% Addendum 76: OP-B-Mass — The λ=1 Normalisation and Spectral-Weight Closure
%              Deriving ln(μ₀)/MU from the CGF of ρ(x) and the
%              Canonical Spectral-Weight Saturation of the Top Quark
% L. F. Vlegels, Independent Researcher
% Copyright: Léon Fernando Vlegels. License: CC BY 4.0. May 2026.

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\title{OP-B-Mass: The $\lambda=1$ Normalisation\\
and Spectral-Weight Closure via the CGF of $\rho$\\[0.5em]
\large Addendum~76 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 left OP-B-Mass precisely open: there is no derivation showing
that the top/charm energy gap satisfies $\Et - \Ec = \ln(\muzero)/\MU$
without invoking Axiom~(MA) (``top is the quark whose mass ratio to charm
equals $\Tr(\Xsec) = \muzero$'').

The present addendum derives this gap from the cumulant generating function
(CGF) $\CGF(t) = \ln\int_0^1 e^{tx}\rho(x)\,dx$ of the TOE monad density
$\rho(x) = 2\pi x + 3\pi^2 x^2 + 16\pi^3 x^3$.

\medskip
\textbf{Main identity (Theorem~\ref{thm:cgf-identity}).}
\begin{equation}
  \frac{\ln\muzero}{\MU} = \frac{\CGF(0)}{\CGF'(0)},
\end{equation}
where $\CGF(0) = \ln\muzero$ and $\CGF'(0) = \MU$ follow directly from the
density without any circular input.  The ratio $\CGF(0)/\CGF'(0)$ is
the canonical $\lambda=1$ spectral-weight gap: the value of the CGF at
the unshifted normalisation point divided by the mean-slope of the CGF
at that point.

\medskip
\textbf{The $\lambda=1$ condition (Theorem~\ref{thm:lambda-one}).}
Let $\CGF_\lambda(t) = \ln\int_0^1 e^{\lambda tx}\rho(x)\,dx$ for
$\lambda \geq 0$.  The top quark occupies the energy $\Et = \Ec + \Delta$
where $\Delta$ is the unique solution of
\begin{equation}\label{eq:lambda-one-abstract}
  e^{\MU\Delta} = \MGF(0) = \int_0^1 \rho(x)\,dx = \muzero,
\end{equation}
i.e.\ $\Delta = \CGF(0)/\CGF'(0) = \ln(\muzero)/\MU$.
Condition~\eqref{eq:lambda-one-abstract} is the \emph{$\lambda=1$
spectral-weight saturation}: the mass lift from charm to top equals the
full normalisation of $\rho$, evaluated at the unshifted ($\lambda=1$)
weight $e^{0 \cdot x} = 1$ applied uniformly over $[0,1]$.

\medskip
\textbf{Status.}  OP-B-Mass is \emph{conditionally closed}: the gap
$\ln(\muzero)/\MU$ is derived purely from $\rho$ via the CGF.
The remaining logical step is identifying the top quark as the
$\lambda=1$ saturation point within $V_1$.  This identification follows
from two pieces already in the corpus: (i) the top is associated with
the bulk idempotent $e_3$ (forced by OP-B-T2, Addendum~52,
Theorem~2.1); (ii) the bulk sector carries total spectral weight
$E_B/\muzero = 4\pi^3/\muzero \approx 0.905$ — the dominant fraction
of $\rho$.  The $\lambda=1$ saturation postulate states that the top,
as the generator of the full $V_1$ trace, saturates the entire
normalisation $\muzero$, not just the bulk fraction.  This is either
a derivable consequence of the $e_3$-placement (which would fully close
OP-B-Mass) or the minimal irreducible input.  Section~\ref{sec:status}
states the precise residual condition.

\medskip
\textbf{Entropy computation (Proposition~\ref{prop:entropy}).}
The differential entropy $H[\bar\rho] = -\int_0^1 \bar\rho(x)\ln\bar\rho(x)\,dx$
of the normalized density $\bar\rho = \rho/\muzero$ satisfies
$H[\bar\rho] = \ln\muzero - \frac{1}{\muzero}\int_0^1\rho\ln\rho$; explicitly
$H[\bar\rho] \approx -0.597$, confirming that $\CGF(0) = \ln\muzero = 4.920$
is the dominant structural scale.  The ratio $\CGF(0)/\CGF'(0)
= \ln\muzero/\MU = 6.201$ is the natural energy unit of the log-mass
coordinate for transitions from the charm to the top.
\end{abstract}

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

We follow Addenda~44--59 throughout.

\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}\\
  E_e     &= \pi,\quad E_b = \pi^2,\quad E_B = 4\pi^3,\label{eq:sec-en}\\
  \Ec     &= E_e + E_b = \pi + \pi^2,\quad
  \Xsec   = \mathrm{diag}(\pi,\pi^2,4\pi^3).\label{eq:Ec-Xsec}
\end{align}

The TOE monad density and mass formula:
\begin{align}
  \rho(x) &= 2\pi x + 3\pi^2 x^2 + 16\pi^3 x^3
            = \sum_{i=1}^3 (n_i+1)\,c_i\,x^{n_i},\label{eq:rho}\\
  \frac{m}{m_e} &= \exp\!\bigl(\MU\,(E - E_e)\bigr),\label{eq:mass-formula}
\end{align}
where $(n_1,n_2,n_3) = (1,2,3)$ and $(c_1,c_2,c_3) = (\pi,\pi^2,4\pi^3)
= (E_e,E_b,E_B)$.

The known moments (Addendum~47):
\begin{equation}\label{eq:moments}
  \mu_k = \int_0^1 x^k \rho(x)\,dx
  = \frac{2\pi}{k+2} + \frac{3\pi^2}{k+3} + \frac{16\pi^3}{k+4}.
\end{equation}
In particular $\mu_0 = \muzero$ and $\mu_1 = \muone$.

\textbf{Status entering this addendum.}
Addendum~52 (OP-B-T2): the bijection $e_1\leftrightarrow\mathrm{edge}$,
$e_2\leftrightarrow\mathrm{boundary}$, $e_3\leftrightarrow\mathrm{bulk}$
is structurally forced.
Addendum~59: Conjecture~B is equivalent to the single condition
$\Et - \Ec = \ln(\muzero)/\MU$, which could not be derived from $\rho$
without Axiom~(MA).  The present addendum derives this gap from the CGF
of $\rho$.

%==========================================================================
\section{The cumulant generating function of $\rho$}
\label{sec:cgf}
%==========================================================================

\subsection{Definition and basic properties}

\begin{definition}[MGF and CGF of $\rho$]\label{def:cgf}
The \emph{moment generating function} (MGF) of $\rho$ is
\begin{equation}\label{eq:MGF}
  \MGF(t) = \int_0^1 e^{tx}\,\rho(x)\,dx, \qquad t \in \R.
\end{equation}
The \emph{cumulant generating function} (CGF) of $\rho$ is
$\CGF(t) = \ln \MGF(t)$.
\end{definition}

\begin{remark}
Note that $\rho$ is not a probability density: $\int_0^1\rho = \muzero \neq 1$.
We do not normalise $\rho$ in the definitions of $\MGF$ and $\CGF$ in order to
preserve the spectral-weight structure; the unnormalised CGF carries more
information than its probability-normalised counterpart.
\end{remark}

\begin{proposition}[CGF at $t=0$]\label{prop:cgf-zero}
\begin{equation}\label{eq:cgf-at-zero}
  \MGF(0) = \int_0^1 \rho(x)\,dx = \muzero, \qquad
  \CGF(0) = \ln\muzero.
\end{equation}
\end{proposition}

\begin{proof}
$e^{0\cdot x} = 1$ for all $x$, so $\MGF(0) = \int_0^1\rho(x)\,dx = \muzero$
by Proposition~2.2 of Addendum~46 (structural, non-circular after Addendum~51).
$\CGF(0) = \ln\MGF(0) = \ln\muzero$.
\end{proof}

\begin{proposition}[CGF slope at $t=0$]\label{prop:cgf-slope}
\begin{equation}\label{eq:cgf-slope-zero}
  \MGF'(0) = \int_0^1 x\,\rho(x)\,dx = \muone, \qquad
  \CGF'(0) = \frac{\MGF'(0)}{\MGF(0)} = \frac{\muone}{\muzero} = \MU.
\end{equation}
\end{proposition}

\begin{proof}
Differentiating under the integral: $\MGF'(t) = \int_0^1 x\,e^{tx}\rho(x)\,dx$,
so $\MGF'(0) = \int_0^1 x\,\rho(x)\,dx = \muone$.
Since $\CGF'(t) = \MGF'(t)/\MGF(t)$, $\CGF'(0) = \muone/\muzero = \MU$.
\end{proof}

\begin{remark}[Intrinsic origin of $\MU$]
Propositions~\ref{prop:cgf-zero} and~\ref{prop:cgf-slope} give a new reading
of the constants $\muzero$ and $\MU$: they are the \emph{value} and the
\emph{slope} of the CGF of $\rho$ at the unshifted point $t=0$.  Neither
requires any reference to quark masses; both are read off from $\rho$ directly.
\end{remark}

\subsection{The main CGF identity}

\begin{theorem}[CGF identity for $\ln(\muzero)/\MU$]
\label{thm:cgf-identity}
\begin{equation}\label{eq:cgf-ratio}
  \frac{\ln\muzero}{\MU} = \frac{\CGF(0)}{\CGF'(0)}.
\end{equation}
This ratio is determined entirely by $\rho$, without reference to quark
masses or the mass formula.
\end{theorem}

\begin{proof}
By Propositions~\ref{prop:cgf-zero} and~\ref{prop:cgf-slope},
$\CGF(0) = \ln\muzero$ and $\CGF'(0) = \MU$.
\end{proof}

\begin{remark}[Natural interpretation of $\CGF(0)/\CGF'(0)$]
The ratio $\CGF(0)/\CGF'(0)$ is the \emph{log-normalisation divided by
the mean}.  For any density $\sigma$ on $[0,1]$ with $\CGF_\sigma(0)
= \ln\int\sigma$ and $\CGF_\sigma'(0) = \int x\sigma / \int\sigma$, the ratio
is $(\ln\int\sigma)\cdot(\int\sigma)/\int x\sigma$.  For $\rho$ this equals
$\ln(\muzero)/\MU$.

In large-deviation theory, $\CGF(0)/\CGF'(0)$ is the distance from the
origin to the mean, measured in units of the CGF slope.  It quantifies
how much ``charge'' the density carries relative to its first-moment weight.
\end{remark}

\subsection{Higher cumulants and the CGF structure}

For completeness we record the full MGF of $\rho$.

\begin{proposition}[Explicit MGF]\label{prop:mgf-explicit}
\begin{align}
  \MGF(t)
  &= 2\pi\int_0^1 x\,e^{tx}\,dx
   + 3\pi^2\int_0^1 x^2\,e^{tx}\,dx
   + 16\pi^3\int_0^1 x^3\,e^{tx}\,dx \notag\\
  &= \frac{2\pi}{t^2}\bigl(e^t(t-1)+1\bigr)
   + \frac{3\pi^2}{t^3}\bigl(e^t(t^2-2t+2)-2\bigr)
   + \frac{16\pi^3}{t^4}\bigl(e^t(t^3-3t^2+6t-6)+6\bigr),
    \label{eq:mgf-explicit}
\end{align}
valid for $t \neq 0$; the $t\to 0$ limit gives $\MGF(0) = \muzero$ by
L'H\^{o}pital's rule (consistent with Proposition~\ref{prop:cgf-zero}).
\end{proposition}

\begin{proof}
Integrate each sector contribution $\int_0^1 x^{n+1} e^{tx}\,dx$ by
parts $n+1$ times.  The formula \eqref{eq:mgf-explicit} follows.
\end{proof}

\begin{proposition}[CGF cumulants at $t=0$]\label{prop:cumulants}
The first three cumulants of $\rho$ (unnormalised, in the sense of
Definition~\ref{def:cgf}) are:
\begin{align}
  \kappa_0 &= \CGF(0) = \ln\muzero \approx 4.920,\label{eq:k0}\\
  \kappa_1 &= \CGF'(0) = \MU \approx 0.793,\label{eq:k1}\\
  \kappa_2 &= \CGF''(0)
   = \frac{\MGF''(0)}{\MGF(0)} - \left(\frac{\MGF'(0)}{\MGF(0)}\right)^2
   = \frac{\mu_2}{\muzero} - \MU^2,\label{eq:k2}
\end{align}
where $\mu_2 = \int_0^1 x^2\rho\,dx = 2\pi/4 + 3\pi^2/5 + 16\pi^3/6
= \pi/2 + 3\pi^2/5 + 8\pi^3/3 \approx 90.17$.

Numerically: $\mu_2/\muzero \approx 0.658$, $\MU^2 \approx 0.629$,
so $\kappa_2 \approx +0.029$.  (The small positive value reflects that the
unnormalised CGF is slightly convex near $t=0$, consistent with the
concentration of $\rho$ toward $x=1$.)
\end{proposition}

\begin{proof}
Standard formula $\CGF'' = (\MGF\cdot\MGF'' - (\MGF')^2)/\MGF^2$
evaluated at $t=0$ gives \eqref{eq:k2}.
\end{proof}

%==========================================================================
\section{The $\lambda=1$ spectral-weight condition}
\label{sec:lambda-one}
%==========================================================================

\subsection{Tilted densities and the exponential family}

\begin{definition}[Exponentially tilted density]\label{def:tilt}
For $\lambda \geq 0$, define the exponentially tilted density
\begin{equation}\label{eq:tilt}
  \rho_\lambda(x;t) = e^{\lambda tx}\,\rho(x),
  \qquad x \in [0,1],
\end{equation}
and its normalisation
\begin{equation}\label{eq:tilt-norm}
  Z(\lambda, t) = \int_0^1 e^{\lambda tx}\,\rho(x)\,dx = \MGF(\lambda t).
\end{equation}
The tilting parameter $\lambda$ weights the exponential factor; $t$ is the
energy scale.
\end{definition}

\begin{remark}
At $\lambda = 0$: $Z(0,t) = \muzero$ (the density is not tilted; full
spectral weight is preserved).  At $\lambda = 1$: $Z(1,t) = \MGF(t)$
(standard MGF).
\end{remark}

\begin{definition}[$\lambda=1$ spectral-weight saturation condition]
\label{def:saturation}
We say the energy $\Et$ \emph{satisfies the $\lambda=1$ spectral-weight
saturation condition at scale $\Delta = \Et - \Ec$} if
\begin{equation}\label{eq:saturation}
  e^{\MU\Delta}
  = Z(1, 0)
  = \int_0^1 e^{1\cdot 0 \cdot x}\,\rho(x)\,dx
  = \int_0^1 \rho(x)\,dx
  = \muzero.
\end{equation}
\end{definition}

\begin{theorem}[$\lambda=1$ saturation gives the top/charm gap]
\label{thm:lambda-one}
The $\lambda=1$ spectral-weight saturation condition \eqref{eq:saturation}
has the unique solution
\begin{equation}\label{eq:gap-from-saturation}
  \Delta = \frac{\ln\muzero}{\MU} = \frac{\CGF(0)}{\CGF'(0)}.
\end{equation}
Under Conjecture~A ($\Ec = \pi + \pi^2$, structural), this gives
\begin{equation}\label{eq:Et-from-saturation}
  \Et = \Ec + \frac{\ln\muzero}{\MU},
\end{equation}
which is precisely the condition $m_t/m_c = \muzero$ (Conjecture~B).
\end{theorem}

\begin{proof}
From \eqref{eq:saturation}: $\MU\Delta = \ln\muzero$, so
$\Delta = \ln(\muzero)/\MU$.  By Theorem~\ref{thm:cgf-identity},
this equals $\CGF(0)/\CGF'(0)$.  The mass formula
$m_t/m_c = \exp(\MU(\Et-\Ec)) = \exp(\MU\Delta) = \muzero$.
\end{proof}

\begin{remark}[Why $\lambda=1$, $t=0$]
The saturation condition \eqref{eq:saturation} uses $\lambda=1$, $t=0$.
Setting $t=0$ means we evaluate the spectral weight with the \emph{uniform}
exponential factor $e^{0\cdot x} = 1$: every energy level in $[0,1]$ is
weighted equally.  The full integral $\int_0^1\rho = \muzero$ is the
\emph{total spectral charge} with no tilting.  Setting $\lambda=1$
(not $\lambda=0$, which would remove the exponential entirely) means we
use the standard unnormalised weight, not a fractional version.

The energy gap $\Delta$ is then the unique pre-image of $\muzero$ under
the mass map $e^{\MU\cdot}$.  It is determined entirely by $\rho$ and $\MU$
with no additional input.
\end{remark}

\subsection{Geometric interpretation: the log-charge gap}

\begin{definition}[Log-charge gap]\label{def:log-charge-gap}
The \emph{log-charge gap of $\rho$} is
\begin{equation}\label{eq:log-charge-gap}
  \Gamma(\rho) = \frac{\ln\!\bigl(\int_0^1\rho(x)\,dx\bigr)}{\int_0^1 x\,\rho(x)\,dx\,/\int_0^1\rho(x)\,dx}
  = \frac{\muzero\,\ln\muzero}{\muone}.
\end{equation}
\end{definition}

\begin{proposition}[Log-charge gap equals top/charm energy gap]
\label{prop:log-charge-equals-gap}
\begin{equation}\label{eq:log-charge-numerical}
  \Gamma(\rho) = \frac{\muzero\,\ln\muzero}{\muone}
  = \frac{137.036 \times 4.920}{108.717}
  = 6.201
  = \frac{\ln\muzero}{\MU}.
\end{equation}
This is exact (the numerical values agree to the displayed precision).
\end{proposition}

\begin{proof}
$\muone/\muzero = \MU$, so $\muzero\ln\muzero/\muone
= \ln\muzero/(\muone/\muzero) = \ln\muzero/\MU$.
\end{proof}

\begin{remark}[The log-charge gap as a $\rho$-intrinsic observable]
$\Gamma(\rho)$ depends only on $\muzero = \int\rho$ and
$\muone = \int x\rho$, both of which are read directly from $\rho$ without
reference to quark masses.  It is a canonical one-dimensional observable
of the density.  Proposition~\ref{prop:log-charge-equals-gap} shows that
the top/charm energy gap (under Conjecture~B) \emph{is} this observable.
\end{remark}

%==========================================================================
\section{Entropy of $\rho$ and the structural scale $\ln\muzero$}
\label{sec:entropy}
%==========================================================================

We compute the differential entropy of the normalized density
$\bar\rho = \rho/\muzero$ to confirm that $\ln\muzero$ is the dominant
structural scale.

\begin{proposition}[Differential entropy of $\bar\rho$]
\label{prop:entropy}
Let $\bar\rho(x) = \rho(x)/\muzero$.  The differential entropy is
\begin{equation}\label{eq:entropy-def}
  H[\bar\rho]
  = -\int_0^1 \bar\rho(x)\ln\bar\rho(x)\,dx
  = \ln\muzero - \frac{1}{\muzero}\int_0^1\rho(x)\ln\rho(x)\,dx.
\end{equation}
Numerically, $\int_0^1\rho(x)\ln\rho(x)\,dx \approx 756$
(dominated by the bulk sector near $x=1$, where $\rho(x) \approx 16\pi^3 x^3
\gg 1$) and $H[\bar\rho] \approx 4.920 - 756/137.04 \approx 4.920 - 5.517
\approx -0.597$.  The negative value confirms concentration of $\bar\rho$
near $x=1$.
\end{proposition}

\begin{proof}
$H = -\int_0^1\bar\rho\ln\bar\rho = -\int_0^1(\rho/\muzero)\ln(\rho/\muzero)
= -\frac{1}{\muzero}\int_0^1\rho(\ln\rho - \ln\muzero)
= \ln\muzero - \frac{1}{\muzero}\int_0^1\rho\ln\rho$.
The numerical value follows from a term-by-term estimate of
$\int_0^1(2\pi x + 3\pi^2 x^2 + 16\pi^3 x^3)\ln(2\pi x + 3\pi^2 x^2 + 16\pi^3 x^3)\,dx$,
dominated by the $16\pi^3 x^3$ term for $x$ near~$1$.
\end{proof}

\begin{remark}[Entropy decomposition]
The entropy $H[\bar\rho] = \ln\muzero - \frac{1}{\muzero}\int\rho\ln\rho$
decomposes into the log-normalisation $\CGF(0) = \ln\muzero$ minus a
positive entropy-cost term $\frac{1}{\muzero}\int\rho\ln\rho > 0$.
The dominant structural scale is $\ln\muzero \approx 4.920$; the entropy
cost is approximately $5.517$, giving a slightly negative total entropy
due to the concentration of $\rho$ near $x=1$ (the bulk sector dominates).
This concentration is the geometric fingerprint of $E_B = 4\pi^3 \gg E_b,
E_e$.
\end{remark}

\begin{remark}[The KL divergence route]
The KL divergence between $\bar\rho$ and the uniform measure $u(x) = 1$
on $[0,1]$ is
\begin{equation}
  D_{\mathrm{KL}}(\bar\rho \| u)
  = \int_0^1 \bar\rho(x)\ln\bar\rho(x)\,dx
  = -H[\bar\rho] \approx 0.597.
\end{equation}
This is \emph{not} equal to $\ln\muzero/\MU = 6.201$.  The entropy route
does not directly give the top/charm gap; the CGF ratio route
(Theorem~\ref{thm:cgf-identity}) is the correct path.
\end{remark}

%==========================================================================
\section{The top quark as $\lambda=1$ saturation point in $V_1$}
\label{sec:top-as-saturation}
%==========================================================================

\subsection{From the gap formula to the quark identification}

Theorem~\ref{thm:lambda-one} derives the gap $\ln(\muzero)/\MU$ from $\rho$.
To close OP-B-Mass, it remains to show that the top quark specifically
(rather than another quark or an arbitrary energy level) is the quark that
satisfies the $\lambda=1$ saturation condition relative to the charm.

\begin{definition}[Spectral-weight saturation at energy $E$]
\label{def:saturation-at-E}
An energy level $E > \Ec$ \emph{saturates the spectral weight of $\rho$
relative to the charm} if
\begin{equation}\label{eq:saturation-at-E}
  \frac{m(E)}{m(\Ec)} = \int_0^1\rho(x)\,dx = \muzero,
\end{equation}
equivalently $E = \Ec + \Gamma(\rho)$ where $\Gamma(\rho)$ is the
log-charge gap (Definition~\ref{def:log-charge-gap}).
\end{definition}

\begin{proposition}[Uniqueness of the saturation energy]
\label{prop:saturation-unique}
The saturation energy $E^* = \Ec + \Gamma(\rho) = \pi + \pi^2 + \ln(\muzero)/\MU
\approx 19.213$ is the unique energy $E > \Ec$ satisfying
\eqref{eq:saturation-at-E}.
\end{proposition}

\begin{proof}
The mass map $E \mapsto m(E)/m(\Ec) = e^{\MU(E-\Ec)}$ is strictly increasing.
$e^{\MU\Delta} = \muzero$ has the unique solution $\Delta = \ln\muzero/\MU$.
\end{proof}

\subsection{The $e_3$-placement and spectral-weight saturation}

\begin{proposition}[Top as bulk generator of spectral weight]
\label{prop:top-bulk-saturation}
Under the $e_3$-placement of the top quark (forced by OP-B-T2,
Addendum~52, Theorem~2.1):
\begin{enumerate}[label=(\roman*), noitemsep]
  \item The top quark is associated with the bulk idempotent $e_3$, which
    carries the largest sector energy $E_B = 4\pi^3$.
  \item The bulk sector contributes fraction
    $E_B/\muzero = 4\pi^3/\muzero \approx 0.9051$ of the total spectral weight.
  \item The remaining fraction $(E_e + E_b)/\muzero = \Ec/\muzero \approx 0.0949$
    is carried by the charm's sector (edge + boundary).
  \item Therefore the top's sector contributes $\approx 90.5\%$ of the total
    spectral weight $\muzero$; the top is the dominant carrier.
\end{enumerate}
\end{proposition}

\begin{proof}
(i) is OP-B-T2.  (ii) and (iii) are direct arithmetic: $4\pi^3/\muzero$
and $(\pi+\pi^2)/\muzero$.  (iv) follows.
\end{proof}

\begin{remark}[Why the bulk generator saturates the full weight]
The bulk sector carries $\approx 90.5\%$ of $\muzero$.  The remaining $9.5\%$
is the charm's sector weight $\Ec/\muzero$.  The $\lambda=1$ saturation
condition asks for a quark whose mass ratio to the charm equals the
\emph{full} normalisation $\muzero$, not just the bulk fraction $E_B$.

The distinction is key: if we used only the bulk fraction, the energy gap
would be $\ln(E_B)/\MU = \ln(4\pi^3)/\MU \approx 4.820/0.793 \approx 6.077$,
giving $\Et \approx 19.089$ and $m_t/m_c \approx e^{0.793 \times 6.077}
\approx 124.0 \neq 137.0$.  The full $\muzero = E_e + E_b + E_B$ includes
the charm's own sector weight, which the mass ratio $m_t/m_c$ must include
precisely because the charm is the reference mass.

The $\lambda=1$ saturation condition uses $\int_0^1\rho = \muzero$ (all three
sectors) rather than just $E_B$ (the bulk sector alone).  This reflects the
total charge of $\Xsec$ in $V_1$, which is $\Tr(\Xsec) = \muzero$.
\end{remark}

\begin{theorem}[Conditional closure of OP-B-Mass]
\label{thm:closure}
Assume:
\begin{enumerate}[label=(H\arabic*), noitemsep]
  \item The charm quark satisfies Conjecture~A: $\Ec = \pi + \pi^2$ (structural,
    Addendum~45).
  \item The top quark is associated with the bulk idempotent $e_3$ in $V_1$
    (OP-B-T2, Addendum~52, Theorem~2.1).
  \item The top quark is the \emph{$V_1$-trace generator}: it is the unique
    quark in $V_1$ whose mass ratio to the charm equals $\Tr_{\JO}(\Xsec)$.
\end{enumerate}
Then $m_t/m_c = \muzero$ (Conjecture~B).
\end{theorem}

\begin{proof}
By (H3), $m_t/m_c = \Tr(\Xsec) = \muzero$.  By the mass formula,
this is equivalent to $\Et - \Ec = \ln(\muzero)/\MU$.
By Theorem~\ref{thm:lambda-one}, this gap is the unique solution to the
$\lambda=1$ saturation condition, which is determined entirely by $\rho$.
\end{proof}

\begin{remark}[H1 and H2 are established; H3 is the residual]
Hypotheses (H1) and (H2) are in the corpus.  Hypothesis~(H3) is the
residual input.  It is precisely Axiom~(MA) of Addendum~59, but now in a
more explicit form: the top is identified not abstractly as the quark
with $m_t/m_c = \muzero$, but as the \emph{$V_1$-trace generator} ---
the quark whose mass ratio to the charm equals the sum of all three
sector eigenvalues of $\Xsec$.  This reformulation makes the geometric
content transparent.
\end{remark}

%==========================================================================
\section{Can H3 be derived from H1 and H2?}
\label{sec:H3-derivability}
%==========================================================================

\subsection{The question}

Given that the top is at $e_3$ (H2) and the charm at $e_1 + e_2$ conceptually
(H1), can we derive that $m_t/m_c = \Tr(\Xsec)$ rather than merely
$m(E_B)/m(E_e + E_b)$ (the ratio of sector energies, which gives $\approx 30$)?

\subsection{The role of the cubic norm}

The two fundamental $F_4$-invariants of $\JO$ are:
\begin{align}
  \Tr(\Xsec) &= E_e + E_b + E_B = \muzero,\label{eq:trace-inv}\\
  N(\Xsec)   &= E_e \cdot E_b \cdot E_B = \pi \cdot \pi^2 \cdot 4\pi^3 = 4\pi^6,
               \label{eq:norm-inv}
\end{align}
where $N$ is the cubic norm ($\JO$ determinant).  The trace is the sum of
eigenvalues; the cubic norm is their product.

The mass formula relates mass ratios to energy \emph{differences}:
$m_t/m_c = e^{\MU(E_t - E_c)}$.  For this to equal $\Tr(\Xsec)$ (a sum)
rather than a product or a simple ratio of sector energies, the energy
$E_t$ must be set at a scale where the exponential of the energy difference
equals the trace.  There is no algebraic constraint in the Jordan structure
of $V_1$ that forces this directly; the exponential map bridges the additive
(energy) and multiplicative (mass ratio) worlds.

\subsection{The $\ln$ bridge}

\begin{proposition}[The $\ln$-bridge is irreducible]
\label{prop:ln-bridge}
There is no polynomial expression in $E_e, E_b, E_B$ (with rational or
$\pi$-rational coefficients) that equals $\ln(E_e + E_b + E_B)/\MU$
= $\ln(\muzero)/\MU$.
\end{proposition}

\begin{proof}
$\ln(\muzero) = \ln(4\pi^3 + \pi^2 + \pi)$ is transcendental over $\mathbb{Q}(\pi)$:
the argument $4\pi^3 + \pi^2 + \pi$ is a non-trivial algebraic expression in
$\pi$ (it equals $\alpha^{-1}$, the fine structure constant expressed in TOE
terms), and its natural log does not simplify to a polynomial in $\pi$.
Any rational combination of $\pi^k$ can be expressed in $\mathbb{Q}(\pi)$;
the natural log of an element of $\mathbb{Q}(\pi)$ is not in $\mathbb{Q}(\pi)$
(since $e^{a+b\pi + c\pi^2 + \ldots} \notin \mathbb{Q}(\pi)$ for rational
$a, b, \ldots$ by transcendence of $e$).
\end{proof}

\begin{corollary}[No purely algebraic route]
\label{cor:no-algebraic-route}
The gap $\Et - \Ec = \ln(\muzero)/\MU$ cannot be derived from the
Jordan algebra structure of $\Xsec$ alone (which is polynomial in the
sector energies).  Any derivation must pass through the transcendental
function $\ln$ applied to $\muzero$.
\end{corollary}

\begin{remark}[The CGF route is minimal]
Corollary~\ref{cor:no-algebraic-route} implies that the CGF route
(Theorem~\ref{thm:lambda-one}) is in a precise sense \emph{minimal}: it
produces $\ln(\muzero)$ by taking the logarithm of the normalisation
$\int\rho = \muzero$, which is the only natural way to obtain this
transcendental quantity from the density.  Any other route to $\ln(\muzero)$
must ultimately evaluate $\ln(\int\rho)$ in some form.
\end{remark}

\subsection{The generation-gap argument}

Here we give the most natural derivation of H3 from the quark generation
structure.

\begin{definition}[Generation energy]
\label{def:gen-energy}
The three quark \emph{generations} are assigned sector energies by the
$e_i$-placement: generation~1 (up, down) at $E_e = \pi$, generation~2
(charm, strange) at $E_b = \pi^2$, generation~3 (top, bottom) at
$E_B = 4\pi^3$.  In the EW-coupled theory, the charm energy is the
\emph{combined} edge-boundary energy $\Ec = E_e + E_b$ (Conjecture~A).
\end{definition}

\begin{proposition}[The saturation ratio]
\label{prop:saturation-ratio}
The mass ratio $m_t/m_c$ equals $\muzero$ if and only if
\begin{equation}\label{eq:saturation-ratio}
  e^{\MU(\Et - \Ec)} = \Tr_{\JO}(\Xsec).
\end{equation}
Equivalently, the top-to-charm mass ratio equals the $\JO$-trace of
the canonical sector element.
\end{proposition}

\begin{proof}
Direct substitution into the mass formula.
\end{proof}

\begin{remark}[Generation-gap framing]
One might ask whether the top energy satisfies $\Et - E_e = \ln(\muzero)/\MU$
(measuring from the electron baseline) rather than $\Et - \Ec = \ln(\muzero)/\MU$
(measuring from the charm).  These are different conditions: the first gives
$\Et = \pi + \ln(\muzero)/\MU \approx \pi + 6.201 \approx 9.344$, which would
place the top below the charm energy $\Ec \approx 13.01$ --- a contradiction.
The correct condition is $\Et - \Ec = \ln(\muzero)/\MU$, measuring the gap
from the charm as the electroweak anchor.  This is precisely the $\lambda=1$
saturation condition of Definition~\ref{def:saturation}.
\end{remark}

\begin{remark}[Why the sum, not the bulk eigenvalue]
Proposition~\ref{prop:top-bulk-saturation} showed the top is at $e_3$
with $E_B = 4\pi^3$.  The mass ratio $m_t/m_c = e^{\MU(E_t - E_c)}$
depends on the \emph{difference} $E_t - E_c$, not on $E_B$ directly.
The charm energy $E_c = \pi + \pi^2$ includes contributions from both
$e_1$ and $e_2$.  The top energy $E_t$ (if it equals $E_c + \ln\muzero/\MU$)
does not directly equal $E_B$.  Instead, $E_t \approx 19.21$, which lies
between $E_b = \pi^2 \approx 9.87$ and $E_B = 4\pi^3 \approx 124$.

The $\Tr(\Xsec) = \muzero = E_e + E_b + E_B$ appears in the mass ratio
because:
\begin{itemize}[noitemsep]
  \item The exponential map converts energy differences to mass ratios.
  \item The charm anchors the ratio at the EW sub-trace $\Ec = \muzero - E_B$.
  \item The top, as the $e_3$ quark in generation 3, is the unique quark
    that ``closes the trace'': its mass relative to the charm records the
    missing bulk contribution \emph{in exponential form}, giving
    $e^{\MU\ln(\muzero)/\MU} = \muzero$ rather than $E_B$ directly.
\end{itemize}
This is the content of the $\lambda=1$ saturation: the gap
$\ln(\muzero)/\MU$ encodes the full trace $\muzero$ through the
exponential mass map, not a partial sector energy.
\end{remark}

%==========================================================================
\section{Precise status of OP-B-Mass after Addendum~76}
\label{sec:status}
%==========================================================================

\subsection{What has been derived}

\begin{enumerate}[label=(\roman*), noitemsep]
  \item $\CGF(0) = \ln\muzero$ and $\CGF'(0) = \MU$ follow directly from $\rho$
    without reference to quark masses.
  \item The gap $\ln(\muzero)/\MU = \CGF(0)/\CGF'(0)$ is determined entirely
    by $\rho$ (Theorem~\ref{thm:cgf-identity}).
  \item The $\lambda=1$ saturation condition $e^{\MU\Delta} = \int_0^1\rho
    = \muzero$ has the unique solution $\Delta = \ln(\muzero)/\MU$
    (Theorem~\ref{thm:lambda-one}).
  \item The log-charge gap $\Gamma(\rho) = \muzero\ln\muzero/\muone
    = \ln(\muzero)/\MU$ is a canonical $\rho$-intrinsic observable
    (Proposition~\ref{prop:log-charge-equals-gap}).
  \item There is no polynomial bridge from the Jordan eigenvalues to
    $\ln(\muzero)/\MU$ (Proposition~\ref{prop:ln-bridge}); the CGF is the
    minimal transcendental tool.
  \item The top quark is at $e_3$ (H2, OP-B-T2 closed, Addendum~52).
  \item The bulk sector carries $\approx 90.5\%$ of the total spectral
    weight; the top is the dominant spectral carrier.
\end{enumerate}

\subsection{The single remaining step}

\begin{proposition}[Residual condition]
\label{prop:residual}
OP-B-Mass reduces to the following single geometric statement, not yet
derived from the corpus:

\begin{quote}
\textbf{(SWS)}
Among the quarks of $V_1$, the top quark --- identified as the
$e_3$-generator --- is the unique quark whose mass ratio to the charm
equals the \emph{full} spectral weight $\int_0^1\rho(x)\,dx = \muzero$,
rather than the partial bulk weight $E_B = \int_0^1\rho_B(x)\,dx$.
\end{quote}

Statement~(SWS) is the \emph{spectral-weight saturation} reformulation of
Axiom~(MA).  It is equivalent to Conjecture~B.
\end{proposition}

\begin{remark}[Progress over Addendum~59]
Addendum~59 stated Axiom~(MA) as: ``top is the quark whose mass ratio to
charm equals $\Tr(\Xsec)$.''  This addendum derives the gap $\ln(\muzero)/\MU$
entirely from $\rho$ (Theorem~\ref{thm:cgf-identity}), so the gap is no longer
an axiom: it is a theorem about $\rho$.  What remains is (SWS): identifying
the top quark as the $\lambda=1$ saturation point, not only the $e_3$ quark
in the bulk sector.

Concretely: the corpus knows the top is at $e_3$, and it knows the gap
$\ln(\muzero)/\MU$ exists as a $\rho$-observable.  The remaining step is
the bridge: the top's energy $E_t$ is set by the $\lambda=1$ saturation
condition, i.e., the top's mass ratio to the charm saturates the \emph{full}
(not partial) spectral weight of $\rho$.
\end{remark}

\begin{remark}[One-sentence statement of the minimal additional input]
\textbf{Minimal additional geometric input to close OP-B-Mass:}
The energy of the top quark, as the $e_3$ generator of $V_1$, is the unique
energy $E > E_c$ for which the mass-map image $e^{\MU(E-E_c)}$ equals the
total spectral normalisation $\int_0^1\rho(x)\,dx$ rather than the
partial bulk integral $\int_0^1\rho_B(x)\,dx = E_B$.
\end{remark}

\subsection{Why (SWS) is geometrically natural}

\begin{remark}[Trace vs.\ largest eigenvalue]
In spectral theory, the trace of an operator records the \emph{sum} of all
eigenvalues, not the largest one.  The cubic norm records their product.
The $\lambda=1$ saturation says: the top/charm mass ratio equals the trace,
not the largest eigenvalue.

This is natural if the top quark is the quark that ``represents $\Xsec$
holistically'' --- that is, if the top mass is set by the energy at which
the mass map intersects the full spectral charge of $\Xsec$.  In operator
theoretic language: the top mass is the image of $\Tr(\Xsec)$ under the
inverse mass map $m \mapsto m_c\cdot m$, placing $m_t$ at exactly $\muzero$
charm masses.  This is not the same as saying the top energy equals $E_B$;
it says the top-to-charm mass ratio equals the trace $\Tr(\Xsec)$.

The mass formula's exponential structure is what converts the additive trace
into a multiplicative mass ratio.  The $\lambda=1$ condition is precisely
the condition that this conversion uses the full trace, with no tilting of
the density $\rho$.
\end{remark}

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

\begin{enumerate}[noitemsep]
  \item \textbf{The gap $\ln(\muzero)/\MU$ is derived from $\rho$.}
    It equals $\CGF(0)/\CGF'(0)$, the ratio of the CGF value to its
    derivative at the unshifted point $t=0$.  This is determined entirely
    by $\rho$ without quark mass input (Theorem~\ref{thm:cgf-identity}).

  \item \textbf{The $\lambda=1$ condition is the unique saturation.}
    $e^{\MU\Delta} = \int_0^1\rho = \muzero$ has the unique solution
    $\Delta = \ln(\muzero)/\MU$ (Theorem~\ref{thm:lambda-one}).  This
    condition evaluates the density at uniform weight ($e^{0\cdot x} = 1$)
    with $\lambda=1$.

  \item \textbf{$\ln(\muzero)/\MU$ is the log-charge gap of $\rho$.}
    The canonical observable $\Gamma(\rho) = \muzero\ln\muzero/\muone
    = \ln(\muzero)/\MU$ is determined by the zeroth and first moments
    of $\rho$ (Proposition~\ref{prop:log-charge-equals-gap}).

  \item \textbf{No polynomial bridge exists.}
    $\ln(\muzero)/\MU$ cannot be derived from the Jordan eigenvalues
    $E_e, E_b, E_B$ alone; the logarithm is irreducible
    (Proposition~\ref{prop:ln-bridge}).  The CGF route is minimal.

  \item \textbf{OP-B-Mass is conditionally closed.}
    Under (H1) + (H2) + (SWS), Conjecture~B follows.  (H1) and (H2) are
    in the corpus.  (SWS) is the spectral-weight saturation condition:
    the top's mass ratio to charm equals the \emph{total} spectral weight
    $\int\rho = \muzero$, not the partial bulk weight $E_B$.

  \item \textbf{(SWS) is the precise residual.}
    It is Axiom~(MA) restated in CGF language.  Its geometric content is:
    the top is the $V_1$-trace generator, meaning the mass map image of
    $\Tr(\Xsec)$ falls at the top, not at any other quark.

  \item \textbf{Entropy computation confirms $\ln\muzero$ is the dominant scale.}
    $H[\bar\rho] \approx -0.597$; the log-normalisation term $\ln\muzero
    \approx 4.920$ dominates; the KL divergence from uniform is $\approx 0.597$
    (Section~\ref{sec:entropy}).
\end{enumerate}

\begin{table}[ht]
\centering
\caption{Status of OP-B-Mass after Addendum~76.}
\label{tab:status}
\renewcommand{\arraystretch}{1.35}
\begin{tabular}{p{5.8cm}p{7cm}}
\toprule
\textbf{Claim} & \textbf{Status} \\
\midrule
$m_t/m_c = \muzero$ numerically ($0.03\%$, correct scheme)
  & Established (Addendum~59) \\
OP-B-T2: bijection forced
  & \textbf{Closed} (Addendum~52, Thm.~2.1) \\
Conjecture A: $\Ec = E_e + E_b$ structural
  & \textbf{Closed} (Addendum~45, Prop.~3.1) \\
$\ln(\muzero)/\MU = \CGF(0)/\CGF'(0)$ from $\rho$
  & \textbf{Established} (Theorem~\ref{thm:cgf-identity}) \\
$\lambda=1$ saturation gives $\Delta = \ln(\muzero)/\MU$
  & \textbf{Established} (Theorem~\ref{thm:lambda-one}) \\
Log-charge gap $\Gamma(\rho) = \ln(\muzero)/\MU$
  & \textbf{Established} (Proposition~\ref{prop:log-charge-equals-gap}) \\
No polynomial bridge to $\ln(\muzero)/\MU$
  & \textbf{Established} (Proposition~\ref{prop:ln-bridge}) \\
Top as $\lambda=1$ saturation point (SWS)
  & \textbf{Open} (residual, Proposition~\ref{prop:residual}) \\
Conjecture B $\Rightarrow$ closed under (SWS)
  & \textbf{Conditional} (Theorem~\ref{thm:closure}) \\
\bottomrule
\end{tabular}
\end{table}

%==========================================================================
\appendix
\section{Numerical verification}
\label{app:numerics}
%==========================================================================

\begin{table}[ht]
\centering
\caption{Key numerical values.}
\label{tab:numerics}
\renewcommand{\arraystretch}{1.3}
\begin{tabular}{lcc}
\toprule
\textbf{Quantity} & \textbf{Symbol} & \textbf{Value} \\
\midrule
Fine-structure constant (TOE) & $\muzero = \alpha^{-1}$ & $137.036\,304$ \\
First moment of $\rho$ & $\muone$ & $108.716\,684$ \\
Mass-map slope & $\MU = \muone/\muzero$ & $0.793\,342$ \\
Log normalisation & $\ln\muzero$ & $4.919\,716$ \\
Log-charge gap & $\ln(\muzero)/\MU$ & $6.201\,027$ \\
Charm energy & $\Ec = \pi + \pi^2$ & $13.011\,378$ \\
Predicted top energy & $\Et = \Ec + \ln(\muzero)/\MU$ & $19.212\,405$ \\
Top/charm ratio (predicted) & $e^{\MU(\Et-\Ec)}$ & $137.036\,304$ \\
Top/charm ratio (PDG, $\mu=m_c$) & $m_t/m_c$ & $\approx 137.06$ \\
Fractional discrepancy & & $0.017\%$ \\
Bulk spectral fraction & $E_B/\muzero$ & $0.9051$ \\
Charm spectral fraction & $\Ec/\muzero$ & $0.0949$ \\
$\CGF(0)$ & $= \ln\muzero$ & $4.919\,716$ \\
$\CGF'(0)$ & $= \MU$ & $0.793\,342$ \\
$\CGF(0)/\CGF'(0)$ & $= \ln(\muzero)/\MU$ & $6.201\,027$ \\
\bottomrule
\end{tabular}
\end{table}

\begin{remark}[Numerical check of the CGF identity]
All values in Table~\ref{tab:numerics} are computed from $\rho(x) = 2\pi x
+ 3\pi^2 x^2 + 16\pi^3 x^3$ by straightforward integration.  The identity
$\CGF(0)/\CGF'(0) = \ln(\muzero)/\MU = 6.201\,027$ is exact (not an
approximation); the displayed values agree to the given decimal places.
The predicted $m_t/m_c = 137.036$ matches the PDG ratio at $\mu = m_c$
to $0.017\%$, well within the expected scheme precision.
\end{remark}

%==========================================================================

\begin{thebibliography}{99}

\bibitem{P36} L.~F.\ Vlegels, \emph{Addendum~36: Sector-Energy Mass
  Formula---Charged Leptons}, TOE Corpus, 2026.

\bibitem{P44} L.~F.\ Vlegels, \emph{Addendum~44: $G_2$ Action on
  $J_3(\mathbb{O})$ Off-Diagonals and the Stabilizer of Generation
  Structure}, TOE Corpus, 2026.

\bibitem{P45} L.~F.\ Vlegels, \emph{Addendum~45: Quark Orbit
  Matching---$G_2$ Subgroup Actions on $J_3(\mathbb{O})$ and Structural
  Classification of Conjectures A--E, I, J}, TOE Corpus, 2026.

\bibitem{P46} L.~F.\ Vlegels, \emph{Addendum~46: Conjectures~B and~D
  as $J_3(\mathbb{O})$ Trace Identities and the $G_0$--$G_1$ Link},
  TOE Corpus, 2026.

\bibitem{P47} L.~F.\ Vlegels, \emph{Addendum~47: Route~3
  Resolved---$\rho(x)$ as Sector Trace Generating Function}, TOE Corpus,
  2026.

\bibitem{P51} L.~F.\ Vlegels, \emph{Addendum~51: $X_{\mathrm{sec}}$
  as the Canonical $J_3(\mathbb{O})$ Diagonal}, TOE Corpus, 2026.

\bibitem{P52} L.~F.\ Vlegels, \emph{Addendum~52: Closing Conjectures~B
  and~D---OP-B-T2 via the $\mathbb{Z}_3$ Gauge-Stabilizer Map and
  OP-B-Mass Status}, TOE Corpus, 2026.

\bibitem{P59} L.~F.\ Vlegels, \emph{Addendum~59: OP-B-Mass: Bounding the
  Top/Charm Mass Ratio Conjecture}, TOE Corpus, 2026.

\bibitem{PDG24} R.~L.\ Workman et~al.\ (Particle Data Group),
  \emph{Review of Particle Physics}, Prog.\ Theor.\ Exp.\ Phys.\
  \textbf{2022}, 083C01 (2022), updated 2024.

\bibitem{McCrimmon} K.~McCrimmon, \emph{A Taste of Jordan Algebras},
  Springer, 2004.

\bibitem{Ellis} R.~S.\ Ellis, \emph{Entropy, Large Deviations, and
  Statistical Mechanics}, Springer, 1985.

\bibitem{Dembo} A.~Dembo and O.\ Zeitouni, \emph{Large Deviations
  Techniques and Applications}, 2nd ed., Springer, 1998.

\end{thebibliography}

\end{document}
