\pdfsuppressptexinfo=-1 \documentclass[11pt]{amsart} \usepackage[margin=1.1in]{geometry} \usepackage{amsmath,amssymb,amsthm,mathtools} \usepackage{enumitem} \usepackage{booktabs} \usepackage{array} \usepackage{longtable} \usepackage[hidelinks]{hyperref} \hypersetup{pdftitle={On the minimal dimension of range-criterion counterexamples to the PPT-squared conjecture}, pdfauthor={Tejasvi Singh Tomar}} \input{supplement-labels} \makeatletter \let\@writetocindents\relax \def\LT@makecaption#1#2#3{\LT@mcol\LT@cols c{\hbox to\z@{\hss\parbox[t]\textwidth{% \normalsize\@tempcnta64 \@makecaption{#1{#2}}{#3}}\hss}}} \makeatother \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{corollary}[theorem]{Corollary} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \numberwithin{equation}{section} \numberwithin{table}{section} \setlength{\emergencystretch}{2em} \setlength{\captionindent}{0pt} \newcommand{\C}{\mathbb{C}} \newcommand{\R}{\mathbb{R}} \newcommand{\Q}{\mathbb{Q}} \newcommand{\PP}{\mathbb{P}} \newcommand{\M}{\mathrm{M}} \newcommand{\Herm}{\mathrm{Herm}} \newcommand{\Tr}{\operatorname{Tr}} \newcommand{\rank}{\operatorname{rank}} \newcommand{\ran}{\operatorname{ran}} \newcommand{\spn}{\operatorname{span}} \newcommand{\Hom}{\operatorname{Hom}} \newcommand{\Sym}{\operatorname{Sym}} \newcommand{\Gr}{\operatorname{Gr}} \newcommand{\Fl}{\operatorname{Fl}} \newcommand{\Ad}{\operatorname{Ad}} \newcommand{\vecop}{\operatorname{vec}} \newcommand{\id}{\mathrm{id}} \newcommand{\cK}{\mathcal{K}} \newcommand{\cS}{\mathcal{S}} \newcommand{\cG}{\mathcal{G}} \newcommand{\cE}{\mathcal{E}} \newcommand{\cF}{\mathcal{F}} \newcommand{\cO}{\mathcal{O}} \newcommand{\ca}{\textsuperscript{\dag}} \newcommand{\ket}[1]{|#1\rangle} \newcommand{\bra}[1]{\langle #1|} \newcommand{\ip}[2]{\langle #1,#2\rangle} \begin{document} \title[Range-criterion counterexamples to PPT$^2$]{On the minimal dimension of range-criterion counterexamples to the PPT$^2$ conjecture} \author{Tejasvi Singh Tomar} \renewcommand{\urladdrname}{{\itshape ORCID}} \urladdr{\href{https://orcid.org/0000-0003-4668-3639}{https://orcid.org/0000-0003-4668-3639}} \subjclass[2020]{81P40; 15A69, 14M15, 14C17, 47L07} \keywords{PPT squared conjecture, entanglement breaking map, range criterion, Kraus space, determinantal variety, Grassmannian, Chern class, Cayley--Bacharach theorem} \begin{abstract} The PPT$^2$ conjecture asserts that the composition of two completely positive and completely copositive (PPT) maps is entanglement breaking. A recently announced counterexample, with PPT maps $\M_{10}\to\M_6\to\M_{10}$, is certified by the range criterion: the range of the composite Choi matrix contains no nonzero product vector. We ask how small a counterexample certified in this way can be. For PPT maps $\Phi\colon\M_m\to\M_k$ and $\Psi\colon\M_k\to\M_n$ with $\Psi\circ\Phi\neq0$ and $\max(m,k,n)\le6$ we prove that the range of the Choi matrix of $\Psi\circ\Phi$ contains a nonzero product vector whenever $k\le5$ or $\min(m,n)\le5$. Hence the range criterion cannot certify a counterexample on $\M_4$ or $\M_5$, and among the types with all three dimensions at most six only $(6,6,6)$ remains open. Since the range is spanned by products of Kraus operators, the question is when such a span can avoid the matrices of rank one. The computer-assisted proof combines rank bounds for PPT maps with Chern class and excess intersection computations; its computed facts are checked in Lean 4, modulo classical, quoted or hand-proved inputs. The announced counterexample lifts through exterior powers a map on $\M_4$ built from a pencil of real matrices (a map of pencil type). When the map of pencil type loses rank transversally at finitely many points, its lifts through any Schur functor, exterior powers included, put product vectors into the kernel of the composite Choi matrix. When the composite is also nonzero, these product vectors form an unextendible product set only for exterior powers, and then one of the three dimensions is at least ten. \end{abstract} \maketitle \section{Introduction}\label{sec:intro} A linear map $\Phi\colon\M_m\to\M_k$ is completely positive if $\id_r\otimes\Phi$ is positive for all $r$, and completely copositive if $\Phi$ composed with the transpose is completely positive. We call $\Phi$ a PPT map if it is both. It is entanglement breaking if its Choi matrix $C_\Phi=\sum_{i,j}E_{ij}\otimes\Phi(E_{ij})$ is separable \cite{HSR03}. Christandl's PPT$^2$ conjecture \cite{Chr12}, in the form of \cite[Conjecture~IV.1]{CMHW19}, asserts that $\Psi\circ\Phi$ is entanglement breaking for all PPT maps $\Phi\colon\M_m\to\M_k$ and $\Psi\colon\M_k\to\M_n$. A PPT map applied to one half of a bipartite state always produces a state with positive partial transpose, which may still be entangled; the conjecture says that two PPT maps applied in turn always produce a separable state. The conjecture holds for maps on $\M_3$ \cite{CMHW19,CYT19} and when the middle algebra is $\M_2$ \cite{Hen26}, and every PPT channel becomes entanglement breaking after finitely many iterations \cite{Par26}. A counterexample was announced recently \cite{OAI26}: a pair of PPT maps on $\M_{10}$ whose composition is not entanglement breaking, and a trace-preserving PPT map on $\M_{21}$ whose square is not entanglement breaking. For maps on $\M_4$ the conjecture is open. The pair in \cite{OAI26} factors through $\M_6$ (Remark~\ref{rem:tenpair}). In our notation it consists of PPT maps $\Phi\colon\M_{10}\to\M_6$ and $\Psi\colon\M_6\to\M_{10}$ such that $Z=C_{\Psi\circ\Phi}$ is nonzero and its range contains no nonzero product vector. A separable positive semidefinite operator is a sum of projections onto product vectors in its range, so $Z$ is entangled by the range criterion \cite{Hor97}. We call such a pair a \emph{range-criterion pair} of type $(m,k,n)=(10,6,10)$ (Definition~\ref{def:rcp}). The authors of \cite{OAI26} do not claim that the dimension ten is minimal. This paper asks how small a range-criterion pair can be. The answer shows in which dimensions a counterexample can be certified as in \cite{OAI26}. The question is algebraic. The range of $C_{\Psi\circ\Phi}$ is spanned by the vectorised products $XY$ of Kraus operators $X$ of $\Psi$ and $Y$ of $\Phi$, and product vectors correspond to matrices of rank one (Lemma~\ref{lem:range}). So we ask when the linear span of these products can avoid the rank-one matrices while both maps stay PPT. Our main result is the following. \begin{theorem}[\ca]\label{thm:main} Let $\Phi\colon\M_m\to\M_k$ and $\Psi\colon\M_k\to\M_n$ be PPT maps with $\Psi\circ\Phi\neq0$ and $\max(m,k,n)\le6$. If $k\le5$ or $\min(m,n)\le5$, then the range of $C_{\Psi\circ\Phi}$ contains a nonzero product vector. \end{theorem} In particular, the range criterion cannot certify a counterexample on $\M_4$ or on $\M_5$, and among the types with all three dimensions at most six only $(6,6,6)$ remains open; Table~\ref{tab:d6} lists its open cases. Theorem~\ref{thm:main} concerns the range criterion only: a counterexample on $\M_d$ with $4\le d\le 9$ whose composite has product vectors in its range is not excluded, and would have to be detected by other means. Results marked with a dagger, like Theorem~\ref{thm:main}, have computer-assisted proofs or depend on one; the computations are exact or carry rigorous error bounds; they include intersection numbers on Grassmannians and related spaces (computed by Schubert calculus or by localisation), ranks of linear systems and certificates for explicit maps. Each computation is described in the proof that uses it and is listed in Table~\ref{tab:verify} in Appendix~\ref{app:verify} of the supplement (numbers with the prefix S refer to the supplement, and numbers without it to this paper). The computed facts are checked in the proof assistant Lean~4, modulo inputs that are classical, quoted from \cite{OAI26} or proved by hand, and Lean derives Theorem~\ref{thm:main} from results that it proves and from nine steps of the proof that it takes as hypotheses, results whose proofs rest on intersection theory or sheaf cohomology, listed in Appendix~\ref{app:verify}. The programs, certificates and Lean code are deposited \cite{Tom26}. Some of our results reach beyond dimension six: there is no range-criterion pair of type $(4,k,4)$, whatever the middle dimension $k$ (Theorem~\ref{thm:44}), and none of type $(4,6,n)$, $(6,4,n)$, $(n,6,4)$ or $(n,4,6)$ with $n\le9$ (Theorems~\ref{o469:thm:469} and~\ref{thm:649ex}). The supplement excludes further types up to dimension nine, among them $(5,5,7)$, $(4,7,5)$, $(4,7,6)$, $(4,8,5)$, $(4,8,6)$ and $(4,9,5)$ and their mirrors (Table~\ref{tab:d7} and Section~\ref{sec:open}). The proofs work with reduced pairs: every PPT pair with $\Psi\circ\Phi\neq0$ can be compressed so that all four marginals $\Phi(I)$, $\Phi^{*}(I)$, $\Psi(I)$ and $\Psi^{*}(I)$ become invertible, without changing whether the composite range contains a product vector (Lemma~\ref{lem:reduce}), and we call the compressed pair reduced. The proof of Theorem~\ref{thm:main} rests on four ingredients. First, after the reduction, a product-free composite range forbids Kraus operators of rank at most two; together with the theorem of Horodecki, Lewenstein, Vidal and Cirac on PPT states of low rank \cite{HLVC00} and a normal form of Chen and Djokovi\'c \cite{CD13}, this bounds the ranks of the Choi matrices of both maps and of their partial transposes from both sides, and leaves $(5,5,5)$ as the only reduced type with all dimensions at most five. Second, at middle dimension five, a nonzero intersection number, the top Chern class of a vector bundle, forces a Kraus operator of $\Psi$ whose kernel lies in the range of a Kraus operator of $\Phi$, so that their product has rank one. Third, at middle dimension six and outer dimension four, the ranges of the rank-three Kraus operators of $\Phi$ form a family of $3$-planes in $\C^6$ that covers $\PP^5$, and an excess intersection count shows that such a family cannot do so when $\rank C_\Psi$ is large; refinements of this count settle the types with middle dimension six, an outer dimension four and all dimensions at most six. Fourth, two facts hold for a single PPT map $\Phi$: the set of vectors $x$ at which $\Phi(xx^{\dagger})$ loses rank is a finite union of linear subspaces, and each product vector in the kernel of its Choi matrix forces certain of its Kraus operators to have ranges orthogonal to the second factor of that product vector. At middle dimension four these two facts show that, for a reduced range-criterion pair, the kernel of the composite Choi matrix contains no nonzero sum of at most two product vectors (Lemma~\ref{lem:nr2}), and intersection counts then exclude the type $(6,4,6)$, the only type with middle dimension four and all dimensions at most six that the rank bounds leave; the partial-transpose argument behind the second fact, with B\'ezout's theorem, also excludes the outer dimensions $(4,4)$ for every middle dimension. For a reduced range-criterion pair of type $(5,6,5)$ the two facts leave exactly two product vectors in the kernel of the composite Choi matrix, up to scalar multiples (Theorem~\ref{thm:565two}), and a congruence modulo $8$ for the Euler characteristic of a sheaf on a projective plane excludes this type. Section~\ref{mp:sec:overview} names the statement that proves each step. In Section~\ref{sec:example} we study the example of \cite{OAI26} and the family of exterior-power lifts to which it belongs. The twenty directions $x$ in which its base map $L\colon\M_4\to\M_4$ loses rank give the twenty product vectors in the kernel of its composite Choi matrix; we show that the quadratic Veronese images of the rank-loss directions and of the kernel vectors of $L(xx^{\dagger})$ at these directions are Gale dual, a consequence of the Cayley--Bacharach theorem, and that the twenty kernel conditions determine the Kraus spaces of both maps (Theorem~\ref{thm:gale}). Within a wider class of lifts, a certificate of the kind used in \cite{OAI26} still needs dimension ten: if the base map loses rank transversally at finitely many points, the product vectors of a Schur-functor lift with nonzero composite form an unextendible product set (a notion that includes the unextendible product bases of \cite{BDMSST99}) only for exterior powers, and then one of the three dimensions is at least ten (Theorem~\ref{thm:schur}). For a general base map on $\M_3$, the member of type $(6,6,6)$ is one product vector short of the range criterion, and for a general pencil in a family of real pencils its factors are rigid: every PPT map that can replace one of them while keeping the ten product vectors in the kernel is a nonnegative multiple of it (Theorem~\ref{thm:rigid}). Another member gives a range-criterion pair of type $(6,10,10)$ (Proposition~\ref{prop:61010}). In Section~\ref{sec:symmetry} we show that symmetry under a compact connected group that acts irreducibly on both outer spaces always puts a product vector into the composite range, in every dimension (Theorem~\ref{thm:compact}); for $\mathrm{SU}(2)$ it suffices that no spin occurs more than twice in one of the outer representations (Theorem~\ref{thm:su2top}\ref{su2top:d}), a condition that is sharp for invariant operators (Corollary~\ref{cor:su2inv}). We also prove that $\Psi\circ\Phi$ is entanglement breaking, as the PPT$^2$ conjecture asserts, for all PPT maps in several covariant classes with outer dimensions up to nine (Theorem~\ref{thm:classes}); the conjecture was settled earlier for two symmetric classes \cite[Proposition~V.1]{CMHW19}, \cite{SN22}. Section~\ref{sec:edge} treats the stronger criterion for edge states \cite{LKCH00}, which also uses the range of the partial transpose, and Section~\ref{sec:open} lists open problems, with examples showing that the types $(6,6,6)$, $(5,6,5)$ and $(5,6,6)$ cannot be excluded by linear algebra and Lemma~\ref{lem:noranktwo} alone. \emph{The supplement.} Section~\ref{mp:sec:overview} states the results of the case analysis that the proof of Theorem~\ref{thm:main} uses, and the main results for larger types, with sketches of their proofs, and Section~\ref{sec:example} states the main further results on lifts in the same way. The proofs and the remaining results are in the supplement, which is deposited with this paper. Section~\ref{sec:numerics} of the supplement collects numerical evidence, which is not used in any proof. \section{Preliminaries}\label{sec:prelim} \subsection{Notation} We write $\M_n$ for the complex $n\times n$ matrices, $\M_{k\times m}$ for the $k\times m$ matrices, $\Herm_n$ for the Hermitian matrices, $A\succeq0$ for positive semidefinite $A$, and $E_{ij}$ for the matrix units. For a matrix $A$, $\bar A$ is the entrywise complex conjugate, $A^{T}$ the transpose and $A^{\dagger}=\bar A^{T}$. For $x\in\C^m$ we put $P_x=xx^{\dagger}$, without normalisation. The transpose map is $T$, and $\Ad_V(X)=VXV^{\dagger}$. The adjoint $\Phi^{*}$ of a linear map $\Phi\colon\M_m\to\M_k$ is defined by $\Tr(\Phi^{*}(B)^{\dagger}A)=\Tr(B^{\dagger}\Phi(A))$. The Choi matrix is $C_\Phi=\sum_{i,j}E_{ij}\otimes\Phi(E_{ij})\in\M_m\otimes\M_k$, with the input factor first. The map $\Phi$ is completely positive if and only if $C_\Phi\succeq0$ \cite{Cho75}, and completely copositive if and only if the partial transpose $C_\Phi^{\Gamma}=C_{T\circ\Phi}$ on the second factor is positive semidefinite. An operator on $\C^p\otimes\C^q$ is separable if it is a sum of operators $P\otimes Q$ with $P,Q\succeq0$. A product vector is a vector $u\otimes v$ with $u,v\neq0$, and a subspace is product-free if it contains no product vector. The local ranks of $\rho\succeq0$ on $\C^p\otimes\C^q$ are the ranks of its two partial traces. For a subspace $\cK\subseteq\M_{k\times m}$ and $x\in\C^m$ we write $\cK x=\{Kx:K\in\cK\}$, and for subspaces $\cK\subseteq\M_{n\times k}$, $\mathcal L\subseteq\M_{k\times m}$ we write $\spn(\cK\mathcal L)$ for the span of all products $XY$ with $X\in\cK$ and $Y\in\mathcal L$. A subspace of matrices is rank-one-free if it contains no matrix of rank one. \subsection{Kraus spaces and the composite range} For $K\in\M_{k\times m}$ put $\vecop(K)=\sum_{i=1}^{m}e_i\otimes Ke_i\in\C^m\otimes\C^k$. Then $\vecop(ba^{T})=a\otimes b$, and the Schmidt rank of $\vecop(K)$ equals $\rank K$. If $\Phi=\sum_r\Ad_{K_r}$ is completely positive, then $C_\Phi=\sum_r\vecop(K_r)\vecop(K_r)^{\dagger}$. Hence the \emph{Kraus space} \[ \cK_\Phi=\{K\in\M_{k\times m}:\vecop(K)\in\ran C_\Phi\}=\spn\{K_r\} \] does not depend on the Kraus decomposition, and $\dim\cK_\Phi=\rank C_\Phi$. Product vectors in $\ran C_\Phi$ correspond to matrices of rank one in $\cK_\Phi$. For $x\in\C^m$ and $u\in\C^m$, $v\in\C^k$, \begin{equation}\label{eq:basic} \ran\Phi(P_x)=\cK_\Phi x,\qquad (u\otimes v)^{\dagger}C_\Phi(u\otimes v)=v^{\dagger}\Phi(P_{\bar u})v . \end{equation} \begin{lemma}\label{lem:range} Let $\Phi\colon\M_m\to\M_k$ and $\Psi\colon\M_k\to\M_n$ be completely positive and $Z=C_{\Psi\circ\Phi}$. \begin{enumerate}[label=(\alph*)] \item\label{range:a} $\ran Z=\vecop\bigl(\spn(\cK_\Psi\cK_\Phi)\bigr)$. \item\label{range:b} For $u\in\C^m$ and $v\in\C^n$ we have $(u\otimes v)^{\dagger}Z(u\otimes v)=\Tr\bigl(\Psi^{*}(P_v)\Phi(P_{\bar u})\bigr)$. Hence $u\otimes v\in\ker Z$ if and only if $v^{\dagger}XY\bar u=0$ for all $X\in\cK_\Psi$ and $Y\in\cK_\Phi$, if and only if $\Phi(P_{\bar u})$ and $\Psi^{*}(P_v)$ have orthogonal ranges. \item\label{range:c} $\ran Z$ contains a product vector if and only if $\spn(\cK_\Psi\cK_\Phi)$ contains a matrix of rank one. \end{enumerate} \end{lemma} \begin{proof} If $\Phi=\sum_r\Ad_{Y_r}$ and $\Psi=\sum_s\Ad_{X_s}$, then $\Psi\circ\Phi=\sum_{r,s}\Ad_{X_sY_r}$, which gives \ref{range:a} by bilinearity. Part \ref{range:b} follows from \eqref{eq:basic} applied to $\Psi\circ\Phi$, from $\ip{u\otimes v}{\vecop(M)}=v^{\dagger}M\bar u$, and from the fact that the trace of a product of two positive semidefinite matrices vanishes exactly when their ranges are orthogonal. Part \ref{range:c} follows from \ref{range:a} and $\vecop(ba^{T})=a\otimes b$. \end{proof} \begin{definition}\label{def:rcp} A \emph{range-criterion pair of type $(m,k,n)$} is a pair of PPT maps $\Phi\colon\M_m\to\M_k$ and $\Psi\colon\M_k\to\M_n$ with $\Psi\circ\Phi\neq0$ such that $\spn(\cK_\Psi\cK_\Phi)$ is rank-one-free, that is, $\ran C_{\Psi\circ\Phi}$ is product-free. \end{definition} A separable operator $Z\neq0$ is a sum $\sum_t p_tp_t^{\dagger}$ of product vectors $p_t$, and each $p_t$ lies in $\ran Z$. So for a range-criterion pair $C_{\Psi\circ\Phi}$ is entangled \cite{Hor97}, and $\Psi\circ\Phi$ is not entanglement breaking \cite{HSR03}. \begin{remark}\label{rem:tenpair} The maps in \cite[(6.5)]{OAI26} are $\Phi_1(A)=ES(A^{T})E^{\dagger}$ and $\Phi_2(B)=R^{*}(E^{\dagger}BE)$, where $S,R\colon\M_{10}\to\M_6$ are PPT and $E\colon\C^6\to\C^{10}$ is an isometry. Hence $\Phi=S\circ T\colon\M_{10}\to\M_6$ and $\Psi=R^{*}\colon\M_6\to\M_{10}$ are PPT with $\Psi\circ\Phi=\Phi_2\circ\Phi_1$, and \cite[Theorem~1.2]{OAI26}, a statement about $\Phi_1$ and $\Phi_2$, implies that $(\Phi,\Psi)$ is a range-criterion pair of type $(10,6,10)$. We recall the construction in Section~\ref{sec:example}. \end{remark} \subsection{Two symmetries and a reduction} \begin{lemma}\label{lem:sym} Let $\Phi\colon\M_m\to\M_k$ and $\Psi\colon\M_k\to\M_n$ be PPT. \begin{enumerate}[label=(\alph*)] \item\label{sym:a} The map $\Phi^{\#}=T\circ\Phi^{*}\circ T\colon\M_k\to\M_m$ is PPT with $\cK_{\Phi^{\#}}=\cK_\Phi^{T}=\{Y^{T}:Y\in\cK_\Phi\}$. The pair $(\Psi^{\#},\Phi^{\#})$ has type $(n,k,m)$ and $\spn(\cK_{\Phi^{\#}}\cK_{\Psi^{\#}})=\spn(\cK_\Psi\cK_\Phi)^{T}$. \item\label{sym:b} The pair $(T\circ\Phi,\Psi\circ T)$ is a PPT pair of type $(m,k,n)$ with the same composition. Moreover $C_{T\circ\Phi}=C_\Phi^{\Gamma}$ and $\rank C_{\Psi\circ T}=\rank C_\Psi^{\Gamma}$. \item\label{sym:c} Both operations preserve the property that all four marginals $\Phi(I)$, $\Phi^{*}(I)$, $\Psi(I)$, $\Psi^{*}(I)$ are positive definite. \end{enumerate} In particular, $(\Phi,\Psi)$ is a range-criterion pair if and only if $(\Psi^{\#},\Phi^{\#})$ is, if and only if $(T\circ\Phi,\Psi\circ T)$ is. \end{lemma} \begin{proof} If $\Phi=\sum_r\Ad_{K_r}$, then $\Phi^{\#}(Y)=\bigl(\sum_rK_r^{\dagger}Y^{T}K_r\bigr)^{T}=\sum_rK_r^{T}Y\bar K_r$, so $\cK_{\Phi^{\#}}=\cK_\Phi^{T}$. The map $T\circ\Phi^{\#}=\Phi^{*}\circ T$ is the adjoint of $T\circ\Phi$, hence completely positive. Part \ref{sym:a} follows since $(XY)^{T}=Y^{T}X^{T}$. In \ref{sym:b}, $C_{\Psi\circ T}=\sum_{i,j}E_{ij}\otimes\Psi(E_{ji})$ is the partial transpose of $C_\Psi$ on the first factor, which is the full transpose of $C_\Psi^{\Gamma}$. Likewise $C_{T\circ\Psi\circ T}=C_\Psi^{T}\succeq0$, so $\Psi\circ T$ is PPT. For \ref{sym:c}, $\Phi^{\#}(I)=\Phi^{*}(I)^{T}$, $(\Phi^{\#})^{*}(I)=\Phi(I)^{T}$, $(T\circ\Phi)(I)=\Phi(I)^{T}$, $(T\circ\Phi)^{*}(I)=\Phi^{*}(I)$, $(\Psi\circ T)(I)=\Psi(I)$ and $(\Psi\circ T)^{*}(I)=\Psi^{*}(I)^{T}$. \end{proof} \begin{lemma}\label{lem:reduce} Let $\Phi\colon\M_m\to\M_k$ and $\Psi\colon\M_k\to\M_n$ be PPT with $\Psi\circ\Phi\ne0$. There are isometries $V_1\colon\C^{m'}\to\C^m$, $V_2\colon\C^{k'}\to\C^k$ and $V_3\colon\C^{n'}\to\C^n$ such that the PPT maps $\tilde\Phi=\Ad_{V_2^{\dagger}}\circ\Phi\circ\Ad_{V_1}$ and $\tilde\Psi=\Ad_{V_3^{\dagger}}\circ\Psi\circ\Ad_{V_2}$ have positive definite marginals $\tilde\Phi(I)$, $\tilde\Phi^{*}(I)$, $\tilde\Psi(I)$, $\tilde\Psi^{*}(I)$, and \[ C_{\tilde\Psi\circ\tilde\Phi}=(\bar V_1\otimes V_3)^{\dagger}\,C_{\Psi\circ\Phi}\,(\bar V_1\otimes V_3),\qquad \ran C_{\Psi\circ\Phi}\subseteq\ran(\bar V_1\otimes V_3). \] In particular $(\Phi,\Psi)$ is a range-criterion pair if and only if $(\tilde\Phi,\tilde\Psi)$ is. \end{lemma} \begin{proof} Write $\Phi=\sum_r\Ad_{A_r}$ and $\Psi=\sum_s\Ad_{B_s}$. Let $P$ be the projection onto the support of $\Phi(I)$ and $Q$ the projection onto the support of $P\Psi^{*}(I)P$, so that $Q\le P$, and let $V_2$ be an isometry onto $\ran Q$. Then $A_r=PA_r$, and $\sum_sPB_s^{\dagger}B_sP=P\Psi^{*}(I)P$ gives $B_sP=B_sQ$. Hence $B_sA_r=B_sV_2V_2^{\dagger}A_r$. The maps $\Ad_{V_2^{\dagger}}\circ\Phi$ and $\Psi\circ\Ad_{V_2}$ have the same composition as $\Phi$ and $\Psi$. Their middle marginals $V_2^{\dagger}\Phi(I)V_2$ and $V_2^{\dagger}P\Psi^{*}(I)PV_2$ are positive definite. Next let $V_1$ be an isometry onto the support of $\Phi^{*}(Q)$ and $V_3$ an isometry onto the support of $\Psi(Q)$. Then $V_2^{\dagger}A_r=V_2^{\dagger}A_rV_1V_1^{\dagger}$ and $B_sV_2=V_3V_3^{\dagger}B_sV_2$. So these compressions leave the middle marginals unchanged and make the outer marginals definite. The Kraus operators of $\tilde\Psi\circ\tilde\Phi$ are $V_3^{\dagger}B_sA_rV_1$, and $B_sA_r=V_3V_3^{\dagger}B_sA_rV_1V_1^{\dagger}$. Since $\vecop(V_3^{\dagger}KV_1)=(V_1^{T}\otimes V_3^{\dagger})\vecop(K)$, this gives both displayed identities. Compositions with $\Ad_V$ preserve the PPT property. Finally, $(\bar V_1\otimes V_3)^{\dagger}$ maps $\ran C_{\Psi\circ\Phi}$ bijectively onto $\ran C_{\tilde\Psi\circ\tilde\Phi}$, and a product vector in $\ran(\bar V_1\otimes V_3)$ has its factors in $\ran\bar V_1$ and $\ran V_3$. So product vectors correspond on both sides. \end{proof} We call a PPT pair \emph{reduced} if all four marginals are positive definite, and we call $(m',k',n')$ in Lemma~\ref{lem:reduce} the \emph{reduced type}. Since $m'\le m$, $k'\le k$ and $n'\le n$, a statement that excludes all reduced types below $(m,k,n)$ excludes the type $(m,k,n)$. For a reduced pair the local ranks of $C_\Phi$ are $(m,k)$ and those of $C_\Psi$ are $(k,n)$. We write \[ a=\rank C_\Psi,\quad b=\rank C_\Phi,\quad S=\spn(\cK_\Psi\cK_\Phi),\quad s=\dim S=\rank C_{\Psi\circ\Phi}, \] so that $a=\dim\cK_\Psi$ and $b=\dim\cK_\Phi$. \subsection{Facts on PPT operators and on spaces of matrices} \begin{lemma}[\cite{HSTT03,HLVC00}]\label{lem:HLVC} Let $\rho\succeq0$ on $\C^p\otimes\C^q$ have positive partial transpose and local ranks $r_1,r_2$. Then $\rank\rho\ge\max(r_1,r_2)$, and if $\rank\rho\le\max(r_1,r_2)$, then $\rho$ is separable. \end{lemma} \begin{lemma}\label{lem:det} Let $0\le r<\min(p,q)$. The matrices of rank at most $r$ in $\M_{q\times p}$ form an irreducible cone of codimension $(p-r)(q-r)$ \cite{BV88}. Hence every subspace of $\M_{q\times p}$ of dimension greater than $(p-r)(q-r)$ contains a nonzero matrix of rank at most $r$. In particular a rank-one-free subspace of $\M_{q\times p}$, equivalently a product-free subspace of $\C^p\otimes\C^q$, has dimension at most $(p-1)(q-1)$ \cite{Par04}. \end{lemma} The second sentence is the projective dimension theorem \cite[Theorem~I.7.2]{Har77}. We also use that positive semidefinite operators with positive partial transpose on $\C^p\otimes\C^q$ with $pq\le6$ are separable \cite{Per96,HHH96}. \subsection{Rank-loss loci of PPT maps} For a completely positive map $\Phi\colon\M_m\to\M_k$ and $r\ge0$ let \[ D_r(\Phi)=\{[x]\in\PP^{m-1}:\rank\Phi(P_x)\le r\}=\{[x]:\dim\cK_\Phi x\le r\}. \] It is Zariski closed, cut out by minors of the matrix $(K_1x,\dots,K_bx)$ for a basis $K_1,\dots,K_b$ of $\cK_\Phi$. \begin{lemma}\label{lem:linear} Let $\Phi\colon\M_m\to\M_k$ be PPT and $r\ge0$. Then $D_r(\Phi)$ is a finite union of projective linear subspaces $\PP(A_1),\dots,\PP(A_p)$, and for each $j$ there is a subspace $W_j\subseteq\C^k$ with $\dim W_j\le r$ and $\cK_\Phi A_j\subseteq W_j$. \end{lemma} \begin{proof} Let $D$ be an irreducible component of $D_r(\Phi)$, let $r_D\le r$ be the maximum of $\dim\cK_\Phi x$ over $[x]\in D$, and let $D^{\circ}\subseteq D$ be the dense open subset where this maximum is attained. On $D^{\circ}$ the map $f(x)=\cK_\Phi x\in\Gr(r_D,\C^k)$ is a morphism. Since $T\circ\Phi$ is completely positive, $(T\circ\Phi)(P_x)=\overline{\Phi(P_x)}$ has range $\cK_{T\circ\Phi}x$, so $\overline{f(x)}=\cK_{T\circ\Phi}x$ is also a morphism on $D^{\circ}$. On the smooth part of $D^{\circ}$ the differential of $f$ is therefore both complex linear and complex antilinear, hence zero. This smooth part is connected, because $D$ is irreducible. So $f$ is constant, equal to some $W$, and since the condition $\cK_\Phi x\subseteq W$ is closed, it holds on all of $D$. By linearity $\cK_\Phi(\spn D)\subseteq W$, so $\PP(\spn D)\subseteq D_r(\Phi)$. This irreducible set contains the component $D$, so it equals $D$. \end{proof} \begin{corollary}\label{cor:blocks} Let $Z\succeq0$ on $\C^m\otimes\C^n$ with $Z^{\Gamma}\succeq0$. \begin{enumerate}[label=(\alph*)] \item\label{blocks:a} $x\otimes y\in\ker Z$ if and only if $x\otimes\bar y\in\ker Z^{\Gamma}$. \item\label{blocks:b} The set of pairs $([x],[y])$ with $x\otimes y\in\ker Z$ is a finite union of products $\PP(A_j)\times\PP(B_j)$ of linear subspaces. \end{enumerate} \end{corollary} \begin{proof} Part \ref{blocks:a} holds because $(x\otimes y)^{\dagger}Z(x\otimes y)=(x\otimes\bar y)^{\dagger}Z^{\Gamma}(x\otimes\bar y)$. For \ref{blocks:b}, $Z=C_F$ for a PPT map $F\colon\M_m\to\M_n$, and by \eqref{eq:basic} $x\otimes y\in\ker Z$ if and only if $y\perp\cK_F\bar x$. If $\dim\cK_F\bar x=r$, then $[\bar x]$ lies in a component $\PP(A)$ of $D_r(F)$ with $\cK_FA\subseteq W$ and $\dim W\le r$ (Lemma~\ref{lem:linear}), so $\cK_F\bar x=W$. Hence the set in question is the union of the products $\PP(\bar A)\times\PP(W^{\perp})$ over all $r$ and all components. \end{proof} A product vector in the kernel of the Choi matrix of a PPT map constrains its Kraus space as follows. \begin{lemma}\label{lem:kernelrank} Let $\Phi\colon\M_m\to\M_k$ be PPT, write $C_\Phi=\sum_{j=1}^{b}\vecop(Y_j)\vecop(Y_j)^{\dagger}$ with $Y_1,\dots,Y_b$ a basis of $\cK_\Phi$, and give $\cK_\Phi$ the inner product in which this basis is orthonormal. If $x\in\C^m$ and $v\in\C^k$ satisfy $v^{\dagger}Yx=0$ for all $Y\in\cK_\Phi$, that is, $\bar x\otimes v\in\ker C_\Phi$, then the orthogonal complement of $\{Y\in\cK_\Phi:Yx=0\}$ is contained in $\{Y\in\cK_\Phi:v^{\dagger}Y=0\}$. \end{lemma} \begin{proof} For $u\in\C^m$, $w\in\C^k$ and $Y\in\cK_\Phi$ we have $\ip{u\otimes w}{\vecop(Y)}=w^{\dagger}Y\bar u$, so $\ip{\bar x\otimes v}{C_\Phi(\bar x\otimes v)}=\sum_j|v^{\dagger}Y_jx|^2=0$. The partial transpose satisfies $\ip{u\otimes w}{C_\Phi^{\Gamma}(u'\otimes w')}=\ip{u\otimes\bar w'}{C_\Phi(u'\otimes\bar w)}$. Hence $\ip{\bar x\otimes\bar v}{C_\Phi^{\Gamma}(\bar x\otimes\bar v)}=0$, and $C_\Phi^{\Gamma}(\bar x\otimes\bar v)=0$ because $C_\Phi^{\Gamma}\succeq0$. For $u\in\C^m$ and $z\in\C^k$ this gives \[ 0=\ip{u\otimes z}{C_\Phi^{\Gamma}(\bar x\otimes\bar v)}=\ip{u\otimes v}{C_\Phi(\bar x\otimes\bar z)}=v^{\dagger}R_z\bar u,\qquad R_z=\sum_j\overline{\bar z^{\dagger}Y_jx}\,Y_j. \] So $v^{\dagger}R_z=0$ for all $z$. In the inner product of $\cK_\Phi$ we have $\ip{R_z}{Y}=\bar z^{\dagger}Yx$ for $Y\in\cK_\Phi$, so the $R_z$ span the orthogonal complement of $\{Y\in\cK_\Phi:Yx=0\}$. \end{proof} For a PPT map $\Phi$, conditions of the form $w^{T}Yx=0$ for all $Y\in\cK_\Phi$ that hold for polynomial families $x(t)$ and $w(t)$ only modulo a power $t^{m_0}$ with $m_0\ge1$ hold exactly for every pair of coefficients $x_a$, $w_b$ with $a+b0$, contradicting $nr$ for every nonzero $x\in\C^m$. \end{corollary} \begin{proof} Let $x\neq0$, put $W=\cK_\Phi x=\ran\Phi(P_x)$, and suppose that $\dim W\le r$. For every $v\in W^{\perp}$, Lemma~\ref{lem:kernelrank} puts the orthogonal complement of $\{Y\in\cK_\Phi:Yx=0\}$ into $\{Y\in\cK_\Phi:v^{\dagger}Y=0\}$, hence into $\{Y\in\cK_\Phi:\ran Y\subseteq W\}$, whose nonzero elements have rank at most $r$. So this complement is zero, $Yx=0$ for all $Y\in\cK_\Phi$, and $x^{\dagger}\Phi^{*}(I)x=\sum_j\|Y_jx\|^2=0$, a contradiction. \end{proof} \begin{corollary}\label{cor:linrank} Let $\Phi\colon\M_m\to\M_k$ be PPT with $\Phi(I)\succ0$, and suppose that $\cK_\Phi$ contains no nonzero matrix of rank at most $r$. If $A\subseteq\C^m$ and $W\subsetneq\C^k$ are subspaces with $\cK_\Phi A\subseteq W$, then $\dim A\le m-r-1$. In particular, every component $\PP(A)$ of $D_{k-1}(\Phi)$ in Lemma~\ref{lem:linear} has $\dim A\le m-r-1$. \end{corollary} \begin{proof} For $x\in A$ and $v\in W^{\perp}$ we have $v^{\dagger}Yx=0$ for all $Y\in\cK_\Phi$. By Lemma~\ref{lem:kernelrank} the orthogonal complement of $\{Y\in\cK_\Phi:Yx=0\}$ lies in $\{Y\in\cK_\Phi:v^{\dagger}Y=0\}$, and intersecting over $v\in W^{\perp}$ it lies in $\cK^W=\{Y\in\cK_\Phi:\ran Y\subseteq W\}$. Taking orthogonal complements, $(\cK^W)^{\perp}\subseteq\{Y\in\cK_\Phi:Yx=0\}$ for every $x\in A$, so the nonzero elements of $(\cK^W)^{\perp}$ have rank at most $m-\dim A$. If $\dim A\ge m-r$, then $(\cK^W)^{\perp}=0$, so every element of $\cK_\Phi$ has range in $W$ and $\ran\Phi(I)\subseteq W\ne\C^k$, a contradiction. \end{proof} The next lemma computes the length of these loci on a projective plane at an isolated point where the rank drops by one or by two. \begin{lemma}\label{lem:pptlen} Let $\Phi\colon\M_m\to\M_k$ be PPT, let $A\subseteq\C^m$ be a three-dimensional subspace and $W\subseteq\C^k$ a subspace with $\cK_\Phi A\subseteq W$. Let $D\subseteq\PP(A)$ be the subscheme defined by the Fitting ideal $\operatorname{Fitt}_0(\operatorname{coker}\sigma)$, that is, by the $(\dim W)$-minors, of the map $\sigma\colon\cK_\Phi\otimes\cO(-1)\to W\otimes\cO$, $Y\otimes x\mapsto Yx$, and let $[x]$ be an isolated point of $D$. If $\dim\cK_\Phi x=\dim W-1$, then $D$ has length one at $[x]$, and if $\dim\cK_\Phi x=\dim W-2$, then it has length three. \end{lemma} \begin{proof} Let $c=\dim W-\dim\cK_\Phi x$ and $\mathcal Y_x=\{Y\in\cK_\Phi:Yx=0\}$. Let $c_0$ be a linear form on $\C^k$ that vanishes on $\cK_\Phi x$ but not on $W$, and let $\xi\in A$ satisfy $c_0(Y\xi)=0$ for all $Y\in\mathcal Y_x$. Then $Y\mapsto c_0(Y\xi)$ factors through $Y\mapsto Yx$, so $c_0(Y\xi)+c_1(Yx)=0$ for all $Y\in\cK_\Phi$ and some linear form $c_1$. Write $c_j(v)=\gamma_j^{T}v$. As $\ip{u\otimes w}{\vecop(Y)}=w^{\dagger}Y\bar u$, the vectors $\bar x\otimes\bar\gamma_0$ and $\bar\xi\otimes\bar\gamma_0+\bar x\otimes\bar\gamma_1$ lie in $\ker C_\Phi$, and Corollary~\ref{cor:jordan}\ref{jor:b} puts $\bar\xi\otimes\bar\gamma_0$ into $\ker C_\Phi$, that is, $c_0(Y\xi)=0$ for all $Y\in\cK_\Phi$. So $c_0$ vanishes on $\cK_\Phi(x+t\xi)\subseteq W$ for all $t$, and the line through $[x]$ and $[\xi]$ lies in $D$ unless $\xi\in\C x$. As $[x]$ is isolated, $\xi\in\C x$. Choose complements $A_0$ of $\C x$ in $A$, $\cK_1$ of $\mathcal Y_x$ in $\cK_\Phi$, and $W_1$ of $\cK_\Phi x$ in $W$, and use the affine coordinates $\xi\in A_0$ near $[x]$. The block of $\sigma(x+\xi)$ from $\cK_1$ to $\cK_\Phi x$ is invertible over the local ring, so row and column operations, which preserve the Fitting ideal, bring $\sigma(x+\xi)$ to the form $\operatorname{diag}(I,S(\xi))$. Here $S(\xi)\colon\mathcal Y_x\to W_1$ vanishes at $\xi=0$, and its linear part is $D_1(\xi)\colon Y\mapsto Y\xi\bmod\cK_\Phi x$, with $W_1\cong W/\cK_\Phi x$. So $D$ is defined near $[x]$ by the $c\times c$ minors of $S$. By the first paragraph, $\xi\mapsto c_0\circ D_1(\xi)$ is injective on $A_0$ for every nonzero linear form $c_0$ on $W/\cK_\Phi x$. If $c=1$, the entries of $S$ therefore have linear parts spanning the dual of $A_0$, they generate the maximal ideal $\mathfrak m$ of the local ring, and the length is one. If $c=2$, the maps $c_0\mapsto(\xi\mapsto c_0\circ D_1(\xi))$ form a pencil of matrices of size $\dim\mathcal Y_x\times2$ and rank two at every $c_0\neq0$, so it has no blocks of type $L_\varepsilon$ and no regular part, and by \cite[Chapter~XII]{Gan59} it is equivalent to $L_2^{T}$ or to $L_1^{T}\oplus L_1^{T}$, padded with zero rows. In both cases the $2\times2$ minors of $D_1(\xi)$ span the quadratic forms in $\xi$, and this span does not depend on the choice of bases. The $2\times2$ minors of $S$ lie in $\mathfrak m^2$ and their initial forms span $\mathfrak m^2/\mathfrak m^3$, so they generate $\mathfrak m^2$ by Nakayama's lemma, and the length is three. \end{proof} Without the PPT property the lengths can be larger. For $\sigma(y)=\left(\begin{smallmatrix}y_0&0&y_2\\y_2&y_1&0\end{smallmatrix}\right)$ on $\PP^2$, the point $[1:0:0]$ is isolated, $\sigma$ has rank one there, and the length is two. In this example the first paragraph of the proof fails for $\xi=e_2$. \begin{corollary}\label{cor:pptlen} In Lemma~\ref{lem:pptlen}, suppose that $\cK_A=\{Y|_A:Y\in\cK_\Phi\}$ has dimension $\dim W+1$, that $D$ is finite, and that $\dim\cK_\Phi x\ge\dim W-2$ for every nonzero $x\in A$. Then $n_1+3n_2=\binom{\dim W+1}{2}$, where $n_c$ is the number of points $[x]\in D$ with $\dim\cK_\Phi x=\dim W-c$. \end{corollary} \begin{proof} Put $w=\dim W$. The scheme $D$ is also defined by the $w$-minors of $\cK_A\otimes\cO(-1)\to W\otimes\cO$, and as it is finite, it has the expected codimension two. By \cite[Theorem~14.4(c)]{Ful98} its length is $c_1^2-c_2$ of $(1-h)^{-(w+1)}$, that is, $(w+1)^2-\binom{w+2}{2}=\binom{w+1}{2}$. Lemma~\ref{lem:pptlen} gives the length at each point. \end{proof} The proof of Lemma~\ref{lem:kernelrank} also shows that the range of $\Phi(P_x)$ is spanned by its intersections with the ranges at any points spanning $x$. \begin{lemma}\label{lem:pptspan} Let $\Phi\colon\M_m\to\M_k$ be PPT, and let $x\in\spn(y_1,\dots,y_r)\subseteq\C^m$. Then $\cK_\Phi x=\sum_i(\cK_\Phi x\cap\cK_\Phi y_i)$. \end{lemma} \begin{proof} Let $U=\cK_\Phi x$, write $x=\sum_i\lambda_iy_i$, and let $\mathcal Y_x^{\perp}$ be the orthogonal complement of $\{Y\in\cK_\Phi:Yx=0\}$ in the inner product of Lemma~\ref{lem:kernelrank}. For every $v$ with $v^{\dagger}U=0$, Lemma~\ref{lem:kernelrank} gives $v^{\dagger}Y=0$ for all $Y\in\mathcal Y_x^{\perp}$, so every element of $\mathcal Y_x^{\perp}$ has range in $U$. The map $Y\mapsto Yx$ sends $\mathcal Y_x^{\perp}$ onto $U$, and $Yx=\sum_i\lambda_iYy_i$ with $Yy_i\in U\cap\cK_\Phi y_i$. \end{proof} \begin{corollary}\label{cor:pptspan} Let $\Phi\colon\M_m\to\M_k$ be PPT with $\Phi^{*}(I)\succ0$, and suppose that $\cK_\Phi$ contains no nonzero matrix of rank at most two. Let $x\in\spn(y_1,y_2,y_3)$ be nonzero and $U=\cK_\Phi x$. Then the three spaces $U\cap\cK_\Phi y_i$ do not all have dimension at most one with two of them equal. \end{corollary} \begin{proof} Otherwise $\dim U\le2$ by Lemma~\ref{lem:pptspan}, contradicting Corollary~\ref{cor:rankgap} with $r=2$. \end{proof} \section{Rank windows and impossibility}\label{sec:windows} This section bounds the Choi ranks of a reduced range-criterion pair from both sides and excludes the reduced types with middle dimension five and outer dimensions at most six. Throughout this section $(\Phi,\Psi)$ is a reduced range-criterion pair of type $(m,k,n)$, and $a$, $b$, $S$, $s$ are as in Section~\ref{sec:prelim}. By Lemma~\ref{lem:sym}, $(T\circ\Phi,\Psi\circ T)$ and $(\Psi^{\#},\Phi^{\#})$ are reduced range-criterion pairs of types $(m,k,n)$ and $(n,k,m)$. Every statement about $\rank C_\Phi$ and $\rank C_\Psi$ below therefore also holds for $\rank C_\Phi^{\Gamma}$ and $\rank C_\Psi^{\Gamma}$, and every statement has a mirror version with the roles of $\Phi$ and $\Psi$ exchanged. \subsection{No Kraus operators of rank at most two} \begin{lemma}\label{lem:noranktwo} Neither $\cK_\Phi$ nor $\cK_\Psi$ contains a nonzero matrix of rank at most two. In particular $m,k,n\ge3$. \end{lemma} \begin{proof} Suppose $Y_0=uw^{T}\in\cK_\Phi$ has rank one. For $X\in\cK_\Psi$ the product $XY_0=(Xu)w^{T}$ has rank at most one and lies in $S$, so $Xu=0$. Then $u^{\dagger}\Psi^{*}(I)u=\sum_s\|X_su\|^2=0$ for the Kraus operators $X_s$ of $\Psi$, which contradicts $\Psi^{*}(I)\succ0$. Suppose $Y_1\in\cK_\Phi$ has rank two, and let $U=\ran Y_1$. If $X\in\cK_\Psi$ satisfies $\rank(X|_U)=1$, then $\rank XY_1=1$, which is excluded. So $\cK_\Psi|_U=\{X|_U:X\in\cK_\Psi\}\subseteq\Hom(U,\C^n)$ is rank-one-free. It is the Kraus space of the map $\Psi_U=\Psi\circ\Ad_{\iota_U}\colon\M_2\to\M_n$, where $\iota_U\colon\C^2\to\C^k$ is an isometry onto $U$. This map is PPT and nonzero, since $\iota_U^{\dagger}\Psi^{*}(I)\iota_U\succ0$. Let $n_U$ be the rank of $\Psi_U(I)$, the output local rank of $C_{\Psi_U}$. The space $\cK_\Psi|_U$ lies in $\Hom(U,\ran\Psi_U(I))\cong\C^2\otimes\C^{n_U}$ and is product-free, so $\rank C_{\Psi_U}=\dim\cK_\Psi|_U\le n_U-1$ by Lemma~\ref{lem:det}. This contradicts Lemma~\ref{lem:HLVC}. The statements for $\cK_\Psi$ follow by applying the above to the mirror pair $(\Psi^{\#},\Phi^{\#})$, whose first factor has Kraus space $\cK_\Psi^{T}$. If $k\le2$ or $m\le2$, every element of $\cK_\Phi$ has rank at most two, so $\cK_\Phi=0$, contradicting $\Phi(I)\succ0$. In the same way $n\ge3$. \end{proof} \subsection{PPT states of low rank} The next lemma gives the lower bounds in the rank windows below. Part \ref{lowrank:a} uses the first half of the normal form in the proof of \cite[Theorem~43]{CD13}. \begin{lemma}\label{lem:lowrank} Let $\rho\succeq0$ on $\C^M\otimes\C^N$ have positive partial transpose and local ranks $M$ and $N$, and suppose that $\ran\rho$ contains no nonzero vector of Schmidt rank at most two. \begin{enumerate}[label=(\alph*)] \item\label{lowrank:a} If $M\ge3$, then $\rank\rho\ge N+2$. \item\label{lowrank:b} If $M\ge5$ and $\rank\rho=N+2$, then $N\ge3M-8$. \end{enumerate} \end{lemma} \begin{proof} By Lemma~\ref{lem:HLVC}, $\rank\rho\ge N$, and $\rank\rho=N$ would make $\rho$ separable, with product vectors in its range. So $\rank\rho=N+p$ with $p\ge1$. We treat $p\in\{1,2\}$ and assume $M\ge p+2$, which holds in both parts. Write $\rho=C^{\dagger}C$ with $C=\sum_{i=0}^{M-1}\bra{i}\otimes C_i$, where the $C_i$ are $(N+p)\times N$ matrices and the rows of $C$ are linearly independent. For $c\in\C^{N+p}$ the vector $C^{\dagger}c=\sum_i\ket i\otimes C_i^{\dagger}c$ lies in $\ran\rho$. Its Schmidt rank is the dimension of the span of the vectors $C_i^{\dagger}c$, so it is at most the number of indices $i$ with $C_i^{\dagger}c\ne0$, and at most the dimension of any subspace containing all $C_i^{\dagger}c$. For $\alpha\in\C^M$ put $C_\alpha=\sum_i\alpha_iC_i$. Local invertible operations and unitaries acting on the left of $C$ preserve all hypotheses and the Schmidt ranks. We use two facts. (P) If $C_\alpha\beta=0$, then $C_j\beta\perp\ran C_\alpha$ for all $j$. Indeed, the partial transpose $\rho^{\Gamma_1}$ on the first factor satisfies $(\bar\alpha\otimes\beta)^{\dagger}\rho^{\Gamma_1}(\bar\alpha\otimes\beta)=\|C_\alpha\beta\|^2$, so $\rho^{\Gamma_1}(\bar\alpha\otimes\beta)=\sum_j\ket j\otimes C_\alpha^{\dagger}C_j\beta$ vanishes. (D) Since $\dim\ker\rho=MN-N-p>(M-1)(N-1)$, the kernel contains a product vector $\alpha\otimes\beta$ (Lemma~\ref{lem:det}), and then $C_\alpha\beta=0$. Let $R$ be the largest value of $\rank C_\alpha$ that is smaller than $N$. Then $R\ge1$, because $C_\alpha=0$ would put $\alpha\otimes\C^N$ into $\ker\rho$, contradicting the local rank $M$. After local operations and a unitary acting on the left of $C$ we may assume that $\rank C_0=R$ and $C_0=\left(\begin{smallmatrix}I_R&0\\0&0\end{smallmatrix}\right)$, with rows split as $R\mid N+p-R$ and columns as $R\mid q$, where $q=N-R$. By (P) applied to $C_0$, for $i\ge1$ we have $C_i=\left(\begin{smallmatrix}C_{i0}&0\\C_{i2}&C_{i3}\end{smallmatrix}\right)$ with $C_{i3}$ of size $(q+p)\times q$. The operator $\sigma=C'^{\dagger}C'$ with $C'=\sum_{i\ge1}\bra i\otimes C_{i3}$ is the compression of $\rho$ to $\spn(e_1,\dots,e_{M-1})\otimes\spn(e_R,\dots,e_{N-1})$, so it has positive partial transpose, its second local rank is $q$, and $\rank\sigma=q+t$ with $0\le t\le p$. For $\xi\in\C^{M}$ with $\xi_0=0$ and small $\epsilon\ne0$, $\rank C_{e_0+\epsilon\xi}=R+\rank C_{\xi3}$, where $C_{\xi3}=\sum_i\xi_iC_{i3}$. By maximality of $R$, every $C_{\xi3}$ has rank $0$ or $q$. Let $\mu$ be the first local rank of $\sigma$, that is, $\mu=\dim\spn\{C_{i3}\}$. If $t=0$, then $\sigma$ is separable with $\rank\sigma=q$ (Lemma~\ref{lem:HLVC}), so $\sigma=\sum_{l=1}^{q}P_{\alpha_l}\otimes P_{\beta_l}$ with $\beta_1,\dots,\beta_q$ linearly independent \cite[Theorem~1]{HLVC00}. After a unitary on the lower rows and local operations we get $C_{i3}=\left(\begin{smallmatrix}\Lambda_i\\0\end{smallmatrix}\right)$ with diagonal $\Lambda_i$ and $(\Lambda_i)_{11}=\delta_{i1}$. Then $C_0e_R=0$, $C_ie_R=0$ for $i\ge2$, and $C_1e_R$ is the first lower unit vector. By (P), this row of $C$ vanishes in every block $C_i$ with $i\ne1$, so it gives a product vector in $\ran\rho$. Hence $t\ge1$. If $\mu>t+1$, then $\sigma$ restricted to $\ran\sigma_1\otimes\C^q$, where $\sigma_1$ is the first marginal, has kernel of dimension $\mu q-q-t>(\mu-1)(q-1)$, which contains a product vector $\xi\otimes\beta$. Then $C_{\xi3}\beta=0$ with $C_{\xi3}\neq0$, which contradicts the dichotomy. Hence $\mu\le t+1$. After a change of basis of $\spn(e_1,\dots,e_{M-1})$ we have $C_{i3}=0$ for $i>\mu$. For such $i$ and every $\beta\in\spn(e_R,\dots,e_{N-1})$, (P) shows that $\ran C_i$ is orthogonal to the column space of $C'$, which has dimension $q+t$ inside the $q+p$ lower rows. In an orthonormal basis of the lower rows adapted to this column space, the first $q+t$ lower rows of $C$ vanish in all blocks except $1,\dots,\mu$. They give range vectors of Schmidt rank at most $\mu$, so $\mu\ge3$ and $t\ge2$. This proves \ref{lowrank:a}, since $p=1$ would give $t\le1$. For \ref{lowrank:b} let $p=2$, so that $t=2$ and $\mu=3$. Then $C_{i2}=0$ and $C_{i3}=0$ for $i\ge4$. The compression $\tau$ of $\rho$ to $\spn(e_0,e_4,\dots,e_{M-1})\otimes\spn(e_0,\dots,e_{R-1})$ equals $D^{\dagger}D$ with $D=(I_R,C_{40},\dots,C_{M-1,0})$. It has positive partial transpose, second local rank $R$ and rank $R$. By \cite[Theorem~1]{HLVC00} it is a sum of $R$ product projections with linearly independent second factors. Comparing the two factorisations of $\tau$ shows that the $C_{i0}$ with $i\ge4$ are simultaneously unitarily diagonalisable, and after a unitary change of the first $R$ coordinates they are diagonal, $C_{i0}=\Delta_i$, while $C_0$ keeps its form. Group the coordinates $0,\dots,R-1$ into blocks $B_1,\dots,B_\kappa$ according to the tuple of diagonal entries $((\Delta_i)_{ss})_{i\ge4}$. If $s\in B_r$ has tuple $d^{(r)}$, then $\alpha=e_i-d^{(r)}_ie_0$ satisfies $C_\alpha e_s=0$. By (P), $C_je_s$ is orthogonal to $\ran C_\alpha$ for all $j$ and all $i\ge4$, and these ranges span the top coordinates outside $B_r$. So in every $C_j$ the column $s$ has no entries in the top rows outside $B_r$. As the blocks partition the top coordinates and the top right block of every $C_j$ is zero, the top rows of $C$ with index in $B_r$ are supported on the columns in $B_r$, and they give range vectors of Schmidt rank at most $|B_r|$. Hence $|B_r|\ge3$ for every $r$. The matrices $C_0,C_4,\dots,C_{M-1}$ are linearly independent, because the first local rank of $\rho$ is $M$, and they lie in the $\kappa$-dimensional space of matrices $\left(\begin{smallmatrix}\Delta&0\\0&0\end{smallmatrix}\right)$ with $\Delta$ diagonal and constant on each block. So $M-3\le\kappa$, and $N-1\ge R=\sum_r|B_r|\ge3(M-3)$. \end{proof} \subsection{Rank windows} \begin{theorem}\label{thm:windows} Let $(\Phi,\Psi)$ be a reduced range-criterion pair of type $(m,k,n)$. Then $m,k,n\ge4$ and \[ \max(m,k)+2\le b\le(k-2)(m-2),\qquad\max(k,n)+2\le a\le(k-2)(n-2). \] The same bounds hold for $\rank C_\Phi^{\Gamma}$ and $\rank C_\Psi^{\Gamma}$. \end{theorem} \begin{proof} By Lemma~\ref{lem:noranktwo}, $m,k,n\ge3$, and $\ran C_\Phi$ contains no vector of Schmidt rank at most two. Lemma~\ref{lem:lowrank}\ref{lowrank:a}, applied with $M=\min(m,k)$ and $N=\max(m,k)$ after exchanging the two tensor factors if $m>k$, gives the lower bound for $b$. If $b>(k-2)(m-2)$, then $\cK_\Phi$ contains a nonzero matrix of rank at most two by Lemma~\ref{lem:det}, contradicting Lemma~\ref{lem:noranktwo}. The bounds for $a$ follow in the same way, and those for the partial transposes from Lemma~\ref{lem:sym}\ref{sym:b}. If $k=3$, the window for $b$ is empty, and the same holds if $m=3$ or $n=3$. \end{proof} \begin{corollary}\label{cor:windows} \begin{enumerate}[label=(\alph*)] \item\label{cw:a} There is no range-criterion pair with $k\le3$, with $\min(m,n)\le3$, with $k=4$ and $\min(m,n)\le5$, or with $\min(m,n)=4$ and $k\le5$. In particular there is none with $\max(m,k,n)\le4$. \item\label{cw:b} A reduced range-criterion pair with $\max(m,k,n)\le5$ has type $(5,5,5)$ and $a,b\in\{7,8,9\}$. \item\label{cw:c} A reduced range-criterion pair with $\max(m,k,n)=6$ has one of the types $(6,4,6)$, $(5,5,6)$, $(6,5,5)$, $(6,5,6)$, $(4,6,4)$, $(4,6,5)$, $(5,6,4)$, $(4,6,6)$, $(6,6,4)$, $(5,6,5)$, $(5,6,6)$, $(6,6,5)$, $(6,6,6)$. \end{enumerate} \end{corollary} \begin{proof} For a reduced type, the window for $b$ is nonempty only if $\max(m,k)+2\le(k-2)(m-2)$. With $k=4$ this forces $m\ge6$, and with $m=4$ it forces $k\ge6$. The same holds for $(k,n)$. Part \ref{cw:a} follows because reduction does not increase dimensions, and \ref{cw:b}, \ref{cw:c} are the reduced types allowed by these conditions. \end{proof} In particular the range criterion never certifies a counterexample through the middle algebra $\M_3$, whatever the outer dimensions. For the pair of \cite{OAI26} the windows read $12\le a,b\le32$, and $a=b=15$. \subsection{Middle dimension five} The next lemma is the topological input. For $1\le d_12n-2$, against the bound $a\le2n-2$ of Theorem~\ref{thm:46n}. At $(4,6,6)$ they give $a\ge9$, and Theorem~\ref{thm:46n} gives $a\le10$. These are the bounds used in Theorem~\ref{thm:466}, which excludes this type. The types $(5,6,4)$ and $(6,6,4)$ are mirrors, and Corollary~\ref{cor:windows} passes from original to reduced types. \subsection{The type \texorpdfstring{$(5,6,5)$}{(5,6,5)}} \begin{proposition}\label{prop:565} Let $(\Phi,\Psi)$ be a reduced range-criterion pair of type $(5,6,5)$. \begin{enumerate}[label=(\alph*)] \item\label{565:a} The set $\cG$ of members is infinite. \item\label{565:b} $S^{\perp}$ contains a nonzero matrix of rank at most two. \end{enumerate} \end{proposition} \emph{Sketch} (proof in Section~S3.6). Here $a=b=9$ by Proposition~\ref{prop:cuts}\ref{cuts:a}. If $\cG$ were finite, the zero scheme of the section $([Y],E)\mapsto Y|_E$ of a bundle of rank $12$ on $\PP(\cK_\Phi)\times\Gr(2,5)$ would be a disjoint union of planes $\PP(M_U)$, counted with multiplicities $m_U\ge1$, and two Chern numbers would give $\sum_Um_U=175$ and $3\sum_Um_U=210$. If $S^{\perp}$ contained no nonzero matrix of rank at most two, then $s=16$, and an exact sequence of vector bundles on $\PP(M_U)\cong\PP^2$ would contradict the Whitney formula. The map $\bar\rho$ in part~\ref{two:b} of the next theorem is defined in its proof, in Section~S3.6 of the supplement. \begin{theorem}\label{thm:565two} Let $(\Phi,\Psi)$ be a reduced range-criterion pair of type $(5,6,5)$. Then $s=16$, and $\ker C_{\Psi\circ\Phi}$ contains exactly two product vectors, $\bar z_1\otimes\bar w_1$ and $\bar z_2\otimes\bar w_2$, up to scalar multiples. Moreover: \begin{enumerate}[label=(\alph*)] \item\label{two:a} The matrices of rank at most two in $\PP(S^{\perp})$ form a finite union of points and lines. Each of these lines is $\PP(\Lambda)$, where $\Lambda=\{A\in S^{\perp}:\ran A\subseteq E,\ AR=0\}$ has dimension two for a plane $E$ and a $3$-space $R$ and contains $z_1w_1^{T}$ or $z_2w_2^{T}$. One of these lines passes through $[z_1w_1^{T}]$ and $[z_2w_2^{T}]$. \item\label{two:b} Let $\{i,j\}=\{1,2\}$ and $W_i=\cK_\Phi z_i$. Then $\dim(M_{W_i}\cap\mathcal Y_{z_j})=2$, and a nonzero $Y\in M_{W_i}$ satisfies $\{M|_{\ker Y}:M\in S\}\neq\Hom(\ker Y,\C^5)$ if and only if $Yz_j=0$. On $\PP(M_{W_i})$, the determinant of the map $\bar\rho$ in the proof vanishes to order nine along $\PP(M_{W_i}\cap\mathcal Y_{z_j})$. \end{enumerate} \end{theorem} \begin{theorem}\label{thm:565no} There is no reduced range-criterion pair of type $(5,6,5)$. \end{theorem} \emph{Sketch} (proofs in Section~S3.6). For a kernel product $(z,w)$, Corollary~\ref{cor:rankgap} and Lemma~\ref{lem:kernelrank} show that $W=\cK_\Phi z$ has dimension three, and they determine $M_W$ and $N_W$. Distinct kernel products have distinct inputs, and three of them would put their outputs on one rational normal curve, which contradicts the description of the spaces $N_W$. So there are at most two. An analysis of the matrices of rank at most two in $\PP(S^{\perp})$ and of the points $[Y]$ of $\PP(M_U)$ at which the restriction map from $S$ to $\Hom(\ker Y,\C^5)$ is not surjective then shows that kernel products exist, that there are exactly two, and that $s=16$. For Theorem~\ref{thm:565no}, the two kernel products give a $2\times2$ map $\Theta$ of vector bundles on a projective plane whose determinant is a constant multiple of $\lambda^8$ for a linear form $\lambda$. Two computations of the Euler characteristic of the cokernel of $\Theta$ give $8d-20$, with $d$ the degree of a line bundle on the line $\{\lambda=0\}$, and $8$, which is impossible. \subsection{The types \texorpdfstring{$(5,6,6)$ and $(6,6,5)$}{(5,6,6) and (6,6,5)}} In the next theorem, $\mathcal A$ is the set of the pairs $([X],[Y])\in\PP(\cK_\Psi)\times\PP(\cK_\Phi)$ with $XY=0$, and $\mathcal R^{\mathrm b}$ is its projection to $\PP(\cK_\Phi)$, as defined in Section~S3.6 of the supplement. \begin{theorem}[\ca]\label{thm:566inc} Let $(\Phi,\Psi)$ be a reduced range-criterion pair of type $(5,6,6)$. Then \[ a+b-18\le\dim\mathcal R_X\le a-6,\qquad\dim\mathcal R^{\mathrm b}\le b-5,\qquad\dim\mathcal A\ge a+2b-27, \] $\dim\mathcal R^{\mathrm b}\le b-6$ if $b\ge10$, $\dim\mathcal R^{\mathrm b}\ge a+b-18$ if $a\le11$, and $s\ge\max(a+6,b+5)$. The mirror statements hold at type $(6,6,5)$. \end{theorem} \emph{Sketch} (proof in Section~S3.6). The lower bounds on $\dim\mathcal R_X$ and $\dim\mathcal R^{\mathrm b}$ come from the incidence count of Lemma~\ref{lem:incidence}, with a section of a bundle of rank $27$ on a variety of dimension $27$. If $\dim\mathcal R_Xn$ are their mirrors. The same ranges hold with $(a,b)$ replaced by $(\rank C_\Psi^{\Gamma},\rank C_\Phi^{\Gamma})$. In the rows that cite only Corollary~\ref{cor:ns2gen}, the ranges follow from the bounds stated after Remark~\ref{rem:747res}, and in the other rows they follow from these bounds and the results cited in the row or in the notes below the table. These bounds also give $b=9$ if $a\ge13$ at $(5,6,7)$ and $b\le12$ if $a\ge13$ at $(6,6,7)$. \begin{table}[htbp] \centering \small \begin{tabular}{@{}lccp{0.52\textwidth}@{}} \toprule reduced type & $a=\rank C_\Psi$ & $b=\rank C_\Phi$ & further restrictions \\ \midrule $(6,4,7)$ & -- & -- & excluded (Thm.~\ref{thm:647no}\ca{}) \\ $(7,4,7)$ & $9$ & $9$ & $s=36$ (Thms.~\ref{thm:747}\ca{} and~\ref{thm:747b}\ca{}) \\ $(5,5,7)$ & -- & -- & excluded (Thm.~\ref{thm:557}\ca{}) \\ $(6,5,7)$ & $10$ & $9$ & Rem.~\ref{rem:657}\ca{}, $s=30$ (Prop.~\ref{prop:657s}\ca{}) \\ $(7,5,7)$ & $10$--$11$ & $10$--$11$ & $(a,b)\ne(11,11)$ (Rem.~\ref{rem:757}\ca{}), $34\le s\le36$ if $(a,b)\neq(10,10)$, and $33\le s\le36$ if $(a,b)=(10,10)$ (Prop.~\ref{prop:757s}\ca{}) \\ $(4,6,7)$ & -- & -- & excluded (Thm.~\ref{thm:467}\ca{}) \\ $(5,6,7)$ & $10$--$15$ & $9$--$11$ & $b=9$ if $a\ge13$, $s\ge a+8$ if $b=9$, and $s\ge a+9$ if $b\ge10$ (Thm.~\ref{thm:567s}\ca{}) \\ $(6,6,7)$ & $10$--$15$ & $10$--$14$ & $b\le12$ if $a\ge13$, $(a,b)\neq(12,14)$; $XY=0$ for some nonzero $X\in\cK_\Psi$, $Y\in\cK_\Phi$ if $a+b\ge23$; $s\ge a+b+5$ if $\ker C_{\Psi\circ\Phi}$ contains no product vector, which holds if $b\ge13$, and $s\ge a+b+3$ if $b\in\{11,12\}$\textsuperscript{a}; $s\ge b+6$ (Prop.~\ref{prop:k6rank6}) \\ $(7,6,7)$ & $10$--$16$ & $10$--$16$ & $a\le15$ if $b\ge13$, $b\le15$ if $a\ge13$, $\{a,b\}=\{10,16\}$ if $\max(a,b)=16$, $s\ge\max(a,b)+7$\textsuperscript{b} \\ $(4,7,4)$ & -- & -- & excluded (Thm.~\ref{thm:44}) \\ $(4,7,5)$ & -- & -- & excluded (Thm.~\ref{thm:475no}\ca{}) \\ $(4,7,6)$ & -- & -- & excluded (Thm.~\ref{thm:476}\ca{}) \\ $(4,7,7)$ & $10$--$11$ & $9$--$10$ & $s\in\{17,18\}$; $(a,s)=(10,18)$ if $b=10$; $s=18$ or $(a,s)=(10,17)$ if $b=9$ (Thm.~\ref{thm:477p}\ca{}) \\ $(5,7,5)$ & $10$--$11$ & $10$--$11$ & $s=16$ if $(a,b)\neq(10,10)$, and $15\le s\le16$ if $(a,b)=(10,10)$\textsuperscript{c} \\ $(5,7,6)$ & $10$--$15$ & $10$--$13$ & $a\le14$ if $b\ge11$, $a=10$ if $b=13$; $s\ge\max(a+2,14)$, $s\ge a+3$ if $a\ge15$, $s\ge a+6$ if $b\ge12$, $s\ge a+7$ if $b=13$\textsuperscript{d} \\ $(5,7,7)$ & $10$--$18$ & $10$--$14$ & $a\le17$ if $b\ge11$, $a\le15$ if $b\ge13$; $s\ge\max(a+2,15)$, $s\ge a+3$ if $a\ge18$, $s\ge a+7$ if $b=12$, $s\ge a+8$ if $b\in\{13,14\}$\textsuperscript{e} \\ $(6,7,6)$ & $10$--$19$ & $10$--$19$ & $a\le15$ if $b\ge13$, $b\le15$ if $a\ge13$ (Cor.~\ref{cor:ns2gen}) \\ $(6,7,7)$ & $10$--$23$ & $10$--$19$ & $a\le18$ if $b\ge13$, $b\le15$ if $a\ge17$ (Cor.~\ref{cor:ns2gen}) \\ $(7,7,7)$ & $10$--$23$ & $10$--$23$ & $a\le18$ if $b\ge17$, $b\le18$ if $a\ge17$ (Cor.~\ref{cor:ns2gen}) \\ \bottomrule \end{tabular} \par\smallskip\parbox{\linewidth}{\raggedright\footnotesize Sources of the marked restrictions. \textsuperscript{a}\,$(6,6,7)$: Thms.~\ref{thm:667corner}\ca{}, \ref{thm:667cong}\ca{}, \ref{thm:667c4}\ca{}, Cors.~\ref{cor:667b15}\ca{}, \ref{cor:667c4}\ca{}, \ref{cor:667b1112}\ca{}. \textsuperscript{b}\,$(7,6,7)$: Bounds after Rem.~\ref{rem:747res}\ca{}, Thms.~\ref{thm:767corner}\ca{} and~\ref{thm:767cong}\ca{}, Prop.~\ref{prop:k6rank6}. \textsuperscript{c}\,$(5,7,5)$: Thms.~\ref{thm:57n}, \ref{thm:57big}, \ref{thm:575c}, \ref{thm:57b12}\ca{}, \ref{thm:57M}, \ref{thm:575nine}\ca{}, \ref{thm:575two}\ca{}, \ref{thm:575x15}\ca{}, \ref{thm:575t16}\ca{} and~\ref{thm:575t14}\ca{}. \textsuperscript{d}\,$(5,7,6)$: Thms.~\ref{thm:57n}, \ref{thm:57big}, \ref{thm:57b12}\ca{}, \ref{thm:57M}, \ref{thm:576}\ca{}, \ref{thm:576e}\ca{}, Prop.~\ref{prop:57b10}, Cor.~\ref{cor:ns2gen}. \textsuperscript{e}\,$(5,7,7)$: Thms.~\ref{thm:57n}, \ref{thm:57big}, \ref{thm:57b12}\ca{}, \ref{thm:57M}, \ref{thm:577f}, \ref{thm:577g}\ca{}, \ref{thm:577h15}\ca{}, \ref{thm:577k}\ca{}, Prop.~\ref{prop:57b10}, Cor.~\ref{cor:ns2gen}.} \caption{The reduced types $(m,k,n)$ with $\max(m,k,n)=7$ and $m\le n$ that Theorem~\ref{thm:windows} allows, and the possible Choi ranks of a range-criterion pair.}\label{tab:d7} \end{table} \begin{theorem}[\ca]\label{thm:467} There is no reduced range-criterion pair of type $(4,6,7)$, $(7,6,4)$, $(4,6,8)$ or $(8,6,4)$. A reduced range-criterion pair of type $(4,6,9)$ has $a=\rank C_\Psi^{\Gamma}=12$ and $s=24$, and one of type $(9,6,4)$ has $b=\rank C_\Phi^{\Gamma}=12$ and $s=24$. Consequently a reduced range-criterion pair of type $(4,6,n)$ has $n\ge9$, and there is no range-criterion pair of type $(4,6,n)$ or $(n,6,4)$ with $n\le8$. \end{theorem} \begin{theorem}[\ca]\label{o469:thm:469} There is no reduced range-criterion pair of type $(4,6,9)$ or $(9,6,4)$. Consequently there is no range-criterion pair of type $(4,6,n)$ or $(n,6,4)$ with $n\le9$. \end{theorem} \emph{Sketch} (proofs in Section~S3.31). At type $(4,6,n)$ with $7\le n\le9$, Proposition~\ref{prop:46nbound}, the rank windows with Lemma~\ref{lem:lowrank}\ref{lowrank:b}, and Lemma~\ref{lem:det} leave ten triples $(n,a,s)$, with $b=8$. On the incidence $\mathcal T\subseteq\PP^7\times\PP^3$ of the pairs $([Y],[z])$ with $Yz=0$, the multiplication map $X\otimes Y\mapsto XY$ has a degeneracy locus whose Thom--Porteous class is computed in exact arithmetic, and the proof leaves only $(a,s)=(12,24)$ at $n=9$. For Theorem~\ref{o469:thm:469}, a bad member $U$ with $n_U=\dim\cK_\Psi U=8$, given by the results that precede its proof in the supplement, carries on $\PP(U)\cong\PP^2$ a map of bundles of ranks nine and eight whose locus of rank at most seven has the class $36h^2\neq0$. So some $x\in U\setminus0$ has $w(x)=\dim\cK_\Psi x\le7$, while the results that precede that proof exclude every value $w(x)\le7$. \begin{theorem}[\ca]\label{thm:477p} Let $(\Phi,\Psi)$ be a reduced range-criterion pair of type $(4,7,7)$. Then $s\in\{17,18\}$ and $a\le11$. If $b=10$, then $(a,s)=(10,18)$, and if $b=9$, then $s=18$ or $(a,s)=(10,17)$. The same holds with $(a,b)$ replaced by $(\rank C_\Psi^{\Gamma},\rank C_\Phi^{\Gamma})$, and for the mirror type $(7,7,4)$ with $a$ and $b$ exchanged. \end{theorem} \emph{Sketch} (proof in Section~S3.31). At type $(4,7,7)$, the rank windows and Lemma~\ref{lem:lowblocks} give $b\in\{9,10\}$ and $a\ge10$, and $s\le18$ by Lemma~\ref{lem:det}. The proof uses the incidence of the proofs of Theorems~\ref{thm:475no} and~\ref{thm:476} and proceeds in twenty steps. It lists the possible triples $(a,s,d)$, where $d$ is an integer attached to the spaces $N_U$ of the members, and it excludes the triples with $d\ge1$ by Thom--Porteous counts, an analysis of annihilators of rank two and of fixed subspaces, and, for $b=9$, a small resolution of the incidence variety. The triples that remain have $d=0$, $s\in\{17,18\}$ and $a\le11$. \subsection{Unextendible product sets} A set of product vectors is an \emph{unextendible product set} if no product vector is orthogonal to all of them. The unextendible product bases of \cite{BDMSST99} are such sets of mutually orthogonal vectors. The certificate of \cite{OAI26} is of this form: the kernel of the composite Choi matrix is spanned by twenty product vectors that form an unextendible product set. \begin{proposition}\label{prop:ups} Let $\Phi\colon\M_m\to\M_k$ and $\Psi\colon\M_k\to\M_n$ be PPT with $Z=C_{\Psi\circ\Phi}\neq0$. \begin{enumerate}[label=(\alph*)] \item\label{ups:a} If $m=n=4$, $k\le7$ and $\ran Z$ is product-free, then $\ker Z$ contains no product vector. \item\label{ups:b} If $m=n=4$, $k\le9$ and $\ran Z$ is product-free, then any two product vectors in $\ker Z$ are proportional. \item\label{ups:c} The kernel of $Z$ contains no unextendible product set if $(m,n)=(4,4)$, if $(m,n)\in\{(4,5),(5,4)\}$ and $k\le9$, or if $(m,k,n)\in\{(4,6,6),(6,6,4),(4,7,6),(6,7,4)\}$. \end{enumerate} \end{proposition} \emph{Sketch} (proof in Section~S3.41). If $\ker Z$ contains an unextendible product set, then $\ran Z$ is product-free, so the pair is a range-criterion pair and may be reduced, and an outer dimension four stays four. At a kernel point $([x],[y])$, that is, with $\bar x\otimes y\in\ker Z$, Lemma~\ref{lem:range}\ref{range:b} gives $r(x)+t(y)\le k$ for $r(x)=\dim\cK_\Phi x$ and $t(y)=\rank\Psi^{*}(P_y)$, while elements of $\cK_\Phi$ that kill several inputs have rank at most two and are excluded by Lemma~\ref{lem:noranktwo}. Counting at one or two kernel points gives parts~\ref{ups:a} and~\ref{ups:b}, and counting over suitable points of an unextendible product set gives part~\ref{ups:c}. By Theorem~\ref{thm:44} the hypotheses of parts~\ref{ups:a} and~\ref{ups:b} never hold, and the types of part~\ref{ups:c} are excluded for every certificate by Theorems~\ref{thm:main}, \ref{thm:44}, \ref{thm:475no}, \ref{thm:476}, \ref{thm:485no} and~\ref{thm:495no}. \subsection{Proof of Theorem~\ref{thm:main}} The results stated above complete the proof of the main theorem. \begin{proof}[Proof of Theorem~\ref{thm:main}] By Lemma~\ref{lem:reduce} we may pass to the reduced pair, whose dimensions are at most $m$, $k$ and $n$. If $k\le5$, Corollary~\ref{cor:windows} leaves the reduced types $(5,5,5)$, $(5,5,6)$, $(6,5,5)$, $(6,5,6)$ and $(6,4,6)$, which Theorems~\ref{thm:incidence} and~\ref{thm:646no} exclude. If $\min(m,n)\le4$, the reduced pair has an outer dimension at most four, which Corollaries~\ref{cor:windows}\ref{cw:a} and~\ref{cor:46n} exclude. If $\min(m,n)\le5$, the reduced pair has middle dimension at most five, an outer dimension at most four, or one of the types $(5,6,5)$, $(5,6,6)$ and $(6,6,5)$, which Theorems~\ref{thm:565no} and~\ref{thm:566no} exclude. \end{proof} \section{The ten-dimensional example and its family}\label{sec:example} The pair of \cite{OAI26} is the exterior-power lift of a map $L\colon\M_4\to\M_4$ of a special kind. This section sets up these lifts and then states how far the construction can be pushed below dimension ten; the proofs of the further results are in Section~S4 of the supplement. Here $m$ and $n$ are the dimensions of the base map $L$, not those of the lifted pair. \subsection{Exterior-power lifts}\label{ssec:lift} Let $M_0,\dots,M_{m-1}$ be real $r\times n$ matrices such that $M_i^{T}M_j$ is symmetric for all $i,j$, and define $L\colon\M_m\to\M_n$ by $L(E_{ij})=M_i^{T}M_j$. We call $L$ a map of \emph{pencil type} and $M(x)=\sum_ix_iM_i$ its pencil. The matrix $C_L$ is the Gram matrix of the columns of the real $r\times mn$ matrix $(M_0,\dots,M_{m-1})$, so $C_L\succeq0$, and $C_L^{\Gamma}=C_L$ because every block $M_i^{T}M_j$ is symmetric. Hence $L$ is PPT and $L\circ T=L$. The Kraus operators $Q_1,\dots,Q_r$ of $L$ are given by $Q_\rho x=(\text{row }\rho\text{ of }M(x))^{T}$, and \[ L(P_x)=M(x)^{T}\,\overline{M(x)},\qquad \rank L(P_x)=\rank M(x). \] A \emph{rank-loss point} of $L$ is a point $[x]\in\PP^{m-1}$ with $\rank M(x)0$ for every $f\neq0$. Since $\bar a$ is a lowest weight vector of $\sigma_1$, for a lowest weight vector $f$ of $\sigma_2$ the vector $\bar a\otimes f$ spans the lowest weight space of $\sigma_1\otimes\sigma_2$, which $Z$ preserves, so $\bar a\otimes f\in\ran Z$. The theorem does not apply to finite groups, which have no Borel subgroup. For reducible outer representations of $\mathrm{SU}(2)$ it extends as long as one of them is multiplicity-free, and more generally, by Theorem~\ref{thm:su2top}\ref{su2top:d} below, as long as no spin occurs more than twice in one of them. The proof uses the following observation. \begin{lemma}\label{lem:corner} Let $Z\succeq0$ on $(\bigoplus_iA_i)\otimes(\bigoplus_jB_j)$, with orthogonal direct sums, and let $Z_{(ij),(i'j')}$ denote the block of $Z$ that maps $A_{i'}\otimes B_{j'}$ to $A_i\otimes B_j$. If $Z^{\Gamma}\succeq0$ and $A_a\otimes B_b\subseteq\ker Z$, then $Z_{(ay),(xb)}=0$ for all $x$ and $y$. \end{lemma} \begin{proof} Positivity of the partial transpose does not depend on the choice of orthonormal bases, so we may use bases adapted to the decompositions. Then the partial transpose maps the block $Z_{(ij'),(i'j)}$ to the block $(Z^{\Gamma})_{(ij),(i'j')}$, transposing it on the second factor. Since $Z_{(ab),(ab)}=0$, the diagonal block $(Z^{\Gamma})_{(ab),(ab)}$ vanishes. As $Z^{\Gamma}\succeq0$, so does the whole block row $(Z^{\Gamma})_{(ab),(xy)}$, which is the partial transpose of $Z_{(ay),(xb)}$. \end{proof} \begin{theorem}\label{thm:su2mf} Let $\sigma_1$ and $\sigma_2$ be unitary representations of $\mathrm{SU}(2)$ on $\C^m$ and $\C^n$, at least one of them multiplicity-free, and let $Z\neq0$ be a positive semidefinite operator on $\C^m\otimes\C^n$ with $Z^{\Gamma}\succeq0$ that commutes with $\sigma_1(g)\otimes\sigma_2(g)$ for all $g$. Then $\ran Z$ contains a product vector. Consequently part~\ref{compact:b} of Theorem~\ref{thm:compact} holds for $K=\mathrm{SU}(2)$ whenever $\sigma_1$ or $\sigma_2$ is multiplicity-free. \end{theorem} \begin{proof} Exchanging the two tensor factors preserves the hypotheses, so we may assume that $\sigma_1$ is multiplicity-free: $\C^m=\bigoplus_iA_i$ with $A_i$ irreducible of spin $s_i$ and the $s_i$ distinct. Let $\C^n=\bigoplus_tB_t$ be the isotypic decomposition of $\sigma_2$, with $B_t\cong V_t\otimes\C^{\mu_t}$ and $V_t$ irreducible of spin $t$. The blocks $A_i\otimes B_t$ are invariant and mutually orthogonal, and we call a block dead if it lies in $\ker Z$. Suppose that $\ran Z$ is product-free. We show by downward induction on $w$ that every block with $s_i+t\ge w$ is dead, which for $w=0$ contradicts $Z\ne0$. Assume that all blocks with $s_i+t>w$ are dead. The isotypic component of spin $w$ of $\C^m\otimes\C^n$ meets a block $A_i\otimes B_t$ only if $s_i+t\ge w$, and for $s_i+t=w$ the intersection is the Cartan component $C_{it}\cong V_w\otimes\C^{\mu_t}$ of the block. Since $Z$ commutes with the action and its range is orthogonal to the dead blocks, it maps $C_w=\bigoplus_{s_i+t=w}C_{it}$ to itself. Let $(i,t)\ne(i',t')$ be two pairs with $s_i+t=s_{i'}+t'=w$. Then $i\neq i'$, and the corner sums $s_i+t'$ and $s_{i'}+t$ add up to $2w$ and differ by $2(s_i-s_{i'})\neq0$. So one of them exceeds $w$, the corresponding corner block $A_i\otimes B_{t'}$ or $A_{i'}\otimes B_t$ is dead, and Lemma~\ref{lem:corner}, together with $Z=Z^{\dagger}$, shows that the block $Z_{(it),(i't')}$ vanishes. Hence $Z$ restricted to $C_w$ is the direct sum of its restrictions to the $C_{it}$, and by Schur's lemma these have the form $1_{V_w}\otimes z_{it}$ with $z_{it}\succeq0$ on $\C^{\mu_t}$. Let $e_i\in A_i$ and $f_t\in V_t$ be highest weight vectors. For $v\in\C^{\mu_t}$ the product vector $e_i\otimes(f_t\otimes v)$ is a highest weight vector in $C_{it}$, and $Z$ maps it to the product vector $e_i\otimes(f_t\otimes z_{it}v)$. So $z_{it}=0$, and $e_i\otimes(f_t\otimes v)\in\ker Z$ for all $v$. The argument in the proof of Theorem~\ref{thm:compact}\ref{compact:a}, applied to the representations on $A_i$ and on $V_t\otimes v$, gives $A_i\otimes(V_t\otimes v)\subseteq\ker Z$. So the block $A_i\otimes B_t$ is dead. The last statement follows as in the proof of Theorem~\ref{thm:compact}\ref{compact:b}. \end{proof} Theorem~\ref{thm:su2mf} leaves open the case where some spin occurs more than once in both outer representations (see Section~\ref{sec:numerics}). Theorem~\ref{thm:su2top}\ref{su2top:d} below settles this case unless every active pair of highest total spin has multiplicity at least three on both sides, and Remark~\ref{rem:tiles} shows that $\ran Z$ can then be product-free. Theorems~\ref{thm:compact} and~\ref{thm:su2mf} say more than the failure of the range criterion for covariant pairs. \begin{corollary}\label{cor:twirl} Let $K$ be a compact group with unitary representations $\sigma_1$, $\sigma_2$ on $\C^m$ and $\C^n$. Suppose that every nonzero positive semidefinite operator $Z$ on $\C^m\otimes\C^n$ with $Z^{\Gamma}\succeq0$ that commutes with all $\sigma_1(g)\otimes\sigma_2(g)$ has a product vector in its range, as in Theorems~\ref{thm:compact} and~\ref{thm:su2mf}. Then no nonzero positive semidefinite operator with positive partial transpose has its range in a product-free subspace of $\C^m\otimes\C^n$ that is invariant under all $\sigma_1(g)\otimes\sigma_2(g)$. \end{corollary} \begin{proof} Let $Z$ be such an operator with range in such a subspace $\mathcal V$. The twirl $\int_K(\sigma_1(g)\otimes\sigma_2(g))Z(\sigma_1(g)\otimes\sigma_2(g))^{\dagger}\,dg$ is nonzero, positive semidefinite and invariant, and its range lies in $\mathcal V$. Its partial transpose is the corresponding twirl of $Z^{\Gamma}$ by $\sigma_1\otimes\bar\sigma_2$, so it is positive semidefinite. This contradicts the hypothesis. \end{proof} So for these groups no composite of PPT maps, covariant or not, has its range in an invariant product-free subspace. \begin{remark}[\ca]\label{rem:topweight} For $\mathrm{SU}(2)$ the Cartan component of $V\otimes V'$ is the component of highest total spin. Let $\Phi$ and $\Psi$ be $\mathrm{SU}(2)$-covariant PPT maps between irreducible representations $V_1$, $V_2$, $V_3$, and let $\Pi$ be the projection onto the Cartan component, so that $w(Z)=\Tr(Z\Pi)/\Tr Z$ is the weight of this component in $Z=C_{\Psi\circ\Phi}$. By Schur's lemma $\Phi(I)$ and $\Psi(I)$ are multiples of the identity, so after normalising both maps $\Tr Z$ is constant, and $w$ is bilinear on the product of the two polytopes of normalised covariant PPT maps of Theorem~\ref{thm:classes}\ref{cl:su2}. Its minimum is attained at a pair of vertices. Table~\ref{tab:topw} lists the exact minima, computed from these vertices. By Theorem~\ref{thm:compact} they are positive. For the dimension-six types in the table the minimum is attained by the pair of extreme rays whose Choi matrices are supported on the components of highest total spin. \end{remark} \begin{table}[ht] \centering \begin{tabular}{@{}lcc@{}} \toprule dimensions of $(V_1,V_2,V_3)$ & minimum of $w(Z)$ & value \\ \midrule $(4,6,5)$ and $(5,6,4)$ & $143/1470$ & $0.097$ \\ $(4,6,6)$ and $(6,6,4)$ & $13/168$ & $0.077$ \\ $(5,6,5)$ & $143/2100$ & $0.068$ \\ $(6,4,6)$ & $143/2268$ & $0.063$ \\ $(5,6,6)$ and $(6,6,5)$ & $13/252$ & $0.052$ \\ $(6,6,6)$ & $26/693$ & $0.038$ \\ all dimensions between $2$ and $9$ & $437/126412$ & $0.0035$ \\ \bottomrule \end{tabular} \caption{Exact minima of the weight of the Cartan component in the composite, over $\mathrm{SU}(2)$-covariant PPT maps between irreducible representations. The last minimum is attained at dimensions $(9,9,9)$.}\label{tab:topw} \end{table} \subsection{Reducible representations}\label{ssec:reducible} For reducible outer representations, covariance still reduces a range-criterion pair to one on the extreme weight spaces of a torus or on the multiplicity spaces of the highest total spin. For subspaces $A\subseteq\C^m$, $K\subseteq\C^k$ and $F\subseteq\C^n$ with isometries $V_A$, $V_K$ and $V_F$ onto them, we call $(\Ad_{V_K^{\dagger}}\circ\Phi\circ\Ad_{V_A},\Ad_{V_F^{\dagger}}\circ\Psi\circ\Ad_{V_K})$ the compression of $(\Phi,\Psi)$ to $(A,K,F)$. Its maps are PPT if $\Phi$ and $\Psi$ are, as in the proof of Lemma~\ref{lem:reduce}, and for $K=\C^k$ the identity in the proof of Lemma~\ref{lem:reduce} shows that the Choi matrix of its composition is the compression of $C_{\Psi\circ\Phi}$ to $\bar A\otimes F$. \begin{theorem}\label{thm:torus} Let $\mathbb T$ be a compact torus with unitary representations $\sigma_1$, $\pi$ and $\sigma_2$ on $\C^m$, $\C^k$ and $\C^n$, let $\Phi\colon\M_m\to\M_k$ and $\Psi\colon\M_k\to\M_n$ be PPT maps that are covariant as in Theorem~\ref{thm:compact}\ref{compact:b}, and let $Z=C_{\Psi\circ\Phi}\neq0$. Let $\C^m=\bigoplus_\mu E_\mu$ and $\C^n=\bigoplus_\nu F_\nu$ be the weight space decompositions of $\sigma_1$ and $\sigma_2$, and call a pair $(\mu,\nu)$ active if $E_\mu\otimes F_\nu\not\subseteq\ker Z$. Fix a linear form $\ell$ on the real span of the characters of $\mathbb T$ that is injective on the weights of $\sigma_1$, on those of $\sigma_2$ and on their sums, and let $w^*$ be the largest value of $\ell(\mu+\nu)$ over the active pairs. \begin{enumerate}[label=(\alph*)] \item\label{torus:a} If $(\mu,\nu)\neq(\mu',\nu')$ are active and $\ell(\mu+\nu)=\ell(\mu'+\nu')=w^*$, then the block of $Z$ from $E_{\mu'}\otimes F_{\nu'}$ to $E_\mu\otimes F_\nu$ vanishes. For every active $(\mu,\nu)$ with $\ell(\mu+\nu)=w^*$, the range of the compression of $Z$ to $E_\mu\otimes F_\nu$ lies in $\ran Z$. \item\label{torus:b} Let $(\Phi,\Psi)$ be a range-criterion pair, and let $(\mu,\nu)$ be active with $\ell(\mu+\nu)=w^*$. Then the compression of $(\Phi,\Psi)$ to $(\bar E_\mu,\C^k,F_\nu)$ is a range-criterion pair, and so is the compression to $(\bar E_\mu,K,F_\nu)$ for some weight space $K$ of $\pi$, of type $(\dim E_\mu,\dim K,\dim F_\nu)$. In particular $\dim E_\mu$, $\dim F_\nu$ and $\dim K$ are at least four, $\dim E_\mu$ and $\dim F_\nu$ are not both four, $\dim E_\mu,\dim F_\nu\ge6$ if $\dim K=4$, and $\dim E_\mu,\dim F_\nu\ge5$ if $\dim K=5$. \item\label{torus:c} If moreover no range-criterion pair has a type that is componentwise at most $(m,k,n)$ and different from it, then $\mathbb T$ acts by scalars on $\C^m$, $\C^k$ and $\C^n$. \end{enumerate} \end{theorem} \begin{proof} As in the proof of Theorem~\ref{thm:compact}\ref{compact:b}, $Z$ commutes with $\sigma_1(t)\otimes\sigma_2(t)$, so it preserves the weight spaces $H_w=\bigoplus_{\mu+\nu=w}E_\mu\otimes F_\nu$ of $\sigma_1\otimes\sigma_2$. \ref{torus:a} If $\mu=\mu'$, then $\ell(\nu)=\ell(\nu')$ and $\nu=\nu'$. So $\mu\neq\mu'$, and as $Z=Z^{\dagger}$ we may assume that $\ell(\mu)>\ell(\mu')$. Then $\ell(\nu')>\ell(\nu)$, so $\ell(\mu+\nu')>w^*$, the pair $(\mu,\nu')$ is not active, and Lemma~\ref{lem:corner} shows that the block vanishes. By the choice of $\ell$, all pairs $(\mu',\nu')$ with $\ell(\mu'+\nu')=w^*$ have the same total weight $w$. On $H_w$ the operator $Z$ vanishes on the spaces $E_{\mu'}\otimes F_{\nu'}$ of the pairs that are not active, and by the first statement its blocks between distinct active pairs vanish. So $Z$ restricted to $H_w$ is the direct sum of the compressions of $Z$ to the spaces $E_\mu\otimes F_\nu$ of the active pairs with $\ell(\mu+\nu)=w^*$. Their ranges lie in $\ran Z$. \ref{torus:b} Let $V_E$ and $V_F$ be isometries onto $\bar E_\mu$ and $F_\nu$, and put $\Phi_E=\Phi\circ\Ad_{V_E}$ and $\Psi_F=\Ad_{V_F^{\dagger}}\circ\Psi$, so that $(\Phi_E,\Psi_F)$ is the compression of $(\Phi,\Psi)$ to $(\bar E_\mu,\C^k,F_\nu)$. The Choi matrix of $\Psi_F\circ\Phi_E$ is the compression of $Z$ to $E_\mu\otimes F_\nu$. It is nonzero as $(\mu,\nu)$ is active, and by \ref{torus:a} its range lies in $\ran Z$ and so contains no product vector. Hence $(\Phi_E,\Psi_F)$ is a range-criterion pair. As $\bar\sigma_1(t)$ acts on $\bar E_\mu$ by a scalar, covariance shows that $\Phi_E(X)$ commutes with $\pi(t)$ for every $X$, so $\Phi_E(X)=\sum_KP_K\Phi_E(X)P_K$, where $K$ runs over the weight spaces of $\pi$. Hence $C_{\Psi_F\circ\Phi_E}$ is the sum over $K$ of the Choi matrices of the compositions $\Psi_F\circ\Ad_{P_K}\circ\Phi_E$, which are positive semidefinite. One of them is nonzero, and its range lies in that of $C_{\Psi_F\circ\Phi_E}$ and so contains no product vector. For this $K$, $\Psi_F\circ\Ad_{P_K}\circ\Phi_E$ is the composition of the compression of $(\Phi,\Psi)$ to $(\bar E_\mu,K,F_\nu)$, which is therefore a range-criterion pair. The bounds follow from Corollary~\ref{cor:windows}\ref{cw:a} and Theorem~\ref{thm:44}. \ref{torus:c} By \ref{torus:b}, $\dim E_\mu=m$, $\dim K=k$ and $\dim F_\nu=n$, so each of $\sigma_1$, $\pi$ and $\sigma_2$ has a single weight. \end{proof} For a compact connected group, Theorem~\ref{thm:torus} applies to a maximal torus. For $\mathrm{SU}(2)$ the highest weight vectors give a finer reduction, to the multiplicity spaces of the highest total spin. \begin{theorem}\label{thm:su2top} Let $\sigma_1$, $\pi$ and $\sigma_2$ be unitary representations of $\mathrm{SU}(2)$ on $\C^m$, $\C^k$ and $\C^n$, let $\Phi\colon\M_m\to\M_k$ and $\Psi\colon\M_k\to\M_n$ be PPT maps that are covariant as in Theorem~\ref{thm:compact}\ref{compact:b}, and let $Z=C_{\Psi\circ\Phi}\neq0$. Let $\C^m=\bigoplus_jA_j$ and $\C^n=\bigoplus_lB_l$ be the isotypic decompositions of $\sigma_1$ and $\sigma_2$, with $A_j\cong\C^{p_j}\otimes V_j$ and $B_l\cong\C^{q_l}\otimes V_l$, where $V_s$ is irreducible of spin $s$. Call $(j,l)$ active if $A_j\otimes B_l\not\subseteq\ker Z$, and let $J$ be the largest value of $j+l$ over the active pairs. For an active pair with $j+l=J$, let $x_j\in V_j$ and $y_l\in V_l$ be highest weight vectors, and put $E_j=\C^{p_j}\otimes x_j\subseteq A_j$ and $F_l=\C^{q_l}\otimes y_l\subseteq B_l$. \begin{enumerate}[label=(\alph*)] \item\label{su2top:a} For distinct active pairs $(j,l)$ and $(j',l')$ with $j+l=j'+l'=J$, the block of $Z$ from $A_{j'}\otimes B_{l'}$ to $A_j\otimes B_l$ vanishes. \item\label{su2top:b} For an active pair with $j+l=J$, the compression $Z_{jl}$ of $Z$ to $E_j\otimes F_l\cong\C^{p_j}\otimes\C^{q_l}$ is nonzero, $Z_{jl}^{\Gamma}\succeq0$, and $\ran Z_{jl}\subseteq\ran Z$. \item\label{su2top:c} If $(\Phi,\Psi)$ is a range-criterion pair, then for every active pair with $j+l=J$ the compression of $(\Phi,\Psi)$ to $(\bar E_j,\C^k,F_l)$ is a range-criterion pair of type $(p_j,k,q_l)$. In particular $p_j,q_l\ge4$ and $\max(p_j,q_l)\ge5$. \item\label{su2top:d} If $\min(p_j,q_l)\le2$ for some active pair with $j+l=J$, then $\ran Z$ contains a product vector. \item\label{su2top:e} If $(\Phi,\Psi)$ is a range-criterion pair and $m,n\le7$, then $Z$ is supported on $A_0\otimes B_0$, and some range-criterion pair has type $(p_0,r,q_0)$, where $r\ge4$ is the multiplicity in $\pi$ of one of its spins. This type is componentwise at most $(m,k,n)$, and equal to it only if $\sigma_1$, $\pi$ and $\sigma_2$ are trivial. \end{enumerate} \end{theorem} \begin{proof} As in the proof of Theorem~\ref{thm:compact}\ref{compact:b}, $Z$ commutes with $\sigma_1(g)\otimes\sigma_2(g)$. \ref{su2top:a} If $j=j'$, then $l=l'$. If $j>j'$, then $l'>l$, so $j+l'>J$, the pair $(j,l')$ is not active, and Lemma~\ref{lem:corner} applies. The case $j0$ and $Z_{jl}\neq0$. The projection onto the highest weight vectors of total spin $J$ commutes with $Z$. Its range is the sum of the spaces $E_{j'}\otimes F_{l'}$ over the pairs with $j'+l'=J$ and of subspaces of the blocks with $j'+l'>J$, which are not active and lie in $\ker Z$. So by \ref{su2top:a}, $Z$ restricted to this range is the direct sum of the operators $Z_{j'l'}$ over the active pairs with $j'+l'=J$, and $\ran Z_{jl}\subseteq\ran Z$. \ref{su2top:c} The operator $Z_{jl}$ is the Choi matrix of the composition of this compression, which consists of PPT maps. By \ref{su2top:b} it is nonzero and its range contains no product vector. The bounds follow from Corollary~\ref{cor:windows}\ref{cw:a} and Theorem~\ref{thm:44}. \ref{su2top:d} By \ref{su2top:b}, $Z_{jl}$ is a nonzero positive semidefinite operator on $\C^{p_j}\otimes\C^{q_l}$ with $Z_{jl}^{\Gamma}\succeq0$, and its range lies in $U\otimes W$, where $U$ and $W$ are the supports of its two marginals, of dimensions $r_1$ and $r_2$ with $\min(r_1,r_2)\le\min(p_j,q_l)\le2$. If $\min(r_1,r_2)=1$, every vector of $U\otimes W$ is a product vector. Otherwise $\min(r_1,r_2)=2$, so $(r_1-1)(r_2-1)=\max(r_1,r_2)-1$, and as $\rank Z_{jl}\ge\max(r_1,r_2)$ by Lemma~\ref{lem:HLVC}, Lemma~\ref{lem:det} applied to $U\otimes W$ gives a product vector in $\ran Z_{jl}$. So $\ran Z_{jl}$ contains a product vector $u\otimes v$, and $(u\otimes x_j)\otimes(v\otimes y_l)$ is a product vector in $\ran Z$. \ref{su2top:e} By \ref{su2top:c}, an active pair with $j+l=J$ has $p_j\ge4$ and $p_j(2j+1)\le m\le7$, so $j=0$, and likewise $l=0$. So $(0,0)$ is the only active pair, and $Z$ is supported on $A_0\otimes B_0$. It is the Choi matrix of the composition of the compression $(\Phi_0,\Psi_0)$ of $(\Phi,\Psi)$ to $(\bar A_0,\C^k,B_0)$. As $\bar\sigma_1$ is trivial on $\bar A_0$, the values of $\Phi_0$ commute with $\pi$, so they lie in $\bigoplus_b\M_{r_b}\otimes I_{V_b}$, where $\C^k=\bigoplus_b\C^{r_b}\otimes V_b$ is the isotypic decomposition of $\pi$. The maps $\pi_b(Y)=(2b+1)^{-1}\Tr_{V_b}(P_bYP_b)$, with $P_b$ the projection onto $\C^{r_b}\otimes V_b$, and $\iota_b(Y)=Y\otimes I_{V_b}$, an operator on $\C^{r_b}\otimes V_b\subseteq\C^k$, are completely positive, and so are $T\circ\pi_b\circ T$ and $T\circ\iota_b\circ T$, so $\pi_b\circ\Phi_0$ and $\Psi_0\circ\iota_b$ are PPT, and $\Psi_0\circ\Phi_0=\sum_b(\Psi_0\circ\iota_b)\circ(\pi_b\circ\Phi_0)$. The Choi matrices of the summands are positive semidefinite, one of them is nonzero, and their ranges lie in $\ran Z$. So $(\pi_b\circ\Phi_0,\Psi_0\circ\iota_b)$ is a range-criterion pair of type $(p_0,r_b,q_0)$ for some $b$, and $r_b\ge4$ by Corollary~\ref{cor:windows}\ref{cw:a}. As $p_0\le m$, $(2b+1)r_b\le k$ and $q_0\le n$, this type is componentwise at most $(m,k,n)$, with equality only if $A_0=\C^m$, $b=0$, $\C^{r_0}=\C^k$ and $B_0=\C^n$. \end{proof} Part~\ref{su2top:d} of Theorem~\ref{thm:su2top} contains Theorem~\ref{thm:su2mf}, as $p_j=1$ for all $j$ if $\sigma_1$ is multiplicity-free, and $q_l=1$ for all $l$ if $\sigma_2$ is. If both representations have repeated spins, part~\ref{su2top:c} reduces a covariant range-criterion pair to one on the multiplicity spaces of the highest total spin, part~\ref{su2top:d} excludes small multiplicities, and part~\ref{su2top:e} shows that a covariant range-criterion pair with $m,n\le7$ yields one whose type is componentwise at most $(m,k,n)$ and different from it, unless all three representations are trivial. Parts~\ref{su2top:a}, \ref{su2top:b} and~\ref{su2top:d} use only that $Z\succeq0$, $Z^{\Gamma}\succeq0$ and that $Z$ commutes with all $\sigma_1(g)\otimes\sigma_2(g)$, so they hold for every such operator $Z\neq0$, as in Theorem~\ref{thm:su2mf}. \begin{remark}\label{rem:tiles} The bound in part~\ref{su2top:d} is sharp for invariant operators. Let $\sigma_1=\sigma_2$ be the sum of three copies of $V_{1/2}$, so that $\C^6\cong\C^3\otimes V_{1/2}$ and $p_{1/2}=q_{1/2}=3$. Let $\rho$ be the projection onto the orthogonal complement of the five product vectors of the Tiles unextendible product basis of $\C^3\otimes\C^3$ \cite{BDMSST99}, and let $Q_1=\frac12(I+F)$, with $F$ the flip, be the projection onto the symmetric tensors in $V_{1/2}\otimes V_{1/2}$, which form its component of spin one. Then $Z=\rho\otimes Q_1$, an operator on $(\C^3\otimes\C^3)\otimes(V_{1/2}\otimes V_{1/2})$, which is $\C^6\otimes\C^6$ after reordering the tensor factors, is positive semidefinite and invariant, and $Z^{\Gamma}=\rho^{\Gamma}\otimes Q_1^{\Gamma}\succeq0$, since $\rho^{\Gamma}=\rho$ as the Tiles vectors are real product vectors, and $Q_1^{\Gamma}=\frac12(I+F^{\Gamma})=\frac12(I+2P)$ with $P$ the projection onto a maximally entangled vector. If $x\otimes y\in\ran Z$, write $x=\sum_ia_i\otimes e_i$ and $y=\sum_jb_j\otimes e_j$ with a basis $e_1,e_2$ of $V_{1/2}$. As $\ran Z$ lies in $\ran\rho\otimes(V_{1/2}\otimes V_{1/2})$, every $a_i\otimes b_j$ lies in $\ran\rho$, which contains no nonzero product vector, so $x\otimes y=0$. So $\ran Z$ contains no product vector, while $\min(p_{1/2},q_{1/2})=3$. \end{remark} \begin{corollary}\label{cor:su2inv} Let $\sigma_1$ and $\sigma_2$ be unitary representations of $\mathrm{SU}(2)$ on $\C^m$ and $\C^n$. The following are equivalent. \begin{enumerate}[label=(\roman*)] \item Every nonzero positive semidefinite operator $Z$ on $\C^m\otimes\C^n$ with $Z^{\Gamma}\succeq0$ that commutes with all $\sigma_1(g)\otimes\sigma_2(g)$ has a product vector in its range. \item $\sigma_1$ or $\sigma_2$ has no spin of multiplicity at least three. \end{enumerate} \end{corollary} \begin{proof} If, say, every spin of $\sigma_1$ has multiplicity at most two, then $\min(p_j,q_l)\le2$ for every active pair, and part~\ref{su2top:d} of Theorem~\ref{thm:su2top}, which holds for every such operator, gives a product vector. Conversely, let $s$ and $t$ be spins of multiplicity at least three in $\sigma_1$ and $\sigma_2$. Then there are isometries $W_1\colon\C^3\otimes V_s\to\C^m$ and $W_2\colon\C^3\otimes V_t\to\C^n$ that intertwine the action of $\mathrm{SU}(2)$ on the second factors with $\sigma_1$ and $\sigma_2$. With $\rho$ as in Remark~\ref{rem:tiles}, let $Z=(W_1\otimes W_2)(\rho\otimes I)(W_1\otimes W_2)^{\dagger}$, where $\rho\otimes I$ acts on $(\C^3\otimes\C^3)\otimes(V_s\otimes V_t)$, which is $(\C^3\otimes V_s)\otimes(\C^3\otimes V_t)$ after reordering the tensor factors. Then $Z\neq0$ is positive semidefinite and invariant, and $Z^{\Gamma}=(W_1\otimes\bar W_2)(\rho^{\Gamma}\otimes I)(W_1\otimes\bar W_2)^{\dagger}\succeq0$, as $\rho^{\Gamma}=\rho$. A product vector in $\ran Z$ has its factors in $\ran W_1$ and $\ran W_2$, as in the proof of Lemma~\ref{lem:reduce}, so it is the image of a product vector in $\ran\rho\otimes(V_s\otimes V_t)$, and the argument of Remark~\ref{rem:tiles}, with bases of $V_s$ and $V_t$, shows that it vanishes. \end{proof} \begin{remark}[\ca]\label{rem:topbound} Part~\ref{su2top:b} has a quantitative form. Let $V_j\otimes V_l$ be as in its proof, with $2j$ and $2l$ between $1$ and $8$. Every nonzero invariant positive semidefinite operator $d=\sum_Lc_LQ_L$ on $V_j\otimes V_l$ with $d^{\Gamma}\succeq0$ satisfies $\Tr(dQ_{j+l})\ge\binom{2j+2l}{2j}^{-1}\Tr d$, and equality occurs. The inequality is a combination with nonnegative rational coefficients of the linear inequalities in the $c_L$ that express $d^{\Gamma}\succeq0$ (Table~\ref{tab:verify}). As $D^{\Gamma}\succeq0$ gives the same inequalities with the operators $z_L^{\Gamma}$ in place of the $c_L$, the operators $z_L$ in the proof of Theorem~\ref{thm:su2top} satisfy \[ (2J+1)z_J^{\Gamma}\succeq\binom{2j+2l}{2j}^{-1}\Bigl(\sum_L(2L+1)z_L\Bigr)^{\Gamma}. \] The composites in Remark~\ref{rem:topweight} are such operators $d$, and the minima in Table~\ref{tab:topw} lie above these bounds. \end{remark} The following corollary shows that a range-criterion pair that is covariant under a nontrivial action of $\mathrm{SU}(2)$ exists exactly when a suitably smaller range-criterion pair exists. \begin{corollary}\label{cor:su2pad} Let $\sigma_1$, $\pi$ and $\sigma_2$ be unitary representations of $\mathrm{SU}(2)$ on $\C^m$, $\C^k$ and $\C^n$, and call a range-criterion pair covariant if it is covariant for them as in Theorem~\ref{thm:compact}\ref{compact:b}. \begin{enumerate}[label=(\alph*)] \item\label{pad:a} If there is a covariant range-criterion pair, then there is a range-criterion pair of a type $(m',k',n')$ with $m'\le m$, $k'\le k$ and $n'\le n$, where $m'\le m-2$ if $\sigma_1$ is nontrivial, $k'\le k-2$ if $\pi$ is nontrivial, and $n'\le n-2$ if $\sigma_2$ is nontrivial. \item\label{pad:b} Conversely, if there is a range-criterion pair of type $(m',k',n')$, and $\sigma_1$, $\pi$ and $\sigma_2$ act trivially on subspaces of dimensions $m'$, $k'$ and $n'$, then there is a covariant range-criterion pair. \item\label{pad:c} If $\max(m,k,n)\le5$ and there is a covariant range-criterion pair, then $\sigma_1$, $\pi$ and $\sigma_2$ are trivial. \item\label{pad:d} If there is a covariant range-criterion pair, $\sigma_1$ is nontrivial and $m\le7$, then $\sigma_1$ is $V_0^{\oplus4}\oplus V_{1/2}$, $V_0^{\oplus5}\oplus V_{1/2}$ or $V_0^{\oplus4}\oplus V_1$, and there is a range-criterion pair of a type $(p_0,k,n')$ with $n'\le n$, where $p_0\in\{4,5\}$ is the multiplicity of $V_0$ in $\sigma_1$. The same holds for $\sigma_2$, with the roles of the outer spaces exchanged. \end{enumerate} \end{corollary} \begin{proof} \ref{pad:a} Theorem~\ref{thm:torus}\ref{torus:b}, applied to a maximal torus $\mathbb T$ of $\mathrm{SU}(2)$, gives a range-criterion pair of type $(\dim E,\dim K,\dim F)$, where $E$, $K$ and $F$ are weight spaces of $\mathbb T$ in $\C^m$, $\C^k$ and $\C^n$. All three dimensions are at least four by Corollary~\ref{cor:windows}\ref{cw:a}. Let $E$ be the weight space of weight $w$ of a nontrivial representation on $\C^d$, with $\dim E\ge2$. An irreducible summand of spin $s$ adds $2s+1$ to $d$, and it adds one to $\dim E$ if $s\ge|w|$ and $s-w$ is an integer, and nothing otherwise. If $w$ is not an integer, every summand that adds to $\dim E$ has $2s+1\ge2$, so $d\ge2\dim E\ge\dim E+2$. If $w$ is an integer, a summand with $s>0$ adds at least two more to $d$ than to $\dim E$. In both cases $\dim E\le d-2$, and applied to $E$, $K$ and $F$ this gives the bounds. \ref{pad:b} Let $(\Phi',\Psi')$ be a range-criterion pair of type $(m',k',n')$, and let $V_1$, $V_2$ and $V_3$ be isometries from $\C^{m'}$, $\C^{k'}$ and $\C^{n'}$ onto subspaces on which $\bar\sigma_1$, $\pi$ and $\sigma_2$ act trivially, where for $\bar\sigma_1$ we take the complex conjugate of a subspace on which $\sigma_1$ acts trivially. The maps $\Phi=\Ad_{V_2}\circ\Phi'\circ\Ad_{V_1^{\dagger}}$ and $\Psi=\Ad_{V_3}\circ\Psi'\circ\Ad_{V_2^{\dagger}}$ are PPT, as in the proof of Lemma~\ref{lem:reduce}, and they are covariant, since $\bar\sigma_1(g)V_1=V_1$, $\pi(g)V_2=V_2$ and $\sigma_2(g)V_3=V_3$ for all $g$. As $V_2^{\dagger}V_2=I$, the Kraus operators of $\Psi\circ\Phi$ are the $V_3KV_1^{\dagger}$ for the Kraus operators $K$ of $\Psi'\circ\Phi'$, so $C_{\Psi\circ\Phi}=(\bar V_1\otimes V_3)C_{\Psi'\circ\Phi'}(\bar V_1\otimes V_3)^{\dagger}\neq0$, and, as in the proof of Lemma~\ref{lem:reduce}, product vectors in the two ranges correspond. \ref{pad:c} By \ref{pad:a}, a nontrivial action gives a range-criterion pair with a dimension at most three, which Corollary~\ref{cor:windows}\ref{cw:a} excludes. \ref{pad:d} By Theorem~\ref{thm:su2top}\ref{su2top:c}, an active pair $(j,l)$ with $j+l=J$ has $p_j\ge4$ and $p_j(2j+1)\le m\le7$, so $j=0$, and the compression there is a range-criterion pair of type $(p_0,k,q_l)$ with $q_l\le n$. The summands of $\sigma_1$ other than the $p_0\ge4$ copies of $V_0$ are nontrivial irreducible representations of total dimension $m-p_0\le3$, and as $\sigma_1$ is nontrivial there is at least one. So they form $V_{1/2}$ or $V_1$, which gives the three representations. The same argument applies to $\sigma_2$. \end{proof} By parts~\ref{pad:b} and~\ref{pad:d}, there are representations $\pi$ and $\sigma_2$ on $\C^k$ and $\C^n$ and a range-criterion pair that is covariant for $V_0^{\oplus4}\oplus V_{1/2}$, $\pi$ and $\sigma_2$ if and only if there is a range-criterion pair of a type $(4,k',n')$ with $k'\le k$ and $n'\le n$. This is the question of outer dimension four in Section~\ref{sec:open}. \begin{remark}[\ca]\label{rem:su2pad6} With Theorem~\ref{thm:main}, part~\ref{pad:c} of Corollary~\ref{cor:su2pad} extends to $\max(m,k,n)\le6$: by part~\ref{pad:a}, a nontrivial action gives a range-criterion pair with all three dimensions at most six and one of them at most four, which Theorem~\ref{thm:main} excludes. Part~\ref{pad:b} shows that this cannot be extended to $\max(m,k,n)\le7$ without deciding types of Table~\ref{tab:d7}. For instance, a range-criterion pair of type $(4,7,7)$ would give one of type $(6,7,7)$ that is covariant for $\sigma_1=V_0^{\oplus4}\oplus V_{1/2}$ and trivial $\pi$ and $\sigma_2$. \end{remark} \subsection{Exact covariant classes} For the following classes we prove the PPT$^2$ conjecture itself, not only the failure of the range criterion. Before stating them, we recall two symmetric classes for which the conjecture was settled earlier: maps on $\M_d\otimes\M_d$ that are covariant under all pairs of unitaries acting on the two tensor factors \cite[Proposition~V.1]{CMHW19}, and maps covariant under the diagonal unitary group, which include the Choi-type maps \cite{SN22}. \begin{theorem}[\ca]\label{thm:classes} In each of the following cases, $\Psi\circ\Phi$ is entanglement breaking for all PPT maps $\Phi\colon\M(V_1)\to\M(V_2)$ and $\Psi\colon\M(V_2)\to\M(V_3)$ that are covariant for the indicated representations $V_1$, $V_2$, $V_3$ of the indicated group. \begin{enumerate}[label=(\alph*)] \item\label{cl:su2} $\mathrm{SU}(2)$, with $V_1,V_2,V_3$ irreducible of dimensions between $2$ and $9$, and also with $V_1=V_3$ of dimension $4$ and $V_2$ irreducible of dimension between $3$ and $10$. \item\label{cl:su2su2} $\mathrm{SU}(2)\times\mathrm{SU}(2)$, with each $V_i$ one of $\C^2\otimes\C^2$, $\C^2\otimes\C^3$, $\C^3\otimes\C^2$, $\C^2\otimes\C^4$, $\C^4\otimes\C^2$, $\C^3\otimes\C^3$ (tensor products of irreducible representations of the two factors). \item\label{cl:g2} $G_2$, with $V_1=V_2=V_3$ its seven-dimensional irreducible representation. \item\label{cl:sn} $S_n$ for $6\le n\le10$, with $V_1=V_2=V_3$ the standard representation, and $S_6$ with $(V_1,V_2,V_3)$ equal to $(\mathrm{std},(4,2),\mathrm{std})$ or $(\mathrm{std},\mathrm{std},(4,2))$. \item\label{cl:s5} $S_5$, with each $V_i$ the irreducible representation $(3,2)$, and also with $(V_1,V_2,V_3)$ equal to $(4,5,4)$, $(4,4,5)$, $(5,4,5)$ or $(4,5,5)$, where $4$ denotes the standard representation and $5$ the representation $(3,2)$. \item\label{cl:pgl} $\mathrm{PGL}(2,q)$ for $q\in\{7,9\}$, acting on the $q+1$ points of $\PP^1(\mathbb F_q)$, with $V_1=V_2=V_3$ the $q$-dimensional standard representation on functions with zero sum. \item\label{cl:bn} The group of signed permutation matrices of size $n$, for $5\le n\le9$, with $V_1=V_2=V_3=\C^n$. \item\label{cl:su3} $\mathrm{SU}(3)$, with $(V_1,V_2,V_3)$ one of $(3,3,3)$, $(3,6,3)$, $(3,8,3)$, $(3,3,6)$, $(3,6,6)$, $(3,8,6)$, $(3,3,8)$, $(3,6,8)$, $(6,3,6)$, $(6,6,6)$, $(6,8,6)$, $(6,3,8)$, $(6,6,8)$, where $3=\C^3$, $6=\Sym^2\C^3$ and $8$ is the adjoint representation, and with the triples obtained from these by passing to conjugate representations and to adjoint maps. \end{enumerate} Each of \ref{cl:su2}, \ref{cl:su2su2}, \ref{cl:s5} and \ref{cl:pgl} includes classes that contain PPT maps that are not entanglement breaking. \end{theorem} \begin{proof} In each case the operators on $V_1\otimes V_2$ that commute with $\bar V_1\otimes V_2$ form a commutative algebra spanned by isotypic projections, the partial transposes of these operators commute with $\bar V_1\otimes\bar V_2$, which again decomposes without multiplicities, and the same holds for $V_2$ and $V_3$. So the Choi matrix of a covariant map is a real combination of isotypic projections with coefficients $\lambda$, complete positivity and complete copositivity are linear inequalities in $\lambda$, and the covariant PPT maps form a polyhedral cone. The Choi matrix of the composition is bilinear in the two coefficient vectors, and separability is preserved under convex combinations. So it suffices to show that the composition of every pair of extreme rays is separable. For each class we computed the extreme rays in exact arithmetic, formed all compositions of pairs, and wrote each composite Choi matrix as an explicit combination, with exact nonnegative weights, of twirls of explicit product vectors. In \ref{cl:su2}--\ref{cl:sn}, \ref{cl:bn} and \ref{cl:su3} the weights and the twirled points are rational. In \ref{cl:s5} they are rational except in the stored certificates of $18$ composites, which use product vectors with non-real coordinates in $\Q(e^{2\pi i/3})$. In \ref{cl:pgl} the group is sharply $3$-transitive, the commutant of $V_1\otimes V_1$ has dimension $q+1$ and is multiplicity-free, the isotypic projections are defined over $\Q(\sqrt2)$ for $q=7$ and over $\Q(\sqrt2,\sqrt5)$ for $q=9$, and the certificates use product vectors with entries in $\mathbb Z[i]$ or $\mathbb Z[e^{2\pi i/3}]$ and weights in these fields. Table~\ref{tab:classes} lists the numbers of composites. In that table, for $\mathrm{SU}(2)$ with dimensions $2$--$9$ the $288$ triples are those with $\dim V_1\le\dim V_3$, and the second count is for the triples with $\dim V_1>\dim V_3$, certified directly; for $\mathrm{SU}(2)\times\mathrm{SU}(2)$ the $68$ triples represent all $216$ triples up to exchanging the two factors of the group and passing to adjoint maps. Non-triviality is witnessed by extreme rays whose Choi matrices violate the realignment criterion, verified in interval arithmetic in \ref{cl:su2} and \ref{cl:su2su2}, and in exact arithmetic in \ref{cl:s5} and \ref{cl:pgl}, where for \ref{cl:pgl} the realigned Choi matrix is diagonal in the isotypic projections. The polytopes of normalised covariant PPT maps have $196$ vertices for $\mathrm{PGL}(2,7)$ and $667$ for $\mathrm{PGL}(2,9)$, of which $139$ and $536$ violate the realignment criterion. For \ref{cl:su2} with outer dimension four an exact Breuer--Hall witness detects an extreme ray for every middle dimension. \end{proof} \begin{table}[ht] \centering \begin{tabular}{@{}lll@{}} \toprule class & triples & composites of extreme pairs certified \\ \midrule $\mathrm{SU}(2)$, dimensions $2$--$9$ & $288$, and adjoints & $226{,}973$ and $148{,}862$ \\ $\mathrm{SU}(2)$, $V_1=V_3=\C^4$ & $8$ & $464$, counted as pairs \\ $\mathrm{SU}(2)\times\mathrm{SU}(2)$ & $68$ & $13{,}363$ \\ $G_2$ & $1$ & $16$ \\ $S_n$ standard, $6\le n\le10$ & $5$ & $22$ each \\ $S_6$ with $(4,2)$ & $2$ & $47$ and $58$ \\ $S_5$, $(3,2)$ and mixed & $5$ & $222$; $24$, $38$, $24$, $114$ \\ $\mathrm{PGL}(2,7)$, standard & $1$ & $9{,}672$ \\ $\mathrm{PGL}(2,9)$, standard & $1$ & $110{,}540$ \\ signed permutations, $5\le n\le9$ & $5$ & $18$ each \\ $\mathrm{SU}(3)$ & $13$, and variants & $95$ \\ \bottomrule \end{tabular} \caption{Exact certificates for Theorem~\ref{thm:classes}.}\label{tab:classes} \end{table} Theorem~\ref{thm:classes}\ref{cl:su2} proves the PPT$^2$ conjecture for $\mathrm{SU}(2)$-covariant maps between irreducible spaces of type $(6,6,6)$, the type that Theorem~\ref{thm:main} leaves open, and of type $(4,3,4)$, for which the conjecture is open in general. The classes in \ref{cl:pgl} are far from trivial, since most of their extreme rays are entangled, yet all compositions are entanglement breaking. Covariant maps between reducible representations split into maps between irreducible ones when distinct pairs of summands share no irreducible constituent. \begin{lemma}\label{lem:covblocks} Let $G$ be a compact group with unitary representations $\rho_1$, $\rho_2$ and $\rho_3$ on $\C^m=\bigoplus_aV_a$, $\C^k=\bigoplus_bW_b$ and $\C^n=\bigoplus_cU_c$, orthogonal decompositions into irreducible subrepresentations. Suppose that for $(a,b)\neq(a',b')$ the representations $\bar V_a\otimes W_b$ and $\bar V_{a'}\otimes W_{b'}$ have no irreducible constituent in common, and likewise $\bar W_b\otimes U_c$ and $\bar W_{b'}\otimes U_{c'}$ for $(b,c)\neq(b',c')$. Let $\Phi\colon\M_m\to\M_k$ and $\Psi\colon\M_k\to\M_n$ be PPT maps with $\Phi(\rho_1(g)X\rho_1(g)^{\dagger})=\rho_2(g)\Phi(X)\rho_2(g)^{\dagger}$ and $\Psi(\rho_2(g)Y\rho_2(g)^{\dagger})=\rho_3(g)\Psi(Y)\rho_3(g)^{\dagger}$ for all $g\in G$. Then $\Phi(X)=\sum_{a,b}P_{W_b}\Phi(P_{V_a}XP_{V_a})P_{W_b}$, the maps $\Phi_{ab}\colon\M(V_a)\to\M(W_b)$ obtained by compression are covariant and PPT, and the same holds for $\Psi$, with maps $\Psi_{bc}\colon\M(W_b)\to\M(U_c)$. The Choi matrix of $\Psi\circ\Phi$ is block diagonal for the orthogonal decompositions $\C^m=\bigoplus_a\bar V_a$, where $\bar V_a=\{\bar v:v\in V_a\}$, and $\C^n=\bigoplus_cU_c$, with the block on $\bar V_a\otimes U_c$ equal to $\sum_bC_{\Psi_{bc}\circ\Phi_{ab}}$. So if every composition $\Psi_{bc}\circ\Phi_{ab}$ is entanglement breaking, then so is $\Psi\circ\Phi$. \end{lemma} \begin{proof} As in the proof of Theorem~\ref{thm:compact}\ref{compact:b}, $C_\Phi$ commutes with $\bar\rho_1(g)\otimes\rho_2(g)$, which preserves the orthogonal subspaces $\bar V_a\otimes W_b$ of $\C^m\otimes\C^k$. By the hypothesis, each isotypic component of this representation lies in one of these subspaces, and $C_\Phi$ preserves the isotypic components, so it is block diagonal for this decomposition. This gives the form of $\Phi$, and the blocks are the Choi matrices of the maps $\Phi_{ab}$, which are PPT as compressions of $\Phi$. The same holds for $\Psi$. Composing, $(\Psi\circ\Phi)(X)=\sum_{a,b,c}P_{U_c}\Psi_{bc}\bigl(\Phi_{ab}(P_{V_a}XP_{V_a})\bigr)P_{U_c}$, which gives the blocks of $C_{\Psi\circ\Phi}$. An operator that is block diagonal for orthogonal decompositions of both tensor factors is separable if its blocks are, and sums of separable operators are separable. \end{proof} \begin{corollary}[\ca]\label{cor:su2blocks} Let $G=\mathrm{SU}(2)$, and suppose that the hypothesis of Lemma~\ref{lem:covblocks} holds and that the summands $V_a$, $W_b$ and $U_c$ have dimension at most nine. Then $\Psi\circ\Phi$ is entanglement breaking for all covariant PPT maps $\Phi$ and $\Psi$ as in that lemma. This applies, for instance, when $\C^m=\C^n$ is the sum of a trivial representation and an irreducible one of spin $3/2$ or $5/2$ and $\C^k$ is irreducible of spin $3$, and when $\C^m=\C^n$ is the sum of a trivial representation and an irreducible one of spin $5/2$ and $\C^k$ is the sum of a trivial representation and an irreducible one of spin $3/2$. \end{corollary} \begin{proof} By Lemma~\ref{lem:covblocks} it suffices that every composition $\Psi_{bc}\circ\Phi_{ab}$ is entanglement breaking. This is clear if one of $V_a$, $W_b$ and $U_c$ has dimension one, and otherwise it is Theorem~\ref{thm:classes}\ref{cl:su2}. In the examples, the Clebsch--Gordan rule shows that the products of distinct pairs of summands have no common constituent. \end{proof} \section{An extended range criterion}\label{sec:edge} If $Z=\sum_tP_{a_t}\otimes P_{b_t}$ is separable, with nonzero $a_t$ and $b_t$, then $Z^{\Gamma}=\sum_tP_{a_t}\otimes P_{\bar b_t}$. So every vector $a_t\otimes b_t$ lies in \[ E(Z)=\{a\otimes b\in\ran Z:\ a\neq0,\ b\neq0,\ a\otimes\bar b\in\ran Z^{\Gamma}\}, \] and $E(Z)$ spans $\ran Z$. This is the range criterion of \cite{Hor97} in the form used for edge states \cite{LKCH00}. A composite with $E(Z)$ not spanning $\ran Z$ would be entangled even if its range contains product vectors. We record its form for composites of PPT maps, show that $E(Z)\neq\emptyset$ for small types, and bound the vectors of $\ker Z^{\Gamma}$ that could make it stronger than the plain criterion. \begin{lemma}\label{lem:edge} Let $\Phi\colon\M_m\to\M_k$ and $\Psi\colon\M_k\to\M_n$ be PPT and $Z=C_{\Psi\circ\Phi}$. \begin{enumerate}[label=(\alph*)] \item\label{edge:a} $\ran Z^{\Gamma}=\vecop\bigl(\spn(\cK_{T\circ\Psi}\cK_\Phi)\bigr)$, and for nonzero $a$ and $b$, $a\otimes b\in E(Z)$ if and only if $ba^{T}\in\spn(\cK_\Psi\cK_\Phi)$ and $\bar ba^{T}\in\spn(\cK_{T\circ\Psi}\cK_\Phi)$. \item\label{edge:b} If $\ker Z^{\Gamma}$ is spanned by product vectors, then $E(Z)$ is the set of all product vectors in $\ran Z$. \item\label{edge:c} Let $A\in\cK_\Phi$, $U=\ran A$, and $\Psi_U=\Psi\circ\Ad_{\iota_U}$ with an isometry $\iota_U$ onto $U$. If $C_{\Psi_U}$ is separable, then $\vecop(\cK_\Psi A)\subseteq\spn E(Z)$. In particular this holds when $\rank A=1$. \end{enumerate} \end{lemma} \begin{proof} \ref{edge:a} $Z^{\Gamma}=C_{(T\circ\Psi)\circ\Phi}$, so Lemma~\ref{lem:range} applies. \ref{edge:b} The vector $a\otimes b$ is orthogonal to $x\otimes y$ if and only if $a\otimes\bar b$ is orthogonal to $x\otimes\bar y$, and by Corollary~\ref{cor:blocks}\ref{blocks:a} the product vectors in $\ker Z^{\Gamma}$ are exactly the $x\otimes\bar y$ with $x\otimes y\in\ker Z$. So for $a\otimes b\in\ran Z$ the vector $a\otimes\bar b$ is orthogonal to all product vectors of $\ker Z^{\Gamma}$, which span it. \ref{edge:c} Write $A=\iota_UA_0$. Then $\cK_{\Psi_U}=\cK_\Psi\iota_U$ and $\cK_{T\circ\Psi_U}=\cK_{T\circ\Psi}\iota_U$. A separable decomposition of $C_{\Psi_U}$ consists of product vectors $w\otimes c$ with $cw^{T}=B\iota_U$ and $\bar cw^{T}=B'\iota_U$ for some $B\in\cK_\Psi$ and $B'\in\cK_{T\circ\Psi}$, and the matrices $cw^{T}$ span $\cK_\Psi\iota_U$. Then $c(A_0^{T}w)^{T}=BA$ and $\bar c(A_0^{T}w)^{T}=B'A$, so $A_0^{T}w\otimes c\in E(Z)$ whenever $A_0^{T}w\neq0$ by \ref{edge:a}, and the matrices $c(A_0^{T}w)^{T}$ span $\cK_\Psi\iota_UA_0=\cK_\Psi A$. \end{proof} Part \ref{edge:b} applies to the pair of \cite{OAI26}, for which $Z^{\Gamma}=Z$ and the kernel is spanned by twenty product vectors: there the extended criterion is the plain one. \begin{proposition}\label{prop:edge} Let $(\Phi,\Psi)$ be a reduced PPT pair of type $(m,k,n)$ with $\Psi\circ\Phi\neq0$. Suppose that for some $r\ge1$ with $rn\le6$ we have $(k-r)(m-r)\le\max(m,k)$. Then $E(Z)\ne\emptyset$. The same holds with $(m,n)$ replaced by $(n,m)$. \end{proposition} \begin{proof} If $b\le\max(m,k)$, then $C_\Phi$ is separable by Lemma~\ref{lem:HLVC}, hence so is $Z$, and $E(Z)$ spans $\ran Z\ne0$. Otherwise $b>(k-r)(m-r)$, and $\cK_\Phi$ contains a nonzero $A$ with $\rank A\le r$, by Lemma~\ref{lem:det} if $r<\min(m,k)$, and trivially if $r\ge\min(m,k)$. Then $C_{\Psi_U}$ is a positive semidefinite operator with positive partial transpose on $\C^{\rank A}\otimes\C^n$, and $\rank A\cdot n\le6$, so it is separable \cite{Per96,HHH96}. Since $\Psi^{*}(I)\succ0$, $\cK_\Psi A\ne0$, and Lemma~\ref{lem:edge}\ref{edge:c} gives $E(Z)\ne\emptyset$. The last statement follows by applying this to the mirror pair $(\Psi^{\#},\Phi^{\#})$ of Lemma~\ref{lem:sym}, which is reduced by Lemma~\ref{lem:sym}\ref{sym:c}: its composite Choi matrix is $Z'=WZW^{\dagger}$, where $W\colon\C^m\otimes\C^n\to\C^n\otimes\C^m$ is the flip, and $Z'^{\Gamma}=W\overline{Z^{\Gamma}}W^{\dagger}$, so $E(Z')=W\,E(Z)$. \end{proof} The hypothesis holds, for example, when $n\le3$ and $\min(m,k)\le3$, for $(4,4,3)$ and $(3,4,4)$, and for $(5,5,2)$, $(6,4,2)$, $(4,6,2)$, $(5,6,2)$, $(6,5,2)$ and their mirrors. For these types the strongest form of the extended criterion, $E(Z)=\emptyset$, can never certify a counterexample. The extended criterion is stronger than the plain one only if $\ker Z^{\Gamma}$ is not spanned by product vectors (Lemma~\ref{lem:edge}\ref{edge:b}). Vectors of $\ker Z^{\Gamma}$ constrain the Kraus spaces as follows. For $w\in\C^m\otimes\C^n$ let $Y(w)\subseteq\C^n$ be the span of the second factors in a decomposition $w=\sum_tx_t\otimes y_t$ with linearly independent $x_t$. It does not depend on the decomposition. \begin{lemma}\label{lem:transfer} Let $\Phi\colon\M_m\to\M_k$ and $\Psi\colon\M_k\to\M_n$ be completely positive, and $Z=C_{\Psi\circ\Phi}$. \begin{enumerate}[label=(\alph*)] \item\label{tr:a} A vector $v\in\C^m\otimes\C^n$ lies in $\ker Z$ if and only if $(I\otimes B^{\dagger})v\in\ker C_\Phi$ for all $B\in\cK_\Psi$. If $\Psi$ is PPT, then $v\in\ker Z^{\Gamma}$ if and only if $(I\otimes B^{\dagger})v\in\ker C_\Phi$ for all $B\in\cK_{T\circ\Psi}$. If $\Phi$ is PPT, then $v\in\ker Z^{\Gamma}$ if and only if $(I\otimes B^{\dagger})v\in\ker C_\Phi^{\Gamma}$ for all $B\in\overline{\cK_\Psi}$. If both maps are PPT, then $v\in\ker Z$ if and only if $(I\otimes B^{\dagger})v\in\ker C_\Phi^{\Gamma}$ for all $B\in\cK_{\Psi\circ T}$. \item\label{tr:b} Let $\cK\subseteq\M_{n\times k}$ be a subspace and $w\in\C^m\otimes\C^n$. Then the vectors $(I\otimes B^{\dagger})w$ with $B\in\cK$ form a subspace of dimension $\dim\cK-\dim\{B\in\cK:B^{\dagger}Y(w)=0\}$, which is at most $k$ times the Schmidt rank of $w$. \end{enumerate} \end{lemma} \begin{proof} \ref{tr:a} If $\Xi=\sum_l\Ad_{B_l}$ is completely positive and $X\succeq0$, then $\ip{v}{(\id\otimes\Xi)(X)v}=\sum_l\ip{(I\otimes B_l^{\dagger})v}{X(I\otimes B_l^{\dagger})v}$. So $v\in\ker(\id\otimes\Xi)(X)$ if and only if $(I\otimes B_l^{\dagger})v\in\ker X$ for all $l$, and, since $B\mapsto(I\otimes B^{\dagger})v$ is conjugate-linear, if and only if $(I\otimes B^{\dagger})v\in\ker X$ for all $B\in\cK_\Xi$. Apply this to $Z=(\id\otimes\Psi)(C_\Phi)$, to $Z^{\Gamma}=C_{(T\circ\Psi)\circ\Phi}=(\id\otimes(T\circ\Psi))(C_\Phi)$, to $Z^{\Gamma}=C_{(T\circ\Psi\circ T)\circ(T\circ\Phi)}=(\id\otimes(T\circ\Psi\circ T))(C_\Phi^{\Gamma})$, where $T\circ\Psi\circ T=\sum_l\Ad_{\bar B_l}$ for $\Psi=\sum_l\Ad_{B_l}$, and to $Z=C_{(\Psi\circ T)\circ(T\circ\Phi)}=(\id\otimes(\Psi\circ T))(C_\Phi^{\Gamma})$. \ref{tr:b} Write $w=\sum_tx_t\otimes y_t$ with linearly independent $x_t$ and $y_t$. The map $B\mapsto\sum_tx_t\otimes B^{\dagger}y_t$ is conjugate-linear, its kernel consists of the $B$ with $B^{\dagger}y_t=0$ for all $t$, and its image lies in $\spn\{x_t\}\otimes\C^k$. \end{proof} For completely positive $\Phi$ and $\Psi$ call $([x],[y])$ a kernel point if $\bar x\otimes y\in\ker Z$, and put $r(x)=\dim\cK_\Phi x$ and $t(y)=\rank\Psi^{*}(P_y)$, as in the proof of Proposition~\ref{prop:ups}. By \eqref{eq:basic}, $\ker C_\Phi$ contains $F(x)=\bar x\otimes(\cK_\Phi x)^{\perp}$, of dimension $k-r(x)$. Since $(u\otimes y)^{\dagger}C_\Psi(u\otimes y)=\bar u^{\dagger}\Psi^{*}(P_y)\bar u$, also $\ker C_\Psi$ contains $G(y)=\overline{(\ran\Psi^{*}(P_y))^{\perp}}\otimes y$, of dimension $k-t(y)$. \begin{proposition}\label{prop:edgecount} Let $\Phi\colon\M_m\to\M_k$ be completely positive and $\Psi\colon\M_k\to\M_n$ PPT, and put $Z=C_{\Psi\circ\Phi}$, $a=\rank C_\Psi$, $a'=\rank C_\Psi^{\Gamma}$ and $b=\rank C_\Phi$. Let $([x_i],[y_i])$, $1\le i\le N$, be kernel points, and let \[ \delta_1=\sum_i(k-r(x_i))-\dim\sum_iF(x_i),\qquad\delta_2=\sum_i(k-t(y_i))-\dim\sum_iG(y_i). \] For $w\in\ker Z^{\Gamma}$ put $W=\{(I\otimes B^{\dagger})w:B\in\cK_{T\circ\Psi}\}$, $\varepsilon_1=\dim\{B\in\cK_{T\circ\Psi}:B^{\dagger}Y(w)=0\}$ and $\varepsilon_2=\dim\bigl(W\cap\sum_iF(x_i)\bigr)$. Then \[ a+b+a'\le(m+n-N)k+\delta_1+\delta_2+\varepsilon_1+\varepsilon_2 . \] \end{proposition} \begin{proof} By Lemma~\ref{lem:transfer}, $W\subseteq\ker C_\Phi$ and $\dim W=a'-\varepsilon_1$, as $\cK_{T\circ\Psi}$ has dimension $\rank C_{T\circ\Psi}=a'$. The spaces $F(x_i)$ also lie in $\ker C_\Phi$, so \[ \sum_i(k-r(x_i))-\delta_1+a'-\varepsilon_1-\varepsilon_2=\dim\Bigl(\sum_iF(x_i)+W\Bigr)\le mk-b . \] Likewise $\sum_i(k-t(y_i))-\delta_2\le kn-a$. Adding the two inequalities and using $r(x_i)+t(y_i)\le k$ (Lemma~\ref{lem:range}\ref{range:b}) gives the claim. \end{proof} For $w=0$ we have $\varepsilon_1=a'$ and $\varepsilon_2=0$, and the inequality is the count $a+b\le(m+n-N)k+\delta_1+\delta_2$. The same happens whenever $w$ lies in the span of the vectors $\bar x_i\otimes\bar y_i$, which lie in $\ker Z^{\Gamma}$ by Corollary~\ref{cor:blocks}\ref{blocks:a}. Indeed, $\bar x_i\otimes B^{\dagger}\bar y_i\in\ker C_\Phi$ for $B\in\cK_{T\circ\Psi}$ by Lemma~\ref{lem:transfer}, and $F(x_i)$ contains every vector of $\ker C_\Phi$ of the form $\bar x_i\otimes u$, so $W\subseteq\sum_iF(x_i)$ and $\varepsilon_1+\varepsilon_2=a'$. \begin{corollary}\label{cor:edgeos} Let $(\Phi,\Psi)$ be a PPT pair of type $(m,k,n)$ with $\Phi(I)\succ0$ and $\Psi^{*}(I)\succ0$, such that $C_\Phi$ and $C_\Psi$ are not separable. If $([x_i],[y_i])$, $1\le i\le m+n-2$, are kernel points, then $\delta_1+\delta_2+\varepsilon_1+\varepsilon_2\ge k+3$ for every $w\in\ker Z^{\Gamma}$. If moreover $\varepsilon_1=0$, then $w$ has Schmidt rank at least two. \end{corollary} \begin{proof} The partial traces of $C_\Phi$ over $\C^m$ and of $C_\Psi$ and $C_\Psi^{\Gamma}$ over $\C^n$ are $\Phi(I)$ and $\overline{\Psi^{*}(I)}$, of rank $k$. The three operators have positive partial transpose, and none of them is separable, as $C_\Psi^{\Gamma}$ is separable only if $C_\Psi$ is. So $a,a',b\ge k+1$ by Lemma~\ref{lem:HLVC}, and Proposition~\ref{prop:edgecount} gives $3k+3\le2k+\delta_1+\delta_2+\varepsilon_1+\varepsilon_2$. If $\varepsilon_1=0$, then $k+1\le a'=\dim W$, which is at most $k$ times the Schmidt rank of $w$ by Lemma~\ref{lem:transfer}\ref{tr:b}. \end{proof} The middle marginals can be made positive definite without changing $Z$, by the first step of the proof of Lemma~\ref{lem:reduce}, and $C_\Phi$ and $C_\Psi$ are not separable when $Z$ is entangled. For such a pair, which is one product vector short in the sense of Lemma~\ref{lem:oneshort} when the kernel points are in general position, Proposition~\ref{prop:edgecount} with $w=0$ only gives $\delta_1+\delta_2\ge a+b-2k\ge2$. So a vector $w\in\ker Z^{\Gamma}$ outside the span of the $\bar x_i\otimes\bar y_i$ forces strongly dependent blocks, a large intersection of $W$ with the blocks $F(x_i)$, or many $B\in\cK_{T\circ\Psi}$ with $B^{\dagger}Y(w)=0$. Proposition~\ref{prop:edge} excludes the strongest form $E(Z)=\emptyset$ of the extended criterion for small types, and Corollary~\ref{cor:edgeos} constrains its weaker form, in which $E(Z)$ does not span $\ran Z$, on strata one product vector short in every type. Finally, if the kernel of the Choi matrix of one factor and that of its partial transpose are nested, then $\ker Z^{\Gamma}$ is determined by $\ker Z$. \begin{proposition}\label{prop:edgenest} Let $\Phi\colon\M_m\to\M_k$ and $\Psi\colon\M_k\to\M_n$ be PPT and $Z=C_{\Psi\circ\Phi}$. \begin{enumerate}[label=(\alph*)] \item\label{nest:a} If $\ker C_\Psi\subseteq\ker C_\Psi^{\Gamma}$ or $\ker C_\Psi^{\Gamma}\subseteq\ker C_\Psi$, then $\ker Z^{\Gamma}=\ker Z$. \item\label{nest:b} If $\overline{\ker C_\Phi}\subseteq\ker C_\Phi^{\Gamma}$ or $\ker C_\Phi^{\Gamma}\subseteq\overline{\ker C_\Phi}$, then $\ker Z^{\Gamma}=\overline{\ker Z}$. \end{enumerate} In both cases, if $\ker Z$ is spanned by product vectors, then $E(Z)$ is the set of all product vectors in $\ran Z$. The hypothesis of \ref{nest:a} holds if $\ker C_\Psi$ is spanned by product vectors $x\otimes y$ with $y$ real, and that of \ref{nest:b} holds if $\ker C_\Phi$ is spanned by product vectors $x\otimes y$ with $x$ real. \end{proposition} \begin{proof} \ref{nest:a} Write $\Phi=\sum_a\Ad_{A_a}$. As $C_\Phi=\sum_a\vecop(A_a)\vecop(A_a)^{\dagger}$ and $\vecop(A_a)=(A_a^{T}\otimes I)\vecop(I_k)$, we have $Z=(\id\otimes\Psi)(C_\Phi)=\sum_a(A_a^{T}\otimes I)C_\Psi(\bar A_a\otimes I)$, and in the same way $Z^{\Gamma}=\sum_a(A_a^{T}\otimes I)C_\Psi^{\Gamma}(\bar A_a\otimes I)$. Suppose that $\ker C_\Psi\subseteq\ker C_\Psi^{\Gamma}$, let $v\in\C^m\otimes\C^n$, and put $w_a=(\bar A_a\otimes I)v$. If $Zv=0$, then all $w_a$ lie in $\ker C_\Psi$, since the terms of $Z$ are positive semidefinite, so they lie in $\ker C_\Psi^{\Gamma}$, and $Z^{\Gamma}v=0$. Conversely, let $Z^{\Gamma}v=0$ and $R=\sum_aw_aw_a^{\dagger}$. Then $R^{\Gamma}$ is the transpose of $((\Phi\circ T)\otimes\id)(vv^{\dagger})$, which is positive semidefinite because $\Phi$ is PPT, and $\Tr(C_\Psi R^{\Gamma})=\Tr(C_\Psi^{\Gamma}R)=\ip{v}{Z^{\Gamma}v}=0$. So $\ran R^{\Gamma}\subseteq\ker C_\Psi\subseteq\ker C_\Psi^{\Gamma}$, hence $\ip{v}{Zv}=\Tr(C_\Psi R)=\Tr(C_\Psi^{\Gamma}R^{\Gamma})=0$, and $Zv=0$. If $\ker C_\Psi^{\Gamma}\subseteq\ker C_\Psi$, the same argument applies to the PPT map $T\circ\Psi$ in place of $\Psi$, since $C_{T\circ\Psi}=C_\Psi^{\Gamma}$ and $C_{(T\circ\Psi)\circ\Phi}=Z^{\Gamma}$. \ref{nest:b} Let $\Gamma_1$ be the partial transpose on the first factor. For Hermitian $X$ we have $X^{\Gamma_1}=\overline{X^{\Gamma}}$, so $\ker X^{\Gamma_1}=\overline{\ker X^{\Gamma}}$. Hence the hypothesis says that $\ker C_\Phi\subseteq\ker C_\Phi^{\Gamma_1}$ or $\ker C_\Phi^{\Gamma_1}\subseteq\ker C_\Phi$, and the claim is that $\ker Z^{\Gamma_1}=\ker Z$. Write $\Psi=\sum_b\Ad_{B_b}$. Then $Z=\sum_b(I\otimes B_b)C_\Phi(I\otimes B_b^{\dagger})$ and $Z^{\Gamma_1}=\sum_b(I\otimes B_b)C_\Phi^{\Gamma_1}(I\otimes B_b^{\dagger})$, and the proof of \ref{nest:a} applies with the vectors $(I\otimes B_b^{\dagger})v$ in place of the $w_a$ and with $\Gamma_1$ in place of $\Gamma$. The operator $S=\sum_b(I\otimes B_b^{\dagger})vv^{\dagger}(I\otimes B_b)$ that replaces $R$ has $S^{\Gamma_1}$ equal to the transpose of $(\id\otimes(\Psi\circ T)^{*})(vv^{\dagger})$, which is positive semidefinite because $\Psi$ is PPT. For the reverse inclusion one uses the PPT map $\Phi\circ T$, with $C_{\Phi\circ T}=C_\Phi^{\Gamma_1}$ and $C_{\Psi\circ\Phi\circ T}=Z^{\Gamma_1}$. In both cases, if $\ker Z$ is spanned by product vectors, then so is $\ker Z^{\Gamma}$, and Lemma~\ref{lem:edge}\ref{edge:b} applies. Finally, by Corollary~\ref{cor:blocks}\ref{blocks:a}, a product vector $x\otimes y\in\ker C_\Psi$ with real $y$ satisfies $x\otimes y=x\otimes\bar y\in\ker C_\Psi^{\Gamma}$, and a product vector $x\otimes y\in\ker C_\Phi$ with real $x$ satisfies $\overline{x\otimes y}=x\otimes\bar y\in\ker C_\Phi^{\Gamma}$. \end{proof} So for a PPT pair whose composite kernel is spanned by product vectors, the extended criterion can be stronger than the plain one only if $\ker C_\Psi$ and $\ker C_\Psi^{\Gamma}$ are incomparable and $\overline{\ker C_\Phi}$ and $\ker C_\Phi^{\Gamma}$ are incomparable. In particular, $\ker C_\Psi$ is then not spanned by product vectors $x\otimes y$ with real $y$, and $\ker C_\Phi$ is not spanned by product vectors $x\otimes y$ with real $x$. \section{Open problems}\label{sec:open} We first show by examples that linear algebra and Lemma~\ref{lem:noranktwo} alone cannot exclude the types $(6,6,6)$, $(5,6,5)$ and $(5,6,6)$, and then list open problems. The type $(6,6,6)$ cannot be excluded using only the conclusions of Lemma~\ref{lem:noranktwo} and the rank windows of Theorem~\ref{thm:windows} together with linear algebra, as the following example shows. Its spaces have dimension $9$, below the bound $a,b\ge10$ of Corollary~\ref{cor:n3}, whose proof uses positivity through Theorem~\ref{thm:n3}. \begin{example}\label{ex:sl2} Let $V=\Sym^5\C^2$, realised as the polynomials of degree at most five in one variable $z$, with $\mathfrak{sl}_2(\C)$ acting by $e=d/dz$, $f=-z^2\,d/dz+5z$ and $h=[e,f]=5-2z\,d/dz$. Let $\mathcal D\subseteq\mathrm{End}(V)$ be the image of the elements of degree at most two of the universal enveloping algebra $U(\mathfrak{sl}_2)$, and identify $\mathrm{End}(V)$ with $\M_6$ by an orthonormal basis for an $\mathrm{SU}(2)$-invariant inner product on $V$. Then $\dim\mathcal D=9$, $I\in\mathcal D$, no nonzero element of $\mathcal D$ has rank at most three, and $\spn(\mathcal D\mathcal D)$ has dimension $25$ and contains no matrix of rank one. No nonzero PPT map has its Kraus space contained in $\mathcal D$. \end{example} \begin{proof} The adjoint action preserves the filtration of $U(\mathfrak{sl}_2)$ by degree, so the image $\mathcal D_d$ of the elements of degree at most $d$ is a subrepresentation of $\mathrm{End}(V)\cong V\otimes V^{*}$, which is the sum of irreducible representations $W_0,W_2,\dots,W_{10}$ with $\dim W_k=k+1$. The weights of $\mathrm{ad}\,h$ on the elements of degree at most $d$ lie in $[-2d,2d]$, and $e^j$, of weight $2j$, is a highest weight vector that acts as $(d/dz)^j\ne0$ for $j\le5$. So $\mathcal D_d=W_0\oplus W_2\oplus\dots\oplus W_{2d}$ for $d\le5$. Hence $\mathcal D=\mathcal D_2$ has dimension $9$ and contains $I$, and $\spn(\mathcal D\mathcal D)=\mathcal D_4$ has dimension $25$. A rank-one operator has a nonzero component in $W_{10}$: under $V^{*}\cong V$, the projection $V\otimes V\to W_{10}\cong\Sym^{10}\C^2$ is, up to a nonzero factor, the multiplication of binary forms, and the product of two nonzero forms is nonzero. A nonzero element of $\mathcal D$ acts as a nonzero differential operator $a(z)\,d^2/dz^2+b(z)\,d/dz+c(z)$, so its kernel on $V$ has dimension at most two and its rank is at least four. Finally, $\vecop(gYg^{-1})=(\bar g\otimes g)\vecop(Y)$ for unitary $g$, so $\vecop(\mathcal D)$ is invariant under $\bar\sigma\otimes\sigma$, where $\sigma$ is the irreducible representation of $\mathrm{SU}(2)$ on $V$, and it is product-free. By Theorem~\ref{thm:compact} and Corollary~\ref{cor:twirl}, no nonzero PPT operator has its range in $\vecop(\mathcal D)$, while the Choi matrix of a nonzero PPT map with Kraus space contained in $\mathcal D$ would be such an operator. \end{proof} \begin{remark}\label{rem:klein} At type $(5,6,5)$ a natural ansatz identifies $\C^6$ with $\Lambda^2\C^4$, so that $\alpha$-planes $v\wedge\C^4$ of the Klein quadric can be members. It requires $\cK_\Phi$ to contain the four-dimensional space $\mathcal Y$ of the maps $u\mapsto v\wedge\ell(u)$, $v\in\C^4$, for a surjection $\ell\colon\C^5\to\C^4$, and $\cK_\Psi$ to contain the space $\mathcal X$ of the maps $\eta\mapsto\nu(w\wedge\eta)$, $w\in\C^4$, for an injection $\nu\colon\Lambda^3\C^4\to\C^5$. By Theorem~\ref{thm:main} this ansatz gives no range-criterion pair. We sketch a direct argument that would give more; its case analysis is not written out, and the remark is not used in any proof. The argument concludes that if $\Phi$ and $\Psi$ are PPT with $\mathcal Y\subseteq\cK_\Phi$, $\mathcal X\subseteq\cK_\Psi$ and $a,b\ge9$, then $s\ge17$ or $S$ contains a matrix of rank one. In outline, suppose that $s\le16$ and that $S$ is rank-one-free. Then the quadrics $u\mapsto e_i\wedge\ell(u)\wedge E(u)\in\Lambda^4\C^4\cong\C$, for $E\in\cK_\Phi$ and a basis $e_i$ of $\C^4$, lie in a space $\mathcal Q$ of quadrics on $\PP^4$ of dimension at most seven, determined by $S^{\perp}$. At a common zero $u$ of $\mathcal Q$ with $\ell(u)\neq0$ this gives $\cK_\Phi u\subseteq\ell(u)\wedge\C^4$, so $[u]\in D_3(\Phi)$, and Lemma~\ref{lem:linear} confines every curve of such points to a line through $[\ker\ell]$. On the other hand $\dim\cK_\Phi/\mathcal Y\ge5$, as $b\ge9$, and $\dim\cK_\Phi/\mathcal Y$ is at most the dimension of the space of linear syzygies of $\mathcal Q$ whose coefficients vanish at $[\ker\ell]$. A case analysis of these syzygies, together with the mirror argument for $\cK_\Psi$, which uses $a\ge9$, shows that this space has dimension at most four, or that $\dim S^{\perp}\le8$, contradicting $s\le16$, or that $\mathcal Q$ has a curve of common zeros that is not such a line. Granted the case analysis, the same ansatz fails for every output dimension, and on the side of $\Psi$ only $\mathcal X\subseteq\cK_\Psi$ is needed: if $\Phi\colon\M_5\to\M_6$ is PPT with $\mathcal Y\subseteq\cK_\Phi$ and $b\ge9$, and $\Psi\colon\M_6\to\M_n$ is completely positive with $\mathcal X\subseteq\cK_\Psi$ for an injection $\nu\colon\Lambda^3\C^4\to\C^n$, $n\ge4$, then $S$ contains a matrix of rank one. Indeed, for $y\in\ran\nu$ and $x\in\C^5$ the bilinear forms $x^{T}Wy$ with $W\in S^{\perp}$ depend only on $W|_{\ran\nu}$, and these restrictions span a space of dimension $14-\dim\mathcal Q$, as $\mathcal Q$ is its annihilator in the $14$-dimensional space of quadrics on $\PP^4$ that vanish at $[\ker\ell]$. If this dimension is at most seven, the forms have a common zero on $\PP(\ran\nu)\times\PP^4$, a variety of dimension seven on which $\cO(1,1)$ is ample, and $yx^{T}\in S$ has rank one. So a rank-one-free $S$ has $\dim\mathcal Q\le6$, and then the case analysis above ends in its first or third alternative, which use neither the output dimension, nor a bound on $s$, nor the mirror argument. Applied to the pair $(\Ad_{G^{-1}}\circ\Psi^{\#},\Phi^{\#}\circ\Ad_G)$, where $G$ is the matrix of the pairing $(\eta,\theta)\mapsto\eta\wedge\theta$ on $\Lambda^2\C^4$, whose composite has Kraus space $S^{T}$, this gives the same conclusion at type $(6,6,5)$ when $\Psi$ is PPT with $\mathcal X\subseteq\cK_\Psi$ and $a\ge9$, and $\mathcal Y\subseteq\cK_\Phi$ for a surjection $\ell\colon\C^6\to\C^4$. \end{remark} \begin{remark}[\ca]\label{rem:565lin} The type $(5,6,5)$ cannot be excluded by linear algebra and Lemma~\ref{lem:noranktwo} alone either. In the notation of Remark~\ref{rem:klein}, let $\ell(x,t)=x$ for $(x,t)\in\C^4\oplus\C=\C^5$, and let $\mathcal Q$ be the span of the six $2\times2$ minors of the matrix with rows $(x_0,x_1,x_2,-x_3)$ and $(x_1,x_2,x_3,t)$, the quadrics through the rational normal curve $\Gamma=\{[1:r:r^2:r^3:-r^4]\}\subseteq\PP^4$. Let $\cK_1$ be the space of linear maps $E\colon\C^5\to\Lambda^2\C^4$ such that the four quadrics $u\mapsto e_i\wedge\ell(u)\wedge E(u)\in\Lambda^4\C^4\cong\C$ lie in $\mathcal Q$, where $e_0,\dots,e_3$ is the standard basis of $\C^4$. Let $G$ be the matrix of the pairing $(\eta,\theta)\mapsto\eta\wedge\theta$ on $\Lambda^2\C^4$ in the basis of the $e_i\wedge e_j$, and let $\cK_2=\{E^{T}G:E\in\cK_1\}$. Then $\dim\cK_1=\dim\cK_2=9$ and $\cK_1\supseteq\mathcal Y$. Also $\cK_2\supseteq\mathcal X$ for the injection $\nu\colon\Lambda^3\C^4\to\C^5$ that sends $\theta$ to $(c_0,c_1,c_2,c_3,0)$, where $e_i\wedge\theta=c_i\,e_0\wedge e_1\wedge e_2\wedge e_3$. The space $\spn(\cK_2\cK_1)$ has dimension $16$ and is rank-one-free, neither $\cK_1$ nor $\cK_2$ contains a nonzero matrix of rank at most two, $\cK_1$ has no common kernel, and $\dim\cK_1u=6$ for general $u$. So $\cK_1$ and $\cK_2$ satisfy the conclusions of Lemma~\ref{lem:noranktwo} and Proposition~\ref{prop:cuts}\ref{cuts:a}, with $s=16$. But $\dim\cK_1u=3$ for $[u]\in\Gamma$, and $\Gamma$ spans $\PP^4$, so by Lemma~\ref{lem:linear} no PPT map has Kraus space $\cK_1$. Moreover, $S$ consists of the matrices $M$ with $e(r)^{T}Me(r)=0$ for all $r$, where $e(r)=(1,r,r^2,r^3,-r^4)$. So a change of sign of the last coordinate on both sides turns $S$ into the kernel of the multiplication $\Sym^4\C^2\otimes\Sym^4\C^2\to\Sym^8\C^2$ in the monomial bases, and $S^{\perp}$ is spanned by the matrices $e(r)e(r)^{T}$ of rank one. The type $(5,6,6)$ has an analogous configuration, with the same input side and a rational normal curve of degree five on the output side: $\cK_\Phi=\cK_1$, and $\cK_\Psi=\{E^{T}G\}$ for the linear maps $E\colon\C^6\to\Lambda^2\C^4$ such that the four quadrics $u\mapsto e_i\wedge\ell'(u)\wedge E(u)$ lie in the span of the nine $2\times2$ minors other than $x_3t_2-t_1^2$ of the matrix with rows $(x_0,x_1,x_2,x_3,t_1)$ and $(x_1,x_2,x_3,t_1,t_2)$, where $\ell'(x,t_1,t_2)=x$ for $(x,t_1,t_2)\in\C^4\oplus\C^2=\C^6$. It has $(a,b)=(12,9)$, inside the ranges of Proposition~\ref{prop:cuts}\ref{cuts:b}, and $s=20=(m-1)(n-1)$, and a change of sign of one coordinate turns $S$ into the kernel of the multiplication $\Sym^4\C^2\otimes\Sym^5\C^2\to\Sym^9\C^2$, a completely entangled subspace of the maximal dimension \cite{Par04}. It satisfies the conclusions of Lemma~\ref{lem:noranktwo}, and Lemma~\ref{lem:linear} excludes it, as $\cK_\Phi u$ and $\cK_\Psi^{T}y$ have rank three along rational normal curves that span $\PP^4$ and $\PP^5$. The stated properties of these configurations were verified in exact arithmetic (Table~\ref{tab:verify}). \end{remark} \begin{enumerate}[label=(\arabic*),leftmargin=*] \item Decide the reduced type $(6,6,6)$, whose open cells Table~\ref{tab:d6} lists; the paragraph before that table and Remarks~\ref{o666t:rem:c4} and~\ref{rem:666open} collect what is known about them. By Example~\ref{ex:sl2}, arguments for $(6,6,6)$ have to use positivity beyond Lemma~\ref{lem:noranktwo}, such as Lemmas~\ref{lem:linear} and~\ref{lem:kernelrank}, which the proofs of Theorems~\ref{thm:565no}, \ref{thm:666zero} and~\ref{thm:566no} use, and Remark~\ref{rem:565lin} shows the same for $(5,6,5)$ and $(5,6,6)$. \item Find a range-criterion pair with $\max(m,k,n)\le9$, or show that none exists. By Theorem~\ref{thm:main} it would need $\max(m,k,n)\ge6$, and $m=k=n=6$ if $\max(m,k,n)=6$; Theorems~\ref{thm:44}, \ref{o469:thm:469} and~\ref{thm:649ex}, Proposition~\ref{prop:count} and Theorem~\ref{thm:schur} exclude further types and constructions. \item Is the composite Choi matrix of the member of type $(6,6,6)$ (Theorem~\ref{thm:rigid}) separable for every pencil? When no six of the ten rank-loss points lie on a conic, Lemma~\ref{lem:oneshort} turns this into a linear program over $252$ atoms. \item Decide whether part~\ref{compact:b} of Theorem~\ref{thm:compact} holds for finite groups (part~\ref{compact:a} appears to fail, Section~\ref{sec:numerics}), and whether there are range-criterion pairs that are covariant under $\mathrm{SU}(2)$. By Theorem~\ref{thm:su2top}\ref{su2top:c}, for such a pair the multiplicities $p_j$ and $q_l$ at the active pairs $(j,l)$ with $j+l=J$ are at least four, and $\max(p_j,q_l)\ge5$, while for invariant operators the conclusion of Theorem~\ref{thm:su2top}\ref{su2top:d} can already fail when these multiplicities are three (Remark~\ref{rem:tiles} and Corollary~\ref{cor:su2inv}). \item The PPT$^2$ conjecture for maps on $\M_d$ with $4\le d\le9$ remains open; for $d\in\{4,5\}$ a counterexample would need a composite whose range contains product vectors (Theorem~\ref{thm:main}). Can the extended criterion of Section~\ref{sec:edge} certify a counterexample with both outer dimensions at least four? The exclusion arguments of Sections~\ref{sec:windows}, \ref{mp:sec:overview} and~\ref{sec:example} assume that the composite range contains no product vector, so they do not apply, and Lemma~\ref{lem:edge}\ref{edge:b} and, on strata one product vector short, Corollary~\ref{cor:edgeos} give conditions that such a certificate must satisfy. \item Decide whether a range-criterion pair can have an outer dimension four. For $\max(m,k,n)\le9$ only the types $(4,7,n)$ and $(4,8,n)$ with $7\le n\le9$, $(4,9,n)$ with $6\le n\le9$, and their mirrors remain, by Theorems~\ref{thm:44}, \ref{thm:467}, \ref{o469:thm:469}, \ref{thm:475no}, \ref{thm:476}, \ref{thm:485no}, \ref{o486b:486no} and~\ref{thm:495no} and Corollary~\ref{cor:windows}\ref{cw:a}; Theorem~\ref{thm:477p} restricts the type $(4,7,7)$, and Corollary~\ref{cor:4kn} and Table~\ref{tab:o4cells} restrict the types $(4,8,n)$ and $(4,9,n)$. \item Decide whether the range of every PPT state on $\C^6\otimes\C^6$ with local ranks $(6,6)$ and rank $10$ contains a nonzero vector of Schmidt rank at most two. 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