# Balanced Measures on Compact Median Algebras

Uri Bader  and  Aviv Taller  Email addresses:  A. Taller<EMAIL_ADDRESS>U. Bader<EMAIL_ADDRESS>Data availability statement : NA

###### Abstract.

We initiate a systematic investigation of group actions on compact median
algebras via the corresponding dynamics on their spaces of measures. We show
that a probability measure which is invariant under a natural push forward
operation must be a uniform measure on a cube and use this to show that every
amenable group action on a locally convex compact median algebra fixes a sub-
cube.

##  1\. Introduction

Median algebras form a common generalization of dendrites and distributive
lattices. While early investigations of these objects mainly dealt with
combinatorial aspects as part of order theory, recently they obtained much
attention due to the observation that CAT(0) cube complexes carry a natural
median space structure, that is, compatible metric and median algebra
structures. A powerful tool in the investigation of a CAT(0) cube complex is
provided by embedding it in a compact median algebra, namely its Roller
compactification, see [Rol98] and also [Fio20] for a further discussion. This
brings to front the class of compact median algebras.

As common for dynamical systems, we study compact median algebras by
investigating their invariant measures. Let $M$ be a second countable compact
median algebra, endowed with a continuous median operator $m:M^{3}\to M$. We
study the operator $\Phi$, called self-median operator, defined on the space
of Borel probability measures of $M$ by the formula

$\Phi:\operatorname{Prob}(M)\to\operatorname{Prob}(M),\quad\Phi(\mu)=m_{*}(\mu^{3}),$

along with its space of invariant measures, $\operatorname{Prob}(M)^{\Phi}$,
which elements we denote balanced measures.

A basic example of a second countable compact median algebra is the cube
$\\{0,1\\}^{I}$, endowed with the product topology and median structure, where
$I$ is a countable index set111In this paper, a countable set could be finite,
in particular empty.. One verifies easily that the uniform measure
$\lambda\in\operatorname{Prob}(\\{0,1\\}^{I})$ is balanced (see the beginning
of §5 for details). Clearly, if $f:N\to M$ is a continuous morphism of second
countable compact median algebras and $\mu$ is a balanced measure on $N$ then
$f_{*}(\mu)$ is a balanced measure on $M$. In particular, for every continuous
median algebra morphism $f:\\{0,1\\}^{I}\to M$, the push forward of the
uniform measure, $f_{*}(\lambda)$, is a balanced measure on $M$. In case $f$
is injective, we say that $f(\\{0,1\\}^{I})$ is a cube in $M$ and
$f_{*}(\lambda)$ is a cubical measure on $M$.

Our main theorem is the following classification result which applies to the
class of second countable, locally open-convex, compact, or for short sclocc,
median algebras (see Definitions 3.1 and 4.1).

######  Theorem A.

Every balanced measure on a sclocc median algebra is cubical.

Theorem A is a direct corollary of the following two propositions.

######  Proposition 1.1.

The support of every balanced measure on a sclocc median algebra is a cube.

######  Proposition 1.2.

The uniform measure is the unique fully supported balanced measure on a cube.

Proposition 1.1 has the following interesting corollary.

######  Corollary B.

Assume that $G$ is a locally compact amenable group which acts on a sclocc
median algebra $M$ by continuous automorphisms. Then $M$ contains a
$G$-invariant cube.

Indeed, by the amenability of $G$, $\operatorname{Prob}(M)^{G}$ is non-empty
and clearly $\Phi$-invariant, thus it contains a balanced measure $\mu$ by the
Tychonoff Fixed Point Theorem and the support of $\mu$ is a $G$-invariant cube
by Proposition 1.1.

Note that one cannot expect a $G$-invariant cube to be pointwise fixed.
Indeed, the group $(\mathbb{Z}/2\mathbb{Z})^{I}$ acts transitively on the cube
$\\{0,1\\}^{I}$ by median automorphisms (see the discussion in the begining of
§5).

The requirements that a topological median algebra is sclocc is satisfied by
the Roller compactification of a (possibly infinite dimensional) CAT(0) cube
complex with countably many vertices, as well as a second countable median
space of finite rank, see Example 4.2. For further discussions on CAT(0) cube
complexes and median spaces, see [Sag95, CFI16, Bow13, Bow20, CDH10, Fio20]
and the references therein.

We note that if $M$ is not assumed second countable then the median operator
$m$ might not be Borel (with respect to the product of the Borel
$\sigma$-algebras on $M^{3}$) even though it is continuous, thus we cannot
define the self median operator $\Phi$ in this generality. Consider the
interval $[0,\omega_{1}]$, where $\omega_{1}$ is the first uncountable
ordinal, and endow it with the order topology and median structure. Even
though this space is not second countable, one verifies easily that $m$ is
Borel, thus $\Phi$ is defined, in this case. Then the Dieudonné measure, which
assigns 1 to Borel sets containing an unbounded closed subset and 0 to their
complements, is a balanced measure which has an empty support. Hereafter, when
discussing measures on compact spaces we will assume that the underlying space
is second countable, thus every measure has a non-empty support.

The structure of this note is as follows. After reviewing median algebras in
the next section, we will focus our attention on the theory of compact median
algebras in §3. We will then prove Proposition 1.1 in §4 and Proposition 1.2
in §5.

Acknowledgments. The authors thank Elia Fioravanti for sharing with them the
observation that a finite rank median space is locally open-convex and
referring them to [Bow20]. The authors wish to thank the anonymous referee for
many valuable comments. This research is supported by ISF Moked 713510 grant
number 2919/19.

##  2\. A review of Median Algebras

A median algebra is a set $M$, equipped with a ternary operation
$m:M\times M\times M\rightarrow M$, called the median operator, that has the
following properties:
$\forall x,y,z,u,v\in M$

  1. (Med 1)

$m(x,y,z)=m(x,z,y)=m(y,x,z)$

  2. (Med 2)

$m(x,x,y)=x$

  3. (Med 3)

$m(m(x,y,z),u,v)=m(x,m(y,u,v),m(z,u,v))$

A median morphism $\varphi:M\rightarrow N$ between two median algebras, is a
map such that $\varphi\circ m=m\circ(\varphi\times\varphi\times\varphi)$
(notice that we may confuse often between the median operators of one object
to another, like we did here).

A basic example of a median algebra is $\\{0,1\\}$, endowed with the standard
median operation. The category of median algebra has arbitrary products, given
by the Cartesian products of the underlying sets and the coordinate-wise
operations. In particular, for any index set $I$, $\\{0,1\\}^{I}$ is a median
algebra. A median algebra that isomorphic to $\\{0,1\\}^{I}$ for some $I$ is
called a cube. A cube in a median algebra $M$ means a median subalgebra of $M$
that is a cube.

For the rest of this section we fix a median algebra $M$. The interval of two
elements $x,y\in M$ is the subset $[x,y]=\\{u\in M\mid m(x,y,u)=u\\}$.

######  Lemma 2.1 (statements (Int 3) and (Int 6) at the beginning of section
2 in [Rol98]).

For every $x,y,z\in M$,

  1. (1)

$y\in[x,z]\quad\implies\quad[x,y]\subset[x,z].$

  2. (2)

$y\in[x,z]\quad\iff\quad[x,y]\cap[y,z]=\\{y\\}.$

A subset $C\subset M$ is called convex, if for every $x,y\in C$, $[x,y]\subset
C$. An important property of convex sets is what usually known as Helly’s
Theorem .

######  Lemma 2.2 ([Rol98, Theorem 2.2]).

If $C_{1},...,C_{n}$ are convex, and for every $i\neq j\ \ C_{i}\cap
C_{j}\neq\varnothing$, then also
$\overset{n}{\underset{i=1}{\cap}}C_{i}\neq\varnothing$.

Let $C\subset M$ be a convex set and $x\in M$. We say that $y\in C$ is the
gate for $x$ in $C$, if $y\in[x,z]$, for every $z\in C$. It follows form Lemma
2.1(2), that if exists, the gate is indeed unique. We say that $C$ is gate-
convex if for every $x\in M$, there exists a gate in $C$. In this case, we
define the gate-projection to $C$, $\pi_{C}:M\rightarrow C$, where
$\pi_{C}(x)$ is the gate of $x$ in $C$. An example for gate-convex set is the
interval $[x,y]$ with gate-projection $\pi_{[x,y]}(z)=m(x,y,z)$, for every
$x,y\in M$. Notice that as a result of Lemma 2.1(2), gate-convex sets are
always convex.

By [Fio20, Proposition 2.1], if $\phi:M\rightarrow M$ is a gate-projection,
then for every $x,y,z\in M$ we have
$\phi(m(x,y,z))=m(\phi(x),\phi(y),\phi(z))$. In particular, gate-projections
map intervals to intervals. In addition, the following is true

######  Lemma 2.3 ([Fio20, Lemma 2.2]).

  1. (1)

If $C_{1}\subset M$ is convex and $C_{2}\subset M$ is gate-convex, the
projection $\pi_{C_{2}}(C_{1})$ is convex. If moreover, $C_{1}\cap
C_{2}\neq\varnothing$, we have $\pi_{C_{2}}(C_{1})=C_{1}\cap C_{2}$.

  2. (2)

If $C_{1},C_{2}\subset M$ are gate-convex and $C_{1}\cap
C_{2}\neq\varnothing$, then $C_{1}\cap C_{2}$ is gate-convex with gate-
projection $\pi_{C_{2}}\circ\pi_{C_{1}}=\pi_{C_{1}}\circ\pi_{C_{2}}$. In
particular, if $C_{2}\subset C_{1}$, then
$\pi_{C_{2}}=\pi_{C_{2}}\circ\pi_{C_{1}}$.

Given a disjoint partition of a median algebra $M$ into non-empty convex
subsets, $M=\mathfrak{h}\sqcup\mathfrak{h}^{*}$, we say that $\mathfrak{h}$
and $\mathfrak{h}^{*}$ are complementary half-spaces in $M$ and we regard the
unordered pair $\mathfrak{w}=\\{\mathfrak{h},\mathfrak{h}^{*}\\}$ as a wall in
$M$. For disjoint subsets $A,B\subset M$, we say that $\mathfrak{h}$ separates
$A$ from $B$ if $A\subset\mathfrak{h}$ and $B\subset\mathfrak{h}^{*}$. We
denote by $\Delta(A,B)$ the collection of half-spaces that separate $A$ from
$B$. By [Rol98, Theorem 2.8], if $A$ and $B$ are convex then $\Delta(A,B)$ is
not empty. Given a wall $\mathfrak{w}$, we say that $\mathfrak{w}$ separates
$A$ and $B$ if $\mathfrak{w}\cap\Delta(A,B)$ is not empty. For $a\notin B$ we
denote $\Delta(a,B):=\Delta(\\{a\\},B)$.

######  Lemma 2.4.

For disjoint non-empty subsets $A,B\subset M$, if $A$ is gate-convex then
there exists $a\in A$ such that $\Delta(A,B)=\Delta(a,B)$.

###### Proof.

Fix $b\in B$ and let $a=\pi_{A}(b)$. Clearly, $\Delta(A,B)\subset\Delta(a,B)$.
We will show the other inclusion. Fix $\mathfrak{h}\in\Delta(a,B)$, and let
$a^{\prime}\in A\cap\mathfrak{h}^{*}$. By the definition of the gate-
projection and the fact that $\mathfrak{h}^{*}$ is convex, we have
$a\in[a^{\prime},b]\subset\mathfrak{h}^{*}$, contradicting $a\in\mathfrak{h}$.
Thus, indeed, $A\subset\mathfrak{h}$ and we conclude that
$\mathfrak{h}\in\Delta(A,B)$. ∎

We denote by $\mathscr{H}$ and by $\mathscr{W}$ the collections of all half-
spaces and all walls in $M$, respectively. There is a natural map
$\mathscr{H}\to\mathscr{W}$, given by
$\mathfrak{h}\mapsto\\{\mathfrak{h},\mathfrak{h}^{*}\\}$. Fix a section
$\sigma:\mathscr{W}\to\mathscr{H}$, that is, a half-space representation for
every wall in $M$. We denote by $\chi_{A}$ the characteristic function of a
set A. For each subset $W\subset\mathscr{W}$ we get a median morphism

$\iota_{W}:M\to\\{0,1\\}^{W},\quad
x\mapsto(\chi_{\sigma(\mathfrak{w})}(x))_{\mathfrak{w}}.$

We say that a set of walls $W$ is separating if $\iota_{W}$ is injective, that
is, the walls in $W$ separate the points of $M$. We say that $W$ is transverse
if $\iota_{W}$ is surjective. Note that these properties are independent of
the choice of the section $\sigma$. Two distinct walls
$\mathfrak{w}_{1},\mathfrak{w}_{2}\in\mathscr{W}$ are said to be transverse if
$\\{\mathfrak{w}_{1},\mathfrak{w}_{2}\\}$ is transverse. We record the
following lemma, which proof is an immediate application of Helly’s Theorem,
Lemma 2.2.

######  Lemma 2.5.

If $W\subset\mathscr{W}$ is finite then it is transverse if and only every
pair of distinct walls in $W$ is transverse.

##  3\. Compact Median Algebras

A Topological median algebra is a median algebra $M$ endowed with a topology
for which the median operator is continuous. In this note we will always
assume that topologies are Hausdorff.

######  Definition 3.1.

A topological median algebra $M$ is said to be locally convex if each of its
points has a basis of convex neighborhoods and it is said to be locally open-
convex if each of its points has a basis of open and convex neighborhoods.

######  Example 3.2.

  1. (1)

The class of locally open-convex median algebras is closed under taking
arbitrary products, hence for every index set I, the cube $\\{0,1\\}^{I}$ is a
locally open-convex compact median algebra, with respect to the product
topology and product median structure.

  2. (2)

The class of locally open-convex median algebras is closed under taking median
subalgebras. Therefore, any closed median subalgebra of a cube, such as the
Roller compactification of a (possibly infinite dimensional) CAT(0) cube
complex, is locally open-convex and compact.

  3. (3)

By [Bow20, Lemma 3.1 and Lemma 3.2], every median space $(X,\rho)$ admits a
pseudo-metric $\sigma$ which open balls are convex. Assume $X$ has a finite
rank. Then by [Bow20, Lemma 6.2], $\sigma$ is bilipschitz equivalent to
$\rho$, thus it separates the points of $X$. It follows that $X$, as well as
all its intervals, are locally open-convex. By [Fio20, Corollary 2.20], the
completion of $X$ has compact intervals and its Roller compactification
$\bar{X}$ is defined in [Fio20, Definition 4.13]. By [Fio20, Definition 4.1],
$\bar{X}$ is a median subalgebra of the product of all intervals in $X$. As in
the previous example, it follows that $\bar{X}$ is locally open-convex and
compact.

For the rest of the section we let $M$ be a locally open-convex and compact
topological median algebra.

We note that the closure of a convex set in $M$ is convex. Indeed, if
$C\subset M$ is convex, we have $C\times C\times M\subset m^{-1}(\bar{C})$
and, as $m^{-1}(\bar{C})$ is closed, we have $\bar{C}\times\bar{C}\times
M\subset m^{-1}(\bar{C})$, thus also $\bar{C}$ is convex. Since $M$ is a
normal topological space, it follows that every point in $M$ also has a basis
of closed convex neighborhoods. We also note that, by [Fio20, Lemma 2.6 and
Lemma 2.7], a convex set in $M$ is gate-convex if and only if it is closed and
the gate-projections to the gate-convex sets are continuous. In particular,
every interval is closed (this is true in fact in every Hausdorff topological
median algebra).

A half-space $\mathfrak{h}$ in $M$ is said to be admissible if it is open and
$\mathfrak{h}^{*}$ has a non empty interior. If further $\mathfrak{h}^{*}$ is
open, we say that $\mathfrak{h}$ is clopen. We denote by $\mathscr{H}^{\circ}$
the collections of all clopen half-spaces and we denote by
$\mathscr{W}^{\circ}$ the collection of all corresponding walls. For a subset
$W\subset\mathscr{W}^{\circ}$, the median morphism
$\iota_{W}:M\to\\{0,1\\}^{W}$ is clearly continuous. By compactness, we get
the following upgrade of Lemma 2.5.

######  Lemma 3.3.

For every $W\subset\mathscr{W}^{\circ}$, $W$ is transverse if and only if
every pair of distinct walls in $W$ is transverse. If moreover $W$ is
separating then $\iota_{W}$ is an isomorphism of topological median algebras,
thus $M$ is a cube.

###### Proof.

The first part follows immediately from Lemma 2.5 by the compactness of $M$.
If further $W$ is separating then $\iota_{W}$ is a continuous isomorphism of
median algebras and it is closed, as $M$ is compact, thus an homeomorphism. ∎

Since in a cube $\mathscr{W}^{\circ}$ is clearly transverse and separating, we
immediately get the following corollary of Lemma 3.3.

######  Corollary 3.4.

$M$ is a cube if and only if $\mathscr{W}^{\circ}$ is separating and every
pair of distinct walls in $\mathscr{W}^{\circ}$ is transverse.

We will use Corollary 3.4 in the next section. In order to apply it we will
use the existence of enough open and admissible half-spaces. These are
provided by the next two results.

######  Proposition 3.5.

Let $U,C\subset M$ be two disjoint non-empty convex sets such that $U$ is open
and $C$ is closed. Then there exists an open half-space
$\mathfrak{h}\in\Delta(U,C)$.

###### Proof.

By Lemma 2.4, there exists a point $c\in C$ such that
$\Delta(U,c)=\Delta(U,C)$. We fix such $c$ and argue to show that there exists
an open half-space $\mathfrak{h}\in\Delta(U,c)$.

We order the collection

$P=\\{V\subset M\mid V\mbox{ is an open-convex set, }U\subset V\mbox{ and
}c\notin V\\}$

by inclusion and note that it is non empty, as $U\in P$. By Zorn’s Lemma, $P$
has a maximal element, as the union of any chain in $P$ forms an upper bound.
We fix such a maximal element $\mathfrak{h}$ and denote
$\mathfrak{h}^{*}=M\setminus\mathfrak{h}$. We argue to show that
$\mathfrak{h}^{*}$ is convex, thus $\mathfrak{h}$ is a half-space.

Assume for the sake of contradiction that $\mathfrak{h}^{*}$ is not convex,
that is, there exist $x,y\in\mathfrak{h}^{*}$ such that
$[x,y]\cap\mathfrak{h}\neq\varnothing$. Fix $z\in[x,y]\cap\mathfrak{h}$ and
set $\omega=\pi_{[x,z]}(c)=m(x,z,c)$. In both cases, $\omega\in\mathfrak{h}$
and $\omega\in\mathfrak{h}^{*}$, we will derive a contradiction to the
maximality of $\mathfrak{h}$ by establishing $\mathfrak{h}\subsetneq V\in P$.

We assume first that $\omega\in\mathfrak{h}^{*}$ and denote
$V\coloneqq\pi_{[\omega,z]}^{-1}(\mathfrak{h}\cap[\omega,z])$. Recalling that
$\pi_{[\omega,z]}$ is a continuous median morphism, $V$ is open and convex.
Applying Lemma 2.3(1) for $C_{1}=\mathfrak{h}$ and $C_{2}=[\omega,z]$ we
conclude that $\mathfrak{h}\subset V$. In particular, $U\subset V$. By Lemma
2.1(1), since $\omega\in[z,x]$, we have $[z,\omega]\subset[z,x]$ and we get by
Lemma 2.3(2),
$\pi_{[z,\omega]}(c)=\pi_{[z,\omega]}\circ\pi_{[z,x]}(c)=\pi_{[z,\omega]}(\omega)=\omega$.
Using our assumption that $\omega\in\mathfrak{h}^{*}$, it follows that
$c\notin V$, thus $V\in P$. Similarly, we use $z\in[x,y]$ to get
$\pi_{[z,\omega]}(y)=\pi_{[z,\omega]}\circ\pi_{[z,x]}(y)=\pi_{[z,\omega]}(z)=z\in\mathfrak{h}\cap[\omega,z]$.
We conclude that $y\in V$, therefore $\mathfrak{h}\subsetneq V$, contradicting
the maximality of $\mathfrak{h}$.

We assume now that $\omega\in\mathfrak{h}$ and denote
$V\coloneqq\pi_{[\omega,c]}^{-1}(\mathfrak{h}\cap[\omega,c])$. Again, $V$ is
clearly open and convex. Applying now Lemma 2.3(1) for $C_{1}=\mathfrak{h}$
and $C_{2}=[\omega,c]$ we conclude that $\mathfrak{h}\subset V$. Since
$\pi_{[\omega,c]}(c)=c\notin\mathfrak{h}$ we have that $c\notin V$, thus $V\in
P$. Again, by Lemma 2.1(1), as $\omega\in[z,c]$, we have
$[\omega,c]\subset[z,c]$ and using Lemma 2.3(2) we get this time
$\pi_{[\omega,c]}(x)=\pi_{[\omega,c]}\circ\pi_{[z,c]}(x)=\pi_{[\omega,c]}(\omega)=\omega$,
which is in $\mathfrak{h}\cap[\omega,c]$ by our assumption that
$\omega\in\mathfrak{h}$. We conclude that $x\in V$, therefore
$\mathfrak{h}\subsetneq V$, contradicting again the maximality of
$\mathfrak{h}$. ∎

######  Proposition 3.6.

Let $C,C^{\prime}\subset M$ be two non-empty disjoint closed convex sets. Then
there exists an admissible half-space $\mathfrak{h}\in\Delta(C,C^{\prime})$.

###### Proof.

Using Lemma 2.4 twice, we find $c\in C$ and $c\in C^{\prime}$ such that
$\Delta(c,c^{\prime})=\Delta(C,C^{\prime})$. Using the fact that $M$ is
normal, we find open-convex neighborhoods $c\in U$ and $c^{\prime}\in V$
having disjoint closures. By Proposition 3.5, there exists an open half-space
$\mathfrak{h}\in\Delta(U,\bar{V})$. As $V\subset\mathfrak{h}^{*}$,
$\mathfrak{h}$ is admissible. We have
$\mathfrak{h}\in\Delta(U,\bar{V})\subset\Delta(c,c^{\prime})=\Delta(C,C^{\prime})$,
thus indeed $\mathfrak{h}\in\Delta(C,C^{\prime})$ ∎

##  4\. The support of a balanced measure

In this section we study the support of balanced measures. Recall that the
support of a measure is the minimal closed set having a null complement and
that every Borel measure on a compact second countable topological space has a
non-empty support.

######  Definition 4.1.

A topological median algebra $M$ is said to be sclocc if it is second
countable, locally open-convex and compact.

In view of Example 3.2, we get the following.

######  Example 4.2.

  1. (1)

For a countable index set I, the cube $\\{0,1\\}^{I}$ is sclocc.

  2. (2)

The Roller compactification of a CAT(0) cube complex with countably many
vertices is sclocc.

  3. (3)

By [Fio20, Theorem 4.14(4)], the Roller compactification of a second
countable, finite rank median space is sclocc.

For the rest of the section we let $M$ be a sclocc median algebra.

We let $\operatorname{Prob}(M)$ be the space of probability Borel measures on
$M$ and recall the definition of the self-median operator

$\Phi:\operatorname{Prob}(M)\to\operatorname{Prob}(M),\quad\Phi(\mu)=m_{*}(\mu^{3}),$

which fixed points are the balanced measures on $M$. The following observation
is trivial, but useful.

######  Lemma 4.3.

For any Borel median algebra morphism $M\to N$, the push forward map
$\operatorname{Prob}(M)\to\operatorname{Prob}(N)$ commutes with the
corresponding self-median operators. In particular, the image of a balanced
measure on $M$ is a balanced measure on $N$.

Another basic lemma is the following.

######  Lemma 4.4.

The balanced measures on $\\{0,1\\}$ are exactly $\delta_{0}$, $\delta_{1}$
and $\frac{1}{2}\delta_{0}+\frac{1}{2}\delta_{1}$.

###### Proof.

Let $\mu$ be a balanced measure on $\\{0,1\\}$ and denote $x=\mu(\\{0\\})$. An
easy calculation gives the equation
$x=\mu(\\{0\\})=m_{*}(\mu^{3})(\\{0\\})=x^{3}+3x^{2}(1-x)$ which solutions are
exactly $0,1$ and $\frac{1}{2}$. ∎

Since every Borel half-space gives a Borel median algebra morphism to
$\\{0,1\\}$ (namely, the characteristic map of the half-space), we get the
following.

######  Corollary 4.5.

For every balanced measure on a median algebra, the measure of any Borel half-
space is either $0,1$ or $\frac{1}{2}$.

For fully supported balanced measures and admissible half-spaces, much more
can be said.

######  Lemma 4.6.

Let $\mu$ be a fully supported balanced measure on the sclocc median algebra
$M$ and let $\mathfrak{h}$ be an admissible half-space. Then $\mathfrak{h}$ is
clopen and $\mu(\mathfrak{h})=\frac{1}{2}$. If $\mathfrak{f}$ is a clopen
half-space corresponding to another wall in $M$ then $\mathfrak{f}$ and
$\mathfrak{h}$ are transverse.

###### Proof.

That $\mu(\mathfrak{h})=\frac{1}{2}$ follows immediately from Corollary 4.5,
by the assumptions that $\mu$ is fully supported and balanced, as by the
admissibility of $\mathfrak{h}$, both $\mathfrak{h}$ and $\mathfrak{h}^{*}$
have positive measures. Fix $x\in\mathfrak{h}$ and use Proposition 3.6 to find
an admissible half-space $\mathfrak{h}^{\prime}\in\Delta(\mathfrak{h}^{*},x)$.
Then also $\mu(\mathfrak{h}^{\prime})=\frac{1}{2}$ and we have
$M=\mathfrak{h}\cup\mathfrak{h}^{\prime}$. It follows that
$\mu(\mathfrak{h}\cap\mathfrak{h}^{\prime})=0$. Since
$\mathfrak{h}\cap\mathfrak{h}^{\prime}$ is open and $\mu$ is fully supported,
we conclude that $\mathfrak{h}\cap\mathfrak{h}^{\prime}=\varnothing$, thus
$\mathfrak{h}^{\prime}=\mathfrak{h}^{*}$ and indeed, $\mathfrak{h}$ is clopen

We now let $\mathfrak{f}$ be a clopen half-space which is not transverse to
$\mathfrak{h}$ and show that $\mathfrak{f}$ and $\mathfrak{h}$ determine the
same wall in $M$. Without loss of the generality we assume
$\mathfrak{f}\cap\mathfrak{h}^{*}=\varnothing$. We have
$\mu(\mathfrak{f}\cap\mathfrak{h})=\mu(\mathfrak{f})=\frac{1}{2}$ and as
$\mu(\mathfrak{h})=\frac{1}{2}$ we get
$\mu(\mathfrak{f}^{*}\cap\mathfrak{h})=0$. Since
$\mathfrak{f}^{*}\cap\mathfrak{h}$ is open and $\mu$ is fully supported, we
conclude that $\mathfrak{f}^{*}\cap\mathfrak{h}=\varnothing$, thus
$\mathfrak{f}=\mathfrak{h}$. This finishes the proof . ∎

We are now ready to prove that the support of a balanced measure is a cube.

###### Proof of Proposition 1.1.

We let $\mu$ be a balanced measure on $M$. We note that the support of $\mu$,
$\operatorname{supp}(\mu)$, is a closed sub-algebra of $M$, and thus, a sclocc
median algebra. Indeed, fixing $x,y,z\in\operatorname{supp}(\mu)$, for every
open neighborhood $O$ of $m(x,y,z)$, there are open neighborhoods $x\in U$,
$y\in V$ and $z\in W$ such that $(x,y,z)\in U\times V\times W\subset
m^{-1}(O)$, and we get $\mu(O)\geq\mu(U)\mu(V)\mu(W)>0$, therefore
$m(x,y,z)\in\operatorname{supp}(\mu)$. By restricting to its support, we
assume as we may that $\mu$ is a fully supported balanced measure on $M$ and
argue to show that $M$ is a cube.

By Corollary 3.4, we need to show that $\mathscr{W}^{\circ}$ is separating and
every pair of distinct walls in $\mathscr{W}^{\circ}$ is transverse.
Proposition 3.6 guarantees that the collection of admissible half-spaces is
separating and, by the first part of Lemma 4.6, this collection coincides with
$\mathscr{W}^{\circ}$. Thus, we get that $\mathscr{W}^{\circ}$ is separating.
By the last part of Lemma 4.6 we also get that every pair of distinct walls in
$\mathscr{W}^{\circ}$ is transverse. Thus, indeed, $M$ is a cube. ∎

##  5\. Balanced measures on cubes

Fix a countable set $I$ and consider the cube $M=\\{0,1\\}^{I}$. It is
convenient to identify $\\{0,1\\}$ with the group $\mathbb{Z}/2\mathbb{Z}$ and
$M$ with the compact group $(\mathbb{Z}/2\mathbb{Z})^{I}$. We denote by
$\lambda$ the Haar measure on $M$,
$\lambda=(\frac{1}{2}\delta_{0}+\frac{1}{2}\delta_{1})^{I}$. It is fully
supported and balanced. To see that it is indeed balanced, recall that
$\lambda$ is the unique probability Borel measure on $M$ that is invariant
under translations, but by Lemma 4.3, also $\Phi(\lambda)$ is invariant under
translations, as translations form median algebra automorphisms.

This section is devoted to the proof of Proposition 1.2, which claims that
$\lambda$ is the unique fully supported balanced measure on $M$. We observe
that it is enough to prove this for finite cubes, that is, in case
$|I|<\infty$, as every cube is the inverse limit of its finite coordinate
projections, which are median algebras surjective morphisms, thus take fully
supported and balanced measure to fully supported and balanced measures by
Lemma 4.3.

Dealing with measures on finite sets, we will identify a measure $\mu$ with
the function $x\mapsto\mu(\\{x\\})$, writing $\mu(x):=\mu(\\{x\\})$. For our
proof, it is beneficial to study a one-parameter class of measures on $M$,
namely, the measures that are invariant under translations by a certain index
two subgroup $K_{0}<M$.

######  Lemma 5.1.

Fix a natural integer $n$ and let $M=(\mathbb{Z}/2\mathbb{Z})^{n}$. Consider
the group homomorphism

$\rho:M\to\mathbb{Z}/2\mathbb{Z},\quad(x_{1},\dots,x_{n})\mapsto\sum x_{i},$

denote its kernel by $K_{0}$ and denote the non-trivial coset of $K_{0}$ by
$K_{1}$. Then the map

$[0,1]\to\operatorname{Prob}(M)^{K_{0}},\quad\omega\mapsto\mu_{t}=\frac{t}{2^{n-1}}\cdot\chi_{K_{0}}+\frac{1-t}{2^{n-1}}\cdot\chi_{K_{1}}$

is bijective and the cubic polynomial

$\phi(t)=t+(-1)^{n}2^{2-n}(t-\frac{1}{2})(t^{2}-t+\frac{1}{4}+(-1)^{n}\frac{3}{4}-(-1)^{n}2^{n-2})$

satisfies the relation, $\Phi(\mu_{t})=\mu_{\phi(t)}$ for every $t\in[0,1]$.

We now provide the proof of Proposition 1.2, based on Lemma 5.1, which proof
we postpone until later.

###### Proof of Proposition 1.2.

We argue to show that $\lambda$ is the unique fully supported balanced measure
on $M=\\{0,1\\}^{I}$. As mentioned above, we may assume that $I$ is finite. We
do so and prove the claim by an induction on $|I|$. The base case $|I|=0$ is
trivial. We note also that the case $|I|=1$ is proven in Lemma 4.4. We now fix
a natural $n>1$, assume $|I|=n$ and that the proposition is known for every
cube $\\{0,1\\}^{J}$ with $|J|<n$. We let $\mu$ be a balanced measure on
$\\{0,1\\}^{I}$ and we argue to prove that for every $x\in\\{0,1\\}^{I}$,
$\mu(x)=\lambda(x)=1/2^{n}$.

We denote by $\\{e_{i}\mid i\in I\\}$ the standard generating set of $M$ and
for every $i\in I$ we consider the obvious projection
$\pi_{i}:M\to\\{0,1\\}^{I\setminus\\{i\\}}$. Fixing $i\in I$ and noticing that
$\pi_{i}$ is a homomorphism of groups, we get by Lemma 4.3 that the push
forward of $\mu$ by $\pi_{i}$ is balanced, thus we conclude by our induction
hypothesis that for every $x\in M$, $\mu(x)+\mu(x+e_{i})=1/2^{n-1}$. It
follows that for evey $i,j\in I$, $\mu(x)=\mu(x+e_{i}+e_{j})$. We denote by
$K_{0}$ the group generated by the set $\\{e_{i}+e_{j}\mid i,j\in I\\}$ and
conclude that $\mu$ is $K_{0}$ invariant.

Noticing that $K_{0}<M$ coincides with the subgroup considered in Lemma 5.1,
it follows that $\mu=\mu_{t}$ for some $t\in[0,1]$. Since $\mu$ is fully
supported, we in fact get that $t\in(0,1)$. Since $\mu$ is balanced, we have
by Lemma 5.1, $\mu_{t}=\Phi(\mu_{t})=\mu_{\phi(t)}$, thus $\phi(t)=t$. We
conclude that $t$ is a root of the polynomial

$(-1)^{n}2^{n-2}(\phi(t)-t)=(t-\frac{1}{2})(t^{2}-t+\frac{1}{4}+(-1)^{n}\frac{3}{4}-(-1)^{n}2^{n-2}).$

We observe that $t=1/2$ is the only root of this polynomial in the region
$(0,1)$. Indeed, for $c\in\mathbb{R}$, the polynomial $t^{2}-t+c$ has a root
in $(0,1)$ iff $0<c\leq 1/4$, which is not satisfied for
$c=\frac{1}{4}+(-1)^{n}\frac{3}{4}-(-1)^{n}2^{n-2})$, since for even $n$ we
have $c\leq 0$ and for odd $n>1$ we have $c\geq 3/2$. We conclude that
$\mu=\mu_{\frac{1}{2}}$, thus indeed, for every $x\in\\{0,1\\}^{I}$,
$\mu(x)=1/2^{n}$. ∎

###### Proof of Lemma 5.1.

The map $t\mapsto\mu_{t}$ is clearly injective and it is onto
$\operatorname{Prob}(M)^{K_{0}}$ as every measure in
$\operatorname{Prob}(M)^{K_{0}}$ (considered as a function on $M$) is constant
on the fibers of $\rho$. Since translation on $M$ are median algebra
automorphisms, we get by Lemma 4.3 that $\Phi$ preserves
$\operatorname{Prob}(M)^{K_{0}}$, thus for every $t\in[0,1]$,
$\Phi(\mu_{t})=\mu_{\psi(t)}$ for some function $\psi:[0,1]\to[0,1]$. We are
left to prove that $\psi=\phi$.

We denote by $0\in M$ the group identity and set $X=m^{-1}(\\{0\\})\subset
M^{3}$. Clearly, $0\in K_{0}$, thus for every $t\in[0,1]$,
$\mu_{t}(0)=t/2^{n-1}$. Applying this to $\psi(t)$, we get that for every
$t\in[0,1]$,

$\mu_{t}^{3}(X)=\mu_{t}^{3}(m^{-1}(\\{0\\}))=m_{*}(\mu_{t}^{3})(\\{0\\})=\Phi(\mu_{t})(\\{0\\})=\mu_{\psi(t)}(0)=\psi(t)/2^{n-1}.$

It is then enough to show that for every $t\in[0,1]$ we have
$\mu_{t}^{3}(X)=\phi(t)/2^{n-1}$, which is what we now proceed to show.

We need to understand the subset $X\subset M^{3}$ and its measure under
$\mu_{t}^{3}$. We use the decomposition $M^{3}=A_{0}\sqcup A_{1}\sqcup
A_{2}\sqcup A_{3}$, where

$A_{i}=\cup\\{K_{\epsilon_{1}}\times K_{\epsilon_{2}}\times
K_{\epsilon_{3}}\mid\epsilon_{1},\epsilon_{2},\epsilon_{3}\in\\{0,1\\},\leavevmode\nobreak\
\epsilon_{1}+\epsilon_{2}+\epsilon_{3}=i\\},$

that is, $A_{i}$ is the subset of $M^{3}$ consists of triples of elements out
of which exactly $i$ are in $K_{1}$. We observe that $\mu_{t}^{3}$, as a
function on $M^{3}$, attains the constant value $t^{3-i}(1-t)^{i}/8^{n-1}$ on
$A_{i}$. Denoting $X_{i}=X\cap A_{i}$ and $a_{i}=|X_{i}|$ we get the formula

(1) $\mu_{t}^{3}(X)=\sum_{i=0}^{3}a_{i}\cdot\frac{t^{3-i}(1-t)^{i}}{8^{n-1}}.$

Our next goal will be to compute the coefficients $a_{i}$. For this we now
emphasize their dependence on $n$, denoting them $a_{i}(n)$. Similarly, we
write $X(n)$ and $X_{i}(n)$ for $X$ and $X_{i}$ correspondingly. Writing
further, $M(n)=\\{0,1\\}^{n}$, we make the identification $M(n)\simeq
M(n-1)\times\\{0,1\\}$ and we identify accordingly also $M(n)^{3}\simeq
M(n-1)^{3}\times\\{0,1\\}^{3}$. As $0\mapsto 0$ under the projection map
$M(n)\to M(n-1)$, we clearly have, using Lemma 4.3, that the image of $X(n)$
under the corresponding map $M(n)^{3}\to M(n-1)^{3}$ is in $X(n-1)$. We denote
by $\pi:X(n)\to X(n-1)$ the corresponding restriction map and consider the
partition $X(n-1)=\sqcup_{j=0}^{3}X_{j}(n-1)$. We fix $i,j\in\\{0,1,2,3\\}$
and for each $x\in X_{j}(n-1)$ count the intersection size of the fiber
$\pi^{-1}(\\{x\\})$ with the sets $X_{i}(n)$, that is, $|\pi^{-1}(\\{x\\})\cap
X_{i}(n)|$. One verifies easily that this size does not depends on $x$, only
on $i,j\in\\{0,1,2,3\\}$ and, denoting it by $s_{i,j}$, we have

$(s_{i,j})=\begin{pmatrix}1&1&0&0\\\ 3&1&2&0\\\ 0&2&1&3\\\
0&0&1&1\end{pmatrix}.$

We therefore obtain the recurrence linear relation

$\begin{pmatrix}a_{0}(n)\\\ a_{1}(n)\\\ a_{2}(n)\\\
a_{3}(n)\end{pmatrix}=\begin{pmatrix}1&1&0&0\\\ 3&1&2&0\\\ 0&2&1&3\\\
0&0&1&1\end{pmatrix}\begin{pmatrix}a_{0}(n-1)\\\ a_{1}(n-1)\\\ a_{2}(n-1)\\\
a_{3}(n-1)\end{pmatrix},\quad\quad\begin{pmatrix}a_{0}(1)\\\ a_{1}(1)\\\
a_{2}(1)\\\ a_{3}(1)\end{pmatrix}=\begin{pmatrix}1\\\ 3\\\ 0\\\
0\end{pmatrix},$

that leads to the explicit formulas

$\displaystyle a_{0}(n)$
$\displaystyle=2^{n}\left((\frac{3}{8}+(-1)^{n}\frac{1}{8})+2^{n}\frac{1}{8}\right),$
$\displaystyle a_{1}(n)$
$\displaystyle=2^{n}\left((\frac{3}{8}-(-1)^{n}\frac{3}{8})+2^{n}\frac{3}{8}\right),$
$\displaystyle a_{2}(n)$
$\displaystyle=2^{n}\left((-\frac{3}{8}+(-1)^{n}\frac{3}{8})+2^{n}\frac{3}{8}\right),$
$\displaystyle a_{3}(n)$
$\displaystyle=2^{n}\left((-\frac{3}{8}-(-1)^{n}\frac{1}{8})+2^{n}\frac{1}{8}\right).$

By substituting these values in equation (1), we get

$\displaystyle\mu_{t}^{3}(X)$
$\displaystyle=\sum_{i=0}^{3}a_{i}\cdot\frac{t^{3-i}(1-t)^{i}}{8^{n-1}}$
$\displaystyle=2^{n}\left((\frac{3}{8}+(-1)^{n}\frac{1}{8})+2^{n}\frac{1}{8}\right)\cdot\frac{t^{3}}{8^{n-1}}$
$\displaystyle+2^{n}\left((\frac{3}{8}-(-1)^{n}\frac{3}{8})+2^{n}\frac{3}{8}\right)\cdot\frac{t^{2}(1-t)}{8^{n-1}}$
$\displaystyle+2^{n}\left((-\frac{3}{8}+(-1)^{n}\frac{3}{8})+2^{n}\frac{3}{8}\right)\cdot\frac{t(1-t)^{2}}{8^{n-1}}$
$\displaystyle+2^{n}\left((-\frac{3}{8}-(-1)^{n}\frac{1}{8})+2^{n}\frac{1}{8}\right)\cdot\frac{(1-t)^{3}}{8^{n-1}}$
$\displaystyle=\frac{t+(-1)^{n}2^{2-n}(t-\frac{1}{2})(t^{2}-t+\frac{1}{4}+(-1)^{n}\frac{3}{4}-(-1)^{n}2^{n-2})}{2^{n-1}}=\frac{\phi(t)}{2^{n-1}},$

thus indeed, $\mu_{t}^{3}(X)=\phi(t)/2^{n-1}$, and this finishes the proof. ∎

## References

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#  SLOCC classification of n qubits invoking the proportional relationships
for spectrums and for standard Jordan normal forms

Dafa Li1,2 1Department of Mathematical Sciences, Tsinghua University, Beijing,
100084, China
2Center for Quantum Information Science and Technology, Tsinghua National
Laboratory for Information Science and Technology (TNList), Beijing, 100084,
China

###### Abstract

We investigate the proportional relationships for spectrums and for SJNFs
(Standard Jordan Normal Forms) of the matrices constructed from coefficient
matrices of two SLOCC (stochastic local operations and classical
communication) equivalent states of $n$ qubits. Invoking the proportional
relationships for spectrums and for SJNFs, pure states of $n$ ($\geq 4$)
qubits are partitioned into 12 groups and 34 families under SLOCC,
respectively. Specially, it is true for four qubits.

##  I Introduction

Quantum entanglement is an essential resource in quantum teleportation,
quantum cryptography, and quantum information and computation Nielsen . A key
task of the entanglement theory is to classify different types of
entanglement. SLOCC classification is very significant because the states in
the same SLOCC class are able to perform the same QIT-tasks Dur Verstraete .
It is well known that two-qubit states were partitioned into two SLOCC
classes, three-qubit states were partitioned into six SLOCC classes, and there
are infinitely many SLOCC classes for $n$ ($\geq 4$) qubits Dur . It is highly
desirable to partition these infinite classes into a finite number of families
according to a SLOCC invariant criterion. In the pioneering work of Verstraete
_et al._ Verstraete , by using their general singular value decomposition
Verstraete _et al._ partitioned pure four-qubit states into nine SLOCC
inequivalent families: $G_{abcd}$, $L_{abc_{2}}$, $L_{a_{2}b_{2}}$,
$L_{ab_{3}}$, $L_{a_{4}}$,$\ L_{a_{2}0_{3\oplus 1}}$, $L_{0_{5\oplus 3}}$,
$L_{0_{7\oplus 1}}$, and $L_{0_{3\oplus 1}0_{3\oplus 1}}$ Verstraete . Since
then, the extensive efforts have contributed to studying entanglement
classification of four qubits Verstraete ; Miyake ; Cao ; LDF07b ; Chterental
; Lamata ; LDFQIC09 ; Borsten ; Viehmann ; Buniy ; Sharma12 .

Recently, considerable efforts have been devoted to find SLOCC invariant
polynomials in the coefficients of states for classifications and measures of
entanglement of$\ n$ qubits Wong ; Luque ; Leifer ; Levay ; LDF07a ;
Osterloh09 ; Viehmann ; Eltschka ; Gour ; LDFPRA13 . It is well known that the
concurrence and the 3-tangle are invariant polynomials of degrees 2 and 4 for
two and three qubits, respectively Coffman . Explicit and simple expresses of
invariant polynomials of degrees 2 for even $n$ qubits LDF07a , 4 for odd $n$
($\geq 4$) qubits LDF07a , 4 for even $n$ ($\geq 4$) qubits LDFPRA13 , were
presented.

Very recently, SLOCC invariant ranks of the coefficient matrices were proposed
for SLOCC classification LDFPRL12 ; LDFPRA12 ; Wang ; Fan ; LDFPRA15 .

In this paper, for two SLOCC equivalent states of $n$ qubits, we show that the
matrices constructed from coefficient matrices of the two states have
proportional spectrums and proportional SJNFs. Invoking the proportional
relationships for spectrums pure states of $n$ ($\geq 4$) qubits are
partitioned into 12 groups under SLOCC, and invoking the proportional
relationships for SJNFs pure states of $n$ ($\geq 4$) qubits are partitioned
into 34 families under SLOCC. Specially, for four qubits, we obtain new SLOCC
classifications.

##  II SLOCC classification of $n$ qubits

###  II.1 The proportional relationships for spectrums and for SJNFs

Let $|\psi\rangle=\sum_{i=0}^{2^{n}-1}a_{i}|i\rangle$ be an $n$-qubit pure
state. It is well known that two $n$-qubit pure states $|\psi\rangle$ and
$|\psi^{\prime}\rangle$ are SLOCC equivalent if and only if there are
invertible local operators $\mathcal{A}_{i}\in GL(2,C)$, $i=1,\cdots,n$, such
that Dur

$|\psi^{\prime}\rangle=\mathcal{A}_{1}\otimes\mathcal{A}_{2}\otimes\cdots\otimes\mathcal{A}_{n}|\psi\rangle.$
(1)

To any state $|\psi\rangle$ of $n$ qubits, we associate a $2^{\ell}$ by
$2^{n-\ell}$ matrix $C_{q_{1}\cdots q_{\ell}}^{(n)}(|\psi\rangle)$ whose
entries are the coefficients $a_{0},a_{1},\cdots,a_{2^{n}-1}$ of the state
$|\psi\rangle$, where $q_{1},\cdots,q_{\ell}$ are chosen as the row bits
LDFPRL12 ; LDFPRA12 . In LDFPRA15 , in terms of the coefficient matrix
$C_{q_{1},...,q_{i}}^{(n)}$ we constructed a $2^{i}$ by $2^{i}$ matrix
$\Omega_{q_{1},...,q_{i}}^{(n)}$

$\displaystyle\Omega_{q_{1},...,q_{i}}^{(n)}(|\psi\rangle)$ (2)
$\displaystyle=$ $\displaystyle
C_{q_{1},...,q_{i}}^{(n)}(|\psi\rangle)\upsilon^{\otimes(n-i)}(C_{q_{1},...,q_{i}}^{(n)}(|\psi\rangle))^{t},$

where $\upsilon=\sqrt{-1}\sigma_{y}$ and $\sigma_{y}\ $is the Pauli operator,
and $C^{t}$ is the transpose of $C$.

From LDFPRA15 , when $q_{1}$ and $q_{2}$ are chosen as the row bits, we can
show that if $n$-qubit states $|\psi^{\prime}\rangle$ and $|\psi\rangle$ are
SLOCC equivalent, then

$\displaystyle\Omega_{q_{1}q_{2}}^{(n)}(|\psi^{\prime}\rangle)$
$\displaystyle=$
$\displaystyle(\Pi_{\ell=3}^{n}\det\mathcal{A}_{q_{\ell}})(\mathcal{A}_{q_{1}}\otimes\mathcal{A}_{q_{2}})\Omega_{q_{1}q_{2}}^{(n)}(|\psi\rangle)(\mathcal{A}_{q_{1}}\otimes\mathcal{A}_{q_{2}})^{t}.$

Let the unitary matrix

$T=\frac{1}{\sqrt{2}}\left(\begin{array}[]{cccc}1&0&0&1\\\ 0&i&i&0\\\
0&-1&1&0\\\ i&0&0&-i\end{array}\right).$ (4)

Let $G_{1}=\ \ T(\mathcal{A}_{q_{1}}\otimes\mathcal{A}_{q_{2}})T^{+}$, where
$T^{+}$ is the Hermitian transpose of $T$. It is easy to check that
$G_{1}G_{1}^{t}=(\det\mathcal{A}_{q_{1}}\ast\det\mathcal{A}_{q_{2}})I$. Let
$S_{q_{1}q_{2}}^{(n)}(\psi^{\prime})=T\Omega_{q_{1}q_{2}}^{(n)}(|\psi^{\prime}\rangle)T^{t}$.
Then, from Eq. (LABEL:g-4) we obtain

$\displaystyle S_{q_{1}q_{2}}^{(n)}(\psi^{\prime})$ (5) $\displaystyle=$
$\displaystyle(\Pi_{\ell=3}^{n}\det\mathcal{A}_{q_{\ell}})\times$
$\displaystyle
T(\mathcal{A}_{q_{1}}\otimes\mathcal{A}_{q_{2}})T^{+}T\Omega_{q_{1}q_{2}}^{(n)}(|\psi\rangle)T^{t}T^{\ast}(\mathcal{A}_{q_{1}}\otimes\mathcal{A}_{q_{2}})^{t}T^{t}$
$\displaystyle=$ $\displaystyle kG_{1}S_{q_{1}q_{2}}^{(n)}(\psi)G_{1}^{-1}$

where $T^{\ast}$ is a conjugate matrix, $T^{t}T^{\ast}=I$, and
$k=\Pi_{\ell=1}^{n}\det\mathcal{A}_{\ell}$. Note that
$S_{q_{1}q_{2}}^{(n)}(\psi^{\prime})$ and $S_{q_{1}q_{2}}^{(n)}(\psi)$ are $4$
by $4$ matrices.

In this paper, we write the direct sum of standard Jordan blocks
$J_{n_{1}}(\lambda_{1})$,$\cdots$, and $J_{n_{j}}(\lambda_{j})$ as
$J_{n_{1}}(\lambda_{1})\cdots J_{n_{j}}(\lambda_{j})$. The Jordan block
$J_{1}(a)$ is simply written as $a$. We define that the two SJNFs
$J_{n_{1}}(\lambda_{1})\cdots J_{n_{j}}(\lambda_{j})$ and
$J_{n_{1}}(k\lambda_{1})\cdots J_{n_{j}}(k\lambda_{j})$, where $k\neq 0$, are
proportional.

Eq. (5) leads to the following theorem 1.

Theorem 1. If the states $|\psi^{\prime}\rangle$ and $|\psi\rangle$ of $n$
qubits satisfy Eq. (1), i.e. the state $|\psi^{\prime}\rangle$ is SLOCC
equivalent to $|\psi\rangle$, then

(1) if $S_{q_{1}q_{2}}^{(n)}(\psi)$ has the spectrum $\lambda_{1}$, $\cdots$,
$\lambda_{4}$, then $S_{q_{1}q_{2}}^{(n)}(\psi^{\prime})$ has the spectrum
$k\lambda_{1}$, $\cdots$, $k\lambda_{4}$, where
$k=\Pi_{\ell=1}^{n}\det\mathcal{A}_{\ell}$.

(2) if $S_{q_{1}q_{2}}^{(n)}(\psi)$ has the SJNF $J_{n_{1}}(\lambda_{1})\cdots
J_{n_{j}}(\lambda_{j})$, then $S_{q_{1}q_{2}}^{(n)}(\psi^{\prime})$ has the
SJNF $J_{n_{1}}(k\lambda_{1})\cdots J_{n_{j}}(k\lambda_{j})$, where
$k=\Pi_{\ell=1}^{n}\det\mathcal{A}_{\ell}$.

We give our argument as follows. Let
$\Gamma=G_{1}S_{q_{1}q_{2}}^{(n)}(\psi)G_{1}^{-1}$. Then,
$S_{q_{1}q_{2}}^{(n)}(\psi^{\prime})=k\Gamma$. Clearly, $\Gamma$ is similar to
$S_{q_{1}q_{2}}^{(n)}(\psi)$. Therefore, $\Gamma$ and
$S_{q_{1}q_{2}}^{(n)}(\psi)$ have the same spectrum and SJNF.

(1). It is clear that if $\Gamma$ has the spectrum $\lambda_{1}$,$\cdots$,
$\lambda_{4}$, then $S_{q_{1}q_{2}}^{(n)}(\psi^{\prime})$ has the spectrum
$k\lambda_{1}$,$\cdots$, $k\lambda_{4}$.

(2). There is an invertible matrix $H$ such that $\Gamma=HJH^{-1}$, where the
SJNF $J=J_{n_{1}}(\lambda_{1})\cdots J_{n_{j}}(\lambda_{j})$. Then,
$k\Gamma=HkJH^{-1}$. It is not hard to see that the SJNF of $kJ$ is
$J_{n_{1}}(k\lambda_{1})\cdots J_{n_{j}}(k\lambda_{j})$.

Example 1. We have the following SLOCC equivalent states of four qubits:
$L_{a_{4}}(a\neq 0)$ and $L_{a_{4}}(a=1)$ LDFQIC09 ; $L_{a_{2}0_{3\oplus
1}}(a\neq 0)$ and $L_{a_{2}0_{3\oplus 1}}(a=1)$ LDFQIC09 ; and
$L_{ab_{3}}^{\ast}(a=0)$ and $L_{ab_{3}}(a=0)$ LDFPRA12 . We list the SJNFs of
$S_{1,2}^{(4)}$ of the states in Table 1.

Table 1:  SJNFs of SLOCC equivalent states  state | $L_{a_{4}}(a\neq 0)$ | $L_{a_{4}}(a=1)$ | $k=a^{2}$
---|---|---|---
SJNF | $J_{4}(a^{2})$ | $J_{4}(1)$ |
State | $L_{a_{2}0_{3\oplus 1}}(a\neq 0)$ | $L_{a_{2}0_{3\oplus 1}}(a=1)$ | $k=a^{2}$
SJNF | $J_{2}(a^{2})J_{2}(0)$ | $J_{2}(1)J_{2}(0)$ |
State | $L_{ab_{3}}^{\ast}(a=0)$ | $L_{ab_{3}}(a=0)$ | $k=1$
SJNF | $0b^{2}J_{2}(0)$ | $0b^{2}J_{2}(0)$ |

Example 2. For four qubits, let
$\zeta_{4}=a(|0\rangle+|15\rangle)+b(|5\rangle+|10\rangle)+|6\rangle$, and
$\zeta_{5}=b(|0\rangle+|15\rangle)+a(|5\rangle+|10\rangle)+|6\rangle$, where
$a\neq b$. The SJNF of $S_{1,2}^{(4)}(\zeta_{4})$ is $J_{2}(b)aa$ while the
SJNF of $S_{1,2}^{(4)}(\zeta_{5})$ is $J_{2}(a)bb$. So, by (2) of Theorem 1
the two states $\zeta_{4}$ and $\zeta_{5}$ are SLOCC inequivalent.

We can rewrite $S_{q_{1}q_{2}}^{(n)}(\psi)$ as

$S_{q_{1}q_{2}}^{(n)}(\psi)=[TC_{q_{1}q_{2}}^{(n)}(|\psi\rangle)]\upsilon^{\otimes(n-2)}[TC_{q_{1}q_{2}}^{(n)}(|\psi\rangle)]^{t}.$
(6)

###  II.2 Partition pure states of $n$ ($\geq 4$) qubits into 12 groups and 34
families

Theorem 1 permits a reduction of SLOCC classification of $n$ ($\geq 4$) qubits
to a classification of $4$ by $4$ complex matrices. For $4$ by $4$ matrices, a
calculation yields 12 types of CPs (characteristic polynomials), 12 types of
spectrums, and 34 types of SJNFs in Table 2. It is easy to see that CPs and
spectrums have the same effect for SLOCC classification.

Note that in Table 2, $\sigma_{i}\neq 0$, $\sigma_{i}\neq\sigma_{j}$ when
$i\neq j$. Next, we give 12 types of CPs of $4$ by $4$ matrices as follows.

CP${}_{1}:(\sigma-\sigma_{1})^{4}$;
CP${}_{2}:(\sigma-\sigma_{1})(\sigma-\sigma_{2})^{3}$;
CP${}_{3}:(\sigma-\sigma_{1})(\sigma-\sigma_{2})(\sigma-\sigma_{3})^{2}$;
CP${}_{4}:(\sigma-\sigma_{1})^{2}(\sigma-\sigma_{2})^{2}$;
CP${}_{5}:\Pi_{i=1}^{4}(\sigma-\sigma_{i})$;
CP${}_{6}:\sigma(\sigma-\sigma_{1})^{3}$;
CP${}_{7}:\sigma(\sigma-\sigma_{1})(\sigma-\sigma_{2})^{2}$;
CP${}_{8}:\sigma\Pi_{i=1}^{3}(\sigma-\sigma_{i})$;
CP${}_{9}:\sigma^{2}(\sigma-\sigma_{1})^{2}$;
CP${}_{10}:\sigma^{2}(\sigma-\sigma_{1})(\sigma-\sigma_{2})$;
CP${}_{11}:\sigma^{3}(\sigma-\sigma_{1})$; CP${}_{12}:\sigma^{4}$.

For each state of $n$ ($\geq 4$) qubits, the spectrum of
$S_{q_{1}q_{2}}^{(n)}\ $must belong to one of the 12 types of the spectrums in
Table 2. Let the states of $n$ ($\geq 4$) qubits, for which spectrums of
$S_{q_{1}q_{2}}^{(n)}$ possess the same type in Table 2, belong to the same
group. Thus, the states of $n$ ($\geq 4$) qubits are partitioned into 12
groups. In light of Theorem 1, the states belonging to different groups must
be SLOCC inequivalent.

For each state of $n$ ($\geq 4$) qubits, the SJNF of $S_{q_{1}q_{2}}^{(n)}$ up
to the order of the standard Jordan blocks must belong to one of the 34 types
of the SJNFs in Table 2. Let the states of $n$ ($\geq 4$) qubits with the same
type of SJNFs of $S_{q_{1}q_{2}}^{(n)}$ in Table 2 up to the order of the
Jordan blocks belong to the same family. Thus, we partition the states of $n$
($\geq 4$) qubits into 34 families. In light of Theorem 1, the states
belonging to different families must be SLOCC inequivalent.

Table 2: 12 types of CPs, 12 types of spectrums, 34 types of the SJNFs for 4 by 4 matrices, and the corresponding states for four qubits. CPi;spectrum | SJNF | state | SJNF | state
---|---|---|---|---
1;$\sigma_{1}\sigma_{1}\sigma_{1}\sigma_{1}$ | $J_{4}(\sigma_{1})\text{ }$ | $\tau_{1}$ | $J_{2}(\sigma_{1})J_{2}(\sigma_{1})$ | $\eta_{1}$
| $J_{3}(\sigma_{1})\sigma_{1}$ | $\theta_{1}$ | $\sigma_{1}\sigma_{1}{}J_{2}(\sigma_{1})$ | $\zeta_{1}$
| $\sigma_{1}\sigma_{1}\sigma_{1}\sigma_{1}{}$ | $G_{1}$ |  |
2;$\sigma_{1}\sigma_{2}\sigma_{2}\sigma_{2}$ | $\sigma_{1}J_{3}(\sigma_{2})$ | $\theta_{2}$ | $\sigma_{1}\sigma_{2}J_{2}(\sigma_{2})$ | $\zeta_{2}$
| $\sigma_{1}\sigma_{2}\sigma_{2}\sigma_{2}{}$ | $G_{2}$ |  |
3;$\sigma_{1}\sigma_{2}\sigma_{3}\sigma_{3}$ | $\sigma_{1}\sigma_{2}J_{2}(\sigma_{3})$ | $\zeta_{3}$ | $\sigma_{1}\sigma_{2}\sigma_{3}\sigma_{3}{}$ | $G_{3}$
4;$\sigma_{1}\sigma_{1}{}\sigma_{2}\sigma_{2}$ | $\sigma_{1}\sigma_{1}{}\sigma_{2}\sigma_{2}{}$ | $G_{4}$ | $\sigma_{1}\sigma_{1}{}J_{2}(\sigma_{2})$ | $\zeta_{4}$
| $J_{2}(\sigma_{1})J_{2}(\sigma_{2})$ | $\eta_{2}$ |  |
5;$\sigma_{1}\sigma_{2}\sigma_{3}\sigma_{4}$ | $\sigma_{1}\sigma_{2}\sigma_{3}\sigma_{4}$ | $G_{5}$ |  |
6;0$\sigma_{1}\sigma_{1}\sigma_{1}$ | $0J_{3}(\sigma_{1})$ | $\theta_{3}$ | $0J_{2}(\sigma_{1})\sigma_{1}$ | $\zeta_{6}$
| $0\sigma_{1}\sigma_{1}\sigma_{1}{}$ | $G_{6}$ |  |
7;0$\sigma_{1}\sigma_{2}\sigma_{2}$ | $0\sigma_{1}J_{2}(\sigma_{2})$ | $\zeta_{7}$ | $0\sigma_{1}\sigma_{2}\sigma_{2}{}$ | $G_{7}$
8;$0\sigma_{1}\sigma_{2}\sigma_{3}$ | $0\sigma_{1}\sigma_{2}\sigma_{3}$ | $G_{8}$ |  |
9;$00\sigma_{1}\sigma_{1}$ | $J_{2}(0)J_{2}(\sigma_{1})$ | $\kappa_{1}$ | $J_{2}(0)\sigma_{1}\sigma_{1}{}$ | $\mu_{2}$
| $00J_{2}(\sigma_{1})$ | $\zeta_{9}$ | $00\sigma_{1}\sigma_{1}{}$ | $\zeta_{8}$
10;$00\sigma_{1}\sigma_{2}$ | $J_{2}(0)\sigma_{1}\sigma_{2}$ | $\mu_{1}$ | $00\sigma_{1}\sigma_{2}$ | $\zeta_{10}$
11;$000\sigma_{1}$ | $J_{3}(0)\sigma_{1}$ | $\xi_{1}$ | $J_{2}(0)0\sigma_{1}$ | $\theta_{4}$
| $000\sigma_{1}$ | $\zeta_{11}$ |  |
12;0000 | $J_{4}(0)$ | $L_{0_{7\oplus 1}}$ | $J_{3}(0)0$ | $\xi_{2}$
| $J_{2}(0)J_{2}(0)$ | $\tau_{2}$ | $J_{2}(0)00$ | $\theta_{5}$
| 0000 | $\zeta_{12}$ |  |

Example 3. For the maximally entangled states $|\Psi_{2}\rangle$,
$|\Psi_{4}\rangle-|\Psi_{6}\rangle$ of five qubits and $|\Xi_{2}\rangle$,
$|\Xi_{4}\rangle-|\Xi_{7}\rangle$ of six qubits Osterloh06 , SJNFs of
$S_{1,2}^{(5)}$ partition $|\Psi_{2}\rangle$,
$|\Psi_{4}\rangle-|\Psi_{6}\rangle$ into three families, and SJNFs of
$S_{1,2}^{(6)}$ partition $|\Xi_{2}\rangle$, $|\Xi_{4}\rangle-|\Xi_{7}\rangle$
into four families. See Table 3.

Table 3:  SJNFs of $S_{1,2}^{(5)}$ and $S_{1,2}^{(6)}$. states | $|\Psi_{2}\rangle$ | $|\Psi_{4}\rangle$ | $|\Psi_{5}\rangle$ | $|\Psi_{6}\rangle$ |
---|---|---|---|---|---
SJNFs | $\pm\frac{1}{2}00$ | $0000$ | $0000$ | $0J_{3}(0)$ |
states | $|\Xi_{2}\rangle$ | $|\Xi_{4}\rangle$ | $|\Xi_{5}\rangle$ | $|\Xi_{6}\rangle$ | $|\Xi_{7}\rangle$
SJNFs | $(\frac{1}{2})(\frac{1}{2})00$ | $0000$ | $0000$ | $00J_{2}(0)$ | $0J_{3}(0)$

##  III SLOCC classification of two, three, and four qubits

###  III.1 SLOCC classification of four qubits

For four qubits, invoking the fact that $T^{+}T^{\ast}=\upsilon^{\otimes 2}$
Eq. (6) reduces to

$S_{q_{1}q_{2}}^{(4)}(\psi)=(TC_{q_{1}q_{2}}^{(4)}T^{+})(TC_{q_{1}q_{2}}^{(4)}T^{+})^{t}.$
(7)

From the above discussion, in light of Theorem 1 pure states of four qubits
are partitioned into 12 groups and 34 families in Table 2. Furthermore, for
each type of spectrums, CPs, and SJNFs in Table 2, we give a state in Table 2
and the appendix for which $S_{1,2}^{(4)}$ has the corresponding type. For
example, $S_{1,2}^{(4)}\ $of the state $\theta_{1}$ has the spectrum
$\sigma_{1}$, $\sigma_{1}$, $\sigma_{1}$, $\sigma_{1}$, the CP
$(\sigma-\sigma_{1})^{4}$, and the SJNF $J_{3}(\sigma_{1})\sigma_{1}$. It is
plain to see that 12 groups and 34 families in Table 2 are both complete for
four qubits.

Here, we make a comparison to Verstraete et al.’s nine families. They showed
that for a complex $n$ by $n$ matrix, there are complex orthogonal matrices
$O_{1}$ and $O_{2}$ such that $R=O_{1}R^{\prime}O_{2}$, where $R^{\prime}$ is
a direct sum of blocks defined in Verstraete . Note that the blocks are not
standard Jordan blocks. The decomposition was called a generalization of the
singular value decomposition and used to partition pure states of four qubits
into nine families Verstraete .

Recently, Chterental and Djoković pointed out an error in Verstraete et al.’s
nine families by indicating that the family$\ L_{ab_{3}}$ is SLOCC equivalent
to the subfamily $L_{abc_{2}}(a=c)$ of the family $L_{abc_{2}}$ Chterental
LDFPRA15 . Thus, the classification for the nine families is incomplete. The
need to redo this classification of four qubits was proposed Chterental .

###  III.2 SLOCC classification of three qubits

For three qubits, Eq. (6) reduces to

$S_{q_{1}q_{2}}^{(3)}(\psi)=[TC_{q_{1}q_{2}}^{(3)}(|\psi\rangle)]\upsilon[TC_{q_{1}q_{2}}^{(3)}(|\psi\rangle)]^{t}.$
(8)

Let
$\lambda^{2}=[(c_{0}c_{7}-c_{1}c_{6})-(c_{2}c_{5}-c_{3}c_{4})]^{2}-4(c_{0}c_{3}-c_{1}c_{2})(c_{4}c_{7}-c_{5}c_{6})$.
Note that $\lambda^{2}$ is just the $3$-tangle. The spectrum of
$S_{1,2}^{(3)}(\psi)$ is $\pm\lambda,0,0$. We list the SJNFs of
$S_{1,2}^{(3)}(\psi)$ and $S_{1,3}^{(3)}(\psi)$ in the Table 4. In light of
Theorem 1, we can distinguish the six SLOCC classes of three qubits.

Table 4: SLOCC classification of three qubits states | SJNF of $S_{1,2}^{(3)}(\psi)$ | SJNF of $S_{1,3}^{(3)}(\psi)$
---|---|---
GHZ | $J_{1}(\pm\frac{1}{2})00$ | $J_{1}(\pm\frac{1}{2})00$
W | $J_{3}(0)0$ | $J_{3}(0)0$
A-BC | $J_{2}(0)J_{2}(0)$ | $J_{2}(0)J_{2}(0)$
B-AC | $J_{2}(0)J_{2}(0)$ | $0000$
C-AB | $0000$ | $J_{2}(0)J_{2}(0)$
$|000\rangle$ | $0000$ | $0000$

###  III.3 SLOCC classification of two qubits

For two qubits, Eq. (6) reduces to

$S_{1,2}^{(2)}(\psi)=[TC_{1,2}^{(2)}(|\psi\rangle)][TC_{1,2}^{(2)}(|\psi\rangle)]^{t}.$
(9)

The spectrum of $S_{1,2}^{(2)}(\psi)$ is 0, 0, 0, $\lambda^{\prime}$, where
$\lambda^{\prime}=2(a_{0}a_{3}-a_{1}a_{2})$. There are two cases for SJNFs.
Case 1. For which $a_{0}a_{3}=a_{1}a_{2}$ (it is a separate state), the SJNF
of $S_{1,2}^{(2)}(\psi)$ is $J_{2}(0)00$. Case 2. For which $a_{0}a_{3}\neq
a_{1}a_{2}$ (it is an entangled state), the SJNF of $S_{1,2}^{(2)}(\psi)$ is
$\lambda^{\prime}000$. Thus, in light of Theorem 1, we can distinguish two-
qubit states into two SLOCC classes.

##  IV SLOCC classification of $n$ qubits under $\mathcal{A}_{i}\in SL(2,C)$

SLOCC classification under $A_{i}\in SL(2,C)$ or the classification under
determinant one SLOCC operations was discussed in previous articles Verstraete
Luque . Note that under $\mathcal{A}_{i}\in SL(2,C)$, $G_{1}\in SO(4,C)$ and
Eq. (5) reduces to

$S_{q_{1}q_{2}}^{(n)}(\psi^{\prime})=G_{1}S_{q_{1}q_{2}}^{(n)}(\psi)G_{1}^{-1}.$
(10)

Thus, Eq. (10) leads to the following theorem.

Theorem 2. If the states $|\psi^{\prime}\rangle$ and $|\psi\rangle$ of $n$
qubits are SLOCC equivalent under $\mathcal{A}_{i}\in SL(2,C)$, then
$S_{q_{1}q_{2}}^{(n)}(\psi^{\prime})$ is orthogonally similar to
$S_{q_{1}q_{2}}^{(n)}(\psi)$. The similarity implies that
$S_{q_{1}q_{2}}^{(n)}(\psi^{\prime})$ and $S_{q_{1}q_{2}}^{(n)}(\psi)$ have
the same CP, spectrum, and SJNF up to the order of the standard Jordan blocks.

Example 4. $L_{ab_{3}}^{\ast}(a=0)$ is SLOCC equivalent to $L_{ab_{3}}(a=0)$
under $A_{i}\in SL(2,C)$ LDFPRA12 . The SJNFs of $S_{1,2}^{(4)}$ are both
$0b^{2}J_{2}(0)$.

Restated in the contrapositive the theorem reads: If two matrices
$S_{q_{1}q_{2}}^{(n)}$ associated with two n-qubit pure states differ in their
CPs, spectrums, or SJNFs, then the two states are SLOCC inequivalent under
$\mathcal{A}_{i}\in SL(2,C)$. From Example 2, by Theorem 2 the two states
$\zeta_{4}$ and $\zeta_{5}$ are SLOCC inequivalent under $A_{i}\in SL(2;C)$
because SJNFs of $S_{1,2}^{(4)}(\zeta_{4})$ and $S_{1,2}^{(4)}(\zeta_{5})$ are
different.

Note that a SLOCC equivalent class may include infinite SLOCC equivalent
classes under $\mathcal{A}_{i}\in SL(2,C)$.

##  V Conclusion

In Theorem 1, we demonstrate that for two SLOCC equivalent states, the
spectrums and SJNFs of the matrices $S_{q_{1}q_{2}}^{(n)}$ have proportional
relationships. Invoking the proportional relationships, we partition pure
states of $n$ ($\geq 4$) qubits into 12 groups and 34 families under SLOCC,
respectively.

In Theorem 2, we deduce that for two equivalent states under determinant one
SLOCC operations, the spectrums, CPs, SJNFs of $S_{q_{1}q_{2}}^{(n)}$ are
invariant. The invariance can be used for SLOCC classification of $n$ qubits
under determinant one SLOCC operations.

To make a comparison, we list the differences between Theorems 1 and 2 in
Table 5.

It is known that SJNF is used to solve a system of linear differential
equations. The classification of SJNFs under SLOCC in this paper seems to be
useful for classifying linear differential systems.

Table 5: Comparison between Theorems 1 and 2 | Theorem 1 | Theorem 2
---|---|---
spect. $\psi$ |  $\lambda_{1}$, $\cdots$, $\lambda_{4}$ |  $\lambda_{1}$, $\cdots$, $\lambda_{4}$
spect. $\psi^{\prime}$ |  $k\lambda_{1}$, $\cdots$, $k\lambda_{4}$ |  $\lambda_{1}$, $\cdots$, $\lambda_{4}$
SJNF $\psi$ | $J_{\ell_{1}}(\lambda_{1})\cdots J_{\ell_{j}}(\lambda_{j})$ | $J_{\ell_{1}}(\lambda_{1})\cdots J_{\ell_{j}}(\lambda_{j})$
SJNF $\psi^{\prime}$ | $J_{\ell_{1}}(k\lambda_{1})\cdots J_{\ell_{j}}(k\lambda_{j})$ | $J_{\ell_{1}}(\lambda_{1})\cdots J_{\ell_{j}}(\lambda_{j})$

Acknowledgement—This work was supported by NSFC (Grant No. 10875061) and
Tsinghua National Laboratory for Information Science and Technology.

##  VI Appendix Corresponding states of four qubits

Using $G_{abcd}$ we obtain the following 8 states.

$G_{1}=G_{abcd}(a=b=c=d\neq 0)$; (we will omit $G_{abcd}$ next);

$G_{2}:abcd\neq$ $0;b=c=d$ but $a\neq b$;

$G_{3}:abcd\neq$ $0$, two of $a,b,c$, and $d$ are equal while the other two
are not equal;

$G_{4}:abcd\neq 0$, $a,b$, $c$, and $d$ consists of two pairs of equal
numbers;

$G_{5}:abcd\neq$ $0,$ $a,b$, $c$, and $d$ are distinct;

$G_{6}:$ only one of $a,b$, $c$, and $d$ is zero and other three are equal;

$G_{7}:$ only one of $a,b$, $c$, and $d$ is zero and only two of them are
equal;

$G_{8}:$ only one of $a,b$, $c$, and $d$ is zero and the other three are
distinct.

Using $L_{abc_{2}}$ we obtain the following 11 states.

$\zeta_{1}=L_{abc_{2}}($ $a=b=c\neq$ $0);$

$\zeta_{2}=L_{abc_{2}}(abc\neq 0$ and one of $a$ and $b$ equals c$)$;

$\zeta_{3}=L_{abc_{2}}(abc\neq 0$ and $a,b,c$ are distinct.$)$;

$\zeta_{4}=a(|0\rangle+|15\rangle)+b(|5\rangle+|10\rangle)+|6\rangle$, where
$a\neq b$;

$\zeta_{6}=L_{abc_{2}}($ only one of $a$ and $b$ is zero while the other is
equal to $c$.$)$;

$\zeta_{7}=L_{abc_{2}}(c\neq 0$ and only one of $a$ and $b$ is zero while the
other is not equal to $c$.$)$;

$\zeta_{8}=L_{abc_{2}}(c=0$ and $a=b\neq 0)$; $\zeta_{9}=L_{abc_{2}}(c\neq 0$
and $a=b=0)$; $\zeta_{10}=L_{abc_{2}}(c=0$ while $ab\neq 0$ and $a\neq b)$;
$\zeta_{11}=L_{abc_{2}}(c=0$ while only one of $a$ and $b$ is zero$)$;
$\zeta_{12}=L_{abc_{2}}(a=b=c=0)$.

Using $L_{a_{2}b_{2}}$ we obtain the following two states.

$\eta_{1}=L_{a_{2}b_{2}}(a=b\neq 0)$; $\eta_{2}=L_{a_{2}b_{2}}(ab\neq 0$ and
$a\neq b)$.

Let
$L_{ab_{3}}^{\prime}=b(|0\rangle+|15\rangle)+\frac{b+a}{2}(|5\rangle+|10\rangle)+\frac{b-a}{2}(|6\rangle+|9\rangle)+\frac{i}{\sqrt{2}}(|1\rangle+|2\rangle-|7\rangle-|11\rangle)$.
Using $L_{ab_{3}}^{\prime}$ we obtain the following five states.

$\theta_{1}=L_{ab_{3}}^{\prime}(a=b\neq 0)$;

$\theta_{2}=L_{ab_{3}}^{\prime}(ab\neq 0$ and $a\neq b)$;

$\theta_{3}=a(|0\rangle+|15\rangle)+\frac{a}{2}(|5\rangle+|10\rangle+|6\rangle+|9\rangle)+\frac{i}{\sqrt{2}}(|1\rangle+|2\rangle-|7\rangle-|11\rangle)$
(obtained from $L_{ab_{3}}^{\prime}(a=0$ but $b\neq 0)$);

$\theta_{4}=L_{ab_{3}}^{\prime}(b=0$ but $a\neq 0)$;
$\theta_{5}=L_{ab_{3}}^{\prime}(a=b=0)$.

Using $L_{a_{4}}$ we obtain the following two states.

$\tau_{1}=L_{a_{4}}(a\neq 0)$; $\tau_{2}=L_{a_{4}}(a=0)$.

Using $L_{a_{2}0_{3\oplus 1}}$ we obtain the following one state.

$\kappa_{1}=L_{a_{2}0_{3\oplus 1}}(a\neq 0)$.

Let $L_{ab0_{3\oplus
1}}=\frac{a+b}{2}(|0\rangle+|15\rangle)+\frac{a-b}{2}(|3\rangle+|12\rangle)+|5\rangle+|6\rangle$.
Using $L_{ab0_{3\oplus 1}}$ we obtain the following two states.

$\mu_{1}=L_{ab0_{3\oplus 1}}(ab\neq 0$ and $a\neq b)$;
$\mu_{2}=L_{ab0_{3\oplus 1}}(a=b\neq 0)$ ;

Let
$\xi=\frac{a}{2}(|0\rangle+|3\rangle+|12\rangle+|15\rangle)+i|1\rangle-i|13\rangle+|10\rangle$.
Using $\xi$ we obtain the following two states.

$\xi_{1}=\xi(a\neq 0)$; $\xi_{2}=\xi(a=0)$.

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