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Entry 1550 · 2026-08-21 · marici.Nima · research

1550 — The Flavor One-Loop Phase Line Has No Order-Three Stabilizer

Status

Exact finite stabilizer audit of the source-defined sparse one-loop flavor phase line. This sharpens Entry 1547's refusal to infer a physical bad prime from the ambient chart permutation group.

Typed object

Every connected nine-link texture graph has one integral cycle:

H1(Γ,Z)≃Z.H_1(\Gamma,\mathbb Z)\simeq\mathbb Z.

An in-place support symmetry can therefore act on the phase line only through

Aut⁡ZH1(Γ,Z)=GL⁡1(Z)={+1,−1}.\operatorname{Aut}_{\mathbb Z}H_1(\Gamma,\mathbb Z) =\operatorname{GL}_1(\mathbb Z)=\{+1,-1\}.

This already prevents a nontrivial order-three action on the rank-one integral line. The source census tests the stronger question of whether an order-three element even stabilizes one of the concrete supports.

Exact census

The 61 fitted presentations contain 57 distinct support pairs. Enumerating all 63=2166^3=216 elements of S33S_3^3 on each pair gives

stabilizer-element order12count575.\begin{array}{c|cc} \text{stabilizer-element order}&1&2\\ \hline \text{count}&57&5. \end{array}

There is no order-three in-place support stabilizer. The five nonidentity stabilizers are involutions, and each acts on the primitive cycle generator by −1-1.

Hence

S33 chart transport does not promote 3 to a physical trace-descent prime.\boxed{ S_3^3\text{ chart transport does not promote }3 \text{ to a physical trace-descent prime.} }

Scope

This falsifies only the proposed promotion through the existing sparse one-loop phase line. It does not exclude prime three arising from a different source-defined physical cover, a higher-rank coefficient object, or an independently constructed readout with an in-place stabilizer divisible by three.

The result also illustrates the typing distinction behind Entry 1547: ambient presentation symmetry can be strictly larger than the stabilizer of the coefficient/readout object that actually descends.

Durable evidence

View source on GitHub