\section{Autocorrelation phase lattices}\label{sec:phase}

Write $C_n=\Z/n\Z$ additively and fix $\zeta_n=e^{2\pi i/n}$.  For
$f:C_n\to\C$, use the Fourier convention
\[
 \widehat f(s)=\sum_{x\in C_n}f(x)\zeta_n^{-sx}.
\]
The order-$k$ autocorrelation is
\[
 M_k(f;t_1,\ldots,t_{k-1})
 =\sum_{x\in C_n}f(x)f(x+t_1)\cdots f(x+t_{k-1}).             \tag{3.1}
\]

Our convention counts the number of signal factors: $M_1(f)=\sum_x f(x)$,
$M_2$ is the usual periodic autocorrelation, and the bispectrum is the Fourier
transform of $M_3$.  This normalization matters when comparing order bounds
across the literature.

Taking the Fourier transform of (3.1) in all shift variables gives, up to the
fixed normalization power of $n$,
\[
 \widehat M_k(f;s_1,\ldots,s_{k-1})
 =\widehat f(-s_1-\cdots-s_{k-1})
   \prod_{j=1}^{k-1}\widehat f(s_j).                         \tag{3.2}
\]
Thus the relevant monomials are indexed by ordered zero-sum $k$-tuples.
Passing from an ordered tuple to its multiplicity vector loses no equation,
because the product is commutative.  This elementary identity is the reason
the monoid $\calB(S)$, rather than merely the set of distinct subset sums,
governs the problem.

We use two results of Casper--Orozco \cite{CasperOrozco2026}.  First, equality
of (3.1) is equivalent to equality of Fourier products over zero-sum
$k$-tuples.  Second, $f$ is rational-valued precisely when its Fourier
transform is Galois equivariant:
\[
 \widehat f(as)=\sigma_a(\widehat f(s))
 \qquad(a\in(\Z/n\Z)^\times).                               \tag{3.3}
\]
The earlier bispectral literature often numbers the order by one fewer Fourier
factor; in particular the convention in \cite{KakaralaIverson1993} is shifted
relative to (3.1).

For clarity, the rationality assertion can also be seen directly.  If
$f(x)\in\Q$, applying $\sigma_a$ termwise to the Fourier sum replaces
$\zeta_n^{-sx}$ by $\zeta_n^{-asx}$ and gives (3.3).  Conversely, if a
Fourier vector is equivariant and is zero on a Galois-stable complement, the
inverse transform
\[
 f(x)=\frac1n\sum_{s\in C_n}\widehat f(s)\zeta_n^{sx}        \tag{3.4}
\]
is fixed by every $\sigma_a$, hence lies in $\Q$.  We will use this converse
in Sections~\ref{sec:signals} and~\ref{sec:galois}.

\begin{lemma}[Exact support from order two]\label{lem:support}
If two real-valued signals have equal second autocorrelations, then their
Fourier transforms have the same zero set.
\end{lemma}

\begin{proof}
Reality gives $\widehat f(-s)=\overline{\widehat f(s)}$.  The zero-sum pair
$(s,-s)$ therefore contributes $|\widehat f(s)|^2$.  Equality of all such
products determines exactly which coefficients vanish.
\end{proof}

\begin{proposition}[Phase-ratio bridge]\label{prop:ratio}
Let $f,g:C_n\to\Q$ have common exact Fourier support $S$, and put
$r_s=\widehat g(s)/\widehat f(s)$.  The autocorrelations agree through order
$d$ if and only if
\[
 r^a=\prod_{s\in S}r_s^{a_s}=1
 \qquad\text{for every }a\in\calL_{\le d}(S).                \tag{3.5}
\]
Translation ratios are exactly the characters of $\Z^S$ that kill $K(S)$.
Thus $\calL_{\le d}(S)=K(S)$ if and only if every complex ratio invisible
through order $d$ is a translation ratio.
\end{proposition}

\begin{proof}
By the Fourier monomial criterion, equality through order $d$ means equality
of the products over all zero-sum tuples of length at most $d$.  A tuple
meeting $C_n\setminus S$ makes both products zero.  On a tuple contained in
$S$, every coefficient is nonzero, so division is legitimate and gives (3.5)
for its multiplicity vector.  The map $a\mapsto r^a$ is a homomorphism
$\Z^S\to\C^\times$, hence it kills the bounded nonnegative rows exactly when
it kills their integral span.

If $g(x)=f(x+y)$, the Fourier shift rule gives
$r_s=\zeta_n^{ys}$.  The quotient map $\Z^S\to\angles S$ has kernel $K(S)$,
so the characters killing $K(S)$ are the characters of $\angles S$.  The
translation ratios have the same cardinality $|\angles S|$ and therefore
exhaust this character group.
\end{proof}

\begin{proposition}[Direct Galois realization]\label{prop:direct-realization}
Let $S\subset C_n$ be Galois-stable and let
$r:S\to\Q(\zeta_n)^\times$ satisfy
\[
 r_{as}=\sigma_a(r_s)\quad(a\in\units n),\qquad
 r^c=1\quad(c\in\calL_{\le d}(S)).                         \tag{3.7}
\]
If $r$ is not a translation ratio, then there are rational-valued signals,
and also integer-valued signals, with exact Fourier support $S$, equal
autocorrelations through order $d$, and not related by translation.
\end{proposition}

\begin{proof}
Set $\widehat f=1$ and $\widehat g=r$ on $S$, and set both transforms to zero
off $S$.  Stability and the first identity in (3.7) make the two Fourier
vectors Galois equivariant, so (3.3)--(3.4) give rational-valued signals.
Every displayed coefficient on $S$ is nonzero, hence the Fourier support is
exactly $S$.  The second identity and Proposition~\ref{prop:ratio} give equal
autocorrelations through order $d$, while the hypothesis on $r$ excludes a
translation.

Choose one positive common denominator $D$ for all values of $f$ and $g$.
Then $Df,Dg$ are integer-valued.  Their order-$k$ autocorrelations are both
multiplied by $D^k$, their exact Fourier supports are unchanged, and a
translation between them would divide by $D$ to a translation between
$f,g$.  Thus the same ratio supplies both arithmetic lower bounds.
\end{proof}

There is a useful exact-sequence version of the same argument.  Applying
$\Hom(-,\C^\times)$ to (2.3) is exact because $\C^\times$ is divisible.  It
gives
\[
 1\longrightarrow\widehat{\angles S}
 \longrightarrow\Hom(A_d(S),\C^\times)
 \longrightarrow\Hom(K(S)/\calL_{\le d}(S),\C^\times)
 \longrightarrow1.                                         \tag{3.6}
\]
The first term is the translation-character subgroup.  The last term is the
residual complex phase ambiguity.  Formula (3.6) also explains why the
integral obstruction itself and the phase group are dual objects, not the
same object.  A free summand in $K/\calL$ produces a copy of $\C^\times$;
a $C_m$ summand produces $\mu_m$.

\begin{example}[Why exact support is a hypothesis]
Suppose a zero-sum tuple contains a frequency at which both Fourier
coefficients vanish.  Its monomial equality is then $0=0$ and imposes no
condition on ratios at the other frequencies.  One cannot divide the tuple
equation and then silently discard the zero.  Lemma~\ref{lem:support} first
identifies the common exact support; Proposition~\ref{prop:ratio} divides only
after restricting to it.
\end{example}

The propositions separate two arithmetic tasks.  The lattice calculation
classifies all complex multiplicative ratios.  On the unit support,
Section~\ref{sec:signals} classifies the complete rational family.  For all
stable $C_{2pq}$ supports, Section~\ref{sec:galois} constructs a
Galois-equivariant ratio whenever the lattice has not yet closed.  That is
enough to determine the exact rational and integer thresholds, without
parametrizing every rational ambiguity torus.

\begin{remark}[Translations versus reflection]
The equivalence throughout the paper is translation only.  Reflection sends
$s$ to $-s$ in frequency and is not quotiented unless it happens to coincide
with a translation on a particular signal.
\end{remark}

The ratio-character method and CRT patching already occur in the classical
cyclic work of Gr\"unbaum--Moore \cite{GrunbaumMoore1995}.  Our later use is a
support-sensitive integral classification, not a claim of ownership of that
method.