# Identifying families of multipartite states with non-trivial local
entanglement transformations

Nicky Kai Hong Li  Institute for Theoretical Physics, University of Innsbruck,
Technikerstr. 21A, 6020 Innsbruck, Austria  Cornelia Spee  Institute for
Theoretical Physics, University of Innsbruck, Technikerstr. 21A, 6020
Innsbruck, Austria  Martin Hebenstreit  Institute for Theoretical Physics,
University of Innsbruck, Technikerstr. 21A, 6020 Innsbruck, Austria  Julio I.
de Vicente  Departamento de Matemáticas, Universidad Carlos III de Madrid,
Avda. de la Universidad 30, E-28911, Leganés (Madrid), Spain  Barbara Kraus
Institute for Theoretical Physics, University of Innsbruck, Technikerstr. 21A,
6020 Innsbruck, Austria  Department of Physics, QAA, Technical University of
Munich, James-Franck-Str. 1, D-85748 Garching, Germany

###### Abstract

The study of state transformations by spatially separated parties with local
operations assisted by classical communication (LOCC) plays a crucial role in
entanglement theory and its applications in quantum information processing.
While transformations of this type among pure bipartite states were
characterized long ago and have a revealing theoretical structure, it turns
out that generic fully entangled pure multipartite states cannot be obtained
from nor transformed to any inequivalent fully entangled state under LOCC.
States with this property are referred to as isolated. Nevertheless,
multipartite states are classified into families, the so-called SLOCC classes,
which possess very different properties. Thus, the above result does not
forbid the existence of particular SLOCC classes that are free of isolation
and therefore display a rich structure regarding LOCC convertibility. In fact,
it is known that the celebrated $n$-qubit GHZ and W states give particular
examples of such classes and in this work, we investigate this question in
general. One of our main results is to show that the SLOCC class of the
3-qutrit totally antisymmetric state is isolation-free as well. Actually, all
states in this class can be converted to inequivalent states by LOCC protocols
with just one round of classical communication (as in the GHZ and W cases).
Thus, we consider next whether there are other classes with this property and
we find a large set of negative answers. Indeed, we prove weak isolation
(i.e., states that cannot be obtained with finite-round LOCC nor transformed
by one-round LOCC) for very general classes, including all SLOCC families with
compact stabilizers and many with non-compact stabilizers, such as the classes
corresponding to the $n$-qunit totally antisymmetric states for $n\geq 4$.
Finally, given the pleasant feature found in the family corresponding to the
3-qutrit totally antisymmetric state, we explore in more detail the structure
induced by LOCC and the entanglement properties within this class.

##  I Introduction

Many quantum technologies that demonstrate advantages over their classical
counterparts rely on multipartite entanglement EntQKD ; Gottesman ;
SecretSharing ; MBC ; MetrologyGHZ ; ByzantineGHZ . Moreover, in the last
years, entanglement theory tools have proven to be instrumental in the
description of interactions and correlations in many body systems reviewmps .
Thus, the resource theory of entanglement is one of the building blocks of
quantum information theory Horodeckis ; ResourceTheory . It aims at
characterizing, classifying, and quantifying entanglement, providing protocols
to harness this resource, and studying how efficient these manipulations can
be. The free operations in this theory are local operations assisted by
classical communication (LOCC), which naturally describe state manipulation
protocols carried out by multiple spatially separated parties. The study of
state transformations under LOCC not only provides practical protocols for
quantum information processing, but it also plays a crucial role in the
foundations of entanglement theory. Since LOCC operations are assumed to be
freely implementable, a state cannot be less useful than any state that can be
obtained from it by this kind of transformation. Hence, LOCC convertibility
defines a meaningful partial order for the entanglement contained in quantum
states and defines a basic monotonicity condition that every entanglement
measure must satisfy. A simple example of an LOCC map is local unitary (LU)
transformations. Given that this is actually an invertible map, all states
related by LU transformations are regarded as equivalent in entanglement
theory.

For bipartite pure states, LOCC transformability is completely determined by a
majorization relation between the Schmidt coefficients of the two states
Nielsen . This is a seminal result in the field. Among other reasons, this
singles out a maximally entangled state which must be the optimal resource for
any task to be performed under the LOCC constraint. Unfortunately, the
situation in the multipartite case is much more involved. First, it is known
that the state space is partitioned into different classes containing all
states that can be interconverted with non-zero probability, i.e., by
stochastic LOCC (SLOCC) 3qubitSLOCC . Therefore, no LOCC transformation can
exist between any two states in different SLOCC classes. To make things more
complicated, except for very specific values of the number of parties and
local dimensions, there exist infinitely many different SLOCC classes
4qubitSLOCC . In addition to this, SLOCC only defines an equivalence relation
and one is still left with the question of characterizing LOCC convertibility
within each SLOCC class in order to obtain a meaningful ordering of states.
Interestingly, general tools based on the stabilizer of the SLOCC class (see
definition below) have been provided GourWallach ; SEPeqn ; ProbStep1 ;
ProbStep2 , which, in principle, allow us to characterize LOCC
transformability MESpaper . However, a second and more drastic difference
between the multipartite and the bipartite cases arises: it turns out that for
most values of the number of parties and local dimension, generic SLOCC
classes have a trivial stabilizer, from which it follows that almost every
fully entangled pure multipartite state is isolated Isolation1 ; Isolation2 ,
i.e., it cannot be obtained from nor transformed to any LU-inequivalent fully
entangled state of the same local dimensions by LOCC operations. Therefore,
except for very specific configurations, entanglement is incomparable among
generic states. In fact, this is the case in $(\mathbb{C}^{d})^{\otimes n}$
for $n>4$ parties with local dimension $d=2$ and for $d>2$ and $n>3$
Isolation1 ; Isolation2 . However, it remains unknown whether this property
extends to the unbalanced case, i.e., for states such that the local
dimensions for each party are not all the same.

The above notwithstanding, in the perspective of identifying the most useful
states for achieving certain tasks that require entanglement, it is still
possible to find SLOCC classes with a non-trivial stabilizer that are of
measure zero with respect to the whole Hilbert space and contain non-isolated
states. Actually, the possible LOCC transformations among the states in the
SLOCC classes of the $n$-qubit GHZ and W states were characterized
respectively in Refs. turgutGHZ  and turgutW . It turns out that these two
classes are isolation-free (i.e., no state therein is isolated). In the last
few years, the aforementioned techniques have enabled the study of LOCC
convertibility in SLOCC classes with a non-trivial stabilizer such as the ones
that contain generic MESpaper  and non-generic 4qubitMES  4-qubit states,
generic 3-qutrit states locc3qutrit , graph states and general stabilizer
states MatthiasStab , matrix product states loccmps1 ; loccmps2 , permutation-
symmetric states of an arbitrary number of parties and local dimension
OurSymmPaper , and states relevant for quantum networks yamasaki ;
NonTrivial4round . Remarkably, no further isolation-free SLOCC class has been
found with the exception of the particular quantum networks studied in Ref.
yamasaki , which only occur in the unbalanced case. Moreover, the results of
Refs. MESpaper  and locc3qutrit  show that generic 4-qubit and 3-qutrit states
are isolated even though the stabilizer is not trivial in these cases. Thus,
despite this active line of investigation, the GHZ and W SLOCC classes stand
out as the unique instances with such feature for the case of all subsystems
having equal dimension. Interestingly, this property seems to go hand in hand
with relevant applications as these states appear in, e.g., quantum
communication SecretSharing ; ByzantineGHZ ; QCommW , quantum metrology
MetrologyGHZ , and entanglement robustness against loss RobustLossW .
Therefore, with the hope of finding multipartite states that are potentially
relevant for new applications, the underlying question that motivates this
work is: being aware that this phenomenon cannot hold generically for almost
all balanced configurations, are there more particular SLOCC classes that are
isolation-free in addition to the few preexisting examples?

One of the main results in this paper is that there is indeed another
isolation-free class, namely the SLOCC class of the 3-qutrit totally
antisymmetric state $|A_{3}\rangle$. Totally antisymmetric states are
fermionic states of total spin zero AdanAntiSymmAppl . They have applications
in solving certain multiparty communication problems such as the Byzantine
agreement ByzantineAgree , $N$ strangers, secret sharing, and liar detection
problems AdanAntiSymmAppl . In more general settings, these states allow the
inversion of unitary InvertUnitary  and isometry operations InvertIsometry
due to their symmetry properties. Given that $|A_{3}\rangle$ has many
applications and has an isolation-free SLOCC class, it is then well motivated
to take a closer look into the entanglement properties of the whole class.
While studying allowed LOCC transformations within the class, we also show
that the maximally entangled set (MES) MESpaper  of this SLOCC class, which is
the minimal set of LU-inequivalent states such that every state in the whole
SLOCC class can be reached by a state from the set with LOCC, is of zero
measure. In other words, the most “useful” set of states constitutes only a
vanishingly small portion of the SLOCC class.

Next, we consider whether the techniques used to prove the lack of isolation
in the 3-qutrit totally antisymmetric case can be extended to other SLOCC
classes and we provide a plethora of no-go results. The way we discover the
new isolation-free SLOCC class is actually by proving the absence of what we
call weakly isolated states. Excluding trivial LU transformations, these are
states that cannot be reached from any other SLOCC-equivalent states with
finite-round LOCC ($\text{LOCC}_{\mathbb{N}}$) and cannot be transformed into
any other SLOCC-equivalent states with LOCC protocols with one round of
classical communication in which a single party acts non-trivially
($\text{LOCC}_{1}$). Since this is a restricted class of LOCC protocols,
obviously, all isolated states are also weakly isolated. Hence, proving that a
SLOCC class does not contain weakly isolated states is enough to conclude that
it is isolation-free. On the other hand, weak isolation does not necessarily
imply isolation. In fact, there are examples of LOCC conversions such that the
input state cannot be transformed into the output state by concatenating
$\text{LOCC}_{1}$ protocols (and, hence, intermediate probabilistic steps are
required) ProbStep1 ; ProbStep2 ; OurSymmPaper . However, this relies on very
particular constructions and it is conceivable that all classes containing
weakly isolated states are not isolation-free. In any case, weak isolation is
a strong indication of SLOCC classes with limited LOCC convertibility
properties. In this part of the manuscript, we prove weak isolation for very
general SLOCC classes including all classes that have a compact stabilizer and
also the ones with a non-compact stabilizer such as all 4-qubit classes and
the classes of $n$-qunit totally antisymmetric states for all $n\geq 4$. All
of these suggest that even among the non-generic SLOCC classes that have a
non-trivial stabilizer, the absence of isolation remains an elusive
phenomenon.

The outline of this paper is as follows. In Sec. II, we first introduce some
definitions and notations that we will use throughout the paper. Then, we
recall some important results that determine transformability under LOCC and
$\text{LOCC}_{\mathbb{N}}$ and whether a state is weakly isolated. Moreover,
we review the reason why the $n$-qubit GHZ and W classes are isolation-free
and some key properties of the totally antisymmetric states. In Sec. III where
we present our main results, we prove that every SLOCC class that contains a
fully entangled state with a finite or compact symmetry group contains weakly
isolated states. The next thing we consider are SLOCC classes with non-compact
infinite symmetry groups. In particular, we show that weak isolation exists in
every SLOCC class that contains an $n$-qudit fully entangled state with a
symmetry group $\\{S^{\otimes n}\mbox{ with }S\in GL(d,\mathbb{C})\\}$ for all
$n\geq 4$. Examples for such classes would be the ones of $n$-qunit totally
antisymmetric states with $d=n\geq 4$. However, remarkably, we prove that
SLOCC classes that contain a $3$-qudit fully entangled state with a symmetry
group $\\{S^{\otimes 3}\mbox{ with }S\in SL(d,\mathbb{C})\\}$ are isolation-
free. The class of $|A_{3}\rangle$ is an example of such a class in the case
$d=3$. Finally, we study the entanglement properties of the $|A_{3}\rangle$
class in Sec. IV. We characterize the set of states in the $|A_{3}\rangle$
class that cannot be obtained from other LU-inequivalent states with
$\text{LOCC}_{\mathbb{N}}$ ($M_{A_{3}}$) and rule out certain LOCC
transformations within $M_{A_{3}}$. We also present a way to decompose and
prepare all the states in $M_{A_{3}}$ with a qutrit, a 2-qubit entangled
state, and two 2-qutrit entangling unitaries.

##  II Preliminaries

Before reviewing some key concepts in entanglement theory we introduce here
our notation. Throughout this paper, the set $\\{1,\ldots,n\\}$ is denoted by
$[n]$ for any $n\in\mathbb{N}$, $\sigma_{x}$ and $\sigma_{z}$ denote the Pauli
$x$ and $z$ matrices, $GL(d,\mathbb{C})$ and $SL(d,\mathbb{C})$ denote the
sets of $d\times d$ complex invertible and special linear matrices,
respectively. As a shorthand notation, we denote a positive definite matrix
$X$ by $X>0$.

In this work, we consider only $n$-qudit fully entangled pure states
$|\psi\rangle\in\bigotimes_{j=1}^{n}\mathbb{C}^{d_{j}}$ where $d_{j}$ is the
local dimension of the $j$-th qudit. All reduced density matrices of fully
entangled states have full rank, i.e.,
$rk(\text{Tr}_{[n]\setminus\\{j\\}}(|\psi\rangle\langle\psi|))=d_{j}$ for all
$j\in[n]$. The SLOCC class of a state $|\psi\rangle$ is defined to be the set
$\\{\bigotimes_{k=1}^{n}g_{k}|\psi\rangle:g_{k}\in
GL(d_{k},\mathbb{C})\;\forall\;k\in[n]\\}$. Two states are SLOCC-equivalent if
they belong to the same SLOCC class. When we study state transformations
within an SLOCC class, it is useful to choose a representative
$|\Psi_{s}\rangle$ (also called a seed state) from an SLOCC class and
characterize its stabilizer. The stabilizer (also called the symmetry group)
of a state $|\psi\rangle$ is the set of local operators
$\mathcal{S}_{\psi}=\\{S=\bigotimes_{k=1}^{n}S^{(k)}:S|\psi\rangle=|\psi\rangle\text{
with }S^{(k)}\in GL(d_{k},\mathbb{C})\;\forall\;k\in[n]\\}$. We say
$|\psi\rangle$ is stabilized by $S$ if $S\in\mathcal{S}_{\psi}$. We also
define the local symmetry group of the $j$-th qudit of $|\psi\rangle$,
$\mathcal{S}_{\psi}^{(j)}=\\{S^{(j)}\in
GL(d_{j},\mathbb{C}):\exists\;S^{(i)}\;\forall\;i\in[n]\setminus\\{j\\}\text{
so that }\bigotimes_{k=1}^{n}S^{(k)}\in\mathcal{S}_{\psi}\\}$ which can, for
all but one site, be assumed to be a subset of $SL(d_{j},\mathbb{C})$
footnote:LocalSymm_SL .

Within the SLOCC class of $|\Psi_{s}\rangle$, we denote the initial and final
states of a transformation as $|\Psi\rangle=g|\Psi_{s}\rangle$ and
$|\Phi\rangle=h|\Psi_{s}\rangle$, respectively, where
$g=\bigotimes_{k=1}^{n}g_{k}$ and $h=\bigotimes_{k=1}^{n}h_{k}$. We will
predominantly work with unnormalized states unless specified otherwise. Since
entanglement is invariant under LU transformations
$\bigotimes_{i=1}^{n}U(d_{i},\mathbb{C})$, the study of state transformations
can be restricted to between LU-inequivalent states. Hence, it is convenient
to consider the LU-invariant operator $G=g^{\dagger}g$ of the state
$g|\Psi_{s}\rangle$. It is also useful to define the set of local singular
operators of a state $|\Psi\rangle$ as
$\mathcal{N}_{\Psi}=\\{N=\bigotimes_{k=1}^{n}N^{(k)}:N|\Psi\rangle=0\\}$.

###  II.1 Necessary condition for LOCC transformability

We will recall here some results regarding LOCC transfromations among pure
multipartite states. Readers who are familiar with results regarding LOCC
transformations can skip Sec. II.1.

Due to the complicated mathematical description of LOCC transformations
SEP>LOCCRef1 ; DonaldHorodecki ; LOCChardEric ; LOCC_Everything , explicit
construction of an LOCC protocol is often infeasible especially when infinite-
round protocols are taken into account. One way to circumvent the difficulties
of studying LOCC transformations is to study state transformations with an
enlarged set of operations, namely the set of separable maps (SEP), that
strictly contains LOCC SEP>LOCCRef1 ; LOCC_Everything  and has much simpler
mathematical characterization despite not having any physical interpretation.
A SEP is a completely positive trace-preserving map for which all the Kraus
operators are local. The necessary and sufficient condition for the existence
of a SEP transformation among pure states is stated in the following theorem
which was proven in Refs. GourWallach ; SEPeqn .

######  Theorem 1 (GourWallach ; SEPeqn ).

The state $g|\Psi_{s}\rangle$ can be transformed to $h|\Psi_{s}\rangle$ via
SEP if and only if there exists a finite set of probabilities $\\{p_{k}\\}$,
symmetries $\\{S_{k}\\}\subseteq\mathcal{S}_{\Psi_{s}}$, and
$N_{q}\in\mathcal{N}_{g\Psi_{s}}$ such that

$\frac{1}{r}\sum_{k}p_{k}S_{k}^{\dagger}HS_{k}+g^{\dagger}\sum_{q}N_{q}^{\dagger}N_{q}g=G,$
(1)

where $r=||h|\Psi_{s}\rangle||^{2}/||g|\Psi_{s}\rangle||^{2}$.

From Theorem 1, we see that the stabilizer plays a crucial role in determining
whether two states from the same SLOCC class are related via SEP. If one can
prove that one state cannot reach another state with SEP, then it also implies
that the transformation is impossible with LOCC. For example, since generic
states only have trivial symmetries, they are isolated under SEP Isolation1 ;
Isolation2 , and hence isolated under LOCC transformations, which is the
reason why we study non-trivial LOCC transformations only among zero-measure
sets of states. Determining SEP transformability with Theorem 1 is often a
non-trivial task since it is hard to fully characterize both the symmetry
group $\mathcal{S}_{\Psi_{s}}$ and the set of operators
$\mathcal{N}_{g\Psi_{s}}$, which annihiliate the state, in general. Moreover,
even if a SEP transformation is proven to exist between two states, it does
not imply that an LOCC protocol exists. Therefore, Theorem 1 serves only as a
necessary condition for the existence of LOCC transformations.

###  II.2 $\text{LOCC}_{\mathbb{N}}$-reachability and weak isolation

A state is isolated if it is not reachable nor convertible via LOCC (excluding
LU transformations). That is, it cannot be obtained by transforming any SLOCC-
equivalent state, nor can it be transformed into any other SLOCC-equivalent
state with LOCC. As pointed out in Sec. II.1, only a necessary condition for
multipartite LOCC transformability is known, so there is no clear way of
determining whether a state is isolated except for generic states that only
have trivial symmetries Isolation1 ; Isolation2 . However, one could prove a
weaker form of isolation by considering a restricted set of
transformations–$\text{LOCC}_{\mathbb{N}}$, which are, from a practical point
of view, more relevant and characterized better. The necessary and sufficient
condition for a state to be reachable from an SLOCC-equivalent state via
$\text{LOCC}_{\mathbb{N}}$ is stated in Theorem 2. Its proof in Ref.
OurSymmPaper  relies on the fact that there exists a final round in any
$\text{LOCC}_{\mathbb{N}}$ protocol.

######  Theorem 2 (OurSymmPaper ).

A state $|\Phi\rangle\propto h|\Psi_{s}\rangle$ is reachable via
$\text{LOCC}_{\mathbb{N}}$, iff there exists $S\in\mathcal{S}_{\Psi_{s}}$ such
that the following conditions hold up to permutations of the particles:

  * (i)

For any $i\geq 2$, $(S^{(i)})^{\dagger}H_{i}S^{(i)}\propto H_{i}$ and

  * (ii)

$(S^{(1)})^{\dagger}H_{1}S^{(1)}\not\propto H_{1}$.

A simple example for an $\text{LOCC}_{\mathbb{N}}$-reachable state would be a
3-qubit state $|\psi\rangle\propto h_{1}\otimes\text{1}^{\otimes
2}|\Psi_{s}\rangle$ with $h_{1}=\frac{1}{2}\text{1}+\sigma_{x}$ and
$\sigma_{z}^{\otimes 3}\in\mathcal{S}_{\Psi_{s}}$ since $\sigma_{z}$ cannot
quasi-commute with
$H_{1}=(\frac{1}{2}\text{1}+\sigma_{x})^{2}=\frac{5}{4}\text{1}+\sigma_{x}$.
In this case, $|\psi\rangle$ can be reached from $|\Psi_{s}\rangle$ with a
one-round LOCC protocol where the first party measures with the operators
$\\{\sqrt{\frac{2}{5}}h_{1}\sigma_{z}^{m}\\}_{m=0,1}$ and parties 2 and 3 each
apply LU $\sigma_{z}^{m}$ depending on the measurement outcome $m$. For more
examples of states that are (not) $\text{LOCC}_{\mathbb{N}}$-reachable, see
e.g., Ref. OurSymmPaper  and Secs. III and IV.

However, to determine whether a state is
$\text{LOCC}_{\mathbb{N}}$-convertible is generally much harder due to, among
other reasons, the existence of protocols that require intermediate
probabilistic steps ProbStep1 ; ProbStep2 ; OurSymmPaper . Instead, if one
asks only for one-round LOCC ($\text{LOCC}_{1}$) convertibility, then one can
use the necessary and sufficient conditions given by Lemma 1 in Ref.
OurSymmPaper . We can define a weaker form of isolation by calling a state
weakly isolated if it is not $\text{LOCC}_{\mathbb{N}}$-reachable and not
$\text{LOCC}_{1}$-convertible. The necessary and sufficient condition for weak
isolation is given by the following lemma OurSymmPaper .

######  Lemma 3 (OurSymmPaper ).

A state $|\Phi\rangle\propto g|\Psi_{s}\rangle$ is weakly isolated if and only
if there exists no $S\in\mathcal{S}_{\Psi_{s}}\setminus\\{\text{1}\\}$ such
that $S^{(i)}$ quasi-commutes with $G_{i}$ [i.e.,
$(S^{(i)})^{\dagger}G_{i}S^{(i)}\propto G_{i}$] for at least $n-1$ sites $i$.

To give an example, a 3-qubit state $|\psi\rangle$ with a symmetry
$\sigma_{z}^{\otimes 3}$, which (quasi-)commutes with $G_{i}=\text{1}$ for all
3 sites, is not (weakly) isolated. It can be transformed into another state
$h_{1}\otimes\text{1}^{\otimes 2}|\psi\rangle$ with $h_{1}\in
GL(2,\mathbb{C})$ such that $H_{1}$ is not diagonal and
$\text{diag}(H_{1})=\text{1}$. The transformation is achieved by having party
1 measure the first qubit of $|\psi\rangle$ with operators
$\\{\frac{1}{\sqrt{2}}h_{1}\sigma_{z}^{m}\\}_{m=0,1}$ and parties 2 and 3
apply a local unitary $\sigma_{z}^{m}$ based on the measurement outcome $m$.
More examples of (non-)weakly isolated states can be found in Ref.
OurSymmPaper  and in Secs. III and IV.

With Lemma 3, one has a relatively simple mathematical criterion to show that
an SLOCC class is isolation-free by proving that every state it contains is
not weakly isolated since any isolated state is also weakly isolated by
definition. In fact, this is exactly how we identify a new isolation-free
class, the SLOCC class of the 3-qutrit totally antisymmetric state (Sec.
III.2). We should also note that weak isolation is only an indicator but not
an implication for full LOCC isolation.

###  II.3 Known isolation-free SLOCC classes

Before starting the search for new isolation-free SLOCC classes, let us
revisit the key properties of the $n$-qubit GHZ and W states,
$|\text{GHZ}_{n}\rangle$ and $|\text{W}_{n}\rangle$, whose SLOCC classes do
not contain any isolated states, i.e., the SLOCC classes are isolation-free.
The symmetry groups of the two states are given by WGHZsymm

$\displaystyle\mathcal{S}_{\text{GHZ}}=\left\\{\bigotimes_{i=1}^{n}P_{z_{i}}\sigma_{x}^{m}:m\in\\{0,1\\},z_{i}\in\mathbb{C}\setminus\\{0\\},\prod_{i=1}^{n}z_{i}=1\right\\},$
(2)
$\displaystyle\mathcal{S}_{\text{W}}=\left\\{\frac{1}{x}\bigotimes_{i=1}^{n}T_{x,y_{i}}:x\in\mathbb{C}\setminus\\{0\\},y_{i}\in\mathbb{C},\sum_{i=1}^{n}y_{i}=0\right\\}$
(3)

where $P_{z_{i}}=diag(z_{i},z_{i}^{-1})$ and
$T_{x,y_{i}}=\begin{pmatrix}1&y_{i}\\\ 0&x\end{pmatrix}$. One can verify that
these two classes do not have any weakly isolated states using the following
observations. Let $H_{i}=\begin{pmatrix}a_{i}&b_{i}\\\
b_{i}^{*}&c_{i}\end{pmatrix}>0$. It holds that
$\sigma_{x}P_{z_{i}}^{\dagger}H_{i}P_{z_{i}}\sigma_{x}\propto H_{i}$ if and
only if $z_{i}=\pm\sqrt{\frac{c_{i}b_{i}}{a_{i}b_{i}^{*}}}$ and
$T_{x,y_{i}}^{\dagger}H_{i}T_{x,y_{i}}\propto H_{i}$ if and only if
$y_{i}=\frac{b_{i}}{a_{i}}(1-x)$ and $|x|=1$. Since $n-1$ parameters
$z_{i}\in\mathbb{C}\setminus\\{0\\}$ ($y_{i}\in\mathbb{C}$) are independent in
$\mathcal{S}_{GHZ}$ ($\mathcal{S}_{W}$), for any $H=\otimes_{i=1}^{n}H_{i}$,
there always exists a non-trivial symmetry from $\mathcal{S}_{GHZ}$
($\mathcal{S}_{W}$) that quasi-commutes with $H$ for $n-1$ sites,
respectively. Therefore, due to Lemma 3, the two SLOCC classes do not have any
state that is (weakly) isolated.

###  II.4 Totally antisymmetric states

We now introduce some properties of totally antisymmetric states, which we
will need when we prove (non-)existence of weakly isolated states in their
SLOCC classes in Sec. III.2 and when we investigate entanglement properties of
the SLOCC class of the 3-qutrit totally antisymmetric state in Sec. IV.

The $n$-qunit totally antisymmetric state is given by

$\displaystyle|A_{n}\rangle$
$\displaystyle=\frac{1}{\sqrt{n!}}\sum_{\pi}sgn(\pi)P_{\pi}(|1,2,\ldots,n\rangle)$
(4)
$\displaystyle=\frac{1}{\sqrt{n!}}\sum_{i_{1},\ldots,i_{n}=1}^{n}\epsilon_{i_{1},\ldots,i_{n}}|i_{1},\ldots,i_{n}\rangle$
(5)

where the sum in Eq. (4) is over all permutations of $n$ elements $\pi\in
S_{n}$, $sgn(\pi)$ is the sign of $\pi$,
$P_{\pi}(|1,2,\ldots,n\rangle)=|\pi(1),\pi(2),\ldots,\pi(n)\rangle$ and
$\epsilon_{i_{1},\ldots,i_{n}}$ is the Levi-Civita symbol. The key property of
$|A_{n}\rangle$ that we will use later is its invariance under permutation up
to a sign. More precisely, it is easy to see that

$P_{\tilde{\pi}}|A_{n}\rangle=sgn(\tilde{\pi})|A_{n}\rangle$ (6)

using the properties of $sgn$:
$sgn(\pi)=sgn(\tilde{\pi}^{-1}\tilde{\pi}\pi)=sgn(\tilde{\pi}^{-1})sgn(\tilde{\pi}\pi)$
and $sgn(\tilde{\pi}^{-1})=sgn(\tilde{\pi})$.

As we saw in Secs. II.1–II.3, in order to determine LOCC transformability and
whether weak isolation exists in an SLOCC class, it is essential to have a
full characterization of the stabilizer of a seed state from that class.
Hence, we need to characterize the symmetry group of $|A_{n}\rangle$, which is
stated in the following observation. Its proof can be found in Appendix A.

######  Observation 4.

The symmetry group of the $n$-qunit totally antisymmetric state
$|A_{n}\rangle$ is $\\{S^{\otimes n}\mbox{ with }S\in SL(n,\mathbb{C})\\}$.

##  III Identifying isolation-free SLOCC classes

In this section, we will first focus on SLOCC classes that contain a fully
entangled state with a compact symmetry group and prove that all these classes
contain weakly isolated states. Then, we move on to consider classes with non-
compact stabilizers and prove that for any local dimension $d$, all 3-qudit
SLOCC classes with a state stabilized by $S^{\otimes 3}$ for all $S\in
SL(d,\mathbb{C})$ are isolation-free. One such example is the SLOCC class of
the 3-qutrit totally antisymmetric state $|A_{3}\rangle$. Finally, we show
that for $n\geq 4$, if a state in an $n$-qudit SLOCC class has all symmetries
of the form $S^{\otimes n}$, then the class contains weak isolation. Examples
of these classes are the ones containing $n$-qunit totally antisymmetric
states $|A_{n}\rangle$. In a slight abuse of notation, here and in the rest of
the paper, we use $\bigotimes_{i=1}^{n}U(d_{i},\mathbb{C})$ to denote the
group with elements $\bigotimes_{i=1}^{n}U_{i}$ where each $U_{i}$ is an
arbitrary unitary matrix in $U(d_{i},\mathbb{C})$ for all $i$ (and similarly
for $\bigotimes_{i=1}^{n}GL(d_{i},\mathbb{C})$).

###  III.1 SLOCC classes with compact stabilizers

We consider here SLOCC classes whose representatives possess a compact
symmetry group. We call a set of $d\times d$ complex matrices closed (compact)
if it is closed (compact) in the Euclidean topology on $\mathbb{C}^{d^{2}}$
unless specified otherwise. As stated in the following theorem, we prove that
all SLOCC classes containing a fully entangled state which is stabilized by a
compact stabilizer contains weakly isolated states.

######  Theorem 5.

Every $n$-qudit SLOCC class that contains a fully entangled state
$|\psi\rangle\in\bigotimes_{i=1}^{n}\mathbb{C}^{d_{i}}$ which has a compact
symmetry group
$\mathcal{S}_{\psi}\subset\bigotimes_{i=1}^{n}GL(d_{i},\mathbb{C})$ contains
weakly isolated states.

In order to prove this, we need the following proposition, which states that a
compact symmetry group is always finite, and it is proven in Ref. GourWallach
.

######  Proposition 6 (GourWallach ).

Let $|\psi\rangle$ be a state with a symmetry group $\mathcal{S}_{\psi}$ that
is compact, then $\mathcal{S}_{\psi}$ is finite.

###### Proof of Theorem 5.

Due to Proposition 6, we are left to prove that any SLOCC class that contains
a fully entangled state $|\psi\rangle$ with a finite symmetry group
$\mathcal{S}_{\psi}$ contains weakly isolated states. Let us first recall that
whenever there exists a state with finitely many symmetries, then there exists
a state in the same SLOCC class with finitely many unitary symmetries
GourWallach . Hence, it remains to show that if a state has only finitely many
unitary symmetries, then there exists weak isolation. In order to show this,
we denote a non-empty finite set of non-trivial $d\times d$ unitary matrices
(i.e., $U\not\propto\text{1}$) by $\mathcal{F}$ and show that there exists
$P>0$ such that $U_{i}^{\dagger}PU_{i}\not\propto P$ for all
$U_{i}\in\mathcal{F}$. This can be easily seen from the fact that all $U_{i}$
have a finite set of eigenspaces. Hence, choosing $P$ as a positive operator
with an eigenvector that is not in any of these finitely many eigenspaces
proves the statement footnote:PosOpNotCommUs . Given that the state could be
transformed if the symmetry commutes with all but one site, it remains to show
that no fully entangled state can have a symmetry of the form
$S^{(j)}\otimes_{k\neq j}\text{1}_{d_{k}}\in\mathcal{S}_{\psi}$ unless it is
trivial, i.e., $S^{(j)}=\text{1}_{d_{j}}$. This last statement follows easily
by considering the Schmidt decomposition in the splitting, system $j$ versus
the rest.∎

Since a unitary symmetry group is also compact (see also Appendix B), the
following corollary follows immediately from Theorem 5.

######  Corollary 7.

Every $n$-qudit SLOCC class that contains a fully entangled state
$|\psi\rangle$ which has a unitary symmetry group
$\mathcal{S}_{\psi}\subseteq\bigotimes_{i=1}^{n}U(d_{i},\mathbb{C})$ contains
weakly isolated states.

The proof of Proposition 6 in Ref. GourWallach  requires some familiarity with
algebraic geometry. However, Corollary 7 can also be proven for
$d_{j}=2\;\forall\;j\in[n]$ without relying on Proposition 6. In Appendix B,
we provide an alternative proof which uses more elementary mathematics and
contains several observations about the general structure of stabilizers that
might be of independent interest to some readers.

To summarize all the ideas behind the proof, weak isolation exists in all
SLOCC classes with compact stabilizers because there are not enough symmetries
in any compact stabilizer to quasi-commute with every possible
$\bigotimes_{i=1}^{n}H_{i}>0$ for $n-1$ qudits and by Lemma 3, this leads to
weak isolation. Although weak isolation does not imply isolation, these
results suggest that it is more promising to investigate SLOCC classes with
states stabilized by a non-compact symmetry group to identify isolation-free
SLOCC classes.

###  III.2 SLOCC classes with non-compact stabilizers

We have seen that non-compact stabilizers are necessary to identify SLOCC
class without weak isolation. In this section we show that tripartite states
with a particular non-compact stabilizer are indeed isolation-free. An example
of such an SLOCC class is given by the totally antisymmetric state. Extending
such stabilizers to more parties leads, however, again to weak isolation, as
we show at the end of this section.

######  Theorem 8.

Let $|\Psi_{s}\rangle$ be a $3$-qudit state with a symmetry group
$\mathcal{S}_{\Psi_{s}}=\\{S^{\otimes 3}\mbox{ with }S\in
SL(d,\mathbb{C})\\}$, with $d\geq 2$. Then, no weakly isolated state exists
within the SLOCC class of $|\Psi_{s}\rangle$. In particular, the SLOCC class
of the 3-qutrit totally antisymmetric state $|A_{3}\rangle$ is isolation-free.

###### Proof.

Using that $\frac{\tilde{g}_{3}^{\otimes
3}}{(\det\tilde{g}_{3})^{3/d}}\in\mathcal{S}_{\Psi_{s}}$, for any
$\tilde{g}_{3}$, we have that any state in the SLOCC class of
$|\Psi_{s}\rangle$ can be written as

$\tilde{g}_{1}\otimes\tilde{g}_{2}\otimes\tilde{g}_{3}|\Psi_{s}\rangle\propto
g_{1}\otimes g_{2}\otimes\text{1}|\Psi_{s}\rangle$ (7)

where $\tilde{g}_{i}\in GL(d,\mathbb{C})$ and
$g_{i}=\tilde{g}_{i}\tilde{g}_{3}^{-1}$ for $i=1,2$. The proportionality
factor in the equation above comes from $\tilde{g}_{3}^{\otimes
3}|\Psi_{s}\rangle=(\det\tilde{g}_{3})^{3/d}|\Psi_{s}\rangle$. To show that
any state of the form of Eq. (7) is not weakly isolated, we define
$G_{i}=g_{i}^{\dagger}g_{i}$ for $i=1,2$ and use Lemma 3. Since the symmetry
group of $|\Psi_{s}\rangle$ is $\\{S^{\otimes 3}\mbox{ with }S\in
SL(d,\mathbb{C})\\}$, there always exists a unitary $S\not\propto\text{1}$
that satisfies $S^{\dagger}\text{1}S\propto\text{1}$ and
$S^{\dagger}G_{i}S\propto G_{i}$ for $i=1$ or 2 by choosing the eigenbasis of
$S$ identical to that of the positive operator $G_{i}$. As the quasi-
commutation relation holds for at least 2 out of 3 sites, any state in form of
Eq. (7) is not weakly isolated by Lemma 3. Since the 3-qutrit totally
antisymmetric state $|A_{3}\rangle$ has the symmetry group $\\{S^{\otimes
3}\mbox{ with }S\in SL(3,\mathbb{C})\\}$ by Observation 4, its SLOCC class has
no (weakly) isolated states. ∎

This makes the SLOCC class of $|A_{3}\rangle$ the third known isolation-free
class that has identical dimension in all local systems. Note however, that
this property does not extend to larger numbers of parties, as we show in the
following theorem, which is proven in Appendix C.

######  Theorem 9.

Let $|\Psi_{s}\rangle$ be an $n$-qudit state $(n\geq 4)$ with a symmetry group
$\mathcal{S}_{\Psi_{s}}\subseteq\\{S^{\otimes n}\mbox{ with }S\in
GL(d,\mathbb{C})\\}$ with $d\geq 2$ arbitrary. Then, there exist weakly
isolated states within the SLOCC class of $|\Psi_{s}\rangle$.

Theorem 9 and the fact that the symmetries of the $n$-qunit totally
antisymmetric states are of the form $S^{\otimes n}$ implies that the SLOCC
classes of the $n$-qunit totally antisymmetric states, $|A_{n}\rangle$, with
$n\geq 4$ contain weakly isolated states.

Hence, in order to find new isolation-free SLOCC classes, one needs to
consider states with non-compact stabilizer different than $S^{\otimes n}$.
However, a more general symmetry group does not preclude isolation. In fact,
we show next that all SLOCC classes of $n$-qubit states with $n\leq 4$, with
the exception of the GHZ and W classes, contain weakly isolated states.

######  Observation 10.

The only isolation-free SLOCC classes for $n\leq 4$ qubits that contain fully
entangled states are the GHZ and W classes.

###### Proof.

Considering only fully entangled state, we have for 2 qubits, only the SLOCC
class of the Bell state $|\Phi^{+}\rangle(=|\text{GHZ}_{2}\rangle)$, and for 3
qubits, the ones of $|\text{GHZ}_{3}\rangle$ and $|\text{W}_{3}\rangle$
3qubitSLOCC . For 4 qubits, despite that there are infinitely many SLOCC
classes with fully entangled states 4qubitSLOCC , the stabilizer for every
class was characterized in Ref. 4qubitMES . By Theorem 5 and Proposition 6, we
only need to show that weak isolation exists in classes with non-compact
infinite stabilizers. One can easily check that for every infinite stabilizer
characterized in Ref. 4qubitMES , one can always find
$\otimes_{i=1}^{4}H_{i}>0$ with which no non-trivial symmetry can quasi-
commute for more than 2 sites. ∎

Notice that such a case-by-case analysis cannot be immediately extended to
more parties. A full characterization of all $n$-qubit SLOCC classes and their
stabilizers for $n>4$ is a formidable problem that, to the best of our
knowledge, remains open so far.

##  IV Entanglement properties of the $|A_{3}\rangle$ SLOCC class

In this section, we investigate the entanglement properties of the SLOCC class
of the 3-qutrit totally antisymmetric state $|A_{3}\rangle$. In particular, we
will identify a superset, $M_{A_{3}}$, of the MES of this SLOCC class. It
comprises of states that are not $\text{LOCC}_{\mathbb{N}}$-reachable and can
collectively reach any SLOCC-equivalent state with LOCC in at most 3 rounds.
Then, we will show that within the $M_{A_{3}}$, LOCC transformations from a
full-measure subset to a zero-measure subset are impossible even with
infinite-round protocols (Lemma 16). Finally, we show that all states in the
$M_{A_{3}}$ can be obtained (up to LU) from applying two 2-qutrit entangling
unitaries on a qutrit and a 2-qubit entangled state.

######  Observation 11.

Any state in the SLOCC class of $|A_{3}\rangle$ is either LU equivalent to or
can be reached from a state in one of the following three forms (up to LU) via
LOCC in at most 3 rounds:

  1. (i)

$|A_{3}\rangle$,

  2. (ii)

$|\psi_{2}(\alpha_{1};\alpha_{2})\rangle\propto\text{diag}(\sqrt{\alpha_{1}},1,1)\otimes\text{diag}(\sqrt{\alpha_{2}},1,1)\otimes\text{1}|A_{3}\rangle$
where $\alpha_{1},\alpha_{2}>0$, $\alpha_{1}\neq\alpha_{2}$ and
$\alpha_{1},\alpha_{2}\neq 1$,

  3. (iii)

$|\psi_{3}(\alpha_{1},\beta_{1};\alpha_{2},\beta_{2})\rangle\propto\text{diag}(\sqrt{\alpha_{1}},\sqrt{\beta_{1}},1)\otimes\text{diag}(\sqrt{\alpha_{2}},\sqrt{\beta_{2}},1)\otimes\text{1}|A_{3}\rangle$
where $\alpha_{1},\alpha_{2},\beta_{1},\beta_{2}>0$,
$\alpha_{1}\neq\beta_{1}\neq\beta_{2}\neq\alpha_{2}$,
$\alpha_{1}\neq\alpha_{2}$,
$\frac{\alpha_{1}}{\beta_{1}}\neq\frac{\alpha_{2}}{\beta_{2}}$ and
$\alpha_{1},\alpha_{2},\beta_{1},\beta_{2}\neq 1$.

Moreover, these states are not reachable via $\text{LOCC}_{\mathbb{N}}$.
Hence, the MES of $|A_{3}\rangle$ is contained in the union set of all these
states, which we call $M_{A_{3}}$.

We prove this observation in Appendix D. Note that within this SLOCC class,
many states can be reached via LOCC. In fact, the MES of the $|A_{3}\rangle$
class is a zero-measure subset of the whole SLOCC class since the MES is
contained in $M_{A_{3}}$ which is parametrized by 4 positive real numbers (see
Observation 11), while general states in the SLOCC class require 8 real
parameters footnote:MA3_zero_measure .

Let us note that some transformations from states in $M_{A_{3}}$ to other
states in the SLOCC class are achievable with non-trivial 3-round LOCC
protocols. We illustrate that with an example: the transformation from
$|A_{3}\rangle$ to $|\psi\rangle\propto
h_{1}\otimes\sqrt{D_{2}}\otimes\text{1}|A_{3}\rangle$ where
$h_{1}=\frac{1}{\sqrt{6}}\begin{pmatrix}\sqrt{27+\sqrt{3}}&\sqrt{3-\sqrt{3}}\\\
\sqrt{3-\sqrt{3}}&\sqrt{3+\sqrt{3}}\end{pmatrix}\oplus 1$ and
$D_{2}=\text{diag}(25,9,1)$. First, party 3 applies a measurement to the third
qutrit of $|A_{3}\rangle$ with operators
$\\{\sqrt{\frac{5}{11}}\text{diag}(\frac{1}{\sqrt{5}},1,1)X^{j}\\}_{j=0,1,2}$
where $X=\sum_{i=0}^{2}|(i+1)\text{mod}\;3\rangle\langle i|$ and parties 1 and
2 each apply $X^{j}$ depending on the measurement outcome $j$ to get
$\text{1}\otimes\text{1}\otimes\text{diag}(\frac{1}{\sqrt{5}},1,1)|A_{3}\rangle\propto\text{diag}(\sqrt{5},1,1)^{\otimes
2}\otimes\text{1}|A_{3}\rangle$. Next, party 2 measures the state with
operators
$\\{\frac{1}{\sqrt{10}}\text{diag}(5,3,1)\tilde{U}^{k}\text{diag}(\frac{1}{\sqrt{5}},1,1)\\}_{k=0,1}$
where $\tilde{U}=1\oplus\begin{pmatrix}0&-1\\\ 1&0\end{pmatrix}$ and parties 1
and 3 each apply $\tilde{U}^{k}$ depending on the measurement outcome $k$.
Finally, party 1 measures with
$\\{\frac{1}{\sqrt{3}}h_{1}\text{diag}(\frac{1}{\sqrt{5}},1,1)Z^{r}\\}_{r=0,1,2}$
and each of the remaining parties apply $Z^{r}$ depending on the measurement
outcome $r$, where $Z=\text{diag}(1,e^{i\frac{2\pi}{3}},e^{i\frac{4\pi}{3}})$.
Despite the fact that there are certain instances [e.g., $h_{1}$ and $D_{2}$
such that $\text{diag}(h_{1}^{\dagger}h_{1})=D_{2}$] where a 2-round protocol
is sufficient, it is worth noting that in this protocol, all three parties act
non-trivially despite $g_{3}=h_{3}$. Other instances for state transformations
that can be achieved with LOCC protocols using more than 2 rounds of
communication in a non-trivial way are presented in Ref. NonTrivial4round .

In the process of characterizing the MES of the $|A_{3}\rangle$ class, we show
that for certain subsets of $M_{A_{3}}$, some states are not related even via
SEP (and, hence, by LOCC). More specifically, we prove in Lemma 16 that none
of the type-(iii) states in $M_{A_{3}}$ (see Observation 11) can be converted
to any type-(ii) states with SEP. In addition, we show that certain type-(ii)
states are not interconvertible via SEP. For more details, please refer to
Lemma 16 in Appendix E. This result together with Observation 11 are
summarized in Fig. 1.

Figure 1: The figure summarizes the results about state transformations within
the SLOCC class of $|A_{3}\rangle$. (a) Any state in the SLOCC class can be
reached (up to LU) from a state in $M_{A_{3}}$ with LOCC using at most 3
rounds of communications, whereas all states in $M_{A_{3}}$ are unreachable
from any other states with $\text{LOCC}_{\mathbb{N}}$. (b) Within $M_{A_{3}}$,
there exist a full-measure subset of states, $\mathcal{D}_{f}$, that cannot
reach states in another zero-measure subset, $\mathcal{D}_{0}$, with SEP and
hence, also not with LOCC. The definitions of sets $\mathcal{D}_{f}$ and
$\mathcal{D}_{0}$ are given in Appendix E.

The proof of these results relies on the satisfiability of a necessary but not
sufficient condition for SEP transformations. For the most general
transformations within $M_{A_{3}}$, arbitrary operators $g$ and $h$
corresponding to states therein and the set $\mathcal{N}_{g|A_{3}\rangle}$
need to be considered. As shown in Appendix E, for the state transformations
presented here, a simple necessary condition for the existence of a SEP
transformation, which is sufficient to show that these transformations do not
exist, can be derived without specifying this set. Hence, we leave the task of
fully characterizing the MES of the $|A_{3}\rangle$ class as an open problem.

Given the simple form of states in $M_{A_{3}}$, one may expect that these
states can be completely characterized by bipartite entanglement across two
different splittings of the three particles. Indeed, we find that all states
in $M_{A_{3}}$ can be generated (up to LU) by entangling a qutrit with a
2-qubit entangled states using two 2-qutrit controlled-unitaries following a
procedure similar to the one proposed by Ref. Decomp3qubits  (see Ref.
SubradianEntAtoms  for a different preparation scheme of $|A_{3}\rangle$). The
preparation procedure of a state that is LU equivalent to the state
$|\psi(\alpha_{1},\alpha_{2},\beta_{1},\beta_{2})\rangle\propto\text{diag}(\sqrt{\alpha_{1}},\sqrt{\alpha_{2}},1)\otimes\text{diag}(\sqrt{\beta_{1}},\sqrt{\beta_{2}},1)\otimes\text{1}|A_{3}\rangle$
footnote:ReorderAlpha2Beta1  is illustrated in Fig. 2 and is outlined as
follows. First, qutrit 1 is prepared in the state
$|\widetilde{+}\rangle=\frac{1}{\sqrt{3}}(|0\rangle+|1\rangle+|2\rangle)$ and
qutrits 2 and 3 are prepared in an entangled state
$|\psi_{s}\rangle=\sqrt{\lambda_{+}}|00\rangle+\sqrt{\lambda_{-}}|11\rangle$
where $\lambda_{\pm}$ depends on $\alpha_{1},\alpha_{2},\beta_{1},\beta_{2}$.
Then, apply two commuting unitaries $U_{1j}=\sum_{k=0}^{2}|k\rangle\langle
k|\otimes(U_{j}^{\dagger}D_{\omega}U_{j})^{k}$ on qutrits 1 and $j$ for
$j=2,3$ where
$D_{\omega}=\text{diag}(1,e^{-i\frac{2\pi}{3}},e^{i\frac{2\pi}{3}})$. The
exact forms of $\lambda_{\pm}$, $U_{2}$ and $U_{3}$ are stated in Appendix F.
With these parametrization,

$U_{13}U_{12}|\widetilde{+}\rangle_{1}|\psi_{s}\rangle_{23}\stackrel{{\scriptstyle\text{LU}}}{{\simeq}}|\psi(\alpha_{1},\alpha_{2},\beta_{1},\beta_{2})\rangle$
(8)

for any $\alpha_{1},\alpha_{2},\beta_{1},\beta_{2}>0$, where the symbol
$\stackrel{{\scriptstyle\text{LU}}}{{\simeq}}$ stands for LU equivalence.

Figure 2: Any state in $M_{A_{3}}$ can be prepared (up to LU) by entangling
qutrit 1 in the state $\frac{1}{\sqrt{3}}(|0\rangle+|1\rangle+|2\rangle)$ to
the entangled qubits 2 and 3 in the state $|\psi_{s}\rangle$ with 2-qutrit
entangling unitaries $U_{12}$ and $U_{13}$.

##  V Conclusion

Although isolation is a generic feature of multipartite states in
$(\mathbb{C}^{d})^{\otimes n}$ for $n>3$ and $d>2$ (and for $n>4$ when $d=2$),
there could still exist zero-measure SLOCC classes with a rich LOCC structure.
In fact, such examples are known (most notably the $n$-qubit GHZ and W SLOCC
classes) but they are scarce. Thus, in this work we considered this question
in general with the hope of finding more isolation-free SLOCC classes. In
order to do so, we investigated whether weak isolation exists in many SLOCC
classes. First, we proved that weakly isolated states are present in all
$n$-qudit SLOCC classes that have fully entangled states with compact
stabilizers for arbitrary local dimensions. Then, we considered $n$-qudit
SLOCC classes that has a fully entangled state stabilized by non-compact
stabilizers restricted to the form $S^{\otimes n}$ for $S\in GL(d,\mathbb{C})$
and any local dimension $d$. We proved that for $n=3$, the SLOCC class is
isolation-free for any $d\geq 2$. We also showed that the SLOCC class of the
3-qutrit totally antisymmetric state $|A_{3}\rangle$ is a special case with
$d=3$, thereby identifying a new isolation-free class. However, we proved that
weakly isolated states exist in $n$-qudit SLOCC classes with non-compact
stabilizers of the form $S^{\otimes n}$ for all $n\geq 4$ and $d\geq 2$.
Examples of such classes are those of $n$-qunit totally antisymmetric states
for all $n\geq 4$. Let us mention here that the $|A_{3}\rangle$ class is
singled out by the fact that this state and the 2-qubit singlet state are the
only absolutely maximally entangled states, i.e., states which are maximally
entangled in any bipartite splitting, in $(\mathbb{C}^{d})^{\otimes n}$ for
all $d,n\geq 2$ that satisfy $U^{\otimes n}|\psi\rangle\propto|\psi\rangle$
for any $U\in U(d,\mathbb{C})$ A3vsAngeq4 . However, it is still unclear
whether these properties are relevant for the absence of isolation.

Finally, we studied the entanglement properties of the new isolation-free
class. We proved that every state in such class can be reached (up to LU) from
a zero-measure subset of states, $M_{A_{3}}$, with LOCC using at most 3
rounds, while the states in $M_{A_{3}}$ are not
$\text{LOCC}_{\mathbb{N}}$-reachable from any other states. Moreover, we
showed that all states in $M_{A_{3}}$ can be prepared with a simple protocol
using only bipartite entanglement resources.

The reason for asking the initial question was that the elements of the MES of
the $n$-qubit GHZ and W SLOCC classes appear in many practical applications
SecretSharing ; ByzantineGHZ ; QCommW ; MetrologyGHZ ; RobustLossW . Thus,
arguably, one might expect to find useful states in other isolation-free
classes due to their exceptionally rich entanglement structure. Indeed, the
newly identified isolation-free class proven in this work contains the state
$|A_{3}\rangle$ in its MES, which also plays an important role in many known
practical protocols AdanAntiSymmAppl ; ByzantineAgree ; InvertUnitary ;
InvertIsometry . In future work, it would be interesting to study whether a
practical scenario could be found, in which the very notion of “isolation-
freeness” translates into some form of advantage. On the more technical level,
it would be desirable to identify more isolation-free classes or to prove that
the examples found thus far exhaust all possibilities. In this regard, it
would be particularly useful to devise new techniques that would make it
possible to prove that weakly isolated SLOCC classes are in fact isolated or
to find a counterexample.

###### Acknowledgements.

We thank Nolan Wallach and Tobias Fritz for helpful discussions regarding
Proposition 6. NKHL, CS, MH, and BK acknowledge financial support from the
Austrian Science Fund (FWF): W1259-N27 (DK-ALM), F 7107-N38 (SFB BeyondC) and
P 32273-N27 (Stand-Alone Project). JIdV acknowledges the following
institutions for financial support: Spanish Ministerio de Ciencia e Innovación
(grant PID2020-113523GB-I00) and Comunidad de Madrid (grant QUITEMAD-
CMS2018/TCS-4342 and the Multiannual Agreement with UC3M in the line of
Excellence of University Professors EPUC3M23 in the context of the V PRICIT).

##  Appendix A Proof of Observation 4

In this appendix, we prove Observation 4 which states the symmetry group of
$n$-qunit totally antisymmetric states for all $n\geq 2$.

######  Observation 4.

The symmetry group of the $n$-qunit totally antisymmetric state
$|A_{n}\rangle$ is $\\{S^{\otimes n}\mbox{ with }S\in SL(n,\mathbb{C})\\}$.

###### Proof.

Using Eq. (6), it is straightforward to see that the existence of a symmetry,
which is not of the form $S^{\otimes n}$ implies that there exists a $B\in
SL(n,\mathbb{C})$ such that footnote:BBinv

$\displaystyle B\otimes B^{-1}\otimes\text{1}^{\otimes
n-2}|A_{n}\rangle=|A_{n}\rangle.$ (9)

It follows from Eq. (9) that

$\displaystyle(B\otimes\text{1}-\text{1}\otimes B)\otimes\text{1}^{\otimes
n-2}|A_{n}\rangle=0.$ (10)

Let $B=\sum_{i,j=1}^{n}B_{i,j}|i\rangle\langle j|$. Projecting the last $n-2$
systems of the state in Eq. (10) onto $|i_{3},\ldots,i_{n}\rangle$ gives

$\displaystyle(B\otimes\text{1}-\text{1}\otimes
B)(|i_{1},i_{2}\rangle-|i_{2},i_{1}\rangle)=0$ (11)

for all $i_{1}\neq i_{2}$. Its overlaps with $\langle i_{1},i_{2}|$ and
$\langle i_{1},i_{1}|$ give the constraints $B_{i_{1},i_{1}}=B_{i_{2},i_{2}}$
and $B_{i_{1},i_{2}}=0$, respectively, for all $i_{1}\neq i_{2}$. Thus, only
$B\propto\text{1}$ can satisfy Eq. (9) and hence, any symmetry of
$|A_{n}\rangle$ must be equal to $S^{\otimes n}$ for some $S\in
GL(n,\mathbb{C})$.

The remaining part of the proof is to show that for any $S\in
GL(n,\mathbb{C})$, $\frac{1}{\det S}S^{\otimes n}$ is a symmetry of
$|A_{n}\rangle$. Consider

$\displaystyle\langle j_{1},\ldots,j_{n}|S^{\otimes n}|A_{n}\rangle$ (12)
$\displaystyle=$
$\displaystyle\;\frac{1}{\sqrt{n!}}\sum_{i_{1},\ldots,i_{n}=1}^{n}\epsilon_{i_{1},\ldots,i_{n}}[S]_{j_{1},i_{1}}[S]_{j_{2},i_{2}}\ldots[S]_{j_{n},i_{n}}$
(13) $\displaystyle=$
$\displaystyle\;\frac{1}{\sqrt{n!}}\epsilon_{j_{1},\ldots,j_{n}}\det S$ (14)

where the last line is obtained from the Leibniz formula for determinants that
uses the Levi-Civita symbol. Due to Eq. (5), it follows that $S^{\otimes
n}|A_{n}\rangle=\det S|A_{n}\rangle$ for all $S\in GL(n,\mathbb{C})$ and
therefore, $S^{\prime}=S/(\det S)^{\frac{1}{n}}\in SL(n,\mathbb{C})$ fulfills
$S^{\prime\otimes n}|A_{n}\rangle=|A_{n}\rangle$. Thus, the symmetry group of
$|A_{n}\rangle$ is $\\{S^{\otimes n}\mbox{ with }S\in SL(n,\mathbb{C})\\}$. ∎

##  Appendix B Proof of Corollary 7

In this appendix, we first prove Corollary 7 using Proposition 12 and then we
provide an alternative proof of Corollary 7 for $n$ qubits, which does not
require Proposition 6. Let us first restate Corollary 7 here.

######  Corollary 7.

Every $n$-qudit SLOCC class that contains a fully entangled state
$|\psi\rangle$ which has a unitary symmetry group
$\mathcal{S}_{\psi}\subseteq\bigotimes_{i=1}^{n}U(d_{i},\mathbb{C})$ contains
weakly isolated states.

The proof follows immediately from Theorem 5, which states that weakly
isolated states exist in every $n$-qudit SLOCC class that contains a fully
entangled state with a compact symmetry group, and the following proposition.

######  Proposition 12.

Let $\mathcal{S}_{\psi}\subseteq\bigotimes_{i=1}^{n}U(d_{i},\mathbb{C})$ be
the symmetry group of a state $|\psi\rangle$. Then, $\mathcal{S}_{\psi}$ is
compact.

###### Proof.

The symmetry group of a state $|\psi\rangle$,
$\mathcal{S}_{\psi}=\\{S\in\bigotimes_{i=1}^{n}U(d_{i},\mathbb{C}):(S-\text{1})|\psi\rangle=0\\}$,
is the intersection of the compact group
$\bigotimes_{i=1}^{n}U(d_{i},\mathbb{C})$ and a Zariski closed set, namely the
set of all solutions to the finite set of complex polynomial equations given
above. As Zariski closed sets are closed in the Euclidean topology
footnote:ZariskiEuclidean  and $\mathcal{S}_{\psi}$ is bounded since it is a
subgroup of a compact group, we have that $\mathcal{S}_{\psi}$ is also
compact. ∎

For the alternative proof restricted to $n$-qubit cases, we will need the
following lemmata. Our first step is to show that every local symmetry group
corresponding to a unitary symmetry group must be closed. Note that
considering the local symmetry group on a single site, $j$, one can always
choose its elements in $SU(d_{j},\mathbb{C})$, i.e.,
$\mathcal{S}_{\psi}^{(j)}=\\{S^{(j)}\in
SU(d_{j},\mathbb{C}):\exists\;S^{(i)}\;\forall\;i\in[n]\setminus\\{j\\}\text{
so that }\bigotimes_{k=1}^{n}S^{(k)}\in\mathcal{S}_{\psi}\\}$. Then, we have
the following lemma.

######  Lemma 13.

Let $\mathcal{S}_{\psi}\subseteq\bigotimes_{i=1}^{n}U(d_{i},\mathbb{C})$ be
the symmetry group of a state $|\psi\rangle$. Then, the local symmetry group
$\mathcal{S}_{\psi}^{(j)}\in SU(d_{j},\mathbb{C})$ is closed for any choice of
$j\in[n]$.

###### Proof.

We will prove the statement by contradiction. By Proposition 12,
$\mathcal{S}_{\psi}$ is compact. Assume that $\mathcal{S}_{\psi}^{(j)}$ is not
closed for some $j$. Then there exists a convergent sequence
$\\{S_{k}^{(j)}\\}_{k\in\mathbb{N}}\subseteq\mathcal{S}_{\psi}^{(j)}$ such
that $\lim_{k\rightarrow\infty}S_{k}^{(j)}\not\in\mathcal{S}_{\psi}^{(j)}$. As
$\mathcal{S}_{\psi}$ is compact, the corresponding sequence
$\\{\bigotimes_{l=1}^{n}S_{k}^{(l)}\\}_{k\in\mathbb{N}}\subseteq\mathcal{S}_{\psi}$
is bounded. By the Bolzano-Weierstrass theorem BolzanoWeierstrass ;
footnote:BW , there exists a convergent subsequence
$\\{\bigotimes_{l=1}^{n}\tilde{S}_{k}^{(l)}\\}_{k\in\mathbb{N}}\subset\\{\bigotimes_{l=1}^{n}S_{k}^{(l)}\\}_{k\in\mathbb{N}}$
and its limit must be an element in $\mathcal{S}_{\psi}$ since this set is
closed. Here, we choose the phase of $\tilde{S}_{k}^{(j)}$ such that
$\tilde{S}_{k}^{(j)}$ is an element of $\\{S_{k}^{(j)}\\}_{k\in\mathbb{N}}$
for each $k$. Since the limit of a convergent sequence
($\\{S_{k}^{(j)}\\}_{k\in\mathbb{N}}$) equals to the limit of any subsequence,
we have
$\lim_{k\rightarrow\infty}\tilde{S}_{k}^{(j)}=\lim_{k\rightarrow\infty}S_{k}^{(j)}$,
which leads to a contradiction as the LHS is an element of
$\mathcal{S}_{\psi}^{(j)}$, whereas the RHS is not. Therefore,
$\mathcal{S}_{\psi}^{(j)}$ is closed for any choice of $j\in[n]$. ∎

Due to the argument in the proof of Theorem 5, it remains to consider SLOCC
classes that have a state $|\psi\rangle$ with an infinite symmetry group
$\mathcal{S}_{\psi}$. A simple counting argument shows that
$\mathcal{S}_{\psi}$ is infinite if and only if the local symmetry group
$\mathcal{S}_{\psi}^{(j)}$ is infinite for at least one $j\in[n]$. Due to
Lemma 13 and the fact that the only infinite closed subgroups of
$SU(2,\mathbb{C})$ are up to unitary conjugation
$\mathcal{G}_{1}=\\{diag(e^{i\theta},e^{-i\theta})\\}_{\theta\in[0,2\pi)}$,
$\mathcal{G}_{2}=\mathcal{G}_{1}\cup\left\\{\begin{pmatrix}0&-e^{i\varphi}\\\
e^{-i\varphi}&0\end{pmatrix}\right\\}_{\varphi\in[0,2\pi)}$, and
$SU(2,\mathbb{C})$ (see Sec. III of Ref. ClosedSubgroupSU2 ), we deduce that
$\mathcal{S}_{\psi}^{(j)}$ is, up to unitary conjugation, one of these three
subgroups of $SU(2,\mathbb{C})$ for at least one $j$.

With the help of the subsequent lemma, we will now show that there exists no
SLOCC class with an infinite unitary local stabilizer. In particular, we show
that any of these local symmetry groups must also contain matrices in
$SL(2,\mathbb{C})\setminus SU(2,\mathbb{C})$. Hence, there exists no SLOCC
class with unitary stabilizer without weak isolation. Without loss of
generality, we can assume that
$\mathcal{S}_{\psi}^{(1)}\in\\{\mathcal{G}_{1},\mathcal{G}_{2},SU(2,\mathbb{C})\\}$.
Since $\mathcal{G}_{1}\subset\mathcal{G}_{2}\subset SU(2,\mathbb{C})$, it
suffices to consider unitary symmetries $\bigotimes_{i=1}^{n}U_{i}$ with
$U_{1}=diag(e^{i\theta},e^{-i\theta})$ for all $\theta\in[0,2\pi)$ for the
rest of the proof footnote:LUequiv .

######  Lemma 14.

If an $n$-qubit state $|\psi\rangle$ has a unitary symmetry
$U(\theta)=e^{-i\varphi}diag(e^{i\theta},e^{-i\theta})\otimes_{i=2}^{n}U_{i}\in
U(2,\mathbb{C})^{\otimes n}$ for $\theta=x\pi$ where
$x\in(0,2)\cap(\mathbb{R}\setminus\mathbb{Q})$, then $|\psi\rangle$ also has a
symmetry $X(z)=w(z)diag(e^{z\theta},e^{-z\theta})\otimes_{i=2}^{n}X_{i}(z)$
where $w(z)\in\mathbb{C}$ and $X_{i}(z)\in SL(2,\mathbb{C})$ for every
$z\in\mathbb{C}$.

Let us first sketch the proof of Lemma 14 here and then give the complete
proof once we prove Lemma 15. The first step is to consider the spectral
decomposition of $U_{j}$, which is a tensor factor of the unitary
$U(\theta)\in U(2,\mathbb{C})^{\otimes n}$, as
$U_{j}=e^{i\alpha_{j}}|v_{j}^{(0)}\rangle\langle
v_{j}^{(0)}|+e^{-i\alpha_{j}}|v_{j}^{(1)}\rangle\langle v_{j}^{(1)}|$ and
determine how $\\{\alpha_{j}\\}_{j=2}^{n}$ depend on $\varphi$ and $\theta$
for $U(\theta)$ to be a symmetry of $|\psi\rangle$. Then, we construct the
operators $X(z)\in GL(2,\mathbb{C})^{\otimes n}$ with
$w(z)=e^{-z\varphi^{\prime}}$ and
$X_{j}(z)=e^{z\alpha^{\prime}_{j}}|v_{j}^{(0)}\rangle\langle
v_{j}^{(0)}|+e^{-z\alpha^{\prime}_{j}}|v_{j}^{(1)}\rangle\langle v_{j}^{(1)}|$
where $z\in\mathbb{C}\setminus i\mathbb{R}$ and
$(\varphi^{\prime},\\{\alpha^{\prime}_{j}\\}_{j=2}^{n})$ are related to
$(\varphi,\\{\alpha_{j}\\}_{j=2}^{n})$ by being solutions to two systems of
linear equations that share the same coefficient matrix (see Lemma 15).
Finally, we show that $X(z)$ is a symmetry of $|\psi\rangle$ by using the fact
that $(\varphi^{\prime},\\{\alpha^{\prime}_{j}\\}_{j=2}^{n})$ solves the
system of linear equations which corresponds to
$X(z)|\psi\rangle=|\psi\rangle$. Note that we will use $\vec{1}_{d}$ to denote
a $d$-dimensional vector with all entries being $1$.

######  Lemma 15.

Let $\theta=x\pi$ where $x\in(0,2)\cap(\mathbb{R}\setminus\mathbb{Q})$. If
there exists a solution $\vec{\alpha}\in\mathbb{R}^{L}$ to the matrix equation
$\mathbf{B}\vec{\alpha}=\vec{\varphi}+\vec{\theta}+2\pi\vec{m}$ where
$\mathbf{B}$ is a $(d_{1}+d_{2})\times L$ matrix with rational entries,
$\vec{\varphi}=\varphi\cdot\vec{1}_{d_{1}+d_{2}}$, $\varphi\in\mathbb{R}$,
$\vec{\theta}=\theta\cdot((-\vec{1}_{d_{1}})\oplus\vec{1}_{d_{2}})$, and
$\vec{m}\in\mathbb{Z}^{d_{1}+d_{2}}$, then there exists a choice for
$\varphi^{\prime}\in\mathbb{R}$ such that the equation
$\mathbf{B}\vec{\alpha^{\prime}}=\vec{\varphi^{\prime}}+\vec{\theta}$ where
$\vec{\varphi^{\prime}}=\varphi^{\prime}\cdot\vec{1}_{d_{1}+d_{2}}$ has a real
solution $\vec{\alpha^{\prime}}\in\mathbb{R}^{L}$.

###### Proof.

Consider a system of equations represented by the matrix equation
$\mathbf{B}\vec{\alpha}=\vec{\varphi}+\vec{\theta}+2\pi\vec{m}$. While
allowing any free choice of $\varphi\in\mathbb{R}$ and
$\vec{m}\in\mathbb{Z}^{d_{1}+d_{2}}$, we determine the necessary and
sufficient conditions for the system of equations to be consistent (and,
hence, admits a real solution). We then choose $\varphi^{\prime}\in\mathbb{R}$
so that the above conditions imply that the system
$\mathbf{B}\vec{\alpha^{\prime}}=\vec{\varphi^{\prime}}+\vec{\theta}$ is
consistent as well. Notice that a system with a matrix of coefficients given
by $\mathbf{B}$ is always consistent if all the rows of this matrix are
linearly independent, so we only need to analyze the situation when this is
not the case.

By performing Gaussian elimination on the augmented matrix corresponding to
$\mathbf{B}\vec{\alpha}=\vec{\varphi}+\vec{\theta}+2\pi\vec{m}$, the system is
consistent if and only if for each row of the resulting matrix where the first
$L$ entries are all zeros, the last entry must also be zero. We will call
these conditions “consistency conditions”. Given that $\mathbf{B}$ is a
rational matrix, they take the form

$x_{1}^{(\lambda)}\varphi+x_{2}^{(\lambda)}\theta+2\pi\sum_{j}y_{j}^{(\lambda)}m_{j}=0$
(15)

where $x_{1}^{(\lambda)},x_{2}^{(\lambda)},y_{j}^{(\lambda)}\in\mathbb{Q}$ for
all $j\in\\{1,\ldots,d_{1}+d_{2}\\}$, $\lambda$ labels the different
consistency conditions, and $m_{j}$ is the $j$-th entry of $\vec{m}$. Notice
that $\theta$ being an irrational multiple of $\pi$ imposes that
$f=\frac{x_{2}^{(\lambda)}}{x_{1}^{(\lambda)}}\in\mathbb{Q}$ is a constant
independent of $\lambda$ for all the consistency conditions for which it does
not hold that $x_{1}^{(\lambda)}=x_{2}^{(\lambda)}=0$ footnote:consistencyCond
.

We consider now the system
$\mathbf{B}\vec{\alpha^{\prime}}=\vec{\varphi^{\prime}}+\vec{\theta}$ and look
for a choice of $\varphi^{\prime}\in\mathbb{R}$ that makes it consistent under
the assumption that all the previous consistency conditions (15) are met. The
consistency conditions are now

$x_{1}^{(\lambda)}\varphi^{\prime}+x_{2}^{(\lambda)}\theta=0$ (16)

where the coefficients $\\{x_{1}^{(\lambda)},x_{2}^{(\lambda)}\\}$ are the
same as before. Observe first that if the previous system had consistency
conditions of the form $x_{1}^{(\lambda)}=x_{2}^{(\lambda)}=0$, these are
trivially met here for any choice of $\varphi^{\prime}$. On the other hand,
all remaining conditions in Eq. (16) are fulfilled by choosing
$\varphi^{\prime}=-f\theta$. ∎

We are now in the position to prove Lemma 14.

###### Proof of Lemma 14.

Let $|\psi\rangle=|0\rangle_{1}|\phi_{0}\rangle_{2\ldots
n}+|1\rangle_{1}|\phi_{1}\rangle_{2\ldots n}$ be an $n$-qubit state where
$|\phi_{0}\rangle,|\phi_{1}\rangle\in(\mathbb{C}^{2})^{\otimes(n-1)}$ are both
unnormalized. Suppose that
$U(\theta)=e^{-i\varphi}diag(e^{i\theta},e^{-i\theta})\otimes_{j=2}^{n}U_{j}\in
U(2,\mathbb{C})^{\otimes n}$ with
$U_{j}=e^{i\alpha_{j}}|v_{j}^{(0)}\rangle\langle
v_{j}^{(0)}|+e^{-i\alpha_{j}}|v_{j}^{(1)}\rangle\langle v_{j}^{(1)}|$, where
$\theta=x\pi$ with $x\in(0,2)\cap(\mathbb{R}\setminus\mathbb{Q})$,
$\varphi,\alpha_{j}\in\mathbb{R}$, and $\\{|v_{j}^{(k)}\rangle\\}_{k=0}^{1}$
are orthonormal for all $j\in\\{2,\ldots,n\\}$, is a symmetry of
$|\psi\rangle$ (i.e., $U(\theta)|\psi\rangle=|\psi\rangle$). This means that

$\bigotimes_{j=2}^{n}\left(\sum_{k=0}^{1}e^{i(-1)^{k}\alpha_{j}}|v_{j}^{(k)}\rangle\langle
v_{j}^{(k)}|\right)|\phi_{l}\rangle=e^{i(\varphi-(-1)^{l}\theta)}|\phi_{l}\rangle$
(17)

has to hold for both $l=0$ and 1. These two equations are satisfied if and
only if

$|\phi_{l}\rangle=\sum_{\vec{k}\in\\{0,1\\}^{n-1}}c_{\vec{k}}^{(l)}\;\bigotimes_{j=2}^{n}|v_{j}^{(k_{j})}\rangle$
(18)

where $c_{\vec{k}}^{(l)}\in\mathbb{C}$ and $c_{\vec{k}}^{(l)}\neq 0$ only for
those $\vec{k}$, for which it holds that
$e^{i\sum_{j=2}^{n}(-1)^{k_{j}}\alpha_{j}}=e^{i(\varphi-(-1)^{l}\theta)}$ for
$l=0$ and 1, which is equivalent to

$\sum_{j=2}^{n}(-1)^{k_{j}}\alpha_{j}=\varphi-(-1)^{l}\theta+2\pi m$ (19)

for some $m\in\mathbb{Z}$. The collection of all constraints on
$\\{\alpha_{j}\\}_{j=2}^{n}$ [in the form of Eq. (19)] associated to all
$c_{\vec{k}}^{(l)}\neq 0$ for $l=0$ and 1 form a system of linear equations of
the form, $\mathbf{B}\vec{\alpha}=\vec{\varphi}+\vec{\theta}+2\pi\vec{m}$
where $\mathbf{B}$ is a $(d_{1}+d_{2})\times(n-1)$ matrix with entries either
$1$ or $-1$, $d_{l+1}$ is the number of $c_{\vec{k}}^{(l)}\neq 0$,
$\vec{\alpha}=(\alpha_{2},\ldots,\alpha_{n})^{T}$, and $\vec{\varphi}$,
$\vec{\theta}$, and $\vec{m}$ are defined in the same way as in Lemma 15.

As $U(\theta)$ is unitary, the equation
$\mathbf{B}\vec{\alpha}=\vec{\varphi}+\vec{\theta}+2\pi\vec{m}$ must have a
real solution. Then, by Lemma 15,
$\mathbf{B}\vec{\alpha^{\prime}}=\vec{\varphi^{\prime}}+\vec{\theta}$ must
also have a real solution $\vec{\alpha^{\prime}}\in\mathbb{R}^{n-1}$, which
implies that footnote:how_to_get_exp_equal

$e^{\sum_{j=2}^{n}(-1)^{k_{j}}z\alpha^{\prime}_{j}}=e^{z(\varphi^{\prime}-(-1)^{l}\theta)}$
(20)

must hold for all $z\in\mathbb{C}$, $\vec{k}\in\\{0,1\\}^{n-1}$ and
$l\in\\{0,1\\}$ such that $c_{\vec{k}}^{(l)}\neq 0$.

Now, let $X(z)=w(z)diag(e^{z\theta},e^{-z\theta})\otimes_{i=2}^{n}X_{i}(z)$
where $w(z)=e^{-z\varphi^{\prime}}$,
$X_{j}(z)=e^{z\alpha^{\prime}_{j}}|v_{j}^{(0)}\rangle\langle
v_{j}^{(0)}|+e^{-z\alpha^{\prime}_{j}}|v_{j}^{(1)}\rangle\langle
v_{j}^{(1)}|$, and $z\in\mathbb{C}$. Note that $X(z)\in
U(2,\mathbb{C})^{\otimes n}$ if and only if $Re(z)=0$. Using the same methods
as above, it is straightforward to see that $X(z)|\psi\rangle=|\psi\rangle$
for all $z\in\mathbb{C}$. Hence, there also exist non-unitary symmetries,
which completes the proof. ∎

We are now in the position to present an alternative proof of Corollary 7 for
$n$ qubits. From the argument above, SLOCC classes that have a state
$|\psi\rangle$ with an infinite symmetry group $\mathcal{S}_{\psi}\subseteq
U(2,\mathbb{C})^{\otimes n}$ means that there exists an LU-equivalent state
$|\phi\rangle$ such that
$\mathcal{S}_{\phi}^{(j)}\in\\{\mathcal{G}_{1},\mathcal{G}_{2},SU(2,\mathbb{C})\\}$
for at least one $j\in[n]$. We will now prove that no such class exists.
Recall that, without loss of generality, we can take
$\mathcal{S}_{\phi}^{(1)}\in\\{\mathcal{G}_{1},\mathcal{G}_{2},SU(2,\mathbb{C})\\}$
and
$U(\theta)=e^{-i\varphi}diag(e^{i\theta},e^{-i\theta})\otimes_{i=2}^{n}U_{i}\in\mathcal{S}_{\phi}$
where $\theta=x\pi$ and $x\in(0,2)\cap(\mathbb{R}\setminus\mathbb{Q})$. By
Lemma 14, the symmetry group $\mathcal{S}_{\phi}$ must also contain $X(z)\in
GL(2,\mathbb{C})^{\otimes n}$ for $z\in\mathbb{C}$ such that $Re(z)\neq 0$,
which are not in $U(2,\mathbb{C})^{\otimes n}$. Hence, these classes with an
infinite stabilizer $\mathcal{S}_{\phi}\subseteq U(2,\mathbb{C})^{\otimes n}$
do not exist. In particular, all $n$-qubit SLOCC classes that have a state
$|\psi\rangle$ with a symmetry group $\mathcal{S}_{\psi}\subseteq
U(2,\mathbb{C})^{\otimes n}$ contain weakly isolated states.

##  Appendix C Proof of Theorem 9

In this appendix, we prove Theorem 9 which is stated in Sec. III.2. We restate
the theorem here for clarity.

######  Theorem 9.

Let $|\Psi_{s}\rangle$ be an $n$-qudit state $(n\geq 4)$ with a symmetry group
$\mathcal{S}_{\Psi_{s}}\subseteq\\{S^{\otimes n}\mbox{ with }S\in
GL(d,\mathbb{C})\\}$ with $d\geq 2$ arbitrary. Then, there exist weakly
isolated states within the SLOCC class of $|\Psi_{s}\rangle$.

###### Proof.

Suppose that the state $|\Psi_{s}\rangle$ has symmetries only of the form
$S^{\otimes n}$. We will show that a weakly isolated state in such an SLOCC
class can be constructed as
$\bigotimes_{i=1}^{n-1}g_{i}\otimes\text{1}|\Psi_{s}\rangle$. Here, the
matrices $g_{i}$ are chosen such that the only matrices $S\in
GL(d,\mathbb{C})$ that satisfy $S^{\dagger}G_{i}S\propto G_{i}$ for $n-1$
sites are $S\propto\text{1}$ (see Lemma 3). We divide this proof into two
cases: (a) $n\geq 5$ and (b) $n=4$. Case (a) has been proven in Ref.
OurSymmPaper  footnote:WI_n>=5 . The main idea there was to choose
$G_{i}\not\propto\text{1}$ for $i=1,2,3$ and $g_{n-1}=\text{1}$ so that for
any $S$ which quasi-commute with $G_{i}$ on $n-1$ sites, it must quasi-commute
with at least two $G_{i}\not\propto\text{1}$ and at least one
$G_{j}=\text{1}$. The last condition implies that $S$ is proportional to a
unitary. Appropriate choices of the eigenbases of
$G_{1},G_{2},G_{3}\not\propto\text{1}$ force $S\propto\text{1}$. More
precisely, $G_{1},G_{2}$ and $G_{3}$ are chosen to be non-degenerate and each
of these operators has an eigenvector that is non-orthogonal to all
eigenvectors of the other two operators. The fact that $S$ has to commute with
at least two of these operators implies that $S\propto\text{1}$. For case (b),
where $n=4$, such a construction is no longer possible, because there exists
$S\not\propto\text{1}$ that can quasi-commute with $G_{1}\not\propto\text{1}$
and $G_{3}=G_{4}=\text{1}$. However, as we will show next, there also exist
weakly isolated states.

Similar to case (a), we show that the state $g_{1}\otimes\ldots\otimes
g_{4}|\Psi_{s}\rangle$, with $g_{i}$ defined below is weakly isolated. Let
$\\{|v^{(1)}_{k}\rangle=|k\rangle\\}_{k=0}^{d-1}$,
$\\{|v^{(2)}_{k}\rangle=U_{F}|k\rangle\\}_{k=0}^{d-1}$ and
$\\{|v^{(3)}_{k}\rangle=\widetilde{U}U_{F}|k\rangle\\}_{k=0}^{d-1}$ (where
$U_{F}=\frac{1}{\sqrt{d}}\sum_{j,k=0}^{d-1}\omega^{jk}|j\rangle\langle k|$,
$\widetilde{U}=|0\rangle\langle
0|+\sum_{k=1}^{d-1}\sqrt{\omega}|k\rangle\langle k|$ and
$\omega=e^{i\frac{2\pi}{d}}$) be three orthonormal bases. One can easily
verify that they are three mutually non-orthogonal bases for any $d\geq 2$. We
choose $g_{i}=\sqrt{G_{i}}$ where $G_{1}=\sum_{j=0}^{d-1}r^{j}|j\rangle\langle
j|$ for any $0<r<1$,
$G_{i}=\sum_{j=0}^{d-1}(1-\epsilon)^{j}|v^{(i)}_{j}\rangle\langle
v^{(i)}_{j}|$ for $i=2,3$ and $0<\epsilon\ll 1$, and $G_{4}=\text{1}$.

To prove that the state is weakly isolated, we must prove that no
$S\not\propto\text{1}$ can satisfy $S^{\dagger}G_{i}S\propto G_{i}$ for any
three sites. We distinguish the cases: (i) site 4 and any two sites from
$\\{1,2,3\\}$ and (ii) sites 1,2 and 3. In case (i), the proof in case (a)
holds directly since the $G_{i}$ defined here correspond to special instances
of the weakly isolated states constructed for $n\geq 5$ (by ignoring $G_{4\leq
i\leq n-1}$ in the construction for $n\geq 5$). In case (ii), we use
Observation 1 in Ref. OurSymmPaper  and $S^{\dagger}G_{1}S\propto G_{1}$ to
get $S\propto\sqrt{G_{1}}^{-1}U\sqrt{G_{1}}$ for some unitary $U$.
Substituting this expression for $S$ into $S^{\dagger}G_{i}S\propto G_{i}$ and
using that traces of both sides of this equation must coincide, we obtain
$[\sqrt{G_{1}}^{-1}G_{i}\sqrt{G_{1}}^{-1},U]=0$. We now define
$Y=\sqrt{G_{1}}^{-1}G_{2}\sqrt{G_{1}}^{-1}$ and
$Z=\sqrt{G_{1}}^{-1}G_{3}\sqrt{G_{1}}^{-1}$ which are both positive definite.
To prove that only $U\propto\text{1}$ can satisfy both $[Y,U]=0$ and
$[Z,U]=0$, we follow the same argument for proving $S\propto\text{1}$ if
$[S,G_{i}]=0$ for $i=1,2$ in case (a). It remains to show that (1) $Y$ and $Z$
have non-degenerate eigenvalues and (2) one of the eigenvectors of $Y$ is non-
orthogonal to all eigenvectors of $Z$. Despite the proof being
straightforward, we present it here for completeness. Since both $G_{1}$ and
$\widetilde{U}$ are diagonal and
$G_{3}=\widetilde{U}G_{2}\widetilde{U}^{\dagger}$, we have that
$Z=\widetilde{U}Y\widetilde{U}^{\dagger}$. In particular, condition (1) is
satisfied if $Y$ is non-degenerate. To prove that $Y$ and $Z$ satisfy
conditions (1) and (2), we begin by writing $Y$ in the computational basis as

$\displaystyle Y$
$\displaystyle=\frac{1}{d}\sum_{j,k,m=0}^{d-1}\frac{(1-\epsilon)^{m}\omega^{m(j-k)}}{\sqrt{r^{j+k}}}|j\rangle\langle
k|$
$\displaystyle=\frac{1-(1-\epsilon)^{d}}{d}\sum_{j,k=0}^{d-1}\frac{1}{\sqrt{r^{j+k}}(1-(1-\epsilon)\omega^{(j-k)})}|j\rangle\langle
k|$
$\displaystyle\propto\sum_{j,k=0}^{d-1}\frac{\epsilon}{\sqrt{r^{j+k}}(1-(1-\epsilon)\omega^{(j-k)})}|j\rangle\langle
k|$ (21) $\displaystyle\equiv H_{0}+\epsilon V(\epsilon)$ (22)

where we use the geometric series in the second line to sum over the index $m$
and define a diagonal matrix
$H_{0}=\sum_{k=0}^{d-1}\frac{1}{r^{k}}|k\rangle\langle k|$ and an off-diagonal
matrix $V(\epsilon)=\sum_{j\neq
k}\frac{1}{\sqrt{r^{j+k}}(1-(1-\epsilon)\omega^{(j-k)})}|j\rangle\langle k|$.
Note that the proportionality factor $\frac{1-(1-\epsilon)^{d}}{\epsilon d}$
omitted in Eq. (21) is well defined for $\epsilon>0$. By setting $\epsilon$
sufficiently small, we can use non-degenerate perturbation theory Sakurai  to
find the eigenvalues and eigenvectors of $H_{0}+\epsilon V(\epsilon)\propto
Y$.

Since $V(\epsilon)$ depends on the perturbation parameter $\epsilon$, we need
to Taylor expand each matrix element around $\epsilon=0$, which results in
$V(\epsilon)=\sum_{m=0}^{\infty}\epsilon^{m}V^{(m)}$ where

$\displaystyle V^{(m)}=\sum_{j\neq
k}\frac{(-1)^{m}\omega^{m(j-k)}}{\sqrt{r^{j+k}}(1-\omega^{j-k})^{m+1}}|j\rangle\langle
k|.$ (23)

The solution to the eigenvalue equation

$\displaystyle(H_{0}+\epsilon V(\epsilon))|e_{p}\rangle=E_{p}|e_{p}\rangle$
(24)

with $E_{p}=\sum_{m=0}^{\infty}\epsilon^{m}E^{(m)}_{p}$ and
$|e_{p}\rangle=\sum_{m=0}^{\infty}\epsilon^{m}|p^{(m)}\rangle$
footnote:pert_converge  is given (up to first order) by
$E^{(0)}_{p}=\frac{1}{r^{p}}$, $|p^{(0)}\rangle=|p\rangle$,
$E^{(1)}_{p}=\langle p^{(0)}|V^{(0)}|p^{(0)}\rangle=0$, and

$\displaystyle|p^{(1)}\rangle=\sum_{j\neq
p}\frac{\sqrt{r}^{j+p}}{(r^{j}-r^{p})(1-\omega^{j-p})}|j\rangle$ (25)

for $p\in\\{0,\ldots,d-1\\}$. Thus, the eigenvalues
$E_{p}=\frac{1}{r^{p}}+\mathcal{O}(\epsilon^{2})$ are non-degenerate if
$0<\epsilon\ll 1$ is sufficiently small footnote:pert . This shows that the
eigenvalues of $Y\propto H_{0}+\epsilon V(\epsilon)$ and $Z$ are also non-
degenerate [satisfying condition (1)].

It remains to show that one of the eigenvectors of $Y$ is non-orthogonal to
all eigenvectors of $Z$. Using the fact that
$Z=\widetilde{U}Y\widetilde{U}^{\dagger}$ and $|e_{p}\rangle$ are the
eigenvectors of $Y$, the task becomes proving that for at least one $q$,
$\langle e_{p}|\widetilde{U}|e_{q}\rangle\neq 0$ for all
$p\in\\{0,\ldots,d-1\\}$. Let
$|e_{p}\rangle=\sum_{j=0}^{d-1}e^{(j)}_{p}|j\rangle$, then

$\displaystyle\langle e_{p}|\widetilde{U}|e_{q}\rangle$
$\displaystyle=e^{(0)*}_{p}e^{(0)}_{q}+\sqrt{\omega}\sum_{j=1}^{d-1}e^{(j)*}_{p}e^{(j)}_{q}$
(26)
$\displaystyle=(1-\sqrt{\omega})e^{(0)*}_{p}e^{(0)}_{q}+\sqrt{\omega}\delta_{p,q}.$
(27)

For $p=q$, this overlap is non-vanishing as $|e^{(0)}_{p}|^{2}$ is real and
$\frac{-\sqrt{\omega}}{1-\sqrt{\omega}}$ is complex. For all $p\neq q$,
$e^{(0)*}_{p}e^{(0)}_{q}\neq 0$ if and only if $e^{(0)}_{p}=\langle
0|e_{p}\rangle\neq 0\;\forall\;p$. From perturbation theory (up to first
order), we find that

$\displaystyle\langle
0|e_{p}\rangle=\begin{cases}1+\mathcal{O}(\epsilon^{2}),\qquad\qquad\qquad\;\text{if
}p=0,\\\
\frac{\epsilon\sqrt{r}^{p}}{(1-r^{p})(1-\omega^{-p})}+\mathcal{O}(\epsilon^{2}),\quad\text{if
}p\neq 0\end{cases}$ (28)

which are non-zero for all $p\in\\{0,\ldots,d-1\\}$ if $0<\epsilon\ll 1$ is
sufficiently small while keeping the eigenvalues $\\{E_{p}\\}$ non-degenerate
footnote:Ep_non-deg . Hence, by choosing $\epsilon$ to be sufficiently small,
conditions (1) and (2) are satisfied, so only $U\propto\text{1}$ can satisfy
$[Y,U]=[Z,U]=0$ $\Rightarrow
S\propto\sqrt{G_{1}}^{-1}U\sqrt{G_{1}}\propto\text{1}$, which completes the
proof for case (b). ∎

##  Appendix D Proof of Observation 11

In this appendix, we will prove Observation 11 which is stated in Sec. IV. We
restate it here for clarity.

######  Observation 11.

Any state in the SLOCC class of $|A_{3}\rangle$ is either LU equivalent to or
can be reached from a state in one of the following three sets of states (up
to LU) via LOCC in at most 3 rounds:

  1. (i)

$|A_{3}\rangle$,

  2. (ii)

$|\psi_{2}(\alpha_{1};\alpha_{2})\rangle\propto\text{diag}(\sqrt{\alpha_{1}},1,1)\otimes\text{diag}(\sqrt{\alpha_{2}},1,1)\otimes\text{1}|A_{3}\rangle$
where $\alpha_{1},\alpha_{2}>0$, $\alpha_{1}\neq\alpha_{2}$ and
$\alpha_{1},\alpha_{2}\neq 1$,

  3. (iii)

$|\psi_{3}(\alpha_{1},\beta_{1};\alpha_{2},\beta_{2})\rangle\propto\text{diag}(\sqrt{\alpha_{1}},\sqrt{\beta_{1}},1)\otimes\text{diag}(\sqrt{\alpha_{2}},\sqrt{\beta_{2}},1)\otimes\text{1}|A_{3}\rangle$
where $\alpha_{1},\alpha_{2},\beta_{1},\beta_{2}>0$,
$\alpha_{1}\neq\beta_{1}\neq\beta_{2}\neq\alpha_{2}$,
$\alpha_{1}\neq\alpha_{2}$,
$\frac{\alpha_{1}}{\beta_{1}}\neq\frac{\alpha_{2}}{\beta_{2}}$ and
$\alpha_{1},\alpha_{2},\beta_{1},\beta_{2}\neq 1$.

Moreover, these states are not reachable via $\text{LOCC}_{\mathbb{N}}$.
Hence, the MES of $|A_{3}\rangle$ is contained in the union set of all these
states, which we call $M_{A_{3}}$.

###### Proof.

Recall from the proof of Theorem 8 that any state $|\phi\rangle$ in the SLOCC
class of $|A_{3}\rangle$ is equal to $h_{1}\otimes
h_{2}\otimes\text{1}|A_{3}\rangle$ with the normalization factor absorbed into
$h_{1}$ or $h_{2}$. Using the singular value decomposition of $h_{2}$,
$h_{2}=V\sqrt{D_{2}}U^{\dagger}$, where $D_{2}$ is positive diagonal, and the
fact that $U^{\otimes 3}$ is a symmetry of $|A_{3}\rangle$, we have that

$|\phi\rangle\stackrel{{\scriptstyle\text{LU}}}{{\cong}}h_{1}U\otimes\sqrt{D_{2}}\otimes\text{1}|A_{3}\rangle.$
(29)

Any of these states can be reached from a state
$\sqrt{D_{1}}\otimes\sqrt{D_{2}}\otimes\text{1}|A_{3}\rangle$ (up to LU) via
$\text{LOCC}_{1}$, where $D_{1}$ is positive diagonal. The transformation is
achieved if party 1 measures with the operators $\mathcal{K}_{D\rightarrow
ND}=\\{\frac{1}{\sqrt{3}}h_{1}U\sqrt{D_{1}}^{-1}Z^{j}\\}_{j=0,1,2}$ and each
of the remaining parties apply $Z^{j}$ depending on the measurement outcome
$j$ of party 1, where
$Z=\text{diag}(1,e^{i\frac{2\pi}{3}},e^{i\frac{4\pi}{3}})$. One can easily
verify that it suffices to take $D_{1}=\text{diag}(U^{\dagger}H_{1}U)$ so that
the operators $\mathcal{K}_{D\rightarrow ND}$ satisfy the completeness
relation for a measurement. Therefore, $|\phi\rangle$ is either LU equivalent
to or can be reached from

$\displaystyle\sqrt{D_{1}}\otimes\sqrt{D_{2}}\otimes\text{1}|A_{3}\rangle$
(30) $\displaystyle\propto$
$\displaystyle\text{\;diag}(\sqrt{\alpha_{1}},\sqrt{\beta_{1}},1)\otimes\text{diag}(\sqrt{\alpha_{2}},\sqrt{\beta_{2}},1)\otimes\text{1}|A_{3}\rangle$

where $\alpha_{i},\beta_{i}>0$ for $i=1,2$. It remains to show that the three
types of states (i)–(iii), which are special cases of Eq. (30), are not
$\text{LOCC}_{\mathbb{N}}$-reachable and all the remaining states of the form
in Eq. (30) can be reached from one of these three types of states with LOCC
in 2 steps.

We now prove that states of types (i)–(iii) are not
$\text{LOCC}_{\mathbb{N}}$-reachable by showing that condition (ii) in Theorem
2 is violated for all these cases. Clearly, for the type-(i) state
($D_{1}=D_{2}=\text{1}$), any $S\in SL(3,\mathbb{C})$ that quasi-commutes with
1 must quasi-commute with $D_{i}$ for all sites $i$. For type-(ii) states, any
$S\in SL(3,\mathbb{C})$ that quasi-commutes with both
$\text{diag}(\alpha,1,1)$ and $\text{diag}(\beta,1,1)$ for $\alpha\neq\beta$
must be of the form $e^{i\theta}\oplus U_{2}$ where $\theta\in\mathbb{R}$ and
$U_{2}\in U(2,\mathbb{C})$ footnote:QCwithDiag . Thus, any $S$ which quasi-
commutes with any two operators of the set
$\\{\text{diag}(\alpha_{1},1,1),\text{diag}(\alpha_{2},1,1),\text{1}\\}$ must
also quasi-commute with the remaining one. Lastly, for type-(iii) states, any
$S\in SL(3,\mathbb{C})$ that quasi-commutes with both
$\text{diag}(\alpha,\beta,1)$ and $\text{diag}(\gamma,\delta,1)$ where
$\frac{\alpha}{\gamma},\frac{\beta}{\delta}$, and 1 take three different
values must be unitary and diagonal footnote:QCwithDiag . Therefore, any $S$
that quasi-commutes with any two operators of the set
$\\{\text{diag}(\alpha_{1},\beta_{1},1),\text{diag}(\alpha_{2},\beta_{2},1),\text{1}\\}$
must also quasi-commute with the remaining one. Thus, all states of types
(i)–(iii) are not $\text{LOCC}_{\mathbb{N}}$-reachable. Since, by definition,
a state is in the MES if and only if it is not reachable via LOCC, the MES of
the $|A_{3}\rangle$ class is contained in the union set of these three types
of states.

Finally, we have to show that all the states in the form of Eq. (30) but not
LU equivalent to type (i), (ii) or (iii) are reachable from one of these three
types of states via LOCC within 2 steps. The first case to consider is when
$\frac{\alpha_{1}}{\beta_{1}}\neq\frac{\alpha_{2}}{\beta_{2}}$ and the set
$\\{\alpha_{1},\alpha_{2},\beta_{1},\beta_{2}\\}$ has at least one element
equal to 1. All states in this case are, up to LU (by exchanging local basis
$|0\rangle\leftrightarrow|1\rangle$ in all three systems) and systems
permutation, equivalent to the state with

$D_{1}=\text{diag}(\alpha,\beta,1)\text{ and }D_{2}=\text{diag}(\delta,1,1).$
(31)

We can further divide this case into the following:

  1. (a)

The state $\sqrt{D_{1}}\otimes\text{1}^{\otimes 2}|A_{3}\rangle$ (with
$\delta=1$) can be obtained from $|A_{3}\rangle$ [type (i)] with a protocol
similar to the one above where party 1 measures with operators
$\\{\frac{1}{\sqrt{\text{Tr}(D_{1})}}\sqrt{D_{1}}X^{j}\\}_{j=0,1,2}$ with
$X=\sum_{i=0}^{2}|(i+1)\text{mod}\;3\rangle\langle i|$ and parties 2 and 3
each apply $X^{j}$ depending on the measurement outcome $j$.

  2. (b)

For $\delta\neq\frac{2\alpha}{\beta+1}$ and $\beta,\delta\neq 1$, the state
can be obtained from a type-(ii) state
$|\psi_{2}(\alpha_{1};\alpha_{2})\rangle$ with
$\alpha_{1}=\frac{2\alpha}{\beta+1}$ and $\alpha_{2}=\delta$ if party 1
measures with operators
$\\{\frac{1}{\sqrt{\beta+1}}\sqrt{D_{1}}\tilde{U}^{j}\text{diag}(\frac{1}{\sqrt{\alpha_{1}}},1,1)\\}_{j=0,1}$
where $\tilde{U}=1\oplus\begin{pmatrix}0&-1\\\ 1&0\end{pmatrix}$ and parties 2
and 3 each apply $\tilde{U}^{j}$ depending on the measurement outcome $j$.

  3. (c)

For $\delta=\frac{2\alpha}{\beta+1}$ and $\beta,\delta\neq 1$, the state can
be obtained from $|A_{3}\rangle$ in 2 steps: (1) Convert $|A_{3}\rangle$ into
the state
$\text{1}\otimes\text{1}\otimes\text{diag}(\sqrt{\frac{\beta+1}{2\alpha}},1,1)|A_{3}\rangle\propto\text{diag}(\sqrt{\frac{2\alpha}{\beta+1}},1,1)^{\otimes
2}\otimes\text{1}|A_{3}\rangle$ with the protocol in (a) but having the roles
of parties 1 and 3 reversed and $D_{1}$ in the measurement operators replaced
by $\text{diag}(\frac{\beta+1}{2\alpha},1,1)$. (2) Party 1 follows protocol
(b) to obtain $\sqrt{D_{1}}\otimes\sqrt{D_{2}}\otimes\text{1}|A_{3}\rangle$.

  4. (d)

For all other values of $\alpha,\beta,\delta>0$, the state is either a
type-(ii) state or can be reached from $|A_{3}\rangle$ with the protocol in
(a) having the roles of parties 1 and 2 (or 3) reversed.

The second case to consider is when
$\frac{\alpha_{1}}{\beta_{1}}\neq\frac{\alpha_{2}}{\beta_{2}}$ and the
collection $\\{\alpha_{1},\alpha_{2},\beta_{1},\beta_{2}\\}$ has no elements
equal to 1. All states belonging to this case are defined by
$D_{1}=\text{diag}(\alpha_{1},\beta_{1},1)$ and
$D_{2}=\text{diag}(\alpha_{2},\beta_{2},1)$. In fact, they all correspond to
the previous case described by Eq. (31) up to LU and permutation of systems,
except for $\alpha_{1}\neq\beta_{1}\neq\beta_{2}\neq\alpha_{2}$ and
$\alpha_{1}\neq\alpha_{2}$, which correspond to the type-(iii) states.

The last case to consider is when
$\frac{\alpha_{1}}{\beta_{1}}=\frac{\alpha_{2}}{\beta_{2}}=\frac{1}{k}$ which
corresponds to the states with $D_{1}=\text{diag}(\alpha,k\alpha,1)$ and
$D_{2}=\text{diag}(\beta,k\beta,1)$. Using the symmetry
$\frac{\sqrt{D_{1}}^{\otimes 3}}{\det\sqrt{D_{1}}}$, it is easy to see that
these states are proportional to
$\text{1}\otimes\text{diag}(1,1,\sqrt{\frac{\alpha}{\beta}})\otimes\text{diag}(1,\frac{1}{\sqrt{k}},\sqrt{\alpha})|A_{3}\rangle$
which are equivalent to the case described by Eq. (31) up to LU and
permutation of systems. This completes the proof. ∎

We have shown that every state in the SLOCC class of $|A_{3}\rangle$ can be
reached from a state in $M_{A_{3}}$ within 3 rounds of LOCC, including the
round which converts a state of the form in Eq. (30) into the most general
state in the SLOCC class described by Eq. (29) with measurement operators
$\mathcal{K}_{D\rightarrow ND}$. We observe that certain state transformations
[e.g., when case (c) is involved] can be achieved with a non-trivial 3-round
protocol. A specific example for that has already been shown in Sec. IV.

##  Appendix E Transformations that are impossible via SEP within $M_{A_{3}}$

In this appendix, we will prove Lemma 16 which provides the full details of
which states in $M_{A_{3}}$ cannot be related by SEP (and hence, by LOCC) as
summarized in Sec. IV. At the end, we will define the sets $\mathcal{D}_{0}$
and $\mathcal{D}_{f}$ that appear in Fig. 1.

######  Lemma 16.

For any type-(ii) state $|\psi_{2}(\alpha_{1};\alpha_{2})\rangle$ in
Observation 11, it holds that: (a) they cannot be reached from any type-(iii)
state and (b) they cannot be reached from nor converted to any type-(ii) state
$|\psi_{2}(\alpha^{\prime}_{1};\alpha^{\prime}_{2})\rangle$ with
$\alpha^{\prime}_{2}\neq\frac{\alpha^{\prime}_{1}(\alpha_{2}-1)+\alpha_{1}-\alpha_{2}}{\alpha_{1}-1}$
by LOCC. Moreover, the state $|A_{3}\rangle$ is in the MES.

###### Proof.

We will prove that the transformations are not achievable even with SEP by
showing that Eq. (1) in Theorem 1 cannot be satisfied. Let the initial state
be
$\sqrt{\Delta^{\prime}}\otimes\sqrt{D^{\prime}}\otimes\text{1}|A_{3}\rangle$
and the target state be
$\sqrt{\Delta}\otimes\sqrt{D}\otimes\text{1}|A_{3}\rangle$ where
$D,D^{\prime},\Delta$ and $\Delta^{\prime}$ are positive diagonal matrices. To
obtain a simple necessary condition for the existence of a SEP transformation,
we compute the overlap of Eq. (1) with $\langle j_{1},j_{2},j_{3}|$ and
$|A_{3}\rangle$ after inserting $\Delta\otimes D\otimes\text{1}$
($\Delta^{\prime}\otimes D^{\prime}\otimes\text{1}$) for $H$ ($G$),
respectively. We get footnote:SEP_A3_singular

$\sum_{i_{1},i_{2},i_{3},k}p_{k}\epsilon_{i_{1},i_{2},i_{3}}\prod_{m=1}^{3}[S_{k}^{\dagger}]_{j_{m},i_{m}}\Delta_{i_{1}}D_{i_{2}}=r\epsilon_{j_{1},j_{2},j_{3}}\Delta^{\prime}_{j_{1}}D^{\prime}_{j_{2}}.$
(32)

For case (a), we will show that any type-(iii) state
$|\psi_{3}(\alpha^{\prime}_{1},\beta^{\prime}_{1};\alpha^{\prime}_{2},\beta^{\prime}_{2})\rangle$
cannot be transformed into any type-(ii) state
$|\psi_{2}(\alpha_{1};\alpha_{2})\rangle$ via SEP. The corresponding Eq. (32)
is equivalent to the following matrix equation

$\begin{pmatrix}\alpha_{1}&-\alpha_{1}&-1&-\alpha_{2}&1&\alpha_{2}\\\
-\alpha_{1}&\alpha_{1}&\alpha_{2}&1&-\alpha_{2}&-1\\\
-\alpha_{2}&\alpha_{2}&1&\alpha_{1}&-1&-\alpha_{1}\\\
1&-1&-\alpha_{2}&-\alpha_{1}&\alpha_{2}&\alpha_{1}\\\
\alpha_{2}&-\alpha_{2}&-\alpha_{1}&-1&\alpha_{1}&1\\\
-1&1&\alpha_{1}&\alpha_{2}&-\alpha_{1}&-\alpha_{2}\end{pmatrix}\cdot\vec{x}=r\begin{pmatrix}\alpha^{\prime}_{1}\beta^{\prime}_{2}\\\
-\alpha^{\prime}_{1}\\\ -\alpha^{\prime}_{2}\beta^{\prime}_{1}\\\
\beta^{\prime}_{1}\\\ \alpha^{\prime}_{2}\\\ -\beta^{\prime}_{2}\end{pmatrix}$
(33)

where the entries of $\vec{x}$ correspond to the non-vanishing terms of the
left hand side of Eq. (32) and are given by
$\sum_{k}p_{k}\prod_{m=1}^{3}[S_{k}^{\dagger}]_{j_{m},i_{m}}$ for
$i_{m},j_{m}\in\\{1,2,3\\}$. Using Gaussian elimination, one can easily check
that Eq. (33) has no solution (i.e., inconsistent) if
$\alpha_{1},\alpha_{2},\alpha^{\prime}_{1},\beta^{\prime}_{1},\alpha^{\prime}_{2},\beta^{\prime}_{2}$
correspond to the parameters of any type-(ii) and type-(iii) states. Hence,
such a transformation is not possible.

For case (b), we will first show that any type-(ii) state
$|\psi_{2}(\alpha^{\prime}_{1};\alpha^{\prime}_{2})\rangle$ cannot be
transformed into any type-(ii) state $|\psi_{2}(\alpha_{1};\alpha_{2})\rangle$
with
$\alpha^{\prime}_{2}\neq\frac{\alpha^{\prime}_{1}(\alpha_{2}-1)+\alpha_{1}-\alpha_{2}}{\alpha_{1}-1}$
via SEP. Then, we will show that it is also impossible to transform
$|\psi_{2}(\alpha_{1};\alpha_{2})\rangle$ into
$|\psi_{2}(\alpha^{\prime}_{1};\alpha^{\prime}_{2})\rangle$ for these
parameters with SEP. For the first direction, we simply set
$\beta^{\prime}_{1}=\beta^{\prime}_{2}=1$ in Eq. (33) and observe that the
equation has no solution if

$\alpha^{\prime}_{2}\neq\frac{\alpha^{\prime}_{1}(\alpha_{2}-1)+\alpha_{1}-\alpha_{2}}{\alpha_{1}-1}.$
(34)

For the reversed direction, one simply has to exchange the parameters
$\alpha_{i}\leftrightarrow\alpha^{\prime}_{i}$ for both $i=1,2$ in Eq. (33)
that has $\beta^{\prime}_{1}=\beta^{\prime}_{2}=1$ and observe that it also
has no solution if Eq. (34) holds by the fact that Eq. (34) is symmetric under
the exchange, $\alpha_{i}\leftrightarrow\alpha^{\prime}_{i}$.

To show that the type-(i) state, $|A_{3}\rangle$, is in the MES, one simply
sets $\alpha_{1}=\alpha_{2}=1$ in Eq. (33) and sees that the matrix equation
is inconsistent when
$\alpha^{\prime}_{1},\beta^{\prime}_{1},\alpha^{\prime}_{2},\beta^{\prime}_{2}$
correspond to the parameters of any type-(ii) or type-(iii) states. Since
$|A_{3}\rangle$ cannot be reached from any other MES candidates (see
Observation 11) via SEP (LOCC), it must be in the MES footnote:A3_in_MES . ∎

We now define the sets $\mathcal{D}_{0}$ and $\mathcal{D}_{f}$ that we mention
in Fig. 1 based on the results in Lemma 16. We first fix a pair of
$(\alpha_{1},\alpha_{2})$ such that $\alpha_{1},\alpha_{2}>0$,
$\alpha_{1}\neq\alpha_{2}$, and $\alpha_{1},\alpha_{2}\neq 1$. As a zero-
measure subset of $M_{A_{3}}$, $\mathcal{D}_{0}$ is defined to contain
$|A_{3}\rangle$, a type-(ii) state $|\psi_{2}(\alpha_{1};\alpha_{2})\rangle$.
The full-measure subset, $\mathcal{D}_{f}$, of $M_{A_{3}}$ is defined to
contain all type-(iii) states and all type-(ii) states
$|\psi_{2}(\alpha_{1}^{\prime};\alpha_{2}^{\prime})\rangle$ such that
$\alpha^{\prime}_{2}\neq\frac{\alpha^{\prime}_{1}(\alpha_{2}-1)+\alpha_{1}-\alpha_{2}}{\alpha_{1}-1}$.

##  Appendix F Decomposition of states in $M_{A_{3}}$

In Sec. IV, we stated that any states
$|\psi(\alpha_{1},\alpha_{2},\beta_{1},\beta_{2})\rangle$
$\propto\text{diag}(\sqrt{\alpha_{1}},\sqrt{\alpha_{2}},1)\otimes\text{diag}(\sqrt{\beta_{1}},\sqrt{\beta_{2}},1)\otimes\text{1}|A_{3}\rangle$
footnote:ReorderAlpha2Beta1  in $M_{A_{3}}$ can be prepared with simple
bipartite entanglement resources. More precisely, one can show that

$\displaystyle
U_{13}U_{12}|\widetilde{+}\rangle_{1}|\psi_{s}\rangle_{23}=V\otimes
U_{2}^{\dagger}\otimes
U_{3}^{\dagger}|\psi(\alpha_{1},\alpha_{2},\beta_{1},\beta_{2})\rangle$ (35)

where
$|\widetilde{+}\rangle_{1}=\frac{1}{\sqrt{3}}(|0\rangle+|1\rangle+|2\rangle)$,
$|\psi_{s}\rangle_{23}=\sqrt{\lambda_{+}}|00\rangle+\sqrt{\lambda_{-}}|11\rangle$,
$U_{1j}=\sum_{k=0}^{2}|k\rangle\langle
k|\otimes(U_{j}^{\dagger}D_{\omega}U_{j})^{k}$ for $j=2,3$,
$D_{\omega}=\text{diag}(1,e^{-i\frac{2\pi}{3}},e^{i\frac{2\pi}{3}})$, and
$V=\frac{1}{\sqrt{3}}\begin{pmatrix}1&1&1\\\ 1&\omega&\omega^{2}\\\
1&\omega^{2}&\omega\end{pmatrix}$ with $\omega=e^{i\frac{2\pi}{3}}$. We now
give the exact forms of $\lambda_{\pm}$, and unitaries $U_{2}$ and $U_{3}$
which we omitted in Sec. IV.

For states $|\psi(\alpha_{1},\alpha_{2},\beta_{1},\beta_{2})\rangle$ with
$\beta_{1}\neq\beta_{2}$, $\lambda_{\pm}=\frac{1\pm\sqrt{2\chi-1}}{2}$ with
$\chi=\frac{\gamma^{2}-2(\alpha_{1}+\alpha_{2}+1)[(\alpha_{1}+\beta_{1})\beta_{2}+\alpha_{2}\beta_{1}]}{\gamma^{2}}$
and
$\gamma=\alpha_{1}+\alpha_{2}+\beta_{1}+\beta_{2}+\alpha_{1}\beta_{2}+\alpha_{2}\beta_{1}$.
The unitary
$U_{2}=\sigma\begin{pmatrix}-\frac{\sqrt{\alpha_{2}\beta_{1}}x_{+,2}}{\mathcal{N}^{(1)}_{+}}&-\frac{\sqrt{\alpha_{2}\beta_{1}}x_{-,2}}{\mathcal{N}^{(1)}_{-}}&\frac{1}{\mathcal{N}^{(1)}_{0}}\sqrt{\frac{\alpha_{1}}{\beta_{1}}}\\\
\frac{\sqrt{\alpha_{1}\beta_{2}}x_{+,1}}{\mathcal{N}^{(1)}_{+}}&\frac{\sqrt{\alpha_{1}\beta_{2}}x_{-,1}}{\mathcal{N}^{(1)}_{-}}&\frac{1}{\mathcal{N}^{(1)}_{0}}\sqrt{\frac{\alpha_{2}}{\beta_{2}}}\\\
\frac{\sqrt{\alpha_{1}\alpha_{2}}(\beta_{1}-\beta_{2})}{\mathcal{N}^{(1)}_{+}}&\frac{\sqrt{\alpha_{1}\alpha_{2}}(\beta_{1}-\beta_{2})}{\mathcal{N}^{(1)}_{-}}&\frac{1}{\mathcal{N}^{(1)}_{0}}\\\
\end{pmatrix}$ with $\sigma=\text{sgn}(\beta_{1}-\beta_{2})$,
$x_{\pm,i}=\gamma\lambda_{\pm}-\alpha_{1}-\alpha_{2}-\beta_{i}$ for
$i\in\\{1,2\\}$,
$\mathcal{N}_{0}^{(m)}=\sqrt{1+\frac{\alpha_{1}}{\beta_{1}^{m}}+\frac{\alpha_{2}}{\beta_{2}^{m}}}$
and
$\mathcal{N}_{\pm}^{(m)}=\sqrt{\alpha_{1}\beta_{2}^{m}x_{\pm,1}^{2}+\alpha_{2}\beta_{1}^{m}x_{\pm,2}^{2}+\alpha_{1}\alpha_{2}(\beta_{1}-\beta_{2})^{2}}$
and
$U_{3}=\begin{pmatrix}-\frac{\sqrt{\alpha_{2}}x_{-,2}}{\mathcal{N}^{(0)}_{-}}&\frac{\sqrt{\alpha_{2}}x_{+,2}}{\mathcal{N}^{(0)}_{+}}&\frac{\sqrt{\alpha_{1}}}{\mathcal{N}^{(0)}_{0}}\\\
\frac{\sqrt{\alpha_{1}}x_{-,1}}{\mathcal{N}^{(0)}_{-}}&-\frac{\sqrt{\alpha_{1}}x_{+,1}}{\mathcal{N}^{(0)}_{+}}&\frac{\sqrt{\alpha_{2}}}{\mathcal{N}^{(0)}_{0}}\\\
\frac{\sqrt{\alpha_{1}\alpha_{2}}(\beta_{1}-\beta_{2})}{\mathcal{N}^{(0)}_{-}}&-\frac{\sqrt{\alpha_{1}\alpha_{2}}(\beta_{1}-\beta_{2})}{\mathcal{N}^{(0)}_{+}}&\frac{1}{\mathcal{N}^{(0)}_{0}}\end{pmatrix}$.

For any state $|\psi(\alpha_{1},\alpha_{2},\beta_{1},\beta_{2})\rangle$ with
$\beta_{1}=\beta_{2}=\beta$ footnote:DecompMA3_b1=b2 , the preparation
procedure is the same as above, except
$\lambda_{+}=\frac{(\alpha_{1}+\alpha_{2}+1)\beta}{\gamma}$,
$\lambda_{-}=1-\lambda_{+}$,
$U_{2}=\begin{pmatrix}-\frac{\sqrt{\alpha_{2}}}{\mathcal{N}^{\prime}}&-\frac{\sqrt{\alpha_{1}\beta}}{\mathcal{M}^{\prime(1)}}&\frac{1}{\mathcal{N}^{(1)}_{0}}\sqrt{\frac{\alpha_{1}}{\beta}}\\\
\frac{\sqrt{\alpha_{1}}}{\mathcal{N}^{\prime}}&-\frac{\sqrt{\alpha_{2}\beta}}{\mathcal{M}^{\prime(1)}}&\frac{1}{\mathcal{N}^{(1)}_{0}}\sqrt{\frac{\alpha_{2}}{\beta}}\\\
0&\frac{\alpha_{1}+\alpha_{2}}{\mathcal{M}^{\prime(1)}}&\frac{1}{\mathcal{N}^{(1)}_{0}}\end{pmatrix}$
and
$U_{3}=\begin{pmatrix}-\frac{\sqrt{\alpha_{1}}}{\mathcal{M}^{\prime(0)}}&\frac{\sqrt{\alpha_{2}}}{\mathcal{N}^{\prime}}&\frac{\sqrt{\alpha_{1}}}{\mathcal{N}^{(0)}_{0}}\\\
-\frac{\sqrt{\alpha_{2}}}{\mathcal{M}^{\prime(0)}}&-\frac{\sqrt{\alpha_{1}}}{\mathcal{N}^{\prime}}&\frac{\sqrt{\alpha_{2}}}{\mathcal{N}^{(0)}_{0}}\\\
\frac{\alpha_{1}+\alpha_{2}}{\mathcal{M}^{\prime(0)}}&0&\frac{1}{\mathcal{N}^{(0)}_{0}}\end{pmatrix}$
with $\mathcal{N}^{\prime}=\sqrt{\alpha_{1}+\alpha_{2}}$ and
$\mathcal{M}^{\prime(m)}=\mathcal{N}^{\prime}\sqrt{\alpha_{1}+\alpha_{2}+\beta^{m}}$.

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  * (42) More precisely, $P$ can be chosen as $P=|v\rangle\langle v|+\text{1}$, where $|v\rangle\in\mathbb{C}^{d}$ is not an eigenvector of any $U_{i}\in\mathcal{F}$. Such a vector always exists as no finite-dimensional vector space over $\mathbb{C}$ is a finite union of proper subspaces (see e.g., Ref. VecSpaceNOTfiniteUnion ).
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  * (44) This can be easily seen as follows. First, due to the symmetry of the state, it is easy to see that any state in the SLOCC class is LU equivalent to $\sqrt{G_{1}}\otimes\sqrt{D_{2}}\otimes\text{1}|A_{3}\rangle$ [see Eq. (29)] where $G_{1}>0$ and $D_{2}=diag(\alpha_{2},\beta_{2},1)>0$. Moreover, using the symmetry $U^{\otimes 3}$ of $|A_{3}\rangle$, where $U=diag(e^{i\theta},e^{i\varphi},e^{-i(\theta+\varphi)})$ with $\theta=-\frac{\arg(\gamma_{1})+\arg(\delta_{1})}{3}$, $\varphi=\frac{2\arg(\gamma_{1})-\arg(\delta_{1})}{3}$, $\gamma_{1}=(G_{1})_{12}$ and $\delta_{1}=(G_{1})_{13}$, leads to a state of the same form as above, but with $G_{1}$ replaced by $UG_{1}U^{\dagger}$, whose entries $(1,2)$ and $(1,3)$ are larger than or equal to zero. Hence, the states are (up to LU) parameterized by 8 parameters.
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  * (47) Note that we exchange the order of $\alpha_{2}$ and $\beta_{1}$ here as opposed to the notation that we use in Observation 11 to denote the states in $M_{A_{3}}$.
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  * (49) The argument we use here to show that $B\otimes B^{-1}\otimes\text{1}^{\otimes n-2}\in\mathcal{S}_{|A_{n}\rangle}$ is the same argument used in Ref. MigdalSymm  (Sec. II) to prove that permutation-symmetric states have symmetries of the form $B\otimes B^{-1}\otimes\text{1}^{\otimes n-2}$.
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  * (56) The considered state might then be LU-equivalent to the initial state.
  * (57) Note that if there exists a consistency condition with $x_{1}^{(\lambda)}=0$ and $x_{2}^{(\lambda)}\neq 0$ while $\theta$ is an irrational multiple of $\pi$, then the system of equations is inconsistent.
  * (58) We obtain Eq. (20) by first multiplying each equation in $\mathbf{B}\vec{\alpha^{\prime}}=\vec{\varphi^{\prime}}+\vec{\theta}$ by a factor $z\in\mathbb{C}$ on both sides, and then exponentiating both sides of each equation.
  * (59) Although the existence of weak isolation was proven for $(n\geq 5)$-qudit SLOCC classes of non-exceptionally symmetric (non-ES) states, which are permutation-symmetric states with only symmetries of the form $S^{\otimes n}$, in Lemma 4 of Ref. OurSymmPaper , the proof also applies to any $n$-qudit SLOCC class that has a state stabilized only by $S^{\otimes n}$ as long as $n\geq 5$.
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  * (61) The perturbation series for $E_{p}$ and $|e_{p}\rangle$ are guaranteed to converge because the matrix $H_{0}+\epsilon V(\epsilon)$ is Hermitian and analytic (i.e., every matrix entry is analytic) in the neighbourhood of $\epsilon=0$ where $\epsilon\in\mathbb{R}$ and by Rellich’s Theorem Rellich ; FriedlandBook , all the eigenvalues and entries of the eigenvectors must also be analytic in the neighbourhood of $\epsilon=0$.
  * (62) F. Rellich, Perturbation Theory of Eigenvalue Problems, Gordon & Breach, New York, 1969.
  * (63) S. Friedland, Matrices: Algebra, Analysis and Applications, World Scientific, 2015.
  * (64) Since the perturbation series of eigenvalue $E_{p}$ converges in $\epsilon$, one can choose $\epsilon$ small enough such that the absolute value of the sum of the $\mathcal{O}(\epsilon^{2})$ terms is strictly less than $\frac{1}{2}(\frac{1}{r}-1)$ for $E_{0}$ and $\frac{1}{2r^{p-1}}(\frac{1}{r}-1)$, which is half the distance between the $(p-1)$-th and the $p$-th unperturbed eigenvalues, for $E_{p}$ where $p\in\\{1,\ldots,d-1\\}$ and $0<r<1$.
  * (65) Since the perturbation series of eigenvector $|e_{p}\rangle$ converges in $\epsilon$, one can choose $\epsilon$ small enough such that the absolute value of the sum of the $\mathcal{O}(\epsilon^{2})$ terms for $\langle 0|e_{p}\rangle$ is strictly smaller than 1 for $|e_{0}\rangle$ and $|\frac{\epsilon\sqrt{r}^{p}}{(1-r^{p})(1-\omega^{-p})}|$ for every $|e_{p}\rangle$ where $p\in\\{1,\ldots,d-1\\}$, while keeping $\\{E_{p}\\}$ non-degenerate footnote:pert .
  * (66) It is easy to see the following: If $S\in SL(d,\mathbb{C})$ quasi-commutes with two $d\times d$ positive definite diagonal matrices $\Lambda$ and $D$ such that $\Lambda\not\propto D$, $S$ must be a direct sum of block matrices that act on the (degenerate) eigenspaces of $\Lambda^{-1}D$. Moreover, for each block in $S$ of which the range lies within the (degenerate) eigenspace of a single eigenvalue of $\Lambda$ or $D$, the block is unitary.
  * (67) When multiplying Eq. (1) by $|A_{3}\rangle$ (which is the seed state $|\Psi_{s}\rangle$ here) where $g=\sqrt{\Delta^{\prime}}\otimes\sqrt{D^{\prime}}\otimes\text{1}$ and $h=\sqrt{\Delta}\otimes\sqrt{D}\otimes\text{1}$, the term $g^{\dagger}\sum_{q}N_{q}^{\dagger}N_{q}g|A_{3}\rangle=0$ because all $N_{q}\in\mathcal{N}_{g\Psi_{s}}$ satisfy $N_{q}g|A_{3}\rangle=0$ by definition.
  * (68) Alternatively, one can see this by showing that $|A_{3}\rangle$ is the only state among all the MES candidates in Observation 11 that has a completely mixed single qutrit reduced density matrix for all 3 bipartite splittings. Applying Nielsen’s theorem Nielsen  to all 3 bipartitions proves that $|A_{3}\rangle$ is indeed not LOCC-reachable.
  * (69) The preparation procedure above does not work for $|\psi(\alpha_{1},\alpha_{2},\beta_{1},\beta_{2})\rangle$ with $\beta_{1}=\beta_{2}$ because one of the columns in $U_{2}$ and $U_{3}$ becomes all zeros when $\beta_{1}=\beta_{2}$.