add example section for ring graph state and its symmetry properties

paper/symmetric_shadow_maps_formal.tex

@@ -17,6 +17,7 @@
 \newtheorem{corollary}{Corollary}
 \newtheorem{remark}{Remark}
 \newtheorem{lemma}{Lemma}
+\newtheorem{example}{Example}
 
 \newcommand{\tr}{\operatorname{tr}}
 \newcommand{\id}{\mathbbm{1}}
@@ -1198,6 +1199,30 @@ Let $\pi\in S_n$ be a permutation of the parties fixing $S$ setwise (i.e.\ $\pi(
 Two families used elsewhere in this note calibrate this gap. The state $W_n$ is invariant under the full party-permutation group $S_n$, and permutation-invariant under any $\pi$ fixing any given $S$ setwise; it therefore block-diagonalizes under Proposition~\ref{prop:block-diagonal} for every cut, yet it is not a stabilizer state (its single-party marginal has eigenvalues $1/n,(n-1)/n$, never the flat spectrum that Lemma~\ref{lem:code-support} forces on a stabilizer-code reduction), so its degeneracies are explained by permutation symmetry alone, with neither Proposition~\ref{prop:sector-symmetry-compatible} nor Lemma~\ref{lem:stabilizer-degeneracy} available. The ring graph state, by contrast, is invariant only under the cyclic subgroup $C_n\le S_n$, which for $n=4$ and $S=\{1,2\}$ already fails to fix $S$ setwise (the generating shift sends $\{1,2\}\mapsto\{2,3\}$), so Proposition~\ref{prop:block-diagonal} does not apply to that cut at all in the permutation channel; its observed degeneracies, as Corollary~\ref{cor:stabilizer-examples} already notes, are accounted for entirely by the stabilizer mechanism instead. Party-permutation symmetry is thus a third selection mechanism, logically independent of both the collective-unitary mechanism of Proposition~\ref{prop:block-diagonal} under the strong hypothesis and the stabilizer mechanism of Lemma~\ref{lem:stabilizer-degeneracy}, and satisfying it need not entail satisfying either of the other two.
 \end{remark}
 
+\begin{example}[Ring graph state: reflections, not rotations, supply the example]
+\label{ex:ring-reflection}
+The cyclic group $C_4=\langle(1234)\rangle$ stabilizing the ring graph state $\rho_\square$ of Section~\ref{sec:qubit-numerics} indeed fixes no nontrivial cut setwise, as anticipated above. But $\mathrm{Aut}(\square)$ is the full dihedral group $D_4$, of order $8$, and its reflections do fix cuts setwise. For $S=\{1,2\}$, $S^c=\{3,4\}$ (the ``adjacent'' cut of Section~\ref{sec:qubit-numerics}), the reflection $\pi=(12)(34)$ is a graph automorphism with $\pi(S)=S$, $\pi(S^c)=S^c$; it is a genuine two-party swap on each side of the cut, hence exactly the situation of Remark~\ref{rem:permutation-separating-example}: the weak hypothesis holds for $\rho_\square$ at this cut, the strong one does not, since swapping qubits $1$ and $2$ is not of the form $U^{(1)}\otimes U^{(2)}$.
+
+Concretely, $R(\pi)$ does not fix $\V_{\{1\}}^{(S)}$: it maps it onto $\V_{\{2\}}^{(S)}$, since $\pi$ relabels qubit $1$ as qubit $2$. So Proposition~\ref{prop:sector-symmetry-compatible}'s hypothesis fails non-vacuously here, exactly as claimed. What Proposition~\ref{prop:block-diagonal} (the weak statement) still guarantees is equivariance of the full block $\widetilde{\mathcal M}_S(\rho_\square)$ under the combined swap on $\V_{\{1\}}^{(S)}\oplus\V_{\{2\}}^{(S)}$ and $\V_{\{3\}}^{(S^c)}\oplus\V_{\{4\}}^{(S^c)}$; restricted to the four singleton-to-singleton blocks this is the concrete prediction
+\begin{equation}
+M_{\{1\}\to\{3\}}(\rho_\square)=M_{\{2\}\to\{4\}}(\rho_\square),
+\qquad
+M_{\{1\}\to\{4\}}(\rho_\square)=M_{\{2\}\to\{3\}}(\rho_\square),
+\label{eq:ring-circulant-prediction}
+\end{equation}
+i.e.\ that the $2\times2$ array of blocks is circulant --- a prediction that follows from equivariance alone, before any stabilizer computation is performed.
+
+A direct check against the stabilizer group $H=\langle K_1,K_2,K_3,K_4\rangle$ of $\rho_\square$ (Lemma~\ref{lem:code-support}) confirms this: in the raw Pauli-string basis, $X_1X_3\in H$ and $X_2X_4\in H$, while $X_1X_4\notin H$ and $X_2X_3\notin H$, so
+\begin{equation*}
+M_{\{1\}\to\{3\}}(\rho_\square)=M_{\{2\}\to\{4\}}(\rho_\square)=1\ (\text{support on the }xx\text{ direction only}),
+\qquad
+M_{\{1\}\to\{4\}}(\rho_\square)=M_{\{2\}\to\{3\}}(\rho_\square)=0,
+\end{equation*}
+matching Eq.~\eqref{eq:ring-circulant-prediction} exactly. This is a case where the weak, cut-factorizing symmetry and the combinatorial stabilizer mechanism of Lemma~\ref{lem:stabilizer-degeneracy} operate on the same numbers from two independent directions: the stabilizer mechanism explains \emph{why} the entries are $0$ or $\pm1$ at all, while the swap-equivariance of Proposition~\ref{prop:block-diagonal} explains, without reference to the stabilizer formalism, why the two nonzero entries must coincide and the two zero entries must coincide. Diagonalizing this $2\times2$ circulant block in the symmetric/antisymmetric basis $(\,e_x^{(1)}\pm e_x^{(2)})/\sqrt2$ gives eigenvalues $1\pm0=1$ in this instance --- here the two channels happen to be degenerate, since the off-diagonal entry vanishes, but the circulant form Eq.~\eqref{eq:ring-circulant-prediction} would hold with generically distinct symmetric/antisymmetric eigenvalues $a\pm b$ for any $\rho$ merely invariant under this same reflection, stabilizer or not.
+
+The ``diagonal'' cut $S=\{1,3\}$, $S^c=\{2,4\}$ of Section~\ref{sec:qubit-numerics} admits the analogous reflection $\pi'=(13)$ (fixing qubits $2,4$ individually and swapping $1,3$ within $S$), giving the same story with $S$ and $S^c$ playing asymmetric roles: here $\pi'$ swaps only within $S$, not within $S^c$, so the weak hypothesis still applies, but now with $R_{S^c}(\pi')$ trivial and $R_S(\pi')$ a nontrivial swap. Equivariance then forces the weaker, one-sided prediction $M_{\{1\}\to T}(\rho_\square)=M_{\{3\}\to T}(\rho_\square)$ for every $T\subseteq\{2,4\}$, again checkable directly against $H$.
+\end{example}
+
 \begin{lemma}[Transitive party symmetry forces cut-value equality]
 \label{lem:transitive-symmetry}
 Suppose all local dimensions coincide, $d_a=d$ for every $a\in P$, and let $H\le S_n$ act on $P$ by permuting parties via the induced permutation unitaries $U_\pi$ on $\mathcal H^{(P)}=(\mathcal H^{(1)})^{\otimes n}$. Suppose $\rho$ is $H$-invariant, $U_\pi\rho U_\pi^\dagger=\rho$ for every $\pi\in H$. Then for every $a\in P$ and $\pi\in H$,