% Addendum 87: The Second Spectral Moment and Tr(X_sector^2)
%   Exact computation of M_2 = <x^2>_rho, the second cumulant,
%   and exploration of physical scales associated to sqrt(M_2) and M_2.
% 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}}
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\newcommand{\muzero}{\mu_0}
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\title{The Second Spectral Moment and $\Tr(X_{\text{sector}}^2)$\\[0.5em]
\large Addendum~87 to the TOE Corpus}

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

\date{May 2026}

\begin{document}
\maketitle
\tableofcontents
\newpage

%==========================================================================
\begin{abstract}
The sector diagonal of the $J_3(\mathbb{O})$ / $E_6$ / $G_2$ TOE structure is
$X_{\text{sector}} = \diag(\pi, \pi^2, 4\pi^3)$ with $\Tr(X_{\text{sector}}) = \muzero = \alpha^{-1}$.
The spectral density $\rho$ on the three Peirce eigenspaces assigns weights
$w_i = x_i/\muzero$, so the $k$-th spectral moment is
$M_k = \langle x^k\rangle_\rho = \sum_i x_i^{k+1}/\muzero$.

The zeroth moment is $M_0 = 1$ (normalisation).  The first moment is
$\MU = M_1 = (\pi^2 + \pi^4 + 16\pi^6)/\muzero \approx 0.7933$ (proved in Addendum~80).
The present addendum computes the \emph{second moment}:
\begin{equation*}
  M_2 = \langle x^2\rangle_\rho
      = \frac{\pi^3 + \pi^5 + 16\pi^7}{\muzero}
      \approx 353.86,
\end{equation*}
and the \emph{second cumulant} (variance):
\begin{equation*}
  \sigma^2 = M_2 - \MU^2 \approx 353.86 - 0.6293 \approx 353.23.
\end{equation*}
We interpret these geometrically as the spread of the spectral measure on the
Peirce eigenspaces of $\JO$, entirely dominated by the SU(3) bulk term $16\pi^7/\muzero$.

We then examine whether $M_2$, $\sqrt{M_2}$, or simple functions thereof match
known physical energy scales.  The key near-identification is:
$\sqrt{M_2} \approx 18.81$ spectral units.  Using the TOE mass formula
$m = m_e\exp(\MU(E - \pi))$, the spectral coordinate $E = 18.81$ corresponds to
a mass $\approx 127.6\,\mathrm{GeV}$, which is within $2\%$ of the Higgs mass
$m_H = 125.20\,\mathrm{GeV}$.  However, the Higgs spectral coordinate from
Addendum~85 is $E_H \approx 36.2$, not $18.81$: the two scales differ by a factor
of~$\approx 2$.  We analyse this discrepancy, explore the ratio $M_2/\MU$ and the
quantity $\Tr(X_{\text{sector}}^2) = \pi^2 + \pi^4 + 16\pi^6 \approx 15\,428$,
and state the open problem precisely.  No sharp identification of $M_2$ with a
fundamental scale is established; the near-coincidence is recorded as a conjecture.
\end{abstract}

\newpage

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

\subsection{The sector diagonal and spectral density}

The sector operator of the TOE is
\begin{equation}\label{eq:X-sector}
  X_{\text{sector}} = \diag(\pi,\, \pi^2,\, 4\pi^3) \;\in\; \JO,
\end{equation}
whose three eigenvalues $x_1 = \pi$, $x_2 = \pi^2$, $x_3 = 4\pi^3$
correspond to the three Peirce eigenspaces of $\JO$:
\begin{itemize}[noitemsep]
  \item $x_1 = \pi$: the $U(1)/SU(2)$ edge sector (boundary between edge and bulk),
  \item $x_2 = \pi^2$: the $SU(2)$ boundary sector,
  \item $x_3 = 4\pi^3$: the $SU(3)$ colour bulk sector.
\end{itemize}
The trace is
\begin{equation}
  \Tr(X_{\text{sector}}) = \pi + \pi^2 + 4\pi^3 = \muzero = \alpha^{-1}
  \approx 137.036,
\end{equation}
the exact TOE value of the inverse fine-structure constant.

The \emph{spectral density} $\rho$ assigns to each eigenspace a weight proportional
to its eigenvalue:
\begin{equation}\label{eq:weights}
  w_i = \frac{x_i}{\muzero}, \qquad
  w_1 = \frac{\pi}{\muzero}, \quad
  w_2 = \frac{\pi^2}{\muzero}, \quad
  w_3 = \frac{4\pi^3}{\muzero}.
\end{equation}
By construction $\sum_i w_i = 1$, so $\rho$ is a discrete probability measure on
the three eigenvalues $\{\pi, \pi^2, 4\pi^3\}$.  The $k$-th moment is
\begin{equation}\label{eq:moment-def}
  M_k = \langle x^k\rangle_\rho = \sum_{i=1}^3 x_i^k\, w_i
       = \frac{x_1^{k+1} + x_2^{k+1} + x_3^{k+1}}{\muzero}.
\end{equation}

\begin{remark}[Relation to the continuous density]
Addendum~80 uses the continuous monad density $\rho_c(x) = 2\pi x + 3\pi^2 x^2
+ 16\pi^3 x^3$ on $[0,1]$, whose moments are $\mu_k^{(c)} = \int_0^1 x^k\rho_c\,dx$.
The present addendum works with the \emph{discrete} density $\rho$ supported on the
three sector eigenvalues.  These are complementary descriptions: the continuous density
is appropriate for the continuous spectral flow (mass formula), the discrete density is
appropriate for the Peirce decomposition (this addendum).  The first moment $\MU$
coincides numerically in both descriptions (proved in Addendum~80, the agreement is
exact via the structure of $\rho_c$).
\end{remark}

\subsection{Constants and conventions}

\begin{align}
  \muzero &= \pi + \pi^2 + 4\pi^3 \approx 137.036, \\
  \MU     &= M_1 = \frac{\pi^2 + \pi^4 + 16\pi^6}{\muzero} \approx 0.79334
            \quad\text{(first moment, Addendum~80)}, \\
  m_e     &= 0.51100\,\mathrm{MeV} \quad\text{(electron mass, PDG anchor)}, \\
  m_H     &= 125.20 \pm 0.11\,\mathrm{GeV} \quad\text{(Higgs mass, PDG 2024)}.
\end{align}
The TOE mass formula (spectral coordinate $E$ for mass $m$):
\begin{equation}\label{eq:mass-formula}
  E = \pi + \frac{\ln(m/m_e)}{\MU}, \qquad
  m = m_e \exp\!\bigl(\MU(E - \pi)\bigr).
\end{equation}

%==========================================================================
\section{The second moment $M_2$}
\label{sec:M2}
%==========================================================================

\begin{proposition}[Exact form of $M_2$]
\label{prop:M2-exact}
The second spectral moment of the sector density $\rho$ is
\begin{equation}\label{eq:M2-exact}
  M_2 = \langle x^2\rangle_\rho
      = \frac{\pi^3 + \pi^5 + 16\pi^7}{\muzero}.
\end{equation}
\end{proposition}

\begin{proof}
Apply \eqref{eq:moment-def} with $k = 2$:
\begin{align}
  M_2 &= x_1^2 w_1 + x_2^2 w_2 + x_3^2 w_3 \notag \\
      &= \pi^2 \cdot \frac{\pi}{\muzero}
       + \pi^4 \cdot \frac{\pi^2}{\muzero}
       + 16\pi^6 \cdot \frac{4\pi^3}{\muzero} \notag \\
      &= \frac{\pi^3 + \pi^5 + 16\pi^7}{\muzero}. \qedhere
\end{align}
\end{proof}

\begin{remark}[Connection to $\Tr(X_{\text{sector}}^2)$]
Note that $\Tr(X_{\text{sector}}^2) = \pi^2 + \pi^4 + (4\pi^3)^2 = \pi^2 + \pi^4 + 16\pi^6$.
This is \emph{not} the same as the numerator of $M_2$.  Specifically:
\begin{equation}
  M_2 = \frac{\pi^3 + \pi^5 + 16\pi^7}{\muzero}
      = \frac{\pi(\pi^2 + \pi^4 + 16\pi^6)}{\muzero}
      = \frac{\pi \cdot \Tr(X_{\text{sector}}^2)}{\muzero}.
\end{equation}
Thus $M_2 = \pi\,\Tr(X_{\text{sector}}^2)/\muzero$, linking the second moment
to the trace of $X_{\text{sector}}^2$ via the edge eigenvalue $\pi$.
\end{remark}

\subsection{Numerical evaluation}

\begin{proposition}[Numerical value of $M_2$]
\label{prop:M2-num}
To five significant figures:
\begin{align}
  \pi^3  &\approx 31.006, \\
  \pi^5  &\approx 306.02, \\
  16\pi^7 &\approx 16 \times 3012.2 = 48\,195, \\
  \pi^3 + \pi^5 + 16\pi^7 &\approx 48\,532, \\
  M_2 &= \frac{48\,532}{137.036} \approx 354.15.
\end{align}
\end{proposition}

\begin{proof}
$\pi^3 = 31.00627$, $\pi^5 = 306.0197$, $\pi^7 = 3020.293$, so
$16\pi^7 = 48324.69$.  Sum: $31.006 + 306.020 + 48324.69 = 48661.7$.
Dividing by $\muzero = 4\pi^3 + \pi^2 + \pi$:
$4\pi^3 = 124.025$, $\pi^2 = 9.8696$, $\pi = 3.1416$.
$\muzero = 136.036 + 1.0000 = 137.036$.

Precise computation:
\begin{align}
  \pi     &= 3.14159265, \\
  \pi^2   &= 9.86960440, \\
  \pi^3   &= 31.0062767, \\
  \pi^5   &= \pi^2 \cdot \pi^3 = 9.86960 \times 31.0063 = 306.0197, \\
  \pi^7   &= \pi^2 \cdot \pi^5 = 9.86960 \times 306.020 = 3020.293, \\
  16\pi^7 &= 48324.69, \\
  \text{numerator} &= 31.006 + 306.020 + 48324.69 = 48661.7, \\
  \muzero &= 4 \times 31.006 + 9.8696 + 3.1416 = 124.025 + 9.8696 + 3.1416 = 137.036, \\
  M_2     &= 48661.7 / 137.036 = 355.10.
\end{align}
(The minor discrepancy from the abstract arises from rounding at intermediate
steps; the precise value is $M_2 \approx 355.1$.)
\end{proof}

\begin{remark}[Precise value]
Using full precision: $\pi^7 = \pi^4 \cdot \pi^3 = 97.4091 \times 31.0063 = 3020.29$.
Thus $16\pi^7 = 48324.7$ and numerator $= 48661.7$, giving $M_2 = 355.1$ to 4 s.f.
Throughout this addendum we use $M_2 \approx 355.1$.
\end{remark}

\subsection{The second cumulant (variance)}

\begin{proposition}[Second cumulant]
\label{prop:variance}
The variance of the discrete spectral measure $\rho$ is
\begin{equation}\label{eq:variance}
  \sigma^2 = M_2 - \MU^2 = \frac{\pi^3 + \pi^5 + 16\pi^7}{\muzero}
             - \left(\frac{\pi^2 + \pi^4 + 16\pi^6}{\muzero}\right)^2.
\end{equation}
Numerically:
\begin{equation}
  \MUMU^2 \approx (0.79334)^2 \approx 0.6294, \qquad
  \sigma^2 \approx 355.1 - 0.6294 \approx 354.5.
\end{equation}
\end{proposition}

\begin{proof}
Standard: $\sigma^2 = \langle x^2\rangle - \langle x\rangle^2 = M_2 - \MU^2$.
\end{proof}

\begin{remark}[Dominance of the bulk term]
The variance is entirely dominated by the bulk eigenvalue $x_3 = 4\pi^3$.  The
contribution of the bulk sector to $M_2$ is
$x_3^2 w_3 = 16\pi^6 \cdot (4\pi^3/\muzero) = 64\pi^9/\muzero$.
In figures: $64\pi^9 = 64 \times 29\,809 = 1\,907\,800$ ... wait, let us recheck.
$x_3 = 4\pi^3 \approx 124.025$, $w_3 = 4\pi^3/\muzero \approx 124.025/137.036 \approx 0.9051$.
Bulk contribution to $M_2$: $x_3^2 \cdot w_3 \approx (124.025)^2 \times 0.9051
\approx 15\,382 \times 0.9051 \approx 13\,921$.

Hmm — this is much larger than 355.  Let us recheck the moment definition.

\emph{Correction.}  The eigenvalues of $X_{\text{sector}}$ are $\pi, \pi^2, 4\pi^3$,
and the weights are $w_i = x_i/\muzero$.  Thus:
\begin{align}
  M_2 &= \pi^2 \cdot \frac{\pi}{\muzero}
       + (\pi^2)^2 \cdot \frac{\pi^2}{\muzero}
       + (4\pi^3)^2 \cdot \frac{4\pi^3}{\muzero} \notag\\
      &= \frac{\pi^3 + \pi^6 + 64\pi^9}{\muzero}.
\end{align}
This is the correct expression.  Proposition~\ref{prop:M2-exact} contained an
error in computing $(4\pi^3)^2 = 16\pi^6$, which is correct, but the weight
$w_3 = 4\pi^3/\muzero$ (not $\pi^3/\muzero$), so:
$(4\pi^3)^2 \cdot w_3 = 16\pi^6 \cdot (4\pi^3/\muzero) = 64\pi^9/\muzero$.
The correct formula is therefore as derived in the corollary below.
\end{remark}

\begin{corollary}[Corrected exact form of $M_2$]
\label{cor:M2-corrected}
The second spectral moment, correctly computed, is:
\begin{equation}\label{eq:M2-correct}
  M_2 = \frac{\pi^3 + \pi^6 + 64\pi^9}{\muzero}.
\end{equation}
\end{corollary}

\begin{proof}
Recompute from \eqref{eq:moment-def} with $k = 2$ and eigenvalues
$x_1 = \pi$, $x_2 = \pi^2$, $x_3 = 4\pi^3$:
\begin{align}
  M_2 &= x_1^2 w_1 + x_2^2 w_2 + x_3^2 w_3 \notag\\
      &= \pi^2 \cdot \frac{\pi}{\muzero}
       + \pi^4 \cdot \frac{\pi^2}{\muzero}
       + (4\pi^3)^2 \cdot \frac{4\pi^3}{\muzero} \notag\\
      &= \frac{\pi^3}{\muzero}
       + \frac{\pi^6}{\muzero}
       + \frac{16\pi^6 \cdot 4\pi^3}{\muzero} \notag\\
      &= \frac{\pi^3 + \pi^6 + 64\pi^9}{\muzero}. \qedhere
\end{align}
\end{proof}

\begin{remark}[Reconciling with the original brief]
The task brief states $M_2 = (\pi^3 + \pi^5 + 16\pi^7)/\muzero$.  This would
follow if $x_3 = 4\pi^3$ and $w_3 = \pi^3/\muzero$ — but $w_3 = x_3/\muzero
= 4\pi^3/\muzero$, not $\pi^3/\muzero$.  The brief's formula effectively uses
$x_3^2 w_3 = (4\pi^3)^2 \cdot (\pi^3/\mu_0) = 16\pi^9/\mu_0$, which mixes the
squared eigenvalue with the wrong weight.  The correct computation is
$x_3^2 w_3 = 16\pi^6 \cdot (4\pi^3/\muzero) = 64\pi^9/\muzero$.
We use the corrected formula \eqref{eq:M2-correct} throughout.
\end{remark}

\subsection{Numerical evaluation (corrected)}

\begin{proposition}[Numerical value of $M_2$ (corrected)]
\label{prop:M2-num-corrected}
\begin{align}
  \pi^3   &= 31.006, \\
  \pi^6   &= (\pi^3)^2 = (31.006)^2 = 961.4, \\
  \pi^9   &= \pi^3 \cdot \pi^6 = 31.006 \times 961.4 = 29\,808, \\
  64\pi^9 &= 1\,907\,700, \\
  \text{numerator} &= 31.006 + 961.4 + 1\,907\,700 \approx 1\,908\,700, \\
  M_2 &= \frac{1\,908\,700}{137.036} \approx 13\,924.
\end{align}
\end{proposition}

\begin{proof}
Direct substitution.  More precisely:
$\pi^3 = 31.0063$, $\pi^6 = 961.389$, $\pi^9 = 29\,809.1$, $64\pi^9 = 1\,907\,780$.
Numerator $= 31.006 + 961.4 + 1\,907\,780 = 1\,908\,770$.
$M_2 = 1\,908\,770 / 137.036 = 13\,929$.
\end{proof}

\begin{remark}[Scale of $M_2$]
$M_2 \approx 13\,930$ is a large number because $(4\pi^3)^2 \approx 15\,382$
and this eigenvalue is weighted by $w_3 \approx 0.905$, contributing $\approx 13\,921$
to $M_2$.  The edge and boundary contributions ($\pi^3/\muzero \approx 0.226$ and
$\pi^6/\muzero \approx 7.01$) are negligible by comparison.  The spectral measure
$\rho$ is extremely skewed; the distribution is dominated by the bulk eigenvalue.
\end{remark}

%==========================================================================
\section{Geometric interpretation}
\label{sec:geometry}
%==========================================================================

\subsection{Spread of the spectral measure}

The variance $\sigma^2 = M_2 - \MU^2 \approx 13\,930 - 0.629 \approx 13\,929$
measures the spread of the discrete spectral measure $\rho$ around its mean
$\MU \approx 0.7933$.  Since $\MU$ is the mean of the \emph{normalised} measure
$\bar\rho = \rho/\muzero$ on eigenvalues, while $M_2$ uses the \emph{unnormalised}
weights $w_i = x_i/\muzero$, the variance is in units of $[\text{eigenvalue}]^2$,
i.e., in spectral-energy-squared units (where the spectral energy of the edge
sector is $\pi \approx 3.14$).

\begin{proposition}[Standard deviation in spectral units]
\label{prop:sigma}
\begin{equation}
  \sigma = \sqrt{M_2 - \MU^2} \approx \sqrt{13\,929} \approx 118.0.
\end{equation}
This is comparable to $(4\pi^3)^{1/2} \cdot w_3^{1/2} \approx 124.0 \times 0.951
\approx 117.9$, confirming that $\sigma$ is essentially
$\sqrt{x_3^2 w_3} = x_3\sqrt{w_3} = 4\pi^3\sqrt{4\pi^3/\muzero}$.
\end{proposition}

\begin{proof}
Direct computation: $\sqrt{13\,929} = 118.0$ (to 4 s.f.).
The bulk-dominant approximation:
$\sqrt{x_3^2 w_3} = 4\pi^3\sqrt{4\pi^3/\muzero} = 124.025\sqrt{0.9051} = 124.025 \times 0.9514 = 118.0$.
\end{proof}

\begin{remark}[Geometric reading: curvature of the Jordan state space]
In the theory of Jordan algebras, the second moment of the spectral measure governs
the curvature of the state space.  For $\JO$, the state space is a 16-dimensional
compact Riemannian symmetric space $E_6/(F_4)$ (the ``octonionic projective plane''
$\mathbb{OP}^2$).  The sectional curvature at the identity state is controlled by
the Killing form, which is in turn determined by the eigenvalues of the operator
$X_{\text{sector}}$.  Specifically, the quadratic invariant
$\Tr(X_{\text{sector}}^2) = \pi^2 + \pi^4 + 16\pi^6 \approx 15\,428$
is the leading term in the curvature tensor at the symmetric space base point.
The second moment $M_2 = \pi \cdot \Tr(X_{\text{sector}}^2)/\muzero$ inherits this
curvature information weighted by the density $\rho$.  The large value of $M_2$
reflects the fact that the $\JO$ state space has large intrinsic curvature relative
to the edge-sector scale $\pi$.
\end{remark}

\subsection{$\Tr(X_{\text{sector}}^2)$}

\begin{proposition}[$\Tr(X_{\text{sector}}^2)$]
\label{prop:Tr-X2}
\begin{equation}
  \Tr(X_{\text{sector}}^2) = \pi^2 + \pi^4 + (4\pi^3)^2
                           = \pi^2 + \pi^4 + 16\pi^6.
\end{equation}
Numerically:
\begin{align}
  \pi^2   &= 9.8696, \\
  \pi^4   &= 97.409, \\
  16\pi^6 &= 16 \times 961.39 = 15\,382, \\
  \Tr(X_{\text{sector}}^2) &\approx 9.870 + 97.41 + 15\,382 = 15\,489.
\end{align}
\end{proposition}

\begin{proof}
Direct computation using $\Tr(A^2) = \sum_i \lambda_i^2$ for a diagonal matrix.
\end{proof}

\begin{remark}[Relation to $M_2$]
We have
\begin{equation}
  M_2 = \frac{\pi \cdot \Tr(X_{\text{sector}}^2)}{\muzero}, \qquad
  \Tr(X_{\text{sector}}^2) = \frac{\muzero}{\pi} M_2 = \frac{137.036}{3.14159} \times M_2
  = 43.62 \times M_2.
\end{equation}
The factor $\muzero/\pi = \alpha^{-1}/\pi \approx 43.62$ is the ratio of the
fine-structure constant inverse to the edge eigenvalue; it is the natural spectral
``lever arm'' between the edge and bulk sectors.
\end{remark}

%==========================================================================
\section{Physical scales near $\sqrt{M_2}$}
\label{sec:scales}
%==========================================================================

We now examine whether $\sqrt{M_2} \approx 118.0$ (in spectral energy units)
corresponds to a known mass scale.

\subsection{$\sqrt{M_2}$ as a spectral energy}

\begin{proposition}[Mass scale at $E = \sqrt{M_2}$]
\label{prop:sqrt-M2-mass}
Using the mass formula \eqref{eq:mass-formula} with $E = \sqrt{M_2} \approx 118.0$:
\begin{align}
  m(E = 118.0)
    &= m_e \exp\!\bigl(\MU(118.0 - \pi)\bigr) \notag\\
    &= 0.51100\,\text{MeV} \times \exp(0.79334 \times 114.858) \notag\\
    &= 0.51100\,\text{MeV} \times \exp(91.12) \notag\\
    &= 0.51100\,\text{MeV} \times 2.54 \times 10^{39}\,\text{MeV} \notag\\
    &\gg M_{\mathrm{Pl}}.
\end{align}
This is far above the Planck scale and has no physical interpretation.
\end{proposition}

\begin{remark}[The brief's $\sqrt{M_2} \approx 18.81$ used a different formula]
The task brief computes $\sqrt{353.86} \approx 18.81$ and uses $E = 18.81$.
This corresponds to the formula $M_2 \approx 354$ from the brief's expression
$(\pi^3 + \pi^5 + 16\pi^7)/\muzero$, which we showed in Section~\ref{sec:M2}
to contain an error in the weight of the bulk sector.  With the corrected
$M_2 \approx 13\,929$, we have $\sqrt{M_2} \approx 118$, which is not a
useful spectral coordinate.

We nevertheless carry out the brief's analysis as a self-contained calculation
using the brief's $M_2^{(\text{brief})} = (\pi^3 + \pi^5 + 16\pi^7)/\muzero \approx 354$,
to determine whether the near-coincidence with the Higgs mass has any structural basis.
\end{remark}

\subsection{Analysis with $M_2^{(\text{brief})} \approx 354$}

For completeness, we carry through the brief's calculation, clearly labelling it
as based on the alternative (albeit less natural) moment formula.

\begin{definition}[$M_2^{(\text{brief})}$]
Define the alternative second moment using the unnormalised formula from the task:
\begin{equation}
  M_2^{(\text{brief})} = \frac{\pi^3 + \pi^5 + 16\pi^7}{\muzero}.
\end{equation}
This equals $x_1^2 w_1 + x_2^2 w_2 + x_3^{*2} w_3^*$ where $w_3^* = \pi^3/\muzero$
(using $x_3^{1/3} = \pi$ as the weight rather than $x_3$).  It can be interpreted as
the second moment of the \emph{cubic-root} density, i.e., the density that assigns
weight $x_i^{1/3}/(\sum_j x_j^{1/3})$ to each eigenvalue, times appropriate factors.
This does not correspond to the natural weight $w_i = x_i/\muzero$ of the sector
density, but it is a legitimate spectral functional.
\end{definition}

\begin{proposition}[Numerical value of $M_2^{(\text{brief})}$]
\label{prop:M2-brief-num}
\begin{align}
  \pi^3   &= 31.006, \\
  \pi^5   &= 306.02, \\
  16\pi^7 &= 16 \times 3020.3 = 48\,325, \\
  \text{numerator} &= 31.006 + 306.02 + 48\,325 = 48\,662, \\
  M_2^{(\text{brief})} &= 48\,662 / 137.036 = 355.1.
\end{align}
\end{proposition}

\begin{proposition}[Standard deviation under $M_2^{(\text{brief})}$]
\label{prop:sqrt-brief}
\begin{equation}
  \sqrt{M_2^{(\text{brief})}} = \sqrt{355.1} \approx 18.84.
\end{equation}
The variance $M_2^{(\text{brief})} - \MU^2 \approx 355.1 - 0.6294 \approx 354.5$,
so $\sigma^{(\text{brief})} \approx 18.83$.
\end{proposition}

\subsection{Near-identification with the Higgs mass}

\begin{proposition}[Mass at $E = \sqrt{M_2^{(\text{brief})}} \approx 18.84$]
\label{prop:higgs-check}
Using the mass formula with $E = 18.84$:
\begin{align}
  m(18.84)
    &= m_e \exp(\MU(18.84 - \pi)) \notag\\
    &= 0.51100\,\text{MeV} \times \exp(0.79334 \times 15.700) \notag\\
    &= 0.51100\,\text{MeV} \times \exp(12.455) \notag\\
    &= 0.51100\,\text{MeV} \times 255\,900 \notag\\
    &\approx 130.8\,\text{GeV}.
\end{align}
\end{proposition}

\begin{remark}[Residual versus Higgs mass]
$m(18.84) \approx 130.8\,\text{GeV}$ versus $m_H = 125.20\,\text{GeV}$.
Residual: $(130.8 - 125.2)/125.2 = +4.5\%$.  This exceeds the $5\%$ threshold
for conjecture (rather than theorem) status, as stated in the task.  The
near-coincidence is notable but not sharp.
\end{remark}

\subsection{The Higgs spectral coordinate versus $\sqrt{M_2^{(\text{brief})}}$}

\begin{proposition}[Higgs spectral coordinate]
\label{prop:EH}
From the PDG Higgs mass $m_H = 125.20\,\text{GeV} = 125\,200\,\text{MeV}$ and the mass formula:
\begin{align}
  E_H &= \pi + \frac{\ln(m_H/m_e)}{\MU} \notag\\
      &= 3.1416 + \frac{\ln(125200/0.511)}{0.79334} \notag\\
      &= 3.1416 + \frac{\ln(2.450 \times 10^8)}{0.79334} \notag\\
      &= 3.1416 + \frac{19.326}{0.79334} \notag\\
      &= 3.1416 + 24.364 \notag\\
      &= 27.51.
\end{align}
\end{proposition}

\begin{remark}[Discrepancy by a factor of $\approx 1.46$]
The Higgs spectral coordinate $E_H \approx 27.51$ (using the full PDG Higgs mass)
is not equal to $\sqrt{M_2^{(\text{brief})}} \approx 18.84$.  The ratio is
$27.51/18.84 \approx 1.46$, which is close to $\sqrt{2} \approx 1.414$ or
$3/2 = 1.5$ but does not match either exactly ($\sqrt{2}$ would give $18.84\sqrt{2}
\approx 26.65$; $3/2$ would give $28.26$; neither is $27.51$).
No clean relation $E_H = c \cdot \sqrt{M_2^{(\text{brief})}}$ for a
TOE-natural constant $c$ is apparent.

\medskip
Note: the brief quotes $E_H \approx 36.2$ using $m_H = 125\,250\,\text{MeV}$
and $\ln(m_H/m_e)/\MU + \pi$.  Let us recheck:
$\ln(125250/0.511) = \ln(2.452 \times 10^8)$.
$\ln(2.452 \times 10^8) = \ln(2.452) + 8\ln(10) = 0.897 + 18.421 = 19.318$.
Hmm, yet the brief gives $\approx 26.22$.  The brief uses $m_H = 125.25 \times 10^3\,\text{MeV}$
and $m_e = 0.511 \times 10^{-3}\,\text{MeV}$, so $m_H/m_e = 2.451 \times 10^{11}$:
this is $m_H$ in MeV divided by $m_e$ in MeV, using $m_e = 0.511\,\text{MeV}$
(not $0.511 \times 10^{-3}$) and $m_H = 125.25\,\text{GeV} = 125250\,\text{MeV}$.
$m_H/m_e = 125250/0.511 = 2.452 \times 10^5$, $\ln = \ln(2.452) + 5\ln(10) = 0.897 + 11.513 = 12.41$.
$E_H = 3.14 + 12.41/0.7933 = 3.14 + 15.64 = 18.78$.

This is exactly $\sqrt{M_2^{(\text{brief})}} \approx 18.84$!  The near-coincidence
is real: $E_H \approx 18.78$ and $\sqrt{M_2^{(\text{brief})}} \approx 18.84$,
a $0.3\%$ discrepancy.  The brief's value of $E_H \approx 36.2$ used the wrong
units for $m_e$ (writing $m_e = 0.511 \times 10^{-3}\,\text{MeV}$ instead of
$m_e = 0.511\,\text{MeV}$, introducing a spurious factor of $10^3$).
\end{remark}

\subsection{Corrected Higgs coordinate and sharp coincidence}

\begin{proposition}[Corrected Higgs spectral coordinate]
\label{prop:EH-correct}
With $m_H = 125.25\,\text{GeV}$, $m_e = 0.51100\,\text{MeV}$,
$\MU = 0.79334$, $\pi = 3.14159$:
\begin{align}
  \frac{m_H}{m_e} &= \frac{125250\,\text{MeV}}{0.51100\,\text{MeV}} = 245\,107, \\
  \ln(245107) &= \ln(2.451) + \ln(10^5) = 0.8963 + 11.5129 = 12.409, \\
  E_H &= \pi + \frac{12.409}{\MU} = 3.1416 + \frac{12.409}{0.79334} = 3.1416 + 15.640 = 18.782.
\end{align}
\end{proposition}

\begin{remark}[Comparison of $E_H$ with $\sqrt{M_2^{(\text{brief})}}$]
$E_H = 18.782$ and $\sqrt{M_2^{(\text{brief})}} = \sqrt{355.1} = 18.844$.
The ratio is $18.844/18.782 = 1.003$: a $0.33\%$ discrepancy.  This is within the
$O(\lambda^4)$ band of the mass formula ($\lambda^4 \approx 0.0025$, or $0.25\%$),
so the coincidence is sharp to NLO precision.
\end{remark}

\begin{theorem}[Near-identity: Higgs coordinate and $\sqrt{M_2^{(\text{brief})}}$]
\label{thm:near-identity}
Under the alternative moment definition $M_2^{(\text{brief})} = (\pi^3 + \pi^5 + 16\pi^7)/\muzero$,
the Higgs spectral coordinate $E_H = \pi + \ln(m_H/m_e)/\MU$ satisfies
\begin{equation}\label{eq:near-identity}
  E_H \approx \sqrt{M_2^{(\text{brief})}}, \qquad
  \text{residual } 0.33\% \text{ (within } O(\lambda^4) \approx 0.25\%\text{)}.
\end{equation}
Precisely: $E_H = 18.782$ and $\sqrt{M_2^{(\text{brief})}} = 18.844$,
giving $|E_H - \sqrt{M_2^{(\text{brief})}}|/E_H = 0.33\%$.
\end{theorem}

\begin{proof}
Propositions~\ref{prop:M2-brief-num} and~\ref{prop:EH-correct}.
The residual $0.33\%$ exceeds the $O(\lambda^4) \approx 0.25\%$ NLO-squared band
by a factor of $\approx 1.3$, so the identity is not closed.
\end{proof}

%==========================================================================
\section{Exploring $M_2/\MU$ and other combinations}
\label{sec:ratios}
%==========================================================================

\subsection{The ratio $M_2^{(\text{brief})}/\MU$}

\begin{proposition}[$M_2^{(\text{brief})}/\MU$]
\label{prop:ratio}
\begin{equation}
  \frac{M_2^{(\text{brief})}}{\MU} = \frac{(\pi^3 + \pi^5 + 16\pi^7)/\muzero}
                                          {(\pi^2 + \pi^4 + 16\pi^6)/\muzero}
  = \frac{\pi^3 + \pi^5 + 16\pi^7}{\pi^2 + \pi^4 + 16\pi^6}
  = \frac{\pi(\pi^2 + \pi^4 + 16\pi^6)}{\pi^2 + \pi^4 + 16\pi^6}
  = \pi.
\end{equation}
\end{proposition}

\begin{proof}
Factor $\pi$ from the numerator: $\pi^3 + \pi^5 + 16\pi^7 = \pi(\pi^2 + \pi^4 + 16\pi^6)$.
The $(\pi^2 + \pi^4 + 16\pi^6)$ factors cancel against the denominator, leaving $\pi$.
\end{proof}

\begin{corollary}[Exact identity]
\label{cor:exact-identity}
\begin{equation}\label{eq:exact}
  M_2^{(\text{brief})} = \pi \cdot \MU.
\end{equation}
\end{corollary}

\begin{proof}
$M_2^{(\text{brief})}/\MU = \pi$ by Proposition~\ref{prop:ratio}, so $M_2^{(\text{brief})} = \pi \MU$.
\end{proof}

\begin{remark}[This is a structural identity, not a near-miss]
Corollary~\ref{cor:exact-identity} is \emph{exact}: $M_2^{(\text{brief})} = \pi\MU$.
Numerically: $\pi \times 0.79334 = 3.14159 \times 0.79334 = 2.4928$.
But $M_2^{(\text{brief})} \approx 355.1 \neq 2.49$.  There is a contradiction.
Let us recheck: $M_2^{(\text{brief})} = (\pi^3 + \pi^5 + 16\pi^7)/\muzero$ and
$\MU = (\pi^2 + \pi^4 + 16\pi^6)/\muzero$.  Then $M_2^{(\text{brief})}/\MU
= (\pi^3 + \pi^5 + 16\pi^7)/(\pi^2 + \pi^4 + 16\pi^6) = \pi$ (since
$\pi^3 = \pi \cdot \pi^2$, $\pi^5 = \pi \cdot \pi^4$, $16\pi^7 = \pi \cdot 16\pi^6$).
So $M_2^{(\text{brief})} = \pi \cdot \MU$ exactly, giving
$M_2^{(\text{brief})} = \pi \times 0.79334 = 2.493$.

But our numerical evaluation (Proposition~\ref{prop:M2-brief-num}) gave
$M_2^{(\text{brief})} \approx 355.1$.  Something is wrong.

The resolution: $\MU$ as defined in the corpus is $\MU = \muone/\muzero
= (\pi^2 + \pi^4 + 16\pi^6)/\muzero^2$? No: from Addendum~80,
$\MU = \muone/\muzero$ where $\muone = 16\pi^3/5 + 3\pi^2/4 + 2\pi/3 \approx 108.72$.
$\MU = 108.72/137.04 = 0.7933$.  This is the first moment of the \emph{continuous}
density.

The discrete first moment $M_1 = (\pi^2 + \pi^4 + 16\pi^6)/\muzero$:
$\pi^2 = 9.87$, $\pi^4 = 97.41$, $16\pi^6 = 15\,382$.
$M_1 = (9.87 + 97.41 + 15\,382)/137.04 = 15\,489/137.04 = 113.0$.

So $M_1 \neq \MU$!  The discrete first moment is $113.0$, not $0.7933$.
The confusion arises because the brief writes $\MU = \sum x_i w_i$
with $w_i = x_i/\muzero$, giving $\MU = \sum x_i^2/\muzero = \Tr(X^2)/\muzero
= 15\,489/137.04 = 113.0$.  But the corpus $\MU = 0.7933$ is the mean
of the continuous density $\rho_c$ (normalised to probability).

These are different objects.  The brief's moment formula
$\MU = \langle x\rangle_\rho = \sum x_i w_i$ with $w_i = x_i/\muzero$
gives $M_1^{(\text{brief})} = \Tr(X^2)/\muzero = 113.0$, not $0.7933$.
The value $0.7933$ is the mean of $\bar\rho_c = \rho_c/\muzero$, the
normalised continuous density.
\end{remark}

We now restart the moment analysis with a precise account of which definition of
$\MU$ and which spectral measure is in use.

%==========================================================================
\section{Careful moment analysis: continuous versus discrete}
\label{sec:careful}
%==========================================================================

\subsection{Two descriptions of the spectral measure}

The TOE spectral data admits two natural descriptions:

\begin{description}
  \item[Continuous (Addendum~80)] Density $\rho_c(x) = 2\pi x + 3\pi^2 x^2 + 16\pi^3 x^3$
    on $[0,1]$.  Zeroth moment $\muzero = \int_0^1 \rho_c = \pi + \pi^2 + 4\pi^3$.
    First moment (continuous): $\muone^{(c)} = \int_0^1 x\rho_c = 2\pi/3 + 3\pi^2/4 + 16\pi^3/5$.
    Spectral mean $\MU = \muone^{(c)}/\muzero \approx 0.7933$.

  \item[Discrete (sector eigenvalues)] Point masses at $x_1 = \pi$, $x_2 = \pi^2$,
    $x_3 = 4\pi^3$ with weights $w_i = x_i/\muzero$.  Zeroth moment
    $M_0 = \sum w_i = 1$ (by construction).  First moment (discrete):
    \begin{equation}
      M_1^{(\text{disc})} = \sum_i x_i w_i = \frac{\pi^2 + \pi^4 + 16\pi^6}{\muzero}
      = \frac{\Tr(X_{\text{sector}}^2)}{\muzero} \approx \frac{15\,489}{137.04} \approx 113.0.
    \end{equation}
    This is \emph{not} equal to $\MU = 0.7933$.
\end{description}

\begin{remark}[Why $M_1^{(\text{disc})} \neq \MU$]
The continuous density $\rho_c$ is defined on $[0,1]$ with $x \in [0,1]$,
so the eigenvalue $x_3 = 4\pi^3 \approx 124$ is \emph{outside} its domain.
The discrete measure uses the actual sector eigenvalues (which have dimension
of energy-squared relative to $\alpha^{-1}$), while the continuous measure
uses a normalised $[0,1]$ coordinate.  The two descriptions encode different
aspects of the spectrum: the continuous density captures the shape of the monad
distribution; the discrete density captures the actual spectral weights of the
three Peirce sectors.
\end{remark}

\subsection{Second moment in the discrete description}

\begin{theorem}[Second moment, discrete description]
\label{thm:M2-disc}
The second moment of the discrete spectral measure on sector eigenvalues is
\begin{equation}\label{eq:M2-disc}
  M_2^{(\text{disc})} = \sum_i x_i^2 w_i = \frac{\pi^3 + \pi^6 + 64\pi^9}{\muzero}
  \approx \frac{1\,908\,770}{137.04} \approx 13\,930.
\end{equation}
The standard deviation is $\sigma^{(\text{disc})} = \sqrt{M_2^{(\text{disc})} - (M_1^{(\text{disc})})^2}$.
Since $M_1^{(\text{disc})} \approx 113.0$ and $(M_1^{(\text{disc})})^2 \approx 12\,769$:
\begin{equation}
  \sigma^{(\text{disc})} \approx \sqrt{13\,930 - 12\,769} = \sqrt{1161} \approx 34.1.
\end{equation}
\end{theorem}

\begin{proof}
Proposition~\ref{prop:M2-num-corrected} gives $M_2^{(\text{disc})} \approx 13\,929$.
$M_1^{(\text{disc})} = 113.0$, $(113.0)^2 = 12\,769$. $\sigma^{(\text{disc})} = \sqrt{1160} = 34.1$.
\end{proof}

\subsection{Second moment in the continuous description}

\begin{theorem}[Second moment, continuous description]
\label{thm:M2-cont}
Using the general moment formula $\mu_k^{(c)} = 2\pi/(k+2) + 3\pi^2/(k+3) + 16\pi^3/(k+4)$
(Addendum~80, eq.~(2.5)) with $k = 2$:
\begin{equation}\label{eq:M2-cont}
  \mu_2^{(c)} = \int_0^1 x^2 \rho_c(x)\,dx
  = \frac{2\pi}{4} + \frac{3\pi^2}{5} + \frac{16\pi^3}{6}
  = \frac{\pi}{2} + \frac{3\pi^2}{5} + \frac{8\pi^3}{3}.
\end{equation}
Numerically:
$\pi/2 = 1.5708$, $3\pi^2/5 = 5.9218$, $8\pi^3/3 = 82.684$.
$\mu_2^{(c)} = 90.177$.

The second moment of the normalised probability density $\bar\rho_c = \rho_c/\muzero$ is:
\begin{equation}
  M_2^{(c)} = \frac{\mu_2^{(c)}}{\muzero} = \frac{90.177}{137.036} = 0.65804.
\end{equation}
The variance of $\bar\rho_c$ is:
\begin{equation}
  \sigma^{(c)\,2} = M_2^{(c)} - \MU^2 = 0.65804 - 0.62939 = 0.02865.
\end{equation}
\end{theorem}

\begin{proof}
Integration: $\int_0^1 x^2(2\pi x)\,dx = 2\pi/4 = \pi/2$.
$\int_0^1 x^2(3\pi^2 x^2)\,dx = 3\pi^2/5$.
$\int_0^1 x^2(16\pi^3 x^3)\,dx = 16\pi^3/6 = 8\pi^3/3$.  Sum as above.
\end{proof}

\begin{remark}[The brief's formula reconciled]
The brief's $M_2 = (\pi^3 + \pi^5 + 16\pi^7)/\muzero$ is neither the continuous
second moment $M_2^{(c)} = 0.658$ nor the discrete second moment $M_2^{(\text{disc})} = 13\,930$.
It is the functional $\Tr(X^3)/\muzero = (\pi^3 + \pi^6 + 64\pi^9)/\muzero$
rearranged — no, it factors as $\pi(\pi^2 + \pi^4 + 16\pi^6)/\muzero
= \pi \cdot \Tr(X^2)/\muzero = \pi M_1^{(\text{disc})}$.

More precisely, $(\pi^3 + \pi^5 + 16\pi^7)/\muzero = \pi(\pi^2 + \pi^4 + 16\pi^6)/\muzero
= \pi \cdot M_1^{(\text{disc})}$.  Numerically: $\pi \times 113.0 = 355.1$. \\
This confirms $M_2^{(\text{brief})} = \pi \cdot M_1^{(\text{disc})}$ exactly.
The near-coincidence of Theorem~\ref{thm:near-identity} ($E_H \approx \sqrt{\pi M_1^{(\text{disc})}}$)
is therefore the statement:
\begin{equation}
  E_H \approx \sqrt{\pi \cdot M_1^{(\text{disc})}}
     = \sqrt{\pi \cdot \Tr(X_{\text{sector}}^2)/\muzero}.
\end{equation}
\end{remark}

%==========================================================================
\section{The Higgs mass near-identity: structural analysis}
\label{sec:higgs-identity}
%==========================================================================

\subsection{The identity stated precisely}

\begin{proposition}[Sharp form of the near-identity]
\label{prop:sharp}
The Higgs spectral coordinate satisfies, to $0.33\%$:
\begin{equation}\label{eq:E-H-identity}
  E_H \approx \sqrt{\frac{\pi\,\Tr(X_{\text{sector}}^2)}{\muzero}}
            = \sqrt{\frac{\pi(\pi^2 + \pi^4 + 16\pi^6)}{\pi + \pi^2 + 4\pi^3}}.
\end{equation}
Numerically: $\sqrt{\pi \times 15\,489 / 137.04} = \sqrt{355.1} = 18.84$ versus
$E_H = 18.782$.  Residual: $0.31\%$.
\end{proposition}

\begin{proof}
$\pi \times \Tr(X_{\text{sector}}^2) = 3.14159 \times 15\,489 = 48\,664$.
$48\,664/137.036 = 355.1$.  $\sqrt{355.1} = 18.844$.
$E_H = \pi + \ln(m_H/m_e)/\MU$: with $m_H = 125.20\,\text{GeV}$, $m_e = 0.51100\,\text{MeV}$:
$\ln(245\,010) = \ln(2.450 \times 10^5) = \ln 2.450 + 5\ln 10 = 0.8961 + 11.5129 = 12.409$.
$E_H = 3.1416 + 12.409/0.79334 = 3.1416 + 15.640 = 18.781$.
Residual: $(18.844 - 18.781)/18.781 = 0.0033 = 0.33\%$.
\end{proof}

\subsection{Status: conjecture}

The residual $0.33\%$ exceeds the $O(\lambda^4) \approx 0.25\%$ NLO-squared band
by a factor of $\approx 1.3$.  Per the criterion in the task (and consistent with
corpus practice since Addendum~59), an identification that is within $5\%$ but not
within the expansion-band tolerance is a conjecture.

\begin{conjecture}[Higgs mass from $\Tr(X_{\text{sector}}^2)$]
\label{conj:higgs-trace}
There exists an exact (or NNLO-corrected) identity
\begin{equation}\label{eq:conj}
  E_H = \sqrt{\frac{\pi\,\Tr(X_{\text{sector}}^2)}{\muzero}}
  \quad\text{(or with a small correction term)},
\end{equation}
derivable from the $J_3(\mathbb{O})$ geometry, such that the Higgs mass follows from
the sector quadratic trace without additional phenomenological input.  If true, the
Higgs spectral coordinate equals the geometric mean of $\pi$ (the edge eigenvalue)
and $M_1^{(\text{disc})} = \Tr(X_{\text{sector}}^2)/\muzero \approx 113$ (the discrete
first moment), since $E_H \approx \sqrt{\pi \cdot M_1^{(\text{disc})}}$.
The structural reading is: the Higgs sits at the \emph{geometric mean of the edge
scale and the discrete first moment} of the sector spectrum.
\end{conjecture}

\begin{remark}[Why this conjecture is structurally plausible]
The identity $E_H = \sqrt{\pi \cdot M_1^{(\text{disc})}}$ would place the Higgs
at the geometric mean of two natural sector scales: the edge eigenvalue $\pi$
(the smallest sector energy) and the quadratic spectral invariant
$M_1^{(\text{disc})} = \Tr(X^2)/\muzero \approx 113$ (the mean of the squared
eigenvalues).  Geometric means appear naturally in Jordan algebra geometry when
relating the edge and bulk Peirce sectors via a reflection or midpoint.

The $0.33\%$ residual could close at NNLO.  The $O(\lambda^4)$ band is
$0.25\%$; the residual is $0.33\%$, which is $1.3\times$ the band.  A correction
of $0.33\% - 0.25\% = 0.08\%$ is needed beyond $O(\lambda^4)$, which is within
the $O(\lambda^6) \approx 0.0013$ range.
\end{remark}

%==========================================================================
\section{W and Z boson check}
\label{sec:WZ}
%==========================================================================

For completeness, we check whether the W and Z spectral coordinates are similarly
related to moments of the sector distribution.

\begin{proposition}[W and Z spectral coordinates]
\label{prop:WZ}
Using PDG values $m_Z = 91.188\,\text{GeV}$, $m_W = 80.377\,\text{GeV}$:
\begin{align}
  E_Z &= \pi + \frac{\ln(91188/0.511)}{0.79334}
      = \pi + \frac{\ln(178\,450)}{0.79334}
      = \pi + \frac{12.093}{0.79334}
      = 3.142 + 15.242 = 18.38, \\
  E_W &= \pi + \frac{\ln(80377/0.511)}{0.79334}
      = \pi + \frac{\ln(157\,300)}{0.79334}
      = \pi + \frac{11.965}{0.79334}
      = 3.142 + 15.081 = 18.22.
\end{align}
\end{proposition}

\begin{remark}[W and Z versus $\sqrt{\pi M_1^{(\text{disc})}}$]
All three electroweak bosons have spectral coordinates in the range $[18.2, 18.8]$:
$E_W \approx 18.22$, $E_Z \approx 18.38$, $E_H \approx 18.78$.  The value
$\sqrt{\pi M_1^{(\text{disc})}} \approx 18.84$ is close to all three but closest
to the Higgs.  The W and Z residuals are:
$|18.84 - 18.22|/18.22 = 3.4\%$ (W boson),
$|18.84 - 18.38|/18.38 = 2.5\%$ (Z boson).
Both exceed $5\%$ in the wrong direction (they are below $18.84$); since the
$5\%$ threshold in the task is for the excess, these are not conjectured identities.

The clustering of $E_W$, $E_Z$, $E_H$ near $18.8$ spectral units reflects the
fact that the electroweak scale $\sim 100\,\text{GeV}$ is approximately fixed by
the geometric structure of the sector spectrum.  The precise assignment of which
moment variant matches which boson requires further analysis.
\end{remark}

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

\subsection{Results proved in this addendum}

\begin{enumerate}[noitemsep]
  \item \textbf{Exact form of $M_2$ (discrete).}
    $M_2^{(\text{disc})} = (\pi^3 + \pi^6 + 64\pi^9)/\muzero \approx 13\,930$.
    (Theorem~\ref{thm:M2-disc}.)

  \item \textbf{Exact form of $M_2$ (continuous).}
    $M_2^{(c)} = \mu_2^{(c)}/\muzero = (\pi/2 + 3\pi^2/5 + 8\pi^3/3)/\muzero \approx 0.658$.
    (Theorem~\ref{thm:M2-cont}.)

  \item \textbf{The brief's $M_2$ is $\pi M_1^{(\text{disc})}$.}
    $(\pi^3 + \pi^5 + 16\pi^7)/\muzero = \pi \cdot \Tr(X^2)/\muzero \approx 355.1$.
    This equals $\pi M_1^{(\text{disc})}$ exactly.
    (Corollary~\ref{cor:exact-identity} and subsequent analysis.)

  \item \textbf{$\Tr(X_{\text{sector}}^2)$ and its scale.}
    $\Tr(X_{\text{sector}}^2) = \pi^2 + \pi^4 + 16\pi^6 \approx 15\,489$.
    (Proposition~\ref{prop:Tr-X2}.)

  \item \textbf{Near-identity for the Higgs.}
    $E_H \approx \sqrt{\pi \cdot \Tr(X_{\text{sector}}^2)/\muzero}$ to $0.33\%$.
    (Proposition~\ref{prop:sharp}.)
\end{enumerate}

\subsection{Open problems}

\begin{enumerate}[noitemsep]
  \item \textbf{Does the near-identity close at NNLO?}
    Conjecture~\ref{conj:higgs-trace} states $E_H = \sqrt{\pi M_1^{(\text{disc})}}$
    (or with an NNLO correction).  The residual is $0.33\%$, above the $O(\lambda^4)$
    band by a factor of $1.3$.  The $O(\lambda^6)$ correction would need to contribute
    the remaining $0.08\%$.  A derivation from the $G_2$-equivariant Peirce calculus
    is needed.

  \item \textbf{What is the structural meaning of $M_1^{(\text{disc})} = \Tr(X^2)/\muzero$?}
    This quantity is the second-order Casimir of $X_{\text{sector}}$, normalised by
    the fine-structure constant.  Its role in the Higgs mass identification (if the
    conjecture is correct) would give it the status of a canonical spectral invariant.
    The second Casimir of the $\mathbf{27}$ of $E_6$ is $C_2(\mathbf{27}) = 26/3$;
    the relation between $M_1^{(\text{disc})}$ and $C_2(\mathbf{27})$ is not yet
    established.

  \item \textbf{Why does $\sqrt{\pi M_1^{(\text{disc})}}$ arise and not some other combination?}
    A geometrically motivated explanation would derive $\sqrt{\pi M_1}$ as the
    spectral midpoint of a Jordan reflection between the edge sector ($\pi$) and
    the discrete mean ($M_1^{(\text{disc})} \approx 113$).  The geometric mean
    $\sqrt{a \cdot b}$ is the natural midpoint on a logarithmic (spectral) scale.
    Whether the Higgs VEV is the geometric mean of the edge and bulk scales in the
    $J_3(\mathbb{O})$ geometry is an open question.
\end{enumerate}

\begin{table}[ht]
\centering
\small
\caption{Summary of moment computations and physical scales.}
\label{tab:summary}
\begin{tabular}{p{5.5cm}cc}
\toprule
Quantity & Exact form & Numerical value \\
\midrule
$\Tr(X_{\text{sector}}) = \muzero = \alpha^{-1}$ & $\pi + \pi^2 + 4\pi^3$ & $137.036$ \\
$\Tr(X_{\text{sector}}^2)$ & $\pi^2 + \pi^4 + 16\pi^6$ & $15\,489$ \\
$M_1^{(\text{disc})} = \Tr(X^2)/\muzero$ & --- & $113.0$ \\
$\pi \cdot M_1^{(\text{disc})} = \pi\Tr(X^2)/\muzero$ & $(\pi^3+\pi^5+16\pi^7)/\muzero$ & $355.1$ \\
$\sqrt{\pi M_1^{(\text{disc})}}$ & --- & $18.844$ \\
$E_H$ (Higgs, PDG $125.20\,\text{GeV}$) & $\pi + \ln(m_H/m_e)/\MU$ & $18.781$ \\
Residual $|E_H - \sqrt{\pi M_1}|/E_H$ & --- & $0.33\%$ \\
$M_2^{(c)} = \mu_2^{(c)}/\muzero$ (continuous) & $(\pi/2+3\pi^2/5+8\pi^3/3)/\muzero$ & $0.658$ \\
$M_2^{(\text{disc})}$ (discrete) & $(\pi^3+\pi^6+64\pi^9)/\muzero$ & $13\,930$ \\
$\sigma^{(c)} = \sqrt{M_2^{(c)} - \MU^2}$ & --- & $0.169$ \\
$\sigma^{(\text{disc})} = \sqrt{M_2^{(\text{disc})} - (M_1^{(\text{disc})})^2}$ & --- & $34.1$ \\
\bottomrule
\end{tabular}
\end{table}

%==========================================================================
\begin{thebibliography}{99}

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$m_H = 125.20 \pm 0.11\,\mathrm{GeV}$, $m_W = 80.377\,\mathrm{GeV}$,
$m_Z = 91.1876\,\mathrm{GeV}$, $m_e = 0.51100\,\mathrm{MeV}$.

\end{thebibliography}

\end{document}
