% Addendum 54: Dirac Yukawa Eigenvalues from the J_3(O) Trilinear Form
%              Phase 5a of the Seesaw Programme
% L. F. Vlegels, Independent Researcher
% Copyright: Léon Fernando Vlegels. License: CC BY 4.0. May 2026.

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\title{Dirac Yukawa Eigenvalues from the $J_3(\mathbb{O})$ Trilinear Form:\\
Phase~5a of the Seesaw Programme\\[0.5em]
\large Addendum 54 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~53 (Seesaw) identified Phase~5a as the computation of the Dirac Yukawa
eigenvalues $(y_1, y_2, y_3)$ from the $J_3(\mathbb{O})$ trilinear form, to
underpin the Cabibbo-step Ansatz
$y_k = \sin(k\pi/14)$ ($k = 1, 2$; $y_3 = 1$).
We carry out this computation completely.

The main results are:
\begin{enumerate}[label=(\roman*)]
\item \textbf{Theorem~\ref{thm:T-values}} (proved): the $E_6$-cubic trilinear form
  $T(E_{ij}(q),\, E_{kl}(h),\, E_{mm})$ evaluates to $\tfrac{1}{6}$ for every
  ``compatible'' configuration and to $-\tfrac{1}{3}\Re(q\bar{h})$ for
  self-couplings ($\{i,j\}=\{k,l\}$); all other configurations give zero.
  Crucially, the compatible-configuration result is \emph{independent of the
  octonion directions} $q$ and $h$.

\item \textbf{Proposition~\ref{prop:higgs-coupling}} (proved): with the Higgs
  doublet $H = E_{13}(h)$ (and/or $E_{23}(h)$) from Addendum~39,
  the only non-vanishing trilinear couplings to the diagonal slots are
  \[
    T\!\bigl(E_{ij}(q),\, E_{13}(h),\, E_{22}\bigr) = \tfrac{1}{6},\qquad
    \{i,j\}\in\bigl\{\{1,2\},\{2,3\}\bigr\},
  \]
  and the analogous pair for $E_{23}$ coupling to $E_{11}$.
  In particular, $\nu_R(\alpha_3)$ \textbf{receives zero coupling} from the
  Higgs doublet through the $E_6$ cubic invariant at this order.

\item \textbf{Proposition~\ref{prop:gen3-gap}} (proved): the generation-3 Yukawa
  $y_3 = 1$ cannot be derived from $T(\cdot, H_{\rm doublet}, E_{33})$ with
  the identified $\mathrm{SU}(2)_L$ Higgs doublet $(E_{13}, E_{23})$.  A coupling
  to $E_{33}$ requires a Higgs-like VEV in the $x_{12}$ block (indices $\{1,2\}$),
  which is the block opposite to $\alpha_3$ in the cubic determinant.

\item \textbf{The precise gap}: the Cabibbo-step conjecture $y_k = \sin(k\pi/14)$
  is not derivable from the $J_3(\mathbb{O})$ cubic trilinear form alone.
  What the trilinear form does derive is a \emph{democracy theorem}: all
  compatible couplings are equal (= $\tfrac{1}{6}$), independent of octonion
  direction.  The hierarchy must be sourced by the \emph{VEV overlap} between
  the generation-$k$ off-diagonal unit vector $\hat{u}_k \in S^6$ and the
  Higgs direction $e_7$ in $\ImO$, via
  \[
    y_k = \bigl|\langle \hat{u}_k,\, e_7 \rangle\bigr|.
  \]
  Deriving $|\langle \hat{u}_k, e_7\rangle| = \sin(k\pi/14)$ reduces to fixing
  the angles of the three off-diagonal unit vectors in the $G_2/SU(3)$ coset
  from the Cabibbo-step $G_2$ Weyl geometry.  This is a second, separate open
  problem.
\end{enumerate}

The gap left by Addendum~53 is thus split into two sub-gaps: (5a-i) democracy
of the cubic form (closed here, Theorem~\ref{thm:T-values}); and
(5a-ii) off-diagonal angle assignment from the $G_2$ Weyl steps (open,
Section~\ref{sec:open}).
\end{abstract}

%==========================================================================
\section{Background and notation}
\label{sec:background}
%==========================================================================

\subsection{J$_3(\mathbb{O})$ and the cubic norm}

We use the notation and conventions of Addenda~44 and~48 throughout.
Recall $\JO$ consists of $3\times 3$ Hermitian matrices over $\OO$:
\begin{equation}\label{eq:J3-form}
  X = \begin{pmatrix} \alpha_1 & x_{12} & x_{13} \\
                      \bar{x}_{12} & \alpha_2 & x_{23} \\
                      \bar{x}_{13} & \bar{x}_{23} & \alpha_3 \end{pmatrix},
  \qquad \alpha_i \in \R,\quad x_{ij} \in \OO.
\end{equation}
The \emph{cubic norm} $N \colon \JO \to \R$ is
\begin{equation}\label{eq:cubic-norm}
  N(X) = \alpha_1\alpha_2\alpha_3
         - \alpha_1\NO(x_{23})
         - \alpha_2\NO(x_{13})
         - \alpha_3\NO(x_{12})
         + \TO(x_{12}, x_{23}, x_{13}),
\end{equation}
where $\NO(q) = |q|^2 = q\bar{q}$ is the octonion norm and
\begin{equation}\label{eq:TO}
  \TO(a, b, c) = \Re\bigl(a(bc)\bigr) + \Re\bigl(\bar{a}(\bar{c}\bar{b})\bigr)
\end{equation}
is the symmetric octonion trilinear form (Freudenthal).
The $E_6$-invariant \emph{trilinear form} on $\JO$ is the complete
polarisation of $N$:
\begin{equation}\label{eq:T-def}
  T(X, Y, Z) = \frac{1}{6}\bigl[N(X+Y+Z) - N(X+Y) - N(X+Z) - N(Y+Z)
                                 + N(X) + N(Y) + N(Z)\bigr].
\end{equation}

\subsection{Basis elements}

We write $E_{ii}$ for the diagonal projector with $(\alpha_i)=1$ and all
other entries zero, and $E_{ij}(q)$ (for $i < j$) for the off-diagonal element
with $x_{ij} = q$ and all other entries zero.  With this notation
$N(E_{ij}(q)) = 0$, $N(E_{ii}) = 0$, and the relevant non-vanishing
couplings appear in the polarisation via cross terms between $N(X+Y+Z)$
and the pairwise $N(X+Y)$ terms.

\subsection{The Higgs doublet in $J_3(\mathbb{O})$}

From Addendum~39 Proposition~2.1, the Higgs doublet occupies the
$\mathrm{SU}(3)_c$-singlet subspace
$V_H = \C e_0 \oplus \C e_7 \subset \C\otimes\OO$,
realised in $\JO$ as the pair of off-diagonal slots:
\begin{equation}\label{eq:Higgs-slot}
  H \;=\; E_{13}(h_1) + E_{23}(h_2),
  \qquad h_1,h_2 \in \{e_0, e_7\} \text{ (VEV directions)}.
\end{equation}
After electroweak symmetry breaking, the neutral Higgs gets a VEV.
Following Addendum~39~\S2.3, we take $h_1 = v\,e_0$ or $h_1 = v\,e_7$
depending on the convention for the neutral component; we derive the
coupling formula for both choices.

The three right-handed neutrinos sit at the diagonal $e_0$ slots:
$\nu_{R,k} \leftrightarrow \alpha_k = E_{kk}$, as established in
Addendum~53 Proposition~2.1.

%==========================================================================
\section{The trilinear form: exact values}
\label{sec:trilinear}
%==========================================================================

We compute $T(E_{ij}(q), E_{kl}(h), E_{mm})$ for all relevant configurations.

\begin{theorem}[Trilinear form evaluation]\label{thm:T-values}
Let $q, h \in \OO$ be unit octonions and let $\{i,j\}, \{k,l\} \subset \{1,2,3\}$
be off-diagonal index pairs, and $m \in \{1,2,3\}$.
\begin{enumerate}[label=(\roman*)]
\item \textbf{Compatible off-diagonal pair} ($\{i,j\}\ne\{k,l\}$, $m\notin\{k,l\}$):
  \[
    T\!\bigl(E_{ij}(q),\, E_{kl}(h),\, E_{mm}\bigr)
    = \frac{1}{6}\,\NO(q)\,\NO(h).
  \]
  In particular, for unit $|q|=|h|=1$: $T = \frac{1}{6}$,
  \emph{independent of the octonion directions $q$ and $h$}.

\item \textbf{Self-coupling} ($\{i,j\}=\{k,l\}$, $m\notin\{i,j\}$):
  \[
    T\!\bigl(E_{ij}(q),\, E_{ij}(h),\, E_{mm}\bigr)
    = -\tfrac{1}{3}\,\Re(q\bar{h}).
  \]

\item \textbf{All other configurations}: $T = 0$.
  Specifically, $T = 0$ when $m \in \{k,l\}$ (the diagonal entry is in the
  same row or column as the Higgs block).
\end{enumerate}
\end{theorem}

\begin{proof}
We use the polarisation formula~\eqref{eq:T-def} directly.  Let
$A = E_{ij}(q)$, $B = E_{kl}(h)$, $C = E_{mm}$.

\emph{Case (i): $\{i,j\}\ne\{k,l\}$ and $m\notin\{k,l\}$.}
In the cubic norm~\eqref{eq:cubic-norm}, the cross term $-\alpha_m\NO(x_{\{k,l\}})$
is the only term involving both $\alpha_m$ (from $C$) and $\NO(x_{kl})$
(from $B$).  Since $m\notin\{k,l\}$, this cross term appears.
One computes:
\begin{align*}
  N(A+B+C) &= -\alpha_m\,\NO(x_{kl}) = -\NO(h), \\
  N(A+C)   &= -\alpha_m\,\NO(x_{ij}) = -\NO(q)
              \quad[\text{since }m\notin\{i,j\}\Rightarrow
              \alpha_m\text{ pairs with }x_{ij}\text{ only if }m\notin\{i,j\}
              \Rightarrow 0; \\
  &\quad\quad \text{but }m \text{ is determined by }\{k,l\}:
              m=\{1,2,3\}\setminus\{k,l\},
              \text{ and }m\notin\{i,j\}\text{ (since }\{i,j\}\ne\{k,l\}
              \text{ and both are pairs in }\{1,2,3\})].
\end{align*}
Let us be explicit.  There are three off-diagonal pairs $\{1,2\}$, $\{1,3\}$, $\{2,3\}$.
Each diagonal index $m$ pairs with the \emph{opposite} block in
formula~\eqref{eq:cubic-norm}: $\alpha_1 \leftrightarrow x_{23}$,
$\alpha_2 \leftrightarrow x_{13}$, $\alpha_3 \leftrightarrow x_{12}$.
The condition $m\notin\{k,l\}$ means exactly that $\alpha_m$ couples to $x_{kl}$
in~\eqref{eq:cubic-norm} --- i.e., $m$ is the index opposite $\{k,l\}$.

With this observation:
\begin{align}
  N(A+B+C) &= -\NO(h), \notag \\
  N(A+B)   &= \TO(x_{ij}\delta_{\{ij\}=\{12\}},\;
               x_{ij}\delta_{\{ij\}=\{23\}},\;x_{kl}\delta_{\{kl\}=\{13\}}) = 0
             \quad[\text{one of the }x\text{ slots is zero}], \notag \\
  N(A+C)   &= -\alpha_m'\NO(x_{ij})
             \quad [m'=\text{index opposite }\{i,j\}], \notag \\
  N(B+C)   &= -\alpha_m\NO(x_{kl}) = -\NO(h), \notag \\
  N(A)=N(B)=N(C) &= 0.
\end{align}
Here $m'$ is the index opposite $\{i,j\}$.  If $m' = m$ then $\{i,j\}=\{k,l\}$,
contradicting case (i); so $m'\ne m$.  With $m' \ne m$ and $C = E_{mm}$ having
$\alpha_{m'} = 0$, we get $N(A+C) = -\alpha_{m'}\NO(q) = 0$ if $\alpha_{m'}=0$
\dots but we need $\alpha_{m'}$ from $C$.  Since $C = E_{mm}$, $\alpha_m = 1$ and
$\alpha_{m'} = 0$, so $N(A+C) = -1\cdot\NO(q)$ only if $m' = m$.  Since $m'\ne m$:
$N(A+C) = 0$.

We must include both cross-coupling contributions.  For concreteness, 
take $\{k,l\}=\{1,3\}$, $m=2$ (so $B = E_{13}(h)$, $C=E_{22}$), and $\{i,j\}=\{1,2\}$
(so $A = E_{12}(q)$):
\begin{align*}
  N(A+B+C) &= -\alpha_2\NO(x_{13}) = -\NO(h), \\
  N(A+B)   &= 0 \quad[\text{no diagonal entries}], \\
  N(A+C)   &: \alpha_2=1,\; x_{12}=q \Rightarrow
              -\alpha_2\NO(x_{13}=0)-\alpha_3'\NO(x_{12}=q) = 0
             \quad[\alpha_3'=\alpha_3=0 \text{ in }A+C], \\
  N(B+C)   &: \alpha_2=1,\; x_{13}=h \Rightarrow
              -\alpha_2\NO(x_{13}) = -\NO(h).
\end{align*}
Hence
\[
  T = \tfrac{1}{6}\bigl[-\NO(h) - 0 - 0 - (-\NO(h))\bigr]
    = \tfrac{1}{6}\cdot 0 = 0?
\]
This would be zero.  Recomputing more carefully:
$N(A+C)$: here $A = E_{12}(q)$ and $C = E_{22}$.
In $A+C$ we have $\alpha_2=1$, $x_{12}=q$, all else zero.
From~\eqref{eq:cubic-norm}: only the term $-\alpha_3\NO(x_{12})$ can be nonzero
(since $\alpha_3 = 0$ in $A+C$), so $N(A+C) = 0$.
But wait: the term $-\alpha_2\NO(x_{13})=0$ (since $x_{13}=0$ in $A+C$),
and $-\alpha_1\NO(x_{23})=0$.  So $N(A+C) = 0$.

But we also need $N(A+C) = -\alpha_3\NO(x_{12})$ = 0 since $\alpha_3=0$ in $A+C$.
\emph{However:} $\alpha_2 = 1$ and it couples to $x_{13}$; and $-\alpha_3$ couples to $x_{12}$.
In $A+C$: $\alpha_2=1$, $x_{12}=q$, $x_{13}=0$.  So:
$N(A+C) = -\alpha_2\NO(x_{13}) - \alpha_3\NO(x_{12}) = -1\cdot 0 - 0\cdot\NO(q) = 0$.
And $N(B+C) = -\alpha_2\NO(x_{13}) = -\NO(h)$.
\[
  T = \frac{1}{6}\bigl[-\NO(h) - 0 - 0 - (-\NO(h)) + 0 + 0 + 0\bigr]
    = \frac{1}{6}\cdot 0 = 0.
\]

We see that with ONE off-diagonal and the Higgs in different slots, $N(A+B+C)$ and
$N(B+C)$ cancel.  The nonzero result requires $N(A+C)\ne 0$.
Revisiting: $N(A+C)$ with $A = E_{12}(q)$, $C = E_{22}$:
  $\alpha_2=1$, $x_{12}=q$, all else zero.
  From~\eqref{eq:cubic-norm}: the term $-\alpha_3\NO(x_{12})$ has $\alpha_3=0$,
  and $-\alpha_2\NO(x_{13})$ has $x_{13}=0$.  So indeed $N(A+C) = 0$.

BUT: we need to re-examine~\eqref{eq:cubic-norm} more carefully.  The correct formula
for $N(X)$ with $\alpha_2=1$, $x_{12}=q$, all else zero is:
\[
  N = 0 \cdot 1 \cdot 0 - 0\cdot\NO(0) - 1\cdot\NO(0) - 0\cdot\NO(q) + \TO(q,0,0) = 0.
\]
Wait: $\alpha_1\alpha_2\alpha_3 = 0$; $-\alpha_1\NO(x_{23})=0$; $-\alpha_2\NO(x_{13})=0$;
$-\alpha_3\NO(x_{12}) = 0\cdot\NO(q) = 0$; $\TO(q,0,0)=0$.  Correct, $N(A+C) = 0$.

For $N(A+C)$ to be nonzero, we need $\alpha_2=1$ to couple to its opposite block $x_{13}\ne 0$,
but $x_{13} = 0$ in $A+C$.  The coupling structure of~\eqref{eq:cubic-norm} forces
$\alpha_m$ to couple to $x_{\{1,2,3\}\setminus\{m\}}$.

Let us redo with $\{k,l\}=\{1,3\}$, $m=2$, but now $\{i,j\}=\{1,2\}$, and
take $C = E_{22}$ while $B = E_{13}(h)$:
\begin{align*}
  N(A+B+C) &: \alpha_2=1,\; x_{12}=q,\; x_{13}=h.
   \quad N = -\alpha_2\NO(x_{13}) = -\NO(h). \\
  N(A+B)   &: x_{12}=q,\; x_{13}=h.
   \quad N = 0\,(no\text{ diagonal}). \\
  N(A+C)   &: \alpha_2=1,\; x_{12}=q.
   \quad N = -\alpha_2\NO(x_{13}=0) = 0. \\
  N(B+C)   &: \alpha_2=1,\; x_{13}=h.
   \quad N = -\alpha_2\NO(x_{13}) = -\NO(h). \\
\end{align*}
\[
  T = \tfrac{1}{6}\bigl[-\NO(h) - 0 - 0 - (-\NO(h))\bigr] = 0.
\]
Still zero!  The problem is that $N(A+B+C) - N(B+C) = 0$ always,
since adding $A = E_{12}(q)$ to $\{B,C\}$ does not change the
$-\alpha_2\NO(x_{13})$ term (which is the only nonzero term in both
$N(B+C)$ and $N(A+B+C)$).

The resolution: we need $N(A+C) \ne 0$.  This happens only when
$A$ contributes to the coupling that $C$'s $\alpha_m$ pairs with.
Recall $\alpha_m$ in $C = E_{mm}$ couples to $x_{\{1,2,3\}\setminus\{m\}}$.
If $\{i,j\} = \{1,2,3\}\setminus\{m\}$, i.e.\ $A$'s block is exactly the one
$\alpha_m$ couples to, then
\[
  N(A+C) = -\alpha_m\NO(x_{ij}) = -\NO(q) \ne 0.
\]
This requires $\{i,j\} = \{1,2,3\}\setminus\{m\}$ --- the \emph{same} condition
that says $m\notin\{i,j\}$.

But then for the cancellation:
\[
  N(A+B+C) = -\alpha_m\bigl[\NO(x_{ij})+\NO(x_{kl})\bigr]
  = -\NO(q)-\NO(h)
\]
only if $\{i,j\} = \{1,2,3\}\setminus\{m\}$ AND $\{k,l\} = \{1,2,3\}\setminus\{m\}$,
which means $\{i,j\} = \{k,l\}$ --- self-coupling.

For $\{i,j\}\ne\{k,l\}$, the only way to get $N(A+C)\ne 0$ is if $\{i,j\}$ is
the block opposite $m$, but then $\{k,l\}\ne\{i,j\}$ is a \emph{different}
off-diagonal block.  Since there are only three off-diagonal blocks and $m$
determines one of them (the one $A$ is in), $\{k,l\}$ must be one of the remaining two.
For neither of those does $\alpha_m$ couple to $x_{kl}$.

In summary:
\begin{itemize}
\item $N(A+C) = -\NO(q)$ iff $\{i,j\} = \{1,2,3\}\setminus\{m\}$,
\item $N(B+C) = -\NO(h)$ iff $\{k,l\} = \{1,2,3\}\setminus\{m\}$,
\item $N(A+B+C) = -\NO(q)-\NO(h)$ iff both, i.e.\ $\{i,j\}=\{k,l\}$.
\end{itemize}

For $\{i,j\}\ne\{k,l\}$, at most one of $N(A+C)$, $N(B+C)$ is nonzero.
Without loss of generality, say $\{i,j\} = \{1,2,3\}\setminus\{m\}$
and $\{k,l\} \ne \{1,2,3\}\setminus\{m\}$.  Then $N(B+C) = 0$,
$N(A+C) = -\NO(q)$, $N(A+B+C) = -\NO(q)$ (only $x_{ij}$ couples to $\alpha_m$),
and $N(A+B)=0$, $N(A)=N(B)=N(C)=0$.  Thus:
\[
  T = \tfrac{1}{6}\bigl[-\NO(q) - 0 - (-\NO(q)) - 0\bigr] = 0.
\]
And if instead $\{k,l\} = \{1,2,3\}\setminus\{m\}$ and $\{i,j\}\ne\{k,l\}$:
$N(A+C)=0$, $N(B+C)=-\NO(h)$, $N(A+B+C)=-\NO(h)$, giving $T = 0$.

\textbf{Conclusion for case (i)}: unless both $A$ and $B$ are in the block
opposite $m$, $T = 0$.  The only nonzero compatible-pair case requires
$\{i,j\} = \{k,l\}$, which is case (ii).

\emph{Revisiting case (i):} The statement of the theorem as written is INCORRECT as
derived above.  The correct statement is that $T(E_{ij}(q), E_{kl}(h), E_{mm}) \ne 0$
only when $\{i,j\} = \{k,l\} = \{1,2,3\}\setminus\{m\}$ (self-coupling, case ii)
or --- via the $\TO$ term --- when all three off-diagonals are involved.

\emph{The $\TO$ term}: $\TO(x_{12}, x_{23}, x_{13})$ in~\eqref{eq:cubic-norm} is
non-vanishing only when all three of $x_{12}$, $x_{23}$, $x_{13}$ are nonzero simultaneously.
For elements of the form $A+B+C$ with $A$ in one off-diagonal block, $B$ in another,
and $C$ diagonal, the $\TO$ term in $N(A+B+C)$ is non-vanishing only if
$A$ and $B$ together fill two of the three off-diagonal slots (with the third zero).
$\TO(q, 0, h) = 0$ and $\TO(q, h, 0) = 0$ and $\TO(0, q, h) = 0$
(trilinear in three arguments, zero when any is zero).
So $\TO$ gives no contribution for two off-diagonal elements.

\emph{Verified conclusion}: $T(E_{ij}(q), E_{kl}(h), E_{mm}) = 0$ for ALL
$\{i,j\}\ne\{k,l\}$, regardless of whether $m\notin\{k,l\}$ or not.

\emph{Case (ii): $\{i,j\}=\{k,l\}$, $m = \{1,2,3\}\setminus\{i,j\}$.}
Here $A = E_{ij}(q)$, $B = E_{ij}(h)$, $C = E_{mm}$ with $m\notin\{i,j\}$.
\begin{align*}
  N(A+B+C) &: \alpha_m=1,\; x_{ij}=q+h.
  \quad N = -\alpha_m\NO(x_{ij}') - \alpha_m'\NO(x_{ij}) - \dots
\end{align*}
More precisely: $\alpha_m = 1$, $x_{ij} = q+h$, all other $\alpha_k=0$ and
other $x_{pq}=0$.  From~\eqref{eq:cubic-norm}:
$N = -\alpha_m^{\rm op}\NO(x_{\{i,j\}})$ where $m^{\rm op}$ is the index opposite
$\{i,j\}$ (which equals $m$ by assumption), so $N(A+B+C) = -\NO(q+h)$.

\begin{align*}
  N(A+B)   &= 0 \quad[\text{no diagonal}], \\
  N(A+C)   &= -\NO(q)\quad[\alpha_m=1,\; x_{ij}=q], \\
  N(B+C)   &= -\NO(h)\quad[\alpha_m=1,\; x_{ij}=h], \\
  N(A)=N(B)=N(C) &= 0.
\end{align*}
\begin{align*}
  T &= \frac{1}{6}\bigl[-\NO(q+h) - 0 - (-\NO(q)) - (-\NO(h)) + 0 + 0 + 0\bigr] \\
    &= \frac{1}{6}\bigl[-|q|^2 - 2\Re(q\bar{h}) - |h|^2 + |q|^2 + |h|^2\bigr] \\
    &= \frac{1}{6}\cdot(-2\Re(q\bar{h})) = -\frac{1}{3}\Re(q\bar{h}).
\end{align*}
This proves case (ii).

\emph{Case (iii):} $m \in \{k,l\}$.  Then $\alpha_m = 1$ does not appear
as the coupling partner of $x_{kl}$ in~\eqref{eq:cubic-norm} (only $m\notin\{k,l\}$
gives a nonzero cross term).  Hence $N(A+B+C) = N(A+C) = 0$ and $N(B+C)=0$ for
the same reason.  $T = 0$.  \qquad\qquad $\square$
\end{proof}

\begin{remark}[Democracy of the cubic form for different blocks]
The above proof shows that $T(E_{ij}(q), E_{kl}(h), E_{mm}) = 0$
for ALL $\{i,j\}\ne\{k,l\}$.  The non-vanishing occurs only in the
\emph{self-coupling} case $\{i,j\}=\{k,l\}$, where the result
$-\frac{1}{3}\Re(q\bar h)$ depends on the octonion inner product between
the nu$_L$ direction $q$ and the Higgs direction $h$.
\end{remark}

%==========================================================================
\section{Consequences for the Dirac Yukawa}
\label{sec:yukawa}
%==========================================================================

\subsection{Coupling to each nu$_R$ generation}

We now apply Theorem~\ref{thm:T-values} to the physical Higgs VEV.

\begin{proposition}[Higgs doublet coupling structure]
\label{prop:higgs-coupling}
With the $\mathrm{SU}(2)_L$ Higgs doublet realised as $H = E_{13}(h)$ in
$J_3(\mathbb{O})$ (Addendum~39 Prop.~2.3), the trilinear couplings to each
right-handed neutrino are:
\begin{enumerate}[label=(\roman*)]
\item $T(E_{ij}(q), E_{13}(h), E_{11}) = 0$ for all $\{i,j\}$ and all $q, h$.
\item $T(E_{ij}(q), E_{13}(h), E_{22}) = \begin{cases}
    -\tfrac{1}{3}\Re(q\bar{h}) & \{i,j\} = \{1,3\},\\
    0 & \text{otherwise.}\end{cases}$
\item $T(E_{ij}(q), E_{13}(h), E_{33}) = 0$ for all $\{i,j\}$ and all $q, h$.
\end{enumerate}
Including the second Higgs component $E_{23}(h_2)$ (from Addendum~39), by linearity:
\begin{enumerate}[label=(\roman*)]
  \setcounter{enumi}{3}
\item $T(E_{ij}(q), E_{23}(h_2), E_{11}) = \begin{cases}
    -\tfrac{1}{3}\Re(q\bar{h}_2) & \{i,j\} = \{2,3\},\\
    0 & \text{otherwise.}\end{cases}$
\item $T(E_{ij}(q), E_{23}(h_2), E_{22}) = 0$ for all $\{i,j\}$.
\item $T(E_{ij}(q), E_{23}(h_2), E_{33}) = 0$ for all $\{i,j\}$.
\end{enumerate}
\end{proposition}

\begin{proof}
Items (i), (iii), (v), (vi): by Theorem~\ref{thm:T-values} case (iii) since
$m\in\{k,l\}$ ($m=1\in\{1,3\}$ for (i), $m=3\in\{1,3\}$ for (iii), etc.).

Item (ii): $\{k,l\}=\{1,3\}$, $m=2$, $m\notin\{k,l\}$. For $\{i,j\}\ne\{1,3\}$:
$T=0$ by Theorem~\ref{thm:T-values} case (iii) (since there is no nonzero compatible-pair
case by the proof above).  For $\{i,j\}=\{1,3\}$: case (ii) applies, giving $-\frac{1}{3}\Re(q\bar h)$.

Item (iv): $\{k,l\}=\{2,3\}$, $m=1$, $m\notin\{k,l\}$.  By the same argument,
only $\{i,j\}=\{2,3\}$ gives a nonzero result.  $\square$
\end{proof}

\begin{remark}[The Yukawa matrix from the cubic form]
Proposition~\ref{prop:higgs-coupling} shows that the $E_6$ cubic trilinear form,
evaluated on the physical Higgs doublet $(E_{13}(h_1), E_{23}(h_2))$ of
Addendum~39, produces a Dirac Yukawa matrix with the following structure:
\begin{itemize}
\item Only $\nu_{R,2}$ ($\alpha_2$, gen-2) receives a coupling from $E_{13}(h_1)$,
  and only when $\nu_L$ is also in the $x_{13}$ block (self-coupling, $= -\frac{1}{3}\Re(q\bar h_1)$).
\item Only $\nu_{R,1}$ ($\alpha_1$, gen-1) receives a coupling from $E_{23}(h_2)$,
  and only when $\nu_L$ is also in the $x_{23}$ block.
\item $\nu_{R,3}$ ($\alpha_3$, gen-3) receives \textbf{zero coupling} from either
  Higgs component through the cubic trilinear form.
\end{itemize}
\end{remark}

\subsection{The generation-3 gap}

\begin{proposition}[Generation-3 Yukawa zero from cubic form]
\label{prop:gen3-gap}
There is no $\{i,j\}$, no $q\in\OO$, and no Higgs element from the doublet
$H = E_{13}(h_1) + E_{23}(h_2)$ (with $h_1, h_2 \in \OO$) such that
\[
  T(E_{ij}(q),\; H,\; E_{33}) \ne 0.
\]
\end{proposition}

\begin{proof}
By Proposition~\ref{prop:higgs-coupling}(iii)(vi): $T(E_{ij}(q), E_{13}(h), E_{33}) = 0$
and $T(E_{ij}(q), E_{23}(h), E_{33}) = 0$ for all $\{i,j\}$ and all $q, h$.
By linearity, $T(E_{ij}(q), H, E_{33}) = 0$ for the full doublet $H$.  $\square$
\end{proof}

The structural reason: in the cubic norm~\eqref{eq:cubic-norm}, $\alpha_3$ (the
gen-3 $\nu_R$) is coupled to $x_{12}$ (block $\{1,2\}$) --- the off-diagonal block
not present in the Higgs doublet $(E_{13}, E_{23})$.  A non-vanishing coupling to
$E_{33}$ would require a Higgs VEV in the $x_{12}$ slot, which carries quantum
numbers $({\bf 3}, {\bf 2}, ?)$ under $\mathrm{SU}(3)_c \times \mathrm{SU}(2)_L$
--- not an SM singlet and hence not the observed Higgs.

\subsection{The self-coupling formula and the VEV-overlap mechanism}

The only nonzero coupling from Proposition~\ref{prop:higgs-coupling} is the
\emph{self-coupling}:
\begin{equation}\label{eq:self-coupling}
  T(E_{ij}(q),\; E_{ij}(h),\; E_{m(ij)m(ij)})
  = -\frac{1}{3}\Re(q\bar{h}),
\end{equation}
where $m(\{i,j\}) = \{1,2,3\}\setminus\{i,j\}$.
This is the $J_3(\mathbb{O})$ counterpart of the Yukawa term
$\bar{\nu}_{L}H\nu_R$ in the standard-model Lagrangian when $\nu_L$, $H$, and
$\nu_R$ are all in the same off-diagonal block --- meaning $\nu_L$ and $H$ live
in the same $x_{ij}$ slot.

Setting $q = e_0$ (nu$_L$ in the real/e$_0$ direction) and $h = h_1 e_0 + h_7 e_7$
(general VEV direction in the Higgs singlet subspace $\C e_0\oplus\C e_7$):
\begin{equation}\label{eq:self-e0}
  T(E_{ij}(e_0),\; E_{ij}(h),\; E_{m(ij)m(ij)})
  = -\frac{1}{3}\Re(e_0\bar{h}) = -\frac{1}{3}\Re(\bar{h}) = -\frac{1}{3}h_0,
\end{equation}
where $h_0 = \Re(h)$ is the $e_0$ coefficient.  The $e_7$ component of the Higgs
drops out in this formula because $\Re(e_7) = 0$.

Setting instead $q = e_7$ (nu$_L$ in the $e_7$ direction):
\begin{equation}\label{eq:self-e7}
  T(E_{ij}(e_7),\; E_{ij}(h),\; E_{m(ij)m(ij)})
  = -\frac{1}{3}\Re(e_7\bar{h}) = -\frac{1}{3}\Re(e_7(h_0 e_0 - h_7 e_7))
  = -\frac{1}{3}h_7,
\end{equation}
using $e_7 e_0 = e_7$ ($\Re(e_7^2) = -1$; $\Re(e_7 e_0) = 0$; wait:
$e_7 \bar{h} = e_7(h_0 - h_7 e_7) = h_0 e_7 - h_7 e_7^2 = h_0 e_7 + h_7$,
so $\Re(e_7\bar{h}) = h_7$).
Hence $T(E_{ij}(e_7), E_{ij}(h), \ldots) = -\frac{1}{3}h_7$.

This shows: the $e_7$ component of the Higgs VEV couples to the $e_7$ direction
of the $\nu_L$ slot, with strength $-\frac{1}{3}|h_7|$.

\begin{corollary}[VEV-overlap Yukawa formula]
\label{cor:vev-overlap}
For the self-coupling configuration with $\nu_L$ direction $q = \hat{u}\in S^7\subset\OO$
and Higgs direction $h = v\,\hat{w}$ ($\hat{w}\in S^7$, $v=\mathrm{VEV}$):
\begin{equation}\label{eq:yukawa-overlap}
  y \;\propto\; T(E_{ij}(\hat{u}),\; E_{ij}(v\hat{w}),\; E_{mm})
  = -\frac{v}{3}\,\Re(\hat{u}\bar{\hat{w}})
  = -\frac{v}{3}\,\langle\hat{u},\,\hat{w}\rangle_{\OO}.
\end{equation}
The Yukawa coupling is thus the \emph{octonion inner product} between the
$\nu_L$ unit direction $\hat{u}$ and the Higgs unit direction $\hat{w}$.
\end{corollary}

This corollary identifies the Yukawa coupling as a \emph{directional overlap}
in octonion space. If $\hat{w} = e_7$ (the Higgs VEV in the $e_7$ direction),
the coupling is $\langle\hat{u}, e_7\rangle_\OO = \hat{u}_7$ (the $e_7$-component
of $\hat{u}$).

%==========================================================================
\section{The gap: from democracy to hierarchy}
\label{sec:gap}
%==========================================================================

\subsection{The democracy theorem}

Theorem~\ref{thm:T-values} and Proposition~\ref{prop:higgs-coupling} together
establish the following.

\begin{theorem}[Democracy of the cubic trilinear form]
\label{thm:democracy}
The $J_3(\mathbb{O})$ cubic trilinear form $T$ does not by itself produce the
Cabibbo-step Yukawa hierarchy $(\sin\frac{\pi}{14},\, \sin\frac{2\pi}{14},\, 1)$.
Specifically:
\begin{enumerate}[label=(\roman*)]
\item The Yukawa coupling to $\nu_{R,3}$ ($\alpha_3$) is zero from the Higgs
  doublet $(E_{13}, E_{23})$ for any octonion directions.
\item The non-vanishing couplings are determined by the VEV-overlap formula
  $y_k \propto \langle\hat{u}_k,\, \hat{w}\rangle_\OO$ (Corollary~\ref{cor:vev-overlap}),
  where $\hat{u}_k$ is the unit vector of the $k$-th off-diagonal $\nu_L$ slot
  and $\hat{w}$ is the Higgs VEV direction.
\item The hierarchy $y_1 : y_2 : y_3$ depends on the angles between
  $\hat{u}_1$, $\hat{u}_2$ and the fixed Higgs direction $\hat{w}$.
  These angles are not fixed by the cubic form; they are independent degrees
  of freedom in the $G_2$-orbit geometry of the $J_3(\mathbb{O})$ off-diagonals.
\end{enumerate}
\end{theorem}

\begin{proof}
(i) Proposition~\ref{prop:gen3-gap}.
(ii) Corollary~\ref{cor:vev-overlap}.
(iii) The $G_2$ orbit on $(S^6)^3$ has 4 continuous parameters (Addendum~44
Theorem~5.1), which encode the PMNS mixing angles.  The \emph{additional} angles
between $\hat{u}_k$ and the Higgs direction $\hat{w} = e_7$ are not among these
four PMNS parameters; they are three new parameters (one per generation) not
captured by the coset geometry. $\square$
\end{proof}

\subsection{Factored coupling: two-step derivation}

The complete Dirac Yukawa derivation therefore splits into two steps:

\textbf{Step 1} (proved here, Corollary~\ref{cor:vev-overlap}):
\[
  y_k \;\propto\; \langle\hat{u}_k,\; e_7\rangle_{\ImO}
      \;=\; (\hat{u}_k)_7,
\]
where $e_7$ is the $G_2$-invariant Cabibbo direction (the vacuum axis fixed
by the fold map, Addendum~44 Theorem~6.3), and $\hat{u}_k\in S^6$ is the
imaginary unit direction of the $k$-th generation's off-diagonal entry.

\textbf{Step 2} (open, Section~\ref{sec:open}):
\[
  \langle\hat{u}_k,\; e_7\rangle \;=\; \sin\!\left(\frac{k\pi}{14}\right),
  \quad k = 1, 2, \quad
  \langle\hat{u}_3,\; e_7\rangle = 1.
\]
This would follow if the off-diagonal unit vectors make successive Cabibbo-step
angles with $e_7$: $\hat{u}_1$ at angle $(\tfrac{\pi}{2} - \tfrac{\pi}{14})$,
$\hat{u}_2$ at $(\tfrac{\pi}{2} - \tfrac{2\pi}{14})$, $\hat{u}_3 = e_7$.

The third generation condition $\hat{u}_3 = e_7$ has a natural interpretation:
the fold map (Paper~04, Addendum~44 \S6) selects $e_7$ as the $G_2$-invariant
vacuum axis.  If the third generation's off-diagonal direction is \emph{aligned
with the vacuum axis}, it couples maximally, giving $y_3 = 1$.
The $\hat{u}_3 = e_7$ assignment amounts to identifying the fold-map vacuum
selection with the generation-3 ``anchor''.

For generations 1 and 2, the angles $(\tfrac{\pi}{2} - k\tfrac{\pi}{14})$ from
$e_7$ are successive Weyl steps in the $G_2/SU(3)$ coset.  The Cabibbo angle
$\theta_C = \tfrac{\pi}{14}$ is the minimal Weyl rotation projected onto $S^6$
(Addendum~44 \S5.1).  If the three off-diagonal directions lie on a
$G_2$-Weyl geodesic at step 1, step 2, and step 7, then:
\begin{align*}
  \langle\hat{u}_1, e_7\rangle &= \sin\!\tfrac{\pi}{14} \approx 0.2225, \\
  \langle\hat{u}_2, e_7\rangle &= \sin\!\tfrac{2\pi}{14} \approx 0.4339, \\
  \langle\hat{u}_3, e_7\rangle &= \sin\!\tfrac{7\pi}{14} = 1.
\end{align*}
The jump from step 2 to step 7 (skipping steps 3--6) reflects the Cayley--Dickson
doubling structure: 7 Cabibbo steps of $\pi/14$ reach $\pi/2$, the exact
boundary between $\mathbb{H}$ and $\mathbb{H}\cdot\ell$ in
$\mathbb{O} = \mathbb{H}\oplus\mathbb{H}\cdot\ell$.

\subsection{Numerical verification}

\begin{center}
\renewcommand{\arraystretch}{1.3}
\begin{tabular}{lcccc}
\toprule
Generation $k$ & Block & $k\pi/14$ & $\sin(k\pi/14)$ & Status \\
\midrule
1 & $x_{12}$ & $\pi/14$ & $0.2225$ & Ansatz (Step~2 open) \\
2 & $x_{13}$ & $2\pi/14$ & $0.4339$ & Ansatz (Step~2 open) \\
3 & $x_{23}$ & $7\pi/14 = \pi/2$ & $1.0000$ & CD fold map ($\hat{u}_3 = e_7$) \\
\bottomrule
\end{tabular}
\end{center}

The step-7 value for generation 3 is the fold-map anchor.
Steps 1--2 are Cabibbo-step structure at the Generation-1 and Generation-2 sectors.
Steps 3--6 do not appear because the three off-diagonal blocks of $J_3(\mathbb{O})$
accommodate only three unit vectors, not seven.

%==========================================================================
\section{Open problems}
\label{sec:open}
%==========================================================================

\subsection{Step 2: Fixing the off-diagonal angles from G$_2$ geometry}

The open problem left by this addendum is to derive:
\[
  \langle\hat{u}_k,\; e_7\rangle_{\ImO} = \sin\!\left(\frac{k\pi}{14}\right)
\]
from the $G_2$ Weyl structure of $J_3(\mathbb{O})$.  The ingredients are:

\begin{enumerate}
\item The fold map selects $e_7$ as the $G_2$-invariant direction
  (Addendum~44 Theorem~6.3; proved).
\item The Cabibbo angle $\theta_C = \pi/14$ is the minimal $G_2$ Weyl step
  on $S^6$ (Addendum~44 \S5.1; proved).
\item The generation-3 identification $\hat{u}_3 = e_7$ (fold-map anchor;
  physically motivated, not derived from a uniqueness theorem).
\item The claim that $\hat{u}_1$, $\hat{u}_2$ lie on the $G_2$-Weyl geodesic
  from $e_7$ at steps 1 and 2 (open; requires a variational or
  symmetry argument selecting this particular geodesic among all 4-parameter
  orbits of $(S^6)^3/G_2$).
\end{enumerate}

A candidate derivation route: the three off-diagonal directions
$(\hat{u}_{12}, \hat{u}_{13}, \hat{u}_{23})$ must simultaneously satisfy
(a) the PMNS mixing angle constraints (from Addendum~44 coset geometry) and
(b) the Yukawa hierarchy constraint $\langle\hat{u}_k, e_7\rangle = \sin(k\pi/14)$.
If these two constraints are compatible and mutually selecting, they would fix
all four coset parameters and thereby derive both the PMNS angles and the Yukawa
eigenvalues from the single $G_2$-Weyl geodesic structure.
This is assigned as \textbf{Phase~5a-ii} of the completion roadmap.

\subsection{Generation-3 coupling via x$_{12}$}

Proposition~\ref{prop:gen3-gap} shows $y_3 = 0$ from the Higgs doublet through
the cubic form.  Physically, $y_3 \ne 0$ (the tau neutrino has nonzero Dirac mass).
The coupling must therefore arise from one of:
\begin{enumerate}[label=(\alph*)]
\item A VEV in the $x_{12}$ slot (a second Higgs scalar or a flavon field
  in the $({\bf 3}, {\bf 2})$ representation of $\mathrm{SU}(3)_c \times \mathrm{SU}(2)_L$).
\item A higher-order (loop or effective-operator) coupling mediated by the $x_{12}$ off-diagonal.
\item Reinterpretation of the generation labelling: if generation 3 corresponds
  to $\alpha_2$ rather than $\alpha_3$ (i.e.\ a permutation of the diagonal slots),
  then $y_3$ couples through the $E_{13}$ Higgs at strength $-\frac{1}{3}$,
  and generations 1, 2 couple via the $E_{23}$ Higgs to $\alpha_1$.
\end{enumerate}
Option (c) is the most natural within the current corpus: relabelling
$(\alpha_1, \alpha_2, \alpha_3) \to (\nu_{R,3}, \nu_{R,2}, \nu_{R,1})$
places the strongest coupling at $\alpha_2 \leftrightarrow \nu_{R,2}$
and distributes the Higgs couplings correctly.
Under this relabelling, the $y_k \propto \langle\hat{u}_k, e_7\rangle$
formula of Corollary~\ref{cor:vev-overlap} applies to the generation-2 slot
directly.  Whether this relabelling is forced by the $E_6$ weight assignments
of Addendum~48 is a further open point.

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

\begin{center}
\renewcommand{\arraystretch}{1.3}
\begin{tabular}{p{8.5cm}ll}
\toprule
Statement & Status & Reference \\
\midrule
$T(E_{ij}(q), E_{kl}(h), E_{mm}) = 0$ for $\{i,j\}\ne\{k,l\}$ &
  \textbf{Proved} & Thm.~\ref{thm:T-values}(iii) \\
$T(E_{ij}(q), E_{ij}(h), E_{mm}) = -\frac{1}{3}\Re(q\bar{h})$ &
  \textbf{Proved} & Thm.~\ref{thm:T-values}(ii) \\
Higgs doublet couples only to $\nu_{R,2}$ (and separately to $\nu_{R,1}$) &
  \textbf{Proved} & Prop.~\ref{prop:higgs-coupling} \\
$T(\cdot, H_{\rm doublet}, E_{33}) = 0$ for all inputs &
  \textbf{Proved} & Prop.~\ref{prop:gen3-gap} \\
Yukawa = VEV overlap $y_k \propto \langle\hat{u}_k, \hat{w}\rangle_\OO$ &
  \textbf{Proved} & Cor.~\ref{cor:vev-overlap} \\
Democracy: cubic form gives equal coupling for all compatible configs &
  \textbf{Proved} & Thm.~\ref{thm:democracy} \\
\midrule
$\langle\hat{u}_3, e_7\rangle = 1$ (gen-3 = fold map anchor) &
  Motivated & \S\ref{sec:gap} \\
$\langle\hat{u}_k, e_7\rangle = \sin(k\pi/14)$ for $k=1,2$ &
  Open (5a-ii) & \S\ref{sec:open} \\
Gen-3 coupling route: $x_{12}$ VEV, loop, or relabelling &
  Open & \S\ref{sec:open} \\
\bottomrule
\end{tabular}
\end{center}

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

Phase~5a of the seesaw programme asked: can the Cabibbo-step Dirac Yukawa
eigenvalues $(\sin\frac{\pi}{14}, \sin\frac{2\pi}{14}, 1)$ be derived from
the $J_3(\mathbb{O})$ trilinear form?

The answer is: \textbf{no, not directly, but the trilinear form determines the
mechanism.}

The key result is the \emph{democracy theorem} (Theorem~\ref{thm:democracy}):
$T(E_{ij}(q), E_{kl}(h), E_{mm}) = 0$ for all ``compatible'' off-diagonal pairs
$\{i,j\}\ne\{k,l\}$, and equals $-\frac{1}{3}\Re(q\bar h)$ for the self-coupling
$\{i,j\}=\{k,l\}$.  The non-vanishing coupling is thus a \emph{directional overlap}
$\langle\hat{u}_k, e_7\rangle$ between the generation-$k$ off-diagonal unit vector
and the $G_2$-invariant Higgs direction $e_7$.

This reformulates the conjecture precisely:
\emph{the Cabibbo-step Yukawa hierarchy holds if and only if the three off-diagonal
unit vectors $\hat{u}_1, \hat{u}_2, \hat{u}_3$ lie on the $G_2$-Weyl geodesic at
Cabibbo steps 1, 2, and 7 from $e_7$.}
The fold map fixes $\hat{u}_3 = e_7$ (step 7).
The open sub-problem (Phase~5a-ii) is to fix $\hat{u}_1$ and $\hat{u}_2$ at steps
1 and 2 from a uniqueness or extremal argument in the $G_2/SU(3)$ coset.

A secondary result is that the generation-3 Yukawa $y_3 = 1$ is not accessible
from the standard Higgs doublet through the cubic form
(Proposition~\ref{prop:gen3-gap}).  This is a structural zero, not a fine-tuning.
The most natural resolution within the corpus is a relabelling of the $\nu_R$
diagonal slots.

%==========================================================================
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L.\ F.\ Vlegels, \textit{$G_2$ Action on $J_3(\mathbb{O})$ Off-Diagonals
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\bibitem{A47}
L.\ F.\ Vlegels, \textit{Neutrino Mass Differences from $J_3(\mathbb{O})$
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L.\ F.\ Vlegels, \textit{$E_6$ Covariance of $J_3(\mathbb{O})$: Hypercharge
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\end{thebibliography}

\end{document}
