# Holographic dual of defect CFT with corner contributions

Xinyu Sun  Institute for Advanced Study, Tsinghua University, Beijing 100084,
China  Shao-Kai Jian<EMAIL_ADDRESS>Department of Physics and Engineering
Physics, Tulane University, New Orleans, Louisiana, 70118, USA

###### Abstract

We study defect CFT within the framework of holographic duality, emphasizing
the impact of corner contributions. We model distinct conformal defects using
interface branes that differ in tensions and are connected by a corner.
Employing the relationship between CFT scaling dimensions and Euclidean
gravity actions, we outline a general procedure for calculating the anomalous
dimensions of defect changing operators at nontrivial cusps. Several
analytical results are obtained, including the cusp anomalous dimensions at
big and small angles. While $1/\phi$ universal divergence appears for small
cusp angles due to the fusion of two defects, more interestingly, we uncover a
bubble phase rendered by a near zero angle cusp, in which the divergence is
absent.

Introduction.—Defect conformal field theory (CFT) appears across various
disciplines in physics [1], ranging from Kondo effect [2, 3] in solid-state
physics to the description of D-branes in the string theory [4]. Fathoming
defect CFT is not only of great importance but also has broad applications.
Consider a general defect CFT on a circle $S^{1}$, hosting two conformal
defects, as illustrated in Fig. 1 (a). Two CFT Hamiltonian, $H_{1}$ and
$H_{2}$, are defined on lattice sites with green and purple colors,
respectively. The two defects of type $a$ and $b$ separating the bulk CFT are
denoted by the red and blue bonds, respectively. The full Hamiltonian reads,

$\begin{split}H_{ab}&=\sum_{r=1}^{L_{1}}h_{1}(r,r+1)+\sum_{r=L_{1}+1}^{L_{1}+L_{2}}h_{2}(r,r+1)\\\
&+\kappa_{a}h_{a}(L_{1}+L_{2},1)+\kappa_{b}h_{b}(L_{1},L_{1}+1)\,.\end{split}$
(1)

Here, $h_{1}$ and $h_{2}$ denote local terms in $H_{1}$ and $H_{2}$ defined in
$r=1,...,L_{1}$ and $r=L_{1}+1,...,L_{1}+L_{2}$. $h_{a}$ and $h_{b}$ with the
strength $\kappa_{a}$ and $\kappa_{b}$ are the local terms of defect type $a$
and $b$, respectively. The total number of sites is $L=L_{1}+L_{2}$, and
$r+L=r$ due to the periodic boundary condition. The angle spanned by these two
defects is $\phi=2\pi L_{1}/L$.

In the low-energy limit, this lattice model under RG will flow to a defect
CFT. The bulk CFT Hamiltonian $H_{1}$ and $H_{2}$ correspond to two CFTs
separated by different conformal defects in the most general case. The ground
state energy has the following form [5],

$E_{\text{GS}}=\epsilon_{\text{bulk}}L+\epsilon_{\text{defect}}+\frac{\Delta_{ab}^{12}(\phi)}{L}+\mathcal{O}(L^{-2})\,,$
(2)

in which $\epsilon_{\text{bulk}}$ and $\epsilon_{\text{defect}}$ denote the
bulk and defect energy density. The third term that is proportional to $1/L$
is universal, and it gives the scaling dimension, $\Delta_{ab}^{12}(\phi)$, of
the defect changing operator with an additional cusp $\phi$. While a partial
list of known examples includes a few minimal models [6, 7, 8, 9, 10, 11, 12,
13, 14, 15, 16, 17, 18], the $O(N)$ model [19, 20, 21, 22, 23, 24], and
supersymmetric Yang-Mills models [25, 26, 27, 28, 29, 30, 31], the solution to
this defect problem is relatively limited, even in 2D. For instance, the
product of two minimal models in the folding trick is no longer a minimal
model, and the conformal boundary condition [32, 33] is not known in general.

Figure 1: Illustration of (a) the defect CFT model and (b) the holographic
model. See the main text for detailed explanations.

In this paper, we propose a bottom-up holographic dual [34, 35, 36] to gain a
general understanding. More explicitly, the CFT is dual to a bulk gravity and
each defect is dual to an interface brane [37, 38, 39, 40], as illustrated in
Fig. 1 (b). $B_{1,2}$ ($M_{1,2}$) denotes the asymptotic boundary (the bulk
region) of the AdS gravity dual to the CFTs. Two branes $W_{a,b}$ are dual to
the defects $a,b$, and they intersect at a corner with internal angles
$\theta_{0}^{1,2}$ in region $M_{1,2}$. It should be noted that the corner
angle $\theta_{0}^{1,2}$ is distinct from the cusp angle $\phi$. Holographic
models for defect CFT have been widely explored [41, 42, 43, 44, 45, 46, 47,
48, 49, 50, 51]. Similar setups either in a single CFT [48] or without a
corner contribution [47] were considered, but we obtain new results as
summarized in the following. We outline a general procedure of solving the
brane solution and getting $\Delta_{ab}^{12}(\phi)$. This gives a universal
methodology to obtain arbitrary $\Delta_{ab}^{12}(\phi)$ numerically in
holography. For some special cases, we can make simplifications to get
analytical results. Specifically, we focus on (a) the defect changing operator
between different conformal defects, $\Delta_{ab}^{11}(\pi)$, without a cusp,
see Eq. (11), and (b) the cusp anomalous dimension for the same kind of defect
between the distinct CFTs $\Delta_{aa}^{12}(\phi)$, see Eq. (12) [Eq. (15)]
for a big (small) cusp angle, or the same CFTs with a nontrivial corner angle
$\Delta_{ab}^{11}(\phi)$, see Eq. (16). More interestingly, we uncover a
bubble-solution phase at near zero cusp angle, $\phi\approx 0$, where the
fusion between defects is unconventional.

Holographic duality and general solutions.— We consider a dual model of Eq.
(1) described by 3D Euclidean gravity action:

$I_{\rm tot}=I_{\rm EH}+I_{T}+I_{\rm surface}+I_{\rm corner}\,.$ (3)

Here $I_{\rm
EH}=-\frac{1}{2}\sum_{i=1,2}\int_{M_{i}}\sqrt{g_{i}}\left(R_{i}+\frac{2}{l_{i}^{2}}\right)$
is the bulk EH action for the two AdS regions $M_{i}$ with $g_{i}$, $R_{i}$,
and $l_{i}$ being the determinant of metric, the Ricci scalar, and the AdS
radius, respectively. The boundaries are $\partial
M_{i}=S_{a}^{i}+S_{b}^{i}+B_{i}$, where $S_{a,b}^{i}$ are the interface branes
and $B_{i}$ is an asymptotic boundary.
$I_{T}=\sum_{\alpha=a,b}\int_{W_{\alpha}}\sqrt{\hat{g}_{\alpha}}\ T_{\alpha}$,
and $I_{\rm surface}=-\sum_{i=1,2}\int_{\partial M_{i}}\sqrt{\hat{g}_{i}}\
K_{i}$ describes the interface brane $W_{a,b}$,
$W_{a,b}=S^{1}_{a,b}=S^{2}_{a,b}$, with $\hat{g}$, $K$ and $T$ denoting the
induced metric, the external curvature, and the corresponding tension,
respectively. The corner contribution joining two interface branes is
described by

$\begin{split}I_{\rm corner}=-\sum_{i=1,2}\int_{S_{a}^{i}\cap
S_{b}^{i}}\sqrt{\hat{g}}\left(\theta_{0}^{i}-\theta^{i}\right).\end{split}$
(4)

with $\theta_{0}^{i}$ being the angle spanning between the tangents of the two
branes at the corner. Note that we omit the corner at the asymptotic boundary
and a counter term $I_{\rm c.t.}$, the detail of which can be found in the
Supplemental Material.

Consider a gravity dual for CFT in a long strip of width $L_{i}$, the AdS bulk
solution is

${\rm d}s_{i}^{2}=(r^{2}-M_{i}l_{i}^{2}){\rm d}\tau^{2}+\frac{l_{i}^{2}{\rm
d}r^{2}}{r^{2}-M_{i}l_{i}^{2}}+r^{2}{\rm d}x^{2},$ (5)

where $l_{1,2}$ is the radius for $AdS_{1,2}$ related to the central charge
$c_{1,2}=\frac{3l_{12}}{2G_{N}}$, where $G_{N}$ is the Newton constant, and
$M_{i}$ is the parameter which is related to the width $L_{i}$. The Euclidean
time is periodic, $\tau=\tau+T_{\text{DCFT}}^{-1}$, $T_{\text{DCFT}}^{-1}\gg
L$. It is worth explaining the physics meaning of $M_{i}$. For a pure AdS
geometry, $M_{i}<0$ determines the perimeter in the compactified coordinate,
$x=x+2\pi/\sqrt{-M_{i}}$. Nevertheless, in our geometry shown in Fig. 1 (b),
two AdS geometries are joined via interface branes, so that the compactified
coordinate is not determined by solely either $M_{1}$ or $M_{2}$, but both of
them. We will need to determine them according to the length $L_{1,2}$ in the
dual CFTs.

The two interface branes dual to the defects are given by
$(x(\sigma),r(\sigma))$ at a constant time slice, with $\sigma$ being a
parameter. Two matching conditions are [47]

$\begin{split}{\rm d}s^{2}|_{S_{\alpha}^{1}}={\rm
d}s^{2}|_{S_{\alpha}^{2}}\,,\quad(K^{1}+K^{2})_{\mu\nu}=T_{\alpha}h_{\mu\nu}\,,\end{split}$
(6)

where $\alpha=a,b$. The first equation states the metric on the brane is
continuous. The second equation relates the tension of the brane $T_{\alpha}$
to the extrinsic curvatures $K^{1,2}$ from the two regions, with $h_{\mu\nu}$
being the induced metric. The second equation gives a constraint on the
tension $T_{\text{min}}\leq T\leq T_{\text{max}}$ with
$T_{\text{min}}=|l_{1}^{-1}-l_{2}^{-1}|$,
$T_{\text{max}}=l_{1}^{-1}+l_{2}^{-1}$. With the corner term, the two
interface branes shall not only satisfy the matching condition (6), but they
will join at a tangent angle $\theta^{1,2}_{0}$. Although there are two
corners in two AdS spacetime, we can prove that two angles are not
independent, and in the following, we require that the angle in $AdS_{1}$ is
$\theta_{0}\equiv\theta_{0}^{1}$. The general procedure to proceed is to solve
the brane trajectory with given $(l_{1,2},T_{a,b},M_{1,2},\theta_{0})$, and
then match the interval $L_{1,2}$ for the two CFTs, which will then lead to an
equation for $M_{1,2}$ and $L_{1,2}$.

Without loss of generality, we assume $l_{1}<l_{2}$. After solving the
equations of motion with matching conditions (6), with a proper coordinate
$\sigma=r^{2}-Ml^{2}$, we have the differential equations
$\dot{x}_{i}^{\alpha}$ for branes trajectories with tension $T$ at each AdS
spacetime, where the explicit form is given in the Supplemental Material. Here
$i=1,2$ label the brane in $AdS_{1,2}$, and $T_{\alpha}=T_{a,b}$ for different
branes. With the condition that the angle at the corner of two branes in
$AdS_{1}$ is $\theta_{0}$, we can solve the coordinate of the crossing point
of two branes with $\sigma=\sigma_{0}$ 111Note that $\sigma\rightarrow\infty$
approaches the intersection between the interface brane and the asymptotic
boundary. Then an equation can be constructed to match $L_{1,2}$. To this end,
it is convenient to define two dimensionless variables
$\gamma=\frac{L_{1}}{L_{2}}=\frac{\phi}{2\pi-\phi}$ and
$\mu=\frac{M_{2}}{M_{1}}$, then, we have

$\gamma=\frac{I_{1}^{\prime}+\pi(1-{\rm
sgn}(I_{1}^{\prime}))}{I_{2}^{\prime}+\frac{\pi}{\sqrt{\mu}}(1-{\rm
sgn}(I_{2}^{\prime}))}\,,$ (7)

where $I_{1,2}^{\prime}$ correspond to two integrals that are related to the
brane trajectory, whose expression can be found in the Supplemental Material.
In the following, we will use $\gamma$ and $\phi$ interchangeably. In general,
Eq. (7) is a complicated integral function about $\mu$ and cannot be solved
analytically. However, in some special cases, we can solve it to get
$\mu(\gamma)$, which is shown below.

Now, let’s discuss the effect of the corner term Eq. (4). In AdS/BCFT, it is
known that a corner term is necessary when tensions are different $T_{a}\neq
T_{b}$, and plays the role of boundary condition changing operator. In our
model, a corner term plays the role of defect changing operator. Specifically,
a nontrivial defect changing operator exists for $\theta_{0}\neq\pi$. To
extract the scaling dimension of defect changing operator, we evaluate the
onshell action:

$I_{\rm tot}=\frac{M_{1}l_{1}L_{1}+M_{2}l_{2}L_{2}}{2T_{\rm DCFT}}\,.$ (8)

Although $I_{\rm tot}$ seems to have a simple form, the complexity is hidden
in $M_{1,2}$ as we need to express them as
$(l_{1,2},T_{a,b},L_{1,2},\theta_{0})$ by solving Eq. (7). As the defect CFT
is dual to the gravitational model with interface branes joined by a corner,
this partition function reveals the information of $\Delta_{ab}^{12}(\phi)$.
In particular,

$I_{\rm tot}=\frac{E_{ab}^{12}(\gamma)}{T_{\rm
DCFT}}\,,~{}\Delta^{12}_{ab}(\phi)=E^{12}_{ab}(\gamma)-\frac{1}{2}(E^{12}_{aa}+E^{12}_{bb})\,,$
(9)

in which $E_{aa}^{12}\equiv E_{aa}^{12}(1)$ is the eigenvalue at $\gamma=1$.
With the procedure outlined above, a general $\Delta^{12}_{ab}(\phi)$ can be
solved numerically. Next, we will discuss several special cases with simpler
analytical expressions.

Defect changing operator.— Consider the defect changing operator between
different conformal defects without a cusp in the same CFT: $\gamma=1$ and
$l_{1}=l_{2}=l$, the geometry has a $\mathbb{Z}_{2}$ symmetry, so we know
$\mu=1$ without solving Eq. (7). Plugging it into Eq. (8), we have $I_{\rm
tot}=-\frac{2l}{T_{\rm DCFT}L}\left[\pi+{\rm
sgn}(\theta_{0}-\pi)(\pi-\arccos{\beta})\right]^{2}$ with

$\beta=\frac{\cos{\theta_{0}}+(lT_{a}/2)(lT_{b}/2)}{\sqrt{(1-(lT_{a}/2)^{2})(1-(lT_{b}/2)^{2})}}\,,$
(10)

and consequently, $E_{ab}^{12}(\gamma=1)=I_{\text{tot}}T_{\text{DCFT}}$. After
subtracting $E_{aa}=-2\pi^{2}l/L$ for $T_{a}=T_{b}$ and $\theta_{0}=\pi$, we
arrive at

$\Delta_{ab}^{11}(\pi)=\frac{2l}{L}(\pi^{2}-\arccos^{2}{\beta}).$ (11)

in which we require $\theta_{0}\leq\pi$ to satisfy
$\Delta_{ab}>\Delta_{aa}=0$.

To better understand the effect of the corner contribution, we set
$T_{a}=T_{b}=T$ which leads to
$\Delta_{ab}|_{T_{\alpha}=T}=\frac{2l}{L}[\pi^{2}-\arccos^{2}{(\frac{4\cos{\theta_{0}}+(Tl)^{2}}{4-(Tl)^{2}})}]$
with the constraint $0<T<T_{c}=2\sin{(\theta_{0}/2)}$. Note that, although
$T_{a}=T_{b}$, it still corresponds to two different defects due to a
nontrivial $\theta_{0}$. The scaling dimension of the defect changing operator
is plotted in Fig. 2 (a). A greater tension $T$ and a larger deviation from
$\theta_{0}=\pi$ will lead to a greater scaling dimension of defect changing
operators. Specially, we have
$\Delta_{ab}|_{T_{\alpha}=T_{c}}=\frac{2l}{L}\pi^{2}$ and
$\Delta_{ab}|_{T_{\alpha}=0}=\frac{2l}{L}(\pi^{2}-\theta_{0}^{2})$.

Cusp anomalous dimension.— Apart from the defect changing operator given
above, interesting conformal data exists even for the same defect: the cusp
anomalous dimension. The same types of defects means $T_{a}=T_{b}=T$ and
$\theta_{0}=\pi$. Then we look at a big angle $\phi\sim\pi$ and small angle
$\phi\sim 0,2\pi$, respectively. Consider $\phi\sim\pi$,
$\gamma=1+\delta_{\gamma}$ and $\delta_{\gamma}\ll 1$. We can check $\mu=1$ is
a solution for $\gamma=1$. Therefore, we can expand $\mu=1+\delta_{\mu}$ with
$\delta_{\mu}\ll 1$, to get
$\delta_{\mu}=\frac{2l_{1}l_{2}T}{l_{1}+l_{2}+l_{1}l_{2}T}\delta_{\gamma}+\mathcal{O}(\delta_{\gamma}^{2})$,
where the expression of the second order $\mathcal{O}(\delta_{\gamma}^{2})$ is
given in the Supplemental Material. With this solution and Eq. (8), the cusp
anomalous dimension near $\phi=\pi$ reads

$\Delta_{aa}^{12}(\phi)=-\frac{l_{1}}{L}\left[\pi\left(\frac{l_{2}}{l_{1}}-1\right)\delta_{\phi}+\frac{2(T-T_{\rm
min})}{T+T_{\rm max}}\delta_{\phi}^{2}\right].$ (12)

where $\delta_{\phi}=\phi-\pi$. When $l_{1}=l_{2}=l$ with a $\mathbb{Z}_{2}$
symmetry, the linear term in Eq. (12) vanishes, and the quadratic term is
negative, consistent with reflection positivity [53, 54, 24].

Next, consider the cusp anomalous dimension near $\phi\rightarrow 0$ or
$\phi\rightarrow 2\pi$ for the same defect: $T_{a}=T_{b}=T$ and
$\theta_{0}=\pi$. Note that these two limits are different when the two CFTs
are distinct $l_{1}\neq l_{2}$. As $\phi\rightarrow 0$ implies
$\gamma\rightarrow 0$, we simplify Eq. (7) as
$\sqrt{\mu}=\frac{2\pi}{B(T,l_{1},l_{2})}\gamma+\mathcal{O}(\gamma^{2})$,
where
$B(T,l_{1},l_{2})=2\pi\Theta\left(l_{2}T-1\right)-\Xi_{1}(T,l_{1},l_{2})$ with

$\Xi_{u_{0}}(T,l_{1},l_{2})=\sqrt{\frac{4v}{-u}}\left[K\left(u_{0},v\right)+\frac{l_{1}^{2}l_{2}^{2}T^{2}-l_{1}^{2}}{l_{2}^{2}}\Pi\left(u_{0},u,v\right)\right],$
(13)

and $u_{0}=1$. We define $v=\frac{(T-T_{\rm max})(T+T_{\rm min})}{(T+T_{\rm
max})(T-T_{\rm min})}$, $u=l_{1}^{2}(T_{\rm max}-T)(T+T_{\rm min})$,
$\Theta(\cdot)$ is the step function, and $K(\cdot,\cdot)$
($\Pi(\cdot,\cdot,\cdot)$) is the incomplete elliptic integral of the first
(third) kind. With this solution and Eq. (8), we can obtain the onshell action
explicitly,

$I_{\rm tot}=-\frac{1}{2T_{\rm DCFT}L}\left(\frac{l\cdot
B(T,l_{1},l_{2})^{2}}{\gamma}+\mathcal{O}(1)\right).$ (14)

The other limit, $\phi\rightarrow 2\pi$, implies instead
$\gamma\rightarrow\infty$, which can be solved similarly. With Eq. (9), the
cusp anomalous dimension is given by

$\Delta_{aa}^{12}(\phi)=\begin{cases}-\frac{l_{2}B(T,l_{2},l_{1})^{2}}{2L}\frac{2\pi}{2\pi-\phi}&\phi\rightarrow
2\pi\\\ -\frac{l_{1}B(T,l_{1},l_{2})^{2}}{2L}\frac{2\pi}{\phi}&\phi\rightarrow
0\end{cases}\,.$ (15)

The cusp anomalous dimension diverges as $1/\phi$ or $1/(2\pi-\phi)$, under
the limit [24, 55]. On the CFT side, the two defects fuse into a trivial
defect. The divergent behavior can be interpreted the Casimir energy and the
prefactor is the fusion coefficient. It is positive or negative for repulsive
and attractive interaction.

Figure 2: (a) Scaling dimension of the defect changing operator with
$l_{1}=l_{2}=l$ and $T_{a}=T_{b}=T$ with different $\theta_{0}$. (b,c) On-
shell action for (b) $\phi\rightarrow 2\pi$ limit and (c) $\phi\rightarrow 0$
limit without corners. (d) On-shell action for $\phi\rightarrow 0$ limit with
a corner. The solid (dash) curves are numerical (analytical) results.

We plot the numerical results obtained by our general procedure and analytical
results in Fig. 2 (b) and (c), in which Fig. 2 (b),(c) show the limit
$\phi\rightarrow 2\pi,0$, respectively. For $\phi\rightarrow 2\pi$, the two
results, i.e., numerical (analytical) results shown by solid (dashed) curves,
agree quite well in Fig. 2 (b). However, for $\phi\rightarrow 0$, some results
agree, but some do not, for instance, the green solid curve with
$2l_{1}=l_{2}=2$, $T=0.75$ shows a significant deviation from the analytical
solution plotted in a dashed line. The reason is that there can emerge a
bubble solution in the limit $\phi\rightarrow 0$, as is discussed next.

Bubble phase and exotic fusion.— As mentioned above, for $\phi\rightarrow 0$,
two defects into one, and lead to a divergent behavior. However, it is not
always the case. In the following, we consider more general solutions, and
also discuss the role of the corner contribution with $\theta_{0}\neq\pi$.

First, we consider the same defect, $T_{a}=T_{b}$, $l_{1}\neq l_{2}$ and
$\theta_{0}=\pi$. Given $B(T,l_{1},l_{2})>0$, the solution
$\sqrt{\mu}=\frac{2\pi}{B(T,l_{1},l_{2})}\gamma$ becomes $2\pi
L_{1}/L_{2}=B(T,l_{1},l_{2})\sqrt{M_{2}/M_{1}}$. It shows that when the length
of one CFT shrinks to zero, $L_{1}\ll L_{2}$, its dual AdS also shrinks to
zero $M_{1}\gg M_{2}$. A schematic plot of the solution is given in Fig. 3 (a)
right panel. However, if $B(T,l_{1},l_{2})\rightarrow 0$, that the length of
one CFT shrinks to zero, $L_{1}\ll L_{2}$, does not imply its dual AdS shrinks
to zero. Instead, we find that at $B(T,l_{1},l_{2})=0$, the solution of Eq.
(7) at $\gamma\rightarrow 0$ is given by a finite
$M_{2}/M_{1}\rightarrow\mu_{0}>0$. A schematic plot of the two geometries
given in Fig. 3 (a), showing a qualitative difference between the two
solutions. The one with $M_{2}/M_{1}\rightarrow 0$ is termed as the no-bubble-
solution phase, and the one with $M_{2}/M_{1}\rightarrow\mu_{0}>0$ is termed
as the bubble-solution phase. Their phase boundary is precisely given by
$B(T,l_{1},l_{2})=0$. We can show that $B(T,l_{1},l_{2})=0$ is possible only
when $l_{1}<l_{2}$. It means that the dual of CFTs with a greater central
charge can support a bubble of CFTs with a smaller central charge, but the
opposite cannot occur. With independent parameters $(l_{1}/l_{2},Tl_{1})$ and
$l_{1}/l_{2}\in[0,1],Tl_{1}\in[1-l_{1}/l_{2},1+l_{1}/l_{2}]$, the phase
diagram is plotted in Fig. 3 (b).

We discuss the implication for the cusp anomalous dimension. For
$B(T,l_{1},l_{2})>0$, namely, in the no-bubble-solution phase, the cusp
anomalous dimension given in Eq. (15) is valid. However, in the bubble-
solution phase, the solution with $\mu=\mu_{0}$ renders a finite onshell
action without the $1/\phi$ divergence, in contrast to Eq. (14). Therefore, we
expect the existence of the bubble-solution phase corresponds to an exotic
fusion of two defects.

We are ready to explain Fig. 2 (b) and (c). For parameters located in the no-
bubble-solution phase, the numerical and analytical results agree, whereas for
parameters located in the bubble-solution phase, specially, for
$l_{2}=2l_{1}=2$, $T=0.75$, the cusp anomalous dimension converges at
$\phi\rightarrow 0$ shown by the solid curve.

Figure 3: (a) Configurations of the bubble-solution phase on the left panel
and no-bubble-solution phase on the right panel, where the blue (orange)
region illustrates $AdS_{1(2)}$. (b) Phase diagram from small cusp angles
$\phi\sim 0$ for $\theta_{0}=\pi$. (c) Phase diagram from small cusp angles
$\phi\sim 0$ with a corner.

Next, we consider a more general case with a corner contribution
$\theta_{0}\neq\pi$ that indicates a nontrivial defect changing operator. We
further assume $l_{1}=l_{2}=l$ and $T_{a}=T_{b}=T$. Then, similar to the
previous discussion, In the limit $\phi\rightarrow 0$, or equivalently
$\gamma\rightarrow 0$, we obtain
$\sqrt{\mu}=\frac{2\pi}{\tilde{B}(T,l,\theta_{0})}\gamma+\mathcal{O}(\gamma^{2})$,
where $\tilde{B}(T,l,\theta_{0})$ is defined in the Supplemental Material. For
$\theta_{0}<\pi$ and $T>1$, it can be simplified to
$\tilde{B}(T,l,\theta_{0})=2\pi-2\Xi_{1}(T,l,l)+\Xi_{u_{0}}(T,l,l)$, where
$u_{0}=[\frac{l^{2}T^{2}+2\cos{\theta_{0}}-2}{(lT-2)(\sqrt{(l^{2}T^{2}+l^{2}T^{2}\cos{\theta_{0}}-2)(\cos{\theta_{0}}-1)}-lT\cos{\theta_{0}})}]^{1/2}$.
Again, $\tilde{B}(T,l,\theta_{0})=0$ defines the phase transition between the
no-bubble-solution phase and the bubble-solution phase. With $lT\in[0,2]$ and
$\cos{\theta_{0}}\in[-1,1-(lT)^{2}/2]$ which is required to have a solution,
the phase diagram is shown in Fig. 3 (c). From the phase diagram, we can find
that there is no bubble-solution phase for $lT<1$.

The asymptotic behavior of the cusp anomalous dimension for $\phi\rightarrow
0$ in the no-bubble-solution phase is,

$\Delta_{ab}(\phi)=-\frac{l\tilde{B}(T,l,\theta_{0})^{2}}{2L}\frac{2\pi}{\phi}\,,$
(16)

which has a $1/\phi$ singularity from the Casimir energy. Finally, we compare
the numerical and analytical results of the anomalous dimension and plot them
in Fig. 2 (d).

It is illustrative to consider $lT=1$. In this case,
$\tilde{B}(1/l,l,\theta_{0})=\theta_{0}-\theta_{0}^{c}$ with
$\theta_{0}^{c}=\frac{2}{\sqrt{3}}\cdot K\left(-\frac{1}{3}\right)$. Hence,
the bubble-solution phase and the no-bubble-solution phase exist for
$\theta_{0}\in(\frac{\pi}{3},\theta_{0}^{c})$ and
$\theta_{0}\in(\theta_{0}^{c},\frac{5\pi}{3})$, respectively.

Concluding remarks.— To illustrate the connection between our results and
concrete CFT models, we consider the defect Ising model
$H=-\frac{1}{2}\sum^{\prime}_{r}[\sigma^{z}(r)+\sigma^{x}(r)\sigma^{x}(r+1)]-\frac{1}{2}\kappa_{a}\sigma^{x}(L_{1}+L_{2})\sigma^{x}(1)-\frac{1}{2}\kappa_{b}\sigma^{x}(L_{1})\sigma^{x}(L_{1}+1)$,
where $\sigma^{\mu}(r)$ denotes the Pauli matrix at site $r$ and the first
summation omits the terms at the two defect bonds. We calculate the defect
changing operator at $\phi=\pi$ (see Supplemental Material)

$\Delta_{ab}^{11}(\pi)=\frac{1}{\pi}\left(\arctan\kappa_{a}-(-1)^{Q}\arctan\kappa_{b}\right)^{2}\,.$
(17)

where $Q=0$ ($1$) denotes the symmetric (antisymmetric) sector. We observe
that in the $Q=1$ sector, when $\kappa_{a}=\kappa_{b}=\kappa$,
$\Delta_{ab}^{11}(\pi)=\frac{4}{\pi}\arctan^{2}\kappa$ still nontrivially
depends on $\kappa$. Note that this happens in the holographic model when
$\theta_{0}\neq\pi$. Hence, the effect of a nontrivial corner term is similar
to the different sector $Q=0,1$. Further investigation on the corner angle
$\theta_{0}$ on the CFT side is important.

We also consider the cusp anomalous dimension of the same types of defects in
a weak defect limit (see Supplemental Material)

$\Delta_{aa}(\phi)=-\frac{1}{2L}\frac{1}{\pi}\left(\frac{\pi-\phi}{\sin{\phi}}-1\right)\delta\kappa^{2}\,,\quad\kappa=1-\delta\kappa\,.$
(18)

It is quadratic near $\phi\sim\pi$ and $1/\phi$ divergent near $\phi\sim 0$
both with a negative prefactor, consistent with our result in the no-bubble-
solution phase.

In the exotic fusion, if two defects are the same with $\theta_{0}=\pi$, the
bubble region will have a smaller AdS radius, and act as a true vacuum [47].
It is exotic because interface and anti-interface do not fuse to a trivial
identity. More interestingly, we find that a nontrivial corner term
$\theta_{0}\neq\pi$ can also support a bubble solution even with $l_{1}=l_{2}$
and $T_{a}=T_{b}$. The exotic fusion on the CFT side is also an interesting
open question.

Acknowledgement. XS acknowledges the support from the Lavin-Bernick Grant
during his visit to Tulane University, where the work was conducted. The work
of SKJ is supported by a startup fund at Tulane University.

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## Supplemental Material

###  .1 General AdS space with defects

In this section, we construct the holographic dual to the defect CFT described
in the main text, and outline a general procedure for solving the geometry. To
have two distinct interface branes, a corner term at their intersection is
necessary. Collecting all terms [47, 48] with $8\pi G_{N}=1$, the Euclidean
action reads

$I_{\rm tot}=I_{\rm EH}+I_{T}+I_{\rm surface}+I_{\rm corner}+I_{\rm c.t.},$
(S1a) where $I_{\rm
EH}=-\frac{1}{2}\sum_{i=1,2}\int_{M_{i}}\sqrt{g_{i}}\left(R_{i}+\frac{2}{l_{i}^{2}}\right),$
(S1b) $I_{T}=\sum_{\alpha=a,b}\int_{W_{\alpha}}\sqrt{\hat{g}_{\alpha}}\
T_{\alpha},$ (S1c) $I_{\rm surface}=-\sum_{i=1,2}\int_{\partial
M_{i}}\sqrt{\hat{g}_{i}}\ K_{i},$ (S1d) $I_{\rm
corner}=\sum_{\alpha=a,b}\int_{B_{1}\cap
B_{2}}\sqrt{\hat{g}}\left(\theta_{\alpha}-\pi\right)-\sum_{i=1,2}\int_{S_{a}^{i}\cap
S_{b}^{i}}\sqrt{\hat{g}}\left(\theta_{0}^{i}-\theta^{i}\right),$ (S1e) $I_{\rm
c.t.}=\sum_{i=1,2}\frac{1}{l_{i}}\int_{B_{i}}\sqrt{\hat{g}_{i}}-\int_{B_{1}\cap
B_{2}}\sqrt{\hat{g}}\left(\theta_{1}+\theta_{2}\right),$ (S1f)

where $\alpha=a,b$ labels different interface branes between two AdS space
labeled by $i=1,2$. Here the boundary of $M_{i}$ is $\partial
M_{i}=S_{a}^{i}+S_{b}^{i}+B_{i}$, where $S_{a,b}^{i}$ are two branes of AdS
bulk $M_{i}$, and $B_{i}$ is its asymptotic boundary. We also denote the
interface brane as $W_{a,b}$ with corresponding tension $T_{a,b}$. Besides, we
specify the direction of each surface that the direction of extrinsic
curvatures, $K_{1,2}$, points outside the AdS bulk. There are also
codimension-2 terms in the action. $B_{1}\cap B_{2}$ corresponds to the
intersection of asymptotic boundaries, with $\theta_{\alpha}$ being the
corresponding angle. $S_{a}^{i}\cap S_{b}^{i}$ corresponds to the corner with
an angle $\theta^{i}$ in the intersection of the two branes labeled by
$i=1,2$. The counter terms are included for completeness [47].

With the saddle point approximation, we have two equations of motion (e.o.m.)
for the branes and the corner on them.

$\left(K^{1}+K^{2}\right)_{\mu\nu}=T_{\alpha}h_{\mu\nu},$ (S2a)
$\theta^{i}=\theta_{0}^{i},$ (S2b)

where $\alpha=a,b$ for two branes which are dual to different defects and have
different tension $T_{\alpha}$, and the e.o.m. from the corner term requires
two branes to form an angle $\theta^{i}_{0}$, which may deviate from $\pi$
corresponding to a nontrivial corner on branes. We will use the first equation
to solve the brane trajectory with undetermined parameters of the AdS
spacetime, and then the second equation (S2b) will become a constraint for the
parameters.

In the following, we use a similar convention of Ref. [47]. The metrics of the
two AdS spacetime are

${\rm d}s^{2}=(r_{i}^{2}-M_{i}l_{i}^{2}){\rm d}\tau^{2}+\frac{l_{i}^{2}{\rm
d}r_{i}^{2}}{r_{i}^{2}-M_{i}l_{i}^{2}}+r_{i}^{2}{\rm d}x_{i}^{2},$ (S3)

where $(\tau,r_{i},x_{i})$ are coordinates for two AdS spacetime labeled by
$i=1,2$, and $M_{i}<0$ corresponds to the cold phase in Ref. [47]. We denote
$\tau_{1}=\tau_{2}=\tau$ by identifying the time for two AdS spaces. With the
parameter $\sigma$ defined on the brane, we have brane coordinates
$(r_{i}=r_{i}(\sigma),x_{i}=x_{i}(\sigma))$ on a constant time slice. Because
of the continuity of metrics on the brane, ${\rm d}s^{2}=f(\sigma){\rm
d}\tau^{2}+g(\sigma){\rm d}\sigma^{2}$, we have

$r_{1}^{2}-M_{1}l_{1}^{2}=r_{2}^{2}-M_{2}l_{2}^{2}\equiv f(\sigma),$ (S4a)
$\frac{l_{1}^{2}\dot{r}_{1}^{2}}{r_{1}^{2}-M_{1}l_{1}^{2}}+r_{1}^{2}\dot{x}_{1}^{2}=\frac{l_{2}^{2}\dot{r}_{2}^{2}}{r_{2}^{2}-M_{2}l_{2}^{2}}+r_{2}^{2}\dot{x}_{2}^{2}\equiv
g(\sigma),$ (S4b)

Then, simplifying the junction condition (S2a) leads to

$\frac{\dot{x}_{1}r_{1}^{2}}{l_{1}}+\frac{\dot{x}_{2}r_{2}^{2}}{l_{2}}=-T_{\alpha}\sqrt{fg}.$
(S5)

We exploit the freedom to choose a proper coordinate $\sigma$:
$\sigma=f(\sigma)$, which leads to

$r_{i}=\sqrt{\sigma+M_{i}l_{i}^{2}}.$ (S6)

With (S4), (S5) and (S6), we can solve the brane trajectory
$x_{i}=x_{i}(\sigma)$. In the following, we drop the label $\alpha$ for
simplicity, and discuss the general property of the brane trajectory. Later,
we will use the constraint from (S2b) to solve the full geometry in which the
label $\alpha$ will be restored. We simplify the function [47] $g(\sigma)$

$g(\sigma)=T^{2}\left[4\frac{r_{1}^{2}r_{2}^{2}}{l_{1}^{2}l_{2}^{2}}-\left(T^{2}\sigma-\frac{r_{1}^{2}}{l_{1}^{2}}-\frac{r_{2}^{2}}{l_{2}^{2}}\right)^{2}\right]^{-1}\equiv\frac{T^{2}}{A\sigma^{2}+2B\sigma+C},$
(S7)

where

$A=\frac{4}{l_{1}^{2}l_{2}^{2}}-\left(T^{2}-\frac{1}{l_{1}^{2}}-\frac{1}{l_{2}^{2}}\right)^{2}=(T^{2}-T_{\rm
min}^{2})(T_{\rm max}^{2}-T^{2}),$ (S8a)
$B=\frac{2}{l_{1}^{2}l_{2}^{2}}\left(M_{1}l_{1}^{2}+M_{2}l_{2}^{2}\right)+\left(M_{1}+M_{2}\right)\left(T^{2}-\frac{1}{l_{1}^{2}}-\frac{1}{l_{2}^{2}}\right)=\left(M_{1}+M_{2}\right)T^{2}-\left(M_{1}-M_{2}\right)T_{0}^{2},$
(S8b) $C=-\left(M_{1}-M_{2}\right)^{2}.$ (S8c)

Here and in the following, we assume AdS radius $l_{1}\leq l_{2}$. Therefore,
the range of the tension is $T_{\rm min}<T<T_{\rm max}$ with $T_{\rm
min}=\frac{1}{l_{1}}-\frac{1}{l_{2}}$ and $T_{\rm
max}=\frac{1}{l_{1}}+\frac{1}{l_{2}}$. Besides, we define
$T_{0}=\sqrt{\frac{1}{l_{1}^{2}}-\frac{1}{l_{2}^{2}}}$ for later convenience.
The denominator in (S7) has two zeros, which we denote as $\sigma_{\pm}$ and
$\sigma_{+}>\sigma_{-}$. Because $A>0$ and $C<0$, we have $\sigma_{+}>0$.

Plugging (S7) into (S4b), we reach the differential equations of the branes

$(\dot{x}_{1})^{2}=l_{1}^{2}\frac{\left[M_{1}-M_{2}+\sigma\left(T^{2}+T_{0}^{2}\right)\right]^{2}}{4(l_{1}^{2}M_{1}+\sigma)^{2}\sigma(A\sigma^{2}+2B\sigma+C)},$
(S9a)
$(\dot{x}_{2})^{2}=l_{2}^{2}\frac{\left[M_{2}-M_{1}+\sigma\left(T^{2}-T_{0}^{2}\right)\right]^{2}}{4(l_{2}^{2}M_{2}+\sigma)^{2}\sigma(A\sigma^{2}+2B\sigma+C)}.$
(S9b)

There is a sign ambiguity that can be fixed by choosing one half the brane.

Supplementary Figure S1: Notations and the conventions of brane. (a) The
geometry of AdS space without the corner on branes. The left and right panels
correspond to $AdS_{1,2}$, and $\psi_{1,2}$ are related to the tension of
branes in (S127). $1$ and $2$ label the two asymptotic boundaries dual to the
two CFTs, and the positive direction of the coordinate $x$ is clockwise. A and
B are two defects that connect two different CFTs. Two branes in each AdS
space correspond to the phases E1 that includes the center $r=0$ and E2 that
excludes the center. In $AdS_{1}$, the blue arrows indicate the positive
direction of $\sigma\in[\sigma_{+},\infty)$. (b) The geometry of AdS space
with a corner on branes. In $AdS_{1}$, the two branes starting from defects A
and B intersect at C with the angle $\theta_{0}$. The tangent vectors at the
intersecting point and its symmetric point are shown with black arrows, which
are consistent with (S17), where $\iota_{A}=1,-1(\iota_{B}=-1,1)$ corresponds
to the half brane close to or further from the defect A (B). For
$\theta_{0}<\pi(>\pi)$, we zoom in the intersecting point with different signs
of $\bm{n}_{a}\times\bm{n}_{b}\cdot\bm{n}_{\tau}$ on the right upper (lower)
panel.

Namely, as shown in Fig. S1 (a), there are two halves of the brane that are
symmetric under reflection in each AdS space. It is easy to see from (S7) that
$\sigma_{+}$ is the symmetric point, which has the smallest $r_{i}$ on the
brane trajectory. Different sign choices in (S9) correspond to different
halves of brane. We take the minus sign for both $\dot{x}_{1}$ and
$\dot{x}_{2}$, so they correspond to the half brane starting from A for the
$AdS_{1}$ and the half brane starting from B for the $AdS_{2}$. We set the
counterclockwise direction as the positive direction of $x$, and from the Fig.
S1 (a), we know that $\dot{x}(\sigma_{+})<0$ $(>0)$ corresponds to an AdS
region with (without) center $r=0$, which is denoted as E1 (E2) phase [47].

With an explicit brane trajectory, we can construct equations for the AdS
spacetime parameters $M_{1,2}$. We will restore the label $\alpha=a,b$ in the
following discussion. In Fig. S1 (b), we consider the branes in $AdS_{1}$,
where the black arrows label the directions of the branes on the right panel.
Let’s denote the intersecting point coordinate as $x_{C}(\sigma_{0})$. Then we
have

$x_{C}=x_{A}+\int_{+\infty}^{\sigma_{+}}\dot{x}_{a}{\rm
d}\sigma+\iota_{A}\int_{\sigma_{+}}^{\sigma_{0}}\dot{x}_{a}{\rm d}\sigma,$
(S10)

where $\iota_{A}=+1$ ($-1$) corresponds to the crossing point locating on the
negative (positive) half of the brane for the brane starting from point A
(here the negative and positive half correspond to the signs of $\dot{x}$).
Remember that we take the minus sign for $\dot{x}$ in (S9). Similarly, for the
point B we have

$x_{C}^{\prime}=x_{B}+\int_{+\infty}^{\sigma_{+}}(-\dot{x}_{b}){\rm
d}\sigma-\iota_{B}\int_{\sigma_{+}}^{\sigma_{0}}(-\dot{x}_{b}){\rm d}\sigma,$
(S11)

where $\iota_{B}=+1$ ($-1$) corresponds to the crossing point on the negative
(positive) half of the brane for the brane starting from point B. Because of
the periodic boundary condition, $x_{C}$ and $x_{C}^{\prime}$ are related by

$x_{C}=x_{C}^{\prime}+2\pi\frac{n}{\sqrt{-M}}.$ (S12)

where $n=1$ ($n=0$) for the E1 (E2) phase with (without) the center. Relating
this to the interval in the dual CFT, we arrive at

$L=x_{A}-x_{B}=\int_{\sigma_{+}}^{+\infty}(\dot{x}^{a}+\dot{x}^{b}){\rm
d}\sigma-\int_{\sigma_{+}}^{\sigma_{0}}(\iota_{A}\ \dot{x}^{a}-\iota_{B}\
\dot{x}^{b}){\rm d}\sigma+\frac{2\pi n}{\sqrt{-M}},$ (S13)

where $a,b$ corresponds to the branes for the $AdS_{i}$ with different
tensions $T_{a,b}$. To summarize, we have for $AdS_{i}$ spacetime

$L_{i}=I_{i}+\frac{2\pi}{\sqrt{-M_{i}}}\cdot\frac{1}{2}(1-{\rm sgn}(I_{i})),$
(S14a) $I_{i}=\int_{\sigma_{+}}^{+\infty}(\dot{x}_{i}^{a}+\dot{x}_{i}^{b}){\rm
d}\sigma-\int_{\sigma_{+}}^{\sigma_{0}}(\iota_{A}\ \dot{x}_{i}^{a}-\iota_{B}\
\dot{x}_{i}^{b}){\rm d}\sigma,$ (S14b)

where we use ${\rm sgn}(I_{i})$ to express $n=0,1$, and $\dot{x}_{1}^{\alpha}$
and $\dot{x}_{2}^{\alpha}$ are

$\dot{x}_{1}^{\alpha}=-l_{1}\frac{M_{1}-M_{2}+\sigma\left(T_{\alpha}^{2}+T_{0}^{2}\right)}{2(l_{1}^{2}M_{1}+\sigma)\sqrt{\sigma(A\sigma^{2}+2B\sigma+C)}},$
(S15a)
$\dot{x}^{\alpha}_{2}=-l_{2}\frac{M_{2}-M_{1}+\sigma\left(T_{\alpha}^{2}-T_{0}^{2}\right)}{2(l_{2}^{2}M_{2}+\sigma)\sqrt{\sigma(A\sigma^{2}+2B\sigma+C)}}.$
(S15b)

Notice $A,B$ and $C$ are now functions of $T=T_{\alpha}$ from Eq. (S8).

It is straightforward to check that ${\rm sgn}(I_{i})=-1$ ($+1$) corresponds
to E1 (E2) phase, i.e.,$n=0$ $(1)$. Therefor, for later convenience, we also
introduce the following notation,

$2\delta_{i,{\rm E1}}=1-{\rm sgn}(I_{i}^{\prime}).$ (S16)

We will use these two notations interchangeably.

Here is a remark: we should be careful about different coordinates for
different AdS spacetime. But in Eq. (S14b) we only introduce one parameter
$\sigma_{0}$. It is because the coordinates of the intersecting point (corner)
on different branes are supposed to be the same due to the junction condition.

In the discussion above, we have introduced parameters $\iota_{A},\iota_{B}$
and $\sigma_{0}$. Now we construct equations to solve their values. As shown
in Fig. S1 (b), the directions of the brane $a$ and $b$ which start from
boundary points A and B are

$\bm{n}_{a}=(-\iota_{A}\
\dot{r}_{a},-\dot{x}_{a}),\quad\quad\bm{n}_{b}=(\iota_{B}\
\dot{r}_{b},\dot{x}_{b}).$ (S17)

We can get an equation from the relation
$\cos{\theta_{0}^{1}}=\bm{n}_{a}\cdot\bm{n}_{b}/|\bm{n}_{a}||\bm{n}_{b}|$.
Additionally, we need another equation to distinguish between
$\theta_{0}^{1}<\pi$ and $\theta_{0}^{1}>\pi$ as illustrated in Fig. S1 (b).
We use $\bm{n}_{a}\times(-\bm{n}_{b})\cdot\bm{n}_{\tau}>0$ ($<0$) for
$\theta_{0}^{1}<\pi$ ($>\pi$), where we regard $\bm{n}_{a}=(0,-\iota_{A}\
\dot{r}_{a},-\dot{x}_{a})$, $\bm{n}_{b}=(0,\iota_{B}\
\dot{r}_{b},\dot{x}_{b})$, and $\bm{n}_{\tau}=(1,0,0)$ as three-dimensional
vectors. These considerations give

$\theta_{0}^{1}=\pi+{\rm
sgn}[\bm{n}_{a}\times\bm{n}_{b}\cdot\bm{n}_{\tau}]\left(\pi-\arccos{\frac{\bm{n}_{a}\cdot\bm{n}_{b}}{|\bm{n}_{a}||\bm{n}_{b}|}}\right).$
(S18)

The equation above is equivalent to

$\cos{\theta_{0}^{1}}=\frac{-\left[\iota_{A}\iota_{B}+\frac{M_{1}-M_{2}+\sigma\left(T_{a}^{2}+T_{0}^{2}\right)}{\sqrt{A_{a}(\sigma-\sigma_{+}^{a})(\sigma-\sigma_{-}^{a})}}\frac{M_{1}-M_{2}+\sigma\left(T_{b}^{2}+T_{0}^{2}\right)}{\sqrt{A_{b}(\sigma-\sigma_{+}^{b})(\sigma-\sigma_{-}^{b})}}\right]}{\sqrt{\left[1+\frac{\left[M_{1}-M_{2}+\sigma\left(T_{a}^{2}+T_{0}^{2}\right)\right]^{2}}{A_{a}(\sigma-\sigma_{+}^{a})(\sigma-\sigma_{-}^{a})}\right]\left[1+\frac{\left[M_{1}-M_{2}+\sigma\left(T_{b}^{2}+T_{0}^{2}\right)\right]}{A_{b}(\sigma-\sigma_{+}^{b})(\sigma-\sigma_{-}^{b})}\right]}}.$
(S19a) ${\rm sgn}{[\theta_{0}^{1}-\pi]}=\iota_{A}\ {\rm
sgn}\left[\frac{M_{1}-M_{2}+\sigma\left(T_{b}^{2}+T_{0}^{2}\right)}{\sqrt{A_{b}(\sigma-\sigma_{+}^{b})(\sigma-\sigma_{-}^{b})}}-\iota_{A}\iota_{B}\frac{M_{1}-M_{2}+\sigma\left(T_{a}^{2}+T_{0}^{2}\right)}{\sqrt{A_{a}(\sigma-\sigma_{+}^{a})(\sigma-\sigma_{-}^{a})}}\right].$
(S19b)

It is easy to show that $(\iota_{A},\iota_{B})$ is unique for a given
$\sigma$. Then, (S19a) can be simplified and reduced to a quadratic equation
for $\sigma$ that $A^{\prime}\sigma^{2}+2B^{\prime}\sigma+C^{\prime}=0$, with

$\begin{split}A^{\prime}=&-(T_{a}^{2}+T_{b}^{2})(T_{0}^{4}+T_{a}^{2}T_{b}^{2})+4T_{a}^{2}T_{b}^{2}\left({l_{2}^{-2}}-{l_{1}^{-2}}\cos^{2}{\theta_{0}^{1}}\right)-2T_{a}T_{b}\cos{\theta_{0}^{1}}(T_{a}^{2}+T_{0}^{2})(T_{b}^{2}+T_{0}^{2}),\\\
B^{\prime}=&-T_{0}^{2}(T_{a}^{2}+T_{b}^{2})(M_{1}-M_{2})-T_{a}T_{b}(2T_{0}^{2}+T_{a}^{2}+T_{b}^{2})(M_{1}-M_{2})\cos{\theta_{0}^{1}}+2T_{a}^{2}T_{b}^{2}(M_{2}-\cos^{2}{\theta_{0}^{1}}M_{1}),\\\
C^{\prime}=&-(M_{1}-M_{2})^{2}(T_{a}^{2}+T_{b}^{2}+2T_{a}T_{b}\cos{\theta_{0}^{1}}).\end{split}$
(S20)

Its solution is nothing but the intersecting point $\sigma_{0}$. To summarize,
we can solve $\sigma_{0}$ according to (S20), and then get $\iota_{A}$ and
$\iota_{B}$ from (S19). Finally, solving (S14) we get $M_{1}$ and $M_{2}$,
i.e., the full geometry is obtained.

For simplicity, we introduce dimensionless parameters [47]:
$\gamma=L_{1}/L_{2}$ and $\mu=M_{2}/M_{1}$. Then (S20) and (S19) can be
simplified with a new variable $s=\sigma/|M_{1}|=-\sigma/M_{1}$. The equation
of $\sigma_{0}$ leads to
$A^{\prime\prime}s^{2}+2B^{\prime\prime}s+C^{\prime\prime}=0$ with

$\begin{split}A^{\prime\prime}=&A^{\prime},\\\
B^{\prime\prime}=&-T_{0}^{2}(T_{a}^{2}+T_{b}^{2})(\mu-1)-T_{a}T_{b}(2T_{0}^{2}+T_{a}^{2}+T_{b}^{2})(\mu-1)\cos{\theta_{0}^{1}}+2T_{a}^{2}T_{b}^{2}(\cos^{2}{\theta_{0}^{1}}-\mu),\\\
C^{\prime\prime}=&-(\mu-1)^{2}(T_{a}^{2}+T_{b}^{2}+2T_{a}T_{b}\cos{\theta_{0}^{1}}).\end{split}$
(S21)

With $s_{\pm}=-\sigma_{\pm}/M_{1}$ and $s_{0}=-\sigma_{0}/M_{1}$,
$\iota_{A},\iota_{B}$ are given by

$\begin{split}\iota_{A}\cdot\iota_{B}={\rm
sgn}\left[-\cos{\theta_{0}^{1}}\prod_{\alpha=a,b}\sqrt{A_{\alpha}(s-s_{+}^{\alpha})(s-s_{-}^{\alpha})+\left[\mu-1+s\left(T_{\alpha}^{2}+T_{0}^{2}\right)\right]^{2}}\left.-\prod_{\alpha=a,b}\left[\mu-1+s\left(T_{\alpha}^{2}+T_{0}^{2}\right)\right]\right]\right|_{s=s_{0}},\end{split}$
(S22a) $\iota_{A}={\rm sgn}{[\theta_{0}^{1}-\pi]}\left.\cdot{\rm
sgn}\left[\frac{\mu-1+s\left(T_{b}^{2}+T_{0}^{2}\right)}{\sqrt{A_{b}(s-s_{+}^{b})(s-s_{-}^{b})}}-\iota_{A}\iota_{B}\frac{\mu-1+s\left(T_{a}^{2}+T_{0}^{2}\right)}{\sqrt{A_{a}(s-s_{+}^{a})(s-s_{-}^{a})}}\right]\right|_{s=s_{0}}.$
(S22b)

Finally, the equation for $\mu$ becomes

$\gamma=\gamma(\mu)\equiv\frac{I_{1}^{\prime}+\pi(1-{\rm
sgn}(I_{1}^{\prime}))}{I_{2}^{\prime}+\frac{\pi}{\sqrt{\mu}}(1-{\rm
sgn}(I_{2}^{\prime}))},$ (S23)

where $I_{1,2}^{\prime}=\sqrt{-M_{1}}\cdot I_{1,2}$. With a tedious
calculation, $I_{1,2}^{\prime}$ can be simplified to

$\begin{split}I_{1}^{\prime}=&(1-\iota_{A})\frac{l_{1}}{\sqrt{As_{+}}}\left.\left[\frac{\mu-1}{l_{1}^{2}}\cdot
K\left(\frac{s_{-}}{s_{+}}\right)-\left(T_{a}^{2}+T_{0}^{2}+\frac{\mu-1}{l_{1}^{2}}\right)\cdot\Pi\left(\frac{l_{1}^{2}}{s_{+}},\frac{s_{-}}{s_{+}}\right)\right]\right|_{a}\\\
+&(1+\iota_{B})\frac{l_{1}}{\sqrt{As_{+}}}\left.\left[\frac{\mu-1}{l_{1}^{2}}\cdot
K\left(\frac{s_{-}}{s_{+}}\right)-\left(T_{b}^{2}+T_{0}^{2}+\frac{\mu-1}{l_{1}^{2}}\right)\cdot\Pi\left(\frac{l_{1}^{2}}{s_{+}},\frac{s_{-}}{s_{+}}\right)\right]\right|_{b}\\\
+&\iota_{A}\frac{l_{1}}{\sqrt{As_{+}}}\left.\left[\frac{\mu-1}{l_{1}^{2}}\cdot
K\left(\sqrt{\frac{s_{+}}{s_{0}}},\frac{s_{-}}{s_{+}}\right)-\left(T_{a}^{2}+T_{0}^{2}+\frac{\mu-1}{l_{1}^{2}}\right)\cdot\Pi\left(\sqrt{\frac{s_{+}}{s_{0}}},\frac{l_{1}^{2}}{s_{+}},\frac{s_{-}}{s_{+}}\right)\right]\right|_{a}\\\
-&\iota_{B}\frac{l_{1}}{\sqrt{As_{+}}}\left.\left[\frac{\mu-1}{l_{1}^{2}}\cdot
K\left(\sqrt{\frac{s_{+}}{s_{0}}},\frac{s_{-}}{s_{+}}\right)-\left(T_{b}^{2}+T_{0}^{2}+\frac{\mu-1}{l_{1}^{2}}\right)\cdot\Pi\left(\sqrt{\frac{s_{+}}{s_{0}}},\frac{l_{1}^{2}}{s_{+}},\frac{s_{-}}{s_{+}}\right)\right]\right|_{b},\end{split}$
(S24a)
$\begin{split}I_{2}^{\prime}=&(1-\iota_{A})\frac{l_{2}}{\sqrt{As_{+}}}\left.\left[\frac{1-\mu}{l_{2}^{2}\mu}\cdot
K\left(\frac{s_{-}}{s_{+}}\right)-\left(T_{a}^{2}-T_{0}^{2}+\frac{1-\mu}{l_{2}^{2}\mu}\right)\cdot\Pi\left(\frac{l_{2}^{2}\mu}{s_{+}},\frac{s_{-}}{s_{+}}\right)\right]\right|_{a}\\\
+&(1+\iota_{B})\frac{l_{2}}{\sqrt{As_{+}}}\left.\left[\frac{1-\mu}{l_{2}^{2}\mu}\cdot
K\left(\frac{s_{-}}{s_{+}}\right)-\left(T_{b}^{2}-T_{0}^{2}+\frac{1-\mu}{l_{2}^{2}\mu}\right)\cdot\Pi\left(\frac{l_{2}^{2}\mu}{s_{+}},\frac{s_{-}}{s_{+}}\right)\right]\right|_{b}\\\
+&\iota_{A}\frac{l_{2}}{\sqrt{As_{+}}}\left.\left[\frac{1-\mu}{l_{2}^{2}\mu}\cdot
K\left(\sqrt{\frac{s_{+}}{s_{0}}},\frac{s_{-}}{s_{+}}\right)-\left(T_{a}^{2}-T_{0}^{2}+\frac{1-\mu}{l_{2}^{2}\mu}\right)\cdot\Pi\left(\sqrt{\frac{s_{+}}{s_{0}}},\frac{l_{2}^{2}\mu}{s_{+}},\frac{s_{-}}{s_{+}}\right)\right]\right|_{a}\\\
-&\iota_{B}\frac{l_{2}}{\sqrt{As_{+}}}\left.\left[\frac{1-\mu}{l_{2}^{2}\mu}\cdot
K\left(\sqrt{\frac{s_{+}}{s_{0}}},\frac{s_{-}}{s_{+}}\right)-\left(T_{b}^{2}-T_{0}^{2}+\frac{1-\mu}{l_{2}^{2}\mu}\right)\cdot\Pi\left(\sqrt{\frac{s_{+}}{s_{0}}},\frac{l_{2}^{2}\mu}{s_{+}},\frac{s_{-}}{s_{+}}\right)\right]\right|_{b},\end{split}$
(S24b)

where we use $(\cdot)|_{a,b}$ to indicate that $s_{\pm}$ and $s_{0}$ are the
functions of $T=T_{a,b}$. $K(y_{0},v)$ and $\Pi(y_{0},u,v)$ are elliptic
integrals and $K(v)=K(1,v)$ and $\Pi(u,v)=\Pi(1,u,v)$. The details of elliptic
integrals and the derivation of equation (S24) are shown in Appendix .6.

We summarize the general procedure to get the solution

  1. 1.

with coefficients given in (S21), solve $s_{0}$;

  2. 2.

plug $s_{0}$ into (S22), to solve $\iota_{A}$ and $\iota_{B}$, which determine
the geometry of branes and the location of the intersection point;

  3. 3.

use $\iota_{A}$ and $\iota_{B}$ to solve (S23) to get $\mu$, which then will
give $M_{1,2}$;

  4. 4.

use $M_{1,2}$ to evaluate the onshell action given in (S116).

From the discussion above, we find that with the constraint
$\theta^{1}=\theta_{0}^{1}$, we can already solve the AdS geometry and the
brane trajectories. It means $\theta_{0}^{1}$ and $\theta_{0}^{2}$ in the
action (S1) are not independent. Hence, we define $\theta_{0}=\theta_{0}^{1}$
for simplicity.

###  .2 Defect changing operator without a cusp

In this section, we derive the scaling dimension of the defect changing
operator given in Eq. (11) in the main text. To this end, we consider the
geometry with $\mathbb{Z}_{2}$ symmetry, $L_{1}=L_{2}$ and $l_{1}=l_{2}$.
Because of the $\mathbb{Z}_{2}$ symmetry, we have $M_{1}=M_{2}=M$ and $\mu=1$
without solving the complicated equations outlined in the last section. Using
$l_{1}=l_{2}=l$ and $T_{0}^{2}=l_{1}^{-2}-l_{2}^{-2}=0$, the coefficients in
(S20) can be simplified as

$\begin{split}A^{\prime}=-T_{a}^{2}T_{b}^{2}(T_{a}^{2}+T_{b}^{2}+2T_{a}T_{b}\cos{\theta_{0}^{1}}-4l^{-2}\sin^{2}{\theta_{0}^{1}}),\qquad
B^{\prime}=2MT_{a}^{2}T_{b}^{2}\sin^{2}{\theta_{0}^{1}},\qquad
C^{\prime}=0,\end{split}$ (S25)

and $\sigma_{0}=-2B^{\prime}/A^{\prime}$.

Next, we need to solve (S22). Firstly, since $r=\sqrt{\sigma+Ml^{2}}\geq 0$,
we must have $\sigma_{0}+Ml^{2}\geq 0$, and it leads to the constraint for
$\cos{\theta_{0}^{1}}$,

$-\mathcal{T}_{a}\mathcal{T}_{b}-\sqrt{(1-\mathcal{T}_{a}^{2})(1-\mathcal{T}_{b}^{2})}<\cos{\theta_{0}^{1}}<-\mathcal{T}_{a}\mathcal{T}_{b}+\sqrt{(1-\mathcal{T}_{a}^{2})(1-\mathcal{T}_{b}^{2})},$
(S26)

where we introduced $\mathcal{T}_{a,b}=l\cdot T_{a,b}/2$ for simplicity. We
assume $T_{a}>T_{b}$ without loss of generality. Using $l_{1}=l_{2}=l$ and
$\mu=1$, the coefficient $A_{\alpha}$ in (S8) can be simplified as
$A_{\alpha}=T_{\alpha}^{2}(4/l^{2}-T_{\alpha}^{2})$, and

$A_{\alpha}s_{\pm}=-\frac{A_{\alpha}}{M}\frac{-B_{\alpha}\pm\sqrt{B_{\alpha}^{2}-A_{\alpha}C_{\alpha}}}{A}=2T_{\alpha}^{2}\pm
2T_{\alpha}^{2}.$ (S27)

Therefore, (S22a) can be simplified to $\iota_{A}\iota_{B}={\rm
sgn}\left[4\cos{\theta_{0}^{1}}-s_{0}\left(\frac{4}{l^{2}}\cos{\theta_{0}^{1}}+T_{a}T_{b}\right)\right]$.
Using $s_{0}=(-\frac{2B^{\prime}}{A^{\prime}})/(-M)$, after a tedious
calculation, we have

$\iota_{A}\iota_{B}=\begin{cases}1,&-\mathcal{T}_{a}\mathcal{T}_{b}-\sqrt{(1-\mathcal{T}_{a}^{2})(1-\mathcal{T}_{b}^{2})}<\cos{\theta_{0}^{1}}<-\frac{\mathcal{T}_{a}}{\mathcal{T}_{b}},\\\
-1,&-\frac{\mathcal{T}_{a}}{\mathcal{T}_{b}}<\cos{\theta_{0}^{1}}<-\mathcal{T}_{a}\mathcal{T}_{b}+\sqrt{(1-\mathcal{T}_{a}^{2})(1-\mathcal{T}_{b}^{2})},\end{cases}$
(S28)

If $\iota_{A}\iota_{B}=-1$, (S22b) leads $\iota_{A}={\rm
sgn}(\theta_{0}^{1}-\pi)$. If $\iota_{A}\iota_{B}=1$, with
$s_{+}^{\alpha}=4/(4l^{-2}-T_{\alpha}^{2})$ and
$s_{0}=4l^{2}\sin^{2}{\theta_{0}^{1}}/[-l^{2}(T_{a}^{2}+T_{b}^{2}+2T_{a}T_{b}\cos{\theta_{0}^{1}})+4\sin^{2}{\theta_{0}^{1}}]$,
we have

$\begin{split}\iota_{A}=&{\rm sgn}{[\theta_{0}^{1}-\pi]}\cdot{\rm
sgn}\left[\left(1-\frac{s_{+}^{a}}{s_{0}}\right)\left(\frac{4}{l^{2}T_{a}^{2}}-1\right)-\left(1-\frac{s_{+}^{b}}{s_{0}}\right)\left(\frac{4}{l^{2}T_{b}^{2}}-1\right)\right]\\\
=&{\rm sgn}{[\theta_{0}^{1}-\pi]}\cdot{\rm
sgn}\left[\frac{(T_{a}^{2}-T_{b}^{2})(T_{a}^{2}+T_{b}^{2}+2T_{a}T_{b}\cos{\theta_{0}^{1}})}{-\sin^{2}{\theta_{0}^{1}}T_{a}^{2}T_{b}^{2}}\right].\end{split}$
(S29)

which leads to

$\begin{cases}\iota_{A}=\iota_{B}=-{\rm
sgn}(\theta_{0}^{1}-\pi),&-\mathcal{T}_{a}\mathcal{T}_{b}-\sqrt{(1-\mathcal{T}_{a}^{2})(1-\mathcal{T}_{b}^{2})}<\cos{\theta_{0}^{1}}<-\frac{\mathcal{T}_{a}}{\mathcal{T}_{b}},\\\
\iota_{A}=-\iota_{B}={\rm
sgn}(\theta_{0}^{1}-\pi),&-\frac{\mathcal{T}_{a}}{\mathcal{T}_{b}}<\cos{\theta_{0}^{1}}<-\mathcal{T}_{a}\mathcal{T}_{b}+\sqrt{(1-\mathcal{T}_{a}^{2})(1-\mathcal{T}_{b}^{2})}.\end{cases}$
(S30)

Finally, we are ready to solve (S14) to get the parameter $M$. Since $s_{-}=0$
and $T_{0}=0$, we have $K(y_{0},0)=\arcsin{y_{0}}$,
$\Pi(y_{0},u,0)=\arctan{\left(y_{0}\sqrt{\frac{1-u}{1-y_{0}^{2}}}\right)}/\sqrt{1-u}$.
Then plugging $A_{\alpha}s_{+}^{\alpha}=4T_{\alpha}^{2}$,
$\frac{l_{1}^{2}}{s_{+}^{\alpha}}=\frac{4-l_{1}^{2}T_{\alpha}^{2}}{4}$ in
(S24a), we obtain

$\begin{split}I_{1}^{\prime}=(1-\iota_{A})\frac{-\pi}{2}+(1+\iota_{B})\frac{-\pi}{2}+\iota_{A}\left[-\arctan{\left(\sqrt{\frac{s_{+}^{a}-l_{1}^{2}}{s_{0}-s_{+}^{a}}}\right)}\right]-\iota_{B}\left[-\arctan{\left(\sqrt{\frac{s_{+}^{b}-l_{1}^{2}}{s_{0}-s_{+}^{b}}}\right)}\right].\end{split}$
(S31)

We discuss two possibilities in (S30). For
$-\mathcal{T}_{a}\mathcal{T}_{b}-\sqrt{(1-\mathcal{T}_{a}^{2})(1-\mathcal{T}_{b}^{2})}<\cos{\theta_{0}^{1}}<-\frac{\mathcal{T}_{a}}{\mathcal{T}_{b}}$,
we can get

$I_{1}^{\prime}=-\pi+{\rm
sgn}(\theta_{0}^{1}-\pi)\arccos{\frac{-(\mathcal{T}_{a}\mathcal{T}_{b}+\cos{\theta_{0}^{1}})}{\sqrt{(1-\mathcal{T}_{a}^{2})(1-\mathcal{T}_{b}^{2})}}}\leq
0.$ (S32)

Then, (S14a) gives

$L_{1}\sqrt{-M_{1}}=I_{i}^{\prime}+2\pi\cdot\frac{1}{2}(1-{\rm
sgn}(I_{i}^{\prime}))=\pi+{\rm
sgn}(\theta_{0}^{1}-\pi)\arccos{\frac{-(\mathcal{T}_{a}\mathcal{T}_{b}+\cos{\theta_{0}^{1}})}{\sqrt{(1-\mathcal{T}_{a}^{2})(1-\mathcal{T}_{b}^{2})}}}=\begin{cases}2\pi-\arccos{\beta},&\theta_{0}^{1}>\pi,\\\
\arccos{\beta},&\theta_{0}^{1}<\pi.\end{cases},$ (S33)

where
$\beta=(\mathcal{T}_{a}\mathcal{T}_{b}+\cos{\theta_{0}^{1}})/\sqrt{(1-\mathcal{T}_{a}^{2})(1-\mathcal{T}_{b}^{2})}$.
Similarly, for
$-\frac{\mathcal{T}_{a}}{\mathcal{T}_{b}}<\cos{\theta_{0}^{1}}<-\mathcal{T}_{a}\mathcal{T}_{b}+\sqrt{(1-\mathcal{T}_{a}^{2})(1-\mathcal{T}_{b}^{2})}$,
we can get

$I_{1}^{\prime}=-\pi+{\rm sgn}(\theta_{0}^{1}-\pi)(\pi-\arccos{\beta})\leq 0,$
(S34)

which leads to the same result as (S33).

Then, the onshell action (S116) reads

$\begin{split}I_{\rm tot}=\frac{M_{1}l_{1}L_{1}+M_{2}l_{2}L_{2}}{2T_{\rm
DCFT}}=-\frac{2l}{T_{\rm DCFT}L}[I_{1}^{\prime}+\pi(1-{\rm
sgn}(I_{1}^{\prime}))]^{2}=-\frac{2l}{T_{\rm DCFT}L}[\pi+{\rm
sgn}(\theta_{0}^{1}-\pi)(\pi-\arccos{\beta})]^{2}.\end{split}$ (S35)

This leads to Eq. (11) in the main text.

###  .3 AdS geometry without a corner on the branes

In this section, we consider the branes without corners, i.e.,
$\theta_{0}^{1}=\theta_{0}^{2}=\pi$ and $T_{a}=T_{b}=T$. Note that a similar
case has been considered in Ref. [47] with a different motivation. We will
derive new results for the cusp anomalous dimension.

####  .3.1 Continuous brane with the same tension

Plugging $\theta_{0}^{1}=\theta_{0}^{2}=\pi$ into (S20), we find that
$A^{\prime},B^{\prime}$ and $C^{\prime}$ are all proportional to
$(T_{a}-T_{b})^{2}$. Hence, $T_{a}=T_{b}$ means the equation for $\sigma$ is
trivial. Plugging $T_{a}=T_{b}$ into (S22a), we get $\iota_{A}\iota_{B}=1$.
While (S22b) is not applicable. With the results above, we know that
$\iota_{A}=\iota_{B}$ and any $\sigma_{0}$ is a solution, which means two
branes starting from A and B will connect smoothly and form a single brane, as
one would expect without corners. Besides, to simplify the problem, we can
take $\iota_{A}=\iota_{B}=1$ and $\sigma_{0}=\sigma_{+}$ ($s_{0}=s_{+}$). Then
(S24) can be simplified to

$\begin{split}I_{1}^{\prime}=\frac{2l_{1}}{\sqrt{As_{+}}}\left[\frac{\mu-1}{l_{1}^{2}}\cdot
K\left(\frac{s_{-}}{s_{+}}\right)-\left(T^{2}+T_{0}^{2}+\frac{\mu-1}{l_{1}^{2}}\right)\cdot\Pi\left(\frac{l_{1}^{2}}{s_{+}},\frac{s_{-}}{s_{+}}\right)\right],\end{split}$
(S36a)
$\begin{split}I_{2}^{\prime}=\frac{2l_{2}}{\sqrt{As_{+}}}\left[\frac{1-\mu}{l_{2}^{2}\mu}\cdot
K\left(\frac{s_{-}}{s_{+}}\right)-\left(T^{2}-T_{0}^{2}+\frac{1-\mu}{l_{2}^{2}\mu}\right)\cdot\Pi\left(\frac{l_{2}^{2}\mu}{s_{+}},\frac{s_{-}}{s_{+}}\right)\right].\end{split}$
(S36b)

####  .3.2 Bubble-solution phase

As we discussed in the main text, the existence of the bubble-solution phase
is that the solution of (S23) at $\gamma\rightarrow 0$ is $\mu_{0}>0$. We
derive the condition for the bubble-solution phase in this subsection. With
the help of (S36), we define the numerator and denominator in (S23),
respectively, as

$F_{\rm num}=I_{1}^{\prime}+\pi(1-{\rm
sgn}(I_{1}^{\prime}))=2\pi\delta_{1,{\rm
E1}}-\frac{2l_{1}}{\sqrt{As_{+}}}\left[\frac{1-\mu}{l_{1}^{2}}\cdot
K\left(\frac{s_{-}}{s_{+}}\right)+\left(T^{2}+T_{0}^{2}+\frac{\mu-1}{l_{1}^{2}}\right)\cdot\Pi\left(\frac{l_{1}^{2}}{s_{+}},\frac{s_{-}}{s_{+}}\right)\right],$
(S37a) $F_{\rm den}=I_{2}^{\prime}+\frac{\pi}{\sqrt{\mu}}(1-{\rm
sgn}(I_{2}^{\prime}))=\frac{2\pi}{\sqrt{\mu}}\delta_{2,{\rm
E1}}-\frac{2l_{2}}{\sqrt{As_{+}}}\left[\frac{\mu-1}{l_{2}^{2}\mu}\cdot
K\left(\frac{s_{-}}{s_{+}}\right)+\left(T^{2}-T_{0}^{2}+\frac{1-\mu}{l_{2}^{2}\mu}\right)\cdot\Pi\left(\frac{l_{2}^{2}\mu}{s_{+}},\frac{s_{-}}{s_{+}}\right)\right].$
(S37b)

We show in Appendix .8 that $\delta_{i,{\rm E1}}$ can be expressed as a
function of $\mu$ explicitly.

Instead of directly solve $\gamma(\mu)=0$, our strategy is to examine the
behavior of

$\gamma(\mu)=\frac{I_{1}^{\prime}+\pi(1-{\rm
sgn}(I_{1}^{\prime}))}{I_{2}^{\prime}+\frac{\pi}{\sqrt{\mu}}(1-{\rm
sgn}(I_{2}^{\prime}))}=\frac{F_{\rm den}}{F_{\rm num}},$ (S38)

as a function of $\mu$, from which we can get the phase boundary of the bubble
solution. The results are summarized in the discussion of (S54) and (S55),
with an illustration of $\gamma(\mu)$ shown in Fig. S2 (a).

Now. we start our analysis by examining the asymptotic behavior of $F_{\rm
num}$ and $F_{\rm den}$, respectively. We first consider the limit
$\mu\rightarrow\infty$ for $F_{\rm num}$. We have $\delta_{1,{\rm E1}}=1$ from
(S121). Then by expanding $s_{-}/s_{+}$ and $l_{1}^{2}/s_{+}$ in the order of
$\mu^{-1}$,

$\frac{s_{-}}{s_{+}}=\frac{(T-T_{\rm max})(T-T_{\rm min})}{(T+T_{\rm
max})(T+T_{\rm min})}+\mathcal{O}(\mu^{-1}),$ (S39a)
$\frac{l_{1}^{2}}{s_{+}}=\frac{1}{\mu}\frac{(T_{\rm max}-T)(T-T_{\rm
min})}{l_{1}^{-2}}+\mathcal{O}(\mu^{-2}),$ (S39b)
$\sqrt{As_{+}}=\mu\cdot(T+T_{\rm max})(T+T_{\rm min})+\mathcal{O}(1),$ (S39c)

and substituting them to the special functions in (S37a), we can obtain after
a simplification

$\begin{split}F_{\rm
num}|_{\mu\rightarrow\infty}=&2\pi+\mathcal{O}(1/\sqrt{\mu})\rightarrow
2\pi>0.\end{split}$ (S40)

Consider the limit $\mu\rightarrow 0$ for $F_{\rm num}$. We have
$\delta_{1,{\rm E1}}=\Theta(Tl_{2}-1)$ from (S121). Hence,

$\begin{split}F_{\rm num}|_{\mu\rightarrow
0}&=2\pi\Theta(Tl_{2}-1)+I_{1}^{\prime}|_{\mu\rightarrow 0},\\\
I_{1}^{\prime}|_{\mu\rightarrow
0}&=-\frac{2l_{1}}{\sqrt{As_{+}}}\left[\frac{1-\mu}{l_{1}^{2}}\cdot
K\left(\frac{s_{-}}{s_{+}}\right)+\left(T^{2}+T_{0}^{2}+\frac{\mu-1}{l_{1}^{2}}\right)\cdot\Pi\left(\frac{l_{1}^{2}}{s_{+}},\frac{s_{-}}{s_{+}}\right)\right].\end{split}$
(S41)

Expanding $s_{-}/s_{+}$ and $\mu l_{2}^{2}/s_{+}$ in the order of $\mu$ gives

$\frac{s_{-}}{s_{+}}=\frac{(T-T_{\rm max})(T+T_{\rm min})}{(T+T_{\rm
max})(T-T_{\rm min})}+\mathcal{O}(\mu),$ (S42a) $\frac{\mu
l_{2}^{2}}{s_{+}}=\mu\cdot\frac{(T_{\rm max}-T)(T+T_{\rm
min})}{l_{2}^{-2}}+\mathcal{O}(\mu^{2}),$ (S42b) $\sqrt{As_{+}}=(T+T_{\rm
max})(T-T_{\rm min})+\mathcal{O}(\mu).$ (S42c)

Then (S36a) can be simplified to

$\begin{split}{I_{1}^{\prime}}|_{\mu\rightarrow 0}=-2[(T-T_{\rm min})(T+T_{\rm
max})]^{-\frac{1}{2}}&\left[l_{1}^{-1}\cdot K\left(\frac{(T-T_{\rm
max})(T+T_{\rm min})}{(T+T_{\rm max})(T-T_{\rm
min})}\right)+l_{1}\left(T^{2}-\frac{1}{l_{2}^{2}}\right)\right.\\\
\times&\left.\Pi\left(l_{1}^{2}(T_{\rm max}-T)(T+T_{\rm min}),\frac{(T-T_{\rm
max})(T+T_{\rm min})}{(T+T_{\rm max})(T-T_{\rm
min})}\right)\right].\end{split}$ (S43)

To discuss its sign, we consider

$\frac{{\rm d}}{{\rm d}T}{I_{1}^{\prime}}|_{\mu\rightarrow
0}=\frac{4T(E(v)-K(v))}{l_{1}(T+T_{\rm min})(T_{\rm max}-T)\sqrt{(T+T_{\rm
max})(T-T_{\rm min})}}.$ (S44)

Note that, for later convenience, we introduce two dimensionless variables,

$v=\frac{(T-T_{\rm max})(T+T_{\rm min})}{(T+T_{\rm max})(T-T_{\rm min})},\quad
u=l_{1}^{2}(T_{\rm max}-T)(T+T_{\rm min}).$ (S45)

Because $E(v)-K(v)$ has the property

$E(v)-K(v)=\int_{0}^{1}\frac{\sqrt{1-vy^{2}}}{\sqrt{1-y^{2}}}{\rm
d}y-\int_{0}^{1}\frac{{\rm
d}y}{\sqrt{(1-y^{2})(1-vy^{2})}}=\int_{0}^{1}\frac{-vy^{2}{\rm
d}y}{\sqrt{(1-y^{2})(1-vy^{2})}}>0,$ (S46)

we conclude that $\frac{{\rm d}}{{\rm d}T}{I_{1}^{\prime}}|_{\mu\rightarrow
0}>0$,

For $T_{\rm min}>l_{2}^{-1}$, the step function is a constant
$\theta(Tl_{2}-1)=1$, so $\frac{{\rm d}}{{\rm d}T}F_{\rm num}|_{\mu\rightarrow
0}=\frac{{\rm d}}{{\rm d}T}{I_{1}^{\prime}}|_{\mu\rightarrow 0}>0$. It is more
subtle for $T_{\rm min}<l_{2}^{-1}$. While the step function has a nontrivial
contribution to $F_{\rm num}|_{\mu\rightarrow 0}$, it can be shown that
$F_{\rm num}|_{\mu\rightarrow 0}$ is actually continuous. Also, there is a
caveat in $\frac{{\rm d}}{{\rm d}T}{I_{1}^{\prime}}|_{\mu\rightarrow 0}$ at
$T=l_{2}^{-1}$: ${I_{1}^{\prime}}|_{\mu\rightarrow 0}$ has a jump at in
$T=l_{2}^{-1}$, but $\lim_{T\rightarrow l_{2}^{-1}+0^{+}}\frac{{\rm d}}{{\rm
d}T}{I_{1}^{\prime}}|_{\mu\rightarrow 0}=\lim_{T\rightarrow
l_{2}^{-1}+0^{-}}\frac{{\rm d}}{{\rm d}T}{I_{1}^{\prime}}|_{\mu\rightarrow
0}$. Hence, we can conclude that $F_{\rm num}|_{\mu\rightarrow 0}$ is a
monotonically increasing function of $T$. Then we need to discuss the behavior
of $F_{\rm num}|_{\mu\rightarrow 0}$ at $T_{\rm min}$ and $T_{\rm max}$. For
$T\rightarrow T_{\rm min}^{+}$, we assume $T=T_{\rm min}\cdot(1+\delta)$ with
$\delta\rightarrow 0^{+}$. Then we can expand the parameters in the elliptic
integrals in (S45) that $v\approx-\frac{2l_{1}}{l_{2}\delta}$,
$u\approx\frac{4l_{1}^{2}}{l_{2}}\left(\frac{1}{l_{1}}-\frac{1}{l_{2}}\right)$.
Using
$K\left(-\frac{a}{\delta}\right)\approx\Pi\left(u,-\frac{a}{\delta}\right)\approx-\ln{\delta}\cdot\sqrt{\delta}/(2\sqrt{a})$
from Appendix .6.1, we arrive at

$\begin{split}I_{1}^{\prime}|_{\mu\rightarrow 0,T\rightarrow T_{\rm
min}}\approx\sqrt{\frac{l_{2}}{l_{1}}-1}\cdot\ln{\delta}\rightarrow-\infty.\end{split}$
(S47)

While, for $T=T_{\rm max}$ in (S43), it is straightforward to get

$\begin{split}I_{1}^{\prime}|_{\mu\rightarrow 0,T\rightarrow T_{\rm
max}}=-\sqrt{\frac{l_{2}}{l_{1}}+1}\cdot\pi.\end{split}$ (S48)

Therefore, we find that, $F_{\rm num}|_{\mu\rightarrow 0}$ will increase as a
function of $T$ monotonically, and

$F_{\rm num}|_{\mu\rightarrow 0,T\rightarrow T_{\rm
min}}\approx\sqrt{\frac{l_{2}}{l_{1}}-1}\ln(T-T_{\rm
min})\rightarrow-\infty,\qquad F_{\rm num}|_{\mu\rightarrow 0,T\rightarrow
T_{\rm max}}\rightarrow\left(2-\sqrt{\frac{l_{2}}{l_{1}}+1}\right)\cdot\pi.$
(S49)

It means that, for $l_{2}>3l_{1}$, $F_{\rm num}|_{\mu\rightarrow 0}$ is less
than zero for any $T$, while, for $l_{2}<3l_{1}$, there is a zero for $F_{\rm
num}|_{\mu\rightarrow 0}$.

Now we discuss the asymptotic behaviors of (S37b). First consider the limit
$\mu\rightarrow\infty$, we have $\delta_{2,{\rm E1}}=\theta(Tl_{1}-1)$ from
(S121). With expansions in (S39), (S36b) can be simplified to be

$\begin{split}F_{\rm
den}|_{\mu\rightarrow\infty}=&\frac{2\pi}{\sqrt{\mu}}\Theta(Tl_{1}-1)+I_{2}^{\prime}|_{\mu\rightarrow\infty},\\\
I_{2}^{\prime}|_{\mu\rightarrow\infty}\approx&\frac{-2}{\sqrt{\mu}}[(T+T_{\rm
min})(T+T_{\rm max})]^{-\frac{1}{2}}\left[l_{2}^{-1}\cdot
K\left(\frac{(T-T_{\rm max})(T-T_{\rm min})}{(T+T_{\rm max})(T+T_{\rm
min})}\right)+l_{2}\left(T^{2}-\frac{1}{l_{1}^{2}}\right)\right.\\\
&\times\left.\Pi\left(l_{2}^{2}(T_{\rm max}-T)(T-T_{\rm min}),\frac{(T-T_{\rm
max})(T-T_{\rm min})}{(T+T_{\rm max})(T+T_{\rm
min})}\right)\right].\end{split}$ (S50)

To proceed, we consider

$\frac{{\rm d}}{{\rm
d}T}(\sqrt{\mu}I_{2}^{\prime}|_{\mu\rightarrow\infty})\approx\frac{4T(E(v^{\prime})-K(v^{\prime}))}{l_{2}(T-T_{\rm
min})(T_{\rm max}-T)\sqrt{(T+T_{\rm max})(T+T_{\rm min})}}>0,$ (S51)

where we use that $v^{\prime}=\frac{(T-T_{\rm max})(T-T_{\rm min})}{(T+T_{\rm
max})(T+T_{\rm min})}<0$ and $E(v^{\prime})-K(v^{\prime})$ has a similar form
to (S46), leading to $E(v^{\prime})-K(v^{\prime})>0$.

Again, there is a subtlety: while the step function has a nontrivial
contribution to $F_{\rm den}|_{\mu\rightarrow\infty}$, it can be shown that
$F_{\rm den}|_{\mu\rightarrow\infty}$ is actually continuous. Also, there is a
caveat in $\frac{{\rm d}}{{\rm
d}T}(\sqrt{\mu}I_{2}^{\prime}|_{\mu\rightarrow\infty})$ at $T=l_{1}^{-1}$:
$\sqrt{\mu}I_{2}^{\prime}|_{\mu\rightarrow\infty}$ has a jump at in
$T=l_{1}^{-1}$, but $\lim_{T\rightarrow l_{1}^{-1}+0^{+}}\frac{{\rm d}}{{\rm
d}T}(\sqrt{\mu}I_{2}^{\prime}|_{\mu\rightarrow\infty})=\lim_{T\rightarrow
l_{1}^{-1}+0^{-}}\frac{{\rm d}}{{\rm
d}T}(\sqrt{\mu}I_{2}^{\prime}|_{\mu\rightarrow\infty})$. Hence, we can
conclude that $F_{\rm den}|_{\mu\rightarrow\infty}$ is a monotonically
increasing function of $T$.

Further, it is straightforward to show that

$F_{\rm den}|_{\mu\rightarrow\infty,T=T_{\rm
min}}\approx\sqrt{1-\frac{l_{1}}{l_{2}}}\cdot\frac{\pi}{\sqrt{\mu}},\qquad
F_{\rm den}|_{\mu\rightarrow\infty,T=T_{\rm
max}}\approx\left(2-\sqrt{1+\frac{l_{1}}{l_{2}}}\right)\frac{\pi}{\sqrt{\mu}}$
(S52)

Since we assume $l_{2}>l_{1}$, we have $F_{\rm den}|_{\mu\rightarrow\infty}>0$
for all $T$.

Consider the limit $\mu\rightarrow 0$, using the expansion (S42) and
$\delta_{2,{\rm E1}}=1$ from (S121), the denominator can be simplified as

$\begin{split}F_{\rm den}|_{\mu\rightarrow
0}=&\frac{2\pi}{\sqrt{\mu}}+\mathcal{O}(1)\rightarrow+\infty.\end{split}$
(S53)

To summarize, at $\mu\rightarrow\infty$, our analysis in (S40) and (S52)
implies

$\lim_{\mu\rightarrow\infty}\frac{F_{\rm num}}{F_{\rm den}}=+\infty.$ (S54)

On the other hand, at $\mu\rightarrow 0^{+}$, our analysis in (S49) and (S53)
indicates

$\frac{F_{\rm num}}{F_{\rm
den}}\approx\begin{cases}\sqrt{\frac{l_{2}}{l_{1}}-1}\ln(T-T_{\rm
min})\cdot\frac{\sqrt{\mu}}{2\pi},\quad&T\approx T_{\rm min}\\\
\left(2-\sqrt{\frac{l_{2}}{l_{1}}+1}\right)\pi\cdot\frac{\sqrt{\mu}}{2\pi},\quad&T=T_{\rm
max}\end{cases}.$ (S55)

This implies that for $l_{2}<3l_{1}$ the right-hand side of (S23) goes to
$0^{-}$ for a small tension $T$, and goes to $0^{+}$ for a large tension $T$.
While, for $l_{2}>3l_{1}$, the right-hand side of (S23) goes to $0^{-}$ for
$\mu\rightarrow 0^{+}$ with any tension $T$.

Supplementary Figure S2: Examples of $\gamma=\gamma(\mu)$ for different phases
without and with a corner on branes. (a) Without a corner, i.e.,
$\theta_{0}=\pi$, we plot $l_{2}=2l_{1}=2$, so $T=0.55$ ($T=1.45$) corresponds
to the (no-)bubble-solution phase, as shown in main text Fig. 3 (b). (b) With
a nontrivial corner contribution, we plot $l_{2}=2l_{1}=2,T=1.25$, so
$\theta_{0}=\pi/2$ ($\theta_{0}=3\pi/4$) corresponds to the (no-)bubble-
solution phase, as shown in main text Fig. 3 (c).

We plot two examples with $l_{2}=2l_{1}=2$ and $T=0.55,1.55$ in Fig. S2 (a).
It shows that for a small tension $T=0.55$, there is a finite solution
$\mu_{0}>0$ for $\gamma\rightarrow 0$, which corresponds to a bubble solution.
While, with a large tension $T=1.45$, we have $\mu\rightarrow 0$ for
$\gamma\rightarrow 0$, which corresponds to the no-bubble-solution phase.

Therefore, for $\gamma\rightarrow\infty$, we expect $\mu\rightarrow\infty$
from (S23), which means there is no bubble-solution phase. For
$\gamma\rightarrow 0^{+}$, there are two possibilities. For a large tension
and $l_{2}<3l_{1}$, we have $\mu\rightarrow 0^{+}$ for $\gamma\rightarrow
0^{+}$, which corresponds to the no-bubble-solution phase. While, for a small
tension with $l_{2}<3l_{1}$ or any tension with $l_{2}>3l_{1}$, we will have a
finite solution $\mu_{0}>0$ with $\gamma\rightarrow 0$ because $\mu\rightarrow
0^{-}$ for $\gamma\rightarrow 0^{+}$. It is the bubble-solution phase.

Based on the discussion above, the condition for the bubble phase is that
$B(T,l_{1},l_{2})\equiv\lim_{\mu\rightarrow 0}F_{\rm num}<0$. Namely, the
phase boundary is determined by

$\begin{split}B(T,l_{1},l_{2})=&2\pi\Theta\left(T^{2}l_{2}^{2}-1\right)-2[(T-T_{\rm
min})(T+T_{\rm max})]^{-\frac{1}{2}}\left[l_{1}^{-1}\cdot
K\left(\frac{(T-T_{\rm max})(T+T_{\rm min})}{(T+T_{\rm max})(T-T_{\rm
min})}\right)\right.\\\
&\left.+l_{1}\left(T^{2}-\frac{1}{l_{2}^{2}}\right)\Pi\left(l_{1}^{2}(T_{\rm
max}-T)(T+T_{\rm min}),\frac{(T-T_{\rm max})(T+T_{\rm min})}{(T+T_{\rm
max})(T-T_{\rm min})}\right)\right].\end{split}$ (S56)

It is interesting to note that this function is the same as the condition of
phase transition in Ref. [45], though the transition is very different from
ours. Defining

$\Xi_{u_{0}}(T,l_{1},l_{2})=\sqrt{\frac{4v}{-u}}\left[K\left(u_{0},v\right)+(l_{1}^{2}T^{2}-l_{1}^{2}/l_{2}^{2})\cdot\Pi\left(u_{0},u,v\right)\right],$
(S57)

with $u,v$ in (S45), we have
$B(T,l_{1},l_{2})=2\pi\Theta\left(l_{2}T-1\right)-\Xi_{1}(T,l_{1},l_{2})$.
Also, we can observe that

$F_{\rm den}|_{\mu\rightarrow\infty}=\frac{B(T,l_{2},l_{1})}{\sqrt{\mu}}.$
(S58)

Since we have shown that $F_{\rm den}|_{\mu\rightarrow\infty}>0$ for any $T$,
it means $B(T,l_{2},l_{1})>0$. Note that we have assumed $l_{2}>l_{1}$, so
$B(T,l_{2},l_{1})$ is distinct from $B(T,l_{1},l_{2})$, the latter can be
either positive and negative.

####  .3.3 On-shell action

We derive the onshell action near big and small cusp angles in this section,
which is used to get the cusp anomalous dimension in the main text. In
general, the derivation of onshell action is given by (S116) in Appendix .7.

We first discuss big cusp angles. At $\phi=\pi$,
$\gamma=\frac{L_{1}}{L_{2}}=1$ we have $\mu=1$, i.e., $M_{1}=M_{2}=M$. From
(S121), we have $\delta_{1,{\rm E1}}=1$ and $\delta_{2,{\rm
E1}}=\Theta(Tl_{1}-1)+\Theta(1-Tl_{1})\Theta(T^{2}-T_{0}^{2})=\Theta(T-T_{0})$.
Therefore, with (S14a), (S130) and (S131), we have

$L_{1}=\frac{2\pi}{\sqrt{-M}}\delta_{1,{\rm
E1}}+\left(-\frac{\pi}{\sqrt{-M}}\right)=\frac{\pi}{\sqrt{-M}},\qquad
L_{2}=\frac{2\pi}{\sqrt{-M}}\delta_{2,{\rm E1}}+\left(-{\rm
sgn}(T^{2}-T_{0}^{2})\frac{\pi}{\sqrt{-M}}\right)=\frac{\pi}{\sqrt{-M}},$
(S59)

which give the consistent result that $L_{1}=L_{2}$. Then, the onshell action
reads

$\begin{split}I_{\rm tot}&=\frac{ML}{2T_{\rm
DCFT}}\cdot\frac{l_{1}+l_{2}}{2}=\frac{l_{1}+l_{2}}{4T_{\rm
DCFT}}\cdot\left(-\frac{4\pi^{2}}{L}\right)=-\frac{\pi^{2}(l_{1}+l_{2})}{T_{\rm
DCFT}}\cdot\frac{1}{L}.\end{split}$ (S60)

Now we can expand the result around $\gamma=1$. Because $\gamma(\mu=1)=1$ for
any $(T,l_{1},l_{2})$, we can consider $\mu=1+\delta_{\mu}$ with
$|\delta_{\mu}|\ll 1$. From (S121), for $l_{1}<l_{2}$ and a small enough
$\delta_{\mu}$, we always have $\delta_{1,{\rm E1}}=1$, and $\delta_{2,{\rm
E1}}=\Theta(-\mu_{2}^{*})+\Theta(\mu_{2}^{*})\Theta(\mu_{2}^{*}-\mu)$, where
$\mu^{\ast}_{1,2}$ are defined in (S117). Expanding $s_{\pm}$ in the order of
$\delta_{\mu}$ gives

$s_{\pm}=\frac{2(T^{2}\pm T^{2})}{(T_{\rm max}^{2}-T^{2})(T^{2}-T_{\rm
min}^{2})}+\frac{((T_{0}^{2}+T^{2})\pm(T_{0}^{2}+T^{2}))\delta_{\mu}}{(T_{\rm
max}^{2}-T^{2})(T^{2}-T_{\rm
min}^{2})}\pm\frac{\delta_{\mu}^{2}}{4T^{2}}+\mathcal{O}(\delta_{\mu}^{3}),$
(S61)

which means $s_{-}/s_{+}=\mathcal{O}(\delta_{\mu}^{2})$. We can plug (S61)
into $F_{\rm num}$ and $F_{\rm den}$ in (S37) and expand them w.r.t.
$\delta_{\mu}$ to get

$F_{\rm
num}\approx\pi+\frac{\pi}{2l_{1}T}\delta_{\mu}-\frac{3\pi(T_{0}^{2}+T^{2})}{16l_{1}T^{3}}\delta_{\mu}^{2}+\mathcal{O}(\delta_{\mu}^{3}),$
(S62a) $F_{\rm
den}=\pi-\left(\frac{1}{l_{2}T}+1\right)\cdot\frac{\pi}{2}\delta_{\mu}+\frac{3\pi}{16}\left(\frac{T_{0}^{2}+3T^{2}}{l_{2}T^{3}}+2\right)\delta_{\mu}^{2}+\mathcal{O}(\delta_{\mu}^{3}).$
(S62b)

Then we can simplify the equation for $\gamma$

$\gamma=\frac{F_{\rm num}}{F_{\rm den}}\approx 1+\frac{T_{\rm
max}+T}{2T}\delta_{\mu}+\frac{(T_{\rm
max}+T)(-3T_{0}^{2}+T(\frac{1}{l_{2}}+\frac{3}{l_{1}})-2T^{2})}{16T^{3}}\delta_{\mu}^{2}+\mathcal{O}(\delta_{\mu}^{3}).$
(S63)

It is worth mentioning that the right-hand side of (S63) is not symmetric by
exchanging $l_{1}$ and $l_{2}$. Defining $\gamma=\delta_{\gamma}+1$, we have

$\begin{split}\delta_{\mu}=\frac{2l_{1}l_{2}T}{l_{1}+l_{2}+l_{1}l_{2}T}\delta_{\gamma}+\frac{l_{1}^{2}l_{2}(-4l_{2}T-l_{2}^{2}T^{2}-4l_{1}T+l_{1}l_{2}T^{2}+2l_{1}l_{2}^{2}T^{3}+3l_{2}(l_{1}+l_{2})(l_{1}^{-2}-l_{2}^{-2}))}{2(l_{1}+l_{2}+l_{1}l_{2}T)^{3}}\delta_{\gamma}^{2}+\mathcal{O}(\delta_{\gamma}^{3})\end{split}$
(S64)

Now, we can calculate the on-shell action with (S116). Comparing (S14a) and
(S37), we know that

$F_{\rm num}=\sqrt{-M_{1}}L_{1}.$ (S65)

Therefore, with $L_{1}+L_{2}=L$, the onshell action is

$\begin{split}I_{\rm tot}&=\frac{M_{1}l_{1}L_{1}+M_{2}l_{2}L_{2}}{2T_{\rm
DCFT}}=\frac{M_{1}l_{1}L_{1}}{2T_{\rm DCFT}}\left(1+\frac{\mu l_{2}}{\gamma
l_{1}}\right)=-\frac{1}{2T_{\rm DCFT}}\left(1+\frac{\mu l_{2}}{\gamma
l_{1}}\right)(1+\gamma^{-1})\frac{l_{1}}{L}\cdot F_{\rm num}^{2}\\\
&\approx\frac{-1}{2T_{\rm
DCFT}L}\left[2(l_{1}+l_{2})\pi^{2}+(l_{2}-l_{1})\pi^{2}\delta_{\gamma}+\frac{l_{1}(l_{1}-l_{2}+l_{1}l_{2}T)}{l_{1}+l_{2}+l_{1}l_{2}T}\pi^{2}\delta_{\gamma}^{2}\right],\end{split}$
(S66)

where the leading term is consistent with (S60). This leads to Eq. (12) in the
main text.

If $l_{1}=l_{2}=l$, the onshell action can be simplified as

$I_{\rm tot}\approx-\frac{1}{2T_{\rm
DCFT}L}\left(4l\pi^{2}+\frac{l^{2}T}{2+lT}\pi^{2}\delta_{\gamma}^{2}\right).$
(S67)

The lowest order of the correction is quadratic, because there is a
$\mathbb{Z}_{2}$ symmetry of exchanging two AdS spacetime for $l_{1}=l_{2}$
and $L_{1}=L_{2}$. Moreover, if $T=0$, the second order correction also
vanishes.

Now, we discuss small cusp angles $\phi\approx 0$ or $\phi\approx 2\pi$. We
consider the expansion around $\gamma=\frac{L_{1}}{L_{2}}\rightarrow 0$ (or
$\gamma=\frac{L_{1}}{L_{2}}\rightarrow\infty$).

For $\gamma=\frac{L_{1}}{L_{2}}\rightarrow\infty$, as discussed in Sec. .3.2,
we have $\mu\rightarrow\infty$. Then from (S40) and (S58), we arrive at

$F_{\rm num}|_{\mu\rightarrow\infty}\approx 2\pi,\quad F_{\rm
den}|_{\mu\rightarrow\infty}\approx\frac{B(T,l_{2},l_{1})}{\sqrt{\mu}},\quad\gamma=\frac{F_{\rm
num}}{F_{\rm
den}}\Big{|}_{\mu\rightarrow\infty}\approx\frac{2\pi}{B(T,l_{2},l_{1})}\sqrt{\mu}+\mathcal{O}(1).$
(S68)

Consequently, the onshell action reads

$\begin{split}I_{\rm tot}&=\frac{M_{1}l_{1}L_{1}+M_{2}l_{2}L_{2}}{2T_{\rm
DCFT}}=-\frac{1}{2T_{\rm DCFT}L}\left(l_{1}+\frac{\mu
l_{2}}{\gamma}\right)(1+\gamma^{-1})\cdot F_{\rm
num}^{2}|_{\mu\rightarrow\infty}=-\frac{1}{2T_{\rm DCFT}L}\left(l_{2}\cdot
B(T,l_{2},l_{1})^{2}\gamma+\mathcal{O}(1)\right).\end{split}$ (S69)

For $\gamma=\frac{L_{1}}{L_{2}}\rightarrow 0$, as discussed in Sec. .3.2,
there are two possibilities. If it is the bubble-solution phase with
$B(T,l_{1},l_{2})<0$, then there is a solution $\mu_{0}>0$ such that
$\gamma(\mu_{0})=0$. Similar to (S69), the onshell action can be written as

$\begin{split}I_{\rm tot}=\frac{M_{1}l_{1}L_{1}+M_{2}l_{2}L_{2}}{2T_{\rm
DCFT}}=-\frac{1}{2T_{\rm
DCFT}L}\left(l_{1}\gamma+l_{2}\mu\right)(1+\gamma)\cdot F_{\rm
den}^{2}|_{\mu=\mu_{0}}.\end{split}$ (S70)

From (S37b), we conclude that $F_{\rm den}^{2}(\mu_{0})$ is also a constant.
On the other hand, if it is a no-bubble-solution phase, we have

$F_{\rm num}|_{\mu\rightarrow 0}\approx B(T,l_{1},l_{2}),\quad F_{\rm
den}|_{\mu\rightarrow
0}\approx\frac{2\pi}{\sqrt{\mu}},\quad\gamma=\frac{F_{\rm num}}{F_{\rm
den}}\Big{|}_{\mu\rightarrow
0}\approx\frac{B(T,l_{1},l_{2})}{2\pi}\sqrt{\mu}.$ (S71)

The onshell action becomes

$\begin{split}I_{\rm tot}&=-\frac{1}{2T_{\rm DCFT}L}\left(\frac{l_{1}\cdot
B(T,l_{1},l_{2})^{2}}{\gamma}+\mathcal{O}(1)\right).\end{split}$ (S72)

The onshell action (S72) and (S69) leads to the cusp anomalous dimension in
Eq. (15) in the main text.

###  .4 AdS geometry with defect on brane

In this section, we consider a nontrivial corner with $\theta_{0}\neq\pi$ on
the brane. In the following, we focus on $l_{1}=l_{2}=l$ and $T_{a}=T_{b}=T$
with a general $\theta_{0}$.

####  .4.1 AdS geometry with defect on brane with general tension

Following the general procedure, we first get the intersecting point from
(S21) with $l_{1}=l_{2}=l$ and $T_{a}=T_{b}=T$ (results not shown here).
Secondly, we simplify the variables $\iota_{A}$ and $\iota_{B}$ in (S22). From
(S22a), for the expression in the sign function to be real, we require
$l^{2}T^{2}+2\cos{\theta_{0}}<2$ for $0<lT<\sqrt{2}$ and
$\cos{\theta_{0}}<2/(l^{2}T^{2})-1$ for $\sqrt{2}<lT<2$. The range for
$\theta_{0}$ in the region $\sqrt{2}<lT<2$ is larger than that in
$0<lT<\sqrt{2}$. Later, we will see that a real solution actually requires
$l^{2}T^{2}+2\cos{\theta_{0}}<2$ for all $0<lT<2$. Using this condition, we
have $\iota_{A}=-\iota_{B}={\rm sgn}(\theta_{0}-\pi){\rm sgn}(l^{2}T^{2}-1)$
for $\mu\rightarrow 0$.

After a tedious calculation, under the limit $\mu\rightarrow 0$, (S24a) can be
simplified as

$\begin{split}I_{1}^{\prime}|_{\mu\rightarrow 0}=&2(1-{\rm
sgn}(\theta_{0}-\pi){\rm
sgn}(l^{2}T^{2}-1))\left(\frac{-K(\frac{lT-2}{lT+2})-(l^{2}T^{2}-1)\cdot\Pi(-lT(lT-2),\frac{lT-2}{lT+2})}{\sqrt{lT(lT+2)}}\right)\\\
+&2{\rm sgn}(\theta_{0}-\pi){\rm
sgn}(l^{2}T^{2}-1)\left(\frac{-K(u_{0},\frac{lT-2}{lT+2})-(l^{2}T^{2}-1)\cdot\Pi(u_{0},-lT(lT-2),\frac{lT-2}{lT+2})}{\sqrt{lT(lT+2)}}\right),\end{split}$
(S73)

where
$u_{0}=\sqrt{\frac{l^{2}T^{2}+2\cos{\theta_{0}}-2}{(lT-2)(\sqrt{(l^{2}T^{2}+l^{2}T^{2}\cos{\theta_{0}}-2)(\cos{\theta_{0}}-1)}-lT\cos{\theta_{0}})}}$.
Here we can see that for $u_{0}$ to be real, the range of parameters is given
by $l^{2}T^{2}+2\cos{\theta_{0}}<2$ for $0<lT<2$. Therefore, we have

$\tilde{B}(T,l,\theta_{0})\equiv F_{\rm num}|_{\mu\rightarrow
0}=I_{1}^{\prime}|_{\mu\rightarrow 0}+\pi(1-{\rm
sgn}(I_{1}^{\prime}|_{\mu\rightarrow 0})),$ (S74)

where $I_{1}^{\prime}|_{\mu\rightarrow 0}$ is given in (S73). Next, we check
the value of $I_{2}^{\prime}|_{\mu\rightarrow 0}$ in the range of $0<lT<2$ and
$-1<\cos{\theta_{0}}<1-l^{2}T^{2}/2$ is finite and negative. It in turn
indicates that

$F_{\rm den}|_{\mu\rightarrow 0}=I_{2}^{\prime}|_{\mu\rightarrow
0}+\frac{\pi}{\sqrt{\mu}}(1-{\rm sgn}(I_{2}^{\prime}|_{\mu\rightarrow
0}))\rightarrow\frac{2\pi}{\sqrt{\mu}}.$ (S75)

We have the behavior of $\gamma(\mu)$ near $\mu\rightarrow 0$

$\gamma(\mu\rightarrow 0)=\frac{F_{\rm num}}{F_{\rm
den}}\Big{|}_{\mu\rightarrow
0}=\frac{\sqrt{\mu}}{2\pi}\cdot\tilde{B}(T,l,\theta_{0}).$ (S76)

Similar to the discussion in Sec. .3.2, the sign of
$\tilde{B}(T,l,\theta_{0})$ determines a phase transition. Namely,
$\tilde{B}(T,l,\theta_{0})>0$ is the no-bubble-solution phase, while
$\tilde{B}(T,l,\theta_{0})<0$ is the bubble-solution phase. We plot two
examples with $l_{1}=l_{2}=1,T=1.25$ and $\theta_{0}=\pi/2,3\pi/4$ in Fig. S2
(b). It shows that for a small corner angle $\theta_{0}=\pi/2$, there is a
finite solution $\mu_{0}>0$ for $\gamma\rightarrow 0$, which corresponds to a
bubble solution. While, with a large corner angle $\theta_{0}=3\pi/4$, we have
$\mu\rightarrow 0$ for $\gamma\rightarrow 0$, which corresponds to the no-
bubble-solution phase. Notice the opposite signs at the limit $\mu\rightarrow
0$.

In the same manner, we can also get the asymptotic behavior of the onshell
action with $\gamma\rightarrow 0$ for the no-bubble-solution phase

$\begin{split}I_{\rm tot}&=-\frac{1}{2T_{\rm
DCFT}L}\left(\frac{l}{\gamma}\cdot\tilde{B}(T,l,\theta_{0})^{2}+\mathcal{O}(1)\right).\end{split}$
(S77)

To further simplify the result, we can prove that for $0<lT<2$,
$I_{1}^{\prime}|_{\mu\rightarrow 0}<0$. Therefore, in the range
$\theta_{0}<\pi$ and $lT>1$, $\tilde{B}(T,l,\theta_{0})$ reads

$\tilde{B}(T,l,\theta_{0})=2\pi-2\ \Xi_{1}(T,l,l)+\ \Xi_{u_{0}}(T,l,l),$ (S78)

where $\Xi_{u_{0}}(T,l_{1},l_{2})$ is defined in (S57), and $u,v$ are the same
as (S45) with $l_{1}=l_{2}=l$. This gives Eq. (16) in the main text.

####  .4.2 Special case of AdS geometry with a corner on branes

In this section, we focus on a special case $lT=1$. Proceeding as before, the
range of $\theta_{0}$ in this special case is $-1<\cos{\theta_{0}}<1/2$. This
range leads to $\iota_{A}=-\iota_{B}={\rm sgn}(\theta_{0}-\pi)$. Actually, it
is what we expect because of the $\mathbb{Z}_{2}$ symmetry in this case, which
maps one brane to another. Then, because we only care about the solution for
$\gamma\rightarrow 0$, we expand $I_{1}^{\prime}$ in (S24) around
$\mu\rightarrow 0$. Most of the derivation is straightforward, but there is an
expansion of the incomplete elliptic integral which is a little complicated.
For $\Pi(y_{0}=\sin{\pi/2-a\mu},u=1-b\mu^{2},v=m)$, we can expand it around
$\mu\rightarrow 0$

$\begin{split}&\Pi(\sin{(\pi/2-a\mu)},1-b\mu^{2},m)=\int_{0}^{1-a^{2}\mu^{2}/2}\frac{{\rm
d}t}{(1-(1-b\mu^{2})t^{2})\sqrt{(1-t^{2})(1-mt^{2})}}\\\
=&\Pi(1,1-b\mu^{2},m)-\int_{0}^{1}\frac{\frac{a^{2}\mu^{2}}{2}{\rm
d}x}{(1-(1-b\mu^{2})(1-\frac{a^{2}\mu^{2}}{2}x)^{2})\sqrt{(1-(1-\frac{a^{2}\mu^{2}}{2}x)^{2})(1-m(1-\frac{a^{2}\mu^{2}}{2}x)^{2})}}\\\
=&\Pi(1,1-b\mu^{2},m)-\frac{\arctan{\sqrt{\frac{a^{2}}{b}}}}{\sqrt{b(1-m)}}\cdot\frac{1}{\mu}+\mathcal{O}(1).\end{split}$
(S79)

Then, using the expansion above, we can simplify the numerator in (S23), and
find that

$\tilde{B}(1/l,l,\theta_{0})\approx\theta_{0}-\theta_{0}^{c},\qquad\theta_{0}^{c}=\frac{2}{\sqrt{3}}\cdot
K\left(-\frac{1}{3}\right).$ (S80)

We can calculate the denominator in (S23) and prove it is positive in the
limit of $\mu\rightarrow 0$. The derivation of the expansion of the
denominator is tedious, and there are also two expansions of equations are
useful:

$\begin{split}\Pi(\mu,-\frac{1}{3})-\Pi(\mu,-\frac{1}{3}+\frac{2\mu}{3})=&\int_{0}^{1}\frac{{\rm
d}t}{(1-\mu
t^{2})\sqrt{(1-t^{2})}}\left(\frac{1}{\sqrt{1+\frac{1}{3}t^{2}}}-\frac{1}{\sqrt{1+\frac{1-2\mu}{3}t^{2}}}\right)\\\
=&\left[\frac{3}{4}\cdot
E\left(-\frac{1}{3}\right)-K\left(-\frac{1}{3}\right)\right]\cdot\mu+\mathcal{O}(\mu^{2}),\end{split}$
(S81)

and similar to (S79), we also have

$\begin{split}&\Pi\left(\sin{(\pi/2-a\mu)},\mu,\frac{2\mu-1}{3}\right)=\int_{0}^{1-a^{2}\mu^{2}/2}\frac{{\rm
d}t}{(1-\mu)t^{2})\sqrt{(1-t^{2})(1-\frac{2\mu-1}{3}t^{2})}}\\\
=&\Pi(1,\mu,\frac{2\mu-1}{3})-\int_{0}^{1}\frac{\frac{a^{2}\mu^{2}}{2}{\rm
d}x}{(1-\mu(1-\frac{a^{2}\mu^{2}}{2}x)^{2})\sqrt{(1-(1-\frac{a^{2}\mu^{2}}{2}x)^{2})(1-\frac{2\mu-1}{3}(1-\frac{a^{2}\mu^{2}}{2}x)^{2})}}\\\
=&\Pi(1,\mu,\frac{2\mu-1}{3})-\frac{\sqrt{3}a}{2}\cdot\mu+\mathcal{O}(\mu^{2}).\end{split}$
(S82)

With these expansions, we find $I_{2}^{\prime}|_{\mu\rightarrow
0}=-\frac{1}{\sqrt{3}}(6E(-1/3)+4K(-1/3))<0$ for $\mu\rightarrow 0$, which
leads to $F_{\rm den}|_{\mu\rightarrow 0}\approx 2\pi/\sqrt{\mu}$.

Therefore, if $\theta_{0}>\theta_{0}^{c}$, the positive numerator leads to the
no-bubble-solution phase with a divergent action which is proportional to
$\gamma^{-1}$. While, for $\theta_{0}<\theta_{0}^{c}$, there is a bubble phase
with a finite action for $\gamma\rightarrow 0$. And remember that here we
focus on $\cos{\theta_{0}}\in(-1,1/2)$ with
$\theta_{0}\in(\frac{\pi}{3},\frac{5\pi}{3})$.

###  .5 Ising model with defects

In this section, we focus on the Ising CFT with periodic boundary condition
and two defects. To be concrete, consider the following Hamiltonian

$H=-\frac{1}{2}\sum^{\prime}_{r}[\sigma^{z}(r)+\sigma^{x}(r)\sigma^{x}(r+1)]-\frac{1}{2}\kappa_{a}\sigma^{x}(L_{1}+L_{2})\sigma^{x}(1)-\frac{1}{2}\kappa_{b}\sigma^{x}(L_{1})\sigma^{x}(L_{1}+1),$
(S83)

where $\sigma^{\mu}(r)$ denotes the Pauli matrix at site $r$ and the first
summation omits the terms at the two defect bonds with strength
$\kappa_{\alpha}$.

The Hamiltonian (S83) can be exactly diagonalized [6], and for the two defects
case, we define $\phi=2\pi\alpha=2\pi p/q$, with coprime integers $p,q$, to
denote the location of two defects. With two defects strength $\kappa_{a}$ and
$\kappa_{b}$, the eigenvalues of (S83) are related to the solution of the
algebraic equation

$(1-\kappa_{a}^{2})(1-\kappa_{b}^{2})\cos{[x(1-2\alpha)]}+(1+\kappa_{a}^{2})(1+\kappa_{b}^{2})\cos{x}+4\kappa_{a}\kappa_{b}(1-2Q)=0,$
(S84)

where $Q=0$ $(1)$ corresponds to the odd (even) sector of (S83) after Jordan-
Wigner transformation. With the solution for $x$, we can calculate the energy
spectrum, particularly the lowest eigenvalue. It is easy to check that (S84)
has a periodicity $2\pi q$. In general, we can define $z={\rm e}^{{\rm
i}x/q}$, then (S84) becomes a polynomial of $z$ of order $2q$:

$\frac{1}{2}(1-\kappa_{a}^{2})(1-\kappa_{b}^{2})(z^{q-2p}+z^{-q+2p})+\frac{1}{2}(1+\kappa_{a}^{2})(1+\kappa_{b}^{2})(z^{q}+z^{-q})+4\kappa_{a}\kappa_{b}(1-2Q)=0,$
(S85)

Therefore, (S85) has $2q$ roots, and if $z_{0}$ is a root, $1/z_{0}$ is also a
root. With the roots $x_{i}\in[0,\tilde{\gamma}(q)\cdot\pi]$ and
$i=1,2,...,\tilde{\gamma}(q)$, the universal $O(\frac{1}{L})$ part of the
lowest eigenvalue is

$E=\frac{2\pi\tilde{\gamma}(q)}{L}\sum_{i=1}^{\tilde{\gamma}(q)}\left[-\frac{1}{12}\left(\frac{1}{2}-6\tilde{\Delta}_{i}^{2}\right)\right],$
(S86a) $\tilde{\Delta}_{i}=\frac{1}{2}-\frac{1}{2\pi\tilde{\gamma}(q)}x_{i},$
(S86b)

where $\tilde{\gamma}(q)=q\ (q/2)$ for the odd (even) $q$.

In the following, we consider defect changing operator for distinct
$\kappa_{a,b}$ at $\phi=\pi$ as well as the cusp anomalous dimension for two
identical defects $\kappa_{a}=\kappa_{b}=\kappa$ at strong and weak defect
limit.

####  .5.1 Defect changing operator in different sectors

When $\phi=\pi$, $\alpha=1/2$, it is straightforward to get the solution of
(S84), and consequently,

$\tilde{\Delta}_{Q}(\kappa_{a},\kappa_{b})=\frac{1}{\pi}\left|\arctan\kappa_{a}-(-1)^{Q}\arctan\kappa_{b}\right|.$
(S87)

Note that there is only one solution we need to consider. Then the scaling
dimension of the defect changing operator is given by

$\Delta_{ab}^{11}(\pi)=\pi\left[\tilde{\Delta}_{Q}(\kappa_{a},\kappa_{b})^{2}-\frac{1}{2}(\tilde{\Delta}_{0}(\kappa_{a},\kappa_{a})^{2}+\tilde{\Delta}_{0}(\kappa_{b},\kappa_{b})^{2})\right]=\frac{1}{\pi}\left(\arctan\kappa_{a}-(-1)^{Q}\arctan\kappa_{b}\right)^{2}.$
(S88)

In the $Q=0$ sector, $\Delta_{ab}^{11}(\pi)$ vanishes when
$\kappa_{a}=\kappa_{b}$ as one would expect. However, in the $Q=1$ sector,
when $\kappa_{a}=\kappa_{b}=\kappa$,
$\Delta_{ab}^{11}(\pi)=\frac{4}{\pi}\arctan^{2}\kappa$ still nontrivially
depends on $\kappa$. Note that this happens in the holographic model when
$\theta_{0}\neq\pi$. Hence, the effect of a nontrivial corner term is similar
to the different sector $Q=0,1$ here.

####  .5.2 Weak defect limit

We discuss the weak defect limit with expansion around $\kappa\approx 1$. To
proceed, we consider $\alpha=p/q=p/(2n+1)$ with $p$ being an even integer and
an integer $n$. Actually, this parametrization does not guarantee $p,q$ are
coprime, but we will find that the result below is still valid. $q=2n+1$ gives
$\tilde{\gamma}(q)=q$. Taking $\kappa=1$, (S84) gives $x=(2m+1)\pi$ for
$m=0,1,...,n$. Once we include a perturbation $\kappa=1-\delta\kappa$, two
degenerate roots at $x=(2m+1)\pi$ for $0\leq m<n$ will split and give rise to
$2n$ roots; while the correction to the solution $x=(2n+1)\pi$ leads to
another root which is less than $(2n+1)\pi$ (the other one that is greater
than $(2n+1)\pi$ is neglected). To the second order in $\delta\kappa$, we set
$x=(2m+1)\pi+a\cdot\delta\kappa+b\cdot\delta\kappa^{2}$ and plug it into
(S84). The vanishing of the coefficient in $\delta\kappa$ and
$\delta\kappa^{2}$ then leads to the equations of $a$ and $b$ as follows,

$a^{2}-2+2\cos[(2m+1)\pi(1-2\alpha)]=0,$ (S89a)
$(1-a^{2}+ab)+\cos[2\pi(\alpha-m+2m\alpha)]+a(2\alpha-1)\sin[2\pi(\alpha-m+2m\alpha)]=0.$
(S89b)

The corresponding solutions are

$a=\pm 2\cos{[\pi(\alpha-m+2m\alpha)]},$ (S90a)
$b=\pm\cos{[\pi(\alpha-m+2m\alpha)]}-(2\alpha-1)\sin{[2\pi(\alpha-m+2m\alpha)]}.$
(S90b)

With these corrections, the energy (S86) becomes

$\begin{split}E&=-\frac{2\pi(2n+1)}{24L}\left\\{\sum_{m=0}^{n-1}\left[2-12\left(2\left(\frac{1}{2}-\frac{(2m+1)\pi}{2\pi(2n+1)}\right)^{2}+2\left(\frac{a_{m}}{2\pi(2n+1)}\right)^{2}\delta\kappa^{2}\right.\right.\right.\\\
&\left.\left.\left.+2\left(\frac{1}{2}-\frac{(2m+1)\pi}{2\pi(2n+1)}\right)\frac{-b_{m}-b_{m}^{\prime}}{2\pi(2n+1)}\delta\kappa^{2}\right)\right]+1-12\left(\frac{2\delta\kappa}{2\pi(2n+1)}\right)^{2}\right\\}+O(\delta\kappa^{3}).\end{split}$
(S91)

where $a_{m},b_{m}$ and $a_{m}^{\prime},b_{m}^{\prime}$ denote the $+$ and $-$
solution in (S90). Performing the summation over $m$, we arrive at the final
result

$E=-\frac{\pi}{12L}\cdot\left[1+\frac{6}{\pi^{2}}\left(\frac{(1-2\alpha)\pi}{\sin{2\alpha\pi}}-1\right)\delta\kappa^{2}\right]+O(\delta\kappa^{3}),$
(S92)

which leads to Eq. (18) in the main text.

####  .5.3 Strong defect limit

We can also consider another limit $\kappa=\delta\kappa\ll 1$. In this limit,
(S84) becomes

$(1+\delta\kappa^{2})^{2}\cos{[x(1-2\alpha)]}+(1+\delta\kappa^{2})^{2}\cos{x}+4\delta\kappa^{2}=0.$
(S93)

Similarly, we consider $\alpha=p/q=p/(2n+1)$ with integer $n$. If $\kappa=0$,
then the zeroth order solution is
$x=(\frac{1}{2}+m_{1})\frac{\pi}{1-\alpha},(\frac{1}{2}+m_{2})\frac{\pi}{\alpha}$
with $0\leq\frac{1}{2}+m_{1}\leq(1-\alpha)q=q-p$ and
$0\leq\frac{1}{2}+m_{2}\leq\alpha q=p$. $m_{1}=0,1,...,q-p-1$ and
$m_{2}=0,1,...,p-1$ constitute the $q$ solutions. To the second order in
$\delta\kappa$, it is straightforward to obtain the correction for the $m_{1}$
type solution
$x=\frac{(2m_{1}+1)\pi}{2(1-\alpha)}+a\delta\kappa+b\delta\kappa^{2}$ with
$a=0$ and $b=\frac{2}{(1-\alpha)\tan{\frac{2m_{1}+1}{4(1-\alpha)}\pi}}$, and
the correction to the $m_{2}$ type solution
$x=\frac{(2m_{2}+1)\pi}{2\alpha}+a\delta\kappa+b\delta\kappa^{2}$ with $a=0$
and $b=\frac{2}{\alpha\tan{\frac{2m_{2}+1}{4\alpha}\pi}}$.

Next we need to substitute the corrected solutions above in (S86), and
performing the summation. The summation cannot be done analytically, but we
obtain the zeroth order result:

$\begin{split}E_{0}&=-\frac{2\pi(2n+1)}{24L}\left\\{\sum_{m_{1}=0}^{q-p-1}\left[1-12\left(\frac{1}{2}-\frac{\frac{2m_{1}+1}{2(1-\alpha)}\pi}{2\pi(2n+1)}\right)^{2}\right]+\sum_{m_{2}=0}^{p-1}\left[1-12\left(\frac{1}{2}-\frac{\frac{2m_{2}+1}{2\alpha}\pi}{2\pi(2n+1)}\right)^{2}\right]\right\\}\\\
&=-\frac{\pi}{12N}\frac{1}{4\alpha(1-\alpha)},\end{split}$ (S94)

where we see the same type of divergence at $\alpha\rightarrow 0$
($\alpha\rightarrow 1$).

## Appendix

###  .6 Special functions and arc equations

####  .6.1 Special functions

In this section, we summarize the elliptic integrals, and discuss some
properties of them. For the complete elliptic integrals of the first, second
and third kind, we have

$K(v)=\int_{0}^{1}\frac{{\rm d}t}{\sqrt{(1-t^{2})(1-vt^{2})}},$ (S95a)
$E(v)=\int_{0}^{1}\frac{\sqrt{1-vt^{2}}{\rm d}t}{\sqrt{1-t^{2}}},$ (S95b)
$\Pi(u,v)=\int_{0}^{1}\frac{{\rm d}t}{(1-ut^{2})\sqrt{(1-t^{2})(1-vt^{2})}}.$
(S95c)

For the incomplete elliptic integrals of the first, second and third kind, we
have

$K(y_{0},v)=\int_{0}^{y_{0}}\frac{{\rm d}t}{\sqrt{(1-t^{2})(1-vt^{2})}},$
(S96a) $E(y_{0},v)=\int_{0}^{y_{0}}\frac{\sqrt{1-vt^{2}}{\rm
d}t}{\sqrt{1-t^{2}}},$ (S96b) $\Pi(y_{0},u,v)=\int_{0}^{y_{0}}\frac{{\rm
d}t}{(1-ut^{2})\sqrt{(1-t^{2})(1-vt^{2})}},$ (S96c)

where $K(v)=K(1,v)$, $E(v)=EK(1,v)$ and $\Pi(u,v)=\Pi(1,u,v)$.

With the definition above, in the following we list some properties of the
complete elliptic integrals, which are used in the main text. For $K(v)$,
changing integral variable $t=\sqrt{1-s^{2}}$, it is easy to show that

$K(v)=\frac{1}{\sqrt{1-v}}K\left(\frac{v}{v-1}\right).$ (S97)

Similarly, for $\Pi(u,v)$, with $t=\sqrt{1-s^{2}}$, we have

$\Pi(u,v)=\frac{1}{(1-u)\sqrt{1-v}}\Pi\left(\frac{u}{u-1},\frac{v}{v-1}\right).$
(S98)

Besides, for the special case $v=0$, we have

$K(y_{0},0)=\arcsin{y_{0}},\qquad\Pi(y_{0},u,0)=\frac{\arctan{\left(y_{0}\sqrt{\frac{1-u}{1-y_{0}^{2}}}\right)}}{\sqrt{1-u}}.$
(S99)

Finally, as mentioned in Ref. [47], with $\delta\ll 1$, the elliptic integrals
have the property that

$K\left(-\frac{a}{\delta}\right)\approx\Pi\left(u,-\frac{a}{\delta}\right)\approx-\frac{\ln{\delta}\cdot\sqrt{\delta}}{2\sqrt{a}}+\mathcal{O}(\sqrt{\delta}).$
(S100)

####  .6.2 Arc equations with special functions

In this section, we show the detailed derivation of (S24). Here, we derive
(S24a) explicitly, and (S24b) can be obtained similarly. For $I_{1}^{\prime}$,
with $s=-\sigma/M_{1}$ and (S14a), we have

$I_{1}^{\prime}=\sqrt{-M_{1}}\cdot
I_{1}=\int_{s_{+}}^{+\infty}(\dot{x}_{1}^{a}+\dot{x}_{1}^{b})(-M_{1})^{-3/2}{\rm
d}s-\int_{s_{+}}^{s_{0}}(\iota_{A}\ \dot{x}_{1}^{a}-\iota_{B}\
\dot{x}_{1}^{b})(-M_{1})^{-3/2}{\rm d}s,$ (S101)

where

$(-M_{1})^{-3/2}\dot{x}_{1}^{\alpha}=-l_{1}\left.\frac{\mu-1+s\left(T_{\alpha}^{2}+T_{0}^{2}\right)}{2(-l_{1}^{2}+s)\sqrt{sA(s-s_{+})(s-s_{-})}}\right|_{\alpha},$
(S102)

in which we take $T=T_{\alpha}$ in the expression of $s_{0}$ and $s_{\pm}$.
Then define $y^{2}=s_{+}/s$, we have

$\begin{split}&\int_{s_{+}}^{s_{0}}\dot{x}_{1}^{\alpha}(-M_{1})^{-3/2}{\rm
d}s=\int_{1}^{\sqrt{s_{+}/s_{0}}}l_{1}\frac{s_{+}(T_{\alpha}^{2}+T_{0}^{2})+(\mu-1)y^{2}}{(s_{+}-l_{1}^{2}y^{2})\sqrt{As_{+}}\sqrt{(1-y^{2})(1-\frac{s_{-}}{s_{+}}y^{2})}}{\rm
d}y\\\
=&\frac{l_{1}}{\sqrt{As_{+}}}\int_{1}^{\sqrt{s_{+}/s_{0}}}\left(\frac{-(\mu-1)l_{1}^{-2}}{\sqrt{(1-y^{2})(1-\frac{s_{-}}{s_{+}}y^{2})}}+\frac{(\mu-1)l_{1}^{-2}s_{+}+s_{+}(T_{\alpha}^{2}+T_{0}^{2})}{(s_{+}-l_{1}^{2}y^{2})\sqrt{(1-y^{2})(1-\frac{s_{-}}{s_{+}}y^{2})}}\right){\rm
d}y\\\
=&\frac{l_{1}}{\sqrt{As_{+}}}\left\\{\frac{\mu-1}{l_{1}^{2}}\left[K\left(\frac{s_{-}}{s_{+}}\right)-K\left(\sqrt{\frac{s_{+}}{s_{0}}},\frac{s_{-}}{s_{+}}\right)\right]-(T_{\alpha}^{2}+T_{0}^{2}+\frac{\mu-1}{l_{1}^{2}})\left[\Pi\left(\frac{l_{1}^{2}}{s_{+}},\frac{s_{-}}{s_{+}}\right)-\Pi\left(\sqrt{\frac{s_{+}}{s_{0}}},\frac{l_{1}^{2}}{s_{+}},\frac{s_{-}}{s_{+}}\right)\right]\right\\}.\end{split}$
(S103)

Plugging (S103) into (S101) and taking $s_{0}\rightarrow+\infty$ for the first
integral in (S101), with $K(0,v)=\Pi(0,u,v)=0$, we can obtain (S24a).

###  .7 On-shell action with a corner

In this section, we derive the on-shell action, following the method in Ref.
[47]. We work in Euclidean signature. The continuity of induced metric on the
interface brane $(0,r=r(\sigma),x=x(\sigma))$ leads ${\rm
d}s^{2}=f(\sigma){\rm d}\tau^{2}+g(\sigma){\rm d}\sigma^{2}$ with $f(\sigma)$
and $g(\sigma)$ given in (S4). For (S5), with $e^{\mu}_{\tau}=(1,0,0)$ and
$e^{\mu}_{x}=(0,\dot{r},\dot{x})$, the normal direction $n$ of the brane for
$AdS_{1}$ reads $n_{\mu}=\tilde{A}(0,-\dot{x},\dot{r})$ with $\tilde{A}>0$ and
$\tilde{A}^{2}=\frac{l^{2}}{\dot{x}^{2}(r^{2}-Ml^{2})+\dot{r}^{2}l^{2}/r^{2}}$.
The extrinsic curvature is then

$K_{\tau\tau}=-\Gamma^{\rho}_{\tau\tau}n_{\rho}=-(Mr-\frac{r^{3}}{l^{2}})A\cdot(-\dot{x})=-\dot{x}\frac{r^{2}}{l}\frac{f}{\sqrt{fg}}.$
(S104)

Now we derive the onshell action (S1). For AdS spacetime with radius $l_{i}$,
we have $R_{i}=-6/l_{i}^{2}$ and $\sqrt{g_{i}}{\rm d}^{3}x=l_{i}r_{i}{\rm
d}r_{i}{\rm d}x_{i}{\rm d}\tau$. The bulk term $I_{\rm EH}$ is

$I_{\rm
EH}=\sum_{i=1,2}\frac{2}{l_{i}^{2}}\int_{M_{i}}\sqrt{g_{i}}=\sum_{i=1,2}\int_{M_{i}}\frac{2r_{i}}{l_{i}}{\rm
d}r_{i}{\rm d}x_{i}{\rm d}\tau.$ (S105)

Defining the vector $\mathbf{n}^{\prime}=(0,r^{2},0)$, by the Stoke’s theorem,
we have $\int_{\Omega}(\nabla\cdot\mathbf{n}^{\prime}){\rm
d}V=\int_{\partial\Omega}\mathbf{n}^{\prime}\cdot{\rm d}\mathbf{S}$. It leads
$\int_{\Omega}2r{\rm d}r{\rm d}x{\rm
d}\tau=\int_{\partial\Omega}r^{2}\hat{\mathbf{n}}^{\prime}\cdot{\rm
d}\mathbf{S}$, where $\hat{\mathbf{n}}^{\prime}$ is the normalized vector of
$\mathbf{n}^{\prime}$ with unit length. Hence, the bulk term can be expressed
as the integral on their surface that

$I_{\rm
EH}=\sum_{i=1,2}\frac{1}{l_{i}}\left(\int_{B_{i}}r_{i}^{2}\hat{\mathbf{n}}_{i}^{\prime}\cdot{\rm
d}\mathbf{S}_{i}+\int_{W}r_{i}^{2}\hat{\mathbf{n}}_{i}^{\prime}\cdot{\rm
d}\mathbf{S}_{i}\right)=\sum_{i=1,2}\frac{1}{l_{i}}\left(\int_{B_{i}}r_{i}^{2}{\rm
d}x_{i}{\rm d}\tau+\int_{W}r_{i}^{2}\hat{\mathbf{n}}_{i}^{\prime}\cdot{\rm
d}\mathbf{S}_{i}\right),$ (S106)

where $W=W_{a}\cup W_{b}$. Using the parameter $\sigma$, the second term in
(S106) can be evaluated to be

$\sum_{i=1,2}\frac{1}{l_{i}}\int_{W}r_{i}^{2}\hat{\mathbf{n}}_{i}^{\prime}\cdot{\rm
d}\mathbf{S}_{i}=-\sum_{i=1,2}\frac{1}{l_{i}}\int_{W}r_{i}^{2}{\rm d}\tau{\rm
d}x_{i}=-\int_{W}\left(\frac{r_{1}^{2}\dot{x}_{1}}{l_{1}}+\frac{r_{2}^{2}\dot{x}_{2}}{l_{2}}\right){\rm
d}\tau{\rm d}\sigma=\sum_{\alpha}T_{\alpha}\int_{W_{\alpha}}\sqrt{fg}\ {\rm
d}\tau{\rm d}\sigma.$ (S107)

where in the last step we used (S5). Besides, with the induced metric on the
brane ${\rm d}s^{2}=f(\sigma){\rm d}\tau^{2}+g(\sigma){\rm d}\sigma^{2}$, the
$I_{T}$ term is

$I_{T}=T\int_{W}\sqrt{fg}\ {\rm d}\sigma{\rm d}\tau.$ (S108)

The extrinsic curvature term $I_{\rm surface}$ can be written as

$\begin{split}I_{\rm surface}=&-\sum_{i=1,2}\int_{B_{i}}\sqrt{\hat{g}_{i}}\
K_{i}-\int_{W}\sqrt{\hat{g}}(K_{1}+K_{2})=-\sum_{i=1,2}\int_{B_{i}}\sqrt{\hat{g}_{i}}\
K_{i}-2T\int_{W}\sqrt{fg}\ {\rm d}\sigma{\rm d}\tau\end{split}$ (S109)

where we used $K_{1}+K_{2}=2T$ for the extrinsic curvature on the brane and
$\int_{W}\sqrt{\hat{g}}=\int_{W}\sqrt{fg}\ {\rm d}\sigma{\rm d}\tau$. Because
the codimension-2 integral on the corner in the action will not contribute to
the onshell result, the total action is

$\begin{split}I_{\rm
tot}&=\sum_{i=1,2}\left[\frac{1}{l_{i}}\int_{B_{i}}r_{i}^{2}{\rm d}x_{i}{\rm
d}\tau-\int_{B_{i}}\sqrt{\hat{g}_{i}}\
K_{i}+\frac{1}{l_{i}}\int_{B_{i}}\sqrt{\hat{g}_{i}}\right].\end{split}$ (S110)

In the following, we calculate the integral in (S110) explicitly. For the
first term in (S110), with the time direction length $T_{\rm DCFT}^{-1}$ and
CFT coordinate $r_{i,\infty}$, we have

$\frac{1}{l_{i}}\int_{B_{i}}r_{i}^{2}{\rm d}x_{i}{\rm
d}\tau=\frac{r_{i,\infty}^{2}T_{\rm DCFT}^{-1}L_{i}}{l_{i}}.$ (S111)

For the second term in (S110), the induced metric on $B$ is ${\rm
d}s^{2}=(r_{\infty}^{2}-Ml^{2}){\rm d}\tau^{2}+r_{\infty}^{2}{\rm d}x^{2}$.
With ${e^{\prime}}_{\tau}^{\mu}=(1,0,0)$ and ${e^{\prime}}_{x}^{\mu}=(0,0,1)$,
${n^{\prime}}_{\mu}=A^{\prime}(0,1,0)$ with
$A^{\prime}=\sqrt{\frac{l^{2}}{r_{\infty}^{2}-Ml^{2}}}$, the extrinsic
curvature becomes

$K_{\tau\tau}=-\Gamma^{\rho}_{\tau\tau}n_{\rho}=-\Gamma^{r}_{\tau\tau}\sqrt{\frac{l^{2}}{r_{\infty}^{2}-Ml^{2}}}=\frac{r_{\infty}}{l}\sqrt{r_{\infty}^{2}-Ml^{2}},~{}K_{xx}=-\Gamma^{\rho}_{xx}n_{\rho}=-\Gamma^{r}_{xx}\sqrt{\frac{l^{2}}{r_{\infty}^{2}-Ml^{2}}}=\frac{r_{\infty}}{l}\sqrt{r_{\infty}^{2}-Ml^{2}},$
(S112)

where
$\Gamma^{r}_{\tau\tau}=\Gamma^{r}_{xx}=Mr_{\infty}-r_{\infty}^{3}/l^{2}$.
Then, we arrive at

$K=g^{\mu\nu}K_{\mu\nu}=K_{\tau\tau}\cdot(r_{\infty}^{2}-Ml^{2})^{-1}+K_{xx}\cdot
r_{\infty}^{-2}=\frac{\frac{2r_{\infty}}{l}-\frac{Ml}{r_{\infty}}}{\sqrt{r_{\infty}^{2}-Ml^{2}}}.$
(S113)

Now, the second term in (S110) is

$\int_{B_{i}}\sqrt{\hat{g}_{i}}\
K_{i}=\int_{B_{i}}\sqrt{(r_{i,\infty}^{2}-M_{i}l_{i}^{2})r_{i,\infty}^{2}}\cdot\frac{\frac{2r_{i,\infty}}{l_{i}}-\frac{M_{i}l_{i}}{r_{i,\infty}}}{\sqrt{r_{i,\infty}^{2}-M_{i}l_{i}^{2}}}{\rm
d}x_{i}{\rm d}\tau=\frac{(2r_{i,\infty}^{2}-M_{i}l_{i}^{2})T_{\rm
DCFT}^{-1}L_{i}}{l_{i}}.$ (S114)

Similarly, the last term in (S110) becomes

$\frac{1}{l_{i}}\int_{B_{i}}\sqrt{\hat{g}_{i}}=\frac{1}{l_{i}}\int_{B_{i}}\sqrt{(r_{i,\infty}^{2}-M_{i}l_{i}^{2})r_{i,\infty}^{2}}{\rm
d}x_{i}{\rm
d}\tau=\frac{\sqrt{(r_{i,\infty}^{2}-M_{i}l_{i}^{2})r_{i,\infty}^{2}}T_{\rm
DCFT}^{-1}L_{i}}{l_{i}}.$ (S115)

Finally, combining (S111), (S114) and (S115), the onshell action reads

$\begin{split}I_{\rm
tot}&=\sum_{i=1,2}\left(\frac{1}{l_{i}}\int_{B_{i}}r_{i}^{2}{\rm d}x_{i}{\rm
d}\tau-\int_{B_{i}}\sqrt{\hat{g}_{i}}\
K_{i}+\frac{1}{l_{i}}\int_{B_{i}}\sqrt{\hat{g}_{i}}\right)\\\
&=\sum_{i=1,2}\left(r_{i,\infty}^{2}-(2r_{i,\infty}^{2}-M_{i}l_{i}^{2})+\sqrt{(r_{i,\infty}^{2}-M_{i}l_{i}^{2})r_{i,\infty}^{2}}\right)\frac{T_{\rm
DCFT}^{-1}L_{i}}{l_{i}}\xrightarrow{r_{i,\infty}\rightarrow\infty}\frac{M_{1}l_{1}L_{1}+M_{2}l_{2}L_{2}}{2T_{\rm
DCFT}}.\end{split}$ (S116)

Due to the cancellation of the integral on the interface brane with the help
of the Stoke’s theorem, it is the same as the result in Ref. [47],

###  .8 Sweeping transition and non-self-intersecting condition without
corners

In this section, we will review and generalize the discussion of the sweeping
transition in Ref. [47], and also discuss the monotonicity of brane solutions.
We will also consider a few special cases.

Firstly, we review the sweeping transition between E1 and E2 phase in Ref.
[47]. For the single brane geometry with brane coordinate
$(r=r(\sigma),x=x(\sigma))$ on a constant time slice, the critical point is
given by that the center $r=0$ is located on the interface brane. With the
reflection symmetry, it means that the critical point corresponds to
$r(\sigma_{+})=\sqrt{\sigma_{+}+Ml^{2}}=0$, in which $\sigma_{+}$ is a
solution of $A\sigma^{2}+2B\sigma+C=0$ with coefficients (S8). Therefore,
using $\sigma_{+}=\sigma_{+}^{c}\equiv-M_{i}l_{i}^{2}$, we can get the
critical $\mu_{i}^{\ast}$

$\mu_{1}^{*}=\frac{l_{1}^{2}}{l_{2}^{2}}-T^{2}l_{1}^{2}=\frac{1-T^{2}l_{2}^{2}}{l_{2}^{2}/l_{1}^{2}},\qquad\mu_{2}^{*}=\frac{1}{l_{2}^{2}/l_{1}^{2}-T^{2}l_{2}^{2}}=\frac{l_{1}^{2}/l_{2}^{2}}{1-T^{2}l_{1}^{2}}.$
(S117)

Remember that we assume $l_{1}<l_{2}$. Now we can determine different phases.
With $\sigma_{+}=(-B+\sqrt{B^{2}-AC})/A$, for different limits we have

$\sigma_{+}\approx\begin{cases}\frac{M_{2}}{-2T/l_{1}+(T^{2}+T_{0}^{2})},&M_{2}\gg
M_{1},\\\ \frac{M_{1}}{-2T/l_{2}+(T^{2}-T_{0}^{2})},&M_{2}\ll
M_{1}.\end{cases}$ (S118)

Then from (S15), for $\dot{x}_{1}$ with $\mu=\frac{M_{2}}{M_{1}}\gg 1$, we
have

$\begin{split}{\rm sgn}(\dot{x}_{1}(\sigma_{+}))=&{\rm
sgn}\left[-l_{1}\frac{M_{1}-M_{2}+\sigma_{+}\left(T^{2}+T_{0}^{2}\right)}{2(l_{1}^{2}M_{1}+\sigma_{+})}\right]\approx{\rm
sgn}\left[\frac{M_{2}}{2\sigma_{+}}-\frac{T^{2}+T_{0}^{2}}{2}\right]=-{\rm
sgn}\left[\frac{(T-T_{\rm min})(T_{\rm max}-T)}{2}\right]=-1,\end{split}$
(S119)

which corresponds to the E1 phase for $AdS_{1}$. Hence, when
$\mu>\mu_{1}^{\ast}$ ($\mu<\mu_{1}^{\ast}$), $AdS_{1}$ is in E1 (E2) phase.
Similarly, from (S15), with $\mu=\frac{M_{2}}{M_{1}}\ll 1$, we have

$\begin{split}{\rm sgn}(\dot{x}_{2}(\sigma_{+}))=&{\rm
sgn}\left[-l_{2}\frac{M_{2}-M_{1}+\sigma_{+}\left(T^{2}-T_{0}^{2}\right)}{2(l_{2}^{2}M_{2}+\sigma_{+})}\right]\approx{\rm
sgn}\left[\frac{M_{1}}{2\sigma_{+}}-\frac{T^{2}-T_{0}^{2}}{2}\right]=-{\rm
sgn}\left[\frac{(T+T_{\rm min})(T_{\rm max}-T)}{2}\right]=-1,\end{split}$
(S120)

which means $AdS_{2}$ is in E1 phase. Hence, when $\mu<\mu_{1}^{\ast}$
($\mu>\mu_{1}^{\ast}$), $AdS_{2}$ is in E1 (E2) phase. Therefore, with the
discussion above, we express $\delta_{i,{\rm E1}}$ explicitly that

$\delta_{1,{\rm E1}}=\Theta(\mu-\mu_{1}^{*}),\qquad\delta_{2,{\rm
E1}}=\Theta(-\mu_{2}^{*})+\Theta(\mu_{2}^{*})\theta(\mu_{2}^{*}-\mu),$ (S121)

where $\Theta(x)$ is the step function that is nonzero only for $x>0$. Because
$\mu_{1}^{\ast}$ changes signs at $T=l_{2}^{-1}$, i.e.,

$\mu_{1}^{\ast}({T<l_{2}^{-1}})>0,\quad\mu_{1}^{\ast}({T>l_{2}^{-1}})<0,$
(S122)

it means that when $T>l_{2}^{-1}$, $AdS_{1}$ is always in the E1 phase. A plot
of the phase diagram is shown in Fig. S3 (c,d,e). Notice that the difference
between (c,d) and (e) comes from whether $T_{\rm min}<l_{2}^{-1}$, i.e., if
$T_{\rm min}<l_{2}^{-1}$, there is a transition for $AdS_{1}$ as shown in
(c,d), otherwise $T_{\rm min}>l_{2}^{-1}$, there are no transitions for
$AdS_{1}$ as shown in (e). On the other hand, the difference between (c) and
(d) is due to the possibility of monotonicity of the gradient $\dot{x}_{i}$ as
we will detail next.

Supplementary Figure S3: (a) The values of special tensions with $l_{1}=1$,
where the blue region corresponds to the valid tension $T_{\rm min}<T<T_{\rm
max}$. There are three regions of $l_{2}$, which have different orders of
$T_{0}$, $T_{\rm min}$ and $l_{2}^{-1}$. The crossing points are
$l_{2}=\sqrt{2}l_{1}$ and $l_{2}=2l_{1}$. Different phase diagrams for
sweeping transition for (b) $l_{2}=l_{1}$, (c) $l_{1}<l_{2}<\sqrt{2}l_{1}$,
(d) $\sqrt{2}l_{1}<l_{2}<2l_{2}$ and (e) $l_{2}>2l_{1}$. The blue, orange and
green region corresponds to different phases. ${\rm[Ei,Ej]}$ means $AdS_{1,2}$
is in ${\rm Ei,Ej}$ phases with $i,j=1,2$. The shadowed regions with
$(a_{1},a_{2})$ indicate the monotonicity of branes. $a_{i}=0,1$ corresponds
to the monotonic or non-monotonic brane in $AdS_{i}$, respectively. Monotonic
branes in both AdS spaces are depicted without shadow lines.

In general, we should start (S15), as $\dot{x}_{i}=0$ gives a non-monotonic
brane solution. It is easy to show the denominator of the right-hand side in
(S15) is non-negative. On the other hand, the numerator a linear function in
$\sigma$, so there is only one transition point where the brane will change
the direction. Setting the numerator of the right-hand side of (S15) to be
zero, we get the zero point, $\sigma_{1,2}$,

$\sigma_{1}=\frac{M_{2}-M_{1}}{T^{2}+T_{0}^{2}},\qquad\sigma_{2}=\frac{M_{1}-M_{2}}{T^{2}-T_{0}^{2}}.$
(S123)

Because the range of the brane parameter is $\sigma\geq\sigma_{+}$, in order
for the zero point to be accessible for a brane solution, it requires

$0<\sigma_{+}/\sigma_{i}<1.$ (S124)

It is easy to check that $\sigma_{i}=\sigma_{+}$ leads $\mu=\mu_{i}^{*}$, and
the following non-monotonicity condition,

$\begin{split}0<\sigma_{+}/\sigma_{1}<1&\Rightarrow 0<\mu<\mu_{1}^{\ast},\\\
0<\sigma_{+}/\sigma_{2}<1&\Rightarrow\begin{cases}0<\mu<\mu_{2}^{\ast}&\quad
T<T_{0}\\\ \mu>\mu_{2}^{\ast}&\quad T_{0}<T<l_{1}^{-1}\end{cases}.\end{split}$
(S125)

Note that the condition for $AdS_{2}$ depends on $T_{0}$. This is the reason
for the difference between Fig. S3 (c) and (d). More explicitly, in Fig. S3
(c,d,e), we use the label $(a_{1},a_{2})$ with $a_{i}=0$ ($a_{i}=1$) to show
if the corresponding phase has (non-)monotonic branes in $AdS_{i}$. For
$T_{\rm min}>l_{2}^{-1}$, there is no non-monotonic brane for $AdS_{1}$, but
there exist non-monotonic branes for $AdS_{2}$ as shown in figure S3 (e). For
$T_{\rm min}<l_{2}^{-1}$, there exist non-monotonic branes for both $AdS_{1}$
and $AdS_{2}$ as shown in Fig. S3 (c) and (d). Fig. S3 (a) summarizes the
relation among the allowed range of tensions, $T_{0}$, and $l_{2}^{-1}$.

The phase diagrams in Fig. S3 seem complicated because the phases E1 and E2
are not directly related to the monotonicity of branes. However, we can
understand it by combining the E1 and E2 phases and the angle between the
brane and the asymptotic boundary. With the metric (S3), near the asymptotic
boundary $r\rightarrow\infty$, we can ignore $M_{i}l_{i}^{2}$. Then, using the
coordinate transformation $r_{i}=l_{i}/z_{i}$, we obtain the Poincare metric

${\rm d}s^{2}=\frac{l_{i}^{2}(-{\rm d}t^{2}+{\rm d}z_{i}^{2}+{\rm
d}x_{i}^{2})}{z_{i}^{2}}.$ (S126)

Then, with the result in Ref. [56, 57], we can solve the brane equation
$x_{i}=\tan{\psi_{i}}z_{i}$ with

$\sin{\psi_{1}}=\frac{l_{1}}{2T}\left(T^{2}+\frac{1}{l_{1}^{2}}-\frac{1}{l_{2}^{2}}\right),\qquad\sin{\psi_{2}}=-\frac{l_{1}}{2T}\left(T^{2}+\frac{1}{l_{2}^{2}}-\frac{1}{l_{1}^{2}}\right).$
(S127)

Here we assume the asymptotic boundary of $AdS_{1}$ is located within
$x_{1}<0$, while the asymptotic boundary of $AdS_{2}$ is within $x_{2}>0$ and
the defect A is located at $x_{i}=0$. The positive direction of the angle
$\psi_{i}$ is defined as pointing outsides the $AdS_{i}$ bulk. Now we can
consider the brane equations. Since $l_{1}<l_{2}$, (S127) indicates
$\sin{\psi_{1}}>0$ and $0<\psi_{1}<\frac{\pi}{2}$. Therefore, we expect that
the E1 phase, which includes the center, will be equivalent to the monotonic
brane for $AdS_{1}$, which is shown in Fig. S4 left upper panel. While, the E2
phase, which excludes the center, will be equivalent to the non-monotonic
brane for $AdS_{1}$, which is shown in Fig. S4 right upper panel. A phase
transition between E1 and E2 shown in figure S4 (a) also leads to a transition
of the monotonicity of branes.

Supplementary Figure S4: Different configurations of branes: Left upper panel:
E1 phase with monotonic brane. Left lower panel: E2 phase with monotonic
brane. Right upper panel: E2 phase with non-monotonic brane. Right lower
panel: E1 phase with non-monotonic brane.

However, for $AdS_{2}$ the situation is more complicated. If $T>T_{0}$ with
$-\frac{\pi}{2}<\psi_{2}<0$, it is similar to $AdS_{1}$ that the E1 (E2) phase
is equivalent to (non-)monotonic branes. However, if $T_{\rm min}<T<T_{0}$
with $0<\psi_{2}<\frac{\pi}{2}$, it is opposite to $AdS_{1}$ that the E2 (E1)
phase is equivalent to (non-)monotonic branes, which is shown in Fig. S4 lower
panels. Therefore, we also have the phase transition for the monotonicity of
branes within the same phase E1 or E2.

Monotonicity is related to self-intersecting branes, because a self-
intersecting brane requires the non-monotonicity from the discussion above.
For example, as the E1 phase shown in Fig. S4 right lower panel, the points
$\dot{x}(\sigma)=0$ may touch each other and make the brane self-intersecting.
Note that the self-intersecting solution may not be physical [58], so it is
important to address it. Numerically, we have checked some non-monotonic
examples, and we expect that all of them will not give self-intersecting
branes.

In the following, we discuss some special cases. (i) $L_{1}=L_{2}=L/2$, i.e.,
$\gamma=1$. In this case, as mentioned in Ref. [47], we always have the
solution $\mu=1$. And with (S117), we have $\mu_{2}^{*}(T_{0})=1$. Therefore,
as shown in Fig. S4 (c), (d) and (e), at $\mu=1$, by increasing the tension
$T$, the AdS geometry changes from [E1,E2] to [E1,E1], at the critical point
$T=T_{0}$. Moreover, there are no non-monotonic branes in the solution. This
result is used in (S59). (ii) $l_{1}=l_{2}=l$. In this case, because
$\mu_{1}^{*}=1-l^{2}T^{2}$ and $\mu_{2}^{*}=1/(1-l^{2}T^{2})$, we have
$\mu_{1}^{*}(T_{\rm min})=\mu_{2}^{*}(T_{\rm min})=1$ with $T_{\rm min}=0$.
The phase diagram is Fig. S3 (b). And similar to the discussion above, we find
that in this phase diagram, if and only if the AdS spacetime is in E2 phase,
it has non-monotonic branes. Moreover, if we also have $L_{1}=L_{2}=L/2$, then
there is only one phase [E1,E1] with monotonic branes.

###  .9 Brane solution for the symmetric case

In this section, we derive the brane solution for $L_{1}=L_{2}=L/2$without
corner terms. In this case, $\gamma=1$, and have $\mu=1$. Then we can solve
the quadratic equation of $\sigma$ with coefficients (S8). Letting
$M_{1}=M_{2}=M$, we have $C=0$, $\sigma_{-}=0$ and
$\sigma_{+}=-\frac{2B}{A}=-4T^{2}M/A$. The differential equations (S15) can be
simplified to

$\dot{x}_{1}=-l_{1}\frac{T^{2}+T_{0}^{2}}{2(l_{1}^{2}M+\sigma)\sqrt{A(\sigma-\sigma_{+})}}=\frac{-l_{1}(T^{2}+T_{0}^{2})}{2(l_{1}^{2}M+\sigma)\sqrt{A\sigma+4T^{2}M}},$
(S128a)
$\dot{x}_{2}=-l_{2}\frac{T^{2}-T_{0}^{2}}{2(l_{2}^{2}M+\sigma)\sqrt{A(\sigma-\sigma_{+})}}=\frac{-l_{2}(T^{2}-T_{0}^{2})}{2(l_{2}^{2}M+\sigma)\sqrt{A\sigma+4T^{2}M}}.$
(S128b)

In the following, we use the integral

$\int_{-b}^{\tilde{\sigma}}\frac{{\rm
d}\sigma}{(\sigma+a)\sqrt{\sigma+b}}=\frac{2}{\sqrt{a-b}}\arctan{\sqrt{\frac{b+\tilde{\sigma}}{a-b}}},$
(S129)

where $-b>-a$ and $\tilde{\sigma}>-b>0>a$. With
$4T^{2}-Al_{1}^{2}=l_{1}^{2}(T^{2}+T_{0}^{2})^{2}>0$, and
$4T^{2}-Al_{2}^{2}=l_{2}^{2}(T^{2}-T_{0}^{2})^{2}>0$, we can apply the
integral above to get the brane equations that

$\begin{split}x_{1}(\tilde{\sigma})=&\int_{\sigma_{+}}^{\tilde{\sigma}}\dot{x}_{1}{\rm
d}\sigma=\frac{-l_{1}(T^{2}+T_{0}^{2})}{2\sqrt{A}}\int_{-\frac{4T^{2}M}{A}}^{\tilde{\sigma}}\frac{{\rm
d}\sigma}{(\sigma+l_{1}^{2}M)\sqrt{\sigma+4T^{2}M/A}}\\\
=&\frac{-l_{1}(T^{2}+T_{0}^{2})}{\sqrt{l_{1}^{2}MA-4T^{2}M}}\arctan{\sqrt{\frac{\tilde{\sigma}+4T^{2}M/A}{l_{1}^{2}M-4T^{2}M/A}}}=\frac{-1}{\sqrt{-M}}\arctan{\sqrt{\frac{\tilde{\sigma}+4T^{2}M/A}{l_{1}^{2}M-4T^{2}M/A}}}.\end{split}$
(S130)

as well as

$\begin{split}x_{2}(\tilde{\sigma})=&\int_{\sigma_{+}}^{\tilde{\sigma}}\dot{x}_{2}{\rm
d}\sigma=\frac{-l_{2}(T^{2}-T_{0}^{2})}{2\sqrt{A}}\int_{-\frac{4T^{2}M}{A}}^{\tilde{\sigma}}\frac{{\rm
d}\sigma}{(\sigma+l_{2}^{2}M)\sqrt{\sigma+4T^{2}M/A}}\\\
=&\frac{-l_{2}(T^{2}-T_{0}^{2})}{\sqrt{l_{2}^{2}MA-4T^{2}M}}\arctan{\sqrt{\frac{\tilde{\sigma}+4T^{2}M/A}{l_{2}^{2}M-4T^{2}M/A}}}=\frac{-{\rm
sgn}(T^{2}-T_{0}^{2})}{\sqrt{-M}}\arctan{\sqrt{\frac{\tilde{\sigma}+4T^{2}M/A}{l_{2}^{2}M-4T^{2}M/A}}}.\end{split}$
(S131)

As discussed in Appendix .8, the brane is always monotonic. Therefore, the
monotonically decreasing function $x_{1}(\tilde{\sigma})$ in (S130)
corresponds to $0<\psi_{1}<\frac{\pi}{2}$. While for $x_{2}(\tilde{\sigma})$,
it relies on the sign of $T^{2}-T_{0}^{2}$, and it is consistent with
$\sin{\psi_{2}}$ because $\sin{\psi_{2}}=-\frac{l_{2}}{2T}(T^{2}-T_{0}^{2})$.
We show the diagram of the geometry in Fig. S5.

Supplementary Figure S5: Brane trajectory in the equipartition case
$L_{1}=L_{2}$. Note that $l_{1}<l_{2}$. $AdS_{1}$ always has $\psi_{1}>0$,
which is shown in the right panel. While $\psi_{2}$ can have different signs
for $AdS_{2}$.

Moreover, with (S130) and (S131), we have
$x_{1}(+\infty)-x_{1}(\sigma_{+})=-\frac{\pi}{2}\cdot\frac{1}{\sqrt{-M}}$ and
$x_{2}(+\infty)-x_{1}(\sigma_{+})=-{\rm
sgn}(T^{2}-T_{0}^{2})\frac{\pi}{2}\cdot\frac{1}{\sqrt{-M}}$. Because of
$L=2\pi/\sqrt{-M}$, the interface brane connects antipodal points at the
asymptotic boundary. This is the same as the result of end-of-the-world brane
[48].