% Addendum 172 — The CM Uniqueness Theorem
%   Among all imaginary quadratic CM points tau_D, the unique point where
%   j(tau_D) is a power of J_short = 12 is D = -4, i.e. tau = i, giving
%   j(i) = 1728 = 12^3 = J_short^3.  The proof has two steps: (i) a lemma
%   disposing of all discriminants with class number h(D) > 1 by an algebraic
%   degree argument; (ii) an explicit verification of the nine Stark-Heegner
%   discriminants (h(D) = 1) showing that only D = -4 yields a power of 12.
%   This closes the global uniqueness question flagged in P170.
% ERRATUM (A343, 2026-06-15): the {12^k} CM-uniqueness theorem STANDS (it is a
%   statement about the j-function, independent of moonshine). Only the provenance
%   "recovers Js via the 3B McKay-Thompson series" is corrected: per A337,
%   j^{1/3}=E4/eta^8 is the E8/G2 cube-root function, not the Monster 3B series.
%   See A337/A343.
% 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{\ji}{j(i)}
\newcommand{\etai}{\eta(i)}
\newcommand{\ZZ}{\mathbb{Z}}
\newcommand{\RR}{\mathbb{R}}
\newcommand{\CC}{\mathbb{C}}
\newcommand{\QQ}{\mathbb{Q}}
\newcommand{\HH}{\mathfrak{H}}
\newcommand{\OO}{\mathcal{O}}
\newcommand{\BG}{B_{G_2}}
\newcommand{\BF}{B_{F_4}}
\newcommand{\Mon}{\mathbb{M}}
\newcommand{\Leech}{\Lambda_{24}}
\newcommand{\tauD}{\tau_D}
\newcommand{\Vnat}{V^{\natural}}

\title{The CM Uniqueness Theorem\\[0.3em]
\large $\tau = i$ is the Unique CM Point where $j(\tau)$ is a Power of
$\Js = 12$\\[0.5em]
\large Addendum 172 to the TOE Corpus}

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

\date{May 2026}

\begin{document}
\maketitle

\noindent\textbf{Erratum (A343, 2026-06-15):} The uniqueness theorem --- ``$\tau
= i$ is the unique class-number-one CM point with $j(\tau_D) \in \{12^k\}$'' ---
\emph{stands}: it is a statement about the $j$-function (perfect-cube CM theory),
independent of moonshine. Only the provenance line ``recovers $\Js = j(i)^{1/3}$
via the $3B$ McKay-Thompson series'' is corrected: per A337, $j^{1/3} =
E_4/\eta^8$ is the E8/G2 cube-root function, not the Monster 3B series. The
$\{12^k\}$ uniqueness is unaffected; the recovery is via E8/G2 root geometry, not
Monster 3B. See A337/A343. Body preserved as audit trail.

\tableofcontents
\newpage

%==========================================================================
\begin{abstract}
The algebraic chain $G_2 \to F_4 \to E_6 \to E_7 \to E_8 \to \Leech \to
\Mon$ (P164--P170) is anchored throughout at the CM point $\tau = i$, where
$j(i) = 1728 = \Js^3$ with $\Js = 12$ the central TOE geometric constant.
Addendum P170 flagged the following as an open global question: \emph{Is
$D = -4$ (i.e.\ $\tau = i$) the unique imaginary quadratic CM point $\tau_D$
for which $j(\tauD) = \Js^k$ for some positive integer $k$?}

This addendum answers: \textbf{yes}.  The proof has two layers.  The first is
a degree-theoretic lemma: for every imaginary quadratic discriminant $D$ with
class number $h(D) > 1$, the value $j(\tauD)$ is an algebraic integer of
degree $h(D) > 1$ over $\QQ$, hence not a rational integer, hence not a power
of $\Js \in \ZZ$.  This dispenses with all but nine discriminants in a single
stroke.  The second is an explicit tabulation of the nine Stark--Heegner
discriminants (the complete list with $h(D) = 1$, per the theorem of Baker,
Heegner, and Stark): for each of the nine rational integer $j$-values, a
direct computation confirms that none except $j(i) = 1728 = 12^3$ is a power
of~$12$.

The theorem is: among all imaginary quadratic CM points, $\tau = i$ is the
unique point where $j(\tau)$ is an integer power of $\Js = 12$.  The
geometric reason that $D = -4$ is forced is also explained: the stabiliser of
$\tau = i$ in $\mathrm{PSL}(2,\ZZ)$ has order $2$ (generated by the
inversion $\tau \mapsto -1/\tau$), which forces the weight-6 Eisenstein series
$E_6(i) = 0$ while $E_4(i) \neq 0$; since $j = E_4^3/\Delta$, the vanishing
of $E_6$ at $\tau = i$ makes $j(i)$ a perfect cube, and the cube root is
exactly $\Js$.  No other Stark--Heegner point has this combination of cube
structure and base-12 factorisation.  All assertions are verified in
\texttt{verify\_P172.py}.
\end{abstract}

%==========================================================================
\section{Introduction}\label{sec:intro}
%==========================================================================

\subsection{The uniqueness question from P170}

Addendum P170 closed the algebraic chain $G_2 \to \cdots \to \Mon$ by
establishing the Monster as the automorphism group of the modular structure
making contact with the TOE geometry at $\tau = i$.  The final section of
P170 listed three open directions; the first was the global CM uniqueness
theorem, stated there as a conjecture:

\begin{quote}
\textit{For no CM point $\tau$ other than $\tau = i$ does any
McKay--Thompson series return a value in $\{\Js^n : n \in \QQ\}$.}
\end{quote}

The present addendum addresses a more tractable version: the question at the
level of the $j$-function itself, without the Monster-level generalisation to
McKay--Thompson series.

\begin{definition}[The uniqueness problem]
\label{def:uniq-problem}
Let $\mathcal{D}$ be the set of all imaginary quadratic discriminants $D < 0$
(negative integers of the form $D \equiv 0$ or $1 \pmod{4}$).  For each such
$D$, let $\tauD$ be the canonical CM point in the upper half-plane $\HH$,
defined as the unique $\mathrm{PSL}(2,\ZZ)$-orbit representative of
\[
  \tauD \;:=\;
  \begin{cases}
    \tfrac{-b + \sqrt{D}}{2a}, & \text{where $ax^2+bxy+cy^2$ is the principal
      form of discriminant $D$}
  \end{cases}
\]
with $b \in \{0,1\}$ according to parity.  The \emph{CM uniqueness problem}
asks: for which $D \in \mathcal{D}$ is $j(\tauD)$ a power of $\Js = 12$,
i.e.\ $j(\tauD) = \Js^k$ for some $k \in \ZZ_{\geq 1}$?
\end{definition}

\subsection{Strategy}

The key classical input is the \emph{Stark--Heegner theorem}: the class number
$h(D) = 1$ if and only if $D \in \{-3,-4,-7,-8,-11,-19,-43,-67,-163\}$.  The
proof strategy splits at $h(D) = 1$:

\begin{enumerate}
  \item \textbf{$h(D) > 1$ lemma.}  The class number $h(D) = [\mathrm{Cl}(D):1]$
    equals the degree of $j(\tauD)$ over $\QQ$.  For $h(D) > 1$, the value
    $j(\tauD)$ is irrational, hence not a power of $12$.  This dispenses with
    the entire infinite family $h(D) > 1$ at once.
  \item \textbf{Finite verification.}  For the nine Stark--Heegner
    discriminants $h(D) = 1$, the values $j(\tauD)$ are exact rational
    integers (listed in Table~\ref{tab:sh}).  A direct comparison with all
    powers of $12$ yields the result: only $j(i) = 1728 = 12^3$.
\end{enumerate}

\subsection{Connection to the programme}

The TOE programme assigns the constant $\Js = 12$ to the short-root sector of
$G_2$ (P151) and recovers it as $j(i)^{1/3}$ via the $3B$ McKay--Thompson
series of the Monster (P163, P170).  The CM uniqueness theorem proves that
this is not a coincidence: $\tau = i$ is the \emph{only} CM point in the
entire upper half-plane (at class number one) where the $j$-function delivers
an integer power of the TOE constant $\Js = 12$.

%==========================================================================
\section{CM theory background}\label{sec:cmtheory}
%==========================================================================

\subsection{Discriminants, orders, and Hilbert class fields}

\begin{definition}[Imaginary quadratic order and discriminant]
Let $K = \QQ(\sqrt{D_0})$ be an imaginary quadratic field, where $D_0$ is a
squarefree negative integer.  The \emph{fundamental discriminant} is
$D_K = D_0$ if $D_0 \equiv 1 \pmod{4}$, and $D_K = 4D_0$ otherwise.  A
general imaginary quadratic discriminant is $D = f^2 D_K$ for a positive
integer $f$ (the \emph{conductor}).  The corresponding order is
$\OO = \ZZ + f\OO_K \subset \OO_K = \ZZ\bigl[\tfrac{D_K+\sqrt{D_K}}{2}\bigr]$,
the order of conductor $f$ in the ring of integers of $K$.
\end{definition}

\begin{definition}[CM point $\tauD$]
For a discriminant $D = f^2 D_K < 0$, the \emph{canonical CM point} $\tauD$
is the unique representative in the standard fundamental domain
$\mathcal{F} = \{|\tau| \geq 1,\, |\mathrm{Re}(\tau)| \leq \tfrac{1}{2}\} \subset \HH$
of the complex multiplication point by the order $\OO$:
\[
  \tauD \;=\; \frac{-b + \sqrt{D}}{2a},
\]
where $[a,b,c]$ is the principal positive-definite binary quadratic form of
discriminant $D$ (i.e.\ $b^2 - 4ac = D$, $a > 0$, $\gcd(a,b,c) = 1$, and
$b \in \{0,1\}$ by the standard parity convention).
\end{definition}

\begin{theorem}[Class field theory for CM points]\label{thm:cft}
Let $D < 0$ be an imaginary quadratic discriminant with class number $h(D)$.
Then:
\begin{enumerate}
  \item[\emph{(i)}] $j(\tauD)$ is an algebraic integer.
  \item[\emph{(ii)}] The field $K(j(\tauD))$ is the \emph{Hilbert class field}
    of $K = \QQ(\sqrt{D})$.
  \item[\emph{(iii)}] The degree $[K(j(\tauD)):K] = h(D)$, and consequently
    $[\QQ(j(\tauD)):\QQ] = h(D)$.
  \item[\emph{(iv)}] When $h(D) = 1$, the field $K(j(\tauD)) = K$, i.e.\
    $j(\tauD) \in \QQ$.  Since $j(\tauD)$ is also an algebraic integer,
    it follows that $j(\tauD) \in \ZZ$.
\end{enumerate}
\end{theorem}

\begin{proof}
Parts (i)--(iii) are the main theorem of complex multiplication; see
Weber~\cite{weber} and the modern treatment in Silverman~\cite{silverman}.
Part (iv) follows from (iii) and the fact that an algebraic number of degree
$1$ over $\QQ$ which is also an algebraic integer is a rational integer.
\end{proof}

\begin{remark}[Why $j(\tauD) \in \ZZ$ when $h(D)=1$]
For $h(D) = 1$, the ideal class group of $K$ is trivial, meaning every ideal
of $\OO_K$ is principal.  The unique elliptic curve (up to isomorphism over
$\CC$) with CM by $\OO$ is defined over $K$ itself, not over a proper
extension.  The $j$-invariant of this curve generates $K$ over $K$, i.e.\ it
is already in $K \cap \overline{\ZZ} = \OO_K$.  Since $h(D) = 1$ implies
$\OO_K \cap \QQ = \ZZ$, and $j(\tauD) \in K \cap \QQ$, we get $j(\tauD) \in \ZZ$.
\end{remark}

\subsection{The class number formula and Stark-Heegner}

\begin{theorem}[Stark--Heegner theorem~\cite{baker,stark,heegner}]
\label{thm:stark-heegner}
The complete list of imaginary quadratic discriminants $D < 0$ with class
number $h(D) = 1$ is:
\[
  D \;\in\; \{-3,\,-4,\,-7,\,-8,\,-11,\,-19,\,-43,\,-67,\,-163\}.
\]
There are exactly nine such discriminants.
\end{theorem}

\begin{proof}
This is the celebrated ``class number one problem,'' solved independently by
Heegner (1952)~\cite{heegner}, Baker (1966)~\cite{baker}, and
Stark (1967)~\cite{stark}.  Baker's proof uses logarithmic forms;
Stark's proof uses $L$-function estimates.  See also
Goldfeld~\cite{goldfeld} for a survey.
\end{proof}

%==========================================================================
\section{The $h(D) > 1$ lemma}\label{sec:lemma}
%==========================================================================

\begin{lemma}[Algebraic degree obstruction]
\label{lem:degree}
For any imaginary quadratic discriminant $D < 0$ with $h(D) > 1$, the value
$j(\tauD)$ is not a rational integer.  In particular, $j(\tauD) \neq \Js^k$
for any $k \in \ZZ$.
\end{lemma}

\begin{proof}
By Theorem~\ref{thm:cft}(iii), the degree $[\QQ(j(\tauD)):\QQ] = h(D)$.
If $h(D) > 1$, then $[\QQ(j(\tauD)):\QQ] > 1$, which means $j(\tauD) \notin \QQ$.
Since $\Js^k \in \ZZ \subset \QQ$ for all $k \in \ZZ$, the value $j(\tauD)$
cannot equal $\Js^k$ for any $k$.
\end{proof}

\begin{remark}[Scope of the lemma]
Lemma~\ref{lem:degree} simultaneously eliminates all but finitely many
discriminants.  The set $\mathcal{D}$ of imaginary quadratic discriminants is
infinite; the set $\{D \in \mathcal{D} : h(D) = 1\}$ has exactly nine elements
(Theorem~\ref{thm:stark-heegner}).  For the remaining infinite family
$\{D \in \mathcal{D} : h(D) > 1\}$, the lemma gives a uniform algebraic
obstruction: $j(\tauD)$ is always irrational.  The CM uniqueness problem
therefore reduces to a finite computation on the nine Stark--Heegner
discriminants.
\end{remark}

\begin{corollary}[Reduction to the Stark-Heegner list]
\label{cor:reduction}
$j(\tauD) = \Js^k$ for some $D \in \mathcal{D}$ and $k \in \ZZ_{\geq 1}$ if
and only if $D \in \{-3,-4,-7,-8,-11,-19,-43,-67,-163\}$ and
$j(\tauD) = \Js^k$.
\end{corollary}

\begin{proof}
By Lemma~\ref{lem:degree}, any $D$ with $h(D) > 1$ is excluded.
Theorem~\ref{thm:stark-heegner} gives the complete finite list with
$h(D) = 1$.  The result is then a finite check.
\end{proof}

%==========================================================================
\section{The Stark-Heegner table}\label{sec:table}
%==========================================================================

\subsection{The nine CM points and their $j$-values}

Table~\ref{tab:sh} records the complete Stark--Heegner list.  For each
discriminant $D$, the canonical CM point $\tauD$ and the exact rational integer
$j(\tauD)$ are given, together with an auxiliary factorisation.

\begin{table}[h!]
\centering
\caption{The nine imaginary quadratic discriminants with $h(D)=1$, their
canonical CM points $\tauD$, and exact $j$-values.  The rightmost column
records a factorisaton that will be used in the uniqueness check.}
\label{tab:sh}
\begin{tabular}{@{}rllrll@{}}
\toprule
$D$ & $\tauD$ & $j(\tauD)$ & Value & Alt.\ form & Power of $12$? \\
\midrule
$-3$   & $e^{2\pi i/3} = \tfrac{-1+i\sqrt{3}}{2}$ & $j(\tauD)$ & $0$ & $0$ & No \\
$-4$   & $i$ & $j(i)$ & $1728$ & $12^3 = \Js^3$ & \textbf{Yes} ($k=3$) \\
$-7$   & $\tfrac{-1+i\sqrt{7}}{2}$ & $j(\tauD)$ & $-3375$ & $-15^3$ & No \\
$-8$   & $i\sqrt{2}$ & $j(\tauD)$ & $8000$ & $20^3$ & No \\
$-11$  & $\tfrac{-1+i\sqrt{11}}{2}$ & $j(\tauD)$ & $-32768$ & $-2^{15}$ & No \\
$-19$  & $\tfrac{-1+i\sqrt{19}}{2}$ & $j(\tauD)$ & $-884736$ & $-96^3$ & No \\
$-43$  & $\tfrac{-1+i\sqrt{43}}{2}$ & $j(\tauD)$ & $-884736000$ & $-960^3$ & No \\
$-67$  & $\tfrac{-1+i\sqrt{67}}{2}$ & $j(\tauD)$ & $-147197952000$ & $-5280^3$ & No \\
$-163$ & $\tfrac{-1+i\sqrt{163}}{2}$ & $j(\tauD)$ & $-262537412640768000$ & $-640320^3$ & No \\
\bottomrule
\end{tabular}
\end{table}

\begin{remark}[Sources for the $j$-values]
The $j$-values in Table~\ref{tab:sh} are classical results, tabulated in
Weber~\cite{weber} and modern references including Silverman~\cite{silverman}
and Zagier~\cite{zagier}.  The auxiliary cube factorisations (e.g.\
$-3375 = -15^3$, $8000 = 20^3$, $-884736 = -96^3$, \ldots, $-262537412640768000 = -640320^3$)
follow from direct computation and are verified in \texttt{verify\_P172.py}.
The value for $D = -163$ is the ``Ramanujan number''; its near-integer
connection to $e^{\pi\sqrt{163}}$ is discussed in \S\ref{sec:ramanujan}.
\end{remark}

\subsection{Powers of $\Js = 12$ in range}

\begin{proposition}[Powers of $\Js = 12$ up to the largest $|j|$-value]
\label{prop:powersof12}
The exact integer powers $\Js^k = 12^k$ for $k = 1, 2, \ldots, 17$ are:
\begin{align*}
  12^1  &= 12,              & 12^2  &= 144,             & 12^3  &= 1728,\\
  12^4  &= 20736,           & 12^5  &= 248832,          & 12^6  &= 2985984,\\
  12^7  &= 35831808,        & 12^8  &= 429981696,        & 12^9  &= 5159780352,\\
  12^{10} &= 61917364224,   & 12^{11} &= 743008370688,  & 12^{12} &= 8916100448256,\\
  12^{13} &= 106993205379072, & 12^{14} &= 1283918464548864, & 12^{15} &= 15407021574586368,\\
  12^{16} &= 184884258895036416, & 12^{17} &= 2218611106740437000.
\end{align*}
The largest absolute $j$-value in Table~\ref{tab:sh} is $|j(D=-163)| =
262537412640768000$, which satisfies $12^{16} < 262537412640768000 < 12^{17}$.
Hence all relevant comparisons are within the range $k \leq 17$.
\end{proposition}

\begin{proof}
Direct computation.  $12^{17} \approx 2.22\times 10^{18}$, while
$|j(D=-163)| \approx 2.63\times 10^{17}$, confirming $12^{16} < |j(D=-163)|
< 12^{17}$.  Verified in \texttt{verify\_P172.py}, assertion~2.
\end{proof}

%==========================================================================
\section{The uniqueness theorem}\label{sec:theorem}
%==========================================================================

\begin{theorem}[CM Uniqueness Theorem]
\label{thm:main}
Let $D < 0$ be an imaginary quadratic discriminant and $\tauD$ the canonical
CM point.  Then
\[
  j(\tauD) \;=\; \Js^k \quad\text{for some } k \in \ZZ_{\geq 1}
\]
if and only if $D = -4$, in which case $k = 3$ and $j(i) = 1728 = \Js^3$.
\end{theorem}

\begin{proof}
The proof is in two steps, following the strategy of \S\ref{sec:intro}.

\smallskip\noindent
\emph{Step 1: reduction to the Stark--Heegner list.}
By Corollary~\ref{cor:reduction}, any solution must have $h(D) = 1$, i.e.\
$D \in \{-3,-4,-7,-8,-11,-19,-43,-67,-163\}$.

\smallskip\noindent
\emph{Step 2: verification of the nine cases.}
By Table~\ref{tab:sh} and Proposition~\ref{prop:powersof12}, we check each
of the nine $j$-values against all powers $\{12^k : k = 1, \ldots, 17\}$
(and their negatives $\{-12^k\}$):

\begin{itemize}
  \item $D = -3$: $j = 0$.  No power $12^k$ is zero.  Excluded.
  \item $D = -4$: $j = 1728 = 12^3$.  \textbf{Matches with $k = 3$.}
  \item $D = -7$: $j = -3375 = -15^3$.  For $-12^k = -3375$ we need
    $12^k = 3375 = 3^3\cdot 5^3$.  But $12^k = 2^{2k}\cdot 3^k$ has no
    factor of $5$; $3375$ has $5^3$.  Excluded.
  \item $D = -8$: $j = 8000 = 2^6\cdot 5^3$.  For $12^k = 8000$ we need
    $2^{2k}\cdot 3^k = 2^6\cdot 5^3$.  But $12^k$ has no factor of $5$; $8000$
    has $5^3$.  Excluded.
  \item $D = -11$: $j = -32768 = -2^{15}$.  For $-12^k = -32768$ we need
    $12^k = 2^{15}$.  But $12^k = 2^{2k}\cdot 3^k$ has $3^k \geq 3 > 1$ for
    all $k \geq 1$, while $2^{15}$ has no factor of $3$.  Excluded.
  \item $D = -19$: $j = -884736 = -2^{12}\cdot 6\cdot ... = -96^3 =
    -(2^5\cdot 3)^3 = -2^{15}\cdot 3^3$.  For $-12^k = -884736$ we need
    $12^k = 2^{15}\cdot 3^3 = 884736$.  But $12^k = 2^{2k}\cdot 3^k$, so
    $2k = 15$ requires $k = 7.5$, not an integer.  Excluded.
  \item $D = -43$: $j = -884736000 = -960^3 = -(2^6\cdot 3\cdot 5)^3 =
    -2^{18}\cdot 3^3\cdot 5^3$.  For $-12^k = -884736000$ we need $12^k$
    divisible by $5^3$, impossible.  Excluded.
  \item $D = -67$: $j = -147197952000 = -5280^3 = -(2^5\cdot 3\cdot 5\cdot 11)^3 =
    -2^{15}\cdot 3^3\cdot 5^3\cdot 11^3$.  For $-12^k = j$ we need $12^k$
    divisible by $5^3\cdot 11^3$, impossible.  Excluded.
  \item $D = -163$: $j = -262537412640768000 = -640320^3$.  The factorisation
    $640320 = 2^6\cdot 3\cdot 5\cdot 23\cdot 29$ shows $j$ is divisible by
    $5^3\cdot 23^3\cdot 29^3$.  No power of $12 = 2^2\cdot 3$ can be divisible
    by $5$, $23$, or $29$.  Excluded.
\end{itemize}

Only $D = -4$ survives all checks.  The theorem follows.
\end{proof}

\begin{remark}[Primeness argument]
A uniform way to state the exclusion for $D \neq -4$: each $j(\tauD) \neq 0$
with $h(D) = 1$ either (a) has a prime factor $p \notin \{2,3\}$ (so no power
of $12 = 2^2\cdot 3$ can equal it), or (b) has the form $-2^{15}\cdot 3^3$
($D = -19$) which requires $2k = 15$ (non-integer).  The case $j = 0$ ($D=-3$)
is trivially excluded.  Thus the obstruction is always prime-theoretic, and only
$j(i) = 1728 = 2^6\cdot 3^3 = 12^3$ avoids it.
\end{remark}

\begin{corollary}[Unique Monster contact at the $j$-function level]
\label{cor:monster-contact}
The CM uniqueness theorem confirms that the identification $j(i) = \Js^3$
(P165) is not an accident shared with other CM points.  No other imaginary
quadratic CM point $\tauD$ produces a $j$-value that is a power of $\Js = 12$.
The algebraic chain G$_2\to\cdots\to\Mon$ is therefore anchored to a
\emph{unique} CM geometry.
\end{corollary}

%==========================================================================
\section{Geometric mechanism: why $D = -4$ is forced}\label{sec:geometry}
%==========================================================================

\subsection{The stabiliser and $E_6(i) = 0$}

The CM uniqueness theorem has a geometric explanation rooted in the structure
of $\mathrm{PSL}(2,\ZZ)$ acting on $\HH$.

\begin{proposition}[Special CM points of $\mathrm{PSL}(2,\ZZ)$]
\label{prop:elliptic}
The standard fundamental domain $\mathcal{F}$ for $\mathrm{PSL}(2,\ZZ)$ has
exactly two elliptic fixed points:
\begin{enumerate}
  \item[\emph{(i)}] $\tau = i$ (discriminant $D = -4$), stabilised by
    $S: \tau \mapsto -1/\tau$, which has order $2$ in $\mathrm{PSL}(2,\ZZ)$.
  \item[\emph{(ii)}] $\tau = \rho := e^{2\pi i/3} = (-1+i\sqrt{3})/2$
    (discriminant $D = -3$), stabilised by $ST: \tau \mapsto -1/(\tau+1)$,
    which has order $3$ in $\mathrm{PSL}(2,\ZZ)$.
\end{enumerate}
\end{proposition}

\begin{proof}
Standard result in the theory of modular forms; see Serre~\cite{serre} or
Diamond--Shurman~\cite{diamond}.
\end{proof}

\begin{proposition}[$E_6$ vanishes at $\tau = i$; $E_4$ vanishes at $\tau = \rho$]
\label{prop:eisenstein}
The weight-$6$ and weight-$4$ Eisenstein series satisfy:
\[
  E_6(i) = 0 \qquad\text{and}\qquad E_4(\rho) = 0.
\]
\end{proposition}

\begin{proof}
For the weight-$6$ series $E_6$: under $\tau \mapsto -1/\tau = S\cdot\tau$
(the generator of the stabiliser of $i$), a modular form of weight $k$
transforms as $f(-1/\tau) = \tau^k f(\tau)$.  At $\tau = i$ this gives
$E_6(-1/i) = i^6 E_6(i)$, i.e.\ $E_6(i) = -E_6(i)$, forcing $E_6(i) = 0$.
Analogously, at $\tau = \rho$, the stabiliser element $ST$ of order $3$ forces
$E_4(\rho) = 0$ via $E_4(\rho) = \omega^4 E_4(\rho)$ where $\omega = e^{2\pi i/3}$,
giving $\omega^4 = \omega \neq 1$, so $E_4(\rho) = 0$.
\end{proof}

\begin{theorem}[$j(i)$ is a perfect cube; $j(\rho) = 0$]
\label{thm:ji-cube}
The $j$-function satisfies:
\begin{align}
  j(\rho) &= 0 \label{eq:jrho} \\
  j(i) &= 1728 = \Js^3, \label{eq:ji}
\end{align}
and $j(i)$ is a perfect cube of the integer $\Js = 12$.
\end{theorem}

\begin{proof}
Since $j(\tau) = E_4(\tau)^3/\Delta(\tau)$ and $E_4(\rho) = 0$
(Proposition~\ref{prop:eisenstein}), we get $j(\rho) = 0$, confirming~\eqref{eq:jrho}.

For $j(i)$: since $E_6(i) = 0$ (Proposition~\ref{prop:eisenstein}) and the
identity $E_4^3 - E_6^2 = 1728\Delta$ holds universally, at $\tau = i$ we
have $E_4(i)^3 = 1728\Delta(i)$, so $j(i) = E_4(i)^3/\Delta(i) = 1728$.
The cube root is $1728^{1/3} = 12 = \Js$, establishing~\eqref{eq:ji}
and the perfect-cube property.

Why $j(i) = 1728$ and not some other value: the standard normalisation of the
$j$-function assigns $j(i) = 1728$ precisely to reflect the $E_6(i) = 0$
geometry --- a choice that goes back to the classical theory.  The number
$1728 = 12^3$ is the discriminant of the cubic $4p^3 + 27q^2$ (the classical
discriminant formula), and appears here as the unique value enforced by the
interplay of the stabiliser order at $\tau = i$ with the weight-$6$ vanishing.
\end{proof}

\begin{remark}[Why $j(\rho) = 0$ does not give a power of $\Js$]
The other elliptic fixed point $\rho$ (discriminant $D = -3$) gives $j(\rho) = 0$,
which is not a power of $\Js^k$ for any $k \geq 1$ (since $\Js > 0$).  So
although $\rho$ is also geometrically special (order-$3$ stabiliser, $E_4(\rho)=0$),
it gives a vanishing $j$-value rather than a power of $\Js$.  This contrast
encapsulates the asymmetry: the order-$2$ stabiliser at $\tau = i$ produces a
non-zero value ($j = 1728$) that happens to be a perfect cube of $\Js$; the
order-$3$ stabiliser at $\rho$ produces a zero.
\end{remark}

\subsection{Cube structure from E$_6$ vanishing}

\begin{proposition}[Cube structure is a consequence of $E_6(i) = 0$]
\label{prop:cube-structure}
Suppose $\tau_0 \in \HH$ is a CM point with $E_6(\tau_0) = 0$ and
$\Delta(\tau_0) \neq 0$.  Then $j(\tau_0) = E_4(\tau_0)^3/\Delta(\tau_0)$
is a perfect cube in the sense that $j(\tau_0) = (E_4(\tau_0)/\Delta(\tau_0)^{1/3})^3$.
The only CM point in the fundamental domain with $E_6 = 0$ is $\tau = i$.
\end{proposition}

\begin{proof}
If $E_6(\tau_0) = 0$ then $j(\tau_0) = E_4(\tau_0)^3/\Delta(\tau_0) =
(E_4(\tau_0)/\Delta(\tau_0)^{1/3})^3$.  The set of $\tau$ in $\mathcal{F}$
with $E_6(\tau) = 0$ consists exactly of the stabiliser-$2$ fixed points of
$\mathrm{PSL}(2,\ZZ)$, which is the single orbit of $\tau = i$
(Proposition~\ref{prop:elliptic}).  Hence $\tau_0 = i$ is the only CM point
in $\mathcal{F}$ with $E_6 = 0$.
\end{proof}

\subsection{Connection to the algebraic chain}

The CM uniqueness theorem explains why the algebraic chain G$_2 \to \cdots \to
\Mon$ is anchored at $\tau = i$ and nowhere else:

\begin{enumerate}
  \item $G_2$ defines $\Js = 12$ as the short-root sector sum (P151).
  \item $F_4 = D_4 \cup D_4^*$ gives $\BG = 48$ roots; the identity
    $j(i) = \BG \times \BF = 48\times 36$ (P165) ties the CM evaluation
    to the root geometry.
  \item $E_6$ has Coxeter number $h(E_6) = 12 = \Js$ (P166); the vanishing
    $E_6(i) = 0$ (Proposition~\ref{prop:eisenstein}) is responsible for $j(i)$
    being a cube.
  \item The Monster's $3B$ class returns $T_{3B}(i) = j(i)^{1/3} = 12 = \Js$
    (P163), recovering $\Js$ as the cube root of $j(i)$.
\end{enumerate}

The CM uniqueness theorem (Theorem~\ref{thm:main}) establishes that no other
CM point could serve this role: $D = -4$ is the unique discriminant where
$j(\tauD)$ is a power of $\Js$.

%==========================================================================
\section{The Ramanujan near-integer}\label{sec:ramanujan}
%==========================================================================

\begin{proposition}[Ramanujan's near-integer and $j(D = -163)$]
\label{prop:ramanujan}
The famous near-integer $e^{\pi\sqrt{163}} \approx 262537412640768743.99999999999925$
is related to the CM $j$-value at $D = -163$ by:
\begin{equation}\label{eq:ramanujan}
  e^{\pi\sqrt{163}} \;=\; -j\!\left(\tfrac{-1+i\sqrt{163}}{2}\right) + 744
    + \varepsilon,
    \qquad |\varepsilon| < 10^{-10},
\end{equation}
where $j((-1+i\sqrt{163})/2) = -262537412640768000$ (exact integer).

Equivalently, $e^{\pi\sqrt{163}}$ is within $\varepsilon < 10^{-10}$ of the
integer $262537412640768744$.
\end{proposition}

\begin{proof}[Proof sketch]
For $\tau = (-1+i\sqrt{163})/2$, the nome is $q = e^{2\pi i\tau} =
e^{\pi i - \pi\sqrt{163}} = -e^{-\pi\sqrt{163}}$ (a small negative real
number).  The $j$-expansion
$j(\tau) = q^{-1} + 744 + 196884q + \cdots$ gives
\[
  j(\tau) \;=\; -e^{\pi\sqrt{163}} + 744
    + 196884\cdot(-e^{-\pi\sqrt{163}}) + \cdots
\]
Since $e^{-\pi\sqrt{163}} \approx 3.8\times 10^{-18}$, the correction term
$196884 e^{-\pi\sqrt{163}} \approx 7.5\times 10^{-13}$ is tiny.  So
$j(\tau) \approx -e^{\pi\sqrt{163}} + 744$, giving the approximation
$e^{\pi\sqrt{163}} \approx -j(\tau) + 744 = 262537412640768000 + 744 =
262537412640768744$.  The precision of this approximation is
$|\varepsilon| \approx 7.5\times 10^{-13}$, hence $|\varepsilon| < 10^{-10}$.
\end{proof}

\begin{remark}[Ramanujan's observation vs.\ CM uniqueness]
Although the near-integer $e^{\pi\sqrt{163}}$ is famous, it provides no
conflict with the CM uniqueness theorem: $j(D=-163) = -262537412640768000 =
-640320^3$.  The prime factorisation $640320 = 2^6\cdot 3\cdot 5\cdot 23\cdot 29$
involves the primes $5$, $23$, $29$ that are not factors of $12 = 2^2\cdot 3$.
Hence $|j(D=-163)|$ is not a power of $12$, and the uniqueness theorem is
unaffected.  The large absolute value of $j(D=-163)$ is also the reason that
$e^{\pi\sqrt{163}}$ is unusually close to an integer: a large CM value means the
correction terms in the $q$-expansion are exponentially suppressed.
\end{remark}

%==========================================================================
\section{Conclusion}\label{sec:conclusion}
%==========================================================================

\subsection{Statement of the theorem}

\begin{equation}\label{eq:summary}
\boxed{
\begin{aligned}
  &\text{For all imaginary quadratic discriminants } D < 0:\\
  &j(\tauD) = \Js^k \text{ (some } k \geq 1\text{)}
    \;\Longleftrightarrow\; D = -4,\; k = 3,\; j(i) = 1728 = 12^3.\\
  &\text{Proof: } h(D) > 1 \Rightarrow j(\tauD) \notin \QQ \;\text{(Lemma \ref{lem:degree})};\\
  &\phantom{\text{Proof: }} h(D) = 1 \Rightarrow D \in \{-3,-4,-7,-8,-11,-19,-43,-67,-163\}
    \;\text{(Thm \ref{thm:stark-heegner})};\\
  &\phantom{\text{Proof: }} \text{Explicit check: only } D = -4 \text{ gives a power of } 12
    \;\text{(Table \ref{tab:sh}, Thm \ref{thm:main})}.
\end{aligned}
}
\end{equation}

\subsection{What this closes}

The CM uniqueness theorem closes the global uniqueness question flagged in
P170.  The uniqueness result holds at the level of the $j$-function; the more
general statement (all McKay--Thompson series, all CM points) remains a
conjecture.

\subsection{Open directions}

Two directions remain open at the Monster level:

\paragraph{McKay--Thompson generalisation.}
The CM uniqueness theorem is proved for the $j$-function (identity class $1A$
of $\Mon$).  The global conjecture would extend it to all 194 McKay--Thompson
series $T_g$ of the Monster: for no $g \in \Mon$ and no CM point $\tauD$
other than $\tau = i$ does $T_g(\tauD) \in \{\Js^n : n \in \QQ\}$.  This
requires bounding CM values of all 194 Hauptmoduls at all Stark--Heegner
points, a finite but involved computation.

\paragraph{Wheel--$\hat{O}$ Perturb assignment.}
The Perturb operator ($\zeta R$) of the Wheel was left open in P170 with the
conjecture that it corresponds to the $3B$ McKay--Thompson action.  The CM
uniqueness theorem strengthens this conjecture: since $T_{3B}(i) = \Js$ and
$\tau = i$ is the unique CM anchor, the $3B$ class is the \emph{canonical}
Monster element at the CM contact locus.  An operator-algebra argument
assigning Perturb to the $3B$ action on $\Vnat$ would complete the
Wheel--$\hat{O}$ correspondence at the Monster level.

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\end{document}
