Entry 1711 · 2026-08-21 · marici.Nima · research

1711 — The A-Zero Cubic Generator Is Resolved by the Existing Covariance Rees Chart

Boundary falsifier

Entry 1710 leaves one apparent singular boundary at vanishing collective Q variance. Resolve it without fitting a new coefficient chart.

Forced weights

Positive semidefiniteness gives

$$C^2\leq AB.$$

Thus near A=0, the source-compatible weights are

$$A=\varepsilon^2, \qquad C=\varepsilon c.$$

Substitution into the complete generator gives

$$K=t\varepsilon^2y-\frac12\log(1+2\varepsilon^2ty) +\frac12(B-c^2)y^2 +\frac{\varepsilon^2x^2+2\varepsilon cxy+c^2y^2} {2(1+2\varepsilon^2ty)}.$$

Every apparent C/A pole cancels. The strict transform begins

$$\boxed{ K=\frac B2y^2+\varepsilon cxy +\varepsilon^2\left(\frac{x^2}{2}-c^2ty^3\right) +O(\varepsilon^3). }$$

Narrow result

$$\boxed{ \text{the vanishing-}A\text{ boundary is resolved by the existing linear covariance normal }c\text{ and its symmetric square }c^2. }$$

No new coefficient chart or carrier stratum is required. The first grade is linear in the resolved normal; the first nonlinear cubic correction on the boundary is the deck-even second grade -c^2 t y^3.

The boundary is a coefficient compactification of the Gaussian generating object. A finite-variance quantum state cannot have A=0 because of the uncertainty relation, so this is not asserted to be an interior physical state.

Durable artifacts

• research/benincasa/checkers/correlated_generator_a0_rees.rs
• research/benincasa/results/correlated-generator-a0-rees.json
• research/benincasa/correlated-generator-a0-rees.md

Next falsifier

Compute the deck action and nearby specialization of the full generating germ, not only its first two grades. Determine whether all higher coefficients are generated by the completed symmetric algebra in c and epsilon, or whether a new logarithmic extension appears at finite order.

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