ExampleConstruction: AI-generatedVerification: AI-generatedprecheck pendingjudge pass (gpt-6-luna)

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.

A nonzero pure-gauge potential with zero curvature

Example

Let $M = R^{2}$ with coordinates $(x, y)$ , let $G = S U (2)$ , and let $P = M \times G$ have the right action $(z, g) h = (z, g h)$ . Use the defining representation on $V = C^{2}$ with its standard positive-definite Hermitian form $h (v, w) = v^{*} w$ , let $E = P \times_{G} C^{2}$ , and take the admitted sector set to be $Σ = {[P]}$ . Set

$T_{3} = - \frac{i}{2} (\begin{array}{cc} 1 & 0 \\ 0 & - 1 \end{array}), u (x, y) = (\begin{array}{cc} e^{- i x / 2} & 0 \\ 0 & e^{i x / 2} \end{array}), A = u^{- 1} d u = T_{3} d x .$

Equip $P$ with the principal form

ω_{(z, g)} (X, ξ) = {Ad}_{g^{- 1}} A_{z} (X) + (θ_{L})_{g} (ξ),

where $θ_{L}$ is the left Maurer–Cartan form. This is a smooth principal connection form with local potential $A$ along the canonical section $s (z) = (z, e)$ . Its Cartan curvature $F = d A + \frac{1}{2} [A \land A]$ vanishes identically although $A$ is nonzero; in this sense $A = u^{- 1} d u$ is a nonzero pure-gauge potential on the trivial bundle.

Example references

Vákár, §1.3.1, principal connection forms

Vákár, §2.5.1, potentials and principal connections

Neeb, Chapter 5 §5.4, trivial-bundle connection forms

Neeb, Chapter 6 §6.1, Cartan curvature formula

Non-Abelian gauge datum and global group choice

Principal connection form

Cartan curvature

The group $S U (2)$ and its Lie algebra

Closed and bounded subsets of Euclidean space are compact

The adjoint action is a group representation

Facts & Assumptions

Given: ${A C}_{ω}$ , the finite-dimensional matrix group $S U (2)$ , the smooth base $M = R^{2}$ , and the matrices and functions displayed in the example.

[A1]

The supplied gauge datum requires a compact noncommutative Lie group, a smooth right principal bundle, and specified smooth unitary representations. Non-Abelian gauge data and bundle sectors

[F1]

$S U (2)$ is an embedded finite-dimensional real Lie group in $U (2)$ with $s u (2) = {X : X^{*} + X = 0, tr X = 0}$ ; its defining inclusion is unitary for the standard Hermitian form. Unitary and special unitary Lie groups

[F2]

A closed bounded subset of finite-dimensional Euclidean space is compact; this Heine–Borel result uses no choice. Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line

[F3]

A smooth right principal connection form reproduces fundamental vectors and is equivariant under the right action; conversely such a form defines its horizontal kernel. Principal connection 1-form from a horizontal distribution

[F4]

For the right action, the local potential is $s^{*} ω$ , and the Cartan curvature is $F = d A + \frac{1}{2} [A \land A]$ with the stated shuffle bracket. Principal curvature, Cartan structure equation, and Bianchi identity, Graded bracket of Lie-algebra-valued differential forms

[F5]

The left Maurer–Cartan form is smooth and is $(θ_{L})_{g} = d (L_{g^{- 1}})_{g}$ ; on matrix groups it evaluates as $g^{- 1} d g$ . Left Maurer--Cartan form

[F6]

The adjoint maps obey ${Ad}_{g h} = {Ad}_{g} {Ad}_{h}$ . Adjoint is a smooth Lie-group representation

[F7]

A smooth principal bundle has a smooth projection and right action and equivariant local product charts. Smooth principal $G$ -bundle

[A2]

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

Verification

Given: $M = R^{2}$ , $G = S U (2)$ , $P = M \times G$ , and $A = T_{3} d x$ .

1.1A1A2F1F2algebra

Regard $M_{2} (C)$ as $R^{8}$ . The conditions $g^{*} g = I$ and $\det g = 1$ are closed equations in the matrix entries, so $S U (2)$ is closed. For a unitary matrix each column has norm one, hence every entry has absolute value at most one; thus the subset is bounded. By [F2], it is compact. The matrices $D = (\begin{array}{cc} i & 0 \\ 0 & - i \end{array})$ and $J = (\begin{array}{cc} 0 & 1 \\ - 1 & 0 \end{array})$ are unitary with determinant one, while $D J = (\begin{array}{cc} 0 & i \\ i & 0 \end{array}) \neq (\begin{array}{cc} 0 & - i \\ - i & 0 \end{array}) = J D$ . Therefore $S U (2)$ is noncommutative. Together with [F1], this verifies the group conditions in [A1]. The declared axiom base is ${A C}_{ω}$ by [A2], and [F2] requires no additional choice.

1.2A1F1algebra

The matrix $u (x, y)$ is unitary and has determinant one because its diagonal entries have modulus one and multiply to one. Differentiating its entries gives $d u = diag (- \frac{i}{2} e^{- i x / 2}, \frac{i}{2} e^{i x / 2}) d x$ , hence $u^{- 1} d u = T_{3} d x$ . The coefficient $T_{3} \neq 0$ , so $A$ is not the zero form. The defining representation is unitary.

2.1F1F3F5F6F7step 1.1step 1.2

The product $P = M \times S U (2)$ has projection to $M$ and right action on its second coordinate; both are smooth, the action is free and transitive on each fibre, and the product chart is equivariant, so [F7] makes it a smooth right principal bundle. By [F5], the displayed $ω$ is smooth. On the fundamental vector generated by $Z \in s u (2)$ , the tangent is $(0, g Z)$ , and $(θ_{L})_{g} (g Z) = Z$ ; thus $ω (Z_{P}) = Z$ . For fixed $h \in S U (2)$ , right translation sends $(X, ξ)$ to $(X, ξ h)$ , and ${Ad}_{(g h)^{- 1}} A (X) + (θ_{L})_{g h} (ξ h) = {Ad}_{h^{- 1}} ({Ad}_{g^{- 1}} A (X) + (θ_{L})_{g} (ξ))$ . The equality uses [F6] for the adjoint product and the matrix formula for $θ_{L}$ from [F5]. Thus $R_{h}^{*} ω = {Ad}_{h^{- 1}} ω$ , and [F3] shows it is a principal connection form. Since $s (z) = (z, e)$ , its pullback is $s^{*} ω = A$ .

3.1F4step 1.2step 2.1∎

The coefficient $T_{3}$ is constant and $d (d x) = 0$ , so $d A = 0$ . For tangent vectors $X, Y$ , the bracket term is $[A (X), A (Y)]$ ; both values are scalar multiples of $T_{3}$ , so this bracket is zero. Therefore $F = d A + \frac{1}{2} [A \land A] = 0$ everywhere by [F4], while step 1.2 showed $A \neq 0$ .

A nonzero pure-gauge potential with zero curvature

Example

Example references

Facts & Assumptions

Verification

Depends on

Dependency tree · 2 levels

Sources