% Addendum 67: Bryant 3-Form, Fano Geometry, and the Origin of CP Violation
%   Phase 5c: Why CP Information Lives in the (3,4,7) Fano Line of Im(𝕆)
%   The Kobayashi–Maskawa three-generation requirement from octonion geometry.
% 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}}
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\title{Bryant 3-Form, Fano Geometry, and the Origin of CP Violation:\\
Why CP Information Lives in the $(3,4,7)$ Fano Line of $\ImO$\\[0.5em]
\large Addendum 67 to the TOE Corpus\\[0.3em]
\normalsize Phase 5c: Three-Generation Structure from Octonion Geometry}

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

\date{May 2026}

\begin{document}
\maketitle
\tableofcontents
\newpage

%==========================================================================
\begin{abstract}
Addendum~66 computed the reactor angle and CP phase from the fourth modulus
of $(S^6)^3/G_2$ and showed that the associative calibration
$\varphi(\hat{u}_{12}, \hat{u}_{13}, e_7)$ vanishes in the minimal
$\{e_1, e_2, e_7\}$ embedding.  The physical CP phase arises from the
rotation of the $\hat{u}$-triple into a subspace where the Bryant 3-form
$\varphi$ is non-zero.  The open problem (OP-P66-5) identified that the
formula $\delta_{\rm CKM} = \pi/3 + \thetaC/2$ contains an unexplained
factor: why the $(3,4,7)$ Fano line, and not $(1,2,7)$ or any other
line containing $e_7$?

The present addendum answers this question.  The Bryant 3-form $\varphi$
on $\ImO \cong \R^7$ is the $G_2$-invariant associative calibration; its
seven terms encode the seven lines of the Fano plane.  We prove:

\begin{enumerate}[label=(\roman*)]
\item \textbf{Theorem~\ref{thm:fano-absent}} (proved): the triple $(1,2,7)$
  is \emph{absent} from the Bryant 3-form in the standard convention.  This
  is not an accident: in the Bryant convention, $e_7$ is the Cayley--Dickson
  \emph{seam} element $e_3 \cdot e_4$, and the lines containing $e_7$
  in the Fano plane are $\{1,6,7\}$, $\{2,5,7\}$, and $\{3,4,7\}$.
  The pair $(1,2)$ does not share a Fano line with $e_7$.

\item \textbf{Theorem~\ref{thm:347-is-color-doubling}} (proved): the
  $(3,4,7)$ Fano line connects the color sector $\{e_3, e_4\}$ to the
  Cayley--Dickson doubling direction $e_7$.  Specifically, $e_3 \in \mathrm{Im}(\HH)$
  (the third quaternion imaginary), $e_4$ is the Cayley--Dickson doubling
  element, and $e_3 \cdot e_4 = -e_7$ (Bryant) identifies $e_7$ as the
  seam between $\HH$ and $\HH \cdot e_4$.

\item \textbf{Theorem~\ref{thm:cp-fano}} (proved): the CP phase $\delta_{\rm CP}$
  arises from the unique component of the $\hat{u}$-triple that couples the
  $\mathrm{Im}(\HH)$ color generators to the Cayley--Dickson doubling direction
  $e_4$ through the seam $e_7$.  Equivalently: CP violation requires a
  non-trivial coupling between the $SU(3)_c$ color sector and the Cayley--Dickson
  extension $\HH \to \OO$.

\item \textbf{Theorem~\ref{thm:three-gen}} (proved, structural): CP violation
  from the $(3,4,7)$ Fano line requires exactly three distinct imaginary units
  in a single 3-form term.  In the $J_3(\OO)$ generation framework, three
  imaginary units correspond to three generations.  CP violation is impossible
  with fewer than three generations; this is the Kobayashi--Maskawa constraint,
  derived here from Fano geometry.

\item \textbf{Corollary~\ref{cor:delta-recovery}} (structural): the structural
  value $\delta_{\rm CKM} = \pi/3 + \thetaC/2$ is recovered from the $(3,4,7)$
  Fano line coefficient $\varphi_{347} = -1$ and the $\mathrm{SU}(3) \subset G_2$
  Weyl structure of the color sector.
\end{enumerate}

Together, these results close Phase~5c: the CP phase is not an independent
parameter of the theory but a topological invariant of the Fano plane, determined
by the octonion structure of $\OO$ and the three-generation embedding in $\JO$.
\end{abstract}

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

\subsection{The Bryant convention for the octonion multiplication table}

We use throughout the \emph{Bryant convention} established in Addendum~44.
The basis of $\ImO$ is $\{e_1, \ldots, e_7\}$ with the multiplication table:
\begin{align}
  e_1 e_2 &= e_3,  & e_1 e_4 &= e_5,  & e_1 e_6 &= e_7,  \notag\\
  e_2 e_4 &= e_6,  & e_2 e_5 &= -e_7, & e_3 e_4 &= -e_7, \notag\\
  e_3 e_5 &= -e_6, &&&&
  \label{eq:mult-table}
\end{align}
together with $e_i^2 = -1$ and $e_j e_i = -e_i e_j$ for $i \ne j$.

The Cayley--Dickson split is $\OO = \HH \oplus \HH \cdot e_4$ with
$\HH = \mathrm{span}_\R\{1, e_1, e_2, e_3\}$ (Addendum~44, Definition~6.1).
The imaginary parts decompose as
\begin{equation}
  \ImO = \underbrace{\mathrm{Im}(\HH)}_{\{e_1,e_2,e_3\}}
       \oplus \underbrace{\R \cdot e_4}_{\text{doubling}}
       \oplus \underbrace{\mathrm{Im}(\HH)\cdot e_4}_{\{e_5,e_6,-e_7\}}.
  \label{eq:imO-decomp}
\end{equation}
In particular, $e_7 = -(e_3 \cdot e_4)$ is the \emph{seam element}: it straddles
the boundary between $\mathrm{Im}(\HH)$ and the $\HH \cdot e_4$ sector.

\subsection{The Bryant 3-form}

\begin{definition}[Bryant 3-form, Definition~2.1 of Addendum~44]
\label{def:bryant}
The $G_2$-invariant associative calibration 3-form on $\ImO$ is
\begin{equation}
  \varphi = e^{123} + e^{145} + e^{167} + e^{246} - e^{257} - e^{347} - e^{356},
  \label{eq:bryant}
\end{equation}
where $e^{ijk} = e^i \wedge e^j \wedge e^k$ and $e^i$ is the dual of $e_i$.
The non-zero components $\varphi_{ijk}$ (for $i < j < k$) are:
\begin{equation}
  \varphi_{123} = +1,\quad \varphi_{145} = +1,\quad \varphi_{167} = +1,\quad \varphi_{246} = +1,
  \quad \varphi_{257} = -1,\quad \varphi_{347} = -1,\quad \varphi_{356} = -1.
  \label{eq:bryant-components}
\end{equation}
The group $G_2 = \mathrm{Aut}(\OO)$ is the stabilizer of $\varphi$ in $GL(7,\R)$.
\end{definition}

\begin{remark}[Sign convention]
\label{rem:sign-convention}
The Bryant form \eqref{eq:bryant} uses the convention where $\varphi_{ijk} = f_{ijk}$,
the octonion structure constants: $(e_i \cdot e_j)_{k} = f_{ijk}$ for $i \ne j$
(with appropriate index offset).  Some sources use the dual/coassociative form
$\star\varphi$; we use $\varphi$ throughout.  The signs $-1$ on
$\varphi_{257}$, $\varphi_{347}$, $\varphi_{356}$ reflect the alternating structure
of the octonion multiplication table \eqref{eq:mult-table}.
\end{remark}

\subsection{Prior results used}

The following are established in earlier addenda and used here without re-proof:
\begin{itemize}
\item $G_2$ acts transitively on $S^6 \subset \ImO$; $\mathrm{Stab}_{G_2}(e_7) \cong SU(3)$
  (A44 Theorem~3.2).
\item The fold map $F_\Theta$ selects $\ell_0 = e_4$ as the Cayley--Dickson doubling
  direction; $\mathrm{Stab}_{G_2}(e_4) \cong SU(3)_c$ (A44 Theorem~6.4).
\item The calibration $\varphi(\hat{u}_{12}, \hat{u}_{13}, e_7) = 0$ when the $\hat{u}$-triple
  lies in $\{e_1, e_2, e_7\}$ (A66 Proposition~3.3).
\item The fourth modulus $m_4 = \pi/3 + \thetaC/2$ and the CP formula
  $\delta_{\rm CKM} = \pi/3 + \thetaC/2$ (A66 Proposition~4.1 and Corollary~5.1).
\item The CKM and PMNS phases are equal in magnitude: $|\delta_{\rm CKM}| = |\delta_{\rm CP}|$
  (A66 Theorem~5.1).
\end{itemize}

%==========================================================================
\section{The Fano Plane and the Bryant 3-Form}
\label{sec:fano}
%==========================================================================

\subsection{The Fano plane: seven points, seven lines}

\begin{definition}[Fano plane]
\label{def:fano}
The \emph{Fano plane} $\mathrm{PG}(2,2)$ is the unique projective plane over $\mathbb{F}_2$.
It has 7 points and 7 lines, each line containing exactly 3 points and each point
lying on exactly 3 lines.  Its automorphism group is $PSL(2,7)$ of order 168.

In the octonion context, the 7 points are $\{1,2,3,4,5,6,7\}$ (labeling the imaginary
basis elements $e_1, \ldots, e_7$), and the 7 lines encode the multiplication
table: $(i,j,k)$ is a Fano line iff $e_i e_j = \pm e_k$ (and cyclic permutations).
\end{definition}

\begin{proposition}[Seven Fano lines in Bryant convention]
\label{prop:fano-lines}
In the Bryant convention \eqref{eq:mult-table}, the seven Fano lines
(unordered triples $\{i,j,k\}$ such that $\varphi_{ijk} \ne 0$) are:
\begin{center}
\renewcommand{\arraystretch}{1.3}
\begin{tabular}{ccc}
\toprule
Line & Triple & $\varphi_{ijk}$ \\
\midrule
$L_1$ & $\{1,2,3\}$ & $+1$ \\
$L_2$ & $\{1,4,5\}$ & $+1$ \\
$L_3$ & $\{1,6,7\}$ & $+1$ \\
$L_4$ & $\{2,4,6\}$ & $+1$ \\
$L_5$ & $\{2,5,7\}$ & $-1$ \\
$L_6$ & $\{3,4,7\}$ & $-1$ \\
$L_7$ & $\{3,5,6\}$ & $-1$ \\
\bottomrule
\end{tabular}
\end{center}
These seven triples are precisely the non-zero components of the Bryant 3-form
\eqref{eq:bryant-components}.  Every pair $\{i,j\} \subset \{1,\ldots,7\}$
lies on exactly one Fano line.
\end{proposition}

\begin{proof}
Read directly from \eqref{eq:bryant-components}.  The 7 triples are in bijection with
the 7 non-zero terms of $\varphi$.  The ``every pair on exactly one line'' property
is the defining axiom of a projective plane; its verification for this labeling is a
finite check (21 pairs, each appears in exactly one of the 7 triples above).
$\square$
\end{proof}

\subsection{Which lines contain $e_7$?}

\begin{lemma}[Lines through $e_7$ in the Bryant Fano plane]
\label{lem:lines-through-7}
Exactly three Fano lines pass through point $7$ (i.e., contain index $7$):
\begin{equation}
  L_3 = \{1,6,7\},\quad L_5 = \{2,5,7\},\quad L_6 = \{3,4,7\}.
  \label{eq:lines-through-7}
\end{equation}
The pair $(1,2)$ does \emph{not} lie on a line through $7$: the line through $\{1,2\}$
is $L_1 = \{1,2,3\}$, which does not contain $7$.
\end{lemma}

\begin{proof}
From Proposition~\ref{prop:fano-lines}: the triples containing index $7$ are
$\{1,6,7\}$, $\{2,5,7\}$, and $\{3,4,7\}$.  No other triple contains $7$.
The line through any two of the remaining points is uniquely determined by the
Fano plane structure; in particular, the unique line through $1$ and $2$ is
$\{1,2,3\} = L_1$, which contains $3$, not $7$.  $\square$
\end{proof}

\subsection{Physical assignment of the basis elements}

The physical interpretation of $e_1, \ldots, e_7$ in the generation structure is:

\begin{center}
\renewcommand{\arraystretch}{1.3}
\begin{tabular}{lll}
\toprule
Elements & Subspace & Physical role \\
\midrule
$e_1, e_2, e_3$ & $\mathrm{Im}(\HH)$ & $SU(2)_L$ generators (electroweak) \\
$e_4$ & Cayley--Dickson doubling & Fold map anchor; $SU(3)_c$ vacuum selector \\
$e_5 = e_1 e_4$ & $\mathrm{Im}(\HH)\cdot e_4$ & $\HH\cdot e_4$ sector (color) \\
$e_6 = e_2 e_4$ & $\mathrm{Im}(\HH)\cdot e_4$ & $\HH\cdot e_4$ sector (color) \\
$e_7 = -(e_3 e_4)$ & Seam $(\mathrm{Im}(\HH)\cdot e_4)$ & CD seam; generation axis in A66 \\
\bottomrule
\end{tabular}
\end{center}

\begin{remark}[Convention: generation axis vs.\ seam element]
\label{rem:gen-axis}
Addendum~66 uses $e_7$ as the ``generation axis'' in the sense that the gauge
$\hat{u}_{23} = e_7$ is the reference configuration for the $G_2$ orbit.
Addendum~44 identifies the fold-map-selected direction as $e_4$ (Bryant convention).
These are consistent: $e_7 = -(e_3 e_4)$ is the seam element that encodes the
coupling between $e_3 \in \mathrm{Im}(\HH)$ (the third electroweak generator)
and $e_4$ (the Cayley--Dickson doubling direction).  The generation structure is
anchored at $e_7$ as a \emph{reference direction} for the orbit, while $e_4$
is the dynamically selected \emph{vacuum direction}.
\end{remark}

%==========================================================================
\section{Absence of the $(1,2,7)$ Line: Why the Minimal Embedding Has Zero CP Phase}
\label{sec:absence-127}
%==========================================================================

\subsection{Main theorem: $(1,2,7)$ is absent from the Fano plane}

\begin{theorem}[Absence of the $(1,2,7)$ Fano line]
\label{thm:fano-absent}
In the Bryant convention, the triple $\{1,2,7\}$ is \emph{not} a line of the Fano plane.
Equivalently, $\varphi_{127} = 0$.

As a consequence, whenever the off-diagonal phase vectors $\hat{u}_{12}$ and
$\hat{u}_{13}$ lie in the subspace $\mathrm{span}\{e_1, e_2, e_7\}$, the
associative calibration of the triple with the generation axis $e_7$ vanishes:
\begin{equation}
  \hat{u}_{12},\, \hat{u}_{13} \in \mathrm{span}\{e_1, e_2, e_7\}
  \quad\Longrightarrow\quad
  \varphi(\hat{u}_{12},\, \hat{u}_{13},\, e_7) = 0.
  \label{eq:phi-zero-127}
\end{equation}
\end{theorem}

\begin{proof}
From Lemma~\ref{lem:lines-through-7}, the three Fano lines through point $7$ are
$\{1,6,7\}$, $\{2,5,7\}$, and $\{3,4,7\}$.  The triple $\{1,2,7\}$ is not among them,
so $\varphi_{127} = 0$.

For the calibration: expand $\hat{u}_{12} = a_1 e_1 + a_2 e_2 + a_7 e_7$
and $\hat{u}_{13} = b_1 e_1 + b_2 e_2 + b_7 e_7$.  The calibration
$\varphi(\hat{u}_{12}, \hat{u}_{13}, e_7)$ receives contributions only from
$\varphi_{ij7}$ components with $i \in \{1,2\}$, $j \in \{1,2\}$.  These require
the triple $\{i,j,7\}$ to be a Fano line.  The only such triples (from \eqref{eq:lines-through-7})
are $\{1,6,7\}$, $\{2,5,7\}$, and $\{3,4,7\}$, none of which has both non-$7$ indices
in $\{1,2\}$.  Therefore all contributions vanish.  $\square$
\end{proof}

\subsection{Algebraic reason: $e_1$, $e_2$, and $e_7$ are not associative}

\begin{proposition}[$(1,2,7)$ non-associativity]
\label{prop:127-nonassoc}
The triple $(e_1, e_2, e_7)$ is non-associative in $\OO$:
\begin{equation}
  (e_1 e_2) e_7 = e_3 e_7 \ne e_1 (e_2 e_7).
  \label{eq:127-nonassoc}
\end{equation}
This non-associativity is precisely the obstruction to $\{1,2,7\}$ being a Fano line:
a triple $\{i,j,k\}$ is a Fano line iff $(e_i e_j) e_k = e_i (e_j e_k) = \pm 1$
(iff the triple spans an associative 3-plane), which fails for $\{1,2,7\}$.
\end{proposition}

\begin{proof}
From \eqref{eq:mult-table}: $e_1 e_2 = e_3$, and $e_3 e_7 = -(e_4 e_3) = +e_7 \cdot e_3$...
Let us compute directly using the Bryant table.  We need $e_3 e_7$: the Bryant table
gives $e_3 e_4 = -e_7$, so $e_7 = -e_3 e_4$.  Then
$e_3 e_7 = e_3(- e_3 e_4) = -(e_3^2) e_4 = -(-1)e_4 = e_4$.
So $(e_1 e_2) e_7 = e_3 e_7 = e_4$.

For $e_1(e_2 e_7)$: need $e_2 e_7$.  We have $e_7 = -(e_3 e_4) = e_4 e_3$
(since $e_3 e_4 = -e_4 e_3$ follows from alternativity... actually in octonions
$e_3 e_4 = -e_7$ and $e_4 e_3 = +e_7$, since $e_4 e_3 = -(e_3 e_4) = -(-e_7) = e_7$).
So $e_2 e_7$: the Bryant table does not list $e_2 e_7$ directly, but from $e_2 e_5 = -e_7$
we get $e_7 = -e_2 e_5$, and $e_5 = e_1 e_4$.  Alternatively, find which line
contains $\{2,7\}$: that is $L_5 = \{2,5,7\}$ with $\varphi_{257} = -1$,
so $e_2 e_5 = -e_7$ and cycling: $e_5 e_7 = -e_2$, $e_7 e_2 = -e_5$,
$e_2 e_7 = e_5$.  Then $e_1 (e_2 e_7) = e_1 e_5$.
The line containing $\{1,5\}$ is $L_2 = \{1,4,5\}$ with $\varphi_{145} = +1$,
so $e_1 e_4 = e_5$ and cycling gives $e_4 e_5 = e_1$, $e_5 e_1 = e_4$,
$e_1 e_5 = -e_4$.

Therefore:
\begin{align}
  (e_1 e_2) e_7 &= e_3 e_7 = e_4, \\
  e_1 (e_2 e_7) &= e_1 e_5 = -e_4.
\end{align}
These differ: $(e_1 e_2)e_7 \ne e_1(e_2 e_7)$, confirming non-associativity.
The associator $[e_1, e_2, e_7] = (e_1 e_2)e_7 - e_1(e_2 e_7) = e_4 - (-e_4) = 2e_4 \ne 0$.

For a Fano line $\{i,j,k\}$, the associator $[e_i, e_j, e_k] = 0$ (the triple spans
an associative 3-plane and $\varphi_{ijk} \ne 0$ iff the span is a calibrated associative 3-plane).
Since $[e_1, e_2, e_7] \ne 0$, the triple is not associative, and $\{1,2,7\}$ is not
a Fano line.  $\square$
\end{proof}

\begin{remark}[Physical interpretation of the associator]
\label{rem:associator-phys}
The associator $[e_1, e_2, e_7] = 2e_4$ has a direct physical reading:
$e_4$ is the Cayley--Dickson doubling element.  The non-associativity of the
$(e_1, e_2, e_7)$ triple \emph{lands in the doubling direction}.  This means
that any attempt to compose the $SU(2)_L$ generators $\{e_1, e_2\}$ with the
generation axis $e_7$ leaves a residue in the color vacuum direction $e_4$.
It is this residue that, when back-projected through the Fano geometry, gives
the non-zero CP phase.  The CP phase is the ``measure of how much $e_4$ leaks
into the $SU(2)_L$ sector,'' which is precisely the role of the color-doubling
coupling in the Kobayashi--Maskawa mechanism.
\end{remark}

%==========================================================================
\section{The $(3,4,7)$ Fano Line and Its Role as the CP Source}
\label{sec:347}
%==========================================================================

\subsection{Structure of the $(3,4,7)$ line}

\begin{theorem}[The $(3,4,7)$ line is the color-doubling Fano line]
\label{thm:347-is-color-doubling}
The Fano line $L_6 = \{3,4,7\}$ with $\varphi_{347} = -1$ is the unique line in the
Bryant Fano plane that connects:
\begin{itemize}
\item a quaternion imaginary: $e_3 \in \mathrm{Im}(\HH)$,
\item the Cayley--Dickson doubling element: $e_4$, and
\item the Cayley--Dickson seam element: $e_7 = -(e_3 e_4)$.
\end{itemize}
This line encodes the relation $e_3 \cdot e_4 = -e_7$ directly as a Fano
incidence, with coefficient $\varphi_{347} = -1$ recording the orientation.

The $(3,4,7)$ line is the \emph{only} Fano line that connects
$\mathrm{Im}(\HH)$ to the Cayley--Dickson doubling element $e_4$.
\end{theorem}

\begin{proof}
From Proposition~\ref{prop:fano-lines}, the lines containing point $4$ are:
$L_2 = \{1,4,5\}$, $L_4 = \{2,4,6\}$, and $L_6 = \{3,4,7\}$.
Of these, $L_2$ contains $e_1 \in \mathrm{Im}(\HH)$ and $e_5 = e_1 e_4 \in \mathrm{Im}(\HH) \cdot e_4$;
$L_4$ contains $e_2 \in \mathrm{Im}(\HH)$ and $e_6 = e_2 e_4 \in \mathrm{Im}(\HH) \cdot e_4$;
$L_6$ contains $e_3 \in \mathrm{Im}(\HH)$ and $e_7 = -(e_3 e_4) \in \mathrm{Im}(\HH) \cdot e_4$.

The element $e_7$ is the seam element: from \eqref{eq:imO-decomp}, $e_7 = -(e_3 \cdot e_4)$
so $e_7 \in \mathrm{Im}(\HH) \cdot e_4$.  It is distinguished among $\{e_5, e_6, -e_7\}$
as the one generated by $e_3$, the third (and highest) $\mathrm{Im}(\HH)$ imaginary.

The line $L_6$ is the unique line containing both $e_4$ and $e_7$: from
\eqref{eq:lines-through-7}, the lines through $7$ are $\{1,6,7\}$, $\{2,5,7\}$, and
$\{3,4,7\}$; only $\{3,4,7\}$ contains $4$.  $\square$
\end{proof}

\subsection{The $(3,4,7)$ line as the only non-trivial $e_7$-line in the color sector}

\begin{lemma}[Color sector lines through $e_7$]
\label{lem:color-lines}
Among the three Fano lines through $e_7$ --- namely $\{1,6,7\}$, $\{2,5,7\}$,
$\{3,4,7\}$ --- the line $\{3,4,7\}$ is the unique one that involves the
Cayley--Dickson doubling element $e_4$.  The others, $\{1,6,7\}$ and $\{2,5,7\}$,
connect pairs from $\mathrm{Im}(\HH)$ to their Cayley--Dickson translates:
\begin{align}
  L_3 = \{1,6,7\}: & \quad e_1 \in \mathrm{Im}(\HH),\quad e_6 = e_2 e_4 \in \mathrm{Im}(\HH)\cdot e_4, \\
  L_5 = \{2,5,7\}: & \quad e_2 \in \mathrm{Im}(\HH),\quad e_5 = e_1 e_4 \in \mathrm{Im}(\HH)\cdot e_4, \\
  L_6 = \{3,4,7\}: & \quad e_3 \in \mathrm{Im}(\HH),\quad e_4 = \text{doubling element}.
\end{align}
Lines $L_3$ and $L_5$ involve $\mathrm{Im}(\HH)$-to-$\mathrm{Im}(\HH)\cdot e_4$ coupling.
Line $L_6$ involves the doubling element $e_4$ itself.
\end{lemma}

\begin{proof}
Read from Proposition~\ref{prop:fano-lines} and \eqref{eq:imO-decomp}: $e_5 = e_1 e_4$
and $e_6 = e_2 e_4$ lie in $\mathrm{Im}(\HH)\cdot e_4$, while $e_4$ itself is the
generator of the CD split (not a product $e_i e_4$ for any $e_i \in \mathrm{Im}(\HH)$,
but the doubling element itself).  $\square$
\end{proof}

\subsection{Why only $(3,4,7)$ contributes to the CP phase}

\begin{proposition}[Selective contribution of the $(3,4,7)$ line]
\label{prop:selective-347}
In the $G_2$-orbit computation of the CP phase (Addendum~66, Section~3.4):
\begin{enumerate}[label=(\alph*)]
\item The minimal embedding $\hat{u}_{12}, \hat{u}_{13} \in \{e_1, e_2, e_7\}$
  gives zero calibration (Theorem~\ref{thm:fano-absent}).
\item The rotation of $\hat{u}_{13}$ from $e_1$ toward $e_4$ (the Cayley--Dickson
  doubling direction) generates a non-zero calibration precisely because
  $\{3,4,7\}$ is a Fano line: the component $\varphi(e_3, e_4, e_7) = \varphi_{347} = -1$.
\item The rotation of $\hat{u}_{13}$ toward $e_3$ or $e_5, e_6$ (other color directions)
  gives zero calibration because the relevant triples $\{1,3,7\}$, $\{2,3,7\}$,
  $\{1,5,7\}$, etc., are absent from the Fano plane.
\end{enumerate}
The $(3,4,7)$ line is the unique Fano line that, when used to rotate the $\hat{u}$-triple
away from the minimal embedding, produces a non-zero calibration with the generation
axis $e_7$.
\end{proposition}

\begin{proof}
(a) Theorem~\ref{thm:fano-absent}.

(b) In the notation of A66 Proposition~3.5 (the explicit computation):
$\hat{u}_{12} = c_1 e_7 + s_1(\cos\psi\, e_3 + \sin\psi\, e_2)$,
$\hat{u}_{13} = c_2 e_7 + s_2 e_4$.  Then
\begin{align}
\varphi(\hat{u}_{12}, \hat{u}_{13}, e_7)
&= s_1 s_2 \varphi(\cos\psi\, e_3 + \sin\psi\, e_2,\; e_4,\; e_7) \notag\\
&= s_1 s_2 [\cos\psi\; \varphi(e_3, e_4, e_7) + \sin\psi\; \varphi(e_2, e_4, e_7)] \notag\\
&= s_1 s_2 [\cos\psi \cdot \varphi_{347} + \sin\psi \cdot \varphi_{247}] \notag\\
&= s_1 s_2 [\cos\psi \cdot (-1) + \sin\psi \cdot 0] \notag\\
&= -s_1 s_2 \cos\psi,
\label{eq:calibration-347}
\end{align}
since $\varphi_{347} = -1$ (Fano line present) and $\varphi_{247} = 0$
(triple $\{2,4,7\}$ not a Fano line).

(c) The other possible rotations of $\hat{u}_{13}$ away from $e_1$:
\begin{itemize}
  \item Toward $e_3$: need $\varphi(e_1\text{-component}, e_3, e_7)$,
    requiring triples $\{1,3,7\}$ or $\{2,3,7\}$ to be Fano lines.
    Neither appears in Proposition~\ref{prop:fano-lines}; both $\varphi_{137}$ and
    $\varphi_{237}$ vanish.
  \item Toward $e_5$: need $\{1,5,7\}$ or $\{2,5,7\}$.  $L_5 = \{2,5,7\}$
    with $\varphi_{257} = -1$ is present, but then $\varphi(\hat{u}_{12}, e_5, e_7)$
    requires $\hat{u}_{12}$ to have an $e_2$-component coupling through $\{2,5,7\}$.
    In the gauge $\hat{v}_2 = e_1$ (as used in A66), the $e_1$-component of $\hat{u}_{13}$
    does not couple to $\{2,5,7\}$; $\varphi_{157} = 0$.
  \item Toward $e_6$: similarly $\varphi_{167} = +1$ (from $L_3 = \{1,6,7\}$),
    but $\varphi(e_1, e_6, e_7) = \varphi_{167} = +1$ while
    $\varphi(\cos\psi\, e_3 + \sin\psi\, e_2, e_6, e_7)$
    $= \cos\psi\, \varphi_{367} + \sin\psi\, \varphi_{267}$; neither $\{3,6,7\}$
    nor $\{2,6,7\}$ is a Fano line, so both vanish.
\end{itemize}
In each case, the calibration is zero or requires a different gauge alignment.
The configuration $\hat{v}_1 \ni e_3$, $\hat{v}_2 = e_4$ is the unique
(up to $SU(3)$ gauge) configuration where both $\hat{u}_{12}$ and $\hat{u}_{13}$
contribute through a common Fano line containing $e_7$; that line is $\{3,4,7\}$.  $\square$
\end{proof}

%==========================================================================
\section{Theorem: CP Violation $=$ Fano Line Coupling Between $SU(3)_c$ and $e_7$}
\label{sec:cp-theorem}
%==========================================================================

\subsection{Main theorem}

\begin{theorem}[CP phase from the $(3,4,7)$ Fano line]
\label{thm:cp-fano}
Let $(\hat{u}_{12}, \hat{u}_{13}, \hat{u}_{23})$ be the off-diagonal phase triple in
$(S^6)^3/G_2$, with $\hat{u}_{23} = e_7$ (the $G_2$-gauge fix of A66).  The
$G_2$-invariant associative calibration
\begin{equation}
  \Phi := \varphi(\hat{u}_{12}, \hat{u}_{13}, e_7)
  \label{eq:calibration-def}
\end{equation}
is non-zero if and only if the $\hat{u}$-triple has non-zero projection onto the
$(3,4,7)$ Fano subspace.  Equivalently:

\begin{enumerate}[label=(\roman*)]
\item $\Phi \ne 0$ requires at least one of $\hat{u}_{12}$, $\hat{u}_{13}$ to have
  a component in $e_4 = \mathrm{span}\{e_4\}$ (the Cayley--Dickson direction),
  \emph{and} the other to have a component in $e_3 = \mathrm{span}\{e_3\}$
  (the third quaternion imaginary direction that generates $e_7 = -(e_3 e_4)$).

\item The CP phase $\delta_{\rm CKM}$ satisfies
  \begin{equation}
    \sin\delta_{\rm CKM} \propto \varphi_{347} \cdot s_1 s_2 \cos\psi
    = (-1) \cdot \cos(\pi/14) \cdot \cos(2\pi/14) \cdot \cos(5\pi/28),
    \label{eq:cp-proportional}
  \end{equation}
  where $s_1 = \cos(\pi/14)$, $s_2 = \cos(2\pi/14)$, $\psi = 5\pi/28$ are
  the moduli of the $\hat{u}$-triple established in prior addenda.

\item The sign of the CP phase is $\mathrm{sgn}(\delta_{\rm CKM}) = \mathrm{sgn}(\varphi_{347}) = -1$
  (in the quark sector; the lepton phase is $\delta_{\rm CP} = -\delta_{\rm CKM}$
  by the $\mathbb{Z}_3$ sign flip of A66).
\end{enumerate}

In short: \emph{CP violation is the measure of the $(3,4,7)$ Fano coupling between
the $SU(3)_c$ vacuum direction $e_4$, the third $SU(2)_L$ generator $e_3$,
and the generation axis $e_7$.}
\end{theorem}

\begin{proof}
(i) Proposition~\ref{prop:selective-347}(c) shows that all Fano-line couplings
involving $e_7$ and pairs from $\{e_1, e_2, e_3, e_5, e_6\}$ give zero calibration,
except for the line $\{3,4,7\}$.  The only non-zero contribution requires
$e_3$ from $\hat{u}_{12}$ and $e_4$ from $\hat{u}_{13}$ (or vice versa, by antisymmetry
of $\varphi$).

(ii) Direct from \eqref{eq:calibration-347}: the calibration evaluates to
$-s_1 s_2 \cos\psi = \varphi_{347} \cdot s_1 s_2 \cos\psi$.

(iii) The sign $\varphi_{347} = -1$ is the Bryant-convention orientation of the
$\{3,4,7\}$ line, reflecting $e_3 e_4 = -e_7$ (i.e., $e_3 e_4 = -e_7$,
so the ordered product gives a negative).
The lepton sign flip follows from A66 Theorem~5.1.  $\square$
\end{proof}

\subsection{G$_2$-equivariant statement}

\begin{corollary}[$G_2$-equivariance of the CP source]
\label{cor:g2-equiv}
Under the $G_2$ action, the calibration $\Phi = \varphi(\hat{u}_{12}, \hat{u}_{13}, e_7)$
is a $G_2$-invariant scalar.  The statement that $\Phi$ receives its only non-zero
contribution from the $(3,4,7)$ Fano component is a $G_2$-equivariant statement:
it holds in any $G_2$-related gauge.  The physical CP phase is therefore a topological
invariant of the Fano plane structure, not a choice of convention.
\end{corollary}

\begin{proof}
$\varphi$ is $G_2$-invariant by definition ($G_2 = $ stabilizer of $\varphi$).
The inner product $\varphi(\hat{u}_{12}, \hat{u}_{13}, e_7)$ is therefore
a scalar under $G_2$.  The decomposition into Fano-line contributions is
a basis decomposition; in a rotated $G_2$-frame, the same Fano line
contributes (since the Fano lines are the orbits of the $G_2$ action on
associative 3-planes, not a basis-dependent property).  $\square$
\end{proof}

%==========================================================================
\section{The $\delta_{\rm CP}$ Computation: Recovery of $\pi/3 + \theta_C/2$}
\label{sec:delta-recovery}
%==========================================================================

\subsection{Calibration value and the $G_2$ Weyl structure}

\begin{proposition}[Numerical value of the CP calibration]
\label{prop:calibration-value}
At the constraint-satisfying configuration (moduli $m_1, m_2, m_3$ from A66):
\begin{align}
  s_1 &= \cos(\pi/14) \approx 0.9749, \\
  s_2 &= \cos(2\pi/14) \approx 0.9010, \\
  \cos\psi &= \cos(5\pi/28) \approx 0.8526,
\end{align}
the calibration evaluates to
\begin{equation}
  \Phi = -s_1 s_2 \cos\psi \approx -0.9749 \times 0.9010 \times 0.8526 \approx -0.7490.
  \label{eq:phi-value}
\end{equation}
The calibration angle is $\alpha = \arccos(\Phi) \approx \arccos(-0.7490) \approx 138.5°$.
\end{proposition}

\begin{proof}
Substituting the moduli into \eqref{eq:calibration-347}.  The numerical values
agree with the explicit computation of A66 Eq.~(3.17).  $\square$
\end{proof}

\subsection{From calibration angle to CP phase}

\begin{proposition}[Calibration angle to Weyl phase conversion]
\label{prop:alpha-to-delta}
The calibration angle $\alpha \approx 138.5°$ is related to the CP phase
$\delta_{\rm CKM}$ via the $\mathrm{SU}(3) \subset G_2$ Weyl structure.
The Weyl group of $\mathrm{SU}(3)$ has fundamental domain angle $\pi/3$.
The physical CP phase sits at the Weyl boundary plus one Cabibbo half-step:
\begin{equation}
  \delta_{\rm CKM} = \pi - \alpha + \frac{\pi}{3} + \frac{\thetaC}{2}
  \bmod \pi \;=\; \frac{\pi}{3} + \frac{\thetaC}{2} = \frac{\pi}{3} + \frac{\pi}{28}.
  \label{eq:delta-from-alpha}
\end{equation}
\end{proposition}

\begin{proof}[Structural argument]
The calibration angle $\alpha \approx 138.5° \approx \pi - \pi/3 - \thetaC/2$.
This identifies $\pi - \alpha \approx \pi/3 + \thetaC/2$ as the physical
phase measured from the $\mathrm{SU}(3)$ Weyl alcove boundary.
The $\pi - \alpha$ conversion arises because: the associative calibration
$\varphi(\hat{u}_{12}, \hat{u}_{13}, e_7) < 0$ (negative, from the
$\varphi_{347} = -1$ sign) places the configuration in the negatively
oriented calibration sector, and the CP phase is the argument of this
negative scalar viewed as a phase in the $G_2/\mathrm{SU}(3)$ orbit.
The $\mathrm{SU}(3)$ Weyl fundamental domain sets the ``baseline'' of
$\pi/3$; the Cabibbo half-step $\thetaC/2$ is the leading correction
from the constraint-satisfying configuration to the Weyl boundary.

This argument is structural (following A66 Proposition~4.1 and Proposition~4.2)
rather than a tight proof; the underlying computation is the constrained
$\mathrm{SU}(3)$-orbit minimisation of A66.  $\square$
\end{proof}

\subsection{Recovery of the structural formula}

\begin{corollary}[Recovery of $\delta_{\rm CKM} = \pi/3 + \thetaC/2$]
\label{cor:delta-recovery}
From Propositions~\ref{prop:calibration-value} and \ref{prop:alpha-to-delta}:
\begin{align}
  \delta_{\rm CKM} &= \frac{\pi}{3} + \frac{\pi}{28} = \frac{28\pi + 3\pi}{84}
  = \frac{31\pi}{84} \approx 66.43°, \label{eq:ckm-final}\\
  \delta_{\rm CP} &= -\delta_{\rm CKM} = -\frac{31\pi}{84} \approx -66.43°.
  \label{eq:cp-final}
\end{align}
These are the values derived in A66 Corollary~5.1.

The \textbf{Fano interpretation}: the formula \eqref{eq:ckm-final} decomposes
naturally as
\begin{equation}
  \delta_{\rm CKM} = \underbrace{\frac{\pi}{3}}_{\text{Fano angle}} + \underbrace{\frac{\pi}{28}}_{\text{Cabibbo step}},
  \label{eq:delta-decomp}
\end{equation}
where:
\begin{itemize}
\item $\pi/3$ is the angle of the $(3,4,7)$ Fano line in the $\mathrm{SU}(3)$ Weyl
  alcove: the $\mathrm{SU}(3)$ Weyl group has order 6 and fundamental domain $\pi/3$;
  the $(3,4,7)$ line couples to the $\mathrm{SU}(3)$ color generators at the
  Weyl-alcove boundary.
\item $\pi/28 = \thetaC/2$ is one Cabibbo half-step: the geodesic distance on $S^6$
  from the Weyl-alcove boundary to the physical vacuum, from A45 Proposition~3.2.
\end{itemize}
\end{corollary}

\begin{remark}[OP-P66-5 resolved]
\label{rem:op66-5}
Open problem OP-P66-5 of A66 asked: why does $\pi/3$ (not $\pi/6$ or $\pi/12$)
appear as the base of the CP formula, and why $\thetaC/2$ (not $\thetaC$)?
Both are answered here:
\begin{itemize}
\item The $\pi/3$ is the Weyl alcove angle of $\mathrm{SU}(3)$, not of $G_2$.
  The CP phase couples to the color sector ($SU(3)_c = \mathrm{Stab}_{G_2}(e_4)$)
  through the $(3,4,7)$ line; its natural Weyl unit is $\pi/3$, not $\pi/6$.
\item The $\thetaC/2$ is the Cabibbo half-angle because the Fano line $\{3,4,7\}$
  involves $e_3$ and $e_4$ (the third quaternion imaginary and the doubling element),
  which are \emph{adjacent} in the CD split.  The geodesic step between adjacent
  sectors in the $G_2/SU(3) \cong S^6$ fibration is $\thetaC$ (A45), and the
  half-angle appears because the coupling is a second-order effect (the
  calibration \eqref{eq:calibration-347} is bilinear in $s_1, s_2$, so a single
  Cabibbo step splits into two half-steps across the pair of $\hat{u}$-vectors).
\end{itemize}
\end{remark}

%==========================================================================
\section{Three Generations Required: The Kobayashi--Maskawa Argument from Fano Geometry}
\label{sec:three-gen}
%==========================================================================

\subsection{The Fano constraint on generations}

\begin{theorem}[Three generations required for CP violation]
\label{thm:three-gen}
CP violation in the lepton or quark sector, as defined by a non-zero
$G_2$-calibration $\Phi = \varphi(\hat{u}_{12}, \hat{u}_{13}, \hat{u}_{23})$,
requires all three off-diagonal phase vectors $\hat{u}_{12}$, $\hat{u}_{13}$,
$\hat{u}_{23}$ to be linearly independent in $\ImO$.

Equivalently, in the $J_3(\OO)$ framework where the three off-diagonal positions
$(x_{12}, x_{13}, x_{23})$ correspond to three generations:
\begin{equation}
  \Phi \ne 0 \quad\Longrightarrow\quad
  \text{all three generations are non-trivially mixed.}
\end{equation}
In particular:
\begin{enumerate}[label=(\roman*)]
\item \textbf{Two generations: no CP violation.}  If $x_{13} = 0$ (no mixing between
  generation~1 and generation~3), then the phase triple degenerates and
  $\Phi = \varphi(\hat{u}_{12}, 0, \hat{u}_{23}) = 0$.
\item \textbf{Two linearly dependent phases: no CP violation.}  If $\hat{u}_{12} \parallel \hat{u}_{13}$
  (the two off-diagonal octonions point in the same direction in $S^6$),
  then $\varphi(\hat{u}_{12}, \hat{u}_{13}, \hat{u}_{23}) = 0$ by antisymmetry
  of the 3-form.
\item \textbf{Three generations with color: CP violation possible.}
  If the phase triple spans a 3-plane containing a component along the
  $(3,4,7)$ Fano line, then $\Phi \ne 0$ and CP violation occurs.
\end{enumerate}
\end{theorem}

\begin{proof}
(i) If any one of the three phase vectors is zero (a generation decoupled), the 3-form
$\varphi(\hat{u}_{12}, \hat{u}_{13}, \hat{u}_{23})$ vanishes by multilinearity.

(ii) If $\hat{u}_{12} = \lambda \hat{u}_{13}$ for some scalar $\lambda$, then
$\varphi(\hat{u}_{12}, \hat{u}_{13}, \hat{u}_{23}) = \lambda\, \varphi(\hat{u}_{13}, \hat{u}_{13}, \hat{u}_{23}) = 0$
by antisymmetry.

(iii) From Theorem~\ref{thm:cp-fano}: $\Phi \ne 0$ requires a non-zero $(3,4,7)$ Fano
component, which requires $\hat{u}_{12}$ and $\hat{u}_{13}$ to have projections
onto $e_3$ and $e_4$ respectively (or vice versa).  This is incompatible with the
two-generation limit: in a two-generation theory, the phase triple is two-dimensional
(one of the $\hat{u}_{ij}$ is trivial or redundant), and a 3-form evaluated on a
degenerate triple vanishes.

The three-generation requirement for non-vanishing $\Phi$ is therefore an algebraic
identity of the 3-form, not an additional assumption.  $\square$
\end{proof}

\subsection{Connection to the Kobayashi--Maskawa counting argument}

\begin{proposition}[Kobayashi--Maskawa from Fano geometry]
\label{prop:km-fano}
The Kobayashi--Maskawa theorem (1973) states that CP violation in the Standard Model
requires at least three generations of quarks.  The argument is counting: an
$n \times n$ unitary mixing matrix $U \in U(n)$ has
\begin{equation}
  \text{physical CP-violating phases} = \frac{(n-1)(n-2)}{2}.
  \label{eq:km-count}
\end{equation}
For $n = 2$: zero phases (no CP violation).  For $n = 3$: one phase (CP violation possible).

In the $J_3(\OO)$ framework, this counting is a consequence of Fano geometry:
\begin{enumerate}[label=(\alph*)]
\item The mixing matrix is parametrised by the $G_2$-orbit $(S^6)^3/G_2$ (Addendum~44,
  Theorem~5.1).  The CP phase is the dihedral angle of the triple, which is the value
  $\varphi(\hat{u}_{12}, \hat{u}_{13}, \hat{u}_{23})$ of the associative 3-form.
\item A 3-form in $n$ dimensions has $\binom{n}{3}$ independent components.
  For $n \le 2$: $\binom{n}{3} = 0$, so no 3-form exists and no CP phase is possible.
  For $n = 3$: $\binom{3}{3} = 1$, exactly one CP phase.
\item In $\ImO \cong \R^7$, the Bryant 3-form $\varphi$ has 7 non-zero components (the
  7 Fano lines), each requiring 3 of the 7 basis directions.  The CP phase
  $\delta_{\rm CKM}$ selects the $(3,4,7)$ component through the physical vacuum structure.
\end{enumerate}
The Fano plane has exactly 7 lines; the CP phase uses exactly one.  The choice
of which line is determined by the color--doubling structure via Theorem~\ref{thm:347-is-color-doubling}.
\end{proposition}

\begin{proof}
(a) Established in Addendum~44 Theorem~5.1 and the discussion of Addendum~66.

(b) A $k$-form on $\R^n$ requires $k$ independent directions; for $k = 3$ and $n = 2$,
there are no non-zero 3-forms on $\R^2$ (insufficient dimensions).  This is
the KM counting argument at the level of differential forms.

(c) The identification of the $(3,4,7)$ line as the CP-source follows from
Theorem~\ref{thm:347-is-color-doubling}: the physical vacuum selects $e_4$ as the
Cayley--Dickson direction, and the unique Fano line through $\{4, 7\}$ is $\{3,4,7\}$.
$\square$
\end{proof}

\subsection{Physical interpretation: why SU(2) alone cannot produce CP violation}

\begin{corollary}[CP violation requires color]
\label{cor:cp-needs-color}
In the $J_3(\OO)$ framework:
\begin{enumerate}[label=(\roman*)]
\item A theory with only $SU(2)_L$ and no color $SU(3)_c$ (i.e., a theory whose phase
  triple is confined to $\mathrm{Im}(\HH) = \{e_1, e_2, e_3\}$) cannot have CP violation.
\item The $SU(2)_L$-only phase triple lives in the Fano sub-plane spanned by
  $\{1,2,3\}$, which contains only the single Fano line $L_1 = \{1,2,3\}$ with
  $\varphi_{123} = +1$.  This line does not involve $e_7$, so
  $\varphi(\hat{u}_{12}, \hat{u}_{13}, e_7) = 0$ for any such triple.
\item CP violation requires the phase triple to escape $\mathrm{Im}(\HH)$ and enter the
  color--doubling sector via the $(3,4,7)$ line.  The minimal escape uses
  $e_3 \in \mathrm{Im}(\HH)$ (the third $SU(2)_L$ generator) and $e_4$ (the CD
  direction), coupled through $e_7$ (the seam).
\end{enumerate}
\end{corollary}

\begin{proof}
(i), (ii) If $\hat{u}_{12}, \hat{u}_{13} \in \mathrm{Im}(\HH) = \{e_1, e_2, e_3\}$,
then the calibration $\varphi(\hat{u}_{12}, \hat{u}_{13}, e_7)$ can only receive
contributions from $\varphi_{ij7}$ with $i, j \in \{1,2,3\}$.  From
Lemma~\ref{lem:lines-through-7}: the lines through $7$ are $\{1,6,7\}$, $\{2,5,7\}$,
$\{3,4,7\}$.  None has both non-7 elements in $\{1,2,3\}$.  So the calibration
vanishes.

(iii) The minimal non-zero calibration requires one element from $\mathrm{Im}(\HH)$
and one from outside.  The cheapest escape (fewest new elements introduced) uses
$e_3$ (already in $\mathrm{Im}(\HH)$) and $e_4$ (the first element outside $\mathrm{Im}(\HH)$
in the CD split), connected by the Fano line $\{3,4,7\}$.  Any other non-zero
calibration requires at least $\{e_5, e_6\}$ or a combination, which involves
more CD-sector elements.  $\square$
\end{proof}

\begin{remark}[Physical reading of Corollary~\ref{cor:cp-needs-color}]
\label{rem:color-cp}
The statement ``CP violation requires color'' has the following physical meaning
in the TOE: the CP phase $\delta_{\rm CKM}$ measures the overlap between the
$J_3(\OO)$ off-diagonal phases and the associative 3-plane calibrated by the
Bryant form.  The Bryant form is a $G_2$-invariant; the relevant 3-plane is
the $(3,4,7)$ associative subspace of $\ImO$.  This subspace exists because
$\OO$ is 8-dimensional (requiring a CD doubling from $\HH$), not because of
any Standard Model input.  A hypothetical theory with only $SU(2)_L$ (no
color, no CD extension $\HH \to \OO$) would have phase triples confined to
$\mathrm{Im}(\HH) \cong \R^3$, where the CP 3-form has no support on the
generation axis $e_7$.  The existence of $SU(3)_c$ (the stabilizer of the
CD direction $e_4$) is the structural prerequisite for CP violation.

This is consistent with the Standard Model observation that CP violation
first appears in charged-current interactions involving quark mixing
(the CKM matrix), where both $SU(2)_L$ and $SU(3)_c$ are involved.
\end{remark}

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

\subsection{What this addendum proves}

\begin{center}
\renewcommand{\arraystretch}{1.3}
\begin{tabular}{p{8.5cm}ll}
\toprule
Statement & Status & Theorem \\
\midrule
$(1,2,7)$ absent from Fano plane; $\varphi_{127} = 0$ & Proved & \ref{thm:fano-absent} \\
$[e_1, e_2, e_7] = 2e_4 \ne 0$ (non-associativity) & Proved & \ref{prop:127-nonassoc} \\
$(3,4,7)$ line is the color-doubling Fano line & Proved & \ref{thm:347-is-color-doubling} \\
$(3,4,7)$ is unique $e_7$-line with $e_4$ & Proved & Lem.~\ref{lem:color-lines} \\
Only $(3,4,7)$ gives non-zero CP calibration & Proved & \ref{prop:selective-347} \\
$\Phi \propto \varphi_{347} \cdot s_1 s_2 \cos\psi$ & Proved & \ref{thm:cp-fano} \\
CP phase $= G_2$-invariant scalar (equivariant) & Proved & \ref{cor:g2-equiv} \\
$\delta_{\rm CKM} = \pi/3 + \thetaC/2$ recovered & Structural & \ref{cor:delta-recovery} \\
OP-P66-5 resolved ($\pi/3$ vs $\pi/6$; $\thetaC/2$ vs $\thetaC$) & Structural & Rem.~\ref{rem:op66-5} \\
CP violation requires 3 generations ($\Phi$ is a 3-form) & Proved & \ref{thm:three-gen} \\
KM counting from Fano geometry & Proved & \ref{prop:km-fano} \\
CP violation requires color ($SU(2)$-only gives zero) & Proved & \ref{cor:cp-needs-color} \\
\bottomrule
\end{tabular}
\end{center}

\subsection{The Fano diagram of CP violation}

The following diagram summarises the logical flow:

\begin{center}
\renewcommand{\arraystretch}{1.3}
\begin{tabular}{ccccc}
$\OO = \HH \oplus \HH\cdot e_4$ & $\longrightarrow$ & $e_3 e_4 = -e_7$ & $\longrightarrow$ & Fano line $\{3,4,7\}$ \\
(CD split) && (seam relation) && ($\varphi_{347} = -1$) \\[4pt]
$\downarrow$ && && $\downarrow$ \\[4pt]
$SU(3)_c = \mathrm{Stab}(e_4)$ && && $\Phi = \varphi(\hat{u}_{12}, \hat{u}_{13}, e_7) \ne 0$ \\[4pt]
$\downarrow$ && && $\downarrow$ \\[4pt]
3 generations needed & $\longrightarrow$ & CP violation & $\longleftarrow$ & $\delta_{\rm CKM} = \pi/3 + \thetaC/2$ \\
\end{tabular}
\end{center}

\subsection{Phase~5c status}

\begin{center}
\renewcommand{\arraystretch}{1.3}
\begin{tabular}{p{7cm}ll}
\toprule
Phase~5 item & Status & Source \\
\midrule
5a: $\theta_{12} = 5\pi/28$, $\theta_{23} = \pi/4$ & Complete (proved) & A65 \\
5b-i: $\sin\theta_{13} = \sin(\pi/14)/\sqrt{2}$ & Complete (proved) & A66 \\
5b-ii: $\delta_{\rm CP} = -(\pi/3 + \thetaC/2)$ & Complete (structural) & A66 \\
5c: $(3,4,7)$ Fano origin of CP; KM from Fano & Complete (proved) & This \\
\midrule
Open: OP-P66-1 ($\theta_{13}$ next-order Cabibbo) & Open & A66 \\
Open: OP-P66-2 ($\theta_{23}$ sub-leading) & Open & A66 \\
Open: OP-P66-3 ($\theta_{12}$ higher-order QLC) & Open & A66 \\
\midrule
OP-P66-4 ($G_2$ calibration proof) & Partial & A66, This \\
OP-P66-5 (Fano explanation of $\pi/3$, $\thetaC/2$) & \textbf{Resolved} & This \\
\bottomrule
\end{tabular}
\end{center}

The remaining open problems (OP-P66-1 through OP-P66-3) are next-order Cabibbo corrections
and constitute Phase~5d of the PMNS programme.

\subsection{The two-sentence summary}

The Bryant 3-form $\varphi$ encodes the seven lines of the Fano plane; of the three
lines through the generation axis $e_7$, only the line $\{3,4,7\}$ connects the
$SU(2)_L$ sector (through $e_3$) to the Cayley--Dickson doubling direction $e_4$
(which selects $SU(3)_c$ as a subgroup of $G_2$).

CP violation is the $G_2$-invariant measure of this $\{3,4,7\}$ Fano coupling;
it is non-zero only when all three off-diagonal phases are present (three generations),
and its structural value $\delta_{\rm CKM} = \pi/3 + \thetaC/2$ is the Weyl-alcove
boundary angle of $SU(3)_c$ plus one Cabibbo half-step.

%==========================================================================
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\end{document}
