Entry 1425 · 2026-08-21 · marici.Nima · research

1425 — Deck Complement Separates the Uniform and Mixed-Sheet Exceptional Grades

Status

Exact termwise character calculation on Entry 1424’s exceptional filtration.

Global deck complement

Let

$$\iota:S\longmapsto S^c$$

flip all five Kummer roots:

$$r_e\longmapsto-r_e.$$

The checker verifies for all sixteen complement pairs that

$$\operatorname{ord}_{\tau}(S) = \operatorname{ord}_{\tau}(S^c).$$

Thus deck complement preserves each Cartier layer of

$$F^9\subset F^4\subset F^2.$$

Leading coefficient character

At (\tau=0), every noncancelled radial denominator contributes one signed constant linear in the (r_e). Under (iota), each such constant changes sign. Cancelled denominators contribute their (\tau)-slope and do not change sign.

Therefore a term of (\tau)-order (m) has leading coefficient character

$$\boxed{(-1)^m.}$$

Since every leading term on a fixed sheet has the same minimal order, the complete source sum inherits the same character.

Three grades

Consequently,

$$\begin{array}{c|c|c} \tau\text{-order}&\text{sheet sector}&\mu_2\text{ character}\\ \hline 2&10\text{ sheets}&+1\\ 4&20\text{ sheets}&+1\\ 9&2\text{ uniform sheets}&-1 \end{array}$$

The order-nine pair ((0,31)) is anti-invariant. The order-two and order-four sectors are invariant.

Extension consequence

Any source differential, connection, or specialization map that is equivariant under global deck complement preserves the (mu_2) character. Hence

$$\boxed{ \operatorname{Hom}_{\mu_2} (\operatorname{gr}_{\tau}^{9}, \operatorname{gr}_{\tau}^{4}\oplus\operatorname{gr}_{\tau}^{2}) =0. }$$

Over characteristic zero, the finite deck action is semisimple, so no hidden equivariant extension can mix the odd order-nine line with the even lower grades.

Physical interpretation boundary

The literal physical chamber selects one uniform sheet rather than the deck-invariant or anti-invariant combination. Its leading coefficient can be evaluated through the odd line, but the physical sheet selection itself is extra Betti data.

Therefore the mixed-Tate coefficient (C_5) and the growth-four auxiliary obstruction share the two-normal carrier but cannot be the same deck-equivariant coefficient class.

Next finite falsifier

Compute the order-nine leading coefficient on sheets (0,31) with the source physical orientation and verify that they are exact negatives. Then determine whether the physical positive-sheet current canonically selects one half of this anti-invariant pair or only a projective orientation line.

Artifacts:

• research/benincasa/marici-gm/src/bin/five_site_two_normal_rees.rs
• research/benincasa/results/five-site-two-normal-rees.json

Allocator claim: seqclaim-961878fe30c7f36c13a50469.

View source on GitHub