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+\documentclass[11pt]{article}
+
+\usepackage[a4paper,margin=1in]{geometry}
+\usepackage[T1]{fontenc}
+\usepackage[utf8]{inputenc}
+\usepackage{lmodern}
+\usepackage{amsmath,amssymb,amsthm,mathtools}
+\usepackage{booktabs}
+\usepackage{graphicx}
+\usepackage{hyperref}
+\usepackage{bbm}
+
+\newtheorem{remark}{Remark}
+
+\newcommand{\tr}{\operatorname{tr}}
+\newcommand{\id}{\mathbbm{1}}
+\newcommand{\R}{\mathbb{R}}
+\newcommand{\V}{\mathcal{V}}
+\newcommand{\norm}[1]{\left\lVert #1 \right\rVert}
+\newcommand{\fro}{\mathrm{F}}
+\newcommand{\GHZ}{\mathrm{GHZ}}
+
+\title{Companion Note on Symmetric Shadow Maps}
+\author{Draft companion note}
+\date{May 2026}
+
+\begin{document}
+
+\maketitle
+
+\begin{abstract}
+This companion note explains the idea behind the symmetric shadow-map criterion in a self-contained way for readers who know, or at least want a quick reminder of, the older local-invariant correlation-sector approach of Aschauer, Calsamiglia, Hein, and Briegel \cite{aschauer}. The main message is simple. The original correlation strengths $L_S$ record only the total quadratic size of each correlation block. The geometric object that was implicit all along is richer: for a chosen source subsystem, one can look at the family of correlation-response vectors induced on the other side. In the multipartite case a single source party casts several such shadows at once, one into each orthogonal correlation sector on the complement. Stacking them produces a direct-sum response operator and its associated response ellipsoid in correlation space. For cut-product states that ellipsoid must collapse to a line; entanglement is witnessed by the failure of that locking. We also explain, in the same informal register, why nothing about this argument is special to a single-party source, why tracing out an unwanted party is a slice of the correlation data rather than a loss of it, and why every construction in the note is best pictured as one flattening of a single underlying correlation cube.
+\end{abstract}
+
+\section{Why another note?}
+
+The formal note states the result in a compressed form. Fix a source party $a$, define the combined shadow map $\mathcal M_a(\rho)$ --- that is, the direct-sum response operator obtained by stacking all one-vs-rest response maps --- and prove that
+\begin{equation}
+\norm{\mathcal M_a(\rho)}_*\le 1
+\end{equation}
+for states separable across $a\mid\bar a$, and then average or maximize over $a$ to obtain permutation-symmetric criteria.
+
+That is the right final statement, but it hides the route by which the construction becomes natural. The purpose of the present note is to reconstruct that route while assuming only familiarity with the older sector-wise quantities
+\begin{equation}
+L_S(\rho)=\norm{C_S(\rho)}_{\fro}^2
+\end{equation}
+from the Aschauer \emph{et al.} framework \cite{aschauer}.
+
+The conceptual progression is:
+\begin{enumerate}
+\item Start from the old quadratic invariants $L_S$.
+\item Notice that they throw away too much directional information.
+\item Replace a single number by a linear response map.
+\item Interpret that map geometrically as an ellipsoid in correlation space.
+\item Observe that in the multipartite case one source party produces several such shadows simultaneously, landing in orthogonal sectors.
+\item Stack those shadows and test whether they remain controlled by one common scalar factor.
+\end{enumerate}
+
+The last step is the origin of the symmetric shadow-map criterion.
+
+There is also a practical reason for spelling the story out this way. The existing correlation-tensor literature is already rich: there are criteria based on full tensors, unfoldings, augmented Bloch tensors, scalar sums over sectors, and even partition-adapted mixed-order block trace norms. So the point of the present construction is not to claim the first criterion of that broad type. The narrower claim is that once a source party $a$ is fixed, one can organize the data canonically by including every nonempty target subset $T\subseteq \bar a$ in one exhaustive direct sum. That source-indexed architecture is the feature to keep in mind when comparing with prior work.
+
+To be completely explicit, the closest prior-art families include bipartite correlation-matrix criteria \cite{devicente,chenwu}, multipartite full-tensor unfoldings and matricizations \cite{hassanjoag,devicentehuber,li2014,jingzhang2023}, nonlinear geometric tensor criteria \cite{laskowski2011}, scalar multi-sector norm criteria \cite{klocklhuber2015}, and recent augmented or partition-adapted mixed-order block criteria \cite{shen2016,sarbicki2020,zhao2020,huang2024extended,liyaoyangfei2025}. In that company, the honest claim here is modest: the novelty is mainly in the canonical source-indexed organization of the response data, not in being the first multipartite trace-norm criterion in the Bloch-tensor world.
+
+There is also a small historical point worth keeping straight. Aschauer \emph{et al.} \cite{aschauer} should be read as an early multipartite criterion based on local-operator expansion coefficients, or equivalently on correlation tensors organized by support sectors. The later Hassan--Joag paper \cite{hassanjoag} marks a different milestone: an explicit multipartite separability criterion framed in Bloch-representation language and based on full-tensor unfoldings. The shadow-map construction sits between those viewpoints: it keeps the older sector decomposition but restores directional information inside and across the sectors.
+
+\section{What was already present in the Aschauer \emph{et al.} framework}
+
+The local-invariant correlation-sector decomposition of Aschauer \emph{et al.} \cite{aschauer} organizes an $n$-qubit state into correlation sectors indexed by subsets $S\subseteq P$. For each nonempty $S$, the tensor $C_S(\rho)$ collects the Pauli correlations with support exactly on $S$, and the scalar
+\begin{equation}
+L_S(\rho)=\norm{C_S(\rho)}_{\fro}^2
+\end{equation}
+measures the squared Euclidean size of that block.
+
+This had two major strengths.
+\begin{itemize}
+\item It was local-unitary invariant.
+\item It admitted a simple convexity argument: for qubits, product states have $L_S=1$, so $L_S>1$ certifies entanglement.
+\end{itemize}
+
+But the compression is severe. Once a whole tensor block has been replaced by one Frobenius norm, the geometry of that block is lost. Two states can have the same $L_S$ although one correlation block is essentially rank one while another is spread across several orthogonal directions.
+
+That is the real motivation for the new construction. The issue is not that this older sector-decomposition paper had the wrong geometry; it is that the geometry was compressed too aggressively.
+
+\section{The basic ellipsoid: response vectors rather than states}
+
+Take first the bipartite two-qubit case. Write the correlation matrix as
+\begin{equation}
+C_{ij}=\tr\!\bigl(\rho\,\sigma_i\otimes\sigma_j\bigr),
+\qquad i,j\in\{x,y,z\}.
+\end{equation}
+Pick a unit vector $u=(u_x,u_y,u_z)\in\R^3$ on the second qubit and form the traceless observable
+\begin{equation}
+\sigma_u=u_xX+u_yY+u_zZ.
+\label{eq:sigma-u-def}
+\end{equation}
+Then the induced vector on the first qubit is
+\begin{equation}
+r(u)=Cu,
+\end{equation}
+with components
+\begin{equation}
+r_i(u)=\sum_j C_{ij}u_j=\tr\!\bigl(\rho\,\sigma_i\otimes\sigma_u\bigr).
+\label{eq:ri}
+\end{equation}
+
+This vector $r(u)$ is not the post-measurement state of the first qubit. It is the list of correlation coefficients obtained when the second qubit is probed in direction $u$. It is therefore better thought of as a \emph{correlation-response vector}.
+
+This matters conceptually. The ellipsoid under discussion is not a steering ellipsoid in state space. It lives in correlation space. One chooses a direction on one side, computes the induced correlation vector on the other, and collects all such vectors. The image of the unit ball under the linear map $u\mapsto Cu$ is then an ellipsoid in $\R^3$.
+
+Its semiaxes are the singular values of $C$. So the singular-value language is not an algebraic add-on. It is simply the natural language of the response ellipsoid.
+
+It is worth being precise about what the different natural norms of this ellipsoid measure, since it is easy to blur them together. Write $s_1,s_2,s_3\ge0$ for the semiaxis lengths, i.e.\ the singular values of $C$. The \emph{volume} of the ellipsoid is proportional to the product $s_1s_2s_3$; that is what a determinant measures, and it is a poor witness here because it vanishes the moment even one semiaxis is zero, no matter how large the other two are. The \emph{Frobenius norm squared}, $s_1^2+s_2^2+s_3^2$, is the sum of the squared semiaxes; this is the old $L_S$ in disguise, and it treats one long axis and three balanced medium axes as interchangeable whenever their total squared length agrees. The \emph{nuclear norm}, $s_1+s_2+s_3$, is the plain sum of the semiaxis lengths, with no squaring and no vanishing on account of a single small or absent axis. It behaves like a perimeter of the ellipsoid's principal directions rather than its volume or its squared size, and that additive, non-vanishing character is exactly what makes it the right quantity for the cut-separable bound: every independent response direction contributes to it in proportion to its own length, never more and never less, and never at the mercy of some other direction happening to vanish.
+
+\section{What a response vector actually measures: conditioning on an outcome}
+\label{sec:conditioning}
+
+The response vector $r(u)$ has a completely concrete operational meaning, and it is worth spelling out, because it explains \emph{why} product states collapse to a line before we even reach that argument in coordinate form.
+
+Write the observable $\sigma_u=u_xX+u_yY+u_zZ$ from Eq.~\eqref{eq:sigma-u-def} in spectral form,
+\begin{equation}
+\sigma_u=P_+-P_-,
+\end{equation}
+where $P_\pm$ projects onto the $\pm1$-eigenspace of $\sigma_u$ on the second qubit. Measuring the second qubit along direction $u$ produces outcome $s=\pm1$ with probability
+\begin{equation}
+p_s=\tr\bigl[(\id\otimes P_s)\rho\bigr],
+\end{equation}
+and leaves the first qubit in the conditional state
+\begin{equation}
+\rho_s=\frac{\tr_2\bigl[(\id\otimes P_s)\rho\bigr]}{p_s},
+\qquad s=\pm1,
+\end{equation}
+with Bloch vector $a_s\in\R^3$. A short calculation using Eq.~\eqref{eq:ri} gives
+\begin{equation}
+r(u)=p_+a_+-p_-a_-.
+\label{eq:conditioning}
+\end{equation}
+In words, $r(u)$ is the probability-weighted difference between the two states the first qubit can be left in, depending on which outcome the measurement on the second qubit produced. If the two possible outcomes never leave the first qubit in different states, $r(u)$ vanishes no matter how sharp that measurement is; if they leave it in very different states, $r(u)$ is correspondingly large. This is a steering-flavored statement without any of steering's usual signalling caveats: nothing physically happens to the first qubit, and $r(u)$ is simply bookkeeping the correlation between the possible outcomes on one side and the possible conditional states on the other.
+
+Eq.~\eqref{eq:conditioning} immediately explains the product-state collapse of the next section. If $\rho=\rho_1\otimes\rho_2$, learning the outcome on the second qubit carries no information about the first: $\rho_+=\rho_-=\rho_1$ for every direction $u$, so $a_+=a_-=a$ and
+\begin{equation}
+r(u)=(p_+-p_-)\,a=\langle\sigma_u\rangle\,a.
+\end{equation}
+The response vector always points along the same fixed direction $a$, the Bloch vector of $\rho_1$; only its length changes as $u$ varies. Genuine entanglement across the cut is precisely the failure of this locking: different measurement directions $u$ on the second qubit can then leave the first qubit in genuinely different conditional states, and the response vector sweeps out more than one dimension as $u$ varies.
+
+For a $d$-dimensional qudit in place of a qubit, the same argument runs with a spectral decomposition into more than two outcomes. Writing $\sigma_u=\sum_k\lambda_kP_k$ for the (now $u$-dependent) eigenbasis of the generic observable $\sigma_u$, with outcome probabilities $p_k$ and conditional Bloch vectors $a_k$, Eq.~\eqref{eq:conditioning} generalizes to an eigenvalue-weighted sum over conditional states rather than a plain difference of two,
+\begin{equation}
+r(u)=\sum_k\lambda_k\,p_k\,a_k.
+\end{equation}
+The product-state argument goes through exactly as before: if the outcome tells you nothing about the other side, every $a_k$ coincides with the fixed Bloch vector of the marginal, and the sum collapses to $\langle\sigma_u\rangle$ times that one fixed vector.
+
+The same reading applies unchanged to the multipartite target-sector maps $M_{a\to T}$ of the formal note: $M_{a\to T}(u)$ is the eigenvalue-weighted sum of the sector-$T$ correlation vectors of the conditional states of the complement, exactly as above but with the single second qubit replaced by the full multi-party target sector $T$.
+
+\section{Why product states collapse to a line}
+\label{sec:collapse-to-line}
+
+In the conditioning language of the previous section, the calculation below is the same statement written out in coordinates: measuring qubit $2$ changes nothing about what one can say about qubit $1$.
+
+Suppose now that
+\begin{equation}
+\rho=\rho_A\otimes\rho_B,
+\end{equation}
+with local Bloch vectors $a,b\in\R^3$. Then the correlation matrix factorizes as
+\begin{equation}
+C=ab^T.
+\end{equation}
+Hence
+\begin{equation}
+r(u)=Cu=a\,(b\cdot u).
+\end{equation}
+Every response vector is parallel to the same fixed vector $a$; only the scalar coefficient changes. Geometrically, the ellipsoid degenerates to a line segment.
+
+This is the cleanest geometric reading of product structure. Product states have only one correlation channel across the cut. The old Frobenius norm sees only the total squared size of that channel. The singular values also see whether there is only one such direction or several independent ones.
+
+That is why the nuclear norm is a stronger refinement. A matrix with one singular value $\sqrt3$ and a matrix with three singular values $1,1,1$ have the same Frobenius norm squared $3$, but they describe very different ellipsoids: one long axis in the first case, three orthogonal axes in the second.
+
+\section{Why the multipartite extension is almost forced}
+
+Once one thinks in terms of response ellipsoids, the multipartite continuation is hard to avoid.
+
+Take three qubits $A,B,C$. If one probes qubit $A$ in a direction $u_A$, then three different responses arise naturally:
+\begin{equation}
+r_B(u_A)\in\R^3,
+\qquad
+r_C(u_A)\in\R^3,
+\qquad
+R_{BC}(u_A)\in\R^9.
+\end{equation}
+These belong to the $AB$, $AC$, and $ABC$ correlation sectors.
+
+The key observation is that these target spaces are orthogonal Hilbert-Schmidt sectors. So there is no compelling reason to study them separately. The first way we wrote this down was as the external direct sum
+\begin{equation}
+\mathcal R_A(u_A)
+=
+\frac{1}{\sqrt3}\bigl(r_B(u_A),r_C(u_A),R_{BC}(u_A)\bigr)
+\in
+\R^3\oplus\R^3\oplus\R^9.
+\end{equation}
+Its image of the unit ball is one combined ellipsoid.
+
+This is the decisive step. The old bipartite ellipsoid was not the endpoint of the story; it was the simplest case of a more general response geometry. In the multipartite case one source party casts several shadows at once, one into each orthogonal sector on the complement. At first this looks like a useful stacking trick. But there is actually a cleaner way to understand it.
+
+For each party $b$, let $\V^{(b)}$ be the real Hilbert-Schmidt space of Hermitian operators on $\mathcal H^{(b)}$, and let $\V_0^{(b)}$ be its traceless subspace, spanned by the generators $\sigma_1^{(b)},\dots,\sigma_{d_b^2-1}^{(b)}$. Then on the complement of $a$ one has the orthogonal decomposition
+\begin{equation}
+\V_0^{(\bar a)}
+=
+\bigoplus_{\emptyset\neq T\subseteq \bar a}^{\perp}\V_T^{(\bar a)},
+\end{equation}
+where $\V_T^{(\bar a)}$ is the span of product basis elements that act nontrivially exactly on the parties in $T$. This is exactly the same kind of invariant-sector decomposition used in the older Aschauer \emph{et al.} picture, now viewed as the codomain decomposition of a response map. So the external direct sum above is just the coordinate form of an internal orthogonal decomposition of the traceless complement space.
+
+In that language the real primary object is the basis-free response operator
+\begin{equation}
+\widetilde{\mathcal M}_a(\rho):\V_0^{(a)}\to \V_0^{(\bar a)},
+\qquad
+\langle Y,\widetilde{\mathcal M}_a(\rho)X\rangle
+=
+\tr\!\bigl(\rho(X\otimes Y)\bigr).
+\end{equation}
+The sector maps are simply its orthogonal projections onto the subspaces $\V_T^{(\bar a)}$. So the combined shadow map is not really a bookkeeping stack after all. It is the full one-vs-rest response operator, written in the orthogonal sector decomposition naturally supplied by the multipartite Bloch space.
+
+This re-interpretation is useful, not just prettier. It explains from the beginning why the target sectors are orthogonal, why Frobenius additivity is just Pythagoras in the codomain, and why collective unitaries on the complement act by orthogonal rotations on the full target space while generally mixing the individual sectors.
+
+For general $n$, choosing a source party $a$ and writing $\bar a=P\setminus\{a\}$, every nonempty subset $T\subseteq\bar a$ gives a map
+\begin{equation}
+M_{a\to T}(\rho):\R^3\to\R^{3^{|T|}}.
+\end{equation}
+Stacking all of them yields the combined shadow map
+\begin{equation}
+\mathcal M_a(\rho)
+=
+\frac{1}{\sqrt{2^{n-1}-1}}
+\bigoplus_{\emptyset\neq T\subseteq\bar a}M_{a\to T}(\rho).
+\end{equation}
+That is the coordinate object used in the formal note. But the more intrinsic viewpoint is that the codomain has one natural ambient Bloch structure first, and the direct sum appears because that ambient space splits orthogonally by support sectors.
+
+\section{What cut separability means geometrically}
+
+To see the geometry, begin with a state that is product across the
+one-vs-rest cut:
+\begin{equation}
+\rho=\rho_A\otimes\sigma_{BC}.
+\end{equation}
+Let $a$ be the Bloch vector of $\rho_A$, and let $b,c,T_{BC}$ denote the one- and two-body correlation data of $\sigma_{BC}$. Then each shadow is driven by the same scalar $a\cdot u_A$:
+\begin{equation}
+r_B(u_A)=(a\cdot u_A)b,
+\qquad
+r_C(u_A)=(a\cdot u_A)c,
+\qquad
+R_{BC}(u_A)=(a\cdot u_A)T_{BC}.
+\end{equation}
+Thus
+\begin{equation}
+\mathcal R_A(u_A)=\frac{a\cdot u_A}{\sqrt3}(b,c,T_{BC}).
+\end{equation}
+
+This is the geometric core of the theorem. Even though the response occupies several different sectors, they are all locked to one common source factor. The combined ellipsoid therefore still collapses to a line segment.
+
+So cut-product structure does not merely bound each sector separately. It imposes a common rank-one organization across \emph{all} sectors at once. That is precisely what separate scalar criteria forget and what the combined shadow map retains.
+
+For mixed states separable across the same cut, convexity preserves the corresponding nuclear-norm bound.
+
+\section{Why the normalization is so clean}
+
+The normalization factor in the definition of $\mathcal M_a$ is not arbitrary. For a state $\sigma_{\bar a}$ on the complement, the old correlation-sum identity gives
+\begin{equation}
+\sum_{\emptyset\neq T\subseteq\bar a}L_T(\sigma_{\bar a})
+=2^{n-1}\tr(\sigma_{\bar a}^2)-1
+\le 2^{n-1}-1.
+\end{equation}
+Thus for a pure cut-product term
+\begin{equation}
+\rho=\rho_a\otimes\sigma_{\bar a},
+\end{equation}
+the stacked target vector has squared norm at most $2^{n-1}-1$, while the source Bloch vector has norm at most $1$. Dividing by $\sqrt{2^{n-1}-1}$ therefore makes the rank-one nuclear norm at most $1$.
+
+This is one of the nicest features of the construction. The geometry and the old purity identity fit together exactly.
+
+\section{What one can already do with the individual summands}
+
+The direct-sum map is the main object in the formal note, but the separate summands are not merely bookkeeping. Each individual sector map already gives a valid cut-separability criterion once it is normalized by the dimension of its own target subsystem.
+
+For a fixed nonempty $T\subseteq\bar a$, define
+\begin{equation}
+\widehat M_{a\to T}(\rho)
+=
+\frac{1}{\sqrt{(d_a-1)(d_T-1)}}M_{a\to T}(\rho),
+\qquad
+d_T:=\prod_{b\in T}d_b.
+\end{equation}
+Then the same rank-one argument used for the full map shows that for every state separable across $a\mid\bar a$,
+\begin{equation}
+\norm{\widehat M_{a\to T}(\rho)}_*\le 1.
+\end{equation}
+So every summand already supplies its own witness.
+
+This is conceptually useful. The full map answers the cut-level question ``how large is the total shadow cast by $a$ onto the whole complement?'' The summands answer the profiling question ``where does that shadow actually land?''
+
+It is therefore natural to name the sector profiles explicitly:
+\begin{equation}
+\Phi_{a\to T}(\rho):=\norm{\widehat M_{a\to T}(\rho)}_*,
+\qquad
+\Phi^{(2)}_{a\to T}(\rho):=\norm{\widehat M_{a\to T}(\rho)}_{\fro}^2.
+\end{equation}
+These are not replacements for the full map. They are a diagnostic profile attached to it.
+
+There is also an exact reconstruction formula. If $\iota_T$ denotes the canonical inclusion of the $T$-sector target space into the full direct sum, 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).
+\end{equation}
+So the full map is literally the dimension-weighted orthogonal assembly of the separately normalized sector maps.
+
+Because the target sectors are orthogonal, the Frobenius norm adds cleanly:
+\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.
+\end{equation}
+So for Frobenius-type criteria the full signal is just the weighted sum of the sector signals.
+
+For the nuclear norm, however, there is no such scalar additivity. The reason is simple: the different sector maps all start from the same source space. So even though the target sectors are orthogonal, the sector shadows can still overlap in their right-singular directions. The full nuclear norm therefore depends not only on the size of each sector shadow but also on how those shadows align as channels out of the source party.
+
+This is exactly why the direct-sum criterion contains more structural information than a mere list of sector sizes.
+
+There is also an important invariance point. If one applies a collective unitary on the whole complement,
+\begin{equation}
+\rho'=(I_a\otimes U_{\bar a})\rho(I_a\otimes U_{\bar a}^{\dagger}),
+\end{equation}
+then the full map for the cut $a\mid\bar a$ changes only by an orthogonal rotation on its target space. So the full nuclear and Frobenius norms are unchanged. The coarse cut signal therefore does exactly what it should do: it depends only on the bipartite split $a\mid\bar a$, not on how one chooses coordinates on the complement.
+
+By contrast, the sector profile is tied to the chosen internal factorization of the complement. A generic collective unitary on $\bar a$ mixes the $T$-sectors with one another. So the numbers $\Phi_{a\to T}(\rho)$ are not invariants of the coarse cut; they are diagnostics of how the cut-level signal is distributed relative to the specific decomposition of $\bar a$ into parties.
+
+\section{Why the nuclear norm is the right refinement}
+
+There is also a simpler Frobenius statement:
+\begin{equation}
+\norm{\mathcal M_a(\rho)}_{\fro}^2
+=
+\frac{1}{2^{n-1}-1}
+\sum_{\emptyset\neq T\subseteq\bar a}L_{\{a\}\cup T}(\rho).
+\end{equation}
+For cut-separable states this is bounded by $1$ as well.
+
+But this is only the old information in a reorganized form. It records total quadratic size, not directional complexity. The nuclear norm is stronger because it distinguishes a shadow that is essentially one-dimensional from a shadow spread over several independent directions.
+
+This is exactly the lesson learned already in the bipartite flattening picture. One dominant singular value means one dominant channel. Several sizable singular values mean several independent channels. The direct-sum shadow criterion imports that lesson into the multipartite setting.
+
+The distinction is visible already in a very simple example. Take three parties and the state
+\begin{equation}
+\rho_{ABC}=\lvert\mathrm{Bell}\rangle_{AB}\!\langle\mathrm{Bell}\rvert\otimes \lvert\psi\rangle_C\!\langle\psi\rvert.
+\end{equation}
+With $A$ as source, the shadow on $B$ is maximal, the shadow on $C$ vanishes, and the shadow on $BC$ is nonzero because the $ABC$ correlations factor into Bell correlations on $AB$ times the local Bloch vector of $C$.
+
+For qubits, the $A\to B$ sector carries no dimension penalty, while both the full map and the $A\to BC$ sector are divided by $\sqrt3$. So the strongest localized signal is the $B$-only shadow, not the combined one. That is not a defect. It simply means that the full map is a balanced witness for the entire cut, while the sectorwise maps are more diagnostic about where the entanglement actually sits. Distributed states such as GHZ-type states are the opposite kind of example: there the combined map benefits from several sectors at once.
+
+This also resolves the apparent white-noise paradox. Suppose $\lvert\psi\rangle_{ABC}$ is a pure state with maximally mixed marginal on $A$. Then some collective unitary on $BC$ converts it to
+\begin{equation}
+\lvert\Phi^+\rangle_{AB}\otimes\lvert\eta\rangle_C.
+\end{equation}
+If one mixes in white noise,
+\begin{equation}
+\rho_p = p\lvert\psi\rangle\!\langle\psi\rvert + (1-p)\frac{\id}{8},
+\end{equation}
+the same collective unitary gives
+\begin{equation}
+\rho'_p=(I_A\otimes U_{BC})\rho_p(I_A\otimes U_{BC}^{\dagger}).
+\end{equation}
+The white-noise part is unchanged, and the full cut map for $A\mid BC$ is merely orthogonally rotated. So the full cut signal and its noise threshold are identical for $\rho_p$ and $\rho'_p$.
+
+What changes is the sector profile. For the concrete Bell-product state
+\begin{equation}
+\lvert\Phi^+\rangle_{AB}\otimes\lvert 0\rangle_C,
+\end{equation}
+one finds
+\begin{equation}
+\Phi_{A\to B}=3,
+\qquad
+\Phi_{A\to C}=0,
+\qquad
+\Phi_{A\to BC}=\sqrt 3,
+\qquad
+\norm{\mathcal M_A}_*=\sqrt 6.
+\end{equation}
+After applying the Bell-basis change
+\begin{equation}
+U_{BC}=\mathrm{CNOT}_{B\to C}(H_B\otimes I_C),
+\end{equation}
+the same full cut value remains
+\begin{equation}
+\norm{\mathcal M_A}_*=\sqrt 6,
+\end{equation}
+but the profile becomes
+\begin{equation}
+\Phi_{A\to B}=1,
+\qquad
+\Phi_{A\to C}=1,
+\qquad
+\Phi_{A\to BC}=\sqrt{\tfrac83}.
+\end{equation}
+So the signal does not move purely into the $A\to BC$ sector. What happens instead is a sharp redistribution from a strongly localized $A\to B$ witness to a mixed profile spread across $A\to B$, $A\to C$, and $A\to BC$, while the full cut singular values stay unchanged.
+
+This makes the white-noise behavior more interesting, not less. The full cut threshold stays
+\begin{equation}
+p>\frac{1}{\sqrt 6},
+\end{equation}
+but the sector thresholds change drastically. Before the collective rotation, the sector $A\to B$ alone already detects for $p>1/3$. After the rotation, the sectors $A\to B$ and $A\to C$ never strictly violate the cut-separable bound at all, because their pure-state values are exactly $1$, and the sector $A\to BC$ only detects for $p>\sqrt{3/8}$. In particular, at $p=0.60$ the full map still witnesses entanglement after the collective rotation, whereas every individual sector misses it.
+
+So the separate summands tell us how the entanglement is organized relative to the chosen internal decomposition of the complement, while the full map tells us how much entanglement there is across the coarse cut itself. That is the right division of labor.
+
+\section{What changes when the source itself has more than one party}
+
+So far the source side has always been a single party $a$. Nothing forces that. The same construction goes through for any cluster of parties $S$ acting jointly as the source, and it is worth seeing why in plain terms before looking at the formal statement.
+
+Suppose the source is two parties, $S=\{A,B\}$, rather than one. The traceless operator space on $AB$ jointly does not need to be treated as one undifferentiated block. It splits, exactly as the complement always did, according to which of $A,B$ are actually ``doing something'' in a given correlation coefficient:
+\begin{itemize}
+\item a coefficient with $A$ nontrivial and $B$ trivial is a plain $A$-marginal effect;
+\item a coefficient with $B$ nontrivial and $A$ trivial is a plain $B$-marginal effect;
+\item a coefficient with both $A$ and $B$ nontrivial is \emph{genuine} $AB$ structure, correlation that cannot be attributed to either party alone.
+\end{itemize}
+These three pieces are orthogonal subspaces of the two-party traceless operator space, in exactly the same sense that the target-side sectors $T\subseteq\bar a$ were orthogonal all along. Nothing new has to be invented to write this down; it is the same tensor-product decomposition applied to the source instead of the target.
+
+Once the source side is graded this way, the response operator naturally becomes \emph{bigraded}: for every source piece $V\in\{A,B,AB\}$ and every target sector $T$ of the complement, there is a block $M_{V\to T}$, and stacking all of them, with the same normalization idea as before, gives the combined bigraduated shadow map $\mathcal M_S$.
+
+The proof that $\|\mathcal M_S(\rho)\|_*\le1$ for states separable across $S\mid S^c$ costs nothing extra. For a product state $\rho_S\otimes\sigma_{S^c}$, every block factorizes the same way regardless of which source piece $V$ it comes from, because the whole point of a product state is that its correlation coefficients factorize block by block, source side and target side alike. Stacking the source blocks just reassembles the ordinary Bloch vector of the two-party state $\rho_S$, treated as one composite $d_S$-dimensional party; the same correlation-sum bound that controlled a single party's Bloch vector controls this composite one without modification. The whole argument is the same proof, run once more with a bigger source alphabet.
+
+\subsection*{Sub-block witnesses come for free}
+
+Once the response operator is graded this finely, one can look at any sub-collection of its blocks and still get a valid witness, simply because throwing away rows and columns of a matrix (formally, compressing it with an orthogonal projection on either side) can only shrink its nuclear norm. In particular, the single block $M_{S\to T}$ that keeps only the fully genuine source piece $V=S$ isolates the correlation attributable to the source cluster acting jointly, with every individual-party contribution filtered out. This block is a valid witness completely on its own: if it alone already exceeds the bound, that already certifies entanglement, and it does so more specifically than the full stacked operator, because it has deliberately discarded everything that a single source party could have produced by itself.
+
+\subsection*{Why this matters: the Smolin state seen at two different resolutions}
+
+The four-qubit Smolin state is the cleanest illustration of why this refinement is worth having. Viewed one party at a time, the shadow criterion sees a clear violation, $\|\mathcal M_A(\rho_{\mathrm{Smo}})\|_*=3/\sqrt7\approx1.13$, correctly flagging that the state is entangled across every one-versus-three cut. But the Smolin state is also a textbook example of a state that is separable across \emph{every} two-versus-two cut; a single-party construction has no way to even ask that question, because its source side is always exactly one party.
+
+The cluster construction can ask it directly. Take $S=\{A,B\}$ and $S^c=\{C,D\}$. Because the Smolin state carries no one- or three-body correlations, every block of $\mathcal M_{AB}$ vanishes except the fully genuine one, $M_{AB\to CD}$. That single surviving block turns out to have three equal singular values, and after the correct normalization for a two-versus-two cut, the resulting nuclear norm comes out to be exactly $1$: sitting precisely on the boundary, not over it. That is not a coincidence or a near-miss; it is the shadow criterion correctly reporting that this particular cut is consistent with separability, exactly as it should for a state that is genuinely separable there. The same tool that clearly detects entanglement at $1\mid3$ resolution correctly stands down at $2\mid2$ resolution, without needing to be told in advance which cut is the interesting one.
+
+\section{Why the symmetric aggregates are natural}
+
+Once one has the family $\{\mathcal M_a\}_{a\in P}$, two permutation-invariant scalars are immediate:
+\begin{equation}
+\Phi_{\mathrm{sym}}(\rho)=\frac{1}{n}\sum_{a\in P}\norm{\mathcal M_a(\rho)}_*,
+\qquad
+\Phi_{\max}(\rho)=\max_{a\in P}\norm{\mathcal M_a(\rho)}_*.
+\end{equation}
+The maximum asks whether at least one one-vs-rest cut has a large combined shadow. The average asks whether the state is simultaneously large across many such cuts.
+
+These are the most natural symmetric descendants of the cut-wise criterion. They do not introduce new structure; they simply package the family of one-vs-rest ellipsoids into permutation-invariant scalars.
+
+For full separability the benchmark is immediate, because a fully separable state is separable across every one-vs-rest cut. For genuine multipartite entanglement one needs a stronger benchmark: the maximum or average over the \emph{biseparable} set. That is why the three-qubit case is special --- there the symmetric average can be optimized explicitly over biseparable states.
+
+\section{How to read the examples}
+
+Two examples capture the point particularly well.
+
+\subsection*{Three-qubit GHZ}
+
+For $\lvert\GHZ_3\rangle$, each source party produces three orthogonal response directions of equal size. So the combined shadow ellipsoid is not a line segment at all; it has three equal axes. This is exactly the kind of pattern the old Frobenius norms blur together but the singular-value picture retains.
+
+The explicit three-qubit biseparable threshold then shows that the symmetric average does more than detect generic entanglement. It can certify genuine tripartite entanglement in this case.
+
+\subsection*{Four-qubit Smolin}
+
+The Smolin state is useful for the opposite reason. Its one-vs-rest combined shadows are still large enough to violate the fully separable bound, but this does \emph{not} imply genuine multipartite entanglement, because the state is separable across every $2\mid 2$ split. So the example is a warning as well as a success: strong one-vs-rest shadow geometry certifies non-full-separability very clearly, but by itself it does not yet solve the full biseparable problem in higher-party systems.
+
+\section{A first four-qubit numerical scan}
+
+The new \texttt{qtensor} package makes it easy to test whether the shadow geometry is restricted to a few hand-picked examples or whether it persists across broader families. The first scan worth doing is over four-qubit graph states.
+
+For the pure states $\GHZ_4$, the line graph, the ring graph, and the star graph, one finds numerically
+\begin{equation}
+\Phi_{\mathrm{sym}}=\Phi_{\max}=\frac{6}{\sqrt 7}\approx 2.268.
+\end{equation}
+This is already striking because these families have rather different pairwise reduced states. For $\GHZ_4$ all two-qubit marginals are diagonal and PPT; for the line and ring graphs some pairs are maximally mixed, while the remaining pairs are still PPT. So the shadow functional is clearly seeing something beyond ordinary pairwise structure.
+
+The stronger surprise is that this value is not special to just those named examples. A numerical scan over all $38$ connected labeled graph states on four qubits gives exactly the same shadow value in every case. In particular, for every such connected graph state,
+\begin{equation}
+\Phi_{\mathrm{sym}}=\Phi_{\max}=\frac{6}{\sqrt 7}.
+\end{equation}
+Under white-noise admixture,
+\begin{equation}
+\rho(p)=p\rho+(1-p)\frac{\id}{16},
+\end{equation}
+the fully separable benchmark is therefore crossed already at
+\begin{equation}
+p>\frac{\sqrt 7}{6}\approx 0.441.
+\end{equation}
+
+There is also a useful systematic pattern beyond the four-qubit scan. For $n=3,4,5$, the families $\GHZ_n$, the line graph state, and the ring graph state all give the same numerical value of $\Phi_{\mathrm{sym}}$ within machine precision. The common values decrease with $n$ --- from $\sqrt 6$ at $n=3$ to $6/\sqrt 7$ at $n=4$ and then to $2.19089\ldots$ at $n=5$ --- but the coincidence across these different graph-like families persists. By contrast, the $W_n$ family sits slightly lower at each scanned $n$, while D\"ur states are already below the fully separable benchmark for $n=4,5$. So the shadow functional appears to single out a fairly rigid kind of global graph-like channel structure.
+
+We also made a first random search over pure four-qubit biseparable states, sampling all inequivalent cuts. The largest sampled value was
+\begin{equation}
+\Phi_{\mathrm{sym}}\approx 2.235,
+\end{equation}
+obtained for a $2\mid 2$ product state, whereas the best sampled $1\mid 3$ values stayed near $1.94$. This is only a numerical hint, not a theorem, but it is informative. It says that if one eventually wants the sharp four-qubit biseparable threshold, the hard competitors are likely to come from $2\mid 2$ cuts. It also says that the connected graph-state value $6/\sqrt 7\approx 2.268$ lies just above the best random biseparable examples we found.
+
+Figure~\ref{fig:noisy-graph-scan} summarizes the numerics for the representative families $\GHZ_4$, line, ring, and star. The upper panel shows that the shadow functional crosses the fully separable threshold at the common value $p=\sqrt 7/6$. The lower panel shows that the minimum eigenvalue of the partial transpose of every two-qubit marginal stays nonnegative throughout that detection region. This does not yet prove genuine multipartite entanglement by the shadow functional alone, because the four-qubit biseparable threshold is not yet known. But it strongly suggests that the criterion is responding to a genuinely multipartite channel structure that pairwise tests fail to reveal.
+
+\begin{figure}[t]
+\centering
+\includegraphics[width=0.92\linewidth]{numerics/figures/noisy_graph_family_scan.png}
+\caption{Numerical scan performed with \texttt{qtensor} for noisy four-qubit graph-state families. Top: the symmetric shadow functional for $\GHZ_4$, line, ring, and star graph states under white noise. Bottom: the smallest eigenvalue of the partial transpose of a two-qubit marginal for the same families. The shadow signal crosses the fully separable threshold at $p=\sqrt 7/6$, while all two-qubit marginals remain PPT in that region.}
+\label{fig:noisy-graph-scan}
+\end{figure}
+
+\section{The single object behind every construction in this note}
+
+Every object introduced so far --- the single-party map $M_a$, its cluster generalization $\mathcal M_S$, and every sub-block restriction of either --- turns out to be a view of one and the same underlying object. It is worth seeing this explicitly, because it explains why none of the earlier bounds needed a separate proof.
+
+\subsection*{A picture: one big data cube, many ways to flatten it}
+
+Imagine collecting every correlation coefficient of $\rho$, for every party, into one giant array with one axis per party. On the axis belonging to party $a$, instead of only the $d_a^2-1$ traceless directions, also keep one extra slot for ``party $a$ was not measured at all'' (the identity direction). This turns the whole collection into a single $n$-axis data cube, one axis per party, where every axis includes a ``not measured'' option.
+
+This cube is the natural common ancestor of everything else in this note. If one slices the cube by declaring a subset $V$ of parties ``measured'' and everyone else ``not measured'', the result is exactly the ordinary correlation tensor $C_V(\rho)$ for that subset. So the sector decomposition used everywhere in this note, on the target side and on the source side alike, is not extra bookkeeping laid on top of the correlation data; it is simply the fact that a cube with several axes naturally organizes itself by which axes are ``in use''.
+
+It is worth being clear that this cube is not something added on top of the Aschauer \emph{et al.} framework from outside. It is, quite literally, the object that framework starts from: the very first equations of \cite{aschauer} expand the state in the full product basis of local generators \emph{together with the identity} on every party, which is exactly this cube, axis for axis. The sector-restricted correlation tensors $C_S(\rho)$ used to build $L_S$ throughout that framework are obtained from it by the same slicing operation described above; they were never a different object, only a restricted view of this one. What the present note adds is not the cube, which was already there, but the observation that looking at more of it at once, and unfolding it into a matrix instead of collapsing it to a scalar, uncovers structure that slicing straight down to $L_S$ throws away.
+
+\subsection*{Product states are the simplest possible cube}
+
+A completely uncorrelated, fully factorized state $\rho=\bigotimes_a\rho_a$ produces the simplest cube there is: one that factors into a separate vector along each axis, so that every entry of the cube is just a product of one number from each party's own vector. This is the tensor analogue of a rank-one matrix, and the formal note makes this exact: the cube factors completely along every axis if and only if the state itself is a full product state.
+
+\subsection*{Unfolding: turning the cube back into a matrix}
+
+The reason this note works with matrices at all, rather than with the cube directly, is that a cube does not have a clean, efficiently computable notion of ``how far from rank one is it''. Matrices do: that is exactly the singular value decomposition used throughout. So the practical move is to \emph{unfold} the cube into a matrix by choosing a bipartition of the parties, gathering all the source-side axes into one long row index and all the target-side axes into one long column index. This is a completely standard operation on multi-axis data, sometimes called a mode unfolding or matricization. Every shadow map in this note, from the single-party $M_a$ to the cluster map $\mathcal M_S$ and every sub-block restriction of either, is exactly such an unfolding of the same underlying cube, for one particular choice of which axes go on which side.
+
+Crucially, unfolding a rank-one cube always produces a rank-one matrix, no matter which bipartition is chosen. That is the single fact underlying every bound proved in this note: separability makes the cube (a convex mixture of) rank one, every unfolding of a rank-one cube is rank one, and a rank-one matrix always has nuclear norm equal to the product of two vector lengths, which the normalization is chosen precisely to keep below $1$. None of the $\le1$ statements in this note, for a single party, for a cluster, or for any sub-block, ever needed its own separate argument; they are the same rank-one fact, looked at through a different choice of which axes get folded into rows and which into columns.
+
+\subsection*{Tracing out a party is cutting a slice, not adding anything up}
+
+One more piece of intuition is worth having explicitly, because it resolves a question that comes up naturally once several parties are in play: if a party is thrown away by tracing it out, is information lost that was only visible while that party was still there?
+
+The answer is no, and the cube picture makes it obvious why. Tracing out a party $E$ corresponds to slicing the cube at the ``not measured'' position on $E$'s axis, nothing more. It is not a sum or an average over that axis; it is simply looking at the one slice of the cube where $E$ never appears. Every correlation coefficient that does not involve $E$ was already sitting in that slice before $E$ was traced out, completely unchanged. What does change is the normalization: a shadow map built directly from the smaller, $E$-free state uses a smaller effective target dimension than the same block would have used inside the full cube, so the very same numbers get rescaled upward once $E$ is genuinely removed rather than merely ignored. This is the difference between a witness that only says ``entangled across $S\mid RE$, somewhere'' and the sharper, more demanding witness that says ``entangled across $S\mid R$, and that entanglement survives even if $E$ is thrown away completely''. The second statement is strictly stronger, and the cube picture shows that it costs nothing beyond correctly bookkeeping which slice one happens to be looking at.
+
+\section{Where the moment hierarchy fits}
+
+There is a natural broader picture behind all this. The full Pauli correlation coefficients determine the density matrix linearly, so the separability problem can be formulated directly in the space of correlation moments. For qubits this becomes a truncated moment problem on a product of Bloch spheres.
+
+From that viewpoint the old criterion $L_S>1$ is a very compressed front-end test, and the shadow-map criterion is a better one: it keeps some of the real geometry while staying simple and explicit. The full moment hierarchy is the exact continuation of the same program, but it is a larger project and does not belong in the core of the present note.
+
+So the right way to think about the shadow maps is not as a competing philosophy, but as a natural intermediate level:
+\begin{center}
+old quadratic summaries \; $\to$ \; geometric response maps \; $\to$ \; exact correlation-moment hierarchy.
+\end{center}
+
+\section{Closing summary}
+
+The shortest summary is this. The Aschauer \emph{et al.} framework already had the right ambient geometry: correlations were organized into invariant sectors. What it compressed too strongly was the internal directional structure of those sectors. The shadow-map construction restores some of that lost structure.
+
+The central geometric statement is equally short. A source party produces response vectors in several orthogonal sectors on the complement. For cut-product states all of those shadows must remain locked to one common scalar factor. Entanglement is witnessed by the failure of that locking.
+
+Two further points round this out, and both turned out to be free once the central statement was understood correctly. First, ``source party'' never had to mean a single party: the same locking argument applies verbatim to any cluster of parties acting jointly as the source, and grading the source side the way the target side was always graded exposes sub-block witnesses, such as the purely genuine cross-cluster block, that a single-party construction cannot even formulate. Second, every one of these objects, single-party or cluster, full map or sub-block, is a different flattening of one and the same correlation cube; the recurring bound is one rank-one fact about that cube, not a family of separately earned results.
+
+That is the idea behind the formal criterion.
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+\end{document}