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Email: \texttt{s.sra@tum.de}.}} \makeatother \hypersetup{colorlinks=true,linkcolor=blue!45!black,citecolor=blue!45!black,urlcolor=red!65!black} \setlist{nosep,leftmargin=*} \renewcommand{\bibnumfmt}[1]{{\color{darkblue}\bf\textsc{#1}}.\hskip 6pt} \newenvironment{aiusage}{\section*{AI usage}}{} \newcommand{\aiacknowledgements}[1][]{% \section*{Acknowledgements} \if\relax\detokenize{#1}\relax\else #1\ \fi The author (SS) thanks the Alexander von Humboldt (AvH) Foundation for their generous support via a Humboldt Professorship in AI.% } \newcommand{\reals}{\mathbb{R}} \newcommand{\N}{\mathbb{N}} \newcommand{\pp}{\mathbb{P}} \newcommand{\E}{\mathbb{E}} \newcommand{\bmOne}{\bm{1}} \newcommand{\C}{\mathbb{C}} \newcommand{\Cc}{\mathcal{C}} \newcommand{\Dc}{\mathcal{D}} \newcommand{\Nc}{\mathcal{N}} \newcommand{\Xc}{\mathcal{X}} \newcommand{\Fc}{\mathcal{F}} \newcommand{\symm}{\mathfrak{S}} \newcommand{\ma}{\bm{A}} \newcommand{\mx}{\bm{X}} \newcommand{\my}{\bm{Y}} \newcommand{\mz}{\bm{Z}} 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before={\par\medskip\noindent}, after={\par\medskip}, colback=yellow!6!white, boxrule=0pt, enhanced jigsaw, top=2pt, left=2pt, right=2pt, bottom=0pt, dashedrounded, #1 } \newtcbox{\tcbmathbox}[1][]{ nobeforeafter, colback=yellow!30!white, boxrule=0pt, enhanced jigsaw, top=0pt, left=1pt, right=1pt, bottom=0pt, borderline horizontal={0.5pt}{0pt}{dashed}, borderline vertical={0.5pt}{0pt}{dashed}, #1 } \hypersetup{pdftitle={Lower bounds for the quantum permanent}, pdfauthor={Suvrit Sra}} \allowdisplaybreaks \setlength{\emergencystretch}{2em} \DeclareMathOperator{\tr}{tr} \DeclareMathOperator{\Ad}{Ad} \DeclareMathOperator{\Id}{Id} \DeclareMathOperator{\rank}{rank} \DeclareMathOperator{\Ent}{Ent} \DeclareMathOperator{\capop}{cap} \newcommand{\R}{\mathbb R} \newcommand{\HS}{\mathrm{HS}} \newcommand{\D}{\mathscr D} \newcommand{\T}{\mathscr T} \newcommand{\fall}[2]{(#1)_{#2}} \keywords{Griffiths positivity, positive maps, quantum permanent, lower bounds, bistochastic} \makeatletter \newcommand{\preprintslink}[2]{% \thanks{First public release: #2, as v1 in the author's preprint repository, \url{https://github.com/suvrit/preprints/tree/main/preprints/#1}; every version is kept there with its timestamp.}} \makeatother \begin{document} \title{Lower bounds for the quantum permanent} \aiauthor \begin{abstract} Griffiths's 1969 conjecture on the positivity of Schur forms reduces to the positivity of the quantum permanent $\Phi_n$ on positive bistochastic maps. This conjecture was refuted by \citet{du2026} with a map for which $\Phi_9<0$. Therefore, we ask: \emph{how negative can $\Phi_n$ be?} We prove a lower bound $\Phi_n\ge-3n/(3n^2+20n+103)$ (for $n\ge 8$), and an asymptotically stronger bound $\Phi_n\ge-1/n+(3e^2-7-o(1))\log n/n^2$. The key to our bounds is the orbital polynomial of the map, which turns $\Phi_n$ into a linear function of a distribution on partitions and positivity into linear constraints. We also construct explicit maps for $n=8, 9$ that improve on Du's example. Finally, if $d_*\in\{4,\ldots,8\}$ is the first dimension for which $\Phi_n<0$, the minimum $\mu_n$ of $\Phi_n$ over positive bistochastic maps satisfies $-\mu_n=\Omega(n^{-d_*})$; we conjecture the matching order $-\mu_n=O(n^{-d_*})$. \end{abstract} \preprintslink{qperm-lower}{24 October 2026} \maketitle \addtocontents{toc}{\protect\setcounter{tocdepth}{-1}} \vspace*{-8pt} \section{Introduction} \label{sec:introduction} Griffiths asked in 1969 whether Griffiths-positive curvature forces positivity of the characteristic forms of a vector bundle \citep[p.~247]{griffiths1969}. \citet[Theorem~1.3]{finski2022} reduced Griffiths's question to the sign of a more accessible object, the quantum permanent of~\citet{gurvits2006} evaluated at a positive linear map, which in turn is just a signed sum of mixed discriminants. For matrices $A_1,\ldots,A_n$, the mixed discriminant $D(A_1,\ldots,A_n)$ \citep{alexandrov1938} is the coefficient of $t_1\cdots t_n$ in $\det(t_1A_1+\cdots+t_nA_n)/n!$. For a complex-linear map $H\colon M_n(\C)\to M_n(\C)$, the \emph{quantum permanent}, also called the double mixed discriminant~\citep[(1.6)]{finski2022}, is \begin{equation*} \Phi_n(H) := \nlsum_{\sigma\in \symm_n}\operatorname{sgn}(\sigma)\, D\bigl(H(E_{1\sigma(1)}),\ldots,H(E_{n\sigma(n)})\bigr), \end{equation*} where $E_{ij}$ are the matrix units and $\symm_n$ is the group of permutations of $\{1,\ldots,n\}$. % We focus on positive linear maps, where $H$ is \emph{positive} if $H(X)\succeq0$ whenever $X\succeq0$; closely related is whether positive $H$ also implies $\Phi_n(H)\ge 0$. More precisely, by Finski's reduction \citep[Theorem~1.3]{finski2022}, the weak form of Griffiths's question over all bundles holds if and only if $\Phi_n(H)\ge0$ for every positive $H$. Finski settled dimension $n=2$ \citep[Proposition~1.4(c)]{finski2022} and Wan $n=3$ \citep[Theorem~0.3]{wan2026}, but Du's recent 9D counterexample \citep[Theorem~2.3]{du2026} dashed hopes for general $n$. His refutation raises two natural questions: \begin{enumerate}[label=(\roman*),leftmargin=4cm,itemsep=0pt] \vspace*{4pt} \it \item Which hypotheses on the map restore the sign? \item How far does the sign fail without them? \end{enumerate} \vskip5pt The companion paper \citep{sra2026qperm} takes up (i). Beyond proving $\Phi_n(H)\ge0$ for every decomposable map (i.e., a sum of a completely positive and a completely copositive map) it studies the positive maps whose every square compression keeps the sign, among other sufficient conditions. The present paper is complementary and takes up (ii). In particular, we ask how far below zero can $\Phi_n$ fall, and which constraints on a positive map control that negative part. The question needs a scale, because $\Phi_n(cH)=c^n\Phi_n(H)$. We focus therefore on \emph{bistochastic} maps, i.e., those where $H(I)=H^*(I)=I$. Our goal is to then find lower bounds on the optimal value of \begin{equation}\label{eq:minimum} \mu_n :=\min_{H\in\mathfrak B_n}\Phi_n(H),\quad\text{where}\quad \mathfrak B_n=\{H:M_n(\C)\to M_n(\C):H \succeq 0 \text{ is bistochastic}\}. \end{equation} In Problem~\eqref{eq:minimum}, the minimum is attained since $\mathfrak B_n$ is compact. We bound $\mu_n$ from below with a technique that rests on a simple observation: $\Phi_n(H)$ depends only on the polynomial $\det H(X)$, and is invariant to its input being conjugated by a unitary (Section~\ref{sec:orbital}). Therefore, it is determined by the \emph{orbital polynomial} $f_H(x):=\E_U\det H(U\operatorname{diag}(x)U^*)$, with $U$ Haar distributed on $U(n)$. For $H\in\mathfrak B_n$, positivity makes the coefficients of $f_H$ nonnegative, and $H(I)=I$ makes them sum to $f_H(1,\ldots,1)=1$, so they form a probability distribution on partitions of $n$, of which $\Phi_n(H)$ is a fixed signed linear function (Proposition~\ref{prop:orbital}). This idea based on orbital polynomials delivers a technique that helps us to turn properties of positive maps into \emph{valid constraints}, i.e., linear inequalities among the coefficients of $f_H$ that hold for every $H\in\mathfrak B_n$. Every set of valid constraints yields a lower bound on $\mu_n$, namely the minimum of the linear function over all distributions that satisfy them. The simplest valid constraint comes from Gurvits's theorem \citep[Theorem~4.1]{gurvits2006}, which bounds the coefficient of $x_1\cdots x_n$ in $f_H$ from below. With it, we obtain the following bound, which we prove in Section~\ref{sec:orbital}. % \begin{dashedbox}[supporting] \begin{prop}[Elementary bound]\label{prop:elementary} Every $H\in\mathfrak B_n$ satisfies \begin{equation}\label{eq:elementary-intro} \Phi_n(H)\ge\frac{(n+1)!}{n^{n+1}}-\frac1n. \end{equation} \end{prop} \end{dashedbox} The positive term in \eqref{eq:elementary-intro} decays exponentially. Our main result uses sharper valid constraints to improve the decay to be polynomial; we note it as Theorem~\ref{thm:main} below. \begin{dashedbox} \begin{theorem}[Main bound]\label{thm:main} Every $H\in\mathfrak B_n$ satisfies \begin{equation}\label{eq:main} \Phi_n(H)\ge-\frac{3n}{3n^2+20n+103}\qquad(n\ge8). \end{equation} For every $\varepsilon>0$ there is a constant $A_\varepsilon$ such that \begin{equation}\label{eq:log-main} \Phi_n(H)\ge-\frac1n+ \frac{(c_{\log}-\varepsilon)\log n-A_\varepsilon}{n^2} \qquad(n\ge21),\qquad c_{\log}=3e^2-7. \end{equation} \end{theorem} \end{dashedbox} The sharper constraints used by Theorem~\ref{thm:main} are also simple to state. Compressions of $H$ to $k$-dimensional subspaces turn the lower bounds known in dimension $k$ into valid constraints (Lemma~\ref{lem:mixed}), and these constraints carry the bound up to higher dimensions (Theorem~\ref{thm:transport}). The spectra of $H(P)$, for projections $P$, give a second family of valid constraints (Section~\ref{sec:projection-spectra}). We prove \eqref{eq:main} in Section~\ref{sec:finite}, starting from bounds in small dimensions that have computational proofs in exact rational arithmetic. In every dimension, these finite certificates also give the single constant $\Phi_n\ge -0.05656\dots$ (Corollary~\ref{cor:uniform-floor-improved}). We prove \eqref{eq:log-main} in Section~\ref{sec:mixed-structure} by feeding each new bound back in as a seed. The spectral constraints add a small gain at every step, and these gains add up to the $\log n$ term. But eventually, this idea hits its limit: we show that recursive programs built from these valid constraints cannot certify a coefficient larger than $3e^2-7$ (Theorem~\ref{thm:optimal-recursive-coefficient}). % Finally, in Section~\ref{sec:negative} we construct negative maps for $n=8, 9$ that improve on Du's construction~\citep{du2026}. Together with \eqref{eq:main} our construction yields the bracket \begin{equation}\label{eq:bracket} \max\left\{\frac{4947}{50000(n)_8}, \frac1{737280\binom n9}\right\}\le-\mu_n \le\frac{3n}{3n^2+20n+103}\qquad(n\ge9), \end{equation} where $(n)_r=n(n-1)\cdots(n-r+1)$ and $(n)_0=1$. Thus $-\mu_n$ decays at a polynomial rate between $n^{-8}$ and $n^{-1}$; the exact order is open, and we formally state it as Conjecture~\ref{conj:universal-order}. \section{Orbital coefficients and their constraints} \label{sec:orbital} The orbital polynomial turns positivity of $H$ into valid constraints on its coefficients, and it expresses $\Phi_n$ as a signed sum of these nonnegative coefficients. Every lower bound in this paper combines the two. To express $\Phi_n$ through $\det H(X)$ alone, we pair this polynomial with the input determinant, which gives \begin{equation}\label{eq:phi} \Gamma_n(H)=\det(\partial_X)\det H(X),\qquad \Phi_n(H)=\frac{\Gamma_n(H)}{n!}, \end{equation} where the entries of $X$ are independent complex variables and \[ \det(\partial_X)=\sum_{\sigma\in \symm_n}\operatorname{sgn}(\sigma) \prod_{i=1}^n\frac{\partial}{\partial X_{i,\sigma(i)}}. \] Since $\det H(X)$ is homogeneous of degree $n$, the derivative in \eqref{eq:phi} is a constant. The unnormalized invariant $\Gamma_n$ is Cayley's first hyperdeterminant of order four of the coefficient tensor of $H$, and it is the quantum permanent of the Choi matrix $C_H=\sum_{i,j}E_{ij}\otimes H(E_{ij})$ in the sense of Gurvits \citep[Definition~1.7]{gurvits2002}, whose Gaussian representation and operator scaling we use throughout \citep{gurvits2004,gurvits2006}. Averaging $\det H(X)$ over unitary conjugates of a diagonal input leaves \eqref{eq:phi} unchanged and yields a symmetric polynomial with coefficients indexed by partitions. To group monomials with the same exponents, write $\lambda\vdash d$ for a partition $\lambda_1\ge\cdots\ge\lambda_\ell>0$ with sum $d$. Its length is $\ell(\lambda)=\ell$, and $m_j(\lambda)$ counts its parts equal to $j$. The monomial symmetric polynomial $m_\lambda$ is the sum of the distinct permutations of $x_1^{\lambda_1}\cdots x_\ell^{\lambda_\ell}$. Trailing zeros are added when a partition is regarded as an exponent vector in $n$ variables. We use \[ e_j(x)=\sum_{|I|=j}\prod_{i\in I}x_i,\qquad e_0=1, \] for the elementary symmetric polynomials; for a matrix, $e_j$ is evaluated on its eigenvalues. In this notation, the orbital polynomial of the introduction expands as \begin{equation*} f_H(x)=\E_U\det H(U\operatorname{diag}(x)U^*) =\sum_{\lambda\vdash n}w_\lambda q_\lambda(x),\qquad q_\lambda(x)=\frac{m_\lambda(x)}{m_\lambda(1^n)}. \end{equation*} Here $1^n=(1,\ldots,1)$, Haar invariance makes $f_H$ symmetric, and the weight $w_\lambda$ is the total coefficient of the monomials of \emph{shape}~$\lambda$. The weights form a probability distribution on the partitions of~$n$. Indeed, for each unitary $U$ with columns $u_1,\ldots,u_n$, the matrices $A_i=H(u_iu_i^*)$ satisfy \[ A_i\succeq0,\qquad \tr A_i=1,\qquad \sum_iA_i=I. \] Rank-one decomposition and Cauchy--Binet show that $\det(\sum_i x_iA_i)$ has nonnegative coefficients, and its value at $1^n$ is $\det I=1$, so \begin{equation*} w_\lambda\ge0,\qquad \sum_{\lambda\vdash n}w_\lambda=1. \end{equation*} The pairing \eqref{eq:phi} sees each monomial of $f_H$ only through the number of variables it involves (Section~\ref{app:radial}), so we group monomials by this number. For a homogeneous degree-$n$ polynomial $f$ with nonnegative coefficients and $f(1^n)=1$, let $w_\ell(f)$ be the total coefficient of monomials supported on exactly $\ell$ variables, and define \begin{equation}\label{eq:coefficient-functional} \mathcal L_n(f)=\sum_{\ell=1}^n \frac{(-1)^{n-\ell}}{\binom n\ell}w_\ell(f). \end{equation} For the symmetric orbital polynomial, $w_\ell(f_H)$ is the sum of $w_\lambda$ over partitions of length $\ell$. Gurvits's theorem \citep[Theorem~4.1]{gurvits2006} also bounds $w_{(1^n)}$ from below. \begin{prop}\label{prop:orbital} For $H\in\mathfrak B_n$, \begin{equation}\label{eq:score} \Phi_n(H)=\mathcal L_n(f_H)=\sum_{\lambda\vdash n}r_\lambda w_\lambda, \qquad r_\lambda=\frac{(-1)^{n-\ell(\lambda)}}{\binom n{\ell(\lambda)}}. \end{equation} Moreover, $w_{(1^n)}\ge n!/n^n$. \end{prop} We call $r_\lambda$ the score of the shape $\lambda$, \eqref{eq:score} the score identity, and the inequality $w_{(1^n)}\ge n!/n^n$ the square-free floor. Proposition~\ref{prop:orbital} is proved in Section~\ref{app:radial}. Since $r_{(1^n)}=1$ and $r_\lambda\ge-1/n$ for every other shape, the score identity and the square-free floor give $\Phi_n(H)\ge(1+1/n)w_{(1^n)}-1/n\ge(n+1)!/n^{n+1}-1/n$, which proves Proposition~\ref{prop:elementary}. \subsection{Compatible compressions} \label{sec:mixed} The elementary bound uses only the normalization of the weights and the square-free floor. Lower bounds known in a smaller dimension $k$ give further valid constraints, once applied to the compressions of $H$ to $k$-dimensional subspaces. Such compressions are positive but need not be bistochastic, and scaling removes the marginal constraints while preserving every nonpositive lower bound. \begin{lemma}[Normalization]\label{lem:normalization} Suppose $\gamma\le0$ and $\Phi_k(L)\ge\gamma$ for every $L\in\mathfrak B_k$. Then every positive $K:M_k(\C)\to M_k(\C)$ satisfies \begin{equation*} \Phi_k(K)\ge\gamma\det K(I). \end{equation*} \end{lemma} The scaling proof is in Section~\ref{app:scaling}. We call a bound $\gamma$ as in Lemma~\ref{lem:normalization} a \emph{seed} in dimension~$k$. The first seeds are $\gamma_2=\gamma_3=0$, obtained from Wan's theorem \citep[Theorem~0.3]{wan2026} by the normalization and limiting argument of the same section. The condition $\gamma\le0$ is essential (Section~\ref{app:scaling}). Lemma~\ref{lem:normalization} bounds one compression at a time. To combine many such bounds, we place them in a single determinant. Let $H:M_n(\C)\to M_n(\C)$ be positive, fix $1\le k\le n$ and an isometry $V:\C^k\to\C^n$, and take finitely many positive semidefinite baseline matrices $B_i\in M_n(\C)$. Set \[ F(x,Z)=\det\left(\sum_i x_iB_i+H(VZV^*)\right). \] The next lemma is the source of all inherited coefficient constraints. \begin{samepage} \begin{lemma}[Mixed determinant inequality]\label{lem:mixed} Under the hypotheses of Lemma~\ref{lem:normalization}, every monomial coefficient of \begin{equation}\label{eq:mixed} \left.\det(\partial_Z)F(x,Z)\right|_{Z=0} -\gamma\left.\partial_s^kF(x,sI_k)\right|_{s=0} \end{equation} is nonnegative whenever $H$ is positive. \end{lemma} \end{samepage} \begin{proof} Write each $B_i$ as a sum of rank-one matrices. Expand the part of $F$ homogeneous of degree $k$ in $Z$ by choosing $n-k$ baseline columns. A choice $S$ contributes \[ x^{\nu(S)}\det(B_S^*B_S)\det K_S(Z),\qquad K_S(Z)=U_S^*H(VZV^*)U_S, \] where $U_S$ is an isometry from $\C^k$ onto the orthogonal complement of their span, $B_S$ has the chosen vectors as columns, and $\nu(S)$ records their baseline multiplicities. Dependent columns contribute zero. Every $K_S$ is positive and every Gram determinant is nonnegative. The two derivatives in \eqref{eq:mixed} give $k!\Phi_k(K_S)$ and $k!\det K_S(I)$, respectively. Lemma~\ref{lem:normalization} makes their difference nonnegative before terms with the same monomial are collected. The compression identity follows by completing the chosen span to an orthonormal basis and expanding along its first $n-k$ coordinates. The expansion is an identity in the entries of $Z$, so $Z$ need not be Hermitian or positive semidefinite. \end{proof} To turn Lemma~\ref{lem:mixed} into inequalities among the weights, we express its two derivatives in the eigenvalue coordinates of $f_H$. On polynomials symmetric in the first $k$ variables, define \begin{equation*} \delta_k(x)=\prod_{ik$, the coefficient of $x_{k+1}\cdots x_n$, divided by $k!$, is \begin{equation}\label{eq:J} J_{n,k,\gamma}:=\sum_{\mu\vdash k} \frac{(k)_s}{(n)_s} \left(\frac{(-1)^j}{\binom kj}-\gamma\right) w_{\mu\cup1^{n-k}}\ge0, \qquad j=k-\ell(\mu),\quad s=k-m_1(\mu). \end{equation} Here $\mu\cup1^{n-k}$ appends $n-k$ parts equal to one to $\mu$; we call the shapes of this form \emph{selected}, since no other shape enters $J_{n,k,\gamma}$. For the shape $(n)$, the coefficient of $x_1^{n-2}$ in the $k=2$, $\gamma_2=0$ inequality gives \begin{equation}\label{eq:R} R_n:=\frac{w_{(n-1,1)}}{n-1}-w_{(n)}\ge0\qquad(n\ge3). \end{equation} The coefficient extractions are verified in Section~\ref{app:rows}. \subsection{Projection spectra} \label{sec:projection-spectra} Lemma~\ref{lem:mixed} uses only positivity of $H$. Positivity and bistochasticity together also constrain the images of projections. For a rank-$a$ projection $P$, they imply $0\preceq H(P)\preceq I$ and $\tr H(P)=a$. Averaged over Haar projections of rank $a$, the characteristic coefficients $e_j(H(P))$ are linear in the weights (Section~\ref{app:finite-certificates}), so these spectra give a second family of valid constraints, used by the finite programs of Section~\ref{sec:finite} and by the logarithmic improvement of Section~\ref{sec:mixed-structure}. The convex hull of the vectors $(e_0(x),\ldots,e_n(x))$ over all spectra $0\le x_i\le1$, $\sum_i x_i=a$, is generated by the characteristic-coefficient vectors of the following spectra: \begin{equation}\label{eq:spectral-generators} (1^a,0^{n-a})\quad\text{and}\quad \left(1^u,\left(\frac{a-u}{v}\right)^v,0^{n-u-v}\right), \quad 0\le u20/3$, monotonicity \eqref{eq:transport-gain} gives $h_n(271/40)>h_n(20/3)$. For $A=20/3$ and $B=103/3$, \[ h_n(A)+\frac n{n^2+An+B} =\frac{(211/9)n+2369/9} {(n(n-1)+A)(n^2+An+B)}>0. \] This inequality proves \eqref{eq:main} for $n\ge24$. For $8\le n\le23$, the seeds of Table~\ref{tab:seeds} and the bound transported from dimension $21$ exceed the right side of \eqref{eq:main}; these finite comparisons are part of the supporting checks. \end{proof} \section{Logarithmic improvement and limits of the coefficient relaxation} \label{sec:mixed-structure} With a fixed seed, Theorem~\ref{thm:transport} gives $\Phi_n\ge-1/n+K/n^2+O(n^{-3})$ for a constant $K$. The logarithmic bound \eqref{eq:log-main} needs $K$ to grow like $\log n$. The mixed rows alone cannot produce this growth, since their relaxation reproduces the transported bound exactly. The projection spectra can, through a small strict gain at every step. \subsection{The mixed relaxation and its projection obstruction} The finite programs use both mixed rows and projection spectra. To separate their roles, we first retain every supplied mixed row but omit the spectral constraints. This smaller set of conditions has an explicit unique minimizer, and its failure to come from a map identifies the information that projection spectra add. Let $\mathcal G$ be a finite collection of seeds $(k,\gamma_k)$ with $\gamma_k\in[-1/k,0]$, including $(2,0)$ and $(3,0)$, and let $k_{\max}$ be its largest dimension. Each seed is a valid lower bound on $\mathfrak B_k$. For $n>k_{\max}$, define $\mathcal R_n(\mathcal G)$ to be the symmetric homogeneous degree-$n$ polynomials $f=\sum_{\lambda\vdash n}w_\lambda q_\lambda$ satisfying \[ w_\lambda\ge0,\qquad \sum_\lambda w_\lambda=1,\qquad (\D_k-\gamma_k\T_k)f\ge_{\mathrm{coeff}}0 \quad((k,\gamma_k)\in\mathcal G). \] All monomial coefficients of the last inequalities are imposed. Put \[ K=\max_{(k,\gamma_k)\in\mathcal G}K(k,\gamma_k),\qquad N=n(n-1). \] The two zero seeds ensure $2\le K\le k_{\max}-1$. \begin{theorem}[Unique mixed-coefficient extremizer]\label{thm:mixed-extremizer} For $n\ge k_{\max}+3$, \[ \min_{f\in\mathcal R_n(\mathcal G)}\mathcal L_n(f)=h_n(K). \] The unique minimizer is \begin{equation}\label{eq:two-shape} f_{n,K}(z)=\frac{K e_n(z)+m_{(2,1^{n-2})}(z)}{N+K} =\frac{e_1(z)e_{n-1}(z)-(n-K)e_n(z)}{N+K}. \end{equation} The minimum and its unique minimizer are unchanged if one also imposes the square-free floor $w_{(1^n)}\ge n!/n^n$ or the scalar polynomial capacity condition $\operatorname{Cap}(f)\ge1$, where \[ \operatorname{Cap}(f)=\inf_{z_1,\ldots,z_n>0} \frac{f(z)}{\prod_i z_i}. \] This scalar capacity is distinct from the map capacity in \eqref{eq:capacity}. \end{theorem} \begin{proof}[Proof idea] Transport gives the lower bound using only the mixed rows. The full coefficient identity \eqref{eq:full-mixed-identity} shows that $f_{n,K}$ satisfies every supplied row. Its two nonzero weights have ratio $K:N$. Strict positivity of all remaining transport residuals, together with $R_n\ge0$ in even dimension, forces this support at equality. The full proof, including scalar capacity and the square-free floor, is in Appendix~\ref{app:relaxation-proofs}. \end{proof} Projection spectra impose a different restriction, which uses both marginals of the map and explains why the preceding minimizer cannot attain a negative value for an actual map. \begin{prop}[Projection obstruction]\label{prop:projection-obstruction} If $H\in\mathfrak B_n$, $n\ge4$, and $f_H=f_{n,K}$ with $K\ge0$, then \begin{equation}\label{eq:projection-K-threshold} K\ge \begin{cases} n,&n\text{ even},\\ n-1,&n\text{ odd}. \end{cases} \end{equation} In particular, $f_{n,K}$ is not realizable when $h_n(K)<0$. \end{prop} \begin{proof} For a Haar rank-$a$ projection $P$, put $Y=H(P)$. Its eigenvalues lie in $[0,1]$ and sum to $a$. Substitution in \eqref{eq:two-shape} gives \begin{equation}\label{eq:two-shape-projection} \E\det(tI+Y)=t^{n-a-1}(t+1)^{a-1} \left(t^2+t+\frac{a(n-a)}{N+K}\right) \qquad(1\le ah_n(K)$. In particular, $\mu_n>h_n(271/40)$ for $n\ge27$. \end{corr} \begin{proof} Every actual map has a feasible orbital polynomial. The unique minimizer of the relaxation is excluded by Proposition~\ref{prop:projection-obstruction}, since $K\le k_{\max}-1s$, coefficientwise nonnegativity of $\D_2f_H$ implies \[ w_{\mu\cup1^{n-s}}\le\beta_{\mu,p}(n) w_{\mu'\cup1^{n-s-1}}, \] where \[ \beta_{\mu,p}(n)=\frac{2p}{(p-1)(n-s)m_p} \begin{cases} (m_2+1)(m_2+2),&p=3,\\ (m_{p-1}+1)(m_2+1),&p\ge4. \end{cases} \] Thus every fixed nonbinary shape has weight $O(n^{-1})$ relative to another fixed nonleading shape. \end{lemma} \begin{proof} For a padded partition $\lambda$, write $a_\lambda=w_\lambda/|\mathcal O_\lambda|$ for its ordinary monomial coefficient. Fix the inactive monomial whose nonunit exponents are $\mu\setminus(p)$ and whose other nonzero exponents are $n-s-1$ ones. The active coefficient of $x^{p-2}y$ in $\D_2f_H$ is \[ 2p\,a_{\mu\setminus(p)\cup(p-1,2)} -(p-1)a_{\mu\setminus(p)\cup(p,1)} -(p+1)a_{\mu\setminus(p)\cup(p+1,0)}\ge0. \] Indeed $\D_2f=\partial_x\partial_yf+ (\partial_yf-\partial_xf)/(x-y)$. For active exponents summing to $p+1$, only the three pairs displayed can contribute to $x^{p-2}y$; direct differentiation gives their coefficients. The count includes $p=3$, when the positive pair is $(2,2)$ and its coefficient is six. Drop the last, nonpositive contribution. The two remaining padded shapes have the same number of zero entries; the positive shape has one fewer entry equal to one. Their orbit-size ratio is exactly the multiplicity factor in the displayed formula, divided by $(n-s)m_p$, which proves the claim. \end{proof} The projection constraints enter each step through a bounded test function $\psi$ on the integers. For a rank-$a$ projection $P$, the eigenvalues of $H(P)$ lie in $[0,1]$ and sum to $a$. Read as success probabilities, they define a Poisson-binomial variable $S$ of mean $a$, and after averaging over Haar projections, $\E\psi(S-a)$ is linear in the weights. If $\E\psi(S-a)\ge0$ for every Poisson-binomial $S$ of integer mean $a$, then $\psi$ is a valid \emph{projection separator}, and this expectation is a valid constraint. The next result shows that the logarithmic coefficient depends on $\psi$ only through its averages on binary shapes, which separates the coefficient estimate from the construction of $\psi$. \begin{prop}[Binary reduction of the logarithmic coefficient]\label{prop:binary-bootstrap} Let $\psi:\mathbb Z\to\mathbb R$ be a bounded symmetric function with $\psi(0)=1$ and $\psi(1)=-1$. Suppose it is a valid projection separator: $\mathbb E\psi(S-a)\ge0$ for every Poisson-binomial random variable $S$ of integer mean $a$. Put \[ B=\max\{1,\sup_t\psi(t)\},\qquad g_r=4^{-r}\sum_{j=-r}^r\binom{2r}{r+j}\psi(j),\qquad M=\sup_{r\ge2}\frac{g_r}{2r-1}. \] Assume $M>0$. Then the mixed coefficient inequalities and projection constraints imply \[ \Phi_n(H)\ge-\frac1n+ \frac{M^{-1}\log n-O(1)}{n^2}. \] It is enough to verify the value of $M$ on finitely many binary shapes: for any integer $L$ with $B/LL$ satisfy $g_r/(2r-1)\le B/L3/M$ and start sufficiently far out with any transported positive seed parameter $K_0$. In each step, use the correction \[ \varepsilon=\min\left\{ \frac{4\Delta}{\alpha+Q_{n,k,K}},\frac A{n^3}, -\frac12h_n(K)\right\}, \] convert the corrected seed back to its transport parameter, and round down to a multiple of $n^{-2}$. The cap preserves a negative seed and gives an upper increment $(A+1)/n$ for all sufficiently large $n$. Rounding loses less than $n^{-2}$, so the parameters stay positive and are $O(\log n)$. The preceding estimates therefore apply; eventually both caps are inactive. The parameter increment is \[ \frac3{Mn}+O(\log n/n^2). \] Summing in steps of three gives $K=M^{-1}\log n+O(1)$. Transport fills the intervening dimensions and proves the claim. \end{proof} \subsection{The parity separator and the logarithmic coefficient} By Proposition~\ref{prop:binary-bootstrap}, a valid separator with ratio $M$ gives the logarithmic coefficient $1/M$, so we want $M$ as small as validity allows. Balanced binary shapes have centered binomial laws, and their parity moments suggest a quadratic separator for which every binary ratio $g_r/(2r-1)$ is the same. The mean-one boundary of the projection hull determines the smallest admissible value of that ratio. \begin{prop}[Bounded separators approaching the limiting coefficient] \label{prop:optimal-clipped-separator} Put $c_{\log}=3e^2-7$ and $M_*=c_{\log}^{-1}$. For every $M>M_*$ there is a bounded even function $\psi:\mathbb Z\to\mathbb R$ such that \[ \psi(0)=1,\qquad \psi(\pm1)=-1,\qquad \mathbb E\psi(S-a)\ge0 \] for every Poisson-binomial random variable $S$ of integer mean $a$. Moreover, for $X_r\sim\operatorname{Bin}(2r,1/2)$, \begin{equation*} \sup_{r\ge2}\frac{\mathbb E\psi(X_r-r)}{2r-1}=M. \end{equation*} \end{prop} \begin{proof} Consider the even function \begin{equation}\label{eq:parity-quadratic-separator} \psi_M(d)= \begin{cases} 1+6Md^2,&d\text{ even},\\ 2M(d^2-1)-1,&d\text{ odd}. \end{cases} \end{equation} Since $M>M_*>1/16$, its values at $|d|\ge2$ are nonnegative and $\psi_M(2)>2$. For $D=X_r-r$, $r\ge2$, the binomial generating function gives \[ \mathbb E D^2=r/2,\qquad \mathbb E(-1)^D=\mathbb E[D^2(-1)^D]=0. \] Since $\psi_M(d)=M(4d^2-1)+(1+M+2Md^2)(-1)^d$, these identities imply \[ \mathbb E\psi_M(X_r-r)=M(2r-1). \] To check validity, the projection-spectrum hull reduces the problem to the centered binomials $X-r$, with $X\sim\operatorname{Bin}(r+s,r/(r+s))$ and $r,s\ge1$, and the deterministic law at zero. The binomial estimates in Section~\ref{app:optimal-logarithm} show that the generators with $r,s\ge2$ are valid whenever $\psi(0)=1$, $\psi(\pm1)=-1$, $\psi(\pm2)\ge2$, and all other values are nonnegative. The remaining generators, up to reflection, are $\operatorname{Bin}(m,1/m)-1$. For the unclipped separator $\psi_M$ their limiting expectation is \[ \lim_{m\to\infty}\mathbb E\psi_M (\operatorname{Bin}(m,1/m)-1) =3M-e^{-2}(1+7M). \] It vanishes at $M=M_*=(3e^2-7)^{-1}$ and is negative for $M0 \qquad(m\ge3,\ M>M_*). \] The generator with $m=2$ has expectation zero. The strict margin permits bounded clipping. For an even integer $L\ge2$, put $\psi(d)=\min\{\psi_M(d),1+6ML^2\}$. The clipped separator retains every value with $|d|\le L$ and the conditions that prove validity for $r,s\ge2$. For the mean-one generators its loss is at most \[ \sum_{j=L+2}^{\infty}\frac{1+6M(j-1)^2}{j!}, \] uniformly in $m$, because $\Pr(\operatorname{Bin}(m,1/m)=j)\le1/j!$. Choose $L$ so that this tail is smaller than the uniform margin. Then the clipped separator is valid for all generators. Clipping can only decrease the binary expectation and leaves the support at $r=2$ unchanged. Thus its supremum ratio is exactly $M$, as claimed. Section~\ref{app:optimal-logarithm} supplies the full finite-binomial verification used above. \end{proof} Applying Proposition~\ref{prop:binary-bootstrap} to the separators of Proposition~\ref{prop:optimal-clipped-separator} gives the logarithmic improvement. \begin{corr}[Logarithmic improvement]\label{cor:optimal-logarithm} For every fixed $0<\alpha<3e^2-7$, the mixed coefficient inequalities and projection constraints imply \[ \Phi_n(H)\ge-\frac1n+ \frac{\alpha\log n-O_\alpha(1)}{n^2}. \] In particular, the minimum $\mu_n$ of \eqref{eq:minimum} satisfies \[ \liminf_{n\to\infty} \frac{n^2(\mu_n+1/n)}{\log n}\ge3e^2-7. \] \end{corr} \begin{proof} Apply Proposition~\ref{prop:binary-bootstrap} to the bounded separator above with $M=1/\alpha$. \end{proof} With $\alpha=c_{\log}-\varepsilon$, Corollary~\ref{cor:optimal-logarithm} gives \eqref{eq:log-main} for all sufficiently large $n$, and increasing $A_\varepsilon$ covers the remaining dimensions $n\ge21$. The proof of Theorem~\ref{thm:main} is now complete. \subsection{Optimality of the logarithmic coefficient} Corollary~\ref{cor:optimal-logarithm} raises the question whether a better separator could beat $3e^2-7$. Within the recursive programs, it cannot. \begin{samepage} \begin{theorem}[Optimal logarithmic coefficient of the relaxation] \label{thm:optimal-recursive-coefficient} Consider recursive coefficient programs using nonnegative normalized weights, the square-free floor, scalar capacity, all projection-spectrum hull constraints, and all proper-dimensional mixed rows. Initialize them with valid nonpositive lower bounds in a finite initial range of dimensions. Every sequence $\gamma_n$ certified recursively by these programs satisfies \begin{equation*} \limsup_{n\to\infty} \frac{n^2(\gamma_n+1/n)}{\log n}\le c_{\log}=3e^2-7. \end{equation*} Together with \eqref{eq:log-main}, this identifies the optimal limiting logarithmic coefficient for the relaxation. \end{theorem} \end{samepage} The proof, given in the supplement, builds feasible profiles from the same mean-one binomial generators, with weights on the shapes $(2^r,1^{n-2r})$ besides the two leading shapes. These profiles limit what the specified recursive programs can certify from finitely many nonpositive initial bounds; they need not arise from maps. Additional compatibility conditions or independent bounds supplied in unbounded dimensions can change the conclusion. The remaining question is whether actual maps can sustain negativity of order $1/n$. \section{Negative maps and polynomial padding} \label{sec:negative} Negative maps give upper bounds on $\mu_n$, and padding carries a negative map in dimension $r$ to every larger dimension $n$, with invariant of order $n^{-r}$. Small dimensions therefore matter most. Our eight-dimensional construction starts from the unnormalized six-dimensional Breuer--Hall map \citetext{\citealp[(2)--(3)]{breuer2006}; \citealp[Theorem~4]{hall2006}}: \[ J=\begin{pmatrix}0&I_3\\-I_3&0\end{pmatrix},\qquad \mathcal B(Y)=\tr(Y)I_6-Y-JY^{\mathsf T}J^*. \] For every $w\in\C^6$, the vectors $w$ and $J\bar w$ are orthogonal and have the same norm. Hence $\mathcal B(ww^*)=\|w\|^2I_6-ww^*-(J\bar w)(J\bar w)^*\succeq0$. The eight-dimensional map is a sum of three positive components. Let $B:\mathbb C^8\to\mathbb C^6$ project onto the first six coordinates, and let $A:\mathbb C^6\to\mathbb C^8$ satisfy $Ae_i=e_i$ for $1\le i\le5$ and $Ae_6=0$. Put $D=\operatorname{diag}(1,1,\sqrt2,1,1,1,1,1)$. Using one-based matrix units, define \[ \begin{split} K_0&=E_{17}+E_{28}+E_{61}+E_{72}+E_{83},\\ L_0&=E_{47}+E_{58}+E_{64}+E_{75}+E_{86},\\ R(X)&=A\mathcal B(BXB^*)A^*+4K_0XK_0^* +4L_0X^{\mathsf T}L_0^*,\\ H_8(X)&=\tfrac18DR(X)D. \end{split} \] The Breuer--Hall term is positive, and the other two terms are congruences of $X$ and $X^{\mathsf T}$. The marginal and determinant calculations are given in Section~\ref{app:witness:eight-family}. The padding below uses the completely positive summand of $H_8$ corresponding to $K_0$, which is $\Ad_Q$, where \[ Q=K_0/\sqrt2, \qquad H_8-\Ad_Q\text{ is positive}. \] Here $DK_0=K_0$, since $K_0$ has zero third row. The map $H_8$ is one member of a two-parameter family, and Du's map, normalized to be bistochastic, is one member of a nine-dimensional family (Sections~\ref{app:witness:eight-family} and~\ref{app:witness:nine-family}). The minima over the two families are as follows. \begin{prop}[Negative families]\label{prop:negative-summary} There are explicit two-parameter families of positive bistochastic maps on $M_9$ and $M_8$, with parameters $(a,b)\in[-1,1]^2$, whose minima are \[ -\frac1{737280}\approx-1.356\cdot10^{-6}\quad\hbox{and}\quad-\frac{11}{6881280}\approx-1.599\cdot10^{-6}, \] respectively. In both families equality holds exactly when $|a|=|b|=1$. The minimizing maps are sums of three positive components. The negative magnitudes exceed those of the normalized Du example and its corresponding compression by factors $21/2$ and $33/5$, respectively. \end{prop} \begin{proof} Propositions~\ref{prop:negative-family-nine} and \ref{prop:negative-family-eight} give the constructions and the calculations. \end{proof} The dimension of a negative map determines the decay exponent of its pinched paddings. For $H_r\in\mathfrak B_r$, its pinched direct sum with an identity map is \[ (H_r\boxplus\Id_m)\begin{pmatrix}X&Z\\W&Y\end{pmatrix} =\begin{pmatrix}H_r(X)&0\\0&Y\end{pmatrix},\qquad m=n-r. \] The normalized determinant differential gives \begin{equation}\label{eq:padding-main} \Phi_n(H_r\boxplus\Id_{n-r})=\frac{\Phi_r(H_r)}{\binom nr}. \end{equation} Indeed, the output determinant factors into the two block determinants, and the input differential contributes $r!(n-r)!/n!$ after normalization. For the eight-dimensional map, extension through the summand $\Ad_Q$ retains off-diagonal entries and improves the padding coefficient to \[ \mu_n\le-\frac{4947}{50000(n)_8}\qquad(n\ge9) \] (Proposition~\ref{app:witness:bound}). The nine-dimensional family gives $\mu_n\le-1/(737280\binom n9)$, completing the bracket \eqref{eq:bracket}. The family minima do not identify $\mu_8$ or $\mu_9$. \section{Discussion} \label{sec:discussion} The padding exponent singles out the first dimension admitting a negative map. Define \begin{equation}\label{eq:first-negative-dimension} d_*:=\min\{d\in\mathbb Z_{\ge1}:\mu_d<0\}. \end{equation} Nonnegativity in dimensions two and three and the eight-dimensional example give $4\le d_*\le8$; its value is unknown. For any fixed $H_*\in\mathfrak B_{d_*}$ with $\Phi_{d_*}(H_*)<0$, the padding identity \eqref{eq:padding-main} gives \[ -\mu_n\ge\frac{-d_*!\Phi_{d_*}(H_*)}{(n)_{d_*}} \qquad(n\ge d_*). \] We conjecture a matching upper bound on $-\mu_n$. \begin{conj}[Order of the universal minimum] \label{conj:universal-order} There is a constant $C>0$, independent of $n$ and $H$, such that for every integer $n\ge d_*$ and every $H\in\mathfrak B_n$, \begin{equation*} \Phi_n(H)\ge-\frac{C}{(n)_{d_*}}. \end{equation*} \end{conj} Together with padding, the conjecture is equivalent to $-\mu_n=\Theta(n^{-d_*})$. Theorem~\ref{thm:optimal-recursive-coefficient} excludes certifying this order with the specified recursive programs initialized by finitely many nonpositive bounds. Determining the unresolved signs of $\mu_4,\ldots,\mu_7$ would fix the exponent in Conjecture~\ref{conj:universal-order}. \subsection*{Other ideas and extensions} Constraints on the map itself, rather than on its orbital weights, give partial evidence for Conjecture~\ref{conj:universal-order}. We collect them, with proofs, in a supplement released with this paper. The conjectured order holds for maps with an entanglement-breaking submap of normalized trace at least $(d_*-1)\log n/n$, and for coherent extensions, which couple a core of fixed dimension to the added coordinates, under spectral restrictions on the coupling. For general coherent extensions, the conjecture requires a noncentral determinant polynomial to be nonnegative, which remains open. For arbitrary maps, $\Phi_n(H)\le-c/n$ with fixed $c>0$ forces $\sum_{i,j}\|H(E_{ij})\|_{\HS}^2=n^2-O(n)$, a contraction-radius bound gives $\mu_2=1/4$, and every negative global minimizer satisfies explicit first-order conditions. The supplement also proves Theorem~\ref{thm:optimal-recursive-coefficient}, and two obstructions for weaker relaxations, showing that the mixed rows alone cannot certify a bound above $-1/(n^2-2)$ and that bounds for individual balanced determinant tuples cannot improve the order $1/n$. \begin{aiusage} % The results in this paper were developed through extensive use of an AI assistant (OpenAI's GPT~Pro). None of the claims were obtained zero-shot: they emerged through sustained back-and-forth discussion, including failed attempts, counterexamples, audits, and repairs. The author also used AI assistants (GPT and Claude) to help polish the writing. \end{aiusage} \aiacknowledgements \begin{thebibliography}{12} \providecommand{\natexlab}[1]{#1} \providecommand{\url}[1]{\texttt{#1}} \expandafter\ifx\csname urlstyle\endcsname\relax \providecommand{\doi}[1]{doi: #1}\else \providecommand{\doi}{doi: \begingroup \urlstyle{rm}\Url}\fi \bibitem[Alexandrov(1938)]{alexandrov1938} A.~D. 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The set $\mathfrak B_n$ is compact, and the minimum in \eqref{eq:minimum} is attained. \end{lemma} \begin{proof} For Hermitian $X$, the unital positive map satisfies $H(X)^2\preceq H(X^2)$. One can see this directly from a spectral decomposition $X=\sum_i x_iP_i$: the matrices $H(P_i)$ form a positive operator-valued partition of identity, so the map $v\mapsto(H(P_i)^{1/2}v)_i$ is an isometry, and compression of $\operatorname{diag}(x_iI)$ gives the asserted inequality. Trace preservation now implies $\|H(X)\|_{\HS}\le\|X\|_{\HS}$. For Hermitian $X,Y$, one has $\|X+iY\|_{\HS}^2=\|X\|_{\HS}^2+\|Y\|_{\HS}^2$, and the same identity holds for their images. Thus the contraction extends to all matrices, so $\mathfrak B_n$ is bounded. Positivity and the two marginal identities are closed conditions, and $\Phi_n$ is a polynomial in the entries of $H$. \end{proof} Write $\Ad_A(X)=AXA^*$. To prove Lemma~\ref{lem:normalization}, we use the covariance rule \begin{equation}\label{eq:covariance} \Phi_k(\Ad_A\circ K\circ\Ad_B) =|\det A|^2|\det B|^2\Phi_k(K). \end{equation} Indeed, the output congruence contributes $|\det A|^2$, while the chain rule for the input congruence contributes $|\det B|^2$. The proof of Lemma~\ref{lem:normalization} rescales $K$ so that both marginals become the identity. The right input scale minimizes the determinant over inputs of unit determinant, which defines the capacity of a positive map, \begin{equation}\label{eq:capacity} \operatorname{cap}(K)=\inf_{X\succ0,\ \det X=1}\det K(X). \end{equation} The determinant constraint removes scalar rescaling of the input. \begin{proof}[Proof of Lemma~\ref{lem:normalization}] First assume $K(X)\succeq c\tr(X)I$ for positive semidefinite $X$, with $c>0$. Minimize $\det K(X)$ over $X\succ0$ with $\det X=1$. The estimate $\det K(X)\ge c^k(\tr X)^k$ bounds the trace on sublevel sets; the determinant constraint then bounds every eigenvalue away from zero. A minimizer $X_0$ exists. Put $C=K(X_0)$. Differentiating $\log\det K(X)$ under $\log\det X=0$ gives $K^*(C^{-1})=\eta X_0^{-1}$. Pairing with $X_0$ gives $k=\eta k$, so $\eta=1$. Define \[ L(Z)=C^{-1/2}K(X_0^{1/2}ZX_0^{1/2})C^{-1/2}. \] The scaled map has $L(I)=L^*(I)=I$ and is positive. Since $\det X_0=1$, covariance \eqref{eq:covariance} yields \[ \Phi_k(K)=\det C\,\Phi_k(L) \ge\gamma\det C\ge\gamma\det K(I), \] where the last step uses $\det C\le\det K(I)$ and $\gamma\le0$. Apply this argument to $K+\varepsilon\tr(\cdot)I$ and let $\varepsilon\downarrow0$ to obtain the general case, including singular $K(I)$. \end{proof} The sign condition in Lemma~\ref{lem:normalization} is essential. The positive map $K(X)=X_{11}I_2$ has $K(I)=I_2$ and $\Phi_2(K)=0$, so $\Phi_2(K)\ge c\det K(I)$ fails for every $c>0$, and no positive constant can replace $\gamma_2$ in the lemma. We use the seed bounds $\gamma_2=\gamma_3=0$. To obtain the second seed, apply Wan's Theorem~0.3 \citep{wan2026} to $3(H+\varepsilon\Delta_3)/(1+\varepsilon)$, where $\Delta_3(X)=\tr(X)I/3$. Its two marginals are $3I$, as required there, and it sends every nonzero positive semidefinite matrix to a positive definite matrix. Homogeneity and continuity give $\Phi_3(H)\ge0$ as $\varepsilon\downarrow0$. For the first seed, the map on $M_3$ that sends a $2+1$ block matrix to $\operatorname{diag}(H(X),y)$ is positive and bistochastic; direct expansion gives its invariant as $\Phi_2(H)/3$. Thus $\Phi_2(H)\ge0$ as well. The reduction-transpose map $R_3(X)=(\tr(X)I-X^{\mathsf T})/2$ is bistochastic, and a brief calculation gives $\Phi_3(R_3)=0$, so $\mu_3=0$; its pinched direct sums with identity maps also give $\mu_k\le0$ for every $k\ge3$. \subsection{Radial determinant calculus and the orbital score} \label{app:radial} The score identity and the radial operator $\D_k$ rest on one computation, the action of $\det(\partial)$ on conjugation-invariant polynomials, which we carry out on Schur polynomials. Define the Schur polynomials in $k$ variables by \[ s_\lambda(x)= \frac{\det(x_i^{\lambda_j+k-j})_{i,j=1}^k} {\det(x_i^{k-j})_{i,j=1}^k}\qquad(\ell(\lambda)\le k). \] The alternating numerator is divisible by the denominator, so this quotient is a polynomial, including at repeated coordinates. For a matrix $X$, $s_\lambda(X)$ and $e_j(X)$ denote the corresponding polynomials of its eigenvalues. In particular, $s_{(1^k)}(X)=\det X$ on $M_k$. We use the complex Fischer inner product, conjugate-linear in the first argument, \[ \langle f,g\rangle_F =\sum_\alpha\alpha!\,\overline{f_\alpha}g_\alpha =\E[\overline{f(G)}g(G)]. \] Here $f_\alpha$ is the coefficient of the monomial with exponent multi-index $\alpha$, and $\alpha!$ is the product of its entrywise factorials. The entries of $G$ are independent circular complex Gaussians with $\E|G_{ij}|^2=1$. For the determinant polynomial $p(X)=\det X$, \begin{equation}\label{app:energy-pairing} \|p\|_F^2=n!,\qquad \Phi_n(H)=\frac{\langle p,p\circ H\rangle_F}{n!}. \end{equation} The first equality follows from the $n!$ determinant monomials; the second is exactly the differential convention in \eqref{eq:phi}. The Gaussian realization is the coefficient pairing used in Gurvits's quantum-permanent calculus \citep[Theorem~3.8 and Appendix~C]{gurvits2004}. A conjugation-invariant holomorphic matrix polynomial is determined by its diagonal restriction: the spectral theorem determines it on Hermitian matrices, and a polynomial vanishing on all Hermitian matrices vanishes identically. We can therefore check the radial identity on the Schur polynomials. Their Fischer norms on $M_k(\C)$ satisfy \[ \langle s_\lambda,s_\mu\rangle_F =\mathbf1_{\lambda=\mu}\prod_{(i,j)\in\lambda}(k+j-i). \] The product runs over the cells $1\le i\le k$, $1\le j\le\lambda_i$, and $\mathbf1_{\lambda=\mu}$ is one when the partitions agree and zero otherwise. Multiplication by $\det$ is adjoint to $\det(\partial)$, while $\det\,s_\mu=s_{\mu+(1^k)}$. Hence both the matrix differential and the alternant operator act by \begin{equation}\label{eq:schur} \D_k s_\lambda =\prod_{i=1}^k(\lambda_i+k-i)s_{\lambda-(1^k)}, \end{equation} with value zero when $\lambda_k=0$. To justify the norm identity, let $d=\sum_i\lambda_i$, let $\chi^\lambda$ be the irreducible character of $\symm_d$ indexed by $\lambda$, meaning the trace of each permutation in that representation, and put $p_\pi(X)=\prod_c\tr(X^{|c|})$, with the product over the cycles of $\pi$. Here $\chi^\lambda(1)$ is the dimension of that representation. The Frobenius formula and Gaussian contraction give \[ s_\lambda(X)=\frac1{d!}\sum_{\pi\in \symm_d}\chi^\lambda(\pi)p_\pi(X), \qquad \langle p_\pi,p_\sigma\rangle_F =\sum_{\tau\in \symm_d}k^{\#\mathrm{cycles}(\pi^{-1}\tau\sigma\tau^{-1})}. \] Character orthogonality reduces the pairing to $\mathbf1_{\lambda=\mu}(d!/\chi^\lambda(1))s_\lambda(1^k)$. For each cell $(i,j)$ of the partition, put $h_{ij}=\lambda_i-j+\lambda'_j-i+1$, where $\lambda'_j$ counts the parts at least $j$. The hook formulas $\chi^\lambda(1)=d!/\prod_{(i,j)}h_{ij}$ and $s_\lambda(1^k)=\prod_{(i,j)}(k+j-i)/h_{ij}$ give the displayed product \citep[Chapter~I, \S\S3, 7]{macdonald1995}. Thus the adjoint calculation proves the matrix differential formula. Differentiating the alternant in the definition of $s_\lambda$ gives the same multiplier for $\D_k$. Since Schur polynomials span symmetric polynomials, the two operators agree on every invariant matrix polynomial. \begin{proof}[Proof of Proposition~\ref{prop:orbital}] Let $\widehat f_H(X)=\E_U\det H(UXU^*)$, whose diagonal restriction is $f_H$. The determinant differential is invariant under conjugation of the input, so averaging does not change its value. For conjugation-invariant matrix polynomials of degree $n$, the radial identity \eqref{eq:schur}, proved above, shows that $\det(\partial_X)/n!$ extracts the coefficient of $s_{(1^n)}$. The monomial-to-Schur coefficient and the normalization are \[ [s_{(1^n)}]m_\lambda =\frac{(-1)^{n-\ell(\lambda)}\ell(\lambda)!}{\prod_jm_j(\lambda)!}, \qquad m_\lambda(1^n)=\frac{n!}{(n-\ell(\lambda))!\prod_jm_j(\lambda)!}. \] Their quotient gives \eqref{eq:score}; the Schur identities follow from the Frobenius character formula \citep[Chapter~I, \S7]{macdonald1995}. The square-free coefficient means $[x_1\cdots x_n]f_H$. For each basis, Gurvits's theorem bounds this coefficient below by $n!/n^n$ \citep[Theorem~4.1]{gurvits2006}. Averaging preserves this inequality. \end{proof} \subsection{The two coefficient rows used by transport} \label{app:rows} For the row $J_{n,k,\gamma}$ in \eqref{eq:J}, only shapes $\mu\cup1^{n-k}$ can leave every inactive variable to the first power. Their active polynomial is $m_\mu/m_{\mu\cup1^{n-k}}(1^n)$; the ratio of its evaluations is \[ \frac{m_\mu(1^k)}{m_{\mu\cup1^{n-k}}(1^n)} =\frac{(k)_s}{(n)_s}. \] On degree-$k$ polynomials, $\D_k/k!$ gives the signed score $(-1)^j/\binom kj$, while $\T_k/k!$ evaluates the polynomial at $1^k$. These evaluations prove every factor in \eqref{eq:J}. For the row $R_n$, take the coefficient of $x_1^{n-2}$ in $\D_2f_H$. Only the active monomials of types $(n)$ and $(n-1,1)$ contribute. The coefficients of $x_1^{n-2}$ in their two-variable radial derivatives are $-n$ and $n$; division by $m_{(n)}(1^n)=n$ and $m_{(n-1,1)}(1^n)=n(n-1)$ proves \eqref{eq:R}. \subsection{Verification of dimension transport} \label{app:transport-proof} \begin{proof}[Proof of Theorem~\ref{thm:transport}] Put $N=n(n-1)$, $a=1-\gamma$, $b=(k-1)(1+k\gamma)$, $D=aN+b$, and $t=(n^2-1)/D$. Define residuals $c_\lambda$ by \begin{equation}\label{eq:transport-certificate} \Phi_n(H)-h_n(K) =tJ_{n,k,\gamma} +\mathbf1_{n\ \mathrm{even}}\bigl(1/n+h_n(K)\bigr)R_n +\sum_{\lambda\vdash n}c_\lambda w_\lambda. \end{equation} Here $\mathbf1_{n\ \mathrm{even}}$ equals one for even $n$ and zero otherwise. More explicitly, if $J_\lambda$ and $R_\lambda$ are the coefficients of $w_\lambda$ in \eqref{eq:J} and \eqref{eq:R}, then \[ c_\lambda=r_\lambda-h_n(K)-tJ_\lambda -\mathbf1_{n\ \mathrm{even}}(1/n+h_n(K))R_\lambda. \] The identity uses $\sum_\lambda w_\lambda=1$. As in the proof idea after Theorem~\ref{thm:transport}, these choices of $t$ and $K$ make the coefficients of $w_{(1^n)}$ and $w_{(2,1^{n-2})}$ in \eqref{eq:transport-certificate} vanish. Since $0\le K\le k-10. \] At $\gamma=-1/k$, the target is $-1/n$. For a selected shape, the residual condition is equivalent to $A_{j,s}(k)\le A_{j,s}(n)$, where \[ A_{j,s}(x)=\frac{(x)_s}{x+1} \left(1+\frac{x(-1)^j}{\binom xj}\right). \] Here $(x)_s=x(x-1)\cdots(x-s+1)$ and $\binom xj=(x)_j/j!$ for real $x$. The two leading cases are constant. For even $j\ge2$, \[ A_{j,s}(x)=\frac{(x)_s}{x+1} +\frac{j!x}{x+1}(x-j)_{s-j}, \] a sum of nonnegative increasing functions on $x\ge s$. For odd $j\ge3$, \[ A_{j,s}(x)=\frac{(x)_s}{x+1} \left(1-\frac{j!}{(x-1)_{j-1}}\right), \] a product of nonnegative increasing functions there. Every unselected score is at least $-1/n$. For even $n$, the $R_n$ term cancels the residual at $(n)$. The other affected shape, $(n-1,1)$, is unselected when $n\ge k+3$, and its final residual is \[ \frac{(n+1)(n-K)}{n(N+K)}>0. \] The case analysis now covers every shape. Affine interpolation proves nonnegativity for all $\gamma\in[-1/k,0]$. For a fixed seed, a finite test can lower the starting dimension. For a selected shape with parameters $(j,s)$, before the parity correction, the residual multiplied by the positive factor $D\,(n)_s$ is the polynomial \begin{equation}\label{eq:transport-polynomial} \begin{split} P_{j,s}(n)={}&(-1)^j j!D\,(n-j)_{s-j} +(an-a-b)(n)_s\\ &-(n^2-1)(k)_s\left(\frac{(-1)^j}{\binom kj}-\gamma\right). \end{split} \end{equation} For $j=s=0$ read $(n)_0=1$; the other pairs satisfy $1\le j0. \] These checks prove the claimed finite test, including the possible boundary case $n_0=k+1$. \end{proof} \subsection{Exact finite certificates} \label{app:finite-certificates} \subsubsection{The rational model} The projection constraints of the program \eqref{eq:lp} come from evaluating $f_H$ at $t1^n+1_{[a]}$, where $1_{[a]}$ has its first $a$ entries equal to one and the rest zero. For $\lambda$ padded with zeros, set \begin{equation}\label{eq:projection-profile} B_{a,j,\lambda}=\binom na^{-1} \sum_{|I|=a}\binom{\sum_{i\in I}\lambda_i}{j}. \end{equation} The expression in \eqref{eq:projection-profile} is the coefficient of $t^{n-j}$ in $q_\lambda(t1^n+1_{[a]})$: a monomial whose distinguished exponents sum to $b$ contributes $t^{n-b}(t+1)^b$. On the map side, the same substitution gives \[ f_H(t1^n+1_{[a]}) =\E_U\det\bigl(tI+H(UP_aU^*)\bigr),\qquad P_a=\operatorname{diag}(1_{[a]}). \] Thus its coefficients are averages of the characteristic coefficients of projection images. Since their vectors lie in the hull of Section~\ref{sec:projection-spectra}, one common convex mixture represents all degrees $j$ for a fixed $a$. If $\mathcal V_{n,a}$ is the finite list in \eqref{eq:spectral-generators}, the spectral constraints are \begin{equation}\label{eq:spectral-lp} \sum_\lambda B_{a,j,\lambda}w_\lambda =\sum_{v\in\mathcal V_{n,a}}e_j(v)\pi_{a,v},\qquad \pi_{a,v}\ge0,\qquad\sum_v\pi_{a,v}=1. \end{equation} It suffices to use $1\le a\le\lfloor n/2\rfloor$ and $2\le j\le n$. The degrees zero and one are fixed by normalization and trace preservation. The complementary constraints follow from $H(I-P)=I-H(P)$, because the coefficients of $\det(tI+I-H(P))$ are linear combinations of those of $\det(tI+H(P))$. Together with the weights, square-free floor and inherited mixed rows, these constraints give the program \eqref{eq:lp}. \subsubsection{Certified seeds} \begin{prop}[Finite certificates]\label{prop:finite} For $4\le n\le21$, every $H\in\mathfrak B_n$ satisfies $\Phi_n(H)\ge\gamma_n$, with $\gamma_n$ as in Table~\ref{tab:seeds}. In dimensions four through seven, $\gamma_n$ is the optimum of the program \eqref{eq:lp} seeded with the preceding values. In dimensions eight through twenty-one, $\gamma_n$ is a downward rounding of an exactly feasible dual objective, and no optimality is asserted. For $k\in\{18,19,20,21\}$, the transported bound $\Phi_n(H)\ge h_n(K(k,\gamma_k))$ holds for every $n\ge k$. \end{prop} \begin{proof} The coefficient formulas \eqref{eq:schur}, \eqref{eq:projection-profile}, and \eqref{eq:spectral-lp} specify the rational data $(A,b,c)$ of \eqref{eq:lp}. As in Section~\ref{sec:finite}, a rational vector $y$ with $A^{\mathsf T}y\le c$ then proves $\Phi_n(H)\ge b^{\mathsf T}y$ by finite checks in exact rational arithmetic. A floating-point optimizer only proposed each dual vector; rational reconstruction and exact feasibility decide whether it certifies a bound. In dimensions four through seven, a rational primal vector with $Ax=b$, $x\ge0$, and the same objective value certifies the optimum. In dimensions eight through twenty-one, the exact dual objective is rounded down to the displayed $\gamma_n$, and the later programs use the rounded value as their seed. For the transported tails, the finite test in the proof of Theorem~\ref{thm:transport} is run from $n_0=k+1$: the inequality \eqref{eq:transport-unselected-test}, and nonnegativity of every coefficient of the shifted polynomials \eqref{eq:transport-polynomial} for the pairs $(j,s)$ with $1\le j0} \frac{d_\lambda+d_\circ R_\lambda}{c_\lambda}. \] The coefficients $d_\lambda$ vanish at the two leading shapes. Since $R_{(n)}=-1$, the adjusted coefficient also vanishes at $(n)$. The maximum is a rational number, attained at $\lambda=(6,1^{18})$. Set $s_\lambda=M c_\lambda-d_\lambda-d_\circ R_\lambda$. Every $s_\lambda$ is nonnegative, and summing the coefficient identities gives \begin{equation}\label{eq:seed24-identity} \begin{split} (\alpha_0+M)(\Phi_n(H)-\gamma^\sharp) =B(H)+MtJ_{n,k,\gamma} +(M\eta+d_\circ)R_n +\sum_{\lambda\vdash n}s_\lambda w_\lambda, \end{split} \end{equation} where $\gamma^\sharp=h+4\Delta/(\alpha_0+M)$. All multipliers on the right are nonnegative, and the exact value satisfies $\gamma^\sharp>-649/22351$. The $145$ spectral inequalities and the $1575$ coefficient identities in \eqref{eq:seed24-identity} are rational checks; they are recorded, with the exact values of $M$ and $\gamma^\sharp$, in the supporting files. For the rounded seed $-649/22351$, all $145$ transport polynomials in \eqref{eq:transport-polynomial} have nonnegative coefficients after the shift $n=25+x$. The unselected-shape test and the parity residual also pass exactly. These checks prove the transported bound for $n\ge25$; the seed itself supplies $n=24$. \end{proof} \subsubsection{Two further consequences} A stronger positivity assumption supplies further zero seeds. A map is $k$-positive if $\Id_{M_k}\otimes H$ is positive, and completely positive if this holds for every $k$. In finite dimensions, complete positivity is equivalent to a representation $H(X)=\sum_a A_aXA_a^*$. The map is $k$-copositive if $H\circ\mathsf T$ is $k$-positive, where $\mathsf T(X)=X^{\mathsf T}$; complete copositivity is defined in the same way. For a map with domain $M_k$, $k$-positivity already implies complete positivity: its Choi matrix is the image of the positive rank-one matrix $\sum_{i,j}E_{ij}\otimes E_{ij}$ under $\Id_{M_k}\otimes H$, and factoring that Choi matrix gives the stated representation. \begin{corr}\label{app:certificates:kpositive} If $H\in\mathfrak B_n$ is $k$-positive or $k$-copositive, $k\ge3$, and $n\ge k+3$, then \[ \Phi_n(H)\ge-\frac{n-k}{n^2-n+k-1}. \] In particular, $(n-3)$-positivity or $(n-3)$-copositivity implies $\Phi_n(H)\ge-3/(n^2-4)$ for $n\ge6$. \end{corr} \begin{proof} The compatible compression $Z\mapsto W^*H(VZV^*)W$ on $M_k$ preserves $k$-positivity and hence is completely positive. Its invariant is nonnegative by the Gaussian determinant formula \citep[Theorem~3.8]{gurvits2004}. In the copositive case, the compression is completely copositive and transposition leaves $\Phi_k$ unchanged. Thus the mixed inequality holds with $\gamma=0$. The coefficient proof of Theorem~\ref{thm:transport} applies with $K=k-1$. \end{proof} The finite seeds also give one constant valid in every dimension. \begin{corr}\label{cor:uniform-floor-improved} For every $n\ge2$ and $H\in\mathfrak B_n$, \[ \Phi_n(H)\ge-\frac{77957}{1378367}\approx-0.05656. \] \end{corr} \begin{proof} Dimensions two and three are nonnegative. Among the seeds in dimensions four through seven, the smallest is $\gamma_5=-77957/1378367$. For $n\ge8$, the right side of \eqref{eq:main} is increasing, since its derivative is $(9n^2-309)/(3n^2+20n+103)^2>0$, and its value at $8$ is $-24/455>\gamma_5$. \end{proof} \section{Proofs for the coefficient relaxations} \label{app:relaxation-proofs} The proofs below supply the estimates behind Section~\ref{sec:mixed-structure}. \subsection{The extremizer of the mixed relaxation} \begin{proof}[Proof of Theorem~\ref{thm:mixed-extremizer}] The coefficient proof of Theorem~\ref{thm:transport} uses only nonnegative normalized weights and the rows $J$ and $R$. It therefore applies to every polynomial in $\mathcal R_n(\mathcal G)$, with a seed attaining $K$, and gives the lower bound. To prove feasibility of \eqref{eq:two-shape} for all rows, split $z=(x,y)$ into $k$ active and $m=n-k$ inactive variables. Write $M_m(y)=m_{(2,1^{m-2})}(y)$ for $m\ge2$, and $M_1=0$. Then \begin{equation}\label{eq:full-mixed-identity} \begin{split} (\D_k-\gamma\T_k)f_{n,K} =\frac{k!}{N+K}\bigl[& (1-\gamma)(K-K(k,\gamma))e_m(y)+(1-\gamma)M_m(y)\\ &+(k+1)(1/k-\gamma)e_1(x)e_{m-1}(y)\bigr]. \end{split} \end{equation} Expand $e_n=e_k(x)e_m(y)$ and $e_{n-1}=e_{k-1}(x)e_m(y)+e_k(x)e_{m-1}(y)$. The required derivatives are \[ \begin{array}{c|cc} p&\D_kp&\T_kp\\\hline e_k&k!&k!\\ e_{k-1}&0&0\\ e_1e_{k-1}&k!&k!k^2\\ e_1e_k&(k+1)(k-1)!e_1&k!(k+1)e_1. \end{array} \] The radial formulas follow from \eqref{eq:schur}; the scalar derivatives follow by substituting $x+t1^k$. Combining them with $e_1(y)e_{m-1}(y)=m e_m(y)+M_m(y)$ proves \eqref{eq:full-mixed-identity}. Its three terms have disjoint active/inactive exponent types. For $\gamma\le0$ and $K\ge0$, every coefficient is nonnegative exactly when $K\ge K(k,\gamma)$. Thus $f_{n,K}$ satisfies every supplied mixed row. Its only weights are \begin{equation}\label{eq:two-shape-weights} w_{(1^n)}=\frac K{N+K},\qquad w_{(2,1^{n-2})}=\frac N{N+K}, \end{equation} whose signed value is $h_n(K)$. For uniqueness, choose $(k,\gamma)\in\mathcal G$ attaining $K$ and use the residual identity \eqref{eq:transport-certificate}. As in that identity, put $D=(1-\gamma)N+(k-1)(1+k\gamma)>0$. At $\gamma=0$ its residuals are strictly positive away from the two shapes in \eqref{eq:two-shape-weights} and the shape $(n)$ in even dimension. In the unselected case this uses the strict inequality $h_n(k-1)<-2/N$ for $n\ge k+3$; the selected cases are the positive expressions in the proof of Theorem~\ref{thm:transport}. At $\gamma=-1/k$ the residuals are nonnegative. Since $K\ge2$, an attaining seed has $\gamma>-1/k$, so affine interpolation preserves strict positivity of the $D$-scaled residuals away from those shapes. Equality therefore removes every positive-residual weight. In even dimension the multiplier of $R_n$ is positive, and $w_{(n-1,1)}=0$; equality in $R_n\ge0$ then removes $w_{(n)}$ as well. The equality $J_{n,k,\gamma}=0$ fixes the remaining weight ratio as $K:N$. Normalization gives \eqref{eq:two-shape-weights}. Finally, for $z_i>0$, \[ \frac{f_{n,K}(z)}{\prod_i z_i} =\frac{K+\sum_{i\ne j}z_i/z_j}{N+K}\ge1, \] by pairing reciprocal ratios; equality holds at $1^n$. Thus $\operatorname{Cap}(f_{n,K})=1$. Also, for $K\ge2$ and $n\ge4$, \[ \frac K{N+K}\ge\frac2{N+2}\ge\frac{n!}{n^n}. \] For the second inequality, check $n=4$ and proceed by induction: the ratio of the left-hand side at consecutive dimensions is at least $1/2$, whereas the ratio for $n!/n^n$ is $(n/(n+1))^n\le1/2$. Hence neither additional condition excludes the unique extremizer. \end{proof} \subsection{Transport residuals and projection coefficients} The recursion of Proposition~\ref{prop:binary-bootstrap} compares projection coefficients with transport residuals, so it needs lower bounds on the residuals and an explicit projection row. \begin{lemma}[A sharper transport residual]\label{lem:sharper-residual} Let $k\ge3$, $n\ge k+3$, $0\le K\le k-1$, and $N=n(n-1)$. Every residual in \eqref{eq:transport-certificate} outside the two leading shapes, except the canceled even-dimensional shape $(n)$, is at least \begin{equation*} \begin{split} \widetilde\rho_{n,k,K} =\frac1{(k-1)(N+K)}\biggl[ &(k-1-K)\frac{(n-3)(n-k)(n+k-2)}{n(n-2)}\\ &+K\left(n-k-2-\frac{2(k-1)}N\right)\biggr]>0. \end{split} \end{equation*} \end{lemma} \begin{proof} At the endpoint $\gamma=-1/k$, the selected-shape residual is \[ c_\lambda=\left(\frac1n+\frac{(-1)^j}{\binom nj}\right) \left(1-\frac{A_{j,s}(k)}{A_{j,s}(n)}\right), \qquad A_{j,s}(x)=\frac{(x)_s}{x+1} \left(1+\frac{x(-1)^j}{\binom xj}\right). \] For nonleading selected shapes $j\ge2$ and $s\ge j+1$. We claim that $A_{j,s}(x)/(x(x-2))$ is nondecreasing for $x\ge s$. For $j=2$ and $s=3$, this ratio is exactly one; larger $s$ append nonnegative increasing factors. For odd $j\ge3$, the ratio is \[ \frac{x-1}{x+1}\prod_{i=3}^{s-1}(x-i) \left(1-\frac{j!}{(x-1)_{j-1}}\right), \] a product of nonnegative increasing functions. For even $j\ge4$ it suffices again to take $s=j+1$. The ratio is then $F(x)+G(x)$, where \[ F(x)=\frac{x-1}{x+1}\prod_{i=3}^{j}(x-i),\qquad G(x)=\frac{j!(x-j)}{(x+1)(x-2)}. \] For $x\ge j+1$, we have \[ \frac FG=\frac{(x-1)_{j-1}}{j!}\ge1,\qquad \frac{F'}F\ge\frac2{x-3},\qquad \frac{G'}G> -\frac2{x-2}. \] Thus $F'+G'>0$, proving the claim. It follows that \[ \frac{A_{j,s}(k)}{A_{j,s}(n)} \le\frac{k(k-2)}{n(n-2)}. \] Since $\binom nj\ge\binom n2$ for the selected range, every such residual is at least \[ \rho_0=\frac{n-3}{n(n-1)} \left(1-\frac{k(k-2)}{n(n-2)}\right). \] Unselected shapes obey the same bound because their score is at least $-2/N$; the corrected $(n-1,1)$ residual is larger than $1/n$. At $\gamma=0$ the existing proof gives the lower bound \[ \rho_1=\frac{n-k}{N+k-1}-\frac2N. \] Both floors are positive when $n\ge k+3$. The products $D c_\lambda$ are affine in $\gamma$. Interpolating the two endpoint floors, with $D_0=(k+1)N/k$, $D_1=N+k-1$, and $\gamma=(K-k+1)/(K+k(k-1))$, gives \[ c_\lambda\ge \frac{(k-1-K)N\rho_0+K D_1\rho_1}{(k-1)(N+K)}. \] Substitution is the stated formula. \end{proof} \begin{lemma}[Growth of the selected residuals]\label{lem:pointwise-growth} For integers $j\ge2$ and $s\ge j+1$, the function \[ A_{j,s}(x)/x^{s-1},\qquad A_{j,s}(x)=\frac{(x)_s}{x+1} \left(1+\frac{x(-1)^j}{\binom xj}\right), \] is nondecreasing for $x\ge s$. Consequently, at the endpoint $\gamma=-1/k$, every selected nonleading shape with $s$ coordinates different from one has residual at least \[ \rho_{0,s}=\frac{n-3}{n(n-1)} \left(1-(k/n)^{s-1}\right). \] Zero padding is included in this count; equivalently, $s$ is the sum of the parts exceeding one. \end{lemma} \begin{proof} For odd $j$, the normalized function factors as \[ \frac{x}{x+1}\prod_{r=1}^{s-1}\left(1-\frac r x\right) \left(1-\frac{j!}{(x-1)_{j-1}}\right), \] a product of nonnegative increasing functions. For even $j$, it suffices to take $s=j+1$, since each additional support coordinate appends a factor $1-r/x$. For $j=2$, the normalized function is $1-2/x$. For $j=4$, it is \[ 1-\frac{11}{x}+\frac{46}{x^2}-\frac{72}{x^3}; \] its derivative has positive numerator $11x^2-92x+216$, whose discriminant is negative. For even $j\ge6$, put $t=x-j\ge1$ and $P(t)=\prod_{r=1}^{j-1}(t+r)$. The normalized function is \[ \frac{t}{t+j+1}\frac{P(t)+j!}{(t+j)^{j-1}}. \] The first factor increases. The derivative numerator of the second is \[ (t+j)P'(t)-(j-1)(P(t)+j!) =\sum_{r=1}^{j-1}(j-r)\prod_{h\ne r}(t+h)-(j-1)j!. \] This numerator increases with $t$; its value at $t=1$ is $j![(j+1)\mathsf H_j-3j+1]$, where $\mathsf H_j=\sum_{r=1}^j1/r$. The bracket is positive at $j=6$, and its successive difference is $\mathsf H_j-2+1/(j+1)>0$, so it is positive for every even $j\ge6$, which proves monotonicity. The endpoint residual formula in Lemma~\ref{lem:sharper-residual} and $\binom nj\ge\binom n2$ now give the stated floor. \end{proof} \begin{lemma}[A coefficientwise projection correction]\label{lem:pointwise-gap} Let $k\ge3$ and let $\gamma=h_k(K)$, $00$: its centered subset sum has absolute value $n/2\ge2$, so $b_{(n)}=C$, and $r_{(n)}=-1/n$. The inequality $R_n\ge0$ replaces its weight by $w_{(n-1,1)}/(n-1)$ and gives the effective coefficients above. Every remaining residual is positive. Finally the transport identity implies $\sum_{\mathcal S_n}c_\lambda w_\lambda\le\delta$, proving the result. \end{proof} \subsection{The finite-binomial separator estimates} \label{app:optimal-logarithm} The proof of Proposition~\ref{prop:optimal-clipped-separator} reduces the validity of $\psi_M$ to the following estimates for binomial laws. \begin{proof}[Finite-binomial estimates for Proposition~\ref{prop:optimal-clipped-separator}] Use the test $\psi_M$ from \eqref{eq:parity-quadratic-separator}. The elementary bounds $71/16$. Thus, for $M\ge M_*$, all values $\psi_M(d)$ with $|d|\ge2$ are nonnegative, and $\psi_M(2)>2$. The projection-spectrum hull reduces validity to the deterministic distribution at zero and the centered binomials $X-r$, where $X\sim\operatorname{Bin}(r+s,r/(r+s))$ and $r,s\ge1$. For these generators, write \[ q=\mathbb P(X=r),\qquad R=\frac{\mathbb P(|X-r|=1)}q =2-\frac1{r+1}-\frac1{s+1}, \] and \[ A_2=\frac{\mathbb P(|X-r|=2)}q =\frac{(r-1)s^2}{r(s+1)(s+2)} +\frac{(s-1)r^2}{s(r+1)(r+2)}. \] For $r,s\ge2$, we have $2A_2\ge R-1$. Indeed, after multiplication by the positive denominator $rs(r+1)(r+2)(s+1)(s+2)$, the difference is \[ \begin{split} P(r,s)={}&2(r-1)s^3(r+1)(r+2) +2(s-1)r^3(s+1)(s+2)\\ &-(rs-1)rs(r+2)(s+2). \end{split} \] For $x=r-2$ and $y=s-2$, its expansion is $P(r,s)=\sum_{i,j=0}^3p_{ij}x^iy^j$, where \[ (p_{ij})_{i,j=0}^3= \begin{pmatrix} 192&320&152&24\\ 320&548&266&42\\ 152&266&129&20\\ 24&42&20&3 \end{pmatrix}. \] Consequently, every function $\psi$ with the prescribed values at $0,\pm1$, with $\psi(\pm2)\ge2$, and with nonnegative values elsewhere satisfies \begin{equation*} \mathbb E\psi(X-r)\ge q(1-R+2A_2)\ge0 \qquad(r,s\ge2). \end{equation*} By symmetry, the remaining generators have $X\sim\operatorname{Bin}(m,1/m)$, centered at one, with $m\ge2$. For $m=2$, the expectation of $\psi_M(X-1)$ is zero. For $m\ge3$, put $a=1-2/m\in[1/3,1)$. Using the probability generating function and its first two derivatives gives \begin{equation}\label{eq:boundary-parity-expectation} \mathbb E\psi_M(X-1) =M(3-4/m)-(1+3M)a^m -2Ma^{m-1}-2M(1-1/m)a^{m-2}. \end{equation} At $M=M_*$, nonnegativity is equivalent to \begin{equation}\label{eq:boundary-exponential-inequality} a^{m-2}\bigl[(3e^2-4)a^2+3a+1\bigr]\le1+2a. \end{equation} Since $m-2=2a/(1-a)$ and $\log a\le2(a-1)/(a+1)$, it suffices to prove \[ (1+2a)\exp\!\left(\frac{4a}{1+a}\right) \ge (3e^2-4)a^2+3a+1. \] Set $t=(1-a)/(1+a)=1/(m-1)\in(0,1/2]$. After multiplication by $(1+t)^2$, this inequality becomes $F(t)\ge0$, where \[ F(t)=e^2\bigl[(3+2t-t^2)e^{-2t}-3+6t-3t^2\bigr] -10t+6t^2. \] The Taylor bound $e^{-2t}\ge1-2t+2t^2-\frac43t^3$ gives \[ F(t)\ge t\bigl[e^2 B(t)-10+6t\bigr],\qquad B(t)=2-2t+2t^2-\frac{14}3t^3+\frac43t^4. \] The identity \[ 7B(t)-10+6t =(1-2t)\left(4+14t^2-\frac{14}3t^3\right)\ge0 \] also implies $B(t)\ge1$. Since $e^2>7$, we obtain $F(t)>0$, which proves \eqref{eq:boundary-exponential-inequality}. Write the right-hand side of \eqref{eq:boundary-parity-expectation} as $M D_m-a^m$, where $D_m$ is independent of $M$. Validity at $M_*$ implies $D_m\ge a^m/M_*$. The function $(1-2/m)^m$ increases for real $m>2$, because \[ \frac{d}{dm}\bigl[m\log(1-2/m)\bigr] =-\log(1+u)+u\ge0,\qquad u=\frac2{m-2}. \] It follows that $a^m\ge1/27$ for $m\ge3$, and therefore \begin{equation*} \mathbb E\psi_M(X-1) \ge\frac{M-M_*}{27M_*}=:\eta_M>0 \qquad(M>M_*,\ m\ge3). \end{equation*} \end{proof} \section{Negative constructions} \label{app:negative-coherent} The constructions below prove Proposition~\ref{prop:negative-summary} and the padding bounds of Section~\ref{sec:negative}. \subsection{An attained minimum in a family extending Du's example} \label{app:witness:nine-family} Du's example uses a six-dimensional positive map and separate terms joining three-dimensional blocks. Combining each pair of joining terms into one congruence introduces off-diagonal terms while preserving positivity. For $X\in M_9$, let $S(X)$ be its leading $6\times6$ block and put $\widetilde{\mathcal B}(X)=\operatorname{diag}(\mathcal B(S(X)),0_3)$. Using three blocks of size three, define \[ A_1=E_{13}\otimes I_3,\quad B_1=E_{31}\otimes I_3,\qquad A_2=E_{23}\otimes I_3,\quad B_2=E_{32}\otimes I_3, \] where these $E_{ij}$ are matrix units in $M_3$. For real $t$ set \[ Q_t^{(j)}(X)=A_jXA_j^*+B_jXB_j^* +t(A_jXB_j^*+B_jXA_j^*),\qquad j=1,2, \] and, with $\mathsf T(X)=X^{\mathsf T}$, define \begin{equation*} G_{a,b}=\tfrac18\widetilde{\mathcal B} +\tfrac12 Q_a^{(1)}+\tfrac12 Q_b^{(2)}\circ\mathsf T. \end{equation*} At $a=b=0$, this is Du's map \citep[(2.12)]{du2026} with parameter $t=1/4$, divided by two. Both marginals of Du's map at that parameter are $2I_9$, so $G_{0,0}$ is its bistochastic normalization. \begin{prop}[Attained nine-dimensional family bound] \label{prop:negative-family-nine} For real $a,b$, the map $G_{a,b}$ is positive and bistochastic exactly when $|a|,|b|\le1$. Its determinant contraction is \begin{equation}\label{eq:nine-family-polynomial} \Gamma_9(G_{a,b})=-\frac3{256} \bigl[4+9(a^2+b^2)+a^4+b^4+16a^2b^2+a^4b^2+a^2b^4\bigr]. \end{equation} Writing $H_9=G_{1,1}$, we have \begin{equation*} \min_{|a|,|b|\le1}\Phi_9(G_{a,b}) =\Phi_9(H_9)=-\frac1{737280}\approx-1.356\cdot10^{-6}. \end{equation*} Equality holds exactly when $|a|=|b|=1$. In comparison, $\Phi_9(G_{0,0})=-1/7741440$, so $H_9$ has $21/2$ times the negative magnitude of the normalized Du example. \end{prop} \begin{proof} For $|t|\le1$, \[ Q_t^{(j)}=\frac{1+t}{2}\Ad_{A_j+B_j} +\frac{1-t}{2}\Ad_{A_j-B_j}. \] This decomposition proves positivity on the square. Let $P_j$ denote projection onto the $j$th three-dimensional block. The Breuer--Hall term contributes $\tfrac12(P_1+P_2)$ to each marginal, and the remaining terms contribute $\tfrac12(P_1+P_2)+P_3$. All terms containing $a$ or $b$ have zero marginals, so both marginals are $I_9$ for every real $a,b$. For necessity, apply the map to the rank-one input with block vector $(e_1,0,e_1)$ and restrict the output to the first coordinate in blocks one and three. The resulting principal matrix is $\tfrac12\left(\begin{smallmatrix}1&a\\a&1\end{smallmatrix}\right)$. The input $(0,e_1,e_1)$ gives the analogous matrix with $b$. Thus positivity requires $|a|,|b|\le1$. For the exact polynomial calculation, write $c_{pqij}=L(E_{ij})_{pq}$ for a map $L:M_n\to M_n$. Determinant expansion and differentiation give \begin{equation}\label{app:witness:contraction} \Gamma_n(L)=\sum_{\sigma,\tau,\upsilon\in \symm_n} \operatorname{sgn}(\sigma)\operatorname{sgn}(\tau) \operatorname{sgn}(\upsilon) \prod_{p=1}^n c_{p,\sigma(p),\tau(p),\upsilon(p)}. \end{equation} Each surviving product uses every input row and column once. Indexing them by $\tau$ and $\upsilon$ makes the input determinant sign $\operatorname{sgn}(\tau)\operatorname{sgn}(\upsilon)$; no additional factorial occurs. Substitution of the integer tensor $8G_{a,b}$ gives \[ \begin{array}{c|r} \text{monomial}&[\text{monomial}]\,\Gamma_9(8G_{a,b})\\ \hline 1&-6291456\\ a^2,\ b^2&-14155776\ \text{each}\\ a^4,\ b^4,\ a^4b^2,\ a^2b^4&-1572864\ \text{each}\\ a^2b^2&-25165824 \end{array} \] All other coefficients vanish. Dividing by $8^9$ proves \eqref{eq:nine-family-polynomial}. An exact computation in the supporting files reproduces the entire coefficient array; a separately ordered contraction checks the normalization and values. The bracket in \eqref{eq:nine-family-polynomial} is strictly increasing in each of $a^2,b^2$ on $[0,1]^2$. Its maximum is $42$, attained exactly at $a^2=b^2=1$. Division by $9!$ proves the claimed minimum and the comparison at $a=b=0$. The same strict increase shows that the normalized Du map $G_{0,0}$ is not a local minimum, since arbitrarily small nonzero $a$ or $b$ decrease its invariant. \end{proof} The attaining map has the simpler expression \[ H_9=\tfrac18\widetilde{\mathcal B} +\tfrac12\Ad_{A_1+B_1} +\tfrac12\Ad_{A_2+B_2}\circ\mathsf T. \] Thus the two pairs of separate terms in the normalized Du map become one completely positive and one completely copositive summand. This is an exact minimum within the displayed family; in the unrestricted problem it gives $\mu_9\le-1/737280$. Pinched identity padding therefore gives \[ \mu_n\le-\frac1{737280\binom n9}\qquad(n\ge9). \] \subsection{Compression and an attained eight-dimensional bound} \label{app:witness:eight-family} A compression need not preserve bistochasticity. For the family above, a diagonal output congruence repairs both marginals explicitly. Let $V:\C^8\to\C^9$ have columns $e_1,\ldots,e_8$, and let $U:\C^8\to\C^9$ have columns $e_1,e_2,e_3,e_4,e_5,e_7,e_8,e_9$. Define \[ C_{a,b}(X)=U^*G_{a,b}(VXV^*)U,\qquad D=\operatorname{diag}(1,1,\sqrt2,1,1,1,1,1), \] \begin{equation*} \mathcal F_{a,b}(X)=D C_{a,b}(X)D,\qquad H_8=\mathcal F_{1,1}. \end{equation*} \begin{prop}[Attained eight-dimensional family bound] \label{prop:negative-family-eight} For real $a,b$, the map $\mathcal F_{a,b}$ is positive and bistochastic exactly when $|a|,|b|\le1$. Its determinant contraction is \begin{equation}\label{eq:eight-family-polynomial} \Gamma_8(\mathcal F_{a,b}) =\frac{8a^4-9a^2-21b^2-34a^2b^2-10}{1024}, \end{equation} and \begin{equation*} \min_{|a|,|b|\le1}\Phi_8(\mathcal F_{a,b}) =\Phi_8(H_8)=-\frac{11}{6881280}\approx-1.599\cdot10^{-6}. \end{equation*} Equality holds exactly when $|a|=|b|=1$. The uncompleted compression has $\Phi_8(\mathcal F_{0,0})=-1/4128768$, so $H_8$ increases its negative magnitude by $33/5$. \end{prop} \begin{proof} Compression and output congruence preserve positivity. To check the normalization, put $F=E_{11}+E_{22}+E_{44}+E_{55}$ in $M_8$. Direct calculation gives, for all real $a,b$, \[ C_{a,b}(I_8)=I_8-\tfrac12E_{33},\qquad C_{a,b}^*(I_8)=I_8-\tfrac18F,\qquad C_{a,b}^*(E_{33})=\tfrac18F. \] Since $D^2=I_8+E_{33}$, both marginals of $\mathcal F_{a,b}$ are $I_8$. For necessity, take the rank-one input $xx^*$ with $x=e_1+e_7$ in $\C^8$ and restrict the output to coordinates $(1,6)$. The resulting principal matrix is $\tfrac12\left(\begin{smallmatrix}1&a\\a&1\end{smallmatrix}\right)$. The input $x=e_4+e_7$ and output coordinates $(4,6)$ give the same matrix with $b$. Thus positivity again requires $|a|,|b|\le1$. The integer tensor $8C_{a,b}$ gives the following complete coefficient array under \eqref{app:witness:contraction}: \[ \begin{array}{c|r} \text{monomial}&[\text{monomial}]\,\Gamma_8(8C_{a,b})\\ \hline 1&-81920\\ a^2&-73728\\ b^2&-172032\\ a^4&65536\\ a^2b^2&-278528 \end{array} \] All other coefficients vanish. The output congruence contributes $|\det D|^2=2$, so multiplication by $2/8^8$ proves \eqref{eq:eight-family-polynomial}. The sharp lower bound follows from \begin{align*} 1024\,8!\left(\Phi_8(\mathcal F_{a,b})+\frac{11}{6881280}\right) ={}&(1-b^2)(21+34a^2)\\ &+27(1-a^2)+8(1-a^2)^2. \end{align*} Every term is nonnegative on the square, and their sum vanishes exactly when $a^2=b^2=1$. At $a=b=0$, the polynomial gives the stated baseline. \end{proof} For the maps $H_8$ and $Q$ defined in Section~\ref{sec:negative}, the determinant contractions needed for the extension reduce to one polynomial: \begin{equation}\label{app:witness:seed-poly} \begin{split} \Phi_8(\Ad_Q+tH_8) ={}&\frac{t^3+5t^4}{1376256} +\frac{41t^5}{6881280}+\frac{t^6}{458752}\\ &\hspace{14mm}-\frac{17t^7+11t^8}{6881280}. \end{split} \end{equation} For completeness, the unnormalized polynomial in \eqref{app:witness:seed-poly} is \[ \begin{split} &\det(\partial_X)\det\bigl(4K_0XK_0^*+tR(X)\bigr)\\ &\qquad=245760t^3+1228800t^4+2015232t^5 +737280t^6-835584t^7-540672t^8. \end{split} \] Multiplication by $2/(8^8\,8!)$ gives \eqref{app:witness:seed-poly}. These integer coefficients are obtained from the same contraction \eqref{app:witness:contraction}, applied to $4\Ad_{K_0}+tR$. An exact computation in the supporting files evaluates the sum in two orders for the coefficient of $t^8$ and reconstructs the polynomial by rational interpolation. \subsection{Extension through the completely positive summand} The summand $\Ad_Q$ permits an extension that keeps some off-diagonal entries. For $n=8+m$, $m\ge1$, and $0\le\alpha\le1$, set \begin{equation*} T_{n,\alpha}\begin{pmatrix}X&Z\\W&Y\end{pmatrix} =\begin{pmatrix}H_8(X)&\sqrt\alpha QZ\\ \sqrt\alpha WQ^*&Y\end{pmatrix}. \end{equation*} This map is positive because \[ T_{n,\alpha}=\Ad_{\sqrt\alpha Q\oplus I_m} +(H_8-\alpha\Ad_Q)_{\mathrm{corner}}, \] where the second term acts on the leading $8\times8$ block and vanishes elsewhere. Its diagonal blocks show that it is bistochastic. To compute its invariant in every dimension, we use an identity for a map supported on a fixed corner. If $P:M_r\to M_r$ is complex linear, $K\in M_r$, and $b_j=[t^j]\Phi_r(\Ad_K+tP)$, then \begin{equation}\label{app:witness:corner} \Phi_n\bigl(\Ad_{K\oplus I_{n-r}}+P_{\mathrm{corner}}\bigr) =\sum_{j=0}^r\frac{(r)_j}{(n)_j}b_j, \qquad n\ge r. \end{equation} For $K=I_r$, expand by the number $j$ of factors from $P$ in the determinant contraction. Every selected index lies in the first $r$ coordinates, while the remaining identity contractions have completion factor $(n-j)!$ in dimension $n$ and $(r-j)!$ in dimension $r$. After division by $n!$ and $r!$, their ratio is $(r)_j/(n)_j$. For invertible $K$, factor out the output congruence by $K\oplus I_{n-r}$ and apply the identity case to $\Ad_{K^{-1}}\circ P$. Both sides acquire the factor $|\det K|^2$; polynomial continuity covers singular $K$. \begin{prop}\label{app:witness:bound} For every integer $n\ge9$, the positive bistochastic map $T_{n,3/(5(n-8))}$ satisfies \[ \Phi_n\bigl(T_{n,3/(5(n-8))}\bigr) \le-\frac{4947}{50000(n)_8}. \] \end{prop} \begin{proof} Write $c_j=[t^j]\Phi_8(\Ad_Q+tH_8)$. Homogeneity gives \[ \Phi_8\bigl(\alpha\Ad_Q+t(H_8-\alpha\Ad_Q)\bigr) =\sum_{k=0}^8c_k\alpha^{8-k}t^k(1-t)^{8-k}. \] Apply \eqref{app:witness:corner} and the finite identity \[ \sum_{j=k}^8(-1)^{j-k}\binom{8-k}{j-k}\frac{(8)_j}{(n)_j} =\frac{(8)_k(m)_{8-k}}{(n)_8}. \] The coefficients in \eqref{app:witness:seed-poly} then give \begin{equation}\label{app:witness:bracket} \begin{split} (n)_8\Phi_n(T_{n,\alpha}) ={}&\frac{\alpha^5(m)_5+25\alpha^4(m)_4}{4096}\\ &+\frac{41\alpha^3(m)_3+45\alpha^2(m)_2}{1024} -\frac{51\alpha m+33}{512}. \end{split} \end{equation} For $u\ge0$, define \[ P_*(u)=\frac{u^5+25u^4+164u^3+180u^2-408u-264}{4096}. \] Since $0\le(m)_j/m^j\le1$ for integers $m\ge1$, the right side of \eqref{app:witness:bracket} at $\alpha=u/m$ is at most $P_*(u)$. Choosing $u=3/5$ keeps $\alpha\le1$ and gives $P_*(3/5)=-4947/50000$. \end{proof} The coefficient can be increased slightly without changing the construction. The polynomial $P_*$ has a unique minimizer $u_*>0$, since \[ 4096P_*'(u)=5u^4+100u^3+492u^2+360u-408 \] is strictly increasing for $u\ge0$ and changes sign on $(0,1)$. Taking $\alpha=u_*/m$ proves \[ \mu_n\le-\frac{C_*}{(n)_8}\quad(n\ge9),\qquad C_*=-P_*(u_*)=0.09894625440\ldots. \] \end{document}