Entry 298 · 2026-08-16 · marici.Benincasa · research

Rees Normalization Line Does Not Exhaust Q

Narrow result

The source-forced correction

$$\widehat\Omega_{111}=\Omega_{111}+\frac{e_6}{8E}$$

has now been expanded through the first $\mathcal Q$-sensitive connection order. Its contribution is

$$\boxed{ \nabla_E\left(\frac{e_6}{8E}\right) = -\frac{e_6}{8E^2} +\frac{e_6}{8sE} +\left(\frac1{8s^2}-\frac1{4p}\right)e_6 +E\left(\frac1{8s^3}+\frac{3s}{8p^2}\right)e_6 +O(E^2), }$$

where $s=x+y$ and $p=xy$.

The linear coefficient is not the source quartic's half-log signature

$$[E^1]\,\frac12\partial_E\log(-\mathcal Q)=\frac1{2p}.$$

Therefore the forced $e_6$ normalization term alone does not realize $\mathcal Q$. If $\mathcal Q$ occurs in the Rees-shifted primitive column, its missing contribution must come from the regularized raw relative connection or from mixing with the other algebraic-kernel direction.

Exact expansion

The invariant line has

$$\nabla_Ee_6=c(E)e_6, \qquad c(E)=-\frac12\partial_E\log H,$$

with

$$H=(E-s)^2J, \qquad J=p^2+2pE^2-2sE^3+2E^4.$$

Hence

$$c(E) = \frac1{s-E} - \frac{2pE-3sE^2+4E^3} {p^2+2pE^2-2sE^3+2E^4}.$$

Coefficient extraction gives

$$\boxed{ c(E) = \frac1s + \left(\frac1{s^2}-\frac2p\right)E + \left(\frac1{s^3}+\frac{3s}{p^2}\right)E^2 +O(E^3). }$$

Substitution into

$$\nabla_E\left(\frac{e_6}{8E}\right) = -\frac{e_6}{8E^2} + \frac{c(E)}{8E}e_6$$

proves the displayed result.

Comparison with Q

Entry 295 established

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

The normalization-line coefficient

$$\frac1{8s^3}+\frac{3s}{8p^2}$$

is not $1/(2p)$, nor a source-independent constant multiple of it on generic $(x,y)$. This comparison is basis-specific only in the declared source-normalized $e_6$ line; it does not yet determine the projection of the full Rees-shifted column.

Classification

$$\begin{array}{c|c} E^{-2}\text{ cancellation}&\text{existing Rees gauge on }\langle e_6\rangle\\ E^{-1},E^0,E^1\text{ correction terms}&\text{Tate/Kummer coefficient data}\\ \mathcal Q\text{ from correction alone}&\text{excluded}\\ \mathcal Q\text{ in full primitive column}&\text{open}\\ \text{new carrier datum}&\text{none} \end{array}$$

This removes one candidate home without modifying the frozen carrier.

Next finite falsifier

Compute the raw two-wall connection after subtracting the universal exact logarithmic lift, through $E^1$. Project its fixed part to

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

Only the sum of that projection and the normalization contribution above may be compared with $\frac12d\log(-\mathcal Q)$. Failure of the frozen relative reduction to generate the missing term is the next finite falsifier.

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