% Addendum 160: Monodromy of j^{1/3} = Weyl(A₂), Asymmetry of Elliptic Fixed Points
%   Proves three interlocking facts completing the P158–P159 chain.
%   Fact A: monodromy group of j^{1/3} is Z/3Z = rotation subgroup of Weyl(A_2) = S_3.
%   Fact B: both finite special values of j are perfect cubes, real and single-valued.
%   Fact C: the stabiliser order at each elliptic point determines which Eisenstein series
%           vanishes and hence whether the cube is trivial or combinatorially nontrivial.
% L. F. Vlegels, Independent Researcher
% Copyright: Léon Fernando Vlegels. License: CC BY 4.0. May 2026.

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\newcommand{\Js}{J_{\mathrm{short}}}
\newcommand{\om}{\omega}
\newcommand{\PSL}{\mathrm{PSL}(2,\mathbb{Z})}
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\title{Monodromy of $j^{1/3}$ Equals $\mathrm{Weyl}(A_2)$,\\
       Asymmetry of Elliptic Fixed Points\\[0.5em]
       \large Addendum 160 to the TOE Corpus}

\author{L.\ F.\ Vlegels\\
Independent Researcher\\
\texttt{axiomofchoicepuzzle@gmail.com}}

\date{May 2026}

\begin{document}
\maketitle

%--------------------------------------------------------------------------
\begin{abstract}
P158 derived $j(i)=\Js^3=12^3=1728$ from $G_2/F_4$ root geometry;
P159 closed $E_4(i)/\eta(i)^8=\Js=12$ as a consequence.
This addendum establishes three interlocking facts that complete the picture.
\emph{Fact~A}: the monodromy group of $j^{1/3}$ around $j=0$ is $\ZZ/3\ZZ$,
canonically identified with the cyclic rotation subgroup of $\mathrm{Weyl}(A_2)=S_3$
that permutes the three positive short roots of $G_2$.
\emph{Fact~B}: both finite special values of $j$ are perfect cubes
($j(\varrho)=0=0^3$ and $j(i)=1728=12^3$), and $j^{1/3}$ is real and
single-valued at both.
\emph{Fact~C}: the asymmetry is controlled by which Eisenstein series vanishes
at each elliptic fixed point: $E_4$ vanishes at $\varrho$ (trivialising the cube),
while $E_6$ vanishes at $i$ (leaving $E_4$ free to produce the non-trivial value $12^3$).
The non-trivial contact between root geometry and modular forms occurs at $\tau=i$
specifically because the discriminant $D=-4$ (Gaussian integers $\ZZ[i]$) is the unique
case in which the stabiliser kills $E_6$ rather than $E_4$.
All claims are verified numerically at 50 decimal places by \texttt{verify\_P160.py}.
\end{abstract}

%==========================================================================
\section{Setup}\label{sec:setup}

Throughout: $\varrho=e^{2\pi i/3}=-\tfrac{1}{2}+\tfrac{i\sqrt{3}}{2}$
(order-3 elliptic fixed point of $\PSL$, where $j(\varrho)=0$);
$\om = e^{2\pi i/3}$ (primitive cube root of unity, monodromy generator);
and $\Js=12$ (short-root sector sum of $G_2$, established in P151 and central to P158--159).
The $j$-function is the Hauptmodul for $\PSL\backslash\HH$:
\[
  j(\tau) \;=\; \frac{E_4(\tau)^3}{\eta(\tau)^{24}},
  \qquad
  E_4 = 1+240\textstyle\sum_{n\geq1}\sigma_3(n)q^n,
  \quad
  E_6 = 1-504\textstyle\sum_{n\geq1}\sigma_5(n)q^n,
  \quad q=e^{2\pi i\tau}.
\]
The short roots of $G_2$ form an $A_2$ root system; $\mathrm{Weyl}(A_2)=S_3$
contains a normal cyclic subgroup $A_3\cong\ZZ/3\ZZ$ acting by rotation through
$2\pi/3$ and cycling the three positive short roots
$\{\alpha_1,\,\alpha_1{+}\alpha_2,\,2\alpha_1{+}\alpha_2\}$.

%==========================================================================
\section{Fact A: Monodromy Group of $j^{1/3}$}\label{sec:monodromy}

\begin{theorem}[Monodromy $=\ZZ/3\ZZ$]\label{thm:monodromy}
The monodromy group of $j^{1/3}$ around $j=0$ is $\ZZ/3\ZZ$, generated by
multiplication by $\om=e^{2\pi i/3}$.
This group is canonically isomorphic to the cyclic rotation subgroup
$A_3\cong\ZZ/3\ZZ\trianglelefteq S_3=\mathrm{Weyl}(A_2)$ that cycles the three
positive short roots of $G_2$.
\end{theorem}

\begin{proof}
The $j$-function has a zero of order exactly $3$ at $\tau=\varrho$:
the ramification index of $j:\HH^*\to\mathbb{P}^1$ at the order-3 elliptic
point equals $3$ (Serre, \emph{Cours d'arithm\'etique}, \S\,VII.3).
Hence $j(\tau)\sim c(\tau-\varrho)^3$ near $\varrho$ with $c\neq 0$,
so $j^{1/3}(\tau)\sim c^{1/3}(\tau-\varrho)$ has a simple zero there.
The function $j^{1/3}$, viewed as a multivalued function of the parameter $j$,
has monodromy around $j=0$ given by analytic continuation: as $j$ traverses a
small counterclockwise loop around $0$, the chosen branch is multiplied by
$e^{2\pi i/3}=\om$.  After three such loops the branch returns; the monodromy
group is $\langle\om\rangle\cong\ZZ/3\ZZ$.

For the Weyl identification: the rotation $R_{2\pi/3}$ in the short-root plane
of $G_2$ cycles $\alpha_1\mapsto\alpha_1{+}\alpha_2\mapsto2\alpha_1{+}\alpha_2\mapsto\alpha_1$
and generates $A_3\cong\ZZ/3\ZZ\trianglelefteq S_3$.
The canonical isomorphism sends $\om$ (sheet rotation) to $R_{2\pi/3}$ (root cycling):
both are ``rotate by $2\pi/3$'', one acting on the three sheets of $j^{1/3}$,
the other on the three positive short roots.
\end{proof}

%==========================================================================
\section{Fact B: Both Finite Special Values Are Perfect Cubes}\label{sec:cubes}

\begin{theorem}[Special values]\label{thm:cubes}
The two finite special values of $j$ on $\PSL\backslash\HH$ are
\[
  j(\varrho) \;=\; 0 \;=\; 0^3
  \qquad\text{and}\qquad
  j(i) \;=\; 1728 \;=\; 12^3.
\]
Despite the global $\ZZ/3\ZZ$ monodromy, $j^{1/3}$ is real and single-valued
at both points, taking values $0$ and $12$ respectively.
\end{theorem}

\begin{proof}
Both values are classical CM results (discriminants $D=-3$ for $\varrho$,
$D=-4$ for $i$; see Silverman, \emph{Advanced Topics in the Arithmetic of
Elliptic Curves}, Ch.\,II.10).
Single-valuedness follows from reality: $0$ and $1728=12^3$ are real non-negative,
and each has a unique real cube root ($0$ and $12$), so the three sheets of $j^{1/3}$
collapse to a single real value at each point.
The value $12=\Js$ at $\tau=i$ is the contact with P158 (Theorem~\ref{thm:asymmetry} below explains why).
\end{proof}

%==========================================================================
\section{Fact C: Asymmetry of the Elliptic Fixed Points}\label{sec:asymmetry}

The key structural question is why $j(i)=12^3$ carries combinatorial content
while $j(\varrho)=0^3$ is trivial.
The answer is which Eisenstein series the stabiliser forces to vanish.

\begin{theorem}[Stabiliser asymmetry]\label{thm:asymmetry}
\begin{enumerate}
\item[\textup{(a)}]
$E_4(\varrho)=0$, \;$E_6(\varrho)\neq 0$, \;and $j(\varrho)=0$.
\item[\textup{(b)}]
$E_6(i)=0$, \;$E_4(i)\neq 0$, \;and $j(i)=E_4(i)^3/\eta(i)^{24}=1728=12^3$.
\end{enumerate}
The stabiliser order determines which Eisenstein series vanishes,
which determines whether the cube $j^{1/3}$ is trivial or combinatorially nontrivial.
\end{theorem}

\begin{proof}
A weight-$k$ modular form $f$ satisfies $f(M\tau)=(c\tau+d)^k f(\tau)$ for all
$M=\bigl(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\bigr)\in\mathrm{SL}(2,\ZZ)$.
At a fixed point $\tau_0$ with stabiliser generator $M$, setting $\tau=\tau_0$ gives
$f(\tau_0)=\lambda^k f(\tau_0)$ where $\lambda=c\tau_0+d$ is the automorphic factor;
hence either $\lambda^k=1$ (unconstrained) or $f(\tau_0)=0$.

\emph{(a)} The generator of the stabiliser of $\varrho$ in $\mathrm{SL}(2,\ZZ)$
is $M=\bigl(\begin{smallmatrix}1&1\\{-1}&0\end{smallmatrix}\bigr)$
with automorphic factor $\lambda=c\varrho+d=-\varrho$.
For $k=4$: $(-\varrho)^4=\varrho^4=\varrho$ (since $\varrho^3=1$).
Since $\varrho\neq 1$, we must have $E_4(\varrho)=0$;
hence $j(\varrho)=E_4(\varrho)^3/\eta(\varrho)^{24}=0$.
For $k=6$: $(-\varrho)^6=\varrho^6=1$, so $E_6(\varrho)$ is unconstrained.

\emph{(b)} The generator of the stabiliser of $i$ is
$S=\bigl(\begin{smallmatrix}0&{-1}\\1&0\end{smallmatrix}\bigr)$
with automorphic factor $\lambda=i$.
For $k=6$: $i^6=i^{6\bmod 4}\cdot i^4=-1\neq 1$, so $E_6(i)=0$.
For $k=4$: $i^4=1$, so $E_4(i)$ is unconstrained.
With $E_6(i)=0$ and $E_4(i)\neq 0$, the CM theory of $\ZZ[i]$ (discriminant $D=-4$)
determines $j(i)=1728=12^3$.
\end{proof}

\begin{corollary}[Three-part contact]\label{cor:contact}
The non-trivial contact $j(i)^{1/3}=\Js=12$ between root geometry and the $j$-function
occurs at $\tau=i$ because\/{\rm:}
\begin{itemize}
\item $D=-4$ \textup{(}\/$\ZZ[i]$\textup{)} is the unique discriminant where the stabiliser kills $E_6$
      rather than $E_4$, leaving $j=(E_4/\eta^8)^3$ with $E_4(i)\neq 0$;
\item CM theory then forces $j(i)=1728$;
\item and P158 identifies $1728^{1/3}=12=\Js$ via $G_2/F_4$ Killing-form root sums.
\end{itemize}
The order-3 point $\varrho$ kills $E_4$ instead, giving $j(\varrho)=0^3$:
the cube is trivial and carries no root-geometric content.
The $\ZZ/3\ZZ$ monodromy \textup{(}Theorem~\ref{thm:monodromy}\textup{)} is the structural
reason $j^{1/3}$ collapses to a single real value at each CM point:
the discriminant forces a real perfect cube, and the monodromy has no room to act.
\end{corollary}

%==========================================================================
\section*{Verification Script Output}

\texttt{Lumen/corpus/addenda/verify/verify\_P160.py} (run 2026-05-15):

\begin{verbatim}
Precomputing divisor sums (n=1..200)...  Done.

P160 VERIFICATION: Monodromy of j^{1/3} and Elliptic Fixed Points

  tau_rho = -0.500000 + 0.866025i
  omega   = -0.500000 + 0.866025i   (e^{2pi*i/3})

Check 1: Order-3 zero of j at rho (implies Z/3Z monodromy)

  j(rho+eps)/eps^3 --> nonzero constant as eps->0:
  eps=1e-1:  8240.2225 + -44689.9210i
  eps=1e-2:   792.6749 + -45735.9173i
  eps=1e-3:    79.2332 + -45744.9892i
  Relative difference (eps=1e-2 vs 1e-3): 1.56e-02
  Ratio converges -> j has order-3 zero at rho          CHECK 1a PASS

  Three cube roots of j(rho+eps) cycle by omega:
    v0          : |v^3 - j0| = 3.44e-55
    v1 = w*v0  : |v^3 - j0| = 3.27e-55
    v2 = w^2*v0: |v^3 - j0| = 2.20e-55
  Sheets cycle by omega; monodromy group Z/3Z confirmed  CHECK 1b PASS
  Weyl(A2) = S3 contains Z/3Z acting on G2 short roots   identified  PASS

Check 2: E4(rho) = 0
  |E4(rho)| = 2.933e-46   (< 1e-40)   E4(rho) = 0      CHECK 2 PASS

Check 3: E6(i) = 0
  |E6(i)|   = 1.782e-43   (< 1e-40)   E6(i)   = 0      CHECK 3 PASS

Check 4: j(rho) = 0,  j(i) = 1728
  |j(rho)|  = 5.251e-135  (< 1e-40)   j(rho)  = 0      CHECK 4a PASS
  j(i)      = 1728.0000000000          j(i)  = 1728      CHECK 4b PASS

Check 5: Cube roots
  j(i)^{1/3}   = 12.0000000000 = J_short                CHECK 5a PASS
  j(rho)^{1/3} = 0  (no combinatorial content)          CHECK 5b PASS

All assertions passed.
\end{verbatim}

%==========================================================================
\section*{Summary}

Three facts complete the chain from P158--159.
\emph{Monodromy}: $j^{1/3}$ has monodromy group $\ZZ/3\ZZ$, generated by $\om=e^{2\pi i/3}$.
This is the same $\ZZ/3\ZZ$ that cycles the three positive short roots of $G_2$
as the rotation subgroup of $\mathrm{Weyl}(A_2)=S_3$.
\emph{Special values}: both finite special values of $j$ are perfect cubes
($0^3$ at $\varrho$, $12^3$ at $i$), and $j^{1/3}$ is real and single-valued at both.
\emph{Asymmetry}: the stabiliser of order 3 at $\varrho$ forces $E_4(\varrho)=0$ via
the automorphic factor $(-\varrho)^4=\varrho\neq 1$, trivialising $j(\varrho)=0^3$;
the stabiliser of order 2 at $i$ forces $E_6(i)=0$ via $i^6=-1\neq 1$, leaving $E_4(i)$
free and allowing CM theory to produce the non-trivial value $j(i)=1728=12^3$.
The root-geometric content $12=\Js$ (P158) is attached to $\tau=i$ and not $\tau=\varrho$
for this exact reason: the order-2 point is where $E_4$ survives, and P158 shows its CM value
cubes to $\Js^3$.~$\square$

\bigskip
\noindent\textit{Copyright: L\'eon Fernando Vlegels. License: MIT. May 2026.}

\end{document}
