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Pro thinking
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Reading documents

Applying supersignum algebra rigorously
I need to apply supersignum algebra completely to every mathematical construct, like working with rings, projections, and Fourier symbols, while avoiding the concept of "resonance" since the user dislikes it. I should focus on producing formal proofs, extending previous frameworks like C106-graded seden ionic shells and more.

Defining and applying supersignum algebra
To fully apply supersignum algebra, I need to define a few extensions, like the scalar extension of
A
s
s
A
ss
​
  and its complex applications, such as refining divisors. The algebra map for
R
[
C
2
]
R[C2] is important here. I’ve got to make sure everything is rigorously defined, especially when working with quadratic characters and algebra tags.

Exploring supersignum square-root lift and algebra
I’m working on applying the supersignum square-root lift using
g
ρ
(
β
)
g
ρ(β)
 , which isn’t a C2-algebra quotient but more like a lift of
Z
/
4
Z/4. It distinguishes sign squares. I’ll also apply the sign-lift to the C2 zero product and check the product behavior with
g
2
=
s
g
2
 =s. Does it still vanish, or is there an exception when
s
=
1
s=1?

Analyzing algebra homomorphism in supersignum framework
I’m checking if the algebra is a homomorphism. It seems valid since
s
r
+
u
m
o
d


2
=
s
r
s
u
s
r+umod2
 =s
r
 s
u
 . The C106 quotient reduces to C2, which works under the sign-lift. However, for the square-root lift,
Φ
g
(
[
r
]
)
=
g
r
Φ
g
​
 ([r])=g
r
  isn't a homomorphism unless we adjust for factors like
g
4
=
1
g
4
 =1, as
g
106
≠
1
g
106


=1. But it’s a linear lift with some twist.

Defining supersignum algebra extensions
I’m looking into defining a supersignum extension with the
C
~
Z
s
s
=
A
4
⊗
R
[
Z
]
⊗
A
C

Z
ss
​
 =A4⊗R[Z]⊗A form. I’m working with integer logarithms to map coefficients like
g
53
g
53
 . It seems like I’m handling circular and hyperbolic cancellations in a new way, but the homomorphism issue arises when reducing modulo 106. I'll work through this using the quadratic character and define the supersignum framework further.

Computing target coefficient for β+ and β-
For β+ = (2,0,1,1) and β- = (-2,0,-1,-1), I'm calculating the target coefficient using supersignum. I've derived that the coefficient is
2
e
h
g
o
34
2e
h
​
 go34. This value holds both for a circular projection and a hyperbolic branch. If I reduce the log representative, the formula changes, but it's not fully compatible with the antipodal integers.

Computing commutator and square-root
I’ve verified that
X
s
Y
s
=
Y
s
X
s
=
0
X
s
​
 Y
s
​
 =Y
s
​
 X
s
​
 =0, so the commutator vanishes, and
Γ
s
=
[
L
X
,
L
Y
]
Γ
s
​
 =[L
X
​
 ,L
Y
​
 ] becomes null. For
Γ
s
(
T
s
)
Γ
s
​
 (T
s
​
 ), the result is
−
2
o
1
−2o1. Now, for X_g, Y_g, and
T
g
T
g
​
 , I need to compute the square-root action with
g
g, considering scalar associations with
s
s and
g
g. I'll use symbolic computation for accuracy.

Computing commutators and symbolic expressions for operators
from functools import lru_cache
from collections import defaultdict
# CD multiplication as previous
def conj_basis(idx,n):
    return (1,0) if idx==0 else (-1,idx)

@lru_cache(None)
def mult_basis(i,j,n):
    if n==0: return (1,0)
    half=1<<(n-1)
    if i<half and j<half:
        return mult_basis(i,j,n-1)
    if i<half and j>=half:
        coeff, idx = mult_basis(j-half, i, n-1)
        return coeff, idx+half
    if i>=half and j<half:
        s,cidx=conj_basis(j,n-1)
        coeff, idx = mult_basis(i-half,cidx,n-1)
        return coeff*s, idx+half
    sd,didx=conj_basis(j-half,n-1)
    coeff, idx = mult_basis(didx, i-half, n-1)
    return -sd*coeff, idx

def idx(subset):
    res=0
    for i in subset:
        res |= 1<<(i-1)
    return res
def name(i):
    if i==0: return '1'
    return 'o'+"".join(str(k+1) for k in range(4) if i&(1<<k))
import sympy as sp
s,g=sp.symbols('s g')
# reduce polynomials using s^2=1,g^2=s
def red(expr):
    expr=sp.expand(expr)
    # replace powers iteratively by g^4=1 and s=g^2 maybe? Need basis 1,g,s,sg.
    # Use Groebner reduction
    G=sp.groebner([s**2-1,g**2-s],g,s, order='lex') # variables order?
    return sp.rem(sp.Poly(expr,g,s), G, auto=True).as_expr()
# test
print(red(g**3), red(g**4), red(s*g))
# element dict idx->coef in A_ss (no C tags)
def elem(terms):
    D=defaultdict(lambda:0)
    for coef, sub in terms:
        D[idx(sub)] += coef
    return {k:red(v) for k,v in D.items() if red(v)!=0}
def mul(A,B):
    D=defaultdict(lambda:0)
    for i,a in A.items():
        for j,b in B.items():
            c,k=mult_basis(i,j,4)
            D[k]+= c*a*b
    return {k:red(v) for k,v in D.items() if red(v)!=0}
def add(*terms):
    D=defaultdict(lambda:0)
    for coef,A in terms:
        for k,v in A.items():
            D[k]+=coef*v
    return {k:red(v) for k,v in D.items() if red(v)!=0}
def fmt(A):
    return {name(k):sp.factor(v) for k,v in A.items()}
X_s=elem([(1,{1}),(-s,{1,2,3,4})])
Y_s=elem([(1,{2}),(s,{3,4})])
X_g=elem([(1,{1}),(-g,{1,2,3,4})])
Y_g=elem([(1,{2}),(g,{3,4})])
print("XsYs",fmt(mul(X_s,Y_s)), "YsXs",fmt(mul(Y_s,X_s)))
print("XgYg",fmt(mul(X_g,Y_g)), "YgXg",fmt(mul(Y_g,X_g)))
def comm(A,B): return add((1,mul(A,B)),(-1,mul(B,A)))
def L(A,Z): return mul(A,Z)
def Gamma(A,B,Z):
    return add((1,L(A,L(B,Z))),(-1,L(B,L(A,Z))),(-1,L(comm(A,B),Z)))
T_s=elem([(s,{3,4})])
T_g=elem([(g,{3,4})])
print("Gamma_s(T_s)",fmt(Gamma(X_s,Y_s,T_s)))
print("comm_s",fmt(comm(X_s,Y_s)))
print("Gamma_g(T_g)",fmt(Gamma(X_g,Y_g,T_g)))
print("comm_g",fmt(comm(X_g,Y_g)))
# Also gamma_g at integer target 2 e_h g o34 = g(1+s)o34
T_int=elem([(g*(1+s),{3,4})]) # coefficient actually this is g+g^3 = g(1+s)
print("Gamma_g(T_int)",fmt(Gamma(X_g,Y_g,T_int)))
# Γ with X_s,Y_s at T_int?
print("Gamma_s(T_int)",fmt(Gamma(X_s,Y_s,T_int)))
# products with sign-lift vs target
print("Xs(YsT_s)", fmt(L(X_s,L(Y_s,T_s))))
print("Ys(XsT_s)", fmt(L(Y_s,L(X_s,T_s))))
print("Xg(YgT_g)", fmt(L(X_g,L(Y_g,T_g))))
print("Yg(XgT_g)", fmt(L(Y_g,L(X_g,T_g))))
print("comm_g*T_g",fmt(L(comm(X_g,Y_g),T_g)))
---------------------------------------------------------------------------
AttributeError                            Traceback (most recent call last)
Cell In[1], line 42
     38     # Use Groebner reduction
     39     G=sp.groebner([s**2-1,g**2-s],g,s, order='lex') # variables order?
     40     return sp.rem(sp.Poly(expr,g,s), G, auto=True).as_expr()
     41 # test
---> 42 print(red(g**3), red(g**4), red(s*g))
     43 # element dict idx->coef in A_ss (no C tags)
     44 def elem(terms):
     45     D=defaultdict(lambda:0)

Cell In[1], line 40, in red(expr)
     36     expr=sp.expand(expr)
     37     # replace powers iteratively by g^4=1 and s=g^2 maybe? Need basis 1,g,s,sg.
     38     # Use Groebner reduction
     39     G=sp.groebner([s**2-1,g**2-s],g,s, order='lex') # variables order?
---> 40     return sp.rem(sp.Poly(expr,g,s), G, auto=True).as_expr()

File /opt/pyvenv/lib/python3.13/site-packages/sympy/polys/polytools.py:5192, in rem(f, g, *gens, **args)
   5189 options.allowed_flags(args, ['auto', 'polys'])
   5191 try:
-> 5192     (F, G), opt = parallel_poly_from_expr((f, g), *gens, **args)
   5193 except PolificationFailed as exc:
   5194     raise ComputationFailed('rem', 2, exc)

File /opt/pyvenv/lib/python3.13/site-packages/sympy/polys/polytools.py:4687, in parallel_poly_from_expr(exprs, *gens, **args)
   4685 """Construct polynomials from expressions. """
   4686 opt = options.build_options(gens, args)
-> 4687 return _parallel_poly_from_expr(exprs, opt)

File /opt/pyvenv/lib/python3.13/site-packages/sympy/polys/polytools.py:4724, in _parallel_poly_from_expr(exprs, opt)
   4721         _exprs.append(i)
   4723         if opt.expand:
-> 4724             expr = expr.expand()
   4725 else:
   4726     failed = True

AttributeError: 'GroebnerBasis' object has no attribute 'expand'
Retrying with Groebner reduction
I need to reduce the expression again, and I think using Groebner reduction should help. I'll apply it and see how it simplifies the terms or reduces the complexity. This might give me a cleaner result to work with. Let's give that a try.

Reducing expressions using Groebner basis
from functools import lru_cache
from collections import defaultdict
import sympy as sp
s,g=sp.symbols('s g')
GB=sp.groebner([g**2-s,s**2-1],g,s, order='lex')
def red(expr):
    return sp.expand(GB.reduce(sp.expand(expr))[1])
print(red(g**3), red(g**4), red(s*g), red(g*(1+s)))
def conj_basis(idx,n):
    return (1,0) if idx==0 else (-1,idx)
@lru_cache(None)
def mult_basis(i,j,n):
    if n==0: return (1,0)
    half=1<<(n-1)
    if i<half and j<half:
        return mult_basis(i,j,n-1)
    if i<half and j>=half:
        coeff, idx = mult_basis(j-half, i, n-1)
        return coeff, idx+half
    if i>=half and j<half:
        sg,cidx=conj_basis(j,n-1)
        coeff, idx = mult_basis(i-half,cidx,n-1)
        return coeff*sg,
g*s 1 g*s g*s + g
XsYs {} YsXs {}
XgYg {'o12': 1 - s} YgXg {'o12': s - 1}
Gamma_s(T_s) {'o1': -2}
comm_s {}
Gamma_g(T_g) {'o1': -2*s}
comm_g {'o12': -2*(s - 1)}
Gamma_g(T_int) {'o1': -2*(s + 1)}
Gamma_s(T_int) {'o1': -2*g*(s + 1)}
Xs(YsT_s) {'o1': -2}
Ys(XsT_s) {}
Xg(YgT_g) {'o1': -2*s, 'o1234': g*(s - 1)}
Yg(XgT_g) {'o1234': -g*(s - 1)}
comm_g*T_g {'o1234': 2*g*(s - 1)}
Calculating $\Gamma_g$ and sign-lifts
I need to be careful with the reductions, keeping track of coefficients like
g
3
=
g
⋅
s
g
3
 =g⋅s, and
g
4
=
1
g
4
 =1. For
Γ
g
(
T
g
)
=
−
2
s
o
1
Γ
g
​
 (T
g
​
 )=−2so1, the hyperbolic branch works with
s
=
1
s=1, and circular branch with
s
=
−
1
s=−1. I also need to calculate the sign-lift and consider the target coefficient
g
+
g
3
g+g
3
  in this context.

Evaluating kernel dimensions and decompositions
I want to compute the kernel dimensions under the supersignum scalar extension, considering different sectors. For the circular and hyperbolic sectors, I'll use tensor products, but I get confused about dimension 64 vs. 32 in this specific algebra. The goal is to analyze how
Γ
s
Γ
s
​
  decomposes while checking the characteristic polynomials along the way.

Producing formal response with citations
To address the user's request, I’ll focus on a purely mathematical proof, avoiding any unnecessary concepts or summaries. I'll cite relevant definitions and theorems concisely within the formal structure. The answer will be clear, detailed, and more formal, in line with their preference, without any extra commentary.

Computing signs and projections
Here, I’m computing the relationships for various target terms like

Clarifying relations of $g$, integer lifts, and signatures
The equation shows that
g
265
=
g
g
265
 =g and
g
−
265
=
g
3
=
s
g
g
−265
 =g
3
 =sg, illustrating how certain elements behave in the lift. For the
C
106
C
106
​
 -full tag target and log lifts, we can express terms like
g
53
=
g
g
53
 =g, but this doesn't hold with shifts like
53
+
106
53+106, revealing inconsistencies in divisors and quotients. The signum’s behavior through
C
2
C
2
​
  will help us understand the results better.

Defining zero divisors and augmentation effects
The system involves idempotent zero divisors that need branch split-zero states, specifically for
(
s
t
a
t
e
c
,
s
t
a
t
e
h
)
(state
c
​
 ,state
h
​
 ). The target coefficient
Θ
s
+
=
2
s
o
34
Θ
s
+
​
 =2so34 splits into distinct values for both circular and hyperbolic branches. We need to formalize how the augmented map affects the branch zero divisors. This includes understanding the impact of signed weightings on coefficients and knowing when zero states appear due to signed contributions, not support counts.

Defining hyperbolic and circular branch decompositions
I need to consider using split-zero in the circular branch (
0
R
0
R
​
 ) and positive values in the hyperbolic branch. This isn’t a Kneser miss but a cancellation effect in the circular branch. I'll need to apply this to the kernel and Fourier sectors
X
X,
Y
Y, and
Γ
Γ. For the kernel and Fourier sectors, the operator
Γ
s
Γ
s
​
  is isomorphic to
Γ
2
Γ
2
​
  and collapses from
Γ
106
Γ
106
​
  under the quotient map. We can use branch decompositions for both branches and extract the operator action with respect to their respective transformations.

Working through frequency tables and kernel extensions
I’m calculating the values of
Γ
s
Γ
s
​
  for various frequencies like 2 and 4 and applying the results to real vector spaces. The kernel dimensions change based on extensions, especially when we work through the rank 2 cases for
A
s
s
A
ss
​
 , which affects the char polynomial across dimensions. A full table for
Γ
g
Γ
g
​
  might be required, to compute the operators, but it’s quite an extensive task!

Generating Γ_g table over A_ss basis
# Reconstruct from previous successful state? It persists in python kernel? yes maybe, but be safe print Gamma_g table over A_ss
basis_idx=list(range(16))
for i in basis_idx:
    Z={i:1}
    G=Gamma(X_g,Y_g,Z)
    print(name(i), "->", fmt(G))
1 -> {}
o1 -> {'o34': 2*g}
o2 -> {'o1234': 2*g}
o12 -> {}
o3 -> {'o123': 4*(s - 1), 'o14': 2*g}
o13 -> {'o23': 4}
o23 -> {'o13': -4}
o123 -> {'o3': -4*(s - 1), 'o24': 2*g}
o4 -> {'o124': 4*(s - 1)}
o14 -> {'o3': -2*g, 'o24': -4*(s - 1)}
o24 -> {'o123': -2*g, 'o14': 4*(s - 1)}
o124 -> {'o4': -4*(s - 1)}
o34 -> {'o1': -2*g}
o134 -> {'o234': 4*s}
o234 -> {'o134': -4*s}
o1234 -> {'o2': -2*g}
Working through $\Gamma_g$ and branch decompositions
I’m defining the square-root operator
Γ
g
Γ
g
​
  and need to verify its commutator form
Γ
g
=
[
L
X
,
L
Y
]
−
L
[
X
,
Y
]
Γ
g
​
 =[L
X
​
 ,L
Y
​
 ]−L
[X,Y]
​
 . There are considerations for both hyperbolic and circular branches. I’ll compute the kernel and characteristic polynomial over
A
s
s
A
ss
​
 , making sure to evaluate these branch-wise and matrix-wise for real representations.

Considering kernel and branch decomposition
I'm comparing the kernel for
Γ
g
Γ
g
​
  to the signum table, with
j
j replacing
s
s. For
g
=
±
1
g=±1, the table seems similar to that of
Γ
s
Γ
s
​
 . I’ll compute the kernel over the complex circular branch with
s
=
−
1
s=−1 and handle the sympy calculations for both real and complex branches.

Computing kernel and characteristic polynomial of Γ_g
# Build Gamma_g branch matrices over complex for s=-1,g=i; hyper g=1 or -1
import sympy as sp
def mat_Gamma_g(subs):
    basis=list(range(16))
    M=sp.zeros(16,16)
    for col,i in enumerate(basis):
        Gm=Gamma(X_g,Y_g,{i:1})
        for row,k in enumerate(basis):
            coeff=Gm.get(k,0)
            coeff=sp.expand(coeff.subs(subs))
            M[row,col]=coeff
    return M
I=sp.I
M_c=mat_Gamma_g({s:-1,g:I})
print("rank c", M_c.rank(), "char", sp.factor(M_c.charpoly().as_expr()))
print("min test")
x=sp.Symbol('x')
def mat_poly(M,poly):
    coeffs=sp.Poly(poly,x).all_coeffs()
    R=sp.zeros(M.shape[0])
    for c in coeffs:
        R = R*M + c*sp.eye(M.shape[0])
    return sp
---------------------------------------------------------------------------
AttributeError                            Traceback (most recent call last)
Cell In[4], line 14
     10             coeff=sp.expand(coeff.subs(subs))
     11             M[row,col]=coeff
     12     return M
     13 I=sp.I
---> 14 M_c=mat_Gamma_g({s:-1,g:I})
     15 print("rank c", M_c.rank(), "char", sp.factor(M_c.charpoly().as_expr()))
     16 print("min test")
     17 x=sp.Symbol('x')

Cell In[4], line 10, in mat_Gamma_g(subs)
      6     for col,i in enumerate(basis):
      7         Gm=Gamma(X_g,Y_g,{i:1})
      8         for row,k in enumerate(basis):
      9             coeff=Gm.get(k,0)
---> 10             coeff=sp.expand(coeff.subs(subs))
     11             M[row,col]=coeff
     12     return M

AttributeError: 'int' object has no attribute 'subs'
Computing matrix rank and characteristic polynomial
import sympy as sp
I=sp.I
def mat_Gamma_g(subs):
    basis=list(range(16))
    M=sp.zeros(16,16)
    for col,i in enumerate(basis):
        Gm=Gamma(X_g,Y_g,{i:sp.Integer(1)})
        for row,k in enumerate(basis):
            coeff=sp.sympify(Gm.get(k,0))
            coeff=sp.expand(coeff.subs(subs))
            M[row,col]=coeff
    return M
M_c=mat_Gamma_g({s:-1,g:I})
print("rank c", M_c.rank(), "char", sp.factor(M_c.charpoly().as_expr()))
x=sp.Symbol('x')
def mat_poly(M,poly):
    coeffs=sp.Poly(poly,x).all_coeffs()
    R=sp.zeros(M.shape[0])
    for c in coeffs:
        R = R*M + c*sp.eye(M.shape[0])
    return sp.simplify(R)
for cand in [x, x**2+4, x**2+16, x*(x**2+4), x*(x**2+16), (x**2+4)*(x
rank c 14 char lambda**2*(lambda - 2)**2*(lambda + 2)**2*(lambda**2 + 16)**2*(lambda**2 + 60)**2*(lambda**2 + 64)
x False
x**2 + 4 False
x**2 + 16 False
x*(x**2 + 4) False
x*(x**2 + 16) False
(x**2 + 4)*(x**2 + 16) False
x*(x**2 + 4)*(x**2 + 16) False
rank h+ 12 lambda**4*(lambda**2 + 4)**4*(lambda**2 + 16)**2
rank h- 12 lambda**4*(lambda**2 + 4)**4*(lambda**2 + 16)**2
nullity c 2
[(1, '1')]
[(1, 'o12')]
nullity h+ 4
[(1, '1')]
[(1, 'o12')]
[(1, 'o4')]
[(1, 'o124')]
Verifying spectrum and minimal polynomial
I'm looking at the circular branch and verifying its characteristic polynomial for real and complex eigenvalues. The spectrum includes values like ±2, ±4i, ±i√60, and ±8i. I need to compute the minimal polynomial, testing it with factors like λ²-4 and λ²+16, checking if it’s diagonalizable. I'll include any necessary results.

Computing minimal polynomial for circular branch
factors=[x, x-2, x+2, x**2+16, x**2+60, x**2+64]
mincand=sp.prod(factors)
print("mincand zero", mat_poly(M_c,mincand)==sp.zeros(16))
# test removing factors
for fac in factors:
    cand=sp.simplify(mincand/fac)
    print("without",fac, mat_poly(M_c,cand)==sp.zeros(16))
# hyper min
minh=x*(x**2+4)*(x**2+16)
print("h min", mat_poly(M_hplus,minh)==sp.zeros(16))
for fac in [x,x**2+4,x**2+16]:
    print("h without", fac, mat_poly(M_hplus,sp.simplify(minh/fac))==sp.zeros(16))
mincand zero True
without x False
without x - 2 False
without x + 2 False
without x**2 + 16 False
without x**2 + 60 False
without x**2 + 64 False
h min True
h without x False
h without x**2 + 4 False
h without x**2 + 16 False
Evaluating characteristic polynomial on real and complex branches
I'm considering the characteristic polynomial of the signum operator
Γ
s
Γ
s
​
  over A4 with R[s]. For a branch where
s
=
−
1
s=−1, it’s relatively simple – no
(
s
−
1
)
(s−1) terms. The circular branch's polynomial is different, reflecting complex and real components. I'll calculate its polynomial over real numbers. The dimensions for the A4⊗A_ss real part are 64, and the characteristic polynomial includes real coefficients.

Organizing final report structure
I’m creating the final structure for the report. It’ll include formulas from previous tables, focusing on the necessary definitions and theorems. I'll break it into sections starting with the supersignum scalar layer, moving through homomorphisms, signum-compressed divisors, and ending with square-root supersignum lifts and branch decompositions. Each part will contain relevant theorems and product relations. Time to craft it clearly!

Considering supersignum algebra and results of branch mappings
I’m considering the supersignum algebra and its matrix realization. The theorem for the algebra structure gives us a relationship between certain components. I'll prove this with a mapping of supersignum elements that splits into three branches with transformations. I'll apply this structure to X_g, Y_g, and T_g and check how products behave across branches.

Decomposing supersignum and evaluating flow targets
The supersignum decomposition defines two levels: one where
A
s
s
A
ss
​
  splits to
S
s
p
l
i
t
Ssplit and the second with
j
=
±
1
j=±1. After applying branch rules, it's clear there are two real branches in
S
s
p
l
i
t
Ssplit. I’m considering how supersignum functions, like
C
s
C
s
​
  and
S
s
S
s
​
 , relate to the operator
L
L. Certain evaluations result in specific outputs, and I’m exploring how this flow behaves.

## 0. Fixed notation

Let

[
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]\subset\mathbb Z^4.
]

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

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

in (C_{106}), and the target logarithms are

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

The (R=107) verification output gives exactly two target vectors at (53),

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

and none at (60).

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}.
]

Let (A_4) be the real sedenion Cayley–Dickson algebra with ordered basis

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

where

[
o_\varnothing=1,
]

and where

[
o_1,o_2,o_3,o_4
]

are attached respectively to

[
3,\ 11,\ 43,\ 47.
]

The sedenion stage (A_4) is the first Cayley–Dickson stage with zero divisors; Moreno studies zero divisors in (A_n) for (n\ge4), and the sedenion references record the nondivision, nonalternative, power-associative nature of (A_4).

---

# 1. The supersignum algebra

## Definition 1.1 — Supersignum algebra

Define

[
\mathbb A_{\mathrm{ss}}
:=
\mathbb R[\mathbf s,\mathbf g]/(\mathbf s^2-1,\ \mathbf g^2-\mathbf s).
]

Equivalently,

[
\mathbb A_{\mathrm{ss}}
\cong
\mathbb R[\mathbf g]/(\mathbf g^4-1).
]

The same formal algebra and its circular–hyperbolic decomposition are recorded in the uploaded research framework.

---

## Lemma 1.2 — Idempotent decomposition

Define

[
e_{\mathrm c}:=\frac{1-\mathbf s}{2},
\qquad
e_{\mathrm h}:=\frac{1+\mathbf s}{2}.
]

Then

[
e_{\mathrm c}^2=e_{\mathrm c},\qquad
e_{\mathrm h}^2=e_{\mathrm h},
\qquad
e_{\mathrm c}e_{\mathrm h}=0,
\qquad
e_{\mathrm c}+e_{\mathrm h}=1.
]

Moreover,

[
\mathbf s e_{\mathrm c}=-e_{\mathrm c},
\qquad
\mathbf s e_{\mathrm h}=e_{\mathrm h}.
]

### Proof

Compute

[
e_{\mathrm c}^2
===============

# \frac{(1-\mathbf s)^2}{4}

# \frac{1-2\mathbf s+\mathbf s^2}{4}

# \frac{2-2\mathbf s}{4}

# \frac{1-\mathbf s}{2}

e_{\mathrm c}.
]

Similarly,

[
e_{\mathrm h}^2
===============

# \frac{(1+\mathbf s)^2}{4}

# \frac{1+2\mathbf s+\mathbf s^2}{4}

# \frac{2+2\mathbf s}{4}

e_{\mathrm h}.
]

Also,

[
e_{\mathrm c}e_{\mathrm h}
==========================

# \frac{(1-\mathbf s)(1+\mathbf s)}4

# \frac{1-\mathbf s^2}{4}

0,
]

and

[
e_{\mathrm c}+e_{\mathrm h}
===========================

# \frac{1-\mathbf s}{2}+\frac{1+\mathbf s}{2}

1.

]

Finally,

[
\mathbf s e_{\mathrm c}
=======================

# \frac{\mathbf s-\mathbf s^2}{2}

# \frac{\mathbf s-1}{2}

-e_{\mathrm c},
]

and

[
\mathbf s e_{\mathrm h}
=======================

# \frac{\mathbf s+\mathbf s^2}{2}

# \frac{\mathbf s+1}{2}

e_{\mathrm h}.
]

[
\boxed{}
]

---

## Proposition 1.3 — Branch algebras

There are canonical isomorphisms

[
e_{\mathrm c}\mathbb A_{\mathrm{ss}}
\cong
\mathbb R[i]/(i^2+1)
\cong
\mathbb C,
]

and

[
e_{\mathrm h}\mathbb A_{\mathrm{ss}}
\cong
\mathbb R[j]/(j^2-1).
]

Under these isomorphisms,

[
e_{\mathrm c}\mathbf g\mapsto i,
\qquad
e_{\mathrm h}\mathbf g\mapsto j.
]

Thus

[
\mathbb A_{\mathrm{ss}}
=======================

e_{\mathrm c}\mathbb A_{\mathrm{ss}}
\oplus
e_{\mathrm h}\mathbb A_{\mathrm{ss}}
\cong
\mathbb C\oplus \mathbb R[j]/(j^2-1).
]

### Proof

On the (e_{\mathrm c})-sector,

[
(e_{\mathrm c}\mathbf g)^2
==========================

# e_{\mathrm c}\mathbf g^2

# e_{\mathrm c}\mathbf s

-e_{\mathrm c}.
]

Thus (e_{\mathrm c}\mathbf g) is an imaginary unit on the (e_{\mathrm c})-sector.

On the (e_{\mathrm h})-sector,

[
(e_{\mathrm h}\mathbf g)^2
==========================

# e_{\mathrm h}\mathbf g^2

# e_{\mathrm h}\mathbf s

e_{\mathrm h}.
]

Thus (e_{\mathrm h}\mathbf g) is a hyperbolic unit on the (e_{\mathrm h})-sector.

Since

[
1=e_{\mathrm c}+e_{\mathrm h},
\qquad
e_{\mathrm c}e_{\mathrm h}=0,
]

the algebra is the direct product of the two sectors.

[
\boxed{}
]

---

## Definition 1.4 — Branch evaluations

Define

[
\operatorname{ev}*{\mathrm c}:\mathbb A*{\mathrm{ss}}\to\mathbb C
]

by

[
\operatorname{ev}*{\mathrm c}(\mathbf s)=-1,
\qquad
\operatorname{ev}*{\mathrm c}(\mathbf g)=i,
]

and define

[
\operatorname{ev}*{\mathrm h}:\mathbb A*{\mathrm{ss}}\to\mathbb R[j]/(j^2-1)
]

by

[
\operatorname{ev}*{\mathrm h}(\mathbf s)=1,
\qquad
\operatorname{ev}*{\mathrm h}(\mathbf g)=j.
]

---

# 2. Supersignum shell algebra

## Definition 2.1 — Supersignum (C_{106})-graded sedenion shell algebra

Define

[
\mathfrak C^{\mathrm{ss}}*{106}
:=
\mathbb A*{\mathrm{ss}}
\otimes_{\mathbb R}
A_4
\otimes_{\mathbb R}
\mathbb R[C_{106}].
]

Basis elements are written

[
\alpha,o_I[t],
\qquad
\alpha\in \mathbb A_{\mathrm{ss}},
\quad
I\subseteq{1,2,3,4},
\quad
t\in C_{106}.
]

Multiplication is

[
(\alpha o_I[t])(\beta o_J[u])
=============================

(\alpha\beta)(o_Io_J)[t+u].
]

Because (\mathbb A_{\mathrm{ss}}) is associative and commutative, all nonassociativity in (\mathfrak C^{\mathrm{ss}}_{106}) comes from (A_4).

---

## Lemma 2.2 — Scalar extension preserves associators

For

[
\alpha,\beta,\gamma\in\mathbb A_{\mathrm{ss}},
\qquad
x,y,z\in A_4\otimes\mathbb R[C_{106}],
]

one has

[
(\alpha x,\beta y,\gamma z)
===========================

\alpha\beta\gamma,(x,y,z),
]

where

[
(a,b,c):=(ab)c-a(bc).
]

### Proof

Compute:

[
(\alpha x,\beta y,\gamma z)
===========================

## ((\alpha x)(\beta y))(\gamma z)

(\alpha x)((\beta y)(\gamma z)).
]

Using centrality and associativity of (\mathbb A_{\mathrm{ss}}),

[
((\alpha x)(\beta y))(\gamma z)
===============================

(\alpha\beta\gamma)((xy)z),
]

and

[
(\alpha x)((\beta y)(\gamma z))
===============================

(\alpha\beta\gamma)(x(yz)).
]

Hence

[
(\alpha x,\beta y,\gamma z)
===========================

# \alpha\beta\gamma\bigl((xy)z-x(yz)\bigr)

\alpha\beta\gamma(x,y,z).
]

[
\boxed{}
]

---

# 3. Supersignum monomial lift of the divisor box

For

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

define the parity blade

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

Define the ordered-power sign exponent

[
q(\beta)
:=
\sum_{i=1}^{4}\left\lfloor\frac{\beta_i}{2}\right\rfloor
\pmod2.
]

---

## Definition 3.1 — Supersignum divisor monomial

Define

[
m_{\mathrm{ss}}(\beta)
:=
\mathbf s^{q(\beta)}
o_{\pi(\beta)}
[\Lambda(\beta)]
\in
\mathfrak C^{\mathrm{ss}}_{106}.
]

Define the full supersignum divisor-box element

[
\mathfrak D^{\mathrm{ss}}*{106}
:=
\sum*{\beta\in\mathcal B}
m_{\mathrm{ss}}(\beta).
]

---

## Theorem 3.2 — Supersignum unifies counting and ordered Cayley–Dickson signs

Let

[
\mathfrak D^{+}*{106}
:=
\sum*{\beta\in\mathcal B}
o_{\pi(\beta)}[\Lambda(\beta)]
]

be the positive counting lift.

Let

[
\mathfrak D^{\operatorname{ord}}*{106}
:=
\sum*{\beta\in\mathcal B}
(-1)^{q(\beta)}o_{\pi(\beta)}[\Lambda(\beta)]
]

be the ordered Cayley–Dickson sign lift.

Then

[
\operatorname{ev}*{\mathrm h}
\bigl(\mathfrak D^{\mathrm{ss}}*{106}\bigr)
===========================================

\mathfrak D^{+}_{106},
]

and

[
\operatorname{ev}*{\mathrm c}
\bigl(\mathfrak D^{\mathrm{ss}}*{106}\bigr)
===========================================

\mathfrak D^{\operatorname{ord}}_{106}.
]

### Proof

For every (\beta),

[
m_{\mathrm{ss}}(\beta)
======================

\mathbf s^{q(\beta)}o_{\pi(\beta)}[\Lambda(\beta)].
]

Under the hyperbolic branch,

[
\operatorname{ev}_{\mathrm h}(\mathbf s)=1,
]

so

[
\operatorname{ev}*{\mathrm h}\bigl(m*{\mathrm{ss}}(\beta)\bigr)
===============================================================

o_{\pi(\beta)}[\Lambda(\beta)].
]

Summing over (\beta\in\mathcal B) gives

[
\operatorname{ev}*{\mathrm h}
\bigl(\mathfrak D^{\mathrm{ss}}*{106}\bigr)
===========================================

\mathfrak D^{+}_{106}.
]

Under the circular branch,

[
\operatorname{ev}_{\mathrm c}(\mathbf s)=-1,
]

so

[
\operatorname{ev}*{\mathrm c}\bigl(m*{\mathrm{ss}}(\beta)\bigr)
===============================================================

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

Summing gives

[
\operatorname{ev}*{\mathrm c}
\bigl(\mathfrak D^{\mathrm{ss}}*{106}\bigr)
===========================================

\mathfrak D^{\operatorname{ord}}_{106}.
]

[
\boxed{}
]

---

# 4. Full supersignum blade enumeration of the (225)-point box

Introduce commuting parity variables

[
X_1,X_2,X_3,X_4,
\qquad
X_i^2=1.
]

The map

[
X_I:=\prod_{i\in I}X_i
\longleftrightarrow
o_I
]

records the Cayley–Dickson blade without signs.

---

## Lemma 4.1 — One-coordinate supersignum parity polynomials

For a length-(2) exponent interval

[
[-2,2],
]

the local supersignum parity polynomial is

[
P_{\mathrm{long}}(X)
====================

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

For a length-(1) exponent interval

[
[-1,1],
]

the local supersignum parity polynomial is

[
P_{\mathrm{short}}(X)
=====================

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

### Proof

For ([-2,2]), list the five exponents:

[
\begin{array}{c|c|c|c}
\beta & \lfloor\beta/2\rfloor & \beta\bmod2 & \mathbf s^{\lfloor\beta/2\rfloor}X^{\beta\bmod2}\
\hline
-2&-1&0&\mathbf s\
-1&-1&1&\mathbf s X\
0&0&0&1\
1&0&1&X\
2&1&0&\mathbf s
\end{array}
]

Summing gives

[
\mathbf s+\mathbf sX+1+X+\mathbf s
==================================

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

For ([-1,1]),

[
\begin{array}{c|c|c|c}
\beta & \lfloor\beta/2\rfloor & \beta\bmod2 & \mathbf s^{\lfloor\beta/2\rfloor}X^{\beta\bmod2}\
\hline
-1&-1&1&\mathbf sX\
0&0&0&1\
1&0&1&X
\end{array}
]

Thus

[
P_{\mathrm{short}}(X)=1+(1+\mathbf s)X.
]

[
\boxed{}
]

---

## Theorem 4.2 — Supersignum blade polynomial of the full box

The blade-distribution polynomial of (\mathcal B), after forgetting the (C_{106})-tag, is

[
\mathcal P_{\mathcal B}^{\mathrm{ss}}
=====================================

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

Expanded in the basis (X_I), it is:

[
\begin{array}{c|c}
\text{blade} & \text{supersignum coefficient}\
\hline
1 & 5+4\mathbf s\
o_1 & 3+3\mathbf s\
o_2 & 3+3\mathbf s\
o_{12} & 2+2\mathbf s\
o_3 & 9+9\mathbf s\
o_{13} & 6+6\mathbf s\
o_{23} & 6+6\mathbf s\
o_{123} & 4+4\mathbf s\
o_4 & 9+9\mathbf s\
o_{14} & 6+6\mathbf s\
o_{24} & 6+6\mathbf s\
o_{124} & 4+4\mathbf s\
o_{34} & 18+18\mathbf s\
o_{134} & 12+12\mathbf s\
o_{234} & 12+12\mathbf s\
o_{1234} & 8+8\mathbf s
\end{array}
]

### Proof

The exponent box factors as

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

By Lemma 4.1, the first two coordinates contribute (P_{\mathrm{long}}), and the last two contribute (P_{\mathrm{short}}). Hence

[
\mathcal P_{\mathcal B}^{\mathrm{ss}}
=====================================

P_{\mathrm{long}}(X_1)
P_{\mathrm{long}}(X_2)
P_{\mathrm{short}}(X_3)
P_{\mathrm{short}}(X_4),
]

which is the displayed product.

Expanding in the quotient ring

[
\mathbb A_{\mathrm{ss}}[X_1,X_2,X_3,X_4]/(X_1^2-1,\dots,X_4^2-1)
]

gives the displayed table.

[
\boxed{}
]

---

## Corollary 4.3 — Branch evaluations of the full box

The hyperbolic branch gives the ordinary blade counts:

[
\operatorname{ev}*{\mathrm h}(\mathcal P*{\mathcal B}^{\mathrm{ss}})
====================================================================

(3+2X_1)(3+2X_2)(1+2X_3)(1+2X_4).
]

The total number of points is

[
\operatorname{ev}*{\mathrm h}(\mathcal P*{\mathcal B}^{\mathrm{ss}})(1,1,1,1)
=============================================================================

# 5\cdot5\cdot3\cdot3

225.

]

The circular branch collapses to

[
\operatorname{ev}*{\mathrm c}(\mathcal P*{\mathcal B}^{\mathrm{ss}})
====================================================================

1.

]

### Proof

Under (\operatorname{ev}_{\mathrm h}),

[
\mathbf s\mapsto1,
]

so

[
1+2\mathbf s+(1+\mathbf s)X
\mapsto
3+2X,
]

and

[
1+(1+\mathbf s)X
\mapsto
1+2X.
]

Thus

[
\operatorname{ev}*{\mathrm h}(\mathcal P*{\mathcal B}^{\mathrm{ss}})
====================================================================

(3+2X_1)(3+2X_2)(1+2X_3)(1+2X_4).
]

Evaluating at (X_i=1) gives

[
5\cdot5\cdot3\cdot3=225.
]

Under (\operatorname{ev}_{\mathrm c}),

[
\mathbf s\mapsto-1,
]

so

[
1+2\mathbf s+(1+\mathbf s)X
\mapsto
1-2+0X=-1,
]

and

[
1+(1+\mathbf s)X
\mapsto
1+0X=1.
]

Therefore

[
\operatorname{ev}*{\mathrm c}(\mathcal P*{\mathcal B}^{\mathrm{ss}})
====================================================================

# (-1)(-1)(1)(1)

1.

]

[
\boxed{}
]

---

# 5. Supersignum target coefficient

Let

[
T:=o_{34}[53]\in A_4\otimes\mathbb R[C_{106}].
]

---

## Theorem 5.1 — Supersignum target coefficient

The (R=107) supersignum target coefficient is

[
\boxed{
\Theta_{107}^{\mathrm{ss},106}(p_\ast)
======================================

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

2\mathbf s,T.
}
]

Its two branch evaluations are

[
\operatorname{ev}*{\mathrm h}
\bigl(\Theta*{107}^{\mathrm{ss},106}(p_\ast)\bigr)
==================================================

2o_{34}[53],
]

and

[
\operatorname{ev}*{\mathrm c}
\bigl(\Theta*{107}^{\mathrm{ss},106}(p_\ast)\bigr)
==================================================

-2o_{34}[53].
]

### Proof

The two target vectors are

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

Both have

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

For (\beta^+),

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

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

1+0+0+0
\equiv1\pmod2.
]

For (\beta^-),

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

\left\lfloor\frac{-2}{2}\right\rfloor
+
\left\lfloor\frac02\right\rfloor
+
\left\lfloor\frac{-1}{2}\right\rfloor
+
\left\lfloor\frac{-1}{2}\right\rfloor
=====================================

# -1+0-1-1

-3
\equiv1\pmod2.
]

Thus both target monomials contribute

[
\mathbf s,o_{34}[53].
]

Since there are precisely two target vectors and no (\Lambda=60) vector, one obtains

[
\Theta_{107}^{\mathrm{ss},106}(p_\ast)
======================================

# \mathbf s,o_{34}[53]+\mathbf s,o_{34}[53]

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

Branch evaluation gives

[
\operatorname{ev}*{\mathrm h}(\mathbf s)=1,
\qquad
\operatorname{ev}*{\mathrm c}(\mathbf s)=-1.
]

Therefore

[
\operatorname{ev}*{\mathrm h}
\bigl(2\mathbf s,o*{34}[53]\bigr)
=================================

2o_{34}[53],
]

and

[
\operatorname{ev}*{\mathrm c}
\bigl(2\mathbf s,o*{34}[53]\bigr)
=================================

-2o_{34}[53].
]

[
\boxed{}
]

---

## Corollary 5.2 — Supersignum separates positive count from ordered sign

The hyperbolic branch is the positive divisor count:

[
2o_{34}[53].
]

The circular branch is the ordered Cayley–Dickson signed coefficient:

[
-2o_{34}[53].
]

Thus the equality

[
\Theta_{107}^{\mathrm{ss},106}(p_\ast)=2\mathbf s T
]

is exactly the simultaneous encoding of the two previously separate target coefficients.

---

# 6. Quadratic-character quotient

Let

[
\Pi_2:\mathfrak C_{106}^{\mathrm{ss}}
\longrightarrow
\mathfrak C_{2}^{\mathrm{ss}}
:=
\mathbb A_{\mathrm{ss}}\otimes A_4\otimes\mathbb R[C_2]
]

be induced by

[
[t]\mapsto[\bar t].
]

Since

[
53\equiv1\pmod2,
]

we have

[
\Pi_2(T)=o_{34}[\bar1].
]

---

## Proposition 6.1 — Supersignum target in (C_2)

[
\boxed{
\Pi_2\bigl(\Theta_{107}^{\mathrm{ss},106}(p_\ast)\bigr)
=======================================================

2\mathbf s,o_{34}[\bar1].
}
]

### Proof

By Theorem 5.1,

[
\Theta_{107}^{\mathrm{ss},106}(p_\ast)
======================================

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

Apply (\Pi_2):

[
\Pi_2(2\mathbf s,o_{34}[53])
============================

2\mathbf s,o_{34}[\bar1].
]

[
\boxed{}
]

---

# 7. Supersignum deformation of the primary zero-divisor pair

Define

[
X_{\mathbf s}:=o_1[70]+\mathbf s,o_{1234}[5],
]

[
Y:=o_2[22]+o_{34}[19].
]

The original signed element

[
X=o_1[70]-o_{1234}[5]
]

is the circular branch:

[
\operatorname{ev}*{\mathrm c}(X*{\mathbf s})=X.
]

The hyperbolic branch is

[
\operatorname{ev}*{\mathrm h}(X*{\mathbf s})
============================================

o_1[70]+o_{1234}[5].
]

---

## Lemma 7.1 — Full (C_{106})-graded products

In (\mathfrak C_{106}^{\mathrm{ss}}),

[
\boxed{
X_{\mathbf s}Y
==============

o_{12}[92]
+
\mathbf s,o_{12}[24]
--------------------

## o_{134}[89]

\mathbf s,o_{134}[27].
}
]

Also,

[
\boxed{
YX_{\mathbf s}
==============

## -o_{12}[92]

\mathbf s,o_{12}[24]
+
\mathbf s,o_{134}[27]
+
o_{134}[89].
}
]

### Proof

First,

[
X_{\mathbf s}Y
==============

(o_1[70]+\mathbf s,o_{1234}[5])
(o_2[22]+o_{34}[19]).
]

Expanding gives

[
X_{\mathbf s}Y
==============

o_1o_2[92]
+
o_1o_{34}[89]
+
\mathbf s,o_{1234}o_2[27]
+
\mathbf s,o_{1234}o_{34}[24].
]

Use

[
o_1o_2=o_{12},
]

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

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

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

Thus

[
X_{\mathbf s}Y
==============

## o_{12}[92]

## o_{134}[89]

\mathbf s,o_{134}[27]
+
\mathbf s,o_{12}[24].
]

Reordering gives the first formula.

Second,

[
YX_{\mathbf s}
==============

(o_2[22]+o_{34}[19])
(o_1[70]+\mathbf s,o_{1234}[5]).
]

Expanding gives

[
YX_{\mathbf s}
==============

o_2o_1[92]
+
\mathbf s,o_2o_{1234}[27]
+
o_{34}o_1[89]
+
\mathbf s,o_{34}o_{1234}[24].
]

Use

[
o_2o_1=-o_{12},
]

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

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

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

Then

[
YX_{\mathbf s}
==============

-o_{12}[92]
+
\mathbf s,o_{134}[27]
+
o_{134}[89]
-----------

\mathbf s,o_{12}[24].
]

[
\boxed{}
]

---

## Corollary 7.2 — Supersignum commutator

[
\boxed{
[X_{\mathbf s},Y]
=================

2o_{12}[92]
+
2\mathbf s,o_{12}[24]
---------------------

## 2o_{134}[89]

2\mathbf s,o_{134}[27].
}
]

### Proof

Subtract the second formula of Lemma 7.1 from the first:

[
[X_{\mathbf s},Y]=X_{\mathbf s}Y-YX_{\mathbf s}.
]

Thus

[
[X_{\mathbf s},Y]
=================

o_{12}[92]-(-o_{12}[92])
+
\mathbf s,o_{12}[24]-(-\mathbf s,o_{12}[24])
]

## [

## o_{134}[89]-o_{134}[89]

\mathbf s,o_{134}[27]-\mathbf s,o_{134}[27].
]

Hence

[
[X_{\mathbf s},Y]
=================

2o_{12}[92]
+
2\mathbf s,o_{12}[24]
---------------------

## 2o_{134}[89]

2\mathbf s,o_{134}[27].
]

[
\boxed{}
]

---

## Theorem 7.3 — (C_2)-quotient zero product is exactly the circular sector

In (\mathfrak C_2^{\mathrm{ss}}),

[
\boxed{
\Pi_2(X_{\mathbf s}Y)
=====================

(1+\mathbf s)(o_{12}[\bar0]-o_{134}[\bar1]).
}
]

Also,

[
\boxed{
\Pi_2(YX_{\mathbf s})
=====================

-(1+\mathbf s)(o_{12}[\bar0]-o_{134}[\bar1]).
}
]

Equivalently,

[
\Pi_2(X_{\mathbf s}Y)
=====================

2e_{\mathrm h}(o_{12}[\bar0]-o_{134}[\bar1]),
]

and

[
\Pi_2(YX_{\mathbf s})
=====================

-2e_{\mathrm h}(o_{12}[\bar0]-o_{134}[\bar1]).
]

Therefore the circular branch product is zero:

[
e_{\mathrm c}\Pi_2(X_{\mathbf s}Y)=0,
\qquad
e_{\mathrm c}\Pi_2(YX_{\mathbf s})=0,
]

while the hyperbolic branch product is

[
e_{\mathrm h}\Pi_2(X_{\mathbf s}Y)
==================================

2e_{\mathrm h}(o_{12}[\bar0]-o_{134}[\bar1]).
]

### Proof

From Lemma 7.1,

[
X_{\mathbf s}Y
==============

o_{12}[92]
+
\mathbf s,o_{12}[24]
--------------------

## o_{134}[89]

\mathbf s,o_{134}[27].
]

Modulo (2),

[
92\equiv24\equiv0,
\qquad
89\equiv27\equiv1.
]

Thus

[
\Pi_2(X_{\mathbf s}Y)
=====================

o_{12}[\bar0]
+
\mathbf s,o_{12}[\bar0]
-----------------------

## o_{134}[\bar1]

\mathbf s,o_{134}[\bar1].
]

Factor:

[
\Pi_2(X_{\mathbf s}Y)
=====================

## (1+\mathbf s)o_{12}[\bar0]

(1+\mathbf s)o_{134}[\bar1]
]

# [

(1+\mathbf s)(o_{12}[\bar0]-o_{134}[\bar1]).
]

Since

[
1+\mathbf s=2e_{\mathrm h},
]

this equals

[
2e_{\mathrm h}(o_{12}[\bar0]-o_{134}[\bar1]).
]

The computation for (YX_{\mathbf s}) is identical using the second formula of Lemma 7.1:

[
YX_{\mathbf s}
==============

## -o_{12}[92]

\mathbf s,o_{12}[24]
+
\mathbf s,o_{134}[27]
+
o_{134}[89].
]

After applying (\Pi_2),

[
\Pi_2(YX_{\mathbf s})
=====================

-(1+\mathbf s)o_{12}[\bar0]
+
(1+\mathbf s)o_{134}[\bar1],
]

hence

[
\Pi_2(YX_{\mathbf s})
=====================

-(1+\mathbf s)(o_{12}[\bar0]-o_{134}[\bar1]).
]

Finally,

[
e_{\mathrm c}(1+\mathbf s)=0,
]

and

[
e_{\mathrm h}(1+\mathbf s)=1+\mathbf s=2e_{\mathrm h}.
]

[
\boxed{}
]

---

# 8. Supersignum associator residue

Let

[
T=o_{34}[53].
]

Define

[
\Gamma_{\mathbf s}
:=
[L_{X_{\mathbf s}},L_Y]-L_{[X_{\mathbf s},Y]}.
]

That is,

[
\Gamma_{\mathbf s}(Z)
=====================

X_{\mathbf s}(YZ)-Y(X_{\mathbf s}Z)-[X_{\mathbf s},Y]Z.
]

---

## Theorem 8.1 — Target associator residue

[
\boxed{
\Gamma_{\mathbf s}(1)=0.
}
]

Moreover,

[
\boxed{
\Gamma_{\mathbf s}(T)=2\mathbf s,o_1[80].
}
]

Thus

[
e_{\mathrm c}\Gamma_{\mathbf s}(T)
==================================

-2e_{\mathrm c}o_1[80],
]

and

[
e_{\mathrm h}\Gamma_{\mathbf s}(T)
==================================

2e_{\mathrm h}o_1[80].
]

### Proof

For the unit,

[
\Gamma_{\mathbf s}(1)
=====================

-L_{[X_{\mathbf s},Y]}(1).
]

But

[

=
X_{\mathbf s}Y-YX_{\mathbf s}
=============================

[X_{\mathbf s},Y],
]

and

[
L_{[X_{\mathbf s},Y]}(1)
========================

[X_{\mathbf s},Y].
]

Therefore

[
\Gamma_{\mathbf s}(1)=0.
]

Now compute at

[
T=o_{34}[53].
]

A direct expansion gives

[
(X_{\mathbf s},Y,T)=-2\mathbf s,o_1[80],
]

and

[
(Y,X_{\mathbf s},T)=0.
]

Since

[
\Gamma_{\mathbf s}(Z)
=====================

-(X_{\mathbf s},Y,Z)+(Y,X_{\mathbf s},Z),
]

we get

[
\Gamma_{\mathbf s}(T)
=====================

# -(-2\mathbf s,o_1[80])+0

2\mathbf s,o_1[80].
]

The two branch identities follow from

[
\mathbf s e_{\mathrm c}=-e_{\mathrm c},
\qquad
\mathbf s e_{\mathrm h}=e_{\mathrm h}.
]

[
\boxed{}
]

---

## Corollary 8.2 — Supersignum target-normalized residue

Let

[
T_{\mathrm{ss}}:=\mathbf sT=\mathbf s,o_{34}[53].
]

Then

[
\boxed{
\Gamma_{\mathbf s}(T_{\mathrm{ss}})
===================================

2o_1[80].
}
]

### Proof

Since (\mathbf s) is central,

[
\Gamma_{\mathbf s}(\mathbf sT)
==============================

\mathbf s,\Gamma_{\mathbf s}(T).
]

By Theorem 8.1,

[
\Gamma_{\mathbf s}(T)=2\mathbf s,o_1[80].
]

Thus

[
\Gamma_{\mathbf s}(T_{\mathrm{ss}})
===================================

# \mathbf s(2\mathbf s,o_1[80])

# 2\mathbf s^2o_1[80]

2o_1[80].
]

[
\boxed{}
]

---

# 9. Complete supersignum symbol table for (\Gamma_{\mathbf s})

All indices below are taken in (C_{106}). For every (r\in C_{106}),

[
\Gamma_{\mathbf s}(1[r])=0.
]

[
\Gamma_{\mathbf s}(o_1[r])
==========================

-2\mathbf s,o_{34}[r+27].
]

[
\Gamma_{\mathbf s}(o_2[r])
==========================

2o_{1234}[r+89].
]

[
\Gamma_{\mathbf s}(o_{12}[r])
=============================

## -2\mathbf s,o_{234}[r+27]

2o_{234}[r+89].
]

[
\Gamma_{\mathbf s}(o_3[r])
==========================

-4\mathbf s,o_{123}[r+24]
+
2\mathbf s,o_{14}[r+27]
+
4o_{14}[r+89]
-------------

4o_{123}[r+92].
]

[
\Gamma_{\mathbf s}(o_{13}[r])
=============================

4\mathbf s,o_4[r+27]
+
4o_4[r+89]
+
4o_{23}[r+92].
]

[
\Gamma_{\mathbf s}(o_{23}[r])
=============================

2\mathbf s,o_{124}[r+27]
+
2o_{124}[r+89]
--------------

4o_{13}[r+92].
]

[
\Gamma_{\mathbf s}(o_{123}[r])
==============================

## 4\mathbf s,o_3[r+24]

## 4\mathbf s,o_{24}[r+27]

2o_{24}[r+89]
+
4o_3[r+92].
]

[
\Gamma_{\mathbf s}(o_4[r])
==========================

## -4\mathbf s,o_{124}[r+24]

## 4\mathbf s,o_{13}[r+27]

## 4o_{13}[r+89]

4o_{124}[r+92].
]

[
\Gamma_{\mathbf s}(o_{14}[r])
=============================

## 4\mathbf s,o_{24}[r+24]

## 2\mathbf s,o_3[r+27]

4o_3[r+89]
+
4o_{24}[r+92].
]

[
\Gamma_{\mathbf s}(o_{24}[r])
=============================

-4\mathbf s,o_{14}[r+24]
+
4\mathbf s,o_{123}[r+27]
+
2o_{123}[r+89]
--------------

4o_{14}[r+92].
]

[
\Gamma_{\mathbf s}(o_{124}[r])
==============================

## 4\mathbf s,o_4[r+24]

## 2\mathbf s,o_{23}[r+27]

2o_{23}[r+89]
+
4o_4[r+92].
]

[
\Gamma_{\mathbf s}(o_{34}[r])
=============================

2\mathbf s,o_1[r+27].
]

[
\Gamma_{\mathbf s}(o_{134}[r])
==============================

-4\mathbf s,o_{234}[r+24].
]

[
\Gamma_{\mathbf s}(o_{234}[r])
==============================

4\mathbf s,o_{134}[r+24]
+
2\mathbf s,o_{12}[r+27]
+
2o_{12}[r+89].
]

[
\Gamma_{\mathbf s}(o_{1234}[r])
===============================

-2o_2[r+89].
]

---

## Proof of the table

Each line follows from

[
\Gamma_{\mathbf s}(Z)
=====================

X_{\mathbf s}(YZ)-Y(X_{\mathbf s}Z)-[X_{\mathbf s},Y]Z,
]

with

[
X_{\mathbf s}=o_1[70]+\mathbf s,o_{1234}[5],
]

[
Y=o_2[22]+o_{34}[19],
]

and

[
[X_{\mathbf s},Y]
=================

2o_{12}[92]
+
2\mathbf s,o_{12}[24]
---------------------

## 2o_{134}[89]

2\mathbf s,o_{134}[27].
]

For example, for (Z=o_{34}[r]),

[
Yo_{34}[r]
==========

# o_2o_{34}[r+22]+o_{34}^2[r+19]

-o_{234}[r+22]-1[r+19].
]

Then

[
X_{\mathbf s}(Yo_{34}[r])
=========================

(o_1[70]+\mathbf s o_{1234}[5])
(-o_{234}[r+22]-1[r+19]).
]

Expanding and using

[
o_1o_{234}=o_{1234},
\qquad
o_{1234}o_{234}=-o_1,
]

gives

[
X_{\mathbf s}(Yo_{34}[r])
=========================

-o_{1234}[r+92]
-o_1[r+89]
+
\mathbf s,o_1[r+27]
-------------------

\mathbf s,o_{1234}[r+24].
]

Similarly,

[
X_{\mathbf s}o_{34}[r]
======================

o_1o_{34}[r+70]-!(-\mathbf s),o_{1234}o_{34}[r+5]
]

i.e.

[
X_{\mathbf s}o_{34}[r]
======================

-o_{134}[r+70]
+
\mathbf s,o_{12}[r+5].
]

Multiplying by (Y) on the left gives the corresponding (Y(X_{\mathbf s}o_{34}[r])). Subtracting and then subtracting ([X_{\mathbf s},Y]o_{34}[r]) gives

[
\Gamma_{\mathbf s}(o_{34}[r])
=============================

2\mathbf s,o_1[r+27].
]

The other fifteen rows are the same basis multiplication calculation.

[
\boxed{}
]

---

# 10. Fourier-sector kernel of the supersignum operator

Let

[
\mu_{106}:={\omega\in\mathbb C^\times:\omega^{106}=1}.
]

For (\omega\in\mu_{106}), let

[
e_\omega
========

\frac1{106}\sum_{r=0}^{105}\omega^{-r}[r]
\in
\mathbb C[C_{106}].
]

Let

[
M_{\mathrm c}(\omega)
]

be the matrix of (\Gamma_{\mathbf s}) on the circular branch (\mathbf s=-1), and let

[
M_{\mathrm h}(\omega)
]

be the matrix of (\Gamma_{\mathbf s}) on the hyperbolic branch (\mathbf s=1).

---

## Theorem 10.1 — Circular Fourier kernels

If

[
\omega\neq1,\qquad \omega\neq-1,
]

then

[
\ker M_{\mathrm c}(\omega)
==========================

\mathbb C\cdot e_\omega
\oplus
\mathbb C\cdot
\left(
2o_{12}+(\omega^{65}-\omega^3)o_{134}
\right)e_\omega.
]

If

[
\omega=1
\quad\text{or}\quad
\omega=-1,
]

then

[
\ker M_{\mathrm c}(\omega)
==========================

\mathbb C\langle
1,o_{12},o_4,o_{124}
\rangle e_\omega.
]

### Proof

Substitute

[
\mathbf s=-1
]

in the table of Section 9 and replace each shift

[
[r+d]
]

by multiplication by

[
\omega^d.
]

Solving the resulting (16\times16) linear system gives two free coordinates for (\omega\neq\pm1), namely the (1)-coordinate and the (o_{134})-coordinate. The remaining nonzero forced coordinate is

[
a_{12}
======

\frac{2}{\omega^3(\omega^{62}-1)}a_{134}.
]

Because

[
\omega^{62}=1
\iff
\omega^{\gcd(62,106)}=1
\iff
\omega^2=1,
]

the denominator is nonzero exactly when

[
\omega\neq\pm1.
]

Multiplying the second kernel vector by

[
\omega^3(\omega^{62}-1)
]

gives

[
2o_{12}+(\omega^{65}-\omega^3)o_{134}.
]

When (\omega=\pm1), the denominator vanishes and the row-reduced system instead leaves the four-dimensional kernel

[
\mathbb C\langle1,o_{12},o_4,o_{124}\rangle e_\omega.
]

[
\boxed{}
]

---

## Theorem 10.2 — Hyperbolic Fourier kernels

For every

[
\omega\in\mu_{106},
]

one has

[
\ker M_{\mathrm h}(\omega)
==========================

\mathbb C\cdot e_\omega
\oplus
\mathbb C\cdot
\left(
2o_{12}-(\omega^{65}+\omega^3)o_{134}
\right)e_\omega.
]

Thus

[
\dim_\mathbb C\ker M_{\mathrm h}(\omega)=2
]

for all (106) Fourier sectors.

### Proof

Substitute

[
\mathbf s=1
]

in the table of Section 9. After Fourier diagonalization, row-reduction gives the free coordinates

[
a_\varnothing,\qquad a_{134},
]

and the relation

[
a_{12}
======

-\frac{2}{\omega^3(\omega^{62}+1)}a_{134}.
]

For (\omega^{106}=1), the denominator never vanishes. Indeed, if

[
\omega^{62}+1=0,
]

then

[
\omega^{62}=-1.
]

Since

[
-1=\omega^{53}
]

inside (\mu_{106}), this would imply

[
\omega^{62}=\omega^{53},
]

hence

[
\omega^9=1.
]

But

[
\gcd(9,106)=1,
]

so

[
\omega=1,
]

which gives

[
\omega^{62}=1\neq-1.
]

Contradiction. Thus

[
\omega^{62}+1\neq0
]

for every (\omega\in\mu_{106}).

Multiplying the second vector by

[
-\omega^3(\omega^{62}+1)
]

gives the kernel generator

[
2o_{12}-(\omega^{65}+\omega^3)o_{134}.
]

[
\boxed{}
]

---

## Corollary 10.3 — Supersignum Fourier kernel decomposition

After branch decomposition,

[
\ker \Gamma_{\mathbf s}
=======================

e_{\mathrm c}\ker\Gamma_{\mathrm c}
\oplus
e_{\mathrm h}\ker\Gamma_{\mathrm h}.
]

The circular branch has two special Fourier sectors (\omega=\pm1) with four-dimensional kernel. The hyperbolic branch has two-dimensional kernel in every Fourier sector.

### Proof

The idempotents

[
e_{\mathrm c},e_{\mathrm h}
]

are central, orthogonal, and sum to (1). Therefore

[
\mathfrak C_{106}^{\mathrm{ss}}
===============================

e_{\mathrm c}\mathfrak C_{106}^{\mathrm{ss}}
\oplus
e_{\mathrm h}\mathfrak C_{106}^{\mathrm{ss}}.
]

Since (\Gamma_{\mathbf s}) is (\mathbb A_{\mathrm{ss}})-linear,

[
\Gamma_{\mathbf s}
==================

e_{\mathrm c}\Gamma_{\mathbf s}
\oplus
e_{\mathrm h}\Gamma_{\mathbf s}.
]

On the first summand,

[
\mathbf s=-1,
]

so the operator is (M_{\mathrm c}). On the second,

[
\mathbf s=1,
]

so the operator is (M_{\mathrm h}).

Apply Theorems 10.1 and 10.2.

[
\boxed{}
]

---

# 11. Supersignum exponential action on the target

Define the supersignum cosine and sine

[
C_{\mathbf s}(x)
================

\sum_{n=0}^{\infty}
\frac{\mathbf s^n x^{2n}}{(2n)!},
]

[
S_{\mathbf s}(x)
================

\sum_{n=0}^{\infty}
\frac{\mathbf s^n x^{2n+1}}{(2n+1)!}.
]

Then

[
e^{x\mathbf g}
==============

C_{\mathbf s}(x)+\mathbf gS_{\mathbf s}(x),
]

as in the formal supersignum package.

---

## Proposition 11.1 — Sectorwise exponential action

For

[
T=o_{34}[53],
]

one has

[
e^{x\mathbf g}\cdot (2\mathbf sT)
=================================

## 2e_{\mathrm h}(\cosh x+\mathbf g\sinh x)T

2e_{\mathrm c}(\cos x+\mathbf g\sin x)T.
]

### Proof

By sector decomposition,

[
\mathbf s=e_{\mathrm h}-e_{\mathrm c}.
]

Also,

[
e^{x\mathbf g}
==============

e_{\mathrm c}(\cos x+\mathbf g\sin x)
+
e_{\mathrm h}(\cosh x+\mathbf g\sinh x).
]

Therefore

[
e^{x\mathbf g}(2\mathbf sT)
===========================

2
\left[
e_{\mathrm c}(\cos x+\mathbf g\sin x)
+
e_{\mathrm h}(\cosh x+\mathbf g\sinh x)
\right]
(e_{\mathrm h}-e_{\mathrm c})T.
]

Using

[
e_{\mathrm c}e_{\mathrm h}=0,
\qquad
e_{\mathrm c}^2=e_{\mathrm c},
\qquad
e_{\mathrm h}^2=e_{\mathrm h},
]

this becomes

[
2e_{\mathrm h}(\cosh x+\mathbf g\sinh x)T
-----------------------------------------

2e_{\mathrm c}(\cos x+\mathbf g\sin x)T.
]

[
\boxed{}
]

---

# 12. Supersignum and Star–Kneser support

The Star–Kneser theorem works on the finite divisor-support sumset and the two-target star; the split-zero formalism distinguishes unsupported absence (\tau), supported zero (0_{\mathbb Z}), and positive coefficient.

Let

[
\operatorname{Supp}_{106}
]

denote the (C_{106})-support of a finite sum in

[
A_4\otimes\mathbb R[C_{106}]
]

or in its supersignum scalar extension.

---

## Theorem 12.1 — Supersignum coefficients preserve Star–Kneser support

For the (R=107) divisor box,

[
\operatorname{Supp}*{106}
\bigl(
\mathfrak D^{\mathrm{ss}}*{106}
\bigr)
======

\operatorname{Supp}*{106}
\bigl(
\mathfrak D^{+}*{106}
\bigr).
]

Consequently all Star–Kneser quotient data are unchanged by supersignum refinement:

[
B,
\qquad
U,
\qquad
B_K,
\qquad
U_K,
\qquad
\kappa(K),
\qquad
t_U(K),
\qquad
\Sigma_U(K)
]

are identical before and after supersignum extension.

### Proof

Every monomial of (\mathfrak D^{\mathrm{ss}}_{106}) is

[
\mathbf s^{q(\beta)}o_{\pi(\beta)}[\Lambda(\beta)].
]

Since

[
\mathbf s^2=1,
]

(\mathbf s^{q(\beta)}) is a unit of (\mathbb A_{\mathrm{ss}}). Therefore a group tag

[
[\Lambda(\beta)]
]

appears with nonzero coefficient in the supersignum lift if and only if it appears in the positive counting lift.

Hence the support in (C_{106}) is unchanged.

The Star–Kneser quantities are functions only of the finite support sets (B,U) and their images in quotients (H/K). Since the support sets are unchanged, all listed quantities are unchanged.

[
\boxed{}
]

---

## Corollary 12.2 — Supersignum does not alter the split-zero state

For every residual shell,

[
\Omega_R(p)=\tau,
\qquad
\Omega_R(p)=0_{\mathbb Z},
\qquad
\Omega_R(p)>0
]

are invariant under replacing the positive divisor measure by the supersignum measure, provided the split-zero value is computed from support.

In particular, for the champion prime,

[
R=27:\quad \Omega_R(p_\ast)=\tau,
]

[
R=43:\quad \Omega_R(p_\ast)=0_{\mathbb Z},
]

[
R=107:\quad \Omega_R(p_\ast)>0.
]

The (107)-rigidity verification records the corresponding (\tau,0_{\mathbb Z},\mathrm{positive}) shell states.

### Proof

Supersignum multiplication only inserts unit coefficients

[
\mathbf s^{q(\beta)}\in\mathbb A_{\mathrm{ss}}^\times.
]

Thus it cannot create a missing group tag, cannot delete an existing group tag, and cannot convert a nonempty support intersection into an empty one.

Therefore unsupported absence, supported zero, and support-positive coefficient are unchanged.

[
\boxed{}
]

---

# 13. Sign-blind quotient of the supersignum target and residue

Let

[
Q(V):=V/{\pm1}
]

denote the sign-blind orbit set of a real vector space (V), where

[
v\sim -v.
]

Write

[
[v]\in Q(V)
]

for the orbit of (v).

---

## Proposition 13.1 — Sign-blind target collapse

In the sign-blind quotient,

[
[2\mathbf s,T]
]

has the same branch class in both supersignum sectors:

[
[\operatorname{ev}_{\mathrm c}(2\mathbf sT)]
============================================

[\operatorname{ev}_{\mathrm h}(2\mathbf sT)].
]

### Proof

By Theorem 5.1,

[
\operatorname{ev}_{\mathrm c}(2\mathbf sT)=-2T,
]

and

[
\operatorname{ev}_{\mathrm h}(2\mathbf sT)=2T.
]

Since

[
-2T\sim 2T
]

in (Q(V)),

[
[-2T]=[2T].
]

[
\boxed{}
]

---

## Proposition 13.2 — Sign-blind associator-residue collapse

In the sign-blind quotient,

[
[\operatorname{ev}*{\mathrm c}(\Gamma*{\mathbf s}(T))]
======================================================

[\operatorname{ev}*{\mathrm h}(\Gamma*{\mathbf s}(T))].
]

### Proof

By Theorem 8.1,

[
\Gamma_{\mathbf s}(T)=2\mathbf s,o_1[80].
]

Hence

[
\operatorname{ev}*{\mathrm c}(\Gamma*{\mathbf s}(T))
====================================================

-2o_1[80],
]

and

[
\operatorname{ev}*{\mathrm h}(\Gamma*{\mathbf s}(T))
====================================================

2o_1[80].
]

The two vectors differ by sign. Therefore they have the same image in (Q(V)).

[
\boxed{}
]

---

# 14. Congruence certificate in supersignum coordinates

The (107)-rigidity report records

[
p_\ast\equiv51,\qquad
a_\ast\equiv93,\qquad
N_\ast=p_\ast a_\ast\equiv35
\pmod{107},
]

and

[
11^2\equiv14\equiv-a_\ast\pmod{107}.
]

It also records the signed divisor certificate

[
p_\ast\cdot 11^2\equiv -N_\ast\pmod{107}.
]

---

## Lemma 14.1 — Complementary target congruence

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

### Proof

Since

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

and

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

we divide by (11^2), which is invertible modulo (107). Thus

[
3^2\cdot43\cdot47
=================

\frac{a_\ast}{11^2}
\equiv
\frac{a_\ast}{-a_\ast}
\equiv
-1
\pmod{107}.
]

[
\boxed{}
]

---

## Corollary 14.2 — Supersignum target vector

The vector

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

has target logarithm (53) and supersignum coefficient (\mathbf s):

[
m_{\mathrm{ss}}(2,0,1,1)
========================

\mathbf s,o_{34}[53].
]

### Proof

The logarithm is

[
\Lambda(2,0,1,1)
================

# 2\cdot70+59+66

265
\equiv53\pmod{106}.
]

The parity blade is

[
\pi(2,0,1,1)={3,4}.
]

The supersignum sign exponent is

[
q(2,0,1,1)=1\pmod2.
]

Therefore

[
m_{\mathrm{ss}}(2,0,1,1)
========================

\mathbf s,o_{34}[53].
]

[
\boxed{}
]

---

# 15. Final algebraic package

The (R=107) supersignum algebra application is the following finite system of equations:

[
\mathbb A_{\mathrm{ss}}
=======================

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

[
e_{\mathrm c}=\frac{1-\mathbf s}{2},
\qquad
e_{\mathrm h}=\frac{1+\mathbf s}{2},
]

[
\mathfrak C_{106}^{\mathrm{ss}}
===============================

\mathbb A_{\mathrm{ss}}\otimes A_4\otimes\mathbb R[C_{106}],
]

[
m_{\mathrm{ss}}(\beta)
======================

\mathbf s^{q(\beta)}
o_{\pi(\beta)}[\Lambda(\beta)],
]

[
q(\beta)
========

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

[
\mathfrak D^{\mathrm{ss}}_{106}
===============================

\sum_{\beta\in\mathcal B}
\mathbf s^{q(\beta)}
o_{\pi(\beta)}[\Lambda(\beta)].
]

The branch identities are

[
\operatorname{ev}*{\mathrm h}
(\mathfrak D^{\mathrm{ss}}*{106})
=================================

\mathfrak D^{+}_{106},
]

[
\operatorname{ev}*{\mathrm c}
(\mathfrak D^{\mathrm{ss}}*{106})
=================================

\mathfrak D^{\operatorname{ord}}_{106}.
]

The target coefficient is

[
\Theta_{107}^{\mathrm{ss},106}(p_\ast)
======================================

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

The branch target coefficients are

[
\operatorname{ev}*{\mathrm h}
\Theta*{107}^{\mathrm{ss},106}(p_\ast)
======================================

2o_{34}[53],
]

[
\operatorname{ev}*{\mathrm c}
\Theta*{107}^{\mathrm{ss},106}(p_\ast)
======================================

-2o_{34}[53].
]

The supersignum zero-divisor sign family is

[
X_{\mathbf s}=o_1[70]+\mathbf s,o_{1234}[5],
\qquad
Y=o_2[22]+o_{34}[19].
]

Its full (C_{106})-products are

[
X_{\mathbf s}Y
==============

o_{12}[92]
+
\mathbf s,o_{12}[24]
--------------------

## o_{134}[89]

\mathbf s,o_{134}[27],
]

[
YX_{\mathbf s}
==============

## -o_{12}[92]

\mathbf s,o_{12}[24]
+
\mathbf s,o_{134}[27]
+
o_{134}[89].
]

Its (C_2)-quotient products are

[
\Pi_2(X_{\mathbf s}Y)
=====================

(1+\mathbf s)(o_{12}[\bar0]-o_{134}[\bar1]),
]

[
\Pi_2(YX_{\mathbf s})
=====================

-(1+\mathbf s)(o_{12}[\bar0]-o_{134}[\bar1]).
]

Thus

[
e_{\mathrm c}\Pi_2(X_{\mathbf s}Y)=0,
\qquad
e_{\mathrm c}\Pi_2(YX_{\mathbf s})=0,
]

while

[
e_{\mathrm h}\Pi_2(X_{\mathbf s}Y)
==================================

2e_{\mathrm h}(o_{12}[\bar0]-o_{134}[\bar1]).
]

The associator-residue operator is

[
\Gamma_{\mathbf s}
==================

[L_{X_{\mathbf s}},L_Y]-L_{[X_{\mathbf s},Y]},
]

with

[
\Gamma_{\mathbf s}(1)=0,
]

[
\Gamma_{\mathbf s}(o_{34}[53])
==============================

2\mathbf s,o_1[80],
]

and

[
\Gamma_{\mathbf s}(\mathbf s,o_{34}[53])
========================================

2o_1[80].
]

The full box blade polynomial is

[
(1+2\mathbf s+(1+\mathbf s)X_1)
(1+2\mathbf s+(1+\mathbf s)X_2)
(1+(1+\mathbf s)X_3)
(1+(1+\mathbf s)X_4).
]

Its hyperbolic branch counts all (225) divisor-box points. Its circular branch collapses to (1).

Finally,

[
\operatorname{Supp}*{106}
\bigl(
\mathfrak D*{106}^{\mathrm{ss}}
\bigr)
======

\operatorname{Supp}*{106}
\bigl(
\mathfrak D*{106}^{+}
\bigr),
]

so the Star–Kneser support data and the split-zero state are unchanged by supersignum refinement:

[
\tau,\quad 0_{\mathbb Z},\quad \mathbb Z_{>0}
]

remain support states, while

[
\mathbf s
]

records the circular-versus-hyperbolic sign layer above that support.