Operator Irreducibility Does Not Select the Flavor Spectrum
The three-state charge and mixing packet and a five-state packet completed by an anomaly-neutral pair both have scalar common commutant. The five-state mixing operator couples the added states to the base sector and is nondegenerate, so the completion is not rejected by the tested basis-invariant irreducibility condition.
Both packets preserve linear charge sum 6, cubic anomaly 36, and primitive portal contrast one. Their current indices are respectively 14 and 16. Scaling the five-state mixing operator preserves irreducibility while leaving a continuous spectral scale.
The threshold-response fiber therefore survives:
Consequence
Scalar-common-commutant irreducibility can reject reducing sectors inside one chosen packet. It does not compare or select among inequivalent irreducible packets. The missing source principle must uniquely select dimension, representations, mixing spectrum, and scale across the full completion class.
Evidence
- Packet: research/flavor/flavor-operator-irreducible-vectorlike-completion-no-go.md
- Checker: research/flavor/checkers/wp833_operator_irreducible_vectorlike_completion_no_go.py
- Generated result: research/flavor/results/wp833_operator_irreducible_vectorlike_completion_no_go.json
- Exact result: 12 of 12 checks passed.
- Ledger-sequence claim: seqclaim-f92f903e4a46e2901702d4af, value 3651.
- Graph admission: ev-000000007842-e3b22c91-b8f3-4d1d-9a1b-06a7d9333434.
Claim boundary
The displayed pair is a finite operator model, not a completed spectral triple or quantum field theory. The theorem rejects scalar-common-commutant irreducibility alone; gauge covariance, real structure, grading, first-order condition, positivity, and a physical action remain separate gates.