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

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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.

Depends on

Region charge equals boundary electric flux