Pipeline-generated

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.

Exact compact U1 rotor plus finite CAR data — Examples

1 · Prerequisites

2 · Summary

These examples give an explicit nonzero state in a fixed-background Gauss sector, a one-fermion counterexample to arbitrary total charge on a closed compact graph, and a charged CAR operator whose expectation vanishes in a physical state. Each verification is conditional on the exact finite CAR supplier stated on the A page; none asserts a continuum limit or an empirical claim.

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

None yet.

5 · Examples, experiments and counterexamples

ExampleConstruction: AI-generatedVerification: AI-generatedverified 2026-10-15 (gpt-6-luna, gpt-6.1-sol, and delegated Codex review/repair agents; see completed run evidence)Open item page →

Explicit physical electric/occupation basis state

Example

Take two vertices $v_{1}, v_{2}$ , one oriented link $e : v_{1} \to v_{2}$ , one fermion mode at each vertex, and background $b_{v_{1}} = 1, b_{v_{2}} = 0$ . Let $ψ$ be the tensor of the zero-electric Fourier vector and the occupation vector with the $v_{1}$ mode occupied and the $v_{2}$ mode empty. This is a concrete nonzero state in the fixed-background physical sector.

Facts & Assumptions

Given: The conditional finite-graph CAR model, one oriented link, and the chosen backgrounds.

[F1]

Gauss operators, gauge group and nonempty physical projection: for the conditional compact-rotor/CAR data, a normalized Fourier/occupation basis vector lies in $H_{p h y s} = ⋂_{x} \ker G_{x}$ exactly when every integer Gauss label $γ_{x}$ is zero, where $γ_{x}$ is outgoing link label minus incoming link label minus site occupation plus $b_{x}$ .

Verification

Given: [F1] and the stated graph, occupation, and link labels.

1.1F1given

The vector $ψ$ is a normalized Fourier/occupation basis vector: its link label is $n_{e} = 0$ , its site occupations are $n_{v_{1}} = 1, n_{v_{2}} = 0$ , and these occupations lie in the finite fermion Fock basis.

2.1F1step 1.1algebra∎

At $v_{1}$ , $γ_{v_{1}} = n_{e} - (n_{v_{1}} - b_{v_{1}}) = 0 - (1 - 1) = 0$ . At $v_{2}$ , $γ_{v_{2}} = - n_{e} - (n_{v_{2}} - b_{v_{2}}) = 0 - (0 - 0) = 0$ . Thus every Gauss operator annihilates $ψ$ , so $ψ \in H_{p h y s}$ and the physical projection is nonzero.

CounterexampleConstruction: AI-generatedVerification: AI-generatedverified 2026-10-15 (gpt-6-luna, gpt-6.1-sol, and delegated Codex review/repair agents; see completed run evidence)Open item page →

Compact gauge model fixes total matter charge

Statement refuted

For every finite oriented graph and every fixed background $b$ , the compact gauge model has a physical state in every total matter-occupation sector $N \in {0, \dots, s ∣ V ∣}$ , independently of $b$ .

Construction

Use two vertices $v_{1}, v_{2}$ , one link $e : v_{1} \to v_{2}$ , one fermion mode per vertex, and $b_{v_{1}} = b_{v_{2}} = 0$ . A state with exactly one occupied fermion mode cannot be physical, whatever link-electric basis label is chosen: summing the two Gauss equations fixes total occupation to zero.

Facts & Assumptions

Given: The finite-graph CAR model on the specified graph, fixed zero background, and the Gauss-law physical subspace.

[F1]

Region charge equals boundary electric flux: on a graph with no external boundary, every physical vector satisfies $\sum_{x} n_{x} = \sum_{x} b_{x}$ .

Counterexample

Given: [F1] and the graph and background in the Construction.

1.1givenF1algebra

Here $G_{v_{1}} = E_{e} - n_{v_{1}}$ and $G_{v_{2}} = - E_{e} - n_{v_{2}}$ , so $G_{v_{1}} + G_{v_{2}} = - (n_{v_{1}} + n_{v_{2}})$ on their common domain; the link flux cancels.

2.1F1step 1.1algebra∎

The zero-background identity from [F1] requires $n_{v_{1}} + n_{v_{2}} = 0$ on every physical vector. The normalized occupation vector with $n_{v_{1}} = 1, n_{v_{2}} = 0$ has total number one, so it is not physical and the asserted every-sector property fails.

ExampleConstruction: AI-generatedVerification: AI-generatedverified 2026-10-15 (gpt-6-luna, gpt-6.1-sol, and delegated Codex review/repair agents; see completed run evidence)Open item page →

Charged expectation vanishes by gauge averaging

Example

Take a two-vertex graph with one link $e : v_{1} \to v_{2}$ , one fermion mode per vertex and zero backgrounds. The zero-electric, zero-occupation vector $Ω$ is physical. The bounded creation operator $O = c_{v_{1}, 1}^{†}$ has nonzero gauge charge at $v_{1}$ , so its expectation in any physical vector, including $Ω$ , vanishes.

Facts & Assumptions

Given: The conditional finite-graph CAR model, the gauge action, and the Axiom of Choice required by the time-dependent Ward supplier.

[A1]

The Axiom of Choice is the principle that every family of nonempty sets has a choice function. The Axiom of Choice

[F1]

Exact regulated gauge and infinitesimal Ward identities: every bounded gauge-charge eigenoperator with nonzero charge has zero expectation in a physical vector; time evolution preserves gauge charge under AC. This current A-page item depends on the still-draft QFT-MF01 CAR supplier through its graph-model dependencies.

Verification

Given: [A1], [F1], and the zero-background two-vertex state described above.

1.1A1F1given

For $Ω$ , $n_{v_{1}} = n_{v_{2}} = 0$ and $E_{e} = 0$ , so both Gauss eigenvalues vanish. The CAR creation operator is bounded and nonzero; the gauge law gives $U (α) O U (α)^{†} = e^{- i α_{v_{1}}} O$ , so its charge vector is $q_{v_{1}} = - 1, q_{v_{2}} = 0$ .

2.1F1step 1.1algebra∎

The nonzero charge in step 1.1 forces $⟨ Ω, O Ω ⟩ = 0$ by the finite-regulator selection rule, even though $O Ω$ is a nonzero one-fermion vector. This demonstrates that the selection rule concerns expectations in the physical sector, not the vanishing of the charged operator itself.

Exact compact U1 rotor plus finite CAR data — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

5 · Examples, experiments and counterexamples

Explicit physical electric/occupation basis state

Example

Facts & Assumptions

Verification

Compact gauge model fixes total matter charge

Statement refuted

Construction

Facts & Assumptions

Counterexample

Charged expectation vanishes by gauge averaging

Example

Facts & Assumptions

Verification

Sources