\section{The Fourier recurrence and the deformations of a quartet}\label{app:fourier}
The notation is that of Lemma~\ref{lem:fourier} and \eqref{eq:linear-pairs}. All indices lie in $\mathbb Z/3$.

Write $D_b=\operatorname{diag}(d_{b,k})$, so that $d_{0,k}=d_{1,k}=1$ and
$d_{2,k}=d_k$. The matrix $W_{b,b+1}$ of \eqref{eq:linear-pairs} is then
\[
 W_{b,b+1}=\operatorname{diag}(w_{b,k})X,\qquad
 w_{b,k}=\bar d_{b,k}\,d_{b+1,k-1}.
\]
Only the part $Y_{ab}X^{-1}$ of each $M_{ab}$ contributes to the equations
of the oriented pairs $(b,b+1)$, which read
\begin{equation}\label{eq:scalar-pairs}
 \sum_k\omega^{(a'-a)k}
 \left(w_{b,k}\,y_{a',b+1,k-1}-w_{b,k+1}\,y_{a,b,k}\right)=0.
\end{equation}

The transform $\widehat y_{b,k}(\mu)$ of the lemma is inverted by
\[
 y_{a,b,k}=\sum_\mu\widehat y_{b,k}(\mu)\,\omega^{\mu a},
\]
Substitute this in \eqref{eq:scalar-pairs}. The
coefficient of $\omega^{-\beta a+(\beta+\mu)a'}$ is
\[
 w_{b,\beta}\,\widehat y_{b+1,\beta-1}(\mu)
 -w_{b,\beta+\mu+1}\,\widehat y_{b,\beta+\mu}(\mu).
\]
The transform in $a,a'$ is invertible, so the pair equations say that every
such coefficient vanishes. With
$k=\beta-1$ this is the recurrence
\[
 \widehat y_{b+1,k}(\mu)=\nu_{b,k}\,\widehat y_{b,k+\mu+1}(\mu),
 \qquad
 \nu_{b,k}=\frac{w_{b,k+\mu+2}}{w_{b,k+1}}.
\]
The numbers $\nu_{b,k}$ are nonzero; they depend on $\mu$.

Two steps of the recurrence express everything through the values at
$b=0$:
\[
 \widehat y_{1,k}(\mu)=\nu_{0,k}\,\widehat y_{0,k+\mu+1}(\mu),\qquad
 \widehat y_{2,k}(\mu)=\nu_{1,k}\,\nu_{0,k+\mu+1}\,\widehat y_{0,k+2\mu+2}(\mu).
\]
The third step returns to $b=0$ and leaves the closure equation
\[
 \widehat y_{0,k}(\mu)=m_{k\mu}\,\widehat y_{0,k}(\mu),
\]
with the multiplier
\[
 m_{k\mu}=\nu_{2,k}\,\nu_{1,k+\mu+1}\,\nu_{0,k+2\mu+2}
 =\frac{d_{k+1}\,d_{k+2\mu+2}}{d_{k+\mu+2}\,d_{k+\mu+1}}.
\]
Using $d_0d_1d_2=1$, the multipliers for $\mu=0,1,2$ are
\[
 1,\qquad d_{k+1}^3,\qquad 1.
\]

Same-support pairs supply nine
unordered equations; the three oriented different-support pairs supply
the other twenty-seven. Reversing a pair conjugates its equation. These are
all thirty-six unordered complex pair constraints.

\subsection{Proof of Lemma~\ref{lem:quartet}}\label{app:quartet-proof}
\begin{proof}
Conjugating by a unitary matrix, we may assume that the first frame is the coordinate basis and
that the second contains the line spanned by $(1,1,1)$. The points of the other three frames are
then the projections onto lines spanned by vectors $(1,x,y)$ with $|x|=|y|=1$.

A first-order deformation of one frame is $q\mapsto[N,q]$ with $N$ anti-Hermitian. Subtracting
one, we may assume that the coordinate basis does not move; the conjugations still available
have $N$ diagonal. Each vector $(1,x,y)$ stays unbiased to the coordinate basis, so only the
phases of $x$ and $y$ vary, at rates
$\alpha$ and $\beta$ say, and a diagonal conjugation shifts $(\alpha,\beta)$ equally for all
vectors; so we may assume $\alpha=\beta=0$ for $(1,1,1)$.

Another vector $(1,x,y)$ of the second frame (a row of the dephased Hadamard matrix)
satisfies $1+x+y=0$, with linearization $x\alpha+y\beta=0$; as $x$
and $y$ are not real multiples of each other, $\alpha=\beta=0$. This step says that the Fourier
matrix of order three has defect zero, in the sense of Sections~3 and~5.3 of Tadej and
\.Zyczkowski\cite{TZ06}. A vector $(1,x,y)$ of the third
or fourth frame is unbiased to the second frame:
\[
 \bar x+x\bar y+y=0.
\]
These three unit numbers form an equilateral triangle, so a real linear relation between them
has equal coefficients; the derivative has coefficients $-\alpha$, $\alpha-\beta$, $\beta$,
equal only when $\alpha=\beta=0$.

So the deformation is a conjugation, and conjugation by $N\in\mathfrak{su}(3)$ vanishes on the
quartet only for $N=0$, because the quartet spans $H_3(\mathbb C)$.
\end{proof}