Inserting the expression (2.37) for $x_{1}^{\prime}$ gives

$\displaystyle
L_{1}=-\int_{\sigma_{+}}^{\infty}\frac{\ell_{1}\,d\sigma}{(\sigma+M_{1}\ell_{1}^{2})}\,\frac{(\lambda^{2}+\lambda_{0}^{2})\,\sigma+M_{1}-M_{2}}{\sqrt{A\sigma(\sigma-\sigma_{+})(\sigma-\sigma_{-})}}\
,$ (B.21)

and likewise for $L_{2}$. The roots $\sigma_{\pm}$ are given by (2.35,3.48).
We assume that we are not in the case [H2, H2] where $M_{1}=M_{2}>0$, nor in
the fringe case $\sigma_{+}=-M_{j}\ell_{j}^{2}$ when the string goes through
an AdS center. These cases will be treated separately.

Separating the integral in two parts, and trading the integration variable
$\sigma$ for $y$, with $y^{2}:=\sigma_{+}/\sigma$, we obtain

$\displaystyle
L_{1}=-\frac{2\ell_{1}}{\sqrt{A\,\sigma_{+}}}\bigg{[}\frac{M_{1}-M_{2}}{M_{1}\ell_{1}^{2}}\int_{0}^{1}\frac{dy}{\sqrt{(1-y^{2})(1-\nu
y^{2})}}$ $\displaystyle\ ,$ (B.22)
$\displaystyle+\Bigl{(}(\lambda^{2}+\lambda_{0}^{2})-\frac{M_{1}-M_{2}}{M_{1}\ell_{1}^{2}}\Bigr{)}\int_{0}^{1}\frac{y^{2}dy}{(1-u_{1}y^{2})\sqrt{(1-y^{2})(1-\nu
y^{2})}}\bigg{]}$ $\displaystyle\ ,$

where $\nu=\sigma_{-}/\sigma_{+}$ and $u_{1}=-M_{1}\ell_{1}^{2}/\sigma_{+}$.
Identifying the elliptic integrals finally gives

$\displaystyle
L_{1}=-\frac{2\ell_{1}}{\sqrt{A\sigma_{+}}}\bigg{[}\frac{M_{1}-M_{2}}{M_{1}\ell_{1}^{\,2}}\Bigl{(}{\rm\bf
K}(\nu)-{\bf\Pi}(u_{1},\nu)\Bigr{)}+(\lambda^{2}+\lambda_{0}^{2}){\bf\Pi}(u_{1},\nu)\bigg{]}\
,$ (B.23)

and a corresponding expression for $L_{2}$

$\displaystyle
L_{2}=-\frac{2\ell_{2}}{\sqrt{A\,\sigma_{+}}}\bigg{[}\frac{M_{2}-M_{1}}{M_{2}\ell_{2}^{\,2}}\,\Bigl{(}{\rm\bf
K}(\nu)-{\bf\Pi}(u_{2},\nu)\Bigr{)}+(\lambda^{2}-\lambda_{0}^{2})\,{\bf\Pi}(u_{2},\nu)\bigg{]}\
,$ (B.24)

with $u_{2}=-M_{2}\ell_{2}^{2}/\sigma_{+}$. The prefactors in (B.23) diverge
when $M_{1}\to 0$ but the singularity is removed by expanding
${\bf\Pi}(u_{1},\nu)$ around $u_{1}=0$. In this limit

$\displaystyle L_{1}(M_{1}=0)$
$\displaystyle=-\frac{2\ell_{1}}{\sqrt{A\,\sigma_{+}}}\bigg{[}\frac{M_{2}}{\sigma_{-}}({\rm\bf
E}(\nu)-{\rm\bf K}(\nu))+(\lambda^{2}+\lambda_{0}^{2}){\rm\bf K}(\nu)\bigg{]}\
,$ (B.25) $\displaystyle L_{2}(M_{2}=0)$
$\displaystyle=-\frac{2\ell_{2}}{\sqrt{A\,\sigma_{+}}}\bigg{[}\frac{M_{1}}{\sigma_{-}}({\rm\bf
E}(\nu)-{\rm\bf K}(\nu))+(\lambda^{2}-\lambda_{0}^{2}){\rm\bf K}(\nu)\bigg{]}\
,$

with ${\rm\bf E}(\nu)$ the complete elliptic integral of the second kind.

The $M_{1}=M_{2}>0$ geometries correspond to the high-temperature phase where
$M_{j}=(2\pi T)^{2}$, $\sigma_{+}=0$ and $\sigma_{-}=-(4\pi T\lambda)^{2}/A$.
The integrals (2.40) simplify to elementary functions in this case. This seems
to be related to the fact that in this case, the membrane metric becomes
AdS2.:

$\displaystyle L_{1}-\Delta_{1}^{\rm
Hor}=-\frac{\ell_{1}(\lambda^{2}+\lambda_{0}^{2})}{\sqrt{A|\sigma_{-}|}}\int_{0}^{\infty}\frac{ds}{(s+a)\sqrt{s+1}}=-\frac{\ell_{1}(\lambda^{2}+\lambda_{0}^{2})}{\sqrt{A|\sigma_{-}|}}\frac{2}{\sqrt{1-a}}{\rm
arctanh}(\sqrt{1-a})\ ,$ (B.26)

with $a=A\ell_{1}^{2}/4\lambda^{2}$. Using the expression (2.35) for $A$, and
going through the samesteps for $j=2$, the expression greatly simplifies :

$\displaystyle L_{1}-\Delta_{1}^{\rm Hor}$ $\displaystyle=-\frac{1}{\pi
T}\,{\rm
tanh}^{-1}\left(\frac{\ell_{1}(\lambda^{2}+\lambda^{2}_{0})}{2\lambda}\right)\
,$ (B.27) $\displaystyle L_{2}-\Delta_{2}^{\rm Hor}$
$\displaystyle=-\frac{1}{\pi T}\,{\rm
tanh}^{-1}\left(\frac{\ell_{2}(\lambda^{2}-\lambda^{2}_{0})}{2\lambda}\right)\
.$

Interestingly, since $\Delta_{2}^{\rm Hor}$ must be positive, $TL_{2}$ is
bounded from below in the range $\lambda<\lambda_{0}$ as discussed in section
2.6.1.

In the high-temperature phase the on-shell action, (B.15), reads

$\displaystyle I_{\rm
gr}^{\rm(high-T)}=4\pi^{2}T\Bigl{[}-\frac{1}{2}(\ell_{1}L_{1}+\ell_{2}L_{2})+\ell_{1}(L_{1}-\Delta_{1}^{\rm
Hor})+\ell_{2}(L_{2}-\Delta_{2}^{\rm Hor})\Bigr{]}\ .$ (B.28)

Using the expressions (B.27) and rearranging the arc-tangent functions gives

$\displaystyle I_{\rm
gr}^{\rm(high-T)}:=\frac{E}{T}-S=-2\pi^{2}T(\ell_{1}L_{1}+\ell_{2}L_{2})-\log\,g_{I}\
,$ (B.29)

where the interface entropy $S=\log\,g_{I}$ is given by (2.31). By this re-
arrangement we are thus able to recover the interface entropy in a somewhat
roundabout way, which nonetheless confirms the correctness of our solutions.

###  B.4 Sweeping is continuous

In this appendix, we show that sweeping transitions are continuous. We focus
for definiteness on the sweeping of the $j=2$ AdS center at zero temperature
(all other cases work out the same). The transition takes place when $\mu$
crosses the critical value $\mu_{2}^{*}$ given by eq. (2.64). Setting
$\mu=\mu_{2}^{*}(1-\delta)$ in expression (2.52) gives

$\displaystyle
f_{2}(\mu)=\frac{\ell_{2}}{\sqrt{A}}\int_{s_{+}}^{\infty}ds\frac{(\lambda^{2}-\lambda_{0}^{2})(s-\mu\ell_{2}^{2})\,+\delta}{(s-\mu\ell_{2}^{2})\sqrt{As(s-s_{+})(s-s_{-})}}\
,$ (B.30)

$\displaystyle=\frac{2\ell_{2}(\lambda^{2}-\lambda_{0}^{2})}{\sqrt{As_{+}}}\,{\bf
K}\bigl{(}\frac{s_{-}}{s_{+}}\bigr{)}\,+\,\frac{\ell_{2}\,\delta}{\sqrt{A}}\,\underbrace{\int_{s_{+}}^{\infty}\frac{ds}{(s-\mu\ell_{2}^{2})\sqrt{s(s-s_{+})(s-s_{-})}}}_{J}\
.$ (B.31)

The first term is continuous at $\delta=0$, but the second requires some care
because the integral $J$ diverges. This is because for small $\delta$

$\displaystyle s_{+}-\mu\ell_{2}^{2}=\
\frac{\delta^{2}}{4\lambda^{2}\mu_{2}^{*}}+\mathcal{O}(\delta^{3})\ ,$ (B.32)

as one finds by explicit computation of the expression (2.53). we set
$\delta=0$, $J$ diverges near the lower integration limit. To bring the
singular behavior to $0$ we perform the change of variable $u^{2}=s-s_{+}$, so
that

$\displaystyle
J=\int_{0}^{\infty}\frac{2du}{(u^{2}+\delta^{2}/4\lambda^{2}\mu_{2}^{*})\sqrt{(u^{2}+s_{+}^{*})(u^{2}+s_{+}^{*}-s_{-}^{*})}}\
,$ (B.33)

where we kept only the leading order in $\delta$, and $s_{\pm}^{*}$ are the
roots at $\mu=\mu_{2}^{*}$. Since $s_{+}^{*}$ and $s_{+}^{*}-s_{-}^{*}$ are
positive and finite, the small-$\delta$ behavior of the integral is (after
rescaling appropriately $u$)

$\displaystyle
J=\frac{4\lambda|\mu_{2}^{*}|}{|\delta|\sqrt{s_{+}^{*}(s_{+}^{*}-s_{-}^{*})}}\underbrace{\int_{0}^{\infty}\frac{du}{u^{2}+1}}_{\pi/2}+{\rm
finite}\ .$ (B.34)

Inserting in expression (B.31) and doing some tedious algebra leads finally to
a discontinuity of the function $f_{2}(\mu)$ equal to
sign$(\delta)\pi/\sqrt{\mu_{2}^{*}}$. This is precisely what is required for
$L_{2}$, (2.50), to be continuous when the red ($j=2$) slice goes from type E1
at negative $\delta$ to type E2 at positive $\delta$.

This, however, does not tell us anything about the continuity of higher
derivatives of $L_{2}(\mu)$. The expansions that one needs to make become
exponentially bigger, and it becomes nigh impossible to keep track of all the
terms, as even Mathematica struggles to perform the simplifications. However,
we made a numerical analysis with several numerical parameters. What this
analysis revealed is that the continuity of the Free energy seems to extend at
least to the third derivative (going deeper produced numerical instabilities
which made it hard to conclude). For this reason, it is safe to say that the
transition is completely smooth in terms of Free energy and its derivatives.
It is thus a phase transition of another nature.

###  B.5 Bubbles exist

We show here that the bubble phenomenon of section 2.8 is indeed realized in a
region of the parameter space of the holographic model.

This is the region of non-degenerate gravitational vacua ($\ell_{2}$ strictly
bigger than $\ell_{1}$) and a sufficiently light domain wall. Specifically, we
will show that for $\lambda$ close to its minimal value, $\lambda_{\rm min}$,
the arc $L_{1}(\mu=0)$ is negative, so the wall self-intersects and $\mu_{0}$
is necessarily finite.

Let $\lambda=\lambda_{\rm min}(1+\delta)$ with $\delta\ll 1$. Setting $\mu=0$
and expanding eqs. (2.53) for small $\delta$ gives

$\displaystyle A=\frac{8\lambda_{\rm
min}^{2}\,\delta}{\ell_{1}\ell_{2}}+\mathcal{O}(\delta^{2}),\quad
s_{+}=\frac{\ell_{2}}{4\lambda_{\rm min}}+\mathcal{O}(\delta),\quad
s_{-}=-\frac{\ell_{1}}{\ 2\lambda_{\rm min}\delta}+\mathcal{O}(1)\ .$ (B.35)

Plugging into (B.23) with $M_{2}=\mu M_{1}\approx 0$ we find :

$\sqrt{|M_{1}|}L_{1}=-\frac{2}{\ell_{1}\sqrt{As_{+}}}\left[{\bf
K}\left(\frac{s_{-}}{s_{+}}\right)+(1-\frac{2\ell_{1}}{\ell_{2}}){\bf\Pi}\left({\ell_{1}^{2}}{s_{+}},\,\frac{s_{-}}{s_{+}}\right)\right]\
,$ (B.36)

where we have only kept leading orders in $\delta$. Now we need the asymptotic
form of the elliptic integrals when their argument diverges

${\bf
K}\left[-\frac{a}{\delta}\right]\approx{\bf\Pi}\left[u,-\frac{a}{\delta}\right]\approx-\frac{\ln(\delta)\sqrt{\delta}}{2\sqrt{a}}+\mathcal{O}(\sqrt{\delta})\
,$ (B.37)

for $\delta\to 0_{+}$ with $u,a$ fixed. Using $a=2\ell_{1}/\ell_{2}$ finally
gives

$\displaystyle\sqrt{|M_{1}|}L_{1}\approx\bigl{(}\frac{\ell_{2}}{\ell_{1}}-1\bigr{)}^{1/2}\ln(\delta)+{\rm
subleading}\ .$ (B.38)

For $\delta\ll 1$ this is negative, proving our claim. Note that we took the
green slice to be of type E2, as follows from our analysis of the sweeping
transitions for light domain walls – see section 2.7.2.

##  Appendix C Steady holographic interfaces

###  C.1 Extrinsic curvature formulas

In this section we collect some formulas for the extrinsic curvature of the
membrane, in the parametrisation :

$\displaystyle
x^{\mu}_{m}(\sigma,\tau)=(\tau+f(\sigma),\sqrt{\sigma-M\ell^{2}},x_{m}(\sigma))\
,$ (C.1)

where the order of coordinates is $(t,r,x)$, and the metric (3.1). The tangent
and normal (co)vector then are :

$\displaystyle t_{\tau}^{\mu}$ $\displaystyle=(1,0,0)\ ,$ (C.2) $\displaystyle
t_{\sigma}^{\mu}$
$\displaystyle=(f^{\prime}(\sigma),\frac{1}{2\sqrt{\sigma-M\ell^{2}}},x_{m}^{\prime}(\sigma))\
,$ $\displaystyle n_{\mu}$
$\displaystyle=\frac{1}{\mathcal{N}}(0,-x_{m}^{\prime}(\sigma),\frac{1}{2\sqrt{\sigma-M\ell^{2}}})\
,$

where
$\mathcal{N}=\sqrt{\frac{(r^{\prime})^{2}}{r^{2}}\left(\frac{\sigma}{h(r)}\right)+h(r)\frac{(x^{\prime})^{2}}{\ell^{2}}}$
is the normalisation of the normal vector. Note that we write
$\sigma-M\ell^{2}=r^{2}(\sigma)\equiv r^{2}$ where useful, and we used the
definition (3.2) for $h(r)$.

Note that the tangent vector is independent from $f(\sigma)$, this is expected
as it does not change the geometry of the membrane embedding. It will however
have an impact on the matching conditions.

Formula (A.11) gives us the Extrinsic curvature :

$\displaystyle K_{\tau\tau}$
$\displaystyle=\frac{rx^{\prime}(\sigma)h(r)}{\ell^{2}\sqrt{\frac{r^{\prime}{}^{2}\sigma}{r^{2}h(r)}+\frac{h(r)x^{\prime
2}}{\ell^{2}}}}=\frac{r^{2}x^{\prime}(\sigma)h(r)}{\ell\sqrt{\sigma}\sqrt{|\hat{g}|}}\
,$ (C.3) $\displaystyle K_{\sigma\tau}$
$\displaystyle=\frac{\frac{rf^{\prime}(\sigma)x^{\prime}(\sigma)h(r)}{\ell^{2}}-\frac{J\ell
x^{\prime}(\sigma)}{2rh(r)}}{\sqrt{\frac{r^{\prime}{}^{2}\sigma}{r^{2}h(r)}+\frac{h(r)x^{\prime
2}}{\ell^{2}}}}=\frac{\frac{r^{2}x^{\prime}h(r)f^{\prime}}{\ell}-\frac{J\ell^{2}x^{\prime}}{2h(r)}}{\sqrt{\sigma}\sqrt{|\hat{g}|}}\
,$

where all $r$ should be seen as function of $\sigma$. Note we used the
shorthand $\hat{g}={\rm det}(\hat{g})$ defined in (3.29). We did not include
the equation for $K_{\sigma\sigma}$ as it is very cumbersome, and is not used
anywhere in the main-text.

Then writing down the traced-reversed Israel condition (2.18). We indeed find
(3.33) and (3.34) as in the main-text. As for the $(\sigma\sigma)$ equation,
it can be shown with alot of handwork that it is indeed satisfied when the
other two are.

###  C.2 Horizon inequalities

In section 3.6 we asserted that BTZ geometries whose ergoregions can be glued
together by a thin brane obey the inequalities

$\displaystyle\sigma_{+}^{{\rm H}1}$ $\displaystyle>\sigma_{+}^{{\rm
H}2}\quad{\rm if}M_{1}>M_{2}\ ,$ (C.4) $\displaystyle\sigma_{+}^{{\rm H}2}$
$\displaystyle<\sigma_{+}^{{\rm H}1}\quad{\rm if}M_{1}<M_{2}\ ,$

where the horizon locations are

$\displaystyle\sigma_{\pm}^{{\rm
H}j}=-\frac{M_{j}\ell_{j}^{2}}{2}\pm\frac{1}{2}\sqrt{M_{j}^{2}\ell_{j}^{4}-J^{2}\ell_{j}^{2}}\
,$ (C.5)

and $J\equiv|J_{1}|=|J_{2}|>0$. This ordering of the outer horizons is
manifest if one expands at the leading order for small $J$. We want to show
that it is valid for all values of $J$.

If as $J$ is cranked up the ordering was at some point reversed, then we would
have $\sigma_{+}^{{\rm H}1}=\sigma_{+}^{{\rm H}2}$, or equivalently

$\displaystyle
M_{2}\ell_{2}^{2}-M_{1}\ell_{1}^{2}=\sqrt{M_{2}^{2}\ell_{2}^{4}-J^{2}\ell_{2}^{2}}-\sqrt{M_{1}^{2}\ell_{1}^{4}-J^{2}\ell_{1}^{2}}\
.$ (C.6)

Squaring twice to eliminate the square roots gives

$J^{2}=\frac{4\ell_{1}^{2}\ell_{2}^{2}(M_{1}-M_{2})(M_{2}\ell_{2}^{2}-M_{1}\ell_{1}^{2})}{(\ell_{2}^{2}-\ell_{1}^{2})^{2}}\
.$ (C.7)

Without loss of generality we assume, as elsewhere in the text, that
$\ell_{1}\leq\ell_{2}$. If $M_{2}>M_{1}$, then automatically
$M_{2}\ell_{2}^{2}>M_{1}\ell_{1}^{2}$ and (C.7) has no solution for real $J$.
In this case the ordering (C.4) cannot be reversed.

If on the other hand $M_{1}>M_{2}$ and $M_{2}\ell_{2}^{2}-M_{1}\ell_{1}^{2}>0$
we need to work harder. Inserting $J^{2}$ from (C.7) back in the original
equation (C.6) gives after rearrangements

$\displaystyle(\ell_{2}^{2}-\ell_{1}^{2})(M_{2}\ell_{2}^{2}-M_{1}\ell_{1}^{2})=$
$\displaystyle\ell_{2}^{2}\bigl{|}(M_{2}\ell_{2}^{2}-M_{1}\ell_{1}^{2})-\ell_{1}^{2}(M_{1}-M_{2})\bigr{|}\
,$ (C.8)
$\displaystyle-\ell_{1}^{2}\left|(M_{2}\ell_{2}^{2}-M_{1}\ell_{1}^{2})-\ell_{2}^{2}(M_{1}-M_{2})\right|\
,$

where the absolute values come from the square roots. This equation is not
automatically obeyed whenever its doubly-squared version is. A solution only
exists if

$M_{1}-M_{2}\leq\frac{\ell_{2}^{2}M_{2}-\ell_{1}^{2}M_{1}}{\ell_{2}^{2}}\Leftrightarrow\frac{M_{1}}{M_{2}}\leq\frac{2\ell_{2}^{2}}{\ell_{2}^{2}+\ell_{1}^{2}}\
.$ (C.9)

Remember now that we only care about solutions with walls in the ergoregion,
for which $M_{1}-M_{2}=\lambda J$, see eq.(3.53). Plugging in (C.7) this gives

$M_{2}=\bigl{[}1-\frac{4\lambda_{0}^{2}\\\
lam^{2}}{\lambda_{0}^{4}+4{\lambda^{2}}/{\ell_{1}^{2}}}\bigr{]}M_{1}\ ,$
(C.10)

with $\lambda_{0}^{2}=(\ell_{2}^{2}-\ell_{1}^{2})/\ell_{1}^{2}\ell_{2}^{2}$,
see eq.(2.30). Consistency with the bound (C.9) for a brane tension in the
allowed range then requires

$\lambda_{\rm min}<\lambda\leq\frac{\ell_{1}\lambda_{0}^{2}}{2}\ ,$ (C.11)

where $\lambda_{\rm min}=(\ell_{2}-\ell_{1})/\ell_{1}\ell_{2}$. As one can
easily check, this implies $\ell_{1}>\ell_{2}$ which contradicts our initial
assumption. We conclude that (C.6) has no solution, and the ordering (C.4)
holds for all $J$.

For completeness, let us also consider the ordering of the inner horizons.
Clearly $\sigma_{+}^{{\rm H}j}>\sigma_{-}^{{\rm H}j}$ always, and for small
$J$ also $\sigma_{+}^{{\rm H}1}>\sigma_{-}^{{\rm H}2}$ and $\sigma_{+}^{{\rm
H}2}>\sigma_{-}^{{\rm H}1}$. To violate these last inequalities we need
$\sigma_{+}^{{\rm H}1}=\sigma_{-}^{{\rm H}2}$ or $\sigma_{+}^{{\rm
H}2}=\sigma_{-}^{{\rm H}1}$ for some finite $J$, or equivalently

$\displaystyle
M_{2}\ell_{2}^{2}-M_{1}\ell_{1}^{2}=\mp\bigl{(}\sqrt{M_{2}^{2}\ell_{2}^{4}-J^{2}\ell_{2}^{2}}+\sqrt{M_{1}^{2}\ell_{1}^{4}-J^{2}\ell_{1}^{2}}\,\bigr{)}\
.$ (C.12)

Squaring twice gives back eq.(C.7) which has no solution if $M_{2}>M_{1}$. But
if $M_{1}>M_{2}$ and $M_{2}\ell_{2}^{2}-M_{1}\ell_{1}^{2}>0$, solutions to
$\sigma_{+}^{{\rm H}2}=\sigma_{-}^{{\rm H}1}$ cannot be ruled out. Indeed,
inserting $J$ from (C.7) in (C.12) with the $+$ sign gives

$\displaystyle(\ell_{2}^{2}-\ell_{1}^{2})(M_{2}\ell_{2}^{2}-M_{1}\ell_{1}^{2})=$
$\displaystyle\ell_{2}^{2}\bigl{|}(M_{2}\ell_{2}^{2}-M_{1}\ell_{1}^{2})-\ell_{1}^{2}(M_{1}-M_{2})\bigr{|}\
,$ (C.13)
$\displaystyle+\ell_{1}^{2}\left|(M_{2}\ell_{2}^{2}-M_{1}\ell_{1}^{2})-\ell_{2}^{2}(M_{1}-M_{2})\right|\
,$

which requires that

$\displaystyle\frac{\ell_{2}^{2}M_{2}-\ell_{1}^{2}M_{1}}{\ell_{2}^{2}}\leq
M_{1}-M_{2}\leq\frac{\ell_{2}^{2}M_{2}-\ell_{1}^{2}M_{1}}{\ell_{1}^{2}}\ .$
(C.14)

These conditions are compatible with $M_{1}-M_{2}=\lambda J$ and $\lambda$ in
the allowed range, so the outer horizon of slice 2 need not always come before
the Cauchy horizon of slice 1.

Finally one may ask if the inner (Cauchy) horizons can join continuously, i.e.
if $\sigma_{-}^{{\rm H}1}=\sigma_{-}^{{\rm H}2}$ is allowed. A simple
calculation shows that this is indeed possible for $\ell_{2}/\ell_{1}<3$, a
critical ratio of central charges that also arose in references [77, 189]. We
don’t know if this is a coincidence, or if some deeper reason lurks below.

###  C.3 Background on flowing funnels

In this appendix, we collect some formulae on the flowing funnels discussed in
section 3.6.3. We start with the most general asymptotically-locally-AdS3
solution in Fefferman-Graham coordinates, generalizing the Banados geometries
to arbitrary boundary metric, see 1.146, but here $g_{(0)}$ will be arbitrary
instead of flat :

$\displaystyle
ds^{2}=\frac{\ell^{2}dz^{2}}{z^{2}}+\frac{1}{z^{2}}g_{\alpha\beta}(x,z)dx^{\alpha}dx^{\beta}\
,$ (C.15)

where $g_{\alpha\beta}$ is a quartic polynomial in $z$ (written here as a
matrix)

$\displaystyle
g(x,z)=g_{(0)}+z^{2}g_{(2)}+\frac{z^{4}}{4}g_{(2)}g_{(0)}^{-1}g_{(2)}\ .$
(C.16)

In this equation $g_{(0)}$ is the boundary metric and $g_{(2)}$ is given by
(see sec.1.7 for more details :

$\displaystyle
g_{(2)\;\alpha\beta}=\frac{\ell^{2}}{2}R_{(0)}g_{(0)\alpha\beta}+{\ell}\langle
T_{\alpha\beta}\rangle\ ,$ (C.17)

where $R_{(0)}$ is the Ricci scalar of $g_{(0)}$, and $\langle
T_{\alpha\beta}\rangle$ the expectation value of the energy-momentum tensor.
This must be conserved, $\nabla^{a}_{(0)}\langle T_{ab}\rangle=0$, and should
obey the trace anomaly equation $g_{(0)}^{ab}\langle
T_{ab}\rangle=-(c/24\pi)R_{(0)}$.

We may take the boundary metric to be that of the Schwarzschild black hole
(this differs from the metric in [170], but since it is not dynamical we are
free to choose our preferred boundary metric),

$\displaystyle ds^{2}_{(0)}=-f(x)\,dt^{2}+\frac{dx^{2}}{f(x)}\qquad{\rm
with}\quad f(x)=\frac{x}{x+a}\ .$ (C.18)

The horizon at $x=0$ has temperature $\Theta_{S}=(4\pi a)^{-1}$. Using the
familiar tortoise coordinates we can write

$\displaystyle ds^{2}_{(0)}=f(x)(-dt^{2}+dx_{*}^{2})\qquad{\rm where}\quad
x_{*}=x+a\log x\ .$ (C.19)

Let $w^{\pm}=x_{*}\pm t$. The expectation value of the energy-momentum tensor
in the black-hole metric can be expressed in terms of $\phi=\log f(x)$ as
follows

$\displaystyle\langle
T_{\pm\pm}\rangle=\frac{\ell}{2}\,\big{[}\partial^{2}_{\pm}\phi-\frac{1}{2}(\partial_{\pm}\phi)^{2}\big{]}+k_{\pm}(w^{\pm})\
,\quad\langle T_{+-}\rangle=-\frac{\ell}{2}\,\partial_{+}\partial_{-}\phi\ ,$
(C.20)

with $k_{\pm}$ arbitrary functions of $w^{\pm}$ that depend on the choice of
state. At $x\gg a$ where the metric is flat, $k_{\pm}$ determines the incoming
and outgoing fluxes of energy. In a stationary solution, these must be
constant. If a heat bath at temperature $\Theta_{+}$ is placed at infinity,
$k_{+}=\pi^{2}\ell\Theta_{+}^{2}$. The function $k_{-}$, on the other hand, is
fixed by requiring that there is no outgoing flux at the Schwarzschild
horizon.

From

$\displaystyle\frac{\ell}{2}\big{[}\partial^{2}_{\pm}\phi-\frac{1}{2}(\partial_{\pm}\phi)^{2}\big{]}=-\frac{\ell(a^{2}+4ax)}{16(x+a)^{4}}\
,$ (C.21)

we deduce $\langle T_{--}\rangle|_{x=0}=0\Longrightarrow
k_{-}=\ell/16a^{2}=\pi^{2}\ell\Theta_{S}^{2}$. The outgoing flux at infinity
is thermalized at the black hole temperature, as expected.

Inserting the expressions (C.17-C.21) in (C.15) and (C.16) gives the flowing-
funnel metric in Fefferman-Graham coordinates. These are however singular
coordinates, not well adapted for calculating the event horizon as shown in
[170]. Following this reference, one can compute the horizon by going to BTZ
coordinates – this is possible because all solutions are locally equivalent in
three dimensions. The change from any metric (C.15) - (C.16) to local BTZ
coordinates has been worked out in ref. [227] (see also [228]) and can be used
to compute the black-funnel shapes. A noteworthy feature is that the funnels
start vertically inwards at $x=0$ [170] making a delta-function contribution
to the area density. Note that figure 3.7 shows two independent flowing
funnels with Schwarzschild temperatures $\Theta_{S}=\Theta_{1}^{\rm eff}$ and
$\Theta_{2}^{\rm eff}$.

##  Appendix D RT surfaces and entanglement entropy

###  D.1 Euclidean construction for single-crossing geodesic

We use fig. 4.5 and aim to express the depicted points as a function of
$\lambda$ (where $O_{2}=(\lambda,0)$), $\varphi$ and $\theta_{i}$. The basic
equation we will use is the law of sines inside various triangles. We will
work with fig. 4.5 as a reference, but if one is careful the formulas are
applicable also to situations not depicted by the figure, for instance, when
$\lambda>0$. When writing something like $O_{1}O_{2}$, we mean the length
between the segment $[O_{1},O_{2}]$. This in particular could become negative
in the formulas we will give, signaling that $O_{1}$ and $O_{2}$ cross. This
is always accompanied by a sign change in the opposing angle so that the law
of sines still works.

By the triangle $(O_{2},O,X)$, we have :

$\displaystyle\frac{O_{2}X}{\sin(\theta_{2})}=\frac{OO_{2}}{\sin(\theta_{2}-\phi)}\Leftrightarrow
O_{2}X=\frac{(-\lambda)\sin(\theta_{2})}{\sin(\theta_{2}-\phi)}\ ,$ (D.1)

where we use $OO_{2}=-\lambda$.

Then we find $OO_{1}$ using the triangle $(O,O_{2},O_{1})$ :

$\displaystyle
OO_{1}=\frac{OO_{2}\sin(\pi-\phi)}{\sin(\phi-\theta_{1})}=\frac{(-\lambda)\sin(\phi)}{\sin(\phi-\theta_{1})}\
.$ (D.2)

The last piece we need is $O_{1}X$, using the triangle $(O_{1},X,O)$ :

$\displaystyle
O_{1}X=\frac{OO_{1}\sin(\pi-\theta_{2}+\theta_{1})}{\sin(\theta_{2}-\phi)}\ .$
(D.3)

With that in mind we can compute the anchor points $\sigma_{1}$, $\sigma_{2}$,
as well as $Ox$ as we should require it to be positive :

$\displaystyle O\sigma_{2}$
$\displaystyle=O_{2}X-OO_{2}=(-\lambda)\left(\frac{\sin(\theta_{2})}{\sin(\theta_{2}-\phi)}-1\right)\
,$ (D.4) $\displaystyle O\sigma_{1}$
$\displaystyle=OO_{1}+O_{1}X=\frac{(-\lambda)\sin(\phi)}{\sin(\phi-\theta_{1})}\left(1+\frac{\sin(\theta_{2}-\theta_{1})}{\sin(\theta_{2}-\phi)}\right)\
,$ $\displaystyle OX$
$\displaystyle=\frac{(-\lambda)\sin(\phi)}{\cos(\theta_{2}-\phi)}\ .$

We can re-express these in terms of the initial angles with the identities
$\theta_{2}=\frac{\pi}{2}+\psi_{2}$ and $\theta_{1}=\psi_{1}+\psi_{2}$.

$\displaystyle O\sigma_{2}$
$\displaystyle=(-\lambda)\left(\frac{\cos(\psi_{2})}{\cos(\psi_{2}-\phi)}-1\right)\
,$ (D.5) $\displaystyle O\sigma_{1}$
$\displaystyle=\frac{(-\lambda)\sin(\phi)}{\sin(\phi-\psi_{2}-\psi_{1})}\left(1+\frac{\cos(\psi_{1})}{\cos(\psi_{2}-\phi)}\right)\
,$ $\displaystyle OX$
$\displaystyle=\frac{(-\lambda)\sin(\phi)}{\cos(\phi-\psi_{2})}\ .$

Then, by replacing $\phi-\psi_{2}\equiv\alpha$, we obtain the equation (4.29)
used in the main-text.

We would like now to compute the length of this geodesic. We will apply the
formula (4.33) to the two pieces composing the geodesic. When the geodesic is
anchored at the boundary, we cut it off at $z_{i}=\epsilon_{i}$. Denoting $R$
the radius of the semi-circle, the initial angle $\theta_{0}$ is given by :

$\displaystyle\frac{\epsilon_{0}}{R}=\sin(\alpha_{0})\approx\alpha_{0}\ .$
(D.6)

Thus because of the presence of the cutoffs, the length of the geodesic does
depend on its radius, unlike what was suggested by the formula (4.33).

The semi-circle going from $\sigma_{2}\rightarrow x$ has radius $R_{2}$ given
by :

$\displaystyle
R_{2}=O_{2}\sigma_{2}=\frac{(-\lambda)\cos(\psi_{2})}{\cos(\alpha)}\ .$ (D.7)

The starting angle for this portion is given by the cutoff $\epsilon_{2}$,
while $\theta_{f}=\phi$. This gives the length $L_{2}$ of this portion :

$\displaystyle
L_{2}=\ell_{2}\ln(\frac{2R_{2}\sin(\alpha+\psi_{2})}{\epsilon_{2}(1+\cos(\alpha+\psi_{2}))})=\ell_{2}\ln(\frac{2R_{2}}{\epsilon_{2}}\tan(\frac{\alpha+\psi_{2}}{2}))\
,$ (D.8)

keeping only the leading order in $\epsilon_{2}$.

The other semi-circle from $X\rightarrow\sigma_{1}$ has radius $R_{1}$ :

$\displaystyle
R_{1}=O_{1}X=\frac{(-\lambda)\cos(\psi_{1})\sin(\alpha+\psi_{2})}{\sin(\alpha-\psi_{1})\cos(\alpha)}\
,$ (D.9)

while the initial and final angles are $\theta_{0}=\alpha-\psi_{1}$ and
$\theta_{f}=\frac{\epsilon_{1}}{R_{1}}$. Then its length $L_{1}$ is :

$\displaystyle
L_{1}=\ell_{1}\ln(\frac{2R_{1}(1+\cos(\alpha-\psi_{1}))}{\epsilon_{1}\sin(\alpha-\psi_{1})})=\ell_{1}\ln(\frac{2R_{1}}{\epsilon_{1}}\frac{1}{\tan(\frac{\alpha-\psi_{1}}{2})})\
.$ (D.10)

The full length of the geodesic is obtained simply by summing the two
contributions, $L=L_{1}+L_{2}$. After some algebra and re-arranging, we obtain
the formula (4.32) used in the main-text.

###  D.2 Euclidean construction for double-crossing geodesic

We use fig. 4.6 to express the depicted points as a function of $\lambda$
(where $O_{2}=(\lambda,0)$), $\varphi$ and the $\theta_{i}$. As in the
previous section, this is done by repeated use of the law of sines.

Beginning with the triangle $(O,O_{1},O_{3})$, we have :

$\displaystyle OO_{2}=\frac{(-\lambda)\sin(\phi)}{\sin(\theta_{1}-\phi)}\ .$
(D.11)

With the triangle $(O,O_{2},O_{3})$ :

$\displaystyle
OO_{3}=OO_{2}\frac{\sin(2\theta_{2}-\theta_{1}-\phi)}{\sin(2\theta_{2}-\phi)}=-\frac{\lambda\sin(\phi)\sin(2\theta_{2}-\theta_{1}-\phi)}{\sin(\theta_{1}-\phi)\sin(2\theta_{2}-\phi)}\
.$ (D.12)

To find $\sigma_{2}$, we need the radius of the semi-circle centered at
$O_{3}$. We use the triangle $(O,O_{3},X_{2})$ :

$\displaystyle
O_{3}X_{2}=OO_{3}\frac{\sin(\theta_{2})}{\sin(\theta_{2}-\phi)}\ ,$ (D.13)

where $OO_{3}$ is given by (D.13). Then $O\sigma_{2}=OO_{3}+O_{3}X_{2}$ :

$\displaystyle O\sigma_{2}$
$\displaystyle=OO_{3}(1+\frac{\sin(\theta_{2})}{\sin(\theta_{2}-\phi)})\ .$
(D.14)

Similarly, $O\sigma_{1}=O_{1}\sigma_{1}-OO_{1}=O_{1}X_{1}-OO_{1}$. We use
triangle $(O,O_{1},X_{1})$ :

$\displaystyle O_{1}X_{1}$
$\displaystyle=(-\lambda)\left(1-\frac{\sin(\theta_{2})}{\sin(\theta_{2}-\phi)}\right)\
,$ (D.15) $\displaystyle O\sigma_{1}$
$\displaystyle=(-\lambda)\left(\frac{\sin(\theta_{2})}{\sin(\theta_{2}-\phi)}\right)\
.$

Collecting the expression and replacing with the angles $\psi_{i}$ :

$\displaystyle O\sigma_{1}$
$\displaystyle=(-\lambda)\left(\frac{\cos(\psi_{2})}{\cos(\psi_{2}-\phi)}-1\right)\
,$ (D.16) $\displaystyle O\sigma_{2}$
$\displaystyle=\frac{(-\lambda)\sin(\phi)}{\sin(\psi_{1}+\psi_{2}-\phi)}\frac{\sin(\phi+\psi_{1}-\psi_{2})}{\sin(\phi-2\psi_{2})}\left(1+\frac{\cos(\psi_{2})}{\cos(\psi_{2}-\phi)}\right)\
,$

which yield the formulas (4.36) when doing the change of variables
$\alpha=\phi-\psi_{2}$. In what follows, we express everythin in terms of
$\alpha$ instead of $\phi$.

For the Length of the geodesics, we must now compute the length of three
separate segments, using the formula (4.33). Some extra angles need to be
determined on fig. 4.6, but we don’t details that here.

Consider first the segment connecting $\sigma_{1}\rightarrow X_{1}$. The
radius is given by $R_{1}$ :

$\displaystyle R_{1}=O_{1}X_{1}=\frac{(-\lambda)\cos(\psi_{2})}{\cos(al)}\ .$
(D.17)

The starting angle $\theta_{0}$ is given by the cutoff, $\epsilon_{1}$, while
the ending $\theta_{f}=\alpha+\psi_{2}$.

This gives a contribution to the length :

$\displaystyle
L_{1}=\ell_{2}\ln\left(\frac{2R_{1}}{\epsilon_{1}}\tan\left(\frac{\alpha+\psi_{2}}{2}\right)\right)\
.$ (D.18)

For the second segment $X_{1}\rightarrow X_{2}$, the Radius of the geodesic is
irrelevant. The initial angle is $\theta_{0}=\pi-(\alpha+\psi_{1})$, final
$\theta_{f}=\psi-(\psi_{1}-\alpha)$. This gives the contribution $L_{3}$ :

$\displaystyle
L_{3}=\ell_{1}\ln\left(\frac{\tan(\frac{\alpha+\psi_{1}}{2})}{\tan(\frac{\psi_{1}-\alpha}{2})}\right)\
.$ (D.19)

For the last segment $L_{2}$, the radius $R_{2}$ is :

$\displaystyle
R_{3}=O_{3}X_{2}=OO_{3}\frac{\sin(\theta_{2})}{\sin(\theta_{2}-\phi)}=\frac{-\lambda\sin(\alpha+\psi_{2})\sin(\alpha+\psi_{1})}{\sin(\psi_{1}-\alpha)\sin(\alpha-\psi_{2})}\frac{\cos(\psi_{2})}{\cos(\alpha)}\
.$ (D.20)

The corresponding length $L_{2}$ is then :

$\displaystyle
L_{3}=\ell_{2}\ln(\frac{2R_{3}}{\epsilon_{2}\tan(\frac{\alpha-\psi_{2}}{2})})\
.$ (D.21)

Adding the contributions of the $L_{i}$ gives the length of the full geodesic.
After re-arranging, we obtain (4.40) used in the main-text.

###  D.3 Peculiar geodesics in the spinning black string geometry

In this section we briefly look at the shape of the different spacelike
geodesics in the spinning string geometry. In particular, we highlight the
fact that there are horizon entering geodesics, but that they cannot have two
anchors on the boundary.

The geodesics are obtained from the equations of sec.4.1, composed with the
change of coordinates (4.54). Consider first the type $K_{+}K_{-}>0$,
described in sec.4.1.1. In poincaré space, this geodesic has two anchor points
on the boundary. However, this does not necessarily translate to Finkelstein
coordinates. To see this, it is useful to consider the inverse of (4.54) (for
$J>0$) :

$\displaystyle r^{2}$
$\displaystyle=r_{+}^{2}+\frac{(w_{+}w_{-})(r_{+}^{2}-r_{-}^{2})}{z_{p}^{2}}\
,$ (D.22) $\displaystyle v$
$\displaystyle=\frac{\ell}{2}\left(\frac{1}{r_{+}-r_{-}}\ln(\frac{w_{+}(r+r_{-})}{r+r_{+}})-\frac{1}{r_{+}+r_{-}}\ln(\frac{w_{-}(r+r_{-})}{r-r_{+}})\right)\
,$ $\displaystyle y$
$\displaystyle=\frac{\ell}{2}\left(\frac{1}{r_{+}-r_{-}}\ln(\frac{w_{+}(r+r_{-})}{r+r_{+}})+\frac{1}{r_{+}+r_{-}}\ln(\frac{w_{-}(r+r_{-})}{r-r_{+}})\right)\
,$

where the $r$ appearing in the RHS should be thought as depending on
$w_{\pm}$, $z_{p}$.

When using Poincaré coordinates to compute geodesics, we must keep in mind the
region of Poincaré which describes Finkelstein space, namely $w_{+}>0$. Thus,
when a geodesics is of the type $K_{+}K_{-}>0$, either it remains in the
region $w_{+}>0$, and thus it is doubly anchored in Finkelstein space (see
fig.4.2), or it reaches the $w_{+}=0$ boundary. In this case, by (D.22), the
geodesics escapes at $(v\rightarrow-\infty,r=r_{+},y\rightarrow-\infty)$. As
such, it follows closely the apparent horizon towards the ”coordinate”
horizon, both in time and space. Note that the divergence is such that
$\frac{v}{y}\approx\frac{r_{+}}{r_{-}}$. In the context of a pure CFT, these
geodesics do not contribute to the RT prescription, as they have only one
anchor point. In ICFT, they might be ”stopped” by the membrane, and upon
crossing to the other side, head back to the boundary, so it is plausible that
they play a role in this case.

By (4.54), the outer horizon of the black hole is mapped to $w_{-}=0$. Thus,
one might argue that it should be possible to enter the horizon of the black
hole and head back to the boundary: a geodesic in Poincaré that dips in the
$w_{-}<0$ region and heads back to the boundary should generate such a horizon
probing geodesic in Finkelstein coordinates. However, from (4.9), the curves
$w_{\pm}(\lambda)$ are obviously monotonic, barring such a possibility. Thus
if a geodesic crosses the region $w_{-}<0$, it will remain in it and thus be
unable to leave the black hole. Its precise fate depends on the geodesic, it
could again escape at infinity before reaching the singularity if it reaches
$w_{+}=0$, otherwise, it will reach $r=0$ as $z_{p}\rightarrow 0$, as seen
from (D.22). Again, these geodesics are irrelevant in the context of a pure
CFT, but might be very important in the context of ICFT. If they intersect the
membrane on their way to the singularity, that could give rise to geodesics of
the type depicted in fig.4.11.

Consider now the case $K_{+}K_{-}<0$, see sec.4.1.2. Now, as explained in the
main text, even in Poincaré we are in presence of two disconnected branches
each with one anchor point on the boundary. Considering them as separate
geodesics, we look at the mapping of one of the two branches to Finkelstein
space. The discussion is very similar, namely, we must pay attention to the
important surfaces $w_{\pm}=0$ in Poincaré space. However, in this case as the
affine parameter $\lambda$ approaches $0$ (we assume we begin with
$\lambda_{0}<0$) we will have $w_{\pm}\rightarrow-{\rm Sign}(w_{\mp})\infty$.
As such, we are assured to either enter the horizon at $w_{-}=0$, or escape at
infinity at $w_{+}=0$. Indeed, as in this case the signs of $\dot{w}_{\pm}$
are opposing, one of the two situations will happen.

We do not comment on the last case $K_{+}K_{-}=0$, as it is irrelevant to the
numerical algorithms. Nothing special happens with respect to the two other
cases, the same reasoning applies.

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