aainstitutetext:  Department of Physics,
Indian Institute of Technology,
Kanpur 208 016, India
bbinstitutetext:  School of Physics and Astronomy,
Sun Yat-sen University,
Zhuhai 519082, China
ccinstitutetext:  Shing-Tung Yau Center and School of Physics,
Southeast University,
Nanjing 210096, China

# Entanglement islands and cutoff branes from path-integral optimization

Ashish Chandra b Zhengjiang Li c Qiang Wen<EMAIL_ADDRESS><EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract

Recently it was proposed that, the AdS/BCFT correspondence can be simulated by
a holographic Weyl transformed CFT2, where the cut-off brane plays the role of
the Karch-Randall (KR) brane Basu:2022crn . In this paper, we focus on the
Weyl transformation that optimizes the path integral computation of the
reduced density matrix for a single interval in a holographic CFT2. When we
take the limit that one of the endpoint of the interval goes to infinity (a
half line), such a holographic Weyl transformed CFT2 matches the AdS/BCFT
configuration for a BCFT with one boundary. Without taking the limit, the
induced cutoff brane becomes a circle passing through the two endpoints of the
interval. We assume that the cutoff brane also plays the same role as the KR
brane in AdS/BCFT, hence the path-integral-optimized purification for the
interval is in the island phase. This explains the appearance of negative
mutual information observed in Camargo:2022mme . We check that, the
entanglement entropy and the balanced partial entanglement entropy (BPE)
calculated via the island formulas, exactly match with the RT formula and the
entanglement wedge cross-section (EWCS), which are allowed to anchor on the
cutoff brane.

##  1 Introduction

Recently, a new rule to compute the entanglement entropy in gravitational
systems, namely the island formula Penington:2019npb ; Almheiri:2019psf ;
Almheiri:2019qdq ; Penington:2019kki ; Almheiri:2019hni , has led to a
remarkable new perspective to understand the long standing puzzle of the black
hole information paradox Hawking:1976ra . The development of the island
formula originates from the celebrated Ryu-Takayanagi (RT) formula Ryu:2006bv
; Hubeny:2007xt , which relates the entanglement entropy in holographic CFT to
the minimal surfaces in the dual AdS bulk. Then the RT formula was refined to
the quantum extremal surface (QES) formula Lewkowycz:2013nqa ;
Faulkner:2013ana ; Engelhardt:2014gca  with the quantum correction taken into
account. In Penington:2019npb ; Almheiri:2019psf  the QES formula was used to
compute the entanglement entropy for the Hawking radiation of an evaporating
black hole, and the result is consistent with the Page curve. Remarkably, the
QES formula suggests that, a region in black hole interior $I$, which we call
the entanglement island, should be considered as a part of the Hawking
radiation $R$. And the entanglement entropy for $R$ is given by the island
formula,

$\displaystyle S_{R}=\text{min}\leavevmode\nobreak\
\text{ext}_{I}\left\\{\frac{\text{Area}(\partial
I)}{4G}+\tilde{S}_{bulk}(R\cup I)\right\\}\,,$ (1)

which contains an optimization for all possible $I$. Soon, it was argued that
when we apply the replica trick in gravitational theory in a path-integral
representation, we should take into account the new geometric configurations
with wormholes in the replica manifold. When the replica wormhole
configuration dominate the path-integral, the entanglement entropy should be
computed by the island formula Penington:2019kki ; Almheiri:2019hni . Note
that, replica wormhole arguments apply to generic gravitational theories and
do not rely on the existence of holography.

Nevertheless, the island formula is still very counter-intuitive compared with
our standard understanding of quantum information theory. There should be some
exotic properties in a system that obeys the island rules. In Basu:2022crn , a
mechanism was proposed for the emergence of the entanglement islands in
quantum systems, which can be either gravitational or non-gravitational. The
mechanism is a result of imposing certain constraints on the system, which
project out large amount of states in the Hilbert space, such that for all the
states remain in the reduced Hilbert space there exists a mapping from the
state of a subset $R$ to the state of another subset $I$. We call this mapping
a coding relation,

$\displaystyle\ket{i}_{I}=f(\ket{j}_{R})\,.$ (2)

We call such systems self-encoded. In other words, the state of $I$ is
determine by the state of $R$ in the reduced Hilbert space. Note that, the
constraints we mentioned above are highly non-trivial. They are not only
imposed in the state of the system, but also imposed on the Hilbert space,
which means that a state in the reduced Hilbert space should always remain in
the reduced space under evolution or action of operators. The constraints have
highly non-local effects, such that spacelike separated degrees of freedom
becomes dependent on each other. Then the formula for entanglement entropy
should be modified correspondingly (see Basu:2022crn  for more details). For
example, given the simple coding-relation eq. 2, the entanglement entropy for
$R$ is given by

$\displaystyle S_{R}=\tilde{S}_{R\cup I}=\frac{\text{Area}(\partial
I)}{4G}+\tilde{S}_{bulk}(R\cup I)\,,$ (3)

where $\tilde{S}_{A}$ represent the von Neumann entropy for the reduced
density matrix $\rho_{A}$, which is computed by tracing out the degrees of
freedom in $\bar{A}$. In the second equality we assumed the region $I$ is
settled in a gravitational background, hence the two terms represents the
gravitational contribution and the corresponding bulk entanglement entropy
Faulkner:2013ana . Not that the compared with (1), the optimization process is
missing in (3), due to the simplicity of the coding relation (2). The specific
coding relation that induces (1) is still an open problem. Nevertheless, after
we carry out the optimization to determine the island region, this coding
relation should reduce to (2) in some way. The explicit reduction of the
Hilbert space and the corresponding coding relation are very interesting
future directions, and they are not necessary in the discussion of this paper.

The AdS/BCFT correspondence is a commonly used context under which the
entanglement islands emerges. These Weyl transformations are special as they
depend on the cutoff scale of theory, which effectively introduces finite
cutoff scale to the island region. They are assumed to be the operations that
vastly reduce the Hilbert space of the theory and hence induces the self-
coding property, as well as the island formula (1) for entanglement entropy
Basu:2022crn . The Weyl transformations will induce a cutoff brane in the
bulk. It was shown in Basu:2022crn ; Basu:2023wmv ; Lin:2023ajt  that, if we
take the cutoff brane as a KR brane where the RT surfaces are allow to anchor,
then we perfectly reproduce the main features of the AdS/BCFT configuration
Takayanagi:2011zk ; Fujita:2011fp .

In this paper, we will consider a special Weyl transformation that optimizes
the path integral computation of the reduced density matrix of an interval in
a holographic CFT2 Caputa:2017urj ; Caputa:2018xuf , and use it to simulate
the AdS/BCFT configuration. This means the path-integral-optimized
purification of the interval is in island phase. We give non-trivial
consistency check for this simulation by computing the entanglement entropy
and the BPE on the field theory side via the island formulas, and find that
the results match with the RT surfaces and EWCSs that are allowed to anchor on
the cutoff brane.

In section 2, we briefly introduce the setup of the AdS3/BCFT2 correspondence
and its simulation via a holographic Weyl transformed CFT2. In section 3, we
introduce the computation of the reduced density matrix for an interval and
the Weyl transformation that optimizes the path integral. Furthermore, we
derive the cutoff brane for the holographic CFT under such Weyl
transformations. In section 4 we assume that the cutoff brane induced by the
Weyl transformations that optimizes the path integral plays the same role as a
KR brane and calculate the entanglement entropies from both sides of the
holography. In section 5, we consider a typical bipartite subregion $AB\equiv
A\cup B$ in the interval and check the correspondence between the EWCS and the
BPE. We give a summary in section 6. In the appendix, we give an explicit
introduction for the balanced partial entanglement entropy (BPE) and its
calculation in island phases. We exhaust the computation for all the phases of
the EWCS and the corresponding BPE.

##  2 AdS/BCFT and holographic Weyl transformed CFT

In the AdS/BCFT correspondence Takayanagi:2011zk  (or equivalently the Karch-
Randall braneworld Karch:2000ct ; Karch:2000gx ), the boundary theory is a $d$
dimensional CFTd with boundaries, and the gravity dual is an AdSd+1 gravity in
$d+1$ dimensions bounded by a co-dimension one KR brane,

$\displaystyle I_{\text{AdS}}$ $\displaystyle=\frac{1}{16\pi
G}\int_{N}\sqrt{-g}(R-2\Lambda)+\frac{1}{8\pi
G}\int_{\mathcal{Q}}\sqrt{-h}(K-T),$ (4)

where $N$ denotes the bulk AdS spacetime, $\mathcal{Q}$ denotes the Karch-
Randall (KR) brane anchored at the boundary of the CFT and $T$ is the tension
of $\mathcal{Q}$. In the bulk, the von Neumann boundary conditions are imposed
on the KR brane. In this paper, we focus on the $d=2$ case and write the bulk
metric in terms of the following coordinates,

$\displaystyle\mathrm{d}s^{2}=$
$\displaystyle\frac{\ell^{2}}{z^{2}}\left(-\mathrm{d}t^{2}+\mathrm{d}x^{2}+\mathrm{d}z^{2}\right)$
(5) $\displaystyle=$
$\displaystyle\mathrm{d}\rho^{2}+\ell^{2}\cosh^{2}\frac{\rho}{\ell}\left(\frac{-\mathrm{d}t^{2}+\mathrm{d}y^{2}}{y^{2}}\right)\,,$
(6)

where $\ell$ is the AdS radius which we will set to be unit in the rest of
this paper, and the two sets of coordinates are related by

$\displaystyle z=y\cosh^{-1}{\rho},\quad x=-y\tanh{\rho}\,.$ (7)

For a BCFT with one boundary settled at $x=0$, the corresponding KR brane
locates at $\rho=\rho_{0}$, where $\rho_{0}$ is a constant determined by the
tension of the brane $\rho_{0}=\text{arctanh}\leavevmode\nobreak\ T$.

On the other hand, let us consider the vacuum state of a holographic CFT2 in
flat background

$ds^{2}=\frac{-dt^{2}+dx^{2}}{\epsilon^{2}},$ (8)

where $\epsilon$ represent the UV cutoff of the CFT. Then we perform a Weyl
transformation characterized by a scalar field $\varphi(x)$, hence the metric
changes to be

$ds^{2}=e^{2\varphi(x)}\left(\frac{-dt^{2}+dx^{2}}{\epsilon^{2}}\right).$ (9)

It can be understood that, the Weyl transformation changes the cutoff scale in
a position-dependent way,

$\displaystyle\epsilon\ \to\ e^{-\varphi(x)}\epsilon\,.$ (10)

The scalar $\varphi(x)$ is chosen to be negative hence the Weyl transformation
enlarges the cutoff scale. Accordingly, the entanglement entropy for a single
interval $A=[a,b]$ is modified to be Caputa:2017urj ; Caputa:2018xuf ;
Camargo:2022mme

$\displaystyle
S_{A}=\frac{c}{3}\log\frac{(b-a)}{\epsilon}-\frac{c}{6}|\varphi(a)|-\frac{c}{6}|\varphi(b)|\,,$
(11)

following the transformation rule of the two-point functions of twist
operators under Weyl transformations. The above result is just the original
entanglement entropy subjecting the absolute value of the scalar field at the
endpoints. Holographically, the constant subjection of the entanglement
entropy was understood as inserting cutoff spheres with radius $|\varphi(x)|$
centered at the endpoints Basu:2022crn . In other words, when computing the
entanglement entropy via the RT formula, we should exclude the portion of the
RT surface inside the cutoff spheres, see Fig.1. Interestingly, the cutoff
sphere centered at $(x,z)=(x_{0},\epsilon)$ in AdS3 is just a circle in flat
background with radius $\alpha$, Wen:2018mev ; Basu:2022crn ,

$\displaystyle(x-x_{0})^{2}+(z-\alpha)^{2}=\alpha^{2},\qquad\alpha=\frac{\epsilon}{2}e^{\absolutevalue{\varphi(x_{0})}}\,.$
(12)

$x_{0}$ Figure 1: Cut-off sphere at $x=x_{0}$. For RT surfaces anchored at
$x_{0}$, the part inside the cutoff sphere is excluded, which has the same
length $\absolutevalue{\phi(x_{0})}$

The particular Weyl transformation, that captures the main features of the
AdS/BCFT with the KR brane settled at $\rho=\rho_{0}$, is given by
Basu:2022crn 111See Suzuki:2022xwv  for an earlier discussion.,

$\varphi(x)=\begin{cases}0,&\text{ if }\quad x>0\\\
-\log\left(\frac{2|x|}{\epsilon}\right)+\kappa,&\ \text{if }\quad
x<0\end{cases}\,,$ (13)

where $\kappa$ is a constant. The cutoff brane Basu:2022crn  is defined as the
common tangent line of all the cutoff spheres, which represent the boundary of
the bulk cutoff region (see Fig.2). In this case the cutoff brane is settled
at

$\displaystyle\rho=\kappa\,,$ (14)

which can be adjust to exactly overlap with the KR brane by setting

$\displaystyle\kappa=\rho_{0}\,.$ (15)

Figure 2: This figure shows the cutoff spheres and cutoff brane on a time
slice for the Weyl transformed CFT with the scalar field (13). The radius of
the cutoff sphere centered at $x=x_{0}$ is given by
$\alpha=-x_{0}e^{-\kappa}$.

To justify the application of the island formula in the Weyl transformed CFT,
we need to give an additional assumption for the setup. For example, we can
assume that the $x<0$ region was coupled to a AdS2 gravity, then entanglement
islands are allowed in this region according to the replica wormhole arguments
Almheiri:2019qdq ; Penington:2019kki . Alternatively, we can also assume that,
a particular coding-relation is induced by the Weyl transformation which
result in the corresponding island formula (1) following the arguments in
Basu:2022crn . Note that, the choice between these two assumptions will not
affect the explicit calculations in this paper. Also note that, the area term
in (1) will not appear since we assume the AdS2 gravity is induced by partial
reduction of the AdS3 spacetime.

The key observation is that, after we identify the island region $I$ by
optimizing (1), then the bulk entanglement entropy $\tilde{S}_{bulk}(R\cup I)$
is given by the RT surface of $R\cup I$ which is always cutoff exactly at the
cutoff brane Basu:2022crn , see Fig.2. In other words, in the Weyl transformed
CFT configuration, the RT surfaces are allowed to anchor on the cutoff brane
in the sense that, they are cut off there. The cutoff brane then plays the
same role as the KR brane in the AdS/BCFT correspondence.

##  3 Weyl transformed CFT from path-integral optimization and AdS/BCFT

In the previous section, we simulated the AdS/BCFT configuration where the KR
brane is settled at $\rho=\rho_{0}$, by the holographic Weyl transformed CFT2
with the scalar field (13) and the constant $\kappa$ adjusted to satisfy
$\kappa=\rho_{0}$. Note that, in this case the Weyl transformation is adjusted
by hand such that the cutoff brane and the KR brane overlap. Since in the
AdS/BCFT correspondence, the location of the KR brane is determined by details
of the theory, including the tension and the boundary conditions of the brane.
It is intriguing to ask what is special about the corresponding Weyl
transformation (13) from the perspective on the field theory side. One of the
main observation in this paper is that, the particular Weyl transformation
(13) that makes the simulation a success is the one that optimizes the path
integral, which implies two key points: 1) it preserves the reduced density
matrix $\rho_{A}$ for an interval $A$ at a particular time, and 2) it
minimizes the path-integral complexity $C_{L}[\phi]$ defined later
Caputa:2017urj ; Caputa:2017yrh .

Let us start with the path integral computation for a quantum state of the
whole system. Consider a 2d CFT on Euclidean flat space
$\mathrm{d}s^{2}=(\mathrm{d}\tau^{2}+\mathrm{d}\xi^{2})/\epsilon^{2}=\delta_{ab}/\epsilon^{2}$,
where $\epsilon$ represents the UV cutoff. Under this metric the ground state
$\ket{\Psi}$ wave functional $\Psi[\tilde{\varphi}(\xi)]$ at $\tau=-\epsilon$
is given by the Euclidean path integral on the half plane,

$\displaystyle\Psi_{\delta_{ab}/\epsilon^{2}}[\tilde{\varphi}(\xi)]=\int\left(\prod_{\xi}\prod_{-\infty<\tau<-\epsilon}D\varphi(\tau,\xi)\right)e^{-S_{CFT}(\varphi)}\cdot\prod_{\xi}\delta(\varphi(-\epsilon,\xi)-\tilde{\varphi}(\xi)).$
(16)

Here the subscript of the wave function $\Psi$ represents the metric of the
Euclidean space where the path integral is performed. One can perform a Weyl
transformation, which is a symmetry of the theory, to the metric,

$\displaystyle
ds^{2}=e^{2\phi(\tau,\xi)}\frac{d\tau^{2}+d\xi^{2}}{\epsilon^{2}},\qquad
e^{2\phi(\tau=-\epsilon,\xi)}=1.$ (17)

where the second equation is the boundary condition for the scalar field that
characterizes the Weyl transformation. Then the state $\Psi$ computed under
the Weyl transformed metric is proportional to (16),

$\displaystyle\Psi_{e^{2\phi}\delta_{ab}/\epsilon^{2}}[(\tilde{\varphi}(\xi))]=e^{C_{L}[\phi]-C_{L}[0]}\Psi_{\delta_{ab}/\epsilon^{2}}[(\tilde{\varphi}(\xi))]\,,$
(18)

where $C_{L}[\phi]$ the Liouville action Polyakov:1981rd ,

$\displaystyle
C_{L}[\phi]=\frac{c}{24\pi}\int_{-\infty}^{\infty}d\xi\int^{-\epsilon}_{-\infty}d\tau\left((\partial_{\xi}\phi)^{2}+(\partial_{\tau}\phi)^{2}+\mu
e^{2\phi}\right).$ (19)

This means the state $\Psi$ is preserved under the Weyl transformation (17).
In Caputa:2017urj ; Caputa:2017yrh , the Liouville action $C_{L}[\phi]$ is
further related to the complexity functional of the quantum state
$\ket{\Psi}$. The path integral optimization then means by computing the path
integral under the Weyl transformation that minimizes the Liouville action
$C_{L}[\phi]$. This can be achieved by solving the equation of motion
$(\partial_{\xi}^{2}+\partial_{\tau}^{2})\phi=e^{2\phi}/\epsilon^{2}$ of
$C_{L}[\phi]$ (or the Liouville equation), which gives us the following simple
solution,

$\displaystyle e^{2\phi}=\frac{\epsilon^{2}}{\tau^{2}}.$ (20)

It is easy to check that the above solution satisfies the boundary condition
in (17). Note that, one can shift the above scalar field by a constant by
choosing a different UV cutoff scale $\epsilon$.

The above optimization procedure was generalized to optimizing the path
integral computations that preserve the state (or the reduced density matrix)
on a single interval $A=[a,b],\,\,t=0$ as described in Caputa:2018xuf . This
involves the Euclidean path integral over a complex plane
$(\eta,\bar{\eta})=(x+it,x-it)$ with the interval $A$ cut open. Interestingly,
one can relate this path integral to the one on a half plane by performing a
conformal transformation which maps the interval $A$ to an infinitely long
line,

$\displaystyle
w=\sqrt{\frac{\eta-a}{b-\eta}}\,,\qquad(\omega,\bar{\omega})=(\xi+i\tau,\xi-i\tau)\,.$
(21)

Then we can optimize the Euclidean path integral on the half $\omega$ plane
whose optimization gives (20). Finally, we may obtain the Weyl transformation
by mapping (20) back to the $(t,x)$ plane. Hence, we get the Weyl
transformation that optimizes the path integral computation for the reduced
density matrix $\rho_{A}$ of the interval $A$. More explicitly, the relation
$\frac{\epsilon^{2}}{\tau^{2}}d\omega d\bar{\omega}=e^{2\phi(x)}d\eta
d\bar{\eta}$ gives us,

$\phi(x)=\left\\{\begin{array}[]{ c l }0&\qquad a<x<b\\\
\log\left[\frac{\epsilon(b-a)}{2(x-a)(x-b)}\right]+\kappa&\qquad x>b\ or\
x<a\end{array}\right\\}\,,$ (22)

where $\kappa$ is a constant which depends on the choice of $\epsilon$.

Now we are ready to derive the cutoff brane induced by the above Weyl
transformation, which is again the common tangent line of all the cutoff
spheres. Firstly, according to (12) the cutoff spheres are circles on a time
slice described by

$\displaystyle(x-x_{0})^{2}+(z-\alpha)^{2}=\alpha^{2},\quad\alpha=\frac{(x_{0}-a)(x_{0}-b)}{b-a}e^{-\kappa}.$
(23)

Then it is easy to check that the common tangent line of all the spheres is
also a part of a circle which passes the two endpoints of the interval (see
Fig.3),

$\displaystyle\text{cutoff
brane:}\quad(x-\frac{a+b}{2})^{2}+(z-z_{0})^{2}=z_{0}^{2}+\frac{(b-a)^{2}}{4},$
(24)

where $z_{0}=\frac{b-a}{4}(e^{\kappa}-e^{-\kappa})$. Changing the constant
$\kappa$ will change the intersection angle between the cutoff brane and the
asymptotic boundary. In Fig.3, we give the cutoff brane for the same interval
with different $\kappa$.

Figure 3: This figure shows the numerical results of to series of cut-off
spheres and the corresponding cutoff brane (dashed line). We have set a=1, b=2
and $\kappa=0,1$ in the left and right figures respectively.

When we send one of the endpoint to infinity by taking the limit $b\to\infty$,
we are back to the BCFT with one boundary discussed in the previous section.
Interestingly, under this limit the cutoff brane (24) becomes,

$\displaystyle z=(a-x)\csch(\kappa)\,.$ (25)

According to the coordinate transformation (7), the above equation is just
$\rho=\kappa$ if we put the other endpoint at the origin by setting $a=0$.
This exactly coincides with the cutoff brane (14), which is adjusted by hand
to match the KR brane in the AdS/BCFT correspondence. Note that, if we choose
an arbitrary smooth function $\phi(x)$ for the Weyl transformation, the shape
of the cutoff brane could be arbitrary. It is remarkable that the Weyl
transformations that optimize the path integral exactly reproduce the KR brane
configurations in the AdS/BCFT correspondence for the BCFT with one boundary.

Then it is intriguing to propose that, the cutoff brane configurations from
path integral optimization can match to the KR branes in more generic AdS/BCFT
setups. If this proposal is right, there should be KR brane configurations as
circles (24) in the Poincaré AdS3 bulk that homologous to the strip BCFT2 with
two boundaries. Since the cutoff brane is connected in the bulk, it belongs to
the confined phase Takayanagi:2011zk ; Fujita:2011fp ; Nozaki:2012qd , rather
than the deconfined phase where the brane is disconnected in the bulk222The
deconfined configurations were used to derive the co-dimension two Wedge
holography Akal:2020wfl .. Unfortunately, such configurations in AdS/BCFT were
not confirmed in the literature yet. There are AdS/BCFT configurations where
the KR brane on a time slice are circles, but the BCFT is not a strip. For
example, in Fujita:2011fp  the authors considered a BCFT on a round disk:
$\tau^{2}+x^{2}\leq L^{2}$ in Euclidean spacetime, and find that the KR brane
in AdS bulk dual is given by a sphere,

$\displaystyle\tau^{2}+x^{2}+(z-\sinh(\rho_{0})L)^{2}-L^{2}\cosh^{2}(\rho_{0})=0,$
(26)

where $\rho_{0}=\text{arctanh}\,T$. For any fixed $\tau$, the KR brane is also
a portion of a circle. Also in Geng:2022tfc , the KR branes connecting the two
asymptotic boundaries in the eternal black hole are circles on a time slice
when mapping to the Poincaré patch (see Fig. 6 in Geng:2022tfc ). Similar
configurations were also proposed in Miao:2017gyt .

##  4 Entanglement islands in holographic Weyl transformed CFT from path
integral optimization

In Takayanagi:2017knl ; Nguyen:2017yqw , it was claimed that the entanglement
of purification (EoP) is the holographic dual of the EWCS in holographic
theories. The optimization for the path integral that computes the reduced
density matrix for an interval was first studied in Caputa:2018xuf  to compute
the entanglement of purification (EoP), and the claim of Takayanagi:2017knl ;
Nguyen:2017yqw  was confirmed in this scenario. Let us consider a bipartite
region in the boundary field theory with a partition $AB\equiv A\cup B$, and
$\rho_{AB}$ is the corresponding reduced density matrix. On a time slice, the
entanglement wedge $\mathcal{W}_{AB}$ of $AB$ is the region enclosed by $AB$
and the RT surface $\mathcal{E}_{AB}$. This allows us to define the EWCS
$\Sigma_{AB}$ as the minimal area cross-section of $\mathcal{W}_{AB}$
separating the regions $A$ and $B$. On the other hand, let
$\ket{\Psi}\in\mathcal{H}_{AA^{\prime}}\otimes\mathcal{H}_{BB^{\prime}}$ be
any purification of $\rho_{AB}$, the EoP ($E_{p}(A:B)$) between $A$ and $B$ is
defined as Terhal:2002riz

$\displaystyle
E_{p}(A:B)=\mathop{\text{min}}\limits_{\ket{\Psi},A^{\prime}}S_{AA^{\prime}}\,,$
(27)

where we take the minimization over all possible purifications of $AB$ and
over all possible partitions of $A^{\prime}B^{\prime}$. The EoP is then given
by the minimal value of $S_{AA^{\prime}}$. The key observation of
Caputa:2018xuf  is that, the purification $\ket{\Psi}$ that gives the
minimized $S_{AA^{\prime}}$ is exactly the holographic Weyl transformed CFT2
with the Weyl transformation that optimizes the path integral computation of
$\rho_{AB}$. More explicitly, it was shown in Caputa:2018xuf  that, in the
path-integral-optimized $\ket{\Psi}$, the minimal $S_{AA^{\prime}}$ over all
the possible partition of the complement $A^{\prime}B^{\prime}$ exactly
matches the length of the EWCS.

For example, let us consider the configuration shown in Fig.4, where the
purification of the mixed state $\rho_{AB}$ is path-integral optimized under
the Weyl transformation characterized by (22). Then we minimize the
entanglement entropy $S_{AA^{\prime}}$ by adjusting the partition point $x=q$
and found that when

$\displaystyle q=\frac{2ab-(a+b)p}{a+b-2p}\,,$ (28)

the entanglement entropy $S_{AA^{\prime}}$ is minimized, hence

$\displaystyle
E_{p}(A:B)=\mathop{\text{min}}\limits_{\ket{\Psi},A^{\prime}}S_{AA^{\prime}}=\frac{c}{6}\log(\frac{2(p-a)(b-p)}{\epsilon(b-a)})\,.$
(29)

Note that the entanglement entropy is calculated following (11).

${\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}A}$${\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}B}$${\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}A^{\prime}}$${\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}B^{\prime}}$${\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}B^{\prime}}$$p$$b$a$q$$\Sigma_{AB}$
Figure 4: The blue circles are the RT surfaces for $AB$ and $AA^{\prime}$. The
Weyl transformation (22) with $\kappa=0$ is performed on
$A^{\prime}B^{\prime}$. When calculating $S_{AA^{\prime}}$, we should subject
a constant term $\frac{c}{6}\absolutevalue{\phi(q)}$, which is exactly the
length of the dashed portion. Then $S_{AA^{\prime}}$ exactly matches the
length of the EWCS $\Sigma_{AB}$.

The authors of Caputa:2018xuf  studied the path-integral-optimized state
$\ket{\Psi}$ in no-island phase333By the time when Caputa:2018xuf  was
published, there was no concept of entanglement islands.. This indicates that
the Hilbert space of the system is factorized as all the degrees of freedom
are independent from each other, and the entanglement entropy for any interval
$[a,b]$ can be defined in a standard way and calculated following the formula
eq. 11. Nevertheless, it was pointed out in Camargo:2022mme  that, under such
a setup there exists negative mutual information in the path-integral-
optimized state $\ket{\Psi}$. Again, let us consider the configuration of
Fig.4 with the partition point given by (28), then we will find the following
mutual information is negative,

$\displaystyle I(A^{\prime}:B^{\prime})$
$\displaystyle=S_{A^{\prime}}+S_{B^{\prime}}-S_{A^{\prime}B^{\prime}}$
$\displaystyle=\frac{c}{3}\log\frac{(a-q)(b-q)}{(b-a)\epsilon}+\frac{c}{3}\phi(q)$
$\displaystyle=-\frac{c}{3}\log 2\,.$ (30)

This means the path-integral-optimized state $\ket{\Psi}$ is not physical.

Following the series of work Basu:2022crn ; Basu:2023wmv ; Lin:2023ajt  where
the holographic CFTs under certain Weyl transformations were understood as a
simulation of the AdS/BCFT correspondence, we consider the possibility that
the path-integrate-optimized state is in island phase and apply the island
formula on this state. This new perspective will eventually solve the above
puzzle of negative mutual information. Let us set $\kappa=0$, hence the Weyl
transformation matches with the one in Caputa:2018xuf , and the cutoff brane
(24) coincide with the RT surface of $AB$. Then we take the cutoff brane as a
KR brane where the RT surfaces for subregions in $AB$ anchor.

The island phase perspective on $\ket{\Psi}$ results in a totally new
understanding on the entanglement structure of this state and the
corresponding geometric dual on the gravity side. Firstly, on the field theory
side the complement $A^{\prime}B^{\prime}$ is the entanglement island of $AB$,
hence $S_{AB}=\tilde{S}_{ABA^{\prime}B^{\prime}}=0$, while on the gravity
side, the RT surface of $AB$ vanishes as it can anchor on the cutoff brane. In
other words, we have $S_{AB}=0$ hence $AB$ is in a pure state.

Secondly, on the field theory side we should apply the island formula (1) to
calculate $S_{A}$,

$\displaystyle S_{A}=$ $\displaystyle\mathop{\text{min
ext}}\limits_{A^{\prime}}\tilde{S}_{AA^{\prime}}$ (31) $\displaystyle=$
$\displaystyle\mathop{\text{min}}\limits_{q}\frac{c}{3}\log\frac{p-q}{\epsilon}+\frac{c}{6}\phi(q)$
(32) $\displaystyle=$
$\displaystyle\frac{c}{6}\log\frac{2(p-a)(b-p)}{\epsilon(b-a)}\,,$ (33)

where in the third equation we plugged in $q=\frac{2ab-(a+b)p}{a+b-2p}$ which
is the solution of the optimization. This solution coincides with
$\eqref{Oppar}$, which the minimizes $S_{AA^{\prime}}$ when calculating the
EoP in Caputa:2018xuf . It was also pointed out in Wen:2021qgx ;
Camargo:2022mme  that, the point $x=q$ satisfying (28) is just the
intersection point between the boundary and the geodesic (see the blue dashed
line in Fig.4) extended from the EWCS $\Sigma_{AB}$. Here, since $AB$ is in a
pure state, the EWCS $\Sigma_{AB}$ coincide with the RT surface
$\mathcal{E}_{A}$ of $A$. On the gravity side, the RT surface of $A$ is just
the geodesic emanating from $P$ and anchors on the cutoff brane vertically,
which coincide with the EWCS. This is consistent with the fact that, when $AB$
is in a pure state, the EoP, or other quantum information quantities claimed
to be dual to the EWCS like the reflected entropy, coincide with $S_{A}$. Also
when $AB$ is in a pure state, the EWCS $\Sigma_{AB}$ coincides with the RT
surface of $A$.

Thirdly, we can take $AB$ is a subregion of the interval. If the cutoff brane
plays the same role as the KR brane, then the RT surface $\mathcal{E}_{AB}$,
entanglement wedge $\mathcal{W}_{AB}$ and the corresponding EWCS $\Sigma_{AB}$
all undergo new phase transitions due to the entanglement islands. On the
other hand, we can calculate the BPE on the field theory side following the
steps in Basu:2023wmv , and find it exactly match with the EWCS. This is a
highly non-trivial test for our proposal that the path-integral-optimized
state $\ket{\Psi}$ should be understood as in the island phase, which is the
main topic of the following section.

All in all, if the path-integral-optimized purification $\ket{\Psi}$ is in the
island phase, where $A^{\prime}B^{\prime}$ is the island region of $AB$, then
according to the discussion in Basu:2022crn  the degrees of freedom in
$A^{\prime}B^{\prime}$ are no longer independent. This put all the general
inequalities satisfied by entanglement entropy into question, as well as the
positivity of the mutual information $I(A^{\prime},B^{\prime})$. In fact even
the concept of the entanglement entropy of $A^{\prime}$ is not well defined,
since $A^{\prime}$ is not independent Basu:2023wmv .

##  5 The correspondence between the EWCS and the BPE

In the articles Wen:2021qgx ; Camargo:2022mme , it was proposed that the so-
called balance partial entanglement entropy (BPE) is the quantum information
quantity that duals to the EWCS. Furthermore, the concept of BPE was
generalized to the island phases in Basu:2023wmv  and its correspondence to
the EWCS was also explicitly checked in the context of AdS/BCFT or the
holographic Weyl transformed CFT2 in Basu:2023wmv ; Lin:2023ajt . See also
Chandrasekaran:2020qtn ; Li:2020ceg ; Afrasiar:2022fid ; Afrasiar:2023jrj ;
Basu:2023jtf ; Basak:2023bnc  for more discussion on the EWCS and its quantum
information dual in holographic configurations with entanglement islands. In
this section, we will study the entanglement wedges $\mathcal{W}_{AB}$ for a
bipartite subregion $AB$ of the path-integral-optimized interval, with the
cutoff brane playing the role of the KR brane in the context of AdS/BCFT. On
the gravity side, we classify the phases of the EWCS of $\mathcal{W}_{AB}$ and
calculate the area of the EWCS in each phase. On the field theory side, we
explicitly calculate the BPE between $A$ and $B$. As expected, we find the
agreement between the EWCS and the BPE in all the phases.

###  5.1 Brief review for the PEE and BPE

Let us first introduce the basic concept of the partial entanglement entropy
(PEE) Wen:2018whg ; Wen:2020ech ; Wen:2019iyq . The PEE is a measure of two-
body correlation between two non-overlapping regions $\mathcal{I}(A,B)$444Note
that, we should not mix between the mutual information $I(A,B)$ and the PEE
$\mathcal{I}(A,B)$.. So far, the fundamental definition for PEE based on the
reduced density matrix is still not established. In some scenarios of
interest, it can be determined by a set of physical requirements Casini:2008wt
; Wen:2019iyq , which include all the properties satisfied by the mutual
information and the additional key property of additivity:

  1. 1.

Additivity: $\mathcal{I}(A,B\cup C)=\mathcal{I}(A,B)+\mathcal{I}(A,C)$;

  2. 2.

Permutation symmetry: $\mathcal{I}(A,B)=\mathcal{I}(B,A)$;

  3. 3.

Normalization: $\mathcal{I}(A,\bar{A})=S_{A}$;

  4. 4.

Positivity: $\mathcal{I}(A,B)>0$;

  5. 5.

Upper boundedness:$\mathcal{I}(A,B)\leq\text{min}\\{S_{A},S_{B}\\}$;

  6. 6.

$\mathcal{I}(A,B)$ should be Invariant under local unitary transformations
inside $A$ or $B$;

  7. 7.

Symmetry: For any symmetry transformation $\mathcal{T}$ under which
$\mathcal{T}A=A^{\prime}$ and $\mathcal{T}B=B^{\prime}$, we have
$\mathcal{I}(A,B)=\mathcal{I}(A^{\prime},B^{\prime})$.

In the above list, $A$, $B$ and $C$ denote non-overlapping regions.

The concept of PEE Wen:2018whg ; Kudler-Flam:2019oru ; Wen:2020ech ;
Wen:2019iyq  originates from the study of the _entanglement contour_ Chen_2014
, which is defined as a function $s_{A}(\textbf{x})$ that characterizes the
contribution to the entanglement entropy of $A$ from each site $\textbf{x}\in
A$. By definition the entanglement contour function should satisfy,

$S_{A}=\int_{A}s_{A}(\textbf{x})d\sigma_{\textbf{x}}\,,$ (34)

where $\sigma_{\textbf{x}}$ is a infinitesimal area element located at the
site x. Subsequently, we can also define the contribution from a subset
$\alpha$ in $A$ to $S_{A}$,

$s_{A}(\alpha)=\int_{\alpha}s_{A}(\textbf{x})d\sigma_{\textbf{x}}\,.$ (35)

The contribution $s_{A}(\alpha)$ is a measure of the correlation between the
subregion $\alpha$ and the complement $\bar{A}$ of A, which is exactly the
two-body correlation we have defined as the PEE,

$\displaystyle\mathcal{I}(\alpha,\bar{A})\equiv s_{A}(\alpha)\,.$ (36)

Following Basu:2023wmv , we call the notation on the left hand side of the
above equation as the two-body correlation representation for the PEE, while
the notation on the right hand side is the contribution representation.
Interestingly, the PEE also follows additive property.

In this paper, we will use a particular construction for the PEE in generic
two-dimensional theories on a line or a circle555See Wen:2018whg ; Wen:2019iyq
; Chen_2014 ; Han:2019scu ; SinghaRoy:2019urc  for other prescriptions to
construct PEE, and Wen:2018mev ; Kudler-Flam:2019oru ; Ageev:2021ipd ;
Rolph:2021nan  for applications of the ALC proposal. . This proposal is
referred to as the _additive linear combination_ (ALC) proposal Wen:2018whg ;
Kudler-Flam:2019oru ; Wen:2019iyq , which claims that the PEE in these
scenarios can be written as a linear combination of subset entanglement
entropies that satisfy the property of additivity. Consider an interval $A$
and divide it into three sub-intervals, namely
$\alpha_{L}\cup\alpha\cup\alpha_{R}$, where $\alpha$ is in the middle and
$\alpha_{L}$ ($\alpha_{R}$) signifies the regions situated to the left (right)
of $\alpha$. Within this arrangement, the ALC proposal of the PEE is given by:

$\displaystyle\emph{The ALC proposal:}\quad
s_{A}(\alpha)=\mathcal{I}(\alpha,\bar{A})=\frac{1}{2}\left(S_{\alpha_{L}\cup\alpha}+S_{\alpha\cup\alpha_{R}}-S_{\alpha_{L}}-S_{\alpha_{R}}\right)\,.$
(37)

It is easy to check that, the above construction satisfies all the seven
requirements in general Wen:2020ech ; Kudler-Flam:2019oru .

The _balanced partial entanglement entropy_ (BPE) is a special PEE that
satisfies a set of balance requirements Wen:2021qgx . In canonical
purification, the definition of the BPE coincide with the reflected entropy
Wen:2021qgx , and the purification independence of the BPE was explored in
Camargo:2022mme . The BPE was proposed to be dual to the EWCS in holographic
theories, and this proposal has undergone rigorous validation across diverse
scenarios including both static Wen:2021qgx  and covariant Wen:2022jxr  setups
of AdS3/CFT2 with or without gravitational anomalies, and in the context of
AdS/BCFT with entanglement islands Basu:2023wmv ; Lin:2023ajt . More
interestingly, in the context of 3-dimensional flat holography Bagchi:2010zz ;
Bagchi:2012cy ; Jiang:2017ecm , on the field theory side the BPE was
calculated in Camargo:2022mme ; Basu:2022nyl , which matches the EWCS explored
in Basu:2021awn .

In normal quantum systems without entanglement islands, the BPE is defined in
the following. In a 2-dimensional theory, we consider a bipartite system
$\mathcal{H}_{A}\otimes\mathcal{H}_{B}$ in a mixed state $\rho_{AB}$, which
can be purified by involving an auxiliary system $A_{1}B_{1}$, so that
$\text{Tr}_{A_{1}B_{1}}\ket{\psi}\bra{\psi}=\rho_{AB}$. The BPE between $A$
and $B$ is expressed as follows

$\displaystyle\text{BPE}(A:B)=\mathcal{I}(A,BB_{1})|_{balanced}=\mathcal{I}(B,AA_{1})|_{balanced}=s_{AA_{1}}(A)|_{balanced}\,.$
(38)

where the partition for the complement $A_{1}B_{1}$ is determined by a set of
balance requirements described below:

Figure 5: This figure is extracted from Basu:2023wmv . Schematics describe the
BPE for two adjacent and disjoint configuration of the intervals $A$ and $B$
located in CFT2 vacuum where the points $q,q_{1}$ and $q_{2}$ indicate the
balance points.

  1. 1.

Balance requirement: For the adjacent configuration as depicted in fig. 5,
this condition can be expressed as a constraint equation in terms of the PEE
for all the possible configurations of the partition $A_{1}B_{1}$ as

$\displaystyle s_{AA_{1}}(A)=s_{BB_{1}}(B)\quad
or\quad\mathcal{I}(A,B_{1})=\mathcal{I}(A_{1},B).$ (39)

However, for non-adjacent case as shown in right part of fig. 5, the
complement of the intervals $A\cup B$ gets disconnected and we require two
balance conditions to obtain the two partition points

$s_{AA_{1}A_{2}}(A_{1})=s_{BB_{1}B_{2}}(B_{1}),\quad
s_{AA_{1}A_{2}}(A)=s_{BB_{1}B_{2}}(B),$ (40)

or

$\mathcal{I}(A_{1},BB_{2})=\mathcal{I}(B_{1},AA_{2}),\quad\mathcal{I}(A,B_{1}B_{2})=\mathcal{I}(B,A_{1}A_{2})\,.$
(41)

Since $S_{A}=S_{B}$, $s_{AA_{1}A_{2}}(A_{2})=s_{BB_{1}B_{2}}(B_{2})$ is
automatically satisfied if the above two conditions are satisfied.

  2. 2.

Minimal requirement: Typically, there exist multiple configurations for
partitioning $A_{1}B_{1}$ that fulfill the balance requirement. The BPE$(A,B)$
is then defined as the minimal $s_{AA_{1}}(A)$ that satisfies the balanced
requirement. In rest of the paper, when we mention the balance requirements,
the minimal requirement is implied.

###  5.2 PEE and BPE in island phases

As described in the article Basu:2022crn , a system in the island phase can be
understood as a self-encoding system. In other words, the state of certain
subsets of the system are totally encoded in the state of their counterpart
subsets of the system. In the typical configurations where entanglement
islands $I$ of the Hawking radiation $R$ emerges666For example the
configuration where a AdS gravity coupled to a CFT bath at the asymptotic
boundary, and the AdS/BCFT correspondence where a gravity is located on the KR
brane., this is equivalent to the statement that, the black hole interior (or
the island $I$) can be reconstructed from the Hawking radiation $R$ after the
Page time. In such a self-encoding system, the dimension of Hilbert space is
vastly reduced and no longer factorizable. Subsequently, the calculation of
entanglement entropy should also be modified to formulas like eq. 3. Given the
self-encoding property, when we compute the PEE between subregions in island
phase, we should also take the contribution from the corresponding island
regions into account. Now we generalize our construction of the PEE and BPE to
configurations with entanglement islands Basu:2023wmv . Here we just list the
basic elements we need to carry out the computations. One should consult
Basu:2023wmv  for more details.

${\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}A}$$A_{2}$
$B_{2}$ $B$ Is($A$) $\underbrace{\hskip 28.45274pt}$Io($AB$) Is($B$)  Figure
6: Ownerless island in one-endpoint example. Here $A_{2},\ B_{2}$ admit no
islands, and there is an ownerless island Io($AB$).

Let us start with a bipartite region $AB$ and the island regions Is$(AB)$,
Is$(A)$ and Is$(B)$, which can be empty set if there is no island. The PEE,
for example $s_{AB}(A)$, should contain the contribution from the island
region, hence we should assign the contribution from $\mathrm{Is}(A)$ to
$s_{AB}(A)$ and similarly assign $\mathrm{Is}(B)$ to $s_{AB}(B)$.
Nevertheless, there are scenarios with regions included in $\mathrm{Is}(AB)$
but outside $\mathrm{Is}(A)\cup\mathrm{Is}(B)$, which should also be assigned
to $s_{AB}(A)$ or $s_{AB}(B)$, which we denote as the _ownerless island_
region in $\mathrm{Is}(AB)$,

$\displaystyle\emph{Ownerless
island:}\qquad\text{Io}(AB)=\mathrm{Is}(AB)/(\mathrm{Is}(A)\cup\mathrm{Is}(B))\,.$
(42)

See fig. 6 for an explicit example. The ownerless island should be further
divided into two parts

$\displaystyle\text{Io}(AB)=\text{Io}(A)\cup\text{Io}(B)\,,$ (43)

which are assigned to $A$ and $B$ respectively. With the additional ownerless
island region, we define the _generalized island_ region associated to $A$ and
$B$ respectively,

$\displaystyle\emph{Generalized
island:}\qquad\mathrm{Ir}(A)=\mathrm{Is}(A)\cup\text{Io}(A),\qquad\mathrm{Ir}(B)=\mathrm{Is}(B)\cup\text{Io}(B)\,,$
(44)

and assign them to $A$ and $B$ respectively when calculating the PEEs. The
explicit partition of the ownerless island region will be given later, when we
discuss the BPE. Also note that, there are simpler scenarios where the
ownerless island region is empty:

  * •

when Is$(AB)$=Is$(A)$=Is$(B)$=$\emptyset$, we have
$\mathrm{Ir}(A)=\mathrm{Ir}(B)=\emptyset$.

  * •

when Is$(AB)$=$\mathrm{Is}(A)\cup\mathrm{Is}(B)\neq\emptyset$, we have
$\text{Io}(A)=\text{Io}(B)=\text{Io}(AB)=\emptyset$,
$\mathrm{Ir}(A)=\mathrm{Is}(A)$ and $\mathrm{Ir}(B)=\mathrm{Is}(B)$.

Now we turn to the computation of the PEEs and denote
$C\equiv\overline{AA_{1}\cup\text{Is}(AA_{1})}$ for convenience. After taking
into account the contributions from the islands, we should have Basu:2023wmv ,

$s_{AA_{1}}(A)=\mathcal{I}(A\cup\text{Ir}(A),C).$ (45)

The key step to compute the PEEs eq. 45 is to write the right hand side of eq.
45 as a linear combination of the PEEs which can be written as
$\mathcal{I}(\gamma,\bar{\gamma})$, i.e. a PEE between a region $\gamma$ and
its complement. The computation for two types of
$\mathcal{I}(\gamma,\bar{\gamma})$ was described in Basu:2023wmv , which is
enough to work out the relevant PEEs eq. 45 further leading to the computation
of the BPE. When $\gamma$ is a single interval, this type of
$\mathcal{I}(\gamma,\bar{\gamma})$ is computed by,

$\textit{Basic proposal}:\leavevmode\nobreak\ \leavevmode\nobreak\
\mathcal{I}(\gamma,\bar{\gamma})=\tilde{S}_{\gamma}=\tilde{\tilde{S}}_{[-a,b]},\qquad\gamma:[-a,b]\,,$
(46)

which are just the two-point functions of twist operators inserted at the
endpoints of $\gamma$. In the holographic Weyl transformed CFT, this is just
calculated by eq. 11. Based on the basic proposal eq. 46, we can derive the
cases for $\gamma=A\cup\text{Ir}(A)$, where $\text{Ir}(A)=[-d,-c]$, $A=[a,b]$
and $a,b,c,d>0$ (see Basu:2023wmv  for the demonstration),

$\displaystyle\mathcal{I}(A\,\text{Ir}(A),\overline{A\,\text{Ir}(A)})=\tilde{S}_{[-d,-c]\cup[a,b]}=\tilde{S}_{[-c,a]}+\tilde{S}_{[-d,b]}\,.$
(47)

The reason we choose the coordinates for the endpoints of the regions is that,
we usually put the gravitational part (or the island regions) of the system at
$x<0$, while put the non-gravitational bath part at $x>0$.

Note that, due to the self-encoding property, one should not understand
$\mathcal{I}(\gamma,\bar{\gamma})$ as the entanglement entropy for $\gamma$
Basu:2023wmv . In the cases where $\gamma$ is not the union of a region and
its entanglement island, the entanglement entropy of $\gamma$ is indeed an
ill-defined concept, as the degrees of freedom inside $\gamma$ and not totally
independent from those outside $\gamma$.

Using the proposals eqs. 46 and 47 and the property of additivity, we can
carry out the computations for the PEE eq. 45. For example, when $A$ and $B$
are adjacent, then $s_{AB}(A)$ is just given by

$\begin{split}s_{AA_{1}}(A)=&\mathcal{I}(A\cup\text{Ir}(A),C)\\\
=&\frac{1}{2}\left[\mathcal{I}(A\text{Ir}(A)A_{1}\text{Ir}(A_{1}),C)+\mathcal{I}(A\text{Ir}(A),A_{1}\text{Ir}(A_{1})C)-\mathcal{I}(A_{1}\text{Ir}(A_{1}),A\text{Ir}(A)C)\right]\\\
=&\frac{1}{2}\left[\tilde{S}_{A\text{Ir}(A)A_{1}\text{Ir}(A_{1})}+\tilde{S}_{A\text{Ir}(A)}-\tilde{S}_{A_{1}\text{Ir}(A_{1})}\right].\end{split}$
(48)

This is our _generalized ALC formula_ Basu:2023wmv  in island phases, which is
just the ALC formula eq. 37 with the replacement
$S_{\gamma}\Rightarrow\tilde{S}_{\gamma\text{Ir}(\gamma)}$ applied to each
term. Similarly, when $A$ is sandwiched by $A_{1}$ and $A_{2}$, we have

$s_{A_{1}AA_{2}}(A)=\frac{1}{2}\left[\tilde{S}_{A_{1}\text{Ir}(A_{1})A\text{Ir}(A)}-\tilde{S}_{A_{1}\text{Ir}(A_{1})}+\tilde{S}_{A\text{Ir}(A)A_{2}\text{Ir}(A_{2})}-\tilde{S}_{A_{2}\text{Ir}(A_{2})}\right].$
(49)

The balanced requirements are then generalized to

$\displaystyle\mathcal{I}(A\text{Ir}(A),BB_{1}\text{Ir}(BB_{1}))=\mathcal{I}(B\text{Ir}(B),AA_{1}\text{Ir}(AA_{1}))\,,$
(50)

for adjacent $A$ and $B$, and generalized to

$\displaystyle\mathcal{I}(A\text{Ir}(A),BB_{1}B_{2}\text{Ir}(BB_{1}B_{2}))$
$\displaystyle=\mathcal{I}(B\text{Ir}(B),AA_{1}A_{2}Ir(AA_{1}A_{2}))$ (51)
$\displaystyle\mathcal{I}(A_{1}\text{Ir}(A_{1}),BB_{1}B_{2}\text{Ir}(BB_{1}B_{2}))$
$\displaystyle=\mathcal{I}(B_{1}\text{Ir}(B_{2}),AA_{1}A_{2}\text{Ir}(AA_{1}A_{2}))\,,$
(52)

for non-adjacent $A$ and $B$ . The assignment of the ownerless island region
$\text{Io}(AB)=\text{Io}(A)\cup\text{Io}(B)$ will be determined by solving the
balance requirements. In Basu:2023wmv , it was found that there are different
solutions for the balance requirements with different assignment for the
ownerless island region. These solutions exactly correspond to different
saddles of the EWCS. Then according to the minimum requirement, we should
choose the assignment that gives us the minimal BPE.

###  5.3 A case study

Here we give a typical example for the study of the EWCS and BPE in the path-
integral-optimized purification of an interval, and assume that the cutoff
brane plays the role of the KR brane and the purification is in island phase.
We consider the path integral optimization for a single interval [a,b], hence
the scalar field that characterizes the Weyl transformation is given by eq. 22
with $\kappa=0$, and the cutoff brane eq. 24 is just a semi-circle. As a case
study, we consider the setup shown in fig. 7, where we considered two adjacent
sub-intervals,

$\displaystyle A:\leavevmode\nobreak\ \leavevmode\nobreak\ [a_{1},p]\qquad
B:\leavevmode\nobreak\ \leavevmode\nobreak\ [p,b_{1}]$ (53)

whose size and position are adjusted such that, the RT surface
$\mathcal{E}_{AB}$ consists of two disconnected pieces of geodesics anchored
on the cutoff brane, and the EWCS $\Sigma_{AB}$ also anchors on the brane.

$A$ $B_{1}$ $B$ $B_{1}$ $A_{1}$ Ir(A)Ir(B)Ir(B)
$\Sigma_{AB}$$a_{1}$$p$$b_{1}$$b$$p_{1}$$a$$q_{2}$$q_{4}$$q_{1}$$q_{3}$Ir($B_{1}$)Ir($B_{1}$)Ir($A_{1}$)Ir($A_{1}$)Ir($A$)Ir($B$)
Figure 7: Two adjacent intervals $AB$ and their complements $A_{1}$ and
$B_{1}$, featuring entanglement island regions
$\text{Ir}(A)=[q_{1},q_{2}],\quad\text{Ir}(B)=(-\infty,q_{1}]\cup[q_{4},\infty),\quad\text{Ir}(A_{1})=[q_{2},q_{3}]\,\,\,\text{and}\,\,\,\,\text{Ir}(B_{1})=[q_{3},a]\cup[b,q_{4}]$
on the entire KR brane, are configured schematically. Here $\text{Ir}(B_{1})$
indicate disconnected entanglement island.

The lengths of the geodesics anchored on the cutoff brane are classified in
appendix A. The cross-section of the entanglement wedge $\mathcal{W}_{AB}$ has
three saddle points, which anchors on either of the two disconnected pieces of
the RT surface, or the cutoff brane, and the EWCS is the saddle point with
minimal area. In the case shown in Fig.7, the EWCS anchors on the cutoff
brane.

Firstly, we check the self-consistency of the simulation by calculating the
entanglement entropy on both sides of holography. In the gravity side,
$S_{AB}$ is calculated by the RT formula, which is the area of the two pieces
of geodesic anchored on the brane described as

$\displaystyle
S_{AB}=\frac{c}{6}\log\left[\frac{2(a_{1}-a)(b-a_{2})}{(b-a)\epsilon}\right]+\frac{c}{6}\log\left[\frac{2(b_{1}-a)(b-b_{2})}{(b-a)\epsilon}\right]$
(54)

On the field theory side, we calculate $S_{AB}$ by island formula and denoting
the island region as $\mathrm{Is}(AB)=(-\infty,q_{2}]\cup[q_{4},\infty)$.
Then, following eq. 47 we have

$\displaystyle S_{AB}=$
$\displaystyle\min_{\mathrm{Is}(AB)}\left(\tilde{S}_{[q_{2},a_{1}]}+\tilde{S}_{[b_{1},q_{4}]}\right)$
(55) $\displaystyle=$
$\displaystyle\frac{c}{6}\log\left[\frac{2(a_{1}-a)(b-a_{2})}{(b-a)\epsilon}\right]+\frac{c}{6}\log\left[\frac{2(b_{1}-a)(b-b_{2})}{(b-a)\epsilon}\right]$
(56)

which coincide with the RT formula. In the second equation we have plugged in
the saddle points for $q_{1}$ and $q_{4}$ expressed as

$\displaystyle q_{2}=\frac{2ab-(a+b)a_{1}}{a+b-2a_{1}},\qquad
q_{4}=\frac{2ab-(a+b)b_{1}}{a+b-2b_{1}}\,.$ (57)

As was shown in Fig.7, these saddles are just the intersecting points between
the boundary and the extended RT surface.

Then we check the correspondence between the EWCS and the BPE. On the gravity
side, the area of the EWCS as shown in Fig.7 for this case can be easily
calculated as follows

$\displaystyle\frac{\text{Area}[\Sigma_{AB}]}{4G}=\frac{c}{6}\log\left[\frac{2(p-a)(b-p)}{(b-a)\epsilon}\right]\,.$
(58)

Prior to calculating the BPE for this phase, we should give an assignment for
the island regions, $\text{Is}(AB)=\text{Ir}(A)\cup\text{Ir}(B)$, where

$\displaystyle\text{Ir}(A)=[q_{1},q_{2}]\,,\qquad\text{Ir}(B)=(-\infty,q_{1}]\cup[q_{4},\infty)$
(59)

where $x=q_{1}$ is the partition point of $\mathrm{Is}(AB)$, which will be
determined by the balance requirements eq. 50. Now we calculate the PEE on
both sides of eq. 50, which are given by

$\displaystyle\mathcal{I}(A\text{Ir}(A),B\text{Ir}(B)B_{1}\text{Ir}(B_{1}))$
$\displaystyle=\frac{1}{2}\Big{(}\tilde{S}_{A\text{Ir}(A)A_{1}\text{Ir}(A_{1})}+\tilde{S}_{A\text{Ir}(A)}-\tilde{S}_{A_{1}\text{Ir}(A_{1})}\Big{)}$
$\displaystyle=\frac{1}{2}(\tilde{S}_{[q_{1},q_{3}]\cup[p_{1},p]}+\tilde{S}_{[q_{1},q_{2}]\cup[a_{1},p]}-\tilde{S}_{[q_{2},q_{3}]\cup[p_{1},a_{1}]})$
$\displaystyle=\frac{1}{2}(\tilde{S}_{[q_{1},p]}+\tilde{S}_{[q_{3},p_{1}]}+\tilde{S}_{[q_{1},p]}+\tilde{S}_{[q_{2},a_{1}]}-\tilde{S}_{[q_{2},a_{1}]}-\tilde{S}_{[q_{3},p_{1}]})$
$\displaystyle=\frac{c}{6}\log\left[\frac{(p-q_{1})^{2}}{\epsilon^{2}}\right]+\frac{c}{6}\phi[q_{1}]\,,$
(60)
$\displaystyle\mathcal{I}(B\text{Ir}(B),A\text{Ir}(A)A_{1}\text{Ir}(A_{1}))$
$\displaystyle=\frac{1}{2}\Big{(}\tilde{S}_{B\text{Ir}(B)B_{1}\text{Ir}(B_{1})}+\tilde{S}_{B\text{Ir}(B)}-\tilde{S}_{B_{1}\text{Ir}(B_{1})}\Big{)}$
$\displaystyle=\frac{1}{2}(\tilde{S}_{(-\infty,q_{1}]\cup[q_{3},p_{1}]\cup[p,\infty)}+\tilde{S}_{(-\infty,q_{1}]\cup[p,b_{1}]\cup[q_{4},\infty)}-\tilde{S}_{[q_{3},p_{1}]\cup[b_{1},q_{4}]})$
$\displaystyle=\frac{1}{2}(\tilde{S}_{[q_{1},p]}+\tilde{S}_{[q_{3},p_{1}]}+\tilde{S}_{[q_{1},p]}+\tilde{S}_{[b_{1},q_{4}]}-\tilde{S}_{[q_{3},p_{1}]}-\tilde{S}_{[b_{1},q_{4}]})$
$\displaystyle=\frac{c}{6}\log\left[\frac{(p-q_{1})^{2}}{\epsilon^{2}}\right]+\frac{c}{6}\phi[q_{1}]\,,$
(61)

where the scalar field is given by (22). Interestingly for this configuration,
the balanced condition is satisfied trivially. However, we need to further
impose the implicit minimal requirement which settles $q_{1}$ to be

$\displaystyle q_{1}$ $\displaystyle=\frac{2ab-(a+b)p}{a+b-2p}\,.$ (62)

Again, this point is also the intersection between the boundary and the
extended EWCS $\Sigma_{AB}$. Finally, we obtain the BPE for this phase as
follows

$\displaystyle\text{BPE}=\mathcal{I}(A\text{Ir}(A),B\text{Ir}(B)B_{1}\text{Ir}(B_{1}))|_{balanced}=\frac{c}{6}\log\left[\frac{2(p-a)(b-p)}{(b-a)\epsilon}\right]\,,$
(63)

where we have plugged in eqs. 22 and 62 in the above equation. As expected,
the BPE eq. 63 precisely match with the area of the EWCS eq. 58.

In the Appendix, we explicitly checked the correspondence between the BPE and
the EWCS for all possible choices for the subregion $AB$ and all possible
phases of the EWCS with $\kappa=0$.

$A$ $B_{1}$ $B$ $B_{1}$ $A_{1}$ Ir(A)Ir(B)Ir(B)
$E_{W}$$a_{1}$$p$$b_{1}$$b$$p_{1}$$a$$q_{2}$$q_{4}$$q_{1}$$q_{3}$Ir($B_{1}$)Ir($B_{1}$)Ir($A_{1}$)Ir($A_{1}$)Ir($A$)Ir($B$)
Figure 8: Two adjacent intervals $AB$ and their complements $A_{1}$ and
$B_{1}$, featuring entanglement island regions
$\text{Ir}(A)=[q_{1},q_{2}],\quad\text{Ir}(B)=(-\infty,q_{1}]\cup[q_{4},\infty),\quad\text{Ir}(A_{1})=[q_{2},q_{3}]\,\,\,\text{and}\,\,\,\,\text{Ir}(B_{1})=[q_{3},a]\cup[b,q_{4}]$
on the entire KR brane, are configured schematically. Here $\text{Ir}(B_{1})$
indicate disconnected entanglement island.

Then we consider the scenarios where $\kappa\neq 0$ in (22), such that the
cutoff brane is described by (24), which is no longer a semi-circle (see
Fig.8). The size, position and the partition of $AB$ are adjusted such that,
the RT surface $\mathcal{E}_{AB}$ and the EWCS $\Sigma_{AB}$ all anchor on the
cutoff brane. The lengths of the geodesics anchored on the cutoff brane are
classified in appendix A. On the gravity side, the area of the EWCS can be
easily calculated as follows

$\displaystyle\frac{\text{Area}[\Sigma_{AB}]}{4G}$
$\displaystyle=\frac{c}{6}\log\left[\frac{2(p-a)(b-p)(\cosh(\kappa)+\sinh(\kappa))}{(b-a)\epsilon}\right]$
(64)
$\displaystyle=\frac{c}{6}\log\left[\frac{2(p-a)(b-p)}{(b-a)\epsilon}\right]+\frac{c}{6}\kappa\,.$
(65)

Then we turn to the calculation for the BPE on the field theory side. Firstly
we give an assignment for the island regions,
$\text{Is}(AB)=\text{Ir}(A)\cup\text{Ir}(B)$, where

$\displaystyle\text{Ir}(A)=[q_{1},q_{2}]\,,\qquad\text{Ir}(B)=(-\infty,q_{1}]\cup[q_{4},\infty)$
(66)

where $x=q_{1}$ is the partition point of $\mathrm{Is}(AB)$, which will be
determined by the balance requirements eq. 50. Now we calculate the PEE on
both sides of eq. 50, which have the same expression as eqs. 60 and 61, with
the only difference that, the $\kappa$ in the scalar field eq. 22 is non-zero.
Again the balanced condition is satisfied trivially and the implicit minimal
requirement which settles $q_{1}$ to be

$\displaystyle q_{1}$ $\displaystyle=\frac{2ab-(a+b)p}{a+b-2p}\,.$ (67)

Finally, we may obtain the BPE for this phase as follows

$\displaystyle\text{BPE}=\mathcal{I}(A\text{Ir}(A),B\text{Ir}(B)B_{1}\text{Ir}(B_{1}))|_{balanced}=\frac{c}{6}\log\left[\frac{2(p-a)(b-p)}{(b-a)\epsilon}\right]+\frac{c}{6}k\,,$
(68)

where we have plugged in eqs. 22 and 67 in the above equation. As expected,
the BPE expressed in eq. 68 precisely match with the area of the EWCS given by
eq. 64.

##  6 Summary and Discussion

In this paper, we considered the special Weyl transformations for a
holographic CFT2, which optimizes the path integral computation for the
reduced density matrix of an interval. Under such Weyl transformations, the
cutoff branes are circles in the AdS bulk passing through the endpoints of the
interval at the boundary. When we take the limit that one of the endpoint goes
to infinity hence the interval becomes a half line, the cutoff branes coincide
with the KR branes in the AdS/BCFT configurations where the BCFT has one
boundary. Without taking the above limit, we only find some AdS/BCFT
configurations, where the KR brane are also circles passing through the
endpoints on the boundary on a time slice Fujita:2011fp ; Miao:2017gyt ;
Geng:2022tfc . Finding the AdS/BCFT configurations that exactly match with the
holographic Weyl transformed CFT we have considered could be an interesting
exploration in the future. Perhaps it is an even more interesting idea to take
the scalar field that characterizes the Weyl transformation dynamical, with
its action just being the Liouville action, whose equivalence to the action of
3d gravity has been extensively discussed Verlinde:1989ua ; Carlip:1991zm ;
Coussaert:1995zp ; Martinec:1998wm ; Rooman:2000zi ; Carlip:2005tz ;
Skenderis:1999nb ; Krasnov:2000zq ; Carlip:2005zn . Then the region under
nontrivial Weyl transformation (or the $\phi\neq 0$ region) is naturally
coupled to a gravity with background geometry that solves the Liouville
equations and the gravitational excitation represented by the perturbation of
the scalar field. This may give a justification for our proposal that the
island formula applies to the Weyl transformed CFT.

Nevertheless, we assume that the cutoff branes induced by the Weyl
transformations that optimize the path integral play the same role as the KR
branes, hence the RT surfaces and the EWCSs can anchor on the brane. And the
boundary state, which is the path-integral optimized purification for the
interval, is in the island phase. The new perspective for the optimized
purification solves the puzzle of negative mutual information in this state.
We calculated the entanglement entropies for subregions of the interval using
the island formula, and find them coincide with the area of the RT surfaces
that are allowed to anchor on the brane. Furthermore, we calculated the BPEs
between two arbitrary non-overlapping subregions of the interval, and find
them coincide with the area of the EWCS.

Our results give a potential new link between the AdS/BCFT correspondence and
the path integral optimization. This is quite important for the simulation of
the AdS/BCFT configurations via the holographic Weyl transformed CFT, since
the Weyl transformation is now determined on the field theory side, rather
than adjusted by hand. Our calculations also provide new evidence for the
correspondence between the BPE and the EWCS. It will be interesting to derive
the BPE/EWCS correspondence based on the scheme to geometrize the PEE in terms
of the bulk geodesics Lin:2023rbd ; Lin:2024dho  and its generalization in
island configurations.

## Acknowledgement

The authors thank Hao Geng and Rongxin Miao for helpful discussions.

##  Appendix A Geodesic Length from Geometrical Analysis

In this section, we give a generic way to derive the lengths for geodesic
chords anchored on the boundary and terminated in the bulk, which we
frequently encounter in this paper. Let us consider two adjacent intervals
$A=[x_{1},x^{\prime}_{2}]$ and $B=[x^{\prime}_{2},x_{2}]$ located in the
vacuum $CFT_{2}$ which constitutes a pure state with the intervals
$A^{\prime}=[x^{\prime}_{1},x_{1}]$ and $B^{\prime}=[x_{2},x^{\prime}_{1}]$,
see Fig.9. When the RT surface of $A^{\prime}A$ and $AB$ are normal to each
other, we can explicitly derive the lengths for the geodesic chords $l_{1}$,
$l_{2}$, $l^{\prime}_{1}$ and $l^{\prime}_{2}$ in Fig.9.

AB${\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}A^{\prime}}$${\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}B^{\prime}}$${\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}B^{\prime}}$$l_{2}$$l_{1}$$l_{1}^{\prime}$$l_{2}^{\prime}$$x_{1}$$x_{2}^{\prime}$$x_{2}$$x_{1}^{\prime}$$q^{\prime}$$C$$C^{\prime}$
Figure 9: Schematics shows the purification of the adjacent intervals
$A=[x_{1},x^{\prime}_{2}]$ and $B=[x^{\prime}_{2},x_{2}]$ located in the
vacuum $CFT_{2}$. The geodesics lengths homologous to the intervals $A\cup
A^{\prime}$ and $A\cup B$ are depicted as $l^{\prime}_{1}+l^{\prime}_{2}$
(black) and $l_{1}+l_{2}$ (blue) respectively and the intersection point of
the corresponding geodesics is labelled as $q^{\prime}$.

To obtain the location of the point $q^{\prime}$ as indicated in Fig.9, we
consider a Euclidean triangle in the $x$-$z$ plane and utilize trivial
Euclidean identities. Consequently, the coordinate of the point $q^{\prime}$
are described as $x=\rho_{2}\cos(\theta^{\prime})$ and
$z=\rho_{2}\sin(\theta^{\prime})$ or $\rho_{1}\sin(\theta)$ using the fig. 10.
We can also re-express the coordinates $\rho_{1}$, $\rho_{2}$ and $y$ in terms
of the interval endpoints and applying cosine law in the triangle results in
following relations,

$\displaystyle\rho_{1}$
$\displaystyle=\frac{x_{2}-x_{1}}{2},\quad\rho_{2}=\frac{x_{2}^{\prime}-x_{1}^{\prime}}{2},\quad
y=\frac{x_{1}+x_{2}-x_{1}^{\prime}-x_{2}^{\prime}}{2}\,,$
$\displaystyle\cos\theta=\frac{\rho_{1}^{2}+y^{2}-\rho_{2}^{2}}{2\rho_{1}y},\quad\cos\theta^{\prime}=\frac{\rho_{2}^{2}+y^{2}-\rho_{1}^{2}}{2\rho_{2}y}\,.$
(69)

$y$ $\rho_{2}$ $\rho_{1}$ $C$$C^{\prime}$$q^{\prime}$$\theta$$\theta^{\prime}$
Figure 10: Euclidean triangle is considered in the $x$-$z$ plane where
$\rho_{1}$ and $\rho_{2}$ are the distances from the point $q^{\prime}$ and
the center of the RT surfaces associated with the intervals $A\cup B$ and
$A^{\prime}\cup A$ respectively. The point $y$ is the distance between the
corresponding centers.

Note that, the two semi-circles depicted in Fig.9 are normal to each other and
therefore we can determine a constraint equation between the endpoint of the
intervals $x_{1},x_{2},x_{1}^{\prime}$ and $x_{2}^{\prime}$ by utilizing right
angle triangle identity $\rho_{1}^{2}+\rho_{2}^{2}=y^{2}$ as follows

$\displaystyle-2x_{1}^{\prime}x_{2}^{\prime}+x_{2}(x_{1}^{\prime}+x_{2}^{\prime})+x_{1}(x_{1}^{\prime}+x_{2}^{\prime}-2x_{2})=0\,.$
(70)

Furthermore, we establish a framework to compute the geodesic length
$l^{\prime}_{1}$, then we further generalize this analysis to other geodesic
segments. In this context, the geodesics equation for the segment
$l^{\prime}_{1}$ is described by $(x-C)^{2}+(z-\epsilon)^{2}=(\rho_{2})^{2}$,
where $C$ is the center for the semi-circle $l^{\prime}_{1}+l^{\prime}_{2}$.
Now we utilize the geodesic length formula in pure $AdS_{3}$ geometry at
constant time slice and integrate over the bulk coordinate $z$ with limits
involving endpoints of the geodesic segment $l^{\prime}_{1}$ as

$\displaystyle\mathcal{L}_{l^{\prime}_{1}}$
$\displaystyle=\int_{\epsilon}^{\rho_{2}\sin(\theta^{\prime})}\differential{z}\frac{\sqrt{1+\frac{\differential{x}}{\differential{z}}^{2}}}{z}\,$
$\displaystyle=\frac{1}{2}\log\left[{\frac{\rho_{2}-\sqrt{\rho_{2}^{2}-z^{2}}}{\rho_{2}+\sqrt{\rho_{2}^{2}-z^{2}}}}\right]\Bigg{|}_{\epsilon}^{\rho_{2}\sin(\theta^{\prime})}\,$
$\displaystyle=\frac{1}{2}\log\left[{\frac{1-\cos\theta^{\prime}}{1+\cos\theta^{\prime}}\frac{4\rho_{2}^{2}}{\epsilon^{2}}}\right]\,.$
(71)

Note that similar analysis can also be followed for the length of the geodesic
segments $l_{1}$, $l_{2}$, and $l^{\prime}_{2}$. For each of the geodesic
length segments shown in Fig.9, the holographic proposal of the entanglement
entropy discussed in Ryu:2006bv  may be used to determine the holographic
entanglement entropy from the (A) in a following way

$\displaystyle\mathcal{L}_{l_{1}}$
$\displaystyle=\frac{1}{2}\log\left[{\frac{(x_{1}-x_{1}^{\prime})(x_{1}-x_{2}^{\prime})(x_{2}-x_{1})^{2}}{(x_{1}^{\prime}-x_{2})(x_{2}-x_{2}^{\prime})\epsilon^{2}}}\right]\,,$
(72) $\displaystyle\mathcal{L}_{l_{2}}$
$\displaystyle=\frac{1}{2}\log\left[{\frac{(x_{2}-x_{1}^{\prime})(x_{2}-x_{2}^{\prime})(x_{2}-x_{1})^{2}}{(x_{1}^{\prime}-x_{1})(x_{1}-x_{2}^{\prime})\epsilon^{2}}}\right]\,,$
(73) $\displaystyle\mathcal{L}_{l_{1}^{\prime}}$
$\displaystyle=\frac{1}{2}\log\left[{\frac{(x_{2}-x_{2}^{\prime})(x_{1}-x_{2}^{\prime})(x_{2}^{\prime}-x_{1}^{\prime})^{2}}{(x_{1}^{\prime}-x_{1})(x_{2}-x_{1}^{\prime})\epsilon^{2}}}\right]\,,$
(74) $\displaystyle\mathcal{L}_{l_{2}^{\prime}}$
$\displaystyle=\frac{1}{2}\log\left[{\frac{(x_{1}-x_{1}^{\prime})(x_{1}^{\prime}-x_{2})(x_{2}^{\prime}-x_{1}^{\prime})^{2}}{(x_{2}-x_{2}^{\prime})(x_{1}-x_{2}^{\prime})\epsilon^{2}}}\right]\,.$
(75)

In the above equations, we have written the lengths for the geodesic chords in
terms of the coordinates $\\{x^{\prime}_{1},x^{\prime}_{2},x_{1},x_{2}\\}$ of
the boundary endpoints, which satisfy (70). In this paper, we utilize the
expression of the corresponding geodesic segments to compute the lengths for
the EWCS and other geodesic chords in various scenarios.

##  Appendix B BPE and EWCS for adjacent intervals

### Adjacent $AB$ with no island

We investigate the BPE and its dual EWCS for two adjacent intervals
$A=[a_{1},p]$ and $B=[p,b_{1}]$ where $A\cup B$ has connected entanglement
wedge. We first consider the simplest scenario where $AB$ admits no island.
Then entanglement island of $A_{1}\cup B_{1}$, the complement of $AB$ in the
interval, occupies the whole island region $(-\infty,a]\cup[b,\infty)$. And
the generalized island region for $A_{1}$ and $B_{1}$ has the following three
possible configurations,

  1. 1.

A1a:
$\text{Ir}(AB)=\emptyset,\quad\text{Ir}(A_{1})=[q_{1},q_{2}],\quad\text{Ir}(B_{1})=(-\infty,q_{1}]\cup[q_{2},a]\cup[b,\infty)$,

  2. 2.

A1b:
$\text{Ir}(AB)=\emptyset,\quad\text{Ir}(A_{1})=\emptyset,\quad\text{Ir}(B_{1})=(-\infty,a]\cup[b,\infty)$.

  3. 3.

A1c:
$\text{Ir}(AB)=\emptyset,\quad\text{Ir}(A_{1})=(-\infty,a]\cup[b,\infty),\quad\text{Ir}(B_{1})=\emptyset$.

The second and third configurations happen when the partition point for
$A_{1}B_{1}$ is close to the two endpoints of $AB$.

#### Phase-A1a

We first describe the computation of the EWCS in the phase-A1a depicted in
Fig.11. To proceed, we need to determine the location of point $q_{1}$. This
can be obtained by constructing the Euclidean triangle by following similar
analysis described in appendix A for the RT surfaces homologous to the
intervals $[q_{1},p]$ and $[a_{1},b_{1}]$ with constraint equation provided in
(70). Thus the value of $q_{1}$ is given by

$\displaystyle q_{1}=\frac{2a_{1}b_{1}-(a_{1}+b_{1})p}{a_{1}+b_{1}-2p}\,.$
(76)

The minimal length of the EWCS can be computed by using (76) and the analysis
of the geodesic segment given in appendix A. Consequently, the expression of
EWCS in this phase is given by,

$\displaystyle E_{W}$ $\displaystyle=\frac{\text{Area}[\text{EWCS}]}{4G}$
$\displaystyle=\frac{c}{6}\log\left[\frac{2(p-a_{1})(b_{1}-p)}{(b_{1}-a_{1})\epsilon}\right]\,.$

${\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}A}$$B_{1}$
${\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}B}$$B_{1}$
$A_{1}$
$E_{W}$$a_{1}$$p$$y$$b_{1}$$b$$p_{1}$$a$Ir($B_{1}$)Ir($B_{1}$)Ir($A_{1}$)Ir($A_{1}$)Ir($B_{1}$)$q_{2}$$q_{1}$
Figure 11: Schematics shows the configuration of two adjacent intervals $AB$
with its complement involving entanglement island region
$\text{Ir}(A_{1})=[q_{1},q_{2}]$ and
$\text{Ir}(B_{1})=(-\infty,q_{1}]\cup[q_{2},a]\cup[b,\infty)$.

We now describe the BPE analysis in this phase. We need to determine the
corresponding partition point by solving the balance requirement,

$\displaystyle\mathcal{I}(A,BB_{1}\text{Ir}(B_{1}))=$
$\displaystyle\mathcal{I}(B,AA_{1}\text{Ir}(A_{1}))\,.$ (77)

The PEEs on both sides of the above equation can be calculated using the
generalized ALC proposal as follows,

$\displaystyle\mathcal{I}(A,BB_{1}\text{Ir}(B_{1}))$
$\displaystyle=\frac{1}{2}\Big{(}\tilde{S}_{AA_{1}\text{Ir}(A_{1})}+\tilde{S}_{A}-\tilde{S}_{A_{1}\text{Ir}(A_{1})}\Big{)}$
$\displaystyle=\frac{1}{2}(\tilde{S}_{[p_{1},p]\cup[q_{1},q_{2}]}+\tilde{S}_{[a_{1},p]}-\tilde{S}_{[p_{1},a_{1}]\cup[q_{1},q_{2}]})$
$\displaystyle=\frac{1}{2}(\tilde{S}_{[q_{2},p_{1}]}+\tilde{S}_{[q_{1},p]}+\tilde{S}_{[a_{1},p]}-\tilde{S}_{[q_{2},p_{1}]}-\tilde{S}_{[q_{1},a_{1}]})$
$\displaystyle=\frac{c}{6}\log\left[\frac{(p-q_{1})(p-a_{1})}{(a_{1}-q)\epsilon}\right]\,,$
(78) $\displaystyle\mathcal{I}(B,AA_{1}\text{Ir}(A_{1}))$
$\displaystyle=\frac{1}{2}\Big{(}\tilde{S}_{BB_{1}\text{Ir}(B_{1})}+\tilde{S}_{B}-\tilde{S}_{B_{1}\text{Ir}(B_{1})}\Big{)}$
$\displaystyle=\frac{1}{2}(\tilde{S}_{(-\infty,q_{1}]\cup[q_{2},p_{1}]\cup[p,\infty)}+\tilde{S}_{[p,b_{1}]}-\tilde{S}_{(-\infty,q_{1}]\cup[q_{2},p_{1}]\cup[b_{1},\infty)})$
$\displaystyle=\frac{1}{2}(\tilde{S}_{[q_{2},p_{1}]}+\tilde{S}_{[q_{1},p]}+\tilde{S}_{[p,b_{1}]}-\tilde{S}_{[q_{2},p_{1}]}-\tilde{S}_{[q_{1},b_{1}]})$
$\displaystyle=\frac{c}{6}\log\left[\frac{(p-q_{1})(b_{1}-p)}{(b_{1}-q_{1})\epsilon}\right]\,.$
(79)

We find that, the $q_{1}$ satisfying (76) solves the balance requirement (77).
Nevertheless, there is no constraint for $p_{1}$. Imposing (76) we get the BPE
for this phase,

$\displaystyle\text{BPE}=\mathcal{I}(A,BB_{1}\text{Ir}(B_{1}))|_{balanced}=\frac{c}{6}\log\left[\frac{2(p-a_{1})(b_{1}-p)}{(b_{1}-a_{1})\epsilon}\right]\,,$
(80)

which exactly matches with the EWCS given by (B).

#### Phase-A1b and Phase-A1c

${\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}A}$${\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}B_{1}}$${\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}B}$${\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}B_{1}}$${\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}A_{1}}$$E_{W}$Ir($B_{1}$)$a_{1}$$p$$q$$b_{1}$$b$$p_{1}$$a$Ir($B_{1}$)Ir($B_{1}$)
Figure 12: The phase-A1b describes entanglement island region
$\text{Ir}(B_{1})=(-\infty,a]\cup[b,\infty)$.

This phase also involves connected entanglement wedge of $AB$ as depicted in
Fig.12. The EWCS in this phase is again given by (B). We now proceed to the
computation of BPE in the phase-A1b, here generalized island region
$\text{Ir}(B_{1})=(-\infty,a]\cup[b,\infty)$, which indicates
$\text{Ir}(AA_{1})=\emptyset$. The generalized balance requirement in this
phase can be written as

$\displaystyle\mathcal{I}(A,BB_{1}\text{Ir}(B_{1}))=$
$\displaystyle\mathcal{I}(B,AA_{1})\,.$ (81)

Again, we can compute the PEEs in the above equation using the generalized ALC
proposal as follows

$\displaystyle\mathcal{I}(A,BB_{1}\text{Ir}(B_{1}))$
$\displaystyle=\frac{1}{2}\Big{(}\tilde{S}_{AA_{1}}+\tilde{S}_{A}-\tilde{S}_{A_{1}}\Big{)}$
$\displaystyle=\frac{1}{2}(\tilde{S}_{[p_{1},p]}+\tilde{S}_{[a_{1},p]}-\tilde{S}_{[p_{1},a_{1}]})$
$\displaystyle=\frac{c}{6}\log\left[\frac{(p-p_{1})(p-a_{1})}{(a_{1}-p_{1})\epsilon}\right]\,,$
(82) $\displaystyle\mathcal{I}(B,AA_{1})$
$\displaystyle=\frac{1}{2}\Big{(}\tilde{S}_{BB_{1}\text{Ir}(B_{1})}+\tilde{S}_{B}-\tilde{S}_{B_{1}\text{Ir}(B_{1})}\Big{)}$
$\displaystyle=\frac{1}{2}(\tilde{S}_{(-\infty,p_{1}]\cup[p,\infty)}+\tilde{S}_{[p,b_{1}]}-\tilde{S}_{(-\infty,p_{1}]\cup[b_{1},\infty)})$
$\displaystyle=\frac{1}{2}(\tilde{S}_{[p_{1},p]}+\tilde{S}_{[p,b_{1}]}-\tilde{S}_{[p_{1},b_{1}]})$
$\displaystyle=\frac{c}{6}\log\left[\frac{(p-p_{1})(b_{1}-p)}{(b_{1}-p_{1})\epsilon}\right]\,.$
(83)

The partition point $x=p_{1}$ can be obtained by solving (81), which has the
same formula as $q_{1}$ in (76). This again leads to the following BPE

$\displaystyle\text{BPE}=\mathcal{I}(A,BB_{1}\text{Ir}(B_{1}))|_{balanced}=\frac{c}{6}\log\left[\frac{2(a_{1}-p)(b_{1}-p)}{\epsilon(a_{1}-b_{1})}\right]\,,$
(84)

which coincides precisely with the EWCS appendix B. One can check that the
$A_{1}A$ with $p_{1}=\frac{2a_{1}b_{1}-(a_{1}+b_{1})p}{a_{1}+b_{1}-2p}$ is at
the critical point to admit an island. This means the discussion in phase-A1b
is just a special case of of phase-A1a. The discussion for the phase-A1c is
similar to this phase.

### Adjacent $AB$ with island

Here we consider the configuration of two adjacent intervals $A=[a_{1},p]$ and
$B=[p,b_{1}]$ where $AB$ admits an island region. Again there are three
possible assignments for the island regions:

  1. 1.

A2a:
$\text{Ir}(A)=\text{Ir}(A_{1})=\emptyset,\quad\text{Ir}(B)=(-\infty,q_{1}]\cup[q_{2},\infty),\quad\text{Ir}(B_{1})=[q_{1},a]\cup[b,q_{2}]$,

  2. 2.

A2b:
$\text{Ir}(A)=[q_{1},q_{2}],\quad\text{Ir}(B)=(-\infty,q_{1}]\cup[q_{4},\infty),\quad\text{Ir}(A_{1})=[q_{2},q_{3}],\quad\text{Ir}(B_{1})=[q_{3},a]\cup[b,q_{4}]$,

  3. 3.

A2c:
$\text{Ir}(B)=\text{Ir}(B_{1})=\emptyset,\quad\text{Ir}(A)=(-\infty,q_{1}]\cup[q_{2},\infty),\quad\text{Ir}(A_{1})=[q_{1},a]\cup[b,q_{2}]$

In the phase-A2a, the interval $AA_{1}$ do not admit an island as shown in
Fig.13. The phase-A2c is symmetric to the phase-A2a. In the phase-A2b, both
$AA_{1}$ and $BB_{1}$ admit an island, and the analysis is given in the main
text.

#### Phase-A2a and phase-A2c

We now discuss the computation of the EWCS in the scenario depicted in Fig.13.
Here we have divided the complement of the adjacent intervals $A\cup B$ into
$A_{1}\cup B_{1}$ with partition point $p_{1}$. Utilizing the constraint (70)
for the RT surfaces homologous to the intervals $[q_{1},a_{1}]$ and
$[p_{1},p]$, we can obtain location of the point $p_{1}$ as

$\displaystyle p_{1}$
$\displaystyle=\frac{2q_{1}a_{1}-p(q_{1}+a_{1})}{q_{1}+a_{1}-2p}\,.$ (85)

In this scenario, the EWCS may be directly obtained from the length of the
geodesic segment described in appendix A as follows

$\displaystyle E_{W}$ $\displaystyle=\frac{\text{Area}[\text{EWCS}]}{4G}$ (86)
$\displaystyle=\frac{c}{6}\log\left[{\frac{2(p-a_{1})(p-q_{1})}{(a_{1}-q_{1})\epsilon}}\right],$

Note that the value of $q_{1}$ can also be obtained in terms of the points
$a$, $a_{1}$ and $b$ by using the constraint relation in (70) as

$\displaystyle q_{1}$ $\displaystyle=\frac{2ab-(a+b)a_{1}}{a+b-2a_{1}}\,.$
(87)

${\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}A}$${\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}B_{1}}$${\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}B}$${\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}B_{1}}$${\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}A_{1}}$Ir(B)Ir(B)
$E_{W}$Ir(B)$a_{1}$$p$$b_{1}$$b$$p_{1}$$a$$q_{1}$$q_{2}$Ir($B_{1}$)Ir($B_{1}$)
Figure 13: Schematics shows the configuration of two adjacent intervals
considered in Weyl $CFT_{2}$ where the intervals $B$ and $B_{1}$ admit
entanglement island region on the KR brane.

Now we carry out the BPE analysis for this phase, where the island assignment
$\text{Ir}(B_{1})=[q_{1},a]\cup[b,q_{2}]$ and
$\text{Ir}(B)=(-\infty,q_{1}]\cup[q_{2},\infty)$ is shown in Fig.13. In this
phase, the balance requirement is given by

$\displaystyle\mathcal{I}(A,B\text{Ir}(B)B_{1}\text{Ir}(B_{1}))=$
$\displaystyle\mathcal{I}(B\text{Ir}(B),AA_{1})\,.$ (88)

We now utilize the generalized ALC proposal to compute PEEs mentioned in the
above balance requirement,

$\displaystyle\mathcal{I}(A,B\text{Ir}(B)B_{1}\text{Ir}(B_{1}))$
$\displaystyle=\frac{1}{2}\Big{(}\tilde{S}_{AA_{1}}+\tilde{S}_{A}-\tilde{S}_{A_{1}}\Big{)}$
$\displaystyle=\frac{1}{2}(\tilde{S}_{[p_{1},p]}+\tilde{S}_{[a_{1},p]}-\tilde{S}_{[p_{1},a_{1}]})$
$\displaystyle=\frac{c}{6}\log\left[\frac{(p-p_{1})(p-a_{1})}{(a_{1}-p_{1})\epsilon}\right]\,,$
(89) $\displaystyle\mathcal{I}(B\text{Ir}(B),AA_{1})$
$\displaystyle=\frac{1}{2}\Big{(}\tilde{S}_{B\text{Ir}(B)B_{1}\text{Ir}(B_{1})}+\tilde{S}_{B\text{Ir}(B)}-\tilde{S}_{B_{1}\text{Ir}(B_{1})}\Big{)}$
$\displaystyle=\frac{1}{2}(\tilde{S}_{(-\infty,p_{1}]\cup[p,\infty)}+\tilde{S}_{(-\infty,q_{1}]\cup[p,b_{1}]\cup[q_{2},\infty)}-\tilde{S}_{[q_{1},p_{1}]\cup[b_{1},q_{2}]})$
$\displaystyle=\frac{1}{2}(\tilde{S}_{[p_{1},p]}+\tilde{S}_{[q_{1},p]}+\tilde{S}_{[b_{1},q_{2}]}-\tilde{S}_{[q_{1},p_{1}]}-\tilde{S}_{[b_{1},q_{2}]})$
$\displaystyle=\frac{c}{6}\log\left[\frac{(p-p_{1})(p-q_{1})}{(p_{1}-q_{1})\epsilon}\right]\,.$
(90)

Solving the balance requirement (88) we get the partition point $p_{1}$, which
is same as (85). Finally, we can obtain the BPE using eqs. 89 and 85,

$\displaystyle\text{BPE}=\mathcal{I}(A,B\text{Ir}(B)B_{1}\text{Ir}(B_{1}))|_{balanced}=\frac{c}{6}\log\left[{\frac{2(p-a_{1})(p-q_{1})}{(a_{1}-q_{1})\epsilon}}\right]\,,$
(91)

which coincide with the EWCS (86).

The discussion for the phase-A2c is similar.

##  Appendix C BPE and EWCS for non-adjacent intervals

In this appendix, we will examine the computation of the BPE and its dual EWCS
for two non-adjacent (disjoint) intervals $AB$.

### Disjoint $AB$ with no island

We first investigate the simplest case of two disjoint intervals
$A=[a_{1},a_{2}]$ and $B=[b_{1},b_{2}]$ with no island region. Here the
complement of $AB$ is described by intervals $A_{1}\cup B_{1}$ and $A_{2}\cup
B_{2}$ with partition points located in Weyl $CFT_{2}$. In this case, we
observe that $A_{1}\cup B_{1}$ may incorporate an entanglement island region,
which can be classified into two distinct phases given by

  1. 1.

D1a:
$\text{Ir}(AB)=\emptyset,\quad\text{Ir}(B_{1})=\left(-\infty,a\right]\cup\left[b,\infty\right)$,

  2. 2.

D1b:
$\text{Ir}(AB)=\emptyset,\quad\text{Ir}(A_{1})=[q_{1},q_{2}],\quad\text{Ir}(B_{1})=(-\infty,q_{1}]\cup\left[q_{2},a\right]\cup[b,\infty)$.

In the initial phase-D1a, $\text{Ir}(B_{1})$ spans entire entanglement island
region. However, the other phases-D1a incorporates the entanglement island
region $\text{Ir}(A_{1})$ and $\text{Ir}(B_{1})$ for $A_{1}\cup B_{1}$. In
both of the above phases, the entanglement wedge of $AB$ is connected
indicated from the RT surface homologous to $AB$.

#### Phase-D1a

To proceed, we need to determine the location of EWCS endpoints situated on
the RT surfaces homologous to the intervals $[a_{1},b_{2}]$ and
$[a_{2},b_{1}]$. These two endpoints may be obtained by constructing Euclidean
triangles as discussed in appendix A for the RT surfaces homologous to the
intervals $[p_{1},p_{2}]$ and $[a_{1},b_{2}]$ for $y_{2}$ subsequently
$[p_{1},p_{2}]$ and $[a_{2},b_{1}]$ for $y_{1}$. Therefore the location of
these endpoints in terms of boundary coordinates are given by

$\displaystyle y_{1}$
$\displaystyle=\frac{\sqrt{a_{2}-p_{1}}\sqrt{p_{1}-b_{1}}\sqrt{a_{2}-p_{2}}\sqrt{b_{1}-p_{2}}}{p_{1}+p_{2}-a_{2}-b_{1}}\,,$
(92) $\displaystyle y_{2}$
$\displaystyle=\frac{\sqrt{a_{1}-p_{1}}\sqrt{a_{1}-p_{2}}\sqrt{p_{1}-b_{2}}\sqrt{b_{2}-p_{2}}}{p_{1}+p_{2}-a_{1}-b_{2}}\,.$

The geodesic length connecting the above bulk points can be computed using the
following length formula

$\displaystyle\mathcal{L}$
$\displaystyle=\int_{y_{1}}^{y_{2}}\differential{z}\frac{\sqrt{1+\frac{\differential{x}}{\differential{z}}^{2}}}{z}\,$
(93)
$\displaystyle=\frac{1}{2}\log\left[{\frac{(p_{2}-p_{1})/2-\sqrt{((p_{2}-p_{1})/2)^{2}-z^{2}}}{(p_{2}-p_{1})/2+\sqrt{((p_{2}-p_{1})/2)^{2}-z^{2}}}}\right]\Bigg{|}_{y_{1}}^{y_{2}}\,.$

Note that, the aforementioned length formula is minimal since, in determining
the EWCS endpoints, we took into consideration that the RT surface homologous
to $[p_{1},p_{2}]$ is perpendicular to the RT surfaces homologous to
$[a_{1},b_{2}]$ and $[a_{2},b_{1}]$. In this context, the EWCS in the phase
can be obtained by utilizing eqs. 93 and 92 as follows

$\displaystyle E_{W}$ $\displaystyle=\frac{\mathcal{L}}{4G}$ (94)
$\displaystyle=\frac{c}{12}\log\left[\frac{\left(a_{2}-p_{1}\right)\left(a_{1}-p_{2}\right)\left(p_{1}-b_{1}\right)\left(b_{2}-p_{2}\right)}{\left(a_{1}-p_{1}\right)\left(a_{2}-p_{2}\right)\left(p_{1}-b_{2}\right)\left(b_{1}-p_{2}\right)}\right]\,.$

where $p_{1}$ and $p_{2}$ are given as follow through solving (70) for the RT
surfaces homologous to $[p_{1},p_{2}]$ and $[a_{1},b_{2}]$, and the RT
surfaces homologous to $[p_{1},p_{2}]$ and $[a_{2},b_{1}]$.

$\displaystyle p_{1}$
$\displaystyle=\frac{a_{1}b_{2}-a_{2}b_{1}-\sqrt{\left(a_{1}-a_{2}\right)\left(b_{1}-b_{2}\right)\left(a_{1}-b_{1}\right)\left(a_{2}-b_{2}\right)}}{a_{1}-a_{2}-b_{1}+b_{2}}\,,$
(95) $\displaystyle p_{2}$
$\displaystyle=\frac{a_{1}b_{2}-a_{2}b_{1}+\sqrt{\left(a_{1}-a_{2}\right)\left(b_{1}-b_{2}\right)\left(a_{1}-b_{1}\right)\left(a_{2}-b_{2}\right)}}{a_{1}-a_{2}-b_{1}+b_{2}}\,.$

${\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}A}$$A_{2}$
${\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}B_{1}}$$B_{2}$
${\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}B}$${\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}B_{1}}$${\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}A_{1}}$$E_{W}$Ir($B_{1}$)$a_{1}$$p_{2}$$b_{2}$$b$$p_{1}$$a$$a_{2}$$b_{1}$$y_{1}$$y_{2}$Ir($B_{1}$)Ir($B_{1}$)
Figure 14: The diagram depicts phases-D1a of two disjoint intervals
$A=[a_{1},a_{2}]$ and $B=[b_{1},b_{2}]$ which are sandwiched by $A_{2}\cup
B_{2}=[a_{2},b_{1}]$ with an entanglement island region spanned by
$\text{Ir}(B_{1})=(-\infty,a]\cup[b,\infty)$ on the full KR brane.

We now compute the BPE. As indicated in Fig.14, where
$\text{Ir}(B_{1})=(-\infty,a]\cup[b,\infty)$ occupies the whole island region.
We need to determine the corresponding partition point via the balance
requirement. In this context, the balance requirements are given by the
following two equations

$\displaystyle\mathcal{I}(A,BB_{1}\text{Ir}(B_{1})B_{2})=$
$\displaystyle\mathcal{I}(B,A_{1}A_{2}A)\,,$ (96)
$\displaystyle\mathcal{I}(A_{1},BB_{1}\text{Ir}(B_{1})B_{2})=$
$\displaystyle\mathcal{I}(B_{1}\text{Ir}(B_{1}),A_{1}A_{2}A)\,.$

The above PEEs can be computed using the generalized ALC proposal as follows

$\displaystyle\mathcal{I}(A,BB_{1}\text{Ir}(B_{1})B_{2})$
$\displaystyle=\frac{1}{2}(\tilde{S}_{AA_{1}}+\tilde{S}_{AA_{2}}-\tilde{S}_{A_{1}}-\tilde{S}_{A_{2}})$
$\displaystyle=\frac{1}{2}\left(\tilde{S}_{\left[p_{1},a_{2}\right]}+\tilde{S}_{\left[a_{1},p_{2}\right]}-\tilde{S}_{\left[p_{1},a_{1}\right]}-\tilde{S}_{\left[a_{2},p_{2}\right]}\right)$
$\displaystyle=\frac{c}{6}\log\left[\frac{(a_{2}-p_{1})(p_{2}-a_{1})}{(a_{1}-p_{1})(p_{2}-a_{2})}\right]\,,$
(97) $\displaystyle\mathcal{I}(B,A_{1}A_{2}A)$
$\displaystyle=\frac{1}{2}(\tilde{S}_{BB_{1}\text{Ir}(B_{1})}+\tilde{S}_{BB_{2}}-\tilde{S}_{B_{1}\text{Ir}(B_{1})}-\tilde{S}_{B_{2}})$
$\displaystyle=\frac{1}{2}\left(\tilde{S}_{\left(-\infty,p_{1}\right]\cup\left[b_{1},\infty\right)}+\tilde{S}_{\left[p_{2},b_{2}\right]}-\tilde{S}_{\left(-\infty,p_{1}\right]\cup\left[b_{2},\infty\right)}-\tilde{S}_{\left[p_{2},b_{1}\right]}\right)\,$
$\displaystyle=\frac{1}{2}\left(\tilde{S}_{\left[p_{1},b_{1}\right]}+\tilde{S}_{\left[p_{2},b_{2}\right]}-\tilde{S}_{\left[p_{1},b_{2}\right]}-\tilde{S}_{\left[p_{2},b_{1}\right]}\right)$
$\displaystyle=\frac{c}{6}\log\left[\frac{(b_{1}-p_{1})(b_{2}-p_{2})}{(b_{2}-p_{1})(b_{1}-p_{2})}\right]\,,$
(98)

and

$\displaystyle\mathcal{I}(A_{1},BB_{1}\text{Ir}(B_{1})B_{2})$
$\displaystyle=\frac{1}{2}(\tilde{S}_{AA_{1}A_{2}}+\tilde{S}_{A_{1}}-\tilde{S}_{AA_{2}})$
$\displaystyle=\frac{1}{2}\left(\tilde{S}_{\left[p_{1},p_{2}\right]}+\tilde{S}_{\left[p_{1},a_{1}\right]}-\tilde{S}_{\left[a_{1},p_{2}\right]}\right)$
$\displaystyle=\frac{c}{6}\log\left[\frac{(p_{2}-p_{1})(a_{1}-p_{1})}{(p_{2}-a_{1})\epsilon}\right]\,,$
(99) $\displaystyle\mathcal{I}(B_{1}\text{Ir}(B_{1}),A_{1}A_{2}A)$
$\displaystyle=\frac{1}{2}(\tilde{S}_{BB_{1}\text{Ir}(B_{1})B_{2}}+\tilde{S}_{B_{1}\text{Ir}(B_{1})}-\tilde{S}_{BB_{2}})$
$\displaystyle=\frac{1}{2}\left(\tilde{S}_{\left(-\infty,p_{1}\right]\cup\left[p_{2},\infty\right)}+\tilde{S}_{\left(-\infty,p_{1}\right]\cup\left[b_{2},\infty\right)}-\tilde{S}_{\left[p_{2},b_{2}\right]}\right)\,$
$\displaystyle=\frac{1}{2}\left(\tilde{S}_{\left[p_{1},p_{2}\right]}+\tilde{S}_{\left[p_{1},b_{2}\right]}-\tilde{S}_{\left[p_{2},b_{2}\right]}\right)$
$\displaystyle=\frac{c}{6}\log\left[\frac{(b_{2}-p_{1})(p_{2}-p_{1})}{(b_{2}-p_{2})\epsilon}\right]\,.$
(100)

From the balance requirement (96), we can compute the partition points $p_{1}$
and $p_{2}$, which are the same as (95). The BPE for this phase-D1a may be
obtained utilizing eqs. 95 and 97 as

BPE $\displaystyle=\mathcal{I}(A,BB_{1}\text{Ir}(B_{1})B_{2})|_{balanced}$
$\displaystyle=\frac{c}{6}\log\left[\frac{b_{2}\left(a_{2}+b_{1}\right)+a_{1}\left(a_{2}+b_{1}-2b_{2}\right)-2a_{2}b_{1}+2\sqrt{\left(a_{1}-a_{2}\right)\left(b_{1}-b_{2}\right)\left(a_{1}-b_{1}\right)\left(a_{2}-b_{2}\right)}}{\left(a_{2}-b_{1}\right)\left(a_{1}-b_{2}\right)}\right]\,.$
(101)

Note that the above result of BPE can be exactly matched with the EWCS (94).

#### Phase-D1b

In this phase-D1b, we follow similar analysis provided in the earlier
phase-D1a to compute the EWCS. Consequently, we proceed to obtain the
endpoints of the EWCS which are given by $y_{1}$ and $y_{2}$ as indicated in
Fig.15

$\displaystyle y_{1}$
$\displaystyle=\frac{\sqrt{a_{1}-q_{1}}\sqrt{a_{1}-p_{2}}\sqrt{q_{1}-b_{2}}\sqrt{b_{2}-p_{2}}}{q_{1}+p_{2}-a_{1}-b_{2}}\,,$
(102) $\displaystyle y_{2}$
$\displaystyle=\frac{\sqrt{a_{2}-q_{1}}\sqrt{a_{2}-p_{2}}\sqrt{q_{1}-b_{1}}\sqrt{b_{1}-p_{2}}}{q_{1}+p_{2}-a_{2}-b_{1}}\,,$

where we used Euclidean triangle construction as discussed in appendix A for
the RT surfaces homologous to the intervals $[q_{1},p_{2}]$, $[a_{1},b_{2}]$
and $[q_{1},p_{2}]$, $[a_{2},b_{1}]$ for calculating the $y_{2}$ and $y_{1}$
respectively. The minimal length at a constant time slice associated to these
endpoints is given by

$\displaystyle\mathcal{L}$
$\displaystyle=\int_{y_{1}}^{y_{2}}\differential{z}\frac{\sqrt{1+\frac{\differential{x}}{\differential{z}}^{2}}}{z}\,$
(103)
$\displaystyle=\frac{1}{2}\log\left[{\frac{(p_{2}-q_{1})/2-\sqrt{((p_{2}-q_{1})/2)^{2}-z^{2}}}{(p_{2}-q_{1})/2+\sqrt{((p_{2}-q_{1})/2)^{2}-z^{2}}}}\right]\Bigg{|}_{y_{1}}^{y_{2}}\,.$

Utilizing eqs. 102 and 103, we may the compute the EWCS for this phase as
follows

$\displaystyle E_{W}$ $\displaystyle=\frac{\mathcal{L}}{4G}$ (104)
$\displaystyle=\frac{c}{12}\log\left[\frac{\left(a_{2}-q_{1}\right)\left(a_{1}-p_{2}\right)\left(q_{1}-b_{1}\right)\left(b_{2}-p_{2}\right)}{\left(a_{1}-q_{1}\right)\left(a_{2}-p_{2}\right)\left(q_{1}-b_{2}\right)\left(b_{1}-p_{2}\right)}\right]\,.$

where $q_{1}$ and $p_{2}$ are given by:

$\displaystyle q_{1}$
$\displaystyle=\frac{a_{1}b_{2}-a_{2}b_{1}-\sqrt{\left(a_{1}-a_{2}\right)\left(b_{1}-b_{2}\right)\left(a_{1}-b_{1}\right)\left(a_{2}-b_{2}\right)}}{a_{1}-a_{2}-b_{1}+b_{2}}\,,$
(105) $\displaystyle p_{2}$
$\displaystyle=\frac{a_{1}b_{2}-a_{2}b_{1}+\sqrt{\left(a_{1}-a_{2}\right)\left(b_{1}-b_{2}\right)\left(a_{1}-b_{1}\right)\left(a_{2}-b_{2}\right)}}{a_{1}-a_{2}-b_{1}+b_{2}}\,.$

Note that $q_{1}$ can also be written in terms of the points $a_{2}$, $b_{1}$
and $p_{2}$ using the constraint condition in (70) for the RT surfaces
homologous to $[q_{1},p_{2}]$ and $[a_{2},b_{1}]$ as

$q_{1}=\frac{2a_{2}b_{1}-p_{2}\left(a_{2}+b_{1}\right)}{a_{2}+b_{1}-2p_{2}}\,.$
(106)

${\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}A}$$A_{2}$
$B_{1}$ $B_{2}$
${\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}B}$$B_{1}$
$A_{1}$
$E_{W}$$a_{1}$$p_{2}$$b_{2}$$b$$p_{1}$$a$$a_{2}$$b_{1}$Ir($B_{1}$)Ir($B_{1}$)Ir($B_{1}$)Ir($A_{1}$)Ir($A_{1}$)Ir($B_{1}$)$q_{2}$$q_{1}$$y_{1}$$y_{2}$
Figure 15: Schematics shows the configuration of two disjoint intervals $AB$
with entanglement island regions $\text{Ir}(A_{1})=[q_{1},q_{2}]$ and
$\text{Ir}(B_{1})=(-\infty,q_{1}]\cup\left[q_{2},a\right]\cup[b,\infty)$ where
the partition points of $A_{2}\cup B_{2}$ and $A_{1}\cup B_{1}$ are labeled as
$p_{2}$ and $p_{1}$ respectively.

In this phase-D1b, the computation of BPE involves entanglement island regions
$\text{Ir}(A_{1})=[q_{1},q_{2}]$ and
$\text{Ir}(B_{1})=(-\infty,q_{1}]\cup\left[q_{2},a\right]\cup[b,\infty)$ shown
in Fig.15. We now calculate the location of partition points $p_{1}$ and
$p_{2}$ from balance requirements

$\displaystyle\mathcal{I}(A,BB_{1}\text{Ir}(B_{1})B_{2})=$
$\displaystyle\mathcal{I}(B,A_{1}\text{Ir}(A_{1})A_{2}A)\,,$ (107)
$\displaystyle\mathcal{I}(A_{1}\text{Ir}(A_{1}),BB_{1}\text{Ir}(B_{1})B_{2})=$
$\displaystyle\mathcal{I}(B_{1}\text{Ir}(B_{1}),A_{1}\text{Ir}(A_{1})A_{2}A)\,.$

The above PEEs may be computed using the generalized ALC proposal as follows

$\displaystyle\mathcal{I}(A,BB_{1}\text{Ir}(B_{1})B_{2})$
$\displaystyle=\frac{1}{2}(\tilde{S}_{AA_{1}\text{Ir}(A_{1})}+\tilde{S}_{AA_{2}}-\tilde{S}_{A_{1}\text{Ir}(A_{1})}-\tilde{S}_{A_{2}})$
$\displaystyle=\frac{1}{2}\left(\tilde{S}_{\left[q_{1},q_{2}\right]\cup\left[p_{1},a_{2}\right]}+\tilde{S}_{\left[a_{1},p_{2}\right]}-\tilde{S}_{\left[q_{1},q_{2}\right]\cup\left[p_{1},a_{1}\right]}-\tilde{S}_{\left[a_{2},p_{2}\right]}\right)\,$
$\displaystyle=\frac{1}{2}\left(\tilde{S}_{\left[q_{1},a_{2}\right]}+\tilde{S}_{\left[q_{2},p_{1}\right]}+\tilde{S}_{\left[a_{1},p_{2}\right]}-\tilde{S}_{\left[q_{1},a_{1}\right]}-\tilde{S}_{\left[q_{2},p_{1}\right]}-\tilde{S}_{\left[a_{2},p_{2}\right]}\right)$
$\displaystyle=\frac{c}{6}\log\left[\frac{\left(a_{2}-q_{1}\right)\left(p_{2}-a_{1}\right)}{\left(a_{1}-q_{1}\right)\left(p_{2}-a_{2}\right)}\right]\,,$
(108) $\displaystyle\mathcal{I}(B,A_{1}\text{Ir}(A_{1})A_{2}A)$
$\displaystyle=\frac{1}{2}(\tilde{S}_{BB_{1}\text{Ir}(B_{1})}+\tilde{S}_{BB_{2}}-\tilde{S}_{B_{1}\text{Ir}(B_{1})}-\tilde{S}_{B_{2}})$
$\displaystyle=\frac{1}{2}\left(\tilde{S}_{\left(-\infty,q_{1}\right]\cup[q_{2},p_{1}]\cup\left[b_{1},\infty\right)}+\tilde{S}_{\left[p_{2},b_{2}\right]}-\tilde{S}_{\left(-\infty,q_{1}\right]\cup[q_{2},p_{1}]\cup\left[b_{2},\infty\right)}-\tilde{S}_{\left[p_{2},b_{1}\right]}\right)\,$
$\displaystyle=\frac{1}{2}\left(\tilde{S}_{\left[q_{1},b_{1}\right]}+\tilde{S}_{\left[q_{2},p_{1}\right]}+\tilde{S}_{\left[p_{2},b_{2}\right]}-\tilde{S}_{\left[q_{1},b_{2}\right]}-\tilde{S}_{\left[q_{2},p_{1}\right]}-\tilde{S}_{\left[p_{2},b_{1}\right]}\right)$
$\displaystyle=\frac{c}{6}\log\left[\frac{\left(b_{1}-q_{1}\right)\left(b_{2}-p_{2}\right)}{\left(b_{2}-q_{1}\right)\left(b_{1}-p_{2}\right)}\right]\,,$
(109)

and

$\displaystyle\mathcal{I}(A_{1}\text{Ir}(A_{1}),BB_{1}\text{Ir}(B_{1})B_{2})$
$\displaystyle=\frac{1}{2}(\tilde{S}_{AA_{1}\text{Ir}(A_{1})A_{2}}+\tilde{S}_{A_{1}\text{Ir}(A_{1})}-\tilde{S}_{AA_{2}})$
$\displaystyle=\frac{1}{2}\left(\tilde{S}_{\left[q_{1},q_{2}\right]\cup\left[p_{1},p_{2}\right]}+\tilde{S}_{\left[q_{1},q_{2}\right]\cup\left[p_{1},a_{1}\right]}-\tilde{S}_{\left[a_{1},p_{2}\right]}\right)\,$
$\displaystyle=\frac{1}{2}\left(\tilde{S}_{\left[q_{1},p_{2}\right]}+\tilde{S}_{\left[q_{2},p_{1}\right]}+\tilde{S}_{\left[q_{1},a_{1}\right]}+\tilde{S}_{\left[q_{2},p_{1}\right]}-\tilde{S}_{\left[a_{1},p_{2}\right]}\right)$
$\displaystyle=\frac{c}{6}\log\left[\frac{(p_{1}-q_{2})^{2}(p_{2}-q_{1})(a_{1}-q_{1})}{\epsilon^{3}(p_{2}-a_{1})}\right]+\frac{c}{6}\phi(q_{1})+\frac{c}{6}\phi(q_{2})\,,$
(110)
$\displaystyle\mathcal{I}(B_{1}\text{Ir}(B_{1}),A_{1}\text{Ir}(A_{1})A_{2}A)$
$\displaystyle=\frac{1}{2}(\tilde{S}_{BB_{1}\text{Ir}(B_{1})B_{2}}+\tilde{S}_{B_{1}\text{Ir}(B_{1})}-\tilde{S}_{BB_{2}})$
$\displaystyle=\frac{1}{2}\left(\tilde{S}_{\left(-\infty,q_{1}\right]\cup[q_{2},p_{1}]\cup\left[p_{2},\infty\right)}+\tilde{S}_{\left(-\infty,q_{1}\right]\cup[q_{2},p_{1}]\cup\left[b_{2},\infty\right]}-\tilde{S}_{\left[p_{2},b_{2}\right]}\right)\,$
$\displaystyle=\frac{1}{2}\left(\tilde{S}_{\left[q_{1},p_{2}\right]}+\tilde{S}_{\left[q_{2},p_{1}\right]}+\tilde{S}_{\left[q_{1},b_{2}\right]}+\tilde{S}_{\left[q_{2},p_{1}\right]}-\tilde{S}_{\left[p_{2},b_{2}\right]}\right)$
$\displaystyle=\frac{c}{6}\log\left[\frac{(p_{1}-q_{2})^{2}(p_{2}-q_{1})(b_{2}-q_{1})}{\epsilon^{3}(b_{2}-p_{2})}\right]+\frac{c}{6}\phi(q_{1})+\frac{c}{6}\phi(q_{2})\,.$
(111)

From the balance requirement (107), the partition points $q_{1}$ and $p_{2}$
are the same as (105). Finally, we may obtain BPE for this phase as follows

BPE $\displaystyle=\mathcal{I}(A,BB_{1}\text{Ir}(B_{1})B_{2})|_{balanced}$
$\displaystyle=\frac{c}{6}\log\left[\frac{b_{2}\left(a_{2}+b_{1}\right)+a_{1}\left(a_{2}+b_{1}-2b_{2}\right)-2a_{2}b_{1}+2\sqrt{\left(a_{1}-a_{2}\right)\left(b_{1}-b_{2}\right)\left(a_{1}-b_{1}\right)\left(a_{2}-b_{2}\right)}}{\left(a_{2}-b_{1}\right)\left(a_{1}-b_{2}\right)}\right]\,.$
(112)

The above result of BPE exactly matches with the EWCS in (104) using the value
of partition point $p_{2}$ and $q_{1}$ given in eqs. 106 and 105.

### Disjoint $AB$ admits island

In this subsection, we consider the configuration of two disjoint intervals
$A=[a_{1},a_{2}]$ and $B=[b_{1},b_{2}]$ where $AB$ admits entanglement island
region. We study the following assignments for the island region,

  1. 1.

D2a:
$\text{Ir}(AA_{1}A_{2})=\emptyset,\quad\text{Ir}(B_{1})=[q_{1},a]\cup[b,q_{2}],\quad\text{Ir}(B)=(-\infty,q_{1}]\cup[q_{2},\infty)$
,

  2. 2.

D2b:
$\text{Ir}(A)=[q_{1},q_{2}],\quad\text{Ir}(B)=(-\infty,q_{1}]\cup[q_{4},\infty),\quad\text{Ir}(A_{1})=[q_{2},q_{3}],\quad\text{Ir}(B_{1})=[q_{3},a]\cup[b,q_{4}]$
,

  3. 3.

D2c:
$\text{Ir}(A)=[q_{2},q_{3}],\quad\text{Ir}(B)=[q_{5},q_{6}],\quad\text{Ir}(A_{1})=[q_{3},q_{4}],\quad\text{Ir}(B_{1})=[q_{4},a]\cup[b,q_{5}],\quad\text{Ir}(A_{2})=[q_{1},q_{2}],\quad\text{Ir}(B_{2})=(-\infty,q_{1}]\cup[q_{6},\infty)$
.

The intervals $A$, $A_{1}$ and $A_{2}$ does not receive any contributions from
the island region in the phase-D2a as shown in Fig.16. The last phase-D2c
involves all entanglement island regions for the all the intervals $A$, $B$,
$A_{1}$, $B_{1}$, $A_{2}$ and $B_{2}$. However in this phase, the entanglement
wedge gets disconnected consequently the EWCS and BPE become zero.

#### Phase-D2a

We use the same approach as in phase-D1a to calculate the EWCS in this phase.
Correspondingly the endpoints $y_{1}$ and $y_{2}$ of the EWCS as shown in
Fig.16, are thus obtained as

$\displaystyle y_{1}$
$\displaystyle=\frac{\sqrt{a_{2}-p_{1}}\sqrt{a_{2}-p_{2}}\sqrt{p_{1}-b_{1}}\sqrt{b_{1}-p_{2}}}{p_{1}+p_{2}-a_{2}-b_{1}}\,,$
(113) $\displaystyle y_{2}$
$\displaystyle=\frac{\sqrt{a_{1}-p_{1}}\sqrt{a_{1}-p_{2}}\sqrt{p_{1}-q_{1}}\sqrt{q_{1}-p_{2}}}{p_{1}+p_{2}-a_{1}-q_{1}}\,.$

At a constant time slice, the minimal length between the endpoints $y_{1}$ and
$y_{2}$ located on the RT surfaces homologous to $[q_{1},a_{1}]$ and
$[a_{2},b_{1}]$ can be computed as follows

$\displaystyle\mathcal{L}$
$\displaystyle=\int_{y_{1}}^{y_{2}}\differential{z}\frac{\sqrt{1+\frac{\differential{x}}{\differential{z}}^{2}}}{z}\,$
(114)
$\displaystyle=\frac{1}{2}\log\left[{\frac{(p_{2}-p_{1})/2-\sqrt{((p_{2}-p_{1})/2)^{2}-z^{2}}}{(p_{2}-p_{1})/2+\sqrt{((p_{2}-p_{1})/2)^{2}-z^{2}}}}\right]\Bigg{|}_{y_{1}}^{y_{2}}\,.$

Finally in this phase, the EWCS can be determined utilizing eqs. 113 and 114
in the following proposal of the EWCS in $AdS_{3}$ geometries

$\displaystyle E_{W}$ $\displaystyle=\frac{\mathcal{L}}{4G}$ (115)
$\displaystyle=\frac{c}{12}\log\left[\frac{\left(a_{2}-p_{1}\right)\left(a_{1}-p_{2}\right)\left(p_{1}-b_{1}\right)\left(q_{1}-p_{2}\right)}{\left(a_{1}-p_{1}\right)\left(a_{2}-p_{2}\right)\left(b_{1}-p_{2}\right)\left(p_{1}-q_{1}\right)}\right]\,.$

where $p_{1}$ and $p_{2}$ are given by:

$\displaystyle p_{1}$
$\displaystyle=\frac{a_{1}q_{1}-a_{2}b_{1}-\sqrt{\left(a_{1}-a_{2}\right)\left(b_{1}-q_{1}\right)\left(a_{1}-b_{1}\right)\left(a_{2}-q_{1}\right)}}{a_{1}-a_{2}-b_{1}+q_{1}}\,,$
(116) $\displaystyle p_{2}$
$\displaystyle=\frac{a_{1}q_{1}-a_{2}b_{1}+\sqrt{\left(a_{1}-a_{2}\right)\left(b_{1}-q_{1}\right)\left(a_{1}-b_{1}\right)\left(a_{2}-q_{1}\right)}}{a_{1}-a_{2}-b_{1}+q_{1}}\,.$

${\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}A}$$A_{2}$
${\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}B_{1}}$${\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}B}$$B_{2}$
${\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}B_{1}}$${\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}A_{1}}$Ir(B)Ir(B)
$E_{W}$Ir(B)$a_{1}$$a_{2}$$p_{2}$$b_{2}$$b$$b_{1}$$p_{1}$$a$$q_{1}$$q_{2}$$y_{1}$$y_{2}$Ir($B_{1}$)Ir($B_{1}$)
Figure 16: The diagram shows the configuration of $AB$ considered in Weyl
$CFT_{2}$ where the intervals $B$ and $B_{1}$ admit entanglement island
regions $\text{Ir}(B_{1})=[q_{1},a_{1}]\cup[b,q_{2}]$ and
$\text{Ir}(B)=(-\infty,q_{1}]\cup[q_{2},\infty)$.

We now compute the BPE in Fig.16. Consequently, the balance requirements are
given by

$\displaystyle\mathcal{I}(A,B\text{Ir}(B)B_{1}\text{Ir}(B_{1})B_{2})=$
$\displaystyle\mathcal{I}(B\text{Ir}(B),A_{1}A_{2}A)\,,$ (117)
$\displaystyle\mathcal{I}(A_{1},B\text{Ir}(B)B_{1}\text{Ir}(B_{1})B_{2})=$
$\displaystyle\mathcal{I}(B_{1}\text{Ir}(B_{1}),A_{1}A_{2}A)\,.$

Utilizing the generalized ALC proposal, the above PEEs may be described as
follows

$\displaystyle\mathcal{I}(A,B\text{Ir}(B)B_{1}\text{Ir}(B_{1})B_{2})$
$\displaystyle=\frac{1}{2}(\tilde{S}_{AA_{1}}+\tilde{S}_{AA_{2}}-\tilde{S}_{A_{1}}-\tilde{S}_{A_{2}})$
$\displaystyle=\frac{1}{2}\left(\tilde{S}_{\left[p_{1},a_{2}\right]}+\tilde{S}_{\left[a_{1},p_{2}\right]}-\tilde{S}_{\left[p_{1},a_{1}\right]}-\tilde{S}_{\left[a_{2},p_{2}\right]}\right)\,$
$\displaystyle=\frac{c}{6}\log\left[\frac{\left(a_{2}-p_{1}\right)\left(p_{2}-a_{1}\right)}{\left(a_{1}-p_{1}\right)\left(p_{2}-a_{2}\right)}\right]\,,$
(118) $\displaystyle\mathcal{I}(B\text{Ir}(B),A_{1}A_{2}A)$
$\displaystyle=\frac{1}{2}(\tilde{S}_{B\text{Ir}(B)B_{1}\text{Ir}(B_{1})}+\tilde{S}_{B\text{Ir}(B)B_{2}}-\tilde{S}_{B_{1}\text{Ir}(B_{1})}-\tilde{S}_{B_{2}})$
$\displaystyle=\frac{1}{2}\left(\tilde{S}_{(-\infty,p_{1}]\cup\left[b_{1},\infty\right)}+\tilde{S}_{\left(-\infty,q_{1}\right]\cup[p_{2},b_{2}]\cup\left[q_{2},\infty\right)}-\tilde{S}_{[q_{1},p_{1}]\cup[b_{2},q_{2}]}-\tilde{S}_{\left[p_{2},b_{1}\right]}\right)\,$
$\displaystyle=\frac{1}{2}\left(\tilde{S}_{\left[p_{1},b_{1}\right]}+\tilde{S}_{\left[q_{1},p_{2}\right]}+\tilde{S}_{\left[b_{2},q_{2}\right]}-\tilde{S}_{\left[q_{1},p_{1}\right]}-\tilde{S}_{\left[b_{2},q_{2}\right]}-\tilde{S}_{\left[p_{2},b_{1}\right]}\right)$
$\displaystyle=\frac{c}{6}\log\left[\frac{\left(b_{1}-p_{1}\right)\left(p_{2}-q_{1}\right)}{\left(b_{1}-p_{2}\right)\left(p_{1}-q_{1}\right)}\right]\,,$
(119)

and

$\displaystyle\mathcal{I}(A_{1},B\text{Ir}(B)B_{1}\text{Ir}(B_{1})B_{2})$
$\displaystyle=\frac{1}{2}(\tilde{S}_{AA_{1}A_{2}}+\tilde{S}_{A_{1}}-\tilde{S}_{AA_{2}})$
$\displaystyle=\frac{1}{2}\left(\tilde{S}_{\left[p_{1},p_{2}\right]}+\tilde{S}_{\left[p_{1},a_{1}\right]}-\tilde{S}_{\left[a_{1},p_{2}\right]}\right)\,$
$\displaystyle=\frac{c}{6}\log\left[\frac{\left(p_{2}-p_{1}\right)\left(a_{1}-p_{1}\right)}{\epsilon\left(p_{2}-a_{1}\right)}\right]\,,$
(120) $\displaystyle\mathcal{I}(B_{1}\text{Ir}(B_{1}),A_{1}A_{2}A)$
$\displaystyle=\frac{1}{2}(\tilde{S}_{B\text{Ir}(B)B_{1}\text{Ir}(B_{1})B_{2}}+\tilde{S}_{B_{1}\text{Ir}(B_{1})}-\tilde{S}_{B\text{Ir}(B)B_{2}})$
$\displaystyle=\frac{1}{2}\left(\tilde{S}_{\left(-\infty,p_{1}\right]\cup\left[p_{2},\infty\right)}+\tilde{S}_{[q_{1},p_{1}]\cup[b_{2},q_{2}]}-\tilde{S}_{\left(-\infty,q_{1}\right]\cup[p_{2},b_{2}]\cup\left[q_{2},\infty\right)}\right)\,$
$\displaystyle=\frac{1}{2}\left(\tilde{S}_{\left[p_{1},p_{2}\right]}+\tilde{S}_{\left[q_{1},p_{1}\right]}+\tilde{S}_{\left[b_{2},q_{2}\right]}-\tilde{S}_{\left[q_{1},p_{2}\right]}-\tilde{S}_{\left[b_{2},q_{2}\right]}\right)$
$\displaystyle=\frac{c}{6}\log\left[\frac{\left(p_{2}-p_{1}\right)\left(p_{1}-q_{1}\right)}{\epsilon\left(p_{2}-q_{1}\right)}\right]\,.$
(121)

Utilizing the balance requirements in (117), we get the solutions of $p_{1}$
and $p_{2}$ that are the same as (116). Finally we obtain the BPE for this
phase-D2 satisfying the balance conditions as follows

BPE
$\displaystyle=\mathcal{I}(A,B\text{Ir}(B)B_{1}\text{Ir}(B_{1})B_{2})|_{balanced}$
(122)
$\displaystyle=\frac{c}{6}\log\left[\frac{q_{1}\left(a_{2}+b_{1}\right)+a_{1}\left(a_{2}+b_{1}-2q_{1}\right)-2a_{2}b_{1}+2\sqrt{\left(a_{1}-a_{2}\right)\left(a_{1}-b_{1}\right)\left(a_{2}-q_{1}\right)\left(b_{1}-q_{1}\right)}}{\left(a_{2}-b_{1}\right)\left(a_{1}-q_{1}\right)}\right]\,.$
(123)

The above result of BPE exactly coincide with the EWCS by putting the value of
$p_{2}$ in (115).

#### Phase-D2b

As depicted in Fig.17, the computation of the bulk EWCS involves endpoints
$y_{2}$ and $y_{1}$ where $y_{2}$ is located on the KR brane. Here $y_{1}$ is
situated on the RT surface homologous to the interval $[a_{2},b_{1}]$. Note
that the minimal length between the corresponding endpoints can be calculated
using the following length formula in the $AdS_{3}$ geometry at a constant
time slice as

$\displaystyle\mathcal{L}$
$\displaystyle=\int_{y_{1}}^{y_{2}}\differential{z}\frac{\sqrt{1+\frac{\differential{x}}{\differential{z}}^{2}}}{z}\,$
(124)
$\displaystyle=\frac{1}{2}\log\left[{\frac{(p_{2}-q_{1})/2-\sqrt{((p_{2}-q_{1})/2)^{2}-z^{2}}}{(p_{2}-q_{1})/2+\sqrt{((p_{2}-q_{1})/2)^{2}-z^{2}}}}\right]\Bigg{|}_{y_{1}}^{y_{2}}\,,$

where the endpoints of the EWCS $y_{1}$ and $y_{2}$ are given by

$\displaystyle y_{1}$
$\displaystyle=\frac{\sqrt{a_{2}-q_{1}}\sqrt{a_{2}-p_{2}}\sqrt{q_{1}-b_{1}}\sqrt{b_{1}-p_{2}}}{q_{1}+p_{2}-a_{2}-b_{1}}\,,$
(125) $\displaystyle y_{2}$
$\displaystyle=\frac{\sqrt{a-q_{1}}\sqrt{a-p_{2}}\sqrt{q_{1}-b}\sqrt{b-p_{2}}}{q_{1}+p_{2}-a-b}\,.$

Utilizing eqs. 125 and 124, we may compute the EWCS for this phase as follows

$\displaystyle E_{W}$ $\displaystyle=\frac{\mathcal{L}}{4G}$ (126)
$\displaystyle=\frac{c}{12}\log\left[\frac{\left(a_{2}-q_{1}\right)\left(a-p_{2}\right)\left(q_{1}-b_{1}\right)\left(b-p_{2}\right)}{\left(a-q_{1}\right)\left(a_{2}-p_{2}\right)\left(q_{1}-b\right)\left(b_{1}-p_{2}\right)}\right]\,.$

where $q_{1}$ and $p_{2}$ are given by:

$\displaystyle q_{1}=$
$\displaystyle\frac{2a_{2}b_{1}-a_{2}p_{2}-b_{1}p_{2}}{a_{2}+b_{1}-2p_{2}}\,,$
(127) $\displaystyle p_{2}=$ $\displaystyle\frac{ab-
a_{2}b_{1}+\sqrt{\left(a-a_{2}\right)\left(b-b_{1}\right)\left(a_{2}-b\right)\left(b_{1}-a\right)}}{a-a_{2}+b-b_{1}}\,.$

$A$ $B_{1}$ $B_{1}$ $A_{1}$ $A_{2}$ $B_{2}$
${\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}B}$Ir(A)Ir(B)Ir(B)
$E_{W}$$a_{1}$$a_{2}$$p_{2}$$b_{1}$$b_{2}$$b$$p_{1}$$a$$q_{2}$$q_{4}$$q_{3}$$q_{1}$$y_{1}$$y_{2}$Ir($B_{1}$)Ir($B_{1}$)Ir($A_{1}$)Ir($A_{1}$)Ir($A$)Ir($B$)
Figure 17: The BPE for this configuration receives contribution from
entanglement island of $A$ and $B$ and other island regions are described as
$\text{Ir}(A_{1})=[q_{2},q_{3}]$ and
$\text{Ir}(B_{1})=[q_{3},a]\cup[b,q_{4}]$. Here the $B_{1}$ has disconnected
entanglement island region.

We now computate the BPE in Fig.17. The balance requirements are given by,

$\displaystyle\mathcal{I}(A\text{Ir}(A),B\text{Ir}(B)B_{1}\text{Ir}(B_{1})B_{2})=$
$\displaystyle\mathcal{I}(B\text{Ir}(B),A\text{Ir}(A)A_{1}\text{Ir}(A_{1})A_{2})\,,$
(128)
$\displaystyle\mathcal{I}(A_{1}\text{Ir}(A_{1}),B\text{Ir}(B)B_{1}\text{Ir}(B_{1})B_{2})=$
$\displaystyle\mathcal{I}(B_{1}\text{Ir}(B_{1}),A\text{Ir}(A)A_{1}\text{Ir}(A_{1})A_{2})\,.$

Above PEEs in the balance conditions can be computed by using generalized ALC
proposal as follows

$\displaystyle\mathcal{I}(A\text{Ir}(A),$ $\displaystyle
B\text{Ir}(B)B_{1}\text{Ir}(B_{1})B_{2})$ $\displaystyle=$
$\displaystyle\frac{1}{2}(\tilde{S}_{A\text{Ir}(A)A_{1}\text{Ir}(A_{1})}+\tilde{S}_{A\text{Ir}(A)A_{2}}-\tilde{S}_{A_{1}\text{Ir}(A_{1})}-\tilde{S}_{A_{2}})\,$
$\displaystyle=$
$\displaystyle\frac{1}{2}\left(\tilde{S}_{\left[p_{1},a_{2}\right]\cup\left[q_{1},q_{3}\right]}+\tilde{S}_{\left[a_{1},p_{2}\right]\cup\left[q_{1},q_{2}\right]}-\tilde{S}_{\left[p_{1},a_{1}\right]\cup\left[q_{2},q_{3}\right]}-\tilde{S}_{\left[a_{2},p_{2}\right]}\right)\,$
$\displaystyle=$
$\displaystyle\frac{1}{2}\left(\tilde{S}_{\left[q_{1},a_{2}\right]}+\tilde{S}_{\left[q_{3},p_{1}\right]}+\tilde{S}_{\left[q_{2},a_{1}\right]}+\tilde{S}_{\left[q_{1},p_{2}\right]}-\tilde{S}_{\left[q_{3},p_{1}\right]}-\tilde{S}_{\left[q_{2},a_{1}\right]}-\tilde{S}_{\left[a_{2},p_{2}\right]}\right)$
$\displaystyle=$
$\displaystyle\frac{c}{6}\log\left[\frac{\left(a_{2}-q_{1}\right)\left(p_{2}-q_{1}\right)}{\epsilon\left(p_{2}-a_{2}\right)}\right]+\frac{c}{6}\phi(q_{1})\,,$
(129) $\displaystyle\mathcal{I}(B\text{Ir}(B),$ $\displaystyle
A\text{Ir}(A)A_{1}\text{Ir}(A_{1})A_{2})$ $\displaystyle=$
$\displaystyle\frac{1}{2}(\tilde{S}_{B\text{Ir}(B)B_{1}\text{Ir}(B_{1})}+\tilde{S}_{B\text{Ir}(B)B_{2}}-\tilde{S}_{B_{1}\text{Ir}(B_{1})}-\tilde{S}_{B_{2}})$
$\displaystyle=$
$\displaystyle\frac{1}{2}\left(\tilde{S}_{\left(-\infty,q_{1}\right]\cup\left[q_{3},p_{1}\right]\cup\left[b_{1},\infty\right)}+\tilde{S}_{\left(-\infty,q_{1}\right)\cup\left[p_{2},b_{2}\right]\cup\left[q_{4},\infty\right)}-\tilde{S}_{\left[b_{2},q_{4}\right]\cup\left[q_{3},p_{1}\right]}-\tilde{S}_{\left[p_{2},b_{1}\right]}\right)$
$\displaystyle=$
$\displaystyle\frac{1}{2}\left(\tilde{S}_{\left[q_{1},b_{1}\right]}+\tilde{S}_{\left[q_{3},p_{1}\right]}+\tilde{S}_{\left[b_{2},q_{4}\right]}+\tilde{S}_{\left[q_{1},p_{2}\right]}-\tilde{S}_{\left[q_{3},p_{1}\right]}-\tilde{S}_{\left[b_{2},q_{4}\right]}-\tilde{S}_{\left[p_{2},b_{1}\right]}\right)$
$\displaystyle=$
$\displaystyle\frac{c}{6}\log\left[\frac{\left(b_{1}-q_{1}\right)\left(p_{2}-q_{1}\right)}{\epsilon(b_{1}-p_{2})}\right]+\frac{c}{6}\phi(q_{1})\,,$
(130)

and

$\displaystyle\mathcal{I}(A_{1}\text{Ir}(A_{1}),$ $\displaystyle
B\text{Ir}(B)B_{1}\text{Ir}(B_{1})B_{2})$ $\displaystyle=$
$\displaystyle\frac{1}{2}(\tilde{S}_{A\text{Ir}(A)A_{1}\text{Ir}(A_{1})A_{2}}+\tilde{S}_{A_{1}\text{Ir}(A_{1})}-\tilde{S}_{A\text{Ir}(A)A_{2}})$
$\displaystyle=$
$\displaystyle\frac{1}{2}\left(\tilde{S}_{\left[p_{2},p_{1}\right]\cup\left[q_{1},q_{3}\right]}+\tilde{S}_{\left[p_{1},a_{1}\right]\cup\left[q_{2},q_{3}\right]}-\tilde{S}_{\left[a_{1},p_{2}\right]\cup\left[q_{1}q_{2}\right]}\right)$
$\displaystyle=$
$\displaystyle\frac{1}{2}\left(\tilde{S}_{\left[q_{1},p_{2}\right]}+\tilde{S}_{\left[q_{3},p_{1}\right]}+\tilde{S}_{\left[q_{2},a_{1}\right]}+\tilde{S}_{\left[q_{3},p_{1}\right]}-\tilde{S}_{\left[q_{2},a_{1}\right]}-\tilde{S}_{\left[q_{1},p_{2}\right]}\right)$
$\displaystyle=$
$\displaystyle\frac{c}{6}\log\left[\frac{(p_{1}-q_{3})^{2}}{\epsilon^{2}}\right]+\frac{c}{6}\phi(q_{3})\,,$
(131) $\displaystyle\mathcal{I}(B_{1}\text{Ir}(B_{1}),$ $\displaystyle
A\text{Ir}(A)A_{1}\text{Ir}(A_{1})A_{2})$ $\displaystyle=$
$\displaystyle\frac{1}{2}(\tilde{S}_{B\text{Ir}(B)B_{1}\text{Ir}(B_{1})B_{2}}+\tilde{S}_{B_{1}\text{Ir}(B_{1})}-\tilde{S}_{B\text{Ir}(B)B_{2}})$
$\displaystyle=$
$\displaystyle\frac{1}{2}\left(\tilde{S}_{\left(-\infty,q_{1}\right]\cup\left[q_{3},p_{1}\right]\cup[p_{2},\infty)}+\tilde{S}_{\left[b_{2},q_{4}\right]\cup\left[q_{3},p_{1}\right]}-\tilde{S}_{\left(-\infty,q_{1}\right]\cup\left[p_{2},b_{2}\right]\cup\left[,q_{4},\infty\right)}\right)$
$\displaystyle=$
$\displaystyle\frac{1}{2}\left(\tilde{S}_{\left[q_{1},p_{2}\right]}+\tilde{S}_{\left[q_{3},p_{1}\right]}+\tilde{S}_{\left[b_{2},q_{4}\right]}+\tilde{S}_{\left[q_{3},p_{1}\right]}-\tilde{S}_{\left[b_{2},q_{4}\right]}-\tilde{S}_{\left[q_{1},p_{2}\right]}\right)$
$\displaystyle=$
$\displaystyle\frac{c}{6}\log\left[\frac{(p_{1}-q_{3})^{2}}{\epsilon^{2}}\right]+\frac{c}{6}\phi(q_{3})\,.$
(132)

As we can observe from the second balance constraint from above which are
trivially satisfied. We may obtain $q_{1}$ from the first balance condition
and it reduces in terms of the partition point $p_{2}$. Therefore, we need to
minimize the BPE for this phase over the partition point $p_{2}$ and it
provides following value of $q_{1}$ and $p_{2}$, which are the same as (127).
The BPE for this phase is given by

BPE
$\displaystyle=\mathcal{I}(A\text{Ir}(A),B\text{Ir}(B)B_{1}\text{Ir}(B_{1})B_{2})|_{balanced}$
(133)
$\displaystyle=\frac{c}{6}\log\left[\frac{(a-b)\left(a_{2}-b_{1}\right){}\left(a_{2}-p_{2}\right){}\left(b_{1}-p_{2}\right){}}{\left(p_{2}\left(a_{2}+b_{1}\right)+a\left(a_{2}+b_{1}-2p_{2}\right)-2a_{2}b_{1}\right){}^{2}\left(a_{2}\left(b-2b_{1}+p_{2}\right)+b_{1}p_{2}+b\left(b_{1}-2p_{2}\right)\right){}}\right]\,.$
(134)

The above result of BPE exactly coincide with the EWCS in (126) by putting the
value of $q_{1}$.

#### Phase-D2c

As displayed in Fig.18, the intervals $A_{2}$ and $B_{2}$ also admit
entanglement islands. In this context, the entanglement wedge in this phase
become disconnected implying the EWCS to be vanished, which indicates a zero
BPE. In this phase the balance requirements are described as follows

$\displaystyle\mathcal{I}(A\text{Ir}(A),B\text{Ir}(B)B_{1}\text{Ir}(B_{1})B_{2}\text{Ir}(B_{2}))=$
$\displaystyle\mathcal{I}(B\text{Ir}(B),A_{1}\text{Ir}(A_{1})A_{2}\text{Ir}(A_{2})A\text{Ir}(A))\,,$
(135)
$\displaystyle\mathcal{I}(A_{1}\text{Ir}(A_{1}),B\text{Ir}(B)B_{1}\text{Ir}(B_{1})B_{2}\text{Ir}(B_{2}))=$
$\displaystyle\mathcal{I}(B_{1}\text{Ir}(B_{1}),A_{1}\text{Ir}(A_{1})A_{2}\text{Ir}(A_{2})A\text{Ir}(A))\,.$

Where

$A$ $B_{1}$ $B_{1}$ $A_{1}$ $A_{2}$ $B_{2}$
${\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}B}$Ir(A)Ir($B_{2}$)Ir(B)
Ir($B_{2}$)
$a_{1}$$a_{2}$$p_{2}$$b_{1}$$b_{2}$$b$$p_{1}$$a$$q_{3}$$q_{5}$$q_{4}$$q_{1}$$q_{6}$$q_{2}$Ir($B_{1}$)Ir($B_{1}$)Ir($A_{1}$)Ir($A_{1}$)Ir($A$)Ir($A_{2}$)Ir($B$)Ir($B_{2}$)Ir($A_{2}$)
Figure 18: The figure illustrates entanglement islands regions for all the
intervals located in Weyl $CFT_{2}$ indicated as $\text{Ir}(A)=[q_{2},q_{3}]$,
$\text{Ir}(B)=[q_{5},q_{6}]$, $\text{Ir}(A_{1})=[q_{3},q_{4}]$,
$\text{Ir}(B_{1})=[q_{4},a]\cup[b,q_{5}]$, $\text{Ir}(A_{2})=[q_{1},q_{2}]$
and $\text{Ir}(B_{2})=(-\infty,q_{1}]\cup[q_{6},\infty)$.

$\displaystyle\mathcal{I}(A\text{Ir}(A)$
$\displaystyle,B\text{Ir}(B)B_{1}\text{Ir}(B_{1})B_{2}\text{Ir}(B_{2}))$
$\displaystyle=\frac{1}{2}(\tilde{S}_{A\text{Ir}(A)A_{1}\text{Ir}(A_{1})}+\tilde{S}_{A\text{Ir}(A)A_{2}\text{Ir}(A_{2})}-\tilde{S}_{A_{1}\text{Ir}(A_{1})}-\tilde{S}_{A_{2}\text{Ir}(A_{2})})$
$\displaystyle=\frac{1}{2}\left(\tilde{S}_{\left[q_{2},q_{4}\right]\cup\left[p_{1},a_{2}\right]}+\tilde{S}_{\left[q_{1},q_{3}\right]\cup\left[a_{1},p_{2}\right]}-\tilde{S}_{\left[q_{3},q_{4}\right]\cup\left[p_{1},a_{1}\right]}-\tilde{S}_{\left[q_{1},q_{2}\right]\cup\left[p_{2},a_{2}\right]}\right)$
$\displaystyle=\frac{1}{2}\left(\tilde{S}_{\left[q_{4},p_{1}\right]}+\tilde{S}_{\left[q_{2},a_{2}\right]}+\tilde{S}_{\left[q_{3},a_{1}\right]}+\tilde{S}_{\left[q_{1},p_{2}\right]}-\tilde{S}_{\left[q_{4},p_{1}\right]}-\tilde{S}_{\left[q_{3},a_{1}\right]}-\tilde{S}_{\left[q_{2},a_{2}\right]}-\tilde{S}_{\left[q_{1},p_{2}\right]}\right)$
$\displaystyle=0\,,$ (136) $\displaystyle\mathcal{I}(B\text{Ir}(B)$
$\displaystyle,A_{1}\text{Ir}(A_{1})A_{2}\text{Ir}(A_{2})A\text{Ir}(A))$
$\displaystyle=\frac{1}{2}(\tilde{S}_{B\text{Ir}(B)B_{1}\text{Ir}(B_{1})}+\tilde{S}_{B\text{Ir}(B)B_{2}\text{Ir}(B_{2})}-\tilde{S}_{B_{1}\text{Ir}(B_{1})}-\tilde{S}_{B_{2}\text{Ir}(B_{2})})$
$\displaystyle=\frac{1}{2}\left(\tilde{S}_{\left[q_{4},p_{1}\right]\cup\left[b_{1},q_{6}\right]}+\tilde{S}_{\left(-\infty,q_{1}\right]\cup\left[p_{2},b_{2}\right]\cup\left[q_{5},\infty\right)}-\tilde{S}_{\left[b_{2},q_{5}\right]\cup\left[q_{4},p_{1}\right]}-\tilde{S}_{\left(-\infty,q_{1}\right]\cup\left[p_{2},b_{1}\right]\cup\left(q_{6},\infty\right]}\right)$
$\displaystyle=\frac{1}{2}\left(\tilde{S}_{\left[b_{1}q_{6}\right]}+\tilde{S}_{\left[q_{4},p_{1}\right]}+\tilde{S}_{\left[b_{2},q_{5}\right]}+\tilde{S}_{\left[q_{1},p_{2}\right]}-\tilde{S}_{\left[b_{2},q_{5}\right]}-\tilde{S}_{\left[q_{4},p_{1}\right]}-\tilde{S}_{\left[b_{1},q_{6}\right]}-\tilde{S}_{\left[q_{1},p_{2}\right]}\right)$
$\displaystyle=0\,.$ (137)

It is obvious that the balanced requirements (135) are satisfied and the BPE
for this phase is just zero.

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