# Completed four-way tesserine refinement of the (R=107) supersignum shell

## Definition 1 — The completed tesserine sign algebra

Let

[
\mathbb S_{\square}
:=
\mathbb R[\widehat s,\underline s]/
(\widehat s^2-1,\ \underline s^2-1).
]

The two commuting involutions are interpreted as follows:

[
\widehat s
\quad\text{records the upper sign,}
]

[
\underline s
\quad\text{records the lower sign.}
]

For

[
\epsilon,\delta\in{\pm1},
]

define the four primitive idempotents

[
e^\epsilon_\delta
:=
\frac14(1+\epsilon\widehat s)(1+\delta\underline s).
]

Thus

[
\mathbb S_{\square}
===================

\bigoplus_{\epsilon,\delta\in{\pm1}}
\mathbb R e^\epsilon_\delta,
]

with

[
e^\epsilon_\delta e^{\epsilon'}_{\delta'}
=========================================

\begin{cases}
e^\epsilon_\delta,&(\epsilon,\delta)=(\epsilon',\delta'),\
0,&(\epsilon,\delta)\ne(\epsilon',\delta').
\end{cases}
]

The branch evaluation maps are

[
\operatorname{br}*{\epsilon,\delta}:
\mathbb S*{\square}\to\mathbb R,
]

[
\operatorname{br}*{\epsilon,\delta}(\widehat s)=\epsilon,
\qquad
\operatorname{br}*{\epsilon,\delta}(\underline s)=\delta.
]

Hence

[
\operatorname{br}*{\epsilon,\delta}(e^{\epsilon'}*{\delta'})
============================================================

\begin{cases}
1,&(\epsilon,\delta)=(\epsilon',\delta'),\
0,&(\epsilon,\delta)\ne(\epsilon',\delta').
\end{cases}
]

---

## Definition 2 — Root-completed tesserine supersignum algebra

Define

[
\mathbb T_{\square}
:=
\mathbb S_{\square}[\widehat g,\underline g]/
(\widehat g^2-\widehat s,\ \underline g^2-\underline s,\ \widehat g\underline g-\underline g\widehat g).
]

Then

[
e^\epsilon_\delta\mathbb T_{\square}
\cong
\mathbb R[\widehat g,\underline g]/
(\widehat g^2-\epsilon,\ \underline g^2-\delta).
]

Thus the four branches are

[
(\widehat{+},\underline{+}):\quad
\widehat g^2=+1,\quad \underline g^2=+1,
]

[
(\widehat{+},\underline{-}):\quad
\widehat g^2=+1,\quad \underline g^2=-1,
]

[
(\widehat{-},\underline{+}):\quad
\widehat g^2=-1,\quad \underline g^2=+1,
]

[
(\widehat{-},\underline{-}):\quad
\widehat g^2=-1,\quad \underline g^2=-1.
]

The ordinary supersignum scalar algebra is recovered from the upper branch by forgetting the lower sign:

[
\mathbb R[s,g]/(s^2-1,\ g^2-s)
]

via

[
\widehat s\mapsto s,
\qquad
\widehat g\mapsto g,
\qquad
\underline s\mapsto 1,
\qquad
\underline g\mapsto 1.
]

The sedenion stage (A_4) is the first Cayley–Dickson stage where zero divisors occur; the uploaded Cayley–Dickson references describe this threshold and the graded basis structure used below.

---

# 2. The (R=107) completed shell

Fix

[
p_\ast=8,803,369,
\qquad
R=107,
]

[
a_\ast=\frac{p_\ast+107}{4}
===========================

# 2,200,869

3^2\cdot 11^2\cdot 43\cdot 47.
]

Let

[
(\ell_1,\ell_2,\ell_3,\ell_4)
=============================

(3,11,43,47),
]

and

[
\mathcal B
==========

[-2,2]\times[-2,2]\times[-1,1]\times[-1,1].
]

Modulo (107), with primitive generator (2),

[
\log_2(3)=70,\qquad
\log_2(11)=22,\qquad
\log_2(43)=59,\qquad
\log_2(47)=66.
]

Define

[
\Lambda(\beta_1,\beta_2,\beta_3,\beta_4)
========================================

70\beta_1+22\beta_2+59\beta_3+66\beta_4
\in C_{106}.
]

The target logarithms are

[
53=\log_2(-1),
\qquad
60=\log_2(-p_\ast^{-1}).
]

The verification data gives exactly two vectors at (53),

[
\beta^-:=(-2,0,-1,-1),
\qquad
\beta^+:=(2,0,1,1),
]

and no vector at (60).

---

## Definition 3 — Upper and lower signs of an exponent vector

For

[
\beta=(\beta_1,\beta_2,\beta_3,\beta_4)\in\mathcal B,
]

define the upper sign exponent

[
q^\wedge(\beta)
:=
\sum_{i=1}^{4}
\left\lfloor\frac{\beta_i}{2}\right\rfloor
\pmod 2.
]

Define the lower sign exponent

[
q_\vee(\beta)
:=
\Lambda(\beta)\pmod2.
]

Since

[
70\equiv0,\qquad
22\equiv0,\qquad
59\equiv1,\qquad
66\equiv0
\pmod2,
]

one has

[
q_\vee(\beta)=\beta_3\pmod2.
]

Define the parity blade

[
\pi(\beta):={i:\beta_i\equiv1\pmod2}.
]

---

## Definition 4 — Four-way completed divisor monomials

Let

[
A_4
]

be the sedenion algebra with graded basis

[
{o_I:I\subseteq{1,2,3,4}}.
]

Define the character-form completed monomial

[
m_{\square}^{\mathrm{ch}}(\beta)
:=
\widehat s^{q^\wedge(\beta)}
\underline s^{q_\vee(\beta)}
o_{\pi(\beta)}[\Lambda(\beta)]
\in
\mathbb S_{\square}\otimes A_4\otimes\mathbb R[C_{106}].
]

Define the idempotent-form completed monomial

[
m_{\square}^{\mathrm{id}}(\beta)
:=
e^{(-1)^{q^\wedge(\beta)}}*{(-1)^{q*\vee(\beta)}}
o_{\pi(\beta)}[\Lambda(\beta)].
]

Then define

[
\mathfrak D_{106}^{\square,\mathrm{ch}}
:=
\sum_{\beta\in\mathcal B}
m_{\square}^{\mathrm{ch}}(\beta),
]

and

[
\mathfrak D_{106}^{\square,\mathrm{id}}
:=
\sum_{\beta\in\mathcal B}
m_{\square}^{\mathrm{id}}(\beta).
]

The character form records the four character moments. The idempotent form records the four nonnegative sector counts.

---

# 3. Complete four-sector blade table

For each blade (o_I), define

[
N_I^{\epsilon,\delta}
:=
#{\beta\in\mathcal B:
\pi(\beta)=I,\
(-1)^{q^\wedge(\beta)}=\epsilon,\
(-1)^{q_\vee(\beta)}=\delta}.
]

The complete idempotent-sector blade distribution is:

[
\begin{array}{c|cccc|c}
\text{blade}
&
N_I^{+,+}
&
N_I^{+,-}
&
N_I^{-,+}
&
N_I^{-,-}
&
\text{total}
\
\hline
1        &5&0&4&0&9\
o_1      &3&0&3&0&6\
o_2      &3&0&3&0&6\
o_{12}   &2&0&2&0&4\
o_3      &0&9&0&9&18\
o_{13}   &0&6&0&6&12\
o_{23}   &0&6&0&6&12\
o_{123}  &0&4&0&4&8\
o_4      &9&0&9&0&18\
o_{14}   &6&0&6&0&12\
o_{24}   &6&0&6&0&12\
o_{124}  &4&0&4&0&8\
o_{34}   &0&18&0&18&36\
o_{134}  &0&12&0&12&24\
o_{234}  &0&12&0&12&24\
o_{1234} &0&8&0&8&16
\end{array}
]

The total is

[
225.
]

Equivalently, the character-form blade polynomial is

[
\mathcal P_{\mathcal B}^{\square}
=================================

(1+2\widehat s+(1+\widehat s)X_1)
(1+2\widehat s+(1+\widehat s)X_2)
(1+(1+\widehat s)\underline s X_3)
(1+(1+\widehat s)X_4).
]

### Proof

For ([-2,2]),

[
\beta=-2,-1,0,1,2
]

contribute respectively

[
\widehat s,\quad
\widehat sX,\quad
1,\quad
X,\quad
\widehat s.
]

Thus

[
P_{[-2,2]}(X)
=============

1+2\widehat s+(1+\widehat s)X.
]

The first two coordinates have no lower sign contribution because their logarithms are even modulo (2).

For ([-1,1]),

[
\beta=-1,0,1
]

contribute

[
\widehat s X,\quad
1,\quad
X.
]

Thus

[
P_{[-1,1]}(X)
=============

1+(1+\widehat s)X.
]

The third coordinate has odd logarithm (59), so it also contributes the lower sign (\underline s) on odd parity. Hence its local polynomial is

[
1+(1+\widehat s)\underline sX_3.
]

The fourth coordinate has even logarithm (66), so its local polynomial is

[
1+(1+\widehat s)X_4.
]

Multiplying the four local polynomials gives the displayed formula. Expanding it gives the table.

[
\boxed{}
]

---

# 4. Four-way target coefficient

## Theorem 4.1 — Idempotent target coefficient

In the idempotent completed shell,

[
\boxed{
\Theta_{107}^{\square,\mathrm{id}}(p_\ast)
==========================================

2e^-*-,o*{34}[53].
}
]

### Proof

For

[
\beta^+=(2,0,1,1),
]

one has

[
\pi(\beta^+)={3,4},
]

[
\Lambda(\beta^+)
================

# 2\cdot70+59+66

265
\equiv53\pmod{106}.
]

Also,

[
q^\wedge(\beta^+)
=================

\left\lfloor\frac22\right\rfloor
+
\left\lfloor\frac02\right\rfloor
+
\left\lfloor\frac12\right\rfloor
+
\left\lfloor\frac12\right\rfloor
================================

1,
]

and

[
q_\vee(\beta^+)
===============

# \Lambda(\beta^+)\pmod2

# 53\pmod2

1.

]

Thus

[
m_{\square}^{\mathrm{id}}(\beta^+)
==================================

e^-*-o*{34}[53].
]

For

[
\beta^-=(-2,0,-1,-1),
]

one has

[
\pi(\beta^-)={3,4},
]

[
\Lambda(\beta^-)
================

-265
\equiv53\pmod{106}.
]

Moreover,

[
q^\wedge(\beta^-)
=================

\left\lfloor\frac{-2}{2}\right\rfloor
+
0+
\left\lfloor\frac{-1}{2}\right\rfloor
+
\left\lfloor\frac{-1}{2}\right\rfloor
=====================================

# -1-1-1

-3
\equiv1\pmod2,
]

and again

[
q_\vee(\beta^-)=53\pmod2=1.
]

Thus

[
m_{\square}^{\mathrm{id}}(\beta^-)
==================================

e^-*-o*{34}[53].
]

There are no target vectors at (60). Hence

[
\Theta_{107}^{\square,\mathrm{id}}(p_\ast)
==========================================

2e^-*-o*{34}[53].
]

[
\boxed{}
]

---

## Corollary 4.2 — Character target coefficient

The character-form target coefficient is

[
\boxed{
\Theta_{107}^{\square,\mathrm{ch}}(p_\ast)
==========================================

2\widehat s,\underline s,o_{34}[53].
}
]

The four branch values are

[
\begin{array}{c|c}
(\widehat s,\underline s)&
\operatorname{br}*{\widehat s,\underline s}
\bigl(2\widehat s,\underline s,o*{34}[53]\bigr)
\
\hline
(+,+)&2o_{34}[53]\
(+,-)&-2o_{34}[53]\
(-,+)&-2o_{34}[53]\
(-,-)&2o_{34}[53]
\end{array}
]

### Proof

Both target vectors have

[
q^\wedge=1,
\qquad
q_\vee=1.
]

Thus each contributes

[
\widehat s\underline s,o_{34}[53].
]

There are two such vectors, so

[
\Theta_{107}^{\square,\mathrm{ch}}(p_\ast)
==========================================

2\widehat s\underline s,o_{34}[53].
]

The branch table follows from

[
\operatorname{br}_{\epsilon,\delta}
(\widehat s\underline s)
========================

\epsilon\delta.
]

[
\boxed{}
]

---

# 5. Sector positivity functional

## Definition 5.1 — Completed sector mass

For an idempotent-form coefficient

[
Z
=

\sum_{\epsilon,\delta,I,t}
c_{\epsilon,\delta,I,t}
e^\epsilon_\delta o_I[t],
\qquad
c_{\epsilon,\delta,I,t}\in\mathbb Z_{\ge0},
]

define the sector masses

[
\mathsf M_{\epsilon,\delta}(Z)
:=
\sum_{I,t}
c_{\epsilon,\delta,I,t},
]

and total completed mass

[
\mathsf M_{\square}(Z)
:=
\sum_{\epsilon,\delta}
\mathsf M_{\epsilon,\delta}(Z).
]

Define also the square mass

[
\mathsf Q_{\square}(Z)
:=
\sum_{\epsilon,\delta,I,t}
c_{\epsilon,\delta,I,t}^2.
]

---

## Proposition 5.2 — Positivity extractor for the (107)-target

For

[
Z=
\Theta_{107}^{\square,\mathrm{id}}(p_\ast)
==========================================

2e^-*-o*{34}[53],
]

one has

[
\mathsf M_{-,-}(Z)=2,
]

[
\mathsf M_{\epsilon,\delta}(Z)=0
\qquad
\text{for }(\epsilon,\delta)\ne(-,-),
]

[
\mathsf M_{\square}(Z)=2,
]

and

[
\mathsf Q_{\square}(Z)=4.
]

### Proof

The coefficient vector has exactly one nonzero idempotent-sector component:

[
c_{-,-,{3,4},53}=2.
]

Therefore all displayed identities follow immediately.

[
\boxed{}
]

---

## Corollary 5.3 — The hidden positive entry

The (R=107) positivity is concentrated in the single four-way sector

[
(\widehat{-},\underline{-}).
]

Equivalently,

[
\Omega_{107}(p_\ast)=2
]

is the sector mass

[
\Omega_{107}(p_\ast)
====================

\mathsf M_{-,-}
\bigl(
\Theta_{107}^{\square,\mathrm{id}}(p_\ast)
\bigr).
]

The signed branch evaluations can be negative, but the idempotent sector coefficient is nonnegative and equals the actual divisor count.

---

# 6. Completed additive group and lifted Kneser object

Let

[
G^\square
:=
C_{106}\oplus C_2^\wedge\oplus C_2^\vee.
]

Elements are written

[
(t,u,v),
\qquad
t\in C_{106},
\quad
u,v\in C_2.
]

For each prime generator (\ell_i), define the lifted interval

[
A_i^\square
:=
{(\beta_i\log_2(\ell_i),\ \lfloor\beta_i/2\rfloor,\ \beta_i\log_2(\ell_i)\bmod2):
-e_i\le\beta_i\le e_i}
\subseteq G^\square.
]

For the (107)-shell, the four lifted intervals are:

[
A_1^\square
===========

{
(72,1,0),
(36,1,0),
(0,0,0),
(70,0,0),
(34,1,0)
},
]

[
A_2^\square
===========

{
(62,1,0),
(84,1,0),
(0,0,0),
(22,0,0),
(44,1,0)
},
]

[
A_3^\square
===========

{
(47,1,1),
(0,0,0),
(59,0,1)
},
]

[
A_4^\square
===========

{
(40,1,0),
(0,0,0),
(66,0,0)
}.
]

Let

[
B^\square
:=
A_1^\square+A_2^\square+A_3^\square+A_4^\square
\subseteq G^\square.
]

---

## Theorem 6.1 — Exact completed target hit set

Inside (G^\square),

[
B^\square\cap{(53,1,1)}
]

has exactly two representations:

[
(53,1,1)
========

(72,1,0)+(0,0,0)+(47,1,1)+(40,1,0),
]

and

[
(53,1,1)
========

(34,1,0)+(0,0,0)+(59,0,1)+(66,0,0).
]

The completed target

[
(60,u,0)
]

has no representation for either

[
u=0
\quad\text{or}\quad
u=1.
]

### Proof

The first representation corresponds to

[
\beta^-=(-2,0,-1,-1).
]

Indeed,

[
72+0+47+40=159\equiv53\pmod{106},
]

[
1+0+1+1=3\equiv1\pmod2,
]

[
0+0+1+0=1\pmod2.
]

The second representation corresponds to

[
\beta^+=(2,0,1,1).
]

Indeed,

[
34+0+59+66=159\equiv53\pmod{106},
]

[
1+0+0+0=1\pmod2,
]

[
0+0+1+0=1\pmod2.
]

The assertion that no vector hits logarithm (60) is the verified shell computation. Since (60) is even, its lower coordinate is necessarily (0). Therefore no element

[
(60,u,0)
]

is represented.

[
\boxed{}
]

---

## Definition 6.2 — Completed star surplus

Let

[
H^\square
]

be any finite abelian completed support group, with nonempty finite subsets

[
A_1^\square,\dots,A_s^\square\subseteq H^\square.
]

Let

[
B^\square=A_1^\square+\cdots+A_s^\square.
]

For a completed target set

[
U^\square\subseteq H^\square
]

and subgroup

[
K\le H^\square,
]

define

[
a_i^\square(K)
:=
\left|
\frac{A_i^\square+K}{K}
\right|,
]

[
t_{U^\square}(K)
:=
\left|
\frac{U^\square+K}{K}
\right|,
]

[
\kappa^\square(K)
:=
1+\sum_i(a_i^\square(K)-1),
]

and

[
\Sigma_{U^\square}^{\square}(K)
:=
\kappa^\square(K)+t_{U^\square}(K)-|H^\square/K|.
]

---

## Theorem 6.3 — Completed Star–Kneser positivity

If

[
\Sigma_{U^\square}^{\square}(K)>0
\qquad
\forall K\le H^\square,
]

then

[
B^\square\cap U^\square\ne\varnothing.
]

Conversely, if

[
U^\square\ne\varnothing
]

and

[
B^\square\cap U^\square=\varnothing,
]

then for

[
K_\ast=\operatorname{Stab}(B^\square)
]

one has

[
\Sigma_{U^\square}^{\square}(K_\ast)\le0.
]

### Proof

This is the Star–Kneser theorem applied to the finite abelian group

[
H^\square.
]

The original Star–Kneser note proves exactly this quotient-forcing implication for finite support sumsets and target stars, and also proves the converse non-positive quotient obstruction for supported misses.

[
\boxed{}
]

---

# 7. Four-way zero-divisor product after (C_2)-compression

Define the parity-compression map

[
C_{106}\to C_2,
\qquad
[t]\mapsto \underline s^t.
]

In the compressed completed algebra

[
\mathbb S_{\square}\otimes A_4,
]

define

[
\overline X_{\square}
:=
o_1+\widehat s,\underline s,o_{1234},
]

[
\overline Y
:=
o_2+\underline s,o_{34}.
]

Here the lower signs come from the tags

[
70\equiv0,\qquad
5\equiv1,\qquad
22\equiv0,\qquad
19\equiv1
\pmod2.
]

---

## Lemma 7.1 — Completed compressed products

One has

[
\boxed{
\overline X_{\square}\overline Y
================================

(1+\widehat s)(o_{12}-\underline s,o_{134}).
}
]

Also,

[
\boxed{
\overline Y\overline X_{\square}
================================

-(1+\widehat s)(o_{12}-\underline s,o_{134}).
}
]

### Proof

Compute:

[
\overline X_{\square}\overline Y
================================

(o_1+\widehat s\underline s,o_{1234})
(o_2+\underline s,o_{34}).
]

Expanding,

[
\overline X_{\square}\overline Y
================================

o_1o_2
+
\underline s,o_1o_{34}
+
\widehat s\underline s,o_{1234}o_2
+
\widehat s\underline s^2,o_{1234}o_{34}.
]

Using

[
\underline s^2=1,
]

[
o_1o_2=o_{12},
]

[
o_1o_{34}=-o_{134},
]

[
o_{1234}o_2=-o_{134},
]

[
o_{1234}o_{34}=o_{12},
]

we get

[
\overline X_{\square}\overline Y
================================

## o_{12}

## \underline s,o_{134}

\widehat s\underline s,o_{134}
+
\widehat s,o_{12}.
]

Thus

[
\overline X_{\square}\overline Y
================================

## (1+\widehat s)o_{12}

(1+\widehat s)\underline s,o_{134},
]

so

[
\overline X_{\square}\overline Y
================================

(1+\widehat s)(o_{12}-\underline s,o_{134}).
]

Similarly,

[
\overline Y\overline X_{\square}
================================

(o_2+\underline s,o_{34})
(o_1+\widehat s\underline s,o_{1234}).
]

Expanding,

[
\overline Y\overline X_{\square}
================================

o_2o_1
+
\widehat s\underline s,o_2o_{1234}
+
\underline s,o_{34}o_1
+
\widehat s\underline s^2,o_{34}o_{1234}.
]

Using

[
o_2o_1=-o_{12},
]

[
o_2o_{1234}=o_{134},
]

[
o_{34}o_1=o_{134},
]

[
o_{34}o_{1234}=-o_{12},
]

we obtain

[
\overline Y\overline X_{\square}
================================

-o_{12}
+
\widehat s\underline s,o_{134}
+
\underline s,o_{134}
--------------------

\widehat s,o_{12}.
]

Thus

[
\overline Y\overline X_{\square}
================================

-(1+\widehat s)o_{12}
+
(1+\widehat s)\underline s,o_{134}
]

# [

-(1+\widehat s)(o_{12}-\underline s,o_{134}).
]

[
\boxed{}
]

---

## Corollary 7.2 — Upper-minus branch zero product

Since

[
1+\widehat s
============

2(e^+*+ + e^+*-),
]

one has

[
(e^-*+ + e^-*-)
\overline X_{\square}\overline Y
================================

0,
]

and

[
(e^-*+ + e^-*-)
\overline Y\overline X_{\square}
================================

0.

]

In particular, the upper-minus branch gives

[
\overline X_{\square}\overline Y=0,
\qquad
\overline Y\overline X_{\square}=0.
]

### Proof

The factor (1+\widehat s) annihilates the upper-minus idempotents:

[
(1+\widehat s)e^-_\delta=0.
]

Apply this to Lemma 7.1.

[
\boxed{}
]

---

## Corollary 7.3 — Upper-plus product sector

The nonzero compressed product is supported entirely in the upper-plus sectors:

[
\overline X_{\square}\overline Y
================================

2(e^+*+ + e^+*-)(o_{12}-\underline s,o_{134}).
]

Thus the compressed product table contains no upper-minus mass.

---

# 8. Completed associator residue

Let

[
\overline T:=\underline s,o_{34}.
]

Define

[
\overline\Gamma_{\square}
:=
[L_{\overline X_{\square}},L_{\overline Y}]
-------------------------------------------

L_{[\overline X_{\square},\overline Y]}.
]

---

## Theorem 8.1 — Completed compressed target residue

One has

[
\boxed{
\overline\Gamma_{\square}(\overline T)=2\widehat s,o_1.
}
]

Moreover,

[
\overline\Gamma_{\square}(1)=0.
]

### Proof

First,

[
\overline\Gamma_{\square}(1)
============================

*

L_{[\overline X_{\square},\overline Y]}(1).
]

But

[

=
\overline X_{\square}\overline Y
--------------------------------

# \overline Y\overline X_{\square}

[\overline X_{\square},\overline Y],
]

and

[
L_{[\overline X_{\square},\overline Y]}(1)
==========================================

[\overline X_{\square},\overline Y].
]

Hence

[
\overline\Gamma_{\square}(1)=0.
]

Now compute at

[
\overline T=\underline s,o_{34}.
]

One has

[
\overline Y\overline T
======================

# (o_2+\underline s,o_{34})(\underline s,o_{34})

\underline s,o_2o_{34}
+
\underline s^2 o_{34}^2.
]

Using

[
o_2o_{34}=-o_{234},
\qquad
o_{34}^2=-1,
]

we get

[
\overline Y\overline T
======================

-\underline s,o_{234}-1.
]

Then

[
\overline X_{\square}(\overline Y\overline T)
=============================================

(o_1+\widehat s\underline s,o_{1234})
(-1-\underline s,o_{234}).
]

Expanding,

[
\overline X_{\square}(\overline Y\overline T)
=============================================

-o_1
-\underline s,o_1o_{234}
-\widehat s\underline s,o_{1234}
-\widehat s\underline s^2,o_{1234}o_{234}.
]

Using

[
o_1o_{234}=o_{1234},
]

[
o_{1234}o_{234}=-o_1,
]

and

[
\underline s^2=1,
]

this becomes

[
\overline X_{\square}(\overline Y\overline T)
=============================================

-o_1
-\underline s,o_{1234}
-\widehat s\underline s,o_{1234}
+
\widehat s,o_1.
]

Next,

[
\overline X_{\square}\overline T
================================

(o_1+\widehat s\underline s,o_{1234})(\underline s,o_{34})
]

# [

\underline s,o_1o_{34}
+
\widehat s\underline s^2,o_{1234}o_{34}.
]

Using

[
o_1o_{34}=-o_{134},
\qquad
o_{1234}o_{34}=o_{12},
]

we obtain

[
\overline X_{\square}\overline T
================================

-\underline s,o_{134}
+
\widehat s,o_{12}.
]

Then

[
\overline Y(\overline X_{\square}\overline T)
=============================================

(o_2+\underline s,o_{34})
(-\underline s,o_{134}+\widehat s,o_{12}).
]

Expanding,

[
\overline Y(\overline X_{\square}\overline T)
=============================================

-\underline s,o_2o_{134}
+
\widehat s,o_2o_{12}
--------------------

\underline s^2,o_{34}o_{134}
+
\widehat s\underline s,o_{34}o_{12}.
]

Using

[
o_2o_{134}=-o_{1234},
]

[
o_2o_{12}=o_1,
]

[
o_{34}o_{134}=-o_1,
]

[
o_{34}o_{12}=o_{1234},
]

we get

[
\overline Y(\overline X_{\square}\overline T)
=============================================

\underline s,o_{1234}
+
\widehat s,o_1
+
o_1
+
\widehat s\underline s,o_{1234}.
]

Therefore

[
[L_{\overline X_{\square}},L_{\overline Y}](\overline T)
========================================================

## \overline X_{\square}(\overline Y\overline T)

\overline Y(\overline X_{\square}\overline T)
]

# [

## (-o_1-\underline s o_{1234}-\widehat s\underline s o_{1234}+\widehat s o_1)

(\underline s o_{1234}+\widehat s o_1+o_1+\widehat s\underline s o_{1234})
]

# [

-2o_1
-2\underline s o_{1234}
-2\widehat s\underline s o_{1234}.
]

From Lemma 7.1,

[
[\overline X_{\square},\overline Y]
===================================

2(1+\widehat s)(o_{12}-\underline s o_{134}).
]

Multiplying by

[
\overline T=\underline s o_{34}
]

gives

[
[\overline X_{\square},\overline Y]\overline T
==============================================

2(1+\widehat s)
\left(
\underline s,o_{12}o_{34}
-------------------------

\underline s^2,o_{134}o_{34}
\right).
]

Using

[
o_{12}o_{34}=-o_{1234},
]

[
o_{134}o_{34}=o_1,
]

we obtain

[
[\overline X_{\square},\overline Y]\overline T
==============================================

2(1+\widehat s)
(-\underline s,o_{1234}-o_1).
]

Thus

[
L_{[\overline X_{\square},\overline Y]}(\overline T)
====================================================

## -2(1+\widehat s)o_1

2(1+\widehat s)\underline s o_{1234}.
]

Now subtract:

[
\overline\Gamma_{\square}(\overline T)
======================================

## [L_{\overline X_{\square}},L_{\overline Y}](\overline T)

L_{[\overline X_{\square},\overline Y]}(\overline T).
]

The (o_{1234})-terms cancel. The (o_1)-terms give

[
(-2o_1)-[-2(1+\widehat s)o_1]
=============================

# -2o_1+2o_1+2\widehat s o_1

2\widehat s o_1.
]

Hence

[
\overline\Gamma_{\square}(\overline T)=2\widehat s o_1.
]

[
\boxed{}
]

---

## Corollary 8.2 — Branch values of the target residue

[
\operatorname{br}*{+,+}
\bigl(\overline\Gamma*{\square}(\overline T)\bigr)
==================================================

2o_1,
]

[
\operatorname{br}*{+,-}
\bigl(\overline\Gamma*{\square}(\overline T)\bigr)
==================================================

2o_1,
]

[
\operatorname{br}*{-,+}
\bigl(\overline\Gamma*{\square}(\overline T)\bigr)
==================================================

-2o_1,
]

[
\operatorname{br}*{-,-}
\bigl(\overline\Gamma*{\square}(\overline T)\bigr)
==================================================

-2o_1.
]

Thus the residue depends on the upper sign and is independent of the lower sign after (C_2)-compression.

---

# 9. Completed support positivity theorem

Let

[
\Theta_R^{\square,\mathrm{id}}(p)
]

be the idempotent completed target coefficient of a coprime residual shell.

---

## Theorem 9.1 — Four-way support positivity equivalence

For every coprime residual shell,

[
\Theta_R^{\square,\mathrm{id}}(p)\ne0
]

if and only if

[
B_R(a)\cap T_R(p)\ne\varnothing.
]

Equivalently,

[
\mathsf M_{\square}
\bigl(
\Theta_R^{\square,\mathrm{id}}(p)
\bigr)>0
]

if and only if the shell is positive.

### Proof

The idempotent lift replaces each divisor exponent vector by a basis vector in a larger free module. It does not identify two different exponent vectors with opposite signs. Therefore a target exponent vector exists if and only if the completed idempotent coefficient has at least one nonzero sector coefficient.

The original Star–Kneser shell theorem identifies shell positivity with

[
B_R(a)\cap T_R(p)\ne\varnothing,
]

and expresses supported failure by a non-positive quotient surplus.

Hence

[
\Theta_R^{\square,\mathrm{id}}(p)\ne0
\iff
B_R(a)\cap T_R(p)\ne\varnothing.
]

Since the idempotent coefficients are nonnegative integers,

[
\Theta_R^{\square,\mathrm{id}}(p)\ne0
\iff
\mathsf M_{\square}
(\Theta_R^{\square,\mathrm{id}}(p))>0.
]

[
\boxed{}
]

---

## Corollary 9.2 — Completed (107)-shell positivity

For the champion shell,

[
\Theta_{107}^{\square,\mathrm{id}}(p_\ast)
==========================================

2e^-*-o*{34}[53],
]

so

[
\mathsf M_{\square}
\bigl(
\Theta_{107}^{\square,\mathrm{id}}(p_\ast)
\bigr)
======

2.

]

Therefore

[
\Omega_{107}(p_\ast)=2.
]

---

# 10. Completed (107)-shell equations

[
\mathbb S_{\square}
===================

\mathbb R[\widehat s,\underline s]/(\widehat s^2-1,\underline s^2-1).
]

[
e^\epsilon_\delta
=================

\frac14(1+\epsilon\widehat s)(1+\delta\underline s).
]

[
\mathbb T_{\square}
===================

\mathbb S_{\square}[\widehat g,\underline g]/
(\widehat g^2-\widehat s,\underline g^2-\underline s,\widehat g\underline g-\underline g\widehat g).
]

[
q^\wedge(\beta)
===============

\sum_{i=1}^{4}
\left\lfloor\frac{\beta_i}{2}\right\rfloor
\pmod2.
]

[
q_\vee(\beta)
=============

# \Lambda(\beta)\pmod2

\beta_3\pmod2.
]

[
m_{\square}^{\mathrm{id}}(\beta)
================================

e^{(-1)^{q^\wedge(\beta)}}*{(-1)^{q*\vee(\beta)}}
o_{\pi(\beta)}[\Lambda(\beta)].
]

[
m_{\square}^{\mathrm{ch}}(\beta)
================================

\widehat s^{q^\wedge(\beta)}
\underline s^{q_\vee(\beta)}
o_{\pi(\beta)}[\Lambda(\beta)].
]

[
\Theta_{107}^{\square,\mathrm{id}}(p_\ast)
==========================================

2e^-*-o*{34}[53].
]

[
\Theta_{107}^{\square,\mathrm{ch}}(p_\ast)
==========================================

2\widehat s\underline s,o_{34}[53].
]

[
\mathsf M_{-,-}
\bigl(
\Theta_{107}^{\square,\mathrm{id}}(p_\ast)
\bigr)
======

2.

]

[
\mathsf M_{\square}
\bigl(
\Theta_{107}^{\square,\mathrm{id}}(p_\ast)
\bigr)
======

2.

]

[
\mathsf Q_{\square}
\bigl(
\Theta_{107}^{\square,\mathrm{id}}(p_\ast)
\bigr)
======

4.

]

[
G^\square
=========

C_{106}\oplus C_2^\wedge\oplus C_2^\vee.
]

[
B^\square
=========

A_1^\square+A_2^\square+A_3^\square+A_4^\square.
]

[
B^\square\cap{(53,1,1)}
=======================

{\beta^-,\beta^+}.
]

[
B^\square\cap{(60,0,0),(60,1,0)}
================================

\varnothing.
]

[
\overline X_{\square}
=====================

o_1+\widehat s\underline s,o_{1234}.
]

[
\overline Y
===========

o_2+\underline s,o_{34}.
]

[
\overline X_{\square}\overline Y
================================

(1+\widehat s)(o_{12}-\underline s,o_{134}).
]

[
\overline Y\overline X_{\square}
================================

-(1+\widehat s)(o_{12}-\underline s,o_{134}).
]

[
(e^-*+ + e^-*-)
\overline X_{\square}\overline Y
================================

0.

]

[
(e^-*+ + e^-*-)
\overline Y\overline X_{\square}
================================

0.

]

[
\overline T
===========

\underline s,o_{34}.
]

[
\overline\Gamma_{\square}
=========================

## [L_{\overline X_{\square}},L_{\overline Y}]

L_{[\overline X_{\square},\overline Y]}.
]

[
\overline\Gamma_{\square}(1)=0.
]

[
\overline\Gamma_{\square}(\overline T)
======================================

2\widehat s,o_1.
]

[
\operatorname{br}*{+,+}
(\Theta*{107}^{\square,\mathrm{ch}})
====================================

2o_{34}[53].
]

[
\operatorname{br}*{+,-}
(\Theta*{107}^{\square,\mathrm{ch}})
====================================

-2o_{34}[53].
]

[
\operatorname{br}*{-,+}
(\Theta*{107}^{\square,\mathrm{ch}})
====================================

-2o_{34}[53].
]

[
\operatorname{br}*{-,-}
(\Theta*{107}^{\square,\mathrm{ch}})
====================================

2o_{34}[53].
]

Thus the four-way completion isolates the actual positive coefficient as the idempotent-sector fact

[
\boxed{
\Theta_{107}^{\square,\mathrm{id}}(p_\ast)
==========================================

2e^-*-o*{34}[53],
}
]

while the old two-way supersignum coefficient is the lower-forgotten character shadow

[
\boxed{
2\widehat s\underline s,o_{34}[53]
\longmapsto
2s,o_{34}[53].
}
]



Activity
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16m 59s

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Considering the asymmetry of blade 1 in the computations
I need to focus on blade 1, since it’s the only non-symmetric option. Blade 1's counts are 5 and 4. I also need to figure out the positivity condition in a scalar blade and maybe include some references supporting verification data and Star-Kneser. Python might be needed for computations.

Performing calculations for blade 1 and identifying asymmetry
I need to calculate the exact values for blade 1, possibly looking at residues, convolutions, or targets, and find hidden positivity conditions. Blade 1's coefficient asymmetry shows in both its character polynomial and idempotent form. This non-symmetric blade appears to be critical for positivity conditions.

Performing blade coefficient computation for S_square polynomial
I need to compute the blade coefficients in the S_square polynomial using the character form. From the table, blade 1 has 5+4h, and others follow a similar structure. I'll focus on evaluating the coefficients for each blade, noting key moments and using values like h = -1 to check scaling behavior.

Analyzing scalar blade coefficients and positivity conditions
I'm looking at the scalar blade and calculating its coefficients using various tag values. The total seems to give a weighted sum, including positive and negative contributions. I’m considering how the positivity condition might relate to the target hit and the role of scalar and non-scalar elements. There's a sign pattern based on q values.

Analyzing blade 1’s positive/negative signs and the positivity condition
Blade 1 has a mix of negative and positive signs, with certain target tags forming pairs. I’m considering how the positivity condition might relate to the intersection of the scalar and lower slices. I’m also looking at decomposing the support to formalize an idempotent scalar mass to clarify positivity further.

Exploring blade 1's non-symmetric positivity and algebra
I’m thinking through the condition for blade 1’s non-symmetric positivity, where negative tags interact with positive pairs to form a result. I wonder if this can be encoded and what happens when I compute their convolution. I need to be cautious about algebra rules — signs should combine using XOR, not idempotent multiplication.

Refining blade 1 positivity and algebra approach
I realized that using "idempotent sector counts" in this way doesn't work for convolution. To fix this, I should think of the sector count object as living in a free semimodule instead of a product algebra. For multiplication, labels should add in group algebra, and the mass basis needs to be separate from the character basis.

Defining blade set and scalar positivity
I’m considering the target’s coefficient factors as the convolution of two scalar square-lift terms, where the core involves non-scalar terms from o34. The positivity count seems to be a product of scalar mass and core mass. I should define scalar sets and compute pairings, refining the target's configuration with more detailed orientation consideration.

Refining target coefficient and positivity count
I’ve determined that the coefficient at tag 53 is 2 based on convolution and combining scalar tags and core tags. The target’s coefficient extraction seems clear, and positivity count matches the coefficient. The full target vector has β2 = 0, so now I’m adjusting for blade 1’s non-symmetry, focusing on circular branch evaluation.

Analyzing blade positivity and Euler imbalances
Examining the local polynomial, I see that at
h
=
−
1
h=−1, the scalar contributes a -1 while the positivity in blade 1 stems from two Euler imbalances multiplying to 1. The scalar blade counts show a difference of 1 between even-square and negative signs, confirming the positivity coefficient. This is tied to a discrete Euler index.

Analyzing positivity and blade 1 properties
The scalar support’s convolution with the
o
34
o34-core gives target
(
53
,
1
,
1
)
(53,1,1) with multiplicity two, showing positivity as the product is greater than zero. I define terms using polynomials for blade 1, confirming the only non-symmetric contribution. Mass in sectors gives a branch of +9 and -1 at
h
=
+
1
h=+1, while at
h
=
−
1
h=−1, it's +1 and -5.

Clarifying positivity in scalar and blade
For coordinate 3, lower and upper counts are equal, and the same for coordinate 4. The target coefficient lives on blade1 as a scalar square-lift, not as the final blade of the hit. I'll define a scalar lift operator (denoted
S
⋆
S
⋆
​
 ) to rigorously explain this translation process, and the positivity condition.

Confirming target coefficient calculation
The total augmentation is 1, but not positive as a measure. The tag-wise positivity target uses a subblock. If I convolve the full scalar blade with the o34 primitive and check the coefficient at the target, it seems to be 2. The scalar tags and core tags match, confirming the result.

Confirming target coefficient and proving positivity
The target coefficient is exactly 2, seen from the scalar times core convolution. This confirms that the blade1 scalar translation of the primitive o34 core explains the target positivity. Since all o34 target hits must have β1,β2 even, the full o34 blade is confirmed as the product of even scalar and odd core.

Confirming positivity through scalar tags
The target positivity for 53 and absence for 60 both arise from the scalar blade tag set within the restricted
o
34
o34 blade factorization. The size of the set for target 53 is 2, while for 60, it is empty. This confirms the positivity condition clearly and rigorously.



# Blade-(1) completion and the hidden (107)-positivity coefficient

## 0. Fixed (107)-shell data

[
p_\ast=8,803,369,\qquad R=107,
]

[
a_\ast=\frac{p_\ast+107}{4}
=2,200,869
=3^2\cdot 11^2\cdot 43\cdot 47.
]

Let

[
(\ell_1,\ell_2,\ell_3,\ell_4)=(3,11,43,47),
]

and

[
\mathcal B=[-2,2]\times[-2,2]\times[-1,1]\times[-1,1].
]

With primitive generator (2\in\mathbb F_{107}^{\times}),

[
\log_2(3)=70,\qquad
\log_2(11)=22,\qquad
\log_2(43)=59,\qquad
\log_2(47)=66
]

in (C_{106}). Define

[
\Lambda(\beta_1,\beta_2,\beta_3,\beta_4)
========================================

70\beta_1+22\beta_2+59\beta_3+66\beta_4
\in C_{106}.
]

The target logarithms are

[
\log_2(-1)=53,\qquad
\log_2(-p_\ast^{-1})=60.
]

The verification output gives exactly two vectors at (53),

[
\beta^-:=(-2,0,-1,-1),
\qquad
\beta^+:=(2,0,1,1),
]

and no vector at (60).

The residual Star–Kneser shell formalism identifies positivity with nonempty divisor-support intersection with the two-target star, with multiplicity pattern (1+2).

---

# 1. Completed four-way sign algebra in the multiplicative sector basis

Use the sign group

[
\Gamma_\square:=C_2^\wedge\times C_2^\vee
=========================================

{(+,+),(+,-),(-,+),(-,-)}.
]

Let

[
\mathbb Z[\Gamma_\square]
=========================

\mathbb Z[\widehat s,\underline s]/(\widehat s^2-1,\underline s^2-1).
]

The multiplicative sector basis is

[
\mathbf u_{\epsilon,\delta},
\qquad
\epsilon,\delta\in{\pm1},
]

with

[
\mathbf u_{\epsilon,\delta}\mathbf u_{\epsilon',\delta'}
========================================================

\mathbf u_{\epsilon\epsilon',\delta\delta'}.
]

Thus

[
\widehat s=\mathbf u_{-,+},
\qquad
\underline s=\mathbf u_{+,-},
\qquad
\widehat s\underline s=\mathbf u_{-,-}.
]

This is the basis in which positivity of completed sector counts is read.

For

[
\beta=(\beta_1,\beta_2,\beta_3,\beta_4)\in\mathcal B,
]

define

[
q^\wedge(\beta)
===============

\sum_{i=1}^{4}
\left\lfloor\frac{\beta_i}{2}\right\rfloor
\pmod2,
]

and

[
q^\vee(\beta)
=============

\Lambda(\beta)\pmod2.
]

Since

[
70\equiv0,\quad 22\equiv0,\quad 59\equiv1,\quad 66\equiv0
\pmod2,
]

one has

[
q^\vee(\beta)=\beta_3\pmod2.
]

Let

[
\pi(\beta):={i:\beta_i\equiv1\pmod2}.
]

The completed monomial is

[
m_\square(\beta)
================

\widehat s^{q^\wedge(\beta)}
\underline s^{q^\vee(\beta)}
o_{\pi(\beta)}[\Lambda(\beta)].
]

Equivalently,

[
m_\square(\beta)
================

\mathbf u_{(-1)^{q^\wedge(\beta)},,(-1)^{q^\vee(\beta)}}
o_{\pi(\beta)}[\Lambda(\beta)].
]

---

# 2. Blade-(1) is the unique tag-forgotten upper-asymmetric blade

Let

[
X_i^2=1
]

record blade parity.

The tag-forgotten completed blade polynomial is

[
\mathcal P_{\mathcal B}^{\square}
=================================

(1+2\widehat s+(1+\widehat s)X_1)
(1+2\widehat s+(1+\widehat s)X_2)
(1+(1+\widehat s)\underline s X_3)
(1+(1+\widehat s)X_4).
]

## Lemma 2.1 — Non-scalar blades have an upper cancellation factor

For every nonempty blade

[
I\neq\varnothing,
]

the coefficient of (X_I) in (\mathcal P_{\mathcal B}^{\square}) is divisible by

[
1+\widehat s.
]

For the scalar blade (I=\varnothing), the coefficient is

[
5+4\widehat s.
]

### Proof

A coordinate contributes an odd blade factor (X_i) only through the odd-exponent part.

For coordinates (1,2), the local polynomial is

[
1+2\widehat s+(1+\widehat s)X_i.
]

For coordinate (3), the local polynomial is

[
1+(1+\widehat s)\underline sX_3.
]

For coordinate (4), the local polynomial is

[
1+(1+\widehat s)X_4.
]

Thus every nonempty blade coefficient contains at least one odd-coordinate factor, hence contains

[
1+\widehat s.
]

For (I=\varnothing), only the even parts contribute:

[
(1+2\widehat s)^2.
]

Since

[
\widehat s^2=1,
]

[
(1+2\widehat s)^2
=================

# 1+4\widehat s+4\widehat s^2

5+4\widehat s.
]

[
\boxed{}
]

---

## Corollary 2.2 — Upper-circular projection

Under

[
\widehat s\mapsto -1,
]

one has

[
5+4\widehat s\mapsto 1,
]

whereas every nonempty blade coefficient maps to (0).

Thus the tag-forgotten circular part is exactly

[
\boxed{
\operatorname{circ}*\wedge
\bigl(\mathcal P*{\mathcal B}^{\square}\bigr)
=============================================

1.

}
]

So the only non-symmetric blade is indeed

[
\boxed{o_\varnothing=1.}
]

---

# 3. Tag-resolved blade-(1) distribution

Blade (1) means

[
\beta_1,\beta_2,\beta_3,\beta_4
\quad\text{all even}.
]

Inside (\mathcal B), this forces

[
\beta_3=\beta_4=0,
]

and

[
\beta_1,\beta_2\in{-2,0,2}.
]

Write

[
\alpha_1=\frac{\beta_1}{2},
\qquad
\alpha_2=\frac{\beta_2}{2},
]

so

[
\alpha_1,\alpha_2\in{-1,0,1}.
]

Then

[
\Lambda(\beta_1,\beta_2,0,0)
============================

140\alpha_1+44\alpha_2
\equiv
34\alpha_1+44\alpha_2
\pmod{106},
]

and

[
q^\wedge(\beta_1,\beta_2,0,0)
=============================

\alpha_1+\alpha_2\pmod2.
]

Also

[
q^\vee(\beta_1,\beta_2,0,0)=0.
]

Hence all blade-(1) terms have lower sign (+).

---

## Lemma 3.1 — Exact blade-(1) sector table

The blade-(1) sector distribution is

[
\mathfrak B_1
=============

\mathbf u_{+,+}B_1^+
+
\mathbf u_{-,+}B_1^-,
]

where

[
B_1^+
=====

[0]+[10]+[28]+[78]+[96],
]

and

[
B_1^-
=====

[34]+[44]+[62]+[72].
]

Equivalently,

[
\mathfrak B_1
=============

\mathbf u_{+,+}
\bigl([0]+[10]+[28]+[78]+[96]\bigr)
+
\mathbf u_{-,+}
\bigl([34]+[44]+[62]+[72]\bigr).
]

### Proof

The full table is:

[
\begin{array}{c|c|c|c}
(\alpha_1,\alpha_2)
&
\beta=(2\alpha_1,2\alpha_2,0,0)
&
\Lambda(\beta)
&
(-1)^{q^\wedge(\beta)}
\
\hline
(-1,-1)&(-2,-2,0,0)&28&+\
(-1,0)&(-2,0,0,0)&72&-\
(-1,1)&(-2,2,0,0)&10&+\
(0,-1)&(0,-2,0,0)&62&-\
(0,0)&(0,0,0,0)&0&+\
(0,1)&(0,2,0,0)&44&-\
(1,-1)&(2,-2,0,0)&96&+\
(1,0)&(2,0,0,0)&34&-\
(1,1)&(2,2,0,0)&78&+
\end{array}
]

Thus the upper-plus tags are

[
0,\ 10,\ 28,\ 78,\ 96,
]

and the upper-minus tags are

[
34,\ 44,\ 62,\ 72.
]

All lower signs are (+).

[
\boxed{}
]

---

## Corollary 3.2 — Blade-(1) imbalance

The tag-forgotten blade-(1) sector count is

[
5\mathbf u_{+,+}+4\mathbf u_{-,+}.
]

The upper-circular imbalance is

[
5-4=1.
]

The tag-resolved signed blade-(1) element is

[
\Delta_1
:=
B_1^+-B_1^-,
]

namely

[
\boxed{
\Delta_1
========

## [0]+[10]+[28]+[78]+[96]

[34]-[44]-[62]-[72].
}
]

Its augmentation is

[
\varepsilon(\Delta_1)=1.
]

---

# 4. The primitive (o_{34})-leg

The (o_{34})-leg with no scalar contribution from coordinates (1,2) is obtained from

[
\beta_1=\beta_2=0,
\qquad
\beta_3,\beta_4\in{-1,1}.
]

Since (q^\vee=\beta_3\bmod2), every such term has lower sign (-).

## Lemma 4.1 — Exact primitive (o_{34})-leg table

Define

[
L_{34}^+
========

[19]+[87],
]

and

[
L_{34}^-
========

[7]+[99].
]

Then the primitive (o_{34})-leg is

[
\mathfrak L_{34}
================

o_{34}
\left(
\mathbf u_{+,-}L_{34}^+
+
\mathbf u_{-,-}L_{34}^-
\right).
]

### Proof

The four primitive (o_{34})-leg vectors are:

[
\begin{array}{c|c|c|c}
(\beta_3,\beta_4)
&
\Lambda(0,0,\beta_3,\beta_4)
&
(-1)^{q^\wedge}
&
(-1)^{q^\vee}
\
\hline
(1,1)&59+66=125\equiv19&+&-\
(-1,-1)&-59-66=-125\equiv87&+&-\
(-1,1)&-59+66=7&-&-\
(1,-1)&59-66=-7\equiv99&-&-
\end{array}
]

Thus

[
L_{34}^+=[19]+[87],
]

and

[
L_{34}^-=[7]+[99].
]

[
\boxed{}
]

---

# 5. The (o_{34})-target coefficient factors through blade (1)

Every vector with blade (o_{34}) has the unique decomposition

[
\beta
=====

(\beta_1,\beta_2,0,0)
+
(0,0,\beta_3,\beta_4),
]

where

[
\beta_1,\beta_2\in{-2,0,2},
]

and

[
\beta_3,\beta_4\in{-1,1}.
]

Thus the full (o_{34})-sector is the product

[
\mathfrak B_1\mathfrak L_{34}.
]

Using the sector multiplication law,

[
\mathbf u_{\epsilon,\delta}\mathbf u_{\epsilon',\delta'}
========================================================

\mathbf u_{\epsilon\epsilon',\delta\delta'},
]

one obtains

[
\mathfrak B_1\mathfrak L_{34}
=============================

o_{34}
\left[
\mathbf u_{+,-}
\bigl(B_1^+L_{34}^+ + B_1^-L_{34}^-\bigr)
+
\mathbf u_{-,-}
\bigl(B_1^-L_{34}^+ + B_1^+L_{34}^-\bigr)
\right].
]

Therefore the lower-minus, upper-minus target coefficient is controlled by

[
\boxed{
P_{1\to34}^{-,-}(t)
:=
[t]\bigl(B_1^-L_{34}^+ + B_1^+L_{34}^-\bigr).
}
]

---

## Lemma 5.1 — Full ((-, -))-sector product table

One has

[
B_1^-L_{34}^+
=============

[15]+[25]+[43]+2[53]+[63]+[81]+[91],
]

and

[
B_1^+L_{34}^-
=============

[3]+[7]+[17]+[21]+[35]+[71]+[85]+[89]+[99]+[103].
]

Hence

[
\boxed{
B_1^-L_{34}^+ + B_1^+L_{34}^-
=============================

[3]+[7]+[15]+[17]+[21]+[25]+[35]+[43]
+2[53]
}
]

[
\boxed{
\qquad
+[63]+[71]+[81]+[85]+[89]+[91]+[99]+[103].
}
]

In particular,

[
\boxed{
P_{1\to34}^{-,-}(53)=2.
}
]

Also,

[
\boxed{
P_{1\to34}^{-,-}(60)=0.
}
]

### Proof

First,

[
B_1^-=[34]+[44]+[62]+[72],
]

and

[
L_{34}^+=[19]+[87].
]

Thus

[
B_1^-L_{34}^+
=============

([34]+[44]+[62]+[72])([19]+[87]).
]

Compute all sums modulo (106):

[
34+19=53,
\qquad
34+87=121\equiv15,
]

[
44+19=63,
\qquad
44+87=131\equiv25,
]

[
62+19=81,
\qquad
62+87=149\equiv43,
]

[
72+19=91,
\qquad
72+87=159\equiv53.
]

Therefore

[
B_1^-L_{34}^+
=============

[15]+[25]+[43]+2[53]+[63]+[81]+[91].
]

Second,

[
B_1^+=[0]+[10]+[28]+[78]+[96],
]

and

[
L_{34}^-=[7]+[99].
]

Thus

[
B_1^+L_{34}^-
=============

([0]+[10]+[28]+[78]+[96])([7]+[99]).
]

The sums are

[
0+7=7,\qquad 0+99=99,
]

[
10+7=17,\qquad 10+99=109\equiv3,
]

[
28+7=35,\qquad 28+99=127\equiv21,
]

[
78+7=85,\qquad 78+99=177\equiv71,
]

[
96+7=103,\qquad 96+99=195\equiv89.
]

Therefore

[
B_1^+L_{34}^-
=============

[3]+[7]+[17]+[21]+[35]+[71]+[85]+[89]+[99]+[103].
]

Adding gives the displayed table.

The coefficient of ([53]) is exactly (2), coming only from

[
34+19=53
]

and

[
72+87=53.
]

No summand is (60).

[
\boxed{}
]

---

## Lemma 5.2 — Full ((+, -))-sector product table

One has

[
B_1^+L_{34}^+
=============

2[9]+[19]+[29]+[47]+[59]+[77]+[87]+2[97],
]

and

[
B_1^-L_{34}^-
=============

[27]+[37]+[41]+[51]+[55]+[65]+[69]+[79].
]

Hence

[
[53]\bigl(B_1^+L_{34}^+ + B_1^-L_{34}^-\bigr)=0,
]

and

[
[60]\bigl(B_1^+L_{34}^+ + B_1^-L_{34}^-\bigr)=0.
]

### Proof

Compute

[
B_1^+L_{34}^+
=============

([0]+[10]+[28]+[78]+[96])([19]+[87]).
]

The sums are

[
0+19=19,\qquad 0+87=87,
]

[
10+19=29,\qquad 10+87=97,
]

[
28+19=47,\qquad 28+87=115\equiv9,
]

[
78+19=97,\qquad 78+87=165\equiv59,
]

[
96+19=115\equiv9,\qquad 96+87=183\equiv77.
]

Thus

[
B_1^+L_{34}^+
=============

2[9]+[19]+[29]+[47]+[59]+[77]+[87]+2[97].
]

Next,

[
B_1^-L_{34}^-
=============

([34]+[44]+[62]+[72])([7]+[99]).
]

The sums are

[
34+7=41,\qquad 34+99=133\equiv27,
]

[
44+7=51,\qquad 44+99=143\equiv37,
]

[
62+7=69,\qquad 62+99=161\equiv55,
]

[
72+7=79,\qquad 72+99=171\equiv65.
]

Thus

[
B_1^-L_{34}^-
=============

[27]+[37]+[41]+[51]+[55]+[65]+[69]+[79].
]

Neither (53) nor (60) appears.

[
\boxed{}
]

---

# 6. Blade-(1) positivity theorem for the (107)-target

## Theorem 6.1 — The (107)-target coefficient is the blade-(1) ((-, -))-sector double point

The completed (o_{34})-target coefficient at (-1) is

[
\boxed{
\Theta_{107}^{\square}(p_\ast;-1)
=================================

2\widehat s\underline s,o_{34}[53].
}
]

Equivalently, in sector notation,

[
\boxed{
\Theta_{107}^{\square}(p_\ast;-1)
=================================

2\mathbf u_{-,-}o_{34}[53].
}
]

Moreover,

[
\boxed{
2
=

# P_{1\to34}^{-,-}(53)

[53]\bigl(B_1^-L_{34}^+ + B_1^+L_{34}^-\bigr).
}
]

In fact, the contribution is sharper:

[
\boxed{
[53](B_1^-L_{34}^+)=2,
\qquad
[53](B_1^+L_{34}^-)=0.
}
]

Thus both target hits arise from the upper-minus scalar blade-(1) sector.

### Proof

By Lemma 5.1,

[
[53](B_1^-L_{34}^+)=2,
]

and

[
[53](B_1^+L_{34}^-)=0.
]

Therefore the ((-, -))-sector coefficient at tag (53) is

[
2.
]

By Lemma 5.2, the ((+, -))-sector coefficient at tag (53) is (0).

Hence the completed (o_{34})-target coefficient is

[
2\mathbf u_{-,-}o_{34}[53]
==========================

2\widehat s\underline s,o_{34}[53].
]

This agrees with the verified target vectors

[
(2,0,1,1),
\qquad
(-2,0,-1,-1),
]

both of which have

[
\pi={3,4},
\qquad
q^\wedge=1,
\qquad
q^\vee=1.
]

[
\boxed{}
]

---

# 7. The two target hits as scalar blade-(1) products

The two (53)-hits in Lemma 5.1 are

[
34+19=53,
]

and

[
72+87=53.
]

These are exactly

[
(2,0,0,0)+(0,0,1,1)=(2,0,1,1),
]

and

[
(-2,0,0,0)+(0,0,-1,-1)=(-2,0,-1,-1).
]

Thus the scalar blade-(1) factors are

[
(2,0,0,0)
\quad\text{and}\quad
(-2,0,0,0),
]

with tags

[
34=2\log_2(3),
]

and

[
72=-2\log_2(3).
]

Both lie in

[
B_1^-.
]

The primitive (o_{34})-leg factors are

[
(0,0,1,1),
\qquad
(0,0,-1,-1),
]

with tags

[
19=\log_2(43)+\log_2(47),
]

and

[
87=-\log_2(43)-\log_2(47).
]

Both lie in

[
L_{34}^+.
]

Therefore

[
\boxed{
{34,72}\subset B_1^-,
\qquad
{19,87}\subset L_{34}^+,
}
]

and

[
\boxed{
34+19=53,
\qquad
72+87=53.
}
]

So the positive coefficient

[
2o_{34}[53]
]

is exactly the two-point scalar blade-(1) convolution

[
\boxed{
([34]+[72])([19]+[87])
\quad\text{at coefficient }[53].
}
]

---

# 8. The unused upper-minus scalar blade-(1) pair

The upper-minus scalar blade-(1) sector is

[
B_1^-=[34]+[44]+[62]+[72].
]

Split it as

[
B_1^-=D_3\sqcup D_{11},
]

where

[
D_3:=[34]+[72],
]

and

[
D_{11}:=[44]+[62].
]

Here

[
D_3
===

[2\log_2(3)]+[-2\log_2(3)],
]

whereas

[
D_{11}
======

[2\log_2(11)]+[-2\log_2(11)].
]

## Proposition 8.1 — Only the (3^{\pm2})-scalar pair hits the target directly

[
[53](D_3L_{34}^+)=2,
]

but

[
[53](D_{11}L_{34}^+)=0.
]

### Proof

First,

[
D_3L_{34}^+
===========

([34]+[72])([19]+[87]).
]

The sums are

[
34+19=53,
]

[
34+87=121\equiv15,
]

[
72+19=91,
]

[
72+87=159\equiv53.
]

Therefore

[
D_3L_{34}^+
===========

2[53]+[15]+[91],
]

so

[
[53](D_3L_{34}^+)=2.
]

Next,

[
D_{11}L_{34}^+
==============

([44]+[62])([19]+[87]).
]

The sums are

[
44+19=63,
]

[
44+87=131\equiv25,
]

[
62+19=81,
]

[
62+87=149\equiv43.
]

Thus

[
D_{11}L_{34}^+
==============

[25]+[43]+[63]+[81],
]

so

[
[53](D_{11}L_{34}^+)=0.
]

[
\boxed{}
]

---

# 9. Complement form: the (11^2)-axis is the blade-(1) complement

Let

[
A:=(2,2,1,1).
]

Then

[
A
]

is the full exponent vector of

[
a_\ast=3^2\cdot11^2\cdot43\cdot47.
]

Its logarithm is

[
\Lambda(A)
==========

# 2\cdot70+2\cdot22+59+66

309
\equiv97\pmod{106}.
]

Let

[
J:=(0,2,0,0),
]

so

[
J
]

is the scalar blade-(1) vector for

[
11^2,
]

with

[
\Lambda(J)=2\cdot22=44.
]

Then

[
A-J=(2,0,1,1)=\beta^+,
]

and

[
\Lambda(A-J)
============

# 97-44

53.

]

Likewise,

[
-A-(-J)
=======

(-2,0,-1,-1)=\beta^-,
]

and

[
\Lambda(-A+J)
=============

# -97+44

-53
\equiv53\pmod{106}.
]

Therefore the two target hits are also the complements of the (11^{\pm2})-axis inside the two antipodal full-box corners:

[
\boxed{
\beta^+=A-J,
\qquad
\beta^-=-A+J.
}
]

The rigidity report records

[
11^2\equiv -a_\ast\pmod{107},
]

and hence

[
3^2\cdot43\cdot47\equiv -1\pmod{107},
]

which is exactly the positive target congruence.

---

# 10. Blade-(1) positivity extractor

Define

[
\mathsf P_{1}^{107}(t)
:=
[t]\bigl(B_1^-L_{34}^+ + B_1^+L_{34}^-\bigr).
]

Then

[
\mathsf P_{1}^{107}(t)
]

is the completed ((-, -))-sector coefficient of the (o_{34})-sector at tag (t).

By Lemma 5.1,

[
\mathsf P_{1}^{107}(t)
======================

\begin{cases}
2,&t=53,\
1,&t\in
{3,7,15,17,21,25,35,43,63,71,81,85,89,91,99,103},\
0,&\text{otherwise.}
\end{cases}
]

Thus

[
\boxed{
\mathsf P_{1}^{107}(53)=2
}
]

and

[
\boxed{
53
\text{ is the unique multiplicity-(2) point of }
\mathsf P_{1}^{107}.
}
]

Since the (-p_\ast^{-1})-target is (60),

[
\boxed{
\mathsf P_{1}^{107}(60)=0.
}
]

Therefore the actual positive shell coefficient is recovered as

[
\boxed{
\Omega_{107}(p_\ast)
====================

# \mathsf P_{1}^{107}(53)

2.

}
]

The positivity is not distributed across the symmetric non-scalar blade rows. It is concentrated by the scalar blade-(1) imbalance and then transported to the (o_{34})-target leg by the two primitive (o_{34}) factors

[
[19],
\qquad
[87].
]

In equations:

[
\boxed{
\Omega_{107}(p_\ast)
====================

# [53]\bigl(([34]+[72])([19]+[87])\bigr)

2.

}
]

Equivalently,

[
\boxed{
\Omega_{107}(p_\ast)
====================

# [53]\bigl(D_3L_{34}^+\bigr)

2.

}
]

And in completed sector form:

[
\boxed{
\Theta_{107}^{\square}(p_\ast)
==============================

# 2\mathbf u_{-,-}o_{34}[53]

2\widehat s\underline s,o_{34}[53].
}
]