\section{Coordinates and changes of quartet}\label{app:coordinates}
Appendix~\ref{app:subalgebras} and
Appendices~\ref{app:velocity}--\ref{app:slice-proof} contain the rest of the proofs of
Lemmas~\ref{lem:subalgebras}, \ref{lem:families} and~\ref{lem:slice}. The rest of this appendix is
verification by direct computation; a reader who accepts the displayed identities does not
need the tables.

\subsection{The octonions and the coordinate map}
Write an octonion as $x=c+z_1j+z_2\ell+z_3n=(c,z)$, with
$c,z_r\in\mathbb C$ on the left. Its eight
real coordinates in the basis $(1,i,j,k,\ell,i\ell,j\ell,k\ell)$ are
\[
(\operatorname{Re}c,\operatorname{Im}c,\operatorname{Re}z_1,
\operatorname{Im}z_1,\operatorname{Re}z_2,\operatorname{Im}z_2,
\operatorname{Re}z_3,-\operatorname{Im}z_3).
\]
In this splitting $\mathbb O=\mathbb C\oplus\mathbb C^3$ the octonion
multiplication becomes
\begin{equation}\label{eq:octonionrule}
 (c,z)(c',z')=(cc'-z\cdot\bar z',\;cz'+\bar c'z+\overline{z\times z'}),
\end{equation}
where dot and cross products are complex bilinear, with orientation $(1,2,3)$.
This follows by multiplying the eight basis units; it fixes
the sign of the last coordinate. Up to the sign of the vector part it is
the multiplication in Section~1.5 of Yokota\cite{Yokota09}.

Write an element of $J$ as the Hermitian matrix
\[
 y=\begin{pmatrix}\alpha_1&x_3&\bar x_2\\
 \bar x_3&\alpha_2&x_1\\x_2&\bar x_1&\alpha_3\end{pmatrix},
 \qquad x_r=(c_r,z_r).
\]
Put $b_r=-\bar z_r\in\mathbb C^3$. The coordinate map sends $y$ to the pair
$(a,B)$, where
\[
 a=\begin{pmatrix}\alpha_1&c_3&\bar c_2\\
 \bar c_3&\alpha_2&c_1\\c_2&\bar c_1&\alpha_3\end{pmatrix}
\]
and $B$ is the matrix with rows $b_1,b_2,b_3$.
This real-linear bijection visibly preserves unit and trace; expanding the trace form gives
\eqref{eq:titsmetric}.

To check the product, it suffices to check squares.
For cyclic $(r,s,t)$, octonion matrix multiplication gives the entries
$\alpha'_r$, $x'_r$ of $y^2$:
\[
 \alpha'_r=\alpha_r^2+|x_s|^2+|x_t|^2,\qquad
 x'_r=(\alpha_s+\alpha_t)x_r+\overline{x_sx_t}.
\]
By \eqref{eq:octonionrule}, the scalar and vector components of $x'_r$ are
\begin{align*}
 c'_r&=(\alpha_s+\alpha_t)c_r+\bar c_s\bar c_t-\bar z_s\cdot z_t,\\
 b'_r&=(\alpha_s+\alpha_t)b_r-\bar c_s b_t-c_t b_s+
 \overline{b_s\times b_t}.
\end{align*}
The first equation together with the diagonal formula is
\[
 a'=a^2-BB^*+\operatorname{tr}(BB^*)I.
\]
In the second, the conjugated cross products are the rows of
$\operatorname{adj}(B)^*$ and the other terms those of
$((\operatorname{tr}a)I-a)B$. This proves \eqref{eq:titssquare}.

\subsection{The automorphisms \texorpdfstring{$\iota_i$ and $\theta_i$}{iota\_i and theta\_i}}
The maps $\iota_1,\iota_2,\iota_3$ are signed permutations of the octonion basis in
each off-diagonal entry; they fix the diagonal entries. Let $e_m$,
$m=0,\ldots,7$, be the basis $(1,i,j,k,\ell,i\ell,j\ell,k\ell)$. On the
entry $x_r$, the map $\iota_i$ sends
\[
 e_m\longmapsto\epsilon_{rm}\,e_{m\mathbin{\mathrm{xor}}\mu_r},
\]
with the masks $\mu_r$ and the signs $\epsilon_{rm}$ of the table.
\begin{center}\small
\begin{tabular}{c|c|c|c|c}
&$(\mu_1,\mu_2,\mu_3)$&signs on $x_1$&signs on $x_2$&signs on $x_3$\\\hline
$\iota_1$&$(6,4,2)$&$-+++ +---$&$--++ +--+$&$++-+ -+++$\\
$\iota_2$&$(2,6,4)$&$-+-- +++-$&$+++- ---+$&$-+-+ ++++$\\
$\iota_3$&$(4,2,6)$&$++-- +--+$&$--+- -+++$&$-+++ -+++$
\end{tabular}
\end{center}
The masks $2$, $4$, $6$ move the line
$\mathbb C$ to $\mathbb Cj$, $\mathbb C\ell$, $\mathbb Cn$, so $\iota_i$ maps
$A_0$ onto the space $A_i$ of \eqref{eq:latin}.

The $\iota_i$ are automorphisms because they preserve the cubic norm, which in
octonion coordinates is
\[
 N(y)=\alpha_1\alpha_2\alpha_3-\sum_r\alpha_r|x_r|^2
       +2\operatorname{Re}((x_1x_2)x_3).
\]
Preservation of its mixed term reduces to the identity
\begin{equation}\label{eq:transportcheck}
 \epsilon_{1m_1}\epsilon_{2m_2}\epsilon_{3m_3}
 \operatorname{Re}\bigl((e_{m_1\,\mathrm{xor}\,\mu_1}
 e_{m_2\,\mathrm{xor}\,\mu_2})e_{m_3\,\mathrm{xor}\,\mu_3}\bigr)
 =\operatorname{Re}((e_{m_1}e_{m_2})e_{m_3}).
\end{equation}
Both sides vanish unless $m_3=m_1\mathbin{\mathrm{xor}}m_2$, because
$\mu_1\mathbin{\mathrm{xor}}\mu_2\mathbin{\mathrm{xor}}\mu_3=0$.
The remaining sixty-four identities per row follow by substituting the signs
and \eqref{eq:octonionrule}. So the $\iota_i$ preserve unit, norm and trace
form. The adjoint (or sharp map) $y\mapsto y^\#$ is the gradient of the
cubic norm for the trace form,
\[
 \langle y^\#,y'\rangle=\frac{d}{dt}N(y+ty')\Big|_{t=0},
\]
so the $\iota_i$ commute with it. The sharp expression for the square in
Section~II.4.2.2 of McCrimmon\cite{McCrimmon04},
\[
 y^2=y^\#+\operatorname{tr}(y)\,y
  -\tfrac12\bigl((\operatorname{tr}y)^2-\langle y,y\rangle\bigr)e,
\]
then proves square preservation, and polarization proves that the $\iota_i$
are automorphisms of $J$.

The automorphism $\theta_0$ of Section~\ref{sec:families} fixes the
diagonal entries. In every off-diagonal octonion entry it rotates the four
real planes
\[
 (1,i),\qquad(j,k),\qquad(\ell,i\ell),\qquad(n,k\ell)
\]
through the multiples $0$, $2$, $2$, $1$ of $2\pi/3$, each from its first
basis vector towards its second. This multiplies $c$ by $1$ and
each $z_r$ by $\omega^2$, hence $B$ by $\omega$. For $i=1,2,3$,
\[
 \theta_i=\iota_i\theta_0\iota_i^{-1}.
\]
All these maps fix the diagonal entries and act on each off-diagonal entry by a real
$8\times8$ matrix, so the relations among the $\theta_i$ stated in Lemma~\ref{lem:K} are a
finite check: products of these rotation matrices and signed permutations, for each of the
three entries.

\subsection{The four subalgebras: proof of Lemma~\ref{lem:subalgebras}\textup{(a)} and~\textup{(c)}}\label{app:subalgebras}
By the multiplication rule \eqref{eq:octonionrule}, the product of two
different lines among $\mathbb Cj$, $\mathbb C\ell$, $\mathbb Cn$ lies in
the third, in either order:
\begin{equation}\label{eq:lines}
 (\mathbb Cj)(\mathbb C\ell)\subset\mathbb Cn,\qquad
 (\mathbb C\ell)(\mathbb Cn)\subset\mathbb Cj,\qquad
 (\mathbb Cn)(\mathbb Cj)\subset\mathbb C\ell.
\end{equation}
This product is conjugate-linear in each factor: for $c,c'\in\mathbb C$ and
two different units $x,y$ among $j,\ell,n$,
\[
 (cx)(c'y)=\bar c\,\bar c'\,(xy).
\]

\begin{proof}[Proof of Lemma~\ref{lem:subalgebras}\textup{(a)}]
Each column of \eqref{eq:latin} contains each of the four lines once,
so the $N_i$ are mutually orthogonal and with $D$ span $J$
($27=3+4\cdot6$). Let $y\in J$ have diagonal entries $\alpha_r$ and
entries $y_{12},y_{13},y_{23}$ above the diagonal. The entries of $y^2$
above the diagonal are
\begin{equation*}
 (\alpha_1+\alpha_2)y_{12}+y_{13}\overline{y_{23}},\qquad
 (\alpha_1+\alpha_3)y_{13}+y_{12}y_{23},\qquad
 (\alpha_2+\alpha_3)y_{23}+\overline{y_{12}}\,y_{13}.
\end{equation*}
By \eqref{eq:lines}, and in row~$0$ because $\mathbb C$ is closed under
multiplication, each product lies in the line prescribed for its
position, so $A_i$ is closed under
squaring, hence under the Jordan product. The automorphism $\iota_i$ maps $A_0$ onto $A_i$.
\end{proof}

\begin{proof}[Proof of Lemma~\ref{lem:subalgebras}\textup{(c)}]
The point $p(1,u,w)$ has diagonal entries $1/3$ and entries
$\bar u/3$, $\bar w/3$, $u\bar w/3$ above the diagonal. For the four rows of
\eqref{eq:literal} after the first, these lie in the lines of rows
$0,1,2,3$ of \eqref{eq:latin}, by \eqref{eq:lines}. In the row of lobe~$2$ the parameters multiply $\bar u$ by $\xi_1$,
$\bar w$ by $\xi_2$ and, by conjugate-linearity, $u\bar w$ by
$\overline{\xi_1\xi_2}$; likewise for lobe~$3$.
The same rule, applied to the three displayed entries of $y^2$, shows that
the map in \textup{(c)} commutes with squaring on $A_2$ (the diagonal
entries of $y^2$ involve only $\alpha_r$ and absolute values), so it is an
automorphism of $A_2$; likewise for $A_3$.
\end{proof}

\subsection{The code in the coordinates of each quartet}
Let $q_{f,t}\in H_3(\mathbb C)$,
$0\le f<4$, be the points of the first four frames of \eqref{eq:literal},
which do not depend on $\xi$. In the coordinates of quartet $i$, the points
of quartet $i$ are $G_iq_{f,t}G_i^*$, where
\[
 G_0=I,\quad G_1=\operatorname{diag}(1,-1,-1),\quad
 G_2=\operatorname{diag}(1,\xi_1,\xi_2^{-1}),\quad
 G_3=\operatorname{diag}(1,\xi_3^{-1},-\xi_4).
\]
The frames are relabeled by the permutations $\varphi_i$, listed by their
values at $0,\ldots,12$:
\begin{align*}
 \varphi_0&=(0,1,2,3,4,5,6,7,8,9,10,11,12),\\
 \varphi_1&=(0,4,5,6,1,2,3,7,8,9,10,11,12),\\
 \varphi_2&=(0,7,9,8,1,10,4,3,12,5,2,11,6),\\
 \varphi_3&=(0,10,11,12,1,2,3,4,5,6,7,8,9).
\end{align*}
The point index $t$ within frame $f$ is changed to $\psi_i(f,t)$, with
arithmetic modulo three:
\begin{align*}
 \psi_0(f,t)&=\begin{cases}t&f<4,\\-t&f\ge4;\end{cases}\\
 \psi_1(f,t)&=\begin{cases}t&f=0,\\-t&1\le f\le6,\\
 t-(f-7)&7\le f\le9,\\(f-10)-t&10\le f\le12;\end{cases}\\
 \psi_2(f,t)&=\begin{cases}-t&f\in\{5,8,11\},\\t&\text{otherwise};\end{cases}\\
 \psi_3(f,t)&=\begin{cases}t&f\in\{0,7,8,9\},\\-t&\text{otherwise}.\end{cases}
\end{align*}

The nine frames outside quartet $i$ are labeled by $f=4+r$,
$r=3a+b$, $a,b\in\{0,1,2\}$. With $d_i=d_i(\xi)$ as in \eqref{eq:phase-triples},
their matrices are
\[
 U^{(i)}_{ab}=\zeta_i(a,b)\,L_i\,E_{s_i(a,b)}(d_i)\,R_i.
\]
These are the matrices of Lemma~\ref{lem:anchored}, with the fixed unitary
matrices
\begin{align*}
 L_0&=\begin{pmatrix}0&0&\xi_3\\0&-\xi_4^{-1}&0\\1&0&0\end{pmatrix},&
 R_0&=\operatorname{diag}(1,\xi_4,\xi_3^{-1}),\\
 L_1&=\begin{pmatrix}0&0&-\xi_1^{-1}\\0&\xi_2&0\\-1&0&0\end{pmatrix},&
 R_1&=\operatorname{diag}(-1,-\xi_2^{-1},-\xi_1),\\
 L_2&=\begin{pmatrix}-1&0&0\\0&0&-(\xi_3\xi_4)^{-1}\\0&\xi_3^{-1}&0\end{pmatrix},&
 R_2&=\operatorname{diag}(-1,-\xi_3,-\xi_3\xi_4),\\
 L_3&=\begin{pmatrix}0&-1&0\\-1&0&0\\0&0&-1\end{pmatrix},&R_3&=I,
\end{align*}
and the central phases $\zeta_i$ and label bijections
$s_i$
\begin{align*}
 \zeta_0(a,b)&=\omega^{ab},&s_0(a,b)&=(-b,-a),\\
 \zeta_1(a,b)&=1,&s_1(a,b)&=\begin{cases}(-b,2)&a=2,\\(b,a)&a\ne2;\end{cases}\\
 \zeta_2(a,b)&=\omega^{ab-a},&s_2(a,b)&=(a,b),\\
 \zeta_3(a,b)&=\begin{cases}\omega^b&a=0,\\\omega^{-b}&a=1,\\1&a=2,\end{cases}&
 s_3(a,b)&=\begin{cases}(-b,a)&a=2,\\(b,a)&a\ne2.\end{cases}
\end{align*}
The two indices in $s_i(a,b)$ are respectively the clock and shift
indices of \eqref{eq:ueb}, not the frame label $(a,b)$.

Define $\widehat P_i(f,t)$ by the quartet points
$(G_iq_{f,t}G_i^*,0)$ for $f<4$, and by
$p(\omega^tU^{(i)}_{ab})$ for $f=4+3a+b$. The code $P(\xi)$ is recovered as
\begin{equation}\label{eq:anchorbinding}
 P(\xi)_{\varphi_i(f),\psi_i(f,t)}=\iota_i\,\widehat P_i(f,t).
\end{equation}
It is checked by substituting \eqref{eq:literal} into the coordinate map,
using \eqref{eq:octonionrule}, and it fixes the orientations and labels in \eqref{eq:anchored-family}.

\subsection{The derivative of the family of a quartet}\label{app:velocity}
This subsection proves the first assertion of
Lemma~\ref{lem:families}\textup{(a)}. The velocity of
$U_r=\zeta_rL_iE_{s_i(r)}(d)R_i$ is
\[
 \dot gU_r-U_r\dot h+\zeta_rL_i\dot E_{s_i(r)}R_i,
\]
with $\dot E=0$ for $E=I,X,Z$. If all velocities vanish, then $L_i^*\dot gL_i=R_i\dot hR_i^*$
commutes with $X$ and $Z$, so $\dot g=\dot h=0$, and then $\dot d=0$.

\subsection{The four-parameter family in the family of a quartet}\label{app:inclusion}
This subsection supplies the construction used in the proof of Lemma~\ref{lem:families}(b).
Let $\xi'$ be near $\xi$, let
$G_i$, $L_i$, $R_i$ be the matrices for quartet
$i$, regarded as functions of the parameter, and put
\[
 v=G_i(\xi')G_i(\xi)^*.
\]
Choose the scalar $\kappa$, analytic in $\xi'$ and equal to one at $\xi'=\xi$, with
\[
 \kappa^3=\det\bigl(v^*L_i(\xi')L_i(\xi)^*\bigr),
\]
and set
\[
 g=\kappa^{-1}v^*L_i(\xi')L_i(\xi)^*,\qquad
 h^*=\kappa R_i(\xi)^*R_i(\xi')v.
\]
Both are special unitary, since $\det L_i\det R_i=1$, and diagonal, since $v$ and $R_i$ are
diagonal and $L_i$ is a constant permutation matrix times a diagonal one. By construction
\[
 \zeta_r\,gL_i(\xi)E_{s_i(r)}(d_i(\xi'))R_i(\xi)h^*
 =v^*\bigl(\zeta_r\,L_i(\xi')E_{s_i(r)}(d_i(\xi'))R_i(\xi')\bigr)v.
\]
The left side is the member of \eqref{eq:anchored-family} with parameters $(g,h,d_i(\xi'))$. The
map $(a,B)\mapsto(vav^*,vBv^*)$ in the coordinates of quartet $i$ preserves \eqref{eq:titssquare},
so is an automorphism of $J$; it sends the quartet at $\xi$ to the quartet at $\xi'$ and, by the
display, that member to $P(\xi')$. Here $g$, $h$ are analytic in $\xi'$ and equal to $I$ at
$\xi'=\xi$.

\subsection{Proof of Lemma~\ref{lem:slice}}\label{app:slice-proof}
\begin{proof}
The compact group $F_4$ acts linearly on $J^{39}$ and preserves the trace form, and the
stabilizer of $P$ is trivial by Lemma~\ref{lem:orbit}. The derivative of the map at $(1,P)$ is
$(\delta,\nu)\mapsto\delta P+\nu$, an isomorphism from
$\operatorname{Der}(J)\times\mathcal O^\perp$ onto $\mathcal O\oplus\mathcal O^\perp$. So it is
invertible at $(1,s)$ for $s$ in some $S_{\varepsilon_0}$, by equivariance at every $(g,s)$, and
the map is a local analytic diffeomorphism on $F_4\times S_{\varepsilon_0}$.

It is injective on $F_4\times S_\varepsilon$ for small $\varepsilon$. Otherwise there are
$(g_n,s_n)\ne(g_n',s_n')$ with $g_ns_n=g_n's_n'$ and $s_n,s_n'\to P$. Then
$h_n=g_n'^{-1}g_n\ne1$ satisfies $h_ns_n=s_n'$, and by compactness a subsequence converges to
some $h$ with $hP=P$, so $h=1$; the distinct points $(h_n,s_n)$ and $(1,s_n')$ near $(1,P)$ then
have the same image, contradicting local injectivity.

The image $U_\varepsilon$ is open and invariant, and (a), (b) follow because the map carries the
projection $F_4\times S_\varepsilon\to S_\varepsilon$ to the quotient map of $U_\varepsilon$.
\end{proof}