\section{The weighted cyclotomic seed and zero-sector ties}
\label{sec:seed}

Assume for this section that $k\geq4$ is even.  Choose an injective map
\[
  s:\A\longrightarrow\mathbb R\setminus\{0\},
  \qquad s(\bar a)=-s(a).
\]
Such a map exists by assigning distinct positive magnitudes to the
complement pairs.  For DNA one may use
$s(A)=-2,s(C)=-1,s(G)=1,s(T)=2$.
The existence, descent, and direct-selector arguments use only real
arithmetic, signs, and injectivity.  Integrality will be imposed separately
in Corollary~\ref{cor:complexity}, where a finite exact encoding and a bit
complexity bound are required.
Put $\zeta=e^{2\pi i/k}$ and, for $w=w_0\cdots w_{k-1}$, define
\[
  P_s(w)=\sum_{j=0}^{k-1}s(w_j)\zeta^{j+1},
  \qquad W_s(w)=\sum_{j=0}^{k-1}s(w_j).
\]
For $u=u_0\cdots u_{k-2}$ define
\[
  \lambda_s(u)=\sum_{j=0}^{k-2}s(u_j)
    \sin\!\left(\frac{2\pi(j+1)}k\right),
  \qquad \chi(u)=\ind[\lambda_s(u)>0].
\]

\begin{lemma}[Weighted covariance]
\label{lem:covariance}
For every length-$k$ word $w$,
\[
  P_s(Rw)=\zeta^{-1}P_s(w),\quad
  P_s(\RC w)=-\overline{P_s(Rw)},\quad
  P_s(R\RC w)=-\overline{P_s(w)}.
\]
Moreover
\[
  \operatorname{Im}P_s(w)=\lambda_s(\tail(w)),\qquad
  \operatorname{Im}P_s(Rw)=\lambda_s(\head(w)),
\]
and $\lambda_s(\RC u)=\lambda_s(u)$ on lower words.
\end{lemma}

\begin{proof}
The rotation identity follows by reindexing and $\zeta^k=1$.  Antisymmetry
of $s$ gives
\[
  P_s(\RC w)
  =-\sum_j s(w_j)\zeta^{-j}
  =-\overline{P_s(Rw)}.
\]
Rotating gives the third identity.  The omitted final letter of
$\tail(w)$ multiplies $\zeta^k=1$ and contributes no imaginary part; apply
the same observation to $Rw$ for the head.  Finally, reindexing the lower
sum, using antisymmetry and
$\sin(2\pi(k-r)/k)=-\sin(2\pi r/k)$, proves lower-word invariance.
\end{proof}

\subsection{The selector}

On each PCR with nonzero embedding select one word by the rule
\begin{enumerate}
  \item select the point on the negative real axis, if it occurs;
  \item otherwise select the unique $w$ satisfying
  $\operatorname{Im}P_s(w)<0<\operatorname{Im}P_s(Rw)$.
\end{enumerate}
On a PCR where $P_s$ vanishes identically, select an arbitrary representative
for now.  Call the resulting set $M_0$ and write $m_0=\ind_{M_0}$.

\begin{lemma}[Full period and uniqueness]
\label{lem:selector-unique}
Every nonzero embedded PCR has full period $k$, and the selector above is
unique.
\end{lemma}

\begin{proof}
If a word in a nonzero PCR had period $d<k$, then
$P_s(R^dw)=\zeta^{-d}P_s(w)=P_s(w)$ would force $P_s(w)=0$.  The nonzero
rotations therefore form a nondegenerate regular $k$-gon.  Since $k$ is
even, a negative-real vertex is unique when present; if no real vertex is
present, clockwise rotation gives exactly one strict lower-to-upper crossing.
\end{proof}

\begin{theorem}[Arbitrary-zero-tie acyclicity]
\label{thm:arbitrary-zero}
Every choice of one representative on each zero-embedding PCR produces an
ordinary MDS $M_0$.
\end{theorem}

\begin{proof}
Set $Q_s=P_s-W_s$.  If
$y=w_1\cdots w_{k-1}a$ follows $w=w_0\cdots w_{k-1}$ and
$\delta=s(a)-s(w_0)$, then
\[
  P_s(y)=\zeta^{-1}P_s(w)+\delta,\qquad
  W_s(y)=W_s(w)+\delta,\qquad
  Q_s(y)=-W_s(w)+\zeta^{-1}P_s(w).
\]
A residual path cannot pass from $\operatorname{Im}P_s\leq0$ to
$\operatorname{Im}P_s>0$: the strict crossing and departure from the
negative ray are selected.  Hence a residual cycle lies wholly in the open
upper or closed lower half-plane.

Write $\theta=2\pi/k$ and $P_s(w)=\rho e^{i\phi}\neq0$.  Along an edge whose
endpoints remain in the open upper region,
\begin{equation}
  \operatorname{Re}Q_s(y)-\operatorname{Re}Q_s(w)
  =2\rho\sin(\theta/2)\sin(\phi-\theta/2)>0.
  \label{eq:q-monotone}
\end{equation}
Along a residual edge in the closed lower region the same expression is
strictly negative: the negative-ray starting point was selected and the
remaining angular interval is $\pi+\theta\leq\phi\leq2\pi$.  Thus a
residual cycle could only consist of edges beginning at $P_s=0$.

If consecutive labels $w$ and $y$ both have zero embedding, then
$P_s(y)=s(a)-s(w_0)=0$.  Injectivity of $s$ gives $a=w_0$, so the edge is a
pure rotation.  Such a cycle is exactly a zero-embedding PCR, from which one
word was selected.  This final contradiction proves acyclicity.  Since
$M_0$ meets every PCR exactly once, it has cardinality $N_q(k)$.
\end{proof}

\subsection{Equivariant zero ties}

Reverse complement permutes the zero PCRs.  On a paired orbit
$\{C,\RC(C)\}$, choose $w\in C$ and $\RC(w)$ in its mate.  On a stable PCR,
choose one of its two reverse-complement-fixed rotations; the
alphabet-general fixed-rotation lemma is proved in
Appendix~\ref{app:technical}.  From now on $M_0$ denotes a seed with these
equivariant zero ties.  The zero-sector part of $m_0$ is then literally
reverse-complement invariant.