# Ruled Ricci surfaces and curves of constant torsion

Alcides de Carvalho, Iury Domingos and Roney Santos  Universidade Federal de
Alagoas
Instituto de Matemática
Campus A. C. Simões, BR 104 - Norte, Km 97, 57072-970, Maceió - AL, Brazil.
<EMAIL_ADDRESS>Universidade Federal de Alagoas
Av. Manoel Severino Barbosa S/N, 57309-005 Arapiraca - AL, Brazil.
<EMAIL_ADDRESS>Universidade de São Paulo
Departamento de Matemática
Rua do Matão, 05508-900, São Paulo - SP, Brazil<EMAIL_ADDRESS>
###### Abstract.

We show that all non-developable ruled surfaces endowed with Ricci metrics in
the three-dimensional Euclidean space may be constructed using curves of
constant torsion and its binormal. This allows us to give characterizations of
the helicoid as the only surface of this kind that admits a parametrization
with plane line of striction, and as the only with constant mean curvature.

###### Key words and phrases:

Ricci surfaces, minimal surfaces, ruled surfaces

###### 2020 Mathematics Subject Classification:

53A10

A. de Carvalho was partially supported by CNPq grant 150285/2023-0 as well as
FAPEAL under the process E:60030.0000000513/2023. I. Domingos was partially
supported by the Brazilian National Council for Scientific and Technological
Development (CNPq), grant 409513/2023-7. R. Santos was partially supported by
grant 2023/14796-3, São Paulo Research Foundation (FAPESP)

##  1\. Introduction

Although very old, the theory of minimal surfaces remains an active and
interesting field in Differential Geometry. These are the surfaces in a
Riemannian 3-manifold that are critical points of the area functional.
Equivalently, a surface is minimal if its mean curvature vanishes identically.

In 1873, Schläfli posed an enduring natural question about isometric
immersions that remains unanswered to this day [12, 13]: can every Riemannian
surface be locally isometrically embedded in the flat space $\mathbb{R}^{3}$?
A partially affirmative response was achieved in 2003 by Han-Hong [6]
contingent upon the behavior of the gradient of the Gauss curvature in the
neighborhood of its zeroes.

In 1895, Ricci-Curbastro raised a related question [10]: under what conditions
does a Riemannian surface support local minimal isometric embedding in the
flat space $\mathbb{R}^{3}$? He gave a partial answer in the same work for the
case of strictly negative Gauss curvature $K$ satisfying $4K=\Delta\log(-K).$

Finally, in 2015, Moroianu-Moroianu [9] given a full positive answer to the
Ricci-Curbastro problem. They showed that a Riemannian surface admits local
isometric embedding in $\mathbb{R}^{3}$ as minimal surface when its Gauss
curvature $K$ is non-positive and satisfies

$K\Delta K-\|\nabla K\|^{2}-4K^{3}=0.$

The concept of Ricci surface was introduced by Moroianu-Moroianu in their work
[9]. Remember that we call Ricci surface a Riemannian surface whose Gauss
curvature satisfies the above equation, which is called Ricci condition.
Furthermore, its metric is referred as Ricci metric. Among other results, they
proved that $K$ does not change sign, and either $K$ vanishes identically or
has only isolated zeroes.

Since Ricci surfaces are an “intrinsic” way to see minimal surfaces of
$\mathbb{R}^{3},$ it is natural the interest in this kind of surfaces. The
theory of Ricci surfaces has attracted considerable interest recently and has
been developed in several works both from the intrinsic point of view [1, 3,
9, 14] and as immersed surfaces in $\mathbb{R}^{3}$ [5].

Our goal in this work is to study ruled Ricci surfaces, namely, ruled surfaces
of $\mathbb{R}^{3}$ whose induced metrics are Ricci. Initially we focus on
their metric classification: we aim to identify the conditions under which a
certain given family of metrics can be classified as Ricci metrics.
Subsequently, we transition to the process of obtain the immersion of these
surfaces. In this direction, we discovered a relation between ruled Ricci
surfaces and curves with constant torsion. This correspondence provides a full
classification of non-developable ruled Ricci surfaces, and shows that this
class of surfaces is as numerous as the class of curves of constant torsion.
As consequences of our results, we obtain that the helicoid is the only
surface of this kind whose mean curvature is constant, and which may be
parametrized with line of striction contained in some plane.

######  Acknowledgements.

We are grateful to L. Ambrozio for your valuable comments on this work.

##  2\. Preliminaries

We will remember some properties of ruled surfaces in order to fix notations.
Let us consider two curves $\alpha,\beta\mathrel{\mathop{\mathchar
58\relax}}I\to\mathbb{R}^{3}$ such that $\beta(t)\neq 0$ for each $t\in I.$ We
call $\Sigma$ a ruled surface if it can be parametrized by the immersion
$X\mathrel{\mathop{\mathchar 58\relax}}I\times J\to\mathbb{R}^{3}$ given as

$X(t,u)=\alpha(t)+u\beta(t).$

One say that the curves $\\{\alpha(t),\beta(t)\\}$ generates $\Sigma,$ and
that $\alpha$ is the line of striction corresponding to $\beta$ when they are
orthogonal.

Before proceeding, we need to set some properties of the curves generating a
ruled surface. As we will see, the non-developable ruled Ricci surfaces we
consider admits parametrization whose line of striction has constant non-zero
speed. So, let us remember some facts about regular curves in
$\mathbb{R}^{3}.$ For more details, we refer the reader to [4].

By considering $\alpha$ regular, let us apply a change of parameters $s=s(t)$
where $s$ is the arc length of $\alpha$ and reparametrize $\Sigma$ as

$X^{\prime}(s,u)=\alpha(s)+u\beta(s),\ (s,u)\in I^{\prime}\times J.$

For our purpose, let us assume for now that $\alpha$ is not necessarily
orthogonal to $\beta.$ At each point $\alpha(s),$ with
$\alpha^{\prime\prime}(s)\neq 0,$ we have associated an orthonormal frame
$\\{T(s),N(s),B(s)\\},$ called Frenet frame, that is formed respectively by
the unit tangent, the unit normal and the unit binormal vectors of $\alpha$ at
$\alpha(s).$ More precisely, we have $\alpha^{\prime}(s)=T(s),$
$\alpha^{\prime\prime}(s)=\kappa(s)N(s)$ for a function
$\kappa\mathrel{\mathop{\mathchar 58\relax}}I^{\prime}\to\mathbb{R}$ called
curvature of $\alpha$ and $B(s)=T(s)\wedge N(s).$ These vector fields obey the
well-known Frenet formulas

$T^{\prime}=\kappa N,\ \ \ N^{\prime}=-{\kappa T-\tau B},\ \ \ B^{\prime}=\tau
N.$

We have omitted the parameter $s$ and the prime means the derivative with
respect to $s.$ The function $\tau\mathrel{\mathop{\mathchar
58\relax}}I^{\prime}\to\mathbb{R}$ is called torsion of $\alpha.$ For our
objective, let’s assume that the zeros of the curvature of $\alpha$ are
isolated.

A standard way to change $\alpha$ with the line of striction
$\widetilde{\alpha}$ of $\beta$ is as follows. We write
$\beta=\beta_{1}T+\beta_{2}N+\beta_{3}B$ and consider

$\widetilde{\alpha}(s)=\alpha(s)-h(s)\beta(s),$

for some function $h\mathrel{\mathop{\mathchar
58\relax}}I^{\prime}\to\mathbb{R}$ which we want to find. By admitting
$\langle\widetilde{\alpha}^{\prime},\beta^{\prime}\rangle=0$ in $I^{\prime},$
a straightforward computation using Frenet formulas gives that

$h(s)=\frac{\beta_{1}^{\prime}(s)-\kappa(s)\beta_{2}(s)}{\|\beta^{\prime}(s)\|^{2}}.$

In particular, if the original curve $\alpha$ is the line of striction of
$\beta,$ we must have

$\beta_{1}^{\prime}(s)=\kappa(s)\beta_{2}(s).$ (1)

From now on, we will not assume that $\alpha$ is necessarily regular.
Regarding the curve $\beta,$ we begin by noting that it can be assumed,
without loss of generality, that its trace is contained in $\mathbb{S}^{2}$
and that $\alpha$ is its line of striction (see [4]). Moreover, to be able to
develop the theory, we need the non-trivial assumption that $\beta^{\prime}$
has no zero in $I.$ In the case where the zeroes of $\beta^{\prime}$ are
isolated, we can divide our surface into pieces such that the theory can be
applied to each of them. However, if the zeroes of $\beta^{\prime}$ have
cluster points, the situation may become complicated and we will avoid it
here. The assumption that $\beta$ is a regular curve in $I$ is usually
expressed by saying that $\Sigma$ is non-cylindrical.

Under all considerations above, unless otherwise stated, from now on we shall
assume that

$X(t,u)=\alpha(t)+u\beta(t)$

is a parametrization of a non-cylindrical ruled surface whose generating
curves are orthogonal and $\|\beta(t)\|=\|\beta^{\prime}(t)\|=1$ for all $t\in
I$. The induced metric in this setting is expressed as

$g=\left(\|\alpha^{\prime}(t)\|^{2}+u^{2}\right)dt^{2}+2\langle\alpha^{\prime}(t),\beta(t)\rangle
dtdu+du^{2}.$

The assumption that $\beta$ is a spherical curve implies that
$\langle\beta(t),\beta^{\prime}(t)\rangle=0$ for all $t\in I$. Therefore, we
get
$\langle\alpha^{\prime}(t),\beta^{\prime}(t)\rangle=\langle\beta(t),\beta^{\prime}(t)\rangle=0,$
and consequently there exists a function $\lambda\mathrel{\mathop{\mathchar
58\relax}}I\to\mathbb{R}^{3}$ such that
$\alpha^{\prime}(t)\wedge\beta(t)=\lambda(t)\beta^{\prime}(t),$ where $\wedge$
denotes the cross product. This function $\lambda$ is called distribuition
parameter of $X.$ Since

$\|\alpha^{\prime}(t)\|^{2}+u^{2}-\langle\alpha^{\prime}(t),\beta(t)\rangle^{2}=\|X_{t}(t,u)\wedge
X_{u}(t,u)\|^{2}=\lambda(t)^{2}+u^{2},$

without loss of generality $g$ can be rewritten as

$g=\left(\|\alpha^{\prime}(t)\|^{2}+u^{2}\right)dt^{2}+2\sqrt{\|\alpha^{\prime}(t)\|^{2}-\lambda(t)^{2}}dtdu+du^{2}.$
(2)

We say that a ruled surface is developable if its distribution parameter
vanishes identicaly. Otherwise, we say that the ruled surface is non-
developable. The relation between the Gauss curvature $K$ and the distribution
parameter $\lambda$ of $X$ is given by the formula

$K(t,u)=-\frac{\lambda(t)^{2}}{\left(\lambda(t)^{2}+u^{2}\right)^{2}}.$

In particular, the distribution parameter of an immersion of a ruled Ricci
surface vanishes either identically or at isolated points due to the property
on the zeroes of the Gauss curvature of a Ricci surface. The first case
trivially implies that the considered ruled surface must be a Ricci surface.
For this reason, we will focus only on the case of non-developable ruled Ricci
surfaces.

##  3\. Ruled Ricci surfaces

As we will see, the Ricci condition imposes strong restrictions to a certain
class of metrics that generalize the one of ruled surfaces. Precisely, on the
case of ruled surfaces, we will show that the distribution parameter and the
speed of the line of striction of a given parametrization of a ruled Ricci
surface are constant and equal, up to sign.

###  3.1. Metric classification

In this section we will obtain necessary and sufficient condition for a class
of metrics that have as particular cases the metrics induced by ruled surfaces
to be Ricci metrics.

######  Lemma 1.

Let $g$ be a non-flat Riemannian metric given as

$g=\left(f(t)^{2}+u^{2}\right)dt^{2}+2\sqrt{f(t)^{2}-\lambda(t)^{2}}dtdu+du^{2},\
(t,u)\in I\times J,$

where $f,\lambda\mathrel{\mathop{\mathchar 58\relax}}I\to\mathbb{R}$ are
functions with $f(t)^{2}\geq\lambda(t)^{2}$. Then, $g$ is a Ricci metric if
and only if $\lambda(t)^{2}=f(t)^{2}=c^{2}$ for some $c>0$ and each $t\in I.$
In particular, we have $g=(c^{2}+u^{2})dt^{2}+du^{2}.$

###### Proof.

Suppose $g$ is a Ricci metric. Firstly, we will show $\lambda(t)^{2}=f(t)^{2}$
in $I.$ To see this, let us admit that $\lambda(t_{0})^{2}\neq f(t_{0})^{2}$
for some $t_{0}\in I.$ In this case, by continuity we must have
$\lambda(t)^{2}\neq f(t)^{2}$ at some interval $I_{0}\subset I$ containing
$t_{0}.$ From now on, for simplicity we will omit the parameter $t$ and use a
prime to denote the derivative with respect to $t.$ Since the curvature $K$ of
$g$ is given as

$K=-\frac{\lambda^{2}}{\left(\lambda^{2}+u^{2}\right)^{2}},$

we get

$K_{t}=-\frac{2\lambda\lambda^{\prime}\left(u^{2}-\lambda^{2}\right)}{\left(\lambda^{2}+u^{2}\right)^{3}}\
\ \ \mbox{and}\ \ \
K_{u}=\frac{4u\lambda^{2}}{\left(\lambda^{2}+u^{2}\right)^{3}}.$

Writing $\delta(t)=\sqrt{f(t)^{2}-\lambda(t)^{2}},$ by means of direct
computations one can see that

$\displaystyle\|\nabla K\|^{2}$
$\displaystyle=\operatorname{trace}\left(\left(\begin{array}[]{cc}f^{2}+u^{2}&\delta\\\
\delta&1\end{array}\right)^{-1}\left(\begin{array}[]{cc}K_{t}^{2}&K_{t}K_{u}\\\
K_{t}K_{u}&K_{u}^{2}\end{array}\right)\right)$
$\displaystyle=\frac{4\lambda^{2}}{\left(\lambda^{2}+u^{2}\right)^{7}}\operatorname{trace}\left(\left(\begin{array}[]{cc}1&-\delta\\\
-\delta&f^{2}+u^{2}\end{array}\right)\left(\begin{array}[]{cc}(\lambda^{\prime})^{2}(u^{2}-\lambda^{2})^{2}&-2u\lambda\lambda^{\prime}(u^{2}-\lambda^{2})\\\
-2u\lambda\lambda^{\prime}(u^{2}-\lambda^{2})&4u^{2}\lambda^{2}\end{array}\right)\right)$

and

$\displaystyle\Delta K$
$\displaystyle=\frac{1}{\sqrt{\lambda^{2}+u^{2}}}\left(\left(\frac{K_{t}}{\sqrt{\lambda^{2}+u^{2}}}-\frac{\delta
K_{u}}{\sqrt{\lambda^{2}+u^{2}}}\right)_{t}+\left(-\frac{\delta
K_{t}}{\sqrt{\lambda^{2}+u^{2}}}+\frac{(f^{2}+u^{2})K_{u}}{\sqrt{\lambda^{2}+u^{2}}}\right)_{u}\right)$
$\displaystyle=\frac{1}{\sqrt{\lambda^{2}+u^{2}}}\left(\left(-\frac{2\lambda\lambda^{\prime}\left(u^{2}-\lambda^{2}\right)+4u\delta\lambda^{2}}{\sqrt{(\lambda^{2}+u^{2})^{7}}}\right)_{t}+\left(\frac{2\delta\lambda\lambda^{\prime}\left(u^{2}-\lambda^{2}\right)+4u(f^{2}+u^{2})\lambda^{2}}{\sqrt{(\lambda^{2}+u^{2})^{7}}}\right)_{u}\right).$

Therefore,

$\begin{split}\|\nabla
K\|^{2}&=\frac{4\lambda^{2}}{(\lambda^{2}+u^{2})^{7}}(a_{4}u^{4}+a_{3}u^{3}+a_{2}u^{2}+a_{1}u+a_{0}),\\\
K\Delta
K&=-\frac{2\lambda^{2}}{\delta(\lambda^{2}+u^{2})^{7}}(b_{4}u^{4}+b_{3}u^{3}+b_{2}u^{2}+b_{1}u+b_{0}),\end{split}$

where $a_{i},b_{i}\mathrel{\mathop{\mathchar
58\relax}}I\to\mathbb{R},\,\,i=0,1,2,3,4$, are

$a_{4}=4\lambda^{2}+(\lambda^{\prime})^{2}$ |  $b_{4}=\delta\left(-8\lambda^{2}-(\lambda^{\prime})^{2}-\lambda\lambda^{\prime\prime}\right)$
---|---
$a_{3}=4\delta\lambda\lambda^{\prime}$ |  $b_{3}=\lambda\left(11\lambda^{2}\lambda^{\prime}-\lambda(f^{2})^{\prime}-9f^{2}\lambda^{\prime}\right)$
$a_{2}=2\lambda^{2}\left(2f^{2}-(\lambda^{\prime})^{2}\right)$ |  $b_{2}=3\delta\lambda^{2}\left(2\lambda^{2}-4f^{2}+3(\lambda^{\prime})^{2}\right)$
$a_{1}=-4\delta\lambda^{3}\lambda^{\prime}$ |  $b_{1}=\lambda^{3}\left(-17\lambda^{2}\lambda^{\prime}-\lambda(f^{2})^{\prime}+19f^{2}\lambda^{\prime}\right)$
$a_{0}=\lambda^{4}(\lambda^{\prime})^{2}$ |  $b_{0}=\delta\lambda^{4}\left(\lambda\lambda^{\prime\prime}+2f^{2}-4(\lambda^{\prime})^{2}\right).$

Hence, the Ricci condition along $I_{0}$ is equivalent to

$c_{4}u^{4}+c_{3}u^{3}+c_{2}u^{2}+c_{1}u+c_{0}=0,$ (3)

where $c_{i}\mathrel{\mathop{\mathchar 58\relax}}I\to\mathbb{R},$
$i=0,1,2,3,4,$ is given as

$\displaystyle c_{4}$
$\displaystyle=2\delta\lambda^{2}\left(\lambda\lambda^{\prime\prime}-(\lambda^{\prime})^{2}\right)$
$\displaystyle c_{3}$
$\displaystyle=2\lambda^{3}\left(-3\lambda^{2}\lambda^{\prime}+f^{2}\lambda^{\prime}+\lambda(f^{2})^{\prime}\right)$
$\displaystyle c_{2}$
$\displaystyle=2\delta\lambda^{4}\left(-4\lambda^{2}+4f^{2}-5(\lambda^{\prime})^{2}\right)$
$\displaystyle c_{1}$
$\displaystyle=2\lambda^{5}\left(9\lambda^{2}\lambda^{\prime}-11f^{2}\lambda^{\prime}+\lambda(f^{2})^{\prime}\right)$
$\displaystyle c_{0}$
$\displaystyle=2\delta\lambda^{6}\left(2\lambda^{2}-2f^{2}+2(\lambda^{\prime})^{2}-\lambda\lambda^{\prime\prime}\right).$

Note that the left side of equation (3) is a polynomial with respect to $u.$
Thus, since the coefficients depending only on $t$, we get $c_{i}=0$ for
$i=0,1,2,3,4.$ Since $g$ is a non-flat Ricci metric, the zeroes of $K$ are
isolated, and thus we can admit without loss of generality that $\lambda\neq
0$ in $I_{0}.$ By $c_{3}=c_{1}=0,$ we get

$0=\left(9\lambda^{2}\lambda^{\prime}-11f^{2}\lambda^{\prime}+\lambda(f^{2})^{\prime}\right)-\left(-3\lambda^{2}\lambda^{\prime}+f^{2}\lambda^{\prime}+\lambda(f^{2})^{\prime}\right)=-12\delta^{2}\lambda^{\prime}.$

Hence, $\lambda^{\prime\prime}=\lambda^{\prime}=0$ and
$c_{0}=-4\delta^{3}\lambda^{6}\neq 0$ in $I_{0}.$ This contradiction shows
that $\delta=0$ in $I.$ Consequently, the metric $g$ is given as

$g=\left(\lambda(t)^{2}+u^{2}\right)dt^{2}+du^{2}$

and for this metric we get

$\displaystyle\|\nabla K\|^{2}$
$\displaystyle=\frac{4\lambda^{2}}{(\lambda^{2}+u^{2})^{7}}\left((4\lambda^{2}+(\lambda^{\prime})^{2})u^{4}+(4\lambda^{4}-2\lambda^{2}(\lambda^{\prime})^{2})u^{2}+\lambda^{4}(\lambda^{\prime})^{2}\right)$
$\displaystyle K\Delta K$
$\displaystyle=\frac{2\lambda^{2}}{(\lambda^{2}+u^{2})^{7}}\left((8\lambda^{2}+(\lambda^{\prime})^{2}+\lambda\lambda^{\prime\prime})u^{4}+(6\lambda^{4}-9\lambda^{2}(\lambda^{\prime})^{2})u^{2}-2\lambda^{6}+4\lambda^{4}(\lambda^{\prime})^{2}-\lambda^{5}\lambda^{\prime\prime}\right).$

Once $\lambda\neq 0$ almost everywhere, the Ricci condition in this case is
equivalent to

$\left((\lambda^{\prime})^{2}-\lambda\lambda^{\prime\prime}\right)u^{4}+5\lambda^{2}(\lambda^{\prime})^{2}u^{2}-\lambda^{4}\left(2(\lambda^{\prime})^{2}-\lambda\lambda^{\prime\prime}\right)=0.$
(4)

Since the left hand of equation (4) is a polynomial in $u$ whose coefficients
are functions of $t,$ as before we conclude that these coefficients vanish.
Therefore, the term of second order of (4) implies that $\lambda^{\prime}$ is
zero in $I,$ which shows that $\lambda$ is constant. This provides a solution
of (4) and shows that $\lambda^{2}=f^{2}$ is constant as claimed.

Reciprocally, assuming $\lambda^{2}=f^{2}$ constant in $I,$ we want to prove
that the Ricci condition holds. This is equivalent to show that (4) holds, and
this can be seen directly. ∎

######  Remark 1.

An interesting problem in classical Differential Geometry is to know when two
surfaces coincide, up to rigid motion of the space. In this direction, the
only minimal isometric immersions of $\mathbb{R}^{2}$, endowed with the Ricci
metric of the above result, into $\mathbb{R}^{3}$ are the ones congruent to
the one-parameter associated family with the helicoid (cf. [7, Theorem 8]).

###  3.2. Ricci condition for ruled surfaces

A direct consequence of the metric classification in Lemma 1 says that the
distribution parameter of a certain parametrization of a ruled Ricci surface
is constant and equals to the speed of its line of striction, up to sign.

######  Corollary 1.

Let $\Sigma$ be a non-developable ruled surface parametrizated by
$X\mathrel{\mathop{\mathchar 58\relax}}I\times J\to\mathbb{R}^{3}$ as

$X(t,u)=\alpha(t)+u\beta(t)$

such that the curves $\alpha\mathrel{\mathop{\mathchar
58\relax}}I\to\mathbb{R}^{3}$ and $\beta\mathrel{\mathop{\mathchar
58\relax}}I\to\mathbb{S}^{2}$ are orthogonal and $\beta$ is parametrized by
arc length. Then, $\Sigma$ is a Ricci surface if and only
$\lambda(t)^{2}=\|\alpha^{\prime}(t)\|^{2}=c^{2}$ for some non-zero real
constant $c$ and each $t\in I,$ and where $\lambda$ is the distribution
parameter of $X.$ In particular, the induced metric of $\Sigma$ is
$g=(c^{2}+u^{2})dt^{2}+du^{2}.$

###### Proof.

Remember that the induced metric $g$ of $\Sigma$ is written as in (2). The
result follows by considering, in Lemma 1, $f(t)=\|\alpha^{\prime}(t)\|$ and
$\lambda$ as the distribution parameter of $X.$ ∎

######  Remark 2.

We point out that in the previous notations for ruled surfaces, since we have
$\langle\alpha^{\prime}(t),\beta(t)\rangle^{2}=\|\alpha^{\prime}(t)\|^{2}-\lambda(t)^{2},$
the condition stated in Corollary 1 for a ruled Ricci surface implies that
$\langle\alpha^{\prime}(t),\beta(t)\rangle$ vanishes in $I.$ However,
$\|\alpha^{\prime}(t)\|^{2}=\lambda(t)^{2}$ in $I$ does not necessarily imply
that the surface is a ruled Ricci surface. Indeed, to see this let us consider
the following parametrization of a family of ruled surfaces given by

$X_{w}(t,u)=(0,0,w(t))+u(\cos(t),\sin(t),0),$

where $w\mathrel{\mathop{\mathchar 58\relax}}I\to\mathbb{R}$ is a smooth
function. We have $\alpha(t)=(0,0,w(t))$ and $\beta(t)=(\cos(t),\sin(t),0).$
Obviously, $\alpha$ is the line of striction of $\beta$ and
$\|\beta(t)\|=\|\beta^{\prime}(t)\|=1$ for any $t\in I.$ Moreover,
$\langle\alpha^{\prime}(t),\beta(t)\rangle=0$ and therefore
$\lambda(t)^{2}=\|\alpha^{\prime}(t)\|^{2}$ in $I.$ But $X_{w}$ is not the
parametrization of a ruled Ricci surface in general. In fact, if we admit that
$w$ is not an affine function, in this case, $\|\alpha^{\prime}\|$ (and
consequently the distribution parameter of $X_{w}$) is not constant and,
according to Corollary 1, $X_{w}$ is not a parametrization of a ruled Ricci
surface.

This class is known as right conoids and includes helicoids, hyperbolic
paraboloids, Plücker conoids, Wallis conicals edges and Whitney umbrellas. As
we will see, except for the helicoid, none of them are Ricci surfaces.

We will use Corollary 1 in order to classify ruled Ricci surfaces in two
aspects depending on the geometry of the line of striction for a certain
parametrization. For the first one, we consider that the surface may be
parametrized in such a way that the line of striction is contained in a
straight line. Remember that the immersion

$X(t,u)=(0,0,at+b)+u(\cos(t),\sin(t),0),\ (t,u)\in I\times J,$

for any constants $a$ and $b$ with $a\neq 0$ gives a parametrization of a
piece of helicoid. We have the following assertion.

######  Theorem 1.

Let $\Sigma$ be a non-developable ruled Ricci surface which can be
parametrized so that its line of striction is contained in a straight line.
Then, $\Sigma$ is a piece of helicoid.

###### Proof.

After rigid motions if necessary, we can parametrize $\Sigma$ as

$X(t,u)=(0,0,w(t))+u(x(t),y(t),z(t)),\ (t,u)\in I\times J,$

where $w,x,y,z\mathrel{\mathop{\mathchar 58\relax}}I\to\mathbb{R}$ are smooth
functions. We can assume that these functions satisfy the conditions
$x^{2}+y^{2}+z^{2}=1$ and
$(x^{\prime})^{2}+(y^{\prime})^{2}+(z^{\prime})^{2}=1,$ and furthermore
$(w^{\prime})^{2}=c^{2}$ for some positive constant $c$ according to Corollary
1. In particular, we have $w(t)=ct+a$ for some constant $a,$ and once
$cz=\langle(0,0,w^{\prime}),(x,y,z)\rangle=0$ then $z=0$. Therefore, the
parametrization $X$ becomes

$X(t,u)=(0,0,ct+a)+u(x(t),y(t),0),$

and consequently its distribution parameter $\lambda$ is given by

$\lambda^{2}=\langle(0,0,w^{\prime})\wedge(x,y,0),(x^{\prime},y^{\prime},0)\rangle^{2}=c^{2}(xy^{\prime}-yx^{\prime})^{2}.$

According to Corollary 1, we have $\lambda^{2}=c^{2}$, which implies
$xy^{\prime}-yx^{\prime}=1,$ up to sign. Since $x^{2}+y^{2}=1$, we must have
$x(t)=\cos\theta(t)$ and $y(t)=\sin\theta(t)$ for some function
$\theta\mathrel{\mathop{\mathchar 58\relax}}I\to\mathbb{R}$, and thus

$1=xy^{\prime}-yx^{\prime}=\cos^{2}(\theta)\theta^{\prime}+\sin^{2}(\theta)\theta^{\prime}=\left(\cos^{2}(\theta)+\sin^{2}(\theta)\right)\theta^{\prime}=\theta^{\prime}.$

Hence, $\theta(t)=t+b$ for some constant $b,$ which proves our assertion. ∎

For the second one classification result, let us assume that the ruled surface
that we consider does not admit parametrization with straight line of
striction. We obtain that the surface is generated by a constant torsion curve
and its binormal. More precisely, we have the following result.

######  Theorem 2.

Let $\Sigma$ be a non-developable ruled surface parametrized by
$X\mathrel{\mathop{\mathchar 58\relax}}I\times J\to\mathbb{R}^{3}$ as

$X(t,u)=\alpha(t)+u\beta(t)$

such that the curves $\alpha\mathrel{\mathop{\mathchar
58\relax}}I\to\mathbb{R}^{3}$ and $\beta\mathrel{\mathop{\mathchar
58\relax}}I\to\mathbb{S}^{2}$ are orthogonal and $\beta$ is parametrized by
arc length. Then, $\Sigma$ is a Ricci surface if and only if $\alpha$ has
contant torsion $\tau_{0}\neq 0$ and $\beta$ is its binormal. Moreover, the
constant distribution parameter $\lambda_{0}$ of $X$ is such that
$(\lambda_{0}\tau_{0})^{2}=1.$

###### Proof.

Suppose $\Sigma$ is a Ricci surface. First, we consider a parametrization
$X\mathrel{\mathop{\mathchar 58\relax}}I\times J\to\mathbb{R}^{3}$ of $\Sigma$
given by $X(t,u)=\alpha(t)+u\beta(t)$ for orthogonal curves
$\alpha,\beta\mathrel{\mathop{\mathchar 58\relax}}I\to\mathbb{R}^{3}$ with
$\|\beta(t)\|=\|\frac{d\beta}{dt}\|=1.$ Initially, we show that after a
reparametrization of $\alpha$ by arc length the surface is generated by
$\alpha$ and its binormal. We know by means of Corollary 1 that $\alpha$ is
regular because $\|\frac{d\alpha}{dt}\|^{2}=c^{2}$ for some $c>0,$ and thus we
can reparametrize it by its arc length $s=s(t).$ Without loss of generality,
we admit $0\in I,$ and thus

$s(t)=\int_{0}^{t}\left\|\frac{d\alpha}{d\sigma}\right\|d\sigma=ct.$

Let $(\beta_{1},\beta_{2},\beta_{3})$ be the coordinate functions of $\beta$
with respect to the Frenet frame $\\{T,N,B\\}$ of $\alpha.$ Since by Corollary
1

$0=\bigg{\langle}\frac{d\alpha}{ds},\beta\bigg{\rangle}=\frac{1}{c}\langle
T,\beta_{1}T+\beta_{2}N+\beta_{3}B\rangle=\frac{\beta_{1}}{c}$

one get $\beta_{1}=0.$ Furthermore, since the curvature of $\alpha$ is
different from zero everywhere it follows from equation (1) that $\beta_{2}$
vanishes along $I.$ Finally, since the trace of $\beta$ is contained in
$\mathbb{S}^{2},$ one conclude that $\beta_{3}^{2}=1.$ This shows that the
immersion $X^{\prime}\mathrel{\mathop{\mathchar 58\relax}}I^{\prime}\times
J^{\prime}\to\mathbb{R}^{3}$ given as

$X^{\prime}(s,u)=\alpha(s)+uB(s)$

is a reparametrization of $\Sigma.$ Now, we assert that the torsion $\tau$ of
$\alpha$ is constant. Indeed, using the Frenet formulas, we obtain

$c^{2}=\lambda^{2}=\bigg{\langle}\frac{d\alpha}{dt}\wedge
B,\frac{dB}{dt}\bigg{\rangle}^{2}=c^{4}\tau^{2}\langle T\wedge
B,N\rangle^{2}=c^{4}\tau^{2},$

which implies that $\tau^{2}=1/c^{2}.$

Conversely, let us suppose that $\Sigma$ is parametrized by an immersion
$X\mathrel{\mathop{\mathchar 58\relax}}I\times J\to\mathbb{R}^{3}$ as

$X(s,u)=\alpha(s)+uB(s),\ \ \ (s,u)\in I\times J$

where $\alpha\mathrel{\mathop{\mathchar 58\relax}}I\to\mathbb{R}^{3}$ is a
regular curve with constant torsion $\tau_{0}\neq 0,$ arc length $s$ and
binormal $B\mathrel{\mathop{\mathchar 58\relax}}I\to\mathbb{S}^{2}.$ In this
case, the arc length $t=t(s)$ of $B$ is given by

$t(s)=\int_{0}^{s}\left\|\frac{dB}{d\sigma}\right\|d\sigma=|\tau_{0}|s,$

where, whitout loss of generality, we have admited $0\in I.$ On one hand, we
have $\langle\frac{d\alpha}{dt},B\rangle=0,$ and consequently
$\lambda^{2}=\|\frac{d\alpha}{dt}\|^{2},$ where $\lambda$ is the distribution
parameter of $X.$ On the other hand,

$\lambda^{2}=\left\langle\frac{d\alpha}{dt}\wedge
B,\frac{dB}{dt}\right\rangle^{2}=\frac{1}{\tau_{0}^{2}}\left\langle T\wedge
B,N\right\rangle^{2}=\frac{1}{\tau_{0}^{2}},$

where $\\{T,N,B\\}$ is the Frenet frame of $\alpha.$ Hence, Corollary 1
implies that $\Sigma$ is a Ricci surface. ∎

######  Remark 3.

We point out that Theorems 1 and 2 suggest a correspondence between curves of
constant torsion and non-developable ruled Ricci surfaces. Furthermore, these
results are a classification for non-developable ruled Ricci surfaces in the
sense that they are the an way to obtain this kind of surfaces.

By combining our results with the fact that plane curves are characterized as
the ones with vanishing torsion, we get the following.

######  Corollary 2.

The only non-developable ruled Ricci surface which may be parametrized with
line of striction contained in a plane is the helicoid.

Note that, in particular, we have proven that all non-developable ruled Ricci
surfaces may be generated by a constant torsion curve and its binormal. In
Theorem 1 the line of striction is a straight line and its binormal is a great
circle, while in Theorem 2 the line of striction has contant non-zero torsion
and its binormal is not contained in a great circle. In both cases, the
spherical curve is the binormal of the line of striction. Furthermore, since
the spherical curve is regular, it is direct to see that Theorem 2 does not
depend on its arc length parameter. This motivates the following.

######  Definition 1.

We call canonical parametrization a parametrization of a non-developable ruled
Ricci surface generated by $\\{\alpha(t),B(t)\\}$ where $\alpha$ has constant
torsion and $B$ is the binormal of $\alpha.$

######  Remark 4.

Let $\alpha\mathrel{\mathop{\mathchar 58\relax}}I\to\mathbb{R}^{3}$ be a
regular curve parametrized by arc length, and let $\\{T,N,B\\}$ be its Frenet
frame. The tangent surface and the normal surface generated respectively by
$\\{\alpha(s),T(s)\\}$ and by $\\{\alpha(s),N(s)\\}$ are developable, which
means that they have zero Gauss curvature. It follows immediately that they
are ruled Ricci surfaces. Furthermore, our Theorem 2 says that the binormal
surface generated by $\\{\alpha(s),B(s)\\}$ is a canonical parametrization of
a non-developable ruled Ricci surface provided that $\alpha$ has constant
torsion.

The next result relates the mean curvature of a non-developable ruled Ricci
surface with the curvature of the line of striction of a canonical
parametrization.

######  Theorem 3.

The mean curvature of a non-developable ruled Ricci surface $\Sigma$ is

$H(t,u)=-\frac{\kappa(t)}{2\sqrt{1+\tau_{0}^{2}u^{2}}},$

where $\kappa$ and $\tau_{0}$ are respectively the curvature and the torsion
of the line of striction in the canonical parametrization of $\Sigma.$ In
particular, the only non-developable ruled Ricci surface with constant mean
curvature is the helicoid.

###### Proof.

If $\Sigma$ is a piece of helicoid, there is noting to do. For the remaining
case, the proof is a long calculation, and for simplicity we will omit the
parameters. By considering the immersion $X\mathrel{\mathop{\mathchar
58\relax}}I\times J\to\mathbb{R}^{3}$ of the ruled Ricci surface $\Sigma$
given by Theorem 2

$X(s,u)=\alpha(s)+uB(s),$

where $s$ and $B$ are the arc length and the binormal of $\alpha,$
respectively, note that in this parametrization the mean curvature of $\Sigma$
is given as

$H=\frac{1}{2}\left(\frac{\langle X_{ss},N_{\Sigma}\rangle}{\langle
X_{s},X_{s}\rangle}+\langle X_{uu},N_{\Sigma}\rangle\right),$

where $N_{\Sigma}$ is the unit normal of $\Sigma.$ If we denote respectively
by $\\{T,N,B\\}$ and $\tau_{0}$ the Frenet frame and the torsion of $\alpha$
we have

$X_{s}=T+\tau_{0}uN\ \ \mbox{and}\ \ X_{u}=B.$

Direct computations shows that

$\langle X_{s},X_{s}\rangle=1+\tau_{0}^{2}u^{2},\ \ X_{ss}=\kappa
N-\tau_{0}u(\kappa T+\tau_{0}B)\ \ \mbox{and}\ \ X_{uu}=0,$

where $\kappa$ is the curvature of $\alpha,$ and that $N_{\Sigma}$ is given by

$N_{\Sigma}=\frac{X_{s}\wedge X_{u}}{\|X_{s}\wedge
X_{u}\|}=\frac{-N+\tau_{0}uT}{\sqrt{1+\tau_{0}^{2}u^{2}}}.$

Consequently, $\langle
X_{ss},N_{\Sigma}\rangle=-\kappa\sqrt{1+\tau_{0}^{2}u^{2}},$ which implies
that

$H=-\frac{\kappa}{2\sqrt{1+\tau_{0}^{2}u^{2}}}.$

Since $\kappa$ does not depend on the parameter of the curve, this shows the
formula we stated. ∎

###  3.3. Constructing ruled Ricci surfaces

It is well-known that for a given curve $B\mathrel{\mathop{\mathchar
58\relax}}I\to\mathbb{S}^{2}$ parametrized by arc length and a constant non-
zero $\tau_{0}$ there exists a curve $\alpha\mathrel{\mathop{\mathchar
58\relax}}I\to\mathbb{R}^{3}$ given by

$\alpha(t)=\frac{1}{\tau_{0}}\int_{t_{0}}^{t}B^{\prime}(\sigma)\wedge
B(\sigma)d\sigma$ (5)

with constant torsion equals $\tau_{0}$ provided that $\langle B(t)\wedge
B^{\prime}(t),B^{\prime\prime}(t)\rangle\neq 0,$ and moreover $B$ is the
binormal of $\alpha.$ Conversely, for a given curve with constant non-zero
torsion, it may be parametrized by formula (5). We refer the reader to [2] and
references therein for details.

We present here explicit examples of ruled Ricci surfaces illustrating the
construction above.

######  Example 1 (Parallel circles).

The first family we will look is associated with closed curves. Let us
consider a family of parallel circles in $\mathbb{S}^{2}$ of radius
$\ell\in(0,1)$ centered at $(0,0,\sqrt{1-\ell^{2}}),$ which is parameterized
by arc length by the map $B_{\ell}\mathrel{\mathop{\mathchar
58\relax}}[0,2\ell\pi]\to\mathbb{S}^{2}$ defined as

$B_{\ell}(t)=\left(\ell\sin\frac{t}{\ell},-\ell\cos\frac{t}{\ell},\sqrt{1-\ell^{2}}\right).$

Associated to this family we produce a family of curves with constant torsion
equals $1.$ A straightforward calculation using (5) shows that the family of
circles $B_{\ell}$ is associated with the family of $\ell$-helix of constant
torsion equals $1$ given by

$\alpha_{\ell}(t)=-\ell\left(\sqrt{1-\ell^{2}}\cos\dfrac{t}{\ell},\sqrt{1-\ell^{2}}\sin\dfrac{t}{\ell},t\right),$

up to translation. Hence, Theorem 2 gives that the map
$X_{\ell}\mathrel{\mathop{\mathchar 58\relax}}[0,2\ell\pi]\times
J\to\mathbb{R}^{3},$ given as $X_{\ell}(t,u)=\alpha_{\ell}(t)+uB_{\ell}(t)$,
is the canonical parametrization of a non-developable ruled Ricci surface for
each $\ell\in(0,1).$ One can see directly that the unit circle in
$\mathbb{S}^{2}$ is associated with a helicoid by making $\ell$ goes to 1, and
that the family degenerates in the vertical straight line when $\ell$ goes to
zero.

Figure 3.1. Non-developable ruled Ricci surfaces associated with circles in
$\mathbb{S}^{2}$ of radius $\frac{1}{2}$, $\frac{3}{4}$ and $1$, respectively.

######  Example 2 (Anti-Salkowski curve).

Now, we will describe a family of ruled Ricci surfaces whose line of striction
is defined at most in an open bounded interval. In [11], Salkowski constructs
a family of curves of constant curvature but non-constant torsion. Later,
Monterde demonstrated in [8] how to construct, from a curve of constant
curvature, another curve of constant torsion. Applying this technique to the
family of curves defined by Salkowski, he was able to generate a new family of
curves with constant torsion equals $1$ but non-constant curvature, and he
referred to this family as anti-Salkowski curves. It may be parametrized by
arc length by $\alpha_{\ell}\mathrel{\mathop{\mathchar
58\relax}}\big{(}-\frac{1}{\ell},\frac{1}{\ell}\big{)}\to\mathbb{R}^{3}$ as

$\alpha_{\ell}(t)=\begin{bmatrix}\cos\theta(t)&-\sin\theta(t)&0\\\
\sin\theta(t)&\cos\theta(t)&0\\\ 0&0&-1\end{bmatrix}\begin{bmatrix}ca(t)\\\
cb(t)\\\ \displaystyle\frac{2\arcsin(\ell t)+\sin(2\arcsin(\ell
t))}{4\ell\sqrt{1+\ell^{2}}}\end{bmatrix}$

for

$a(t)=-2+3\ell^{2}t^{2}+3\ell^{4}t^{2},$ |  $\displaystyle c=\frac{\ell}{1-2\ell^{2}-3\ell^{4}},$
---|---
$b(t)=\sqrt{1+\ell^{2}}\left(1+3\ell^{2}\right)t\sqrt{1-\ell^{2}t^{2}},$ |  $\displaystyle\theta(t)=\frac{\sqrt{1+\ell^{2}}\arcsin(\ell t)}{\ell},$

where $\ell$ is positive and different of $\frac{1}{\sqrt{3}}.$ Its binormal
is

$B_{\ell}(t)=\frac{1}{\sqrt{1+\ell^{2}}}\begin{bmatrix}\cos\theta(t)&-\sin\theta(t)&0\\\
\sin\theta(t)&\cos\theta(t)&0\\\ 0&0&1\end{bmatrix}\begin{bmatrix}d(t)\\\
\ell^{2}t\\\ \ell t\end{bmatrix}$

for

$d(t)=-\sqrt{1+\ell^{2}}\sqrt{1-\ell^{2}t^{2}}.$

Since the torsion of $\alpha$ is $1,$ then $t$ is also the arc length of
$B_{\ell}$. Hence, from Theorem 2 the map $X_{\ell}\mathrel{\mathop{\mathchar
58\relax}}\big{(}-\frac{1}{\ell},\frac{1}{\ell}\big{)}\times
J\to\mathbb{R}^{3},$ given as $X_{\ell}(t,u)=\alpha_{\ell}(t)+uB_{\ell}(t),$
is the canonical parametrization of a non-developable ruled Ricci surface.
These surfaces are non-complete since $t$ lies at most in bounded interval.
Furhtermore, once $\theta(t)\to t$ as $\ell\to 0,$ we must have that
$B_{\ell}(t)$ becomes a geodesic of $\mathbb{S}^{2}$ when $\ell$ goes to zero,
which means that the family converges to the helicoid in this case.

Figure 3.2. Spherical curves and non-developable ruled Ricci surfaces for
anti-Salkowski curves with $\ell=0.1$, $\ell=\frac{1}{3}$ and $\ell=0.57$,
respectively.

######  Example 3 (Borderline spherical curve).

The last ruled Ricci surfaces we present is such that is associated with a
curve on the unit sphere whose maximal interval of definition is the real
line. Let $B\mathrel{\mathop{\mathchar 58\relax}}\mathbb{R}\to\mathbb{S}^{2}$
be the spherical curve parameterized by arc length as

$B(t)=(\tanh(t)\cos(t),\tanh(t)\sin(t),\operatorname{sech}(t)).$

Notice that $B$ acumulates in the equator as $t\to\pm\infty.$ Direct
computations shows that

$\displaystyle
B^{\prime}(t)=\operatorname{sech}^{2}(t)(\cos(t),\sin(t),0)+\tanh(t)(-\sin(t),\cos(t),-\operatorname{sech}(t)).$

Hence, it follows from formula (5) that the curve

$\alpha(t)=(-\cos(t)\operatorname{sech}(t),-\sin(t)\operatorname{sech}(t),\tanh(t)-t)$

has constant torsion equals 1 and the map $X\mathrel{\mathop{\mathchar
58\relax}}\mathbb{R}^{2}\to\mathbb{R}^{3},$ given as $X(t,u)=\alpha(t)+uB(t),$
provides the canonical parametrization of a complete ruled Ricci surface.

Figure 3.3. Borderline spherical curve and its associated non-developable
ruled Ricci surface.

## References

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# Entanglement of formation and monogamy of multi-party quantum entanglement

Jeong San Kim<EMAIL_ADDRESS>Department of Applied Mathematics and
Institute of Natural Sciences, Kyung Hee University, Yongin-si, Gyeonggi-do
446-701, Korea

###### Abstract

We provide a sufficient condition for the monogamy inequality of multi-party
quantum entanglement of arbitrary dimensions in terms of entanglement of
formation. Based on the classical-classical-quantum(ccq) states whose quantum
parts are obtained from the two-party reduced density matrices of a three-
party quantum state, we show the additivity of the mutual information of the
ccq states guarantees the monogamy inequality of the three-party pure state in
terms of EoF. After illustrating the result with some examples, we generalize
our result of three-party systems into any multi-party systems of arbitrary
dimensions.

###### pacs:

03.67.Mn, 03.65.Ud

##  I Introduction

Quantum entanglement is a non-classical nature of quantum mechanics, which is
a useful resource in many quantum information processing tasks such as quantum
teleportation, dense coding and quantum cryptography tele ; qkd ; rev .
Because of its important roles in the field of quantum information and
computation theory, there has been a significant amount of research focused on
quantification of entanglement in bipartite quantum systems. Entanglement of
formation(EoF) is the most well-known bipartite entanglement measure with an
operational meaning that asymptotically quantifies how many bell states are
needed to prepare the given state using local quantum operations and classical
communications bdsw . Although EoF is defined in any bipartite quantum systems
of arbitrary dimension, its definition for mixed states is based on ‘convex-
roof extension’, which takes the minimum average over all pure-state
decompositions of the given state. As such an optimization is hard to deal
with, analytic evaluation of EoF is known only in two-qubit systems wootters
and some restricted cases of higher-dimensional systems so far.

In multi-party quantum systems, entanglement shows a distinct behavior that
does not have any classical counterpart; if a pair of parties in a multi-party
quantum system is maximally entangled, then they cannot be entanglement, not
even classically correlated, with the rest parties. This restriction of
sharing entanglement in multi-party quantum systems is known as the monogamy
of entanglement(MoE) T04 ; KGS . MoE plays an important role such as the
security proof of quantum key distribution in quantum cryptography qkd ; qkd2
and the $N$-representability problem for fermions in condensed-matter physics
anti .

Mathematically, MoE can be characterized using monogamy inequality; for a
three-party quantum state $\rho_{ABC}$ and its two-party reduced density
matrices $\rho_{AB}$ and $\rho_{AC}$,

$\displaystyle E\left(\rho_{A(BC)}\right)\geq
E\left(\rho_{AB}\right)+E\left(\rho_{AC}\right)$ (1)

where $E\left(\rho_{XY}\right)$ is an entanglement measure quantifying the
amount of entanglement between subsystems $X$ and $Y$ of the bipartite quantum
state $\rho_{XY}$. Inequality (1) shows the mutually exclusive nature of
bipartite entanglement $E\left(\rho_{AB}\right)$ and $E\left(\rho_{AC}\right)$
shared in three-party quantum systems so that their summation cannot exceeds
the total entanglement $E\left(\rho_{A(BC)}\right)$.

Using tangle CKW  as the bipartite entanglement measure, Inequality (1) was
first shown to be true for all three-qubit states, and generalized for multi-
qubit systems as well as some cases of higher-dimensional quantum systems OV ;
KDS . However, not all bipartite entanglement measures can characterize MoE in
forms of Inequality (1), but only few measures are known so far satisfying
such monogamy inequality KSRenyi ; KimT ; KSU . Although EoF is the most
natural bipartite entanglement measure with the operational meaning in quantum
state preparation, EoF is known to fail in characterizing MoE as the monogamy
inequality in (1) even in three-qubit systems; there exists quantum states in
three-qubit systems violating Inequality (1) if EoF is used as the bipartite
entanglement measure. Thus, a natural question we can ask is ‘On what
condition does the monogamy inequality hold in terms of the given bipartite
entanglement measure?’

Here, we provide a sufficient condition that monogamy inequality of quantum
entanglement holds in terms of EoF in multi-party, arbitrary dimensional
quantum systems. For a three-party quantum state, we first consider the
classical-classical-quantum(ccq) states whose quantum parts are obtained from
the two-party reduced density matrices of the three-party state. By evaluating
quantum mutual information of the ccq states as well as their reduced density
matrices, we show that the additivity of the mutual information of the ccq
states guarantees the monogamy inequality of the three-party quantum state in
terms of EoF. We provide some examples of three-party pure state to illustrate
our result, and we generalize our result of three-party systems into any
multi-party systems of arbitrary dimensions.

This paper is organized as follows. First we briefly review the definitions of
classical and quantum correlations in bipartite quantum systems and recall
their trade-off relation in three-party quantum systems. After providing the
definition of ccq states as well as their mutual information between classical
and quantum parts, we establish the monogamy inequality of three-party quantum
entanglement in arbitrary dimensional quantum systems in terms of EoF
conditioned on the additivity of the mutual information for the ccq states. We
also illustrate our result of monogamy inequality in three-party quantum
systems with some examples, and we generalize our result of entanglement
monogamy inequality into multi-party quantum systems of arbitrary dimensions.
Finally, we summarize our results.

##  II Correlations in bipartite quantum systems

For a bipartite pure state ${\left|\psi\right\rangle}_{AB}$, its entanglement
of formation(EoF) is defined by the entropy of a subsystem,

${E}_{\bf f}\left({\left|\psi\right\rangle}_{AB}\right)=S(\rho_{A}),$ (2)

where
$\rho_{A}=\mbox{$\mathrm{tr}$}_{B}{\left|\psi\right\rangle}_{AB}{\left\langle\psi\right|}$
is the reduced density matrix of ${\left|\psi\right\rangle}_{AB}$ on subsystem
$A$, and $S\left(\rho\right)=-\mbox{$\mathrm{tr}$}\rho\ln\rho$ is the von
Neumann entropy of the quantum state $\rho$. For a bipartite mixed state
$\rho_{AB}$, its EoF is defined by the minimum average entanglement

$E_{\bf f}\left(\rho_{AB}\right)=\min\sum_{i}p_{i}E_{\bf
f}({\left|\psi_{i}\right\rangle}_{AB}),$ (3)

over all possible pure state decompositions of
$\rho_{AB}=\sum_{i}p_{i}|\psi_{i}\rangle_{AB}\langle\psi_{i}|$.

For a probability ensemble $\mathcal{E}=\\{p_{i},\rho_{i}\\}$ realizing a
quantum state $\rho$ such that $\rho=\sum_{i}p_{i}\rho_{i}$, its Holevo
quantity is defined as

$\displaystyle\chi\left(\mathcal{E}\right)=S\left(\rho\right)-\sum_{i}p_{i}S\left(\rho_{i}\right).$
(4)

Given a bipartite quantum state $\rho_{AB}$, each measurement
$\\{M^{x}_{B}\\}$ applied on subsystem $B$ induces a probability ensemble
$\mathcal{E}=\\{p_{x},\rho_{A}^{x}\\}$ of the reduced density matrix
$\rho_{A}=\mbox{$\mathrm{tr}$}_{A}\rho_{AB}$ in the way that
$p_{x}=\mbox{$\mathrm{tr}$}[(I_{A}\otimes M_{B}^{x})\rho_{AB}]$ is the
probability of the outcome $x$ and
$\rho^{x}_{A}=\mbox{$\mathrm{tr}$}_{B}[(I_{A}\otimes{M_{B}^{x}})\rho_{AB}]/p_{x}$
is the state of system $A$ when the outcome was $x$. The one-way classical
correlation (CC) HV  of a bipartite state $\rho_{AB}$ is defined by the
maximum Holevo quantity

$\displaystyle{\mathcal{J}}^{\leftarrow}(\rho_{AB})$
$\displaystyle=\max_{\mathcal{E}}\chi\left(\mathcal{E}\right)$ (5)

over all possible ensemble representations $\mathcal{E}$ of $\rho_{A}$ induced
by measurements on subsystem $B$.

The following proposition shows a trade-off relation between classical
correlation and quantum entanglement(measured by CC and EoF, respectively)
distributed in three-party quantum systems.

######  Proposition 1.

KW  For a three-party pure state ${\left|\psi\right\rangle}_{ABC}$ with
reduced density matrices
$\rho_{AB}=\mbox{$\mathrm{tr}$}_{C}{\left|\psi\right\rangle}_{ABC}{\left\langle\psi\right|}$,
$\rho_{AC}=\mbox{$\mathrm{tr}$}_{B}{\left|\psi\right\rangle}_{ABC}{\left\langle\psi\right|}$
and
$\rho_{A}=\mbox{$\mathrm{tr}$}_{BC}{\left|\psi\right\rangle}_{ABC}{\left\langle\psi\right|}$,
we have

$\displaystyle S(\rho_{A})={\mathcal{J}}^{\leftarrow}(\rho_{AB})+E_{\bf
f}\left(\rho_{AC}\right).$ (6)

##  III Classical-Classical-Quantum(CCQ) States

In this section, we consider a four-party ccq states obtained from a bipartite
state $\rho_{AB}$, and provide detail evaluations of their mutual information.
Without loss of generality, we assume that any bipartite state as a two-
qu$d$it state by taking $d$ as the dimension of larger dimensional subsystem.

For a two-qudit state $\rho_{AB}$, let us consider a spectral decomposition

$\displaystyle\rho_{B}=\sum_{i=0}^{d-1}\lambda_{i}{\left|e_{i}\right\rangle}_{B}{\left\langle
e_{i}\right|}$ (7)

of the reduced density matrix $\rho_{B}=\mbox{$\mathrm{tr}$}_{A}\rho_{AB}$.
Let

$\displaystyle\mathcal{E}_{0}=\\{\lambda_{i},\sigma_{A}^{i}\\}_{i}$ (8)

be the probability ensemble of $\rho_{A}=\mbox{$\mathrm{tr}$}_{B}\rho_{AB}$
from the measurement $\\{{\left|e_{i}\right\rangle}_{B}{\left\langle
e_{i}\right|}\\}_{i=1}^{d-1}$ on subsystem $B$ of $\rho_{AB}$, in a way that

$\displaystyle\lambda_{i}=\mbox{$\mathrm{tr}$}\left[(I_{A}\otimes{\left|e_{i}\right\rangle}_{B}{\left\langle
e_{i}\right|})\rho_{AB}\right]$ (9)

and

$\displaystyle\sigma_{A}^{i}=\frac{1}{\lambda_{i}}\mbox{$\mathrm{tr}$}_{B}\left[(I_{A}\otimes{\left|e_{i}\right\rangle}_{B}{\left\langle
e_{i}\right|})\rho_{AB}\right].$ (10)

Based on the eigenvectors $\\{{\left|e_{j}\right\rangle}_{B}\\}$ of
$\rho_{B}$, we also consider the $d$-dimensional Fourier basis elements

$|\tilde{e}_{j}\rangle_{B}=\frac{1}{\sqrt{d}}\sum_{k=0}^{d-1}\omega_{d}^{jk}{\left|e_{k}\right\rangle}_{B}$
(11)

for each $j=0,\ldots,d-1$, where $\omega_{d}=e^{\frac{2\pi i}{d}}$ is the
$d$th-root of unity. Let

$\displaystyle\mathcal{E}_{1}=\\{\frac{1}{d},\tau_{A}^{j}\\}_{j},$ (12)

be the probability ensemble of $\rho_{A}=\mbox{$\mathrm{tr}$}_{B}\rho_{AB}$
obtained by measuring subsystem $B$ in terms of the Fourier basis
$\\{{\left|\tilde{e}_{j}\right\rangle}_{B}{\left\langle\tilde{e}_{j}\right|}\\}_{j=1}^{d-1}$
where

$\displaystyle\frac{1}{d}=\mbox{$\mathrm{tr}$}\left[(I_{A}\otimes{\left|\tilde{e}_{j}\right\rangle}_{B}{\left\langle\tilde{e}_{j}\right|})\rho_{AB}\right]$
(13)

and

$\displaystyle\tau_{A}^{j}=d\mbox{$\mathrm{tr}$}_{B}\left[(I_{A}\otimes|\tilde{e}_{j}\rangle_{B}\langle\tilde{e}_{j}|)\rho_{AB}\right].$
(14)

Now we define the generalized $d$-dimensional Pauli operators based on the
eigenvectors of $\rho_{B}$ as

$\displaystyle
Z=\sum_{j=0}^{d-1}\omega_{d}^{j}{\left|e_{j}\right\rangle}_{B}{\left\langle
e_{j}\right|},~{}X=\sum_{j=0}^{d-1}{\left|e_{j+1}\right\rangle}_{B}{\left\langle
e_{j}\right|}=\sum_{j=0}^{d-1}\omega_{d}^{-j}|\tilde{e}_{j}\rangle_{B}\langle\tilde{e}_{j}|,$
(15)

and consider a four-qudit ccq state $\Gamma_{XYAB}$

$\displaystyle\Gamma_{XYAB}=\frac{1}{d^{2}}\sum_{x,y=0}^{d-1}{\left|x\right\rangle}_{X}{\left\langle
x\right|}\otimes{\left|y\right\rangle}_{Y}{\left\langle
y\right|}\otimes\left[(I_{A}\otimes X^{x}_{B}Z^{y}_{B})\rho_{AB}(I_{A}\otimes
Z^{-y}_{B}X^{-x}_{B})\right],$ (16)

for some $d$-dimensional orthonormal bases $\\{{\left|x\right\rangle}_{X}\\}$
and $\\{{\left|y\right\rangle}_{Y}\\}$ of the subsystems $X$ and $Y$,
respectively. From Eqs. (9), (10), (13), (14) and (15), the reduced density
matrices of $\Gamma_{XYAB}$ are obtained as

$\displaystyle\Gamma_{XAB}=\frac{1}{d}\sum_{x=0}^{d-1}$
$\displaystyle{\left|x\right\rangle}_{X}{\left\langle
x\right|}\otimes\left[\left(I_{A}\otimes
X^{x}_{B}\right)\left(\sum_{i=0}^{d-1}\sigma_{A}^{i}\otimes\lambda_{i}{\left|e_{i}\right\rangle}_{B}{\left\langle
e_{i}\right|}\right)\left(I_{A}\otimes X_{B}^{-x}\right)\right],$ (17)
$\displaystyle\Gamma_{YAB}=$
$\displaystyle\frac{1}{d}\sum_{y=0}^{d-1}{\left|y\right\rangle}_{Y}{\left\langle
y\right|}\otimes\left[\left(I_{A}\otimes
Z_{B}^{y}\right)\left(\sum_{j=0}^{d-1}\tau_{A}^{j}\otimes\frac{1}{d}|\tilde{e}_{j}\rangle_{B}\langle\tilde{e}_{j}|\right)\left(I_{A}\otimes
Z_{B}^{-y}\right)\right],$ (18)

and

$\Gamma_{AB}=\rho_{A}\otimes\frac{I_{B}}{d},$ (19)

where $I_{A}$ and $I_{B}$ are $d$-dimensional identity operators of subsystems
$A$ and $B$, respectively.

Before we move to the next section, we evaluate the mutual information of the
ccq state in Eq. (16) as well as the reduced density matrices in Eqs. (17) and
(18); the classical parts of the four-qudit ccq state $\Gamma_{XYAB}$ in Eq.
(16) is

$\displaystyle\Gamma_{XY}=\frac{1}{d^{2}}\sum_{x,y=0}^{d-1}{\left|x\right\rangle}_{X}{\left\langle
x\right|}\otimes{\left|y\right\rangle}_{Y}{\left\langle y\right|},$ (20)

which is the maximally mixed state in $d^{2}$-dimensional quantum system,
therefore its von Neumann entropy is

$\displaystyle
S\left(\Gamma_{XY}\right)=-\sum_{x,y=0}^{d-1}\frac{1}{d^{2}}\ln\left(\frac{1}{d^{2}}\right)=2\ln
d.$ (21)

We also note that Eq. (19) leads us to

$\displaystyle S\left(\Gamma_{AB}\right)=S(\rho_{A})+\ln d.$ (22)

From the joint entropy theorem joint1 ; joint , we have

$\displaystyle S(\Gamma_{XYAB})=$ $\displaystyle 2\ln
d+\frac{1}{d^{2}}\sum_{x,y=0}^{d-1}S\left((I_{A}\otimes
X^{x}_{B}Z^{y}_{B})\rho_{AB}(I_{A}\otimes Z^{-y}_{B}X^{-x}_{B})\right)=2\ln
d+S\left(\rho_{AB}\right),$ (23)

where the second equality is due to the unitary invariance of von Neumann
entropy. Thus Eqs. (21), (22) and (23) give us the mutual information of the
four-qudit ccq state $\Gamma_{XYAB}$ with respect to the bipartition between
$XY$ and $AB$ as

$\displaystyle{I}\left(\Gamma_{XY:AB}\right)=\ln d+S(\rho_{A})-S(\rho_{AB}).$
(24)

For the von Neumann entropy of $\Gamma_{XAB}$ in Eq. (17), we have

$\displaystyle S\left(\Gamma_{XAB}\right)=$ $\displaystyle\ln
d+\frac{1}{d}\sum_{x=0}^{d-1}S\left(\left(I_{A}\otimes
X^{x}_{B}\right)\left(\sum_{i=0}^{d-1}\sigma_{A}^{i}\otimes\lambda_{i}{\left|e_{i}\right\rangle}_{B}{\left\langle
e_{i}\right|}\right)\left(I_{A}\otimes X_{B}^{-x}\right)\right)$
$\displaystyle=$ $\displaystyle\ln
d+S\left(\sum_{i=0}^{d-1}\sigma_{A}^{i}\otimes\lambda_{i}{\left|e_{i}\right\rangle}_{B}{\left\langle
e_{i}\right|}\right)$ $\displaystyle=$ $\displaystyle\ln
d+H(\Lambda)+\sum_{i=1}^{d-1}\lambda_{i}S\left(\sigma_{A}^{i}\right),$ (25)

where the first equality is from the joint entropy theorem, the second
equality is due to the unitary invariance of von Neumann entropy and the last
equality is the joint entropy theorem together with

$H(\Lambda)=-\sum_{i}\lambda_{i}\ln\lambda_{i},$ (26)

which is the shannon entropy of the spectrum $\Lambda=\\{\lambda_{i}\\}$ of
$\rho_{B}$ in Eq. (7). Thus we can rewrite the von Neumann entropy of
$\Gamma_{XAB}$ as

$\displaystyle S\left(\Gamma_{XAB}\right)=\ln
d+S\left(\rho_{B}\right)+\sum_{i=1}^{d-1}\lambda_{i}S\left(\sigma_{A}^{i}\right).$
(27)

Because the classical parts of $\Gamma_{XAB}$ is the $d$-dimensional maximally
mixed state
$\Gamma_{X}=\frac{1}{d}\sum_{x=0}^{d-1}{\left|x\right\rangle}_{X}{\left\langle
x\right|}$, we have the mutual information of $\Gamma_{XAB}$ with respect to
the bipartition between $X$ and $AB$ as

$\displaystyle{I}(\Gamma_{X:AB})=$ $\displaystyle
S(\Gamma_{X})+S(\Gamma_{AB})-S(\Gamma_{XAB})=\ln
d-S(\rho_{B})+\chi(\mathcal{E}_{0}).$ (28)

For the von Neumann entropy of $\Gamma_{YAB}$ in Eq. (18), we have

$\displaystyle S\left(\Gamma_{XAB}\right)=$ $\displaystyle\ln
d+\frac{1}{d}\sum_{y=0}^{d-1}S\left(\left(I_{A}\otimes
Z_{B}^{y}\right)\left(\sum_{j=0}^{d-1}\tau_{A}^{j}\otimes\frac{1}{d}|\tilde{e}_{j}\rangle_{B}\langle\tilde{e}_{j}|\right)\left(I_{A}\otimes
Z_{B}^{-y}\right)\right)$ $\displaystyle=$ $\displaystyle\ln
d+S\left(\sum_{j=0}^{d-1}\tau_{A}^{j}\otimes\frac{1}{d}|\tilde{e}_{j}\rangle_{B}\langle\tilde{e}_{j}|\right)$
$\displaystyle=$ $\displaystyle 2\ln
d+\frac{1}{d}\sum_{j=1}^{d-1}S\left(\tau_{A}^{j}\right),$ (29)

where the first and third equalities are due to the joint entropy theorem and
the second equality is from the unitary invariance of von Neumann entropy.
Thus the mutual information of $\Gamma_{YAB}$ with respect to the bipartition
between $Y$ and $AB$ is

$\displaystyle{I}(\Gamma_{Y:AB})=$ $\displaystyle
S(\Gamma_{Y})+S(\Gamma_{AB})-S(\Gamma_{YAB})$ $\displaystyle=$
$\displaystyle\ln d+S(\rho_{A})+\ln d-2\ln
d-\frac{1}{d}\sum_{j=1}^{d-1}S\left(\tau_{A}^{j}\right)$ $\displaystyle=$
$\displaystyle\chi(\mathcal{E}_{1}).$ (30)

##  IV Monogamy inequality of multi-party entanglement in terms of EoF

It is known that quantum mutual information is superadditive for any ccq state
Kim16T

$\Xi_{XYAB}=\frac{1}{d^{2}}\sum_{x,y=0}^{d-1}{\left|x\right\rangle}_{X}{\left\langle
x\right|}\otimes{\left|y\right\rangle}_{Y}{\left\langle
y\right|}\otimes\sigma^{xy}_{AB},$ (31)

that is,

${I}\left(\Xi_{XY:AB}\right)\geq{I}\left(\Xi_{X:AB}\right)+{I}\left(\Xi_{Y:AB}\right).$
(32)

Here we show that the additivity of quantum mutual information for ccq states
guarantees the monogamy inequality of three-party quantum entanglement in
therms of EoF.

######  Theorem 1.

For any three-party pure state ${\left|\psi\right\rangle}_{ABC}$ with its two-
qudit reduced density matrices
$\mbox{$\mathrm{tr}$}_{C}{\left|\psi\right\rangle}_{ABC}{\left\langle\psi\right|}=\rho_{AB}$
and
$\mbox{$\mathrm{tr}$}_{B}{\left|\psi\right\rangle}_{ABC}{\left\langle\psi\right|}=\rho_{AC}$,
we have

$\displaystyle E_{\bf f}\left({\left|\psi\right\rangle}_{A(BC)}\right)\geq$
$\displaystyle E_{\bf f}\left(\rho_{AB}\right)+E_{\bf
f}\left(\rho_{AC}\right),$ (33)

conditioned on the additivity of quantum mutual information for the ccq states
in Eq. (16) obtained by $\rho_{AB}$ and $\rho_{AC}$.

###### Proof.

Let us first consider the four-qudit ccq state $\Gamma_{XYAB}$ of the form in
Eq. (16) obtained by the two-qudit reduced density matrix $\rho_{AB}$ of
${\left|\psi\right\rangle}_{ABC}$. From Eqs. (24), (28) and (30), the
additivity condition of quantum mutual information for $\Gamma_{XYAB}$

$\displaystyle{I}\left(\Gamma_{XY:AB}\right)={I}\left(\Gamma_{X:AB}\right)+{I}\left(\Gamma_{Y:AB}\right)$
(34)

can be rewritten as

$\displaystyle\chi(\mathcal{E}_{0})+\chi(\mathcal{E}_{1})=S\left(\rho_{A}\right)+S\left(\rho_{B}\right)-S\left(\rho_{AB}\right)={I}\left(\rho_{AB}\right),$
(35)

where $\mathcal{E}_{0}$ and $\mathcal{E}_{1}$ are the probability ensembles of
$\rho_{A}$ in Eqs. (8) and (12), respectively.

Because $\mathcal{E}_{0}$ and $\mathcal{E}_{1}$ can be obtained from measuring
subsystem $B$ of $\rho_{AB}$ by the rank-1 measurement
$\\{{\left|e_{i}\right\rangle}_{B}{\left\langle e_{i}\right|}\\}_{i=1}^{d-1}$
and
$\\{{\left|\tilde{e}_{j}\right\rangle}_{B}{\left\langle\tilde{e}_{j}\right|}\\}_{j=1}^{d-1}$,
respectively, the definition of CC in Eq. (5) leads us to

$\displaystyle{\mathcal{J}}^{\leftarrow}(\rho_{AB})\geq\chi(\mathcal{E}_{0}),~{}{\mathcal{J}}^{\leftarrow}(\rho_{AB})\geq\chi(\mathcal{E}_{1}),$
(36)

therefore

$\displaystyle{\mathcal{J}}^{\leftarrow}(\rho_{AB})\geq\frac{1}{2}\left(\chi(\mathcal{E}_{0})+\chi(\mathcal{E}_{1})\right)=\frac{1}{2}{I}\left(\rho_{AB}\right).$
(37)

By considering the ccq state $\Gamma_{XYAC}$ from the two-qudit reduced
density matrix $\rho_{AC}$ of ${\left|\psi\right\rangle}_{ABC}$, we can
analogously have

$\displaystyle{\mathcal{J}}^{\leftarrow}(\rho_{AC})\geq$
$\displaystyle\frac{1}{2}{I}\left(\rho_{AC}\right).$ (38)

As the trade-off relation of Eq. (6) in Proposition 1 is universal with
respect to the subsystems, we also have

$\displaystyle S(\rho_{A})={\mathcal{J}}^{\leftarrow}(\rho_{AC})+E_{\bf
f}\left(\rho_{AB}\right),$ (39)

for the given two-qudit state ${\left|\psi\right\rangle}_{ABC}$, therefore

$\displaystyle E_{\bf f}\left(\rho_{AB}\right)+E_{\bf
f}\left(\rho_{AC}\right)=$ $\displaystyle
2S(\rho_{A})-\left({\mathcal{J}}^{\leftarrow}(\rho_{AB})+{\mathcal{J}}^{\leftarrow}(\rho_{AC})\right).$
(40)

Now Inequalities (37), (38) as well as Eq. (40) lead us to

$\displaystyle E_{\bf f}\left(\rho_{AB}\right)+E_{\bf
f}\left(\rho_{AC}\right)\leq$ $\displaystyle
2S(\rho_{A})-\frac{1}{2}\left({I}(\rho_{AB})+{I}(\rho_{AC})\right)$
$\displaystyle=$ $\displaystyle
2S(\rho_{A})-\frac{1}{2}\left(S(\rho_{A})+S(\rho_{B})-S(\rho_{AB})+S(\rho_{A})+S(\rho_{C})-S(\rho_{AC})\right)$
$\displaystyle=$ $\displaystyle S(\rho_{A})$ $\displaystyle=$ $\displaystyle
E_{\bf f}\left({\left|\psi\right\rangle}_{A(BC)}\right),$ (41)

where the second equality is due to $\rho_{AC}=\rho_{B}$ and
$\rho_{AB}=\rho_{C}$ for three-party pure state
${\left|\psi\right\rangle}_{ABC}$. ∎

To illustrate Theorem 1, let us first consider three-qubit GHZ state GHZ ,

$\displaystyle{\left|GHZ\right\rangle}_{ABC}=\frac{1}{\sqrt{2}}\left({\left|000\right\rangle}_{ABC}+{\left|111\right\rangle}_{ABC}\right),$
(42)

with its reduced density matrices

$\displaystyle\rho_{AB}=\frac{1}{2}\left({\left|00\right\rangle}_{AB}{\left\langle
00\right|}+{\left|11\right\rangle}_{AB}{\left\langle
11\right|}\right),~{}\rho_{A}=\frac{1}{2}\left({\left|0\right\rangle}_{A}{\left\langle
0\right|}+{\left|1\right\rangle}_{A}{\left\langle 1\right|}\right)$ (43)

and

$\displaystyle\rho_{B}=\frac{1}{2}\left({\left|0\right\rangle}_{B}{\left\langle
0\right|}+{\left|1\right\rangle}_{B}{\left\langle 1\right|}\right).$ (44)

The eigenvalues of $\rho_{B}$ are $\lambda_{0}=\lambda_{1}=\frac{1}{2}$ with
corresponding eigenvectors
${\left|e_{0}\right\rangle}_{B}={\left|0\right\rangle}_{B}$ and
${\left|e_{1}\right\rangle}_{B}={\left|1\right\rangle}_{B}$ respectively. Thus
the ensemble of $\rho_{A}$ induced by measuring subsystem $B$ of $\rho_{AB}$
in terms of the eigenvectors of $\rho_{B}$, that is,
$\\{M_{B}^{0}={\left|0\right\rangle}_{B}{\left\langle
0\right|},M_{B}^{1}={\left|1\right\rangle}_{B}{\left\langle 1\right|}\\}$ is

$\displaystyle\mathcal{E}_{0}=\\{\lambda_{0}=\frac{1}{2},\sigma_{A}^{0}={\left|0\right\rangle}_{A}{\left\langle
0\right|},\lambda_{1}=\frac{1}{2},\sigma_{A}^{1}={\left|1\right\rangle}_{A}{\left\langle
1\right|}\\}.$ (45)

Because the Fourier basis elements of subsystem $B$ with respect to the
eigenvectors of $\rho_{B}$ are

$\displaystyle|\tilde{e}_{0}\rangle_{B}=\frac{1}{2}\left({\left|0\right\rangle}_{B}+{\left|1\right\rangle}_{B}\right),~{}|\tilde{e}_{1}\rangle_{B}=\frac{1}{2}\left({\left|0\right\rangle}_{B}-{\left|1\right\rangle}_{B}\right),$
(46)

the ensemble of $\rho_{A}$ induced by measuring subsystem $B$ of $\rho_{AB}$
in terms of the Fourier basis in Eq. (46) is

$\displaystyle\mathcal{E}_{1}=\\{\frac{1}{2},\tau_{A}^{0}=\frac{1}{2}I_{A},\frac{1}{2},\tau_{A}^{1}=\frac{1}{2}I_{A}\\}.$
(47)

Now we consider the additivity of mutual information of the ccq state
$\Gamma_{XYAB}$ obtained from $\rho_{AB}$ in Eq. (43). Due to Eq. (24), the
mutual information of $\Gamma_{XYAB}$ between $XY$ and $AB$ is

$\displaystyle I\left(\Gamma_{XY:AB}\right)=$ $\displaystyle\ln
2+S(\rho_{A})-S(\rho_{AB})=\ln 2$ (48)

because $S(\rho_{A})=S(\rho_{AB})=\ln 2$ from Eqs. (43). For the mutual
information of $\Gamma_{XAB}$ between $X$ and $AB$, Eq. (28) leads us to

$\displaystyle I\left(\Gamma_{X:AB}\right)=$ $\displaystyle\ln
2-S(\rho_{B})+\chi(\mathcal{E}_{0})=\ln 2$ (49)

where the second equality is from $S(\rho_{B})=\ln 2$ and

$\displaystyle\chi(\mathcal{E}_{0})=S(\rho_{A})-\frac{1}{2}S({\left|0\right\rangle}_{A}{\left\langle
0\right|})-\frac{1}{2}S({\left|1\right\rangle}_{A}{\left\langle 1\right|})=\ln
2,$ (50)

for the ensemble $\mathcal{E}_{0}$ in Eq. (45). For the mutual information of
$\Gamma_{YAB}$ between $Y$ and $AB$, Eq. (30) leads us to

$\displaystyle I\left(\Gamma_{Y:AB}\right)=$
$\displaystyle\chi(\mathcal{E}_{1})=S(\rho_{A})-\frac{1}{2}S\left(\frac{1}{2}I_{A}\right)-\frac{1}{2}S\left(\frac{1}{2}I_{A}\right)=0$
(51)

where the second equality is due to the ensemble $\mathcal{E}_{1}$ in Eq.
(47).

From Eqs. (48), (49) and (51), we note that the mutual information of the ccq
state $\Gamma_{XYAB}$ obtained from $\rho_{AB}$ is additive as in Eq. (34).
Moreover, the symmetry of GHZ state assures that the same is also true for the
reduced density matrix
$\rho_{AC}=\mbox{$\mathrm{tr}$}_{B}{\left|GHZ\right\rangle}_{ABC}{\left\langle
GHZ\right|}$. Thus Theorem 1 guarantees the monogamy inequality of the three-
qubit GHZ state in Eq. (42) in terms of EoF. In fact, we have

$\displaystyle E_{\bf
f}\left({\left|GHZ\right\rangle}_{A(BC)}\right)=S(\rho_{A})=\ln 2$ (52)

whereas the two-qubit reduced density matrices $\rho_{AB}$ and $\rho_{AC}$ are
separeble. Thus

$\displaystyle E_{\bf f}\left(\rho_{AB}\right)=E_{\bf
f}\left(\rho_{AC}\right)=0,$ (53)

and this implies the monogamy inequality in (33).

Let us consider another example of three-qubit state; three-qubit W-state is
defined as W

$\displaystyle{\left|W\right\rangle}_{ABC}=\frac{1}{\sqrt{3}}\left({\left|100\right\rangle}_{ABC}+{\left|010\right\rangle}_{ABC}+{\left|001\right\rangle}_{ABC}\right).$
(54)

The two-qubit reduced density matrix of ${\left|W\right\rangle}_{ABC}$ on
subsystem $AB$ is obtained as

$\displaystyle\rho_{AB}=\frac{2}{3}{\left|\psi^{+}\right\rangle}_{AB}{\left\langle\psi^{+}\right|}+\frac{1}{3}{\left|00\right\rangle}_{AB}{\left\langle
00\right|}$ (55)

where

$\displaystyle{\left|\psi^{+}\right\rangle}_{AB}=\frac{1}{\sqrt{2}}\left({\left|01\right\rangle}_{AB}+{\left|10\right\rangle}_{AB}\right)$
(56)

is the two-qubit Bell state, and the one-qubit reduced density matrices of
$\rho_{AB}$ are

$\displaystyle\rho_{A}=\frac{2}{3}{\left|0\right\rangle}_{A}{\left\langle
0\right|}+\frac{1}{3}{\left|1\right\rangle}_{A}{\left\langle
1\right|},~{}\rho_{B}=\frac{2}{3}{\left|0\right\rangle}_{B}{\left\langle
0\right|}+\frac{1}{3}{\left|1\right\rangle}_{B}{\left\langle 1\right|}.$ (57)

From to the spectral decomposition of $\rho_{B}$ in Eq. (57) with the
eigenvalues $\lambda_{0}=\frac{2}{3},\lambda_{1}=\frac{1}{3}$ and
corresponding eigenvectors
${\left|e_{0}\right\rangle}_{B}={\left|0\right\rangle}_{B}$ and
${\left|e_{1}\right\rangle}_{B}={\left|1\right\rangle}_{B}$, respectively, it
is straightforward to check that the ensemble of $\rho_{A}$ induced from
measuring subsystem $B$ of $\rho_{AB}$ by the eigenvectors of $\rho_{B}$ is

$\displaystyle\mathcal{E}_{0}=\\{\lambda_{0}=\frac{2}{3},\sigma_{A}^{0}=\frac{1}{2}I_{A},\lambda_{1}=\frac{1}{3},\sigma_{A}^{1}={\left|0\right\rangle}_{A}{\left\langle
0\right|}\\}.$ (58)

Because the Fourier basis of subsystem $A$ is the same as Eq. (46), it is also
straightforward to obtain the ensemble of $\rho_{A}$ induced by measuring
subsystem $B$ of $\rho_{AB}$ in terms of the Fourier basis,

$\displaystyle\mathcal{E}_{1}=\\{\frac{1}{2},\tau_{A}^{j}\\}_{j=1,2},$ (59)

where

$\displaystyle\tau_{A}^{0}=$
$\displaystyle\frac{1}{3}\left(2{\left|0\right\rangle}_{A}{\left\langle
0\right|}+{\left|0\right\rangle}_{A}{\left\langle
1\right|}+{\left|1\right\rangle}_{A}{\left\langle
0\right|}+{\left|1\right\rangle}_{A}{\left\langle
1\right|}\right),~{}\tau_{A}^{1}=\frac{1}{3}\left(2{\left|0\right\rangle}_{A}{\left\langle
0\right|}-{\left|0\right\rangle}_{A}{\left\langle
1\right|}-{\left|1\right\rangle}_{A}{\left\langle
0\right|}+{\left|1\right\rangle}_{A}{\left\langle 1\right|}\right).$ (60)

For the mutual information of the ccq state $\Gamma_{XYAB}$ obtained from
$\rho_{AB}$ in Eq. (55), Eq. (24) together with Eqs. (55) and (57) lead us to

$\displaystyle I\left(\Gamma_{XY:AB}\right)=$ $\displaystyle\ln
2+S(\rho_{A})-S(\rho_{AB})=\ln 2.$ (61)

For the mutual information of $\Gamma_{XAB}$ between $X$ and $AB$, Eq. (28)
leads us to

$\displaystyle I\left(\Gamma_{X:AB}\right)=$ $\displaystyle\ln
2-S(\rho_{B})+\chi(\mathcal{E}_{0}),$ (62)

where Eq. (58) impies

$\displaystyle\chi(\mathcal{E}_{0})=S(\rho_{A})-\frac{2}{3}S\left(\frac{1}{2}I_{A}\right)-\frac{1}{3}S\left({\left|0\right\rangle}_{A}{\left\langle
0\right|}\right)=S(\rho_{A})-\frac{2}{3}\ln 2.$ (63)

Due to Eq. (57), we have $S(\rho_{A})=S(\rho_{B})$, therefore Eqs. (64) and
(63) lead us to

$\displaystyle I\left(\Gamma_{X:AB}\right)=\frac{1}{3}\ln 2.$ (64)

For the mutual information of $\Gamma_{YAB}$ between $Y$ and $AB$, Eq. (30)
leads us to

$\displaystyle I\left(\Gamma_{Y:AB}\right)=$
$\displaystyle\chi(\mathcal{E}_{1})=S(\rho_{A})-\frac{1}{2}S\left(\tau_{A}^{0}\right)-\frac{1}{2}S\left(\tau_{A}^{1}\right),$
(65)

where the second equality is from the ensemble $\mathcal{E}_{1}$ in Eq. (59).
Here we note that $\tau_{A}^{0}$ and $\tau_{A}^{1}$ in Eq. (60) have the same
eigenvalues, that is $\mu_{0}=\frac{3+\sqrt{5}}{6}$ and
$\mu_{0}=\frac{3-\sqrt{5}}{6}$, therefore we have
$S\left(\tau_{A}^{0}\right)=S\left(\tau_{A}^{1}\right)$. From the spectral
decomposition of $\rho_{A}$ in Eq. (57), we have

$\displaystyle S\left(\rho_{A}\right)=\ln 3-\frac{2}{3}\ln 2$ (66)

and this turns Eq. (65) into

$\displaystyle I\left(\Gamma_{Y:AB}\right)=\ln 3-\frac{2}{3}\ln
2-S\left(\tau_{A}^{0}\right).$ (67)

From Eqs. (61), (64) and (67), we have

$\displaystyle
I\left(\Gamma_{XY:AB}\right)-I\left(\Gamma_{X:AB}\right)-I\left(\Gamma_{Y:AB}\right)=\frac{4}{3}\ln
2-\ln 3+S\left(\tau_{A}^{0}\right),$ (68)

where

$\displaystyle\ln 2\approx 0.693147,~{}\ln 3\approx 1.098612,$ (69)

and

$\displaystyle
S\left(\tau_{A}^{0}\right)=-\mu_{0}\ln\mu_{0}-\mu_{1}\ln\mu_{1}\approx
0.381264.$ (70)

Thus we have

$\displaystyle
I\left(\Gamma_{XY:AB}\right)-I\left(\Gamma_{X:AB}\right)-I\left(\Gamma_{Y:AB}\right)\approx
0.206848>0,$ (71)

which implies the nonadditivity of mutual information for the ccq state
$\Gamma_{XYAB}$ obtained from $\rho_{AB}$ in Eq. (55). We also note that the
symmetry of W state in Eq. (54) would imply the nonadditivity of mutual
information for the ccq state $\Gamma_{XYAC}$ obtained from the two-qubit
reduced density matrix $\rho_{AC}$ of W state in Eq. (54).

As the additivity of mutual information in Theorem 1 is only a sufficient
condition for monogamy inequality in terms of EoF, nonadditivity does not
directly imply violation of Inequality (33) for the W state in Eq. (54).
However, we note that $\rho_{AB}$ in Eq. (55) is a two-qubit state, therefore
its EoF can be analytically evaluated as wootters

$\displaystyle E_{\bf f}\left(\rho_{AB}\right)\approx 0.3812.$ (72)

Moreover, the symmetry of the W state assures that the EoF of
$\rho_{AC}=\mbox{$\mathrm{tr}$}_{B}{\left|W\right\rangle}_{ABC}{\left\langle
W\right|}$ is the same,

$\displaystyle E_{\bf f}\left(\rho_{AC}\right)\approx 0.3812,$ (73)

whereas

$\displaystyle E_{\bf
f}\left({\left|W\right\rangle}_{A(BC)}\right)=S(\rho_{A})\approx 0.6365.$ (74)

As Eqs. (72), (73) and (74) imply the violation of Inequality (33), W state in
Eq. (54) can be considered as an example for the contraposition of Theorem 1;
violation of monogamy inequality in (33) implies nonadditivity of quantum
mutual information for the ccq state.

Now, we generalize Theorem 1 for multi-party quantum states of arbitrary
dimension.

######  Theorem 2.

For any multi-party quantum state $\rho_{A_{1}A_{2}\cdots A_{n}}$ with two-
party reduced density matrices $\rho_{A_{1}A_{i}}$ for $i=2,\cdots,n$, we have

$\displaystyle E_{\bf f}\left(\rho_{A_{1}(A_{2}\cdots A_{n})}\right)\geq$
$\displaystyle\sum_{i=2}^{n}E_{\bf f}\left(\rho_{A_{1}A_{i}}\right),$ (75)

conditioned on the additivity of quantum mutual information for the ccq states
in the form of Eq. (16) obtained by $\rho_{A_{1}A_{i}}$ for $i=2,\cdots,n$.

###### Proof.

We first prove the theorem for any three-party mixed state $\rho_{ABC}$, and
inductively show the validity of the theorem for any $n$-party quantum state
$\rho_{A_{1}A_{2}\cdots A_{n}}$. For a three-party mixed state $\rho_{ABC}$,
let us consider an optimal decomposition of $\rho_{ABC}$ realizing EoF with
respect to the bipartition between $A$ and $BC$, that is,

$\displaystyle\rho_{ABC}=\sum_{i}p_{i}{\left|\psi_{i}\right\rangle}_{ABC}{\left\langle\psi_{i}\right|},$
(76)

with

$\displaystyle E_{\bf f}\left(\rho_{A(BC)}\right)=\sum_{i}p_{i}E_{\bf
f}\left({\left|\psi_{i}\right\rangle}_{A(BC)}\right).$ (77)

From Theorem 1, each pure state ${\left|\psi_{i}\right\rangle}_{ABC}$ of the
decomposition (76) satisfies

$\displaystyle E_{\bf f}\left({\left|\psi_{i}\right\rangle}_{A(BC)}\right)\geq
E_{\bf f}\left(\rho^{i}_{AB}\right)+E_{\bf f}\left(\rho^{i}_{AC}\right)$ (78)

with
$\rho^{i}_{AB}=\mbox{$\mathrm{tr}$}_{C}{\left|\psi_{i}\right\rangle}_{ABC}{\left\langle\psi_{i}\right|}$
and
$\rho^{i}_{AC}=\mbox{$\mathrm{tr}$}_{B}{\left|\psi_{i}\right\rangle}_{ABC}{\left\langle\psi_{i}\right|}$,
therefore,

$\displaystyle E_{\bf f}\left(\rho_{A(BC)}\right)=$
$\displaystyle\sum_{i}p_{i}E_{\bf
f}\left({\left|\psi_{i}\right\rangle}_{A(BC)}\right)\geq\sum_{i}p_{i}E_{\bf
f}\left(\rho^{i}_{AB}\right)+\sum_{i}p_{i}E_{\bf
f}\left(\rho^{i}_{AC}\right).$ (79)

For each $i$ and the two-party reduced density matrices $\rho_{AB}^{i}$, let
us consider their optimal decompositions for EoF, that is,

$\displaystyle\rho_{AB}^{i}=$
$\displaystyle\sum_{j}r_{ij}{\left|\mu_{j}^{i}\right\rangle}_{AB}{\left\langle\mu_{j}^{i}\right|}~{},$
(80)

such that

$\displaystyle E_{\bf f}\left(\rho_{AB}^{i}\right)=$
$\displaystyle\sum_{j}r_{ij}E_{\bf
f}\left({\left|\mu_{j}^{i}\right\rangle}_{AB}\right).$ (81)

Now we have

$\displaystyle\sum_{i}p_{i}E_{\bf
f}\left(\rho^{i}_{AB}\right)=\sum_{i,j}p_{i}r_{ij}E_{\bf
f}\left({\left|\mu_{j}^{i}\right\rangle}_{AB}\right)\geq E_{\bf
f}\left(\rho_{AB}\right),$ (82)

where the inequality is due to

$\displaystyle\rho_{AB}=\sum_{i}p_{i}\rho_{AB}^{i}=$
$\displaystyle\sum_{i,j}p_{i}r_{ij}{\left|\mu_{j}^{i}\right\rangle}_{AB}{\left\langle\mu_{j}^{i}\right|}$
(83)

and the definition of EoF.

For each $i$, we also consider an optimal decomposition of $\rho_{AC}$

$\displaystyle\rho_{AC}^{i}=$
$\displaystyle\sum_{l}s_{il}{\left|\nu_{l}^{i}\right\rangle}_{AC}{\left\langle\nu_{l}^{i}\right|},$
(84)

for each $i$, such that

$\displaystyle E_{\bf f}\left(\rho_{AC}^{i}\right)=$
$\displaystyle\sum_{l}s_{il}E_{\bf
f}\left({\left|\nu_{l}^{i}\right\rangle}_{AC}\right).$ (85)

We can analogously have

$\displaystyle\sum_{i}p_{i}E_{\bf f}\left(\rho^{i}_{AC}\right)\geq$
$\displaystyle E_{\bf f}\left(\rho_{AC}\right),$ (86)

due to

$\displaystyle\rho_{AC}=\sum_{i}p_{i}\rho_{AC}^{i}=$
$\displaystyle\sum_{i,l}p_{i}s_{il}{\left|\nu_{l}^{i}\right\rangle}_{AC}{\left\langle\nu_{l}^{i}\right|},$
(87)

and the definition of EoF. From Inequalities (79), (82) and (86), we have

$\displaystyle E_{\bf f}\left(\rho_{A(BC)}\right)\geq E_{\bf
f}\left(\rho_{AB}\right)+E_{\bf f}\left(\rho_{AC}\right),$ (88)

which proves the theorem for three-party mixed states.

For general multi-party quantum system, we use the mathematical induction on
the number of parties $n$; let us assume Inequality (75) is true for any
$k$-party quantum state, and consider an $k+1$-party quantum state
$\rho_{A_{1}A_{2}\cdots A_{k+1}}$ for $k\geq 3$. By considering
$\rho_{A_{1}A_{2}\cdots A_{k+1}}$ as a three-party state with respect to the
tripartition $A_{1}$, $A_{2}\cdots A_{k}$ and $A_{k+1}$, Inequality (88) leads
us to

$\displaystyle E_{\bf f}\left(\rho_{A_{1}(A_{2}\cdots A_{k+1})}\right)\geq$
$\displaystyle E_{\bf f}\left(\rho_{A_{1}(A_{2}\cdots A_{k})}\right)+E_{\bf
f}\left(\rho_{A_{1}A_{k+1}}\right).$ (89)

As $\rho_{A_{1}A_{2}\cdots A_{k}}$ in Inequality (89) is a $k$-party quantum
state, the induction hypothesis assures that

$\displaystyle E_{\bf f}\left(\rho_{A_{1}(A_{2}\cdots A_{k})}\right)\geq
E_{\bf f}\left(\rho_{A_{1}A_{2}}\right)+\cdots+E_{\bf
f}\left(\rho_{A_{1}A_{k}}\right).$ (90)

Now Inequalities (89) and (90) lead us to the monogamy inequality

$\displaystyle E_{\bf f}\left(\rho_{A_{1}(A_{2}\cdots A_{k+1})}\right)\geq$
$\displaystyle E_{\bf f}\left(\rho_{A_{1}A_{2}}\right)+\cdots+E_{\bf
f}\left(\rho_{A_{1}A_{k+1}}\right),$ (91)

which completes the proof. ∎

##  V Discussion

We have considered possible conditions for monogamy inequality of multi-party
quantum entanglement in terms of EoF, and shown that the additivity of mutual
information of the ccq states implies the monogamy inequality of three-party
quantum entanglement in terms of EoF. We have also provided examples of three-
qubit GHZ and W states to illustrate our result in three-party case, and
generalized our result into any multi-party systems of arbitrary dimensions.

Most monogamy inequalities of quantum entanglement deal with bipartite
entanglement measures based on the minimization over all possible pure state
ensembles. As analytic evaluation of such entanglement measure is generally
hard especially in higher dimensional quantum systems more than qubits, the
situation becomes far more difficult in investigating and establishing
entanglement monogamy of multi-party quantum systems of arbitrary dimensions.
The sufficient condition provided here deals with the quantum mutual
information of the ccq states to guarantee the monogamy inequality of
entanglement in terms of EoF in arbitrary dimensions. As the sufficient
condition is not involved with any minimization process, our result can
provide a useful methodology to understand the monogamy nature of multi-party
quantum entanglement in arbitrary dimensions. We finally remark that it would
be an interesting future task to investigate if the condition provided here is
also necessary.

## Acknowledgments

This work was supported by Basic Science Research
Program(NRF-2020R1F1A1A010501270) and Quantum Computing Technology Development
Program(NRF-2020M3E4A1080088) through the National Research Foundation of
Korea(NRF) grant funded by the Korea government(Ministry of Science and ICT).

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