% Addendum 69: OP-B-Mass via the F₄ Cubic Norm ---
%              Jordan Exponential, Cubic Norm Identity, and the
%              Remaining Step to Conjecture B Closure
% L. F. Vlegels, Independent Researcher
% Copyright: Léon Fernando Vlegels. License: CC BY 4.0. May 2026.

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\newcommand{\R}{\mathbb{R}}
\newcommand{\C}{\mathbb{C}}
\newcommand{\OO}{\mathbb{O}}
\newcommand{\JO}{J_3(\OO)}
\newcommand{\muzero}{\mu_0}
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\title{OP-B-Mass via the $F_4$ Cubic Norm:\\
Jordan Exponential, the Identity
$\ln N(\exp_J(\Xsec)) = \Tr(\Xsec)$,\\
and the Remaining Step to Closure\\[0.5em]
\large Addendum~69 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~59 identified the $F_4$-representation route --- specifically, the
cubic norm and trace as the two fundamental $F_4$-invariants of $\JO$ --- as
the most structurally natural remaining path to close OP-B-Mass (Conjecture~B:
$m_t/m_c = \Tr(\Xsec) = \muzero$).  The present addendum carries out this
route in full and reaches a precise verdict.

\medskip
\textbf{Main algebraic identity (Theorem~\ref{thm:cubic-norm-log}).}
The Jordan exponential $\exp_J(\Xsec) \in \JO$ satisfies:
\begin{equation}
  \ln \NJ\!\bigl(\exp_J(\Xsec)\bigr) = \Tr_{\JO}(\Xsec) = \muzero.
\end{equation}
This identity is exact, algebraic, and $F_4$-natural: the cubic norm $\NJ$
and the trace $\Tr$ are the two fundamental $F_4$-invariant polynomials
on $\JO$, and the identity above expresses a canonical relation between them
via the Jordan exponential on the diagonal subspace $V_1$.

\medskip
\textbf{$F_4$ character evaluation (Theorem~\ref{thm:character}).}
The character of the $26$-dimensional representation of $F_4$ (the unique
irreducible constituent of $\JO \otimes_\R \C$ complementary to the
identity summand) evaluated at $\Xsec$ as a Cartan element does not equal
$\muzero$ or $m_t/m_c$.  The character is a sum over short roots and zero
weights; it does not factor as a ratio that recovers the top-to-charm mass
ratio.

\medskip
\textbf{$F_4$ heat kernel (Proposition~\ref{prop:heat-kernel}).}
The $F_4$ heat kernel $Z(\beta, \Xsec) = \Tr_{\mathbf{26}}\,e^{-\beta \Xsec}$
evaluated at $\beta = 1$ does not equal $\muzero$.  At $\beta = 1/\MU$ it
reduces to a sum of exponentials with no clean closed form in terms of the
sector energies.

\medskip
\textbf{What the cubic norm identity does accomplish.}
The identity $\ln \NJ(\exp_J(\Xsec)) = \muzero = \Tr(\Xsec)$ shows that the
mass ratio target $\muzero$ arises as the log-cubic-norm of a canonical $F_4$
object.  Combined with Conjecture~A ($\Ec = \pi + \pi^2$, structural), the
cubic norm identity reduces OP-B-Mass to a single residual claim: that the
top quark energy $\Et$ satisfies $\MU(\Et - \Ec) = \ln \NJ(\exp_J(\Xsec))$, i.e.,
the top-to-charm energy gap \emph{equals} the log-cubic-norm of the Jordan
exponential of the canonical sector element.  This is the irreducible
remaining step.

\medskip
\textbf{Precise remaining gap (Theorem~\ref{thm:residual-gap}).}
After all routes explored in Addenda~44--69, the single remaining obstacle
to Conjecture~B is: derive the identification
\[
  \MU \cdot (\Et - \Ec) = \ln \NJ\!\bigl(\exp_J(\Xsec)\bigr)
\]
from the placement of the top quark at the bulk diagonal position $e_3 \in V_1$
and the $F_4$-invariant structure of $\JO$, without using $\Et$ as input.
The cubic norm identity removes all \emph{numerical coincidence} from the
problem: the right-hand side is $\muzero$ by an exact algebraic theorem.
The gap is now \emph{purely structural}: why does the top quark energy gap
above the charm equal the log-cubic-norm of the Jordan exponential of the
canonical sector element?
\end{abstract}

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

We follow the conventions of Addenda~44--59.

\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: $E_e = \pi$, $E_b = \pi^2$, $E_B = 4\pi^3$.
The canonical sector element:
\begin{equation}
  \Xsec = \mathrm{diag}(\pi,\,\pi^2,\,4\pi^3) \in V_1 \subset \JO,
  \qquad \Tr(\Xsec) = \muzero.
\end{equation}
The charm energy (Conjecture~A, Addendum~45, Proposition~3.1; structural):
\begin{equation}
  \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 (Addendum~52, Theorem~2.1; Addendum~59, Theorem~5.1):
\begin{equation}\label{eq:gap-form}
  \Et - \Ec = \frac{\ln\muzero}{\MU} = \frac{\ln(\Tr(\Xsec))}{\MU}.
\end{equation}

The two fundamental $F_4$-invariant polynomials on $\JO$ (cf.\ McCrimmon~\cite{McCrimmon},
Springer~\cite{Springer63}):
\begin{itemize}[noitemsep]
  \item \textbf{Trace}: $\Tr: \JO \to \R$, linear, $F_4$-invariant.
  \item \textbf{Cubic norm}: $\NJ: \JO \to \R$, degree-$3$ polynomial,
    $F_4$-invariant.
\end{itemize}
For a diagonal element $X = \mathrm{diag}(a,b,c)$:
\begin{equation}\label{eq:diag-invariants}
  \Tr(X) = a + b + c, \qquad \NJ(X) = abc.
\end{equation}
The cubic norm is the Jordan-algebraic analogue of the determinant: for
$X \in \JO$, $\NJ(X) = \tfrac{1}{6}(\Tr X)^3 - \tfrac{1}{2}\Tr(X)\,\Tr(X^2) + \tfrac{1}{3}\Tr(X^3)$,
which reduces to $abc$ on the diagonal.

%==========================================================================
\section{The Jordan exponential and its cubic norm}
\label{sec:jordan-exp}
%==========================================================================

\subsection{Jordan exponential on $V_1$}

\begin{definition}[Jordan exponential on $V_1$]
\label{def:jordan-exp}
For $X = \mathrm{diag}(a,b,c) \in V_1 \subset \JO$, the \emph{Jordan
exponential} is defined by the functional calculus in the Jordan algebra:
\begin{equation}\label{eq:jordan-exp-def}
  \exp_J(X) = \sum_{n=0}^{\infty} \frac{X^{\circ n}}{n!}
\end{equation}
where $X^{\circ n}$ denotes the $n$-th Jordan power.  On the diagonal
subspace $V_1$, Jordan powers are ordinary powers (since $V_1$ is
associative and commutative as a sub-algebra):
\begin{equation}
  \exp_J\!\bigl(\mathrm{diag}(a,b,c)\bigr)
  = \mathrm{diag}\!\bigl(e^a,\,e^b,\,e^c\bigr).
\end{equation}
\end{definition}

For $\Xsec = \mathrm{diag}(\pi, \pi^2, 4\pi^3)$:
\begin{equation}\label{eq:expJ-Xsec}
  \exp_J(\Xsec)
  = \mathrm{diag}\!\bigl(e^\pi,\,e^{\pi^2},\,e^{4\pi^3}\bigr).
\end{equation}

\begin{remark}[$\exp_J(\Xsec)$ as an $F_4$-orbit representative]
The element $\exp_J(\Xsec) \in \JO$ has trace
$\Tr(\exp_J(\Xsec)) = e^\pi + e^{\pi^2} + e^{4\pi^3}$,
which is not a simple quantity.  However, its $F_4$-invariant cubic norm
has a remarkably clean form, as the next theorem shows.
\end{remark}

\subsection{The cubic norm identity}

\begin{theorem}[Cubic norm of the Jordan exponential]
\label{thm:cubic-norm-log}
Let $\Xsec = \mathrm{diag}(\pi, \pi^2, 4\pi^3) \in V_1 \subset \JO$.  Then:
\begin{equation}\label{eq:cubic-norm-main}
  \NJ\!\bigl(\exp_J(\Xsec)\bigr) = e^{\muzero} = e^{\Tr(\Xsec)}.
\end{equation}
Equivalently:
\begin{equation}\label{eq:log-cubic-norm}
  \ln \NJ\!\bigl(\exp_J(\Xsec)\bigr) = \Tr(\Xsec) = \muzero.
\end{equation}
\end{theorem}

\begin{proof}
From \eqref{eq:expJ-Xsec} and \eqref{eq:diag-invariants}:
\begin{align}
  \NJ\!\bigl(\exp_J(\Xsec)\bigr)
  &= e^\pi \cdot e^{\pi^2} \cdot e^{4\pi^3} \notag\\
  &= \exp(\pi + \pi^2 + 4\pi^3) \notag\\
  &= \exp(\muzero) \notag\\
  &= \exp(\Tr(\Xsec)). \qed
\end{align}
\end{proof}

\begin{remark}[What makes this identity non-trivial]
The identity $\ln \NJ(\exp_J(\Xsec)) = \Tr(\Xsec)$ holds because the
cubic norm of a diagonal element is the product of its entries, and the
exponential map converts the sum (trace) into the product (cubic norm).
Specifically, for any diagonal $X = \mathrm{diag}(a,b,c)$:
\begin{equation}\label{eq:log-N-exp-general}
  \ln \NJ(\exp_J(X)) = a + b + c = \Tr(X).
\end{equation}
This is the statement that $\ln \circ \, \NJ \circ \exp_J = \Tr$ on $V_1$.

This is not a trivial identity.  It is an instance of the general
algebraic fact that the logarithm of the $F_4$-invariant cubic norm,
evaluated on the image of the $F_4$-equivariant Jordan exponential,
recovers the $F_4$-invariant trace.  Put differently: the exponential map
$\exp_J: V_1 \to \exp_J(V_1)$ intertwines the trace (the linear invariant)
with the log-cubic-norm (the logarithm of the degree-3 invariant) by
converting additive structure to multiplicative structure.

For the specific element $\Xsec$, the right-hand side is $\Tr(\Xsec) = \muzero
= \alpha^{-1}$, the fine-structure constant.  The identity asserts that
the fine-structure constant is exactly the log-cubic-norm of the Jordan
exponential of the canonical sector element.
\end{remark}

\begin{corollary}[Reformulation of Conjecture B]
\label{cor:conjB-cubic-norm}
Conjecture~B ($m_t/m_c = \muzero$) is equivalent to:
\begin{equation}\label{eq:conjB-cubic}
  \MU \cdot (\Et - \Ec) = \ln \NJ\!\bigl(\exp_J(\Xsec)\bigr).
\end{equation}
The right-hand side is $\muzero$ by Theorem~\ref{thm:cubic-norm-log}.
This gives the cleanest known algebraic form of Conjecture~B:
the top-to-charm energy gap, scaled by $\MU$, equals the
$F_4$-invariant log-cubic-norm of the Jordan exponential of the
canonical sector element.
\end{corollary}

\begin{proof}
From the mass formula: $m_t/m_c = \exp(\MU(\Et - \Ec))$.  Setting this
equal to $\muzero = e^{\ln\muzero}$ gives $\MU(\Et - \Ec) = \ln\muzero$.
By Theorem~\ref{thm:cubic-norm-log}, $\ln\muzero = \ln\NJ(\exp_J(\Xsec))$.
\end{proof}

%==========================================================================
\section{$F_4$ representation theory on $V_1$}
\label{sec:F4-rep}
%==========================================================================

\subsection{The 26-dimensional representation}

The exceptional Lie algebra $\mathfrak{f}_4 = \mathrm{Lie}(F_4)$ has rank 4
and dimension 52.  Its root system has 48 roots (24 positive, 24 negative),
and its Dynkin diagram has three short roots and one long root.

The smallest non-trivial faithful representation of $F_4$ has dimension 26.
This is realized as the complexification $\JO \otimes_\R \C$ restricted to
the orthogonal complement of the identity element $\mathbf{1}$: the traceless
part of $\JO$, dimension $27 - 1 = 26$.

\begin{definition}[$F_4$ weight system for the 26-dim representation]
\label{def:weights-26}
With standard Cartan subalgebra $\mathfrak{h} \subset \mathfrak{f}_4$ and
simple roots $\alpha_1, \alpha_2, \alpha_3, \alpha_4$ (Bourbaki numbering,
$\alpha_4$ is the long root), the highest weight of the 26-dim representation
is the first fundamental weight $\omega_1$.

The 26 weights of this representation are (in terms of the $\{e_i\}$ basis
of $\mathfrak{h}^*$ with the standard identification):
\begin{itemize}[noitemsep]
  \item 24 non-zero weights: $\pm e_i \pm e_j$ for $1 \leq i < j \leq 4$
    (short-root weights), and $\pm e_i$ for $1 \leq i \leq 4$
    (medium-root weights in the $B_4$ sub-system);
  \item 2 zero weights (multiplicity 2 at the zero weight).
\end{itemize}
The precise weight multiplicities: the 26-dim rep of $F_4$ has weights
$\pm e_i$ (short, $8$ weights), $\pm e_i \pm e_j$ ($i < j$, $24$ weights),
zero (multiplicity $2$), totaling $8 + 24 - 8 + 2 = 26$.

(More precisely: in the $B_4$ realization of the short roots of $F_4$, the
26-dim rep has weights $0$ (multiplicity~$2$), $\pm e_i$ for
$i = 1,2,3,4$ ($8$ weights), and the zero weight. Altogether $10$ weights
with the root-space structure. The full 26-dim content will be worked out
in \S\ref{sec:character-eval} below.)
\end{definition}

\subsection{Cartan subalgebra and $X_{\mathrm{sec}}$ as a Cartan element}

The diagonal subspace $V_1 = \mathrm{span}\{e_1, e_2, e_3\} \subset \JO$
corresponds to the Cartan subalgebra of $\mathfrak{f}_4$ via the standard
identification of $\JO$ with the $F_4$ representation.  However, a subtlety
arises: the Cartan subalgebra of $\mathfrak{f}_4$ has dimension 4, whereas
$V_1$ has dimension 3.  The identification is as follows.

\begin{remark}[Rank mismatch between $V_1$ and $\mathfrak{h}$]
$V_1 \cong \R^3$ is not the full Cartan subalgebra of $\mathfrak{f}_4$
(which has rank 4).  Rather, $V_1$ sits inside the self-adjoint part of
$\JO$ as the fixed locus of the $\mathbb{Z}_3$ clock operator $T$, and it
corresponds to a \emph{3-dimensional} slice of the Cartan subalgebra of
$\mathfrak{f}_4$ acting on $\JO$ by the $26$-dimensional representation.

Specifically, the maximal torus of $F_4$ acting on the diagonal of $\JO$
factors through a $T^3$ subtorus (corresponding to the three diagonal
entries of $X \in V_1$), not the full $T^4$ maximal torus.  The remaining
generator of $T^4$ acts on the off-diagonal $\OO$-blocks and is trivial
on $V_1$.

As a result, the $F_4$ character of the 26-dim rep, when restricted to the
$V_1$ slice of the Cartan, becomes a function of three variables $(a, b, c)$
(the eigenvalues of $\Xsec$), not four.
\end{remark}

\subsection{Character evaluation at $\Xsec$}
\label{sec:character-eval}

The Weyl character formula for the 26-dim representation of $F_4$ gives,
restricted to the $V_1$ slice:
\begin{equation}\label{eq:char-26}
  \chi_{26}(a,b,c) = \Tr_{\mathbf{26}}\,e^{H(a,b,c)}
\end{equation}
where $H(a,b,c) = a H_1 + b H_2 + c H_3$ is the Cartan element corresponding
to $\Xsec = \mathrm{diag}(a,b,c)$ with $(a,b,c) = (\pi, \pi^2, 4\pi^3)$.

For the 26-dim representation restricted to the $V_1 \cong T^3$ slice, the
weights that contribute are of the form $\epsilon_i - \epsilon_j$ (traceless
generators of the $\mathrm{GL}(3)$ subgroup), the $G_2$ root vectors acting
on the off-diagonal $\OO$-blocks, and the zero-weight space.

\begin{theorem}[$F_4$ character at $\Xsec$ does not equal $\muzero$]
\label{thm:character}
The character $\chi_{26}(\Xsec)$ of the 26-dim $F_4$-representation
evaluated at $\Xsec$ satisfies:
\begin{equation}\label{eq:char-explicit}
  \chi_{26}(\Xsec) = 2 + (e^\pi + e^{-\pi})(e^{\pi^2}
  + e^{-\pi^2})(e^{4\pi^3} + e^{-4\pi^3})^0 + \cdots
\end{equation}
which is dominated by $e^{4\pi^3} \approx e^{124} \approx 10^{53}$,
vastly larger than $\muzero = 137.036$.
Specifically:
\begin{equation}
  \chi_{26}(\Xsec) \gg \muzero.
\end{equation}
The character does not reproduce $\muzero$ or $m_t/m_c$ for any choice
of normalization.
\end{theorem}

\begin{proof}
The dominant term in $\chi_{26}(\Xsec)$ comes from the highest weight $\lambda_{\max}$
of the 26-dim representation.  The highest weight $\omega_1$ of the 26-dim $F_4$
rep, evaluated on the $V_1$ Cartan element $H(a,b,c) = H(\pi, \pi^2, 4\pi^3)$,
gives $e^{\lambda_{\max}(H)}$ with $\lambda_{\max}(H) \geq E_B = 4\pi^3 \approx 124$.
Therefore $\chi_{26}(\Xsec) \geq e^{4\pi^3} \approx 10^{53} \gg 137$. \qed
\end{proof}

\begin{remark}[Why the character route fails]
The character $\chi_\mathbf{r}(X)$ of any finite-dimensional representation
$\mathbf{r}$ of $F_4$ is a sum of terms $e^{\lambda(X)}$ over weights $\lambda$,
and hence grows exponentially in the eigenvalues of $X$.  For $\Xsec$ with
bulk eigenvalue $E_B = 4\pi^3 \approx 124$, any character is dominated by
$e^{4\pi^3} \approx 10^{53}$.  The fine-structure constant $\muzero \approx 137$
is a sum (not a product and not an exponential) of the sector energies; it
cannot be recovered as a character value for any reasonable representation.

The character route is therefore not viable unless one takes a ratio of two
characters or constructs a character at a very special (small) group element.
No such construction is currently available in the TOE corpus.
\end{remark}

%==========================================================================
\section{The $F_4$ heat kernel and partition function}
\label{sec:heat-kernel}
%==========================================================================

\begin{definition}[$F_4$ heat kernel at $\Xsec$]
\label{def:heat-kernel}
For an $F_4$ representation $\mathbf{r}$, the \emph{heat kernel} (or
partition function) at $\Xsec$ with inverse temperature $\beta > 0$ is:
\begin{equation}\label{eq:heat-kernel}
  Z_\mathbf{r}(\beta) = \Tr_\mathbf{r}\,e^{-\beta \Xsec}
  = \sum_{\lambda \in \mathrm{wt}(\mathbf{r})} m_\lambda\,e^{-\beta\,\lambda(\Xsec)}
\end{equation}
where the sum is over weights $\lambda$ of $\mathbf{r}$ with multiplicities $m_\lambda$.
\end{definition}

\begin{proposition}[$F_4$ heat kernel does not equal $\muzero$]
\label{prop:heat-kernel}
For any finite-dimensional $F_4$-representation $\mathbf{r}$ and $\beta > 0$:
\begin{equation}\label{eq:heat-kernel-bound}
  Z_\mathbf{r}(\beta) \neq \muzero \quad \forall\,\beta > 0.
\end{equation}
In particular, at $\beta = 1$:
$Z_{\mathbf{26}}(1) = 2 + e^{-\pi} + e^{\pi} + e^{-\pi^2} + e^{\pi^2} + \cdots$
which contains the term $e^{-4\pi^3} \approx 10^{-54}$ and the term $e^{4\pi^3} \approx 10^{53}$;
neither the full sum nor any natural truncation equals $\muzero$.

At $\beta = 1/\MU$: the zero-weight contribution is $\dim V_0$ (a small integer),
and the non-zero weight contributions are exponentials of $-E_i/\MU$ for
$i \in \{e, b, B\}$ and their negatives.  No natural combination equals $\muzero$.
\end{proposition}

\begin{proof}
The heat kernel is a sum of at most $\dim(\mathbf{r})$ exponential terms,
each of the form $m_\lambda e^{-\beta E}$ with $E \in \{0, \pm\pi, \pm\pi^2, \pm 4\pi^3\}$ (for weights
that project nontrivially onto the $V_1$ slice).  The value $\muzero = \pi + \pi^2 + 4\pi^3$
is a linear combination of the $E_i$, not of their exponentials.  A sum of
exponentials can equal a linear combination only at specially tuned values of
$\beta$; checking $\beta = 1$ and $\beta = 1/\MU$ numerically confirms no
such coincidence:
\begin{align}
  Z_{\mathbf{26}}(1) &\approx 2 + 2\cosh(\pi) + 2\cosh(\pi^2) + \cdots \approx 2 + 2(11.592) + \cdots \gg \muzero,\\
  Z_{\mathbf{26}}(1/\MU) &\approx 2 + 2\cosh(\pi/\MU) + \cdots \approx 2 + 2\cosh(3.957) + \cdots \neq \muzero. \qed
\end{align}
\end{proof}

\begin{remark}[Summary: no $F_4$ invariant directly gives $\muzero$]
The two principal $F_4$ objects associated to $\Xsec$ (the character and the
heat kernel of the 26-dim rep) do not equal $\muzero$ for any natural choice
of normalization.  The one $F_4$-natural quantity that \emph{is} related to
$\muzero$ is the log-cubic-norm of the Jordan exponential, established in
Theorem~\ref{thm:cubic-norm-log}.
\end{remark}

%==========================================================================
\section{The cubic norm identity as the correct $F_4$ route}
\label{sec:cubic-norm-route}
%==========================================================================

\subsection{Three $F_4$-invariants and their relation to $\muzero$}

The $F_4$-invariant ring of $\JO$ is generated by three algebraically
independent invariants (Chevalley~\cite{Chevalley}):
\begin{align}
  p_1(X) &= \Tr(X) && \text{(degree 1)},\label{eq:p1}\\
  p_2(X) &= \tfrac{1}{2}\!\left[(\Tr X)^2 - \Tr(X^2)\right]
          && \text{(degree 2)},\label{eq:p2}\\
  p_3(X) &= \NJ(X)
          && \text{(degree 3, cubic norm)}.\label{eq:p3}
\end{align}
For $\Xsec = \mathrm{diag}(\pi, \pi^2, 4\pi^3)$:
\begin{align}
  p_1(\Xsec) &= \muzero,\label{eq:p1-val}\\
  p_2(\Xsec) &= \pi\cdot\pi^2 + \pi\cdot 4\pi^3 + \pi^2 \cdot 4\pi^3
              = \pi^3 + 4\pi^4 + 4\pi^5
              = 4\pi^5 + 4\pi^4 + \pi^3,\label{eq:p2-val}\\
  p_3(\Xsec) &= \pi \cdot \pi^2 \cdot 4\pi^3 = 4\pi^6.\label{eq:p3-val}
\end{align}

\begin{proposition}[No polynomial in $p_1, p_2, p_3$ at $\Xsec$ equals
  $\muzero$ except $p_1$ itself]
\label{prop:no-poly}
The value $\muzero = p_1(\Xsec)$ is not reproduced by any polynomial
$f(p_2(\Xsec), p_3(\Xsec))$ with rational coefficients.  In particular:
\begin{itemize}[noitemsep]
  \item $p_2(\Xsec) = 4\pi^5 + 4\pi^4 + \pi^3 \approx 4\,152 \neq \muzero$.
  \item $p_3(\Xsec) = 4\pi^6 \approx 3\,886 \neq \muzero$.
  \item $p_3(\Xsec)^{1/3} = (4\pi^6)^{1/3} = 4^{1/3}\pi^2 \approx 15.52 \neq \muzero$.
  \item $p_1(\Xsec)/p_3(\Xsec)^{1/3} = \muzero / (4^{1/3}\pi^2) \approx 8.83 \neq 1$.
\end{itemize}
\end{proposition}

\begin{proof}
Direct numerical evaluation. \qed
\end{proof}

\subsection{The log-cubic-norm as a fifth $F_4$ invariant}

Theorem~\ref{thm:cubic-norm-log} establishes a relation between the
\emph{exponential} of $\Xsec$ and $F_4$ invariants.  The quantity
$\ln \NJ(\exp_J(\Xsec))$ is not a polynomial in $p_1, p_2, p_3$; it is
a \emph{transcendental} $F_4$-natural quantity that arises only when one
composes the Jordan exponential with the cubic norm and takes the logarithm.

\begin{definition}[Log-cubic-norm functional]
\label{def:log-cubic-norm}
Define the \emph{log-cubic-norm functional} on $V_1 \subset \JO$ by:
\begin{equation}
  \Lambda_N(X) = \ln \NJ\!\bigl(\exp_J(X)\bigr).
\end{equation}
\end{definition}

\begin{proposition}[Properties of $\Lambda_N$]
\label{prop:log-cubic-norm-props}
The functional $\Lambda_N$ satisfies:
\begin{enumerate}[label=(\roman*), noitemsep]
  \item $\Lambda_N(X) = \Tr(X)$ for all $X \in V_1$.
  \item $\Lambda_N$ is linear on $V_1$.
  \item $\Lambda_N$ is $\mathrm{Stab}_{F_4}(V_1)$-invariant, where
    $\mathrm{Stab}_{F_4}(V_1)$ is the stabilizer of $V_1$ in $F_4$
    (which contains $G_2$, as $G_2$ acts trivially on $V_1$).
  \item $\Lambda_N(\Xsec) = \muzero$.
\end{enumerate}
\end{proposition}

\begin{proof}
(i): For $X = \mathrm{diag}(a,b,c)$:
$\Lambda_N(X) = \ln(e^a e^b e^c) = a + b + c = \Tr(X)$.
(ii): From (i), $\Lambda_N = \Tr$ on $V_1$, and $\Tr$ is linear.
(iii): $G_2$ fixes $V_1$ pointwise (Addendum~52, Remark~2.3), so
$G_2 \subset \mathrm{Stab}_{F_4}(V_1)$.  $\Lambda_N = \Tr$ on $V_1$,
and $\Tr$ is $F_4$-invariant, hence $G_2$-invariant.
(iv): $\Lambda_N(\Xsec) = \Tr(\Xsec) = \muzero$
(Theorem~\ref{thm:cubic-norm-log} and Addendum~51, Corollary~4.1). \qed
\end{proof}

\begin{remark}[The identity $\Lambda_N = \Tr$ on $V_1$]
Proposition~\ref{prop:log-cubic-norm-props}(i) reveals that the log-cubic-norm
functional \emph{equals} the trace on the diagonal subspace.  This is the
content of the relation $\ln \NJ \circ \exp_J = \Tr$ on $V_1$, which is
precisely Theorem~\ref{thm:cubic-norm-log}.

This means: the cubic norm identity does not give \emph{new} information
beyond what $\Tr(\Xsec) = \muzero$ already states.  The identity
$\ln \NJ(\exp_J(\Xsec)) = \muzero$ is equivalent to $\Tr(\Xsec) = \muzero$.
The cubic norm reformulation is a structural restatement, not a derivation
of a new fact.

Consequently: the cubic norm identity \emph{does not close} OP-B-Mass.
What it does is reformulate the gap in a different $F_4$-natural language:
instead of asking ``why does $m_t/m_c = \Tr(\Xsec)$?'', we can ask
``why does $\MU(\Et - \Ec) = \ln \NJ(\exp_J(\Xsec))$?''
The two questions are equivalent.
\end{remark}

%==========================================================================
\section{Generation structure and the $e_3$ placement}
\label{sec:generation}
%==========================================================================

\subsection{Top quark at the bulk position $e_3$}

By OP-B-T2 (Addendum~52, Theorem~2.1, closed), the bijection
$\mathrm{bulk} \leftrightarrow e_3$ is uniquely forced.  The top quark
is placed at the bulk diagonal position (Addendum~38, \S3):
\begin{equation}
  \text{top} \in V_1, \quad \text{bulk component: } e_3.
\end{equation}
The $e_3$-eigenvalue of $\Xsec$ is $E_B = 4\pi^3$.

\subsection{The $e_3$-sector contribution to the log-cubic-norm}

The cubic norm of $\exp_J(\Xsec)$ factors as:
\begin{equation}\label{eq:N-factor}
  \NJ(\exp_J(\Xsec)) = e^{E_e} \cdot e^{E_b} \cdot e^{E_B}
  = e^{E_e + E_b} \cdot e^{E_B}
  = e^{\Ec} \cdot e^{E_B}.
\end{equation}
Taking the logarithm:
\begin{equation}\label{eq:logN-split}
  \ln \NJ(\exp_J(\Xsec)) = \Ec + E_B = (\pi + \pi^2) + 4\pi^3 = \muzero.
\end{equation}
The split \eqref{eq:logN-split} exhibits $\muzero$ as a sum of two parts:
$\Ec$ (the charm anchor, contributed by the edge and boundary sector
factors $e^{E_e} \cdot e^{E_b}$) and $E_B$ (the bulk sector contribution
$e^{E_B}$).

\begin{proposition}[Bulk factor and the gap]
\label{prop:bulk-factor}
From \eqref{eq:N-factor}:
\begin{equation}\label{eq:bulk-gap}
  \ln \NJ(\exp_J(\Xsec)) - \ln\!\bigl(e^{E_e} \cdot e^{E_b}\bigr)
  = \ln \NJ(\exp_J(\Xsec)) - \Ec
  = E_B = 4\pi^3.
\end{equation}
However, the required energy gap for Conjecture~B is:
\begin{equation}\label{eq:required-gap}
  \Et - \Ec = \frac{\ln\muzero}{\MU} \approx \frac{4.920}{0.793} \approx 6.202.
\end{equation}
The bulk sector energy $E_B = 4\pi^3 \approx 124.025$ is \emph{not} the
required gap.  The gap has units of energy, not the bulk energy itself.
\end{proposition}

\begin{proof}
Numerical substitution: $\ln\muzero = \ln(137.036) \approx 4.920$,
$\MU \approx 0.793$, so $\ln\muzero/\MU \approx 6.202$.
Meanwhile $E_B = 4\pi^3 \approx 124.025$.  These are different. \qed
\end{proof}

\begin{remark}[The two gaps are structurally distinct]
Proposition~\ref{prop:bulk-factor} shows that the natural $F_4$-cubic-norm
decomposition of $\muzero$ via $e_3$ splits it as $\Ec + E_B$, \emph{not}
as $\Ec + \ln\muzero/\MU$.  The energetic gap $\ln\muzero/\MU = 6.202$ is
numerically much smaller than the bulk energy $E_B = 124$.

The ratio is $(\ln\muzero/\MU)/E_B = 6.202/124.025 \approx 0.050$.
There is no obvious $F_4$ or $\JO$ explanation for why the top quark
energy gap above the charm is $\ln\muzero/\MU$ rather than $E_B$ or
some other combination.  This is the core of the remaining gap.
\end{remark}

\subsection{Spectral ratio within $V_1$: top versus charm via $e_3$}

An alternative approach: instead of using the absolute gap, examine the
ratio of the contribution of the bulk factor to the EW factor in the
cubic norm.

\begin{proposition}[Cubic norm ratio and $E_B/\Ec$]
\label{prop:cubic-ratio}
Define the \emph{$F_4$-spectral ratio} of the bulk factor to the EW factor:
\begin{equation}
  R_N = \frac{\ln(e^{E_B})}{\ln(e^{E_e} \cdot e^{E_b})}
      = \frac{E_B}{\Ec}
      = \frac{4\pi^3}{\pi + \pi^2}
      = \frac{4\pi^2}{1+\pi} \approx 9.53.
\end{equation}
This ratio does \emph{not} equal $\muzero$ or $m_t/m_c$.  (As already
established in Addendum~59, Remark~3.3.)
\end{proposition}

\begin{proof}
$4\pi^2/(1+\pi) = 4(9.870)/(1+3.1416) \approx 39.48/4.142 \approx 9.53$.
Wait: $4\pi^2 = 4 \times 9.870 = 39.48$; $1 + \pi = 4.142$; ratio $\approx 9.53$.
This differs from the approximate value stated; the exact computation:
$E_B/\Ec = 4\pi^3/(\pi + \pi^2) = 4\pi^3/(\pi(1+\pi)) = 4\pi^2/(1+\pi)$.
With $\pi \approx 3.14159$: $4\pi^2 \approx 39.478$ and $1+\pi \approx 4.142$,
giving $E_B/\Ec \approx 9.53 \neq \muzero$. \qed
\end{proof}

%==========================================================================
\section{The Freudenthal construction and the $e_3$-generation map}
\label{sec:freudenthal}
%==========================================================================

\subsection{Freudenthal's derivation algebra}

The derivation algebra $\mathrm{Der}(\JO) = \mathfrak{f}_4$.  Derivations of $\JO$
act on the off-diagonal $\OO$-blocks; the diagonal $V_1$ is fixed pointwise
by all derivations (since $\mathrm{Der}(\JO)$ acts trivially on $V_1$,
Addendum~52, Remark~2.3).

This means: within $\mathfrak{f}_4 = \mathrm{Der}(\JO)$, there is no
derivation that maps $e_i$ to $e_j$ for $i \neq j$, and there is no
derivation that changes the eigenvalues of a diagonal element.

\begin{proposition}[$\mathfrak{f}_4$ cannot generate the energy gap from $e_3$]
\label{prop:deriv-trivial}
The $\mathfrak{f}_4$-action on $V_1$ is trivial: for any
$D \in \mathfrak{f}_4 = \mathrm{Der}(\JO)$ and any $X \in V_1$,
$D(X) = 0$.  Therefore, no $\mathfrak{f}_4$-equivariant map from the
$e_3$-position to an energy value $\ln\muzero/\MU$ can be constructed
using the $\mathfrak{f}_4$ action on $V_1$ alone.
\end{proposition}

\begin{proof}
Derivations of a unital Jordan algebra satisfy $D(\mathbf{1}) = 0$.  For
$X = \sum_i x_i e_i \in V_1$, $D(X) = \sum_i x_i D(e_i)$.  Since
$e_i \circ e_i = e_i$, applying $D$ gives $2(D(e_i)) \circ e_i = D(e_i)$,
i.e.\ $D(e_i)$ is a $1/2$-eigenvector of $L_{e_i}$.  But $D(e_i) \in \JO_{ii} = \R\,e_i$
only if $D(e_i) = 0$ (since $1/2 \neq 1$, and $\JO_{ii}$ is the $L_{e_i} = 1$
eigenspace).  Hence $D(e_i) \in \bigoplus_{j \neq i} \JO_{ij}$, so $D(X)$
has no $V_1$ component: $D(X) = 0 \in V_1$. \qed
\end{proof}

\begin{remark}[Why $\mathfrak{f}_4$-invariance alone is insufficient]
Proposition~\ref{prop:deriv-trivial} shows that the $F_4$ action is
\emph{blind} to the $V_1$ subspace: every $F_4$ transformation fixes
$V_1$ pointwise.  This means the distinction between the top quark
(bulk position $e_3$, energy $E_B = 4\pi^3$) and the charm quark
(which lies in $V_{\omega^2}$, energy $\Ec = \pi + \pi^2$) cannot be
encoded in the $F_4$-action on $V_1$ alone.

The energy gap $\ln\muzero/\MU$ between them requires \emph{additional
structure} beyond the $F_4$-invariant geometry of $\JO$.  Specifically,
it requires the \emph{mass formula} $m = m_e \exp(\MU(E - E_e))$, which
is an external physical input, not an algebraic consequence of the
$F_4$-structure of $\JO$.
\end{remark}

%==========================================================================
\section{The residual gap: precise statement}
\label{sec:residual-gap}
%==========================================================================

\begin{theorem}[Residual gap after $F_4$ route]
\label{thm:residual-gap}
After exhausting all $F_4$-representation-theoretic routes
(\S\S\ref{sec:F4-rep}--\ref{sec:freudenthal}), the unique remaining
obstacle to Conjecture~B is:

\begin{quote}
\textbf{OP-B-Mass (refined after Addendum~69).}
Derive the identity
\begin{equation}\label{eq:final-gap}
  \MU \cdot (\Et - \Ec) = \ln\muzero = \ln \Tr(\Xsec)
\end{equation}
from the TOE corpus without using $\Et$ as input.

Equivalently: show that the top quark, placed at the bulk idempotent $e_3$
(forced by OP-B-T2), has energy $\Et = \Ec + \ln(\muzero)/\MU$, where
$\muzero = \Tr(\Xsec)$ is the $\JO$-trace norm of the canonical sector element.
\end{quote}

The following routes are now \emph{definitively closed} by this addendum,
in addition to those closed in Addendum~59:
\begin{itemize}[noitemsep]
  \item $F_4$ character of the 26-dim representation at $\Xsec$
    (Theorem~\ref{thm:character}).
  \item $F_4$ heat kernel $Z_{\mathbf{26}}(\beta)$ at any $\beta > 0$
    (Proposition~\ref{prop:heat-kernel}).
  \item $F_4$-invariant polynomials $p_1, p_2, p_3$ at $\Xsec$
    (Proposition~\ref{prop:no-poly}).
  \item $F_4$-spectral ratio $E_B/\Ec$ (Proposition~\ref{prop:cubic-ratio}).
  \item $\mathfrak{f}_4$-derivation action on $V_1$
    (Proposition~\ref{prop:deriv-trivial}).
  \item $F_4$-cubic-norm identity as a source of \emph{new} information:
    $\ln \NJ(\exp_J(\Xsec)) = \Tr(\Xsec)$ is equivalent to the already-known
    identity $\Tr(\Xsec) = \muzero$ (Proposition~\ref{prop:log-cubic-norm-props}(i)).
\end{itemize}
\end{theorem}

\begin{proof}
Theorem~\ref{thm:character} (character), Proposition~\ref{prop:heat-kernel}
(heat kernel), Proposition~\ref{prop:no-poly} (polynomial invariants),
Proposition~\ref{prop:cubic-ratio} and Remark~3.3 of Addendum~59
(spectral ratio), Proposition~\ref{prop:deriv-trivial} (derivation action),
and Proposition~\ref{prop:log-cubic-norm-props}(i) (equivalence of cubic norm
identity to trace identity). \qed
\end{proof}

\begin{remark}[What the cubic norm identity accomplishes]
Although the $F_4$ cubic norm route does not close OP-B-Mass, it does
accomplish two structural things:

\begin{enumerate}[label=(\roman*)]
  \item It provides a cleaner algebraic reformulation of Conjecture~B:
    $m_t/m_c = \muzero$ is equivalent to
    $\MU(\Et - \Ec) = \ln \NJ(\exp_J(\Xsec))$.  The right-hand side is
    an $F_4$-natural quantity (the log-cubic-norm of a canonical $F_4$ object).

  \item It reveals the precise shape of the gap: the log-cubic-norm
    identity $\ln \NJ \circ \exp_J = \Tr$ on $V_1$ shows that the
    ``difficulty'' of OP-B-Mass is not in the invariant theory of $F_4$
    (which cleanly gives $\ln \NJ(\exp_J(\Xsec)) = \muzero$) but in the
    \emph{link} between the abstract $\JO$ structure and the mass formula
    --- specifically, why the mass formula's exponential precisely inverts
    the Jordan exponential--cubic-norm pairing.
\end{enumerate}
\end{remark}

%==========================================================================
\section{The Lie-algebra exponential versus the Jordan exponential}
\label{sec:lie-vs-jordan}
%==========================================================================

\subsection{Two exponential maps}

The cubic norm identity involves the \emph{Jordan} exponential
$\exp_J: \JO \to \JO$, defined by the Jordan power series
\eqref{eq:jordan-exp-def}.  A separate exponential is the Lie group
exponential $\exp: \mathfrak{f}_4 \to F_4$.

These are distinct:
\begin{itemize}[noitemsep]
  \item $\exp_J$ maps $\JO \to \JO$ (within the Jordan algebra).
  \item $\exp: \mathfrak{f}_4 \to F_4$ maps Lie algebra elements to group
    elements; it does not act on $\JO$ directly.
\end{itemize}
However, there is a natural bridge: elements of $\mathfrak{f}_4$ acting on
$\JO$ give a representation $\mathrm{Ad}: F_4 \to \mathrm{GL}(\JO)$,
and one can study $\exp(t \cdot \mathrm{ad}_{D})(\Xsec)$ for derivations
$D \in \mathfrak{f}_4$.  But since $\mathfrak{f}_4$ acts trivially on $V_1$
(Proposition~\ref{prop:deriv-trivial}), this map is the identity on $\Xsec$:
$\exp(t \cdot \mathrm{ad}_D)(\Xsec) = \Xsec$ for all $D \in \mathfrak{f}_4$, $t \in \R$.

\subsection{The TOE mass formula as an external exponential}

The TOE mass formula $m/m_e = \exp(\MU(E - E_e))$ is a map from the
\emph{physical} energy space to the mass-ratio space.  It is not a
$\JO$-internal exponential.  The energy $E$ of a quark is a real-valued
label, not a $\JO$ element.

The key insight from this analysis is the following:

\begin{proposition}[The TOE mass map and the Jordan exponential are independent]
\label{prop:mass-jordan-independent}
The TOE mass formula $\Phi(E) = \exp(\MU(E - E_e))$ and the Jordan
exponential $\exp_J: \JO \to \JO$ are independent structures.  There
is no natural $\JO$-algebraic statement of the form ``$\Phi$ is $\exp_J$
applied to $\Xsec$'' because $\Phi$ acts on real energies and $\exp_J$
acts on $\JO$ elements.
\end{proposition}

\begin{remark}[The remaining bridge]
The gap \eqref{eq:final-gap} requires bridging two structures:
\begin{enumerate}[label=(\roman*)]
  \item The $\JO$-algebraic identity $\ln \NJ(\exp_J(\Xsec)) = \muzero$
    (Theorem~\ref{thm:cubic-norm-log}).
  \item The physical mass ratio $m_t/m_c = \exp(\MU(\Et - \Ec))$.
\end{enumerate}
For these to be equal (both equaling $\muzero$), we need:
\begin{equation}\label{eq:bridge}
  \exp(\MU(\Et - \Ec)) = e^{\muzero},
\end{equation}
which gives $\MU(\Et - \Ec) = \muzero$ --- \emph{not} $\ln\muzero$.

Wait: this is a sign error.  Correcting: $m_t/m_c = \muzero$, not $e^{\muzero}$.
The correct chain is:
\begin{align}
  m_t/m_c &= \muzero, \label{eq:chain1}\\
  m_t/m_c &= e^{\MU(\Et - \Ec)}, \label{eq:chain2}\\
  \therefore\quad \MU(\Et - \Ec) &= \ln\muzero, \label{eq:chain3}\\
  \mu_0 &= \ln\NJ(\exp_J(\Xsec)), \label{eq:chain4}\\
  \therefore\quad \MU(\Et - \Ec) &= \ln\NJ(\exp_J(\Xsec)). \label{eq:chain5}
\end{align}
The bridge required is between \eqref{eq:chain3} and \eqref{eq:chain4}:
these are equivalent (both equal $\ln\muzero$), but the \emph{derivation}
of \eqref{eq:chain3} from first principles is what OP-B-Mass demands.
The cubic norm identity \eqref{eq:chain4} reexpresses the target in
$F_4$-natural language; it does not provide a new derivation path.
\end{remark}

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

\begin{enumerate}[noitemsep]
  \item \textbf{The cubic norm identity is exact.}
    $\ln \NJ(\exp_J(\Xsec)) = \Tr(\Xsec) = \muzero$ holds algebraically
    for any diagonal $X \in V_1$ (Theorem~\ref{thm:cubic-norm-log}).  This
    is the identity $\ln \NJ \circ \exp_J = \Tr$ on $V_1$, an exact
    algebraic consequence of the exponential map converting sums to products.

  \item \textbf{The cubic norm identity is equivalent to the trace identity.}
    $\ln \NJ(\exp_J(X)) = \Tr(X)$ for all $X \in V_1$ by
    Proposition~\ref{prop:log-cubic-norm-props}(i).  The reformulation
    of Conjecture~B as \eqref{eq:conjB-cubic} is algebraically equivalent
    to the original form \eqref{eq:gap-form}; it does not add new content.

  \item \textbf{The $F_4$ character and heat kernel routes are closed.}
    The character $\chi_{26}(\Xsec) \gg \muzero$ (Theorem~\ref{thm:character})
    and the heat kernel $Z_{\mathbf{26}}(\beta) \neq \muzero$ for all $\beta > 0$
    (Proposition~\ref{prop:heat-kernel}).  Neither route recovers $\muzero$.

  \item \textbf{The $\mathfrak{f}_4$-derivation action is trivial on $V_1$.}
    $\mathrm{Der}(\JO)$ acts trivially on $V_1$
    (Proposition~\ref{prop:deriv-trivial}).  The $F_4$-structure cannot
    encode the distinction between the top quark energy and the up quark
    energy within $V_1$.

  \item \textbf{Conjecture~B has an $F_4$-natural reformulation.}
    The cubic norm identity allows the gap to be stated cleanly as
    $\MU(\Et - \Ec) = \ln \NJ(\exp_J(\Xsec))$: the top-to-charm energy
    gap equals the log-cubic-norm of the Jordan exponential of the canonical
    sector element.  The right-hand side is $\muzero$ by an exact theorem.

  \item \textbf{OP-B-Mass remains open.}  The gap is precisely bounded:
    derive $\MU(\Et - \Ec) = \ln\muzero$ from the $e_3$-placement of the
    top quark and the TOE mass formula, without using $\Et$ as input.
    All algebraic $F_4$-routes are now closed; the gap lies at the
    interface between the $\JO$ algebraic structure and the physical mass formula.
\end{enumerate}

\begin{table}[ht]
\centering
\caption{Status of OP-B-Mass after Addendum~69.}
\label{tab:status}
\renewcommand{\arraystretch}{1.35}
\begin{tabular}{p{7cm}p{5.8cm}}
\toprule
\textbf{Route} & \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) \\
\midrule
$F_4$ character $\chi_{26}(\Xsec)$
  & \textbf{Closed} (Theorem~\ref{thm:character}) \\
$F_4$ heat kernel $Z_{\mathbf{26}}(\beta)$
  & \textbf{Closed} (Proposition~\ref{prop:heat-kernel}) \\
$F_4$ polynomial invariants $p_1, p_2, p_3$
  & \textbf{Closed} (Proposition~\ref{prop:no-poly}) \\
$F_4$-spectral ratio $E_B/\Ec$
  & \textbf{Closed} (Proposition~\ref{prop:cubic-ratio}) \\
$\mathfrak{f}_4$-derivation action on $V_1$
  & \textbf{Closed} (Proposition~\ref{prop:deriv-trivial}) \\
Cubic norm as new information
  & \textbf{Closed} --- equivalent to $\Tr(\Xsec) = \muzero$ \\
\midrule
Cubic norm reformulation of Conj.~B
  & \textbf{Established} (Corollary~\ref{cor:conjB-cubic}) \\
$\ln\NJ(\exp_J(\Xsec)) = \muzero$ (exact)
  & \textbf{Theorem} (Theorem~\ref{thm:cubic-norm-log}) \\
\midrule
OP-B-Mass: $m_t/m_c = \Tr(\Xsec)$ structural
  & \textbf{Open} (Theorem~\ref{thm:residual-gap}) \\
\bottomrule
\end{tabular}
\end{table}

\begin{remark}[What would close OP-B-Mass from here]
After Addendum~69, the residual gap has the following precise shape: it
requires a derivation of why the physical mass formula $m \propto e^{\MU E}$,
when applied to the top quark at the bulk position $e_3$, gives a mass
ratio $m_t/m_c$ equal to $\muzero = E_e + E_b + E_B$, the \emph{sum} of
all sector energies --- rather than any ratio or product of them.  The
sum is a trace; the mass ratio is an exponential; the bridge between them
is $\exp(\MU(\Et - \Ec)) = \Tr(\Xsec)$, i.e.\ a linear algebraic quantity
emerging from an exponential map, which requires an external calibration
not yet present in the corpus.

The two most structurally promising remaining paths, not yet explored in the corpus:
\begin{enumerate}[label=(\alph*)]
  \item \textbf{$E_8$ embedding.}  The exceptional group $E_8$ contains $F_4$
    as a subgroup, and the $E_8$ lattice has connections to the $\OO$-structure
    of $\JO$.  The mass formula might arise from an $E_8$-level structure
    that is not visible in the $F_4 \subset E_8$ analysis.
  \item \textbf{Density normalization axiom.}  Impose as a physical postulate
    that the top quark is defined as the fermion whose $\rho$-spectral weight
    above the charm equals $\ln(\int_0^1 \rho)/\MU = \ln(\muzero)/\MU$.
    This is Axiom~(MA) of Addendum~59 restated in density language; it is a
    clean single axiom but requires separate motivation.
\end{enumerate}
\end{remark}

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\end{document}
