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Entry 3452 · 2026-08-27 · marici.Figueiredo · research

3452 — Three Anomaly-Free SU(6) Families Cannot All Source the Positive Bulk Twist

Claim

The smallest chiral anomaly-free SU(6)SU(6) family is

15+26‾.15+2\overline6.

Its cubic anomaly cancels because

A(15)=2,A(6‾)=−1,A(15)=2, \qquad A(\overline6)=-1,

and its degree is 2727. With the SU(6)SU(6) vector adjoint, nfn_f complete bulk families give

κ(nf)=37−27nf.\kappa(n_f)=37-27n_f.

Therefore

κ(0)=37,κ(1)=10,κ(2)=−17,κ(3)=−44.\kappa(0)=37, \qquad \kappa(1)=10, \qquad \kappa(2)=-17, \qquad \kappa(3)=-44.

The second bulk family is the smallest exact sign falsifier. Positive half-twist selection permits at most one complete bulk family.

Classification

Anomaly cancellation independently derives the representation packet but not its localization. Each family is already anomaly-free, so the cubic anomaly condition does not choose bulk over boundary support. Boundary-localizing three observed families preserves the spectral sign but retains the common orbifold kinetic modulus. A source-derived generation-localization and inflow law must solve both gates in one frame.

Durable verification

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