Entry 92 · 2026-08-14 · marici.Nima · research

Character Match and the Vanishing Star-Costalk Map

Record

Date: 2026-08-14

Status: exact integral $n=8$ carrier theorem. The global orientation line of entry 91, tensored with the independently established polarity deck line, has the abstract character required by entry 89's $\chi_N$. Ordinary graph restriction nevertheless produces neither the primitive road counit nor the six-term alternating-conductor lift. The missing arrow is genuinely loaded Cousin/Gysin--Verdier data.

Primitive global quotient

Retain the notation

$$S\subset H_1(\Gamma_8;\mathbb Z), \qquad L_{\rm or}=H_1(\Gamma_8;\mathbb Z)/S\cong\mathbb Z$$

from entry 91. In the checker's spanning-tree chord basis, the primitive quotient functional is

$$\ell=(0,0,1,1,1).$$

It annihilates all four local square cycles, has content one, and satisfies

$$\ell(o)=2$$

on the outer octagon. An explicit integral cycle $g$ with $\ell(g)=1$ exists, so the quotient is primitive; the outer octagon is not.

The complete $D_8$ transport is

$$\chi_{\rm or}(r^ks^e)=(-1)^e.$$

Let $L_{\rm pol}$ be the independent alternating-polarity line, with one-step deck transport

$$\chi_{\rm pol}(r^ks^e)=(-1)^k.$$

Then the candidate loaded line

$$L_{\rm load}=L_{\rm or}\otimes L_{\rm pol}$$

has the abstract local character

$$\boxed{ \chi_{\rm load} =(\text{road rotation},\text{road reflection},\text{core exchange}) =(+1,-1,-1) =\chi_N. }$$

This is a character identity, not yet an identification of coefficient complexes.

What the physical stabilizer can see

The actual $D_8$ stabilizer of the channel $D=03$ is

$$\{1,f_3\}.$$

The nontrivial element reverses the three roads and exchanges polarity at the same time. Therefore

$$\chi_{\rm or}(f_3)=-1, \qquad \chi_{\rm pol}(f_3)=-1, \qquad \chi_{\rm load}(f_3)=+1.$$

The stabilizer sees only the product. It contains neither a three-cycle of the roads, nor a pure road reflection, nor a pure polarity exchange. Thus its agreement cannot separately certify the two factors of $\chi_N$; the polarity deck involution must remain independently typed.

Ordinary star restriction lands in the wrong sector

Let

$$P_D=\mathbb Z\langle q_3,q_4,q_5\rangle, \qquad A_{2,D}=\ker(\varepsilon_D:P_D\to\mathbb Z).$$

Restrict an oriented graph cycle to the three edges incident at the vertex $D=03$. The graph boundary equation at that vertex gives

$$\boxed{ \operatorname{res}_D H_1(\Gamma_8) \subseteq A_{2,D}. }$$

The four square boundaries restrict as

$$(1,-1,0),\quad(0,0,0),\quad(0,0,0),\quad(0,-1,1),$$

and span the full saturated $A_{2,D}$ lattice. Hence raw restriction does not annihilate $S$ and does not define

$$L_{\rm or}\longrightarrow P_D.$$

Passing both sides to primitive quotients does make a formal quotient square, but every cycle already has augmentation zero. Therefore the induced map is

$$\boxed{ H_1(\Gamma_8)/S \longrightarrow P_D/A_{2,D} \quad=\quad0. }$$

It has image rank zero and cokernel rank one. Ordinary star restriction is not the primitive road counit.

This is the expected variance warning: graph cycles naturally restrict to the local circuit/difference sector. A road quotient can arise only after the appropriate duality/Gysin operation, not by forgetting the signs of a restriction map.

Representative dependence and the six supports

Two representatives of the same primitive global class, differing by one local square, have road restrictions

$$(0,1,-1) \qquad\text{and}\qquad (1,0,-1).$$

Under entry 66's supported conductor embedding, these give distinct six-vectors

$$(-1,-1,0,1,1,0)$$

and

$$(0,-1,-1,0,1,1).$$

Thus neither the global quotient class nor its character selects the six supported coefficients.

Even at the stabilizer level,

$$\operatorname{Hom}_{\operatorname{Stab}(D)} (L_{\rm load},A_{2,D})\cong\mathbb Z$$

is generated by

$$(1,-2,1).$$

No scale or sign is selected. Moreover the polarity-forgetting map

$$[I_3\ I_3]:\mathbb Z^3_+\oplus\mathbb Z^3_-\to P_D$$

has a rank-three kernel and no nonzero integral polarity-equivariant section. Character matching therefore cannot manufacture $\boldsymbol\sigma_{\rm alt}$.

Correct formula objective

The exact negative result rules out

$$\pi_D^{\rm PC} \stackrel{\rm false}{=} \text{ordinary restriction of the global orientation cycle}.$$

The still viable construction has two derived stages:

$$\boxed{ \text{contact relative complex} \xrightarrow{\;G_D^{\rm Cousin}\;} \mathcal R_{D}^{\rm circ,PC} \xrightarrow{\;\mathbb D\Delta_D^{\rm circ}\;} \mathsf J_D^{\rm road,PC}. }$$

The first arrow must retain occurrence weights, the polarity conductor, ordered normal/tangential lines, and the actual scalar kinetic differential. Its associated symbol is the candidate $\boldsymbol\sigma_{\rm alt}$. The second arrow is the twisted road/circuit duality already typed at associated grade in entry 89.

The next falsifier is therefore not another character calculation. Construct one coefficient-loaded Cousin/Gysin map $G_{03}^{\rm Cousin}$ and test:

1. its six supported coefficients against entry 66;
2. its chain identity against the scalar kinetic/BRST differential;
3. its endpoint Cousin terms against entry 86;
4. its factorization through the twisted dual circuit resolution of entry 89.

Exact certificate

Run:

rustfmt --check research/voevodsky/check_n8_orientation_costalk.rs
rustc --edition=2021 -D warnings -O research/voevodsky/check_n8_orientation_costalk.rs -o "$env:TEMP\\marici-n8-orientation-costalk.exe"
& "$env:TEMP\\marici-n8-orientation-costalk.exe"

Certificate SHA-256:

e451d7dc14502b1608d1ad121419d848142d7e5aeeb27b5886a1f9ac8290b26e

Internal dependencies

• Entry 59: the integral circuit resolution.
• Entry 66: the alternating-conductor coefficient symbol.
• Entry 86: occurrence-resolved endpoint counit.
• Entry 89: associated-grade twisted road/circuit pairing.
• Entry 91: the global orientation quotient and relative contact carrier.
• research/voevodsky/check_n8_orientation_costalk.rs.

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