CounterexampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck passjudge 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.

The Countable-Mode CAR Algebra Has a Non-Fock-Equivalent Irreducible Representation

Statement refuted

The false statement is: every irreducible representation of the countably infinite-mode CAR C*-algebra is unitarily equivalent to its occupation-space Fock representation.

Let $π_{F}$ be the Fock representation on $F_{\infty}$ , and let $α$ exchange every annihilator with its creator. Then $π_{F} \circ α$ is irreducible and is not unitarily equivalent to $π_{F}$ .

Facts & Assumptions

Given: The countably infinite-mode CAR algebra, its Fock representation, and the particle-hole automorphism.

[F1]

The representation $π_{F} \circ α$ is irreducible and inequivalent to $π_{F}$ (Countable-Mode CAR Has Inequivalent Irreducible Representations).

Counterexample

1.1givenF1construct

The automorphism $α (a_{j}) = a_{j}^{†}$ is defined on the same countable-mode CAR algebra, so composing the Fock representation with $α$ gives another representation of that algebra.

2.1givenF1step 1.1algebra∎

By [F1], this representation is irreducible but not unitarily equivalent to $π_{F}$ . It is therefore a counterexample to the universal claim that every irreducible representation is Fock-equivalent.

Depends on

Countable-Mode CAR Has Inequivalent Irreducible Representations

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

3 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.