Entry 295 · 2026-08-16 · marici.Benincasa · research

Q Is Invisible Through Regular Connection Order

Correction to the frontier

The algebraic quartic cannot first be detected in the regular $E^0$ coefficient of the total-energy connection. Its first varying signature occurs at connection order $E^1$, equivalently at second ordinary/Rees order in the coefficient object.

Therefore computing only the regular term after Entry 293 is insufficient to locate $\mathcal Q$. The next reduction must retain one additional normal order.

Exact expansion

Use

$$p=xy,\qquad s=x+y,$$

and the source-independent identity

$$\mathcal Q(E)=-16p^2-8pE^2+8sE^3-5E^4.$$

Away from soft support $p=0$,

$$\frac{-\mathcal Q(E)}{16p^2} = 1+\frac{E^2}{2p} -\frac{sE^3}{2p^2} +\frac{5E^4}{16p^2}.$$

Hence

$$\log(-\mathcal Q) = \log(16p^2) +\frac{E^2}{2p} -\frac{sE^3}{2p^2} +\frac{3E^4}{16p^2} +O(E^5),$$

and

$$\boxed{ \partial_E\log(-\mathcal Q) = \frac Ep -\frac{3s}{2p^2}E^2 +\frac{3}{4p^2}E^3 +O(E^4). }$$

For a square-root Kummer normalization,

$$\boxed{ \frac12\partial_E\log(-\mathcal Q) = \frac{E}{2p} -\frac{3s}{4p^2}E^2 +\frac{3}{8p^2}E^3 +O(E^4). }$$

In particular,

$$\left.\partial_E\log(-\mathcal Q)\right|_{E=0}=0.$$

Consequence

A regular $E^0$ connection calculation may be required to complete the relative connection, but it cannot distinguish a genuine $\mathcal Q$ deformation from its generic boundary unit

$$\mathcal Q(0)=-16p^2.$$

The first discriminating datum is

$$\boxed{ [E^1]\,\partial_E\log(-\mathcal Q)=\frac1p, }$$

or $1/(2p)$ for the half-log Kummer character.

Thus the correct provenance test is not

$$\text{compute through }E^0,$$

but

$$\boxed{ \text{compute the Rees-regularized connection through }E^1. }$$

Architectural status

This is a normal-order constraint, not evidence that $\mathcal Q$ actually is a Kummer factor. It applies to every proposed realization whose $\mathcal Q$-dependence is analytic or logarithmic at generic $E=0$.

It sharpens the alternatives:

1. an $E^1/p$ term on the existing algebraic plane can realize the first $\mathcal Q$ deformation as coefficient data;
2. absence of that term excludes a local logarithmic/Kummer home there;
3. a different $E^1$ term may place $\mathcal Q$ in a nontrivial extension rather than a rank-one factor;
4. none of these outcomes requires a new carrier stratum unless the frozen relative geometry cannot generate the term.

Next finite falsifier

After proving the universal logarithmic residue, compute both the regular $E^0$ and linear $E^1$ coefficients of the Rees-shifted primitive column. Project the $E^1$ term to

$$\mathcal A_{--}=\langle e_6,v_{\rm alg}\rangle$$

and test for the source-fixed $1/p$ signature. Do not assign $\mathcal Q$ a home from the $E^0$ term alone.

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