1algebraconstruct

3.1

The statement is therefore false; the minus sign in the library convention is essential. For abelian groups both signs give the zero bracket, which is why a nonabelian witness is required. The witness is the four-dimensional open matrix group ${GL}_{2} (R)$ and has no endpoint or degenerate issue. ${A C}_{ω}$ is inherited through [A1]; the explicit matrix calculation itself is finite and choice-free.

A1F1step 2.1discharge-construct: nonzero $2$-by-$2$ matrix bracket witness∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

32 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

John M. Lee, Introduction to Smooth Manifolds, 2nd ed. (standard reference, not scraped)
Pavel Etingof, MIT 18.745 Lie Groups and Lie Algebras I (standard reference, not scraped)