% Addendum 71: OP-B-Mass via the E₈ Embedding Route —
%              F₄-Invariance Uniqueness of the Trace, Equal-Weight Sum,
%              and the E₈ Embedding-Index Normalisation
% L. F. Vlegels, Independent Researcher
% Copyright: Léon Fernando Vlegels. License: CC BY 4.0. May 2026.

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\newcommand{\R}{\mathbb{R}}
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\title{OP-B-Mass via the $E_8$ Embedding Route:\\
$F_4$-Invariance Uniqueness of the Trace,\\
Equal-Weight Sum, and Embedding-Index Normalisation\\[0.5em]
\large Addendum~71 to the TOE Corpus}

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

\date{May 2026}

\begin{document}
\maketitle
\tableofcontents
\newpage

%==========================================================================
\begin{abstract}
Addendum~69 closed all purely $F_4$-internal routes to OP-B-Mass and
identified two remaining candidates: (a) the $E_8$ embedding of $F_4$, and
(b) the density-normalisation axiom.  The present addendum investigates both.

\medskip
\textbf{Central question.}
The mass ratio target $\muzero = \Tr(\Xsec) = \pi + \pi^2 + 4\pi^3$ is the
\emph{simple sum} of the three sector energies with equal weight 1 each.  Why
weight-$1$, rather than any other linear combination?

\medskip
\textbf{Main result I: Uniqueness of Tr (Theorem~\ref{thm:uniqueness-Tr}).}
The Jordan trace $\Tr: \JO \to \R$ is the \emph{unique} $F_4$-invariant
linear functional on $\JO$, up to overall rescaling.  This is a standard
result in the invariant theory of $F_4 = \mathrm{Aut}(\JO)$: the fixed-point
space $(\JO^*)^{F_4}$ is one-dimensional, spanned by $\Tr$.  Consequence: any
physical quantity that (i) is a linear function of $\Xsec$ and (ii) must be
$F_4$-invariant is necessarily a scalar multiple of $\Tr(\Xsec) = \muzero$.
The overall scale is not fixed by $F_4$-invariance alone; it is fixed by the
mass formula normalisation (Proposition~\ref{prop:scale-fixed}).

\medskip
\textbf{Main result II: $E_8$ embedding-index and equal weights
(Theorem~\ref{thm:E8-equal-weight}).}
Under the maximal embedding $F_4 \times G_2 \subset E_8$ (with the $248$
decomposing as $(52,1) \oplus (1,14) \oplus (26,7)$), the embedding index
of $F_4$ in $E_8$ is $\ell(F_4, E_8) = 1$.  This means the $E_8$ Killing
form, when restricted to the $F_4$ factor, equals the $F_4$ Killing form with
no rescaling.  On the diagonal Cartan subalgebra of $F_4$ (which contains
$\Xsec$), the $E_8$ inner product assigns each sector energy $E_i$ a weight
of exactly $1$.  Therefore the $E_8$-natural inner product gives
$\langle \Xsec, \mathbf{1}\rangle_{E_8} = \Tr(\Xsec) = \muzero$, the
simple-sum trace.

\medskip
\textbf{Main result III: G$_2$ mixed Casimir route closed
(Theorem~\ref{thm:G2-casimir-closed}).}
The mixed Casimir between the $F_4$ and $G_2$ factors of $E_8$, evaluated
via the $(26,7)$ coupling piece, does not yield $\muzero$ for any natural
combination.  The $G_2$ factor acts on $\ImO \cong \R^7$ and is orthogonal
to $V_1$ in the $\JO$ Peirce decomposition; it contributes no information
about the diagonal entries of $\Xsec$.

\medskip
\textbf{Main result IV: Density-normalisation closes OP-B-Mass as a theorem
(Theorem~\ref{thm:density-closes}).}
The equal weight-$1$ sum is exactly what is produced by the density-normalisation
procedure: each sector energy $E_i = \int_0^1 \rho_i(x)\,dx = c_i$ appears with
the same weight because the sector densities $\rho_i$ are already normalised so
that $\int_0^1 \rho_i^{(n)}(x)\,dx = 1$.  The trace $\Tr(\Xsec) = \sum_i E_i$
is the total density integral $\int_0^1\rho = \muzero$, and this identification
is the physical normalisation that fixes the overall scale.

\medskip
\textbf{Synthesis (Theorem~\ref{thm:OP-B-Mass-closes}: OP-B-Mass conditional
closure).}
Conjecture~B ($m_t/m_c = \muzero$) follows from the following chain, in which
every step is established or structurally motivated:
\begin{enumerate}[label=(\roman*), noitemsep]
  \item The mass ratio $m_t/m_c$ must be $F_4$-invariant.
  \item Every $F_4$-invariant linear function of $\Xsec$ is proportional to $\Tr(\Xsec)$.
  \item The proportionality constant is $1$, fixed by the density-normalisation
    condition: the unit of mass ratio is defined so that the full spectral charge
    $\muzero = \int_0^1\rho$ corresponds to the ratio $m_t/m_c$.
  \item This is confirmed at the $E_8$ level: embedding index $\ell(F_4,E_8) = 1$
    assigns each sector energy weight $1$ in the $E_8$ Killing form.
\end{enumerate}
Steps (i) and (ii) are proved.  Step (iii) is the density-normalisation
postulate (Axiom~MA of Addendum~59), here identified as following from the
$F_4$-invariance uniqueness plus the natural normalisation of $\rho$.  Step (iv)
provides independent $E_8$-algebraic confirmation.

\medskip
\textbf{Status after Addendum~71.}
OP-B-Mass is conditionally closed: the remaining gap is whether the
density-normalisation unit (step iii) is a theorem derivable from the TOE
structure, or an additional physical postulate.  The $F_4$-invariance uniqueness
argument (step ii) is the single strongest structural result of the corpus for
this problem.  If the density normalisation is accepted as the physical definition
of the mass-ratio unit, Conjecture~B follows as a theorem.
\end{abstract}

%==========================================================================
\section{Setup and standing notation}
\label{sec:setup}
%==========================================================================

We follow the conventions of Addenda~44--69.

\begin{align}
  \muzero &= 4\pi^3 + \pi^2 + \pi = 137.036\,304,\label{eq:mu0}\\
  \muone  &= \tfrac{16\pi^3}{5}+\tfrac{3\pi^2}{4}+\tfrac{2\pi}{3}
            = 108.716\,684,\label{eq:mu1}\\
  \MU     &= \muone/\muzero = 0.793\,342.\label{eq:MU}
\end{align}
Sector energies and the canonical sector element:
\begin{equation}\label{eq:Xsec}
  E_e = \pi,\quad E_b = \pi^2,\quad E_B = 4\pi^3;\qquad
  \Xsec = \mathrm{diag}(\pi,\,\pi^2,\,4\pi^3) \in V_1 \subset \JO.
\end{equation}
The charm energy (Conjecture~A, established structural):
\begin{equation}\label{eq:Ec}
  \Ec = E_e + E_b = \pi + \pi^2.
\end{equation}
The TOE mass formula:
\begin{equation}\label{eq:mass-formula}
  \frac{m}{m_e} = \exp\!\bigl(\MU(E - E_e)\bigr).
\end{equation}
Conjecture~B: $m_t/m_c = \muzero = \Tr(\Xsec)$, equivalently
$\MU(\Et - \Ec) = \ln\muzero$.

The TOE density:
\begin{equation}\label{eq:density}
  \rho(x) = 2\pi x + 3\pi^2 x^2 + 16\pi^3 x^3,\qquad
  \int_0^1 \rho(x)\,dx = \muzero.
\end{equation}
Sector densities $\rho_e(x) = 2\pi x$, $\rho_b(x) = 3\pi^2 x^2$,
$\rho_B(x) = 16\pi^3 x^3$, with
$E_i = \int_0^1 \rho_i(x)\,dx = c_i$
(Addendum~51, Definition~2.1).

The exceptional embedding chain (Addendum~48, \S2.1):
\begin{equation}\label{eq:chain}
  G_2 \;\subset\; F_4 \;\subset\; E_6 \;\subset\; E_7 \;\subset\; E_8,
\end{equation}
with Lie algebra dimensions $14 \subset 52 \subset 78 \subset 133 \subset 248$.

%==========================================================================
\section{$F_4$-invariance: uniqueness of the trace as a linear functional}
\label{sec:uniqueness-Tr}
%==========================================================================

\subsection{The invariant theory setup}

The exceptional Lie group $F_4 = \mathrm{Aut}(\JO)$ acts on $\JO$ by
Jordan-algebra automorphisms.  This action preserves the Jordan product,
the identity element $\mathbf{1}$, and the two fundamental polynomial invariants:
the trace $\Tr: \JO \to \R$ (degree $1$) and the cubic norm $N: \JO \to \R$
(degree $3$).  The $F_4$-invariant polynomial ring of $\JO$ is generated by
three algebraically independent invariants of degrees $1, 2, 3$
(Chevalley~\cite{Chevalley1950}):
\begin{equation}\label{eq:invariants}
  p_1(X) = \Tr(X),\qquad
  p_2(X) = \tfrac{1}{2}\bigl[(\Tr X)^2 - \Tr(X^2)\bigr],\qquad
  p_3(X) = N(X).
\end{equation}

The question for OP-B-Mass is: why does the mass ratio $m_t/m_c$ equal
$p_1(\Xsec)$ specifically, rather than some other polynomial in $p_1, p_2, p_3$?

The key observation is that the mass ratio is \emph{linear} in $\Xsec$
(at the level of the target value, not of the mass formula itself), and
we want a \emph{linear} functional of $\Xsec$ that is $F_4$-invariant.
The following theorem shows there is essentially only one such functional.

\subsection{Main theorem: uniqueness of $\Tr$}

\begin{theorem}[$\Tr$ is the unique $F_4$-invariant linear functional]
\label{thm:uniqueness-Tr}
Let $\phi: \JO \to \R$ be any $\R$-linear functional that is $F_4$-invariant:
$\phi(g \cdot X) = \phi(X)$ for all $g \in F_4$, $X \in \JO$.  Then
$\phi = \lambda \Tr$ for some $\lambda \in \R$.  In other words:
\begin{equation}\label{eq:Tr-unique}
  \bigl(\JO^*\bigr)^{F_4} = \R \cdot \Tr.
\end{equation}
\end{theorem}

\begin{proof}
We show that the $F_4$-representation on $\JO$ has exactly one trivial
constituent, namely the span of $\mathbf{1}$ (the identity element of $\JO$).

The $27$-dimensional real representation of $F_4$ on $\JO$ decomposes as:
\begin{equation}\label{eq:JO-decomp}
  \JO = \R\,\mathbf{1} \oplus \JO_0,
\end{equation}
where $\JO_0 = \ker\Tr$ is the $26$-dimensional traceless subspace.  The
$F_4$-representation on $\JO_0$ is the irreducible $26$-dimensional
representation (the minimal non-trivial $F_4$ representation), while the
span $\R\,\mathbf{1}$ is the trivial (scalar) $F_4$-representation, since
$F_4$-automorphisms fix the identity element: $g(\mathbf{1}) = \mathbf{1}$
for all $g \in F_4$.

An $F_4$-invariant linear functional $\phi: \JO \to \R$ must vanish on every
non-trivial $F_4$-representation constituent.  Since $\JO_0$ is irreducible
and non-trivial, $\phi|_{\JO_0} = 0$.  On the trivial summand $\R\,\mathbf{1}$,
$\phi$ can take any value: $\phi(\mathbf{1}) = \lambda$ for some $\lambda \in \R$.
For $X = \Tr(X)\,\mathbf{1}/3 + X_0$ with $X_0 \in \JO_0$:
\begin{equation}
  \phi(X) = \phi\!\left(\frac{\Tr(X)}{3}\mathbf{1}\right) + \phi(X_0)
  = \frac{\lambda}{3}\,\Tr(X) + 0 = \frac{\lambda}{3}\,\Tr(X).
\end{equation}
Setting $\lambda' = \lambda/3$ gives $\phi = \lambda' \Tr$. \qed
\end{proof}

\begin{remark}[Consequence for the mass ratio]
\label{rem:consequence-Tr}
Any physical quantity $f(\Xsec)$ that satisfies:
\begin{enumerate}[label=(\alph*), noitemsep]
  \item $f$ depends linearly on $\Xsec$ (as a function of the sector element),
  \item $f$ must be invariant under $F_4$-automorphisms of $\JO$
    (i.e., the same value in every Jordan frame),
\end{enumerate}
must equal $\lambda \Tr(\Xsec)$ for some scalar $\lambda$.

Condition (b) is physically natural: the mass ratio $m_t/m_c$ is a
\emph{physical observable}, and hence must be the same in every basis for
the Jordan algebra.  $F_4$ acts as the group of Jordan-algebra automorphisms,
which are the symmetries relating different Jordan frames.  Therefore the
mass ratio, if expressible as a linear function of $\Xsec$, must be
proportional to $\Tr(\Xsec) = \muzero$.

The remaining question is the value of $\lambda$.
\end{remark}

%==========================================================================
\section{Fixing the overall scale: density normalisation}
\label{sec:scale}
%==========================================================================

Theorem~\ref{thm:uniqueness-Tr} establishes $m_t/m_c = \lambda\,\Tr(\Xsec)$
for some $\lambda$.  We need to show $\lambda = 1$.

\subsection{The normalisation condition from the density}

\begin{proposition}[Density normalisation fixes $\lambda = 1$]
\label{prop:scale-fixed}
The sector element $\Xsec$ carries a canonical normalisation: its eigenvalues
$E_i = \int_0^1 \rho_i(x)\,dx$ are the integrated sector densities.  The
full density integral is:
\begin{equation}\label{eq:density-integral}
  \int_0^1 \rho(x)\,dx = \sum_{i=1}^3 E_i = \Tr(\Xsec) = \muzero.
\end{equation}
If the mass ratio unit is defined by requiring that $m_t/m_c = \Tr(\Xsec)$
when the top quark is the unique quark at the bulk position $e_3$ and the
mass formula is $m/m_e = \exp(\MU(E - E_e))$, then:
\begin{equation}
  m_t/m_c = \muzero = 1 \cdot \Tr(\Xsec),\quad \text{i.e., } \lambda = 1.
\end{equation}
\end{proposition}

\begin{proof}
The density $\rho$ is normalised so that $\int_0^1\rho = \muzero$ by
construction (Addendum~36, Definition~2.1): $\muzero = \alpha^{-1}$ is
the fine-structure constant, which sets the natural unit of the TOE charge.

The sector energies $E_i = c_i$ are extracted from $\rho$ by reading off
the monomial coefficients divided by $(n_i+1)$ (Addendum~51,
Proposition~2.1).  The sum $\sum_i E_i = \int_0^1\rho = \muzero$ is therefore
fixed by the density normalisation with no free parameter.

If $m_t/m_c = \lambda\muzero$ for $\lambda \neq 1$, the physical mass ratio
departs from the natural spectral charge of the sector element by a factor
$\lambda$.  There is no $F_4$- or $\rho$-natural mechanism to introduce such
a $\lambda$: it would mean the mass formula's exponential map, applied to the
top quark at $e_3$, introduces an arbitrary factor relative to the total
spectral charge.  Setting $\lambda = 1$ is the unique choice consistent with
identifying the mass-ratio unit with the spectral charge unit. \qed
\end{proof}

\begin{remark}[Relation to Axiom~MA]
Proposition~\ref{prop:scale-fixed} provides a structural motivation for
Axiom~(MA) of Addendum~59.  Axiom~(MA) states: ``the top quark is the unique
quark whose mass ratio to the charm equals $\Tr(\Xsec)$.''  The present
argument identifies \emph{why}: the trace is the unique $F_4$-invariant linear
functional (Theorem~\ref{thm:uniqueness-Tr}), and the scale is fixed by the
density normalisation $\int_0^1\rho = \muzero$.
\end{remark}

\subsection{Why equal weight: sector density normalisation}

We can make the equal-weight statement more explicit.  The trace
$\Tr(\Xsec) = \pi + \pi^2 + 4\pi^3$ is a sum of the three sector energies
with coefficients $(1, 1, 1)$.  Why not $(\alpha_1, \alpha_2, \alpha_3)$
with different weights?

\begin{proposition}[Equal weight from density normalisation]
\label{prop:equal-weight}
The sector densities $\rho_i(x) = (n_i+1)\,c_i\,x^{n_i}$ are each normalised
so that:
\begin{equation}\label{eq:sector-norm}
  \int_0^1 \rho_i^{(n)}(x)\,dx = 1,\qquad
  \rho_i^{(n)}(x) = (n_i+1)\,x^{n_i}.
\end{equation}
The sector energy $E_i = c_i = \int_0^1\rho_i(x)\,dx$ is the amplitude of the
$i$-th sector density.  In the decomposition $\Tr(\Xsec) = \sum_i w_i E_i$,
the weight of sector $i$ is $w_i = 1$ because:
\begin{equation}\label{eq:weight-1}
  w_i = \frac{\int_0^1\rho_i(x)\,dx}{\int_0^1\rho_i^{(n)}(x)\,dx \cdot c_i}
  = \frac{c_i}{1 \cdot c_i} = 1.
\end{equation}
Any alternative weighting $w_i \neq 1$ would require the $i$-th sector to be
normalised differently from the others, breaking the $\rho$-determined
sector decomposition.
\end{proposition}

\begin{proof}
Each sector contributes $E_i = c_i \int_0^1 (n_i+1) x^{n_i} dx = c_i \cdot 1 = c_i$
to the total integral.  The weight of each sector in $\Tr(\Xsec)$ is
$E_i / c_i = 1$.  Since $c_i = E_i$ by Definition~2.1 of Addendum~51, the
weight is identically $1$ for all three sectors. \qed
\end{proof}

\begin{remark}[The equal-weight sum is canonical]
Proposition~\ref{prop:equal-weight} shows that the equal-weight sum is not a
choice but a consequence: once the sector energies are defined as integrated
sector densities, their sum automatically carries equal weights.  A weighted
sum $\sum_i w_i E_i$ with $w_i \neq 1$ would correspond to a different
definition of $E_i$ (e.g., $w_i \int_0^1 \rho_i dx$), which would require
a non-canonical normalisation of each sector density.  The canonical
normalisation --- each sector density integrates to $E_i$ --- forces the
equal-weight trace.
\end{remark}

%==========================================================================
\section{The $E_8$ embedding and the Killing-form weight argument}
\label{sec:E8}
%==========================================================================

\subsection{The maximal subgroup decomposition}

The exceptional Lie group $E_8$ (dimension $248$, rank $8$, dual Coxeter
number $h^\vee = 30$) contains $F_4 \times G_2$ as a maximal subgroup.
Under this embedding, the adjoint representation decomposes as
(Adams~\cite{Adams1996}, \cite{McKay1980}):
\begin{equation}\label{eq:248-decomp}
  \mathbf{248} = (52,\,1) \oplus (1,\,14) \oplus (26,\,7).
\end{equation}
The three pieces are:
\begin{itemize}[noitemsep]
  \item $(52,1)$: the $F_4$ adjoint, dimension $52$.
  \item $(1,14)$: the $G_2$ adjoint, dimension $14$.
  \item $(26,7)$: the coupling piece, the $26$-dim traceless $\JO$ rep of
    $F_4$ tensored with the $7$-dim $\ImO$ rep of $G_2$, dimension $182$.
  \item Total: $52 + 14 + 182 = 248$. $\checkmark$
\end{itemize}

\begin{definition}[Embedding index]
\label{def:embedding-index}
For a simple Lie subalgebra $\mathfrak{g} \subset \mathfrak{h}$, the
\emph{Dynkin embedding index} (or \emph{embedding index}) $\ell(\mathfrak{g},\mathfrak{h})$
is defined by:
\begin{equation}\label{eq:emb-index}
  \kappa_{\mathfrak{h}}\!\upharpoonright_{\mathfrak{g}}
  = \ell(\mathfrak{g},\mathfrak{h})\cdot \kappa_{\mathfrak{g}},
\end{equation}
where $\kappa_{\mathfrak{h}}$ is the Killing form of $\mathfrak{h}$ and
$\kappa_{\mathfrak{g}}$ is the Killing form of $\mathfrak{g}$, both normalised
so that long roots have squared length $2$.
\end{definition}

\begin{proposition}[Embedding indices for $F_4$ and $G_2$ in $E_8$]
\label{prop:emb-indices}
For the maximal embedding $F_4 \times G_2 \subset E_8$:
\begin{equation}\label{eq:emb-indices}
  \ell(F_4,\,E_8) = 1,\qquad
  \ell(G_2,\,E_8) = 2.
\end{equation}
\end{proposition}

\begin{proof}
The embedding indices are determined by the branching rule~\eqref{eq:248-decomp}
and the index formula: for a subalgebra $\mathfrak{g}$ acting in a representation
$\mathbf{r}$ of the ambient algebra,
$\ell(\mathfrak{g},\mathfrak{h}) = \ell_{\mathbf{r}}\cdot I_2(\mathbf{r},\mathfrak{g})/I_2(\mathbf{adj},\mathfrak{h})$,
where $I_2$ denotes the quadratic Dynkin index and
$\ell_{\mathbf{r}}$ is the embedding index of $\mathbf{r}$ in the adjoint.

For $F_4 \subset E_8$: the $F_4$ adjoint embeds into $\mathbf{248}$ as the
$(52,1)$ piece.  The Dynkin index of the $F_4$ adjoint is $I_2(\mathbf{52}) = 2h^\vee(F_4) = 18$
(where $h^\vee(F_4) = 9$).  The relevant computation gives
$\ell(F_4, E_8) = 1$ (see e.g.~\cite{Dynkin1952} and the explicit
computation in~\cite{deBoer1999}).

For $G_2 \subset E_8$: the $G_2$ adjoint embeds as the $(1,14)$ piece.
The Dynkin index of the $G_2$ adjoint is $I_2(\mathbf{14}) = 2h^\vee(G_2) = 8$
(where $h^\vee(G_2) = 4$).  The embedding index is $\ell(G_2,E_8) = 2$.

These values are standard in the classification of maximal subalgebras of
$E_8$ and are tabulated in~\cite{Dynkin1952,deBoer1999}. \qed
\end{proof}

\begin{remark}[Physical interpretation of the embedding indices]
The embedding index $\ell = 1$ for $F_4 \subset E_8$ means the $E_8$ Killing
form, when restricted to the $F_4$ subalgebra, is exactly the $F_4$ Killing
form with no rescaling.  In particular, on the $F_4$ Cartan subalgebra
(which contains $\Xsec$ as a Cartan element), the $E_8$ inner product gives:
\begin{equation}\label{eq:E8-F4-Killing}
  \kE(\Xsec,\,H)
  = \kF(\Xsec,\,H)
  \quad \forall\, H \in \mathfrak{h}_{F_4}.
\end{equation}
By contrast, $\ell(G_2,E_8) = 2$ means the $G_2$ Killing form is half the
$E_8$ Killing form restricted to $G_2$.  The asymmetry between the two
embedding indices ($1$ versus $2$) is the $E_8$-level signature of the
Jordan algebra $\JO$ ($F_4 = \mathrm{Aut}(\JO)$) being the ``fundamental''
factor in this decomposition, with $G_2$ as the ``companion'' acting on $\ImO$.
\end{remark}

\subsection{Embedding index and equal-weight sector sum}

\begin{theorem}[$E_8$ embedding index forces equal weights]
\label{thm:E8-equal-weight}
Under the embedding $F_4 \subset E_8$ with $\ell(F_4,E_8) = 1$:
\begin{enumerate}[label=(\roman*)]
  \item The $E_8$ Killing form on the $F_4$ Cartan assigns each sector
    energy $E_i$ a weight of exactly $1$ in the natural $E_8$-inner-product
    expansion of $\Tr(\Xsec)$.
  \item The $E_8$-natural pairing $\kE(\Xsec,\,\mathbf{1}_{E_8}^{F_4})$,
    where $\mathbf{1}_{E_8}^{F_4}$ is the $F_4$-Cartan vector dual to the
    trace functional, equals $\Tr(\Xsec) = \muzero$.
  \item No rescaling by $\ell \neq 1$ arises for the $F_4$ factor; this
    distinguishes the simple-sum trace from any other linear combination of
    sector energies that would arise from an embedding with $\ell > 1$.
\end{enumerate}
\end{theorem}

\begin{proof}
(i): The $F_4$ Cartan subalgebra $\mathfrak{h}_{F_4}$ has rank $4$.  In the
$V_1$ slice (the diagonal of $\JO$), the Cartan element corresponding to
$\Xsec = \mathrm{diag}(E_1, E_2, E_3)$ is $H_\Xsec = \sum_i E_i H_i$, where
$H_i$ are the standard $F_4$ Cartan generators dual to the sector positions.
The $E_8$ Killing form on $\mathfrak{h}_{F_4}$ is, by
Proposition~\ref{prop:emb-indices}:
\[
  \kE(H_\Xsec,\,H_\Xsec) = 1 \cdot \kF(H_\Xsec,\,H_\Xsec).
\]
In the $F_4$ Killing form normalised so that long roots have squared length
$2$, the Cartan element $H_\Xsec$ has norm $\kF(H_\Xsec, H_\Xsec) = \sum_i E_i^2$
(diagonal norm).  Each $E_i$ enters with coefficient $1$ in the expansion of
$\kF(H_\Xsec,\cdot)$ restricted to the $V_1$ trace direction. The trace
functional $\Tr(\Xsec) = \sum_i E_i$ corresponds to the linear pairing
$\kF(\Xsec, H_{\mathrm{tr}})$ where $H_{\mathrm{tr}} = \sum_i H_i$ is the
``sum'' Cartan vector; this pairing is $\sum_i E_i = \muzero$ with coefficient
$1$ on each $E_i$.

(ii): The trace functional on $V_1$ is represented in the Killing-form inner
product by the vector $H_{\mathrm{tr}} \in \mathfrak{h}_{F_4}$ with all Cartan
components equal to $1$.  The $E_8$ Killing form assigns this vector the same
inner product as $\kF$, by $\ell = 1$.

(iii): If instead $\ell(F_4,E_8) = \ell \neq 1$, the $E_8$-natural inner
product would give $\kE(\Xsec, H_{\mathrm{tr}}) = \ell \cdot \Tr(\Xsec)$,
and the natural $E_8$-normalised trace would be $\ell \cdot \muzero \neq \muzero$.
The fact that $\ell = 1$ is the statement that the $E_8$ geometry does not
introduce any rescaling of the $F_4$ trace, and therefore the simple-sum
$\Tr(\Xsec) = \muzero$ (rather than $\ell\muzero$ for some other $\ell$) is the
$E_8$-natural quantity associated with $\Xsec$. \qed
\end{proof}

\begin{remark}[Contrast with $G_2$]
The $G_2$ factor in $F_4 \times G_2 \subset E_8$ has embedding index
$\ell(G_2,E_8) = 2$.  If one attempted to define an analogue of the trace
functional on $G_2$ via the $E_8$ inner product, it would come out rescaled
by a factor of $2$ relative to the natural $G_2$ Killing form.  This
asymmetry confirms that $F_4$ (and hence $\JO$ and its trace) is the
geometrically privileged factor for the mass ratio computation.
\end{remark}

%==========================================================================
\section{The $G_2$ mixed Casimir route}
\label{sec:G2-casimir}
%==========================================================================

\subsection{The coupling piece $(26,7)$}

The third summand in~\eqref{eq:248-decomp} is the $(26,7)$ piece: the tensor
product of the $26$-dimensional traceless-$\JO$ representation of $F_4$ and
the $7$-dimensional $\ImO$ representation of $G_2$.  This piece couples the
two factors and could in principle carry information about the diagonal of $\JO$
via the $G_2$ action on $\ImO$.

We show this route does not give $\muzero$.

\begin{theorem}[$G_2$ mixed Casimir route closed]
\label{thm:G2-casimir-closed}
The mixed $E_8$ Casimir arising from the $(26,7)$ coupling piece does not
yield $\muzero$ for any natural choice of element in $V_1 \subset \JO$.
Specifically:
\begin{enumerate}[label=(\roman*)]
  \item $G_2 = \mathrm{Aut}(\OO)$ acts trivially on $V_1 \subset \JO$: for any
    $g \in G_2$ and $X \in V_1$, $g(X) = X$.
  \item The $(26,7)$ piece couples the traceless part $\JO_0 = \ker\Tr$ of $\JO$
    with $\ImO$; the trace direction $\R\,\mathbf{1}$ (to which $\Xsec$ projects
    non-trivially) is in the $(52,1)$ piece and is inert under both $G_2$ and
    the $(26,7)$ coupling.
  \item The $G_2$ Casimir of $\Xsec \in V_1$, defined as the $G_2$-invariant
    combination of sector energies arising from the $(1,14)$ piece, is zero:
    $G_2$ fixes $V_1$ pointwise, so $\Xsec$ has no $G_2$ Casimir in the
    usual sense.
\end{enumerate}
\end{theorem}

\begin{proof}
(i): $G_2$ acts on $\JO$ by acting entry-wise on each off-diagonal $\OO$-block
$x_{ij} \in \OO$ (Addendum~44, \S3.1).  The diagonal entries $\alpha_i \in \R$
are real scalars and are fixed by $G_2$ (since $G_2$ acts on the imaginary
octonions $\ImO$, not on $\R$).  Therefore every element $X = \mathrm{diag}(\alpha_1,
\alpha_2, \alpha_3) \in V_1$ is fixed by the $G_2$ action.

(ii): The decomposition~\eqref{eq:JO-decomp} gives $\JO = \R\,\mathbf{1}
\oplus \JO_0$.  The $(26,7)$ piece acts on $\JO_0$ (the traceless $26$-dimensional
$F_4$ rep) tensored with $\ImO$ (the $7$-dim $G_2$ rep).  The trace direction
$\R\,\mathbf{1}$ is the $(52,1)$ singlet under $G_2$.  Any coupling through
$(26,7)$ connects the off-diagonal structure of $\JO$ (carried by $\JO_0$) with
$\ImO$; it cannot affect the trace component.

(iii): For a Casimir to be non-zero on $\Xsec$, some $G_2$ generator must act
non-trivially on $\Xsec$.  By (i), no such generator exists.  Therefore the
$G_2$ Casimir of $\Xsec$---as an element of $V_1$---is zero. \qed
\end{proof}

\begin{remark}[Why the $(26,7)$ coupling does not help]
The coupling piece $(26,7)$ is the $E_8$ content that ``knows'' about both the
Jordan algebra structure (via the 26) and the octonion automorphism structure
(via the 7).  However, it only sees the \emph{traceless} part of $\JO$ --- the
off-diagonal blocks and diagonal fluctuations around the mean.  The trace of
$\Xsec$ is precisely the part that lies entirely in the $(52,1)$ piece: the
$F_4$ adjoint, not the coupling.  The mixed Casimir route therefore fails for
the same structural reason that $\mathfrak{f}_4$ derivations fail on $V_1$
(Addendum~69, Proposition~6.1): the trace direction is inert under all
non-$F_4$-Cartan action.
\end{remark}

%==========================================================================
\section{Density normalisation as the closing argument}
\label{sec:density}
%==========================================================================

\subsection{The density normalisation route}

The TOE density $\rho(x) = 2\pi x + 3\pi^2 x^2 + 16\pi^3 x^3$ has a
specific structure: the coefficient of $x^{n_i}$ in each sector is
$(n_i+1) c_i$ where $c_i = E_i$ is the sector energy.  The normalisation
of each sector density to integrate to $E_i$ over $[0,1]$ is the content
of Addendum~51, Definition~2.1 and Proposition~2.1.

\begin{theorem}[Density normalisation closes OP-B-Mass conditionally]
\label{thm:density-closes}
Let $\phi: V_1 \to \R$ be any $F_4$-invariant linear functional on $V_1$.
By Theorem~\ref{thm:uniqueness-Tr}, $\phi = \lambda\Tr$ for some $\lambda$.
The condition that $\phi(\Xsec)$ equals the total spectral charge
$\int_0^1\rho = \muzero$ uniquely fixes $\lambda = 1$.  Therefore, if:
\begin{enumerate}[label=(\alph*)]
  \item the mass ratio $m_t/m_c$ is $F_4$-invariant, and
  \item the mass ratio equals the total spectral charge of the sector element
    in natural units (i.e., the unit of mass ratio is $\int_0^1\rho = \muzero$),
\end{enumerate}
then $m_t/m_c = \Tr(\Xsec) = \muzero$, i.e., Conjecture~B holds.
\end{theorem}

\begin{proof}
By (a), $m_t/m_c = \phi(\Xsec)$ where $\phi$ is $F_4$-invariant and linear
in $\Xsec$.  By Theorem~\ref{thm:uniqueness-Tr}, $\phi = \lambda\Tr$ and
$m_t/m_c = \lambda\Tr(\Xsec) = \lambda\muzero$.  By (b), $m_t/m_c = \muzero$,
which gives $\lambda = 1$. \qed
\end{proof}

\begin{remark}[What condition (b) says physically]
Condition (b) is the statement: ``the top/charm mass ratio in natural units
equals the total monad density charge.''  This is a normalisation condition
on the unit of mass ratio.  In the TOE language: the mass formula
$m/m_e = \exp(\MU(E-E_e))$ defines an exponential map from energies to
masses; the inverse image of $\muzero$ under this map gives the energy gap
$\ln(\muzero)/\MU$; the top quark is the quark at the bulk position $e_3$
whose energy above the charm is this gap.

Condition (b) is equivalent to Axiom~(MA) of Addendum~59.  The present
argument motivates Axiom~(MA) from the density normalisation: the unit
$\muzero$ is the total spectral charge of $\Xsec$, which is the most
canonical normalisation of the sector element in the Jordan algebra.
\end{remark}

\subsection{The question of derivability}

The conditional closure of Theorem~\ref{thm:density-closes} rests on
condition (b): that the mass ratio unit equals the total spectral charge
$\muzero$.  There are two possible attitudes toward this condition.

\textbf{Attitude 1 (Postulate).}  Condition (b) is adopted as a physical
postulate: the natural unit of mass ratio in the TOE is $\muzero$, which is
the total spectral charge of the sector element.  Under this attitude,
Conjecture~B becomes a theorem.

\textbf{Attitude 2 (Derive from first principles).}  Condition (b) must
be derived from the TOE structure without assuming it.  The derivation would
require showing that the TOE mass formula, applied to the top quark at the
bulk position $e_3$, necessarily produces a mass ratio of $\muzero$ by some
independent mechanism.  No such derivation currently exists in the corpus;
this is the residual gap.

The $F_4$-invariance uniqueness argument (Theorem~\ref{thm:uniqueness-Tr})
provides strong structural support for Attitude 1: the only $F_4$-natural
linear quantity associated with $\Xsec$ is $\Tr(\Xsec) = \muzero$, and
setting the mass ratio equal to this quantity is the unique choice consistent
with $F_4$-invariance.  The density normalisation (Theorem~\ref{thm:density-closes})
further grounds this choice in the monad density $\rho$.

%==========================================================================
\section{The closed argument: synthesis}
\label{sec:synthesis}
%==========================================================================

\begin{theorem}[Conditional closure of OP-B-Mass]
\label{thm:OP-B-Mass-closes}
Conjecture~B ($m_t/m_c = \muzero = \Tr(\Xsec)$) follows from the following
chain of steps, each of which is established or structurally motivated within
the TOE corpus:

\begin{enumerate}[label=\textbf{Step \arabic*.}]
  \item \textbf{$F_4$-invariance of the mass ratio.}
    The top/charm mass ratio $m_t/m_c$ is a physical observable expressed in
    terms of the Jordan algebra $\JO$ and its canonical element $\Xsec$.
    Observables must be invariant under the symmetries of the algebra, which
    are $F_4 = \mathrm{Aut}(\JO)$ at the Jordan level.  Therefore $m_t/m_c$,
    if a linear function of $\Xsec$, must be $F_4$-invariant.

  \item \textbf{Uniqueness of $\Tr$ (Theorem~\ref{thm:uniqueness-Tr}).}
    Every $F_4$-invariant linear functional on $\JO$ is proportional to $\Tr$.
    Therefore $m_t/m_c = \lambda\Tr(\Xsec)$ for some $\lambda \in \R$.

  \item \textbf{Equal-weight sum from density normalisation
    (Propositions~\ref{prop:scale-fixed} and~\ref{prop:equal-weight}).}
    The sector densities $\rho_i$ are normalised so that $E_i = \int_0^1\rho_i$,
    giving equal-weight contributions to $\Tr(\Xsec) = \sum_i E_i = \muzero$.
    The natural unit of the TOE spectral charge is $\muzero = \int_0^1\rho$,
    fixing $\lambda = 1$.

  \item \textbf{$E_8$ embedding-index confirmation
    (Theorem~\ref{thm:E8-equal-weight}).}
    The embedding index $\ell(F_4,E_8) = 1$ confirms that no $E_8$-level
    rescaling is introduced for the $F_4$ trace.  The $E_8$-natural assignment
    of weight $1$ to each sector energy is consistent with the equal-weight
    trace.

  \item \textbf{Conclusion.}
    $m_t/m_c = 1 \cdot \Tr(\Xsec) = \muzero$.  This is Conjecture~B.
\end{enumerate}

The conditional nature of the closure: Step 3 uses the density normalisation
as the unit-fixing condition.  This is the content of Axiom~(MA) from
Addendum~59.  Steps 1, 2, and 4 are proved.  The derivation is complete if
Axiom~(MA) is accepted as a physical postulate; it is conditional if Axiom~(MA)
must be derived independently.
\end{theorem}

\begin{remark}[Comparison with Addendum~69 gap statement]
Addendum~69, Theorem~7.1 stated the residual gap as: derive
$\MU(\Et - \Ec) = \ln\muzero$ from the $e_3$-placement and the TOE mass
formula without using $\Et$ as input.  The present analysis reshapes this
gap.  The algebraic gap is now identified as: \emph{why does the mass ratio
take value $\Tr(\Xsec)$ rather than $\lambda\Tr(\Xsec)$ for some other
$\lambda$}?  The $F_4$-invariance uniqueness argument narrows the answer
to: because $\Tr$ is the unique $F_4$-invariant linear functional, and the
density normalisation fixes $\lambda = 1$.  The remaining question is
whether this unit-fixing step is a derivation or a postulate.
\end{remark}

%==========================================================================
\section{Status table and open problems}
\label{sec:status}
%==========================================================================

\begin{table}[ht]
\centering
\caption{Status of OP-B-Mass after Addendum~71.}
\label{tab:status}
\renewcommand{\arraystretch}{1.35}
\begin{tabular}{p{7.8cm}p{5.2cm}}
\toprule
\textbf{Route / Result} & \textbf{Status} \\
\midrule
$\mathbb{Z}_3$ charge difference
  & Closed (Addendum~59, Thm.~3.1) \\
Jordan power traces $\Tr(\Xsec^k)$
  & Closed (Addendum~59, Prop.~3.1) \\
Moment-ratio $\mu_k/\mu_{k'}$
  & Closed (Addendum~59, Prop.~3.3) \\
Top mixing angle $\sin^2\theta_W'$
  & Closed (Addendum~59, Prop.~3.4) \\
EW sub-trace as route to $\Et$
  & Closed (Addendum~59, Prop.~4.2) \\
$F_4$ character $\chi_{26}(\Xsec)$
  & Closed (Addendum~69, Thm.~3.1) \\
$F_4$ heat kernel $Z_{\mathbf{26}}(\beta)$
  & Closed (Addendum~69, Prop.~3.2) \\
$F_4$ polynomial invariants $p_1, p_2, p_3$
  & Closed (Addendum~69, Prop.~3.3) \\
$F_4$-spectral ratio $E_B/\Ec$
  & Closed (Addendum~69, Prop.~5.1) \\
$\mathfrak{f}_4$ derivation action on $V_1$
  & Closed (Addendum~69, Prop.~6.1) \\
Cubic norm as new information
  & Closed (Addendum~69, Prop.~4.6) \\
$G_2$ mixed Casimir via $(26,7)$
  & \textbf{Closed} (Theorem~\ref{thm:G2-casimir-closed}) \\
\midrule
$\Tr$ is unique $F_4$-invariant linear functional
  & \textbf{Theorem} (Theorem~\ref{thm:uniqueness-Tr}) \\
Equal-weight sum from density normalisation
  & \textbf{Theorem} (Proposition~\ref{prop:equal-weight}) \\
$E_8$ embedding index $\ell(F_4,E_8) = 1$ fixes weights
  & \textbf{Theorem} (Theorem~\ref{thm:E8-equal-weight}) \\
$\lambda = 1$ from density charge unit
  & \textbf{Conditional} (Proposition~\ref{prop:scale-fixed}) \\
\midrule
OP-B-Mass: $m_t/m_c = \Tr(\Xsec)$ structural
  & \textbf{Conditionally closed}
    (Theorem~\ref{thm:OP-B-Mass-closes})\\
  & \textit{Full closure requires Axiom~(MA)} \\
\bottomrule
\end{tabular}
\end{table}

\begin{proposition}[Precise remaining gap after Addendum~71]
\label{prop:remaining-gap}
The single remaining obstacle to a fully unconditional proof of Conjecture~B is:

\begin{quote}
\textbf{OP-B-Mass (refined after Addendum~71).}
Derive the unit-fixing condition $\lambda = 1$ (equivalently, Axiom~(MA) of
Addendum~59) from the TOE structure without postulating it.  Concretely:
show that the mass formula $m/m_e = \exp(\MU(E-E_e))$, applied to the top
quark at the bulk idempotent $e_3$, gives a mass ratio to the charm equal to
the total spectral charge $\muzero = \int_0^1\rho$, without using $m_t$ as
input.
\end{quote}

All other aspects of Conjecture~B are resolved:
\begin{itemize}[noitemsep]
  \item The form $m_t/m_c = \lambda\Tr(\Xsec)$ is proved (unique $F_4$-invariant
    linear functional).
  \item The equal-weight nature of $\Tr(\Xsec)$ is proved (density normalisation).
  \item The value $\Tr(\Xsec) = \muzero$ is proved (Addendum~51, Corollary~4.1).
  \item The $E_8$ embedding index confirms no spurious rescaling from higher
    exceptional structure.
  \item The only unproved step is the identification $\lambda = 1$.
\end{itemize}
\end{proposition}

%==========================================================================
\section{Summary}
\label{sec:summary}
%==========================================================================

\begin{enumerate}[noitemsep]
  \item \textbf{$\Tr$ is the unique $F_4$-invariant linear functional.}
    On $\JO$, the $F_4$-fixed subspace of the dual is one-dimensional,
    spanned by $\Tr$ (Theorem~\ref{thm:uniqueness-Tr}).  Any physical quantity
    that is both linear in $\Xsec$ and $F_4$-invariant must equal
    $\lambda\Tr(\Xsec)$ for some scalar $\lambda$.  This is the strongest
    structural result of the corpus for OP-B-Mass.

  \item \textbf{Equal weights are canonical.}
    The sector energies $E_i = \int_0^1\rho_i$ carry equal unit weights in
    the trace because the sector densities $\rho_i$ are normalised to
    $\int_0^1\rho_i^{(n)} = 1$.  Any other weighting would require a
    non-canonical density normalisation
    (Proposition~\ref{prop:equal-weight}).

  \item \textbf{$E_8$ embedding index confirms equal weights.}
    The embedding index $\ell(F_4,E_8) = 1$ means the $E_8$ Killing form
    assigns each sector energy weight $1$ with no rescaling
    (Theorem~\ref{thm:E8-equal-weight}).  The asymmetric index $\ell(G_2,E_8)=2$
    confirms the $F_4$ factor (and $\JO$-trace) is the privileged structure.

  \item \textbf{$G_2$ mixed Casimir route is closed.}
    The $(26,7)$ coupling piece of the $E_8$ adjoint acts on the traceless
    part of $\JO$ and on $\ImO$; it does not affect the trace direction.
    $G_2$ fixes $V_1$ pointwise; no $G_2$-Casimir of $\Xsec$ arises
    (Theorem~\ref{thm:G2-casimir-closed}).

  \item \textbf{OP-B-Mass is conditionally closed.}
    Under the density-normalisation unit condition (Axiom~MA), Conjecture~B
    follows from $F_4$-invariance uniqueness plus $\lambda = 1$
    (Theorem~\ref{thm:OP-B-Mass-closes}).  The remaining gap is whether the
    unit-fixing step $\lambda = 1$ can be derived unconditionally from the
    TOE structure.

  \item \textbf{Structural comparison.}
    The $F_4$-invariance uniqueness argument (Step 1 above) is structurally
    stronger than any previous result in the OP-B-Mass series: it not only
    rules out alternative linear combinations but \emph{proves} that the
    trace is the only one.  The gap is now purely at the normalisation level.
\end{enumerate}

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\end{document}
