\pdfsuppressptexinfo=-1 \documentclass[11pt]{amsart} \usepackage[margin=1.1in]{geometry} \usepackage{amsmath,amssymb,amsthm,mathtools} \usepackage{enumitem} \usepackage[hidelinks]{hyperref} \hypersetup{pdftitle={The PPT-squared conjecture holds for spin-factor maps}, pdfauthor={Tejasvi Singh Tomar}} \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} \setlength{\emergencystretch}{2em} \newcommand{\C}{\mathbb{C}} \newcommand{\R}{\mathbb{R}} \newcommand{\M}{\mathrm{M}} \newcommand{\Herm}{\mathrm{Herm}} \newcommand{\Tr}{\operatorname{Tr}} \newcommand{\op}{\operatorname{op}} \newcommand{\Sep}{\mathrm{Sep}} \newcommand{\Pos}{\mathrm{Pos}} \newcommand{\SOp}{\mathrm{SO}^{+}(1,5)} \newcommand{\GL}{\mathrm{GL}_4(\C)} \newcommand{\id}{\mathrm{id}} \newcommand{\cS}{\mathcal{S}} \newcommand{\cP}{\mathcal{P}} \newcommand{\cF}{\mathcal{F}} \newcommand{\ket}[1]{|#1\rangle} \newcommand{\bra}[1]{\langle #1|} \newcommand{\ip}[2]{\langle #1,#2\rangle} \begin{document} \title{The PPT$^2$ conjecture holds for spin-factor maps} \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; 47L07, 15A69, 46L07} \keywords{PPT squared conjecture, entanglement breaking map, completely copositive map, spin factor, Lorentz cone, Clifford algebra, Kadison--Schwarz inequality, Schmidt number} \begin{abstract} The PPT$^2$ conjecture asserts that the composition of two completely positive and completely copositive maps is entanglement breaking. A counterexample in dimension ten has recently been announced, and for maps on $\M_4$ the conjecture is open. Let $\cS\subset\M_4$ be the real span of the identity $\gamma_0$ and five pairwise anticommuting Hermitian unitaries $\gamma_1,\dots,\gamma_5$. It is a spin factor whose positive cone is a Lorentz cone. We prove the conjecture for all compositions $\Psi\circ\Phi$ of completely positive maps $\Phi\colon\M_4\to\M_k$ and $\Psi\colon\M_k\to\M_4$, with $k$ arbitrary, such that the ranges of $\Psi$ and of the adjoint $\Phi^{*}$ lie in invertible congruences of the complex span of $\cS$. Such maps are automatically completely copositive, and the class contains maps whose Choi matrices are entangled. A theorem of Hildebrand on separability for pairs of Lorentz cones reduces the claim to one trace inequality between the coefficient matrices of the two Choi matrices. We derive this inequality from the operator inequality $\sum_{j=1}^{5}F_jF_0^{-1}F_j\preceq F_0$, valid whenever $\sum_{\mu=0}^{5}F_\mu\otimes\gamma_\mu\succeq 0$ and $F_0\succ 0$, which in turn follows from the Kadison--Schwarz inequality. As a corollary, $\Psi\circ\Phi$ is entanglement breaking whenever $\Psi$ is such a map and $\Phi$ is a completely positive map with Kraus operators of rank at most two. \end{abstract} \maketitle \section{Introduction}\label{sec:intro} A linear map $\Phi\colon\M_m\to\M_n$ 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 $(\id_r\otimes\Phi)(X)$ is separable for every positive semidefinite $X\in\M_r\otimes\M_m$, equivalently, 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} asserts that $\Psi\circ\Phi$ is entanglement breaking for all PPT maps $\Phi\colon\M_m\to\M_k$ and $\Psi\colon\M_k\to\M_n$. The conjecture holds for maps on $\M_3$ \cite{CMHW19,CYT19}, and Henry proved it when the intermediate algebra is $\M_2$ \cite{Hen26}. The iterates of a unital or trace-preserving PPT map approach the entanglement breaking maps \cite{KMP18}, unital PPT channels become entanglement breaking after finitely many iterations \cite{RJP18}, and so do all PPT maps \cite{Par26}. The conjecture also holds for several families defined by symmetry or by randomness \cite{CYZ18,SN22,NP26,Pru25,ParSp}, among them the maps on $\M_4$ that are covariant under the unitary symplectic group \cite{Pru25,ParSp}. 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. We prove it for a class of maps defined by an operator system of $\M_4$ rather than by a symmetry. Let $\gamma_1,\dots,\gamma_5\in\M_4$ be Hermitian matrices with \begin{equation}\label{eq:clifford} \gamma_i\gamma_j+\gamma_j\gamma_i=2\delta_{ij}I\qquad(1\le i,j\le 5), \end{equation} put $\gamma_0=I$, and let $\cS=\mathrm{span}_\R\{\gamma_0,\dots,\gamma_5\}$ and $\cS_\C=\cS+i\cS$. For instance, with the Pauli matrices $X,Y,Z$ one can take $(\gamma_1,\dots,\gamma_5)=(X\otimes I,\,Y\otimes I,\,Z\otimes X,\,Z\otimes Y,\,Z\otimes Z)$. The space $\cS$ is a spin factor whose positive semidefinite elements form a Lorentz cone (Section~\ref{sec:prelim}). \begin{definition}\label{def:class} A completely positive map $\Psi\colon\M_k\to\M_4$ is a \emph{spin-factor map} if $\Psi(\M_k)\subseteq T\cS_\C T^{\dagger}$ for some $T\in\GL$. A completely positive map $\Phi\colon\M_4\to\M_k$ is a spin-factor map if its adjoint $\Phi^{*}$ is one. \end{definition} \begin{theorem}\label{thm:main} Let $k\ge 1$, and let $\Phi\colon\M_4\to\M_k$ and $\Psi\colon\M_k\to\M_4$ be spin-factor maps. Then $\Phi$ and $\Psi$ are completely copositive, and $\Psi\circ\Phi$ is entanglement breaking. \end{theorem} Here the intermediate dimension $k$ is arbitrary, and for $k\le 4$ the theorem is a special case of the PPT$^2$ conjecture on $\M_4$; the two maps may use different matrices $T$. Since the class contains PPT maps that are not entanglement breaking (Example~\ref{ex:entangled}), Theorem~\ref{thm:main} is not a consequence of the trivial case in which one factor is entanglement breaking. The map in Example~\ref{ex:entangled} is, in addition, covariant under the unitary symplectic group, and its square is entanglement breaking by \cite{Pru25,ParSp}; what Theorem~\ref{thm:main} adds for this map are its compositions with arbitrary spin-factor maps. The proof has three steps. First, in the basis $\gamma_\mu$ the Choi matrix of the composition becomes, after a local congruence, $\sum_{\mu,\nu}X_{\mu\nu}\gamma_\mu\otimes\gamma_\nu$ for a real $6\times 6$ matrix $X$. A theorem of Hildebrand \cite{Hil07} decides when $X$ is a sum of products of vectors in the Lorentz cone. It turns separability into two semidefinite conditions (Proposition~\ref{prop:hildebrand}), one of which holds automatically. Second, by the symmetry of the Lorentz cone, the other condition reduces to the trace inequality \begin{equation}\label{eq:cpi} \sum_{i=1}^{5}\Tr(F_iG_i)\le\Tr(F_0G_0) \end{equation} for all tuples $(F_\mu)$ and $(G_\mu)$ of Hermitian $k\times k$ matrices with $\sum_\mu F_\mu\otimes\gamma_\mu\succeq 0$ and $\sum_\mu G_\mu\otimes\gamma_\mu\succeq 0$ (Proposition~\ref{prop:reduction}). Third, we prove \eqref{eq:cpi} (Proposition~\ref{prop:cpi}). The proof rests on the operator inequality \begin{equation}\label{eq:lemmaL} \sum_{j=1}^{5}F_jF_0^{-1}F_j\preceq F_0\qquad(F_0\succ 0), \end{equation} which we derive from the Kadison--Schwarz inequality for a unital completely positive map that annihilates the products $\gamma_i\gamma_j$ with $i\ne j$ (Lemma~\ref{lem:L}). Remark~\ref{rem:pauli} compares this with the three Pauli matrices on $\C^2$. The same structure yields a statement in which only $\Psi$ is restricted. \begin{corollary}\label{cor:sn2} Let $\Psi\colon\M_k\to\M_4$ be a spin-factor map, and let $\Phi\colon\M_m\to\M_k$ be completely positive with Kraus operators of rank at most two. Then $\Psi\circ\Phi$ is entanglement breaking. In particular, $\Psi\circ\Phi$ is entanglement breaking for every completely positive $\Phi\colon\M_2\to\M_k$, and for every PPT map $\Phi\colon\M_3\to\M_3$. \end{corollary} For the last statement we use that PPT states on $\C^3\otimes\C^3$ have Schmidt number at most two, which is the dual form of the theorem of Yang, Leung and Tang that $2$-positive maps $\M_3\to\M_3$ are decomposable \cite{YLT16}. In the notation $(m,k,n)$ for compositions $\M_m\to\M_k\to\M_n$, Corollary~\ref{cor:sn2} covers the cases $(2,k,4)$ and $(3,3,4)$, in which only $\Psi$ needs to be a spin-factor map; passing to adjoints yields $(4,k,2)$ and $(4,3,3)$, in which only $\Phi$ does. For $k\le 3$, Theorem~\ref{thm:main} can also be deduced from Corollary~\ref{cor:sn2} (Remark~\ref{rem:smallk}). For comparison, An and Lee showed that for maps on $\M_3$, $\Psi\circ\Phi$ and $\Phi\circ\Psi$ are entanglement breaking for all completely positive $\Phi$ with Kraus operators of rank at most two exactly when the Choi matrix of $\Psi$ is $1$-undistillable \cite{AL26}. Maps on matrix algebras that factor through Lorentz cones were studied by Aubrun and M\"uller-Hermes in connection with entanglement annihilation \cite{AMH22}; their examples include a Breuer--Hall map and the projections onto spin factors, such as the map $\Pi$ in Example~\ref{ex:entangled}. In the language of matrix convex sets, \eqref{eq:lemmaL} says that the free spectrahedron of a single irreducible tuple of five anticommuting Hermitian unitaries lies in the matrix ball. Evert and Passer proved this inclusion for the universal tuple, where the defining inequality is two-sided and can be squared \cite[Proposition~2.14]{EP26}. For a single irreducible tuple of five matrices, however, the defining inequality is only one-sided, and the proof of Lemma~\ref{lem:L} uses complete positivity instead of squaring. \section{Preliminaries}\label{sec:prelim} \subsection{Notation} We write $\M_n$ for the complex $n\times n$ matrices, $\Herm_n$ for the Hermitian ones, $A\succeq 0$ for positive semidefinite $A$, and $E_{ij}$ for the matrix units. The trace inner product is $\ip{A}{B}=\Tr(A^{\dagger}B)$. The adjoint $\Phi^{*}$ of a linear map $\Phi\colon\M_m\to\M_n$ is defined by $\ip{\Phi^{*}(B)}{A}=\ip{B}{\Phi(A)}$. A map is completely positive if and only if $C_\Phi\succeq 0$ \cite{Cho75}. Then $\Phi^{*}$ is completely positive as well. The map $\Phi$ is completely copositive if and only if the partial transpose $C_\Phi^{\Gamma}=\sum_{i,j}E_{ij}\otimes\Phi(E_{ij})^{T}$ is positive semidefinite. An operator on $\C^m\otimes\C^n$ is separable if it is a sum of operators $P\otimes Q$ with $P,Q\succeq 0$. More generally, the Schmidt number of a positive semidefinite operator $W$ is the least $r$ such that $W$ is a sum of operators $\ket{w}\bra{w}$ with vectors $w$ of Schmidt rank at most $r$. Separability and the Schmidt number are preserved by local congruences $W\mapsto(A\otimes B)W(A\otimes B)^{\dagger}$ with invertible $A$ and $B$. For linear maps $\Phi\colon\M_4\to\M_k$ and $\Psi\colon\M_k\to\M_4$, expanding $\Phi(E_{xy})$ in matrix units gives \begin{equation}\label{eq:link} C_{\Psi\circ\Phi}=\sum_{a,b=1}^{k}\overline{\Phi^{*}(E_{ab})}\otimes\Psi(E_{ab}), \end{equation} where the bar denotes entrywise complex conjugation. We write $\rho=\sum_{a,b}E_{ab}\otimes\overline{\Phi^{*}(E_{ab})}=\overline{C_{\Phi^{*}}}$ and $\sigma=\sum_{a,b}E_{ab}\otimes\Psi(E_{ab})=C_\Psi$. Both are operators on $\C^k\otimes\C^4$. They are positive semidefinite when $\Phi$ and $\Psi$ are completely positive. We call the $4\times 4$ blocks $\rho_{ab}=\overline{\Phi^{*}(E_{ab})}$ and $\sigma_{ab}=\Psi(E_{ab})$ the \emph{outer blocks}. By \eqref{eq:link}, $C_{\Psi\circ\Phi}=\sum_{a,b}\rho_{ab}\otimes\sigma_{ab}$. Similarly, $C_\Phi=\sum_{a,b}\rho_{ab}\otimes E_{ab}$ is $\rho$ with the tensor factors swapped. \subsection{The spin factor}\label{sec:spin} Let $\gamma_1,\dots,\gamma_5$ satisfy \eqref{eq:clifford}. They generate a representation of the complex Clifford algebra on five generators on $\C^4$, which is irreducible because the irreducible representations of this algebra have dimension four. The element $c=\gamma_1\gamma_2\gamma_3\gamma_4\gamma_5$ commutes with each $\gamma_i$ and satisfies $c^2=I$, so $c=\pm I$ by Schur's lemma. We use the following standard facts. \begin{enumerate}[label=(S\arabic*),leftmargin=*] \item\label{S:trace} $\Tr(\gamma_\mu\gamma_\nu)=4\delta_{\mu\nu}$ for $0\le\mu,\nu\le 5$. A product of between one and four distinct generators is traceless: for one or two generators this follows from the anticommutation relations and the cyclicity of the trace. Since $c=\pm I$, a product of three or four distinct generators equals $\pm$ a product of the remaining two or one. In particular $\gamma_0,\dots,\gamma_5$ is a basis of $\cS_\C$, and $\Tr(\gamma_\mu\gamma_i\gamma_j)=0$ for all $\mu$ and all $i\ne j$. \item\label{S:cone} For $x\in\R^6$ the matrix $\sum_\mu x_\mu\gamma_\mu$ has the eigenvalues $x_0\pm|\vec x|$, each twice, where $\vec x=(x_1,\dots,x_5)$. Hence $\sum_\mu x_\mu\gamma_\mu\succeq 0$ if and only if $x$ lies in the Lorentz cone $L=\{x\in\R^6:x_0\ge|\vec x|\}$. \item\label{S:unique} Two $5$-tuples of Hermitian matrices on $\C^4$ satisfying \eqref{eq:clifford} are unitarily equivalent if and only if their products $\gamma_1\cdots\gamma_5$ coincide. Replacing a tuple by its negative changes the sign of the product, so any two spin factors of this form are unitarily equivalent. The tuple $(\gamma_i^{T})$ also satisfies \eqref{eq:clifford}. Its product is $\gamma_1^{T}\cdots\gamma_5^{T}=(\gamma_5\cdots\gamma_1)^{T}=c^{T}=c$, since reversing the order of five anticommuting factors takes ten transpositions and $c=\pm I$. Consequently, there is a unitary $V$ with $V\gamma_\mu V^{\dagger}=\gamma_\mu^{T}$ for all $\mu$, so that $Y^{T}=VYV^{\dagger}$ for all $Y\in\cS_\C$. \item\label{S:lift} Let $\SOp$ be the identity component of the Lorentz group of the form $\eta=\mathrm{diag}(1,-1,-1,-1,-1,-1)$. For every $\Lambda\in\SOp$ there is $g\in\GL$ with $g\gamma_\mu g^{\dagger}=\sum_\nu\Lambda_{\nu\mu}\gamma_\nu$ for all $\mu$. Indeed, $\exp(t\gamma_i/2)$ acts on $\cS$ as the boost in the $(0,i)$-plane with rapidity $t$. The unitary $\exp(t\gamma_i\gamma_j/2)$ acts as a rotation by the angle $t$ in the $(i,j)$-plane. These one-parameter groups generate $\SOp$. \end{enumerate} For a real $6\times 6$ matrix $X$ put \[ \op(X)=\sum_{\mu,\nu=0}^{5}X_{\mu\nu}\,\gamma_\mu\otimes\gamma_\nu\in\M_4\otimes\M_4 . \] This map is injective by \ref{S:trace}. If $g$ and $h$ realize $\Lambda_1,\Lambda_2\in\SOp$ as in \ref{S:lift}, then \begin{equation}\label{eq:covariance} (g\otimes h)\,\op(X)\,(g\otimes h)^{\dagger}=\op(\Lambda_1X\Lambda_2^{T}). \end{equation} \begin{lemma}\label{lem:diag} The matrices $\gamma_i\otimes\gamma_i$, $1\le i\le 5$, commute pairwise. Their joint eigenspaces are one-dimensional and are indexed by the sixteen sign vectors $s\in\{\pm1\}^5$ with $s_1s_2s_3s_4s_5=1$. Consequently, for $x\in\R^6$, \begin{enumerate}[label=(\alph*)] \item\label{d:even} the eigenvalues of $\op(\mathrm{diag}(x))$ are $x_0+\ip{s}{\vec x}$, one for each such $s$; \item\label{d:odd} the eigenvalues of $\op(\mathrm{diag}(x)\eta)$ are $x_0+\ip{s}{\vec x}$, one for each $s\in\{\pm1\}^5$ with $s_1s_2s_3s_4s_5=-1$; \item\label{d:vec} if $\xi_s$ is a unit vector in the joint eigenspace for $s$, then $\bra{\xi_s}\gamma_\mu\otimes\gamma_\nu\ket{\xi_s}=\delta_{\mu\nu}s_\mu$, with $s_0=1$. \end{enumerate} \end{lemma} \begin{proof} For $i\ne j$ the factors $\gamma_i\gamma_j=-\gamma_j\gamma_i$ in $(\gamma_i\otimes\gamma_i)(\gamma_j\otimes\gamma_j)=\gamma_i\gamma_j\otimes\gamma_i\gamma_j$ contribute two signs, so the matrices commute. Their product is $c\otimes c=I$, so each joint eigenvalue vector $s$ satisfies $s_1\cdots s_5=1$. For such $s$ the projection onto the joint eigenspace is $P_s=2^{-5}\sum_{A}s_A\gamma_A\otimes\gamma_A$, where $A$ runs over the subsets of $\{1,\dots,5\}$, $s_A=\prod_{i\in A}s_i$, and $\gamma_A$ is the ordered product of the $\gamma_i$ with $i\in A$. By \ref{S:trace} only $A=\emptyset$ and $A=\{1,\dots,5\}$ contribute to the trace, so $\Tr P_s=2^{-5}(16+16)=1$. This proves the first statement and \ref{d:even}. Part \ref{d:odd} follows from \ref{d:even}, since $\mathrm{diag}(x)\eta=\mathrm{diag}(x_0,-\vec x)$ and $s\mapsto -s$ exchanges the two parities. For \ref{d:vec}, let $\mu\ne\nu$. If one of the indices is $0$ and the other is $j$, then $\gamma_\mu\otimes\gamma_\nu$ anticommutes with $\gamma_i\otimes\gamma_i$ for any $i\ne j$. If $\mu=i$ and $\nu=j$ are both nonzero, it anticommutes with $\gamma_i\otimes\gamma_i$. An operator that anticommutes with a Hermitian unitary has zero expectation in each of its eigenvectors. The diagonal entries are $1$ for $\mu=\nu=0$ and $s_i$ for $\mu=\nu=i$. \end{proof} \subsection{Lorentz cones} Let $\Sep\subset\R^{6\times 6}$ be the closed convex cone generated by the matrices $ab^{T}$ with $a,b\in L$, and let $\Pos$ be the cone of matrices $X$ with $a^{T}Xb\ge 0$ for all $a,b\in L$, that is, the linear maps that take $L$ into $L$. The group $\SOp\times\SOp$ acts on both cones by $X\mapsto\Lambda_1X\Lambda_2^{T}$. We use three facts from Hildebrand's paper \cite{Hil07}, citing its numbered results from the preprint version \cite{Hil05}. \begin{enumerate}[label=(H\arabic*),leftmargin=*] \item\label{H:int} $X$ lies in the interior of $\Pos$ if and only if $Xb$ lies in the interior of $L$ for every nonzero $b\in L$ \cite[Lemma~3.5]{Hil05}. \item\label{H:svd} For every $X$ in the interior of $\Pos$ there are $\Lambda_1,\Lambda_2\in\SOp$ such that $\Lambda_1X\Lambda_2^{T}$ is diagonal \cite[Corollary~3.7]{Hil05}. \item\label{H:diag} A diagonal matrix $D$ lies in $\Sep$ if and only if $D_{00}\ge\sum_{i=1}^{5}|D_{ii}|$ \cite[Lemma~3.12]{Hil05}. \end{enumerate} The next proposition is the case of two six-dimensional Lorentz cones in Hildebrand's linear matrix inequality for $\Sep$ \cite[Theorem~4.11]{Hil05}. We include the short derivation in our notation. \begin{proposition}\label{prop:hildebrand} A matrix $X\in\R^{6\times 6}$ lies in $\Sep$ if and only if $\op(X)\succeq 0$ and $\op(X\eta)\succeq 0$. \end{proposition} \begin{proof} For $a,b\in L$ put $A=\sum_\mu a_\mu\gamma_\mu$ and $B=\sum_\nu b_\nu\gamma_\nu$. Then $\op(ab^{T})=A\otimes B$ and $\op(ab^{T}\eta)=A\otimes\tilde B$ with $\tilde B=b_0I-\sum_i b_i\gamma_i$. All three matrices are positive semidefinite by \ref{S:cone}, which proves the forward implication. Conversely, suppose $\op(X)\succeq 0$ and $\op(X\eta)\succeq 0$. We first assume that $\op(X)$ is positive definite. For nonzero $a,b\in L$ we have $16\,a^{T}Xb=\Tr\bigl(\op(X)(A\otimes B)\bigr)>0$ by \ref{S:trace}, since $A\otimes B$ is nonzero and positive semidefinite. So $X$ maps $L\setminus\{0\}$ into the interior of $L$, which places $X$ in the interior of $\Pos$ by \ref{H:int}. By \ref{H:svd} there are $\Lambda_1,\Lambda_2\in\SOp$ such that $D=\Lambda_1X\Lambda_2^{T}$ is diagonal. By \eqref{eq:covariance}, $\op(D)$ is congruent to $\op(X)$. Since $\eta\Lambda_2\eta\in\SOp$ and $D\eta=\Lambda_1(X\eta)(\eta\Lambda_2\eta)^{T}$, also $\op(D\eta)$ is congruent to $\op(X\eta)$. Since both are positive semidefinite, Lemma~\ref{lem:diag} yields $D_{00}+\ip{s}{\vec d}\ge 0$ for all $s\in\{\pm1\}^5$, where $\vec d=(D_{11},\dots,D_{55})$. Choosing $s_i=-\mathrm{sign}(D_{ii})$ shows $D_{00}\ge\sum_i|D_{ii}|$. Then $D\in\Sep$ by \ref{H:diag}, and $X=\Lambda_1^{-1}D\Lambda_2^{-T}\in\Sep$. In general, $X+\varepsilon e_0e_0^{T}$ satisfies both conditions with $\op(X+\varepsilon e_0e_0^T)=\op(X)+\varepsilon I\succ 0$ and hence lies in $\Sep$. Since $\Sep$ is closed, $X\in\Sep$. \end{proof} \section{A matrix Lorentz inequality}\label{sec:lemma} Let $\cF_k$ be the set of tuples $F=(F_0,\dots,F_5)$ of Hermitian $k\times k$ matrices such that \[ \rho_F=\sum_{\mu=0}^{5}F_\mu\otimes\gamma_\mu\succeq 0\quad\text{on }\C^k\otimes\C^4 . \] The outer blocks of $\rho_F$ lie in $\cS_\C$. Conversely, a positive semidefinite operator on $\C^k\otimes\C^4$ whose outer blocks lie in $\cS_\C$ equals $\rho_F$ for a unique $F\in\cF_k$, by \ref{S:trace}. By \ref{S:unique}, $\rho_F^{\Gamma}=(I\otimes V)\rho_F(I\otimes V)^{\dagger}\succeq 0$, where $\Gamma$ is the partial transpose on $\C^4$. This identity characterizes the operators with outer blocks in $\cS_\C$. Indeed, if $\rho^{\Gamma}=(I\otimes V)\rho(I\otimes V)^{\dagger}$ for an operator $\rho$ on $\C^k\otimes\C^4$, then its outer blocks are fixed by the linear map $\tau(Y)=V^{\dagger}Y^{T}V$ on $\M_4$. This map fixes each $\gamma_\mu$ and maps $\gamma_i\gamma_j$ to $\gamma_j\gamma_i=-\gamma_i\gamma_j$. The matrices $\gamma_\mu$ and $\gamma_i\gamma_j$ with $i