\section{Exact meaning has a translation-period cost}
\label{sec:period}
The minimum-error theorem concerns a fixed product alphabet. A separate obstruction applies to every covariant phase alphabet \eqref{eq:phaseword}, including compressed alphabets.

\begin{definition}[Semantic period subgroup]
For any function $f:G\to\mathcal Y$, let
\[
 H_f=\{h\in G:f(x+h)=f(x)\text{ for every }x\in G\}.
\]
This is a subgroup: the identity is a period, periods add, and a period's inverse is a period. Its annihilator is
$H_f^\perp=\{a\in\widehat G:\chi_a(h)=1\text{ for every }h\in H_f\}$.
\end{definition}

\begin{theorem}[Semantic period closure]
\label{thm:period}
For the phase words \eqref{eq:phaseword}, the following are equivalent:
\begin{enumerate}
\item A measurement reads $f(x)$ with certainty for every $x$.
\item The Gram function satisfies $g(h)=0$ for all $h\notin H_f$.
\item The spectral distribution $p$ is invariant under translation by $H_f^\perp$.
\end{enumerate}
Any such alphabet distinguishes every coset of $H_f$, and its minimum possible active dimension is
\begin{equation}
 D_{\min}=|G/H_f|.
 \label{eq:periodcost}
\end{equation}
\end{theorem}
\begin{proof}
Perfectly distinguishable pure states carrying different output labels must be orthogonal. If $h\notin H_f$, some $x$ has $f(x+h)\ne f(x)$; covariance makes that pair's overlap $g(h)$, so $g(h)=0$. Conversely, if different semantic labels occur at $x,y$, then $y-x\notin H_f$. Condition 2 makes all cross-label spans orthogonal, and projection onto them reads the label exactly.

Finite Fourier inversion says that $g$ is supported on $H_f$ precisely when $p$ is constant on each coset of $H_f^\perp$. Every nonempty spectral support therefore contains at least $|H_f^\perp|=|G/H_f|$ characters. Uniform $p$ on $H_f^\perp$ attains the bound: its Gram function is the indicator of $H_f$. All distinct $H_f$ cosets are orthogonal, proving the stronger coset assertion.
\end{proof}

\begin{corollary}[A binary sparse predicate can require the full alphabet]
For $G=\F_p^n$ and $0\le k<n$, the predicate $f(x)=\1\{\wt(x)\le k\}$ has $H_f=\{0\}$. Exact all-input decision by a phase alphabet therefore requires $D=p^n$.
\end{corollary}
\begin{proof}
Take nonzero $h$ of weight $s$. Choose $x$ supported outside $\supp h$ with weight $\min(k,n-s)$. Then $\wt(x)\le k$, but $\wt(x+h)=s+\min(k,n-s)>k$. Thus no nonzero $h$ is a period.
\end{proof}
This is an obstruction for a specified covariant encoder. A sender who already knows the requested predicate may send its single classical output bit. The theorem does not prohibit that different encoding task.

\section{TACHY / coherent refusal of sparse repairs}
\label{sec:sparse}

\subsection{The standard support principle, made explicit}
Let $\mathcal L$ be any legal set of pure words. Put $V_{\mathcal L}=\Span\{\ket{\psi_u}:u\in\mathcal L\}$. If an effect $0\preceq E\preceq I$ obeys $\bra{\psi_u}E\ket{\psi_u}=0$ on every legal word, positivity gives $E\ket{\psi_u}=0$. Hence
\begin{equation}
 E\preceq E_*:=I-P_{V_{\mathcal L}}.
 \label{eq:maxrefusal}
\end{equation}
The projector $E_*$ is optimal simultaneously for every input and every prior. This is the established support geometry behind unambiguous state filtering\ \cite{filter,booleanfilter}. The new work below identifies its exact algebraic content for the sparse-repair language and for hereditary product grammars.

\subsection{Sparse repair becomes a zero-marginal spectral trade}
Let $B\in\F_p^{r\times m}$ have full row rank, let $0\le k<m$, and use the legal language $C_k(B)$ of \eqref{eq:repairlanguage}. For a set $\mathcal A\subseteq\F_p^r$ of $D$ distinct characters and positive weights $\mu_a$ summing to one, prepare
\begin{equation}
 \ket{\psi_s}=\sum_{a\in\mathcal A}\sqrt{\mu_a}\,\omega_p^{a\cdot s}\ket a,
 \qquad \omega_p=e^{2\pi i/p}.
 \label{eq:sparseword}
\end{equation}

\begin{theorem}[Spectral-trade characterization]
\label{thm:trade}
A nonzero sound refusal effect exists exactly when there is a nonzero complex coefficient array $(c_a)_{a\in\mathcal A}$ satisfying
\begin{equation}
 \sum_{\substack{a\in\mathcal A\\(B^Ta)|_S=b}}c_a=0
 \quad\text{for every }S\subseteq[m],\ |S|\le k,
 \quad b\in\F_p^S.
 \label{eq:trademarginals}
\end{equation}
Equivalently, extend $c$ to a function on $\F_p^m$ by $c(B^Ta)=c_a$ and zero elsewhere. Every marginal on at most $k$ coordinates vanishes.
\end{theorem}
\begin{proof}
A nonzero sound effect has a nonzero range vector $v$ orthogonal to every legal word. Put $c_a=\overline{v_a}\sqrt{\mu_a}$. For any repair $r$ supported on $S$,
\[
 0=\sum_a c_a\omega_p^{a\cdot Br}
 =\sum_{b\in\F_p^S}\left(\sum_{a:(B^Ta)|_S=b}c_a\right)\omega_p^{b\cdot r_S}.
\]
The characters of $\F_p^S$ are linearly independent, so every parenthesized marginal vanishes. Conversely, \eqref{eq:trademarginals} makes the character sum zero on all legal syndromes. The normalized vector with coordinates proportional to $\overline{c_a}/\sqrt{\mu_a}$ defines a sound rank-one effect. Full row rank makes $B^T$ injective, so the extended coefficient function is well-defined; character independence ensures the effect is nonzero on at least one syndrome.
\end{proof}

\begin{lemma}[Zero-marginal support bound]
\label{lem:tradebound}
If a nonzero complex function on a finite product set has every marginal on at most $k$ coordinates equal to zero, its support has size at least $2^{k+1}$.
\end{lemma}
\begin{proof}
For $k=0$, zero total sum requires at least two nonzero entries. For $k\ge1$, some coordinate has at least two nonempty support slices; otherwise there is only one support point. Every nonempty slice has all marginals on at most $k-1$ other coordinates zero, because those marginals are slices of the original marginals that include the selected coordinate. Induction gives at least $2^k$ points in each of at least two slices.
\end{proof}

\begin{corollary}[Sharp coherent-alphabet barrier]
\label{cor:modebound}
Any nontrivial sound refusal of $C_k(B)$ by \eqref{eq:sparseword} requires
\begin{equation}
 \boxed{D\ge2^{k+1}.}
 \label{eq:modebound}
\end{equation}
\end{corollary}
The extremal zero-marginal mechanism is closely related to classical trade bounds\ \cite{trades}; it is not claimed as a newly invented combinatorial fact. The contribution is its exact interpretation as a sparse-semantic quantum resource law. The bound concerns Hilbert dimension: it means at least $k+1$ qubits of active dimension, not necessarily $2^{k+1}$ separate physical components. Registers, oracle circuitry, and preparation depth must be counted separately.

If $B$ is rank deficient, restrict the syndrome space to $\im B$ and identify characters with the same restriction. Duplicate character labels can be combined by a fixed isometry. The correct dimension is the number of distinct active restricted characters.

\subsection{A matching affine query cube}
Let $t=k+1$. Suppose $\im B^T$ contains nonzero vectors $B^Ta_1,\ldots,B^Ta_t$ with pairwise disjoint coordinate supports. They are linearly independent. Choose an offset $a_0$ and use
\[
 a_\epsilon=a_0+\sum_{j=1}^t\epsilon_ja_j,
 \qquad \epsilon\in\{0,1\}^t.
\]
These are $2^t$ distinct characters, including for odd $p$. For odd $p$ the set is a binary query cube, not an entire $\F_p$-affine subspace.

\begin{theorem}[Optimal cube detector and amplitudes]
\label{thm:cube}
For arbitrary positive query weights $\mu_\epsilon$, the sound-refusal subspace is one-dimensional and
\begin{align}
 Z&=\sum_\epsilon\mu_\epsilon^{-1},&
 v_\epsilon&=\frac{(-1)^{|\epsilon|}}{\sqrt{\mu_\epsilon Z}},&
 E_*&=\ket v\bra v,\nonumber\\
 \Qref^\mu(s)&=\frac{\prod_{j=1}^t|1-\omega_p^{a_j\cdot s}|^2}{Z}.
 \label{eq:cubesuccess}
\end{align}
Uniform weights maximize this probability simultaneously at every input with nonzero numerator, giving
\begin{equation}
 \Qref(s)=\prod_{j=1}^{k+1}\sin^2\!\left(\frac{\pi(a_j\cdot s)}p\right).
 \label{eq:uniformcube}
\end{equation}
\end{theorem}
\begin{proof}
The alternating character sum factors as
\[
 \sum_\epsilon(-1)^{|\epsilon|}\omega_p^{a_\epsilon\cdot s}
 =\omega_p^{a_0\cdot s}\prod_j(1-\omega_p^{a_j\cdot s}).
\]
Any repair on at most $t-1$ coordinates misses one of the disjoint supports $\supp(B^Ta_j)$, so one factor vanishes. This proves soundness.

For uniqueness, select one nonzero coordinate from each $B^Ta_j$. A repair on any $t-1$ selected coordinates can vary the corresponding $t-1$ query values arbitrarily while leaving the last value zero. Character independence forces the coefficient sum over each binary index to vanish with the others fixed. Thus every sound coefficient array is a scalar multiple of $(-1)^{|\epsilon|}$. Normalize the resulting direction to get \eqref{eq:cubesuccess}. Cauchy--Schwarz gives $Z\ge(2^t)^2$, with equality exactly for uniform weights. The identity $|1-e^{i\theta}|^2=4\sin^2(\theta/2)$ yields \eqref{eq:uniformcube}.
\end{proof}

\begin{proposition}[A general amplitude compiler]
For any nonzero algebraic witness $c$ satisfying \eqref{eq:trademarginals}, discard its zero coefficients and take $\supp(c)$ as the active alphabet. Write $F_c(s)=\sum_ac_a\chi_a(s)$. Among rank-one detectors compiled from this fixed witness, the optimal spectral weights are
\[
 \mu_a=\frac{|c_a|}{\|c\|_1},\qquad
 \Qref(s)=\frac{|F_c(s)|^2}{\|c\|_1^2}.
\]
\end{proposition}
\begin{proof}
Normalization of the detector costs $\sum_a|c_a|^2/\mu_a$. Weighted Cauchy--Schwarz bounds this below by $\|c\|_1^2$, with equality at the displayed weights. The numerator is unchanged.
\end{proof}

\subsection{A complete four-mode example}
Over $\F_2$, take
\begin{equation}
 B=\begin{pmatrix}
 1&0&1&0&0&0\\0&1&1&0&0&0\\
 0&0&0&1&0&1\\0&0&0&0&1&1
 \end{pmatrix},\qquad k=1.
 \label{eq:blockB}
\end{equation}
Each two-coordinate syndrome block is produced by at most one repair column. Thus $d_B(s)$ is the number of nonzero blocks. Choose $a_1=(1,0,0,0)$ and $a_2=(0,0,1,0)$, and uniform weights on their four-point query cube.

The syndrome $1111$ has distance two and is refused with probability one. The syndrome $0101$ also has distance two, but its chosen phase word equals the legal zero word and cannot be refused by any sound detector on this alphabet. Select instead a uniformly random nonzero covector in each two-coordinate block. For every nonzero block, its selected parity is one with probability $2/3$; every nonrepairable syndrome is therefore refused with probability $4/9$ per run. There are no false refusals in the ideal model.

The legal predicate has trivial translation period, so exact deterministic all-input reading requires sixteen dimensions by Theorem~\ref{thm:period}. Four dimensions suffice for this selective probabilistic refusal. This is a separation between semantic contracts; classical randomized fingerprints can achieve comparable testing in this structured example.

\subsection{Gauge compatibility and the cost of the oracle}
For the admissible source transformation $B'=PBQ^{-1}$ and $s'=Ps$, with $P$ invertible and $Q$ monomial, use $a'=P^{-T}a$. Then
\[
 a'\cdot s'=a\cdot s,\qquad B'^Ta'=Q^{-T}B^Ta.
\]
Characters, support disjointness, and the zero-marginal constraints transport correctly. Dense residual-coordinate mixing can change sparsity and is excluded from this invariance.

A one-query preparation assumes a coherent operation $O_s\ket a=\chi_a(s)\ket a$. Realizing it may involve many interactions with the source. If electronics first reconstruct $s$ and then program the phases, the quantum state is an encoding of an already known syndrome. The mathematical readout theorem does not turn that route into an independent sensing advantage.