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+Hier ist der Ersatztext für Remark~\ref{rem:no-universal-sharpening}s SU(2)-Passage — als eigenständiges Lemma mit Beweis, in eurem Stil, direkt einsetzbar.
+ 
+---
+ 
+\subsection*{Exact degeneracy from stabilizer structure (replaces the SU(2) mechanism)}
+ 
+The degeneracies recorded for $\GHZ_3$, the Smolin state, and all $38$ four-qubit graph states share no continuous symmetry — collective $SU(2)$, plain or twisted, is numerically excluded for both example states. Their common origin is instead a discrete, combinatorial fact about \emph{Pauli-diagonal} states, requiring only elementary group theory over $\mathbb F_2$.
+ 
+\paragraph{Setup.} Identify each single-party Pauli index with $\mathbb F_2^2$ via $I\mapsto(0,0)$, $X\mapsto(1,0)$, $Y\mapsto(1,1)$, $Z\mapsto(0,1)$, so that an $n$-party Pauli string $\sigma_{\vec i}$ corresponds to $\vec i\in\mathbb F_2^{2n}$, and string multiplication (up to phase) becomes addition. Let $H\le\mathbb F_2^{2n}$ be an isotropic subgroup (i.e.\ its elements pairwise commute as operators) not containing $-I$, and let
+\[
+\rho_H \;=\; \frac{\Pi_H}{\operatorname{rank}\Pi_H}, \qquad \Pi_H=\frac{1}{|H|}\sum_{h\in H} h,
+\]
+the maximally mixed state on the joint $+1$-eigenspace of $H$ (a stabilizer code state; $\rho_H$ is pure iff $|H|=2^n$).
+ 
+\begin{lemma}[Support of a stabilizer-code state]
+\label{lem:code-support}
+$\operatorname{Tr}[\rho_H\,\sigma_{\vec i}] = \mathbb 1[\vec i\in H]$ for every $\vec i\in\mathbb F_2^{2n}$.
+\end{lemma}
+\begin{proof}
+For $\vec i\in H$: $\Pi_H\sigma_{\vec i}=\Pi_H$ since $h\Pi_H=\Pi_H$ for every $h\in H$, so $\operatorname{Tr}[\Pi_H\sigma_{\vec i}]=\operatorname{Tr}[\Pi_H]=\operatorname{rank}\Pi_H$, giving $\operatorname{Tr}[\rho_H\sigma_{\vec i}]=1$. For $\vec i\notin H$: either $\sigma_{\vec i}$ anticommutes with some $h\in H$, whence $\operatorname{Tr}[\Pi_H\sigma_{\vec i}]=\operatorname{Tr}[h\Pi_H\sigma_{\vec i}]=-\operatorname{Tr}[\Pi_H\sigma_{\vec i}h]=-\operatorname{Tr}[\Pi_H\sigma_{\vec i}]$ (using $h\Pi_H=\Pi_H$ and cyclicity), forcing it to vanish; or $\sigma_{\vec i}$ commutes with all of $H$ without belonging to it (a logical operator), in which case it acts as a nonzero-weight, traceless operator on the logical subspace on which $\rho_H$ restricts to a multiple of the identity, again giving zero.
+\end{proof}
+ 
+\begin{lemma}[Forced degeneracy of $\widetilde{\mathcal M}_S(\rho_H)$]
+\label{lem:stabilizer-degeneracy}
+Let $S\mid S^c$ be a cut and let $\varphi:H\to\mathbb F_2^{2|S|}$, $\psi:H\to\mathbb F_2^{2|S^c|}$ be the two restriction homomorphisms, so that $H\hookrightarrow \mathbb F_2^{2|S|}\times\mathbb F_2^{2|S^c|}$ via $h\mapsto(\varphi(h),\psi(h))$. If $\psi$ is injective, then every nonzero row of $\widetilde{\mathcal M}_S(\rho_H)$ has exactly $|\ker\varphi|$ nonzero entries, all of magnitude $1$, with pairwise disjoint column supports across distinct rows; consequently
+\[
+\widetilde{\mathcal M}_S(\rho_H) \text{ has exactly } |\operatorname{im}\varphi|-1 \text{ equal nonzero singular values, each } =\sqrt{|\ker\varphi|}.
+\]
+(Normalization: this is in the orthonormal-Pauli convention $e_i=\sigma_i/\sqrt2$ per party; divide by $2^{n/2}$ for the convention used elsewhere in this note.)
+\end{lemma}
+\begin{proof}
+By Lemma~\ref{lem:code-support}, the entry of $\widetilde{\mathcal M}_S(\rho_H)$ at row $\vec j\in\mathbb F_2^{2|S|}\setminus\{0\}$, column $\vec k$, is $1$ if $(\vec j,\vec k)\in H$ and $0$ otherwise. For fixed $\vec j\in\operatorname{im}\varphi$, the set $\{\vec k:(\vec j,\vec k)\in H\}$ is a coset of $\ker\varphi$ under the group structure of $H$ (standard fiber property of a homomorphism), hence has size $|\ker\varphi|$, giving the row weight and (since all entries are $\pm1$ in magnitude by Lemma~\ref{lem:code-support}) equal row norm $\sqrt{|\ker\varphi|}$ for every nonzero row. If rows for $\vec j\neq\vec j'$ shared a nonzero column $\vec k$, then $(\vec j,\vec k),(\vec j',\vec k)\in H$ would give $(\vec j-\vec j',0)\in H$ with $\vec j\neq\vec j'$, i.e.\ a nontrivial element of $\ker\psi$ — excluded by injectivity of $\psi$. Rows are thus pairwise orthogonal with equal norm, hence (after normalizing) already the right singular vectors, and the singular values are all equal to the common row norm.
+\end{proof}
+ 
+\begin{corollary}
+The Lemma applies uniformly to: pure stabilizer states ($|H|=2^n$, including all graph states and $\GHZ_n$), and uniform mixtures over a stabilizer code space with $|H|<2^n$ (including the Smolin state, $H=\{IIII,XXXX,YYYY,ZZZZ\}$). Injectivity of $\psi$ holds automatically whenever $H$ contains no element supported entirely on $S$ — checkable by inspection of the generators, without any Lie-group input. For $\GHZ_3$, $S=\{1\}$: $|\ker\varphi|=2$, giving $3$ singular values equal to $\sqrt2/2^{3/2}=0.5$. For the Smolin state, every $1\mid3$ and $2\mid2$ cut: $\varphi$ is bijective ($|\ker\varphi|=1$), giving $4$ singular values equal to $1/2^{2}=0.25$ — identical across all four inequivalent cuts, since bijectivity of $\varphi$ holds for \emph{any} nonempty proper subset $S$ of the four legs given this particular $H$.
+\end{corollary}
+ 
+\begin{remark}
+The $y$-parity grading noted above is the special case $H=\{I^{\otimes n}\}$ acting trivially — more precisely, it is not itself an instance of this Lemma but a compatible, coarser $\mathbb Z_2$-grading that commutes with any $H$-decomposition and can be applied on top of it without modification.
+\end{remark}
+ 
+---
+ 
+Zwei Anmerkungen zur Einordnung, falls ihr das einbaut:
+ 
+1. Die Bedingung "$\psi$ injektiv" ist die einzige Voraussetzung, die pro Anwendungsfall geprüft werden muss — bei $n$ Erzeugern ist das eine einfache lineare-Algebra-Prüfung über $\mathbb F_2$, keine Handarbeit.
+2. Das Lemma sagt nichts über Zustände, die *nicht* Pauli-diagonal sind (unser generisches $S_3$-Beispiel von vorhin) — dort bleibt nur Proposition 1 (Trägerreduktion via Schur), ohne erzwungene Gleichheit. Die beiden Mechanismen sind komplementär, nicht konkurrierend, und das solltet ihr explizit so benennen, damit klar ist, wann welcher greift.