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% Addendum 390 (the dihedral structure; extends 385/386/387/389, P18/P06).
% The family/CPT group is D3 = S3 = the A2 Weyl group = Z3 rotations (families) + Z2 reflection (C=P),
% CP=s^2=id, reflection fixed locus = the void (387). Unifies families+C/P+void+A2 Weyl; corrects 389
% (S^3/D3 is an orbifold, not free). But the reflection deficits OVERSHOOT the H0 window (F=pi->H0=38),
% so D3 does NOT rescue H0; the tension stays structural.
% Author: Leon Fernando Vlegels. Text CC BY 4.0. Verifier: verify_P390.py (9/9 pass).
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\title{Addendum 390: The Dihedral Family Group\\
\large $D_3=S_3=$ the $A_2$ Weyl group unifies families, $C/P$, and the void ---\\
but its reflection deficits overshoot $H_0$\\[0.3em]
\normalsize (Extends Addenda 385, 386, 387, 389; Papers 18, 06)}
\author{L\'eon Fernando Vlegels}
\date{2026-06-19}

\begin{document}
\maketitle

\begin{abstract}
Addendum~389 found the family quotient $L(3,1)=S^3/\mathbb Z_3$ to be a \emph{free} action (smooth, the
horizon factor $F=2\pi$, no conical deficit), making the $H_0$ over-prediction structural. But $\mathbb
Z_3$ is only the rotation subgroup: Addenda~385/386 showed the corpus also carries a $\mathbb Z_2$
reflection --- the conjugation $\omega\leftrightarrow\bar\omega$ that realises \emph{both} $C$ and $P$.
The rotations and that reflection together generate the dihedral group. \textbf{The structure.} With
$r=T_{\rm cycle}$ (the $\mathbb Z_3$ family generator) and $s$ the $C=P$ reflection, the relations
$r^3=s^2=1$, $srs^{-1}=r^{-1}$ generate $\Dthree$ (order~6) $=S_3=$ the Weyl group of $A_2$/SU(3). So the
family/CPT discrete symmetry is $\Dthree$: the three rotations are the three fermion families (free,
389), and the three reflections are $C=P$ (with $CP=s^2=\mathrm{id}$, recovering Addendum~386). The
reflections' fixed locus is the real/antipodal locus --- the void at $\Theta=\pi$ (Addendum~387), which
is dark energy. So $\Dthree$ unifies four threads in one group: the families, $C$ and $P$, the void /
dark energy, and the $A_2$/SU(3) Weyl group. \textbf{It corrects 389:} $S^3/\mathbb Z_3$ is free, but the
\emph{full} $S^3/\Dthree$ is not --- the reflections fix the void, so it is an orbifold; 389's smoothness
held only for the rotations. \textbf{But it does not rescue $H_0$.} The data want $F\in[5.587,6.066]$ (a
$3$--$11\%$ deficit); the reflection orbifold deficits are far larger --- a $\mathbb Z_2$ cone (deficit
$\pi$) gives $F=\pi$, $H_0=37.9$; the $\Dthree$ fundamental domain gives $F=2\pi/3$, $H_0=25.3$ --- both
far below the window. The available deficits are quantised too coarsely (0 or $\sim\!\pi$), with nothing
in the $3$--$11\%$ range. So $\Dthree$ deepens rather than rescues 389's negative: the $H_0$ tension
stays structural. A structural win and an honest $H_0$ negative. All claims pinned by
\texttt{verify\_P390.py} (9/9).
\end{abstract}

\section{The family/CPT group is dihedral}
\label{sec:d3}

\begin{theorem}[$\Dthree=S_3=$ the $A_2$ Weyl group]
\label{thm:d3}
Let $r=T_{\rm cycle}$ be the $\mathbb Z_3$ family generator (Paper~18; the three families, Paper~06) and
$s$ the reflection $\omega\mapsto\omega^2=\bar\omega$ that realises $C=P$ (Addenda~385/386). Then
$r^3=s^2=1$ and $srs^{-1}=r^{-1}$, so $\langle r,s\rangle=\Dthree$, the dihedral group of order~6, which
is $S_3$ and the Weyl group of $A_2$/SU(3). The three rotations are the three families; the three
reflections are $C=P$. Since $C=P=s$, the composite $CP=s^2=\mathrm{id}$, recovering Addendum~386's
$CP=\mathrm{id}$ from the group structure: $C$ and $P$ are the same reflection, so their product is the
identity.
\end{theorem}

\begin{corollary}[$\Dthree$ unifies four threads]
\label{cor:unify}
The dihedral group ties together what had been separate results. The \emph{three families} are its
rotations ($\mathbb Z_3$). $C$ and $P$ are its reflections (Addenda~385/386). The reflections' fixed
locus --- the real/antipodal locus, $\omega=\pm1$, the void at $\Theta=\pi$ --- is dark energy
(Addendum~387). And $\Dthree\cong S_3$ is the Weyl group of $A_2$/SU(3), the corpus's gauge-family
structure. Families, $C/P$, the void, and the Weyl group are one group.
\end{corollary}

\section{The correction to 389, and the $H_0$ verdict}
\label{sec:h0}

\begin{proposition}[$S^3/\Dthree$ is an orbifold]
\label{prop:orbifold}
Addendum~389's claim that $S^3/\mathbb Z_3=L(3,1)$ is a \emph{free} action (hence smooth, $F=2\pi$) is
true of the \emph{rotation} subgroup only. The full $S^3/\Dthree$ is not free: the reflections fix a
locus (the real/antipodal locus, the void at $\Theta=\pi$), so $S^3/\Dthree$ is an \emph{orbifold} with a
singular locus there. So 389's smoothness was incomplete --- it covered the families, not the full
family/CPT group.
\end{proposition}

\begin{theorem}[The reflection deficits overshoot $H_0$]
\label{thm:overshoot}
Although $S^3/\Dthree$ has a singular locus (Proposition~\ref{prop:orbifold}), the conical deficits it
provides are far too large for the $H_0$ window. The data want $F\in[5.587,6.066]$ (a $3$--$11\%$ deficit
below $2\pi$). A $\mathbb Z_2$ reflection cone has deficit $\pi$, giving $F=\pi$ and $H_0=37.9$; the
$\Dthree$ fundamental domain on the circle gives $F=2\pi/3$ and $H_0=25.3$. Both lie far below the data
($67$--$73$). The available structural deficits are quantised too coarsely --- $0$ (the $\mathbb
Z_3$-free case, $F=2\pi$, $H_0=75.8$) or $\sim\!\pi$ (a reflection cone, $H_0\sim38$) --- with nothing in
the $3$--$11\%$ range. The dihedral does not fine-tune $F$; it jumps.
\end{theorem}

\begin{remark}[A structural win, an honest $H_0$ negative]
\label{rmk:net}
So the dihedral both deepens and confirms Addendum~389. It deepens it: $S^3/\Dthree$ is an orbifold, not
the smooth space 389 described, and the corpus's discrete family/CPT symmetry is richer ($\Dthree$, not
$\mathbb Z_3$) and more unified (Corollary~\ref{cor:unify}) than had been recorded. It confirms it: the
orbifold structure available overshoots the $H_0$ window rather than fine-tuning it
(Theorem~\ref{thm:overshoot}), so $H_0=75.8$ and the $\sim\!3\sigma$ tension remain structural. The
dihedral's value is the structural unification of families, $C/P$, the void, and the $A_2$ Weyl group ---
not an $H_0$ rescue. The $H_0$ fix, if any, still lies deeper: the $a_0=cH_0/2\pi$ relation, the chart
anchoring, or the closure hypothesis (Addendum~370).
\end{remark}

\section*{Status}
The family/CPT discrete group is the dihedral $\Dthree=S_3=$ the $A_2$/SU(3) Weyl group: the three
rotations are the three fermion families ($\mathbb Z_3$, $T_{\rm cycle}$) and the three reflections are
$C=P$ (Addenda~385/386), with $CP=s^2=\mathrm{id}$ recovered from the group (Theorem~\ref{thm:d3}). The
reflections' fixed locus is the void / dark energy (Addendum~387), so $\Dthree$ unifies families, $C/P$,
the void, and the Weyl group in one structure (Corollary~\ref{cor:unify}). This corrects Addendum~389:
$S^3/\mathbb Z_3$ is free, but the full $S^3/\Dthree$ is an orbifold (Proposition~\ref{prop:orbifold}).
But the reflection orbifold deficits overshoot the $H_0$ window --- $F=\pi$ gives $H_0=37.9$, $F=2\pi/3$
gives $H_0=25.3$, both far below the data --- with no $3$--$11\%$ deficit available
(Theorem~\ref{thm:overshoot}). So $\Dthree$ is a structural win and an honest $H_0$ negative
(Remark~\ref{rmk:net}): the tension stays structural and the fix lies deeper. No published number
changes. Verifier: \texttt{verify\_P390.py}, 9/9 pass.

\end{document}
