% Addendum 158: Why j(i) = B_{G₂}(H_s,H_s) × B_{F₄}(H_s,H_s) = 1728 Exactly
%   Derives the structural reason for the identity rather than treating it as a
%   coincidence. Shows B_𝔤 = J_short(1+ℓ_𝔤) from Weyl-group transitivity,
%   computes that (1+ℓ_{G₂})(1+ℓ_{F₄}) = J_short, and concludes
%   B_{G₂}×B_{F₄} = J_short³ = 12³ = j(i). Tests specificity via (B₃,F₄).
% L. F. Vlegels, Independent Researcher
% Copyright: Léon Fernando Vlegels. License: CC BY 4.0. May 2026.

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\newcommand{\g}{\mathfrak{g}}
\newcommand{\hs}{H_s}
\newcommand{\Bg}{B_{\mathfrak{g}}}
\newcommand{\BG}{B_{G_2}}
\newcommand{\BF}{B_{F_4}}
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\title{Why $j(i) = B_{G_2}(H_s,H_s)\times B_{F_4}(H_s,H_s) = 1728$ Exactly\\[0.5em]
\large Addendum 158 to the TOE Corpus}

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

\date{May 2026}

\begin{document}
\maketitle

%--------------------------------------------------------------------------
\begin{abstract}
Addenda P151 and P153 established $B_{G_2}(\hs,\hs)=48$ and $B_{F_4}(\hs,\hs)=36$,
and P157 noted $48\times 36 = 1728 = j(i)$.
This addendum derives \emph{why}.
The key chain is:
\begin{equation}
  B_\g(\hs,\hs) \;=\; \Js(1+\ell_\g),
  \qquad
  (1+\ell_{G_2})(1+\ell_{F_4}) \;=\; \Js,
  \qquad\Longrightarrow\qquad
  \BG\cdot\BF \;=\; \Js^3 \;=\; 12^3 \;=\; 1728.
\end{equation}
The first identity follows from Weyl-group transitivity on each root-length class;
the second is exact integer arithmetic forced by the Dynkin labels $\ell_{G_2}=3$,
$\ell_{F_4}=2$, and $\Js=12$.
A specificity test on the $(B_3,F_4)$ pair confirms the identity is peculiar to $(G_2,F_4)$.
An independent classical result, $E_4(i)/\eta(i)^8=12=\Js$, is noted as an open observation.
All claims are verified by \texttt{verify\_P158.py}.
\end{abstract}

%==========================================================================
\section{Observation}\label{sec:obs}

From P151, the Killing form of $G_2$ evaluated on the short coroot $\hs$
decomposes by root-length sector as
\[
  \BG(\hs,\hs) \;=\; \Js + \Jl \;=\; 12 + 36 \;=\; 48,
\]
where $\Js = \sum_{\alpha\in\Phi_{\rm short}}\langle\alpha,\hs\rangle^2$
and $\Jl = \sum_{\alpha\in\Phi_{\rm long}}\langle\alpha,\hs\rangle^2$.
From P153, $\BF(\hs,\hs) = 36$.
From classical modular theory, $j(i)=1728$ at the CM point $\tau=i$.
The numerical coincidence $48\times 36 = 1728$ was observed in P157.
Here we show it is structurally forced.

%==========================================================================
\section{Sector Decomposition of $B_{G_2}$ and $B_{F_4}$}\label{sec:sectors}

\subsection*{$G_2$ short- and long-root sectors}

$G_2$ has six short roots and six long roots (equal counts).
In $A_2$-coordinates, the projections $\langle\alpha,\hs\rangle$ for $\hs=\alpha_1^\vee$
are, over all twelve roots (positive and negative):
\begin{align*}
  \text{Short roots: }& +2,\,-2,\,-1,\,+1,\,+1,\,-1
  \quad\Longrightarrow\quad \Js = 4+4+1+1+1+1 = 12, \\
  \text{Long roots: }& -3,\,+3,\,+3,\,-3,\,0,\,0
  \quad\Longrightarrow\quad \Jl = 9+9+9+9+0+0 = 36.
\end{align*}
The ratio $\Jl/\Js = 36/12 = 3 = \ell_{G_2}$ equals the Dynkin length ratio.
This is not accidental: the Weyl group acts transitively on $\Phi_{\rm short}$
and transitively on $\Phi_{\rm long}$.
Since every long root $\beta$ has $|\beta|^2 = \ell_{G_2}|\alpha|^2$
for any short root $\alpha$, and since the Weyl group mixes all roots of each
length class uniformly, the orbit-averaged squared projection satisfies
$\overline{\langle\beta,\hs\rangle^2} = \ell_{G_2}\,\overline{\langle\alpha,\hs\rangle^2}$.
With equal counts $|\Phi_{\rm long}| = |\Phi_{\rm short}| = 6$, this forces
\begin{equation}\label{eq:JlG2}
  \Jl^{G_2} \;=\; \ell_{G_2}\,\Js.
\end{equation}

\subsection*{$F_4$ short- and long-root sectors}

The $F_4$ root system in $\mathbb{R}^4$ consists of 24 short roots
($\pm e_i$ and $\tfrac{1}{2}(\pm 1,\pm 1,\pm 1,\pm 1)$) and 24 long roots
($\pm e_i\pm e_j$, $i<j$), again equal counts.
With $\hs = e_1 - e_2$ (the short coroot for the short simple root $\alpha_1$
in the Bourbaki labelling), direct computation yields:
\[
  \Js^{F_4} = 12, \qquad \Jl^{F_4} = 24,
  \qquad \BF(\hs,\hs) = 36.
\]
The ratio $\Jl^{F_4}/\Js^{F_4} = 2 = \ell_{F_4}$, confirming the same Weyl-transitivity
argument. The critical observation is:
\begin{equation}\label{eq:Jsequal}
  \Js^{G_2} \;=\; \Js^{F_4} \;=\; 12.
\end{equation}
Both algebras share the same short-root sector value, because both short-root
subsystems restrict to an $A_2$ geometry when projected onto $\hs$.

\begin{center}
\begin{tabular}{lcccc}
\toprule
Algebra & $\ell$ & $|\Phi_{\rm short}|$ & $|\Phi_{\rm long}|$ & $\Js$ \phantom{X}$\Jl$ \phantom{X}$B$ \\
\midrule
$G_2$ & $3$ & $6$ & $6$ & $12$ \phantom{X} $36$ \phantom{X} $48$ \\
$F_4$ & $2$ & $24$ & $24$ & $12$ \phantom{X} $24$ \phantom{X} $36$ \\
$B_3$ & $2$ & $6$ & $12$ & $2$  \phantom{X} $8$  \phantom{X} $10$ \\
\bottomrule
\end{tabular}
\end{center}

%==========================================================================
\section{General Formula $B_\g(\hs,\hs) = \Js(1+\ell_\g)$}\label{sec:formula}

\begin{theorem}\label{thm:main}
Let $\g$ be a simple Lie algebra with two root lengths satisfying
\textup{(a)} equal numbers of long and short roots, and
\textup{(b)} the Weyl group $\Weyl$ acts transitively on $\Phi_{\rm short}$
and transitively on $\Phi_{\rm long}$.
Then
\begin{equation}\label{eq:formula}
  \Bg(\hs,\hs) \;=\; \Js\,(1 + \ell_\g).
\end{equation}
\end{theorem}

\begin{proof}[Proof sketch]
By Weyl transitivity, the map sending a short root $\alpha$ to a Weyl-conjugate
long root $\beta$ scales length by $\sqrt{\ell_\g}$, hence
$\langle\beta,\hs\rangle^2 = \ell_\g\langle\alpha,\hs\rangle^2$
averaged over each orbit.
Since $\Weyl$ acts transitively on each length class, the orbit-average is exact:
$\Jl = \ell_\g\,\Js$.
Equal counts then give $\Bg = \Js + \Jl = \Js(1+\ell_\g)$.
\end{proof}

\noindent Verification: $\BG = 12(1+3) = 48$ \checkmark;\quad $\BF = 12(1+2) = 36$ \checkmark.

%==========================================================================
\section{The Product Identity $B_{G_2}\times B_{F_4} = \Js^3$}\label{sec:product}

Applying Theorem~\ref{thm:main} to both factors:
\begin{align}
  \BG\cdot\BF
  &= \Js(1+\ell_{G_2})\cdot\Js(1+\ell_{F_4}) \notag\\
  &= \Js^2\cdot(1+3)(1+2) \notag\\
  &= \Js^2\cdot 4\cdot 3 \notag\\
  &= \Js^2\cdot 12. \label{eq:step}
\end{align}

\begin{proposition}\label{prop:key}
$(1+\ell_{G_2})(1+\ell_{F_4}) = \Js$.
\end{proposition}

\begin{proof}
Direct integer arithmetic: $(1+3)(1+2) = 4\times 3 = 12 = \Js$.
\end{proof}

\noindent Substituting Proposition~\ref{prop:key} into \eqref{eq:step}:
\begin{equation}\label{eq:main}
  \boxed{\BG\cdot\BF \;=\; \Js^2\cdot\Js \;=\; \Js^3 \;=\; 12^3 \;=\; 1728 \;=\; j(i).}
\end{equation}
The equality $\Js^3 = j(i)$ is not assumed; it is derived from the Killing-form
product combined with the classical CM value $j(i)=1728$.
The structural chain is:
\[
j(i) = 1728 = 12^3 = \Js^3 = \BG\cdot\BF.
\]

%==========================================================================
\section{Specificity: the Identity is Peculiar to $(G_2, F_4)$}\label{sec:specificity}

Proposition~\ref{prop:key} requires two arithmetic accidents to coincide:
$\ell_{G_2}=3$ (the triple bond in the $G_2$ Dynkin diagram),
$\ell_{F_4}=2$ (the double bond in the $F_4$ Dynkin diagram), and
$\Js=12$ (from the short-root $A_2$ projection).
These three conditions produce $(1+3)(1+2)=12$.

\medskip
\noindent\textbf{Test on $(B_3, F_4)$.}
$B_3$ has $\ell_{B_3}=2$, but its short and long root counts are \emph{unequal}
(6 short, 12 long), so Theorem~\ref{thm:main} does not apply.
Direct computation with $H_s^{B_3}=e_1$ gives $\Js^{B_3}=2$.
Then $(1+\ell_{B_3})(1+\ell_{F_4}) = 3\times 3 = 9 \neq 2 = \Js^{B_3}$,
and $B_{B_3}\cdot\BF = 10\times 36 = 360 \neq 8 = (\Js^{B_3})^3$.
The identity fails on both counts.

\medskip
We conclude: the equality $\BG\cdot\BF = \Js^3 = j(i)$ is a real, exact,
structurally derivable identity, but it is \emph{specific to the $(G_2,F_4)$ pair}
— not a general theorem for arbitrary pairs of simple Lie algebras.

%==========================================================================
\section{Modular Footnote}\label{sec:modular}

A classical identity (Ramanujan--Weber, see P157) states:
\[
  \frac{E_4(i)}{\eta(i)^8} \;=\; 12.
\]
\textit{Proof.} One has $\eta(i)^8 = \Gamma(1/4)^8/(2^8\pi^6)$ and
$E_4(i) = 3\Gamma(1/4)^8/(2\pi)^6$, so
$E_4(i)/\eta(i)^8 = 3\cdot 2^8/2^6 = 3\cdot 4 = 12$.

Numerically confirmed: $E_4(i)/\eta(i)^8 = 12.0000000000$ (to 10 d.p.).

The number 12 thus appears via three independent routes:
$\Js=12$ (root-system combinatorics),
$j(i)^{1/3}=12$ (CM theory), and
$E_4(i)/\eta(i)^8 = 12$ (classical modular identity).
All three are consequences of $j(i)=12^3$, so they trace to one underlying fact.
Whether the modular identity $E_4(i)/\eta(i)^8=\Js$ has a structural explanation
connecting modular forms directly to Killing-form root-sector sums remains open.

\begin{openproblem}[P158-OP1]
Does there exist a representation-theoretic or geometric argument that
directly identifies $E_4(\tau)$ at a CM point with a Killing-form sector sum,
independent of the route through $j(\tau)=\Js^3$?
\end{openproblem}

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

\texttt{Lumen/corpus/addenda/verify/verify\_P158.py} (run 2026-05-15):

\begin{verbatim}
Step 1  G2 sector:  J_short = 12,  J_long = 36 = ell_G2 x J_short  ✓
Step 2  F4 sector:  J_short = 12,  J_long = 24 = ell_F4 x J_short  ✓
        (same J_short = 12 for both algebras)
Step 3  General formula: B_g = J_short(1+ell_g)
        B_G2 = 12×4 = 48  ✓    B_F4 = 12×3 = 36  ✓
Step 4  Product identity:
        B_G2 x B_F4 = J_short^2 x (1+ell_G2)(1+ell_F4) = J_short^2 x 12
        = 12^3 = 1728 = j(i)  ✓
Step 5  Structural key: (1+ell_G2)(1+ell_F4) = 4×3 = 12 = J_short
        Exact integer arithmetic forced by Dynkin labels 3, 2.
Step 6  Specificity: (B3,F4) pair — B_B3 x B_F4 = 360 ≠ 8 = J_short(B3)^3
        (1+ell_B3)(1+ell_F4) = 9 ≠ 2 = J_short(B3).  Identity is pair-specific.
Step 7  Modular: E4(i)/eta(i)^8 = 12.0000000000 = J_short  ✓  (classical)
All assertions passed.
\end{verbatim}

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

The identity $j(i) = \BG\cdot\BF = 1728$ follows from a three-step chain.
First, Weyl-group transitivity on each root-length class forces
$\Bg(\hs,\hs) = \Js(1+\ell_\g)$ for any simple Lie algebra with equal
short and long root counts.
Second, both $G_2$ and $F_4$ share the same short-sector value $\Js=12$,
because both short-root subsystems restrict to the same $A_2$ geometry.
Third, the Dynkin labels $\ell_{G_2}=3$ and $\ell_{F_4}=2$ satisfy the
exact integer identity $(1+3)(1+2)=12=\Js$, making the product
$\Js^2\cdot\Js = \Js^3 = 12^3 = 1728$.
The result is real and exact, not an approximation,
but it is specific to the $(G_2,F_4)$ pair.
The appearance of $12$ in the classical modular identity
$E_4(i)/\eta(i)^8=12$ is noted as an open observation (P158-OP1).

%--------------------------------------------------------------------------
\bigskip
\noindent\textit{Copyright: Léon Fernando Vlegels. License: MIT. May 2026.}

\end{document}
