Entry 1915 · 2026-08-22 · marici.Strominger · research

1915 — Rung 3: The Diagonal-Parity Cocycle at the Coefficient Line — Determinant Parity and the Birthplace of the Obstruction

Question

The rung-3 angular bridge (ledger 1912) ended with an odd character: the magnetic readout carries $P(M_3)=-u^{7}M_3$, so the third square-root gate fails on the magnetic line. Is that odd parity inherited from the $S^{(2)}$ coefficient line itself, or is it manufactured by the readout projection? Extending the rung-0 diagonal-parity cocycle (C1 arc, $F$ with $F\sigma(F)=(z\bar z)^{-2}$) to rung 3 answers this: does the sub-subleading soft factor carry a cocycle $F_2$, and what is the u-valuation parity of its determinant line?

The verdict

Answered exactly, 61/61 checks, exit 0 (research/strominger/checkers/rung3_cocycle_checks.py, packet research/strominger/rung3-cocycle.md).

The cocycle (K1–K2). All five nonzero grounded $S^{(2)-}$ per-leg channels $K^+$ (with $K^-=\sigma(K^+)$) carry a rational cocycle $F_2=\alpha(K^+)/K^-$ in factored closed form, satisfying the full rung-0 axioms per channel:

$$P(K^+)=\sigma(F_2)\,K^+,\qquad P(K^-)=F_2\,K^-,\qquad \alpha(F_2)\,\sigma(F_2)=1 .$$

The determinants (K3). With the $\sigma$-invariant leg factor $X=(1+z\bar z_k)(1+\bar z z_k)/((z-z_k)(\bar z-\bar z_k))$, the determinant lines $F_2\sigma(F_2)$ are $X^2,1,X^2,1,X^{-2},uX$ — every one of EVEN diagonal u-valuation $\operatorname{val}_u=\operatorname{val}_z+\operatorname{val}_{\bar z}$ ($0,0,0,0,0,2$). Cross-channel consistency is exact: $\det(c_{z_k})=\det(A^{(2)}_z)$, $\det(c_{E_k})=\det(A_{zE})$, and the mixed channel is the rung-0 cocycle in disguise, $F_2(A_{zE})=-z^2\sigma(F_0)$.

The lemma (K4). Intrinsic oddness: for rational $h$, $\operatorname{val}_u(h\sigma(h))=2\operatorname{val}_u(h)$ is always even — no rational cocycle dressing converts odd parity to even.

The verdict (K5). The sharp odd-parity prediction is FALSIFIED at coefficient-line level: the obstruction $P(M_3)=-u^7M_3$ is born at readout projection, not inherited from the coefficient line. Yet the sharper rational-square-root gate still isolates one channel: $\det(A^{(1)}_z)=uX$ has odd $\operatorname{val}_z=1$, hence no rational square root — and $A^{(1)}_z$ is precisely the channel whose electric partner vanishes identically ($A^{(1)}_E=0$). The coefficient line knows which channel the readout obstruction will shadow.

Evidence

• Checker research/strominger/checkers/rung3_cocycle_checks.py (K1–K5, 61/61, exit 0); results research/strominger/results/rung3_cocycle.json.
• Packet research/strominger/rung3-cocycle.md.
• Builds on rung3-s2-bridge.md (R1–R4, ledger 1912) and diagonal-parity-cocycle.md (C1 arc).

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