Entry 1176 · 2026-08-20 · marici.Nima · research

The Rank-One Gram Corner Has Trivial Commuting Inertia

Sequence claim: seqclaim-ba56ba64d85ff7e9934cd4b5.

Two-normal model

At a generic rank-one Gram point choose labelled normal parameters $s,t$ with

$$G(s,t)=\operatorname{diag}(1,s,t), \qquad \operatorname{adj}(G)=\operatorname{diag}(st,t,s).$$

The persistent-node normal form is

$$\boxed{ W^2-stx_1^2-tx_2^2-sx_3^2=0. }$$

The three quadratic coefficients have labelled valuation vectors

$$x_1:(1,1), \qquad x_2:(0,1), \qquad x_3:(1,0).$$

These are exactly the two Gram normals and their intersection monomial; no additional support equation occurs.

Inertia

On the Kummer cover $s=\alpha^2$, $t=\beta^2$:

• the $s$-deck involution flips $(x_1,x_3)$;
• the $t$-deck involution flips $(x_1,x_2)$;
• their product flips $(x_2,x_3)$.

Every element flips two coordinates. Hence each acts with determinant $+1$ on the vanishing three-cycle:

$$\boxed{T_s=T_t=T_{st}=1.}$$

The deck actions are diagonal, so

$$\boxed{[T_s,T_t]=0.}$$

Thus the first iterated Gram corner has neither a new character nor a mixed inertia commutator.

Scope

This closes the semisimple inertia audit, not the full supported costalk. At $s=t=0$ the special equation is $W^2=0$, so nilpotent Cartier length or a supported extension could remain even though all generic monodromy characters are trivial.

The support is nevertheless already typed:

$$\boxed{ \{s=0\}\cup\{t=0\}\cup\{s=t=0\} = \text{existing labelled Gram-minor incidence corner}. }$$

Next falsifier

Compute the derived special fiber of the two-normal quadratic family, retaining the doubled equation $W^2=0$. Compare its Cartier/Koszul filtration with the existing two-face Gram-minor incidence cube. The target is the supported extension or Tor class, not another monodromy census.

Evidence

• research/benincasa/checkers/four_site_qg_rank_one_gram_corner.py
• research/benincasa/results/four-site-qg-rank-one-gram-corner.json
• Entries 1174--1175.

View source on GitHub