\section{Computations for the rank-five family}\label{sec:ppt3-app-family} This appendix contains the proofs of the statements of Section~\ref{sec:ppt3-family}, whose notation it uses. \begin{proof}[Proof of Theorem~\ref{thm:ppt3-family}, PPT and rank claims] Cyclic permutations of the qubits leave $\rho_t$ invariant, so its three partial transposes are unitarily equivalent. The unnormalised numerator of $\rho_t^{T_A}$ commutes with the global parity $Z^{\otimes3}$, so it splits into even and odd parity blocks. With $\beta=3+\omega$ and $\gamma=t+3\omega^2$ ($|\beta|^2=7$, $|\gamma|^2=t^2-3t+9$), in the respective basis orders $(000,011,101,110)$ and $(001,010,100,111)$ these blocks of $7(t+3)\rho_t^{T_A}$ are \[ \text{even: }\begin{pmatrix} t & -t & \beta & \bar\beta\\ -t & t+3 & 0 & 0\\ \bar\beta & 0 & t+3 & \gamma\\ \beta & 0 & \bar\gamma & t+3 \end{pmatrix},\qquad \text{odd: }\begin{pmatrix} 4 & \beta & -t & \bar\gamma\\ \bar\beta & 4 & -t & \gamma\\ -t & -t & 4 & 0\\ \gamma & \bar\gamma & 0 & 3t \end{pmatrix}, \] and their leading principal minors are \begin{equation}\label{eq:ppt3-minors} \begin{aligned} \text{even: }& t,\quad 3t,\quad (t+3)(3t-7),\quad 3(8t^2-15t-36),\\ \text{odd: }& 4,\quad 9,\quad 3(12-t^2),\quad -3(t-3)(3t^2+4t-48). \end{aligned} \end{equation} The even determinant is nonnegative exactly when $t\ge t_-$. For $3\le t<2\sqrt3$, the first three leading minors of each block are positive, while the odd determinant is nonnegative exactly when $t\le t_+$. Since $30$. Setting $U=2z$, $O=s+z^2$ gives exactly $q_s(z)$ in \eqref{eq:ppt3-curve}; it remains nonzero at $z=0$ and $z^2=-s$. At the projective point $(0,1)^T$ (i.e.\ $z=\infty$), the upper $BC$ row must vanish, which forces $U=W=E=O=0$ and leaves only $V$, the ray of $v$. This proves the full cut-product classification. For Appendix~\ref{sec:ppt3-kernel}, $R^\perp$ is spanned by the three symmetry-sector vectors omitted from $R$, namely $\bar e$, $\bar o$ and $3\ket{000}+\ket{011}+\ket{101}+\ket{110}$: the $3\times8$ matrix $K$ whose rows are their bras, \[ K_0=[0,0,0,1,0,\omega,\omega^2,0],\quad K_1=[0,\omega^2,\omega,0,1,0,0,0],\quad K_2=[3,0,0,1,0,1,1,0] \] has rank three and $KB=0$, so $R=\ker K$. \emph{(b).} A rescaling removes the parameter $s$: with $D_s=\operatorname{diag}(1,s^{3/2},s^{1/2},s,s^{1/2})$, \[ q_s(\sqrt s\,z)=\sqrt s\,D_sq_1(z),\qquad H_s=s^3D_s^{-1}H_5D_s^{-1}, \] so $q_s(\sqrt s z)\dg H_sq_s(\sqrt s z)=s^4\,q_1(z)\dg H_5q_1(z)$, and the right side of \eqref{eq:ppt3-sos} at $(s,\sqrt s z)$ is $s^4$ times its value at $(1,z)$. It suffices to take $s=1$, which is \eqref{eq:ppt3-h5}. Put $d=\operatorname{Re}(z^2)=x^2-y^2$, so $r+d=2x^2$. The squared moduli of the entries of $q_1(z)$ are \[ 4r,\quad r^2(r^2+9+6d),\quad r^2+1-2d,\quad r(r^2+1+2d),\quad r^2+1+2d, \] and $H_5$ is diagonal, so \begin{align*} q_1\dg H_5q_1&=-48r+r^2(r^2+9+6d)+11(r^2+1-2d)-6r(r^2+1+2d)+14(r^2+1+2d)\\ &=r^4-6r^3+34r^2-54r+25+6d(r-1)^2 . \end{align*} Since $(r-1)^2(r^2-4r+25)=r^4-6r^3+34r^2-54r+25$ and $12x^2=6(r+d)$, this equals $(r-1)^2\bigl[(r-5)^2+12x^2\bigr]$. This proves \eqref{eq:ppt3-sos}. The cyclic permutation of qubits fixes $u_s,v,w$ and acts on $e,o$ by phases, which commute with the diagonal $H_s$, so the form controls the other cuts as well. The point at infinity contributes $H_{vv}=1$. Finally, every positive pure-state decomposition of an operator supported on $R_s$ uses vectors in $R_s$: its expectation on $R_s^\perp$ vanishes, and each positive summand must vanish there separately. Thus the nonnegative curve values also control mixtures across different cuts. \end{proof} \begin{proof}[Proof of Corollary~\ref{cor:ppt3-chirality}] \emph{(a).} The weights $t,3t,3,3,1$ of $\rho_t$ in \eqref{eq:ppt3-rho} are positive for every $t>0$, so $\operatorname{ran}\rho_t=R_1$. A local unitary is an invertible local filter, so (a) is a special case of (b), proved next. \emph{(b).} By Proposition~\ref{prop:ppt3-witness}(a) the $A|BC$-product rays of $R_s$ are $[v]$ and $[B_sq_s(z)]=[(\ket0+z\ket1)\otimes\ket{\phi_s(z)}]$, $z\in\mathbb C$, and substituting \eqref{eq:ppt3-curve} shows that the $2\times2$ coefficient matrix of $\ket{\phi_s(z)}$ has determinant $-3(z^2+s)^2$. So $\ket{\phi_s(z)}$ is a $B|C$ product exactly when $z=\pm i\sqrt s$, where $B_sq_s(\pm i\sqrt s)=\pm2i\sqrt s\,(\ket0\pm i\sqrt s\ket1)^{\otimes3}$. Hence $R_s$ contains exactly three fully product rays, $P_\infty=[\ket1^{\otimes3}]$ and $P_\pm=[(\ket0\pm i\sqrt s\ket1)^{\otimes3}]$; they span $\operatorname{span}(u_s,v,w)$, because $(\ket0\pm i\sqrt s\ket1)^{\otimes3}=u_s\pm i\sqrt s\,(w-sv)$. The conjugate $\overline{R_s}$ contains the conjugate rays, the same three with $P_+$ and $P_-$ exchanged. An invertible local filter $F$ that maps $R_s$ onto $R_s$ or onto $\overline{R_s}$ permutes the three rays. Each ray is $[\chi^{\otimes3}]$, so $F_A$, $F_B$ and $F_C$ all induce the same permutation of the three distinct points $[\chi]\in\mathbb{CP}^1$, and since a M\"obius map is determined by its action on three points, $F=cF_0^{\otimes3}$. Such an $F$ commutes with the cyclic shift $\Pi$, so it maps the $\omega^2$-eigenspace of $\Pi$ into itself. That eigenspace is $\operatorname{span}(e,o)\subset R_s$, whereas $\overline{R_s}$ meets it in $0$: $\overline{R_s}\ni\bar e,\bar o$ lies in the $\omega$-eigenspace and the other three generators of $\overline{R_s}$ are $\Pi$-invariant. So $F(R_s)\ne\overline{R_s}$. If a state $\rho$ with range $R_s$ satisfied $\rho=F\bar\rho F\dg$ (or the same with $Y^{\otimes3}\bar\rho Y^{\otimes3}$, a local-unitary image of $\bar\rho$), then there would be an invertible local filter $F'$ ($F^{-1}$, respectively $Y^{\otimes3\dagger}F^{-1}$) with $F'(R_s)=\overline{R_s}$, which the argument excludes. For the stabiliser, each of the six permutations of the three points is induced by a unique M\"obius map, and $F_\pi^{\otimes3}$ permutes $P_\infty,P_\pm$, hence preserves $\operatorname{span}(u_s,v,w)$, and as it commutes with $\Pi$ it preserves the $\omega^2$-eigenspace $\operatorname{span}(e,o)$ of $\Pi$ (which is all of that eigenspace in $(\mathbb C^2)^{\otimes3}$). So the stabiliser is $\{[F_\pi]^{\otimes3}\}\cong S_3$. \end{proof} \begin{proof}[Proof of Corollary~\ref{cor:ppt3-extreme6}] Write the numerator of $\sigma_6$ as $t\,uu\dg+b\,vv\dg+3\,ww\dg+3\,ee\dg+oo\dg$; it is the numerator of $\rho_t$ with the weight $3t$ of $vv\dg$ replaced by $b$. The vector $v=\ket{111}$ enters $\sigma_6^{T_A}$ only through the $(111,111)$ entry, the bottom diagonal entry of the odd block, and equals $b$ there. Hence the even block, and the three leading principal minors of the odd block, are those of $\rho_t^{T_A}$ in \eqref{eq:ppt3-minors}: at $t=t_-$ the even minors $t,\ 3t,\ (t+3)(3t-7)$ are positive and, since $8t^2-15t-36=0$, the even determinant vanishes, so the even block is positive semidefinite of rank three. The odd minors $4,\ 9,\ 3(12-t^2)$ are positive ($t_-<2\sqrt3$), and the odd determinant is affine in $b$, \[ \det(\text{odd block})=3(12-t^2)\,b+3(5t^2+24t-144), \] which equals $-3(t-3)(3t^2+4t-48)$ of \eqref{eq:ppt3-minors} at $b=3t$. It vanishes at $b=(5t^2+24t-144)/(t^2-12)$, and this equals $8(9-t)/5$ modulo $8t^2-15t-36=0$, since $5(5t^2+24t-144)-8(9-t)(t^2-12)=8t^3-47t^2+24t+144$ is divisible by $8t^2-15t-36$ (the quotient is $t-4$ and the remainder is $0$). Hence at $b=8(9-t)/5$ the odd block is positive semidefinite of rank three as well. Thus each partial transpose of $\sigma_6$ has rank six. The supported witness has expectation \[ \tr(H_5X_6) =\frac{b-12t+29}{4t+b+21} =\frac{217-68t}{5(4t+b+21)}<0, \] because $t>13/4>217/68$. This proves GME by Proposition~\ref{prop:ppt3-witness}. For extremality, $\sigma_6$ is extreme in the set of fully-PPT states exactly when the only Hermitian $D$ with range in $\operatorname{ran}\sigma_6$ and $D^{T_c}k=0$ for every kernel vector $k$ of $\sigma_6^{T_c}$, at each cut $c$, is a multiple of $\sigma_6$ (the supported-face criterion of~\cite{GIM}). Write $D=BX_DB\dg$ with $X_D$ Hermitian. The conditions $D^{T_c}k=0$ are a real-linear system on the $25$-dimensional real space of such $X_D$. To compute its rank exactly, complexify: let $X_D$ range over all complex $5\times5$ matrices (25 complex unknowns) and impose both $D^{T_c}k=0$ and $k\dg D^{T_c}=0$ for the kernel vectors $k$, at each cut. The two sets are adjoint to each other for Hermitian $D$, so the solution space is closed under $X_D\mapsto X_D\dg$, and is the complexification of the real solution space on Hermitian $X_D$ (split $X_D$ into its Hermitian and anti-Hermitian parts). Real nullity therefore equals complex nullity. The complexified system has rank $24$ over $\mathbb Q(\sqrt{17},i\sqrt3)$, which contains the entries of $\sigma_6$ and of the kernel vectors, so its nullspace, and hence the real nullspace, is exactly the scalar direction of \eqref{eq:ppt3-sigma6}. \end{proof}