\section{Outline of the proof}\label{sec:outline}
Fix a nonresonant parameter $\xi$ and write $P=P(\xi)$.

\subsection{Step 1: one family for each quartet}
\emph{Each quartet~$i$ gives a family $M_i$ of tight codes through $P$ with
ten parameters modulo $F_4$, containing the four-parameter family.}

Let $\theta_i$ be the automorphism of order
three that fixes $A_i$ and multiplies its orthogonal complement
$M_3(\mathbb C)$ by a primitive cube root of unity. Points having
overlap $1/3$ with every point of quartet~$i$ are given by matrices in
$SU(3)$, their frames are the orbits of $\theta_i$, and two frames are
mutually unbiased exactly when their matrices $U$, $V$ are orthogonal,
$\operatorname{tr}(U^*V)=0$ (Lemma~\ref{lem:dictionary}). So the other nine
frames are nine pairwise orthogonal matrices in $SU(3)$. The
shift-and-multiply construction\cite{VW00} supplies such matrices with
$8+8+2$ parameters; eight of them are automorphisms of $J$
(Lemma~\ref{lem:stabilizer}), which leaves ten. Up to an automorphism, each $P(\xi')$ with $\xi'$ near $\xi$ is a member of
this family (Lemma~\ref{lem:families}), so the family contains the four-parameter family.

\subsection{Step 2: first order}
\emph{Modulo $F_4$, the first-order deformations of $P$ form a space
$\overline{\mathcal T}$ of dimension $28$, the sum of the tangent spaces of
the four families.}

With quartet~$i$ held fixed, the linearized problem is solved by a Fourier
transform over $\mathbb Z/3$ (Lemma~\ref{lem:fourier}). Every solution is
tangent to the family of quartet~$i$ unless an entry of the triple $d_i(\xi)$
of \eqref{eq:phase-triples} has cube one, which makes $\xi$ resonant.

The $\theta_i$ preserve the code and generate a group
$K\simeq(\mathbb Z/3)^2$ (Lemma~\ref{lem:K}). The deformations fixed by
$\theta_i$ are those just found, and counting characters of $K$ gives
\[
 \overline{\mathcal T}=\overline{\mathcal T}^K\oplus W_0\oplus W_1\oplus
 W_2\oplus W_3,\qquad 28=4+4\cdot6
\]
(Proposition~\ref{prop:kernel}, Table~\ref{tab:dimensions}). The first
summand is tangent to the four-parameter family. Each $W_i$ is a complex three-space; with the first summand it makes up
the tangent space of $M_i$.

A slice transverse to the $F_4$-orbit of $P$ has analytic coordinates
\[
 (\lambda,u,\eta)\in\mathbb R^4\times(\mathbb C^3)^4\times\mathbb R^{973},
 \qquad u=(u_0,u_1,u_2,u_3),
\]
in which $M_i$ is the subspace where only $\lambda$ and the triple $u_i$ may
be nonzero (Proposition~\ref{prop:simultaneous}).

\subsection{Step 3: second order}
\emph{Once $\eta$ is removed, the quadratic part of the equations vanishes on a first-order
deformation only if at most one $u_i$ is nonzero.}

The equations
\[
 xy=yz=zx=0
\]
have zero derivative at the origin, yet their solution set is the three
coordinate axes (Figure~\ref{fig:axes}), the analogues of the four
families.

\begin{conceptfigure}
\includegraphics{figures/fig3-axes.pdf}
\captionof{figure}{The model $xy=yz=zx=0$; the dashed direction is tangent to
no solution curve.}\label{fig:axes}
\end{conceptfigure}

The implicit function theorem removes $\eta$. The remaining equations vanish
on the four families and have no linear term. A
\emph{stress} at a solution is a linear functional on the residuals of the
frame equations that annihilates their linearization there. Composed with
the frame equations it gives a function, its \emph{energy}, which is a
quadratic form in the displacement because the frame equations have degree
two
(Definition~\ref{def:stress}).

Proposition~\ref{prop:finite} gives twenty-four complex stresses at
$P$. On
transverse directions their energies are the components of quadratic maps,
two for each quartet (Appendix~\ref{app:quadrics}). Every monomial of these
maps multiplies entries of two different triples $u_j$. The quartic identity \eqref{eq:quartic},
verified by expansion, writes the sum of the six products
$\|u_i\|^2\|u_j\|^2$ as a sum of terms each containing such a component, as
\[
 x^2y^2+y^2z^2+z^2x^2=(xy)^2+(yz)^2+(zx)^2
\]
does for the model. So these energies all vanish only if at most one triple
is nonzero, and the tangent cone of the solution set is the model
\eqref{eq:axes}. Part of \eqref{eq:quartic} is the multiplicativity of the octonion norm:
up to signs and conjugations, and after adding products of entries of one triple, the first
maps of the four quartets together give the off-diagonal entries of the square of an element of
$J$ with zero diagonal, as computer algebra verifies (Appendix~\ref{app:quadrics}).

\subsection{Step 4: all orders}
\emph{Near $P$ in the slice the solutions are exactly the points of the four
families \textup{(}Proposition~\ref{prop:exhaustion}\textup{)}; this proves
Theorem~\ref{thm:real}.}

A tangent cone does not determine a solution set: $y(y-x^2)=0$ has the
branches $y=0$ and $y=x^2$, but its lowest-order term $y^2$ vanishes only on
their common tangent (Figure~\ref{fig:exhaustion}). Solutions could likewise
leave a family at higher order.

\begin{conceptfigure}
\includegraphics{figures/fig4-exhaustion.pdf}
\captionof{figure}{The curve $y(y-x^2)=0$ and the zero set of its
lowest-order term $y^2$.}\label{fig:exhaustion}
\end{conceptfigure}

Lemma~\ref{lem:stress} excludes this. It asks for a stress at every point of
each family, continuous in the point. Some directions leaving the
family are excluded already by the mixed second-order terms; on the others,
the energy of the stress at $P$ must be positive. The proof compares a supposed further solution with the nearest
point of the nearest family and rescales the difference to unit size; the
stress there removes the linear term and leaves a positive energy, a
contradiction.

Such stresses exist because, for each quartet, three of the twenty-four have
energies involving only its twelve points, fixed along its family; the
positivity is
\eqref{eq:kernelstress}, a consequence of \eqref{eq:quartic}. The toy curve
has none: the derivative leaving $y=0$ is $-x^2$, so a continuous multiplier
annihilating it vanishes at the origin.

The same argument bounds the distance to the solution set by the square root
of the residual (Theorem~\ref{thm:bound}).

\subsection{Step 5: the unlabeled space and the original code}
\emph{Every neighborhood of the original code in $\mathcal M$ contains an
open ten-dimensional manifold \textup{(}Theorem~\ref{thm:patches}\textup{)},
whence Corollary~\ref{cor:no24} by invariance of dimension.}

By Theorem~\ref{thm:real}, points of $M_i$ off the four-parameter family have
ten-dimensional manifold neighborhoods in $Y$. Forgetting the labels divides
by $\mathfrak S_{39}$; this is a local homeomorphism where the stabilizer is
locally constant, which gives open patches in $\mathcal M$
(Lemma~\ref{lem:finite-patch}). Nonresonant parameters tend to $\xi=1$.