Datasets:

PureOne
/

NIRAEL-Quantum-Instant-Language

\section{EVE / the phase alphabet and its meanings}
\label{sec:foundations}

\subsection{Two exact mechanisms retained from the source manuscripts}
AELHYEUM IV\ \cite{sourcehush} starts with a binary cycle dictionary $x\mapsto Cx$, whose vertex-boundary syndrome vanishes, and a noisy optical dictionary $F(x)$. Its averaged Bhattacharyya overlap has the form
\begin{equation}
 w(h)=\frac1N\sum_{x\in\F_2^n}
 \exp\!\left[-\frac{\|F(x)-F(x+h)\|^2}{4\sigma^2}\right],\qquad N=2^n.
 \label{eq:hush}
\end{equation}
The Gaussian kernel is positive definite; averaging its translated feature inner products gives a positive-definite function on the binary group. Thus its normalized Walsh spectrum is nonnegative. An independent parity sensor of overlap $\beta_a$ multiplies $w(h)$ by $\beta_a^{a\cdot h}$, and the resulting multiplicative risk objective has diminishing returns. These are the precise spectral and design mechanisms retained here.

AELORYN III\ \cite{sourcerefusal} begins with local finite-field transports and residuals,
\[
 r_e=z_v-T_ez_u,\qquad
 D_C=(I-H_C)z_0=\sum_{e\in C}P_{C,e}r_e.
\]
The suffix transports $P_{C,e}$ express every residual in the same base fiber. Stacking the selected loop coordinates gives $Br=s$ over $\F_p$. The repair language and its distance are
\begin{equation}
 C_k(B)=\{Br:\wt(r)\le k\},\qquad
 d_B(s)=\min_{Br=s}\wt(r).
 \label{eq:repairlanguage}
\end{equation}
An exact classical sketch $Ls$ may certify $s\notin C_k(B)$ without reconstructing $r$. We retain this one-sided meaning and ask which coherent character alphabets can express it.

\subsection{A phase word is an actual normalized state}
Let $G$ be a finite abelian group with dual group $\widehat G$. For a probability distribution $p$ on its characters, define
\begin{equation}
 \ket{\psi_x(p)}=\sum_{a\in\widehat G}\sqrt{p(a)}\chi_a(x)\ket a,
 \qquad x\in G.
 \label{eq:phaseword}
\end{equation}
The active Hilbert dimension is $D=|\supp p|$, and the Gram function is
\begin{equation}
 \braket{\psi_x(p)}{\psi_y(p)}
 =g(y-x),\qquad g(h)=\sum_a p(a)\chi_a(h).
 \label{eq:gramphase}
\end{equation}
For the real binary kernel in \eqref{eq:hush}, set
$p(a)=N^{-1}\sum_h w(h)(-1)^{a\cdot h}$. Positivity gives $p(a)\ge0$, and $w(0)=1$ gives $\sum_a p(a)=1$. Equation~\eqref{eq:phaseword} therefore realizes that kernel as a quantum-state Gram geometry.

This is a mathematical realization, not a claim that classical noisy optical samples have already prepared the coherent state. A device must implement the amplitudes and relative phases. Symmetric phase ensembles and their decoding have extensive prior art\ \cite{eldar,piveteau,buzet}.

\begin{proposition}[Spectral fusion]
The product family $\ket{\psi_x(p)}\otimes\ket{\psi_x(q)}$ is isometric, on its word span, to the phase family with spectral distribution $p*q$. For a binary parity token of overlap $\beta\in[0,1]$ and mask $a$, appending the token changes
\[
 p\longmapsto \frac{1+\beta}{2}p+\frac{1-\beta}{2}\tau_a p,
 \qquad (\tau_a p)(u)=p(u+a).
\]
\end{proposition}
\begin{proof}
Product-state overlaps multiply. The Fourier transform of the convolution $p*q$ is the product of the two Fourier transforms, so both word families have identical Gram matrices. The linear map taking one family to the other preserves inner products and extends to an isometry of their spans. A binary token has Gram factor $1$ when $a\cdot h=0$ and $\beta$ otherwise, whose spectral distribution is $\frac{1+\beta}{2}\delta_0+\frac{1-\beta}{2}\delta_a$.
\end{proof}
The two factors must be available as separate prepared systems. The isometry does not manufacture a second copy of an unknown state.

\subsection{Three semantic contracts}
\begin{definition}[Quantum instant language]
A language instance consists of an input set $X$, an encoding $x\mapsto\rho_x$, a declared semantic task, a calibrated measurement, and a classical output rule. We distinguish:
\begin{enumerate}
\item \textbf{Minimum-error reading:} output $f(x)$ with maximum average success under a stated prior.
\item \textbf{Sound refusal:} output \texttt{REFUSE} with probability zero on every legal word; all remaining outcomes mean \texttt{UNDECIDED}.
\item \textbf{Exact reading:} output $f(x)$ correctly with probability one on every input.
\end{enumerate}
\end{definition}
An undecided outcome supplies no assertion of legality. A source prior can improve minimum-error decisions, but cannot relax the universal soundness condition on a refusal detector.

\begin{figure}[ht]
\centering
\begin{tikzpicture}[node distance=7mm, every node/.style={font=\small\sffamily},
 box/.style={draw=teal,fill=mist,rounded corners=0mm,minimum height=12mm,text width=37mm,align=center},
 arr/.style={-{Latex[length=2mm]},draw=teal}]
\node[box,text width=65mm] (word) {Prepared phase word\\or product quantum alphabet};
\node[box,below=10mm of word] (refuse) {One-sided\\sound refusal};
\node[box,left=of refuse] (guess) {Minimum-error\\Boolean reading};
\node[box,right=of refuse] (exact) {Exact deterministic\\semantic reading};
\node[box,below=of guess] (raven) {RAVEN\\Exact affine boundary};
\node[box,below=of refuse] (tachy) {TACHY\\Support, trade, grammar};
\node[box,below=of exact] (period) {Period subgroup\\Dimension cost};
\draw[arr] (word.south)--(refuse.north);
\draw[arr] (word.south)--++(0,-4mm)-|(guess.north);
\draw[arr] (word.south)--++(0,-4mm)-|(exact.north);
\draw[arr] (guess)--(raven);\draw[arr] (refuse)--(tachy);\draw[arr] (exact)--(period);
\end{tikzpicture}
\caption{The declared semantic contract determines the measurement problem and the applicable structural theorem. Each branch uses the state-model hypotheses specified in its corresponding section.}
\end{figure}

\section{RAVEN / an exact local--global boundary}
\label{sec:rigidity}

\subsection{The precise recent question}
Gupta et al.\ \cite{gupta} study a canonical receiver: measure each pure-state symbol by its equal-prior Helstrom measurement, obtain a bitstring $y$, then return $f(y)$. Their Theorem 4 leaves finitely many possible interior overlaps at which a non-affine $f$ might still be globally optimal. The same qualification remains in the located July 2026 theses\ \cite{zheng,mohan}. The following result eliminates it by replacing a dense Gram matrix with its logarithmic generator.

\begin{theorem}[Exact affine boundary, including heterogeneous overlaps]
\label{thm:rigidity}
For each coordinate $i$, let $\ket{\phi_{i,0}},\ket{\phi_{i,1}}$ be known normalized pure states with
\[
 0<s_i=|\braket{\phi_{i,0}}{\phi_{i,1}}|<1.
\]
The input $x$ is uniform on $\F_2^n$ and is encoded as $\ket{\Phi_x}=\bigotimes_i\ket{\phi_{i,x_i}}$. Let $P_{\rm can}$ be the success probability of independent equal-prior Helstrom measurements followed by $f(y)$, and let $P_{\rm glob}$ be the optimal success probability over all joint measurements. Then
\begin{equation}
 P_{\rm can}=P_{\rm glob}
 \quad\Longleftrightarrow\quad
 f(x)=b_0\oplus\bigoplus_{i=1}^n b_ix_i.
 \label{eq:exactboundary}
\end{equation}
For every non-affine $f$, the inequality is strict for \emph{every} $(s_1,\ldots,s_n)\in(0,1)^n$. For affine $f$,
\begin{equation}
 P_{\rm can}=P_{\rm glob}
 =\frac12\left(1+\prod_{i:b_i=1}\sqrt{1-s_i^2}\right).
 \label{eq:affinesuccess}
\end{equation}
The empty product is one.
\end{theorem}

\begin{proof}
\textbf{1. Put the ensemble in polar coordinates.}
Rephase each local pair so its overlap is positive, and restrict to its two-dimensional span. Write $V_i$ for the matrix with those two states as columns, $V=\bigotimes_iV_i$, and
\[
 G=V^*V=\bigotimes_i(I+s_iX),\qquad R=G^{1/2},\qquad V=UR.
\]
Here $X$ is the Pauli flip and $U$ is a tensor product of local polar unitaries. In the $U$ basis the local square-root measurement is the computational basis. For an equal-prior pure-state pair it is the Helstrom measurement: explicitly, if $R_i=\begin{psmallmatrix}a_i&b_i\\b_i&a_i\end{psmallmatrix}$, then
$R_i ZR_i=(a_i^2-b_i^2)Z=\sqrt{1-s_i^2}\,Z$.

Set $N=2^n$, $d_x=(-1)^{f(x)}$, and $D=\operatorname{diag}(d_x)$. The canonical binary observable in this basis is $D$. The difference of the subnormalized hypotheses is
\begin{equation}
 H=U^*\!\left[\frac1N\sum_xd_x\ket{\Phi_x}\bra{\Phi_x}\right]U
 =\frac1N RDR.
 \label{eq:H}
\end{equation}
The Helstrom variational formula gives $P_{\rm glob}=(1+\|H\|_1)/2$ and $P_{\rm can}=(1+\Tr(DH))/2$. Since $R$ is invertible, $H$ is nonsingular. The unique optimal signed observable is $\operatorname{sign}(H)$. Canonical optimality therefore forces $D=\operatorname{sign}(H)$ and hence
\begin{equation}
 [D,RDR]=0\quad\Longleftrightarrow\quad[R,DRD]=0.
 \label{eq:commutatorroot}
\end{equation}

\textbf{2. Take a principal logarithm.}
Both $R$ and $DRD$ are positive definite. Commuting positive definite matrices have commuting principal logarithms. Let $t_i=\operatorname{atanh}(s_i)>0$ and let $X_i$ flip label coordinate $i$. Since
\[
 I+s_iX=\sqrt{1-s_i^2}\exp(t_iX),
\]
we obtain
\begin{equation}
 \log R=\frac14\sum_i\log(1-s_i^2)I+\frac12 A,
 \qquad A=\sum_i t_iX_i.
 \label{eq:loggenerator}
\end{equation}
Thus \eqref{eq:commutatorroot} implies $[A,DAD]=0$. A numerical condition at one overlap tuple has become a structural condition on a weighted Boolean cube.

\textbf{3. Inspect one square.}
For $i\ne j$, write $x^i=x\oplus e_i$ and $y=x\oplus e_i\oplus e_j$. Only two length-two cube paths connect $x$ and $y$, giving
\begin{equation}
 [A,DAD]_{x,y}
 =t_it_j(d_y-d_x)(d_{x^i}+d_{x^j}).
 \label{eq:square}
\end{equation}
If a square has an odd number of negative signs, one diagonal has opposite endpoint signs and the other has equal endpoint signs. The displayed entry for the first diagonal is then nonzero. Therefore every square has even sign parity:
\begin{equation}
 f(x)\oplus f(x^i)\oplus f(x^j)\oplus f(x^{ij})=0.
 \label{eq:secondderivative}
\end{equation}
Each first Boolean derivative is invariant under every coordinate flip and is consequently constant. Integrating these constants gives $f(x)=f(0)\oplus\bigoplus_i[f(e_i)\oplus f(0)]x_i$.

\textbf{4. Prove sufficiency and the success formula.}
For affine $f$, $D$ is, up to a sign, a product of coordinate $Z_i$ operators. Hence $DAD=\sum_i(-1)^{b_i}t_iX_i$ commutes with $A$. It follows that $DRD$ commutes with $R$ and $(DRD)R$ is positive definite. Thus $DH>0$, so $D=\operatorname{sign}(H)$ and the canonical receiver is optimal. Each local output has correctness bias $\sqrt{1-s_i^2}$; independent parity biases multiply, proving \eqref{eq:affinesuccess}.
\end{proof}

\begin{boundary}
\textbf{Receiver scope.} The theorem concerns the fixed rule ``bitwise equal-prior Helstrom, then $f(y)$.'' The next corollary extends its equality classification to arbitrary classical postprocessing of these same outputs. Neither result classifies different local measurement bases or adaptive receivers. The hypotheses are a uniform input prior and interior pure-state overlaps; other alphabets and priors require separate statements.
\end{boundary}

\begin{corollary}[Classical postprocessing cannot restore global equality]
\label{cor:postprocess}
Allow any deterministic or randomized classical decision rule after the same product of local Helstrom measurements. Its best success equals $P_{\rm glob}$ if and only if $f$ is affine. If a postprocessing rule attains global equality, its signed decision observable must coincide with the canonical $D$.
\end{corollary}
\begin{proof}
Any postprocessed observable is diagonal in the $U$ basis, with diagonal entries in $[-1,1]$. If it is globally optimal, uniqueness of the optimum for nonsingular $H$ makes it $J=\operatorname{sign}(H)$, a diagonal involution. Thus $J$ commutes with both $D$ and $H$.

We show $J=D$. The matrix $DH=N^{-1}(DRD)R$ is a product of two positive definite matrices and therefore has strictly positive real spectrum. Since it commutes with $J$, both of its restrictions to the $J$ eigenspaces have positive spectrum. On the $J=+1$ subspace, $H_+$ is positive definite. The product $D_+H_+$ is similar to $H_+^{1/2}D_+H_+^{1/2}$; its positive spectrum and Sylvester's inertia law force $D_+=I$. On the $J=-1$ subspace, write $H_-=-K$ with $K>0$. Positivity of the spectrum of $-D_-K$ similarly forces $D_-=-I$. Consequently $D=J$. Theorem~\ref{thm:rigidity} now applies.
\end{proof}
This corollary does not say $f(y)$ is always the best classical rule. It says that a better rule cannot close the entire global gap for a non-affine function in this model.

\begin{figure}[ht]
\centering\includegraphics[width=\textwidth]{figures/local_global_boundary.pdf}
\caption{Synthetic exact-matrix evaluations for AND on two bits and majority on three bits. The global advantage is positive at every interior overlap; it can approach zero at a boundary. The theorem concerns the entire open interval, not only the plotted points.}
\end{figure}

\section{A quantitative law for nonlinear semantic curvature}
\label{sec:gap}
The logarithmic argument gives more than absence of exceptional points. It identifies an explicit measure of the nonlinearity responsible for the gap.

Define the mixed Boolean derivative $\partial_{ij}f(x)$ as the XOR in \eqref{eq:secondderivative}, and set
\begin{equation}
 \kappa_{\mathbf s}(f)=\sum_{i<j}t_i^2t_j^2\Pr_x\{\partial_{ij}f(x)=1\},
 \quad
 \lambda_-=\prod_i(1-s_i),\quad \lambda_+=\prod_i(1+s_i).
 \label{eq:curvature}
\end{equation}
This is a weighted density of non-affine squares. All weights are strictly positive, so it vanishes exactly for affine functions.

\begin{theorem}[Quantitative curvature gap]
\label{thm:gap}
Under the hypotheses of Theorem~\ref{thm:rigidity},
\begin{equation}
 \boxed{\quad P_{\rm glob}-P_{\rm can}
 \ \ge\ \frac{\lambda_-^2}{32\lambda_+}\,\kappa_{\mathbf s}(f).\quad}
 \label{eq:gapbound}
\end{equation}
\end{theorem}

\begin{proof}
Write $\delta=P_{\rm glob}-P_{\rm can}$, $S=\operatorname{sign}(H)$, and $M=\|H\|_{\rm op}$. Since $[S,H]=0$, the Frobenius triangle inequality and $H^2\preceq M|H|$ give
\begin{align}
 \|[D,H]\|_F^2
 &\le4\Tr\bigl((D-S)^2H^2\bigr)\nonumber\\
 &\le4M\Tr\bigl((D-S)^2|H|\bigr)=16M\delta.
 \label{eq:commgap}
\end{align}
For the last equality, expand $(D-S)^2$ and use $S|H|=H$ and cyclicity of trace.

We need an elementary logarithm estimate. If $X\succeq\alpha I>0$, diagonalizing $X$ shows, for any matrix $B$,
\[
 \|[\log X,B]\|_F\le\alpha^{-1}\|[X,B]\|_F,
\]
because $|\log u-\log v|\le|u-v|/\alpha$ for $u,v\ge\alpha$. Apply this twice to $X=R$ and $Y=DRD$, both with minimum eigenvalue $\alpha=\sqrt{\lambda_-}$:
\[
 \|[\log R,\log(DRD)]\|_F
 \le\lambda_-^{-1}\|[R,DRD]\|_F.
\]
Equations~\eqref{eq:H} and \eqref{eq:loggenerator} then imply
\begin{equation}
 \|[D,H]\|_F\ge\frac{\lambda_-}{4N}\|[A,DAD]\|_F.
 \label{eq:logbound}
\end{equation}
Each odd-sign square contributes two oriented entries of magnitude $4t_it_j$ to \eqref{eq:square}. Each geometric square is counted four times by a uniformly ranged $x$. Therefore
\begin{equation}
 \|[A,DAD]\|_F^2=8N\kappa_{\mathbf s}(f).
 \label{eq:normcurvature}
\end{equation}
Finally $M\le\|R\|^2/N=\lambda_+/N$. Combining \eqref{eq:commgap}--\eqref{eq:normcurvature} yields \eqref{eq:gapbound}.
\end{proof}

\subsection{Meaning and limitations of the bound}
The bound turns a combinatorial obstruction into a certified positive probability advantage over the canonical rule. It is not asserted as a bound against optimally postprocessed local outputs. It is conservative: the product factor $\lambda_-^2/\lambda_+$ can be very small in large dimension or near indistinguishable symbols. For two-bit AND at $s=0.5$, the computed gap is approximately $0.002243649$, while the lower bound is approximately $0.0000790324$. The theorem supplies strictness and an explicit floor; it does not claim the best constant or the optimal asymptotic dependence.

The proof method is portable as a research tool: where a positive Gram kernel has a sparse logarithmic generator, an optimality condition may be tested through local configurations of that generator. Extending this idea beyond the present binary product kernel is a concrete future question, not an established theorem of this manuscript.

	\section{EVE / the phase alphabet and its meanings}
	\label{sec:foundations}

	\subsection{Two exact mechanisms retained from the source manuscripts}
	AELHYEUM IV\ \cite{sourcehush} starts with a binary cycle dictionary $x\mapsto Cx$, whose vertex-boundary syndrome vanishes, and a noisy optical dictionary $F(x)$. Its averaged Bhattacharyya overlap has the form
	\begin{equation}
	w(h)=\frac1N\sum_{x\in\F_2^n}
	\exp\!\left[-\frac{\\|F(x)-F(x+h)\\|^2}{4\sigma^2}\right],\qquad N=2^n.
	\label{eq:hush}
	\end{equation}
	The Gaussian kernel is positive definite; averaging its translated feature inner products gives a positive-definite function on the binary group. Thus its normalized Walsh spectrum is nonnegative. An independent parity sensor of overlap $\beta_a$ multiplies $w(h)$ by $\beta_a^{a\cdot h}$, and the resulting multiplicative risk objective has diminishing returns. These are the precise spectral and design mechanisms retained here.

	AELORYN III\ \cite{sourcerefusal} begins with local finite-field transports and residuals,
	\[
	r_e=z_v-T_ez_u,\qquad
	D_C=(I-H_C)z_0=\sum_{e\in C}P_{C,e}r_e.
	\]
	The suffix transports $P_{C,e}$ express every residual in the same base fiber. Stacking the selected loop coordinates gives $Br=s$ over $\F_p$. The repair language and its distance are
	\begin{equation}
	C_k(B)=\{Br:\wt(r)\le k\},\qquad
	d_B(s)=\min_{Br=s}\wt(r).
	\label{eq:repairlanguage}
	\end{equation}
	An exact classical sketch $Ls$ may certify $s\notin C_k(B)$ without reconstructing $r$. We retain this one-sided meaning and ask which coherent character alphabets can express it.

	\subsection{A phase word is an actual normalized state}
	Let $G$ be a finite abelian group with dual group $\widehat G$. For a probability distribution $p$ on its characters, define
	\begin{equation}
	\ket{\psi_x(p)}=\sum_{a\in\widehat G}\sqrt{p(a)}\chi_a(x)\ket a,
	\qquad x\in G.
	\label{eq:phaseword}
	\end{equation}
	The active Hilbert dimension is $D=\|\supp p\|$, and the Gram function is
	\begin{equation}
	\braket{\psi_x(p)}{\psi_y(p)}
	=g(y-x),\qquad g(h)=\sum_a p(a)\chi_a(h).
	\label{eq:gramphase}
	\end{equation}
	For the real binary kernel in \eqref{eq:hush}, set
	$p(a)=N^{-1}\sum_h w(h)(-1)^{a\cdot h}$. Positivity gives $p(a)\ge0$, and $w(0)=1$ gives $\sum_a p(a)=1$. Equation~\eqref{eq:phaseword} therefore realizes that kernel as a quantum-state Gram geometry.

	This is a mathematical realization, not a claim that classical noisy optical samples have already prepared the coherent state. A device must implement the amplitudes and relative phases. Symmetric phase ensembles and their decoding have extensive prior art\ \cite{eldar,piveteau,buzet}.

	\begin{proposition}[Spectral fusion]
	The product family $\ket{\psi_x(p)}\otimes\ket{\psi_x(q)}$ is isometric, on its word span, to the phase family with spectral distribution $p*q$. For a binary parity token of overlap $\beta\in[0,1]$ and mask $a$, appending the token changes
	\[
	p\longmapsto \frac{1+\beta}{2}p+\frac{1-\beta}{2}\tau_a p,
	\qquad (\tau_a p)(u)=p(u+a).
	\]
	\end{proposition}
	\begin{proof}
	Product-state overlaps multiply. The Fourier transform of the convolution $p*q$ is the product of the two Fourier transforms, so both word families have identical Gram matrices. The linear map taking one family to the other preserves inner products and extends to an isometry of their spans. A binary token has Gram factor $1$ when $a\cdot h=0$ and $\beta$ otherwise, whose spectral distribution is $\frac{1+\beta}{2}\delta_0+\frac{1-\beta}{2}\delta_a$.
	\end{proof}
	The two factors must be available as separate prepared systems. The isometry does not manufacture a second copy of an unknown state.

	\subsection{Three semantic contracts}
	\begin{definition}[Quantum instant language]
	A language instance consists of an input set $X$, an encoding $x\mapsto\rho_x$, a declared semantic task, a calibrated measurement, and a classical output rule. We distinguish:
	\begin{enumerate}
	\item \textbf{Minimum-error reading:} output $f(x)$ with maximum average success under a stated prior.
	\item \textbf{Sound refusal:} output \texttt{REFUSE} with probability zero on every legal word; all remaining outcomes mean \texttt{UNDECIDED}.
	\item \textbf{Exact reading:} output $f(x)$ correctly with probability one on every input.
	\end{enumerate}
	\end{definition}
	An undecided outcome supplies no assertion of legality. A source prior can improve minimum-error decisions, but cannot relax the universal soundness condition on a refusal detector.

	\begin{figure}[ht]
	\centering
	\begin{tikzpicture}[node distance=7mm, every node/.style={font=\small\sffamily},
	box/.style={draw=teal,fill=mist,rounded corners=0mm,minimum height=12mm,text width=37mm,align=center},
	arr/.style={-{Latex[length=2mm]},draw=teal}]
	\node[box,text width=65mm] (word) {Prepared phase word\\or product quantum alphabet};
	\node[box,below=10mm of word] (refuse) {One-sided\\sound refusal};
	\node[box,left=of refuse] (guess) {Minimum-error\\Boolean reading};
	\node[box,right=of refuse] (exact) {Exact deterministic\\semantic reading};
	\node[box,below=of guess] (raven) {RAVEN\\Exact affine boundary};
	\node[box,below=of refuse] (tachy) {TACHY\\Support, trade, grammar};
	\node[box,below=of exact] (period) {Period subgroup\\Dimension cost};
	\draw[arr] (word.south)--(refuse.north);
	\draw[arr] (word.south)--++(0,-4mm)-\|(guess.north);
	\draw[arr] (word.south)--++(0,-4mm)-\|(exact.north);
	\draw[arr] (guess)--(raven);\draw[arr] (refuse)--(tachy);\draw[arr] (exact)--(period);
	\end{tikzpicture}
	\caption{The declared semantic contract determines the measurement problem and the applicable structural theorem. Each branch uses the state-model hypotheses specified in its corresponding section.}
	\end{figure}

	\section{RAVEN / an exact local--global boundary}
	\label{sec:rigidity}

	\subsection{The precise recent question}
	Gupta et al.\ \cite{gupta} study a canonical receiver: measure each pure-state symbol by its equal-prior Helstrom measurement, obtain a bitstring $y$, then return $f(y)$. Their Theorem 4 leaves finitely many possible interior overlaps at which a non-affine $f$ might still be globally optimal. The same qualification remains in the located July 2026 theses\ \cite{zheng,mohan}. The following result eliminates it by replacing a dense Gram matrix with its logarithmic generator.

	\begin{theorem}[Exact affine boundary, including heterogeneous overlaps]
	\label{thm:rigidity}
	For each coordinate $i$, let $\ket{\phi_{i,0}},\ket{\phi_{i,1}}$ be known normalized pure states with
	\[
	0<s_i=\|\braket{\phi_{i,0}}{\phi_{i,1}}\|<1.
	\]
	The input $x$ is uniform on $\F_2^n$ and is encoded as $\ket{\Phi_x}=\bigotimes_i\ket{\phi_{i,x_i}}$. Let $P_{\rm can}$ be the success probability of independent equal-prior Helstrom measurements followed by $f(y)$, and let $P_{\rm glob}$ be the optimal success probability over all joint measurements. Then
	\begin{equation}
	P_{\rm can}=P_{\rm glob}
	\quad\Longleftrightarrow\quad
	f(x)=b_0\oplus\bigoplus_{i=1}^n b_ix_i.
	\label{eq:exactboundary}
	\end{equation}
	For every non-affine $f$, the inequality is strict for \emph{every} $(s_1,\ldots,s_n)\in(0,1)^n$. For affine $f$,
	\begin{equation}
	P_{\rm can}=P_{\rm glob}
	=\frac12\left(1+\prod_{i:b_i=1}\sqrt{1-s_i^2}\right).
	\label{eq:affinesuccess}
	\end{equation}
	The empty product is one.
	\end{theorem}

	\begin{proof}
	\textbf{1. Put the ensemble in polar coordinates.}
	Rephase each local pair so its overlap is positive, and restrict to its two-dimensional span. Write $V_i$ for the matrix with those two states as columns, $V=\bigotimes_iV_i$, and
	\[
	G=V^*V=\bigotimes_i(I+s_iX),\qquad R=G^{1/2},\qquad V=UR.
	\]
	Here $X$ is the Pauli flip and $U$ is a tensor product of local polar unitaries. In the $U$ basis the local square-root measurement is the computational basis. For an equal-prior pure-state pair it is the Helstrom measurement: explicitly, if $R_i=\begin{psmallmatrix}a_i&b_i\\b_i&a_i\end{psmallmatrix}$, then
	$R_i ZR_i=(a_i^2-b_i^2)Z=\sqrt{1-s_i^2}\,Z$.

	Set $N=2^n$, $d_x=(-1)^{f(x)}$, and $D=\operatorname{diag}(d_x)$. The canonical binary observable in this basis is $D$. The difference of the subnormalized hypotheses is
	\begin{equation}
	H=U^*\!\left[\frac1N\sum_xd_x\ket{\Phi_x}\bra{\Phi_x}\right]U
	=\frac1N RDR.
	\label{eq:H}
	\end{equation}
	The Helstrom variational formula gives $P_{\rm glob}=(1+\\|H\\|_1)/2$ and $P_{\rm can}=(1+\Tr(DH))/2$. Since $R$ is invertible, $H$ is nonsingular. The unique optimal signed observable is $\operatorname{sign}(H)$. Canonical optimality therefore forces $D=\operatorname{sign}(H)$ and hence
	\begin{equation}
	[D,RDR]=0\quad\Longleftrightarrow\quad[R,DRD]=0.
	\label{eq:commutatorroot}
	\end{equation}

	\textbf{2. Take a principal logarithm.}
	Both $R$ and $DRD$ are positive definite. Commuting positive definite matrices have commuting principal logarithms. Let $t_i=\operatorname{atanh}(s_i)>0$ and let $X_i$ flip label coordinate $i$. Since
	\[
	I+s_iX=\sqrt{1-s_i^2}\exp(t_iX),
	\]
	we obtain
	\begin{equation}
	\log R=\frac14\sum_i\log(1-s_i^2)I+\frac12 A,
	\qquad A=\sum_i t_iX_i.
	\label{eq:loggenerator}
	\end{equation}
	Thus \eqref{eq:commutatorroot} implies $[A,DAD]=0$. A numerical condition at one overlap tuple has become a structural condition on a weighted Boolean cube.

	\textbf{3. Inspect one square.}
	For $i\ne j$, write $x^i=x\oplus e_i$ and $y=x\oplus e_i\oplus e_j$. Only two length-two cube paths connect $x$ and $y$, giving
	\begin{equation}
	[A,DAD]_{x,y}
	=t_it_j(d_y-d_x)(d_{x^i}+d_{x^j}).
	\label{eq:square}
	\end{equation}
	If a square has an odd number of negative signs, one diagonal has opposite endpoint signs and the other has equal endpoint signs. The displayed entry for the first diagonal is then nonzero. Therefore every square has even sign parity:
	\begin{equation}
	f(x)\oplus f(x^i)\oplus f(x^j)\oplus f(x^{ij})=0.
	\label{eq:secondderivative}
	\end{equation}
	Each first Boolean derivative is invariant under every coordinate flip and is consequently constant. Integrating these constants gives $f(x)=f(0)\oplus\bigoplus_i[f(e_i)\oplus f(0)]x_i$.

	\textbf{4. Prove sufficiency and the success formula.}
	For affine $f$, $D$ is, up to a sign, a product of coordinate $Z_i$ operators. Hence $DAD=\sum_i(-1)^{b_i}t_iX_i$ commutes with $A$. It follows that $DRD$ commutes with $R$ and $(DRD)R$ is positive definite. Thus $DH>0$, so $D=\operatorname{sign}(H)$ and the canonical receiver is optimal. Each local output has correctness bias $\sqrt{1-s_i^2}$; independent parity biases multiply, proving \eqref{eq:affinesuccess}.
	\end{proof}

	\begin{boundary}
	\textbf{Receiver scope.} The theorem concerns the fixed rule ``bitwise equal-prior Helstrom, then $f(y)$.'' The next corollary extends its equality classification to arbitrary classical postprocessing of these same outputs. Neither result classifies different local measurement bases or adaptive receivers. The hypotheses are a uniform input prior and interior pure-state overlaps; other alphabets and priors require separate statements.
	\end{boundary}

	\begin{corollary}[Classical postprocessing cannot restore global equality]
	\label{cor:postprocess}
	Allow any deterministic or randomized classical decision rule after the same product of local Helstrom measurements. Its best success equals $P_{\rm glob}$ if and only if $f$ is affine. If a postprocessing rule attains global equality, its signed decision observable must coincide with the canonical $D$.
	\end{corollary}
	\begin{proof}
	Any postprocessed observable is diagonal in the $U$ basis, with diagonal entries in $[-1,1]$. If it is globally optimal, uniqueness of the optimum for nonsingular $H$ makes it $J=\operatorname{sign}(H)$, a diagonal involution. Thus $J$ commutes with both $D$ and $H$.

	We show $J=D$. The matrix $DH=N^{-1}(DRD)R$ is a product of two positive definite matrices and therefore has strictly positive real spectrum. Since it commutes with $J$, both of its restrictions to the $J$ eigenspaces have positive spectrum. On the $J=+1$ subspace, $H_+$ is positive definite. The product $D_+H_+$ is similar to $H_+^{1/2}D_+H_+^{1/2}$; its positive spectrum and Sylvester's inertia law force $D_+=I$. On the $J=-1$ subspace, write $H_-=-K$ with $K>0$. Positivity of the spectrum of $-D_-K$ similarly forces $D_-=-I$. Consequently $D=J$. Theorem~\ref{thm:rigidity} now applies.
	\end{proof}
	This corollary does not say $f(y)$ is always the best classical rule. It says that a better rule cannot close the entire global gap for a non-affine function in this model.

	\begin{figure}[ht]
	\centering\includegraphics[width=\textwidth]{figures/local_global_boundary.pdf}
	\caption{Synthetic exact-matrix evaluations for AND on two bits and majority on three bits. The global advantage is positive at every interior overlap; it can approach zero at a boundary. The theorem concerns the entire open interval, not only the plotted points.}
	\end{figure}

	\section{A quantitative law for nonlinear semantic curvature}
	\label{sec:gap}
	The logarithmic argument gives more than absence of exceptional points. It identifies an explicit measure of the nonlinearity responsible for the gap.

	Define the mixed Boolean derivative $\partial_{ij}f(x)$ as the XOR in \eqref{eq:secondderivative}, and set
	\begin{equation}
	\kappa_{\mathbf s}(f)=\sum_{i<j}t_i^2t_j^2\Pr_x\{\partial_{ij}f(x)=1\},
	\quad
	\lambda_-=\prod_i(1-s_i),\quad \lambda_+=\prod_i(1+s_i).
	\label{eq:curvature}
	\end{equation}
	This is a weighted density of non-affine squares. All weights are strictly positive, so it vanishes exactly for affine functions.

	\begin{theorem}[Quantitative curvature gap]
	\label{thm:gap}
	Under the hypotheses of Theorem~\ref{thm:rigidity},
	\begin{equation}
	\boxed{\quad P_{\rm glob}-P_{\rm can}
	\ \ge\ \frac{\lambda_-^2}{32\lambda_+}\,\kappa_{\mathbf s}(f).\quad}
	\label{eq:gapbound}
	\end{equation}
	\end{theorem}

	\begin{proof}
	Write $\delta=P_{\rm glob}-P_{\rm can}$, $S=\operatorname{sign}(H)$, and $M=\\|H\\|_{\rm op}$. Since $[S,H]=0$, the Frobenius triangle inequality and $H^2\preceq M\|H\|$ give
	\begin{align}
	\\|[D,H]\\|_F^2
	&\le4\Tr\bigl((D-S)^2H^2\bigr)\nonumber\\
	&\le4M\Tr\bigl((D-S)^2\|H\|\bigr)=16M\delta.
	\label{eq:commgap}
	\end{align}
	For the last equality, expand $(D-S)^2$ and use $S\|H\|=H$ and cyclicity of trace.

	We need an elementary logarithm estimate. If $X\succeq\alpha I>0$, diagonalizing $X$ shows, for any matrix $B$,
	\[
	\\|[\log X,B]\\|_F\le\alpha^{-1}\\|[X,B]\\|_F,
	\]
	because $\|\log u-\log v\|\le\|u-v\|/\alpha$ for $u,v\ge\alpha$. Apply this twice to $X=R$ and $Y=DRD$, both with minimum eigenvalue $\alpha=\sqrt{\lambda_-}$:
	\[
	\\|[\log R,\log(DRD)]\\|_F
	\le\lambda_-^{-1}\\|[R,DRD]\\|_F.
	\]
	Equations~\eqref{eq:H} and \eqref{eq:loggenerator} then imply
	\begin{equation}
	\\|[D,H]\\|_F\ge\frac{\lambda_-}{4N}\\|[A,DAD]\\|_F.
	\label{eq:logbound}
	\end{equation}
	Each odd-sign square contributes two oriented entries of magnitude $4t_it_j$ to \eqref{eq:square}. Each geometric square is counted four times by a uniformly ranged $x$. Therefore
	\begin{equation}
	\\|[A,DAD]\\|_F^2=8N\kappa_{\mathbf s}(f).
	\label{eq:normcurvature}
	\end{equation}
	Finally $M\le\\|R\\|^2/N=\lambda_+/N$. Combining \eqref{eq:commgap}--\eqref{eq:normcurvature} yields \eqref{eq:gapbound}.
	\end{proof}

	\subsection{Meaning and limitations of the bound}
	The bound turns a combinatorial obstruction into a certified positive probability advantage over the canonical rule. It is not asserted as a bound against optimally postprocessed local outputs. It is conservative: the product factor $\lambda_-^2/\lambda_+$ can be very small in large dimension or near indistinguishable symbols. For two-bit AND at $s=0.5$, the computed gap is approximately $0.002243649$, while the lower bound is approximately $0.0000790324$. The theorem supplies strictness and an explicit floor; it does not claim the best constant or the optimal asymptotic dependence.

	The proof method is portable as a research tool: where a positive Gram kernel has a sparse logarithmic generator, an optimality condition may be tested through local configurations of that generator. Extending this idea beyond the present binary product kernel is a concrete future question, not an established theorem of this manuscript.