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\title{Symmetric Shadow Maps and Multipartite Correlation Criteria}
\author{Draft formal note}
\date{June 2026}

\begin{document}

\maketitle

\begin{abstract}
Multipartite correlation-tensor criteria for entanglement are well
established, including full-tensor unfoldings and cut-aware block
trace-norm bounds. We introduce a bigraduated shadow map
$\mathcal M_S(\rho)$ that stacks all correlation-response maps from a
source cluster $S$ into every sector of the complement, and show that
$\norm{\mathcal M_S(\rho)}_*\le1$ for states separable across $S\mid S^c$
--- a bound that recovers, rather than supersedes, existing
correlation-matrix and realignment criteria. Its value lies in the
structure the bigraduation exposes: every sub-block and sector profile
is automatically a valid witness, though we show these profiles are not
invariant under unitaries on the coarse cut; the map itself is one
matricization of a single augmented Bloch tensor, so the bound is
inherited rather than separately proved at each level; and when $\rho$
carries an exact or twirled symmetry, the map block-diagonalizes further
by representation-theoretic isotype, giving a second, combinatorial
(stabilizer-support) mechanism, together accounting for degeneracies
observed numerically throughout. We give an explicit three-qubit
biseparable threshold, detect the two-qutrit Tiles bound entangled state
despite it being numerically PPT, and show all $38$ connected
four-qubit graph states saturate a common value $6/\sqrt7$, robust to
white noise past $p\approx0.441$, undetected by any two-qubit marginal.
The four-qubit biseparable threshold remains open.
\end{abstract}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\section{From the old criterion to a new object}
\label{sec:old-criterion}

Let $P=\{1,\dots,n\}$ be a set of parties with local Hilbert spaces $\mathcal H^{(a)}$ of finite dimensions $d_a$. Following the notation of the local-invariant correlation-sector decomposition introduced by Aschauer \emph{et al.} \cite{aschauer}, choose for each party $a$ traceless Hermitian generators
\begin{equation*}
\sigma^{(a)}_1,\dots,\sigma^{(a)}_{d_a^2-1}
\end{equation*}
together with the identity $\sigma_0^{(a)}=\id$, normalized by
\begin{equation}
\tr\!\left(\sigma^{(a)}_i\sigma^{(a)}_j\right)=d_a\,\delta_{ij}
\label{eq:generator-orthogonality}
\end{equation}
for $i,j\in\{0,1,\dots,d_a^2-1\}$, so that the orthogonality relation already includes the identity generator. For each nonempty subset $S\subseteq P$ we denote by
\begin{equation}
C_S(\rho)=\bigl(c_{i_a:a\in S}\bigr)_{1\le i_a\le d_a^2-1}
\label{eq:corr-tensor-def}
\end{equation}
the $S$-body correlation tensor of the state $\rho$, where
\begin{equation}
c_{i_a:a\in S}=\tr\!\left(\rho\bigotimes_{a\in S}\sigma_{i_a}^{(a)}\right).
\label{eq:corr-tensor-entries}
\end{equation}
The old correlation strengths are
\begin{equation*}
L_S(\rho)=\norm{C_S(\rho)}_{\fro}^2.
\end{equation*}
They are local-unitary invariants and admit a clean convexity-based entanglement criterion, but they retain only the total quadratic size of each tensor block.

The point of the present note is that the old geometric picture has a natural multipartite continuation. Instead of compressing each block to one scalar, we keep the full family of source-indexed correlation-response maps produced by a chosen source party and stack the responses landing in the different orthogonal sectors of the complement. In this sense the main contribution is best viewed as a new object architecture rather than a new functional applied to a familiar full correlation tensor. The general construction and cut-separable bound below work for arbitrary finite local dimensions; the later biseparable benchmark and all numerical examples then specialize to qubits, where the constants and geometry are especially transparent.

A substantial literature already studies separability criteria in the Bloch-representation and correlation-tensor language, including bipartite correlation-matrix criteria \cite{devicente,chenwu}, multipartite unfoldings and matricizations of the full correlation tensor \cite{hassanjoag,devicentehuber,li2014,jingzhang2023}, nonlinear geometric tensor criteria \cite{laskowski2011}, scalar multi-sector norm criteria \cite{klocklhuber2015}, and more recent extended or partition-adapted mixed-order block constructions \cite{shen2016,sarbicki2020,zhao2020,huang2024extended,liyaoyangfei2025}. This note is intended to sit explicitly within that landscape rather than outside it.

Historically, it is useful to distinguish two nearby lines of work. The local-invariant sector decomposition of Aschauer \emph{et al.} \cite{aschauer} already gave an early multipartite entanglement criterion in terms of the coefficients of a local operator expansion; that expansion is carried out, from the first equations of that paper onward, in the full product basis of local generators together with the identity on each party, so the full tensor $\mathcal C(\rho)$ used in Section~\ref{sec:tensor-viewpoint} below is already present there, and the sector-restricted tensor $C_S(\rho)$ used for $L_S$ is obtained from it by exactly the restriction to nonidentity indices already carried out in that paper. Later work such as Hassan and Joag \cite{hassanjoag} made the Bloch-representation terminology explicit and developed criteria from full-tensor unfoldings. The present construction belongs to that broader correlation-tensor lineage, but its organization is closest in spirit to, and its starting tensor is literally the one already used in, \cite{aschauer}.

It is worth stating carefully what is and is not being claimed here. The recent correlation-tensor literature already contains powerful criteria based on full tensors, matricizations, Bloch tensors, and cut-aware mixed-order block trace norms. In particular, recent generalized-Bloch constructions already study multipartite cut-aware mixed-order block trace-norm criteria \cite{liyaoyangfei2025}. Thus the intended novelty claim is narrow: not the first multipartite block trace-norm criterion of this general kind, but the specific direct-sum organization obtained by fixing one source party $a$ and stacking 
\emph{all} response maps
\begin{equation*}
M_{a\to T},
\qquad
\emptyset\neq T\subseteq \bar a,
\end{equation*}
into one canonical operator $\mathcal M_a(\rho)$.

The following table makes this division explicit, separating what is already available in the cited literature from what this note adds on top of it.

\begin{table}[h]
\centering
\small
\begin{tabular}{@{}p{0.46\textwidth}p{0.46\textwidth}@{}}
\toprule
\textbf{Already established elsewhere} & \textbf{Reference} \\
\midrule
A single collapsed correlation-matrix / realignment-type nuclear-norm bound for one bipartite cut & \cite{devicente,chenwu,devicentehuber,hassanjoag} \\
Cut-aware, mixed-order block trace-norm criteria built from an extended correlation tensor & \cite{liyaoyangfei2025} \\
Averaging a bipartition-indexed trace norm over many bipartitions to obtain a genuine multipartite criterion & \cite{liyaoyangfei2025}, Theorem~5 \\
\bottomrule
\end{tabular}
\vspace{0.8em}
\begin{tabular}{@{}p{0.46\textwidth}p{0.46\textwidth}@{}}
\toprule
\textbf{New in this note} & \textbf{Where} \\
\midrule
Simultaneous bigraduation of the same cut matrix by \emph{both} source sector $V$ and target sector $T$, rather than one collapsed block & Definition~\ref{def:bigraduated} \\
Sub-block and sector-profile witnesses that this bigraduation makes available automatically, with no separate proof & Corollary~\ref{cor:sub-block}, profiles $\Phi_{a\to T}$, $\Phi^{(2)}_{a\to T}$ \\
Sector profiles are demonstrably \emph{not} invariant under unitaries acting collectively on the coarse complement, even though the full cut norm is & Proposition following Corollary~\ref{cor:projection-bound}, Bell-product/CNOT example \\
Every shadow map, at every graduation level, is one matricization of a single augmented Bloch tensor, so the $\le1$ bound is inherited rather than reproved at each level & Theorem~\ref{thm:one-fact}, Section~\ref{sec:tensor-viewpoint} \\
An explicit, fully proved three-qubit biseparable threshold in closed form & Eq.~\eqref{eq:bisep-threshold} \\
Two rigorously distinguished degeneracy mechanisms (continuous isotypic symmetry vs.\ discrete stabilizer support) explaining numerically observed singular-value degeneracies & Section~\ref{sec:symmetry-blocks} \\
A qutrit PPT-entangled benchmark and a systematic scan of $38$ four-qubit graph states and $n=3,4,5$ state families & qutrit benchmark in Section~\ref{sec:tensor-viewpoint}, Section~\ref{sec:qubit-numerics} \\
\bottomrule
\end{tabular}
\caption{What this note recovers from the existing correlation-tensor literature (top) versus what it contributes on top of that baseline (bottom). The headline bound $\norm{\mathcal M_S(\rho)}_*\le1$ itself belongs to the top half; the substantive claims of the note are the structural and numerical items in the bottom half.}
\label{tab:novelty}
\end{table}

A guiding thread throughout the note is that this architecture is not tied to a single source party. Sections~\ref{sec:response-maps}--\ref{sec:multiparty-sources} show that fixing a source \emph{cluster} $S\subseteq P$ instead of a single party costs nothing in the proof: the rank-one mechanism behind the cut-separable bound is agnostic to how many parties sit on the source side. Section~\ref{sec:tensor-viewpoint} then makes precise in what sense this is not a coincidence: every construction in this note---the single-party map $\mathcal M_a$, its cluster generalization $\mathcal M_S$, and every sub-block compression of either---is a matricization or sub-block restriction of one and the same full correlation tensor, and the bound $\le 1$ is a single rank-one fact about that tensor, inherited unchanged through each linear operation. The single-party case is simply the version of this fact that is cheapest to state first.

As a concrete demonstration that this stacked structure detects entanglement invisible to the standard PPT test, Section~\ref{sec:tensor-viewpoint} exhibits a two-qutrit state built from the Tiles unextendible product basis of \cite{bennettUPB}: it is numerically PPT to machine precision, so the Peres--Horodecki criterion is silent on it \cite{peres,horodeckiPPT}, yet the shadow-map witness detects its entanglement outright. We flag this example here because it is, in our view, the single clearest piece of evidence in the note that the construction has practical bite beyond reorganizing known bounds.

\subsection*{Notation guide}

The construction accumulates several closely related maps and scalars as it is refined step by step; Table~\ref{tab:notation} collects the main ones for reference, in the order they are introduced.

\begin{table}[h]
\centering
\small
\begin{tabular}{@{}lp{0.72\textwidth}@{}}
\toprule
Symbol & Meaning \\
\midrule
$C_S(\rho)$, $L_S(\rho)$ & sector correlation tensor and its squared Frobenius norm (Aschauer \emph{et al.} \cite{aschauer}) \\
$\widetilde{\mathcal M}_a(\rho)$ & intrinsic (basis-free) one-vs-rest response map, source party $a$ \\
$M_{a\to T}(\rho)$ & unnormalized response block landing in target sector $T\subseteq\bar a$ \\
$\mathcal M_a(\rho)$ & combined, normalized shadow map for a single source party $a$ (Definition~\ref{def:combined-shadow}) \\
$\widehat M_{a\to T}(\rho)$ & sector shadow map, normalized independently of the other sectors \\
$\Phi_{a\to T}(\rho)$, $\Phi^{(2)}_{a\to T}(\rho)$ & sector nuclear and Frobenius shadow profiles, $\norm{\widehat M_{a\to T}}_*$ and $\norm{\widehat M_{a\to T}}_\fro^2$ \\
$\Phi_a^{(\le k)}(\rho)$, $\Phi_a^{(\ge k)}(\rho)$ & nuclear norm after projecting onto grouped target sectors of order $\le k$ or $\ge k$ \\
$\mathcal M_S(\rho)$ & bigraduated shadow map for a source \emph{cluster} $S$ (Definition~\ref{def:bigraduated}); reduces to $\mathcal M_a(\rho)$ when $S=\{a\}$ \\
$M_{V\to T}(\rho)$ & bigraduated block from source sector $V\subseteq S$ to target sector $T\subseteq S^c$ \\
$\Phi_{\mathrm{sym}}(\rho)$, $\Phi_{\max}(\rho)$ & source-aggregated functionals, average and maximum of $\norm{\mathcal M_a(\rho)}_*$ over $a\in P$ \\
$\mathcal C(\rho)$ & full order-$n$ Bloch tensor of which every map above is a matricization or slice (Definition~\ref{def:full-tensor}) \\
$A_\lambda$ & reduced shadow map on the multiplicity space of isotype $\lambda$, once $\rho$ carries a compatible symmetry (Section~\ref{sec:symmetry-blocks}) \\
\bottomrule
\end{tabular}
\caption{Main notation, in order of introduction. $\Phi_{\mathrm{sym}}$, $\Phi_{\max}$, and the biseparable threshold are qubit-specific; everything above them in the table is defined for arbitrary finite local dimensions.}
\label{tab:notation}
\end{table}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\section{One-vs-rest response maps (shadow maps)}
\label{sec:response-maps}

Fix a party $a\in P$, and write $\bar a=P\setminus\{a\}$. For each party $b$, let $\V^{(b)}$ be the real Hilbert space of Hermitian operators on $\mathcal H^{(b)}$, equipped with the Hilbert-Schmidt inner product, and let
\begin{equation*}
\V_0^{(b)}:=\vrspan\{\sigma_1^{(b)},\dots,\sigma_{d_b^2-1}^{(b)}\}
\end{equation*}
be the traceless local subspace. Then the complement operator space
\begin{equation*}
\V^{(\bar a)}:=\bigotimes_{b\neq a}\V^{(b)}
\subseteq \mathrm{Herm}(\mathcal H^{(\bar a)})
\end{equation*}
has the orthogonal decomposition
\begin{equation}
\V^{(\bar a)}
=
\V_{\emptyset}^{(\bar a)}
\oplus
\bigoplus_{\emptyset\neq T\subseteq \bar a}^{\perp}\V_T^{(\bar a)},
\label{eq:complement-bloch-decomposition}
\end{equation}
where $\V_{\emptyset}^{(\bar a)}=\vrspan\{\id_{\bar a}\}$ and, for nonempty $T\subseteq \bar a$,
\begin{equation*}
\V_T^{(\bar a)}
:=
\vrspan\Bigl\{
\bigotimes_{b\neq a}\sigma^{(b)}_{i_b}
:
i_b\neq 0 \iff b\in T
\Bigr\}.
\end{equation*}
Thus
\begin{equation*}
\V_0^{(\bar a)}
=
\bigoplus_{\emptyset\neq T\subseteq \bar a}^{\perp}\V_T^{(\bar a)}
\end{equation*}
is the traceless complement space, resolved into orthogonal correlation sectors indexed by the nonempty subsets of $\bar a$.

The natural one-vs-rest response object is therefore the intrinsic linear map
\begin{equation*}
\widetilde{\mathcal M}_a(\rho):\V_0^{(a)}\to \V_0^{(\bar a)}
\end{equation*}
defined by
\begin{equation}
\langle Y,\widetilde{\mathcal M}_a(\rho)X\rangle
=
\tr\!\bigl(\rho(X\otimes Y)\bigr),
\qquad
X\in \V_0^{(a)},
\quad
Y\in \V_0^{(\bar a)}.
\label{eq:intrinsic-response}
\end{equation}
For each nonempty subset $T\subseteq \bar a$, let
\begin{equation*}
P_T:\V_0^{(\bar a)}\to \V_T^{(\bar a)}
\end{equation*}
be the orthogonal projection. The sector response maps are then simply the projected components
\begin{equation*}
\widetilde M_{a\to T}(\rho):=P_T\widetilde{\mathcal M}_a(\rho).
\end{equation*}
After choosing orthonormal bases in $\V_0^{(a)}$ and in each $\V_T^{(\bar a)}$, these become the coordinate maps used below. We will use the shorter term \emph{shadow maps} for this family. We emphasize that this usage is unrelated to the classical-shadows measurement protocols of \cite{huangkuengpreskill2020}: the shadow maps of this note are linear response operators built directly from the correlation tensor of $\rho$, not estimators reconstructed from randomized single-copy measurements.

\begin{definition}
\label{def:combined-shadow}
For each party $a$, define the combined shadow space
\begin{equation*}
\mathcal W_a:=\bigoplus_{\emptyset\neq T\subseteq\bar a}\R^{\prod_{b\in T}(d_b^2-1)}
\end{equation*}
and the normalized direct-sum response operator
\begin{equation}
\mathcal M_a(\rho)
:=
\frac{1}{\sqrt{(d_a-1)(d_{\bar a}-1)}}
\bigoplus_{\emptyset\neq T\subseteq\bar a}M_{a\to T}(\rho).
\label{eq:combined-map}
\end{equation}
where
\begin{equation*}
d_{\bar a}:=\prod_{b\neq a} d_b.
\end{equation*}
Under the orthogonal decomposition in Eq.~\eqref{eq:complement-bloch-decomposition}, this is just the matrix representation of the intrinsically defined map $\widetilde{\mathcal M}_a(\rho)$ in sector-adapted orthonormal coordinates. We refer to this direct-sum response operator as the \emph{combined shadow map}. The image of the unit ball in $\R^{d_a^2-1}$ under $\mathcal M_a(\rho)$ is the corresponding response ellipsoid in correlation space, which we also call the combined shadow ellipsoid of the party $a$.
\end{definition}

For qubits this reduces to the earlier normalization, since $(d_a-1)(d_{\bar a}-1)=2^{n-1}-1$. In particular, for three qubits this direct sum is
\begin{equation*}
\mathcal W_A=\R^3\oplus\R^3\oplus\R^9,
\end{equation*}
corresponding to the $B$, $C$, and $BC$ response sectors.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\section{Cut-separable states}
\label{sec:cut-separable}
\subsection*{The cut-separable bound and its refinements}

The key fact is easiest to see first for states that are product across $a\mid\bar a$: then the full response operator is rank one, and the sector maps are simply its orthogonal components. The general cut-separable case follows by convexity.

\begin{theorem}
\label{thm:cut-bound}    
Let $\rho$ be separable across the cut $a\mid\bar a$. Then
\begin{equation}
\norm{\mathcal M_a(\rho)}_*\le 1.
\label{eq:cut-bound}
\end{equation}
Consequently,
\begin{equation*}
\norm{\mathcal M_a(\rho)}_*>1
\qquad\Longrightarrow\qquad
\rho\text{ is entangled across }a\mid\bar a.
\end{equation*}
\end{theorem}

\begin{proof}
It is enough to begin with a product state across the cut,
\begin{equation*}
\rho=\rho_a\otimes\sigma_{\bar a}.
\end{equation*}
Let $r^{(a)}\in\V_0^{(a)}$ be the Bloch vector of $\rho_a$, defined by
\begin{equation*}
\langle X,r^{(a)}\rangle=\tr(\rho_a X),
\qquad X\in \V_0^{(a)},
\end{equation*}
and let $v_{\bar a}\in\V_0^{(\bar a)}$ be the traceless Bloch vector of $\sigma_{\bar a}$, defined analogously. Then
\begin{equation*}
\widetilde{\mathcal M}_a(\rho)=v_{\bar a}(r^{(a)})^T
\end{equation*}
is rank one, and in sector-adapted coordinates this becomes the direct sum of the component maps. Equivalently, for each nonempty $T\subseteq\bar a$ let
\begin{equation*}
v_T:=C_T(\sigma_{\bar a})\in\R^{\prod_{b\in T}(d_b^2-1)}.
\end{equation*}
Then tensor factorization gives
\begin{equation*}
C_{\{a\}\cup T}(\rho)=r^{(a)}\otimes v_T,
\end{equation*}
and hence
\begin{equation*}
M_{a\to T}(\rho)=v_T\,(r^{(a)})^T.
\end{equation*}
Therefore the combined map is rank one:
\begin{equation*}
\mathcal M_a(\rho)=
\frac{1}{\sqrt{(d_a-1)(d_{\bar a}-1)}}
\left(\bigoplus_{\emptyset\neq T\subseteq\bar a}v_T\right)(r^{(a)})^T.
\end{equation*}
Its nuclear norm is the product of the Euclidean norms of the two factors,
\begin{equation*}
\norm{\mathcal M_a(\rho)}_*=
\frac{\norm{r^{(a)}}}{\sqrt{(d_a-1)(d_{\bar a}-1)}}
\left(\sum_{\emptyset\neq T\subseteq\bar a}\norm{v_T}^2\right)^{1/2}.
\end{equation*}
By the original correlation-strength bound for one-party sectors, one has
\begin{equation*}
\norm{r^{(a)}}^2=d_a\tr(\rho_a^2)-1\le d_a-1.
\end{equation*}
Moreover,
\begin{equation*}
\sum_{\emptyset\neq T\subseteq\bar a}\norm{v_T}^2
=
\sum_{\emptyset\neq T\subseteq\bar a}L_T(\sigma_{\bar a})
=d_{\bar a}\tr(\sigma_{\bar a}^2)-1
\le d_{\bar a}-1,
\end{equation*}
using the correlation-sum identity from the Aschauer \emph{et al.} framework for the $(n-1)$-party state $\sigma_{\bar a}$. Thus Eq.~\eqref{eq:cut-bound} holds for every product state across the cut.

Now let $\rho$ be mixed and separable across the cut,
\begin{equation*}
\rho=\sum_l p_l\,\rho_{a,l}\otimes\sigma_{\bar a,l}.
\end{equation*}
The map $\rho\mapsto\mathcal M_a(\rho)$ is linear, and the nuclear norm is convex, so
\begin{equation*}
\norm{\mathcal M_a(\rho)}_*
\le
\sum_l p_l\,\norm{\mathcal M_a(\rho_{a,l}\otimes\sigma_{\bar a,l})}_*
\le \sum_l p_l=1.
\end{equation*}
This proves the theorem.
\end{proof}

\begin{corollary}
\label{cor:projection-bound}
Let $\Pi$ be any orthogonal projection on $\V_0^{(\bar a)}$, and let $\Pi\mathcal M_a(\rho)$ denote the corresponding projected map in any orthonormal coordinates adapted to the decomposition of $\V_0^{(\bar a)}$. If $\rho$ is separable across the cut $a\mid\bar a$, then
\begin{equation}
\norm{\Pi\mathcal M_a(\rho)}_*\le 1.
\label{eq:projection-bound}
\end{equation}
Consequently, every orthogonally selected target subspace of $\V_0^{(\bar a)}$ yields a valid cut witness.
\end{corollary}

\begin{proof}
Orthogonal projection is contractive for the operator norm and hence for singular values. Therefore
\begin{equation*}
\norm{\Pi\mathcal M_a(\rho)}_*\le \norm{\mathcal M_a(\rho)}_*.
\end{equation*}
The claim follows from Theorem~\ref{thm:cut-bound}.
\end{proof}

\begin{remark}
Equation~\eqref{eq:projection-bound} produces a whole hierarchy of weaker but natural cut witnesses. Besides the individual sectors $\Pi=P_T$, one can project onto grouped sector subspaces. For instance, for $1\le k\le |\bar a|$ let
\begin{equation*}
\Pi_a^{(\le k)}:=\sum_{\substack{\emptyset\neq T\subseteq \bar a\\ |T|\le k}}P_T,
\qquad
\Pi_a^{(\ge k)}:=\sum_{\substack{\emptyset\neq T\subseteq \bar a\\ |T|\ge k}}P_T,
\end{equation*}
and define
\begin{equation*}
\Phi_a^{(\le k)}(\rho):=\norm{\Pi_a^{(\le k)}\mathcal M_a(\rho)}_*,
\qquad
\Phi_a^{(\ge k)}(\rho):=\norm{\Pi_a^{(\ge k)}\mathcal M_a(\rho)}_*,
\end{equation*}
Then every state separable across $a\mid\bar a$ satisfies
\begin{equation*}
\Phi_a^{(\le k)}(\rho)\le 1,
\qquad
\Phi_a^{(\ge k)}(\rho)\le 1,
\end{equation*}
and one has the monotone chains
\begin{equation*}
\Phi_a^{(\le 1)}(\rho)\le \Phi_a^{(\le 2)}(\rho)\le \cdots \le \Phi_a^{(\le |\bar a|)}(\rho)=\norm{\mathcal M_a(\rho)}_*,
\end{equation*}
\begin{equation*}
\Phi_a^{(\ge |\bar a|)}(\rho)\le \Phi_a^{(\ge |\bar a|-1)}(\rho)\le \cdots \le \Phi_a^{(\ge 1)}(\rho)=\norm{\mathcal M_a(\rho)}_*,
\end{equation*}
Thus the combined shadow map is the top element of a nested family of projection-based witnesses rather than an isolated stacked object.
\end{remark}

\begin{corollary}
For each nonempty subset $T\subseteq\bar a$, let
\begin{equation*}
d_T:=\prod_{b\in T} d_b
\end{equation*}
and define the normalized sector shadow map
\begin{equation*}
\widehat M_{a\to T}(\rho)
:=
\frac{1}{\sqrt{(d_a-1)(d_T-1)}}M_{a\to T}(\rho).
\end{equation*}
If $\rho$ is separable across the cut $a\mid\bar a$, then
\begin{equation*}
\norm{\widehat M_{a\to T}(\rho)}_*\le 1.
\end{equation*}
Consequently,
\begin{equation*}
\norm{\widehat M_{a\to T}(\rho)}_*>1
\qquad\Longrightarrow\qquad
\rho\text{ is entangled across }a\mid\bar a.
\end{equation*}
The same normalization also gives the weaker sectorwise Frobenius bound
\begin{equation*}
\norm{\widehat M_{a\to T}(\rho)}_{\fro}^2\le 1.
\end{equation*}
\end{corollary}

\begin{proof}
For a product state across the cut,
\begin{equation*}
\rho=\rho_a\otimes\sigma_{\bar a},
\end{equation*}
the proof of Theorem~\ref{thm:cut-bound} gives
\begin{equation*}
M_{a\to T}(\rho)=v_T\,(r^{(a)})^T,
\end{equation*}
where $r^{(a)}$ is the one-body correlation vector of $\rho_a$ and
\begin{equation*}
v_T=C_T(\sigma_{\bar a}).
\end{equation*}
Hence
\begin{equation*}
\norm{M_{a\to T}(\rho)}_* = \norm{r^{(a)}}\,\norm{v_T}.
\end{equation*}
Now $C_T(\sigma_{\bar a})$ depends only on the reduced state $\sigma_T$, so
\begin{equation*}
\norm{v_T}^2 = L_T(\sigma_T)
\le
\sum_{\emptyset\neq U\subseteq T}L_U(\sigma_T)
= d_T\tr(\sigma_T^2)-1
\le d_T-1.
\end{equation*}
Together with $\norm{r^{(a)}}^2\le d_a-1$, this proves the product-state case. The mixed separable case again follows by linearity and convexity, and the Frobenius statement follows from $\norm{X}_{\fro}\le \norm{X}_*$.
\end{proof}

\begin{definition}
For each nonempty subset $T\subseteq\bar a$, define the sector nuclear shadow profile by
\begin{equation*}
\Phi_{a\to T}(\rho):=\norm{\widehat M_{a\to T}(\rho)}_*,
\end{equation*}
and the corresponding sector Frobenius profile by
\begin{equation*}
\Phi^{(2)}_{a\to T}(\rho):=\norm{\widehat M_{a\to T}(\rho)}_{\fro}^2.
\end{equation*}
\end{definition}

\begin{remark}
The combined map is exactly the dimension-weighted orthogonal assembly of its normalized sectors. Let
\begin{equation*}
\iota_T:
\R^{\prod_{b\in T}(d_b^2-1)}\hookrightarrow\mathcal W_a
\end{equation*}
denote the canonical inclusion of the $T$-sector. Then
\begin{equation}
\mathcal M_a(\rho)
=
\sum_{\emptyset\neq T\subseteq\bar a}
\sqrt{\frac{d_T-1}{d_{\bar a}-1}}\;\iota_T\,\widehat M_{a\to T}(\rho).
\label{eq:weighted-sector-reconstruction}
\end{equation}
Since the target sectors are orthogonal,
\begin{equation*}
\mathcal M_a(\rho)^T\mathcal M_a(\rho)
=
\sum_{\emptyset\neq T\subseteq\bar a}
\frac{d_T-1}{d_{\bar a}-1}
\widehat M_{a\to T}(\rho)^T\widehat M_{a\to T}(\rho).
\end{equation*}
Taking traces gives the weighted Frobenius additivity formula
\begin{equation}
\norm{\mathcal M_a(\rho)}_{\fro}^2
=
\sum_{\emptyset\neq T\subseteq\bar a}
\frac{d_T-1}{d_{\bar a}-1}
\norm{\widehat M_{a\to T}(\rho)}_{\fro}^2,
\label{eq:frobenius-sector-additivity}
\end{equation}
so the previous Frobenius criterion is just the corresponding weighted sum of the sectorwise Frobenius criteria. By contrast,
\begin{equation*}
\norm{\mathcal M_a(\rho)}_*
=
\tr\sqrt{
\sum_{\emptyset\neq T\subseteq\bar a}
\frac{d_T-1}{d_{\bar a}-1}
\widehat M_{a\to T}(\rho)^T\widehat M_{a\to T}(\rho)},
\end{equation*}
which depends not only on the sizes of the individual sector maps but also on how their right-singular directions align in the common source space. Thus the full nuclear-norm signal is not, in general, a linear combination of the individual sector nuclear norms. In the cut-product case all sector maps share one common right factor, so the full map is again rank one and the proof of Theorem~\ref{thm:cut-bound} reduces to one Euclidean bound on the stacked target vector.
\end{remark}

\subsection*{Sector profiles are not cut invariants}
\begin{proposition}
Fix a source party $a$, and let
\begin{equation*}
\rho'=(U_a\otimes U_{\bar a})\rho(U_a^{\dagger}\otimes U_{\bar a}^{\dagger})
\end{equation*}
for unitaries $U_a$ on $\mathcal H^{(a)}$ and $U_{\bar a}$ on $\mathcal H^{(\bar a)}$. Then there exist orthogonal transformations
\begin{equation*}
O_a(U_a)\in O(d_a^2-1),
\qquad
O_{\bar a}(U_{\bar a})\in O(d_{\bar a}^2-1)
\end{equation*}
such that the full response map for the cut $a\mid\bar a$ transforms by
\begin{equation}
\mathcal M_a(\rho')=O_{\bar a}(U_{\bar a})\,\mathcal M_a(\rho)\,O_a(U_a)^T.
\label{eq:two-sided-collective-covariance}
\end{equation}
Consequently,
\begin{equation*}
\norm{\mathcal M_a(\rho')}_* = \norm{\mathcal M_a(\rho)}_*,
\qquad
\norm{\mathcal M_a(\rho')}_{\fro} = \norm{\mathcal M_a(\rho)}_{\fro}.
\end{equation*}
The sector profiles $\Phi_{a\to T}(\rho)$ and $\Phi^{(2)}_{a\to T}(\rho)$ are therefore not invariants of the coarse cut $a\mid\bar a$ under general collective complement unitaries; they are invariants only under unitaries that preserve the chosen internal factorization of $\bar a$ into parties.
\end{proposition}

\begin{proof}
Choose orthonormal bases of traceless Hermitian operators on $\mathcal H^{(a)}$ and $\mathcal H^{(\bar a)}$, normalized by
\begin{equation*}
\tr(\sigma_i\sigma_j)=d_a\,\delta_{ij},
\qquad
\tr(\tau_i\tau_j)=d_{\bar a}\,\delta_{ij}.
\end{equation*}
The intrinsic response map transforms by the adjoint actions on source and target Bloch spaces:
\begin{equation*}
\widetilde{\mathcal M}_a(\rho')=\mathrm{Ad}_{U_{\bar a}}\,\widetilde{\mathcal M}_a(\rho)\,\mathrm{Ad}_{U_a}^{\,T}.
\end{equation*}
In the chosen orthonormal bases these adjoint actions are represented by real orthogonal matrices $O_a(U_a)$ and $O_{\bar a}(U_{\bar a})$, giving Eq.~\eqref{eq:two-sided-collective-covariance}. Left and right multiplication by orthogonal matrices preserve both nuclear and Frobenius norms, so the norm equalities follow.

The sector maps arise only after choosing the product operator basis on $\mathcal H^{(\bar a)}$ determined by the internal decomposition of $\bar a$ into parties and then splitting that basis into orthogonal summands. A general collective unitary on $\bar a$ need not preserve those summands, so it reshuffles the sector profile even though the full cut norm is unchanged.
\end{proof}

\begin{remark}
The failure of sector invariance is already visible in the simplest three-qubit Bell-product example. Let
\begin{equation*}
\rho
=
\lvert\Phi^+\rangle_{AB}\!\langle\Phi^+\rvert\otimes \lvert 0\rangle_C\!\langle 0\rvert,
\qquad
\lvert\Phi^+\rangle=
\frac{\lvert 00\rangle+\lvert 11\rangle}{\sqrt2},
\end{equation*}
with $A$ as source. A direct calculation gives
\begin{equation*}
\Phi_{A\to B}(\rho)=3,
\qquad
\Phi_{A\to C}(\rho)=0,
\qquad
\Phi_{A\to BC}(\rho)=\sqrt 3,
\qquad
\norm{\mathcal M_A(\rho)}_*=\sqrt 6.
\end{equation*}
Now apply the collective unitary
\begin{equation*}
U_{BC}=\mathrm{CNOT}_{B\to C}(H_B\otimes \id_C),
\qquad
\rho'=(I_A\otimes U_{BC})\rho(I_A\otimes U_{BC}^{\dagger}).
\end{equation*}
Then Eq.~\eqref{eq:two-sided-collective-covariance} implies
\begin{equation*}
\norm{\mathcal M_A(\rho')}_*=\norm{\mathcal M_A(\rho)}_*=\sqrt 6,
\end{equation*}
while the sector profile becomes
\begin{equation*}
\Phi_{A\to B}(\rho')=1,
\qquad
\Phi_{A\to C}(\rho')=1,
\qquad
\Phi_{A\to BC}(\rho')=\sqrt{\frac83}.
\end{equation*}
So a collective unitary on $BC$ does not push the signal purely into the highest-order sector. Instead it redistributes a strongly localized $A\to B$ witness into a mixed profile spread across $A\to B$, $A\to C$, and $A\to BC$, while leaving the full $A\mid BC$ witness unchanged.

This persists under white noise. For
\begin{equation*}
\rho_p=p\rho+(1-p)\frac{\id}{8},
\qquad
\rho'_p=p\rho'+(1-p)\frac{\id}{8},
\end{equation*}
the full-map threshold is the same in both forms,
\begin{equation*}
\norm{\mathcal M_A(\rho_p)}_*>1
\iff
\norm{\mathcal M_A(\rho'_p)}_*>1
\iff
p>\frac{1}{\sqrt 6}.
\end{equation*}
But the sector thresholds differ sharply: before the collective rotation the sector $A\to B$ already detects for $p>1/3$, whereas after the rotation the sectors $A\to B$ and $A\to C$ never strictly violate the cut-separable bound and the sector $A\to BC$ only detects for $p>\sqrt{3/8}$. In particular, at $p=0.60$ one has $\norm{\mathcal M_A(\rho'_p)}_*>1$ while all three individual sectors still satisfy $\Phi_{A\to T}(\rho'_p)\le 1$.
\end{remark}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\section{Multi-party sources: bigraduated shadow maps}
\label{sec:multiparty-sources}

The construction of Section~\ref{sec:response-maps} singles out one party $a$ as source and treats the entire complement $\bar a$ as target. Nothing in the argument in fact requires $|S|=1$ on the source side; the same object exists for any source cluster $\emptyset\neq S\subsetneq P$, with complement $S^c:=P\setminus S$. Making this explicit exposes a layer of internal structure on the source side that Definition~\ref{def:combined-shadow} discards by construction, and it costs nothing beyond re-reading the definitions and the proof of Theorem~\ref{thm:cut-bound} with $a$ replaced by $S$.

\subsection*{Source sector decomposition}

For nonempty $S\subseteq P$, set $\V^{(S)}:=\bigotimes_{a\in S}\V^{(a)}$ and apply exactly the decomposition of Eq.~\eqref{eq:complement-bloch-decomposition}, now with $S$ in the role previously played by $\bar a$:
\begin{equation}
\V^{(S)}
=
\V_\emptyset^{(S)}
\oplus
\bigoplus_{\emptyset\neq V\subseteq S}^\perp \V_V^{(S)}.
\label{eq:source-bloch-decomposition}
\end{equation}
This is not a new construction, only the source-side instance of the same orthogonal sector decomposition already used for the complement. The traceless source space is accordingly
\begin{equation*}
\V_0^{(S)}=\bigoplus_{\emptyset\neq V\subseteq S}^\perp \V_V^{(S)},
\end{equation*}
graded by which parties within $S$ are active. For $S=\{a\}$ the only nonempty $V\subseteq S$ is $V=S$ itself, so this decomposition is trivial for a singleton source; the bigraduation below is genuinely new structure only once $|S|\ge 2$.

\subsection*{Bigraduated shadow maps}

\begin{definition}[Combined bigraduated shadow map]
\label{def:bigraduated}
For nonempty $S\subsetneq P$ with complement $S^c$, the intrinsic source-to-complement response operator
\begin{equation*}
\widetilde{\mathcal M}_S(\rho):\V_0^{(S)}\to\V_0^{(S^c)},
\qquad
\langle Y,\widetilde{\mathcal M}_S(\rho)X\rangle=\tr\!\bigl(\rho(X\otimes Y)\bigr),
\end{equation*}
is given by the same formula as Eq.~\eqref{eq:intrinsic-response}, with $a$ replaced by $S$. Both $\V_0^{(S)}$ and $\V_0^{(S^c)}$ carry an orthogonal sector decomposition, Eq.~\eqref{eq:source-bloch-decomposition} on the source side and Eq.~\eqref{eq:complement-bloch-decomposition} on the target side, so the coordinate representation of $\widetilde{\mathcal M}_S(\rho)$ is naturally \emph{bigraded} by source sector $V\subseteq S$ and target sector $T\subseteq S^c$ simultaneously. Writing $\iota_V:\V_V^{(S)}\hookrightarrow\V_0^{(S)}$ for the canonical inclusion of a source sector and $P_T:\V_0^{(S^c)}\to\V_T^{(S^c)}$ for the orthogonal projection onto a target sector, define
\begin{equation*}
M_{V\to T}(\rho):=P_T\,\widetilde{\mathcal M}_S(\rho)\,\iota_V.
\end{equation*}
The \emph{combined bigraduated shadow map} is the normalized direct sum of all such blocks,
\begin{equation}
\mathcal M_S(\rho)
:=
\frac{1}{\sqrt{(d_S-1)(d_{S^c}-1)}}
\bigoplus_{\substack{\emptyset\neq V\subseteq S\\ \emptyset\neq T\subseteq S^c}}
M_{V\to T}(\rho),
\qquad
d_S:=\prod_{a\in S}d_a,
\label{eq:bigraduated-map}
\end{equation}
i.e.\ the matrix representation of $\widetilde{\mathcal M}_S(\rho)$, normalized exactly as in Eq.~\eqref{eq:combined-map}, in sector-adapted orthonormal coordinates on both sides.
\end{definition}

For $S=\{a\}$, Definition~\ref{def:bigraduated} reduces exactly to Definition~\ref{def:combined-shadow}: the source-side decomposition then has only the single summand $V=S$, so the bigraduation collapses to the ordinary target-only graduation of $\mathcal M_a$. Definition~\ref{def:combined-shadow} is thus the singleton case of this construction, not a separate object introduced in parallel to it.

\begin{remark}[Computational cost]
\label{rem:computational-cost}
The total number of scalar entries in $\mathcal M_S(\rho)$ is $(d_S-1)(d_{S^c}-1)$, exactly as for the collapsed cut matrix, so evaluating the norm $\norm{\mathcal M_S(\rho)}_*$ in Theorem~\ref{thm:cluster-cut} costs a single singular value decomposition of that size and is no more expensive than the realignment-type bounds it recovers. The bigraduation of Definition~\ref{def:bigraduated} does not change this cost; it only reorganizes the same entries into $(2^{|S|}-1)(2^{|S^c|}-1)$ combinatorial blocks $M_{V\to T}$. Consequently, using the full map $\mathcal M_S(\rho)$ as a single witness remains cheap, but exploiting the sub-block witnesses of Corollary~\ref{cor:sub-block} exhaustively --- inspecting every source sector against every target sector separately, rather than only the handful used in the examples below --- requires examining up to $O(2^{|S|+|S^c|})=O(2^n)$ individual blocks in the worst case. Section~\ref{sec:symmetry-blocks} shows that when $\rho$ carries a compatible symmetry, this exponential proliferation of combinatorial blocks collapses instead onto a typically much smaller number of representation-theoretic multiplicity spaces, giving a tractable route to the same sub-block information without enumerating every sector by hand.
\end{remark}

\begin{theorem}[Cluster cut-separable bound]
\label{thm:cluster-cut}
Let $\rho$ be separable across the cut $S\mid S^c$. Then
\begin{equation}
\norm{\mathcal M_S(\rho)}_*\le1.
\label{eq:cluster-cut-bound}
\end{equation}
Consequently, $\norm{\mathcal M_S(\rho)}_*>1$ implies that $\rho$ is entangled across $S\mid S^c$.
\end{theorem}

\begin{proof}
The proof of Theorem~\ref{thm:cut-bound} goes through with $a\to S$ and $\bar a\to S^c$ without modification. For a product state $\rho=\rho_S\otimes\sigma_{S^c}$, let $r^{(S)}\in\V_0^{(S)}$ be the traceless Bloch vector of the $|S|$-party reduced state $\rho_S$ -- the same object as $r^{(a)}$ in the proof of Theorem~\ref{thm:cut-bound}, now for the composite system $S$ treated as a single $d_S$-dimensional party -- and let $v_{S^c}\in\V_0^{(S^c)}$ be defined analogously from $\sigma_{S^c}$. Tensor factorization gives $\widetilde{\mathcal M}_S(\rho)=v_{S^c}(r^{(S)})^T$, which is rank one, and
\begin{equation*}
\norm{r^{(S)}}^2=d_S\tr(\rho_S^2)-1\le d_S-1,
\qquad
\norm{v_{S^c}}^2=d_{S^c}\tr(\sigma_{S^c}^2)-1\le d_{S^c}-1,
\end{equation*}
by the same correlation-sum identity used in the proof of Theorem~\ref{thm:cut-bound}, now applied to the $|S|$-party state $\rho_S$ and the $|S^c|$-party state $\sigma_{S^c}$ rather than to single-party marginals. Hence $\norm{\mathcal M_S(\rho)}_*\le1$ for every product state across the cut, and convexity of the nuclear norm extends the bound to mixtures exactly as in the proof of Theorem~\ref{thm:cut-bound}.
\end{proof}

\begin{corollary}[Sub-block witnesses]
\label{cor:sub-block}
Let $\mathcal V\subseteq\{V:\emptyset\neq V\subseteq S\}$ and $\mathcal T\subseteq\{T:\emptyset\neq T\subseteq S^c\}$ be any nonempty families, and let $P_{\mathcal V}$, $P_{\mathcal T}$ be the orthogonal projections onto $\bigoplus_{V\in\mathcal V}\V_V^{(S)}$ and $\bigoplus_{T\in\mathcal T}\V_T^{(S^c)}$ respectively. If $\rho$ is separable across $S\mid S^c$, then
\begin{equation}
\norm{P_{\mathcal T}\,\mathcal M_S(\rho)\,P_{\mathcal V}}_*\le\norm{\mathcal M_S(\rho)}_*\le1,
\label{eq:sub-block-bound}
\end{equation}
with no separate proof required.
\end{corollary}

\begin{proof}
Orthogonal projections are contractions for the operator norm, and the nuclear norm satisfies $\norm{AXB}_*\le\norm{A}_{\mathrm{op}}\norm{X}_*\norm{B}_{\mathrm{op}}$ for linear maps $A,B$ of compatible size. Taking $A=P_{\mathcal T}$ and $B=P_{\mathcal V}$ gives Eq.~\eqref{eq:sub-block-bound}; the claim then follows from Theorem~\ref{thm:cluster-cut}. This is the two-sided extension of Corollary~\ref{cor:projection-bound}, which only ever compressed the target side.
\end{proof}

In particular, taking $\mathcal V=\{S\}$ isolates the single block $M_{S\to T}$, which carries only the correlation attributable to the full cluster $S$ acting jointly rather than to any proper sub-cluster of $S$ -- a witness targeted specifically at ``genuine $S$'' structure landing in the sector $T$.

\subsection*{Example: the Smolin state under two different cuts}
\label{ex:smolin}

Consider the four-qubit Smolin state
\begin{equation*}
\rho_{\mathrm{Smo}}=\frac1{16}\left(\id^{\otimes 4}+X^{\otimes 4}+Y^{\otimes 4}+Z^{\otimes 4}\right),
\end{equation*}
where here and in what follows for qubits $X,Y,Z$ denote the Pauli matrices, i.e.\ $\sigma_1=X$, $\sigma_2=Y$, $\sigma_3=Z$ in the generator convention fixed formally in Section~\ref{sec:tensor-viewpoint} below. We evaluate it under two cuts side by side.

\emph{The $2\mid2$ cut.} Take the cluster $S=\{A,B\}$, $S^c=\{C,D\}$. Every block $M_{V\to T}$ with $V\neq S$ or $T\neq S^c$ vanishes identically, because the Smolin state carries no one- or three-body correlations. Only $M_{S\to S^c}$ survives, as the $9\times9$ matrix diagonal on the aligned Pauli directions $(x,x)\to(x,x)$, $(y,y)\to(y,y)$, $(z,z)\to(z,z)$ with unit coefficients and zero elsewhere. After the normalization $1/\sqrt{(d_S-1)(d_{S^c}-1)}=1/3$ this gives three singular values of $1/3$ each, so
\begin{equation*}
\norm{\mathcal M_{AB}(\rho_{\mathrm{Smo}})}_*=1
\quad\text{exactly --- saturating, not violating, the bound of Theorem~\ref{thm:cluster-cut}.}
\end{equation*}

\emph{The $1\mid3$ cut.} Take instead a single source party $a$, so $\bar a$ is the remaining three-qubit cluster. Again only the full four-body sector survives, giving three orthogonal response directions with unnormalized singular value $1$ each. The normalization is now $1/\sqrt{(d_a-1)(d_{\bar a}-1)}=1/\sqrt{1\cdot7}=1/\sqrt7$, so
\begin{equation*}
\norm{\mathcal M_a(\rho_{\mathrm{Smo}})}_* = \frac{3}{\sqrt7}\approx1.134>1,
\end{equation*}
strictly violating the bound of Theorem~\ref{thm:cut-bound}.

So the same state sits exactly at the boundary for every $2\mid2$ cut while clearly violating the $1\mid3$ bound --- the two cut types are correctly told apart within one framework. The refinement from singleton to cluster sources costs nothing in the proof yet makes this distinction available at all, since the singleton construction of Definition~\ref{def:combined-shadow} cannot even pose the $2\mid2$ question. Section~\ref{sec:qubit-numerics} below returns to the $1\mid3$ value in the source-aggregated language of $\Phi_{\mathrm{sym}}$ and $\Phi_{\max}$, and adds the white-noise robustness threshold.


%% \subsection*{Example: the Smolin state under a $2\mid2$ cut}
%% 
%% Consider the four-qubit Smolin state
%% \begin{equation*}
%% \rho_{\mathrm{Smo}}=\frac1{16}\left(\id^{\otimes 4}+X^{\otimes 4}+Y^{\otimes 4}+Z^{\otimes 4}\right)
%% \end{equation*}
%% (examined again from the single-party viewpoint in Section~\ref{sec:qubit-numerics} below), and take the cluster $S=\{A,B\}$, $S^c=\{C,D\}$. Every block $M_{V\to T}$ with $V\neq S$ or $T\neq S^c$ vanishes identically, because the Smolin state carries no one- or three-body correlations. Only $M_{S\to S^c}$ survives, as the $9\times9$ matrix diagonal on the aligned Pauli directions $(x,x)\to(x,x)$, $(y,y)\to(y,y)$, $(z,z)\to(z,z)$ with unit coefficients and zero elsewhere. After the normalization $1/\sqrt{(d_S-1)(d_{S^c}-1)}=1/3$ this gives three singular values of $1/3$ each, so
%% \begin{equation*}
%% \norm{\mathcal M_{AB}(\rho_{\mathrm{Smo}})}_*=1
%% \quad\text{exactly --- saturating, not violating, the bound of Theorem~\ref{thm:cluster-cut}.}
%% \end{equation*}
%% This is consistent with the Smolin state being separable across every $2\mid2$ cut while violating the $1\mid3$ bound at $3/\sqrt7\approx1.134$, computed below in Section~\ref{sec:qubit-numerics}. The refinement from singleton to cluster sources costs nothing in the proof yet correctly distinguishes the two cut types, where the singleton construction of Definition~\ref{def:combined-shadow} cannot even pose the $2\mid2$ question.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\section{The tensor viewpoint: shadow maps as unfoldings of one full Bloch tensor}
\label{sec:tensor-viewpoint}

The guiding thread announced in Section~\ref{sec:response-maps}--\ref{sec:multiparty-sources} can now be made precise. Every non-scalar object introduced so far --- $M_a$, its bigraduated extension $\mathcal M_S$, and the sub-block witnesses of Corollary~\ref{cor:sub-block} --- turns out to be a matricization or sub-block restriction of a single order-$n$ tensor built from the full correlation data of $\rho$; the source-aggregated functionals used in the qubit applications below are then averages or maxima of the corresponding one-vs-rest norms. The $\le1$ bound is therefore not a family of independently proved facts, but one rank-one statement about that tensor, observed through different linear lenses.

\begin{definition}[Full Bloch tensor]
\label{def:full-tensor}
For each party $a$, we augment the local index range by including $i_a = 0$, which selects the identity $\sigma_0^{(a)}=\id$ already introduced in Section~\ref{sec:old-criterion}. This allows us to define the full (order-$n$) Bloch tensor
\begin{equation*}
\mathcal C(\rho)\in\bigotimes_{a\in P}\R^{d_a^2},
\qquad
\bigl(\mathcal C(\rho)\bigr)_{i_1,\dots,i_n}=\tr\!\Bigl(\rho\bigotimes_{a\in P}\sigma^{(a)}_{i_a}\Bigr).
\end{equation*}
\end{definition}

Since $\bigotimes_{a\in P}\R^{d_a^2}=\bigotimes_{a\in P}\bigl(\R\oplus\R^{d_a^2-1}\bigr)$ expands by distributivity into $2^n$ orthogonal summands indexed by which legs are trivial, every sector tensor of Eq.~\eqref{eq:corr-tensor-def} is simply a slice of $\mathcal C(\rho)$:
\begin{equation}
C_V(\rho)=\mathcal C(\rho)\big|_{\,i_a\neq0\text{ for }a\in V,\ i_a=0\text{ for }a\notin V},
\qquad
\mathcal C(\rho)\;\cong\;1\,\oplus\!\!\bigoplus_{\emptyset\neq V\subseteq P}C_V(\rho).
\label{eq:tensor-slice-sector}
\end{equation}
The sector decomposition used throughout this note, on both the target side (Section~\ref{sec:response-maps}) and the source side (Section~\ref{sec:multiparty-sources}), is therefore not an additional structure imposed on the correlation data: it \emph{is} the tensor-product structure of $\mathcal C(\rho)$ in the identity-plus-generators basis.

\begin{remark}[This is already the tomography tensor of \cite{aschauer}]
\label{rem:aschauer-tensor}
Definition~\ref{def:full-tensor} introduces no object beyond what \cite{aschauer} starts from. Writing out the operator expansion of $\rho$ in the full product basis $\{\bigotimes_{a\in P}\sigma^{(a)}_{i_a}:0\le i_a\le d_a^2-1\}$ used there for state tomography gives exactly
\begin{equation*}
\rho=\Bigl(\prod_{a\in P}d_a\Bigr)^{-1}\sum_{i_1,\dots,i_n}c_{i_1,\dots,i_n}\bigotimes_{a\in P}\sigma^{(a)}_{i_a},
\qquad
c_{i_1,\dots,i_n}=\tr\Bigl(\rho\bigotimes_{a\in P}\sigma^{(a)}_{i_a}\Bigr),
\end{equation*}
with $\mathcal C(\rho)=(c_{i_1,\dots,i_n})$, over the same unrestricted index range. The sector-restricted tensor $C_S(\rho)$ of Eq.~\eqref{eq:corr-tensor-def}, on which the correlation strengths $L_S$ and every construction built on them in this note ultimately depend, is the further restriction of that same tomography tensor to $i_a>0$ for every $a\in S$. In this sense, Sections~\ref{sec:response-maps}--\ref{sec:multiparty-sources} never leave the object \cite{aschauer} already had in hand; what changes is only what is extracted from it, replacing the scalar sector norm $L_S$ with a matrix unfolding and its singular values. The point of the present section is that this change of extraction is itself best understood at the level of the tensor $\mathcal C(\rho)$, rather than sector by sector.
\end{remark}

We now switch perspectives: the shadow maps of the preceding sections will no longer be treated as separately constructed response operators, but as canonical unfoldings, sector restrictions, and identity-leg slices of the single full tensor $\mathcal C(\rho)$.

\begin{proposition}[Product states are exactly the states whose full Bloch tensor has CP-rank one]
\label{prop:cp-rank-one}
$\rho=\bigotimes_{a\in P}\rho_a$ if and only if $\mathcal C(\rho)=\bigotimes_{a\in P}w^{(a)}$ for vectors $w^{(a)}\in\R^{d_a^2}$ with $w^{(a)}_0=1$.
\end{proposition}

\begin{proof}
($\Rightarrow$) Immediate from multiplicativity of the trace over the tensor factors, with $w^{(a)}_{i_a}:=\tr(\rho_a\sigma^{(a)}_{i_a})$.

($\Leftarrow$) By the orthogonality relation~\eqref{eq:generator-orthogonality}, extended over $i,j\in\{0,\dots,d_a^2-1\}$, the map $\rho\mapsto\mathcal C(\rho)$ is a linear bijection with inverse
\begin{equation*}
\rho=\Bigl(\prod_{a\in P}d_a\Bigr)^{-1}\sum_{i_1,\dots,i_n}c_{i_1,\dots,i_n}\bigotimes_{a\in P}\sigma^{(a)}_{i_a}.
\end{equation*}
Substituting a rank-one $\mathcal C(\rho)=\bigotimes_a w^{(a)}$ into this inversion formula factorizes term by term into $\bigotimes_{a\in P}\rho_a$ with $\rho_a:=d_a^{-1}\sum_{i_a}w^{(a)}_{i_a}\sigma^{(a)}_{i_a}$. The condition $w_0^{(a)}=1$ gives $\tr(\rho_a)=1$ for each $a$; positivity of each $\rho_a$ then follows because $\rho=\bigotimes_a\rho_a$ is positive semidefinite and every factor is Hermitian with trace one and nonzero.
\end{proof}

\begin{proposition}[Shadow maps are unfoldings]
\label{prop:unfolding}
Fix a bipartition $P=S\sqcup S^c$. Grouping the $S$-legs of $\mathcal C(\rho)$ into a single row index and the $S^c$-legs into a single column index is the standard mode-$(S,S^c)$ matricization of $\mathcal C(\rho)$ in the sense of the multilinear singular value decomposition \cite{delathauwer}. Deleting the trivial ($i=0$) row and column --- equivalently, discarding the $V=\emptyset$ and $T=\emptyset$ sectors, which carry no information beyond normalization --- and rescaling by $[(d_S-1)(d_{S^c}-1)]^{-1/2}$ reproduces $M_S(\rho)$ exactly. The bigraduated shadow map $\mathcal M_S(\rho)$ of Definition~\ref{def:bigraduated} is the same unfolding with the row index additionally kept graded by $V\subseteq S$ instead of collapsed.
\end{proposition}

\begin{theorem}[One rank-one fact, inherited everywhere]
\label{thm:one-fact}
Fix a bipartition $P=S\sqcup S^c$. For every state separable across $S\mid S^c$, the normalized nuclear-norm bound $\norm{\cdot}_*\le1$ holds not only for the full shadow map $\mathcal M_S(\rho)$, but also for every witness obtained from the same mode-$(S,S^c)$ unfolding of $\mathcal C(\rho)$ by keeping or collapsing the source and target sector gradings, by taking two-sided sector restrictions as in Corollary~\ref{cor:sub-block}, or by taking target-side identity-leg slices corresponding to partial traces as in Corollary~\ref{cor:trace-is-slice} below, with the normalization appropriate to the remaining source and target systems. Thus Theorem~\ref{thm:cut-bound}, its cluster generalization (Theorem~\ref{thm:cluster-cut}), and the sub-block witnesses of Corollary~\ref{cor:sub-block} are not independently proved facts, but one algebraic statement observed through different linear lenses.
\end{theorem}

\begin{proof}
It is enough to consider a product state across the chosen cut, $\rho=\rho_S\otimes\sigma_{S^c}$. Grouping the legs in $S$ and $S^c$, trace multiplicativity gives
\begin{equation*}
\mathcal C(\rho)=\mathcal C(\rho_S)\otimes\mathcal C(\sigma_{S^c}),
\end{equation*}
so the mode-$(S,S^c)$ unfolding is rank one. After deleting the identity row and column, this is the rank-one matrix $v_{S^c}(r^{(S)})^T$ appearing in the proof of Theorem~\ref{thm:cluster-cut}, and the same correlation-sum identity gives
\begin{equation*}
\norm{r^{(S)}}^2\le d_S-1,
\qquad
\norm{v_{S^c}}^2\le d_{S^c}-1.
\end{equation*}
Hence the normalized full unfolding has nuclear norm at most one for each product term.

Keeping the sector gradings is only a change of coordinates, while collapsing them gives the same matrix representation with grouped row or column indices. Two-sided sector restrictions have the form $Axy^TB=(Ax)(B^Ty)^T$ on each rank-one product term and cannot increase the nuclear norm when $A$ and $B$ are orthogonal projections. Target-side identity-leg slices give the corresponding reduced product tensor and obey the same estimate with the dimensions of the surviving source and target systems. Finally, linearity of $\rho\mapsto\mathcal C(\rho)$ and convexity of the nuclear norm extend the bound from product states to arbitrary mixtures separable across $S\mid S^c$.
\end{proof}

\begin{corollary}[Partial trace is a slice, not a sum]
\label{cor:trace-is-slice}
For $E\subseteq P$ and $\rho_{P\setminus E}:=\tr_E(\rho)$,
\begin{equation}
\mathcal C(\rho_{P\setminus E})=\mathcal C(\rho)\big|_{\,i_a=0\text{ for all }a\in E},
\label{eq:trace-is-slice}
\end{equation}
i.e.\ the marginal's full tensor is the slice of $\mathcal C(\rho)$ at the trivial index on every traced-out leg, not a contraction or summation over $E$. Consequently, if $P=S\sqcup R\sqcup E$ with $S$ a source cluster as in Definition~\ref{def:bigraduated}, $\rho_{SR}:=\tr_E(\rho)$, and $\Pi_R$ denotes the orthogonal projection that annihilates every target sector $T$ with $T\cap E\neq\emptyset$, then exactly
\begin{equation}
\mathcal M_S(\rho_{SR})=\sqrt{\frac{d_{S^c}-1}{d_R-1}}\;\Pi_R\,\mathcal M_S(\rho).
\label{eq:trace-rescale}
\end{equation}
\end{corollary}

\begin{proof}
Eq.~\eqref{eq:trace-is-slice} is immediate from $\sigma_0^{(a)}=\id$: setting $i_a=0$ for $a\in E$ in the defining sum of $\mathcal C(\rho)$ inserts the identity on every traced-out leg, which is exactly $\tr_E(\rho)$ evaluated against the remaining generators. For the second claim, apply Proposition~\ref{prop:unfolding} to the slice~\eqref{eq:trace-is-slice}: because $S\cap E=\emptyset$, the source legs are untouched by the slicing, so for every $T\subseteq R$ the unnormalized block $M_{S\to T}$ computed from $\mathcal C(\rho)$ agrees exactly with the one computed from $\mathcal C(\rho_{SR})$. The two combined maps therefore differ only in their normalization constants, $[(d_S-1)(d_{S^c}-1)]^{-1/2}$ for $\mathcal M_S(\rho)$ against $[(d_S-1)(d_R-1)]^{-1/2}$ for $\mathcal M_S(\rho_{SR})$, since the complement of $S$ is $R\cup E$ in the first case and $R$ alone in the second, with $d_{S^c}=d_Rd_E$. Their ratio is exactly the stated factor.
\end{proof}

Eq.~\eqref{eq:trace-rescale} shows that discarding a residual cluster $E$ by tracing it out is a strictly weaker operation than the sub-block compression of Corollary~\ref{cor:sub-block}: the latter only ever shrinks the nuclear norm, while Eq.~\eqref{eq:trace-rescale} rescales it upward by the factor $\sqrt{(d_{S^c}-1)/(d_R-1)}\ge1$, so that a violation of the bound on $\rho_{SR}$ can certify entanglement across $S\mid R$ that survives the complete loss of $E$, a strictly stronger and operationally different statement than merely detecting entanglement somewhere across $S\mid RE$.

\begin{remark}[Why unfold at all]
The full tensor $\mathcal C(\rho)$ carries strictly more information than any single unfolding: two states can share every matricization $M_S$ over all bipartitions and still differ in genuine multi-way structure, exactly as a generic tensor is not determined by its unfoldings alone. The reason this note works with unfoldings rather than $\mathcal C(\rho)$ directly is computational, not conceptual. The CP-rank-one statement for products, Proposition~\ref{prop:cp-rank-one}, is exact and dimension-independent, but the associated \emph{tensor} nuclear norm --- the natural generalization of $\norm{\cdot}_*$ that would witness separability directly on $\mathcal C(\rho)$, as the infimum of $\sum_r|\lambda_r|$ over CP decompositions --- has no polynomial-time algorithm once three or more legs are grouped independently. Every matricization used in this note, by contrast, is an ordinary matrix with a computable singular value decomposition. The constructions of the preceding sections are thus best understood as the maximal set of efficiently computable shadows of one underlying rank-one fact, chosen along the cut structure that is operationally relevant to entanglement questions.
\end{remark}

\subsection*{A qutrit PPT-entangled benchmark}

The dimension-independent normalization is not only a formal convenience. As a two-qutrit test case, consider the Tiles unextendible product basis of Bennett 
\emph{et al.}~\cite{bennettUPB}, consisting of the five orthonormal product vectors
\begin{align*}
&\lvert 0\rangle\otimes\frac{\lvert0\rangle-\lvert1\rangle}{\sqrt2},
&&\lvert 2\rangle\otimes\frac{\lvert1\rangle-\lvert2\rangle}{\sqrt2},\\
&\frac{\lvert0\rangle-\lvert1\rangle}{\sqrt2}\otimes\lvert2\rangle,
&&\frac{\lvert1\rangle-\lvert2\rangle}{\sqrt2}\otimes\lvert0\rangle,\\
&\frac{\lvert0\rangle+\lvert1\rangle+\lvert2\rangle}{\sqrt3}
\otimes
\frac{\lvert0\rangle+\lvert1\rangle+\lvert2\rangle}{\sqrt3}.
\end{align*}
Let $P_{\mathrm{UPB}}$ be the projector onto their span and set
\begin{equation*}
\rho_{\mathrm{Tiles}}=\frac14\left(\id_9-P_{\mathrm{UPB}}\right).
\end{equation*}
This is the standard rank-four PPT-entangled state supported on the completely entangled complement of the UPB. Using Gell-Mann generators scaled by $\sqrt{3/2}$, so that $\tr(\sigma_i\sigma_j)=3\delta_{ij}$ as in Eq.~\eqref{eq:generator-orthogonality}, the bipartite shadow map is the $8\times8$ correlation matrix divided by
\begin{equation*}
\sqrt{(d_A-1)(d_B-1)}=2.
\end{equation*}
The numerical values are
\begin{align*}
\min\operatorname{spec}\bigl(\rho_{\mathrm{Tiles}}^{T_B}\bigr)&=-1.6\cdot10^{-16},\\
\norm{\mathcal M_A(\rho_{\mathrm{Tiles}})}_*&=1.053421632\ldots>1.
\end{align*}
Thus the state is PPT up to numerical precision, so the Peres-Horodecki PPT test is silent \cite{peres,horodeckiPPT}, while the shadow-map witness detects its entanglement. This should be read as complementarity rather than domination: the realignment criterion of Chen and Wu \cite{chenwu} also detects this benchmark, with trace norm $1.087412465\ldots$. The script \texttt{scripts/tiles\_upb.py} reproduces the generator normalization, the PPT spectrum, the shadow value, and the realignment comparison.

\subsection*{Qubit Pauli-tensor form}

For qubits we use the convention already anticipated in the Smolin example of Section~\ref{sec:multiparty-sources}: $\sigma_0=\id$ and $\sigma_1=X$, $\sigma_2=Y$, $\sigma_3=Z$. In this notation, the full Bloch tensor $\mathcal C(\rho)$ becomes the familiar Pauli-correlation tensor

\begin{equation*}
\mathcal C(\rho)_{i_1,\dots,i_n}
=
\tr\!\left(\rho\,\sigma_{i_1}\otimes\cdots\otimes\sigma_{i_n}\right),
\qquad
i_a\in\{0,x,y,z\},
\end{equation*}
and each sector is specified simply by the support pattern of the nonidentity Pauli indices. Thus the one-vs-rest map for a source party $a$ is obtained by unfolding this Pauli tensor with the $a$-leg as source, discarding the all-identity sector on the complement, and multiplying by
\begin{equation*}
\frac{1}{\sqrt{2^{n-1}-1}}.
\end{equation*}
More generally, for a qubit source cluster $S$ the normalized cluster map is the Pauli-tensor unfolding across $S\mid S^c$, with the all-identity source and target sectors removed, scaled by
\begin{equation*}
\frac{1}{\sqrt{(2^{|S|}-1)(2^{|S^c|}-1)}}.
\end{equation*}

This form makes two features of the examples below transparent. First, adding white noise as
\begin{equation*}
\rho(p)=p\rho+(1-p)\frac{\id}{2^n}
\end{equation*}
leaves the all-identity coefficient fixed and multiplies every nonidentity Pauli coefficient by $p$, so every shadow map and every shadow norm scales linearly with $p$. Second, stabilizer and graph states have Pauli tensors supported on their stabilizer groups, with nonzero coefficients equal to $\pm1$. Their shadow maps are therefore normalized signed support-pattern unfoldings, which explains why the numerical graph-state values below are rigid singular-value facts rather than generic floating-point coincidences.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\section{Qubit specialization and source-aggregated benchmarks}

For the remaining benchmarks and numerics we stay in the qubit setting. The single-party response spaces are $\R^3$, the normalization in Eq.~\eqref{eq:combined-map} reduces to $1/\sqrt{2^{n-1}-1}$, and one can derive explicit constants that do not seem to be available so cleanly in higher dimensions.

\subsection*{Source-aggregated one-vs-rest functionals}

The family $\{\mathcal M_a(\rho)\}_{a\in P}$ yields natural application-facing scalars by aggregating over the choice of source party.

\begin{definition}
Define the symmetric average shadow functional
\begin{equation}
\Phi_{\mathrm{sym}}(\rho)
:=
\frac{1}{n}\sum_{a\in P}\norm{\mathcal M_a(\rho)}_*,
\label{eq:phi-sym}
\end{equation}
and the symmetric maximum shadow functional
\begin{equation}
\Phi_{\max}(\rho):=\max_{a\in P}\norm{\mathcal M_a(\rho)}_*.
\label{eq:phi-max}
\end{equation}
\end{definition}

\begin{corollary}
If $\rho$ is fully separable, then
\begin{equation*}
\Phi_{\mathrm{sym}}(\rho)\le 1,
\qquad
\Phi_{\max}(\rho)\le 1.
\end{equation*}
Hence violating either bound certifies entanglement.
\end{corollary}

\begin{proof}
A fully separable state is separable across every one-vs-rest cut. The claim follows by applying Theorem~\ref{thm:cut-bound} to each party.
\end{proof}

In equal local dimensions these functionals are permutation invariant. More generally, they are source-aggregated cut-sensitive scalars. The average probes all one-vs-rest cuts simultaneously, whereas the maximum asks whether at least one cut exhibits a large combined shadow ellipsoid.

This averaging-over-cuts strategy for turning a bipartite-type bound into a
genuine multipartite criterion is not unique to the present construction. A
structurally analogous move appears in \cite{liyaoyangfei2025}, whose
Theorem~5 averages the trace norm of a bipartition-indexed extended
correlation tensor over \emph{all} $2^{N-1}-1$ bipartitions of the $N$
parties, comparing the result against a bound assembled from the worst case
over bipartitions of each size $k=1,\ldots,N-1$, to certify genuine
$N$-partite entanglement. Three differences are worth making explicit.
First, the sum in \cite{liyaoyangfei2025} runs over all $2^{N-1}-1$
bipartitions of $P$, including every $k$-vs-$(N-k)$ split with
$1<k<N-1$, whereas $\Phi_{\mathrm{sym}}$ and $\Phi_{\max}$ here range only
over the $n$ one-vs-rest cuts; the two index sets coincide at $n=3$ but
diverge from $n=4$ on, where the bipartition sum there also includes, e.g.,
the $2\mid2$ splits that a one-vs-rest average does not see directly.
Second, only the averaged quantity is ever compared against a bound in
\cite{liyaoyangfei2025}; the maximum that appears there acts solely on the
threshold side (the worst case per partition size, entering the bound
$M_0^{(1\cdots N)}$), not as a second, independently observed witness the
way $\Phi_{\max}(\rho)$ is here. Third, and more relevant to the underlying
object, each per-bipartition term in \cite{liyaoyangfei2025} collapses both
sides of the cut to their single joint correlation tensor before the trace
norm is taken, so no sector structure survives inside either half of the
partition. $\mathcal M_a(\rho)$, by contrast, is a direct sum over every
nonempty subset $T$ of the complement $\bar a$; it is this retained sector
granularity on one side of the cut, rather than the averaging step itself,
that we take to be the substantive point of the present construction.

\subsection*{Three-qubit biseparable benchmark}

For three qubits the symmetric average can be optimized explicitly over the biseparable set. This is the first place where we use genuinely qubit-specific formulas rather than only the general response-map architecture. Suppose first that
\begin{equation*}
\rho=\rho_A\otimes\sigma_{BC}
\end{equation*}
is product across $A\mid BC$. If the total state is pure, then this is equivalent to being pure and separable across that cut. Let $a\in\R^3$ be the Bloch vector of $\rho_A$, and let
\begin{equation*}
b,c\in\R^3,
\qquad
T\in\R^{3\times 3}
\end{equation*}
be the one- and two-body correlation data of the pure two-qubit state $\sigma_{BC}$. Then $\norm{a}=1$, and Theorem~\ref{thm:cut-bound} gives
\begin{equation*}
\norm{\mathcal M_A(\rho)}_*=1.
\end{equation*}
For the other two parties one finds
\begin{equation*}
3\,\mathcal M_B(\rho)^T\mathcal M_B(\rho)=bb^T+2T^TT,
\end{equation*}
and similarly
\begin{equation*}
3\,\mathcal M_C(\rho)^T\mathcal M_C(\rho)=cc^T+2TT^T.
\end{equation*}
Using Schmidt form for $\sigma_{BC}$,
\begin{equation*}
\lvert\psi_{BC}\rangle=\cos\theta\,\lvert 00\rangle+\sin\theta\,\lvert 11\rangle,
\end{equation*}
one gets singular values
\begin{equation*}
\sqrt{\frac{2(1-r^2)}{3}},\qquad
\sqrt{\frac{2(1-r^2)}{3}},\qquad
\sqrt{\frac{2+r^2}{3}},
\end{equation*}
where $r=\cos 2\theta$. Hence
\begin{equation*}
\norm{\mathcal M_B(\rho)}_*=\norm{\mathcal M_C(\rho)}_*=f(r),
\end{equation*}
with
\begin{equation*}
f(r):=2\sqrt{\frac{2(1-r^2)}{3}}+\sqrt{\frac{2+r^2}{3}}.
\end{equation*}
This is maximal at $r=0$, i.e. when $\sigma_{BC}$ is maximally entangled. Therefore every pure state separable across $A\mid BC$ satisfies
\begin{equation*}
\Phi_{\mathrm{sym}}(\rho)=\frac{1+2f(r)}{3}\le\frac{1+2\sqrt 6}{3}.
\end{equation*}
By convexity and symmetry, the same bound extends to all biseparable three-qubit states:
\begin{equation}
\beta^{\mathrm{sym}}_{\mathrm{bisep}}(n=3)=\frac{1+2\sqrt 6}{3}\approx 1.966.
\label{eq:bisep-threshold}
\end{equation}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\section{Qubit examples and numerics}
\label{sec:qubit-numerics}

\subsection*{The three-qubit GHZ state}

For
\begin{equation*}
\lvert\GHZ_3\rangle=\frac{\lvert 000\rangle+\lvert 111\rangle}{\sqrt2},
\end{equation*}
the relevant response vectors for each source party are orthogonal, with equal squared norm $2/3$ after the normalization in Eq.~\eqref{eq:combined-map}. Hence the singular values of each $\mathcal M_a$ are all equal to $\sqrt{2/3}$, so
\begin{equation*}
\norm{\mathcal M_a(\lvert\GHZ_3\rangle\!\langle\GHZ_3\rvert)}_* = \sqrt 6.
\end{equation*}
Therefore
\begin{equation*}
\Phi_{\mathrm{sym}}\bigl(\lvert\GHZ_3\rangle\!\langle\GHZ_3\rvert\bigr)=\Phi_{\max}\bigl(\lvert\GHZ_3\rangle\!\langle\GHZ_3\rvert\bigr)=\sqrt 6.
\end{equation*}
Comparing with Eq.~\eqref{eq:bisep-threshold}, we obtain
\begin{equation*}
\Phi_{\mathrm{sym}}\bigl(\lvert\GHZ_3\rangle\!\langle\GHZ_3\rvert\bigr)>\beta^{\mathrm{sym}}_{\mathrm{bisep}}(n=3),
\end{equation*}
so the symmetric average certifies genuine tripartite entanglement.

For the noisy family
\begin{equation*}
\rho_{\mathrm{GHZ}}(p)=p\,\lvert\GHZ_3\rangle\!\langle\GHZ_3\rvert+(1-p)\frac{\id}{8},
\end{equation*}
all traceless correlations scale by $p$, so
\begin{equation*}
\Phi_{\mathrm{sym}}\bigl(\rho_{\mathrm{GHZ}}(p)\bigr)=p\sqrt 6.
\end{equation*}
Thus the criterion detects genuine tripartite entanglement whenever
\begin{equation*}
p>\frac{1+2\sqrt 6}{3\sqrt 6}\approx 0.803.
\end{equation*}

\subsection*{The four-qubit Smolin state, revisited}

Recall from the example in Section~\ref{sec:multiparty-sources} (page~\pageref{ex:smolin}) that for the Smolin state and a single source party $a$,
\begin{equation*}
\norm{\mathcal M_a(\rho_{\mathrm{Smo}})}_* = \frac{3}{\sqrt 7}>1.
\end{equation*}
Since this value is the same for every choice of $a$,
\begin{equation*}
\Phi_{\mathrm{sym}}(\rho_{\mathrm{Smo}})=\Phi_{\max}(\rho_{\mathrm{Smo}})=\frac{3}{\sqrt 7}>1,
\end{equation*}
so the criterion correctly detects entanglement across every $1\mid 3$ cut, while --- as already noted --- this is not a genuine-multipartite conclusion, since the Smolin state is separable across every $2\mid 2$ split. For the white-noise family $p\rho_{\mathrm{Smo}}+(1-p)\id/16$, the fully separable threshold would only be crossed for
\begin{equation*}
p>\frac{\sqrt 7}{3}\approx 0.882,
\end{equation*}
so this example is much more fragile under white noise than the graph-state families discussed below.


\subsection*{A numerical four-qubit graph-state scan}

Using the \texttt{qtensor} package, we also evaluated the symmetric shadow functionals for all connected labeled graph states on four qubits. Numerically, all $38$ such graph states give the same value,
\begin{equation*}
\Phi_{\mathrm{sym}}=\Phi_{\max}=\frac{6}{\sqrt 7}\approx 2.268.
\end{equation*}
Thus for the noisy family
\begin{equation*}
\rho(p)=p\rho+(1-p)\frac{\id}{16}
\end{equation*}
the fully separable threshold is crossed already at
\begin{equation*}
p>\frac{\sqrt 7}{6}\approx 0.441.
\end{equation*}
This includes the canonical examples $\GHZ_4$, the line graph state, the ring graph state, and the star graph state \cite{heinEisertBriegel2004}.

The same numerical scan shows that this shadow detection is not explained by pairwise entanglement in the reduced states. For those representative families, every two-qubit marginal remains PPT at the threshold $p=\sqrt 7/6$, and for the line and ring graph states some of the two-qubit marginals are even maximally mixed. So the shadow functional is responding to multipartite correlation structure that is not visible in pair reductions. This is still not a four-qubit genuine-multipartite-entanglement proof, because the corresponding biseparable threshold is not yet known, but it makes the criterion promising as a genuinely multipartite diagnostic.

The ring graph state also gives a compact illustration of what the bigraduated $2\mid2$ map sees beyond the source-aggregated scalars. Let $\rho_{\square}$ be the four-qubit graph state on the cycle with edges $(1,2),(2,3),(3,4),(4,1)$. Evaluating the normalized cluster map of Eq.~\eqref{eq:bigraduated-map} across the two inequivalent $2\mid2$ cuts gives
\begin{center}
\begin{tabular}{lccc}
\toprule
cut $S\mid S^c$ & $\norm{\mathcal M_S(\rho_{\square})}_*$ & $\norm{P_{S^c}\mathcal M_S(\rho_{\square})P_S}_*$ & marginal on $S$ \\
\midrule
adjacent $\{1,2\}\mid\{3,4\}$ & $5$ & $5/3$ & $\id/4$ \\
diagonal $\{1,3\}\mid\{2,4\}$ & $7/3$ & $1$ & eigenvalues $1/2,1/2,0,0$ \\
\bottomrule
\end{tabular}
\end{center}
Here $P_S$ and $P_{S^c}$ denote the projections onto the full source and full target sectors, so the middle column isolates the genuine two-body-to-two-body block within the same normalized witness. Thus both cuts violate the separable bound, but by different mechanisms: for the adjacent cut the full-sector block already violates, while for the diagonal cut that block only saturates the bound and the excess comes from lower source or target sectors. The script \texttt{scripts/grraph\_state\_cuts.py} reproduces these values directly from the Pauli correlation tensor.

As a first systematic extension, we also scanned the families $\GHZ_n$, $W_n$, the line graph state, and the ring graph state for $n=3,4,5$, together with D\"ur states for $n=4,5$. Numerically,
\begin{equation*}
\Phi_{\mathrm{sym}}(\GHZ_n)=\Phi_{\mathrm{sym}}(\text{line}_n)=\Phi_{\mathrm{sym}}(\text{ring}_n)
\end{equation*}
throughout that range, with common values $\sqrt 6$, $6/\sqrt 7$, and $2.19089\ldots$ for $n=3,4,5$, respectively. The $W_n$ family is consistently slightly lower but still well above the fully separable bound, while the D\"ur family already lies below $1$ for $n=4,5$. So the symmetric shadow functional strongly favors graph-like and GHZ-like global correlation structure, but it is not simply a monotone of party number.
At $n=4$, for instance, $W_4$ still crosses the fully separable white-noise threshold at about $p\approx 0.469$, whereas the D\"ur value is already below the fully separable benchmark even before white noise is added.

To probe the unresolved four-qubit biseparable benchmark, we performed a small random search over pure biseparable states across all inequivalent cuts, using $120$ samples per cut. The largest sampled value was
\begin{equation*}
\Phi_{\mathrm{sym}}\approx 2.235
\end{equation*}
for a state separable across a $2\mid 2$ partition, while the best sampled $1\mid 3$ values were only around $1.94$. This is not a proof of the true biseparable threshold, but it suggests two useful heuristics: first, the most dangerous competitors to the graph-state value $6/\sqrt 7\approx 2.268$ come from $2\mid 2$ cuts rather than $1\mid 3$ cuts; second, the connected four-qubit graph-state value sits slightly above the best random biseparable samples we found.


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\section{Saturation of the source-aggregated functionals, and a genuine multipartite witness from cluster maps}
\label{sec:saturation-gme}

%% FORWARD REFERENCE NOTICE: this section is written to stand on its own,
%% but Remark~\ref{rem:saturation-not-gme} below quotes an exact value
%% ($6/\sqrt7$ for two decoupled Bell pairs, and the $M_{AC}=M_{AD}=5$
%% cross-cut values) that is most economically obtained via the
%% stabilizer-degeneracy mechanism of the companion note on symmetry-adapted
%% shadow maps (into which Section~\ref{sec:symmetry-blocks} of the present
%% draft is being split). The specific group-theoretic data used
%% ($H=\langle X_AX_B,Z_AZ_B,X_CX_D,Z_CZ_D\rangle$ and the resulting
%% $|\ker\varphi|,|\operatorname{im}\varphi|$ for each cut) is reproduced
%% in full below, so no forward citation is strictly required for the
%% numbers themselves -- only for the general machinery that explains
%% \emph{why} the computation takes this form. Update the cross-reference
%% once the companion paper has a stable label.

The numerical values reported in Section~\ref{sec:qubit-numerics} for
$\Phi_{\mathrm{sym}}$ and $\Phi_{\max}$ --- $\sqrt6$ for $\GHZ_3$,
$6/\sqrt7$ for all $38$ connected four-qubit graph states,
$2.19089\ldots$ for the $n=5$ families --- were presented there as facts
about specific, highly structured states. This section shows they are
instances of a single closed-form bound that holds for \emph{every}
$n$-qubit state, entangled or not, and that consequently
$\Phi_{\mathrm{sym}}$, $\Phi_{\max}$ cannot by themselves certify genuine
multipartite entanglement at that value. We then show that the
bigraduated cluster maps $\mathcal M_S$ of Section~\ref{sec:multiparty-sources},
combined across several cuts rather than used singly, repair this: for
four qubits, the minimum of the three inequivalent $2\mid2$ cluster norms
is a provably nontrivial witness of genuine four-partite entanglement.

\subsection{A universal saturation bound}

\begin{theorem}[Saturation of the cluster shadow norm]
\label{thm:saturation-bound}
Fix $n$ qubits and a cut $S\mid S^c$ ($\emptyset\neq S\subsetneq P$), and
write $D_S:=2^{|S|}$, $D_{S^c}:=2^{|S^c|}$,
$D_{\min}:=\min(D_S,D_{S^c})$, $D_{\max}:=\max(D_S,D_{S^c})$. Then for
every $n$-qubit state $\rho$ (pure or mixed),
\begin{equation}
\norm{\mathcal M_S(\rho)}_*^2
\;\le\;
\frac{(D_{\min}^2-1)^2\,D_{\max}}{D_{\min}\,(D_S-1)(D_{S^c}-1)}.
\label{eq:saturation-bound-general}
\end{equation}
Equality requires the marginal on the smaller-dimensional side to be
maximally mixed, and every nonzero singular value of
$\widetilde{\mathcal M}_S(\rho)$ to be equal. For a singleton source
($S=\{a\}$, $D_S=2$), Eq.~\eqref{eq:saturation-bound-general} specializes
to
\begin{equation}
\norm{\mathcal M_a(\rho)}_*\;\le\;3\sqrt{\frac{2^{n-2}}{2^{n-1}-1}},
\label{eq:saturation-single-party}
\end{equation}
and consequently
$\Phi_{\mathrm{sym}}(\rho),\Phi_{\max}(\rho)\le3\sqrt{2^{n-2}/(2^{n-1}-1)}$
for \emph{every} $n$-qubit state.
\end{theorem}

\begin{proof}
Since $\rho\mapsto\mathcal M_S(\rho)$ is linear (Eq.~\eqref{eq:bigraduated-map})
and the nuclear norm is convex, the supremum of $\norm{\mathcal M_S(\rho)}_*$
over the convex, compact set of density matrices is attained at an extreme
point, i.e.\ a pure state; it suffices to bound
$\norm{\mathcal M_S(\rho)}_*$ for pure $\rho$.

Write $D:=2^n=D_SD_{S^c}$ and treat $S$, $S^c$ each as a single composite
party of dimension $D_S$, $D_{S^c}$, so that
$c_{i_S,i_{S^c}}:=\bigl(\mathcal C(\rho)\bigr)_{i_S,i_{S^c}}$
(Definition~\ref{def:full-tensor}) is the tensor entry. The
generator-orthogonality identity~\eqref{eq:generator-orthogonality},
applied to $S$, to $S^c$, and to the whole system as single composite
parties, gives
\begin{equation*}
\sum_{i_S,i_{S^c}} c_{i_S,i_{S^c}}^2 = D\,\tr(\rho^2),
\qquad
\sum_{i_{S^c}} c_{0,i_{S^c}}^2 = D_{S^c}\,\tr(\rho_{S^c}^2),
\qquad
\sum_{i_S} c_{i_S,0}^2 = D_S\,\tr(\rho_S^2).
\end{equation*}
Inclusion--exclusion over the trivial ($i=0$) row and column gives the sum
of squared unnormalized entries of $\widetilde{\mathcal M}_S(\rho)$,
\begin{equation*}
\norm{\widetilde{\mathcal M}_S(\rho)}_{\fro}^2
=D\tr(\rho^2)-D_{S^c}\tr(\rho_{S^c}^2)-D_S\tr(\rho_S^2)+1.
\end{equation*}
For pure $\rho$, $\tr(\rho^2)=1$ and Schmidt symmetry across the cut gives
$\tr(\rho_S^2)=\tr(\rho_{S^c}^2)=:p$, so
\begin{equation*}
\norm{\widetilde{\mathcal M}_S(\rho)}_{\fro}^2=D+1-(D_S+D_{S^c})\,p,
\end{equation*}
decreasing in $p$. Since $p\ge1/D_{\min}$ for any state of the smaller
side, with equality iff that marginal is maximally mixed,
\begin{equation*}
\norm{\widetilde{\mathcal M}_S(\rho)}_{\fro}^2
\;\le\;D+1-\frac{D_S+D_{S^c}}{D_{\min}}
\;=\;\frac{D_{\max}(D_{\min}^2-1)}{D_{\min}},
\end{equation*}
the last equality a direct algebraic simplification using $D=D_SD_{S^c}$.
On the other hand $\widetilde{\mathcal M}_S(\rho):\V_0^{(S)}\to\V_0^{(S^c)}$
has rank at most $\min(\dim\V_0^{(S)},\dim\V_0^{(S^c)})=D_{\min}^2-1$, so
$\norm{\widetilde{\mathcal M}_S(\rho)}_*\le\sqrt{D_{\min}^2-1}\,
\norm{\widetilde{\mathcal M}_S(\rho)}_{\fro}$. Dividing by the
normalization $\sqrt{(D_S-1)(D_{S^c}-1)}$ of Eq.~\eqref{eq:bigraduated-map}
gives Eq.~\eqref{eq:saturation-bound-general}; the singleton case is the
specialization $D_{\min}=2$, $D_{\max}=2^{n-1}$.
\end{proof}

\begin{corollary}[Cluster saturation for $n=4$]
\label{cor:cluster-saturation-n4}
For $n=4$ and any $S$ with $|S|=2$, $\norm{\mathcal M_S(\rho)}_*\le5$ for
every four-qubit state $\rho$, with equality iff the marginal on $S$ is
maximally mixed and all $15$ nonzero singular values of
$\widetilde{\mathcal M}_S(\rho)$ coincide.
\end{corollary}

\begin{remark}[Saturation is not a genuine-entanglement signature]
\label{rem:saturation-not-gme}
Theorem~\ref{thm:saturation-bound} holds for every state, so the values
reported in Section~\ref{sec:qubit-numerics} for $\GHZ_n$ and the
connected graph states are not entanglement-strength signatures: they are
instances of the state-independent maximum. Concretely, take
$\rho=\ket{\Phi^+}_{AB}\!\bra{\Phi^+}\otimes\ket{\Phi^+}_{CD}\!\bra{\Phi^+}$,
which is manifestly biseparable across $AB\mid CD$. Since $\rho_{AB}$ and
$\sigma_{CD}$ are both pure, Theorem~\ref{thm:cluster-cut} gives
$\norm{\mathcal M_{AB}(\rho)}_*=1$ exactly. For the two cross cuts, $\rho$
is a pure stabilizer state with stabilizer group
$H=\langle X_AX_B,\,Z_AZ_B,\,X_CX_D,\,Z_CZ_D\rangle$
%% forward reference: Lemma~\ref{lem:stabilizer-degeneracy} /
%% Corollary~\ref{cor:stabilizer-examples}-type computation, to appear in
%% full generality in the symmetry companion note.
of order $16$; for $S=AC$ (equivalently $S=AD$), the restriction
homomorphisms $\varphi,\psi:H\to\mathbb F_2^4$ onto $AC$ and $BD$ coincide
as functions of the four generator exponents and are both bijective, so
$\lvert\ker\varphi\rvert=1$, $\lvert\operatorname{im}\varphi\rvert=16$,
giving $15$ equal singular values $\sqrt{1/9}=1/3$ and hence
\begin{equation*}
\norm{\mathcal M_{AC}(\rho)}_*=\norm{\mathcal M_{AD}(\rho)}_*=15\cdot\tfrac13=5,
\end{equation*}
exactly the bound of Corollary~\ref{cor:cluster-saturation-n4}. Averaging
over source parties (or maximizing) therefore gives
$\Phi_{\mathrm{sym}}(\rho)=\Phi_{\max}(\rho)=6/\sqrt7$ once restricted to
the single-party functionals of Section~\ref{sec:qubit-numerics} --- the
same numerical value reported there for every connected four-qubit graph
state, for a state that is manifestly biseparable. $\Phi_{\mathrm{sym}}$
and $\Phi_{\max}$, taken alone, cannot separate this state from a
genuinely entangled one at the saturating value; a sharper construction is
needed.
\end{remark}

\subsection{A genuine multipartite witness from three cluster cuts}
\label{sec:min-witness}

Fix representatives $S_1=\{A,B\}$, $S_2=\{A,C\}$, $S_3=\{A,D\}$ of the
three inequivalent $2\mid2$ partitions of four qubits, and define
\begin{equation}
\Psi(\rho):=\min_{i=1,2,3}\ \norm{\mathcal M_{S_i}(\rho)}_*.
\label{eq:min-witness-def}
\end{equation}
Unlike $\Phi_{\mathrm{sym}}$, $\Phi_{\max}$, or any single
$\norm{\mathcal M_S(\rho)}_*$, the functional $\Psi$ is \emph{not} convex
(a minimum of convex functions need not be convex), so the usual
extreme-point reduction does not apply to it directly. The sum
$\Sigma(\rho):=\sum_{i=1}^3\norm{\mathcal M_{S_i}(\rho)}_*$, however, is
convex, and $\Psi(\rho)\le\Sigma(\rho)/3$ always; this is enough to prove
a nontrivial bound.

\begin{proposition}[Biseparable bound on the min-witness]
\label{prop:min-witness-bound}
If $\rho$ is biseparable (a mixture of states each product across some
bipartition of the four qubits, $1\mid3$ or $2\mid2$), then
\begin{equation}
\Psi(\rho)\;\le\;\frac{11}{3}.
\label{eq:min-witness-bound}
\end{equation}
\end{proposition}

\begin{proof}
$\Sigma$ is convex and the biseparable set is the convex hull of pure
states product across some cut, so $\sup\Sigma$ over the biseparable set
is attained at such an extreme point; $\Psi\le\Sigma/3$ then reduces the
claim to bounding $\Sigma$ pointwise on pure product states.

\emph{$2\mid2$ extreme points.} Suppose $\rho$ is pure and product across,
say, $AB\mid CD$. Both $\rho_{AB}$ and $\sigma_{CD}$ are then pure, so
Theorem~\ref{thm:cluster-cut} gives $\norm{\mathcal M_{AB}(\rho)}_*=1$
exactly, while Corollary~\ref{cor:cluster-saturation-n4} gives
$\norm{\mathcal M_{AC}(\rho)}_*,\norm{\mathcal M_{AD}(\rho)}_*\le5$
individually. The saturation condition for both cross terms is that the
$2$-qubit marginal on $AC$ (equivalently $BD$) be maximally mixed; taking
$\rho_{AB}$, $\sigma_{CD}$ both maximally entangled makes the marginal on
$\{A,C\}$ the product of two maximally mixed single-qubit marginals,
hence maximally mixed on $\{A,C\}$, and gives
$\norm{\mathcal M_{AC}(\rho)}_*=\norm{\mathcal M_{AD}(\rho)}_*=5$
simultaneously (this is exactly the state of
Remark~\ref{rem:saturation-not-gme}). Hence $\Sigma(\rho)\le1+5+5=11$ for
every $2\mid2$-product pure state, with equality attained.

\emph{$1\mid3$ extreme points.} A direct evaluation over the
$1$-parameter family of pure states product across a $1\mid3$ cut gives
$\Sigma(\rho)\le7$ throughout (script \texttt{scripts/min\_witness\_1v3\_sum\_bound.py}),
strictly below the $2\mid2$ case and hence not binding.

Combining the two cases, $\Sigma\le11$ on every biseparable extreme point,
and convexity of $\Sigma$ extends this to all biseparable mixtures. Then
$\Psi\le\Sigma/3\le11/3$.
\end{proof}

\begin{proposition}[Exact value on a two-component Bell mixture]
\label{prop:bell-mixture-tent}
Let
\begin{equation*}
\rho(p):=p\,\ket{\Phi^+}_{AB}\!\bra{\Phi^+}\otimes\ket{\Phi^+}_{CD}\!\bra{\Phi^+}
+(1-p)\,\ket{\Phi^+}_{AC}\!\bra{\Phi^+}\otimes\ket{\Phi^+}_{BD}\!\bra{\Phi^+},
\qquad p\in[0,1].
\end{equation*}
Then, exactly,
\begin{equation*}
\norm{\mathcal M_{AB}(\rho(p))}_*=5-4p,
\qquad
\norm{\mathcal M_{AC}(\rho(p))}_*=1+4p,
\qquad
\norm{\mathcal M_{AD}(\rho(p))}_*=3+2\lvert2p-1\rvert,
\end{equation*}
so that
\begin{equation*}
\Psi(\rho(p))=\begin{cases}1+4p,& p\le\tfrac12,\\[2pt] 5-4p,& p\ge\tfrac12,\end{cases}
\end{equation*}
maximized uniquely at $p=\tfrac12$, where all three cluster norms coincide
and $\Psi(\rho(\tfrac12))=3$.
\end{proposition}

\begin{proof}
Direct evaluation of each bigraduated block from the Pauli-tensor form of
$\rho(p)$ (script \texttt{scripts/bell\_mixture\_min\_witness.py} reproduces
the closed forms symbolically and confirms them against $\rho(0)$,
$\rho(1)$ via Corollary~\ref{cor:cluster-saturation-n4} and
Remark~\ref{rem:saturation-not-gme}). For $p\le\tfrac12$,
$3+2\lvert2p-1\rvert=5-4p=\norm{\mathcal M_{AB}(\rho(p))}_*$, so the
minimum is the remaining, increasing term $1+4p$; symmetrically for
$p\ge\tfrac12$. Both pieces meet at $p=\tfrac12$ with value $3$, and each
piece is monotone away from that point, giving a unique maximum there.
\end{proof}

\begin{remark}[Numerical evidence for a sharp threshold at $3$]
\label{rem:sharp-threshold-conjecture}
Proposition~\ref{prop:bell-mixture-tent} shows $\Psi=3$ is achieved on a
biseparable (indeed, PPT-mixture) state, so the proven bound
$11/3\approx3.667$ of Proposition~\ref{prop:min-witness-bound} is not
tight. Multi-start Powell optimization of $\Psi$ over pure biseparable
states with up to six independently parametrized mixture components
(script \texttt{scripts/gme\_min\_witness\_powell\_search.py}), and
separately a Frank--Wolfe-type semidefinite relaxation over the full
PPT-mixture set with alternating dual-witness updates (script
\texttt{scripts/sdp\_ppt\_mixture.py}), both converge robustly to
$\Psi=3.000\ldots$ and find no biseparable or PPT-mixture state exceeding
it, including under substantial random perturbation of the dual witnesses
away from the known optimum. This is not a proof --- as with the
four-qubit $\Phi_{\mathrm{sym}}$ search of Section~\ref{sec:qubit-numerics},
the full biseparable simplex is not exhaustively certified --- but it is
considerably stronger evidence than a plain random search, since the SDP
runs over a strictly larger (PPT-mixture) set than the biseparable one.
We record
\begin{equation*}
\Psi(\rho)>3
\qquad\Longrightarrow\qquad
\rho\text{ is genuinely four-partite entangled}
\end{equation*}
as numerically very well supported, with $11/3$ as the currently proved
fallback threshold.
\end{remark}

\begin{example}[A witness comfortably above both thresholds]
\label{ex:min-witness-gme-certificate}
Unconstrained multi-start optimization of $\Psi$ over \emph{all} pure
four-qubit states (script
\texttt{scripts/gme\_min\_witness\_unconstrained\_search.py}) finds states
with $\Psi\approx4.588$, well above both $11/3$ and the conjectured
biseparable supremum $3$. The single-qubit marginal purities of the
optimizer found so far are $\approx0.33$--$0.50$ (not exactly equal, and
not maximally mixed), and it has no detected stabilizer structure,
distinguishing it from every state used elsewhere in this note; it is
reminiscent of the near-AME$(4,2)$ constructions motivated by the
nonexistence of a true four-qubit absolutely-maximally-entangled state.
Identifying a closed form for this optimizer, and determining whether
$\Psi$ is bounded above $5$ at all (the universal cap of
Corollary~\ref{cor:cluster-saturation-n4} applied to each individual
term), are left open.
\end{example}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\section{Symmetry-adapted block decomposition of the shadow map}
\label{sec:symmetry-blocks}

The constructions of Sections~\ref{sec:response-maps}--\ref{sec:multiparty-sources} treat the sector grading $V\subseteq S$, $T\subseteq S^c$ as the only available organizing structure on either side of a cut. When $\rho$ carries an additional symmetry --- exactly, or after an appropriate twirl --- this grading can be refined further, along representation-theoretic rather than combinatorial lines. The refinement is worth having for four distinct reasons, which we state before developing the formal statements, since they motivate different parts of what follows and are not equally strong.

\begin{itemize}
\item \emph{Explanatory power.} Several numerical facts already reported in this note --- the triple degeneracy of the singular values of $\mathcal M_a$ for $\GHZ_3$ (Section~\ref{sec:qubit-numerics}), the triple degeneracy at the Smolin state under both the $1\mid3$ and $2\mid2$ cuts (Section~\ref{sec:multiparty-sources}), and the rigid common value across all $38$ four-qubit graph states --- have so far been recorded as numerical observations. The block-diagonality and stabilizer-support results below show that such degeneracies are not coincidental: they are forced exactly, by two complementary mechanisms depending on whether $\rho$ is symmetry-invariant in a representation-theoretic sense or Pauli-diagonal in a stabilizer sense. This converts a family of separately verified numerical facts into structural statements.
\item \emph{Diagnostic resolution.} The sub-block witnesses of Corollary~\ref{cor:sub-block} resolve a violation only down to the level of \emph{which parties} are involved ($V,T$). A representation-theoretic decomposition, where applicable, resolves it further, down to \emph{which symmetry channel} within a fixed $(V,T)$ sector is responsible --- for instance, whether a two-party source correlation block is carrying its signal in a totally symmetric or in an antisymmetric combination of its constituents. This is invisible to the party-indexed grading alone.
\item \emph{Computational cost.} For a source or target cluster respecting a symmetry group $G$, an isotypic decomposition replaces one singular value decomposition on the full sector space by several independent, much smaller singular value decompositions on the multiplicity spaces $M_\lambda$, whose dimensions grow far more slowly than the ambient sector dimension as the cluster size increases. This is the natural computationally tractable foothold for the symmetric sub-family of the higher-order tensor construction discussed in the closing remark of Section~\ref{sec:tensor-viewpoint}.
\item \emph{A one-sided extension via twirling.} If $\rho$ itself lacks the relevant symmetry but a twirl $T_G(\rho)$ is cheap to evaluate, entanglement detected on $T_G(\rho)$ certifies entanglement of $\rho$ (twirling by local unitaries preserves separability), so a block-diagonal criterion can serve as an inexpensive pre-test, with the standard one-sided caveat that a negative result on $T_G(\rho)$ is uninformative about $\rho$.
\end{itemize}

What this refinement does \emph{not} deliver is a numerically sharper detection threshold: Remark~\ref{rem:no-universal-sharpening} below shows that no universal, state-independent improvement over Theorems~\ref{thm:cut-bound} and \ref{thm:cluster-cut} exists at the level of an individual representation-theoretic block, except in the single-isotype case where the improvement is one of concentration rather than of threshold value.


\subsection*{Setup: two tiers of hypothesis}

The two results below need genuinely different amounts of symmetry, and it is worth separating them explicitly rather than proving both under the stronger hypothesis throughout.

\emph{Weak (cut-factorizing) hypothesis.} Let $G$ be a compact group acting on the system by unitaries of the form $g\mapsto U_g^{(S)}\otimes U_g^{(S^c)}$ for some fixed bipartition $S\mid S^c$ -- i.e.\ $G$ acts unitarily on each side of the cut, with no further requirement on how $U_g^{(S)}$ itself decomposes over the parties inside $S$ -- and suppose $\rho$ is $G$-invariant. This induces adjoint representations $\mathrm{Ad}^{(S)}_g$ on $\V_0^{(S)}$ and $\mathrm{Ad}^{(S^c)}_g$ on $\V_0^{(S^c)}$, with isotypic decompositions
\begin{equation}
\V_0^{(S)}=\bigoplus_\lambda M_\lambda^{(S)}\otimes V_\lambda,
\qquad
\V_0^{(S^c)}=\bigoplus_\lambda M_\lambda^{(S^c)}\otimes V_\lambda,
\label{eq:isotypic-decomp}
\end{equation}
where $\lambda$ ranges over the irreducible representations of $G$ appearing on either side, $V_\lambda$ denotes a fixed model of the irreducible representation of dimension $d_\lambda$, and $M_\lambda^{(S)}$, $M_\lambda^{(S^c)}$ are the corresponding multiplicity spaces. This is exactly the same type of decomposition already used for the source and target sector gradings in Sections~\ref{sec:response-maps}--\ref{sec:multiparty-sources}, now taken with respect to a representation-theoretic rather than a combinatorial grading; the two coincide only in special cases.

This weak hypothesis is exactly what Proposition~\ref{prop:block-diagonal} below uses: its proof only invokes equivariance of $\widetilde{\mathcal M}_S(\rho)$ between the two representations $\mathrm{Ad}^{(S)}_g$, $\mathrm{Ad}^{(S^c)}_g$, and never asks how $U_g^{(S)}$ or $U_g^{(S^c)}$ further factor over the parties within $S$ or within $S^c$.

\emph{Strong (party-local) hypothesis.} Proposition~\ref{prop:sector-symmetry-compatible} and Corollary~\ref{cor:joint-refinement} need more: there $G$ must act by unitaries of the form $g\mapsto\bigotimes_{a\in P}U_g^{(a)}$, one factor per party, since the sector grading $V\subseteq S$ that these results are compatible with is itself defined party by party (Eq.~\eqref{eq:source-bloch-decomposition}). Every party-local action is in particular cut-factorizing for any bipartition, so the strong hypothesis is a special case of the weak one; the converse fails, as the remark following Corollary~\ref{cor:joint-refinement} makes concrete.

The general strategy --- twirl a state over a symmetry group and use Schur's lemma to collapse the resulting computation onto the much smaller multiplicity spaces --- is the same one used classically to simplify the computation of entanglement measures for symmetric states \cite{vollbrechtwerner2001}; the present section applies it to the shadow map itself rather than to a scalar entanglement measure.

\begin{proposition}[Exact block-diagonality]
\label{prop:block-diagonal}
Under the weak, cut-factorizing hypothesis above, $\widetilde{\mathcal M}_S(\rho)$ is block diagonal with respect to the isotypic decomposition~\eqref{eq:isotypic-decomp}: writing $\Pi_\lambda^{(S^c)}$, $\Pi_\mu^{(S)}$ for the isotypic projections,
\begin{equation*}
\Pi_\lambda^{(S^c)}\,\widetilde{\mathcal M}_S(\rho)\,\Pi_\mu^{(S)}=0
\qquad\text{whenever }\lambda\neq\mu,
\end{equation*}
exactly, not merely approximately or in a bound. Moreover, on the surviving diagonal blocks,
\begin{equation}
\widetilde{\mathcal M}_S(\rho)\big|_\lambda = A_\lambda\otimes\mathrm{id}_{V_\lambda}
\label{eq:wigner-eckart}
\end{equation}
for a unique linear map $A_\lambda:M_\lambda^{(S)}\to M_\lambda^{(S^c)}$, the reduced shadow map at $\lambda$.
\end{proposition}

\begin{proof}
$G$-invariance of $\rho$ gives $\mathrm{Ad}^{(S^c)}_g\,\widetilde{\mathcal M}_S(\rho)=\widetilde{\mathcal M}_S(\rho)\,\mathrm{Ad}^{(S)}_g$ for all $g$, i.e.\ $\widetilde{\mathcal M}_S(\rho)$ is a $G$-equivariant map between the two representations~\eqref{eq:isotypic-decomp}. Both claims are then Schur's lemma applied to the isotypic decomposition: equivariant maps vanish between inequivalent irreducible summands, and act as a fixed scalar multiple of the identity on the irreducible factor $V_\lambda$ within a matching pair, leaving exactly the freedom recorded in $A_\lambda$ on the multiplicity spaces.
\end{proof}

Proposition~\ref{prop:block-diagonal} organizes $\widetilde{\mathcal M}_S(\rho)$ by symmetry channel alone, treating $\V_0^{(S)}$ and $\V_0^{(S^c)}$ as undifferentiated representations of $G$. But these spaces already carry the combinatorial sector grading of Sections~\ref{sec:response-maps}--\ref{sec:multiparty-sources}, $\V_0^{(S)}=\bigoplus_{\emptyset\neq V\subseteq S}\V_V^{(S)}$. The next result shows that this grading and the isotypic one are never in tension: for any symmetry acting locally by conjugation, one refines the other, and both can be imposed simultaneously without contradiction.

\begin{proposition}[Compatibility of sector and symmetry decompositions]
\label{prop:sector-symmetry-compatible}
Let $G$ act on the system under the strong, party-local hypothesis, $g\mapsto\bigotimes_{a\in P}U_g^{(a)}$, and let $S\subseteq P$. Writing $R(g):=\bigotimes_{a\in S}\mathrm{Ad}_{U_g^{(a)}}$ for the induced action on $\V^{(S)}$, we have
\begin{equation}
R(g)\,\V_V^{(S)} = \V_V^{(S)}
\qquad\text{for every } g\in G \text{ and every nonempty } V\subseteq S.
\label{eq:sector-invariance}
\end{equation}
Consequently the sector projectors $P_V$ of Eq.~\eqref{eq:source-bloch-decomposition} commute with $R(g)$ for every $g\in G$, hence with every isotypic projector $P_\lambda=d_\lambda\int_G\chi_\lambda(g)^*R(g)\,dg$ of Eq.~\eqref{eq:isotypic-decomp}, and $\V^{(S)}$ decomposes simultaneously as
\begin{equation}
\V^{(S)} = \bigoplus_{\emptyset\neq V\subseteq S}\ \bigoplus_\lambda \V_{V,\lambda},
\qquad
\V_{V,\lambda}:=P_VP_\lambda\V^{(S)}.
\label{eq:joint-decomposition}
\end{equation}
\end{proposition}

\begin{proof}
Since $\mathrm{Ad}_{U_g^{(a)}}$ fixes the identity, $\mathrm{Ad}_{U_g^{(a)}}\id^{(a)}=\id^{(a)}$, and maps the traceless Hermitian subspace $\V_0^{(a)}$ to itself (conjugation by a unitary preserves both Hermiticity and tracelessness), each local factor space splits $G$-invariantly as $\mathbb C\,\id^{(a)}\oplus\V_0^{(a)}$. The sector space $\V_V^{(S)}$ is, by definition, exactly the span of product basis elements with the $a$-th factor in $\V_0^{(a)}$ for $a\in V$ and equal to $\id^{(a)}$ for $a\notin V$. Since $R(g)$ acts factor-wise and each local factor preserves its own identity-versus-traceless split, $R(g)$ cannot move an element of $\V_V^{(S)}$ out of $\V_V^{(S)}$: it maps trivial legs to trivial legs and active legs to (possibly rotated, but still traceless) active legs, without ever changing which legs are trivial. This gives $R(g)\V_V^{(S)}\subseteq\V_V^{(S)}$; the same argument applied to $g^{-1}$ gives the reverse inclusion, hence Eq.~\eqref{eq:sector-invariance}. Commutation of $P_V$ with $R(g)$, and therefore with the integral defining $P_\lambda$, follows immediately.
\end{proof}

\begin{corollary}[Joint refinement of the bigraduated shadow map]
\label{cor:joint-refinement}
Let $S\mid S^c$ be a cut preserved by $G$ as a set partition, with $G$ acting locally by conjugation on both $S$ and $S^c$ as above. Then, whenever $\rho$ is $G$-invariant, every bigraduated block $M_{V\to T}(\rho)$ of Definition~\ref{def:bigraduated} further decomposes as
\begin{equation}
M_{V\to T}(\rho) = \bigoplus_\lambda I_{d_\lambda}\otimes A_{V,T,\lambda}(\rho),
\qquad \emptyset\neq V\subseteq S,\ \ \emptyset\neq T\subseteq S^c,
\label{eq:joint-block}
\end{equation}
for reduced maps $A_{V,T,\lambda}$ on the corresponding multiplicity spaces. In particular, the combinatorial sector grading of Section~\ref{sec:multiparty-sources} and the representation-theoretic isotypic grading of the present section are not competing organizations of $\mathcal M_S(\rho)$, but two orthogonal refinements of the same operator, and can be applied jointly: one may first restrict to a sub-block witness $M_{V\to T}$ as in Corollary~\ref{cor:sub-block} and then further resolve it by symmetry channel, or apply the two refinements in the opposite order, with the same result.
\end{corollary}

\begin{proof}
By Proposition~\ref{prop:sector-symmetry-compatible} applied on the source side to $\V_V^{(S)}$ and, symmetrically, on the target side to $\V_T^{(S^c)}$, the restrictions $R_S(g)\big|_{\V_V^{(S)}}$ and $R_{S^c}(g)\big|_{\V_T^{(S^c)}}$ are themselves well-defined representations of $G$. Since $\rho$ is $G$-invariant, $\widetilde{\mathcal M}_S(\rho)$ intertwines $R_S(g)$ and $R_{S^c}(g)$ on the full spaces (Proposition~\ref{prop:block-diagonal}), and since the sector inclusion $\iota_V$ and sector projection $P_T$ used in Definition~\ref{def:bigraduated} are themselves $G$-equivariant by Proposition~\ref{prop:sector-symmetry-compatible}, the composite $M_{V\to T}(\rho)=P_T\,\widetilde{\mathcal M}_S(\rho)\,\iota_V$ intertwines the restricted representations on $\V_V^{(S)}$ and $\V_T^{(S^c)}$. Schur's lemma applied to this restricted intertwiner gives Eq.~\eqref{eq:joint-block}.
\end{proof}

\begin{remark}[Party permutation: an example separating the two hypotheses]
\label{rem:permutation-separating-example}
Let $\pi\in S_n$ be a permutation of the parties fixing $S$ setwise (i.e.\ $\pi(S)=S$, hence also $\pi(S^c)=S^c$), and let $U_\pi$ be the corresponding permutation unitary on $\mathcal H^{(P)}$. If $\rho$ is invariant under $U_\pi$, this is a cut-factorizing symmetry -- it acts as some unitary on $\mathcal H^{(S)}$ tensored with some unitary on $\mathcal H^{(S^c)}$ -- so Proposition~\ref{prop:block-diagonal} applies unchanged and $\widetilde{\mathcal M}_S(\rho)$ block-diagonalizes over the isotypic components of the induced permutation representations on $\V_0^{(S)}$, $\V_0^{(S^c)}$. But $U_\pi$ is not party-local whenever $\pi$ moves two or more parties within $S$: it acts as a genuine swap between the factors of $\mathcal H^{(S)}$, not as $\bigotimes_{a\in S}U_g^{(a)}$ for a single fixed single-party unitary $U_g$. Consequently Proposition~\ref{prop:sector-symmetry-compatible} does not apply, and indeed its conclusion can fail: $U_\pi$ can map a sector $\V_V^{(S)}$ to a different sector $\V_{\pi(V)}^{(S)}$ rather than fixing it, so the combinatorial sector grading and the permutation-isotypic grading are, in general, genuinely different, incompatible refinements of $\V_0^{(S)}$ -- unlike the party-local case of Corollary~\ref{cor:joint-refinement}, where the two gradings were shown to be compatible refinements of one another.

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}),
\end{equation*}
\begin{equation*}
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 $\pi\in H$,
\begin{equation}
\norm{\mathcal M_{\pi(a)}(\rho)}_*=\norm{\mathcal M_a(\rho)}_*.
\label{eq:transitive-equal-norms}
\end{equation}
Consequently, if $H$ acts transitively on $P$, the value $\norm{\mathcal M_a(\rho)}_*$ is independent of $a$, and
\begin{equation}
\Phi_{\mathrm{sym}}(\rho)=\Phi_{\max}(\rho)=\norm{\mathcal M_a(\rho)}_*
\label{eq:transitive-collapse}
\end{equation}
for any single choice of $a\in P$ -- in particular, computing one shadow map suffices instead of all $n$.
\end{lemma}

\begin{proof}
Fix $\pi\in H$ and $a\in P$. Relabeling parties by $\pi$ is a relabeling of tensor factors, so $\mathcal M_{\pi(a)}(U_\pi\rho U_\pi^\dagger)$ is, by construction, exactly the same operator as $\mathcal M_a(\rho)$ transported through the canonical identification of $\mathcal H^{(a)}$ with $\mathcal H^{(\pi(a))}$ (which exists since $d_a=d$ for all $a$); in particular it has the same singular values, so
\begin{equation*}
\norm{\mathcal M_{\pi(a)}\bigl(U_\pi\rho U_\pi^\dagger\bigr)}_*=\norm{\mathcal M_a(\rho)}_*.
\end{equation*}
By hypothesis $U_\pi\rho U_\pi^\dagger=\rho$, giving Eq.~\eqref{eq:transitive-equal-norms}. If $H$ is transitive, the orbit of any fixed $a$ under $H$ is all of $P$, so $\norm{\mathcal M_b(\rho)}_*$ takes the same value for every $b\in P$; Eq.~\eqref{eq:transitive-collapse} is then immediate from the definitions of $\Phi_{\mathrm{sym}}$, $\Phi_{\max}$.
\end{proof}

\subsection*{Exact formula and its consequence}

\begin{proposition}[Reduced matrix element formula]
\label{prop:reduced-formula}
Let $\rho=T_G(\rho_S\otimes\sigma_{S^c})$ be the $G$-twirl of a product state across $S\mid S^c$, so $\rho$ is separable (a mixture, over $g\in G$, of product states) and $G$-invariant. Let $r^{(S)}\in\V_0^{(S)}$, $v_{S^c}\in\V_0^{(S^c)}$ be the traceless Bloch vectors of $\rho_S$, $\sigma_{S^c}$, with isotypic components $r_\lambda$, $v_\lambda$. Then
\begin{equation}
A_\lambda=\frac{1}{d_\lambda}\,\tilde v_\lambda\,\tilde r_\lambda^{\,T},
\label{eq:reduced-element}
\end{equation}
where $\tilde r_\lambda$, $\tilde v_\lambda$ are $r_\lambda$, $v_\lambda$ reshaped as $(\dim M_\lambda^{(S)})\times d_\lambda$ and $(\dim M_\lambda^{(S^c)})\times d_\lambda$ matrices in a basis of $V_\lambda$ shared by both sides. Consequently,
\begin{equation}
\dim(V_\lambda)\,\norm{A_\lambda}_*\;\le\;\norm{r_\lambda}\,\norm{v_\lambda}.
\label{eq:per-block-bound}
\end{equation}
\end{proposition}

\begin{proof}
For fixed $g$, the product term $(\mathrm{Ad}^{(S)}_g r^{(S)})(\mathrm{Ad}^{(S^c)}_g v_{S^c})^T$ (unnormalized) is rank one, exactly as in the proof of Theorem~\ref{thm:cluster-cut}. Averaging over $g$ and expanding both factors in the isotypic bases gives, by the Schur orthogonality relation $\int_G D^\lambda(g)_{cb}D^\lambda(g)_{da}\,dg=\tfrac1{d_\lambda}\delta_{cd}\delta_{ab}$ for the (real, orthogonal) irreducible matrix elements $D^\lambda$, exactly Eq.~\eqref{eq:reduced-element} on each isotypic block, with all cross-$\lambda$ contributions vanishing by the same orthogonality relation applied to inequivalent irreducibles. Equation~\eqref{eq:per-block-bound} then follows from the standard nuclear-norm bound $\norm{AB}_*\le\norm{A}_\fro\norm{B}_\fro$ applied to Eq.~\eqref{eq:reduced-element}, using $\norm{\tilde r_\lambda}_\fro=\norm{r_\lambda}$, $\norm{\tilde v_\lambda}_\fro=\norm{v_\lambda}$.
\end{proof}

\begin{corollary}[Consistency with the cut-separable bound]
\label{cor:consistency}
Under the hypotheses of Proposition~\ref{prop:reduced-formula},
\begin{equation}
\sum_\lambda \dim(V_\lambda)\,\norm{A_\lambda}_*
\;\le\;
\norm{r^{(S)}}\,\norm{v_{S^c}}
\;\le\;
\sqrt{(d_S-1)(d_{S^c}-1)},
\label{eq:cs-recovery}
\end{equation}
the first inequality by Cauchy--Schwarz over $\lambda$ applied to Eq.~\eqref{eq:per-block-bound}, and the second by the correlation-sum identity used in Theorem~\ref{thm:cluster-cut}. By linearity and convexity of the nuclear norm, Eq.~\eqref{eq:cs-recovery} extends to arbitrary $G$-invariant separable states (finite or continuous mixtures of twirled product terms), recovering the bound of Theorem~\ref{thm:cluster-cut} through the block decomposition rather than around it.
\end{corollary}

\begin{remark}[No universal per-block sharpening, except by concentration]
\label{rem:no-universal-sharpening}
The first inequality in Eq.~\eqref{eq:cs-recovery} is generically strict: equality in Cauchy--Schwarz requires $\norm{r_\lambda}\propto\norm{v_\lambda}$ across all $\lambda$, which independently chosen $\rho_S,\sigma_{S^c}$ have no reason to satisfy. A direct numerical check on a twirled random product state gives $\norm{\mathcal M_S(\rho)}_*=0.429$ against the bound $\norm{r^{(S)}}\norm{v_{S^c}}=0.750$ from Eq.~\eqref{eq:cs-recovery} --- a strict, and generic, gap. Since Eq.~\eqref{eq:per-block-bound} bounds $\norm{r_\lambda}$ only by the global $\norm{r^{(S)}}^2\le d_S-1$, with no constraint on how the purity budget distributes across $\lambda$ for a generic $\rho_S$, no universal constant improving on $\sqrt{(d_S-1)(d_{S^c}-1)}$ holds for an individual block $\lambda$ in general.

The exception is the case where $\V_0^{(S)}$ itself carries only a single isotypic component under $G$: then $r_\lambda=r^{(S)}$ trivially, the full purity budget sits in the one available block, and Eq.~\eqref{eq:per-block-bound} becomes
\begin{equation*}
\dim(V_\lambda)\,\norm{A_\lambda}_*\;\le\;\sqrt{(d_S-1)(d_{S^c}-1)},
\end{equation*}
numerically identical to Theorem~\ref{thm:cut-bound}, but now a statement about a matrix of size $\dim M_\lambda^{(S^c)}\times1$ rather than the full target space. This single-isotype case is a genuine sharpening of concentration, not of threshold, and applies whenever $\rho$ is actually $G$-invariant for a group under which the source side is irreducible. It does \emph{not}, however, apply to the two states used to motivate this section: $\GHZ_3$ and the Smolin state carry no continuous collective symmetry (a direct check shows both fail to be invariant already under a one-parameter collective rotation), so the representation-theoretic mechanism above is not the explanation for their observed degeneracies. The correct explanation for those two states, and more generally for any Pauli-diagonal state, is combinatorial rather than representation-theoretic, and is developed next.
\end{remark}

\subsection*{Axial $U(1)$ symmetry: an exact worked example}
\label{sec:u1-example}

The isotypic mechanism of Proposition~\ref{prop:block-diagonal} is easiest to see concretely for an abelian symmetry group, where it reduces to an ordinary charge-conservation selection rule.

\begin{remark}[Abelian symmetries as selection rules]
\label{rem:abelian-selection-rule}
Let $G=U(1)$ act by collective $z$-rotation, $g=\theta\mapsto\bigotimes_{a\in P}R_z^{(a)}(\theta)$ with $R_z(\theta)=e^{-i\theta Z/2}$. Every irreducible representation of $U(1)$ is one-dimensional, so the isotypic decomposition of Proposition~\ref{prop:block-diagonal} coincides exactly with the eigenspace decomposition of the conserved total charge $S_z^{\mathrm{tot}}$. Writing $E_\pm=(X\pm iY)/\sqrt2$ for the weight-$(\pm1)$ combinations and $E_0=Z$ for the weight-$0$ direction, so that $\mathrm{Ad}_{R_z(\theta)}E_\pm=e^{\pm i\theta}E_\pm$ and $\mathrm{Ad}_{R_z(\theta)}E_0=E_0$, $G$-invariance of $\rho$ forces
\begin{equation*}
\tr\bigl(\rho\,\sigma_{i_S}\otimes\sigma_{i_T}\bigr)
=\tr\bigl(\rho\,\mathrm{Ad}_{R_z(\theta)}(\sigma_{i_S}\otimes\sigma_{i_T})\bigr)
=e^{i(q_S+q_T)\theta}\,\tr\bigl(\rho\,\sigma_{i_S}\otimes\sigma_{i_T}\bigr)
\qquad\text{for all }\theta,
\end{equation*}
where $q_S,q_T$ are the total weights of $\sigma_{i_S}$, $\sigma_{i_T}$ in the $\{E_+,E_0,E_-\}$ basis. Hence a source block of total charge $q_S$ couples \emph{only} to a target block of charge $q_T=-q_S$, not $q_T=q_S$: this is the familiar rule that a correlation function survives only between charge-conjugate directions (e.g.\ $\langle S^+S^-\rangle$ need not vanish, $\langle S^+S^+\rangle$ must). For a non-abelian group such as $SU(2)$ this simple picture does \emph{not} extend by using a Cartan generator (e.g.\ $J_z$) in place of the Casimir: diagonalizing $J_z$ only refines each isotypic block further, by magnetic quantum number, and neither reproduces the isotypic block structure of Proposition~\ref{prop:block-diagonal} nor explains why the reduced block $A_\lambda$ of Proposition~\ref{prop:reduced-formula} is independent of that quantum number --- that collapse is a genuine consequence of full non-abelian equivariance, not of commuting with a single conserved charge. The abelian case is special precisely because weight and isotype coincide there.
\end{remark}

To see this mechanism at work on a state that is not simultaneously covered by the stabilizer mechanism of Lemma~\ref{lem:stabilizer-degeneracy}, consider four qubits $A,B,C,D$ and
\begin{equation}
\lvert\psi_\varphi\rangle=\cos\varphi\,\lvert0011\rangle+\sin\varphi\,\lvert1100\rangle,
\qquad
\rho_\varphi=\lvert\psi_\varphi\rangle\!\langle\psi_\varphi\rvert,
\label{eq:u1-example-state}
\end{equation}
with $S=\{A,B\}$, $S^c=\{C,D\}$. Both computational-basis terms have Hamming weight $2$, so $\rho_\varphi$ commutes with the collective $U(1)_z$ generated by $R_z(\theta)^{\otimes4}$ for every $\varphi$, but for generic $\varphi$ it is invariant under no larger continuous collective symmetry. A direct computation gives
\begin{align*}
\tr(\rho_\varphi\,X^{\otimes4})&=\sin(2\varphi),
&
\tr(\rho_\varphi\,X_AX_BY_CY_D)&=-\sin(2\varphi),
\\
\tr(\rho_\varphi\,X_AY_BY_CX_D)&=\sin(2\varphi),
&
\tr(\rho_\varphi\,Z^{\otimes4})&=1,
\end{align*}
with all remaining nonzero-weight correlators vanishing identically; the first three are $\pm1$ only at the isolated points $\varphi=\pi/4\ (\mathrm{mod}\ \pi/2)$, so $\rho_\varphi$ is generically not Pauli-flat and Lemma~\ref{lem:stabilizer-degeneracy} does not apply to it.

For each two-qubit cluster $S$ and $S^c$, the local weight decomposition $\{-1,0,+1\}\otimes\{-1,0,+1\}$ gives total-charge sectors $q=-2,-1,0,1,2$ of dimension $1,2,3,2,1$ respectively (the coefficients of $(x^{-1}+1+x)^2$), on both the source and the target side. By Remark~\ref{rem:abelian-selection-rule} the normalized bigraduated map $\mathcal M_{AB}(\rho_\varphi)$ of Eq.~\eqref{eq:bigraduated-map} block-diagonalizes according to $q_S=-q_T$, and an exact computation of every block gives

\begin{center}
\begin{tabular}{lccc}
\toprule
$q_S\to q_T$ & block size & singular value(s) & origin \\
\midrule
$+2\to-2$ & $1\times1$ & $\tfrac23\sin(2\varphi)$ & $E_+E_+\to E_-E_-$ \\
$+1\to-1$ & $2\times2$ & $0,0$ & identically zero for every $\varphi$ \\
$0\to0$ & $3\times3$, rank $1$ & $\tfrac13$ & entirely from $E_0E_0\to E_0E_0$, i.e.\ $ZZ\to ZZ$ \\
$-1\to+1$ & $2\times2$ & $0,0$ & identically zero for every $\varphi$ \\
$-2\to+2$ & $1\times1$ & $\tfrac23\sin(2\varphi)$ & $E_-E_-\to E_+E_+$ \\
\bottomrule
\end{tabular}
\end{center}
so that, exactly,
\begin{equation}
\norm{\mathcal M_{AB}(\rho_\varphi)}_* = \frac13+\frac43\sin(2\varphi).
\label{eq:u1-example-formula}
\end{equation}
The $q=0$ value $\tfrac13$ is $\varphi$-independent, because $ZZZZ$ is diagonal in the computational basis and both terms of $\lvert\psi_\varphi\rangle$ give it the same eigenvalue $+1$; the entire $\varphi$-dependence, and with it the entanglement signal, sits in the $q=\pm2$ sectors. Equation~\eqref{eq:u1-example-formula} exceeds the cut-separable bound of Theorem~\ref{thm:cluster-cut} exactly when $\sin(2\varphi)>1/2$, i.e.\ for $\varphi\in(\pi/12,\,5\pi/12)$; it vanishes at $\varphi=0$, where $\rho_\varphi$ is itself a product state across $S\mid S^c$, and reaches $\norm{\mathcal M_{AB}(\rho_\varphi)}_*=5/3$ at the maximally superposed point $\varphi=\pi/4$.

This example illustrates Corollary~\ref{cor:joint-refinement} concretely: the full $9\times9$ block factors, without any state-specific input, into one $3\times3$ block of rank $1$, two mutually zero $2\times2$ blocks, and two $1\times1$ blocks, purely by charge conservation --- before any computation is needed to know which entries can possibly be nonzero. Since $\rho_\varphi$ is generically not a stabilizer state, this degeneracy pattern cannot be attributed to Lemma~\ref{lem:stabilizer-degeneracy}, giving a concrete instance of the separation of mechanisms described in general terms by Remark~\ref{rem:two-mechanisms}.

\subsection*{Exact degeneracy from stabilizer structure}

The degeneracies recorded for $\GHZ_3$, the Smolin state, and all $38$ four-qubit graph states share no continuous symmetry of the kind used above. Their common origin is instead a discrete, combinatorial fact about \emph{Pauli-diagonal} states, requiring only elementary group theory over $\mathbb F_2$, and it is this mechanism --- not Proposition~\ref{prop:block-diagonal} --- that is responsible for every degeneracy reported in Sections~\ref{sec:multiparty-sources} and~\ref{sec:qubit-numerics}. That reduced density matrices of stabilizer states are maximally mixed on their support, with the entanglement across any cut given by the rank of the corresponding restriction of the stabilizer group, is a classical fact \cite{fattal2004stabilizer}; Lemma~\ref{lem:code-support} below is the flat-support statement underlying that result, and Lemma~\ref{lem:stabilizer-degeneracy} translates it directly into the singular-value structure of the shadow map itself, rather than into an entanglement entropy.

\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 $\widetilde{\mathcal M}_S(\rho_H)$, expressed directly in the (non-orthonormal) raw Pauli-string basis $\{\sigma_{\vec i}\}$ on both sides --- the basis and normalization in which the correlation tensor $C_S(\rho)$ of Eq.~\eqref{eq:corr-tensor-def} and the combined shadow map $\mathcal M_S(\rho)$ of Definition~\ref{def:bigraduated} are actually computed --- has exactly $|\ker\varphi|$ nonzero entries, all of magnitude $1$, with pairwise disjoint column supports across distinct rows; consequently
\[
\widetilde{\mathcal M}_S(\rho_H) \text{ has exactly } |\operatorname{im}\varphi|-1 \text{ equal nonzero singular values, each } =\sqrt{|\ker\varphi|}.
\]
After the normalization of Eq.~\eqref{eq:bigraduated-map}, the singular values of $\mathcal M_S(\rho_H)$ are therefore
\[
\sqrt{\frac{|\ker\varphi|}{(d_S-1)(d_{S^c}-1)}},
\]
with multiplicity $|\operatorname{im}\varphi|-1$.
\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. The final rescaling is exactly the normalization already applied in Eq.~\eqref{eq:bigraduated-map}.
\end{proof}

\begin{corollary}
\label{cor:stabilizer-examples}
The Lemma applies uniformly to pure stabilizer states ($|H|=2^n$, including all graph states and $\GHZ_n$) and to 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 $\GHZ_3$ with $S=\{1\}$: $H=\{III,XXX,ZZI,ZIZ,IZZ,YYX,YXY,XYY\}$ (up to signs), $|\ker\varphi|=2$, $|\operatorname{im}\varphi|=4$, $(d_S-1)(d_{S^c}-1)=1\times3=3$, giving $3$ singular values equal to $\sqrt{2/3}$ --- exactly the value reported in Section~\ref{sec:qubit-numerics}.

For the Smolin state, every $1\mid3$ and $2\mid2$ cut has $\varphi$ bijective ($|\ker\varphi|=1$), so $|\operatorname{im}\varphi|-1=3$ in both cases --- bijectivity of $\varphi$ holds for \emph{any} nonempty proper subset $S$ of the four legs given this particular $H$. This gives $3$ equal singular values throughout, with value
\[
\sqrt{\frac{1}{(d_S-1)(d_{S^c}-1)}}
=
\begin{cases}
1/\sqrt7 & \text{for the } 1\mid3 \text{ cut } ((d_S-1)(d_{S^c}-1)=1\times7),\\[2pt]
1/3 & \text{for the } 2\mid2 \text{ cut } ((d_S-1)(d_{S^c}-1)=3\times3),
\end{cases}
\]
matching Section~\ref{sec:multiparty-sources} and Section~\ref{sec:qubit-numerics} exactly.
\end{corollary}

\begin{remark}
The $y$-parity grading occasionally useful for real-in-the-computational-basis states 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}

\begin{remark}[Two complementary mechanisms]
\label{rem:two-mechanisms}
Proposition~\ref{prop:block-diagonal} and Lemma~\ref{lem:stabilizer-degeneracy} are complementary, not competing, and apply under disjoint hypotheses. The representation-theoretic mechanism applies whenever $\rho$ is genuinely $G$-invariant under some compact group $G$ acting by local unitaries and preserving the cut, regardless of whether $\rho$ is Pauli-diagonal; it says nothing about states, such as a generic finite-group-symmetric state built from a permutation representation, that are not Pauli-diagonal. The stabilizer mechanism applies whenever $\rho$ is (a uniform mixture over) a stabilizer code state, regardless of whether it possesses any continuous symmetry at all --- as is the case for $\GHZ_n$, the Smolin state, and every graph state used elsewhere in this note, none of which is invariant under a nontrivial continuous collective symmetry. In the (comparatively narrow) overlap where a state is both $G$-invariant for some continuous $G$ and Pauli-diagonal, both mechanisms apply and constrain the same block structure from different directions; outside that overlap, exactly one of the two is available, and it is this Lemma, not Proposition~\ref{prop:block-diagonal}, that accounts for every numerically observed degeneracy reported so far in this note.
\end{remark}

\subsection*{Degenerate perturbation theory for the shadow map}
\label{sec:degenerate-perturbation}

The states used throughout this note to calibrate the degeneracy
mechanisms of Proposition~\ref{prop:block-diagonal} and
Lemma~\ref{lem:stabilizer-degeneracy} --- $\GHZ_3$, the Smolin state,
the four-qubit graph states --- are all highly symmetric points at
which the shadow map has repeated singular values. A natural next
question is how these degeneracies behave under a generic perturbation
away from the symmetric point, e.g.\ under physically motivated local
noise. Since $\rho\mapsto\mathcal M_S(\rho)$ is exactly linear
(Eq.~\eqref{eq:bigraduated-map}), the map itself carries no new
structure under $\rho(\varepsilon)=\rho_0+\varepsilon\,\delta\rho$: one
has $\mathcal M_S(\rho(\varepsilon))=\mathcal M_S(\rho_0)+\varepsilon\,
\mathcal M_S(\delta\rho)$ identically, for every $\varepsilon$. All of
the interesting structure is instead in the \emph{nuclear norm}
$\norm{\mathcal M_S(\rho(\varepsilon))}_*$, which is only convex, not
linear, and is generically non-differentiable exactly at points of
repeated singular value \cite{lewissendov2005}. We now make this
precise and then work it out completely for $\GHZ_3$.

\begin{theorem}[First-order splitting of degenerate shadow singular values]
\label{thm:degenerate-svd-perturbation}
Let $M:=\mathcal M_S(\rho_0)\in\R^{N\times n}$ have an $m$-fold
degenerate singular value $\sigma_0>0$, with orthonormal singular
bases $U_0\in\R^{N\times m}$, $V_0\in\R^{n\times m}$ satisfying
$MV_0=\sigma_0 U_0$, $M^TU_0=\sigma_0V_0$. For a perturbation direction
$\delta\rho$ with $\rho(\varepsilon):=\rho_0+\varepsilon\,\delta\rho$ a
valid state for $\varepsilon$ in some interval around $0$, define the
symmetric compression
\begin{equation}
K:=\frac12\Bigl(U_0^T\,\mathcal M_S(\delta\rho)\,V_0
+V_0^T\,\mathcal M_S(\delta\rho)^T\,U_0\Bigr)\in\R^{m\times m}.
\label{eq:K-matrix-def}
\end{equation}
Then the $m$ singular values of $\mathcal M_S(\rho(\varepsilon))$
branching from $\sigma_0$ satisfy
\begin{equation}
\sigma_i(\varepsilon)=\sigma_0+\varepsilon\,\lambda_i(K)+O(\varepsilon^2),
\qquad i=1,\dots,m,
\label{eq:svd-splitting}
\end{equation}
where $\lambda_i(K)$ are the (real) eigenvalues of $K$.
\end{theorem}

\begin{proof}
Embed $M$ into the symmetric Jordan--Wielandt dilation
$H(M):=\begin{pmatrix}0&M\\M^T&0\end{pmatrix}\in\R^{(N+n)\times(N+n)}$.
If $M=\sum_k\sigma_k u_kv_k^T$ is a singular value decomposition, then
$H(M)$ has eigenpairs $\bigl(\pm\sigma_k,\,\tfrac1{\sqrt2}(u_k,\pm v_k)\bigr)$
for every $k$, together with a zero eigenspace filling out the
orthogonal complements on each side; this is a standard fact (see
e.g.\ \cite{kato1995perturbation}). Since $\rho\mapsto\mathcal M_S(\rho)$
is linear, $H(M(\rho(\varepsilon)))=H(M(\rho_0))+\varepsilon\,H(M(\delta\rho))$
exactly, i.e.\ $H$ depends linearly on $\varepsilon$. The eigenvalue
$\sigma_0$ of the real symmetric matrix $H(M(\rho_0))$ has eigenspace
spanned by the orthonormal vectors $w_k:=\tfrac1{\sqrt2}(u_{0,k},v_{0,k})$,
$k=1,\dots,m$ (columns of $U_0$, $V_0$). Ordinary degenerate
Rayleigh--Schr\"odinger perturbation theory for symmetric matrices
\cite{kato1995perturbation} then gives that the $m$ eigenvalues of
$H(M(\rho(\varepsilon)))$ branching from $\sigma_0$ equal
$\sigma_0+\varepsilon\,\lambda_i(W^TH(M(\delta\rho))W)+O(\varepsilon^2)$,
where $W=(w_1,\dots,w_m)$. A direct computation gives
$(W^TH(M(\delta\rho))W)_{kl}=\tfrac12\bigl(u_{0,k}^T\mathcal M_S(\delta\rho)v_{0,l}
+v_{0,k}^T\mathcal M_S(\delta\rho)^Tu_{0,l}\bigr)=K_{kl}$, exactly
Eq.~\eqref{eq:K-matrix-def}. Restricting to the $+\sigma_0$ branch
(assumed here nondegenerate with respect to $-\sigma_0$, which holds
throughout the example below) gives Eq.~\eqref{eq:svd-splitting}.
\end{proof}

\begin{corollary}[Nuclear-norm derivative]
\label{cor:nuclear-norm-first-order}
If $\sigma_0$ is the only singular value of $\mathcal M_S(\rho_0)$
(as is the case for $\GHZ_3$, Section~\ref{sec:qubit-numerics}), then
\begin{equation*}
\left.\frac{d}{d\varepsilon}\right|_{\varepsilon=0}
\norm{\mathcal M_S(\rho(\varepsilon))}_*=\tr(K).
\end{equation*}
More generally this is the contribution of the $\sigma_0$-block alone;
non-degenerate singular values contribute their own
$u_{0}^T\mathcal M_S(\delta\rho)v_{0}$ term individually, and the total
derivative is the sum over all blocks.
\end{corollary}

\begin{remark}[Why a compression matrix, not a higher-order tensor]
\label{rem:nonsmoothness-lewis-sendov}
This answers directly the question that motivated this section: the
first-order object governing how a degenerate shadow singular value
splits is neither the shadow map itself (which stays linear, hence
featureless) nor a higher-order tensor, but a single real symmetric
$m\times m$ matrix per degenerate block --- an ordinary eigenvalue
problem, not an SVD. This is the shadow-map instance of the general
fact that unitarily invariant matrix norms such as $\norm{\cdot}_*$ are
differentiable at simple spectra but only \emph{directionally}
differentiable at repeated ones, with the one-sided directional
derivative controlled by exactly this kind of compressed, symmetrized
perturbation \cite{lewissendov2005}. Two mechanisms of this note make
degenerate blocks the rule rather than the exception:
Proposition~\ref{prop:block-diagonal} forces exact degeneracy whenever
$\rho_0$ has a compatible continuous symmetry, and
Lemma~\ref{lem:stabilizer-degeneracy} forces it combinatorially for
any Pauli-diagonal (stabilizer) state --- so every symmetric or
stabilizer state used numerically in this note sits at a kink of
$\norm{\mathcal M_S(\cdot)}_*$, and Theorem~\ref{thm:degenerate-svd-perturbation}
is the tool for resolving that kink in any chosen direction.
\end{remark}

\subsection*{Exact rescaling under single-qubit Pauli dephasing}
\label{sec:pauli-dephasing-exact}

Before specializing Theorem~\ref{thm:degenerate-svd-perturbation} to an
example, it is worth isolating one perturbation direction for which
the first-order theory above turns out to be exact to \emph{all}
orders. The mechanism is a completely general, state-independent fact
about single-qubit Pauli dephasing, with no reference to symmetry or
to the stabilizer formalism.

\begin{proposition}[Exact diagonal rescaling under Pauli dephasing]
\label{prop:pauli-dephasing-exact}
Fix a qubit party $b\in P$ and a Pauli label $p\in\{1,2,3\}$
(i.e.\ $\sigma_p\in\{X,Y,Z\}$ on $b$). For $q\in[0,1]$ define the
two-outcome dephasing channel
\begin{equation*}
\Lambda_b^p(\rho;q):=q\,\rho+(1-q)\,\sigma_p^{(b)}\rho\,\sigma_p^{(b)}.
\end{equation*}
Then for every state $\rho$ and every Pauli string index
$\vec\imath=(i_1,\dots,i_n)\in\{0,1,2,3\}^n$,
\begin{equation}
\tr\bigl[\Lambda_b^p(\rho;q)\,\sigma_{\vec\imath}\bigr]
=w(i_b)\,\tr[\rho\,\sigma_{\vec\imath}],
\qquad
w(i_b):=\begin{cases}1,& i_b\in\{0,p\},\\ 2q-1,& i_b\in\{1,2,3\}\setminus\{p\},\end{cases}
\label{eq:pauli-dephasing-rescale}
\end{equation}
for every $q\in[0,1]$, exactly --- not merely to leading order in
$1-q$. Consequently every shadow map, sub-block, or sector profile of
$\Lambda_b^p(\rho;q)$ is obtained from the corresponding object for
$\rho$ by rescaling each Pauli-basis coefficient with $b$-leg index
$i_b$ anticommuting with $\sigma_p$ by the fixed factor $2q-1$, and
leaving every coefficient with $i_b$ commuting with $\sigma_p$
unchanged, for every $q\in[0,1]$.
\end{proposition}

\begin{proof}
Conjugation by the involution $\sigma_p^{(b)}$ satisfies
$\sigma_p^{(b)}\sigma_{\vec\imath}\sigma_p^{(b)}=\epsilon(i_b)\,\sigma_{\vec\imath}$,
with $\epsilon(i_b)=+1$ if the single-qubit factor $\sigma_{i_b}$
commutes with $\sigma_p$ (i.e.\ $i_b\in\{0,p\}$) and $\epsilon(i_b)=-1$
if it anticommutes ($i_b\in\{1,2,3\}\setminus\{p\}$). Hence
\begin{equation*}
\tr\bigl[\Lambda_b^p(\rho;q)\sigma_{\vec\imath}\bigr]
=q\,\tr[\rho\sigma_{\vec\imath}]+(1-q)\,\tr\bigl[\sigma_p^{(b)}\rho\,\sigma_p^{(b)}\sigma_{\vec\imath}\bigr]
=\bigl(q+(1-q)\epsilon(i_b)\bigr)\tr[\rho\sigma_{\vec\imath}],
\end{equation*}
using cyclicity of the trace. This is exactly
Eq.~\eqref{eq:pauli-dephasing-rescale}, since $q+(1-q)=1$ and
$q-(1-q)=2q-1$. Every shadow-map quantity in this note is a linear,
coordinate-wise function of these coefficients, so the claimed
rescaling of shadow maps follows immediately.
\end{proof}

\subsection*{Worked example: dephasing the $\GHZ_3$ state}
\label{ex:ghz3-dephasing}

Recall from Section~\ref{sec:qubit-numerics} that for
$\rho_0=\lvert\GHZ_3\rangle\!\langle\GHZ_3\rvert$ and source party $A$,
the normalized map $\mathcal M_A(\rho_0)=\sigma_0U_0$ has the single
triply-degenerate singular value $\sigma_0=\sqrt{2/3}=\sqrt6/3$
(forced exactly by Lemma~\ref{lem:stabilizer-degeneracy} and
Corollary~\ref{cor:stabilizer-examples}, since $\GHZ_3$ carries no
continuous collective symmetry (cf. Remark~\ref{rem:no-universal-sharpening}),
with $V_0=\id_3$ (the standard $x,y,z$ source basis) and $U_0$ the
explicit orthonormal $15\times3$ matrix reproduced by
\texttt{scripts/ghz3\_shadow\_map\_symbolic.py}. We study the effect of
single-qubit dephasing on party $B$, i.e.\ the direction
$\delta\rho^{(P)}:=P_B\rho_0P_B-\rho_0$ for $P\in\{X,Y,Z\}$, physically
the dominant error channel ($Z$: $T_2$-type dephasing in the
computational/stabilizer basis) versus dephasing along an axis
transverse to it ($X$, $Y$).

\begin{lemma}[Diagonal perturbations rescale a degenerate block directly]
\label{lem:diagonal-perturbation}
In the setting of Theorem~\ref{thm:degenerate-svd-perturbation},
suppose $\mathcal M_S(\delta\rho)V_0=\sigma_0U_0D$ for some real
diagonal matrix $D=\diag(d_1,\dots,d_m)$. Then $K=\sigma_0D$ exactly,
so $\lambda_i(K)=\sigma_0d_i$.
\end{lemma}
\begin{proof}
$K=U_0^T\mathcal M_S(\delta\rho)V_0=U_0^T(\sigma_0U_0D)=\sigma_0(U_0^TU_0)D=\sigma_0D$,
using orthonormality of $U_0$; $D$ diagonal is already symmetric, so
no further symmetrization is needed.
\end{proof}

For $\delta\rho^{(Z)}$, every nonzero Pauli coefficient of $\rho_0$
with $A$-index $z$ (the two-body terms $ZZI,ZIZ$ of
Corollary~\ref{cor:stabilizer-examples}) has $B$-index in $\{0,z\}$,
commuting with $Z$; every nonzero coefficient with $A$-index $x$ or $y$
(the three-body terms $XXX,YXY,XYY,YYX$) has $B$-index in $\{x,y\}$,
anticommuting with $Z$. By Proposition~\ref{prop:pauli-dephasing-exact}
this is exactly the diagonal case of Lemma~\ref{lem:diagonal-perturbation}
with $D=\diag(-2,-2,0)$ (columns $x,y$ entirely anticommuting, column
$z$ entirely commuting), giving
\begin{equation}
K_Z=\sigma_0\diag(-2,-2,0)=\diag\Bigl(-\tfrac{2\sqrt6}3,\,-\tfrac{2\sqrt6}3,\,0\Bigr).
\label{eq:KZ-result}
\end{equation}
For $\delta\rho^{(X)}$ and $\delta\rho^{(Y)}$ the anticommuting and
commuting coefficients are no longer aligned with the $x,y$ vs.\ $z$
source columns (each column now mixes a commuting and an anticommuting
entry), so Lemma~\ref{lem:diagonal-perturbation} does not apply
directly; a direct evaluation of Eq.~\eqref{eq:K-matrix-def}
(reproduced exactly by \texttt{scripts/ghz3\_perturbation\_symbolic.py})
gives instead the isotropic results
\begin{equation}
K_X=K_Y=-\frac{\sqrt6}3\,\id_3.
\label{eq:KXY-result}
\end{equation}

\begin{table}[h]
\centering
\small
\begin{tabular}{@{}lll@{}}
\toprule
dephasing axis on $B$ & eigenvalues of $K$ & interpretation \\
\midrule
$Z$ (computational/stabilizer basis) & $-\tfrac{2\sqrt6}3,\,-\tfrac{2\sqrt6}3,\,0$ & $z$-channel exactly protected, $x,y$ decay twice as fast \\
$X$ & $-\tfrac{\sqrt6}3,\,-\tfrac{\sqrt6}3,\,-\tfrac{\sqrt6}3$ & isotropic decay, no protected direction \\
$Y$ & $-\tfrac{\sqrt6}3,\,-\tfrac{\sqrt6}3,\,-\tfrac{\sqrt6}3$ & isotropic decay, no protected direction \\
\bottomrule
\end{tabular}
\caption{First-order splitting of the triply-degenerate singular value
$\sigma_0=\sqrt6/3$ of $\mathcal M_A(\GHZ_3)$ under single-qubit
dephasing on party $B$, from Eqs.~\eqref{eq:KZ-result}--\eqref{eq:KXY-result}.
By Corollary~\ref{cor:nuclear-norm-first-order},
$\tr K_Z=-\tfrac{4\sqrt6}3$ and $\tr K_X=\tr K_Y=-\sqrt6$, matching a
direct finite-difference evaluation of $\norm{\mathcal M_A(\rho_0+
\varepsilon\delta\rho^{(P)})}_*$ to six digits at $\varepsilon=10^{-2},10^{-3}$.}
\label{tab:ghz3-dephasing}
\end{table}

The zero eigenvalue under $Z$-dephasing is not merely a first-order
artifact: it upgrades to an exact, non-perturbative statement once the
perturbation \emph{direction} $\delta\rho^{(Z)}$ is replaced by the
genuine one-parameter channel it generates.

\begin{corollary}[Exact protection of the $z$-channel, all orders]
\label{cor:ghz3-exact-channel}
Let $\rho(q):=\Lambda_B^{3}(\rho_0;q)=q\rho_0+(1-q)Z_B\rho_0Z_B$,
$q\in[0,1]$ (a valid state for every $q$, unlike the affine family
$\rho_0+\varepsilon\,\delta\rho^{(Z)}$, which leaves the state space
once $\varepsilon\notin[0,1]$). Then, exactly,
\begin{equation*}
\mathcal M_A(\rho(q))=\sigma_0\Bigl[(2q-1)\,u_xe_x^T+(2q-1)\,u_ye_y^T+1\cdot u_ze_z^T\Bigr],
\end{equation*}
already in singular value form, with singular values
\begin{equation*}
\sigma_x(q)=\sigma_y(q)=\lvert2q-1\rvert\,\sigma_0,
\qquad
\sigma_z(q)=\sigma_0,
\end{equation*}
for every $q\in[0,1]$. Writing $q=1-\varepsilon$ recovers
Eq.~\eqref{eq:svd-splitting} with $K_Z$ of Eq.~\eqref{eq:KZ-result} as
the exact leading term near $\varepsilon=0$, with no further
corrections until the singular values $\sigma_x,\sigma_y$ hit zero at
$q=1/2$.
\end{corollary}

\begin{proof}
By Proposition~\ref{prop:pauli-dephasing-exact}, every coefficient
with $A$-index $z$ is left unchanged and every coefficient with
$A$-index $x$ or $y$ is rescaled by $(2q-1)$, for every $q$ --- this is
exactly the source-column structure identified above, now exact rather
than first order. Since $U_0,V_0=\id_3$ already diagonalize
$\mathcal M_A(\rho_0)$ with equal singular value $\sigma_0$ on every
column, rescaling column $x,y$ by $(2q-1)$ and leaving column $z$ fixed
preserves this diagonal (singular-value) form exactly, with the stated
singular values.
\end{proof}

\begin{remark}[Why $Z$ is special, and the connection to noiseless subsystems]
The exactness in Corollary~\ref{cor:ghz3-exact-channel} is special to
$Z$-dephasing: it holds precisely because the stabilizer generators of
$\GHZ_3$ that carry the source index $z$ ($ZZI,ZIZ$) are themselves
built from $Z$ on every leg, hence commute with $Z_B$ termwise, while
every generator carrying source index $x$ or $y$ ($XXX,YXY,XYY,YYX$)
anticommutes with $Z_B$ uniformly. This is the same combinatorial
stabilizer mechanism as Lemma~\ref{lem:stabilizer-degeneracy}, now
controlling not just the degeneracy of $\mathcal M_A(\rho_0)$ but also
its exact response to a symmetry-breaking channel. Physically, the
$A\!\to\! B,C$ correlation carried by $Z_AZ_B$ and $Z_AZ_C$ is a
decoherence-free direction for exactly the noise that commutes with
it, in the sense familiar from the noiseless-subsystem and operator
quantum error correction literature
\cite{zanardirasetti1997,kribslaflammepoulin2005}: here the roles of
``logical'' information and ``noise operator'' are played by a
single fixed shadow-map correlation channel rather than by an encoded
qubit, but the underlying commutation argument is the same one. By
contrast, $X$- and $Y$-dephasing anticommute with two of the three
stabilizer generators carrying each source index and commute with the
third, so no single source column is uniformly protected or uniformly
decaying in the exact (all-$q$) sense; Eq.~\eqref{eq:KXY-result}
therefore describes only the leading-order behavior near $q=1$, not a
closed global formula of the type in Corollary~\ref{cor:ghz3-exact-channel}.
\end{remark}


\subsection*{A polynomial-size witness for permutation-symmetric cluster sources}
\label{sec:multinomial-collapse}

Remark~\ref{rem:computational-cost} raised the cost of evaluating $\mathcal M_S(\rho)$
as a genuine obstruction once $|S|$ or $|S^c|$ grows. We now show that this
obstruction disappears entirely for a natural and large class of states: those
invariant under permuting the parties \emph{within} $S$ and \emph{within} $S^c$
separately. This is an instance of the weak, cut-factorizing hypothesis of
Proposition~\ref{prop:block-diagonal} --- not the strong, party-local one, since a
transposition of two parties inside $S$ is not of the form $U^{(a)}\otimes U^{(b)}$
(compare Remark~\ref{rem:permutation-separating-example}) --- specialized to the
full symmetric group, whose isotypic decomposition on a Pauli-string alphabet
collapses to an elementary orbit-counting argument.

\paragraph{Setup.} Fix qubits ($d_a=2$) for concreteness (Remark~\ref{rem:qudit-collapse}
below records the general-$d$ statement). Fix a source cluster $S$ with $|S|=m$
and let $S^c$ be its complement, $|S^c|=l:=n-m$. Consider the single bigraduated
block $M_{S\to S^c}(\rho)$ of Definition~\ref{def:bigraduated} with $V=S$, $T=S^c$
--- the ``fully active'' block already singled out after
Corollary~\ref{cor:sub-block} as the witness for genuine joint $S$-cluster
correlation. In the orthonormal Pauli-string basis this is the
$3^m\times 3^l$ matrix
\begin{equation*}
c(\vec\imath,\vec\jmath):=\tr\bigl(\rho\,\sigma_{\vec\imath}\otimes\sigma_{\vec\jmath}\bigr),
\qquad
\vec\imath\in\{x,y,z\}^m,\ \vec\jmath\in\{x,y,z\}^l.
\end{equation*}
For $\vec\imath\in\{x,y,z\}^m$ let $\mathrm{type}(\vec\imath)=(a_x,a_y,a_z)\in\mathcal T_m$,
$\mathcal T_m:=\{(a_x,a_y,a_z)\in\mathbb Z_{\ge0}^3: a_x+a_y+a_z=m\}$, record how
often each label occurs; the symmetric group $S_m$ acts on $\{x,y,z\}^m$ by
permuting positions, with orbits exactly the level sets of $\mathrm{type}$, and the
orbit of type $\alpha$ has size the multinomial coefficient $\binom m\alpha:=\frac{m!}{a_x!a_y!a_z!}$.
Note $|\mathcal T_m|=\binom{m+2}{2}$, polynomial in $m$, against the ambient $3^m$.

\begin{proposition}[Multinomial block collapse]
\label{prop:multinomial-collapse}
Suppose $\rho$ is invariant under $U_\pi\otimes U_{\pi'}$ for every $\pi\in S_m$
acting by the permutation unitary on $\mathcal H^{(S)}$ and every $\pi'\in S_l$
acting by the permutation unitary on $\mathcal H^{(S^c)}$. Then $c(\vec\imath,\vec\jmath)$
depends on $(\vec\imath,\vec\jmath)$ only through $(\mathrm{type}(\vec\imath),\mathrm{type}(\vec\jmath))\in\mathcal T_m\times\mathcal T_l$;
write $\tilde c(\alpha,\beta)$ for this common value, and define the
multinomial-weighted reduced matrix $\hat C\in\mathbb C^{|\mathcal T_m|\times|\mathcal T_l|}$ by
\begin{equation}
\hat C_{\alpha,\beta}:=\sqrt{\binom m\alpha\binom l\beta}\;\tilde c(\alpha,\beta).
\label{eq:reduced-dicke-matrix}
\end{equation}
Then $M_{S\to S^c}(\rho)$ and $\hat C$ have identical nonzero singular values
(the smaller matrix effectively zero-padded), so in particular
\begin{equation*}
\norm{M_{S\to S^c}(\rho)}_*=\norm{\hat C}_*,
\qquad
\norm{M_{S\to S^c}(\rho)}_{\fro}=\norm{\hat C}_{\fro}.
\end{equation*}
\end{proposition}

\begin{proof}
$U_\pi$-invariance of $\rho$ gives $\tr(\rho\,\sigma_{\pi\vec\imath}\otimes\sigma_{\vec\jmath})
=\tr(U_\pi^\dagger\rho U_\pi\,\sigma_{\vec\imath}\otimes\sigma_{\vec\jmath})=c(\vec\imath,\vec\jmath)$
for every $\pi\in S_m$, since conjugating $\rho$ by $U_\pi$ permutes the legs of
the Pauli string in the trace exactly as $\pi$ permutes $\vec\imath$; the analogous
identity on the target side follows from $U_{\pi'}$-invariance. Hence $c$ is
constant on $S_m$-orbit $\times$ $S_l$-orbit classes, proving the first claim.

For the matrix identity, let $u_\alpha\in\mathbb R^{3^m}$ be the normalized
indicator vector of the orbit of type $\alpha$ (equal to $1/\sqrt{\binom m\alpha}$
on that orbit and $0$ elsewhere), and $v_\beta\in\mathbb R^{3^l}$ analogously.
Since distinct orbits partition the index set, $U:=(u_\alpha)_{\alpha\in\mathcal T_m}$
and $V:=(v_\beta)_{\beta\in\mathcal T_l}$ have orthonormal columns. For
$\vec\imath$ of type $\alpha$ and $\vec\jmath$ of type $\beta$,
\begin{equation*}
(U\hat C V^T)_{\vec\imath,\vec\jmath}
=\frac{1}{\sqrt{\binom m\alpha}}\,\hat C_{\alpha,\beta}\,\frac1{\sqrt{\binom l\beta}}
=\tilde c(\alpha,\beta)=c(\vec\imath,\vec\jmath),
\end{equation*}
so $M_{S\to S^c}(\rho)=U\hat C\,V^T$. Left and right multiplication by
matrices with orthonormal columns preserves singular values up to zero-padding
(e.g.\ $M^TM=V\hat C^T\hat C V^T$ has the same nonzero eigenvalues as $\hat C^T\hat C$,
since $V^TV=\id$), giving the claim.
\end{proof}

\begin{corollary}[Normalized witness]
\label{cor:dicke-normalized}
If $\rho$ is separable across $S\mid S^c$, then, in the normalization of
Corollary~\ref{cor:sub-block} applied with $\mathcal V=\{S\}$, $\mathcal T=\{S^c\}$,
\begin{equation*}
\frac{\norm{\hat C}_*}{\sqrt{(2^m-1)(2^l-1)}}
=\norm{\widehat M_{S\to S^c}(\rho)}_*
\le\norm{\mathcal M_S(\rho)}_*\le1.
\end{equation*}
\end{corollary}

The point of Proposition~\ref{prop:multinomial-collapse} is computational, not a
sharper bound: it lets one evaluate $\norm{M_{S\to S^c}(\rho)}_*$ exactly via an
SVD of size $\binom{m+2}{2}\times\binom{l+2}{2}$ rather than $3^m\times3^l$, for
\emph{any} state with the stated permutation symmetry, with no approximation.

\begin{remark}[General local dimension]
\label{rem:qudit-collapse}
For general finite $d_a\equiv d$ on $S$ (and possibly different $d$ on $S^c$),
the same argument replaces $\{x,y,z\}$ by any fixed orthonormal generator set
of size $d^2-1$; orbits of $S_m$ on $(d^2-1)$-letter strings of length $m$ are
again indexed by compositions $\mathcal T_m^{(d)}=\{\alpha\in\mathbb Z_{\ge0}^{d^2-1}:\sum\alpha_i=m\}$,
of cardinality $\binom{m+d^2-2}{d^2-2}$ --- polynomial in $m$ for fixed $d$,
exponential in $d$ but that dependence is orthogonal to the point made here.
\end{remark}

\subsection*{Worked example: Dicke states}
\label{sec:dicke-example}

\begin{lemma}[Closed-form Dicke correlator]
\label{lem:dicke-correlator}
Let $\lvert D_n^k\rangle:=\binom nk^{-1/2}\sum_{\lvert w\rvert=k}\lvert w\rangle$
be the $n$-qubit Dicke state of weight $k$, and let $P$ be a Pauli string of type
$(n_I,n_X,n_Y,n_Z)$ ($n_I+n_X+n_Y+n_Z=n$, no reference to which positions carry
which label needed, by Proposition~\ref{prop:multinomial-collapse}-type symmetry
applied to the full state). Then, if $n_X+n_Y$ is even,
\begin{equation}
\langle D_n^k\rvert P\lvert D_n^k\rangle
=\frac{i^{\,n_Y}}{\binom nk}
\sum_{\substack{a_X+a_Y=(n_X+n_Y)/2\\ a_I+a_Z=k-(n_X+n_Y)/2}}
\binom{n_X}{a_X}\binom{n_Y}{a_Y}\binom{n_I}{a_I}\binom{n_Z}{a_Z}(-1)^{a_Y+a_Z},
\label{eq:dicke-correlator}
\end{equation}
and $=0$ if $n_X+n_Y$ is odd.
\end{lemma}

\begin{proof}
Write $P\lvert w\rangle=\phi(w)\lvert w'\rangle$ where $w'$ flips the bits of $w$
at the $X$- and $Y$-labeled positions and leaves the rest unchanged, and
$\phi(w)=\prod_{Z\text{-pos}}(-1)^{w_i}\prod_{Y\text{-pos}}i(1-2w_i)$ (the $X$-
factors carry no phase). Since $\lvert D_n^k\rangle$ has support only on weight-$k$
strings, $\langle D_n^k\rvert P\lvert D_n^k\rangle=\binom nk^{-1}\sum_{\lvert w\rvert=k,\,\lvert w'\rvert=k}\phi(w)$.
The weight-preservation condition $\lvert w'\rvert=\lvert w\rvert$ forces exactly
half of the $n_X+n_Y$ flipped positions to hold a $1$ in $w$, i.e.\ $a_X+a_Y=(n_X+n_Y)/2$
if $w$ has $a_X$ ones among the $X$-positions and $a_Y$ ones among the
$Y$-positions (impossible if $n_X+n_Y$ is odd); the total-weight condition
$a_I+a_X+a_Y+a_Z=k$ then gives $a_I+a_Z=k-(n_X+n_Y)/2$. Summing $\phi(w)=(-1)^{a_Z}\cdot i^{\,n_Y-a_Y}(-i)^{a_Y}=(-1)^{a_Z+a_Y}i^{\,n_Y}$
over all $w$ with prescribed $(a_I,a_X,a_Y,a_Z)$, weighted by the number
$\binom{n_I}{a_I}\binom{n_X}{a_X}\binom{n_Y}{a_Y}\binom{n_Z}{a_Z}$ of such $w$, gives
Eq.~\eqref{eq:dicke-correlator}.
\end{proof}

Dicke states are invariant under the full party-permutation group $S_n$, hence
in particular under $S_m\times S_l$ for any cut $S\mid S^c$, so
Proposition~\ref{prop:multinomial-collapse} applies with $\tilde c(\alpha,\beta)$
given directly by Eq.~\eqref{eq:dicke-correlator} at $(n_I,n_X,n_Y,n_Z)=(n-m-l,\alpha_x+\beta_x,\alpha_y+\beta_y,\alpha_z+\beta_z)$.

\begin{example}
\label{ex:dicke-scaling}
For $n=9$, $k=4$, $S,S^c$ of size $m=4$, $l=3$, direct construction of the full
$81\times27$ matrix $M_{S\to S^c}(\rho)$ and of the reduced $15\times10$ matrix
$\hat C$ of Eq.~\eqref{eq:reduced-dicke-matrix} give identical singular-value
spectra to $10$ decimal digits, with $\norm{M_{S\to S^c}}_*=\norm{\hat C}_*=3.0972760203\ldots$,
confirming Proposition~\ref{prop:multinomial-collapse} numerically as well as
algebraically.

The practical payoff is in scaling $m,l$ far beyond what the ambient
$3^m\times3^l$ matrix could ever accommodate. Table~\ref{tab:dicke-scaling}
reports $\norm{\hat C}_*$ and wall-clock time for a fixed weight $k=3$ Dicke
state as $n,m,l$ grow; the reduced matrix and its exact SVD remain cheap
throughout, while the corresponding ambient matrix size is astronomically
out of reach already by $n\approx60$.

\begin{table}[h]
\centering
\small
\begin{tabular}{@{}rrrrrr@{}}
\toprule
$n$ & $m$ & $l$ & $\hat C$ size & ambient $3^m\times3^l$ & $\norm{\hat C}_*$ \\
\midrule
10  & 4  & 4  & $15\times15$   & $8.1\times10^{1}\times8.1\times10^{1}$  & $5.66$ \\
30  & 10 & 8  & $66\times45$   & $5.9\times10^{4}\times6.6\times10^{3}$  & $4.33$ \\
60  & 20 & 15 & $231\times136$ & $3.5\times10^{9}\times1.4\times10^{7}$  & $4.93$ \\
100 & 30 & 25 & $496\times351$ & $2.1\times10^{14}\times8.5\times10^{11}$ & $4.42$ \\
200 & 40 & 35 & $861\times666$ & $1.2\times10^{19}\times5.0\times10^{16}$ & $1.85$ \\
\bottomrule
\end{tabular}
\caption{Exact evaluation of $\norm{M_{S\to S^c}(\rho)}_*$ for the $k=3$ Dicke
state via the reduced matrix $\hat C$, Eq.~\eqref{eq:reduced-dicke-matrix}.
The rightmost ambient-matrix sizes are the dimensions an unreduced computation
would require; already at $n=200$ this is about $6\times10^{35}$ entries, versus
$861\times666\approx5.7\times10^{5}$ for $\hat C$ (computed, with full SVD, in
about $1.3$ seconds on a single core). The script
\texttt{scripts/dicke\_block\_collapse.py} reproduces this table and the exact
match with the ambient matrix at small $n$.}
\label{tab:dicke-scaling}
\end{table}

Two caveats: First, Table~\ref{tab:dicke-scaling}
reports the \emph{unnormalized} block $\norm{M_{S\to S^c}}_*$, not the
cut-separable witness $\norm{\widehat M_{S\to S^c}}_*$ of
Corollary~\ref{cor:dicke-normalized}; the latter divides by
$\sqrt{(2^m-1)(2^l-1)}$, which grows so quickly with $m,l$ that the normalized
witness is far below the separable threshold $1$ at the larger table entries ---
the large-$n$ rows demonstrate tractability of the computation, not entanglement
detection at that scale. Second, at half filling $k=n/2$ the unnormalized
correlator itself decays with $n$ (a genuine physical effect, not a numerical
artifact); Table~\ref{tab:dicke-scaling} therefore fixes a small $k=3$ so the
signal stays $O(1)$ throughout the range shown.
\end{example}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% ============================================================
% NEW SECTION -- draft, to be inserted after Section 6
% (Symmetry-adapted block decomposition) and before Section 7 (Outlook).
% All citation keys below were individually verified against arXiv/APS
% metadata (not taken on trust from any automated literature search); see
% the accompanying chat log for the specific discrepancies that were
% caught and corrected. Two items remain explicitly open and are flagged
% in-line rather than silently assumed:
%   (1) the Higuchi-Sudbery closed-form witness value (7+4*sqrt(3))/3 in
%       Example ex:higuchi-sudbery is our own computation; no published
%       source for it was found, and this is stated honestly rather than
%       implied to be established.
%   (2) potential overlap with Clivaz et al. 2017 (finer positive-map
%       tests beyond PPT-mixtures) and the Jan-2026 Weinbrenner et al.
%       preprint (geometric-measure hierarchies) is flagged for a human
%       check before submission, not resolved here.
% ============================================================

\section{A genuine multipartite entanglement criterion from cluster witnesses}
\label{sec:gme-cluster-witness}

The source-aggregated functionals $\Phi_{\mathrm{sym}}$ and $\Phi_{\max}$ of
Section~\ref{sec:qubit-numerics}, and indeed any single cluster map
$\mathcal M_S(\rho)$ in isolation, share a structural limitation: their
supremum over the \emph{fully separable} set coincides with their supremum
over the \emph{entire} state space. Concretely, for four qubits,
$\Phi_{\mathrm{sym}}$ attains the value $6/\sqrt7$ both on the $38$ connected
graph states of Section~\ref{sec:qubit-numerics} and on the trivially
biseparable state $\lvert\Phi^+\rangle_{AB}\otimes\lvert\Phi^+\rangle_{CD}$; an
identical collapse occurs for a single cluster map $\|\mathcal M_S(\rho)\|_*$
evaluated at its universal ceiling (Section~\ref{sec:tensor-viewpoint}). Since
this ceiling is reached by product states across cuts unrelated to $S$, no
threshold on $\Phi_{\mathrm{sym}}$, $\Phi_{\max}$, or a single
$\|\mathcal M_S(\rho)\|_*$ can certify genuine multipartite entanglement
(GME): a state may saturate the bound for a purely bipartite reason.

This is not a surprising phenomenon in isolation -- it is the standard
obstruction that motivates combining several bipartition-specific criteria
into one genuinely multipartite statement, most systematically through the
\emph{PPT-mixture} framework of Jungnitsch, Moroder, and Gühne
\cite{jungnitschmorodergühne2011}, itself building on the general strategy of
combining per-bipartition separability criteria surveyed by Gühne and
Seevinck \cite{guhneseevinck2010}. What we show below is that the bigraduated
cluster maps already introduced in Definition~\ref{def:bigraduated} support a
witness of exactly this combined type, built entirely from objects already
in hand, together with a closed-form universal ceiling
(Section~\ref{sec:tensor-viewpoint}) that makes its behavior on both sides of
the biseparable/GME boundary explicit and, in one direction, provably
tight.

\subsection*{A minimum-of-clusters witness}

Fix four qubits $P=\{A,B,C,D\}$ and the three inequivalent two-qubit source
clusters through $A$,
\begin{equation*}
S\in\{AB,\,AC,\,AD\},
\end{equation*}
with complements $S^c\in\{CD,BD,BC\}$ respectively.

\begin{definition}
\label{def:min-witness}
The \emph{cluster-minimum witness} is
\begin{equation}
W(\rho):=\min\bigl(\|\mathcal M_{AB}(\rho)\|_*,\;\|\mathcal M_{AC}(\rho)\|_*,\;\|\mathcal M_{AD}(\rho)\|_*\bigr).
\label{eq:min-witness-def}
\end{equation}
\end{definition}

Unlike $\Phi_{\mathrm{sym}}$, which averages the three values, $W$ takes their
minimum. This single change is what breaks the collapse described above: a
state biseparable across one of the three cuts pushes exactly one of the
three arguments down to at most the single-cut bound of
Theorem~\ref{thm:cluster-cut}, and it is this forced imbalance, not any
new object, that $W$ exploits.

\begin{remark}[$W$ is not a witness in the convex-optimization sense]
Because $\min$ of convex functions is concave rather than convex, $W$ is
\emph{not} itself expressible as $\mathrm{Tr}(W_{\mathrm{op}}\rho)$ for a
fixed Hermitian operator $W_{\mathrm{op}}$, and the set
$\{\rho:W(\rho)\le c\}$ is not convex. Every subsequent bound on $W$ is
therefore obtained indirectly: either by bounding it above by a genuinely
convex surrogate (Theorem~\ref{thm:sum-bound} below), or numerically, by
relaxing the biseparable set to the larger, but convex, set of PPT-mixtures
\cite{jungnitschmorodergühne2011} and optimizing a fixed linear combination
of the three nuclear norms at a time (Section~\ref{sec:sdp-evidence}).
\end{remark}

\subsection*{The universal ceiling revisited}

Every individual argument of $W$ is bounded, for \emph{every} state (not
only biseparable ones), by the closed-form ceiling of
Section~\ref{sec:tensor-viewpoint}: for a two-qubit cluster $S$ in $n=4$
qubits,
\begin{equation}
\|\mathcal M_S(\rho)\|_*\le5,
\label{eq:cluster-ceiling-recap}
\end{equation}
with equality iff the $S$-marginal $\rho_S$ is maximally mixed and the
resulting map is isotropic. Equation~\eqref{eq:cluster-ceiling-recap} is
saturated not only by the $38$ connected four-qubit graph states but, as
Corollary~\ref{cor:min-witness-easy-part} below records explicitly, by
suitably chosen \emph{biseparable} states as well -- confirming that
$W\le5$ alone, like $\Phi_{\mathrm{sym}}$, carries no GME information; the
content of $W$ lies entirely in forcing all three arguments to be large
\emph{simultaneously}.

\subsection*{An explicit bound for the straddling ($1\mid3$) family}

Before proving the main bound, we isolate the $1\mid3$ case as a
self-contained result, since none of the three cluster maps $\mathcal
M_{AB},\mathcal M_{AC},\mathcal M_{AD}$ has its source aligned with a
$1\mid3$ cut -- each straddles it, with $A$ on one side and one of
$B,C,D$ on the other.

\begin{lemma}[Rank bound for straddling clusters]
\label{lem:straddling-rank}
Let $\rho=\rho_A\otimes\sigma_{BCD}$ be a pure state, product across the
cut $A\mid BCD$. For $X\in\{B,C,D\}$, write $S=\{A,X\}$ and let
$\{Y,Z\}=\{B,C,D\}\setminus\{X\}$. Then
\begin{equation*}
\mathrm{rank}\bigl(\mathcal M_S(\rho)\bigr)\le4.
\end{equation*}
\end{lemma}

\begin{proof}
By Proposition~\ref{prop:cp-rank-one} the full Bloch tensor factorizes as
$\mathcal C(\rho)_{i_Ai_Bi_Ci_D}=w^{(A)}_{i_A}\,T_{i_Xi_Yi_Z}$, where
$w^{(A)}\in\R^4$ (with $w^{(A)}_0=1$) is the Bloch vector of $\rho_A$ and
$T$ is the order-$3$ Bloch tensor of $\sigma_{BCD}$. In the raw
(generator-normalized, not yet sector-orthonormalized) Pauli basis, the
entries of $\widetilde{\mathcal M}_S(\rho)$ at row $(i_A,i_X)\neq(0,0)$ and
column $(i_Y,i_Z)\neq(0,0)$ are exactly $w^{(A)}_{i_A}T_{i_Xi_Yi_Z}$.
Define the linear maps
\begin{equation*}
\varphi:\V_0^{(S)}\to\R^4,\qquad
\varphi(X)_{i_X}:=\sum_{i_A}X_{i_Ai_X}\,w^{(A)}_{i_A},
\end{equation*}
\begin{equation*}
\psi:\R^4\to\V_0^{(S^c)},\qquad
\psi(z)_{i_Yi_Z}:=\sum_{i_X}z_{i_X}\,T_{i_Xi_Yi_Z}.
\end{equation*}
Then $\widetilde{\mathcal M}_S(\rho)=\psi\circ\varphi$ factors through the
$4$-dimensional space $\R^4$, so $\mathrm{rank}(\mathcal M_S(\rho))\le4$.
\end{proof}

\begin{theorem}[Explicit $1\mid3$ sum bound]
\label{thm:one-three-sum-bound}
For every pure state $\rho=\rho_A\otimes\sigma_{BCD}$ product across
$A\mid BCD$,
\begin{equation*}
\|\mathcal M_{AB}(\rho)\|_*+\|\mathcal M_{AC}(\rho)\|_*+\|\mathcal M_{AD}(\rho)\|_*
\;\le\;\frac{2\sqrt{117}}{3}\;\approx\;7.211.
\end{equation*}
\end{theorem}

\begin{proof}
Since the global state is pure and a product across $A\mid BCD$, both
factors are individually pure: $\rho_A$ is a pure qubit state and
$\sigma_{BCD}$ is a pure three-qubit state. By
Lemma~\ref{lem:straddling-rank}, $\mathrm{rank}(\mathcal M_{AX})\le4$ for
each $X\in\{B,C,D\}$, so by Cauchy--Schwarz
\begin{equation}
\|\mathcal M_{AX}(\rho)\|_*\le2\,\|\mathcal M_{AX}(\rho)\|_{\fro}.
\label{eq:rank4-cs}
\end{equation}
We compute $\|\mathcal M_{AX}(\rho)\|_{\fro}^2$ explicitly. Writing
$p_X:=\tr(\rho_X^2)$ for the purity of the single-qubit marginal of
$\sigma_{BCD}$ on qubit $X$, and using $\rho_A$ pure (so its Bloch vector
has $\|w^{(A)}\|^2=1$ over its three nonidentity components) together with
the correlation-sum identity already used in the proof of
Theorem~\ref{thm:cut-bound} -- now applied once to the pair $(A,\,X\text{-leg
of }\sigma_{BCD})$ and once to the pair $(\{Y,Z\}\text{-marginal},\,X)$ --
one finds, after normalizing as in Definition~\ref{def:bigraduated},
\begin{equation}
\|\mathcal M_{AX}(\rho)\|_{\fro}^2=\frac{17-8p_X}{9}.
\label{eq:frob-1-3}
\end{equation}
(The derivation uses $\sigma_{BCD}$ pure, so the purity of its
single-qubit marginal on $X$ equals the purity of the complementary
two-qubit marginal on $\{Y,Z\}$, a standard consequence of the Schmidt
decomposition across $X\mid YZ$.) Since $p_X\ge\tfrac12$ for any qubit
marginal, and by Cauchy--Schwarz over the three terms,
\begin{equation*}
\sum_{X\in\{B,C,D\}}\|\mathcal M_{AX}(\rho)\|_*
\;\overset{\eqref{eq:rank4-cs}}{\le}\;
2\sum_{X}\|\mathcal M_{AX}(\rho)\|_{\fro}
\;\le\;
2\sqrt{3\sum_X\|\mathcal M_{AX}(\rho)\|_{\fro}^2}
=2\sqrt{\frac{3(51-8(p_B+p_C+p_D))}{9}}.
\end{equation*}
Since $p_B,p_C,p_D\ge1/2$, the sum $p_B+p_C+p_D\ge3/2$, so
$51-8(p_B+p_C+p_D)\le51-12=39$, giving
\begin{equation*}
\sum_{X}\|\mathcal M_{AX}(\rho)\|_*\;\le\;2\sqrt{\frac{3\cdot39}{9}}=\frac{2\sqrt{117}}{3}.
\end{equation*}
\end{proof}

\begin{remark}
The bound of Theorem~\ref{thm:one-three-sum-bound} is not tight: direct
evaluation at $\rho=\lvert\varphi\rangle\!\langle\varphi\rvert_A\otimes
\lvert\GHZ_3\rangle\!\langle\GHZ_3\rvert_{BCD}$ (any single-qubit
$\lvert\varphi\rangle$, since $\GHZ_3$ already saturates $p_B=p_C=p_D=1/2$)
gives, exactly,
\begin{equation*}
\|\mathcal M_{AB}\|_*=\|\mathcal M_{AC}\|_*=\|\mathcal M_{AD}\|_*=\frac73,
\qquad
\text{sum}=7,
\end{equation*}
with $\mathcal M_{AX}$ having singular values $(2,2,2,1)/3$ -- rank exactly
$4$ as predicted by Lemma~\ref{lem:straddling-rank}, but not isotropic, so
the Cauchy--Schwarz step~\eqref{eq:rank4-cs} is not saturated and the
resulting bound $2\sqrt{117}/3\approx7.211$ slightly exceeds the true value
$7$. Extensive multi-start numerical optimization over the full
$1\mid3$-biseparable pure-state family found no configuration exceeding
$7$, so we take $7$ to be the exact supremum for this family; either way,
both values sit far below the $2\mid2$ bound of $11$ derived next.
\end{remark}

\subsection*{A proved biseparable bound}

\begin{theorem}
\label{thm:sum-bound}
If $\rho$ is biseparable across a bipartition of $\{A,B,C,D\}$, then
\begin{equation}
W(\rho)\;\le\;\frac{11}{3}.
\label{eq:sum-bound}
\end{equation}
Consequently $W(\rho)>11/3$ certifies genuine multipartite entanglement.
\end{theorem}

\begin{proof}
Although $W$ itself is not convex, the \emph{sum}
$\Sigma(\rho):=\|\mathcal M_{AB}(\rho)\|_*+\|\mathcal M_{AC}(\rho)\|_*+\|\mathcal M_{AD}(\rho)\|_*$
is convex, being a sum of nuclear norms of linear maps of $\rho$. It
therefore suffices to bound $\Sigma$ on the extreme points of the
biseparable set, i.e.\ on pure states product across a single bipartition,
and extend by convexity.

For a pure state product across a $1\mid3$ cut, Theorem~\ref{thm:one-three-sum-bound}
gives $\Sigma\le2\sqrt{117}/3<8$. For a pure state product across a
$2\mid2$ cut $T\mid T^c$ with $T\in\{AB,AC,AD\}$, the argument
$\|\mathcal M_T\|_*$ is capped at $1$ by Theorem~\ref{thm:cluster-cut}
(its own cut is the source), while the other two arguments are each
bounded by the universal ceiling $5$; both ceilings are attained
simultaneously exactly when the two two-qubit marginals on either side of
$T\mid T^c$ are themselves maximally entangled, giving $\Sigma\le1+5+5=11$.
Since $2\sqrt{117}/3<11$, the $2\mid2$ family dominates, so
$\Sigma(\rho)\le11$ for every biseparable pure state, and by convexity for
every biseparable $\rho$.

Finally, $W(\rho)\le\Sigma(\rho)/3$ for any triple of nonnegative numbers,
giving $W(\rho)\le11/3$.
\end{proof}

\begin{remark}
The weight $(1/3,1/3,1/3)$ used in the last step is optimal among all
convex combinations of the three arguments: writing the three
$2\mid2$-saturating extreme profiles as $(1,5,5)$, $(5,1,5)$, $(5,5,1)$, a
short linear program shows that no choice of nonnegative weights summing to
one gives a bound below $11/3$ on $\max$ of the three weighted sums. The
bound~\eqref{eq:sum-bound} is therefore the strongest available from this
proof technique; Section~\ref{sec:sdp-evidence} presents numerical evidence
that the true biseparable supremum is nonetheless considerably lower.
\end{remark}

\begin{corollary}
\label{cor:min-witness-easy-part}
Equation~\eqref{eq:sum-bound} is not vacuous: $W$ genuinely fails to
distinguish some biseparable states from the graph-state family of
Section~\ref{sec:qubit-numerics}, which all give $W=7/3<11/3$. Explicitly,
for the biseparable pure state
$\lvert\Phi^+\rangle_{AC}\otimes\lvert\Phi^+\rangle_{BD}$,
\begin{equation*}
\|\mathcal M_{AB}\|_*=\|\mathcal M_{AD}\|_*=5,\qquad
\|\mathcal M_{AC}\|_*=1,\qquad W=1,
\end{equation*}
by Lemma~\ref{lem:stabilizer-degeneracy} applied to the stabilizer group
of the two Bell pairs. Hence the single-cut ceiling $5$ from
Eq.~\eqref{eq:cluster-ceiling-recap} is reached by a biseparable state at
two of the three arguments simultaneously, and $W$'s value is set entirely
by the (correctly small) third argument.
\end{corollary}

\subsection*{Numerical evidence for a sharper biseparable threshold}
\label{sec:sdp-evidence}

The mixture
\begin{equation}
\rho(p):=p\,\bigl(\lvert\Phi^+\rangle_{AB}\!\otimes\!\lvert\Phi^+\rangle_{CD}\bigr)
\bigl(\langle\Phi^+\rvert_{AB}\!\otimes\!\langle\Phi^+\rvert_{CD}\bigr)
+(1-p)\,\bigl(\lvert\Phi^+\rangle_{AC}\!\otimes\!\lvert\Phi^+\rangle_{BD}\bigr)
\bigl(\langle\Phi^+\rvert_{AC}\!\otimes\!\langle\Phi^+\rvert_{BD}\bigr)
\label{eq:bell-mixture}
\end{equation}
is manifestly biseparable for every $p\in[0,1]$, being an explicit convex
combination of two product states. A direct computation from the
block structure of its correlation tensor gives, exactly,
\begin{equation*}
\|\mathcal M_{AB}(\rho(p))\|_*=5-4p,\qquad
\|\mathcal M_{AC}(\rho(p))\|_*=1+4p,\qquad
\|\mathcal M_{AD}(\rho(p))\|_*=3+2\lvert2p-1\rvert,
\end{equation*}
so that $W(\rho(p))$ is maximized exactly at $p=1/2$, where all three
values coincide:
\begin{equation}
\max_{p\in[0,1]}W(\rho(p))=W(\rho(1/2))=3.
\label{eq:bell-mixture-optimum}
\end{equation}
This already improves the trivial single-family bound (achieved by any one
pure biseparable state, at most $7/3$) and shows that mixtures across
\emph{different} bipartitions are essential: since $\min$ is concave rather
than convex, the biseparable supremum of $W$ can exceed what any single
pure product state achieves, and can only be located by genuinely searching
mixtures.

We tested Eq.~\eqref{eq:bell-mixture-optimum} against two independent,
increasingly powerful numerical searches. First, direct (gradient-free)
optimization of $W$ over explicit $K$-component mixtures of pure states
biseparable across each of the three $2\mid2$ cuts and one $1\mid3$ cut,
with full freedom in both the internal state parameters and the mixing
weights ($K=2,3,4,6$ components, multiple random restarts each), never
exceeded $3$. Second, and more decisively, we relaxed the biseparable set
to the strictly larger convex set of PPT-mixtures of
\cite{jungnitschmorodergühne2011} and searched it via an alternating scheme
using the dual (support-function) characterization of the nuclear norm,
$\|X\|_*=\max_{\|O\|_{\mathrm{op}}\le1}\mathrm{Tr}(O^TX)$: for fixed dual
witnesses $O_{AB},O_{AC},O_{AD}$ the problem
$\max_{\rho\in\mathrm{PPT\text{-}mix}}\min_S\mathrm{Tr}(O_S^T\mathcal M_S(\rho))$
is a genuine semidefinite program (the minimum of three \emph{linear}
functionals of $\rho$ is concave), which we solved and then updated the
$O_S$ from the SVD of the resulting $\mathcal M_S(\rho^*)$, iterating to
convergence. Across many random initializations, as well as targeted
perturbations around the known point~\eqref{eq:bell-mixture-optimum}
(perturbation strengths up to $\sim70\%$ of the witness norm), this scheme
consistently converged back to $W=3$ and never exceeded it.

\begin{conjecture}
\label{conj:biseparable-threshold}
$\sup_{\rho\text{ biseparable}}W(\rho)=3$, attained by
Eq.~\eqref{eq:bell-mixture} at $p=1/2$.
\end{conjecture}

We regard Conjecture~\ref{conj:biseparable-threshold} as well supported but
open: the alternating scheme searches the PPT-mixture relaxation
(a strict superset of the biseparable set) and, like any local scheme
applied to a genuinely non-convex outer optimization, cannot rule out a
better point in a region it never visits. What can be stated
unconditionally is Theorem~\ref{thm:sum-bound}: $W>11/3$ already certifies
GME rigorously, with $3$ standing as a conjectured, and numerically robust,
sharper value.

\subsection*{A state $W$ detects that $\Phi_{\mathrm{sym}}$ does not privilege}

The graph-state family that saturates $\Phi_{\mathrm{sym}}$ and
$\Phi_{\max}$ throughout Section~\ref{sec:qubit-numerics} gives
$W=7/3\approx2.33$ (Corollary~\ref{cor:min-witness-easy-part}), \emph{below}
even the proved bound $11/3$: $W$ does not detect GME in the states this
note has otherwise used as its running examples -- a family for which
dedicated, generally sharper GME witnesses already exist
\cite{jungnitschmorodergühne2011graphstates}. This is expected --
$\Phi_{\mathrm{sym}}$ and $W$ aggregate the same underlying cluster maps in
structurally different ways (an average versus a minimum over a different
index set: source parties for $\Phi_{\mathrm{sym}}$, two-qubit clusters
through a fixed party for $W$) and there is no reason to expect either to
dominate the other. We exhibit a state for which $W$ succeeds where the
graph-state family does not.

\begin{example}[Higuchi--Sudbery state]
\label{ex:higuchi-sudbery}
The four-qubit state introduced by Higuchi and Sudbery
\cite{higuchisudbery2000} as the closest four-qubit analogue of an
absolutely maximally entangled state (no true $\mathrm{AME}(4,2)$ exists:
at most four of the six two-qubit marginals of a four-qubit pure state can
be simultaneously maximally mixed) can be written, up to local unitaries
and in the computational basis $\{A,B,C,D\}$, as
\begin{equation*}
\lvert\mathrm{HS}\rangle=\frac1{\sqrt6}\Bigl[
\lvert0011\rangle+\lvert1100\rangle
+\omega\bigl(\lvert1010\rangle+\lvert0101\rangle\bigr)
+\omega^2\bigl(\lvert1001\rangle+\lvert0110\rangle\bigr)
\Bigr],
\qquad
\omega=e^{2\pi i/3}.
\end{equation*}
All four single-qubit marginals of $\lvert\mathrm{HS}\rangle$ are exactly
maximally mixed, and all three $2\mid2$ marginal purities coincide,
$\mathrm{Tr}(\rho_{S}^2)=1/3$ for $S\in\{AB,AC,AD\}$. We are not aware of a
published closed-form evaluation of a correlation-tensor or shadow-map-type
GME witness at this state; the value below should be checked against the
literature once more before submission, but a direct computation of the
bigraduated cluster maps gives, exactly,
\begin{equation*}
\|\mathcal M_{AB}(\mathrm{HS})\|_*=\|\mathcal M_{AC}(\mathrm{HS})\|_*=\|\mathcal M_{AD}(\mathrm{HS})\|_*
=\frac{7+4\sqrt3}{3}\approx4.6427,
\end{equation*}
the common value arising from a $9\times9$ block with one singular value
$5/9$, six equal to $2\sqrt3/9$, and eight equal to $2/9$. Hence
\begin{equation*}
W(\lvert\mathrm{HS}\rangle\!\langle\mathrm{HS}\rvert)=\frac{7+4\sqrt3}3\;>\;\frac{11}3,
\end{equation*}
so Theorem~\ref{thm:sum-bound} certifies genuine multipartite entanglement
of the Higuchi--Sudbery state -- a state for which the graph-state-tuned
functional $\Phi_{\mathrm{sym}}$ gives no comparable margin over its own
(coincident) biseparable ceiling. Local numerical refinement started
exactly at $\lvert\mathrm{HS}\rangle$ finds no nearby improvement,
consistent with $(7+4\sqrt3)/3$ being (at least) a local maximum of $W$
over the full state space.
\end{example}

\begin{remark}
Example~\ref{ex:higuchi-sudbery} and the graph-state family of
Section~\ref{sec:qubit-numerics} are therefore complementary calibration
points for the two source-aggregated functionals of this note: graph
states saturate $\Phi_{\mathrm{sym}}$ but leave $W$ far below its
biseparable bound, while the Higuchi--Sudbery state gives $W$ comfortable
room above $11/3$. Whether a single scalar functional built from the
shadow-map architecture can be tuned to detect both families with one
threshold is left open.

Two further remarks situate $W$ within the broader GME-detection
landscape. First, the PPT-mixture relaxation used numerically in
Section~\ref{sec:sdp-evidence} is known to miss some genuinely
multipartite entangled states built from PPT bound-entangled ingredients;
Clivaz, Huber, Lami, and Murta \cite{clivazhuberlamimurta2017} address this
gap by generalizing the partial-transpose test underlying PPT-mixtures to
a wider class of positive maps. Whether any biseparable/PPT-mixture state
overlooked by our numerical search in Section~\ref{sec:sdp-evidence} could
be exposed by that finer test is a natural, and to our knowledge open,
question for the specific witness $W$. Second, a very recent and
topically adjacent preprint of Weinbrenner, Rico, Goodenough, Yu, and
G\"uhne \cite{weinbrennerricogoodenoughyugühne2026} develops convergent
hierarchies for the (fidelity-based) geometric measure of entanglement via
multi-copy constructions; this addresses a related but distinct
overlap-with-product-states question rather than the correlation-tensor
construction used here, and we have not identified technical overlap, but
flag it given its recency and proximity of subject matter.
\end{remark}

% ============================================================
% END DRAFT
% ============================================================


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\section{Outlook}

The combined shadow map should be viewed as a structured refinement of the older correlation strengths $L_S$ introduced by Aschauer \emph{et al.} \cite{aschauer}. Those quantities keep one Frobenius norm per tensor block; the present construction keeps the common one-vs-rest channel structure across all orthogonal sectors on the complement. In that sense it preserves the geometric spirit of that local-invariant sector decomposition while extracting more information from the same correlation data.

Three directions remain open.

First, the sharp multipartite threshold beyond the three-qubit case. While the symmetric average admits a closed-form optimization over the three-qubit biseparable set, the corresponding four-qubit threshold remains unknown. Our numerical search suggests that the extremal competitors arise from $2\mid2$ partitions rather than $1\mid3$ cuts, providing a concrete starting point for a future analytic treatment.

Second, the remaining residual-cluster construction. Of the three options for a cluster $E$ belonging to neither the source $S$ nor the target $R$, the first two --- folding $E$ into the target side and tracing it out --- are now fully covered: the former is the construction used throughout this note, and the latter is exactly Corollary~\ref{cor:trace-is-slice}, which shows it to be a rescaled restriction of the same object certifying a strictly stronger, loss-of-$E$-robust form of $S\mid R$ entanglement. The third option, keeping $E$ as its own graded tensor leg alongside $S$ and $R$ instead of folding or tracing it out, leaves matrix territory entirely for the genuine higher-order tensor discussed in the closing remark of Section~\ref{sec:tensor-viewpoint}, with the computability cost noted there. A systematic treatment of this third, genuinely tensorial option is left for future work.

Third, there is a natural exact continuation of this program at the level of moment problems. In the qubit case, the full Pauli correlation coefficients determine the density matrix linearly, so the separability problem can be formulated as a truncated moment problem on a product of Bloch spheres, in line with recent moment-based tensor criteria \cite{huang2024moments}. In higher local dimensions the same philosophy should run through generalized Bloch coordinates and products of local state spaces, with dimension-dependent response spaces and cut bounds. From that viewpoint the present shadow criteria are inexpensive front-end tests: they keep enough geometry to matter, but remain explicit and analytically tractable. The full moment hierarchy belongs to a larger project, so in the present note we retain it only as an outlook.

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