\section{First-order deformations and local coordinates}\label{sec:linear}
Fix a parameter $\xi$ and write $P=P(\xi)$, with points $p_{f,t}$ labeled as in \eqref{eq:literal}.
The first-order deformations of $P$ form the space
\[
 \mathcal T=\ker DF_P\subset J^{39}.
\]
It contains the tangent space $\mathcal O$ of the orbit $F_4P$, and
$\overline{\mathcal T}=\mathcal T/\mathcal O$ consists of the first-order deformations modulo
$F_4$. Table~\ref{tab:dimensions} collects the dimensions.

\input{sections/tab-dimensions}

\subsection{Solving the linearized matrix equations}
Hold one quartet fixed. By Lemma~\ref{lem:dictionary} the nine frames outside it are nine
special unitary matrices $U_{ab}$ with
\[
 \operatorname{tr}(U_{ab}^*U_{a'b'})=0\qquad\bigl((a,b)\ne(a',b')\bigr).
\]
We linearize these conditions at the matrices $E_{ab}(d)$ of \eqref{eq:ueb}. Every
first-order variation of $E_{ab}(d)$ inside $SU(3)$ has the unique form
\[
 \dot U_{ab}=X^bD_bM_{ab}Z^a,\qquad M_{ab}^*=-M_{ab},\quad
 \operatorname{tr}M_{ab}=0.
\]
Differentiating the trace conditions gives
\begin{equation}\label{eq:linear-pairs}
 \operatorname{tr}\left(Z^{a'-a}
 [W_{bb'}M_{a'b'}-M_{ab}W_{bb'}]\right)=0,
 \qquad W_{bb'}=D_b^*X^{b'-b}D_{b'}.
\end{equation}
Both real and imaginary parts are required. The matrices $W_{bb'}$ occur only here and in Appendix~\ref{app:fourier}; they are not the transverse spaces $W_i$.

\begin{lemma}[First-order solutions of the matrix problem]\label{lem:fourier}
Let $d\in T^2$. \leannote{unrestricted kernel}{Proofs/PairingA3Matrix.lean}{15}The real vector space of all
solutions $(M_{ab})$ of \eqref{eq:linear-pairs}, with no normalization imposed, is isomorphic to
\[
 \mathbb R^6\oplus\mathbb C^6\oplus
 \mathbb C^{\{j:d_j^3=1\}}.
\]
In particular its dimension is $18+2r$, where $r$ is the number of entries
of $d$ with cube one; this number is $0$, $1$ or $3$.
\end{lemma}

\begin{proof}
The matrix $W_{bb}$ is the identity, and $W_{bb'}$ has zero diagonal for $b'\ne b$. So the
equations with $b=b'$ involve only the diagonal entries of the matrices $M_{ab}$, and the
others only the off-diagonal entries.

\emph{Diagonal entries.} The equations with $b=b'$ say that the purely imaginary diagonal of
$M_{ab}$ does not depend on $a$: two real parameters for each $b$.

\emph{Off-diagonal entries.} Write the off-diagonal part of $M_{ab}$ uniquely as
\[
 Y_{ab}X^{-1}-(Y_{ab}X^{-1})^*,\qquad Y_{ab}=\operatorname{diag}(y_{ab,k}),
\]
and take the Fourier transform in $a$,
\[
 \widehat y_{b,k}(\mu)=\tfrac13\sum_a y_{ab,k}\omega^{-\mu a}.
\]
The equations with $b'=b+1$ become a three-step scalar recurrence in $b$. By
Appendix~\ref{app:fourier}, going once round it multiplies the initial value
$\widehat y_{0,k}(\mu)$ by $1$, $d_{k+1}^3$, $1$ for $\mu=0,1,2$. Hence six complex initial values are free, and the remaining one for index $k$ is free exactly
when its multiplier is one.

The inverse Fourier transform and the displayed reconstruction of $M_{ab}$ give both directions of
the isomorphism. Two entries of $d$ with cube one force the third, because the product of the
three entries is one.
\end{proof}

The exact solutions $gE_{ab}(d)h^*$ of Section~\ref{sec:families} have $18=8+8+2$ parameters, with injective derivative (Lemma~\ref{lem:embedded};
Lemma~\ref{lem:families}(a)). So when no entry of $d$ has cube one, every first-order solution
with one quartet held fixed is tangent to this family.

\subsection{The finite symmetry}
The calculations for the four quartets are combined by a group $K\simeq(\mathbb Z/3)^2$ of automorphisms of $J$
that map $P$ to itself, each followed by a relabeling within frames that keeps the
labeled configuration fixed. The relabeling cannot be omitted, because the
thirty-nine points span $J$ (Lemma~\ref{lem:span}), so only the identity fixes them all.

\begin{lemma}[The symmetry $K$]\label{lem:K}
\leannote{action in quartet coordinates}{Proofs/A3ElimAlg.lean}{97}For $i=0,1,2,3$ let $\theta_i$ be the map of Section~\ref{sec:families}, which is
$(a,B)\mapsto(a,\omega B)$ in the coordinates of
quartet $i$. It does not depend on $\xi$.
\begin{enumerate}
\item[\textup{(a)}] Each \leannote{finite symmetry}{Proofs/A3Automorphisms.lean}{520}$\theta_i$ is an
automorphism of $J$ of order three. It fixes the twelve points of quartet $i$ and permutes the
three points of every other frame of $P$ cyclically.
\item[\textup{(b)}] The four maps commute and generate a group $K\simeq(\mathbb Z/3)^2$, with
\[
 \theta_2=\theta_0^2\theta_1^2,\qquad\theta_3=\theta_0^2\theta_1.
\]
Its four subgroups of order three are $K_i=\langle\theta_i\rangle$. The maps $\theta_0$ and
$\theta_1$ fix the points of frame~$0$; on the other frames they send $p_{f,t}$ to $p_{f,t-m}$,
with indices modulo three, where $m$ depends only on the lobe containing frame $f$:
\[
\begin{array}{c|cccc}
 \text{lobe}&0&1&2&3\\\hline
 m\text{ for }\theta_0&0&1&1&1\\
 m\text{ for }\theta_1&1&0&2&1
\end{array}
\]
\item[\textup{(c)}] For $k\in K$ let $\tilde k$ be the map of $J^{39}$ that applies $k$ to each
point and then undoes its permutation of the labels:
\[
 (\tilde ky)_{f,t'}=k\,y_{f,t}\qquad\text{when }kp_{f,t}=p_{f,t'}.
\]
Then $\tilde k$ fixes the labeled configuration $P$, is orthogonal for the
trace form, and satisfies
\[
 \tilde k(gy)=(kgk^{-1})(\tilde ky)\qquad(g\in F_4,\ y\in J^{39}).
\]
It preserves the space $\mathcal T$ of first-order deformations, the orbit tangent space
$\mathcal O$ and its orthogonal complement $\mathcal O^\perp$.
\end{enumerate}
\end{lemma}

\begin{proof}
(a) The map $\theta_i$ is right multiplication of $B$ by the scalar matrix $\omega I$, which
lies in $SU(3)$. By
Lemma~\ref{lem:stabilizer} it is an automorphism of $J$ fixing quartet $i$ pointwise. By
Lemma~\ref{lem:dictionary} every other frame of $P$ is one of its orbits.

(b) By (a), every element of the group generated by the four maps permutes the points of each
frame cyclically. Cyclic permutations of three objects commute, so a commutator fixes all
thirty-nine points and, as these span $J$ (Lemma~\ref{lem:span}), is the identity; likewise an element is determined by
the permutations it induces. Substitution in \eqref{eq:literal} gives the table, and shows that
the two sides of each displayed relation induce the same permutations. The group has order
nine, since $\theta_1$ moves points of quartet~$0$ and
so is not a power of $\theta_0$. Its four subgroups of order three are the $K_i$, distinct
because $\theta_i$ fixes the points of quartet $i$ and no others.

(c) By construction $\tilde k$ fixes $P$. Automorphisms preserve the trace form, and the relabeling
permutes the equations \eqref{eq:frame-equations}, so $F\circ\tilde k$ is $F$ followed by an
invertible linear map; differentiating at $P$ shows that the first-order deformations are
preserved. The displayed identity holds because the relabeling commutes with the action of
$F_4$. Differentiated, it gives $\tilde k(\delta P)=(k\delta k^{-1})P$
for every derivation $\delta$ of $J$. So the orbit tangent space is preserved and, by
orthogonality, so is its orthogonal complement.
\end{proof}

From now on $K$ acts on $J^{39}$ through the maps $\tilde k$, and so on the first-order
deformations, on the orbit tangent space and on the quotient $\overline{\mathcal T}$.

\subsection{Combining the four linearized calculations}

\begin{lemma}[Real representations of $K$]\label{lem:repK}
Let $V$ be a real vector space of finite dimension on which $K$ acts linearly. Write $V^K$ and
$V^{K_i}$ for the vectors fixed by $K$ and by $K_i$, and $V_{[i]}$ for the vectors fixed by
$K_i$ whose average over $K$ is zero.
\begin{enumerate}
\item[\textup{(a)}] \emph{Splitting.} The space $V$ and its fixed spaces split as
\begin{equation}\label{eq:isotypic}
 V=V^K\oplus V_{[0]}\oplus V_{[1]}\oplus V_{[2]}\oplus V_{[3]},\qquad
 V^{K_i}=V^K\oplus V_{[i]},
\end{equation}
and consequently
\begin{equation}\label{eq:averaging}
 \dim V=\sum_{i=0}^3\dim V^{K_i}-3\dim V^K.
\end{equation}
\item[\textup{(b)}] \emph{Quotients.} The averages $\pi_i$ and $\pi_K$ over $K_i$ and over $K$
are the projections onto $V^{K_i}$ and $V^K$, and $V_{[i]}$ is the image of $\pi_i-\pi_K$. For
an invariant subspace $V'$,
\[
 (V/V')^{K_i}=V^{K_i}/V'^{K_i},\qquad (V/V')^K=V^K/V'^K.
\]
\item[\textup{(c)}] \emph{Complex structure.} The group $K$ acts on $V_{[i]}$ through
$K/K_i\simeq\mathbb Z/3$, and a generator $\theta$ of this quotient fixes no nonzero vector of
$V_{[i]}$. So $\theta^2+\theta+1=0$ on $V_{[i]}$, which is thereby a complex vector space with
$\theta$ as multiplication by $\omega$ (the other generator gives the conjugate structure).
\end{enumerate}
\end{lemma}

\begin{proof}
(a) This is the isotypic decomposition: each of the eight nontrivial characters
$K\to\{1,\omega,\omega^2\}$ has one of the four subgroups $K_i$ as kernel, each $K_i$ is the
kernel of a conjugate pair, and the two isotypic components of the pair in $V\otimes\mathbb C$
form the complexification of $V_{[i]}$. Taking dimensions gives \eqref{eq:averaging}.

(b) Averaging over a finite group projects onto its fixed vectors, and it commutes with
equivariant maps, in particular with the quotient map from $V$ to $V/V'$.

(c) A vector of $V_{[i]}$ fixed by $\theta$ is fixed by $K$, so it equals its average over $K$,
which is zero. The image of $\theta^2+\theta+1$ consists of such vectors, as $\theta^3=1$.
\end{proof}

\begin{lemma}[The complex quartet]\label{lem:quartet}
The first-order deformations of a complex quartet inside its subalgebra $H_3(\mathbb C)$ are
exactly the infinitesimal conjugations $q\mapsto[N,q]$ with $N\in\mathfrak{su}(3)$. Each complex
quartet thus contributes exactly eight tangent directions.
\end{lemma}

The proof is in Appendix~\ref{app:quartet-proof}.

\begin{lemma}[First-order deformations fixed by one subgroup]\label{lem:TH}
In the coordinates of quartet $i$, the fixed space $\mathcal T^{K_i}$ is the direct sum of
the first-order deformations of the quartet inside $H_3(\mathbb C)$ and the
first-order solutions of the matrix problem for the nine remaining frames.
Consequently
\[
 \dim\mathcal T^{K_i}=26+2r_i,
\]
where $r_i$ is the number of entries of the triple $d_i(\xi)$ with cube one.
\end{lemma}

\begin{proof}
Let $U_r$, for $r=0,\ldots,8$, be the matrices of Lemma~\ref{lem:anchored} representing the nine
frames outside quartet $i$; the points of frame $r$ are $p(\omega^tU_r)$, permuted cyclically by
$\theta_i$. Let $v$ be a first-order deformation fixed by $\tilde\theta_i$. Its components at
the quartet are fixed by $\theta_i$, so lie in $H_3(\mathbb C)$; its components at frame $r$ are
carried to one another by $\theta_i$, so they are
\[
 (\dot a_r,\ \omega^tV_r/3),\qquad t=0,1,2.
\]
The proof of Lemma~\ref{lem:dictionary} holds at first order, step by step.
\begin{enumerate}
\item[\textup{(1)}] The frame sums give $3\dot a_r=0$.
\item[\textup{(2)}] By \eqref{eq:titssquare}, linearized idempotency is then the linearized
condition for $U_r$ to lie in $SU(3)$:
\[
 U_rV_r^*+V_rU_r^*=0,\qquad\operatorname{tr}(U_r^*V_r)=0.
\]
\item[\textup{(3)}] By \eqref{eq:central-overlap}, the linearized overlap equations between two
different frames $r$ and $s$ say that the real parts of $z$ and $\omega z$ vanish, where
$z=\operatorname{tr}(V_rU_s^*+U_rV_s^*)$. So $z=0$, that is,
\[
 \operatorname{tr}(V_r^*U_s+U_r^*V_s)=0.
\]
The overlap equations inside one frame follow from step~\textup{(2)}.
\item[\textup{(4)}] The overlap of a quartet point with a point outside the quartet varies by
one third of the trace of the variation of the quartet point; this
vanishes by the trace equation, so the two kinds of variation are not coupled.
\end{enumerate}
The only divisions are by $3$ and $\omega-1$, so no solution is lost. The equations of steps \textup{(2)} and
\textup{(3)} are the linearized matrix problem at the matrices $U_r$, the rest are the linearized frame
equations of the quartet inside $H_3(\mathbb C)$, and conversely solutions of the two
reconstruct a first-order deformation fixed by $\tilde\theta_i$.

The first summand has dimension eight by Lemma~\ref{lem:quartet}. For the second, the map
$V\mapsto\zeta_rL_iVR_i$ carries the linearized matrix problem at the matrices
$E_{ab}(d_i(\xi))$ to the one at the matrices $U_r$, so Lemma~\ref{lem:fourier} gives dimension
$18+2r_i$.
\end{proof}

\begin{lemma}[First-order deformations fixed by $K$]\label{lem:TK}
$\dim\mathcal T^K=8$.
\end{lemma}

\begin{proof}
Work in the coordinates of quartet~$0$. By Lemmas~\ref{lem:TH} and~\ref{lem:quartet} a
first-order deformation fixed by $K_0$ consists of an infinitesimal conjugation of the quartet
by some $N\in\mathfrak{su}(3)$ and a first-order solution $(V_r)$ of the matrix problem. It
remains to impose invariance under $\tilde\theta_1$. In these coordinates
\[
 \theta_1(a,B)=(gag^*,gBh^*),\qquad
 g=\operatorname{diag}(1,\omega^2,\omega),\quad
  h=\operatorname{diag}(\omega,\omega^2,1).
\]
This is a finite check. By Appendix~\ref{app:coordinates}, in each off-diagonal entry
$\theta_1=\iota_1\theta_0\iota_1^{-1}$ rotates the planes $(1,i)$, $(j,k)$, $(\ell,i\ell)$,
$(n,k\ell)$ through multiples of $2\pi/3$, and the coordinate map turns these rotations into the
factors $g$ and $h$.

On the quartet, $\tilde\theta_1$ replaces $N$ by $gNg^*$. As $g$ has three distinct entries, the
fixed $N$ are diagonal, which leaves two dimensions.

On the other nine frames, write the solution in the form of Lemma~\ref{lem:fourier}:
\[
 V_r=\zeta_rL_0\,X^bD_bM_{ab}Z^a\,R_0,\qquad(a,b)=s_0(r).
\]
The automorphism $\theta_1$ maps each of the nine frames to itself, so
$gU_rh^*=\omega^mU_r$ for some $m$ depending on $r$. Hence $\theta_1$ sends the point
$p(\omega^tU_r)$ to $p(\omega^{t+m}U_r)$ and the variation $(0,\omega^tV_r/3)$ at it to
$(0,\omega^tgV_rh^*/3)$. The relabeling in $\tilde\theta_1$ cancels the factor $\omega^m$, so
$\tilde\theta_1$ replaces $V_r$ by $\omega^{-m}gV_rh^*$. Since $Z^aR_0$ and $h$ are diagonal,
they commute, and the identity for $U_r$ becomes
\[
 g\,\zeta_rL_0X^bD_b\,h^*=\omega^m\zeta_rL_0X^bD_b.
\]
Therefore
\[
 \omega^{-m}gV_rh^*
 =\omega^{-m}\bigl(g\,\zeta_rL_0X^bD_b\,h^*\bigr)\bigl(hM_{ab}h^*\bigr)Z^aR_0
 =\zeta_rL_0\,X^bD_b\,(hM_{ab}h^*)Z^a\,R_0,
\]
and $\tilde\theta_1$ replaces $M_{ab}$ by $hM_{ab}h^*$. As $h$ has
three distinct entries, the fixed solutions have every $M_{ab}$ diagonal, a space of dimension
six by the proof of Lemma~\ref{lem:fourier}. The dimension is $2+6=8$.
\end{proof}

\begin{lemma}[The orbit tangent space]\label{lem:orbit}
The labeled stabilizer of $P(\xi)$ in $F_4$, that is, the group of automorphisms fixing each of
the thirty-nine points, is trivial, and
\[
 \dim\mathcal O=52,\qquad\dim\mathcal O^{K_i}=16,\qquad\dim\mathcal O^K=4.
\]
\end{lemma}

\begin{proof}
The thirty-nine points span $J$ (Lemma~\ref{lem:span}), so the stabilizer is trivial. Hence $\delta\mapsto\delta P$ is an
isomorphism from the derivations of $J$ onto $\mathcal O$, and $\dim\mathcal O=52$. By
Lemma~\ref{lem:K}(c) it carries conjugation by $k$ to $\tilde k$, so
$\mathcal O^{K_i}$ and
$\mathcal O^K$ correspond to the Lie algebras of the centralizers of $\theta_i$ and of $K$.

In the coordinates of quartet $i$, the maps
\[
 (a,B)\longmapsto(gag^*,gBh^*),\qquad g,h\in SU(3),
\]
preserve \eqref{eq:titssquare}, because $\operatorname{adj}(gBh^*)^*=g\operatorname{adj}(B)^*h^*$.
They are therefore automorphisms of $J$; they commute with $\theta_i$ and form a connected group
of dimension sixteen, the identity component of the centralizer, because $\theta_i$ is conjugate in $F_4$ to $\theta_0$, which up to the conventions for the octonion basis is Yokota's $w$ or its inverse (Section~\ref{sec:families}), and the centralizer of $w$ in $F_4$ is $(SU(3)\times SU(3))/(\mathbb Z/3)$
by Theorem~2.12.2 of Yokota\cite{Yokota09}. Thus $\mathcal O^{K_i}$ has dimension sixteen.

For $\mathcal O^K$ take $i=0$. By the proof of Lemma~\ref{lem:TK}, $\theta_1$ is one of these
maps, with both matrices diagonal with distinct entries. A derivation
$(a,B)\mapsto([N,a],NB-BN')$ in the Lie algebra of this group commutes with $\theta_1$ exactly
when $N$ commutes with the first of them and $N'$ with the second, that is, when both are
diagonal. So $\mathcal O^K$ has dimension $2+2=4$.
\end{proof}

\begin{proposition}[First-order deformations and their splitting]\label{prop:kernel}
At every parameter $\xi$,
\begin{equation}\label{eq:fullkernel}
 \dim\mathcal T=\sum_i\dim\mathcal T^{K_i}-3\dim\mathcal T^K=80+2\sum_ir_i.
\end{equation}
In particular $\xi$ is nonresonant, that is $\dim\mathcal T=80$, exactly when every $r_i$ is zero:
when no entry of the four triples \eqref{eq:phase-triples} has cube one.
At nonresonant parameters, $\overline{\mathcal T}=\mathcal T/\mathcal O$ satisfies
\begin{equation}\label{eq:Qsplit}
 \overline{\mathcal T}=\overline{\mathcal T}^K\oplus\bigoplus_iW_i,\qquad
 \overline{\mathcal T}^{K_i}=\overline{\mathcal T}^K\oplus W_i,\qquad
 \dim\overline{\mathcal T}^K=4,\quad\dim W_i=6.
\end{equation}
Here $W_i=\overline{\mathcal T}_{[i]}$ is the $i$th isotypic part in the notation of
Lemma~\ref{lem:repK}, the image of $\pi_i-\pi_K$. The sum is orthogonal for every $K$-invariant
inner product, and each $W_i$ is a complex vector space of dimension three, on which $K$ acts
through $K/K_i$ by cube roots of unity.
\end{proposition}

\begin{proof}
Apply Lemma~\ref{lem:repK}(a). For $V=\mathcal T$, formula \eqref{eq:averaging} is the first
equality of \eqref{eq:fullkernel},
and Lemmas~\ref{lem:TH} and~\ref{lem:TK} give the second:
$\sum_i(26+2r_i)-3\cdot8=80+2\sum_ir_i$. For $V=\overline{\mathcal T}$, \eqref{eq:isotypic} is
the splitting in \eqref{eq:Qsplit}. At a nonresonant parameter the left side of
\eqref{eq:fullkernel} is $80$, so every
$r_i$ is zero, and by Lemma~\ref{lem:repK}(b) and Lemma~\ref{lem:orbit}
\[
 \dim\overline{\mathcal T}^{K_i}=26-16=10,\qquad\dim\overline{\mathcal T}^K=8-4=4,
\]
so each $W_i$ has dimension six. Distinct isotypic parts are orthogonal for an invariant inner
product, and the complex structure is that of Lemma~\ref{lem:repK}(c).
\end{proof}

\subsection{Coordinates for the four families}
The next lemma is the slice theorem for compact group actions at a point with
trivial stabilizer; Appendix~\ref{app:slice-proof} has a proof.

\begin{lemma}[The slice]\label{lem:slice}
Let $S=P+\mathcal O^\perp$ and, for $\varepsilon>0$, let $S_\varepsilon$ be the set of $s\in S$
with $\|s-P\|<\varepsilon$ in the norm of the trace form. For $\varepsilon$ small enough the map
$(g,s)\mapsto gs$ is an equivariant analytic diffeomorphism from $F_4\times S_\varepsilon$
onto an open invariant neighborhood $U_\varepsilon$ of the orbit of $P$ in $J^{39}$; orbits,
invariance and equivariance are those of $F_4$. In particular:
\begin{enumerate}
\item[\textup{(a)}] two points of $S_\varepsilon$ lie in the same orbit only if they are
equal;
\item[\textup{(b)}] every invariant subset $\Sigma$ of $U_\varepsilon$ equals
$F_4(\Sigma\cap S_\varepsilon)$, and $s\mapsto F_4s$ is a homeomorphism from
$\Sigma\cap S_\varepsilon$ onto $\Sigma/F_4$.
\end{enumerate}
\end{lemma}

A point of $U_\varepsilon$ is $gs$ for a unique point $s$ of the slice $S_\varepsilon$, its
\emph{slice representative}. The slice representative depends analytically
on the point, and the derivative of this dependence at $P$ is the orthogonal projection $\Pi$
of $J^{39}$ onto $\mathcal O^\perp$. This projection identifies $\overline{\mathcal T}$ with
$\mathcal T\cap\mathcal O^\perp$, compatibly with the action of $K$
\textup{(}Lemma~\ref{lem:K}(c)\textup{)}.

\begin{lemma}[The four families in the slice]\label{lem:families}
\begin{enumerate}
\item[\textup{(a)}] The eighteen-parameter map \eqref{eq:anchored-family} has injective derivative
at $(I,I,d_i(\xi))$, and the velocities in the eight parameters $h$ are exactly the velocities of
the family that lie in $\mathcal O$ \textup{(}Lemma~\ref{lem:stabilizer}\textup{)}. Consequently:
\begin{itemize}
\item the slice representatives of the members with $h=I$ and $(g,d)$ near $(I,d_i(\xi))$ form a
ten-dimensional analytic submanifold $M_i$ of $S$ through $P$, the family of quartet $i$;
\item the union of the $F_4$-orbits of the points of $M_i$ is a submanifold of dimension
$62=52+18-8$.
\end{itemize}
\item[\textup{(b)}] The four derivatives $\partial P/\partial\xi_j$ are independent modulo
$\mathcal O$, and the slice representatives of the codes $P(\xi')$, $\xi'$ near $\xi$, form a
four-dimensional submanifold $M_*$. Moreover $M_*\subset M_i$ for every $i$.
\item[\textup{(c)}] Every point of $M_i$ is fixed by the subgroup $K_i$ of $K$. At nonresonant
parameters the tangent space of $M_i$ at $P$ is $\overline{\mathcal T}^{K_i}$, and that of
$M_*$ is $\overline{\mathcal T}^K$.
\end{enumerate}
\end{lemma}

\begin{proof}
(a) The injectivity is the argument with the entries $I$, $X$, $Z$ used for
Lemma~\ref{lem:embedded}; it is written out in Appendix~\ref{app:velocity}. For the velocities
in $h$: the family does not move the quartet, so a velocity of it in $\mathcal O$ is $\delta P$
for a derivation $\delta$ annihilating the quartet, and by Lemma~\ref{lem:stabilizer},
differentiated, these are exactly the velocities in $h$. Hence the map sending $(g,d)$ to the
slice representative of the member with $h=I$ has injective derivative at $(I,d_i(\xi))$, and
the image of a small neighborhood is $M_i$. By Lemma~\ref{lem:slice} the union of the orbits of
its points is the image of $F_4\times M_i$, of dimension $52+10$.

(b) Suppose that a real combination of the four derivatives lies in the orbit tangent space,
so that it is $\delta P$ for a derivation $\delta$. The parameters do not enter the first seven
frames of \eqref{eq:literal}, so $\delta$ annihilates quartet~$0$; by
Lemma~\ref{lem:stabilizer}, differentiated, it is then $(a,B)\mapsto(0,BN)$ for some $N$ in
$\mathfrak{su}(3)$. It also annihilates a point $p(U)$ of frame~$4$, so $UN=0$ and $N=0$. Thus
the combination is zero. Its coefficients then vanish, because the first two parameters occur
in different entries of frame~$7$, and the last two in different entries of frame~$10$. So the
map sending a parameter $\xi'$ to the slice representative of $P(\xi')$ has injective derivative
at $\xi$, and its image near $P$ is $M_*$.

By Lemma~\ref{lem:subalgebras}(c) a change of the parameter moves the last two lobes by
automorphisms of their subalgebras that multiply the off-diagonal entries by unit complex
numbers. For $\xi'$ near $\xi$,
Appendix~\ref{app:inclusion} constructs diagonal matrices $g$ and $h$ in $SU(3)$, analytic in
$\xi'$ and equal to the identity at $\xi'=\xi$, and an automorphism of $J$ that carries the
member with parameters $(g,h,d_i(\xi'))$ to $P(\xi')$. Replacing $h$ by the identity
moves that member by an automorphism \textup{(}Lemma~\ref{lem:stabilizer}\textup{)}. So $P(\xi')$
lies in the orbit of the member with parameters $(g,I,d_i(\xi'))$, and these are near the
parameters of $P$; hence the slice representative of $P(\xi')$ lies in $M_i$.

(c) Every member of \eqref{eq:anchored-family} is fixed by $\tilde\theta_i$: the quartet lies in
the subalgebra fixed by $\theta_i$, and each other frame is one of its orbits, labeled as at
$P$. Let $s$ be the slice representative of such a member $y$. By Lemma~\ref{lem:K}(c),
$\tilde\theta_i$ maps $S_\varepsilon$ to itself and maps $s$ into the orbit of
$\tilde\theta_iy=y$; by Lemma~\ref{lem:slice}(a) it fixes $s$. So $K_i$ fixes every point of $M_i$.

The points of $M_i$ are zeros of $F$, so the tangent space of $M_i$ at $P$ lies in
$\mathcal T\cap\mathcal O^\perp$ and is fixed by $K_i$; it is contained in
$\overline{\mathcal T}^{K_i}$, and both have dimension ten at a nonresonant parameter. By (b) the
points of $M_*$ are fixed by two of the $K_i$, hence by $K$, and the same argument with dimension
four applies.
\end{proof}

\begin{proposition}[Straightening the four families]\label{prop:simultaneous}
\leannote{simultaneous chart}{Proofs/A5Main.lean}{131}At every nonresonant parameter there are analytic coordinates
\[
 (\lambda,u_0,u_1,u_2,u_3,\eta)\in\mathbb R^4\times(\mathbb C^3)^4\times\mathbb R^{973}
\]
on a neighborhood of $P$ in the slice, each $\mathbb C^3$ being regarded as a real vector
space of dimension six, in which each family $M_i$ is the coordinate subspace $\eta=0$, $u_j=0$
for $j\ne i$. The four-parameter family $M_*$ is $u_0=\cdots=u_3=0$, $\eta=0$; it is the
intersection of any two of the $M_i$.
\end{proposition}

\begin{proof}
Choose analytic parametrizations $\varphi_*(\lambda)$ of $M_*$ and $\varphi_i(\lambda,u_i)$ of
$M_i$, with $\lambda\in\mathbb R^4$ and $u_i\in\mathbb C^3$, such that
$\varphi_i(\lambda,0)=\varphi_*(\lambda)$ (take a linear complement to the tangent space of
$M_*$ in that of $M_i$ and apply the inverse function theorem). In the affine space $S$ define
\[
 \Phi(\lambda,u_0,u_1,u_2,u_3)=\sum_i\varphi_i(\lambda,u_i)-3\varphi_*(\lambda).
\]
Its derivative at the origin is injective by \eqref{eq:Qsplit}. Add a linear coordinate
$\eta\in\mathbb R^{973}$ on a complement of the image of this derivative in $\mathcal O^\perp$.
The resulting map is an analytic chart of the $1001$-dimensional slice near $P$. On the
coordinate subspace named in the proposition it equals $\varphi_i$, so $M_i$ is that
subspace.
Because the chart is injective, any two of the four families meet exactly in $M_*$.
\end{proof}