\section{PPT mixtures and the remaining boundary}\label{sec:ppt3-relaxations}

The examples separate full PPT from biseparability. We explain why PPT-mixture tests miss them, show that the supported
witness method applies to every minimal-rank example, and distinguish a remaining question about maximally mixed marginals.

\subsection{Why PPT-mixture tests cannot detect the states}

A \emph{PPT mixture} is a convex combination of states, each PPT
across one of the three cuts~\cite{JMG,NMG}. Every biseparable
state is a PPT mixture, but the converse fails~\cite{HK}.

\begin{corollary}[PPT mixtures and outer relaxations]\label{cor:ppt3-ppt-mixture}
\cls{proved; the level-two detection is the exact-rational certificates of Section~\ref{sec:ppt3-noise}}
Every fully-PPT GME state in Theorems~\ref{thm:ppt3-family}
and~\ref{thm:ppt3-orbit} and Corollaries~\ref{cor:ppt3-extreme6},
\ref{cor:ppt3-extreme5} and~\ref{cor:ppt3-robust} is a
PPT mixture. No PPT-mixture witness can detect any of them.
The level-two symmetric-extension certificates of
Section~\ref{sec:ppt3-noise} do detect the first family and
the states $\rho_*$, $\sigma$ and $\tilde\rho$ of Corollaries~\ref{cor:ppt3-noise-reused},
\ref{cor:ppt3-extreme5} and~\ref{cor:ppt3-robust}; we do not assert a level-two certificate for
every member of the local-filter surface.
\end{corollary}
\begin{proof}
Every fully-PPT state is itself a one-term PPT mixture. The GME and level-two witness claims
were proved in the preceding sections.
\end{proof}

\begin{remark}[rank-four PPT pieces do not always suffice]\label{rem:ppt3-splitting}
\cls{proved}
The family of Theorem~\ref{thm:ppt3-family} also rules out a proposed universal decomposition
of fully-PPT rank-five states into positive operators
$\rho=\sigma_A+\sigma_B+\sigma_C$ with
$\sigma_c^{T_c}\succeq0$ and $\rank\sigma_c\le4$.
By the $2\times4$ low-rank theorem~\cite{KCKL,HLVC},
each summand would be separable across its indicated cut,
making $\rho$ biseparable.
\end{remark}

\subsection{Every minimal-rank example has the same geometry}\label{sec:ppt3-moment-cone}

The supported-witness method extends beyond our chosen vectors. For a fully-PPT GME state of minimal rank, the kernel contains no cut-product
vector; this forces the product rays at each cut to form a rational quartic curve spanning the support.
The three curves determine all its biseparable states and all supported witnesses.

\begin{theorem}[three images of one quartic moment cone]\label{thm:ppt3-universal-moments}
\cls{proved}
Let $R\subset(\mathbb C^2)^{\otimes3}$ be a five-dimensional subspace whose orthogonal complement contains no nonzero vector
that is a product across any of the three cuts; by Lemma~\ref{lem:ppt3-product-kernel} (Appendix~\ref{sec:ppt3-app-moment}) this holds for the range of every
rank-five fully-PPT GME state. Let $B$ be an $8\times5$ basis matrix of $R$, put $\nu(a)=(a_0^4,a_0^3a_1,a_0^2a_1^2,
a_0a_1^3,a_1^4)^T$ for $a\in\mathbb C^2$, and let $\mathcal M_4=\operatorname{cone}\{\nu(a)\nu(a)\dg\}$ be the spin-two
(quartic) moment cone. For each cut $c$ there is an invertible $5\times5$ matrix $T_c$ such that the cut-$c$ product
rays in $R$ are exactly the rays of $BT_c\nu(a)$, $a\neq0$. Consequently, for Hermitian $X$,
\begin{equation}\label{eq:ppt3-universal-bs}
 BXB\dg\ \text{is biseparable}\iff X\in\sum_{c=A,B,C}T_c\mathcal M_4T_c\dg ,
\end{equation}
every biseparable operator on $R$ is a sum of at most $25$ cut-product projectors, and a Hermitian $5\times5$ matrix $H$
satisfies $\tr(HX)\ge0$ for every biseparable $BXB\dg$ if and only if
\begin{equation}\label{eq:ppt3-universal-dual}
 \nu(a)\dg T_c\dg HT_c\,\nu(a)\ge0\qquad\text{for all }a\in\mathbb C^2,\ c=A,B,C .
\end{equation}
\end{theorem}

\emph{Idea of the proof.} A fully-PPT state whose kernel contains a product vector is separable across that cut, so the
kernel of a minimal-rank example is product-free. The product vectors $a\otimes b$ in $R$ are then the solutions of a
$3\times4$ linear pencil in $a$ of constant rank three, which is forced to be a single Kronecker block; its kernel is
cubic in $a$, so the product vectors trace a rational normal quartic spanning $R$. Appendix~\ref{sec:ppt3-app-moment}.

Thus, on the support of a minimal-rank example, biseparability is decided by where three copies of one universal cone
sit relative to each other, and genuine multipartite entanglement is certified exactly by a $5\times5$ Hermitian $H$
whose three quartic forms~\eqref{eq:ppt3-universal-dual} are nonnegative while $\tr(HX)<0$. Such an $H$ exists for
every GME state supported on $R$, since the cone in~\eqref{eq:ppt3-universal-bs} is closed. The witnesses $H_s$ and $H_0$ of
Sections~\ref{sec:ppt3-family} and~\ref{sec:ppt3-surface} are instances, with the quartic forms factorised by hand.
Testing the nonnegativity of a quartic form in one complex variable is a small problem: a real polynomial of degree eight
in two variables.

\subsection{The flat-marginal question}\label{sec:ppt3-flat}

A three-qubit state is \emph{flat} if every proper marginal
is maximally mixed. Flat states are a natural place to look for further examples, because every flat state is automatically fully
PPT (shown below), so only biseparability is in question. Our examples settle existence of fully-PPT GME, but leave the following
conjecture open:
\begin{conjecture}[flat rank-five states]\label{conj:ppt3-flat}
\cls{open; repository name Conjecture R}
Every flat three-qubit state of rank five is biseparable.

\end{conjecture}
Every flat three-qubit state is fully PPT, with rank profile $(r,r,r,r)$, $r=\rank\rho$. Indeed
$\rho=I/8+T$ with $T$ a real combination of three-body Pauli products. Partial transposition on qubit $c$ followed by
conjugation with $\sigma_y$ on that qubit negates every Pauli factor on $c$, and the antiunitary
$J=\sigma_y^{\otimes3}K$ ($K$ complex conjugation) negates all three factors, so
\[
 \sigma_y^{(c)}\rho^{T_c}\sigma_y^{(c)}=I/8-T=J\rho J^{-1}.
\]
Each partial transpose is therefore isospectral to $\rho$. Profiles are unchanged by invertible local filtering.

\begin{remark}[rank profile is not flatness]\label{rem:ppt3-flat-profile}
\cls{proved}
The first family has profile $(5,8,8,8)$ in the PPT interval
and $(5,7,7,7)$ at its endpoints, while
Corollary~\ref{cor:ppt3-extreme6} has profile $(5,6,6,6)$.
None is locally filter-equivalent to a flat rank-five state.
The second family and its extreme member \emph{do} have profile
$(5,5,5,5)$, yet they are GME. Thus that profile by itself
cannot imply biseparability; Conjecture~\ref{conj:ppt3-flat} must use the
flat-marginal condition, not only the ranks.
\end{remark}

\subsection{Relation to earlier work}\label{sec:ppt3-prior}

Fully-PPT genuine multipartite entanglement is known in systems larger than three qubits. For states supported on the totally
symmetric subspace, any cut-product vector in the support
is fully product, so biseparability equals full separability
there. This connects the symmetric multipartite PPT-entangled
constructions for four or more qubits~\cite{TG,TAHKSL,ATSL} to GME;
mere permutation invariance, without support in
the symmetric subspace, is not sufficient for this inference.
Fully-PPT GME examples also exist for three qutrits
and higher local dimensions~\cite{HS,PM,SB,SBZL}; for~\cite{SBZL} the state is the normalised projector onto the
complement of a genuinely unextendible product basis.

For three qubits, Ac\'in, Bru\ss, Lewenstein and Sanpera gave an early construction of fully-PPT three-qubit edge states, with rank
profiles up to $(7,7,7,7)$, including an X-shaped family~\cite{ABLS}, of which Kampermann et al.\ generated a pseudo-bound-entangled NMR version~\cite{KPBS}.
By Han and Kye, every X-shaped fully-PPT multi-qubit state is biseparable~\cite[Cor.~5.4]{HanK}; these NMR
states are therefore not GME, and, since biseparability and PPT are invariant under local unitaries, the GME states of this
paper are not X-shaped in any product basis. Explicit three-qubit PPT entangled states of low rank are also given
in~\cite{HaKye15}.
A PPT state is an edge state across a cut if no product vector of its range has its partial conjugate in the range of the partial
transpose. The states of the $(5,5,5,5)$ surface are edge states across every cut (Corollary~\ref{cor:ppt3-extreme5}). The interior states $\rho_t$ are not: their partial transposes are invertible
and the range contains the product vector $\ket{111}$.

Ha and Kye gave a three-qubit PPT mixture that is not biseparable, but their rank-seven example was established
PPT across a selected cut, not all three~\cite{HK}. This already refuted the conjecture of Novo, Moroder and G\"uhne that
PPT mixtures coincide with biseparable states for three qubits~\cite{NMG}. The states here are a stronger counterexample: a
single-term PPT mixture that is PPT across all three cuts. By Observation~3 of~\cite{NMG}, a permutation-invariant
three-qubit state is biseparable if and only if it is a PPT mixture, so no state of this paper is permutation invariant, in
line with Corollary~\ref{cor:ppt3-symmetrisation}.
Garberg, Irgens and Myrheim found fully-PPT states
numerically with many rank profiles, including $(5,5,5,5)$ and extreme states, without a GME
certificate~\cite{GIM}. Their phrase ``genuine tripartite entanglement'' for rank-$(4,4,4,4)$ states means entangled while
separable across every bipartition, which is not GME in the present sense.
Hari Krishnan, Ranjan and Banik~\cite{KRB} provided
fully-PPT three-qubit states outside
$\bigcap_c\mathrm{Sep}_c$, the intersection of the
three cut-separable sets, by cyclically symmetrising an embedded $2\times4$ PPT state and subtracting low-rank terms.
This does not imply GME:
the biseparable set is the larger convex hull
$\mathrm{BS}=\operatorname{conv}(\bigcup_c\mathrm{Sep}_c)$.
Our supported witness excludes the \emph{convex hull},
not merely the intersection. T\'oth and V\'ertesi found numerically a three-qubit state that is PPT across all
bipartitions and nevertheless useful for metrology~\cite{TV}; it is entangled but was not shown to be GME.
Singh and Banik noted that no explicit three-qubit state that is PPT across all three cuts and GME was known,
and showed that, if one exists, it can be taken extreme with all four ranks at least five and the smallest equal to five
or six~\cite{SB}; they did not settle existence. Our extreme states satisfy these restrictions.

To our knowledge these are the first explicit
\emph{three-qubit} examples proved both fully PPT and GME;
the claim does not extend to more parties
or higher local dimensions.