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\keywords{Positive linear map, quantum permanent, mixed discriminant,
entanglement-breaking map, decomposable map, Breuer--Hall map, Chern form,
Schur form}
\hypersetup{pdftitle={Weak positivity of Schur forms under decomposable curvature},
pdfauthor={Suvrit Sra}}

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\begin{document}
\title[Weak positivity of Schur forms under decomposable curvature]
{Weak positivity of Schur forms\\ under decomposable curvature}
\aiauthor

\begin{abstract}
Griffiths's 1969 conjecture predicts positivity of Schur forms for
Griffiths-positive vector bundles. Finski reduced its weak form to a
sign question for the double mixed discriminant $\Phi_n(H)$ of a
positive linear map. Du's rank-nine counterexample shows that
$\Phi_n(H)\ge0$ fails in general.
We prove $\Phi_n(H)\ge0$ for every decomposable map $H$, a sum of a
completely positive and a completely copositive map, using a quaternionic
Gaussian determinant representation. Combined with Finski's incidence formula,
this inequality gives weak nonnegativity of every Schur form under decomposable
curvature, without splitting assumptions. Beyond the decomposable
cone, nonnegativity of $\Phi_k(H_{V,U})$ on every square
compression is equivalent to $\Phi_n(H+S)\ge0$ for every
entanglement-breaking map $S$. This class contains explicit indecomposable maps, but is
nonconvex in every size $n\ge9$. We prove the nonconvexity by expressing Du's map
as an average of four members, and the obstruction persists when all
compression inequalities are strict.
\end{abstract}
\preprintslink{qperm}{8 October 2026}
\maketitle

\setcounter{tocdepth}{1}
\tableofcontents

\section{Introduction}
\label{sec:introduction}

Griffiths asked in 1969 whether positivity of the curvature of a
Hermitian holomorphic vector bundle forces positivity of its
characteristic forms \citep[p.~247]{griffiths1969}. In the Schur basis,
the weak form of the question asks whether every Schur form
$s_\lambda(E,h)$ is weakly positive under Griffiths-positive curvature.
The corresponding numerical statement for Schur classes of ample
bundles is the theorem of \citet{fulton_lazarsfeld1983}.
Du's counterexample shows that the statement for forms fails: for every
$r\ge9$, the bundle $\mathcal O(1)^{\oplus r}$ over $\mathbb P^r$
carries a strictly Griffiths-positive metric whose top Chern form is
negative on a nonempty open set \citep{du2026}. This counterexample leaves the problem
of identifying curvature conditions that do ensure pointwise
positivity.

Such restricted cases have a longer history. Bott and Chern proved
nonnegativity of Chern forms under the condition now called Bott--Chern
nonnegativity \citep[Lemma~5.3 and (5.5)]{bottchern1965}, and Li
extended the conclusion to Schur forms
\citep[Proposition~3.1]{li2021}. Finski formulated the equivalent
positive-map problem, proved Schur-form positivity under Nakano and
dual Nakano positivity, and introduced decomposable curvature
\citep{finski2022}. Wan subsequently proved weak positivity for
strongly decomposable curvature of type~I and positivity for type~II
\citep[Theorems~1.2 and~1.4]{wan2024}. These hypotheses impose
splittings on the tangent or fiber space. We consider ordinary
decomposable curvature, with no such splitting.

Let $(E,h)$ have rank $r$ over a complex manifold of dimension $n$.
In Finski's operator dictionary, the curvature at $x$ is a map
$H_x:M_n(\mathbb C)\to M_r(\mathbb C)$
(Section~\ref{sec:applications}). We call it \emph{decomposable} if
\begin{equation}\label{eq:intro-decomp}
 H_x(X)=\sum_a A_aXA_a^*+\sum_b B_bX^{\mathsf T}B_b^*,
\end{equation}
a sum of a completely positive and a completely copositive map, with no
orthogonality or splitting imposed between the two families.

\emph{\textbf{Schur forms under decomposable curvature.}}
Theorem~\ref{thm:decomposable-schur} proves that every Schur form is
weakly nonnegative at a point where the curvature is decomposable.
Thus a negative Schur-form evaluation requires indecomposable
curvature. The proof uses Finski's quotient stability and incidence
formula \citep[Proposition~2.22 and Theorem~3.2]{finski2022}: a Schur
form is the fiber integral of the top Chern form of an auxiliary
quotient bundle, whose curvature is again decomposable. It therefore
suffices to control the top Chern form in every dimension.

The coefficient of a top Chern form is a determinant invariant of the
curvature map. For a complex-linear map $H:M_n(\mathbb C)\to M_n(\mathbb C)$,
put
\begin{equation}\label{eq:intro-functional}
 \Gamma_n(H)=\det(\partial_X)\det H(X),
 \qquad \Phi_n(H)=\frac{\Gamma_n(H)}{n!},
\end{equation}
where $X=(x_{ij})$ has algebraically independent entries and
$\partial_X=(\partial/\partial x_{ij})$. For $n=r$, the curvature
convention in Section~\ref{sec:applications} gives
\[
 c_n(E,h)|_x=\Gamma_n(H_x)\,dV_x,
\]
with $dV_x$ the coordinate volume form. Griffiths semipositivity is
equivalent to positivity of $H_x$, meaning $H_x(X)\succeq0$ for
$X\succeq0$.
 With the unnormalized
mixed-discriminant convention, $\Gamma_n(H)$ is Gurvits's quantum
permanent $\QP(C_H)$ of the Choi matrix of $H$
\citep[Definition~1.7]{gurvits2002}. It is also Cayley's first
hyperdeterminant of order four of the coefficient tensor. The normalized
invariant $\Phi_n$ is
the double mixed discriminant in Finski's formulation
\citep[(1.6), (3.47)--(3.48)]{finski2022}. His Theorem~1.3 identifies
the universal weak Schur-form problem with the inequality
$\Phi_n(H)\ge0$ for positive $H$.

Wan proved this inequality for all positive maps in size three \citep{wan2026};
sizes one and two follow as well. Du's example corresponds to a positive
\emph{bistochastic}
map $H_9$ on $M_9$, meaning $H_9(I)=H_9^*(I)=I$, with
$\Gamma_9(H_9)=-3/64$; Proposition~\ref{prop:du-self-contained} gives a
self-contained determinant calculation for it.

\emph{\textbf{Quaternionic representation.}}
The sign needed for Theorem~\ref{thm:decomposable-schur} follows from
a Gaussian determinant formula.
For a decomposable map \eqref{eq:intro-decomp} with $n=r$, put
$A(g)=\sum_ag_aA_a$ and $B(h)=\sum_bh_bB_b$, where all $g_a,h_b$
are independent standard complex Gaussians. Theorem~\ref{thm:decomposable}
gives
\[
 \Gamma_n(H)=\E\det
 \begin{pmatrix}A(g)&-B(h)\\\overline{B(h)}&\overline{A(g)}\end{pmatrix}
 \ge0.
\]
The block determinant is nonnegative for each realization. For $B=0$,
the formula is Gurvits's identity $\Gamma_n(H)=\E|\det A(g)|^2$
\citep[Theorem~3.8]{gurvits2004}. The Gaussian matrices $A(g)$ and $B(h)$ are
independent, and the entries of each may have arbitrary covariance. Equivalently, the formula evaluates
$\QP(P+Q^{\mathsf T_{\rm in}})$ for arbitrary $P,Q\succeq0$, where
$\mathsf T_{\rm in}$ transposes the input tensor factor.

Beyond the decomposable cone, the sign need not survive compression
or addition. Proposition~\ref{prop:obstruction} gives positive maps with
$\Gamma_n(H)>0$ but $\Gamma_k(U^*H(V\,\cdot\,V^*)U)<0$, and an
entanglement-breaking map $S$ for which $\Gamma_n(H+tS)<0$ for large
$t$. This failure motivates imposing the sign condition on every compression.

We ask which positive perturbations preserve determinant positivity, and
why that stability stops short of arbitrary convex mixing. For
isometries $V,U:\mathbb C^k\to\mathbb C^n$, chosen independently, write
$H_{V,U}(X)=U^*H(VXV^*)U$ for the \emph{square compression} of $H$,
and define
\begin{equation}\label{eq:intro-K}
 \cK_n=\{H:M_n\to M_n\text{ positive}:\
       \Gamma_k(H_{V,U})\ge0\text{ for every }k,V,U\}.
\end{equation}
We call this property \emph{hereditary determinant positivity}. Since
compressions of decomposable maps are decomposable,
Theorem~\ref{thm:decomposable} places the decomposable cone in $\cK_n$.
An \emph{entanglement-breaking} map is one of the form
$S(X)=\sum_{\ell=1}^N(b_\ell^*Xb_\ell)a_\ell a_\ell^*$.

\emph{\textbf{Entanglement-breaking stability.}} Theorem~\ref{thm:eb-maximality}
proves
\begin{equation}\label{eq:intro-maximal}
 H\in\cK_n
 \quad\Longleftrightarrow\quad
 \Gamma_n(H+S)\ge0\quad\text{for every entanglement-breaking }S.
\end{equation}
For the forward implication, Proposition~\ref{prop:eb-addition}
expands $\Gamma_n(H+S)$ over subsets of the rank-one summands of $S$.
Each term is a complementary compression invariant of $H$ multiplied
by two nonnegative Gram determinants. The converse exposes a negative
compression as the leading coefficient of the polynomial
$t\mapsto\Gamma_n(H+tS)$ for a suitable $S$. The class is closed under square compression,
adjoints, independent congruences, and further entanglement-breaking
addition (Proposition~\ref{prop:K-closure}), and a coordinate version
of the theorem characterizes stability under all nonnegative classical
additions (Proposition~\ref{prop:classical-maximality}). Two
consequences are given by Corollary~\ref{cor:low-rank-eb}. For positive
$H$, Wan's theorem gives
\[
 \min\{\rank H(I),\rank H^*(I)\}\le3
 \quad\Longrightarrow\quad H\in\cK_n.
\]
A positive map $K$ on $M_4$ with $\Gamma_4(K)\ge0$, embedded in $M_n$ by
congruence, also belongs to $\cK_n$, by congruence covariance for its
size-four compressions. Thus the single inequality $\Gamma_4(K)\ge0$
certifies the embedded map in every dimension $n\ge4$.

\emph{\textbf{Hereditary determinant positivity is not convex.}}
Theorem~\ref{thm:K-nonconvex} constructs
\[
 A_0,A_1,A_2,A_3\in\cK_9,\qquad
 H_9=\frac14\sum_{j=0}^3A_j,\qquad \Gamma_9(H_9)=-\frac3{64}.
\]
Three summands are embedded four-dimensional Breuer--Hall maps;
the fourth is decomposable. Padding gives nonconvexity for every
$n\ge9$. The summands can even lie in $\operatorname{int}\cK_n$
while their average has negative invariant
(Corollary~\ref{cor:strict-nonconvex}).
 Consequently, no convex cone
invariant under simultaneous unitary conjugation can contain all
decomposable maps and a padded four-dimensional Breuer--Hall map while
keeping $\Gamma_n\ge0$ (Corollary~\ref{cor:no-convex-gluing}).
Convex subclasses with a fixed input or output support remain
available (Proposition~\ref{prop:fixed-support-convex}).

On the bundle side, every Chern form is weakly nonnegative wherever all
square compressions of the curvature map have nonnegative invariant
(Proposition~\ref{prop:chern-compression-sum}). The regularized maps
of Corollary~\ref{cor:strict-nonconvex} give strictly Griffiths-positive
curvature tensors, each with every Chern form strictly positive in the
weak sense, whose average has a negative top Chern form
(Corollary~\ref{cor:chern-nonconvexity}). Even simultaneous strict
pointwise positivity of all Chern forms is thus not a convex condition
on Griffiths-positive curvature.

Two certificates for single maps are proved in the appendices. A
Hilbert--Schmidt sector condition, two linear matrix inequalities on
$H$, gives $\operatorname{Re}\Gamma_n(H)\ge0$ and, for positive
bistochastic $H$, the bound $\Phi_n(H)\ge n^{-n}$
(Theorem~\ref{thm:hs-sector}). An exact moment formula gives
$\Gamma_{2m}(B_{2m})>0$ for the skew-unitary Breuer--Hall maps in every
even size $2m\ge4$ (Theorem~\ref{thm:bh-all-even}); its size-four value
enters the proof of Theorem~\ref{thm:K-nonconvex}. These two results are
certificates for single maps, not membership tests for $\cK_n$:
Proposition~\ref{prop:obstruction} exhibits a positive map whose invariant
becomes negative after an entanglement-breaking addition, although the map
is positive semidefinite as a Hilbert--Schmidt operator.

\subsection{Related work}
\label{sec:related-work}

\smallskip\noindent\emph{\textbf{Quantum permanents and Gaussian representations.}}
Gurvits introduced the quantum permanent and its matroidal
specialization \citep{gurvits2002,gurvits2004}; his Gaussian polynomial
norm, Kraus-pencil determinant formula, mixed determinant correlation,
and Hilbert--Schmidt square representation are the calculus of
Section~\ref{sec:preliminaries}. Barvinok's quaternionic Gaussian
estimators give nonnegative determinant representations of permanents
and of real positive-semidefinite mixed discriminants
\citep{barvinok1999}, and the sign of the complex-adjoint quaternionic
determinant is classical \citep{cohen2000}.
Theorem~\ref{thm:decomposable} establishes the determinant
representation for an arbitrary CP-plus-coCP map, by identifying its Wick expansion with $\Gamma_n$.

\smallskip\noindent\emph{\textbf{Minor sums and stability under additions.}}
The addition expansions of Section~\ref{sec:eb-blocks} specialize the
hyperdeterminantal minor-summation formula of Luque--Thibon
\citep{luquethibon2003}, recorded with proof in
\citep[Proposition~3.9]{amanov2023}; for zero background, the
entanglement-breaking expansion is Gurvits's matroidal permanent. We put
these sums in positive-map form, with complementary subspaces and
nonnegative Gram weights, and prove the converse: hereditary determinant positivity
is equivalent to stability under every entanglement-breaking addition.
The nonconvexity theorem
combines the Breuer--Hall and Robertson maps
\citep{breuer2006,hall2006,chruscinski2008} with Du's rank-nine
construction \citep{du2026}; Du's negative map is the input, and its
assembly from four members of $\cK_9$ and the resulting convex obstruction are
the conclusions. Mapping cones in the sense of \citet{girard2021} are
also closed under completely positive compositions; the obstruction of
Corollary~\ref{cor:no-convex-gluing} uses only simultaneous unitary conjugation.

\smallskip\noindent\emph{\textbf{Additional certificates.}}
Positive maps fixing the diagonal algebra were studied by Kye and
Kim--Kye \citep{kye1995,kimkye1994}; Proposition~\ref{prop:phase-permanent}
is the phase-permanent formula for their invariant, and Marcus's
inequality \citep{marcus1963} gives its lower bound. The sector
certificate of Appendix~\ref{sec:hs-sector} applies the tensor-phase
calculus of \citet{wang2020} to the determinant vector in Gaussian Fock
space; the Hilbert--Schmidt positive-semidefinite square case is in
\citep[Appendix~C]{gurvits2004}. The Breuer--Hall theorem of
Appendix~\ref{sec:bh} uses the characteristic-polynomial duality of
\citet{forrester2024}, attributed there to \citet{kulkarni2025}; the
three-exponential moment formula and the sign of its odd moments are the
further steps.

\smallskip\noindent\emph{\textbf{Existence of Griffiths-positive metrics.}}
Whether every ample bundle admits a Griffiths-positive metric is a
separate question. Du and Xie construct a rank-two ample bundle on an
abelian surface admitting no smooth Griffiths-semipositive Hermitian
metric \citep[Theorem~1.1]{duxie2026}. Their obstruction concerns
existence, whereas the rank-nine construction starts with a strictly
Griffiths-positive metric and yields a negative top Chern form.

\subsection{What is not proved}
\label{sec:not-proved}

The results leave open the unrestricted size-four inequality,
convexity of the bistochastic slice of $\cK_n$, and the sign on the
cone of sums of a $2$-positive and a $2$-copositive map.
Section~\ref{sec:scope} discusses these questions. For Schur forms,
the theorem applies to decomposable curvature; the other subclasses
of $\cK_n$ give Chern-form positivity, and extending them to all Schur
forms requires further closure properties of the auxiliary quotient
curvature (Section~\ref{sec:applications}).

\subsection{Organization}

Section~\ref{sec:preliminaries} fixes conventions, records Gurvits's
Gaussian identities, and proves Theorem~\ref{thm:decomposable}.
Section~\ref{sec:eb-blocks} develops the perturbation calculus and
proves Theorem~\ref{thm:eb-maximality} with its consequences.
Section~\ref{sec:convex-geometry} proves the nonconvexity theorem and
the convex obstruction. Section~\ref{sec:applications} contains the
curvature applications, and
Section~\ref{sec:scope} the boundary cases and open problems.
Appendices~\ref{sec:hs-sector}--\ref{app:separation} prove the sector
and Breuer--Hall certificates and separate the sufficient classes.
Appendix~\ref{app:ferrers} records explicit CP curvature tensors
that extract individual Schur coefficients.

\section{Determinant calculus and the decomposable cone}
\label{sec:preliminaries}

To prove Schur-form nonnegativity, we first establish the sign of
$\Gamma_n$ on decomposable maps. We fix its normalization through
Gurvits's Gaussian identities, prove the quaternionic representation,
and then collect the compression and scaling facts needed for
entanglement-breaking additions.

\subsection{Positive maps and conventions}

We use the Hilbert--Schmidt inner product
\(\langle X,Y\rangle_{\mathrm{HS}}=\tr(X^*Y)\), conjugate-linear in
its first variable. A positive complex-linear map is Hermiticity
preserving. Its Hilbert--Schmidt adjoint is positive as well, by
self-duality of the positive-semidefinite matrix cone.

For \(H:M_p\to M_q\), the Choi matrix is
\[
 C_H=\sum_{i,j=1}^p E_{ij}\otimes H(E_{ij}).
\]
A completely positive map is a map of the form
\(H(X)=\sum_\ell A_\ell XA_\ell^*\), with \(A_\ell\) of size \(q\times p\).
Equivalently, \(C_H\succeq0\) \citep{choi1975}.
A completely copositive map has this form after input transposition.
A decomposable map is a sum of a completely positive and a completely
copositive map. These definitions apply in rectangular dimensions.

An entanglement-breaking (EB) map is a map with the representation
\[
 S(X)=\sum_{\ell=1}^N(b_\ell^*Xb_\ell)a_\ell a_\ell^*.
\]
Equivalently, it is completely positive with a rank-one Kraus
representation; this equivalence is the finite-dimensional measure-and-prepare
characterization of \citet[Theorem~4]{horodecki2003}. We use its conic
version, without requiring trace preservation. Every such map is
decomposable.

For isometries \(V:\mathbb C^k\to\mathbb C^p\) and
\(U:\mathbb C^k\to\mathbb C^q\), define the \emph{square
compression}
\begin{equation}\label{eq:compression-definition}
 H_{V,U}(X)=U^*H(VXV^*)U.
\end{equation}
Here ``square'' means only that the two compressed spaces have
the same dimension. No relation between the input and output subspaces
is imposed. Coordinate sets \(I\subseteq[p]\) and \(J\subseteq[q]\)
of equal size give \(H_{I,J}\) by the corresponding coordinate embeddings.
Set \(\Gamma_0=1\), and give the determinant and permanent of the empty
matrix the value one.

\subsection{Apolar and Gaussian identities}

Gurvits's Gaussian polynomial isometry
\citep[Theorem~3.8, Appendix~C]{gurvits2004} turns determinant
differentiation into a polynomial inner product. This isometry gives both the
correlation formula for $\Gamma_n$ and a square formula for
Hilbert--Schmidt positive-semidefinite maps.

Let \(G\) be a matrix with independent circular complex Gaussian entries,
normalized by \(\E|G_{ij}|^2=1\). If
\(p(z)=\sum_\alpha p_\alpha z^\alpha\) and
\(q(z)=\sum_\alpha q_\alpha z^\alpha\), their Fischer Hermitian product is
\begin{equation}\label{eq:fischer-product}
 \langle p,q\rangle_F
 =\sum_\alpha\alpha!\,\overline{p_\alpha}q_\alpha
 =\E[\overline{p(G)}q(G)].
\end{equation}
The final equality follows from
\(\E[\bar z^a z^b]=\delta_{ab}a!\) in one complex variable and
independence. Consequently
\begin{equation}\label{eq:apolar}
 \Gamma_n(H)=\langle\det,\det\circ H\rangle_F
 =\E[\overline{\det G}\det H(G)].
\end{equation}
The last expression is a correlation of two polynomials evaluated on the same random
matrix. It is not in general a squared norm.

\begin{lemma}[Fock-space calculus]\label{lem:fock-calculus}
On a fixed homogeneous polynomial space, let
\(\mathcal U_Lp=p\circ L\). Then
\[
 \mathcal U_L^*=\mathcal U_{L^*},
 \qquad
 \mathcal U_B\mathcal U_C=\mathcal U_{C\circ B}.
\]
In particular, for any complex-linear \(L:M_n\to M_n\),
\begin{equation}\label{eq:fock-square}
 \Gamma_n(L\circ L^*)=\E|\det L(G)|^2.
\end{equation}
For a Hermiticity-preserving \(H\), the number \(\Gamma_n(H)\) is real,
and \(\Gamma_n(H^*)=\Gamma_n(H)\).
\end{lemma}
\begin{proof}
Monomial orthogonality identifies the homogeneous polynomial space with
a symmetric tensor power; the induced adjoint is the symmetric tensor
power of \(L^*\). Alternatively the adjoint identity follows directly
on products of linear forms from \eqref{eq:fischer-product}.
The order of composition reverses because the action is by pullback.
Thus
\[
 \Gamma_n(LL^*)
 =\langle\det,\mathcal U_{L^*}\mathcal U_L\det\rangle_F
 =\|\mathcal U_L\det\|_F^2.
\]
In an orthonormal basis of Hermitian matrices, the matrix of a
Hermiticity-preserving map is real. The determinant polynomial is also
real in these coordinates: it is real on all real coordinate vectors,
so its coefficients are real. These two facts prove reality of the correlation.
The adjoint identity gives
\(\Gamma_n(H^*)=\overline{\Gamma_n(H)}\).
\end{proof}

For later normalization checks, let
\(\Delta_n(X)=\tr(X)I_n/n\). Direct differentiation gives
\begin{equation}\label{eq:calibrations}
 \Gamma_n(\Id)=\Gamma_n(\mathsf T)=n!,
 \qquad \Gamma_n(\Delta_n)=\frac{n!}{n^n}.
\end{equation}

Input and output congruences change $\Gamma_n$ by nonnegative
determinant factors. We use this relative invariance of the quantum
permanent \citep[(18)]{gurvits2004} when changing compression frames
and normalizing maps.

\begin{lemma}[Covariance and pinched sums]\label{lem:covariance}
For square matrices \(A,B\), with \(\Ad_A(X)=AXA^*\),
\begin{align}
 \Gamma_n(\Ad_A\circ H\circ\Ad_B)
 &=|\det A|^2|\det B|^2\Gamma_n(H),\label{eq:filter}\\
 \Gamma_n(\mathsf T\circ H)
 &=\Gamma_n(H\circ\mathsf T)=\Gamma_n(H).\label{eq:transpose}
\end{align}
For the pinched sum
\[
 (H\boxplus K)\begin{pmatrix}X&Y\\Z&W\end{pmatrix}
 =\begin{pmatrix}H(X)&0\\0&K(W)\end{pmatrix},
\]
one has
\begin{equation}\label{eq:directsum}
 \Gamma_{a+b}(H\boxplus K)=\Gamma_a(H)\Gamma_b(K).
\end{equation}
\end{lemma}
\begin{proof}
The output congruence in \eqref{eq:filter} contributes \(|\det A|^2\).
For the input congruence, the chain rule replaces the derivative matrix
by the corresponding left and right matrix factors, contributing
\(|\det B|^2\). First take invertible factors and then use polynomial
continuity for singular factors. Transposing the entries of \(X\)
preserves \(\det(\partial_X)\); transposing the output preserves its
determinant. For \eqref{eq:directsum}, the output determinant factors.
Every surviving input differential permutation preserves the two
blocks, and hence the differential factors as well, without a
multinomial coefficient.
\end{proof}

\subsection{The decomposable cone}

For a completely positive map, Gurvits's identity expresses
$\Gamma_n$ as a squared Gaussian determinant. To include a completely
copositive summand, we combine two independent Kraus pencils in a
quaternionic block matrix. Its determinant is nonnegative: it is the
square of the Study determinant in the convention of
\citet[Sections~3 and~7]{cohen2000}. Barvinok used related quaternionic
estimators for permanents and real positive-semidefinite mixed
discriminants \citep[Sections~2.3 and~3.3]{barvinok1999}. Here the
matrix entries of each Gaussian pencil may have arbitrary covariance.

\begin{theorem}[Quaternionic determinant representation]
\label{thm:decomposable}
Suppose
\[
 H(X)=\sum_a A_aXA_a^*+\sum_b B_bX^{\mathsf T}B_b^*.
\]
For independent standard circular complex Gaussians \(g_a,h_b\), put
\(A(g)=\sum_a g_aA_a\) and \(B(h)=\sum_b h_bB_b\). Then
\begin{equation}\label{eq:quaternionic}
 \Gamma_n(H)=
 \E\det\begin{pmatrix}
 A(g)&-B(h)\\
 \overline{B(h)}&\overline{A(g)}
 \end{pmatrix}\ge0.
\end{equation}
\end{theorem}
\begin{proof}
Write the displayed block matrix as \(M\). Its top \(n\) rows are
holomorphic in \(g,h\), and its bottom \(n\) rows are antiholomorphic.
Wick contraction pairs each top row \(\alpha\) with a bottom row
\(\tau(\alpha)\), for some \(\tau\in S_n\).
Nonzero covariances require the two paired columns to lie in opposite
column blocks. A compatible determinant permutation is specified by
\(i,j\in S_n\) and a subset \(S\subseteq[n]\): if \(\alpha\notin S\),
the top row uses column \(i(\alpha)\) and the paired bottom row uses
column \(n+j(\alpha)\); if \(\alpha\in S\), those column blocks are
exchanged. The determinant sign is
\[
 (-1)^{|S|}\operatorname{sgn}(i)\operatorname{sgn}(j)
 \operatorname{sgn}(\tau).
\]
The top-right block supplies another \((-1)^{|S|}\).
Summing over the choices of \(S\) therefore gives
\begin{equation}\label{eq:quaternionic-contraction}
 \E\det M
 =\sum_{\tau,i,j\in S_n}
 \operatorname{sgn}(\tau)\operatorname{sgn}(i)\operatorname{sgn}(j)
 \prod_{\alpha=1}^n
 H(E_{i(\alpha),j(\alpha)})_{\alpha,\tau(\alpha)}.
\end{equation}
Here a term with \(\alpha\notin S\) uses
\(\sum_a(A_a)_{\alpha,i(\alpha)}
\overline{(A_a)_{\tau(\alpha),j(\alpha)}}\), while a term with
\(\alpha\in S\) uses
\(\sum_b(B_b)_{\alpha,j(\alpha)}
\overline{(B_b)_{\tau(\alpha),i(\alpha)}}\), as required.
Setting the input permutation equal to \(j\circ i^{-1}\) identifies
\eqref{eq:quaternionic-contraction} with the full Leibniz expansion of
\(\det(\partial_X)\det H(X)\). This identification proves the identity with its
normalization.

For the sign, put
\[
 J_0=\begin{pmatrix}0&-I_n\\I_n&0\end{pmatrix}.
\]
One has \(MJ_0=J_0\bar M\), so \(M\) commutes with the antiunitary
\(v\mapsto J_0\bar v\), whose square is \(-I\).
Nonreal eigenvalues occur with their conjugates. The generalized
eigenspace for a real eigenvalue is invariant under this antiunitary
and has even complex dimension. Indeed, on a \(d\)-dimensional such
space its matrix \(C\) satisfies \(C\bar C=-I_d\); taking determinants
forces \(d\) even. Each real eigenvalue therefore has even algebraic
multiplicity. It follows that \(\det M\ge0\), including singular and
nondiagonalizable cases.
\end{proof}

For $B_b=0$ the theorem reduces to the completely positive identity of
\citet[Theorem~3.8]{gurvits2004},
\begin{equation}\label{eq:cp-gaussian}
 \Gamma_n\left(\sum_a A_a(\,\cdot\,)A_a^*\right)
 =\E\left|\det\left(\sum_a g_aA_a\right)\right|^2.
\end{equation}
Although the block determinant is nonnegative, the formula does not
imply that all mixed coefficients
of a sum of a completely positive and a completely copositive map
are nonnegative.

\subsection{Rank-one resolutions and permanents}

\begin{lemma}[Hermitian Kadison inequality]\label{lem:kadison}
If \(T\) is positive and unital and \(X=X^*\), then
\(T(X^2)\succeq T(X)^2\). A positive bistochastic map is a contraction
for the Hilbert--Schmidt norm.
\end{lemma}
\begin{proof}
Write \(X=\sum_j\lambda_jP_j\), put \(A_j=T(P_j)\), and define
\(V\xi=(A_j^{1/2}\xi)_j\). Since \(\sum_jA_j=I\), \(V\) is an
isometry. For \(D=\diag(\lambda_jI)\),
\[
 T(X^2)-T(X)^2=V^*D(I-VV^*)DV\succeq0.
\]
If $T$ preserves trace, then for Hermitian $X$,
\[
 \|T(X)\|_{\mathrm{HS}}^2
 =\tr T(X)^2\le\tr T(X^2)=\tr X^2
 =\|X\|_{\mathrm{HS}}^2.
\]
For $X=A+iB$ with $A,B$ Hermitian, apply this bound to both terms in
$\|T(X)\|_{\mathrm{HS}}^2
=\|T(A)\|_{\mathrm{HS}}^2+\|T(B)\|_{\mathrm{HS}}^2$.
\end{proof}

Positive maps that fix the diagonal algebra, and their Hadamard
representations, were studied by \citet{kye1995}; indecomposable
examples in dimension at least four appear already in
\citet{kimkye1994}. For this class the invariant is an average of
permanents of positive semidefinite matrices with unit diagonal, which
gives the lower bound one.

\begin{prop}[Phase-permanent representation]
\label{prop:phase-permanent}
If \(T:M_n\to M_n\) is positive and \(T(E_{ii})=E_{ii}\), then
\[
 T(E_{ij})\in\operatorname{span}\{E_{ij},E_{ji}\},
 \qquad
 \Gamma_n(T)=\E_\theta\per T(v_\theta v_\theta^*)\ge1,
\]
where the \(\theta_i\) are independent uniform phases and
\(v_\theta=(e^{i\theta_i})_i\).
\end{prop}
\begin{proof}
Apply Lemma~\ref{lem:kadison} to \(E_{ii}+tX\), for Hermitian \(X\).
The constant term in its positive difference is zero. Both signs of
arbitrarily small real \(t\) force
\[
 T(E_{ii}X+XE_{ii})=E_{ii}T(X)+T(X)E_{ii}.
\]
Complex linearity extends this identity to all \(X\). Applying it to
matrix units yields the asserted support. We can therefore write
\[
 (T(Z))_{ij}=a_{ij}z_{ij}+b_{ij}z_{ji}\quad(i\ne j),
 \qquad (T(Z))_{ii}=z_{ii}.
\]

Expand the permanent by an output permutation \(\sigma\).
On a nontrivial cycle, encode a forward or reversed choice of the input
variable by \(\varepsilon_i\in\{1,-1\}\).
The exponent of \(\theta_i\) is
\(\varepsilon_i-\varepsilon_{\sigma^{-1}(i)}\).
Phase averaging retains exactly the choices of a constant orientation
on each cycle. In \eqref{eq:apolar}, a selected input monomial pairs
with \(\det G\) exactly when every input row and every input column
is used once. This requirement is the same cycle condition. The input permutation
reverses the cycles on which the reversed orientation was chosen.
Reversing a cycle preserves its sign, so the two determinant signs
multiply to \(+1\). Every retained Gaussian monomial is squarefree
and has moment one. The cycle analysis thus proves the exact permanent identity,
including two-cycles and fixed points.

The next step is Marcus's permanent analogue of Hadamard's inequality
\citep{marcus1963}; the short Gaussian norm proof is included.
For a positive-semidefinite matrix \(C\),
\(\per C\ge\prod_iC_{ii}\). To verify this inequality directly, realize \(C\)
as the covariance matrix of complex Gaussian linear forms \(L_i\).
Wick contraction gives
\(\per C=\|\prod_iL_i\|_F^2\).
If \(L(z)=\sum_ja_jz_j\) and
\(D=\sum_j\bar a_j\partial_{z_j}\), then the adjoint relation and
\([D,L]=\sum_j|a_j|^2\) give
\[
 \|Lf\|_F^2=\|Df\|_F^2+\|a\|_2^2\|f\|_F^2
 \ge\|a\|_2^2\|f\|_F^2.
\]
Iterating proves the permanent inequality.
For $C_\theta=T(v_\theta v_\theta^*)$, positivity and the fixed
diagonal give $C_\theta\succeq0$ and $(C_\theta)_{ii}=1$. Hence
$\Gamma_n(T)=\E_\theta\per C_\theta\ge1$.
\end{proof}

\subsection{Strict normalization}

Sign statements for $\Gamma_n$ can often be reduced to bistochastic
maps. The next lemma is the strictly positive case of operator scaling,
whose capacity and attainment formulation is due to Gurvits
\citep[Theorem~4.7]{gurvits2004}. We include the proof because a
boundary map need not admit an exact invertible scaling. Gurvits's use of
``indecomposable'' in that scaling theorem denotes a rank-increasing
condition, not indecomposability in the CP-plus-coCP sense used here.

\begin{lemma}[Bistochastic normalization]\label{lem:balancing}
Every strictly positive map \(H:M_n\to M_n\) admits a bistochastic
normalization by invertible input and output congruences.
Here strict positivity means \(H(X)\succ0\) for every
nonzero \(X\succeq0\). These congruences preserve the sign of \(\Gamma_n\).
\end{lemma}
\begin{proof}
Compactness of the unit spheres gives \(\delta>0\) such that
\(H(X)\succeq\delta\tr(X)I\) for \(X\succeq0\).
Minimize \(\log\det H(X)\) over \(X\succ0\), \(\det X=1\).
The lower bound
\[
 \log\det H(X)\ge n\log(\delta\tr X)
\]
is coercive on this constraint. A bounded trace and determinant one
bound every eigenvalue above and away from zero, so a minimizer
\(X_0\) exists. The Lagrange equation is
\[
 H^*(H(X_0)^{-1})=\lambda X_0^{-1}.
\]
Pairing with \(X_0\) gives \(\lambda=1\). Thus
\[
 \widehat H(Z)=H(X_0)^{-1/2}
 H(X_0^{1/2}ZX_0^{1/2})H(X_0)^{-1/2}
\]
satisfies $\widehat H(I)=\widehat H^*(I)=I$. Since $\det X_0=1$,
the covariance identity \eqref{eq:filter} gives
\[
 \Gamma_n(\widehat H)
 =\frac{\Gamma_n(H)}{\det H(X_0)},
 \qquad \det H(X_0)>0.
\]
\end{proof}

A positive map on the boundary need not admit an invertible
bistochastic normalization. Universal sign statements can nonetheless
be reduced to the strictly positive bistochastic case by first adding
\(\epsilon\tr(\,\cdot\,)I\), normalizing, and finally using polynomial
continuity as \(\epsilon\downarrow0\).

We use Wan's theorem \citep[Theorem~0.3]{wan2026} in the following form:
\begin{equation}\label{eq:wan-small}
 H:M_k\to M_k\text{ positive},\quad k\le3
 \quad\Longrightarrow\quad\Gamma_k(H)\ge0.
\end{equation}
The non-strict case follows by trace augmentation and continuity.
Sizes one and two also follow by pinched identity padding into size three.
\section{Perturbation calculus and hereditary determinant positivity}
\label{sec:eb-blocks}

To determine which additions preserve the sign of $\Gamma_n$, we
expand it in rank-one perturbations. Each coefficient is a determinant
invariant of a complementary compression, multiplied by nonnegative
Gram factors. These coefficients prove both directions of the
characterization of $\cK_n$. The coordinate version gives block
constructions with classical additions. Both expansions specialize the
hyperdeterminantal minor-summation formula
\citep{luquethibon2003,amanov2023}.

\subsection{Complementary compressions and entanglement-breaking additions}

When the background map is zero, the following formula reduces to
Gurvits's matroidal permanent \citep[Definition~1.1]{gurvits2002}.
For a nonzero background, each remaining factor is the invariant of
a complementary input/output compression. We include a direct proof
so that the arbitrary nonorthogonal rank-one terms and the factorial
normalization are explicit.

\begin{prop}[Entanglement-breaking addition formula]
\label{prop:eb-addition}
Let
\[
 S(X)=\sum_{a=1}^N(b_a^*Xb_a)a_aa_a^*
\]
be entanglement breaking.  For $L\subseteq[N]$ of cardinality $m\le n$,
put $A_L=[a_a]_{a\in L}$ and $B_L=[b_a]_{a\in L}$.  If both have rank
$m$, choose isometries $U_L,V_L$ onto the orthogonal complements of
their respective column spaces.  Then
\begin{equation}\label{eq:eb-addition}
 \Gamma_n(H+S)=
 \sum_{\substack{L\subseteq[N]\\|L|\le n}}
 \det(A_L^*A_L)\det(B_L^*B_L)
 \Gamma_{n-|L|}(H_{V_L,U_L}).
\end{equation}
Rank-deficient subsets contribute zero.  The empty subset contributes
$\Gamma_n(H)$, and subsets of cardinality $n$ use $\Gamma_0=1$.
The formula holds for every complex-linear $H$, with no positivity
assumption on it.
\end{prop}
\begin{proof}
Introduce distinct parameters $t_a$ in the sum defining $S$.  The output
determinant is affine in each $t_a$, since its associated matrix has
rank one.  For a fixed subset $L$ its coefficient is
\begin{equation}\label{eq:eb-output-coefficient}
 \det(A_L^*A_L)\det(U_L^*H(X)U_L)
 \prod_{a\in L}(b_a^*Xb_a).
\end{equation}
To verify this coefficient, choose a unitary basis whose first $m$ coordinates span
the columns of $A_L$.  The coefficient using all $m$ rank-one terms
exhausts these rows and columns.  Its factor is the squared volume
$\det(A_L^*A_L)$ times the complementary determinant.  If the columns
are dependent, the coefficient is zero.

For every homogeneous polynomial $f$ of degree $n-m$, the matching
input identity is
\begin{equation}\label{eq:eb-input-identity}
 \det(\partial_X)\left[
 \prod_{a\in L}(b_a^*Xb_a)f(X)\right]
 =\det(B_L^*B_L)\det(\partial_Y)f(V_LYV_L^*).
\end{equation}
For independent columns, choose a unitary input basis whose first $m$
coordinates span them.  All $m$ linear factors depend on the leading
$m\times m$ block.  In a surviving determinant differential, every one
of its $m$ input-row derivatives must differentiate these factors;
their columns consequently exhaust the same block.  The first
differential factor is
\[
 \det(\partial_Z)\prod_{a=1}^m(r_a^*Zr_a)
 =|\det[r_1\ \cdots\ r_m]|^2=\det(B_L^*B_L),
\]
by direct alternating expansion.  The remaining differential acts on
the complementary compression of $f$.  If the columns are dependent,
their support has dimension less than $m$, so this allocation of the
$m$ row derivatives is impossible and both sides vanish.  This case analysis proves
\eqref{eq:eb-input-identity}.  Apply it to
\eqref{eq:eb-output-coefficient} and sum over $L$.  This summation also proves
that no factorial is missing.
\end{proof}

\begin{theorem}[Characterization by entanglement-breaking additions]
\label{thm:eb-maximality}
Let $H:M_n\to M_n$ be complex linear. The following are equivalent:
\begin{enumerate}
 \item Every square compression $H_{V,U}$, in every size
 $1\le k\le n$, has $\Gamma_k(H_{V,U})\ge0$.
 \item $\Gamma_n(H+S)\ge0$ for every entanglement-breaking map $S$.
\end{enumerate}
Either condition implies that $H$ is positive.
\end{theorem}
\begin{proof}
The first condition implies the second by
Proposition~\ref{prop:eb-addition}.  Conversely, fix a size-$k$
compression, with $k<n$, and choose orthonormal bases
$a_1,\ldots,a_m$ and $b_1,\ldots,b_m$ of its output and input
complements, where $m=n-k$.  Set
\[
 S(X)=\sum_{j=1}^m(b_j^*Xb_j)a_ja_j^*.
\]
Both Gram matrices are identities, so \eqref{eq:eb-addition} gives
\[
 p(t):=\Gamma_n(H+tS),\qquad
 \deg p\le m,\qquad [t^m]p(t)=\Gamma_k(H_{V,U}).
\]
The second condition gives $p(t)\ge0$ for $t\ge0$, hence
$p\in\mathbb R[t]$ and $[t^m]p\ge0$. For $k=n$, take $S=0$
and use unitary covariance.
Finally, the size-one inequalities are
$u^*H(vv^*)u\ge0$ for all unit vectors $u,v$. They imply
$H(vv^*)\succeq0$, and spectral decomposition then proves positivity
of $H$.
\end{proof}

The characterization reduces stability under all EB additions to
compression inequalities. The next results give closure properties and
ways to verify those inequalities from low-dimensional maps.

\begin{prop}[Closure properties of $\cK_n$]\label{prop:K-closure}
The class $\cK_n$ of \eqref{eq:intro-K} is closed. For $H\in\cK_n$,
$c\ge0$, $A,B\in M_n$, and an entanglement-breaking map $S$,
\[
 cH,\ H^*,\ \mathsf T\circ H,\ H\circ\mathsf T,\
 \Ad_A\circ H\circ\Ad_B,\ H+S\ \in\cK_n.
\]
For isometries $U,V:\mathbb C^k\to\mathbb C^n$, one also has
$H_{V,U}\in\cK_k$. Every coefficient of
$\Gamma_n(H+\sum_a t_aS_a)$ is nonnegative when the $S_a$ are
rank-one entanglement-breaking summands.
\end{prop}
\begin{proof}
Closedness and nonnegative scaling follow from the defining polynomial
inequalities. Iterated compression satisfies
$(H_{V,U})_{V',U'}=H_{VV',UU'}$. Moreover,
$(H+S)_{V,U}=H_{V,U}+S_{V,U}$ and $S_{V,U}$ is entanglement breaking.
Theorem~\ref{thm:eb-maximality} therefore proves addition stability.
 The coefficient assertion follows from
\eqref{eq:eb-addition} with a separate parameter on each summand.

Adjoints interchange the input and output isometries, and
Lemma~\ref{lem:fock-calculus} preserves the real value of the invariant.
For invertible $A,B$, factor $BV=V_0R$ and $A^*U=U_0L$,
where $V_0,U_0$ are isometries and $R,L\in M_k$ are invertible. Then
\[
 (\Ad_A\circ H\circ\Ad_B)_{V,U}
 =\Ad_{L^*}\circ H_{V_0,U_0}\circ\Ad_R,
\]
so \eqref{eq:filter} multiplies its invariant by
$|\det L|^2|\det R|^2$.
Singular congruences follow by continuity. Whole transposition changes
one compression plane to its complex conjugate and preserves its
invariant by \eqref{eq:transpose}.
\end{proof}

\begin{corr}[Low-dimensional certificates]
\label{cor:low-rank-eb}
Each of the following positive maps $H:M_n\to M_n$ belongs to
$\cK_n$. Consequently, $H+S\in\cK_n$ for every entanglement-breaking
map $S$:
\begin{enumerate}
 \item $\min\{\operatorname{rank}H(I),\operatorname{rank}H^*(I)\}\le3$.
 \item $H(X)=A K(B^*XB)A^*$, where $A,B$ are $n\times d$ matrices,
 $d\le4$, and $K:M_d\to M_d$ is positive.  If $d=4$, additionally
 assume that $\Gamma_4(K)\ge0$ has been established.
\end{enumerate}
\end{corr}
\begin{proof}
In the first case, if $H(I)$ has rank at most three, order bounds show
that all outputs are supported on its range.  Every size-$k$
compression with $k>3$ then has identically zero determinant.
If instead $H^*(I)$ has rank at most three, let $P$ be its support.
For $x\in\ker P$, positivity and
\[
 \tr H(xx^*)=x^*H^*(I)x=0
\]
give $H(xx^*)=0$.  For any $y$ and complex $z$, positivity of
$H((x+zy)(x+zy)^*)$ forces the term linear in $z$ to vanish. Thus
$H(xy^*)=H(yx^*)=0$, and $H(X)=H(PXP)$.
  A size-$k$ compression with $k>3$ therefore ignores
a nonzero input direction after a unitary change of basis.  Its
determinant differential is zero.  All size-$k$ compressions with
$k\le3$ are ordinary positive maps covered by Wan.

For the second case, compressions of size $k\le3$ again follow from
Wan, and those with $k>d$ have deficient output rank.  At $k=d=4$,
put $A_k=U^*A$ and $B_k=V^*B$.  They are square, and filtering gives
\[
 \Gamma_4(H_{V,U})
 =|\det A_k|^2|\det B_k|^2\Gamma_4(K)\ge0.
\]
Singular factors give zero.  In each case the signs of all square
compressions have been established, so Theorem~\ref{thm:eb-maximality} applies.
\end{proof}

\begin{rmk}
We call the map $K$ in the second case a \emph{core} of $H$. The
four-dimensional clause applies to a core with $\Gamma_4(K)\ge0$; it does
not assume the unrestricted size-four inequality. It applies, for example, to
decomposable cores, to cores satisfying the sector condition of
Appendix~\ref{sec:hs-sector}, and to the four-dimensional Breuer--Hall
map. A Hilbert--Schmidt PSD map in higher dimension need not have
nonnegative square compressions (Proposition~\ref{prop:obstruction}),
so the corollary does not license EB addition to that entire class.
\end{rmk}

Corollary~\ref{cor:low-rank-eb} already places indecomposable maps in
$\cK_n$. The next family, built from Choi's map, gives strictly positive
bistochastic examples with a lower bound on $\Phi_n$.

\begin{prop}[A strictly positive indecomposable family]
\label{prop:choi-eb-family}
Define the normalized Choi map \citep{choi_biquadratic1975}
$C:M_3\to M_3$ by
\[
 2C(X)=\begin{pmatrix}
 x_{11}+x_{33}&-x_{12}&-x_{13}\\
 -x_{21}&x_{22}+x_{11}&-x_{23}\\
 -x_{31}&-x_{32}&x_{33}+x_{22}
 \end{pmatrix}.
\]
For $n\ge3$, let $K_0$ be its pinched direct sum with diagonal pinching
on the remaining $n-3$ coordinates, and put
\[
 H_\epsilon=(1-\epsilon)K_0+\epsilon\Delta_n,
 \qquad \Delta_n(X)=\frac{\tr X}{n}I.
\]
For $0<\epsilon<n/(n+14)$, the map $H_\epsilon$ is strictly positive,
bistochastic, and indecomposable, and
\[
 \Phi_n(H_\epsilon)\ge(\epsilon/n)^n>0.
\]
\end{prop}
\begin{proof}
The map $C$ is unital and trace preserving.  For $X=xx^*$, put
$a=|x_1|^2$, $b=|x_2|^2$, $c=|x_3|^2$.  The two-by-two principal
minors of $2C(xx^*)$ are
\[
 a^2+ac+bc,\qquad ab+bc+c^2,\qquad ab+ac+b^2.
\]
Its determinant is $a^2b+b^2c+c^2a-3abc\ge0$ by AM--GM.  All principal
minors are nonnegative, proving ordinary positivity.

Let $\Omega=\sum_{i=1}^3|ii\rangle$, with indices read cyclically, and set
\[
 W=|\Omega\rangle\langle\Omega|
 +\frac12\sum_i|i,i+1\rangle\langle i,i+1|
 +2\sum_i|i+1,i\rangle\langle i+1,i|.
\]
The matrix $W$ is PSD.  Its partial transpose has diagonal entries one and, on
each off-diagonal unordered pair, the PSD matrix
$\left(\begin{smallmatrix}1/2&1\\1&2\end{smallmatrix}\right)$, up to
interchanging the two basis vectors.  Thus $W$ is PPT.  Direct contraction
with the Choi matrix gives
\[
 \tr(C_CW)=-\frac34,\qquad \tr W=\frac{21}{2}.
\]
Embed this witness in the first three input and output coordinates.
Its pairing with $C_{H_\epsilon}$ is
\[
 -\frac34(1-\epsilon)+\frac{21}{2n}\epsilon<0
 \quad\Longleftrightarrow\quad
 \epsilon<\frac{n}{n+14}.
\]
A decomposable Choi matrix has nonnegative pairing with every PPT
matrix, proving indecomposability.

Both summands in $H_\epsilon$ are bistochastic, and
$H_\epsilon(X)\succeq(\epsilon/n)\tr(X)I$ for $X\succeq0$.
Its core $(1-\epsilon)C$, embedded on three input and output
coordinates, has marginal rank three.  All remaining terms are EB.
Corollary~\ref{cor:low-rank-eb} and the exact addition formula prove the
sign.  More explicitly the EB remainder is classical, with every
entry of its matrix at least $\epsilon/n$.  Its permanent contribution
alone is at least $n!(\epsilon/n)^n$, proving the quantitative bound.
\end{proof}

\subsection{Coordinate expansions and rectangular blocks}

The coordinate identity below is the order-four specialization of the
hyperdeterminantal minor-summation formula
\citep[Proposition~3.9]{amanov2023} to a diagonal perturbation. For the
tensor of \(C_D\), the surviving input index sets coincide in pairs, as
do the output sets; the permutation signs square to one and the
complementary hyperdeterminant becomes \(\per D[J^c,I^c]\). The
maximality statement for classical additions, that is, for the maps
\(C_D\) below with \(D\) entrywise nonnegative, follows immediately
afterward.

\begin{prop}[Exact classical addition identity]
\label{prop:classical-addition}
For a scalar $n\times n$ matrix $D$, put
\[
 C_D(X)=\operatorname{diag}(D\operatorname{diag}X).
\]
For every complex-linear $H:M_n\to M_n$, without positivity assumptions,
\begin{equation}\label{eq:classical-addition}
 \Gamma_n(H+C_D)=
 \sum_{\substack{I,J\subseteq[n]\\|I|=|J|}}
 \Gamma_{|I|}(H_{I,J})\,\per D[J^c,I^c].
\end{equation}
The permanent of an empty matrix is one.
\end{prop}
\begin{proof}
Expanding the output determinant by the entries of $C_D(X)$ gives
\[
 \det(H(X)+C_D(X))=
 \sum_{J\subseteq[n]}\det H(X)[J,J]
 \prod_{a\notin J}\left(\sum_bD_{ab}x_{bb}\right).
\]
Every input derivative assigned to the second factor must be diagonal.
If $I$ is the set of input derivative indices assigned to the first factor,
homogeneity forces $|I|=|J|$.  The input permutation fixes $I^c$ and permutes
$I$.  Assigning the complementary distinct diagonal derivatives to the
factors from $C_D$ gives $\per D[J^c,I^c]$.  The remaining alternating
differential gives $\Gamma_{|I|}(H_{I,J})$.  These assignments exhaust the Leibniz
expansion and prove the identity, without an additional factorial.
\end{proof}

\begin{prop}[Exact maximality for classical additions]
\label{prop:classical-maximality}
For a complex-linear map $H:M_n\to M_n$, the following are equivalent:
\begin{enumerate}
 \item $\Gamma_{|I|}(H_{I,J})\ge0$ for every pair of equally sized
 coordinate sets $I,J\subseteq[n]$.
 \item $\Gamma_n(H+C_D)\ge0$ for every entrywise nonnegative matrix $D$.
\end{enumerate}
\end{prop}
\begin{proof}
The forward implication follows from
Proposition~\ref{prop:classical-addition}.  Conversely, fix $I,J$ of
cardinality $k<n$, and let $D_0$ be a partial permutation matrix mapping
the input coordinates $I^c$ bijectively to the output coordinates $J^c$.
Since $\per D_0[J^c,I^c]=1$, the expansion
\eqref{eq:classical-addition} gives
\[
 p(t):=\Gamma_n(H+tC_{D_0}),\qquad
 \deg p\le n-k,\qquad [t^{n-k}]p(t)=\Gamma_k(H_{I,J}).
\]
As before, $p(t)\ge0$ for $t\ge0$ implies $p\in\mathbb R[t]$
and $[t^{n-k}]p\ge0$. For $k=n$, take $D=0$; for $k=0$, use
$\Gamma_0=1$.
\end{proof}

\begin{lemma}[Weighted rank-one resolutions]
\label{lem:weighted-resolution}
Suppose $K:M_m\to M_m$ is positive and
$K(E_{ii})=\alpha_iE_{ii}$ with $\alpha_i\ge0$.  Then
\[
 K(E_{ij})\in\operatorname{span}\{E_{ij},E_{ji}\},
 \qquad
 \Gamma_{|I|}(K_{I,I})\ge\prod_{i\in I}\alpha_i.
\]
If $I,J$ have the same cardinality and $I\ne J$, then
$\Gamma_{|I|}(K_{I,J})=0$.
\end{lemma}
\begin{proof}
If $\alpha_i=0$, positivity on
$(e_i+ze_j)(e_i+ze_j)^*$, for arbitrarily small complex $z$, shows that
$u^*K(E_{ij})u=0$ for every output vector $u$.  Polarization gives
$K(E_{ij})=K(E_{ji})=0$.  Moreover $K(I)=\operatorname{diag}\alpha$,
so the order bounds on images of Hermitian matrices show that every
output is supported on the positive-$\alpha$ coordinates.

On these coordinates, output congruence by
$\operatorname{diag}(\alpha_i^{-1/2})$ gives a positive map $T$ fixing
each $E_{ii}$.  Proposition~\ref{prop:phase-permanent} gives the support restriction
and, applied to each principal compression,
$\Gamma_{|I|}(T_{I,I})\ge1$.  The output
congruence factor is $\prod_{i\in I}\alpha_i$, proving the bound when
all these numbers are positive.  If a number vanishes, the relevant
determinant vanishes and the bound still holds.

Finally, $K(M_I)$ is supported in $I$.  Compressing it to a different
set $J$ of the same cardinality leaves support in $I\cap J$, of smaller
dimension.  Its determinant is identically zero.
\end{proof}

\begin{theorem}[Rectangular block enlargement]
\label{thm:rectangular-blocks}
Let $A_s\subseteq[n]$ be pairwise disjoint nonempty input coordinate sets,
and $B_s\subseteq[n]$ pairwise disjoint nonempty output coordinate sets.
They need not exhaust the coordinates.  Put $a_s=|A_s|$, $b_s=|B_s|$,
and let $K_s:M_{a_s}\to M_{b_s}$ be positive. Define $H_0$ by
\[
 H_0(X)[B_s,B_s]=K_s(X[A_s,A_s]),\qquad
 H_0(X)_{ij}=0\quad\text{if }(i,j)\notin\bigcup_s(B_s\times B_s).
\]
Suppose each block satisfies at least one of:
\begin{enumerate}
 \item $K_s$ is decomposable, with arbitrary rectangular dimensions;
 \item $\min(a_s,b_s)\le3$;
 \item $a_s=b_s$ and $K_s(E_{ii})=\alpha_iE_{ii}$ in its designated
 local bases, with $\alpha_i\ge0$;
 \item $a_s=b_s=4$ and $\Gamma_4(K_s)\ge0$ has already been certified.
\end{enumerate}
Then, for every entrywise nonnegative $D$,
\begin{equation}\label{eq:block-per-lower}
 \Gamma_n(H_0+C_D)\ge\per D\ge0.
\end{equation}
The nonnegative conclusion is preserved by independent input and output
congruences, global input or output transposition, and limits.
\end{theorem}
\begin{proof}
In a coordinate compression, put $I_s=I\cap A_s$ and $J_s=J\cap B_s$.
If either $I$ or $J$ uses an unused coordinate, its $\Gamma$ vanishes.
The output determinant factors over the blocks, with its $s$-th factor
of degree $|J_s|$ in the variables of $A_s$.  Every surviving input
differential permutation preserves each $I_s$.  Its degree in that
same variable group is $|I_s|$.  Consequently the invariant vanishes
unless $|I_s|=|J_s|=:k_s$ for every $s$.  In the remaining cases,
\begin{equation}\label{eq:block-compression-factor}
 \Gamma_{|I|}((H_0)_{I,J})
 =\prod_s\Gamma_{k_s}((K_s)_{I_s,J_s}).
\end{equation}
There is no multinomial coefficient, since the differential factors
act on disjoint groups of variables.  Reordering blocks simultaneously
in the rows and columns introduces no determinant sign.

Each factor is nonnegative.  In the first case compression preserves
decomposability.  In the second $k_s\le3$, so the compressed ordinary
positive map is covered by Wan. The third follows from
Lemma~\ref{lem:weighted-resolution}. In the fourth case the proper square
compressions have size at most three and the full compression is the
certified map $K_s$ itself. Apply
Proposition~\ref{prop:classical-addition}; every permanent is nonnegative,
and the $I=J=\varnothing$ term is $\per D$.

For nonsingular congruences the exact covariance is
\[
 \Gamma_n(\Ad_L\circ H\circ\Ad_R)
 =|\det L|^2|\det R|^2\Gamma_n(H).
\]
Singular cases follow by continuity.  Determinant transposition leaves
$\Gamma_n$ unchanged, and polynomial continuity proves the assertion
about limits.
\end{proof}

\begin{corr}[Sharper bound for a weighted resolution]
\label{cor:weighted-classical}
If $H_0(E_{ii})=\alpha_iE_{ii}$ as in
Lemma~\ref{lem:weighted-resolution}, then
\begin{align}
 \Gamma_n(H_0+C_D)
 &=\sum_{I\subseteq[n]}\Gamma_{|I|}((H_0)_{I,I})
                    \per D[I^c,I^c]\\
 &\ge\per(D+\operatorname{diag}\alpha).
\end{align}
If $H_0+C_D$ is bistochastic, then $D+\operatorname{diag}\alpha$ is
doubly stochastic and
\[
 \Phi_n(H_0+C_D)\ge n^{-n}.
\]
\end{corr}
\begin{proof}
The classical addition identity and the weighted-resolution lemma give
\[
 \Gamma_n(H_0+C_D)
 \ge\sum_{I\subseteq[n]}\Bigl(\prod_{i\in I}\alpha_i\Bigr)
       \per D[I^c,I^c]
 =\per(D+\operatorname{diag}\alpha).
\]
Bistochasticity gives
$D\mathbf1+\alpha=D^{\mathsf T}\mathbf1+\alpha=\mathbf1$,
so van der Waerden's inequality
\citep{egorychev1981,falikman1981} gives the final bound.
\end{proof}

\begin{rmk}
The bound $n^{-n}$ is specific to the last corollary; a size-three
reduction block can have zero invariant.  For general blocks,
\eqref{eq:block-per-lower} implies
$\Phi_n\ge(\epsilon/n)^n$ whenever $D_{ij}\ge\epsilon/n$.
The theorem requires pinched input blocks.  It does not permit
uncontrolled coherent couplings between them, or arbitrary mixing
across incompatible block structures.
\end{rmk}

\section{Convex geometry and obstructions}
\label{sec:obstructions}
\label{sec:convex-geometry}

Hereditary determinant positivity permits arbitrary entanglement-breaking
additions. It does not permit arbitrary sums of members of $\cK_n$.
We first identify closed convex subclasses with fixed support, and then
exhibit an explicit obstruction to combining all such certificates into
a unitary-invariant convex enlargement of decomposability.

\subsection{Closed convex subclasses with prescribed support}

Fix a proper subspace \(E\subset\mathbb C^n\), with orthogonal
projection \(P_E\), let \(\EB_n\) be the cone of entanglement-breaking
maps on \(M_n\), and set
\[
 \mathfrak F_E^{\rm out}
 =\{K:M_n\to M_n\text{ positive}:K(I)=P_EK(I)P_E\},
 \qquad
 \mathfrak C_E^{\rm out}=\mathfrak F_E^{\rm out}+\EB_n.
\]
Define the input analogues by taking Hilbert--Schmidt adjoints.

\begin{prop}[Fixed-support convex subclasses]
\label{prop:fixed-support-convex}
If \(\dim E\le3\), then \(\mathfrak C_E^{\rm out}\) and
\(\mathfrak C_E^{\rm in}\) are closed convex cones contained in
\(\cK_n\). For \(\dim E=3\), they contain indecomposable maps.
They do not, in general, contain all decomposable maps.
\end{prop}
\begin{proof}
The set \(\mathfrak F_E^{\rm out}\) is a closed convex face of the
positive-map cone. Indeed, order bounds on Hermitian inputs give
\[
 K(I)=P_EK(I)P_E
 \quad\Longleftrightarrow\quad
 K(X)=P_EK(X)P_E\quad\text{for every }X.
\]
For $K\in\mathfrak F_E^{\rm out}$ with $K=K_1+K_2$ and $K_i$ positive,
the inequalities
$0\preceq K_i(I)\preceq K(I)$ put both $K_i$ in
$\mathfrak F_E^{\rm out}$. The sum with $\EB_n$ is convex.

To prove closedness of the sum, use the linear functional
\(\ell(T)=\tr T(I)\). It is strictly positive on every nonzero
positive map: \(\ell(T)=0\) forces \(T(I)=0\), and order bounds
then force \(T=0\). Consequently the set of positive maps with
\(\ell(T)\le M\) is compact in finite dimension. If
\(K_j+S_j\) converges with \(K_j\in\mathfrak F_E^{\rm out}\) and
\(S_j\in\EB_n\), both \(\ell(K_j)\) and \(\ell(S_j)\) are
nonnegative and bounded. A subsequence therefore converges in each
closed cone, proving closedness of their sum. Closedness of \(\EB_n\)
follows equivalently from the compact convex base of separable Choi
matrices.

The low-dimensional certificate in Corollary~\ref{cor:low-rank-eb}
proves membership in \(\cK_n\). Embedding the Choi map on three input
coordinates and on \(E\) gives an indecomposable member, as witnessed
by the embedded PPT matrix in Proposition~\ref{prop:choi-eb-family}.
Taking adjoints gives the input statements.

For example, the identity map does not belong to
\(\mathfrak C_E^{\rm out}\) when \(E\) is proper and \(n>1\).
If \(\Id=K+S\) with both summands positive, then
\(0\preceq K(xx^*)\preceq xx^*\) for every \(x\).
Linearity applied to \((e_i+ze_j)(e_i+ze_j)^*\) implies
\(K=c\Id\) for a scalar \(0\le c\le1\).
Its prescribed proper output support forces \(c=0\), leaving
\(S=\Id\), which is not entanglement breaking. Thus these cones are
convex subclasses, not cones containing the whole decomposable cone.
\end{proof}

The support is fixed before taking sums. The union over supports,
or the convex hull of maps certified on different supports, is not
covered by this proposition.

\subsection{An exact rank-nine calculation}

The nonconvexity theorem splits Du's rank-nine map into members of
$\cK_9$. We first write that map as $P_t$, a Breuer--Hall part plus a
decomposable part, and recompute its invariant.

For even $n$ and a skew-symmetric unitary $J$, the Breuer--Hall map is
\begin{equation}\label{eq:bh-definition}
 B_n(X)=\tr(X)I_n-X-JX^{\mathsf T}J^*.
\end{equation}
Since $\langle v,J\bar v\rangle=0$ and $\|J\bar v\|=\|v\|$,
\[
 B_n(vv^*)=\|v\|^2I_n-vv^*-(J\bar v)(J\bar v)^*\succeq0,
\]
so $B_n$ is positive \citep{breuer2006,hall2006}.
Write a size-nine matrix in three size-three diagonal blocks \(X,Y,Z\),
and denote its leading size-six block by
\[
 S=\begin{pmatrix}X&W\\ V&Y\end{pmatrix},
 \qquad
 J=\begin{pmatrix}0&I_3\\-I_3&0\end{pmatrix}.
\]
Define
\begin{align*}
 \widetilde B_6(\mathcal X)&=\diag(B_6(S),0_3),\\
 Q(\mathcal X)&=\diag(Z,Z^{\mathsf T},X+Y^{\mathsf T}),\\
 P_t&=t\widetilde B_6+Q.
\end{align*}
The first map is positive by the rank-one test for \(B_6\).
The map \(Q\) is decomposable: its CP part is
\(\diag(Z,0_3,X)\), and its coCP part is
\(\diag(0_3,Z^{\mathsf T},Y^{\mathsf T})\).

\begin{prop}[The rank-nine coefficient, after Du]
\label{prop:du-self-contained}
For every \(t\ge0\), the map \(P_t\) is positive and
\[
 \Gamma_9(P_t)=-1536t^3.
\]
In particular,
\begin{equation}\label{eq:du-H9}
 H_9=\frac12P_{1/4}
 \quad\text{is bistochastic, with}\quad
 \Gamma_9(H_9)=-\frac3{64},\qquad
 \Phi_9(H_9)=-\frac1{7741440}.
\end{equation}
\end{prop}
\begin{proof}
We give the determinant computation for the construction of
\citep{du2026}. Put \(A=X+Y^{\mathsf T}\) and
\(R(A)=\tr(A)I_3-A\). For a column \(k\in\mathbb C^3\), write
\[
 \widehat k=\begin{pmatrix}
 0&-k_3&k_2\\k_3&0&-k_1\\-k_2&k_1&0
 \end{pmatrix}.
\]
Define \(k,l\) by \(\widehat k=W^{\mathsf T}-W\) and
\(\widehat l=V^{\mathsf T}-V\). With \(M=Z+tR(A)\), the top
output block is
\[
 \begin{pmatrix}M&t\widehat k\\t\widehat l&M^{\mathsf T}\end{pmatrix}.
\]
A three-row determinant expansion, equivalently a six-dimensional
Pfaffian expansion, gives the polynomial identity
\begin{equation}\label{eq:du-block-square}
 \det\begin{pmatrix}M&\widehat k\\\widehat l&M^{\mathsf T}\end{pmatrix}
 =\bigl(\det M+k^{\mathsf T}Ml\bigr)^2.
\end{equation}
Consequently
\[
 \det P_t(\mathcal X)
 =\det A\,\bigl(\det(Z+tR(A))
           +t^2 k^{\mathsf T}(Z+tR(A))l\bigr)^2.
\]
The output uses none of the input cross blocks joining \(S\) to \(Z\).
The input determinant differential therefore factors as
\(\det(\partial_S)\det(\partial_Z)\). The latter selects degree three
in \(Z\) from the degree-six squared factor, so homogeneity supplies
exactly \(t^3\). It suffices to evaluate \(t=1\).

Set \(D=\det(\partial_A)\) and
\(L=\sum_{i,j}\partial_{k_i}\partial_{A_{ij}}\partial_{l_j}\).
On functions of \(A,k,l\), the chain rule and
\eqref{eq:du-block-square} give
\[
 \det(\partial_S)=(D+L)^2.
\]
In fact \(\partial_X=\partial_A\),
\(\partial_Y=\partial_A^{\mathsf T}\),
\(\partial_W=-\widehat{\partial_k}\), and
\(\partial_V=-\widehat{\partial_l}\); the two minus signs cancel.
For three-by-three matrix variables the Leibniz formula gives
\[
 \det(\partial_M)(\det M)^2=24\det M,
 \qquad
 \det(\partial_M)(\det M\,M_{ij})=8M_{ij}.
\]
After differentiating in \(Z\) and setting \(Z=0\), the remaining
polynomial is
\[
 24\det A\det R(A)+16\det A\,k^{\mathsf T}R(A)l.
\]
Thus its final constant after \((D+L)^2\) is
\begin{equation}\label{eq:du-reduced-contraction}
 24D^2\bigl(\det A\det R(A)\bigr)
 +32D\sum_{i,j}\partial_{A_{ij}}
             \bigl(\det A\,R(A)_{ij}\bigr).
\end{equation}
The same three-row expansion gives
\[
 D(\det A\det R(A))=15\det R(A),
 \qquad
 \sum_{i,j}\partial_{A_{ij}}(\det A\,R(A)_{ij})
       =\tr(A)e_2(A)-9\det A.
\]
In particular the first term of \eqref{eq:du-reduced-contraction} vanishes.
Since \(\det R(A)=\tr(A)e_2(A)-\det A\),
\(D\det R(A)=0\), and \(D\det A=6\), the second differential in
\eqref{eq:du-reduced-contraction} equals \(6-54=-48\).
The final constant is therefore \(24\cdot0+32(-48)=-1536\).
All identities are in ordinary commuting polynomial variables.

Finally,
\[
 P_t(I_9)=P_t^*(I_9)
   =\diag((1+4t)I_6,2I_3).
\]
At \(t=1/4\), division by two is exactly bistochastic. Degree-nine
homogeneity gives
\(-1536/(4^3\,2^9)=-3/64\), proving the normalization.
\end{proof}

Both summands \(\widetilde B_6\) and \(Q\) have zero size-nine
invariant. The first has a zero output block. For the second,
\(\det Q(\mathcal X)=(\det Z)^2\det(X+Y^{\mathsf T})\), whose
degree in \(Z\) exceeds the three available input derivatives.
The next result strengthens this elementary nonconvexity obstruction
by ensuring that \emph{every} square compression of each summand
used in the construction has a nonnegative invariant.

\subsection{Rank-four localization and nonconvexity of $\cK_n$}

The family is due to Breuer and Hall \citep{breuer2006,hall2006}; its
four-dimensional member is unitarily equivalent to Robertson's map
\citep[Theorem~1]{chruscinski2008}. The block identity below writes
\(B_{2m}\) as a sum of embedded copies of \(B_4\). It lets us assemble
Du's negative map from embedded four-dimensional cores, each in
$\cK_9$, which is what the nonconvexity theorem needs.

For the next lemma use adjacent symplectic pairs, so
\(J_{2m}=\diag(J_2,\ldots,J_2)\), where
\(J_2=\left(\begin{smallmatrix}0&1\\-1&0\end{smallmatrix}\right)\).
Let \(V_{ij}:\mathbb C^4\to\mathbb C^{2m}\) embed the \(i\)-th and
\(j\)-th pairs, in that order.

\begin{lemma}[Localization of the Breuer--Hall family]
\label{lem:bh-localization}
For every \(m\ge2\),
\begin{equation}\label{eq:bh-localization}
 B_{2m}(X)
 =\sum_{1\le i<j\le m}
       V_{ij}B_4(V_{ij}^*XV_{ij})V_{ij}^*.
\end{equation}
\end{lemma}
\begin{proof}
Write \(X=(X_{ij})\) in two-by-two blocks.
Using $A+J_2A^{\mathsf T}J_2^*=\tr(A)I_2$, we obtain
\[
 B_{2m}(X)_{ij}=
 \begin{cases}
  \displaystyle\sum_{\ell\ne i}\tr(X_{\ell\ell})I_2,&i=j,\\
  -X_{ij}-J_2X_{ji}^{\mathsf T}J_2^*,&i\ne j.
 \end{cases}
\]
Each off-diagonal block and each diagonal summand occurs in exactly
one embedded $B_4$, proving \eqref{eq:bh-localization}.
\end{proof}

\begin{theorem}[Nonconvexity of hereditary determinant positivity]
\label{thm:K-nonconvex}
For every \(n\ge9\), the class \(\cK_n\) is nonconvex. In size nine,
four explicit members have equal-weight average \(H_9\) from
\eqref{eq:du-H9}. The slice
\(\{H\in\cK_9:\tr H(I)=9\}\) is also nonconvex, and a negative
convex combination within it is bistochastic.
\end{theorem}
\begin{proof}
Use the symplectic pairs
\(\operatorname{span}(e_i,e_{i+3})\), \(i=1,2,3\), in the first
six coordinates of \(\mathbb C^9\). Let
\(W_{ij}:\mathbb C^4\to\mathbb C^9\) have columns
\((e_i,e_{i+3},e_j,e_{j+3})\), and set
\[
 C_{ij}(X)=W_{ij}B_4(W_{ij}^*XW_{ij})W_{ij}^*.
\]
The normalization in Theorem~\ref{thm:bh-all-even}, or its size-four
calculation, gives \(\Gamma_4(B_4)=24\).
Corollary~\ref{cor:low-rank-eb} therefore places every \(C_{ij}\) in
\(\cK_9\). The map
\(Q\) is decomposable and hence belongs to \(\cK_9\) as well.

Lemma~\ref{lem:bh-localization} gives
\(\widetilde B_6=C_{12}+C_{13}+C_{23}\). Therefore the four maps
\begin{equation}\label{eq:four-safe-maps}
 A_0=2Q,\qquad A_{12}=\tfrac12C_{12},\qquad
 A_{13}=\tfrac12C_{13},\qquad A_{23}=\tfrac12C_{23}
\end{equation}
belong to \(\cK_9\), but their average is
\[
 \frac14(A_0+A_{12}+A_{13}+A_{23})
 =\frac12Q+\frac18\widetilde B_6=H_9.
\]
Proposition~\ref{prop:du-self-contained} proves its negative invariant.

For a common trace normalization, put
\(G_0=\tfrac34Q\) and \(G_{ij}=\tfrac98C_{ij}\).
Indeed \(\tr Q(I)=12\) and \(\tr C_{ij}(I)=8\), so all four maps
have \(\tr G(I)=9\). Their convex combination is
\begin{equation}\label{eq:trace-normalized-nonconvex}
 H_9=\frac23G_0+\frac19(G_{12}+G_{13}+G_{23}).
\end{equation}

For \(n>9\), pad each map in \eqref{eq:four-safe-maps} by zero and
add the same diagonal pinching map on coordinates \(10,\ldots,n\).
The first result is decomposable. Each of the other three is a
four-dimensional core with nonnegative $\Gamma_4$ plus an entanglement-breaking map,
so Corollary~\ref{cor:low-rank-eb} places it in \(\cK_n\).
Their average is the pinched sum of \(H_9\) and diagonal pinching
on the remaining coordinates. Diagonal pinching has
invariant one. Equation~\eqref{eq:directsum} therefore gives
\(\Gamma_n=-3/64\) for the average.
\end{proof}

\begin{corr}[Strict compression positivity does not restore convexity]
\label{cor:strict-nonconvex}
For each \(n\ge9\), there are four strictly positive maps in the
interior of \(\cK_n\) whose average has negative \(\Gamma_n\).
Thus the nonconvexity is not confined to deficient-rank boundary maps.
\end{corr}
\begin{proof}
Let \(A_0,\ldots,A_3\) be the maps constructed in the preceding
proof, including the common tail pinching when \(n>9\).
For \(\epsilon>0\), set \(A_j^\epsilon=A_j+\epsilon\Delta_n\).
Every size-\(k\) compression satisfies
\[
 \Gamma_k((A_j^\epsilon)_{V,U})
 \ge k!(\epsilon/n)^k>0.
\]
Indeed, its added map is \((\epsilon/n)\tr(\,\cdot\,)I_k\),
and the terms of the EB-addition formula that use only the added map
give the displayed bound. Also \(A_j^\epsilon(xx^*)\succeq
(\epsilon/n)\|x\|^2I_n\).
These strict bounds, continuity, and compactness of the finitely many
input/output Grassmannians imply that each \(A_j^\epsilon\) lies in
the interior of \(\cK_n\), relative to Hermiticity-preserving maps.
The average is the preceding negative map plus \(\epsilon\Delta_n\).
Its invariant remains negative for sufficiently small positive
\(\epsilon\), by polynomial continuity.
\end{proof}

\begin{corr}[No unitary-invariant convex gluing of all rank-four cores]
\label{cor:no-convex-gluing}
Fix \(n\ge9\). There is no convex cone of positive maps on \(M_n\)
which simultaneously
\begin{enumerate}
 \item contains all decomposable maps;
 \item contains the standard zero-padded four-dimensional Breuer--Hall
       map and is invariant under simultaneous unitary conjugation
       $H\mapsto\Ad_U\circ H\circ\Ad_{U^*}$; and
 \item has \(\Gamma_n\ge0\) on every member.
\end{enumerate}
\end{corr}
\begin{proof}
Extend each isometry $W_{ij}$ from the construction above to a
unitary on $\mathbb C^n$, after zero padding when $n>9$.
Simultaneous conjugation by these unitaries
sends the standard padded $B_4$ to $C_{12},C_{13},C_{23}$.
Conicity and convexity permit their positive linear combinations with
\(Q\). In size nine one such combination is \(H_9\). In larger size add the
entanglement-breaking diagonal pinching on the remaining coordinates,
which is already in the decomposable cone. The resulting map has
\(\Gamma_n=-3/64\) by the proof of
Theorem~\ref{thm:K-nonconvex}, contradicting the third condition.
\end{proof}

\subsection{A single-map certificate need not be hereditary}

A certificate for $\Gamma_n(H)\ge0$ alone need not survive
entanglement-breaking addition, even when $H$ is positive semidefinite
as a Hilbert--Schmidt operator. This failure is why
Theorem~\ref{thm:eb-maximality} asks for the sign on every compression.

\begin{prop}\label{prop:obstruction}
Suppose $K:M_s\to M_s$ is positive and bistochastic, with
$\Gamma_s(K)<0$. There are a positive bistochastic map $R:M_{2s}\to M_{2s}$, positive
semidefinite as a Hilbert--Schmidt operator, and an entanglement-breaking
map $D$ such that $\Gamma_{2s}(R+tD)<0$ for all sufficiently large $t$.
\end{prop}
\begin{proof}
For $s$-by-$s$ blocks define
\[
 R\begin{pmatrix}X&Y\\Z&W\end{pmatrix}
 =\frac12\begin{pmatrix}X+K(W)&Y\\Z&W+K^*(X)\end{pmatrix},
 \qquad
 D\begin{pmatrix}X&Y\\Z&W\end{pmatrix}
 =\begin{pmatrix}\operatorname{diag}(W)&0\\0&0\end{pmatrix}.
\]
Writing $R=(\Id+\mathcal S)/2$, the map $\mathcal S$ is positive and
bistochastic. It is Hilbert--Schmidt selfadjoint, with diagonal-block
operator matrix $\left(\begin{smallmatrix}0&K\\K^*&0\end{smallmatrix}\right)$,
and is zero on off-diagonal input blocks. Lemma~\ref{lem:kadison} gives
its Hilbert--Schmidt contraction property. Hence
$\|\mathcal S\|\le1$ and $R\succeq0$ in Hilbert--Schmidt order.

The map $D$ has rank-one Kraus operators $|i\rangle\langle s+i|$, so is
entanglement breaking. The polynomial
$p(t)=\Gamma_{2s}(R+tD)$ has degree at most $s$, because $D$ has
only $s$ nonzero output rows. The leading output-determinant coefficient is
\[
 [t^s]\det(R+tD)(\mathcal X)
 =2^{-s}\left(\prod_{i=1}^s W_{ii}\right)\det\bigl(K^*(X)+W\bigr).
\]
It uses no cross-block input variables. The full input determinant
differential therefore factors into those in $X$ and $W$. Its $s$
derivatives in $X$ annihilate all terms using $W$ in the second determinant.
The complementary diagonal differentiation gives one, not $s!$. Therefore
\begin{equation}\label{eq:badleading}
 [t^s]p(t)=2^{-s}\Gamma_s(K^*)
 =2^{-s}\Gamma_s(K)<0.
\end{equation}
The adjoint equality is Lemma~\ref{lem:fock-calculus}. A polynomial with a negative leading coefficient
is negative for all sufficiently large positive $t$.
\end{proof}

For $K=H_9$, \eqref{eq:badleading} gives
\[
 [t^9]\Gamma_{18}(R+tD)=-\frac3{32768}.
\]
An explicit, deliberately conservative parameter is
$t=2^{33}(18!)^3$. Indeed every matrix-unit coefficient of the contraction
$R$ has modulus at most one, and every coefficient of $D$ is zero or one.
The full triple-permutation expression has $(18!)^3$ terms. Thus, with
$p(t)=\sum_{k=0}^9 a_kt^k$,
\[
 \sum_{k<9}|a_k|\le B:=2^{18}(18!)^3,
 \qquad
 p(t)\le t^8\left(-\frac{3t}{32768}+B\right)=-2Bt^8<0.
\]
This bound gives a finite parameter without locating the smallest admissible $t$.

Thus adding an entanglement-breaking map, and hence a completely
positive or decomposable map, to a Hilbert--Schmidt positive-semidefinite
map can make the invariant negative. The cross compressions of $R$ are
$K/2$ and $K^*/2$, so $R\notin\cK_{2s}$.

\section{Applications}
\label{sec:applications}
\label{sec:geometric-implications}
\label{sec:geometry}

We now express Chern forms through the determinant invariants of
curvature compressions. Hereditary determinant positivity then controls every Chern
form, and decomposability permits the incidence construction needed for
all Schur forms. Throughout this section, nonnegativity is pointwise
for the Chern--Weil forms of the specified metric; strict positivity
will be stated explicitly.

\subsection{From hereditary determinant positivity to Chern forms}
\label{sec:chern-forms}

Let $(E,h)$ have rank $r$ over a complex manifold of dimension $n$.
At a point $x$, choose holomorphic coordinates and a unitary fiber frame,
and write its normalized Chern curvature as
\begin{equation}\label{eq:curvature-map-convention}
 \Omega=\frac{i}{2\pi}R^h
       =i\sum_{a,b=1}^n H_x(E_{ab})\,dz_a\wedge d\bar z_b,
 \qquad H_x:M_n(\mathbb C)\longrightarrow M_r(\mathbb C).
\end{equation}
Hermitian symmetry of curvature makes $H_x$ Hermiticity preserving.
Griffiths semipositivity is exactly ordinary positivity of $H_x$, since
its evaluation on a tangent vector $v$ and fiber vector $\xi$ is
$\xi^*H_x(vv^*)\xi\ge0$.  With
\[
 dV_k=\bigwedge_{a=1}^k(i\,dz_a\wedge d\bar z_a),
 \qquad \Gamma_k(T)=\det(\partial_X)\det T(X)=k!\Phi_k(T),
\]
the top-form convention is
\begin{equation}\label{eq:top-chern-gamma}
 c_k(E,h)|_x=\Gamma_k(H_x)dV_k
 \quad\text{when }n=r=k.
\end{equation}
In particular, there is no further factorial multiplying $\Gamma_k$.
This convention agrees with \citet[Theorem~3.4]{finski2022} after converting volume
conventions, and with the explicit normalization in
\citet[\S2, (2.1)--(2.6)]{du2026}.

Every Hermiticity-preserving map $H:M_n\to M_r$ can be realized as
the curvature map at the origin of
\[
 h_H(z)=I_r-2\pi H(zz^*),
\]
which is a positive definite metric on a sufficiently small ball.
Since
\[
 h_H(0)=I_r,\qquad dh_H(0)=0,\qquad
 \partial_{z_a}\partial_{\bar z_b}h_H(0)=-2\pi H(E_{ab}),
\]
its curvature at zero satisfies \eqref{eq:curvature-map-convention}.
  Consequently a negative
$\Gamma_n$ for a positive square map is an actual pointwise
top-Chern obstruction, not a mismatch between representations of
the curvature tensor.

Here is a direct verification of \eqref{eq:top-chern-gamma} that also treats lower Chern forms.
For isometries $V:\mathbb C^k\to\mathbb C^n$ and
$U:\mathbb C^k\to\mathbb C^r$, put
\[
 H_{V,U}(X)=U^*H_x(VXV^*)U.
\]
The image of $V$ is the tangent $k$-plane $L$.

\begin{prop}[Chern forms as sums of compression invariants]
\label{prop:chern-compression-sum}
For $k\le\min(n,r)$ and any orthonormal fiber basis,
\begin{equation}\label{eq:chern-compression-sum}
 c_k(E,h)|_L=
 \left(\sum_{\substack{J\subseteq\{1,\ldots,r\}\\|J|=k}}
 \Gamma_k(H_{V,U_J})\right)dV_L,
\end{equation}
where $U_J$ is the coordinate inclusion.  Equivalently, for invariant
probability measure on the complex Grassmannian,
\begin{equation}\label{eq:chern-compression-average}
 c_k(E,h)|_L=
 \binom{r}{k}\left(\int_{\mathrm{Gr}_{\mathbb C}(k,r)}
 \Gamma_k(H_{V,U})\,d\mu(U)\right)dV_L.
\end{equation}
Consequently, if all square compressions of $H_x$ have
nonnegative $\Gamma$, every Chern form is weakly nonnegative at $x$.
\end{prop}
\begin{proof}
The coefficient of $t^k$ in $\det(I+t\Omega)$ is the sum of the
$k$-by-$k$ principal determinants of $\Omega$.  For a square map $T$,
expand the determinant of
$i\sum_{a,b}T(E_{ab})\,dz_a\wedge d\bar z_b$.
A nonzero term uses every holomorphic and antiholomorphic index once.
If their assignments to output rows are $\rho,\eta\in S_k$ and the
output determinant permutation is $\sigma\in S_k$, its coefficient of
$dV_k$ is
\[
 \operatorname{sgn}(\sigma)\operatorname{sgn}(\rho)
 \operatorname{sgn}(\eta)
 \prod_{a=1}^k T(E_{\rho(a),\eta(a)})_{a,\sigma(a)}.
\]
These products are exactly the terms of $\det(\partial_X)\det T(X)$:
the input derivative permutation is $\eta\rho^{-1}$.
This matching proves \eqref{eq:top-chern-gamma} and
\eqref{eq:chern-compression-sum}.

For an orthonormal $k$-frame $U$, write $u_J=\det U[J,[k]]$.
Unitary invariance and Cauchy--Binet give
\[
 \int\overline{u_J}u_K\,d\mu(U)
 =\frac{\delta_{JK}}{\binom rk},
 \qquad \sum_{|J|=k}|u_J|^2=1.
\]
The entries of $\Omega$ have even degree and commute, so
\[
 \int\det(U^*\Omega U)\,d\mu(U)
 =\sum_{J,K}\det\Omega[J,K]
       \int\overline{u_J}u_K\,d\mu(U)
 =\binom rk^{-1}\sum_J\det\Omega[J,J].
\]
The last display proves \eqref{eq:chern-compression-average}.
Weak nonnegativity means precisely nonnegativity of the restriction
to each tangent $k$-plane, so the final assertion follows.
\end{proof}

The preceding proposition proves only weak nonnegativity.  It does not
assert that the Hermitian form on the entire exterior power is positive
semidefinite, or that the differential form is strongly nonnegative.
In top degree the distinction disappears: a negative multiple of the
complex volume form violates all these positivity notions.

Besides decomposable curvature, the proposition applies to the
rectangular versions of the low-dimensional classes in
Corollary~\ref{cor:low-rank-eb}: maps with input or output marginal
rank at most three, and maps of the form
\begin{equation}\label{eq:geometric-certified-core}
 H_x(X)=A K(B^*XB)A^*+S(X),
 \qquad A:\mathbb C^d\to\mathbb C^r,\quad
 B:\mathbb C^d\to\mathbb C^n,
\end{equation}
where $S$ is entanglement breaking, $K:M_d\to M_d$ is positive,
$d\le4$, and $\Gamma_4(K)\ge0$ when $d=4$. The proof of that
corollary applies to every square compression in these
rectangular dimensions.

The sector and Hilbert--Schmidt criteria for a single square map,
and the coordinate block criteria, give the Chern-form sign on the
corresponding plane.  For the top Chern form of a rank-$r$ bundle,
the certificate must hold on every tangent $r$-plane.  For a lower
Chern form, one needs the square input/output compressions in
\eqref{eq:chern-compression-sum}.

\subsection{From decomposability to all Schur forms}

We use the usual Chern-root convention $s_\lambda$ for Schur
polynomials.  Finski uses the conjugate indexing
\[
 P_a(c)=\det(c_{a_i-i+j})=s_{a'}.
\]
In particular, $c_k=s_{(1^k)}=P_{(k)}$.  The distinction matters when
comparing Chern--Weil evaluations with Schur coefficients in
representation theory.

Call the curvature \emph{decomposable} at a point if
its rectangular map in \eqref{eq:curvature-map-convention} has the form
\begin{equation}\label{eq:rectangular-decomposable-curvature}
 H_x(X)=\sum_\alpha A_\alpha X A_\alpha^*
       +\sum_\beta B_\beta X^{\mathsf T}B_\beta^*.
\end{equation}
The matrices and their number may depend on the point; no smooth
choice of a decomposition is required.  This definition is the operator form
of Finski's decomposable nonnegativity
\citep[Proposition 2.18]{finski2022}.

The quotient stability in the next lemma is Finski's
\citep[Proposition~2.22]{finski2022}. We restate it in the present Kraus
conventions, together with the elementary pullback and direct-sum
operations, to identify precisely the input needed below.

\begin{lemma}[Operations needed for the incidence construction]
\label{lem:decomposable-geometric-closure}
Decomposable curvature is preserved by holomorphic
pullback, orthogonal direct sums, adjoining a flat bundle, and taking
a holomorphic quotient with its induced metric.
\end{lemma}
\begin{proof}
For a tangent pullback with derivative $L$,
\[
 H_x(LXL^*)
 =\sum_\alpha(A_\alpha L)X(A_\alpha L)^*
  +\sum_\beta(B_\beta\bar L)X^{\mathsf T}(B_\beta\bar L)^*.
\]
Output direct sums and zero blocks preserve the same Kraus form.

For a quotient, identify its fiber with the orthogonal complement
of the kernel, and let $P$ be the corresponding output isometry.
In local unitary frames, the normalized second-fundamental-form
contribution is
\[
 i\sum_\nu c_\nu\wedge\bar c_\nu^{\mathsf T},
 \qquad c_\nu=\sum_a \ell_{\nu a}\,dz_a,
\]
absorbing the positive factor $1/(2\pi)$ into the columns.
With $L_\nu=[\ell_{\nu1}\ \cdots\ \ell_{\nu n}]$, the quotient map is
\[
 H^Q(X)=P^*H_x(X)P+\sum_\nu L_\nu X L_\nu^*.
\]
The added map is completely positive, so the quotient retains
the form \eqref{eq:rectangular-decomposable-curvature}.
This computation also proves directly the quotient assertion of
\citet[Proposition~2.22]{finski2022}.  Transposing all matrix
conventions merely interchanges the two summand types.
\end{proof}

The closure lemma allows us to apply the determinant theorem to the
auxiliary quotient in Finski's incidence formula. Fiber integration
then transfers its top Chern-form sign to the required Schur form.

\begin{theorem}[Schur forms for ordinary decomposable curvature]
\label{thm:decomposable-schur}
Let $(E,h)$ be a Hermitian holomorphic vector bundle of rank $r$ over
a complex manifold of dimension $n$. If its curvature
is decomposable at a point $x$, every Schur form
$s_\lambda(E,h)$ is weakly nonnegative at $x$.  In particular,
curvature that is decomposable at every point implies weak nonnegativity of
all Schur forms throughout the base.
\end{theorem}
\begin{proof}
The empty partition gives the form $1$, while $|\lambda|>n$ or
$\operatorname{length}(\lambda)>r$ gives the zero form.  Otherwise
let $1\le k=|\lambda|\le n$, and write $a=\lambda'$ in Finski's
notation, padding it by zeros.  Put $V_0=\mathbb C^{r+k}$ and
\[
 F=\operatorname{Hom}(V_0,E),\qquad
 Y=\mathbb P(F\oplus\mathcal O),\qquad
 Q=\pi^*(F\oplus\mathcal O)/\mathcal O_Y(-1).
\]
Thus $F$ is the orthogonal direct sum of $r+k$ copies of $E$, and
$Q$ has rank $N=r(r+k)$.  Fix the flag of subspaces in $V_0$
with dimensions $r+i-a_i$, as in the Kempf--Laksov construction,
and let
\[
 \Omega_a(E)_x=\{u\in\Hom(V_0,E_x):
                 \dim(\ker u\cap V_i)\ge i,\ 1\le i\le k\}.
\]
Its projective compactification is
$Z_a=\mathbb P(\Omega_a(E)\oplus\mathcal O)\subset Y$.
The fibers of $Z_a$ have complex dimension $N-k$.

Finski's identity of differential forms, rather than merely classes,
is
\begin{equation}\label{eq:finski-incidence-pushforward}
 s_\lambda(E,h)=P_a(c(E,h))
       =\pi_*\bigl(c_N(Q,h^Q)\wedge[Z_a]\bigr).
\end{equation}
Here the expression on the right means integration over the regular
loci of the analytic incidence fibers; equivalently it is the
pushforward current.  The resulting current is the smooth form on
the left \citep[Theorem 3.2]{finski2022}.

Lemma~\ref{lem:decomposable-geometric-closure} shows that the
curvature of $Q$ is decomposable at every point above $x$.  Its
restriction to any tangent $N$-plane there is therefore a
decomposable square map.  The
decomposable-map determinant theorem gives its $\Gamma_N\ge0$.
Equation~\eqref{eq:top-chern-gamma} proves that $c_N(Q,h^Q)$ is
weakly nonnegative at those points.

Finally, fix a tangent $k$-plane on the base.  At a regular point of
an incidence fiber, combine that plane with the $(N-k)$-dimensional
vertical tangent space.  The integrand defining the restriction of
\eqref{eq:finski-incidence-pushforward} is the evaluation of the
weakly nonnegative $(N,N)$-form on this decomposable $N$-plane,
with the complex orientation.  It is nonnegative.  The singular
strata have zero measure for integration over the analytic cycle,
and its multiplicities are positive.  Fiber integration is therefore
nonnegative, proving the theorem.  Only the original curvature at
$x$ enters these evaluations, so no decomposition on a neighborhood
of $x$ is required.
\end{proof}

The proof does not give strictness from strict Griffiths
positivity on the boundary of the decomposable cone.

The same proof applies whenever the curvature of $Q$ in
\eqref{eq:finski-incidence-pushforward} has $\Gamma_N\ge0$ on every
tangent $N$-plane along the regular incidence locus. Under this
condition, the corresponding Schur form is weakly nonnegative.

\subsection{Nonconvexity of simultaneous pointwise Chern positivity}

The obstruction in Section~\ref{sec:convex-geometry} has a pointwise
geometric consequence stronger than the negativity of a single top
Chern form under ordinary positivity.

\begin{corr}\label{cor:chern-nonconvexity}
In complex dimension and bundle rank nine, there exist four strictly
Griffiths-positive curvature tensors, on the same tangent and fiber
spaces, such that every Chern form of each tensor is strictly weakly
positive, whereas the top Chern form of their average is negative.
The same conclusion holds in every common dimension and rank at least
nine.
\end{corr}
\begin{proof}
Use the four maps \(A_j^\epsilon\) from
Corollary~\ref{cor:strict-nonconvex}, with \(\epsilon\) sufficiently
small to retain the negative invariant of their average.
Assign to each map the curvature tensor in
\eqref{eq:curvature-map-convention}. Its strict product-vector bound
is exactly strict Griffiths positivity. For every tangent $k$-plane $L=\operatorname{im}V$,
\[
 \sum_{|J|=k}\Gamma_k((A_j^\epsilon)_{V,U_J})
 \ge\binom nk k!(\epsilon/n)^k>0,
 \qquad 1\le k\le n,
\]
so \eqref{eq:chern-compression-sum} gives strict weak positivity.
The average tensor has top coefficient
$\Gamma_n(\frac14\sum_j A_j^\epsilon)<0$.
 By the local metric realization in Section~\ref{sec:chern-forms},
the four tensors are genuine curvature tensors at a point, not merely
formal arrays.
\end{proof}

The conclusion concerns averaging curvature tensors in fixed frames.
Averaging Hermitian metrics need not average their curvatures, and
the corollary makes no assertion about all Schur forms of the four
tensors.

\subsection{What the rank-nine obstruction changes}

For $H=H_9$, the metric $h_H$ of Section~\ref{sec:chern-forms} realizes the negative invariant in
\eqref{eq:du-H9} as a negative top Chern form. A small positive trace
addition makes the curvature strictly Griffiths positive and retains
the negative sign. Du's construction realizes such negativity globally
on $\mathcal O(1)^{\oplus r}$ over $\mathbb P^r$ for every $r\ge9$
\citep[Theorem~1.1 and Propositions~3.1--3.2]{du2026}.

This negative top Chern form disproves the unrestricted pointwise implication from
Griffiths positivity to all Schur forms, since the top Chern form is
itself a Schur form.  It leaves intact both the structural
hypotheses in Theorem~\ref{thm:decomposable-schur} and numerical
Schur positivity for ample bundles.  For example,
$c_r(\mathcal O_{\mathbb P^r}(1)^{\oplus r})=c_1(\mathcal O(1))^r$
has integral one, independently of the chosen metric.  A negative
density on one open set does not contradict that positive integral.

\section{Boundary cases and open problems}
\label{sec:scope}

This section records the boundary of what the preceding theorems prove:
an exact equality case for the normalized size-four problem, the
size-four form of the maximality theorem, and the open problems that
the counterexamples do not decide.

\subsection{A zero in the normalized rank-four problem}

Gurvits's skew-symmetric size-three example has $\Gamma_3=0$
\citep[Example~2.6]{gurvits2002}; padding that example by a scalar block
gives an equality case, which Appendix~\ref{sec:hs-sector} uses to test
the sector certificates:
\begin{equation}\label{eq:T0}
 T_0(X)=\diag\left(
 \frac{\tr(A)I_3-A^{\mathsf T}}2,x_{44}\right),
 \qquad A=X_{[3],[3]}.
\end{equation}
It is positive, unital, and trace preserving. In fact the first block
is completely positive: its Choi matrix is \((I_9-F_3)/2\), where
\(F_3\) is the flip. Formula~\eqref{eq:cp-gaussian} realizes its
invariant $\Gamma_3$ as the expected squared determinant of a
skew-symmetric Gaussian matrix of size three. That determinant vanishes
identically. Pinched factorization therefore gives
\[
 \Gamma_4(T_0)=\Phi_4(T_0)=0.
\]
The two Choi blocks are orthogonal for the Frobenius inner product, and
\[
 \|C_{T_0}\|_F^2
 =\frac14\tr(I_9-F_3)^2+1
 =\frac{18-6}{4}+1=4.
\]

Since $T_0$ is also a positive map on $M_3$ plus an
entanglement-breaking scalar block, Corollary~\ref{cor:low-rank-eb} puts
it in $\cK_4$, with $\Gamma_4(T_0)=0$. On the traceless diagonal direction
\(\diag(1,-1,0,0)\), \(T_0\) acts by \(-1/2\).
It fails the Hilbert--Schmidt sector condition, so the strict
bistochastic lower bound from that theorem does not apply.

\subsection{The size-four boundary of the maximality theorem}

\begin{corr}\label{cor:rank-four-EB}
The class \(\cK_3\) is the entire positive-map cone, and
\[
 \cK_4=\{H:M_4\to M_4\text{ positive}:\Gamma_4(H)\ge0\}.
\]
For any positive \(H:M_4\to M_4\) and any entanglement-breaking \(S\),
every nonconstant coefficient of \(\Gamma_4(H+tS)\) is nonnegative.
In particular,
\begin{equation}\label{eq:rank-four-EB-lower}
 \Gamma_4(H+tS)\ge\Gamma_4(H)+t^4\Gamma_4(S)
 \qquad(t\ge0).
\end{equation}
\end{corr}
\begin{proof}
All compressions of size at most three are covered by
\eqref{eq:wan-small}. Full-size isometries preserve the sign by
\eqref{eq:filter}, which proves the descriptions of \(\cK_3,\cK_4\).
In \eqref{eq:eb-addition}, every nonempty subset compresses \(H\)
to size at most three. Its Gram factors and its invariant
are nonnegative. Homogeneity identifies the coefficient of \(t^4\)
with \(\Gamma_4(S)\), proving the final inequality.
\end{proof}

The corollary does not determine the constant coefficient
\(\Gamma_4(H)\). In particular, a nonnegative value after depolarizing
addition does not yield a nonnegative value before that addition.

\subsection{Open problems}

The first open question is the sign of $\Gamma_4$ on all positive
maps, equivalently whether $\cK_4$ is the full positive cone. In higher
dimensions, Theorem~\ref{thm:K-nonconvex} leaves the convexity of the
bistochastic slice unresolved: the constructed summands have a
bistochastic average but are not individually bistochastic. Another
question is which convex cones can extend decomposability while
preserving the sign; Corollary~\ref{cor:no-convex-gluing} excludes
those containing the padded $B_4$ and closed under simultaneous
unitary conjugation.

For the nondecomposable subclasses of $\cK_n$, extending Chern-form
positivity to all Schur forms requires control of the quotient
curvatures in \eqref{eq:finski-incidence-pushforward}. Their
second-fundamental-form contribution is completely positive but need
not be entanglement breaking, so the addition theorem does not settle
this extension.

Finally, the examples do not decide the larger cone consisting of sums
of a \(2\)-positive and a \(2\)-copositive map. A map is \(2\)-positive
if \(\Id_{M_2}\otimes H\) is positive, and \(2\)-copositive if its
input-transposed version is \(2\)-positive.
Breuer--Hall maps are atomic, meaning that they lie outside this sum
cone \citep[Theorem~3]{chruscinski2008}. The leading size-six compression
of \(H_9\) is \(B_6/8\); since the sum cone is preserved by completely
positive embeddings and compressions, \(H_9\) is atomic as well.
Thus the displayed size-nine counterexample does not settle the sign
problem on that cone. 

\appendix
\section{A Hilbert--Schmidt sector certificate}
\label{sec:hs-sector}

Addition of matrix phases under tensor products is part of the
sectorial-matrix calculus of \citet[Theorem~11.1]{wang2020}. We apply
that mechanism, with a singular-support endpoint argument, to the
determinant vector in Gaussian Fock space. The result is an explicit
sufficient condition for \(\operatorname{Re}\Gamma_n(H)\ge0\) and a
bistochastic lower bound; the Hilbert--Schmidt positive-semidefinite
square case is in \citet[Appendix~C]{gurvits2004}.

All adjoints in this appendix refer to the Hilbert--Schmidt inner
product on $M_n(\mathbb C)$. A positive-semidefinite superoperator
need not have a positive-semidefinite Choi matrix. We use the
normalization and Fischer product from
\eqref{eq:intro-functional} and \eqref{eq:fischer-product}.

\begin{theorem}[Hilbert--Schmidt sector certificate]
\label{thm:hs-sector}
Let $n\ge2$ and let $H:M_n(\mathbb C)\to M_n(\mathbb C)$ be complex
linear. Put
\[
 S=\frac{H+H^*}{2},\qquad A=\frac{H-H^*}{2i},\qquad
 t_n=\tan\frac{\pi}{2n}.
\]
Suppose the following two Hilbert--Schmidt operator inequalities hold:
\begin{equation}
 -t_nS\preceq A\preceq t_nS.
 \label{eq:hs-sector-lmi}
\end{equation}
Then $\operatorname{Re}\Gamma_n(H)\ge0$. If $H$ is Hermiticity
preserving, then $\Gamma_n(H)$ is real and $\Phi_n(H)\ge0$.
If in addition $H(I)=H^*(I)=I$, then
\begin{equation}
 \operatorname{Re}\Phi_n(H)\ge n^{-n}.
 \label{eq:hs-sector-lower}
\end{equation}
Thus an ordinary positive bistochastic map satisfying
\eqref{eq:hs-sector-lmi} satisfies $\Phi_n(H)\ge n^{-n}$.
\end{theorem}

\begin{proof}
Adding the two positive inequalities in \eqref{eq:hs-sector-lmi} gives
$S\succeq0$. They also imply $A\ker S=0$. Indeed, for $v\in\ker S$,
the positive operators $t_nS\pm A$ both have zero quadratic form at $v$,
so both annihilate $v$. On the support of $S$ define
\[
 K=S^{-1/2}AS^{-1/2}.
\]
Then $K=K^*$, $\|K\|\le t_n$, and, extending $K$ by zero on the
kernel,
\begin{equation}
 H=S^{1/2}(I+iK)S^{1/2}.
 \label{eq:hs-sector-factor}
\end{equation}
For every integer $0\le k\le n$, the normal operator
$(I+iK)^{\otimes k}$ has eigenvalues
\[
 \prod_{j=1}^k(1+i\kappa_j),\qquad |\kappa_j|\le t_n.
\]
For $k\le n$,
\[
 \left|\arg\prod_{j=1}^k(1+i\kappa_j)\right|
 \le\sum_{j=1}^k|\arctan\kappa_j|
 \le\frac{k\pi}{2n}\le\frac\pi2.
\]
Congruence by $(S^{1/2})^{\otimes k}$ therefore proves
\begin{equation}
 \operatorname{Re}H^{\otimes k}\succeq0,
 \qquad 0\le k\le n.
 \label{eq:hs-sector-tensors}
\end{equation}

On homogeneous Gaussian Fock space, use the pullback operators
$\mathcal U_Lf=f\circ L$ from Lemma~\ref{lem:fock-calculus}.
The symmetric-tensor model transfers \eqref{eq:hs-sector-tensors} to
nonnegative Hermitian part of $\mathcal U_H$ in every homogeneous
degree $k\le n$. Equivalently, diagonalize $K$ and apply the preceding
product-of-eigenvalues calculation to monomials, then use the
palindromic factorization
\[
 \mathcal U_H=
 \mathcal U_{S^{1/2}}\mathcal U_{I+iK}\mathcal U_{S^{1/2}}.
\]
Its outside factors are selfadjoint, so
\[
 \operatorname{Re}\Gamma_n(H)
 =\langle\det,(\operatorname{Re}\mathcal U_H)\det\rangle_F\ge0.
\]

For Hermiticity-preserving $H$, reality follows from
Lemma~\ref{lem:fock-calculus}.

For the quantitative assertion, decompose
\[
 G=zI/\sqrt n+G_0,\qquad
 \det G=\sum_{k=0}^n z^{n-k}q_k(G_0),
\]
where $G_0$ is the standard Gaussian on the traceless Hilbert--Schmidt
subspace, independent of $z$. Bistochasticity makes the scalar and
traceless subspaces invariant for both $H$ and $H^*$. The traceless
restriction has the same sector bounds. Orthogonality of the powers
of $z$ gives
\[
 \Gamma_n(H)=\sum_{k=0}^n(n-k)!
 \left\langle q_k,
   \mathcal U_{H|_{\tr=0}}q_k\right\rangle.
\]
Each summand has nonnegative real part, and $q_0=n^{-n/2}$. Thus
\[
 \operatorname{Re}\Gamma_n(H)\ge n!|q_0|^2=\frac{n!}{n^n},
\]
which proves \eqref{eq:hs-sector-lower}.
\end{proof}

For fixed $n$, condition \eqref{eq:hs-sector-lmi} defines a convex
cone by two linear matrix inequalities. It contains the Hilbert--Schmidt
PSD cone and allows a nonzero skewadjoint part. Intersecting it with the
ordinary positive-map cone gives a sufficient class without requiring
complete positivity. It does not assert that arbitrary decomposable
summands may be added to a map in this class.

\begin{rmk}[Recovery of the square formula]
For $H=H^*\succeq0$, Lemma~\ref{lem:fock-calculus} gives
\[
 \Gamma_n(H)=\E|\det H^{1/2}(G)|^2.
\]
Here $H^{1/2}$ is the Hilbert--Schmidt operator square root and need
not preserve positive matrices. If $H(I)=I$, selfadjointness gives
$H^*(I)=I$, so the theorem yields $\Phi_n(H)\ge n^{-n}$.
\end{rmk}

\subsection{A sharper phase certificate for normal operators}

\begin{prop}[Normal spectral-support certificate]
\label{prop:normal-phase}
Suppose $H$ is normal as a Hilbert--Schmidt operator. Choose an
orthonormal eigenbasis $E_j$, with eigenvalues $\lambda_j$, and write
\[
 \det\left(\sum_j z_jE_j\right)
 =\sum_{|\alpha|=n}c_\alpha z^\alpha.
\]
Then
\begin{equation}
 \Gamma_n(H)=\sum_{|\alpha|=n}
 \alpha!|c_\alpha|^2\lambda^\alpha.
 \label{eq:normal-phase-exact}
\end{equation}
Consequently it suffices that
\begin{equation}
 \operatorname{Re}\lambda^\alpha\ge0
 \quad\hbox{whenever }c_\alpha\ne0.
 \label{eq:normal-phase-condition}
\end{equation}
For Hermiticity-preserving $H$ this condition implies $\Phi_n(H)\ge0$; for
bistochastic Hermiticity-preserving $H$ it implies
$\Phi_n(H)\ge n^{-n}$.
\end{prop}

\begin{proof}
Use the same standard Gaussian coefficients $z_j$ in both determinants.
The coefficient of $z^\alpha$ in $\det H(G)$ is
$c_\alpha\lambda^\alpha$. Gaussian monomial orthogonality proves
\eqref{eq:normal-phase-exact}, and
\eqref{eq:normal-phase-condition} makes every summand's real part
nonnegative. The support condition depends only on the nonzero multihomogeneous
components across distinct eigenspaces, so it is unchanged by a unitary
basis change within one eigenspace. In the bistochastic case choose
$E_0=I/\sqrt n$ and
complete it within the invariant traceless subspace. The pure scalar
term has coefficient $n^{-n/2}$ and contributes $n!/n^n$.
\end{proof}

Condition~\eqref{eq:normal-phase-condition} is a finite criterion on the support of a determinant polynomial
and the phases of eigenvalue products, rather than an assumption on
the sign of the sum in \eqref{eq:normal-phase-exact}. It can admit
negative eigenvalues. For example, transposition is $+1$ on symmetric
matrices and $-1$ on skew-symmetric matrices. Transposition invariance
of the determinant excludes odd total degree in the skew variables,
so every nonzero eigenvalue product is $+1$.

For the equality map $T_0$, put $D=\diag(1,1,-2,0)$. Then
\[
 T_0(D)=-D/2,\qquad T_0(E_{44})=E_{44},\qquad
 \det(tD+zE_{44})=-2t^3z.
\]
The nonzero $t^3z$ term has eigenvalue product $(-1/2)^3<0$,
so $T_0$ fails \eqref{eq:normal-phase-condition}, consistently with
$\Phi_4(T_0)=0$.

\section{The entire skew-unitary Breuer--Hall family}
\label{sec:bh}

The Breuer--Hall maps \citep{breuer2006,hall2006,chruscinski2008} are
positive and indecomposable in every even size at least four. We prove
that their invariant $\Gamma_n$ is positive in every even size, through an
exact three-exponential moment formula and a sign argument for its odd
moments. The moment formula comes from a skew-Gaussian duality of
Forrester and Kulkarni--Kawabata--Ryu by a polynomial specialization in
the coupling parameter; no Gaussian law is continued to negative
variance.

\begin{theorem}\label{thm:bh-all-even}
Let $n=2m$ and let $J\in M_n(\mathbb C)$ be skew-symmetric and unitary.
Define
\[
 B_n(X)=\tr(X)I_n-X-JX^{\mathsf T}J^*.
\]
If $X_1,X_2,X_3$ are independent exponential random variables of mean one,
then
\begin{equation}\label{eq:bh-main-moment}
 \Gamma_{2m}(B_{2m})=
 \E\left(X_1^2+X_2^2+X_3^2
       -2X_1X_2-2X_1X_3-2X_2X_3\right)^m.
\end{equation}
This value is zero for $m=1$ and strictly positive for every $m\ge2$.
For $n\ge4$, the normalized map $\widehat B_n=B_n/(n-2)$ is positive,
bistochastic, and indecomposable, and
\begin{equation}\label{eq:bh-normalized}
 \Phi_n(\widehat B_n)
 =\frac{\Gamma_n(B_n)}{n!(n-2)^n}>0,
 \qquad \|C_{\widehat B_n}\|_F^2=\frac{3n}{n-2}.
\end{equation}
\end{theorem}

The next identity is a specialization of the characteristic-polynomial
duality of \citet[Proposition~1.2, equation~(1.13)]{forrester2024},
attributed there to \citet{kulkarni2025}. In that formula set $N=m$,
$k=1$ and identify the self-dual matrix as $JA$. Then
$\det(tI-JA)=\det(A+tJ)$ and the dual real matrix is $2\times2$,
with variance-one entries. We give a finite algebraic proof of this
specialization to fix every sign and the common Gaussian coefficients.

\begin{lemma}[Skew Gaussian duality]\label{lem:bh-duality}
Let $n=2m$ and let $J_0$ have adjacent diagonal blocks
$\left(\begin{smallmatrix}0&1\\-1&0\end{smallmatrix}\right)$.
Let $A$ be a complex skew-symmetric matrix with independent standard
complex Gaussian entries above the diagonal, normalized by
$\E|A_{ij}|^2=1$. Let $G$ be a real $2\times2$ matrix with independent
$N(0,1)$ entries. For every $t\in\mathbb C$,
\begin{equation}\label{eq:bh-duality}
 \E_A|\det(A+tJ_0)|^2
 =\E_G\det(GG^{\mathsf T}+|t|^2I_2)^m.
\end{equation}
\end{lemma}

\begin{proof}
We use a finite Grassmann algebra, so all exponentials and expectations
below are coefficientwise finite algebraic operations. Introduce odd
variables $x_{i,a},y_{i,b}$, where $1\le i\le n$ and $a,b\in\{1,2\}$.
The extraction order is
\begin{equation}\label{eq:bh-extraction-order}
 x_{1,1}\cdots x_{n,1}\;
 x_{1,2}\cdots x_{n,2}\;
 y_{1,1}\cdots y_{n,1}\;
 y_{1,2}\cdots y_{n,2}.
\end{equation}
Put
\[
 U_{ij}=\sum_{a=1}^2x_{i,a}x_{j,a},\qquad
 V_{ij}=\sum_{b=1}^2y_{i,b}y_{j,b}.
\]
Since a determinant of a skew matrix is its Pfaffian squared, the left
side of \eqref{eq:bh-duality} is the coefficient of
\eqref{eq:bh-extraction-order} in
\[
 \E_A\exp\!\left(
 \sum_{i<j}(A_{ij}+t(J_0)_{ij})U_{ij}
 +\sum_{i<j}(\overline{A_{ij}}+\bar t(J_0)_{ij})V_{ij}\right).
\]
The quadratic Grassmann elements are even and commute. The complex
Gaussian identity, with variance one, gives
\[
 \E_A\exp\!\left(\sum_{i<j}(A_{ij}U_{ij}+\bar A_{ij}V_{ij})\right)
 =\exp\!\left(\sum_{i<j}U_{ij}V_{ij}\right).
\]
If $S_{ab}=\sum_i x_{i,a}y_{i,b}$, one interchange of odd variables
and the vanishing of repeated odd variables give
\[
 \sum_{i<j}U_{ij}V_{ij}=-\frac12\sum_{a,b=1}^2S_{ab}^2.
\]
The variance-one real Gaussian identity therefore yields
\[
 \exp\!\left(-\frac12\sum_{a,b}S_{ab}^2\right)
 =\E_G\exp\!\left(i\sum_{a,b}g_{ab}S_{ab}\right).
\]

The remaining exponent factors over the $m$ pairs
$(i,j)=(2p-1,2p)$ specifying $J_0$. Regrouping
\eqref{eq:bh-extraction-order} into these pairs exchanges chunks of two
odd variables, and hence introduces no sign. For one pair the extraction
order is
\[
 x_{i,1},x_{j,1},x_{i,2},x_{j,2},
 y_{i,1},y_{j,1},y_{i,2},y_{j,2},
\]
and the exponent is
\[
 t\sum_a x_{i,a}x_{j,a}
 +\bar t\sum_b y_{i,b}y_{j,b}
 +i\sum_{a,b}g_{ab}
       (x_{i,a}y_{i,b}+x_{j,a}y_{j,b}).
\]
The top coefficient can be checked according to the number of factors
containing an entry of $G$:
\[
 \begin{array}{c|c}
 \text{number of such factors}&\text{contribution}\\ \hline
 0&|t|^4\\
 2&|t|^2(g_{11}^2+g_{12}^2+g_{21}^2+g_{22}^2)\\
 4&(g_{11}g_{22}-g_{12}g_{21})^2.
 \end{array}
\]
Their sum is exactly $\det(GG^{\mathsf T}+|t|^2I_2)$.
Every pair uses the \emph{same} four coefficients $g_{ab}$. Consequently
the product is the $m$th power inside one common expectation, proving
\eqref{eq:bh-duality}.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:bh-all-even}]
First take $J=J_0$ as in Lemma~\ref{lem:bh-duality}.
For a real parameter $s\ge0$, define
\[
 T_s(X)=\tr(X)I_n-JX^{\mathsf T}J^*+sX.
\]
Write $\Omega=\sum_i e_i\otimes e_i$, $P=|\Omega\rangle\langle\Omega|$,
and let $F$ denote swap. Set
$F_J=(I\otimes J)F(I\otimes J^*)$. Then
\[
 C_{T_s}=I-F_J+sP\succeq0.
\]
Conjugating its output factor by $J^*$ gives
\[
 I-F+s|(I\otimes J)\Omega\rangle
          \langle(I\otimes J)\Omega|.
\]
The sign $J^*=-J$ has no effect on the last rank-one operator. This operator is
the covariance matrix of $A+\sqrt{s}\,gJ$, where $A$ is the skew Gaussian
matrix in the lemma and $g$ is an independent standard complex Gaussian.
The exact CP determinant identity therefore gives
\begin{equation}\label{eq:bh-polynomial-family}
 \Gamma_n(T_s)=\E_{A,g}|\det(A+\sqrt{s}\,gJ)|^2
 =\E_{G,E}\det(GG^{\mathsf T}+sEI_2)^m,
\end{equation}
where $E=|g|^2$ is exponential of mean one and independent of $G$.
There is no extra factorial: the CP identity is normalized by
$\Gamma_n(\Id)=\E|g|^{2n}=n!$.

Both sides of \eqref{eq:bh-polynomial-family} are polynomials in $s$ of
degree at most $2m$. Equality for all $s\ge0$ is consequently a polynomial
identity, valid at $s=-1$. As $T_{-1}=B_n$, the identity at $s=-1$ proves
\begin{equation}\label{eq:bh-wishart}
 \Gamma_n(B_n)=\E_{G,E}\det(GG^{\mathsf T}-EI_2)^m.
\end{equation}
The probability law in \eqref{eq:bh-wishart} is still the original
positive product law of four real Gaussians and one exponential.

The real variables
\[
 p=\frac{g_{11}+g_{22}}{\sqrt2},\quad
 q=\frac{g_{12}-g_{21}}{\sqrt2},\quad
 r=\frac{g_{11}-g_{22}}{\sqrt2},\quad
 u=\frac{g_{12}+g_{21}}{\sqrt2}
\]
are independent $N(0,1)$ variables. Put
$X_1=(p^2+q^2)/2$ and $X_2=(r^2+u^2)/2$. These two variables are independent
mean-one exponentials, and
\[
 \|G\|_F^2=2(X_1+X_2),\qquad \det G=X_1-X_2.
\]
Substitute these identities in \eqref{eq:bh-wishart}, and rename $E$ as
$X_3$, to obtain \eqref{eq:bh-main-moment}.

It remains to prove its sign. Define
\[
 \lambda(x_1,x_2,x_3)=\sum_i x_i^2-2\sum_{i<j}x_ix_j.
\]
For the three independent exponentials,
$S=X_1+X_2+X_3$ is independent of $U=(X_1,X_2,X_3)/S$.
The vector $U$ is uniform on the equilateral simplex
\[
 \Delta=\{u\in\mathbb R^3:u_i\ge0,\ \textstyle\sum_i u_i=1\},
\]
and $\E S^{2m}=(2m+2)!/2$. It therefore suffices to determine the sign
of the integral of $\lambda^m$ over $\Delta$.

Set $c=\frac13(1,1,1)$, the simplex center.
In Euclidean polar coordinates write $u=c+\rho e_\theta$ with
$0\le\rho\le R(\theta)$. Then
\[
 \lambda(u)=2\rho^2-\frac13.
\]
On an edge $u_k=0$, with $\{i,j,k\}=\{1,2,3\}$,
\[
 b(\theta):=2R(\theta)^2-\frac13
 =\lambda(u)=(u_i-u_j)^2\in[0,1].
\]
This boundary value is nonconstant in $\theta$.
 Direct radial
integration gives
\begin{equation}\label{eq:bh-radial}
 \int_\Delta\lambda(u)^m\,dA
 =\frac1{4(m+1)}\int_0^{2\pi}
       \bigl[b(\theta)^{m+1}-(-1/3)^{m+1}\bigr]\,d\theta.
\end{equation}
For $m=1$, direct exponential moments give
$\E\lambda(X_1,X_2,X_3)=3\cdot2-2\cdot3=0$.
Equation~\eqref{eq:bh-radial} therefore implies
$\E_\theta b(\theta)^2=1/9$, where $\theta$ is uniform on $[0,2\pi]$.
If $m$ is even, the integrand in \eqref{eq:bh-radial} is strictly
positive. If $m\ge3$ is odd, strict Jensen applied to the nonconstant
variable $b^2$ gives
\[
 \E_\theta b^{m+1}
 >\bigl(\E_\theta b^2\bigr)^{(m+1)/2}
 =(1/3)^{m+1}.
\]
These two cases prove the asserted strict positivity. For $m=1$, the map itself
is zero, consistently with the moment calculation.

Positivity follows from the rank-one calculation following
\eqref{eq:bh-definition}.
Also $B_n(I_n)=B_n^*(I_n)=(n-2)I_n$, proving the stated bistochastic
normalization.

The Choi matrix is $C_{B_n}=I-P-F_J$. Consider
$W=I+P+F_J\succeq0$. Its partial transpose on the output factor is
\[
 W^{\mathsf T_{\rm out}}=I+F+|(I\otimes J)\Omega\rangle
                  \langle(I\otimes J)\Omega|\succeq0.
\]
Here $P^{\mathsf T_{\rm out}}=F$ and, for real $J$,
$F_J^{\mathsf T_{\rm out}}=(I\otimes J)P(I\otimes J^{\mathsf T})$.
Since $F_J\Omega=-\Omega$ and $P^2=nP$,
\[
 (P+F_J)^2=I+(n-2)P,\qquad
 \tr(C_{B_n}W)=-n(n-2)<0\quad(n\ge4).
\]
A decomposable Choi matrix pairs nonnegatively with every PPT positive
semidefinite operator. The negative pairing above therefore proves indecomposability. Expanding
$\tr(I-P-F_J)^2$, using $\tr F_J=n$ and $PF_J=-P$, gives
\[
 \|C_{B_n}\|_F^2=3n(n-2),
\]
which proves the Choi-norm formula in \eqref{eq:bh-normalized}.

Finally, every complex skew-unitary matrix has the form
$J=UJ_0U^{\mathsf T}$ for a unitary $U$. One obtains this form by
choosing orthonormal pairs for the antiunitary $J$ followed by complex
conjugation, whose square is $-I$. Direct substitution gives
\[
 B_J=\Ad_U\circ B_{J_0}\circ\Ad_{U^*}.
\]
These unitary changes preserve the determinant invariant, positivity,
bistochasticity, indecomposability, and the Choi norm. Thus all the conclusions
hold for the original complex skew-unitary $J$.
\end{proof}

\smallskip\noindent\emph{\textbf{Exact controls and scope.}}
Independent sparse-Pfaffian expansions give
\[
\begin{array}{c|rrrrr}
n&2&4&6&8&10\\\hline
\Gamma_n(B_n)&0&24&384&31104&2488320.
\end{array}
\]
In particular, the bistochastic rank-four and rank-six maps have
squared Choi norms $6$ and $9/2$, and invariant values
$\Phi_4=1/16$ and $\Phi_6=1/7680$, respectively. The theorem covers skew-unitary Breuer--Hall maps;
it asserts no closure under arbitrary addition and no result for
arbitrary skew contractions. Pinched direct sums preserve the product
of the $\Gamma$ invariants, and therefore furnish positive indecomposable
examples in odd dimensions as well.

\smallskip\noindent\emph{\textbf{The moment theorem goes beyond the normal phase criterion.}}
For even $n\ge6$, use adjacent-pair $J$ and put
$D=\operatorname{diag}(1,1,1,1,-2,-2,0,\ldots,0)$.
Then $\tr D=0$, $JD^{\mathsf T}J^*=D$, and
$\widehat B_n(D)=-2D/(n-2)$, whereas $\widehat B_n(I_n)=I_n$.
But
\[
 \det(zI_n+tD)=z^{n-6}(z+t)^4(z-2t)^2
\]
has coefficient $-4$ at $t^3z^{n-3}$. After orthonormal rescaling this
coefficient remains nonzero. The determinant therefore has a nonzero
odd-degree component in the negative Hilbert--Schmidt eigenspace, so
the sufficient condition in Proposition~\ref{prop:normal-phase} fails.
Existence of this homogeneous component is independent of the eigenbasis
inside that eigenspace. The positive moment theorem thus controls
cancellations that the termwise phase condition cannot certify.

The same calculation excludes the Jordan orbits of the
Hilbert--Schmidt PSD class, with the terminology of
Appendix~\ref{app:separation}. A Jordan extension of the polar partial
isometry would fix $I_n$ and send $D$ to $-D$, forcing the displayed
determinant to be even in $t$. For $n=4$, the corresponding exclusion
is the Clifford argument in Section~\ref{subsec:bh4-clifford}.
\section{Strict separation of the sufficient classes}
\label{app:separation}

The comparison classes in this appendix are decomposable maps,
Hilbert--Schmidt positive-semidefinite positive maps, and positive maps
preserving an orthogonal rank-one resolution together with their
depolarizing regularizations \((1-\beta)T+\beta\Delta_n\),
\(0\le\beta\le1\). A Jordan symmetry means a unitary conjugation or its composition
with transposition. Every reference to a filtered class means
independent invertible input and output congruences, with whole
input/output transposition also allowed.

\subsection{An independent rank-four Breuer--Hall control}
\label{subsec:bh4-clifford}

Let $X,Y,Z$ be the Pauli matrices, and set
\[
 s_1=X\otimes I_2,\quad s_2=Y\otimes I_2,\quad
 s_3=Z\otimes X,\quad s_4=Z\otimes Y,\quad s_5=Z\otimes Z.
\]
These five matrices are pairwise anticommuting traceless Hermitian involutions, with
$s_1s_2s_3s_4s_5=-I_4$. Let $J=s_2s_4=iX\otimes Y$, a real skew
orthogonal matrix, and define the normalized Breuer--Hall map
\[
 B(A)=\frac{\tr(A)I_4-A-JA^{\mathsf T}J^*}{2}.
\]
The involution $T_J(A)=JA^{\mathsf T}J^*$ fixes
$I_4,s_1,\ldots,s_5$. Its fixed space has dimension six: multiplication
by $J$ identifies it with the skew matrices. Hence those six matrices
span the fixed space. The Hilbert--Schmidt eigenvalues of $B$ are $1$
on $I_4$, $-1$ on the five $s_i$, and zero on the complementary space.

The matrices $I_4/2,s_1/2,\ldots,s_5/2$ are orthonormal. The Clifford
relations yield
\begin{equation}
 \det\frac{z_0I_4+\sum_{i=1}^5z_i s_i}{2}
 =\frac1{16}\left(z_0^2-\sum_{i=1}^5z_i^2\right)^2.
 \label{eq:bh4-clifford-det}
\end{equation}
This polynomial is unchanged when all five $z_i$ change sign. Thus
every nonzero eigenvalue product in
\eqref{eq:normal-phase-exact} is positive, verifying
\eqref{eq:normal-phase-condition}. More explicitly,
\eqref{eq:bh4-clifford-det} is the orthogonal Fock projection of the
full determinant onto these six coordinates. Its squared norm gives
\[
 \Gamma_4(B)=\frac{24+80+120+160}{256}=\frac32,
 \qquad \Phi_4(B)=\frac1{16}.
\]
This computation is an independent control for the all-even-dimensional
Breuer--Hall argument.

The map is positive and bistochastic. For $\|x\|=1$,
\[
 B(xx^*)=\tfrac12P_{\{x,J\bar x\}^{\perp}},
 \qquad \rank B(xx^*)=2,
\]
since $x\perp J\bar x$. Here $P_F$ denotes the orthogonal projection
onto $F$.

The PPT witness in the proof of Theorem~\ref{thm:bh-all-even}
has pairing $-4$ with the Choi matrix of this normalized map, proving
indecomposability.

The same map lies outside the Hilbert--Schmidt PSD class even after
input/output Jordan symmetries. On its support, its
polar partial isometry fixes $I_4$ and negates every $s_i$. A unitary
conjugation with this action would negate their scalar product
$-I_4$, a contradiction. A transpose-conjugation gives the same
contradiction, because reversing five anticommuting generators has
sign $(-1)^{10}=1$. A putative Jordan factor in a PSD factorization
would have to extend this polar partial isometry. Finally, its
rank-two pure outputs and failure of strict positivity exclude both
the projection-preserving class and its positive depolarizing
regularizations considered here.

\subsection{A strict sector example outside three comparison classes}
\label{subsec:sector-explicit-example}

\begin{lemma}[Uniqueness of strict bistochastic scaling]
\label{lem:strict-filter-uniqueness}
Let $T:M_n\to M_n$ be strictly positive and bistochastic. If
$\Ad_L\circ T\circ\Ad_R$ is also bistochastic, with $L,R$ invertible,
then $L=cU$ and $R=c^{-1}V$ for a positive scalar $c$ and unitaries
$U,V$.
\end{lemma}
\begin{proof}
Strict positivity and compactness give
$T(X)\succeq\eta\tr(X)I$ for $X\succeq0$ and some $\eta>0$.
Duality gives the same bound for $T^*$, so $T^*$ is strict as well.
Put $A=L^*L$ and $B=RR^*$. The two normalizations are exactly
\[
 T(B)=A^{-1},\qquad T^*(A)=B^{-1}.
\]
If $B$ is nonscalar, strict positivity applied to
$B-\lambda_{\min}(B)I$ and
$\lambda_{\max}(B)I-B$ gives
\[
 \lambda_{\min}(B)I\prec T(B)\prec\lambda_{\max}(B)I,
 \qquad \operatorname{cond}(A)<\operatorname{cond}(B).
\]
The matrix $A$ must then also be nonscalar: otherwise unitality of
$T^*$ in the second equation would make $B$ scalar. Applying the
same argument to $T^*$ gives the contradictory strict inequality
$\operatorname{cond}(B)<\operatorname{cond}(A)$. Thus $B=bI$ and
$A=b^{-1}I$, which proves the assertion.
\end{proof}

\begin{corr}[Filtered PSD maps reduce to Jordan orbits after balancing]
\label{cor:filtered-hs-psd}
A strictly positive bistochastic map that is an invertible input/output
congruence of an ordinary positive Hilbert--Schmidt PSD map is
input/output unitarily equivalent to a bistochastic Hilbert--Schmidt
PSD map. Allowing input/output transposes gives the corresponding
Jordan equivalence.
\end{corr}
\begin{proof}
Invertible congruences preserve strict positivity, so the original
Hilbert--Schmidt PSD map $Q$ is strict. By the strict balancing result
in Lemma~\ref{lem:balancing}, choose positive definite $A,B$ such that
$T=\Ad_A\circ Q\circ\Ad_B$ is bistochastic. Since $Q=Q^*$,
\[
 T^*=\Ad_{BA^{-1}}\circ T\circ\Ad_{B^{-1}A}
\]
is also bistochastic. Lemma~\ref{lem:strict-filter-uniqueness} implies
$A^{-1}B^2A^{-1}=cI$ for some $c>0$. Uniqueness of the positive
square root gives $B=\sqrt c\,A$, and hence
\[
 T=c\,\Ad_A\circ Q\circ\Ad_A\succeq0
\]
as a Hilbert--Schmidt operator. Comparing any other strictly positive
bistochastic filter image with $T$ and applying the lemma again
reduces its filters to scalar unitaries, whose scalar factors cancel.
Transposition commutes through congruences by conjugating their
matrices, proving the last assertion.
\end{proof}

The following construction gives a strictly positive example separating
the sector theorem itself from the comparison classes. Let $K:M_s\to M_s$
be any positive bistochastic indecomposable map; below we take $s=3$
and the normalized Choi map $C$ of
Proposition~\ref{prop:choi-eb-family}. Put
$n=2s+3$. First define on $M_{2s}$
\[
 \mathcal S\begin{pmatrix}A&C\\D&E\end{pmatrix}
 =\begin{pmatrix}K(E)&0\\0&K^*(A)\end{pmatrix},
 \qquad R=\frac{\Id+\mathcal S}{2}.
\]
As in the proof of Proposition~\ref{prop:obstruction}, both maps are
positive and bistochastic, and $\mathcal S$ is a selfadjoint
Hilbert--Schmidt contraction. Hence $R\succeq0$ in Hilbert--Schmidt
order. These properties use only positivity and bistochasticity of $K$.

The map $R$ sends no nonzero rank-one input to rank one. Otherwise,
for $P=vv^*$, the positive summands $P$ and $\mathcal S(P)$ would
share the line of $v$. The second is nonzero and block diagonal,
forcing $v$ into one block. It would then be supported in the
opposite block from $v$, a contradiction.

Let $P_3$ be a three-cycle permutation matrix, and put
\[
 \epsilon=\frac1{n+2},\qquad
 M=(1-\epsilon)I_3+\epsilon P_3,
 \qquad T_M(A)=\operatorname{diag}(M\operatorname{diag}A).
\]
The map $T_M$ is a CP bistochastic diagonal channel. With $\boxplus$ denoting
a pinched direct sum, set
\begin{equation}
 H_0=R\boxplus T_M,
 \qquad H=\frac{\Id_n+H_0}{2}.
 \label{eq:sector-explicit-map}
\end{equation}
The resulting map is positive and bistochastic. On the first
$M_{2s}$ block, it acts as the Hilbert--Schmidt PSD operator
$(\Id+R)/2$. On off-diagonal blocks and on off-diagonal entries in
the final $M_3$ block, it acts as $\Id/2$. Its sole nonselfadjoint
block is the classical diagonal subspace, where its matrix is
\[
 B_0=(I_3+M)/2=(1-a)I_3+aP_3,
 \qquad a=\frac1{2(n+2)}.
\]
The eigenvalues of that normal matrix are
\[
 1,\quad u\pm iv,
 \qquad u=\frac{4n+5}{4(n+2)},\quad
 v=\frac{\sqrt3}{4(n+2)}.
\]
Consequently $S\succeq I/2$ on the full Hilbert--Schmidt space, and
\[
 \frac{|v|}{u}=\frac{\sqrt3}{4n+5}
 <\frac1n<\tan\frac{\pi}{2n}.
\]
This bound verifies \eqref{eq:hs-sector-lmi} on every block, so
Theorem~\ref{thm:hs-sector} gives $\Phi_n(H)\ge n^{-n}$.

Embedding an input in the lower $s\times s$ block within $M_{2s}$
and compressing the output to its upper $s\times s$ block recovers
$K/4$. Since decomposability survives CP embeddings and compressions,
$H$ is indecomposable.

The Hilbert--Schmidt operator $H$ is invertible. Its polar unitary
equals the identity outside the classical diagonal subspace and the
polar unitary of $B_0$ on that subspace. Put
\[
 \rho=\sqrt{1-3a+3a^2}\in(0,1).
\]
Dividing the two nonconstant Fourier eigenvalues of $B_0$ by $\rho$
shows that its polar unitary is a real circulant matrix
\[
 U_{B_0}=\alpha I_3+\beta P_3+\gamma P_3^2,
 \qquad \gamma=\frac{1-1/\rho}{3}<0.
\]
It sends a coordinate PSD projection to a diagonal matrix with a
negative entry, and so is not a positive Jordan automorphism.
If independent Hilbert--Schmidt-unitary Jordan automorphisms
$J_1,J_2$ transformed $H$ to a PSD operator $Q$, then
\[
 H=J_1^*QJ_2^*
 =(J_1^*J_2^*)(J_2QJ_2^*)
\]
would be its polar decomposition with a positive Jordan unitary
factor. Uniqueness gives a contradiction. Hence $H$ lies outside
the Hilbert--Schmidt PSD class and its stated Jordan orbits.

Finally, $T_M$ sends every nonzero rank-one input to rank at least
two: the support of $M(|v_i|^2)_i$ has at least two elements, since
$0<\epsilon<1$ and $P_3$ has no fixed point. Together with the rank
property of $R$, this support bound implies that neither $H_0$ nor $H$ sends a
nonzero rank-one input to rank one. On the other hand, $H$ is not
strictly positive: a rank-one input supported in the first $2s$
coordinates has output supported there. Positive depolarizing
regularizations of projection-preserving maps are strict, whereas
their unregularized versions preserve designated rank-one outputs.
Thus neither possibility describes $H$.

Taking $s=3$ and $K=C$ gives an example on $M_9$, with
$\epsilon=1/11$, $a=1/22$, and
\[
 \frac{|v|}{u}=\frac{\sqrt3}{41},\qquad
 \gamma=\frac{1-22/\sqrt{421}}3<0.
\]
Now make it strictly positive by setting
\begin{equation}
 H_\delta=(1-\delta)H+\delta\Delta_9,
 \qquad \delta=\frac1{10}.
 \label{eq:sector-strict-nine}
\end{equation}
The sector inequalities are convex and $\Delta_9$ is Hilbert--Schmidt
PSD, so $H_\delta$ remains in the certified sector. It is bistochastic,
satisfies $H_\delta(X)\succeq\tr(X)I_9/90$ for $X\succeq0$, and obeys
\[
 \Phi_9(H_\delta)\ge9^{-9}.
\]

The witness in Proposition~\ref{prop:choi-eb-family} has trace $21/2$
and pairs with $C_C$ to give $-3/4$. Embed it in the lower-input,
upper-output three-dimensional corner used above. That corner of
$H$ is $C/4$, so its pairing with $C_{H_\delta}$ is exactly
\[
 -\frac{3(1-\delta)}{16}+\frac{7\delta}{6}
 =\frac{-9+65\delta}{48}.
\]
It is negative for $\delta<9/65$ and equals $-5/96$ at $\delta=1/10$.
Thus the strictly positive map in \eqref{eq:sector-strict-nine} is
indecomposable.

On the reducing decomposition $M_9=\mathbb C I\oplus\{X:\tr X=0\}$,
write $H=1\oplus H_{\mathrm{tr}}$. Then
\[
 H_\delta=1\oplus(1-\delta)H_{\mathrm{tr}}.
\]
Since $H$ is invertible and $1-\delta>0$, the polar unitary is
unchanged; in particular, it retains the negative coefficient $\gamma$.
 The earlier uniqueness-of-polar-decomposition
argument excludes the Hilbert--Schmidt PSD class and its stated
Jordan orbits for $H_\delta$ as well. Corollary~\ref{cor:filtered-hs-psd}
strengthens this exclusion to all invertible input/output congruences
of ordinary positive Hilbert--Schmidt PSD maps, including transposes.

To exclude all depolarized projection-preserving maps, define the
minimum over unit product vectors by
\[
 \mu(L)=\min_{\|x\|=\|y\|=1}y^*L(xx^*)y.
\]
The unregularized $H$ has $\mu(H)=0$: use an input supported in the
first six coordinates and an output test vector in the last three.
Hence $\mu(H_\delta)=\delta/9$ exactly. If $T$ is positive and sends
one orthogonal rank-one resolution to another, then $\mu(T)=0$,
attained by testing a designated rank-one output in a perpendicular
direction. Consequently
\[
 \mu\bigl((1-\beta)T+\beta\Delta_9\bigr)=\beta/9
 \qquad(0\le\beta\le1).
\]
A representation of $H_\delta$ in this form would force
$\beta=\delta$ and then $T=H$, contradicting the fact that $H$ sends
no nonzero rank-one input to rank one. The same argument covers the
input/output unitary and transposition symmetries, which preserve
the depolarizing map and these product-vector minima.

This argument also excludes arbitrary invertible filters of the earlier
projection-preserving families. An unregularized projection-preserving
map is not strict, and invertible filters cannot change that property.
A positively depolarized member is already strictly positive and
bistochastic. Lemma~\ref{lem:strict-filter-uniqueness} reduces any
bistochastic filter image of it to unitary input/output changes,
which were just excluded. Positive scalar multiples can be absorbed
into the filters, and transposes commute through them.

Thus \eqref{eq:sector-strict-nine} is a fully specified strictly
positive bistochastic map on $M_9$ outside the union of all three
comparison classes even after their invertible input/output filtering and
Jordan symmetries. Only the intermediate $H_0$ is a pinched direct
sum; $H$ retains off-diagonal identity terms, and its sign and that of
$H_\delta$ are certified here by the sector theorem.
\section{Ferrers tests for Schur coefficients}
\label{app:ferrers}

We record explicit completely positive curvature tensors that extract
individual Schur coefficients. This construction complements the
Schur--Griffiths cone identification
\citep[Appendix~A, Proposition~A.3]{fulton_lazarsfeld1983} and Li's
sum-of-squares formula \citep[Propositions~3.1 and~8.2]{li2021}.
It uses the Frobenius character formula and exterior Cauchy
decomposition; see \citet[Sections~2--5]{tamvakis2004}.

Fix $d,r,m\ge1$, let $U=\mathbb C^r$ and $W=\mathbb C^m$, and identify
$U\otimes W$ with $M_{r\times m}$ with its standard Hermitian product.
For $A_1,\ldots,A_d\in M_{r\times m}$, put
\begin{equation}\label{eq:ferrers-curvature}
 \Omega=i\sum_{j,k=1}^d A_jA_k^*\,dz_j\wedge d\bar z_k,
 \qquad dV_d=\bigwedge_{j=1}^d(i\,dz_j\wedge d\bar z_j).
\end{equation}
For homogeneous symmetric polynomials $p$ of degree $d$, define
$L_\Omega(p)$ by $p(\Omega)=L_\Omega(p)dV_d$. The map
$H(E_{jk})=A_jA_k^*$ is CP, with Kraus matrices $K_s e_j=A_j e_s$;
conversely, every CP map $M_d\to M_r$ has this form for some $m$.

For $\mu\vdash d$, let $\chi_\mu$ be the irreducible character of $S_d$
of dimension $f^\mu$. On $(U\otimes W)^{\otimes d}$, let $T_\pi^U$
permute only the $U$ factors and set
\begin{align*}
 \Pi_\mu^U&=\frac{f^\mu}{d!}
    \sum_{\pi\in S_d}\chi_\mu(\pi^{-1})T_\pi^U,\\
 w&=A_1\wedge\cdots\wedge A_d
   =\frac1{\sqrt{d!}}\sum_{\sigma\in S_d}
       \operatorname{sgn}(\sigma)
       A_{\sigma(1)}\otimes\cdots\otimes A_{\sigma(d)}.
\end{align*}
Thus $\Pi_\mu^U$ is the orthogonal projection onto the row
$\mu$-isotypic component, and the exterior embedding is isometric.

\begin{lemma}[Schur-form projector identity]
\label{lem:schur-projector}
For the curvature tensor in \eqref{eq:ferrers-curvature},
\begin{equation}\label{eq:schur-projector}
 L_\Omega(s_\mu)=\frac{d!}{f^\mu}
       \langle w,\Pi_\mu^U w\rangle
   =\frac{d!}{f^\mu}\|\Pi_\mu^U w\|^2.
\end{equation}
In particular, every Schur form is weakly nonnegative for CP curvature.
\end{lemma}

\begin{proof}
For $\pi\in S_d$, let $p_\pi$ be the product of power sums indexed
by its cycle lengths. Expanding the traces in $p_\pi(\Omega)$ and
alternating the holomorphic and antiholomorphic indices gives
\[
 L_\Omega(p_\pi)=d!\langle w,T_{\pi^{-1}}^U w\rangle.
\]
Indeed, the two alternations contribute
$\operatorname{sgn}(\sigma)\operatorname{sgn}(\tau)$, and the
normalization of $w$ contributes $1/d!$; the factors $i^d$ cancel
against $dV_d$. Since the even-degree entries of $\Omega$ commute,
the Frobenius identity gives
\begin{align*}
 s_\mu&=\frac1{d!}\sum_{\pi\in S_d}\chi_\mu(\pi)p_\pi,\\
 L_\Omega(s_\mu)&=\sum_{\pi\in S_d}\chi_\mu(\pi)
       \langle w,T_{\pi^{-1}}^U w\rangle
       =\frac{d!}{f^\mu}\langle w,\Pi_\mu^U w\rangle.
\end{align*}
The norm identity follows because $\Pi_\mu^U$ is an orthogonal
projection. It preserves the exterior power because it commutes
with simultaneous row and column permutations.
\end{proof}

\begin{samepage}
\begin{theorem}[Ferrers coefficient tests]
\label{thm:ferrers-coefficient-test}
Let $\lambda\vdash d$ have at most $r$ rows and at most $m$ columns.
Index the tangent coordinates by
\[
 D_\lambda=\{(a,s):1\le a\le r,\ 1\le s\le\lambda_a\},
\]
and choose $A_{(a,s)}=E_{as}\in M_{r\times m}$ in
\eqref{eq:ferrers-curvature}. The resulting CP curvature tensor
$\Omega_\lambda$ satisfies
\begin{equation}\label{eq:ferrers-isolation}
 L_{\Omega_\lambda}(s_\mu)
       =\delta_{\lambda\mu}\frac{d!}{f^\lambda}
       \qquad(\mu\vdash d).
\end{equation}
Consequently, for $p=\sum_{\mu\vdash d,\,\ell(\mu)\le r}c_\mu s_\mu$,
\begin{equation}\label{eq:ferrers-coefficient-extraction}
 L_{\Omega_\lambda}(p)=\frac{d!}{f^\lambda}c_\lambda.
\end{equation}
\end{theorem}
\end{samepage}

\begin{proof}
The wedge $w_\lambda$ of the matrix units indexed by $D_\lambda$ has
norm one and row and column weights $(\lambda,\lambda')$. Every row
or column raising operator annihilates it: moving a cell upward or
leftward produces a repeated exterior factor. Hence $w_\lambda$ is a
highest weight vector in the $\lambda$ summand of the exterior Cauchy
decomposition
\[
 \bigwedge^d(U\otimes W)
   \simeq\bigoplus_{\substack{\nu\vdash d\\\ell(\nu)\le r,\ \nu_1\le m}}
          S_\nu U\otimes S_{\nu'}W.
\]
Thus $\Pi_\mu^U w_\lambda=\delta_{\mu\lambda}w_\lambda$;
Lemma~\ref{lem:schur-projector} and linearity give the two identities.
\end{proof}

The factor $d!/f^\lambda$ is the product of the hook lengths of
$\lambda$. The corresponding map is
\[
 H_\lambda:M_d\longrightarrow M_r,\qquad
 H_\lambda(X)_{ab}
  =\sum_{s\le\min(\lambda_a,\lambda_b)}x_{(a,s),(b,s)}.
\]

\begin{corr}[Exact separation of the Schur cone]
\label{cor:cp-separates-schur-cone}
A homogeneous symmetric polynomial with real coefficients in $r$
variables is Schur-positive if and only if its Chern--Weil form is
weakly nonnegative on every CP curvature tensor. It suffices to test
the Ferrers tensors of its degree with $m=d$. The universal CP
quantifier may also be restricted to maps with positive-definite
Choi matrices.
\end{corr}

\begin{proof}
Lemma~\ref{lem:schur-projector} proves sufficiency of Schur positivity;
\eqref{eq:ferrers-coefficient-extraction} detects every negative
coefficient. Adding $\varepsilon\operatorname{tr}(\,\cdot\,)I_r$ to
a detecting map makes its Choi matrix positive definite, while
polynomial continuity preserves the negative evaluation for
sufficiently small $\varepsilon>0$.
\end{proof}

\begin{aiusage}  %
  The results in this paper were developed through extensive use of an AI assistant GPT-6 Astra
  (Pro). The choice of the questions worth answering was fully made by the author, as was the selection
  of the subset of results worth retaining. This choice of material evolved over many months, including
  up to the recent disproof of $\Phi_9\ge0$. None of the claims were obtained zero-shot: they emerged
  through sustained back-and-forth discussion, including failed attempts, counterexamples, audits,
  and repairs. The writing, polishing, and presentation were done with GPT-6 as well as Claude's (Fable
  5.1 and Opus 5.5) assistance. 
\end{aiusage}

\aiacknowledgements

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\end{thebibliography}

\end{document}