Six-point obstruction to the proposed Stirling-filtration bridge

Mode: ATTACK
Date: 7 September 2026
Status: exact stopping result for the selected bridge; not a result about the Dong–Stieberger rank conjecture

Result

The filtration proposed in the preceding HUNT is not the six-point open–closed collinear filtration. More precisely, at an exact generic five-hard-particle kinematic point:

1. no diagonal chamber rescaling carries the cumulative physical filtration to the opposite dual of the Varchenko–Gelfand Heaviside filtration;
2. no type-(A_2) Varchenko matrix does so, for any values of its three hyperplane weights; and
3. twisting either the physical or arrangement chamber basis by the sign character does not repair the failure.

This refutes the specific canonical-intertwiner proposal selected by the HUNT. It does not refute the Stirling rank formula, Dong and Stieberger’s determinant conjecture, or the possibility that another universal object explains them.

1. The two filtrations at six points

At five points the test behaves as intended. Taking

$$(s_{12},s_{13},s_{23})=(2,3,-5)$$

gives ((s_{1P},s_{2P},s_{3P})=(-5,3,2)). Dong and Stieberger’s rows are

$$G_5=(2,-3),\qquad R_5=(4,9),$$

and the natural rescaling (D_5=(1/2,1/3)) sends (G_5D_5) to ((1,-1)), exactly the opposite first Heaviside piece. The failure below therefore first appears at the nontrivial three-level flag; it is not a mismatch in the two-chamber convention.

Six-point conventions

Use the chamber/column order

$$234,243,324,342,423,432.$$

For the braid arrangement on the labels (2,3,4), let

$$Q=(\mathbf 1,h_{23},h_{24},h_{34}),$$

where (h_{ab}) indicates that (a) precedes (b). Explicitly,

$$\begin{aligned} \mathbf 1&=(1,1,1,1,1,1)^t,\\ h_{23}&=(1,1,0,0,1,0)^t,\\ h_{24}&=(1,1,1,0,0,0)^t,\\ h_{34}&=(1,0,1,1,0,0)^t. \end{aligned}$$

The first two opposite Heaviside pieces are therefore

$$T_1=\ker Q^t,\qquad T_2=\mathbf 1^\perp.$$

Dong and Stieberger’s two ordinary EYM rows give (F_6^1), while adjoining their three quadratic rows gives (F_6^2). These definitions use the cumulative row spaces, so the calculation is independent of the displayed choice of basis for the quadratic block.

2. An exact generic specialization

Choose

$$(s_{12},s_{13},s_{14},s_{23},s_{24},s_{34})=(2,3,5,7,11,-28).$$

Their sum is zero, hence (P=-(p_1+p_2+p_3+p_4)) is massless, and

$$(s_{1P},s_{2P},s_{3P},s_{4P})=(-10,-20,18,12).$$

Every factor in the exact six-point reconstruction determinant

$$\frac14\prod_{a=1}^4s_{aP}^2 (s_{12}+s_{34})(s_{13}+s_{24})(s_{14}+s_{23})$$

is nonzero. Thus this lies away from the degeneracy divisor on which a filtration test could give a false failure.

The degree-one and degree-at-most-two row matrices specialize to

$$G=\begin{pmatrix} 12&-18&0&0&-18&-20\\ 0&0&12&20&18&20 \end{pmatrix},$$

$$R=\begin{pmatrix} 72&126&0&0&-72&-200\\ 0&0&72&260&-162&-300\\ 48&162&-36&-200&0&0 \end{pmatrix}.$$

The stacked matrix ([G;R]) has rank five.

3. No diagonal chamber rescaling

Suppose (D=(d_1,,d_6)) carries both cumulative filtration levels to their arrangement counterparts. Since (F_6^2D=T_2=), its diagonal vector must satisfy

$$\begin{pmatrix}G\\R\end{pmatrix}d=0.$$

The kernel is one-dimensional. Normalizing its last coordinate to one gives

$$d=\left(\frac{10}{13},\frac{20}{39},-\frac{10}{7}, \frac67,-\frac{10}{9},1\right)^t.$$

But the first-level requirement (F_6^1DT_1) fails already on one Heaviside generator:

$$GD h_{23}=\binom{20}{-20}\ne0.$$

Hence no diagonal matrix–not merely no preferred ansatz–identifies the two filtrations at this generic point.

4. No Varchenko pairing

Give the hyperplanes (23),(24),(34) arbitrary weights (x,y,z). In the same chamber order the full type-(A_2) Varchenko matrix is

$$V(x,y,z)= \begin{pmatrix} 1&z&x&xy&yz&xyz\\ z&1&xz&xyz&y&xy\\ x&xz&1&y&xyz&yz\\ xy&xyz&y&1&xz&z\\ yz&y&xyz&xz&1&x\\ xyz&xy&yz&z&x&1 \end{pmatrix}.$$

If (F_6^1V=T_1), even the first row (g) of (G) must satisfy (gVQ=0). Treat the four monomials ((xy),x,y,1) as linear unknowns for fixed (z). Those four equations become

$$M(z)\begin{pmatrix}xy\\x\\y\\1\end{pmatrix}=0,$$

where

$$M(z)=\begin{pmatrix} -22z-13&-18z-13&-13z-18&-13z-22\\ -10z-10&-10&-3z-18&-3z-12\\ -19z-10&6-9z&-19z-9&-3z-3\\ 6-28z&6-18z&-19z&6-19z \end{pmatrix}.$$

Exact elimination gives

$$\det M(z)=-64(z-1)(z+1)(189z^2-608z+189).$$

Thus a solution can occur only at these four values of (z). Away from (z=-1), an adjugate kernel vector has Segre defect

$$A D-B C=-64(z+1)^2 \bigl(7551z^4+29650z^3-70178z^2+29650z+7551\bigr).$$

The quartic is coprime to

$$(z-1)(189z^2-608z+189),$$

so the required identity (A D=BC), expressing that the first coordinate is the product (xy), fails at all three remaining determinant roots. At (z=-1), direct solution gives

$$(A,B,C)=\left(-1,\frac95,-\frac95\right), \qquad A-BC=\frac{56}{25}\ne0.$$

Therefore no values of the three Varchenko weights work, even over the algebraic closure. The exact Gröbner basis of the four equations is simply ((1)).

Conjugating (V) by the sign character only changes the signs of its three weights and is already covered. For completeness, the attached audit also tests the one-sided sign twists; all four left/right choices again have unit Gröbner ideal.

5. Consequence and stopping boundary

The Stirling dimensions remain striking, but they do not arise from the opposite Heaviside filtration through either canonical map proposed in the HUNT. The first violated classical condition is concrete: after the unique rescaling forced by the second cumulative level, an ordinary EYM row is not orthogonal to (h_{23}).

Repairing this by an arbitrary element of (GL_6(K)) would be vacuous: any two flags with the same dimensions are linearly isomorphic. The next meaningful step is therefore a new HUNT for a different intrinsic object, perhaps one that retains the kinematic local system rather than replacing it by constant chamber functions. This route stops here; no larger multiplicity census is warranted.

Source and credit

The physical rows, six-point determinant, and Stirling-rank conjecture are from Jin Dong and Stephan Stieberger, Subleading Collinear Limits of Yang–Mills Amplitudes from Gravity, arXiv:2609.00153v1. The chamber filtration and Varchenko form are classical arrangement-theoretic objects. The negative comparison above is an FPRD Lab calculation and has not been externally reviewed.