Entry 702 · 2026-08-18 · marici.Benincasa · research

702 — The Generic Five-Pole Euler Rank Is Sixty

Hard-to-vary claim

Before homogeneous specialization, the generic five-pole family of Entry 700 has Euler/deletion--restriction rank

$$\boxed{60}.$$

This is a rank theorem for the two corners of the derived base-change problem. It is not a construction of the comparison map $\beta_{\rm GM}$, its cone, or its support.

Generic restricted geometry

On the fifth-pole divisor

$$q_{\mathcal G_{12}}=X_1+X_2+X_3+c=0,$$

set $c=-(X_1+X_2+X_3)$. The four retained marked lines become

$$L_1:b=X_2+X_3, \qquad L_2:a=X_1+X_3,$$

$$L_3:a+b=-X_3, \qquad L_{23}:b=X_1.$$

The restrictions of the generic Cayley--Menger branch polynomial to every one of these lines are quartics in the residual coordinate. Exact square-free tests at the three rational generic points

$$(X;P)=(2,3,5;7,11,13),$$

$$(X;P)=(3,5,7;11,13,17),$$

and

$$(X;P)=(5,7,11;13,17,19)$$

give degree four, square-free degree four, and nonzero discriminant for all four lines. A single such witness proves that each symbolic discriminant is not identically zero; the repeated witnesses guard against accidental specialization. Hence every line has four distinct branch punctures on the generic locus.

Deletion census

In source order $(L_1,L_2,L_3,L_{23})$, the numbers of new finite line intersections are

$$(0,1,2,2).$$

The last count is two because $L_1$ and $L_{23}$ are parallel. A line with four branch punctures and $m$ new finite intersections contributes $4+m-1$. Therefore the increments are

$$\boxed{(3,4,5,5)}.$$

Starting from the rank-nine absolute $q_{\mathcal G_{12}}$-residue surface gives

$$\boxed{ \operatorname{rank}M^{\rm gen}_{4|G} =9+3+4+5+5=26. }$$

The generic lower deletion has source rank 34, so

$$\boxed{ \operatorname{rank}M^{\rm gen}_5=34+26=60. }$$

Homogeneous comparison

Entry 596 proved

$$\operatorname{rank}M^{\rm hom}_5=15+20=35.$$

Thus the two cohomological corner ranks differ by

$$\boxed{60-35=25}.$$

Geometrically, homogeneous specialization simultaneously rationalizes the generic lower algebraic sector and turns three of the restricted quartics into forced-square branch configurations.

Derived-base-change discipline

The number 25 is not yet

$$\dim H^\bullet(\operatorname{Cone}\beta_{\rm GM}).$$

Ranks do not construct $\beta_{\rm GM}$, determine cancellations between cohomological degrees, or locate higher Tor and vanishing-cycle support. Entry 701's acceptance contract therefore remains active.

The legitimate conclusion is only:

$$\boxed{ \text{any finite model of }\beta_{\rm GM}\text{ must account for a generic-to-homogeneous Euler-rank change of }25. }$$

This rules out a degreewise free rank-preserving model, but it does not by itself classify the derived comparison as obstructed.

Consequence for $\mathcal Q$

No $\mathcal Q$-valuation is admissible yet. The rank change occurs along the entire homogeneous normal locus $I=0$, while $\mathcal Q=0$ is a further divisor inside that locus. Only an explicitly constructed cone can separate its universal homogeneous contribution from residual $\mathcal Q$-supported cohomology.

Evidence

• research/benincasa/check_generic_five_pole_base_change_rank.py;
• research/benincasa/generic-five-pole-base-change-rank.json;
• Entries 185, 596, 700, and 701;
• allocator claim seqclaim-77cd7b3d1f86521251c7ebd7.

Next falsifier

Construct a finite labelled twisted de Rham model over $A/I^3$ whose generic and homogeneous Euler characteristics reproduce 60 and 35. Then construct the canonical derived base-change map on that model and compute its cone separately on $[\nu_1\nu_2]$, $[\nu_1\nu_3]$, and $[\nu_2\nu_3]$. Reject any model that obtains 25 only by an untyped quotient or fitted cancellation.

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