% DRAFT -- resolved Addendum 255 after TASK_022/TASK_023/TASK_024 synthesis.
% Status: Codex/Read staged draft, not promoted to corpus canon.
%
% Verdict:
%   Joint A: DERIVED.
%   Hopf-fiber-to-L(3,1): DERIVED.
%   Strong triality C3 = lens Z3: NEGATIVE / PROVABLY INDEPENDENT.
%
% L. F. Vlegels, Independent Researcher
% Copyright: Leon Fernando Vlegels. License: CC BY 4.0. May 2026.

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\title{DRAFT Addendum 255: Two Primitive Threes\\
and the Derived Hopf--Lens Bridge\\[0.4em]
\large Quaternionic Hopf Closure, Lens-Space Families,\\
and the Independence of Triality}

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

\date{May 2026}

\begin{document}
\maketitle

\begin{abstract}
This addendum resolves the TASK~022 bridge question.  The quaternionic Hopf
side closes: P01's Cayley--Dickson split $\OO=\HH\oplus\HH\ell$ supplies an
$\HH^2$ coordinate model, and the projection
$S^7_{\HH^2}\to\HP^1\simeq S^4$ has $S^3$ fibers.  The fiber over $[1:0]$
is exactly the embedded unit-quaternion $S^3\subset S^7$ already present in
P01.

The Hopf-fiber-to-lens step also closes.  On a unit quaternion
$q=z_1+z_2j$, left multiplication by
$\epsilon=e^{2\pi i/3}$ acts as
$(z_1,z_2)\mapsto(\epsilon z_1,\epsilon z_2)$, exactly the P01 generator of
$L(3,1)=S^3/\ZZ_3$.

The stronger proposed identification does not close.  The triality subgroup
$C_3<\Out(\Spin(8))$ and the lens deck group $\ZZ_3<\Diff(S^3)$ are
provably independent as typed objects.  The lens generator moves
$1\in S^3$ to $\epsilon$, while every element of $G_2=\Aut(\OO)$ fixes
$1$ and the real scalars.  Therefore the lens generator is not in $G_2$ and
cannot be obtained through $Z(SU(3))\subset SU(3)\subset G_2$.  The canon
must state two primitive threes: the lens-space topological $Z_3$ and the
representation-theoretic triality $C_3$.  A44's cyclic permutation of the
three off-diagonal entries of $\JO$ is a third, distinct order-three object.
\end{abstract}

\tableofcontents
\newpage

\section{Typed Objects}

\begin{definition}[Cayley--Dickson coordinate model]
P01 defines the octonions by the Cayley--Dickson split
\[
  \OO=\HH\oplus\HH\ell
\]
at \texttt{01\_Paper\_PerfectStableSphere.tex:805--812}.  Define
\[
  \Phi:\HH^2\to\OO,\qquad \Phi(a,b)=a+b\ell .
\]
Let
\[
  S^7_{\HH^2}
  =
  \{(a,b)\in\HH^2:\lVert a\rVert^2+\lVert b\rVert^2=1\}.
\]
\end{definition}

\begin{remark}[Module action, not octonion multiplication]
The right $\HH$ action used for the Hopf fibration is the module action
\[
  (a+b\ell)\cdot_R h := ah+(bh)\ell .
\]
It is not literal octonion multiplication by $h$.  Under octonion
multiplication the second component is conjugated in the Cayley--Dickson
rule; for example $a=0$, $b=1$, $h=i$ gives $(0,i)$ for the module action
and $(0,-i)$ for the octonion product.  This typing prevents a false
associativity claim.
\end{remark}

\section{Joint A: Quaternionic Hopf Closure}

\begin{theorem}[Quaternionic Hopf closure]
After choosing the P01 Cayley--Dickson split, the unit octonion sphere is the
total space of the quaternionic Hopf fibration
\[
  S^3\longrightarrow S^7_{\HH^2}\longrightarrow \HP^1\simeq S^4 .
\]
\end{theorem}

\begin{proof}
The Cayley--Dickson norm gives
\[
  \lVert a+b\ell\rVert^2=\lVert a\rVert^2+\lVert b\rVert^2,
\]
so $\Phi$ identifies $S^7_{\HH^2}$ with
$S^7_{\OO}=\{x\in\OO:\lVert x\rVert=1\}$.  Define
\[
  \pi_{\HH}(a,b)=[a:b]\in\HP^1 .
\]
The right unit-quaternion action
\[
  (a,b)\cdot u=(au,bu),\qquad \lVert u\rVert=1,
\]
has orbits equal to the fibers of $\pi_{\HH}$, so each fiber is $S^3$.
The base $\HP^1$ is diffeomorphic to $S^4$.
\end{proof}

\begin{corollary}[P01 embedded sphere is a Hopf fiber]
The fiber over $[1:0]$ is
\[
  \{(u,0):\lVert u\rVert=1\}\cong S^3 .
\]
Under $\Phi$, this is exactly P01's embedded unit-quaternion sphere
$S^3\subset S^7$ from
\texttt{01\_Paper\_PerfectStableSphere.tex:897--917}.
\end{corollary}

\section{The Lens Generator on the Hopf Fiber}

\begin{theorem}[Fiber-to-lens closure]
Let $q=z_1+z_2j$ be a unit quaternion, with $z_1,z_2\in\CC$.  Let
$\epsilon=e^{2\pi i/3}\in\CC\subset\HH$.  Left multiplication by $\epsilon$
acts on the Hopf fiber $S^3$ by
\[
  q\mapsto \epsilon q
  \quad\Longleftrightarrow\quad
  (z_1,z_2)\mapsto(\epsilon z_1,\epsilon z_2).
\]
This is the P01 generator of $L(3,1)=S^3/\ZZ_3$.
\end{theorem}

\begin{proof}
Since $\epsilon\in\CC\subset\HH$ acts on the left,
\[
  \epsilon(z_1+z_2j)=\epsilon z_1+\epsilon z_2j.
\]
P01 records the $p=3,q=1$ lens generator as
$(z_1,z_2)\mapsto(\omega z_1,\omega z_2)$ at
\texttt{01\_Paper\_PerfectStableSphere.tex:993--998}.  Taking
$\omega=\epsilon$ gives the same action.  The action is free: if
$\epsilon q=q$, then $(\epsilon-1)q=0$; since $\HH$ is a division algebra and
$\epsilon\ne 1$, this forces $q=0$, impossible on $S^3$.
\end{proof}

\section{Triality Is Independent}

\begin{proposition}[The lens generator is not in $G_2$]
The lens generator $L_\epsilon:S^3\to S^3$, $q\mapsto\epsilon q$, is not the
restriction of any element of $G_2=\Aut(\OO)$.
\end{proposition}

\begin{proof}
The point $1\in S^3\subset\HH$ is sent by the lens generator to
$\epsilon\ne 1$.  But $G_2=\Aut(\OO)$ fixes the real unit.  P01 defines
$G_2=\Aut(\OO)$ at
\texttt{01\_Paper\_PerfectStableSphere.tex:814--818}; A44 explicitly records
that $G_2=\Aut(\OO)$ fixes $e_0=1$ and the real subspace at
\texttt{44\_Addendum\_G2ActionJ3.tex:324--328}.  Hence no $g\in G_2$ can
agree with $L_\epsilon$ on $S^3$.
\end{proof}

\begin{corollary}[No $SU(3)$-center route]
The lens deck transformation is not obtained through
$Z(SU(3))\subset SU(3)\subset G_2$.
\end{corollary}

\begin{proof}
Every element of $SU(3)$ in the corpus route is an element of $G_2$, and so
fixes $1$.  The lens generator does not fix $1$.
\end{proof}

\begin{proposition}[Triality and lens $Z_3$ are independent]
The cyclic triality subgroup $C_3<\Out(\Spin(8))$ and the lens deck group
$\ZZ_3<\Diff(S^3)$ are not the same structure in the current canon.
\end{proposition}

\begin{proof}
Triality is used in P01/P05 as a representation-theoretic order-three
structure supporting the bulk coefficient $16$; see
\texttt{01\_Paper\_PerfectStableSphere.tex:371--374} and
\texttt{05\_Paper\_GeometryOfReality.tex:590--596}.  The lens group acts on
points of the Hopf fiber $S^3$.  No corpus-defined map
\[
  C_3<\Out(\Spin(8))\longrightarrow \Diff(S^3)
\]
sends the triality generator to $L_\epsilon$, and the $G_2/SU(3)$ route is
blocked by the preceding proposition.
\end{proof}

\section{The Third Order-Three Structure}

A44 contains a separate $\ZZ_3$ acting on the off-diagonal slots of $\JO$:
\[
  \sigma:(x_{12},x_{13},x_{23})\mapsto(x_{23},x_{12},x_{13}).
\]
This is recorded at
\texttt{44\_Addendum\_G2ActionJ3.tex:396--404} and summarized at
\texttt{44\_Addendum\_G2ActionJ3.tex:437--439}.  It acts on Jordan-algebra
slots, not on the Hopf fiber $S^3$, and it is not the triality subgroup used
for the P01/P05 bulk coefficient.

\section{Canonical Axioms}

\begin{axiom}[LS: lens-space topology]
The boundary Hopf fiber $S^3$ carries the free order-three action generated
by left multiplication by $\epsilon=e^{2\pi i/3}$.  The quotient is
$L(3,1)=S^3/\ZZ_3$, and this is the topological source of the three
fermion-family covering.
\end{axiom}

\begin{axiom}[TR: triality]
The octonionic bulk carries the representation-theoretic order-three
triality of $\Spin(8)$, permuting the vector and two half-spinor
representations.  In the current corpus this supports the bulk coefficient
$16=8+8$.
\end{axiom}

\begin{remark}[Do not merge the threes]
The canon may relate LS, TR, and the A44 $\JO$ permutation structurally, but
it may not identify them without an explicit typed map.  The $G_2/SU(3)$
route cannot supply that map for the lens generator because $G_2$ fixes
$1\in\OO$ while the lens generator moves it.
\end{remark}

\section{Status Table}

\begin{center}
\begin{tabular}{lll}
\toprule
Claim & Status & Canon action \\
\midrule
Quaternionic Hopf from P01 split & DERIVED & Add to A255 as theorem. \\
P01 embedded $S^3$ is Hopf fiber & DERIVED & Add as corollary. \\
Fiber $C_3$ gives $L(3,1)$ & DERIVED & Add as theorem. \\
Triality $C_3$ equals lens $\ZZ_3$ & NEGATIVE & State independence. \\
A44 $\JO$ permutation $\ZZ_3$ & DISTINCT & Catalogue separately. \\
\bottomrule
\end{tabular}
\end{center}

\end{document}

