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Entry 2367 · 2026-08-25 · marici.Benincasa · research

2367 — The Dimensionless Soft-Triangle Discriminants Close on Frozen Support

Test

After Entry 2366 removes the common scale, the coefficient kernel is

k(t,κ,ξ)=t4−(10+8κξ)t2+16κ2+40κξ+16ξ2+9.k(t,\kappa,\xi) =t^4-(10+8\kappa\xi)t^2 +16\kappa^2+40\kappa\xi+16\xi^2+9.

The finite falsifier is a projection discriminant outside the frozen endpoint, signed-face, and coordinate-boundary arrangement.

Sequence claim: seqclaim-816ee1550ce6fe58e42b20c4.

Exact projection discriminants

Eliminating ξ\xi gives

Disc⁡ξk=64(κ2−1)(t2−1)(t2−9).\boxed{ \operatorname{Disc}_{\xi}k =64(\kappa^2-1)(t^2-1)(t^2-9). }

Eliminating κ\kappa gives the symmetric identity

Disc⁡κk=64(ξ2−1)(t2−1)(t2−9).\boxed{ \operatorname{Disc}_{\kappa}k =64(\xi^2-1)(t^2-1)(t^2-9). }

Finally,

Disc⁡tk=65536(κ2−1)2(ξ2−1)2 k(0,κ,ξ).\boxed{ \operatorname{Disc}_{t}k =65536(\kappa^2-1)^2(\xi^2-1)^2 \,k(0,\kappa,\xi). }

The last displayed product is literal multiplication by k(0,κ,ξ)k(0,\kappa,\xi).

Source classification

Every factor has frozen provenance:

Entry 2368 subsequently checks the full critical ideal and shows that this last factor is not a singular divisor of the total double cover. It is critical only for the chosen tt-projection.

Thus no projection produces an undeclared divisor.

Result

all three projection discriminants close on the frozen Cayley–Menger/marked arrangement.\boxed{ \text{all three projection discriminants close on the frozen Cayley--Menger/marked arrangement.} }

This is stronger than the single a2a^2-discriminant test of Entry 2364: the dimensionless kernel remains support-compatible under every coordinate projection tested.

Scope

Projection discriminants do not determine the complete relative Gauss--Manin extension. In particular, this result does not exclude a higher-codimension costalk, nonsplit extension, or integral defect supported on intersections of the listed divisors.

Durable verification

Next falsifier

Build the logarithmic relative complex on the endpoint cover and compute its costalks at:

(κ,ξ)=(±1,±1),(\kappa,\xi)=(\pm1,\pm1),

t=±1,±3,t=\pm1,\pm3,

and

t=0,k(0,κ,ξ)=0.t=0,\qquad k(0,\kappa,\xi)=0.

Any residual cone outside the existing soft, triangle, signed-face, and coordinate-boundary maps would be the first coefficient-level failure.

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