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\title{Corrected Interaction Calculus for the Residual Erd\H{o}s--Straus Obstruction\\
\large Weighted Snowflake Defects, Split-Zero Positivity, Niemeier Glue, and Modular Support Resolution}
\author{The Clankers}
\date{May 2026}

\begin{document}
\maketitle

\begin{abstract}
This note replaces the object-list presentation of the residual Erd\H{o}s--Straus framework by a single interacting calculus.  For a hard prime class $p\equiv 289^k\pmod {840}$ and a residual shell $R\equiv3\pmod4$, the fixed-shell divisor equation is first projectively normalized, producing the invariant target $-1$.  The missing structure is then restored by the six-sector coordinate
\[
        C_6\cong C_3\times C_2,
\]
so that the two normalized targets are not merely an unordered pair but the weighted snowflake defect
\[
        \mathfrak D_k=E_{0,-}+2E_{-k,-(-1)^k}\in \Z[C_3\times C_2].
\]
The coefficient $2$ is forced by the two $p$-origins in divisors of $p^2a^2$.  Certificate existence is proved equivalent to positivity of the split-zero pullback count
\[
        \Omega_R(p)=\widetilde{\mathfrak c}_R(a;-1)+2\widetilde{\mathfrak c}_R(a;-p^{-1}).
\]
The target defect has full Fourier support for every $k$, so a persistent obstruction cannot come from target degeneracy; it can only come from bounded positive-mass failure of the signed divisor box.  The champion prime $p_*=8{,}803{,}369$ realizes the strict ladder
\[
        (\tauzero,\tauzero)\longrightarrow(0_{\Z},0_{\Z})\longrightarrow(2,0_{\Z})
\]
for $R=27,43,107$.  The same full-arm sign reversal is resolved by the modular critical arm $744\leftrightarrow747\leftrightarrow750$: the raw $A_8^3$ eta sheet vanishes on all three points, while Fricke and Eisenstein channels separate the negative endpoint, the supported-zero hub, and the positive endpoint.  We compute the nonlinear $A_8^3$ support resolvent, the linear lambency-six $A_5^4D_4$ resolvents, and the finite transport matrices $P_{9\to6}^{(i)}$ over $\F_{107}$, including their characteristic polynomials.  The note also records the exact $A_5^4D_4$ glue datum of order $72$ and the non-split $GL_2(3)$ symmetry, only insofar as these objects act in the positivity calculus.
\end{abstract}

\tableofcontents

\section{The interacting object}

The residual obstruction is the composite
\[
\boxed{
\begin{gathered}
\text{divisor box}\longrightarrow
\text{projective target}\longrightarrow
\text{weighted snowflake/gerbe defect}\\
\longrightarrow
\text{split-zero support value}\longrightarrow
\text{positivity}
\end{gathered}}
\]
The point is not that the projective target, gerbe sectors, split-zero semiring, eta sheets, and theta envelopes exist as separate objects.  The point is that each layer controls exactly one map in the following pullback-and-evaluation diagram.

\begin{center}
\begin{tikzpicture}[node distance=13mm and 20mm,>=Latex, every node/.style={font=\small}]
\node[draw,rounded corners,align=center,minimum width=29mm,minimum height=9mm] (box) {$\mathcal B_R(a)$\\signed divisor box};
\node[draw,rounded corners,align=center,right=of box,minimum width=31mm,minimum height=9mm] (unit) {$(\Z/R\Z)^\times$\\residue torus};
\node[draw,rounded corners,align=center,right=of unit,minimum width=32mm,minimum height=9mm] (target) {$\{-1,-p^{-1}\}$\\two targets};
\node[draw,rounded corners,align=center,below=of target,minimum width=36mm,minimum height=9mm] (def) {$\mathfrak D_k\in\Z[C_3\times C_2]$\\weighted defect};
\node[draw,rounded corners,align=center,below=of def,minimum width=32mm,minimum height=9mm] (split) {$G(\Z_{\ge0})$\\split-zero value};
\node[draw,rounded corners,align=center,below=of split,minimum width=32mm,minimum height=9mm] (pos) {$\Omega_R(p)>0$\\certificate};
\draw[->] (box) -- node[above] {$\lambda_R$} (unit);
\draw[<-] (target) -- node[above] {target inclusion} (unit);
\draw[->] (target) -- node[right] {$C_3\times C_2$ sectorization} (def);
\draw[->] (def) -- node[right] {weighted support pairing} (split);
\draw[->] (split) -- node[right] {positive part} (pos);
\end{tikzpicture}
\end{center}

The arithmetic assertion is the existence of a shell for which the last object is positive.  All preceding layers constrain how positivity may appear; none of them is a substitute for positivity.

\begin{theorem}[Total interacting obstruction theorem]\label{thm:master}
Let
\[
        g:=289,
        \qquad
        S_{840}=\{1,121,169,289,361,529\}=\langle g\rangle\subset (\Z/840\Z)^\times.
\]
Let $p\equiv g^k\pmod {840}$ be in the hard six-square sector.  For every shell $R\equiv3\pmod4$, set
\[
        a=a_R(p)=\frac{p+R}{4},\qquad N=pa.
\]
Assume first that $(N,R)=1$.  Define the signed divisor box
\[
        \mathcal B_R(a):=
        \left\{
        \prod_{q^{e_q}\Vert a}q^{\beta_q}\bmod R:
        -e_q\le \beta_q\le e_q
        \right\}
        \subset (\Z/R\Z)^\times.
\]
Let
\[
        \widetilde{\mathfrak c}_R(a;t)\in G(\Z_{\ge0})=\Z_{\ge0}\sqcup\{\tauzero\}
\]
be the split-zero count of exponent vectors in $\mathcal B_R(a)$ hitting $t$, with value $\tauzero$ when $t$ is not even in the subgroup generated by the prime divisors of $a$, value $0_\Z$ when it is in that subgroup but not in the bounded box, and the ordinary positive count otherwise.  Then the shell $R$ gives an Erd\H{o}s--Straus certificate with first denominator $a$ if and only if
\[
        \Omega_R(p):=
        \widetilde{\mathfrak c}_R(a;-1)+2\widetilde{\mathfrak c}_R(a;-p^{-1})
        \in \Z_{>0}.
\]
Moreover, under $C_6\cong C_3\times C_2$, the target object is the weighted defect
\[
        \mathfrak D_k=E_{0,-}+2E_{-k,-(-1)^k},
\]
and every Fourier coefficient of $\mathfrak D_k$ is nonzero.  Hence the residual obstruction is exactly bounded positive-mass failure of $\mathcal B_R(a)$ against a full-support target defect.
\end{theorem}

The rest of the note proves this theorem, computes the finite defect spectrum, and shows how the modular $744\leftrightarrow747\leftrightarrow750$ arm resolves the same zero/support issue.

\section{Fixed shells and projective normalization}

\subsection{The divisor identity}

Let $p$ be prime and choose a shell
\[
        a=\frac{p+R}{4},\qquad R=4a-p,
        \qquad N=pa.
\]
Then
\[
        \frac4p-\frac1a=\frac{R}{N}.
\]
Thus the fixed-shell completion problem is
\[
        \frac{R}{N}=\frac1b+\frac1c.
\]

\begin{lemma}[Divisor certificate]\label{lem:divisor}
Assume $(R,N)=1$.  The equation
\[
        \frac{R}{N}=\frac1b+\frac1c
\]
has a solution in positive integers $b,c$ if and only if there exists a divisor $d\mid N^2$ such that
\[
        d\equiv -N\pmod R.
\]
Given such $d$, the denominators are
\[
        b=\frac{d+N}{R},
        \qquad
        c=\frac{N^2/d+N}{R}.
\]
\end{lemma}

\begin{proof}
Multiplying $R/N=1/b+1/c$ by $Nbc$ gives
\[
        Rbc=N(b+c).
\]
Equivalently,
\[
        R^2bc-RNb-RNc+N^2=N^2,
\]
so
\[
        (Rb-N)(Rc-N)=N^2.
\]
Set $d=Rb-N$.  Then $d\mid N^2$ and $d\equiv -N\pmod R$.  Conversely, if $d\mid N^2$ and $d\equiv -N\pmod R$, then $b=(d+N)/R$ is integral.  Since $(R,N)=1$, $d\equiv -N$ is invertible modulo $R$, and
\[
        \frac{N^2}{d}\equiv N^2(-N)^{-1}\equiv -N\pmod R,
\]
so $c=(N^2/d+N)/R$ is integral.  Substitution into $(Rb-N)(Rc-N)=N^2$ gives the equation.  Positivity is immediate.
\end{proof}

For non-coprime shells one replaces $(R,N)$ by the reduced pair $(R_0,N_0)=(R/g,N/g)$ with $g=(R,N)$ and applies the same lemma to $R_0/N_0$.  The champion shells used below are coprime, so no reduction is hidden in the numerical tables.

\subsection{Why the invariant target is $-1$}

The congruence in \Cref{lem:divisor} is scale-dependent:
\[
        d\equiv -N\pmod R.
\]
Since $N$ is invertible modulo $R$, multiplying by $N^{-1}$ gives
\[
        dN^{-1}\equiv -1\pmod R.
\]
Thus the projective target is $-1$.  This is the same algebraic pattern as the projective Springborn reduction: a raw scale-dependent obstruction is divided by a common projective scale, leaving a smaller invariant obstruction.  In the present problem, however, this removes only the sign coordinate.  It does not remember the three-arm coordinate of the residual six-sector.

\subsection{The three $p$-origins and the forced coefficient $2$}

Let
\[
        a=\prod_q q^{e_q}.
\]
Every divisor $d\mid N^2=p^2a^2$ has the form
\[
        d=p^iu,
        \qquad i\in\{0,1,2\},
        \qquad u\mid a^2.
\]
Writing
\[
        u a^{-1}=\prod_{q^{e_q}\Vert a}q^{\beta_q},
        \qquad -e_q\le \beta_q\le e_q,
\]
the normalized condition $dN^{-1}\equiv -1$ becomes
\[
        p^{i-1}u a^{-1}\equiv -1.
\]
Therefore the signed $a$-ratio must hit
\[
        \begin{array}{c|c}
        i&u a^{-1}\text{ target}\\
        \hline
        0&-p\\
        1&-1\\
        2&-p^{-1}.
        \end{array}
\]
The signed box is inversion-invariant: $x\in\mathcal B_R(a)$ if and only if $x^{-1}\in\mathcal B_R(a)$, with the same number of exponent vectors.  Since $(-p)^{-1}=-p^{-1}$, the $i=0$ and $i=2$ channels combine and force multiplicity $2$.

\begin{proposition}[Two-channel shell coefficient]\label{prop:two-channel}
For a coprime shell,
\[
        R\text{ succeeds}
        \quad\Longleftrightarrow\quad
        \widetilde{\mathfrak c}_R(a;-1)+2\widetilde{\mathfrak c}_R(a;-p^{-1})\in\Z_{>0}.
\]
\end{proposition}

\begin{proof}
The preceding table gives the three target channels.  The two channels $-p$ and $-p^{-1}$ have equal bounded counts by inversion of exponent vectors.  Therefore the ordinary certificate count equals
\[
        \mathfrak c_R(a;-1)+2\mathfrak c_R(a;-p^{-1}).
\]
Replacing ordinary counts by split-zero counts preserves positivity and separates unsupported absence from supported zero.  Hence positivity of the displayed split-zero expression is equivalent to existence of a divisor certificate.
\end{proof}

\section{Split-zero support values}

The coefficient semiring is
\[
        G(\Z_{\ge0})=\Z_{\ge0}\sqcup\{\tauzero\}.
\]
Here $\tauzero$ is the unsupported external zero, while $0_\Z$ is the supported arithmetic zero.  Addition uses $\tauzero$ as the additive identity:
\[
        \tauzero+n=n,
        \qquad
        n,m\in\Z_{\ge0},
\]
and multiplication uses $\tauzero$ as an absorber.  The three states used in the obstruction are
\[
        \tauzero
        \quad<\quad
        0_\Z
        \quad<\quad
        \Z_{>0}.
\]
For a target $t$ and shell $(R,a)$, let
\[
        \Gamma_R(a):=\langle q\bmod R:q\mid a\rangle\le (\Z/R\Z)^\times.
\]
Define
\[
\widetilde{\mathfrak c}_R(a;t)=
\begin{cases}
\tauzero,&t\notin\Gamma_R(a),\\[1mm]
0_\Z,&t\in\Gamma_R(a)\text{ but }t\notin\mathcal B_R(a),\\[1mm]
\#\left\{(\beta_q):\prod q^{\beta_q}\equiv t\pmod R\right\},&t\in\mathcal B_R(a).
\end{cases}
\]
This is the precise location of positivity: not at the projective target, not at the snowflake, not at the eta sheet, but at the split-zero evaluation of the arithmetic box against the weighted target.

\section{The modulo-$840$ support fibration and the glue datum}

The modulo-$840$ support set is
\[
        V_{840}:=\{r\in(\Z/840\Z)^\times:r\equiv1\pmod {24}\}.
\]
By the Chinese remainder theorem,
\[
        V_{840}\cong (\Z/5\Z)^\times\times(\Z/7\Z)^\times,
\]
so $|V_{840}|=4\cdot6=24$.  The two active quadratic characters are
\[
        \chi_5(r)=\left(\frac r5\right),
        \qquad
        \chi_7(r)=\left(\frac r7\right).
\]
The fibration
\[
        V_{840}\twoheadrightarrow C_2^2,
        \qquad
        r\mapsto(\chi_5(r),\chi_7(r))
\]
has four fibers of size six.  The principal square fiber is
\[
        S_{840}=\{1,121,169,289,361,529\}=\langle289\rangle\cong C_6.
\]
The cyclic order is
\[
        1,\\ 289,\\ 361,\\ 169,\\ 121,\\ 529.
\]

\begin{center}
\begin{tikzpicture}[>=Latex, every node/.style={font=\small}]
\node[draw,rounded corners,align=center,minimum width=80mm,minimum height=10mm] (v) {$V_{840}=\{r\in(\Z/840\Z)^\times:r\equiv1\pmod{24}\}$\\$24$ residual classes};
\node[draw,rounded corners,below=13mm of v,minimum width=82mm,minimum height=12mm,align=center] (q) {$C_2^2$ character base\\$(\chi_5,\chi_7)=(+,+),(+,-),(-,+),(-,-)$};
\node[draw,rounded corners,below=13mm of q,minimum width=78mm,minimum height=10mm,align=center] (s) {$S_{840}=\langle289\rangle\cong C_6$\\principal six-class fiber};
\draw[->] (v) -- node[right] {$r\mapsto((r/5),(r/7))$} (q);
\draw[->] (q) -- node[right] {select $(+,+)$} (s);
\end{tikzpicture}
\end{center}

Each six-point fiber has augmentation lattice $A_5$.  The four fibers give $A_5^4$.  The $C_2^2$ base contributes the marked $D_4$ discriminant component.  The resulting root frame is
\[
        R_{\ES}=A_5^4D_4.
\]
The glue calculation is included because it is the exact reason the residual object sits in the lambency-six umbral row rather than in a decorative ADE analogy.

\subsection{The explicit isotropic subgroup}

For $A_5$,
\[
        D(A_5)\cong \Z/6\Z,
        \qquad
        q_{A_5}(k)=\frac{5k^2}{6}\pmod {2\Z}.
\]
For $D_4$,
\[
        D(D_4)\cong \F_2^2,
        \qquad q_{D_4}(0)=0,
        \qquad q_{D_4}(y)=1\quad(y\ne0).
\]
Thus
\[
        D(R_{\ES})\cong (\Z/6\Z)^4\oplus\F_2^2
\]
with
\[
        q(x_1,x_2,x_3,x_4;y)=\frac56\sum_{i=1}^4x_i^2+q_D(y)\pmod {2\Z}.
\]
Its order is
\[
        6^4\cdot4=5184=72^2.
\]
Let $Q=\F_2^2=\{0,a,b,a+b\}$.  Define $H_2\subset \F_2^4\oplus Q$ by
\[
\begin{array}{c|c}
 u&y\\
\hline
0000&0\\
1111&0\\
1100&a\\
0011&a\\
1010&b\\
0101&b\\
1001&a+b\\
0110&a+b.
\end{array}
\]
Let
\[
        H_3:=\Span_{\F_3}\{(1,1,1,0),(1,2,0,1)\}\subset \F_3^4.
\]
Then $|H_3|=9$, and every nonzero word of $H_3$ has Hamming weight $3$: indeed a word is
\[
        \alpha(1,1,1,0)+\beta(1,2,0,1)
        =(\alpha+\beta,\alpha+2\beta,\alpha,\beta),
\]
and if $(\alpha,\beta)\ne(0,0)$ exactly one coordinate is zero.

Embed $\F_2\oplus\F_3$ into $\Z/6\Z$ by CRT,
\[
        k_i=3u_i+2v_i\pmod 6,
\]
and define
\[
        H_{\ES}:=\{((3u_1+2v_1,\ldots,3u_4+2v_4),y):(u,y)\in H_2,\ v\in H_3\}.
\]
Equivalently, $H_{\ES}$ is generated by
\[
\begin{array}{rcl}
        g_1&=&(3,3,3,3;0,0),\\
        g_2&=&(3,3,0,0;1,0),\\
        g_3&=&(3,0,3,0;0,1),\\
        g_4&=&(2,2,2,0;0,0),\\
        g_5&=&(2,4,0,2;0,0).
\end{array}
\]

\begin{theorem}[Glue theorem]\label{thm:glue}
The subgroup $H_{\ES}$ is isotropic of order $72$.  The corresponding overlattice
\[
        N_{\ES}:=\{x\in R_{\ES}^{\vee}:x\bmod R_{\ES}\in H_{\ES}\}
\]
is even unimodular of rank $24$ and has root system exactly $A_5^4D_4$.
\end{theorem}

\begin{proof}
The CRT map is a coordinatewise bijection $\F_2\oplus\F_3\to\Z/6\Z$, so
\[
        |H_{\ES}|=|H_2||H_3|=8\cdot9=72.
\]
For $k=3u+2v$ one has in $\Q/2\Z$
\[
        \frac{5k^2}{6}=\frac56(3u+2v)^2
        \equiv \frac32u+\frac43v^2,
\]
since the cross-term is even.  Therefore
\[
        q((3u+2v),y)=\frac32\wt(u)+\frac43\wt(v)+q_D(y).
\]
For $(u,y)\in H_2$, the possibilities are $(0,0)$, $u=1111,y=0$, or $\wt(u)=2,y\ne0$.  The binary-plus-$D_4$ contribution is respectively
\[
        0,
        \qquad
        \frac32\cdot4=6,
        \qquad
        3+1=4,
\]
all even.  For $v\in H_3$, either $v=0$ or $\wt(v)=3$, so the ternary contribution is either $0$ or $\frac43\cdot3=4$.  Hence $q$ vanishes on $H_{\ES}$.

Even overlattices of an even lattice correspond to isotropic subgroups of its discriminant form.  Since $H_{\ES}$ is isotropic, $N_{\ES}$ is even.  Its discriminant group has order
\[
        |D(N_{\ES})|=\frac{|D(R_{\ES})|}{|H_{\ES}|^2}=\frac{72^2}{72^2}=1,
\]
so $N_{\ES}$ is unimodular.

It remains to check that no roots are introduced by nonzero glue cosets.  In the discriminant coset $k\in\Z/6\Z$ of $A_5$, the minimum norm is
\[
        \mu_A(k)=\frac{k(6-k)}6.
\]
Thus the coordinate possibilities are
\[
\begin{array}{c|c|c|c}
 u_i&v_i&k_i&\mu_A(k_i)\\
\hline
0&0&0&0\\
1&0&3&3/2\\
0&\ne0&2\text{ or }4&4/3\\
1&\ne0&1\text{ or }5&5/6.
\end{array}
\]
The allowed cases from $H_2$ and $H_3$ give the lower bounds
\[
\begin{array}{c|c|c|c}
 \wt(u)&\wt(v)&\wt(\supp u\cap\supp v)&\text{minimum}\\
\hline
2&0&0&2(3/2)+1=4\\
4&0&0&4(3/2)=6\\
0&3&0&3(4/3)=4\\
2&3&1&(3/2)+2(4/3)+(5/6)+1=6\\
2&3&2&(4/3)+2(5/6)+1=4\\
4&3&3&(3/2)+3(5/6)=4.
\end{array}
\]
Every nonzero glue coset has minimum at least $4$, while roots have norm $2$.  Hence all norm-$2$ vectors lie in the zero glue coset, and the root system remains $A_5^4D_4$.
\end{proof}

The glue-preserving finite group is the non-split double cover
\[
        1\to\{\pm I\}\to GL_2(3)\to PGL_2(3)\cong S_4\to1.
\]
It projects to the permutation action on the four $A_5$ fibers and retains the central sign/glue datum invisible in the projective quotient.  This is the finite symmetry used by the weighted defect calculus below.

\section{The six-class snowflake and the weighted target defect}

The cyclic fiber admits the product decomposition
\[
        C_6\cong C_3\times C_2,
        \qquad
        g^k\longmapsto(k\bmod3,(-1)^k).
\]
The $A_8^3$ discriminant realization writes
\[
        v_k=(-1)^k3e_{k\bmod3}\in D(A_8^3),
\]
so
\[
        S_{840}\cong\{\pm3e_0,\pm3e_1,\pm3e_2\}.
\]
The projective target $-1$ is the distinguished hub
\[
        h=v_3=-3e_0.
\]
If $p\equiv g^k\pmod {840}$, then the second normalized target is
\[
        -p^{-1}\longleftrightarrow v_{3-k}.
\]
Thus the target edge is
\[
        E_k=\{v_3,v_{3-k}\}.
\]
But the actual coefficient is not the unweighted edge.  It is the weighted defect coming from \Cref{prop:two-channel}.

\begin{definition}[Weighted target defect]
Let $E_{r,\epsilon}$ denote the basis element of $\Z[C_3\times C_2]$ with arm $r\in\Z/3\Z$ and sign $\epsilon\in\{+,-\}$.  Define
\[
        \mathfrak D_k:=E_{0,-}+2E_{-k,-(-1)^k}.
\]
\end{definition}

The six explicit matrices, with rows $r=0,1,2$ and columns $+,-$, are
\[
\begin{array}{c|c}
 k&\mathfrak D_k\\
\hline
0&\begin{pmatrix}0&3\\0&0\\0&0\end{pmatrix}\\[5mm]
1&\begin{pmatrix}0&1\\0&0\\2&0\end{pmatrix}\\[5mm]
2&\begin{pmatrix}0&1\\0&2\\0&0\end{pmatrix}\\[5mm]
3&\begin{pmatrix}2&1\\0&0\\0&0\end{pmatrix}\\[5mm]
4&\begin{pmatrix}0&1\\0&0\\0&2\end{pmatrix}\\[5mm]
5&\begin{pmatrix}0&1\\2&0\\0&0\end{pmatrix}.
\end{array}
\]

The champion prime satisfies
\[
        p_*=8{,}803{,}369\equiv169\equiv g^3\pmod {840},
\]
so
\[
        \mathfrak D_3=2E_{0,+}+E_{0,-}.
\]
It lies on one arm and flips endpoint sign.

\subsection{Why neither projection is sufficient}

Define the arm marginal
\[
        A_k(r)=\sum_{\epsilon=\pm}\mathfrak D_k(r,\epsilon)
\]
and the sign marginal
\[
        B_k(\epsilon)=\sum_{r=0}^{2}\mathfrak D_k(r,\epsilon).
\]
Then
\[
\begin{array}{c|c|c}
 k&A_k&B_k\\
\hline
0&(3,0,0)&(0,3)\\
1&(1,0,2)&(2,1)\\
2&(1,2,0)&(0,3)\\
3&(3,0,0)&(2,1)\\
4&(1,0,2)&(0,3)\\
5&(1,2,0)&(2,1).
\end{array}
\]

\begin{theorem}[Both coordinates are necessary]\label{thm:projections}
The sign projection $B_k$ does not determine $\mathfrak D_k$.  The arm projection $A_k$ does not determine $\mathfrak D_k$.  The product grading $C_3\times C_2$ is the minimal grading that separates the six weighted defects.
\end{theorem}

\begin{proof}
The sign projection loses arm data because
\[
        B_0=B_2=B_4=(0,3),
\]
while
\[
        A_0=(3,0,0),\quad A_2=(1,2,0),\quad A_4=(1,0,2).
\]
The arm projection loses sign data because
\[
        A_0=A_3=(3,0,0),
\]
while
\[
        B_0=(0,3),\quad B_3=(2,1).
\]
Together the ordered pair $(A_k,B_k)$ separates the six displayed matrices.  Hence neither the projective $C_2$ sheet nor the gerbe $C_3$ arm coordinate is sufficient alone.
\end{proof}

This is the formal reason the Springborn/projective reduction and the gerbe sectorization must be composed rather than listed separately.

\subsection{Mixed arm-sign defect}

The independent tensor predicted by the two marginals is
\[
        \mathfrak D_k^{\mathrm{ind}}(r,\epsilon)=\frac{A_k(r)B_k(\epsilon)}{3}.
\]
Define the mixed defect
\[
        \mathfrak M_k:=\mathfrak D_k-\mathfrak D_k^{\mathrm{ind}}.
\]

\begin{theorem}[Mixed defect classification]\label{thm:mixed}
One has
\[
        \mathfrak M_k=0
        \quad\Longleftrightarrow\quad
        k\notin\{1,5\}.
\]
For the genuinely mixed sectors,
\[
\mathfrak M_1=\begin{pmatrix}
-2/3&2/3\\
0&0\\
2/3&-2/3
\end{pmatrix},
\qquad
\mathfrak M_5=\begin{pmatrix}
-2/3&2/3\\
2/3&-2/3\\
0&0
\end{pmatrix}.
\]
\end{theorem}

\begin{proof}
Substitute each of the six explicit matrices into
\[
        \mathfrak M_k=\mathfrak D_k-A_k\otimes B_k/3.
\]
For $k=0,2,3,4$, the matrix is supported in a single row or single column, and the tensor reconstruction is exact.  For $k=1,5$, direct subtraction gives the two displayed rank-one signed rectangles.
\end{proof}

Thus the six classes split into loop defects ($k=0$), cross-arm same-sign defects ($k=2,4$), same-arm opposite-sign defects ($k=3$), and genuinely mixed arm-sign defects ($k=1,5$).  The champion is the same-arm opposite-sign sector.

\section{Fourier nondegeneracy and all characteristic polynomials}

Let
\[
        \omega=e^{2\pi i/3}.
\]
Characters of $C_3\times C_2$ are indexed by
\[
        (u,v)\in\Z/3\Z\times\Z/2\Z
\]
and evaluate
\[
        (r,\epsilon)\mapsto \omega^{ur}\epsilon^v.
\]

\begin{theorem}[Full Fourier support]\label{thm:fourier}
For every $k$,
\[
        \widehat{\mathfrak D_k}(u,v)=(-1)^v\left(1+2(-1)^{kv}\omega^{-uk}\right).
\]
In particular,
\[
        \widehat{\mathfrak D_k}(u,v)\ne0
\]
for every character $(u,v)$.
\end{theorem}

\begin{proof}
The hub endpoint $E_{0,-}$ contributes $(-1)^v$.  The second endpoint $E_{-k,-(-1)^k}$ contributes
\[
        2\omega^{-uk}(-(-1)^k)^v
        =2\omega^{-uk}(-1)^v(-1)^{kv}.
\]
Adding gives the formula.  The parenthetical factor cannot vanish because its two summands have absolute values $1$ and $2$.
\end{proof}

\begin{corollary}[No target-mode obstruction]
The target defect never loses a Fourier mode.  Therefore a shell failure is not caused by a missing target character.  It is caused by the arithmetic box $\mathcal B_R(a)$ failing to deliver bounded positive mass to the full-support target defect.
\end{corollary}

For completeness, we also record the characteristic polynomial of the six translation/circulant defect operators.  Let $C(\mathfrak D_k)$ be multiplication by the group-ring element $e_3+2e_{3-k}$ on $\Z[C_6]$.

\begin{theorem}[Characteristic polynomials of all six defect operators]\label{thm:char-defects}
The determinants and characteristic polynomials over $\Z$ are
\[
\begin{array}{c|c|c}
 k&\det C(\mathfrak D_k)&\chi_k(X)\\
\hline
0&-729&(X-3)^3(X+3)^3\\
1&63&(X-3)(X-1)(X^2+3)(X^2+4X+7)\\
2&-81&(X-3)(X+3)(X^2+3)^2\\
3&27&(X-3)^3(X-1)^3\\
4&-81&(X-3)(X+3)(X^2+3)^2\\
5&63&(X-3)(X-1)(X^2+3)(X^2+4X+7).
\end{array}
\]
In particular every determinant is nonzero modulo $107$.
\end{theorem}

\begin{proof}
The eigenvalues of a circulant operator are the Fourier values
\[
        \zeta_6^{3m}+2\zeta_6^{(3-k)m},
        \qquad m=0,\ldots,5.
\]
Multiplying $(X-\lambda_m)$ over the six Fourier modes gives the displayed polynomials.  The determinant is the product of the same eigenvalues.  Each listed determinant is prime to $107$.
\end{proof}

\section{Champion positivity with all layers active}

Let
\[
        p_*=8{,}803{,}369.
\]
Then
\[
        p_*\equiv169\equiv g^3\pmod {840},
\]
so the target defect is
\[
        \mathfrak D_3=2E_{0,+}+E_{0,-}.
\]
For each shell below, the ordered support state is
\[
        \mathcal C_R(p_*)=
        \bigl(\widetilde{\mathfrak c}_R(a_R(p_*);-1),
        \widetilde{\mathfrak c}_R(a_R(p_*);-p_*^{-1})\bigr).
\]

\begin{center}
\begin{tabular}{c|c|c|c}
\toprule
$R$&$a_R(p_*)$&support state&meaning\\
\midrule
$27$&$7\cdot314407$&$(\tauzero,\tauzero)$&no subgroup descent\\
$43$&$379\cdot5807$&$(0_\Z,0_\Z)$&subgroup descent, bounded miss\\
$107$&$3^2\cdot11^2\cdot43\cdot47$&$(2,0_\Z)$&positive certificate\\
\bottomrule
\end{tabular}
\end{center}

\begin{theorem}[Champion support ladder]\label{thm:champion}
The shells $R=27,43,107$ for $p_*$ realize
\[
        (\tauzero,\tauzero)
        \longrightarrow
        (0_\Z,0_\Z)
        \longrightarrow
        (2,0_\Z).
\]
Hence positivity first appears at $R=107$, and it appears in the $-1$ channel.
\end{theorem}

\begin{proof}
For $R=27$,
\[
        a=2{,}200{,}849=7\cdot314407.
\]
The subgroup generated by the two primes modulo $27$ is
\[
        \{1,4,7,10,13,16,19,22,25\}.
\]
The target pair is
\[
        -1\equiv26,
        \qquad
        -p_*^{-1}\equiv17\pmod {27},
\]
and neither target lies in the subgroup.  Thus the state is $(\tauzero,\tauzero)$.

For $R=43$,
\[
        a=2{,}200{,}853=379\cdot5807.
\]
The target pair is
\[
        -1\equiv42,
        \qquad
        -p_*^{-1}\equiv41\pmod {43}.
\]
Both targets lie in the generated subgroup, but the bounded box is
\[
        \mathcal B_{43}(a)=\{1,2,8,16,22,27,32,35,39\},
\]
so neither target is hit.  Thus the state is $(0_\Z,0_\Z)$.

For $R=107$,
\[
        a=2{,}200{,}869=3^2\cdot11^2\cdot43\cdot47.
\]
The targets are
\[
        -1\equiv106,
        \qquad
        -p_*^{-1}\equiv86\pmod {107}.
\]
The signed exponent box has two exponent vectors hitting $106$,
\[
        (-2,0,-1,-1),
        \qquad
        (2,0,1,1),
\]
with coordinates ordered by $(3,11,43,47)$, and has no exponent vector hitting $86$.  Hence the state is $(2,0_\Z)$.
\end{proof}

The strictness of this ladder is the operational meaning of split zero: $R=27$ has no support descent; $R=43$ has support but no bounded mass; $R=107$ has bounded positive mass.

\section{The critical arm $744\leftrightarrow747\leftrightarrow750$}

The completed $A_8$ arm is
\[
        -3e_0,\\ 0_0,\\ +3e_0.
\]
It is represented by
\[
        744,
        \qquad
        747,
        \qquad
        750.
\]
Define
\[
        v(n)=\frac{n-747}{3}.
\]
Then
\[
        v(744)=-1,
        \qquad
        v(747)=0,
        \qquad
        v(750)=1.
\]
The involution
\[
        n\mapsto1494-n
\]
is exactly $v\mapsto-v$.

\begin{center}
\begin{tabular}{c|c|c|c}
\toprule
index&$744$&$747$&$750$\\
\midrule
arm state&$-3e_0$&$0_0$&$+3e_0$\\
$v$&$-1$&$0$&$1$\\
role&negative endpoint&supported-zero hub&positive endpoint\\
\bottomrule
\end{tabular}
\end{center}

The important point is that $747$ is not absence.  It is the supported-zero hub of the arm.

\subsection{The $A_8^3$ raw-zero sheet}

Set
\[
        T_9(\tau)=\frac{\eta(\tau)^3}{\eta(9\tau)^3}=q^{-1}P_9(q).
\]
Jacobi's eta-cube dissection gives
\[
        P_9(q)=H(q^3)-3q
\]
for some $H(q^3)\in\Z[[q^3]]$.  Therefore
\[
        T_9=q^{-1}H(q^3)-3.
\]
All positive nonconstant exponents of $T_9$ are of the form $3m-1$, i.e. congruent to $2\pmod3$.  Since $744,747,750$ are divisible by $3$,
\[
        [q^{744}]T_9=[q^{747}]T_9=[q^{750}]T_9=0.
\]
The raw $A_8^3$ eta sheet sees the whole completed arm as zero.  The support states are recovered only after Fricke and Eisenstein completion.

Define
\[
        I_9:=\frac{27}{T_9},
        \qquad
        J_9:=T_9+3+\frac{27}{T_9},
        \qquad
        Y_9:=T_9-\frac{27}{T_9},
\]
and
\[
        f_9:=-q\frac{d}{dq}\log T_9.
\]
Then
\[
        f_9=\frac{-3E_2(\tau)+27E_2(9\tau)}{24},
\]
so
\[
        [q^n]f_9=3\sigma_1(n)-27\sigma_1(n/9),
\]
with the second term omitted if $9\nmid n$.

Modulo $107$, the critical-arm values are
\[
\begin{array}{c|ccccc}
 n&T_9&I_9&J_9&Y_9&f_9\\
\hline
744&0&35&35&72&89\\
747&0&31&31&76&45\\
750&0&42&42&65&52.
\end{array}
\]
Thus
\[
        J_9(744)=35=N_*,
        \qquad
        Y_9(744)=-35=-N_*,
\]
\[
        f_9(744)=89=93^2=a_*^2,
\]
\[
        J_9(747)^2=31^2\equiv105\equiv -2,
\]
and
\[
        J_9(750)^2=42^2\equiv52=f_9(750),
        \qquad
        f_9(750)^{-1}=35=N_*.
\]
The hub identity $J_9(747)^2=-2$ records the two-edge support state at $747\equiv -2\pmod {107}$.

\subsection{The nonlinear $A_8^3$ support resolvent}

Since $Y_9=-J_9$ on this arm, the pair $(J_9,Y_9)$ is rank one.  The nonlinear coordinate $J_9^2$ is required to separate the hub.  Define
\[
        \mathcal R_9(n)=(J_9(n),J_9(n)^2,f_9(n)).
\]
For $n=744,747,750$,
\[
        M_9=
        \begin{pmatrix}
        35&48&89\\
        31&105&45\\
        42&52&52
        \end{pmatrix}.
\]

\begin{theorem}[$A_8^3$ support determinant]\label{thm:M9}
\[
        \det M_9\equiv103\equiv -4\pmod {107}.
\]
\end{theorem}

\begin{proof}
Expanding along the first row,
\[
\det M_9=35(105\cdot52-45\cdot52)-48(31\cdot52-45\cdot42)+89(31\cdot52-105\cdot42).
\]
Reduction modulo $107$ gives $103\equiv -4$.
\end{proof}

\section{Lambency-six moonshine support resolution}

The residual lambency-six eta quotient is
\[
        T_6(\tau)=\frac{\eta(\tau)^5\eta(3\tau)}{\eta(2\tau)\eta(6\tau)^5}.
\]
It has expansion
\[
        T_6=q^{-1}-5+6q+4q^2-3q^3-12q^4-8q^5+\cdots.
\]
Define
\[
        J_6:=T_6+5+\frac{72}{T_6},
        \qquad
        Y_6:=T_6-\frac{72}{T_6},
\]
\[
        F_6:=-2\eta(6\tau)^4T_6,
        \qquad
        f_6:=-q\frac{d}{dq}\log T_6.
\]
The logarithmic derivative is
\[
        f_6=\frac{-5E_2(\tau)+2E_2(2\tau)-3E_2(3\tau)+30E_2(6\tau)}{24},
\]
and the coefficient formula is
\[
        [q^n]f_6=
        5\sigma_1(n)-2\sigma_1(n/2)+3\sigma_1(n/3)-30\sigma_1(n/6),
\]
with terms omitted when the denominator does not divide $n$.

On the same critical arm, modulo $107$,
\[
\begin{array}{c|cccccc}
 n&T_6&F_6&J_6&Y_6&f_6\\
\hline
744&34&19&16&52&67\\
747&83&77&1&58&48\\
750&101&106&2&93&21.
\end{array}
\]
Consequently
\[
        J_6(747)=1,
        \qquad
        Y_6(750)=93=a_*,
\]
\[
        F_6(750)=106=-1,
        \qquad
        f_6(750)=21=S(3).
\]
The lambency-six sheet separates the hub and endpoint linearly.

Define
\[
        M_{6,1}:=
        \begin{pmatrix}
        J_6(744)&Y_6(744)&F_6(744)\\
        J_6(747)&Y_6(747)&F_6(747)\\
        J_6(750)&Y_6(750)&F_6(750)
        \end{pmatrix}
        =
        \begin{pmatrix}
        16&52&19\\
        1&58&77\\
        2&93&106
        \end{pmatrix}
\]
and
\[
        M_{6,2}:=
        \begin{pmatrix}
        J_6(744)&F_6(744)&f_6(744)\\
        J_6(747)&F_6(747)&f_6(747)\\
        J_6(750)&F_6(750)&f_6(750)
        \end{pmatrix}
        =
        \begin{pmatrix}
        16&19&67\\
        1&77&48\\
        2&106&21
        \end{pmatrix}.
\]

\begin{theorem}[Lambency-six support determinants]\label{thm:M6}
\[
        \det M_{6,1}\equiv82\equiv -25\pmod {107},
        \qquad
        \det M_{6,2}\equiv25\pmod {107}.
\]
\end{theorem}

\begin{proof}
Direct determinant expansion of the two displayed matrices gives $82$ and $25$ modulo $107$ respectively.
\end{proof}

\section{Finite transport from $A_8^3$ to $A_5^4D_4$}

The $A_8^3$ sheet is raw-zero and requires the nonlinear support coordinate $J_9^2$.  The $A_5^4D_4$ lambency-six sheet separates linearly.  Their finite relation on the critical arm is encoded by transport matrices
\[
        P_{9\to6}^{(1)}:=M_{6,1}M_9^{-1},
        \qquad
        P_{9\to6}^{(2)}:=M_{6,2}M_9^{-1}
\]
over $\F_{107}$.

\begin{theorem}[Transport matrices and characteristic polynomials]\label{thm:transport}
Modulo $107$,
\[
        P_{9\to6}^{(1)}=
        \begin{pmatrix}
        32&67&44\\
        98&72&41\\
        51&81&71
        \end{pmatrix},
\]
with
\[
        \det P_{9\to6}^{(1)}\equiv33,
\]
and
\[
        \chi_{P_{9\to6}^{(1)}}(X)=X^3+39X^2+18X-33.
\]
This polynomial is irreducible over $\F_{107}$.  Also
\[
        P_{9\to6}^{(2)}=
        \begin{pmatrix}
        3&28&46\\
        106&64&81\\
        38&47&84
        \end{pmatrix},
\]
with
\[
        \det P_{9\to6}^{(2)}\equiv74,
\]
and
\[
        \chi_{P_{9\to6}^{(2)}}(X)=X^3-44X^2-28X+33
        =(X+17)(X^2+46X+46).
\]
\end{theorem}

\begin{proof}
Since $\det M_9=-4$ modulo $107$, the inverse $M_9^{-1}$ exists.  Multiplying the displayed matrices gives the two stated transport matrices.  The determinants follow from
\[
        \det P_{9\to6}^{(1)}=\frac{-25}{-4}=25\cdot4^{-1}\equiv25\cdot27\equiv33
\]
and
\[
        \det P_{9\to6}^{(2)}=\frac{25}{-4}=-25\cdot4^{-1}\equiv74.
\]
The characteristic polynomials are obtained by evaluating $\det(XI-P)$ for the two matrices.  The factorization of the second polynomial is direct.  For the first, substituting all $107$ field elements gives no root, and a cubic over a field is reducible if and only if it has a root; hence it is irreducible.
\end{proof}

This is the finite computation linking the $A_8^3$ raw-zero support completion to the $A_5^4D_4$ lambency-six moonshine completion.  It is not a symbolic arrow; it is an invertible $\F_{107}$-linear transport between the three resolved support states.

\section{Gerbe and paramodular envelope as functorial lifts of the same defect}

The gerbe construction is used here only for one purpose: it supplies a natural six-sector module basis indexed by
\[
        \Z_3\times\Z_2.
\]
Under
\[
        C_6\cong C_3\times C_2,
\]
the snowflake sector $E_{r,\epsilon}$ is the gerbe sector $\Theta_{r,\epsilon}$.  The actual object is therefore
\[
        \mathfrak D_k=\Theta_{0,-}\oplus2\Theta_{-k,-(-1)^k}.
\]
The full Fourier support theorem shows that this defect is representation-complete on the six-sector slice.

The genus-two theta/paramodular envelope lifts the same weighted defect from sectors to binary quadratic coefficient indices.  If $L$ is an even unimodular rank-$24$ lattice and $T$ is a positive semidefinite half-integral binary quadratic form, write $r_L(E_{r,\epsilon};T)$ for the number of lattice pairs in sector $E_{r,\epsilon}$ with Gram index $T$.  Define
\[
        r_L(\mathfrak D_k;T):=
        r_L(E_{0,-};T)+2r_L(E_{-k,-(-1)^k};T).
\]

\begin{proposition}[No mode loss under the binary-quadratic envelope]
For each fixed $T$, the map
\[
        \mathfrak D\mapsto r_L(\mathfrak D;T)
\]
is a linear functional on $\Z[C_3\times C_2]$.  Since $\mathfrak D_k$ has nonzero Fourier coefficient in every character, the envelope does not kill any target mode on the defect side.  Any vanishing in $r_L(\mathfrak D_k;T)$ is therefore a property of the lattice-pair coefficient at $T$, not a degeneracy of the target defect.
\end{proposition}

\begin{proof}
Linearity is immediate from the definition.  Fourier inversion on $C_3\times C_2$ decomposes the pairing into modewise products of coefficient modes and target modes.  By \Cref{thm:fourier}, all target modes are nonzero.
\end{proof}

This is the exact sense in which the gerbe and paramodular objects interact with the arithmetic obstruction: both are lifts of the same weighted target defect.  They are not independent analogies.

\section{The remaining arithmetic theorem}

The preceding results reduce the deep-tail problem to one positivity statement.

\begin{theorem}[Equivalent positivity formulation]\label{thm:remaining}
For hard primes in the six-square sector, the Erd\H{o}s--Straus assertion is equivalent to the following reachability statement:

For every prime $p\equiv g^k\pmod {840}$ in $S_{840}$, there exists a shell $R\equiv3\pmod4$ such that
\[
        \Omega_R(p)=
        \widetilde{\mathfrak c}_R(a_R(p);-1)+2\widetilde{\mathfrak c}_R(a_R(p);-p^{-1})
        \in\Z_{>0}.
\]
Equivalently, the signed divisor box $\mathcal B_R(a_R(p))$ intersects the weighted target defect
\[
        \mathfrak D_k=E_{0,-}+2E_{-k,-(-1)^k}
\]
with positive split-zero mass.
\end{theorem}

\begin{proof}
The standard congruence identities and the $R=3,7$ sieve reduce the remaining primes to the six-square sector.  For a fixed shell, \Cref{prop:two-channel} proves equivalence of shell success and positivity of $\Omega_R(p)$.  Existence of an Erd\H{o}s--Straus decomposition is equivalent to success of at least one shell.  The target-edge statement is the $C_3\times C_2$ sectorization of the two normalized targets.
\end{proof}

\section{Verification recurrences}

All coefficient calculations in this note are finite.  The verification script accompanying the note implements the following recurrences.

Let
\[
        T_h=q^{-1}P_h(q),
        \qquad
        P_h(q)=\prod_{n\ge1}(1-q^n)^{E_h(n)}.
\]
For $T_6$,
\[
        E_6(n)=5-\mathbf 1_{2\mid n}+\mathbf 1_{3\mid n}-5\mathbf 1_{6\mid n}.
\]
For $T_9$,
\[
        E_9(n)=3-3\mathbf 1_{9\mid n}.
\]
If
\[
        P_h(q)=\sum_{n\ge0}a_nq^n,
        \qquad a_0=1,
\]
and
\[
        C_h(m)=\sum_{d\mid m}dE_h(d),
\]
then
\[
        na_n=-\sum_{m=1}^{n}C_h(m)a_{n-m}.
\]
The inverse series
\[
        P_h(q)^{-1}=\sum_{n\ge0}b_nq^n,
        \qquad b_0=1,
\]
is computed by
\[
        b_n=-\sum_{j=1}^{n}a_jb_{n-j}.
\]
Then
\[
        \frac{72}{T_6}=72qP_6(q)^{-1},
        \qquad
        \frac{27}{T_9}=27qP_9(q)^{-1}.
\]
All congruences in the tables are taken in $\F_{107}$.

\section{Conclusion}

The corrected object is not a list of structures.  It is the positivity evaluation
\[
        \mathcal B_R(a)
        \longrightarrow
        \{-1,-p^{-1}\}
        \longrightarrow
        \mathfrak D_k
        \longrightarrow
        G(\Z_{\ge0})
        \longrightarrow
        \Z_{>0}.
\]
Projective normalization creates the invariant sign target $-1$.  The $C_3\times C_2$ snowflake restores the arm/sign coordinates.  The coefficient $2$ in $\mathfrak D_k$ is forced by divisor origins.  Split-zero support distinguishes no subgroup descent, supported bounded miss, and genuine positive mass.  The target defect has full Fourier support, so failure cannot be blamed on missing target modes.  The champion prime demonstrates the entire ladder explicitly.  The $744,747,750$ modular arm shows the same phenomenon in the eta sheets: raw zero is resolved only after Fricke/Eisenstein completion, and the transport matrices compute the finite passage from the $A_8^3$ support resolution to the lambency-six $A_5^4D_4$ moonshine resolution.

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\end{document}
