††institutetext: Kavli Institute for Theoretical Sciences (KITS),
University of Chinese Academy of Sciences (UCAS), Beijing 100190, China

# Islands in Generalized Dilaton Theories

Jia Tian<EMAIL_ADDRESS>
###### Abstract

In this work we systematically study the island formula in the general
asymptotically flat eternal black holes in generalized dilaton gravity
theories or in higher dimensional spherical black holes. Under some reasonable
and mild assumptions we prove that the island always appears barely outside of
the horizon in the late time of Hawking radiation so that the information
paradox is resolved. In particular, we find proper island in Liouville black
hole which solves the puzzle of Li:2021lfo .

##  1 Introduction

Over the last five years there is a dramatic acceleration of progress on
quantum aspects of black holes Bousso:2022ntt . The most exciting achievement
is the successful derivation of Page curve Page:1993wv  which resolves the
long-standing information paradox Hawking:1975vcx  by proposing a new rule,
the island formula Penington:2019npb ; Almheiri:2019psf ; Almheiri:2019hni ,
for computing the entanglement entropy of the Hawking radiations. The island
formula mimics the quantum extremal surface (QES) prescription
Engelhardt:2014gca  of the generalized entanglement entropy:

$\displaystyle S_{R}=\text{min}\Big{\\{}\text{ext}\left[\frac{A(\partial
I)}{4G_{N}}+S_{\text{semi-cl}}[\text{Rad}\cup I]\right]\Big{\\}},$ (1)

where $I$ is the island which is a codimension-one region and $A(\partial I)$
is the area of its boundary $\partial I$, the QES. The surprising fact about
this formula is that its right-hand side only depends on semi-classical
physics and importantly the island is not added by hand but its existence can
be justified by the replica tricks of the gravitational Euclidean path
integral Penington:2019kki ; Almheiri:2019qdq . Another way to understand the
island formula is by combining the AdS/BCFT correspondence and the brane world
holography Sully:2020pza ; Chen:2020uac ; Chen:2020hmv ; Suzuki:2022xwv .

The island formula has been successfully applied to black holes in various of
gravitational theories Krishnan:2020fer ; Caceres:2020jcn ; Geng:2021wcq ;
Geng:2021iyq ; Gautason:2020tmk ; Hartman:2020swn ; Hollowood:2020cou ;
Goto:2020wnk ; Chen:2020jvn ; Wang:2021mqq ; Almheiri:2019psy ;
Hashimoto:2020cas ; Wang:2021woy ; Yu:2021cgi ; Ahn:2021chg ;
Karananas:2020fwx ; Lu:2021gmv ; Krishnan:2020oun ; Alishahiha:2020qza ;
Anegawa:2020ezn ; He:2021mst . However there is also a notable counterexample
Li:2021lfo  where it was claimed that the island formula can not save the
information paradox of Liouville black hole. More puzzlingly this claim seems
to be inconsistent with the systematic analysis of QES in general
D-dimensional asymptotically flat (or AdS) eternal black hole performed in
He:2021mst . In this work, we solve this puzzle by finding a new Liouville
black hole solution with the help of the general construction of classical
solutions of generalized dilaton theories. We also conduct a systematic
analysis of islands in asymptotically flat eternal black holes. Our general
results agree with the ones in He:2021mst .

##  2 2D Generalized dilaton Gravity Theory

Because of their remarkable solvability, 2D GDTs as toy models of quantum
gravity serve as a laboratory for studying properties of black holes. The
general action of 2D GDTs is

$\displaystyle
S=\frac{1}{2\pi}\int\operatorname{d}^{2}x\sqrt{-g}\left[XR-U(X)(\nabla
X)^{2}-2V(X)\right],$ (2)

which depends on the metric $g_{\mu\nu}$ and the scalar field $X$. The scalar
field $X$ is usually related to the dilaton $\phi$ via $X=e^{-2\Phi}$. In
terms of $\Phi$, the action is in a more familiar form

$\displaystyle
S_{\text{dil}}=\frac{1}{2\pi}\int\operatorname{d}^{2}x\sqrt{-g}e^{-2\Phi}\left[R-\tilde{U}(\Phi)(\nabla\Phi)^{2}-2\tilde{V}(\Phi)\right]$
(3)

with the identifications

$\displaystyle\tilde{U}=4e^{-2\Phi}U,\quad\tilde{V}=e^{2\Phi}V.$ (4)

All the classical solutions of (2) can be found in a closed form with the help
of the first-order formalism of GDT. We will show our convention and review
the construction of classic solutions in the Appendix A. The general solution
of (2) is given by111here we only consider the interesting linear dilaton
vacua. Grumiller:2002nm ; Grumiller:2006rc

$\displaystyle\operatorname{d}s^{2}$ $\displaystyle=$ $\displaystyle
2e^{Q}\operatorname{d}v(\operatorname{d}X+(w(X)-C_{0})\operatorname{d}v)$ (5)
$\displaystyle=$ $\displaystyle
2\operatorname{d}v\operatorname{d}\tilde{X}+\xi(\tilde{X})\operatorname{d}v^{2},$
(6)

where we have introduced

$\displaystyle\operatorname{d}\tilde{X}=\operatorname{d}Xe^{Q},\quad\xi(\tilde{X})=2e^{Q}(w-C_{0}),$
(7) $\displaystyle Q=\int^{X}U(y)\operatorname{d}y,$ (8) $\displaystyle
w=\int^{X}e^{Q}V(y)\operatorname{d}y.$ (9)

The solution is parameterized by a constant $C_{0}$ which is usually related
to the mass of the black hole. We can transform the metric to the diagonal
gauge by introducing the coordinates $x^{0}=r,x^{1}=r$, thus
$\operatorname{d}\tilde{X}=\dot{\tilde{X}}\operatorname{d}t+\tilde{X}^{\prime}\operatorname{d}r$,
$\operatorname{d}v=\dot{v}\operatorname{d}t+v^{\prime}\operatorname{d}r$ and
setting

$\displaystyle\tilde{X}=\tilde{X}(r),\quad\tilde{X}^{\prime}+\xi
v^{\prime}=0.$ (10)

The resulting metric is

$\displaystyle\operatorname{d}s^{2}=\xi(\dot{v}^{2}\operatorname{d}t^{2}-{v^{\prime}}^{2}\operatorname{d}r^{2}).$
(11)

If we further set $\tilde{X}=r$ and $\dot{v}=1$, we can get

$\displaystyle\operatorname{d}s^{2}=\xi\operatorname{d}t^{2}-\frac{1}{\xi}\operatorname{d}r^{2}\,,$
(12)

which is in the Schwarzschild gauge if the solution describes a black hole. In
this paper we will focus on asymptotically flat solutions so we take the
ansatz that that $\xi$ approaches to some constant

$\displaystyle\lim_{r\rightarrow\infty}\xi=-\xi_{0}^{2},$ (13)

where $\xi_{0}^{2}$ should be positive to ensure the correct signature. Then
we rescale the coordinates as

$\displaystyle t=\frac{t^{\prime}}{\xi_{0}},\quad r=\xi_{0}r^{\prime}$ (14)

such that

$\displaystyle\operatorname{d}s^{2}=\frac{\xi}{\xi_{0}^{2}}{\operatorname{d}t^{\prime}}^{2}-\frac{\xi_{0}^{2}}{\xi}{\operatorname{d}r^{\prime}}^{2},\quad\lim_{r^{\prime}\rightarrow\infty}\operatorname{d}s^{2}=-{\operatorname{d}t^{\prime}}^{2}+{\operatorname{d}r^{\prime}}^{2}.$
(15)

The horizon of the black hole is at
$r_{H}^{\prime},\quad\xi(r_{H}^{\prime})=0$. The temperature and the entropy
of the black hole are Iyer:1994ys

$\displaystyle
T_{\text{BH}}=\partial_{r^{\prime}}{\xi/\xi_{0}^{2}}\,\Big{|}_{r^{\prime}=r^{\prime}_{H}},\quad
S_{\text{BH}}=2X(r^{\prime}_{H}).$ (16)

##  3 General results

###  3.1 Setting up the calculation

Our goal is to compute the entanglement entropy of the Hawking radiation of
the eternal black hole with the island formula (1). For 2d GDT (2), the first
term is given by the value of dilaton field at the position of the island
boundary $2X(\partial I)$ Iyer:1994ys . $S_{\text{semi-cl}}[\text{Rad}\cup I]$
is semi-classical entanglement entropy of Hawking radiation in the region
$\text{Rad}\cup I$. For eternal black holes, we first introduce Kruskal
coordinates

$\displaystyle\operatorname{d}s^{2}=-e^{2\rho(y^{+},y^{-})}\operatorname{d}y^{+}\operatorname{d}y^{-}$
(17)

to cover the whole regions then we choose Rad to be two symmetric intervals
$[y_{L\infty},y_{-a}]\cup[y_{a},y_{R\infty}]$, where the coordinates of the
end points are

$\displaystyle y_{L\infty}=(y_{a}^{0},-\infty),\quad
y_{R\infty}=(y_{a}^{0},\infty),\quad y_{-a}=(y_{a}^{0},-y_{a}^{1}),\quad
y_{a}=(y_{a}^{0},y_{a}^{1}).$ (18)

With this symmetric choice, it is reasonable to expect that the island222We
assume that the one-interval island configuration dominates. $[y_{-d},y_{d}]$
also enjoys this symmetry

$\displaystyle y_{-d}=(y_{d}^{0},-y_{d}^{1}),\quad
y_{d}=(y_{d}^{0},y_{d}^{1}).$ (19)

The Kruskal coordinates are related to two copies of Schwarzschild coordinates
via

$\displaystyle e^{lx_{R}^{+}}=ly_{R}^{+},\quad
e^{-lx_{R}^{-}}=-ly_{R}^{-},\quad e^{-lx_{L}^{+}}=-ly_{L}^{+},\quad
e^{lx_{L}^{-}}=ly_{L}^{-},$ (20) $\displaystyle y^{+}_{R}\geq 0,\quad
y^{-}_{R}\leq 0,\quad y^{+}_{L}\leq 0,\quad y^{-}_{L}\geq 0\,,$ (21)

where $y_{R(L)}^{\pm}=y^{0}_{R(L)}\pm y^{1}_{R(L)}$ labels the position in the
right (left) patch of Kruskal spacetime and $l$ is some convenient constant
and

$\displaystyle x^{\pm}=t^{\prime}\pm x^{*},\quad
x^{*}=\int\frac{\operatorname{d}r^{\prime}}{-\xi/\xi_{0}^{2}}.$ (22)

From the transformations (about $x_{R}$) we find that

$\displaystyle ly^{+}\operatorname{d}x^{+}=\operatorname{d}y^{+},\quad-
ly^{-}\operatorname{d}x^{-}=\operatorname{d}y^{-}\,$ (23) $\displaystyle
e^{2\rho}=\frac{\xi(y^{+}y^{-})}{\xi_{0}^{2}l^{2}y^{+}y^{-}}.$ (24)

We will model the Hawking radiation with a probe conformal field theory with a
central charge $c<<1/G_{N}$ so can use the semi-classical formula to compute
the entanglement entropy $S_{\text{semi-cl}}(\text{Rad}\cup I)$.

###  3.2 Entanglement entropy without islands

In the early time of Hawking radiation, there are very few Hawking quanta in
the interior of black holes so the island configuration can not be supported.
So as more and more Hawking quanta escape to the infinity the asymptotic
(Schwarzschild) observer should see a growing entanglement entropy. Assuming
that the initial state of the quantum field is pure then

$\displaystyle S_{\text{semi-cl}}[\text{Rad}]$ $\displaystyle=$
$\displaystyle\frac{c}{6}\log\left({|(y_{a}-y_{-a})^{+}(y_{a}-y_{-a})^{-}|e^{\rho(y_{a})}e^{\rho(y_{-a})}}\right)$
$\displaystyle=$
$\displaystyle\frac{c}{6}\log\left(|2y_{a}^{1}|^{2}e^{\rho(y_{a}^{+},y_{a}^{-})}e^{\rho(y_{a}^{-},y_{a}^{+})}\right),$
$\displaystyle=$
$\displaystyle\frac{c}{3}\log\left(2\cosh(lt^{\prime}_{a})\right)+\frac{c}{6}\log\left(-\frac{\xi}{\xi_{0}^{2}l^{2}}\right),$
(26)

thus indeed as shown in the Page curve the growth of early entanglement
entropy is general. Without introducing islands, the entanglement entropy will
exceed the Bekenstein-Hawking entropy before the black hole completely
evaporates. However the Bekenstein-Hawking entropy should be the upper bound
of entanglement entropy. This contradiction is the well known information
paradox.

###  3.3 Entanglement entropy with islands

In the late time of black hole evaporation, as we have seen that we have to
include the islands. But as stressed in Almheiri:2020cfm , the island region
is not included by hand but its appearance is a result of evaluating
gravitational path integral around non-trivial saddles. As a result, the black
hole evaporation is unitary such that the entanglement entropy will vanish in
the end as the Page curve shows. This implies the position of the $\partial I$
is very close to the horizon i.e. $y_{d}^{-}=y_{d}^{0}-y_{d}^{1}\approx 0$
which means that $y_{d}^{1}$ is very large. Therefore we can approximate the
entanglement entropy $S_{\text{Rad}}([y_{-a},y_{-d}]\cup[y_{d},y_{a}])$ with
$2S_{\text{Rad}}([y_{d},y_{a}])$ thus

$\displaystyle
S_{\text{island}}=\frac{4}{G_{N}}X(y_{d})+\frac{c}{3}\log\left(|(y_{a}-y_{d})^{+}(y_{a}-y_{d})^{-}|e^{\rho(y_{a})}e^{\rho(y_{d})}\right),$
(27)

where we have added back the Newton’s constant $G_{N}$ to indicate that in the
semi-classical limit, the first term should be much larger than the second
term. Taking this approximation as an ansatz and then solving $y_{d}$ by
extremizing the generalized entanglement entropy (27) we can show indeed this
approximation is correct.

Differentiating (27) with respect to $y_{d}^{\pm}$ gives two extremal
conditions

$\displaystyle\frac{4}{G_{N}}\frac{\operatorname{d}X}{\operatorname{d}y_{d}^{+}}+\frac{c}{3}\left(\frac{1}{y_{d}^{+}-y_{a}^{+}}+\frac{\operatorname{d}\rho}{\operatorname{d}y_{d}^{+}}\right)=0,$
(28)
$\displaystyle\frac{4}{G_{N}}\frac{\operatorname{d}X}{\operatorname{d}y_{d}^{-}}+\frac{c}{3}\left(\frac{1}{y_{d}^{-}-y_{a}^{-}}+\frac{\operatorname{d}\rho}{\operatorname{d}y_{d}^{-}}\right)=0.$
(29)

Recalling that $\tilde{X}=r$ first term can be evaluated as

$\displaystyle\frac{\operatorname{d}X}{\operatorname{d}d_{R}^{\pm}}=\frac{\operatorname{d}X}{\operatorname{d}\tilde{X}}\frac{\operatorname{d}r}{\operatorname{d}y_{d}^{\pm}}=e^{-Q}\frac{\xi_{0}\operatorname{d}r^{\prime}}{\operatorname{d}y_{d}^{\pm}}=-e^{-Q}\frac{\xi\operatorname{d}x^{*}}{\xi_{0}\operatorname{d}y_{d}^{\pm}}=-e^{-Q}\frac{\xi}{2l\xi_{0}y_{d}^{\pm}}$
(30)

where $\xi(z)$ should be understood as function of $y_{d}^{+}y_{d}^{-}\equiv
z$. Using the expression (24) the last term can be computed as

$\displaystyle\frac{\operatorname{d}\rho}{\operatorname{d}y_{d}^{\pm}}=\frac{1}{2y_{d}^{\pm}}\left(\frac{\xi^{\prime}}{\xi}\frac{\operatorname{d}z}{y_{d}^{\pm}}-\frac{1}{y_{d}^{\pm}}\right)=\frac{1}{2y_{d}^{\pm}}\left(\frac{\xi^{\prime}z}{\xi}-1\right)$
(31)

Then the two equations (28) and (29) can be written as

$\displaystyle\frac{1}{3}\frac{1}{y_{d}^{\pm}-y_{a}^{\pm}}-\frac{2\xi
e^{-Q}}{cG_{N}l\xi_{0}}\frac{1}{y_{d}^{\pm}}+\frac{1}{3}\frac{\xi-z\xi^{\prime}}{2\xi}\frac{1}{y_{d}^{\pm}}=0,$
(32)

from which we can obtain the relation

$\displaystyle\frac{y_{d}^{+}}{y_{a}^{+}-y_{d}^{+}}=\frac{y_{d}^{-}}{y_{a}^{-}-y_{d}^{-}},\quad\rightarrow\quad
y_{a}^{+}y_{d}^{-}=y_{a}^{-}y_{d}^{+},\quad\text{or}\quad\frac{y_{a}^{+}}{y_{a}^{-}}=\frac{y_{d}^{+}}{y_{d}^{-}}.$
(33)

In the semi-classical limit, $cG_{N}<<1$ thus the last term in (32) can be
neglected so we only need consider the equations

$\displaystyle\frac{z}{y_{d}^{-}y_{a}^{+}-z}=-\frac{6\xi
e^{-Q}}{cG_{N}l\xi_{0}},\quad\frac{z}{y_{d}^{+}y_{a}^{-}-z}=-\frac{6\xi
e^{-Q}}{cG_{N}l\xi_{0}}$ (34)

from which we can derive an single equation for $z$

$\displaystyle z\left(1-\frac{\epsilon}{\xi
e^{-Q}}\right)^{2}=y_{a}^{+}y_{a}^{-},\quad\epsilon\equiv\frac{cG_{N}l\xi_{0}}{6},$
(35)

where we will also assume that the ”effective” coupling $\epsilon$ is small.
Solving it and using the relation(33) we can solve $y_{d}^{\pm}$.

In the end we need to transfer to the Schwarzschild coordinate because it
describes the asymptotic observer and it turns out it is more convenient to do
the computation with the Schwarzschild coordinates. Let the positions $y_{a}$
and $y_{d}$ are $(t_{a},a)$ and $(t_{d},d)$ in the Schwarzschild coordinates
such that

$\displaystyle\frac{1}{l}Ye^{lt_{a}}=y_{a}^{+},\quad\frac{1}{l}Ye^{-lt_{a}}=-y_{a}^{-},\quad
Y=e^{ly^{*}},\quad
y^{*}=\int^{a}\frac{\operatorname{d}r^{\prime}}{-\xi/\xi_{0}^{2}},$ (36)
$\displaystyle\frac{1}{l}De^{lt_{d}}=y_{d}^{+},\quad\frac{1}{l}De^{-lt_{d}}=-y_{d}^{-},\quad
D=e^{ld^{*}},\quad
d^{*}=\int^{d}\frac{\operatorname{d}r^{\prime}}{-\xi/\xi_{0}^{2}},$ (37)

thus the entanglement entropy (27) can be written as

$\displaystyle
S_{\text{island}}=\frac{4X(d)}{G_{N}}+\frac{c}{3}\log\frac{1}{l^{2}}\left(Y^{2}+D^{2}-2YD\cosh
l(t_{a}-t_{d})\right)+\frac{c}{3}\left(\rho(d)+\rho(a)\right).$ (38)

Varying with respect to $t_{d}$ implies $t_{d}=t_{a}$ therefore the extremal
of $S_{island}$ is time-independent as expected. Assuming the temperature of
the black hole (15) is not zero so it can evaporate then $\xi$ has a single
zero at $r^{\prime}=r^{\prime}_{H}$. It implies that $d^{*}$ has a logarithm
singularity at $d=r_{H}^{\prime}$ so we can rewrite

$\displaystyle d^{*}=f(d)+\frac{1}{2l}\log(d-r^{\prime}_{H})$ (39)

where $f(d)$ is regular at $r^{\prime}_{H}$. Since we have assumed that the
island is very close to the horizon $d\approx r^{\prime}_{H}$ or equivalently
$z\approx 0$ thus we have the approximation

$\displaystyle
e^{2ld^{*}}=e^{2lf(d)}(d-r^{\prime}_{H})=-\frac{1}{l^{2}}z,\quad\rightarrow$
(40) $\displaystyle
d=r^{\prime}_{H}-e^{-2lf(r^{\prime}_{H})}\frac{1}{l^{2}}z+\mathcal{O}(z^{2}).$
(41)

If we also assume that $e^{-Q}$ is regular and not vanishing at
$d=r^{\prime}_{H}$ then

$\displaystyle\xi e^{-Q}=(d-r^{\prime}_{H})g(d),$ (42)

where $g(d)$ is a regular at $r^{\prime}_{H}$. Thus (35) becomes

$\displaystyle z(1-\frac{\epsilon
l^{2}e^{2lf(r^{\prime}_{H})}}{zg(d)})^{2}\approx
z+\frac{1}{z}\left(\frac{\epsilon
l^{2}e^{2lf(r^{\prime}_{H})}}{g(r^{\prime}_{H})}\right)^{2}-2\frac{\epsilon
l^{2}e^{2lf(r^{\prime}_{H})}}{g(r^{\prime}_{H})}=y_{a}^{+}y_{a}^{-}\equiv
y^{2}.$ (43)

There is indeed one solution which satisfies our ansatz $z\approx 0$:

$\displaystyle
z=\frac{\beta^{2}}{y^{2}}+\mathcal{O}(\beta^{3}),\quad\beta=\frac{\epsilon
l^{2}e^{2lf(r^{\prime}_{H})}}{g(r^{\prime}_{H})},$ (44) $\displaystyle
y_{d}^{+}=-\frac{\beta}{y_{a}^{-}},\quad
y_{d}^{-}=-\frac{\beta}{y_{a}^{+}}\,,$ (45)

which leads to

$\displaystyle S_{\text{island}}$ $\displaystyle=$
$\displaystyle\frac{4X(r^{\prime}_{H})}{G_{N}}+\frac{c}{3}\log\left(|y_{a}^{+}y_{a}^{-}|e^{\rho(y_{a}^{\pm})}\right),$
(46) $\displaystyle=$ $\displaystyle 2S_{\text{BH}}+S_{\text{matter}}$ (47)

where $S_{\text{matter}}$ is the quantum correction of order
$\mathcal{O}(G_{N}^{0})$ due to the presence of matter fields. This is the
main result of this paper: under some reasonable assumptions for a general
asymptotically flat eternal black hole in GDT we can find island such that
(generalized) entanglement entropy of Hawking radiation follows Page curve
which resolves the information paradox.

## Examples

##  4 CGHS model

The most well-studied GDT which admits an asymptotically flat black hole
solution is the Callan-Giddings-Harvery-Strominger model (CGHS model)
Callan:1992rs . Islands in this model have been found in Gautason:2020tmk ;
Anegawa:2020ezn . In this section, we will rederive the island with our
general procedures to confirm the validity of our general analysis. The action
of the CGHS model is

$\displaystyle
S=\frac{1}{2\pi}\int\operatorname{d}^{2}x\sqrt{-g}\left[e^{-2\phi}(R+4(\nabla\phi)^{2}+4\lambda^{2})\right]$
(48)

.

###  4.1 The geometry

Comparing (48) with (3) we recognize that

$\displaystyle U=-\frac{1}{X},\quad V=-2\lambda^{2}X\,.$ (49)

Therefore according to our general discussion, we can compute the following
data

$\displaystyle e^{Q}=\frac{1}{X},\quad X=\exp{\tilde{X}},\quad
w=-2\lambda^{2}X=-2\lambda^{2}e^{\tilde{X}}$ (50)
$\displaystyle\xi=-2C_{0}e^{-\tilde{X}}-4\lambda^{2},\quad\xi_{0}=2\lambda.$
(51)

So the Schwarzschild metric is

$\displaystyle\operatorname{d}s^{2}=-\left(\frac{C_{0}}{2\lambda^{2}}e^{-2\lambda
r^{\prime}}+1\right){\operatorname{d}t^{\prime}}^{2}+\frac{{\operatorname{d}r^{\prime}}^{2}}{\left(\frac{C_{0}}{2\lambda^{2}}e^{-2\lambda
r^{\prime}}+1\right)}.$ (52)

It is easy to find that the horizon and curvature singularity are located at

$\displaystyle
r^{\prime}_{H}=-\frac{1}{2\lambda}\log\left(-\frac{2\lambda^{2}}{C_{0}}\right),\quad
r^{\prime}_{s}=-\infty.$ (53)

Therefore we should take $\lambda^{2}>0$ and $C_{0}\equiv-C<0$ to get the
black hole geometry. To summarize, the classical solution describes a
asymptotically flat black hole with metric and dilaton

$\displaystyle\operatorname{d}^{2}s=-(1-\frac{C}{2\lambda^{2}}e^{-2\lambda
r^{\prime}})\operatorname{d}^{2}t^{\prime}+\frac{1}{(1-\frac{C}{2\lambda^{2}}e^{-2\lambda
r^{\prime}})}\operatorname{d}^{2}r^{\prime},$ (54) $\displaystyle
X=\exp\left(2\lambda r^{\prime}\right).$ (55)

The temperature of this black hole can be found using the equation

$\displaystyle
T=-\frac{1}{4\pi}\partial_{r^{\prime}}\sqrt{-\frac{g_{t^{\prime}t^{\prime}}}{g_{r^{\prime}r^{\prime}}}}\,|_{r=r_{H}}=\frac{\lambda}{2\pi},$
(56)

and the entropy is given by the Wald formula

$\displaystyle S=2X|_{r^{\prime}=r^{\prime}_{H}}=2e^{2\lambda
r^{\prime}}|_{r^{\prime}=r^{\prime}_{H}}=\frac{C}{\lambda^{2}}.$ (57)

Introducing the new variable (22)

$\displaystyle
x^{*}=\int\frac{\operatorname{d}r^{\prime}}{1-\frac{C}{2\lambda^{2}}e^{-2\lambda
r^{\prime}}}=\frac{\log\left(2\lambda^{2}e^{2r^{\prime}\lambda}-C\right)}{2\lambda},$
(58)

we can obtain the Kruskal coordinates $y^{\pm}$ as for example through

$\displaystyle e^{\lambda x^{+}}=\lambda y^{+},\quad e^{-\lambda
x^{-}}=-\lambda y^{-}$ (59)

thus the metric and dilaton become

$\displaystyle\operatorname{d}^{2}s=-\frac{\operatorname{d}y^{+}\operatorname{d}y^{-}}{C-\lambda^{2}y^{+}y^{-}},\quad
e^{2\rho}=\frac{1}{C-\lambda^{2}y^{+}y^{-}},$ (60) $\displaystyle
X=\frac{1}{2\lambda^{2}}\left(C-\lambda^{2}y^{+}y^{-}\right).$ (61)

In the Kruskal coordinates, the horizon is located at $y^{+}y^{-}=0$ and the
singularity is located at $y^{+}y^{-}=C$. As we have shown in the general
discussion, without including the island the entanglement entropy is given by
the general formula (26). Let us focus on the derivation of the island.

###  4.2 The derivation of island

In the late time, the entanglement entropy with island is given by:

$\displaystyle
S_{\text{island}}=\frac{1}{G_{N}}\frac{4}{2\lambda^{2}}\left(C-\lambda^{2}y_{d}^{+}y_{d}^{-}\right)+\frac{c}{3}\log\left(\frac{|(y_{a}-y_{d})^{+}(y_{a}-y_{d})^{-}|}{\sqrt{C-\lambda^{2}y_{a}^{+}y_{a}^{-}}\sqrt{C-\lambda^{2}y_{d}^{+}y_{d}^{-}}}\right).$
(62)

Taking derivative with respect to $y_{d}^{-}$ and $y_{d}^{+}$ we obtain the
equations

$\displaystyle\frac{\lambda^{2}y_{d}^{+}}{6(C-\lambda^{2}y_{d}^{+}y_{d}^{-})}+\frac{1}{3(y_{d}^{-}-y_{a}^{-})}-\frac{2y_{d}^{+}}{cG_{N}}=0,$
(63)
$\displaystyle\frac{\lambda^{2}y_{d}^{-}}{6(C-\lambda^{2}y_{d}^{+}y_{d}^{-})}+\frac{1}{3(y_{d}^{+}-y_{a}^{+})}-\frac{2y_{d}^{-}}{cG_{N}}=0,$
(64)

The exact solutions can be straightforwardly obtained but the exact solutions
are very complicated. To exact the useful information we again take the semi-
classical limit $G_{N}\rightarrow 0$. In this limit, we find that the non-
trivial solutions are

$\displaystyle y_{d}^{+}=-\frac{cG_{N}}{6y_{a}^{-}},\quad
y_{d}^{-}=-\frac{cG_{N}}{6y_{a}^{+}}$ (65)

and corresponding extremal entanglement entropy in the Schwarzschild
coordinate is

$\displaystyle
S_{\text{island}}=2\frac{1}{G_{N}}\frac{C}{\lambda^{2}}+\frac{c}{3}\log\left(\frac{|y_{a}^{+}y_{a}^{-}|}{\sqrt{C\left(C-\lambda^{2}y_{a}^{+}y_{a}^{-}\right)}}\right)$
(66)

which is time-independent and coincides with the results in Anegawa:2020ezn ;
Gautason:2020tmk .

Alternatively, we can apply our general result (35):

$\displaystyle z(1-\frac{\epsilon}{\xi e^{-Q}})^{2}=y^{2}\quad\rightarrow\quad
z(1-\frac{\epsilon}{2\lambda^{2}z})^{2}=y^{2},\quad\epsilon=\frac{cG_{N}\lambda^{2}}{3}$
(67)

which has two solutions

$\displaystyle
z_{1}=y^{2}+\frac{\epsilon}{\lambda^{2}}-\frac{\epsilon^{2}}{4\lambda^{4}y^{2}}+\mathcal{O}(\epsilon^{3}),\quad
z_{2}=\frac{\epsilon^{2}}{4y^{2}\lambda^{4}}+\mathcal{O}(\epsilon^{3}).$ (68)

The solution $z_{1}$ leads to the trivial solution while the solution $z_{2}$
leads to (65).

##  5 Liouville gravity

A particular generalization of CGHS model is the one with exponential
potential. The action is

$\displaystyle S=\frac{1}{2\pi}\int
d^{2}x\sqrt{-g}\left[RX+\sum_{i}4\alpha_{i}^{2}e^{\beta_{i}X}\right].$ (69)

There are some interesting reasons to consider such exponential potentials. It
is shown in Cruz:1997nj  that this kind of model admits extra (conformal)
symmetries. If we add $2X$ in the potential, this kind of models as
deformations of JT gravity is shown to have a matrix model dual Witten:2020wvy
. So it means that

$\displaystyle U(X)=0,\quad V(X)=-2\sum_{i}\alpha_{i}^{2}e^{\beta_{i}X}\,.$
(70)

For simplicity, let us take $k=1$ and the model is called the Liouville
gravity. Surprisingly, it is claimed in Li:2021lfo  that island formula can
not resolve the information paradox of Liouville gravity based on the black
hole solution which is found in Cruz:1997nj ; Mann:1993rf . In this section,
we will use the general solution of GDT to derive a different solution such
that the island formula successfully resolves the information paradox.

###  5.1 The geometry

Given the potentials (70) we can compute the following data

$\displaystyle Q=0,\quad X=\tilde{X}=r,\quad w=-\frac{2\alpha^{2}e^{\beta
X}}{\beta},$ (71)
$\displaystyle\xi=2(w-C_{0}^{L})=-2(\frac{2\alpha^{2}e^{\beta
X}}{\beta}+C_{0}^{L}),\quad\xi_{0}=\sqrt{2C_{0}^{L}},$ (72)

where we choose $\beta<0$. The corresponding Schwarzschild metric is

$\displaystyle\operatorname{d}s^{2}=-(1+\frac{2\alpha^{2}e^{\beta\sqrt{2C_{0}^{L}}r^{\prime}}}{C_{0}^{L}\beta}){\operatorname{d}t^{\prime}}^{2}+\frac{1}{1+\frac{2\alpha^{2}e^{\beta\sqrt{2C_{0}^{L}}r^{\prime}}}{C_{0}^{L}\beta}}{\operatorname{d}r^{\prime}}^{2}.$
(73)

Therefore the horizon is at

$\displaystyle
r^{\prime}_{H}=\frac{1}{\beta}\log\left(\frac{-C_{0}^{L}\beta}{2\alpha^{2}}\right),$
(74)

and the Ricci scalar is

$\displaystyle R=-4\beta e^{\beta\sqrt{2C_{0}^{L}}r^{\prime}}\alpha^{2}.$ (75)

So to ensure asymptotic flatness we can set

$\displaystyle\beta<0,\quad C_{0}^{L}>0,\quad\alpha^{2}>0,$ (76)

Note that the metric (73) is same as (54) if we identity

$\displaystyle\beta=-\frac{2\lambda}{\sqrt{2C_{0}^{L}}},\quad\alpha^{2}=\frac{C}{2\lambda}\sqrt{\frac{C_{0}^{L}}{2}}.$
(77)

If we also set $C_{0}^{L}=2\lambda^{2}$ such that
$\alpha^{2}=\frac{C}{2},\beta=-1$ then the dilaton is given by

$\displaystyle X=\tilde{X}=2\lambda r^{\prime},$ (78)

thus the dilaton of the Liouville black hole is the Logarithm of the one of
the CGHS black hole. Below we will keep $C_{0}^{L}$ general. Similarly the
geometry in the Kruskal coordinates are333We have double checked that the
solutions indeed solve the equations of motion in the second order formalism.

$\displaystyle\operatorname{d}^{2}s=-\frac{\operatorname{d}y^{+}\operatorname{d}y^{-}}{C-\lambda^{2}y^{+}y^{-}},$
(79) $\displaystyle
X=-\frac{1}{\beta}\log\left[\frac{1}{2\lambda^{2}}\left(C-\lambda^{2}y^{+}y^{-}\right)\right].$
(80)

Since the geometry is same the entanglement entropy without island is also
same as one in CGHS black hole. Let us focus on the entanglement entropy of
the Hawking radiation in the presence of possible islands.

###  5.2 The derivation of island

In the late time, the entanglement entropy with island is given by:

$\displaystyle
S_{\text{island}}=\frac{4}{G_{N}}\frac{\sqrt{2C_{0}^{L}}}{2\lambda}\log\left[\frac{1}{2\lambda^{2}}\left(C-\lambda^{2}y_{d}^{+}y_{d}^{-}\right)\right]+\frac{c}{3}\log\left(\frac{|(y_{a}-y_{d})^{+}(y_{a}-y_{d})^{-}|}{\sqrt{C-\lambda^{2}y_{a}^{+}y_{a}^{-}}\sqrt{C-\lambda^{2}y_{d}^{+}y_{d}^{-}}}\right).$

The extremal conditions are

$\displaystyle-\frac{4}{G_{N}}\frac{\sqrt{2C_{0}^{L}}}{2\lambda}\frac{\lambda^{2}y_{d}^{+}}{C-\lambda^{2}y_{d}^{+}y_{d}^{-}}+\frac{c}{3(y_{d}^{-}-y_{a}^{-})}+\frac{cy_{d}^{+}}{6(C-\lambda^{2}y_{d}^{+}y_{d}^{-})}=0,$
(82)
$\displaystyle-\frac{4}{G_{N}}\frac{\sqrt{2C_{0}^{L}}}{2\lambda}\frac{\lambda^{2}y_{d}^{-}}{C-\lambda^{2}y_{d}^{+}y_{d}^{-}}+\frac{c}{3(y_{d}^{+}-y_{a}^{+})}+\frac{cy_{d}^{-}}{6(C-\lambda^{2}y_{d}^{+}y_{d}^{-})}=0\,.$
(83)

These equations are quadratic so can be easily solved. In the limit
$G_{N}\rightarrow 0$ the two solutions behave as

$\displaystyle
y_{d}^{-}=\frac{cG_{N}C\beta}{12\lambda^{2}y_{a}^{+}}=-\frac{cG_{N}C}{6\sqrt{2C_{0}^{L}}y_{a}^{+}\lambda},\quad
y_{d}^{+}=\frac{cG_{N}C\beta}{12\lambda^{2}y_{a}^{-}}=-\frac{cG_{N}C}{6\sqrt{2C_{0}^{L}}y^{-}_{a}\lambda},$
(84) $\displaystyle y_{d}^{-}=y^{-}_{a}+\mathcal{O}(G_{N}),\quad
y_{d}^{+}=y^{+}_{a}+\mathcal{O}(G_{N}),.$ (85)

Let us try to derive these solutions directly from our general result (35):

$\displaystyle z(1-\frac{\epsilon}{\xi
e^{-Q}})^{2}=y^{2}-\quad\rightarrow\quad
z\left((1+\frac{\epsilon}{2C_{0}^{L}})-\frac{\epsilon
C}{2C_{0}^{L}\lambda^{2}z}\right)^{2}=y^{2},$ (86)

with

$\displaystyle\epsilon=\frac{cG_{N}\lambda\sqrt{2C_{0}^{L}}}{6}.$ (87)

The equation (86) is also quadratic with solutions to be

$\displaystyle
z_{1}=\frac{C^{2}\epsilon^{2}}{4{C_{0}^{L}}^{2}\lambda^{4}y^{2}}+\mathcal{O}(\epsilon^{3}),\quad
z_{2}=y^{2}-\frac{\epsilon(\lambda^{2}y^{2}-C)}{C_{0}^{L}\lambda^{2}}+\mathcal{O}(\epsilon^{2})$
(88)

which will correspond to (84) and (85), respectively. The first solution (84)
is the non-trivial one which gives the generalized entanglement entropy

$\displaystyle
S_{\text{island}}=2\frac{1}{G_{N}}\frac{\sqrt{2C_{0}^{L}}}{\lambda}\log\left(\frac{C}{2\lambda^{2}}\right)+\frac{c}{3}\log\left(\frac{|y_{a}^{+}y_{a}^{-}|}{\sqrt{C\left(C-\lambda^{2}y_{a}^{+}y_{a}^{-}\right)}}\right).$
(89)

Thus we have derived the Page curve for the Liouville black hole. The reason
why we succeed is that we have derived another black hole solution whose
parameters are opposite to those in the solutions used in Li:2021lfo  or
derived in Cruz:1997nj ; Mann:1993rf . Let us revisit the black geometry which
is used in Li:2021lfo .

###  5.3 The other black geometry

To get that solution we start from (73) and reverse the radial coordinate

$\displaystyle r^{\prime}\rightarrow-r^{\prime},$ (90)

such that the metric becomes

$\displaystyle\operatorname{d}s^{2}=-(1+\frac{2\alpha^{2}e^{-\beta\sqrt{2C_{0}^{L}}r^{\prime}}}{C_{0}^{L}\beta}){\operatorname{d}t^{\prime}}^{2}+\frac{1}{1+\frac{2\alpha^{2}e^{-\beta\sqrt{2C_{0}^{L}}r^{\prime}}}{C_{0}^{L}\beta}}{\operatorname{d}r^{\prime}}^{2}.$
(91)

The position of the event horizon and the Ricci scalar are

$\displaystyle
r_{H}^{\prime}=\frac{1}{\sqrt{2C_{0}^{L}}\beta}\log\left(-\frac{2\alpha^{2}}{C_{0}^{L}\beta}\right),\quad
R=-4\beta\alpha^{2}e^{-\sqrt{2C_{0}^{L}}\beta r^{\prime}}.$ (92)

So requiring the asymptotic flatness at $r^{\prime}\rightarrow\infty$ forces
the choice

$\displaystyle\beta>0,$ (93)

and having a well-defined horizon forces the choice

$\displaystyle\alpha^{2}<0.$ (94)

With these choices, the solution in the Kruskal coordinates are still given by
(79)

$\displaystyle\operatorname{d}^{2}s=-\frac{\operatorname{d}y^{+}\operatorname{d}y^{-}}{C-\lambda^{2}y^{+}y^{-}},$
(95) $\displaystyle
X=-\frac{1}{\beta}\log\left[\frac{1}{2\lambda^{2}}\left(C-\lambda^{2}y^{+}y^{-}\right)\right].$
(96)

but with different identification

$\displaystyle\beta=\frac{2\lambda}{\sqrt{2C_{0}^{L}}},\quad\alpha^{2}=-\frac{C}{2\lambda}\sqrt{\frac{C_{0}^{L}}{2}}.$
(97)

But in this black hole solution the position of the island is at

$\displaystyle
y_{d}^{-}=\frac{cG_{N}C\beta}{12\lambda^{2}y_{a}^{+}}=\frac{cG_{N}C}{6\sqrt{2C_{0}^{L}}y_{a}^{+}\lambda},\quad
d^{+}=\frac{cG_{N}C\beta}{12\lambda^{2}y_{a}^{-}}=\frac{cG_{N}C}{6\sqrt{2C_{0}^{L}}y_{a}^{-}\lambda},$
(98)

which is in the left Kruskal patch. This contradicts the assumption that
$y_{d}$ is in the right patch and this is why Li:2021lfo  claims the failure
of island formula. However we have shown this is only because a ”wrong”
solution is used.

To summarize, island can save the information paradox of Liouville gravity.

##  6 ab-family

In this section, we consider a large family of dilaton gravity theories which
has the following potentials

$\displaystyle U(X)=-\frac{a}{X},\quad V(X)=-\frac{B}{2}X^{a+b}.$ (99)

In general, there are two free parameters and sometimes this family is called
the ab-family Katanaev:1996ni . From our general discussion, the classical
solution is

$\displaystyle\operatorname{d}^{2}s=2X^{-a}\operatorname{d}X\operatorname{d}v-X^{-2}(2C_{0}+\frac{BX^{b+1}}{b+1})\operatorname{d}^{2}v.$
(100)

Because we are interested in the asymptotically flat black hole solutions
$\lim_{X\rightarrow\infty}R=0$ we will choose Katanaev:1996ni

$\displaystyle b=a-1,\quad a\in(0,1).$ (101)

This choice can be understood form the behavior of Ricci scalar 444here we
consider the case $b\neq 1$.

$\displaystyle R=-2aC_{0}X^{a-2}+\frac{bB(a-b-1)}{b+1}X^{a+b-1},$ (102)

by noticing that $C_{0}$ is related to mass of black hole therefore the
solution with $C_{0}=0$ should be the Minkowski spacetime. Following the
general discussion we compute

$\displaystyle Q=-a\log X,\quad
X=(1-a)^{\frac{1}{1-a}}\tilde{X}^{\frac{a}{1-a}},\quad
w=-\frac{B}{2a}(1-a)^{\frac{a}{1-a}}\tilde{X}^{\frac{a}{1-a}},$ (103)
$\displaystyle\xi=-\frac{B}{a}-\frac{2C_{0}^{f}}{(1-a)^{\frac{a}{1-a}}\tilde{X}^{\frac{a}{1-a}}},\quad\xi_{0}=\sqrt{\frac{B}{a}}.$
(104)

Thus the corresponding Schwarzschild metric is

$\displaystyle\operatorname{d}s^{2}=-(1-\frac{1}{2\lambda{r^{\prime}}^{\frac{a}{1-a}}}){\operatorname{d}t^{\prime}}^{2}+\frac{1}{1-\frac{1}{2\lambda{r^{\prime}}^{\frac{a}{1-a}}}}{\operatorname{d}r^{\prime}}^{2},$
(105)

where

$\displaystyle\frac{1}{2\lambda}=-\frac{2C_{0}^{f}a}{B}\left(\sqrt{\frac{B}{a}}(1-a)\right)^{\frac{a}{a-1}},\quad
C_{0}^{f}<0,\quad\lambda>0.$ (106)

So the horizon is located at

$\displaystyle r^{\prime}_{H}=(2\lambda)^{\frac{a-1}{a}}.$ (107)

Next we can transform it the conformal gauge by introducing

$\displaystyle
x^{*}=\int\frac{\operatorname{d}r^{\prime}}{1-\frac{1}{2\lambda{r^{\prime}}^{\frac{a}{1-a}}}}=2(a-1)\lambda{r^{\prime}}^{\frac{1}{1-a}}\,_{2}F_{1}\left(1,\frac{1}{a};1+\frac{1}{a};2{r^{\prime}}^{\frac{a}{1-a}}\lambda\right)+c_{1},$
(108)

where $c_{1}$ is a constant which can be chosen for our convenience. The
hypergeometric function generally can not be inverted to write
$r^{\prime}(x^{*})$ as a function of $x^{*}$ explicitly. However for the
special case of $a=1/2$, we can invert the function with product logarithm:

$\displaystyle x^{*}=r^{\prime}+\frac{\log(2\lambda
r^{\prime}-1)}{2\lambda}+c_{1},\quad\quad\rightarrow$ (109) $\displaystyle
r^{\prime}=\frac{1}{2\lambda}+\frac{W_{0}\left(e^{2\lambda(x^{*}-c_{1})-1}\right)}{2\lambda},$
(110)

where $W_{0}$ is the principle branch of the Lambert $W$ function or product
logarithm. Thus it is natural to introduce the Kruskal coordinates as

$\displaystyle e^{\pm\lambda x^{\pm}}=\pm\lambda y^{\pm},\quad
c_{1}=-\frac{1}{2\lambda},$ (111)
$\displaystyle\operatorname{d}s^{2}=-e^{2\rho}\operatorname{d}y^{+}\operatorname{d}y^{-},\quad
e^{2\rho}=\frac{1}{e^{W_{0}(-\lambda^{2}y^{2})}-\lambda^{2}y^{2}},$ (112)
$\displaystyle
X=\sqrt{\frac{B}{2}}\frac{1}{4\lambda}\left(1+W_{0}(-\lambda^{2}y^{2})\right),\quad
y^{2}=y^{+}y^{-}.$ (113)

The equation for determining the position of island becomes

$\displaystyle z\left(\frac{\epsilon
2^{1/4}\sqrt{\lambda}\sqrt{W_{0}\left(\lambda^{2}(-z)\right)+1}}{B^{5/4}W_{0}\left(\lambda^{2}(-z)\right)}+1\right)^{2}=y^{2},\quad\epsilon=\frac{cG_{N}\lambda}{6}\sqrt{\frac{B}{a}}.$
(114)

Assuming $\epsilon\rightarrow 0$ we can expand the left-hand side to the first
order of $z$ then the we will obtain two solutions

$\displaystyle
z_{1}=y^{2}+\frac{2^{1/4}\alpha(2-3y^{2}\lambda^{2})}{B^{5/4}\lambda^{3/2}}+\mathcal{O}(\epsilon^{2}),\quad
z_{2}=\frac{\beta^{2}}{y^{2}},\quad\beta^{2}=\frac{\sqrt{2}\epsilon^{2}}{B^{5/2}\lambda^{3}}.$
(115)

Thus the physical solution is

$\displaystyle y_{d}^{+}=-\frac{\beta}{y_{a}^{-}},\quad
y_{d}^{-}=-\frac{\beta}{y_{a}^{+}}\,,$ (116)

which leads to

$\displaystyle
S_{\text{island}}=\sqrt{\frac{B}{2}}\frac{1}{G_{N}\lambda}+\frac{c}{3}\log\frac{|y_{a}^{+}y_{a}^{-}|}{\sqrt{e^{W_{0}(-\lambda^{2}y^{2})}-\lambda^{2}y^{2}}}+\mathcal{O}(G_{N}).$
(117)

For generic $a$, we observe that $x^{*}$ can be always decomposed into

$\displaystyle
x^{*}=f(r^{\prime})+\frac{1-a}{a}(2\lambda)^{\frac{a-1}{a}}\log(r^{\prime}-r^{\prime}_{H}),$
(118)

where $f(r^{\prime})$ is regular at $r^{\prime}=r^{\prime}_{H}$ as we expect
in (39). Here we omit the further analysis since it is very similar to the
general result.

##  7 Reissner-Nordstrom

Islands in charged black hole have been studied in Wang:2021woy ;
Karananas:2020fwx ; Ahn:2021chg ; Yu:2021cgi . Even though they considered
4-dimensional black holes, effectively and technically the model is still
2-dimensional after a dimensional reduction of the two sphere. Therefore we
can also study them with our general procedure. To support a Reissner-
Nordstrom black hole, the simplest choice of potentials are

$\displaystyle U(X)=-\frac{1}{2X},\quad V(X)=-\lambda^{2}+\frac{A}{X}$ (119)

which lead to the following data

$\displaystyle e^{Q}=\frac{1}{\sqrt{X}},\quad X=\frac{\tilde{X}^{2}}{4},\quad
w=-\frac{2(A+\lambda^{2}X)}{\sqrt{X}}=-\lambda^{2}\tilde{X}-\frac{4A}{\tilde{X}},$
(120)
$\displaystyle\xi=-4\lambda^{2}-\frac{4C_{0}^{R}}{\tilde{X}}-\frac{16A}{\tilde{X}^{2}},\quad\xi_{0}=2\lambda.$
(121)

Thus the metric and dilaton are

$\displaystyle\operatorname{d}s^{2}=-(1+\frac{C_{0}^{R}}{2r^{\prime}\lambda^{3}}+\frac{A}{r^{\prime}\lambda^{4}}){\operatorname{d}t^{\prime}}^{2}+\frac{1}{1+\frac{C_{0}^{R}}{2r^{\prime}\lambda^{3}}+\frac{A}{r^{\prime}\lambda^{4}}}{\operatorname{d}r^{\prime}}^{2},$
(122) $\displaystyle X=\lambda^{2}{r^{\prime}}^{2}.$ (123)

Comparing with the standard Reissner-Nordstrom we can identify the following
parameters

$\displaystyle C_{0}^{R}=-4M\lambda^{3},\quad A=\lambda^{4}Q_{c}^{2},$ (124)
$\displaystyle M=\frac{r_{+}+r_{-}}{2},\quad Q_{c}=\sqrt{r_{+}r_{-}},$ (125)

where $M$ and $Q_{c}$ are the mass and charge of the black hole and $r_{\pm}$
are positions of the outer $(+)$ and inner $(-)$ horizons. Next we introduce
new variable

$\displaystyle
x^{*}=\int\frac{\operatorname{d}r^{\prime}}{1-\frac{2M}{r^{\prime}}+\frac{Q_{c}^{2}}{{r^{\prime}}^{2}}}=r^{\prime}+\frac{r_{+}^{2}\log(r^{\prime}-r_{+})-r_{-}^{2}\log(r^{\prime}-r_{-}^{2})}{r_{+}-r_{-}},$
(126)
$\displaystyle\exp(2x^{*})=e^{2r^{\prime}}(r^{\prime}-r_{+})^{\frac{2r_{+}^{2}}{r_{+}-r_{-}}}(r^{\prime}-r_{-})^{-\frac{2r_{-}^{2}}{r_{+}-r_{-}}},$
(127)

to get the conformal metric

$\displaystyle\operatorname{d}s^{2}=-H(r^{\prime})\operatorname{d}x^{+}\operatorname{d}x^{-},\quad
H(r^{\prime})=1-\frac{2M}{r^{\prime}}+\frac{Q_{c}^{2}}{{r^{\prime}}^{2}}\,.$
(128)

The Kruskal coordinates can be defined as

$\displaystyle e^{lx^{+}}=ly^{+},\quad e^{-lx^{-}}=-ly^{-},\quad
l=\frac{r_{+}-r_{-}}{2r_{+}^{2}}.$ (129)

In order to use (35) to solve the position of the island, we first express $d$
in terms of $z$. Using (127) we can directly get

$\displaystyle-l^{2}z=e^{2dl}(d-r_{+})(d-r_{-})^{-\frac{r_{-}^{2}}{r_{+}^{2}}},$
(130)

which leads to

$\displaystyle
d=r_{+}-e^{-2lr_{+}}l^{2}(r_{+}-r_{-})^{\frac{r_{-}^{2}}{r_{+}^{2}}}z+\mathcal{O}(z^{2})\,.$
(131)

Recall that we expect that $z\rightarrow 0$. Substituting (131) into (35) with
help of (120) and (121) we end up an equation of $z$. The non-trivial solution
is

$\displaystyle
z=\frac{\beta^{2}}{y^{2}},\quad\beta^{2}=\frac{\epsilon^{2}e^{4lr_{+}}r_{+}^{2}(r_{+}-r_{-})^{-\frac{2r_{-}^{2}}{r_{+}^{2}}-2}}{16l^{4}\lambda^{6}y^{2}},\quad\epsilon=\frac{cG_{N}l\lambda}{3},$
(132) $\displaystyle y_{d}^{+}=-\frac{\beta}{y_{a}^{-}},\quad
y_{d}^{-}=-\frac{\beta}{y_{a}^{+}}\,.$ (133)

###  7.1 Other Charged dilaton Black Hole I

We can also consider island in other charged dilaton black hole. In
Ahn:2021chg , the charged dilaton black hole has the metric

$\displaystyle\operatorname{d}s^{2}=-r^{2}\left(1-\frac{2M}{r^{2}}+\frac{Q_{c}^{2}}{4r^{4}}\right)\operatorname{d}t^{2}+\left(1-\frac{2M}{r^{2}}+\frac{Q_{c}^{2}}{4r^{4}}\right)^{-1}\operatorname{d}r^{2}+r^{2}(\operatorname{d}x^{2}+dy^{2}).$
(134)

The effective 2D model is

$\displaystyle\operatorname{d}s^{2}=-H(r)\operatorname{d}t^{2}+r^{2}H(r)^{-1}\operatorname{d}r^{2},\quad
X=r^{2},$ (135)

where

$\displaystyle
H(r)=r^{2}\left(1-\frac{2M}{r^{2}}+\frac{Q_{c}^{2}}{4r^{4}}\right).$ (136)

To transform to the Schwarzschild metric let us introduce

$\displaystyle dr^{\prime}=2r\operatorname{d}r,\quad\rightarrow,\quad
r^{\prime}=r^{2},$ (137)

such that

$\displaystyle\operatorname{d}s^{2}=-H(r^{\prime}){\operatorname{d}t^{\prime}}^{2}+H(r^{\prime})^{-1}{\operatorname{d}r^{\prime}}^{2},\quad
t^{\prime}=t/2,$ (138) $\displaystyle
H(r^{\prime})=4r^{\prime}-8M-\frac{Q_{c}^{2}}{r^{\prime}}=\frac{4(r-r_{+})(r-r_{-})}{r^{\prime}},\quad
r_{\pm}=M\pm\frac{\sqrt{4M^{2}-Q_{c}^{2}}}{2}.$ (139)

The geometry is asymptotically flat. The outer event horizon and curvature
singularity are located at $r^{\prime}=r_{+}$ and $r^{\prime}=0$ respectively.
The solution can be embedded into dilaton gravity by choosing the possible
potentials to be

$\displaystyle U(X)=0,\quad V(X)=-2+\frac{Q_{c}^{2}}{2X^{2}},\quad C_{0}=4M.$
(140)

When $Q_{c}=0$, the geometry (138) reduces to Rindler patch. From the
potential we obtain

$\displaystyle Q=0,\quad X=\tilde{X}=r^{\prime},\quad
w=-2X-\frac{Q_{c}^{2}}{2X},\quad\xi=8M-\frac{Q_{c}^{2}}{X}-4X.$ (141)

Next we introduce new variable

$\displaystyle
x^{*}=\int\frac{\operatorname{d}r^{\prime}}{H(r^{\prime})}=\frac{r_{+}}{4(r_{+}-r_{-})}\log(r^{\prime}-r_{+})-\frac{r_{-}}{4(r_{+}-r_{-})}\log(r^{\prime}-r_{-}),$
(142) $\displaystyle
e^{2lx^{*}}=(r^{\prime}-r_{-})^{-\frac{r_{-}}{r_{+}}}(r^{\prime}-r_{+}),\quad\frac{1}{2l}=\frac{r_{+}}{4(r_{+}-r_{-})}.$
(143)

Following the general procedure we find that in the late time the position of
the island is

$\displaystyle d=r_{+}-l^{2}(r_{+}-r_{-})^{\frac{r_{-}}{r_{+}}}z,$ (144)
$\displaystyle
z=\frac{\beta^{2}}{y^{2}},\quad\beta^{2}=\frac{\epsilon^{2}r_{+}^{2}(r_{+}-r_{-})^{-\frac{2(r_{-}+r_{+})}{r_{+}}}}{16l^{4}},\quad\epsilon=\frac{cG_{N}l}{6},$
(145) $\displaystyle y_{d}^{+}=-\frac{\beta}{y_{a}^{-}},\quad
y_{d}^{-}=-\frac{\beta}{y_{a}^{+}}\,,$ (146)

which coincides with the results in Ahn:2021chg .

###  7.2 Other Charged dilaton Black Hole II

In Yu:2021cgi , the charged dilaton black hole has the metric

$\displaystyle\operatorname{d}s^{2}=-W(r)\operatorname{d}t^{2}+W^{-1}\operatorname{d}r^{2}+R(r)^{2}\operatorname{d}\Omega^{2},$
(147)

with the function

$\displaystyle
W(r)=\left(1-\frac{r_{+}}{r}\right)\left(1-\frac{r_{-}}{r}\right)^{n},\quad
R^{2}=r^{2}\left(1-\frac{r_{-}}{r}\right)^{1-n},\quad n\in[0,1).$ (148)

The effective 2D model is

$\displaystyle\operatorname{d}s^{2}=-W(r)\operatorname{d}t^{2}+W^{-1}\operatorname{d}r^{2},\quad
X=r^{2}\left(1-\frac{r_{-}}{r}\right)^{1-n}\equiv f(r).$ (149)

The corresponding 2d dilaton potentials can be

$\displaystyle
e^{Q}=\frac{\operatorname{d}f^{-1}(X)}{\operatorname{d}X}\equiv{f^{-1}}^{\prime},\quad
U(X)=\frac{\operatorname{d}\ln({f^{-1}}^{\prime})}{\operatorname{d}X},$ (150)
$\displaystyle
V(X)=-\frac{1}{2}e^{-Q(X)}\frac{\operatorname{d}\left(e^{-Q(X)}W(f^{-1}(X))\right)}{\operatorname{d}X}.$
(151)

In general, $f(r)$ is hard to invert but to solve the island the explicit
expressions of the potentials are not needed. We only need the following
quantity

$\displaystyle\xi
e^{-Q}=-W\frac{dX}{\operatorname{d}r}=-\frac{(2r-(1+n)r_{-})(r-r_{+})}{r},$
(152)

which appears in (35) and the relation between $r$ and $z$:

$\displaystyle x^{*}$ $\displaystyle=$
$\displaystyle\int\frac{\operatorname{d}r}{W(r)}=\frac{r^{n}}{(r-r_{-})^{n-1}}+(nr_{-}+r_{+})B_{1-\frac{r_{-}}{r}}(1-n,0)$
(153) $\displaystyle-$
$\displaystyle\left(\frac{r_{+}}{r_{+}-r_{-}}\right)^{n}B_{t}(1-n,0),\quad
t=\frac{r_{+}}{r_{+}-r_{-}}\left(1-\frac{r_{-}}{r}\right),$

where $B_{\alpha}(a,b)$ is the incomplete beta function. Note that

$\displaystyle\lim_{t\rightarrow 1}B_{t}(1-n,0)=-\log(t-1)$ (154)

thus let us denote $x^{*}$ as

$\displaystyle
x^{*}=\mathcal{R}+\left(\frac{r_{+}}{r_{+}-r_{-}}\right)^{n}\log(r-r_{+}),$
(155) $\displaystyle
e^{2lx^{*}}=e^{2l\mathcal{R}}(r-r_{+}),\quad\frac{1}{2l}=\left(\frac{r_{+}}{r_{+}-r_{-}}\right)^{n}.$
(156)

It implies that

$\displaystyle d=r_{+}-{l^{2}}e^{-2l\mathcal{R}(r_{+})}z+\mathcal{O}(z^{2}).$
(157)

Substituting into (35) we can solve

$\displaystyle z=\frac{\beta^{2}}{y^{2}},\quad\beta=\frac{\epsilon
r_{+}}{e^{-2l\mathcal{R}(r_{+})}(2r_{+}-r_{-}-nr_{-})},\quad\epsilon=\frac{cG_{N}}{6},$
(158) $\displaystyle y_{d}^{+}=-\frac{\beta}{y_{a}^{-}},\quad
y_{d}^{-}=-\frac{\beta}{y_{a}^{+}}\,,$ (159)

which is consistent with the result in Yu:2021cgi  while our method is much
simpler.

##  8 Kaluza-Klein black holes

Our last example is the 4-dimensional Kaluza-Klein black hole. The island of
this black hole is studied in Lu:2021gmv . The metric of a non-rotating KK
black hole in 4d asymptotically flat spacetime is

$\displaystyle\operatorname{d}s^{2}=-W(r)\operatorname{d}t^{2}+\frac{\operatorname{d}r^{2}}{W(r)}+H^{1/2}r^{2}\operatorname{d}\Omega^{2},$
(160)

where

$\displaystyle W(r)=\frac{f(r)}{\sqrt{H(r)}},\quad
f(r)=1-\frac{r_{h}}{r},\quad H(r)=1+\frac{Q_{c}}{r}.$ (161)

When $Q=0$, (160) is just the 4 dimensional Schwarzschild black hole and the
corresponding dilaton gravity model has the potentials

$\displaystyle U(X)=-\frac{1}{2X},\quad V(X)=-\lambda^{2}.$ (162)

To embed the solution (160) into dilaton gravity we can first identify

$\displaystyle X=H^{1/2}r^{2}\equiv f(\tilde{X}),\quad r=\tilde{X}.$ (163)

Supposing that $f(\tilde{X})$ is invertible, we can solve
$\tilde{X}=f^{-1}(X)$. The potentials of the corresponding GDT can be found
with (150) and (151). For the solution (160), the results are very involved.
The results in leading order of $Q$ are

$\displaystyle U(X)=-\frac{1}{2X}+\frac{3Q^{2}}{32X^{2}},\quad
V(X)=-1+\frac{r_{h}Q}{4X}+\frac{Q^{2}(2\sqrt{X}-3r_{h})}{8X^{3/2}}.$ (164)

To compute the generalized entanglement entropy we can directly use the 2d
dilaton gravity solutions

$\displaystyle\operatorname{d}s^{2}=-W(r)\operatorname{d}t^{2}+\frac{\operatorname{d}r^{2}}{W(r)},$
(165) $\displaystyle X=\sqrt{1+\frac{Q_{c}}{r}}r^{2},\quad
W(r)=(1-\frac{r_{h}}{r})(1+\frac{Q_{c}}{r})^{-\frac{1}{2}},$ (166)

which are already in the Schwarzschild coordinates. To derive the island first
we compute

$\displaystyle\xi
e^{-Q}=-W\frac{dX}{\operatorname{d}r}=-\frac{(4r+3Q_{c})}{2(r+Q_{c})}(r-r_{h}),$
(167)

and

$\displaystyle x^{*}$ $\displaystyle=$
$\displaystyle\int\frac{\operatorname{d}r}{W(r)}=\sqrt{r(Q_{c}+r)}+(Q_{c}+r_{h})\sinh^{-1}\left(\
\sqrt{\frac{r}{Q_{c}}}\right)$ $\displaystyle-$
$\displaystyle\sqrt{r_{h}(Q_{c}+r)}\log\left(1+\sqrt{\frac{{r(Q_{c}+r_{h})}}{{r_{h}(Q_{c}+r)}}}\right)+\sqrt{r_{h}(Q_{c}+r)}\log\left(1-\sqrt{\frac{{r(Q_{c}+r_{h})}}{{r_{h}(Q_{c}+r)}}}\right)$
$\displaystyle=$
$\displaystyle\mathcal{R}+\sqrt{r_{h}(Q_{c}+r)}\log(r-r_{h}),$ (169)

where $\mathcal{R}$ is again regular at $r=r_{h}$. Then we can solve $d$ in
terms of $z$:

$\displaystyle
e^{2lx^{*}}=e^{2l\mathcal{R}}(r-r_{h}),\quad\frac{1}{2l}=\sqrt{r_{h}(Q_{c}+r)},$
(170) $\displaystyle d=r_{h}-l^{2}e^{-2l\mathcal{R}(r_{h})}z.$ (171)

Substituting into (35) we find

$\displaystyle
z=\frac{\beta^{2}}{y^{2}},\quad\beta=\frac{2\epsilon(r_{h}+Q_{c})}{e^{-2l\mathcal{R}(r_{h})}(3Q_{c}+4rh)},\quad\epsilon=\frac{cG_{N}}{6},$
(172) $\displaystyle y_{d}^{+}=-\frac{\beta}{y_{a}^{-}},\quad
y_{d}^{-}=-\frac{\beta}{y_{a}^{+}}\,.$ (173)

##  9 Conclusion and Discussion

In this work we have studied island formula (1) in the general asymptotically
flat eternal black holes in GDT. Under some reasonable and mild assumptions we
prove that the island always appears barely outside of the horizon in the late
time of Hawking radiation so that the information paradox is resolve, in
particular, in the Liouville gravity theory in which it was reported in
Li:2021lfo  that the island proposal failed. We find that failure is due to
the use of a “wrong” black hole solution. With the help of general
construction of classical solutions of GDT we find a different black hole
solution where the island appears as expected. We further apply our general
analysis to a large family of GDT and several 4-dimensional black holes
including different charged dilaton black holes and the KK black hole. It
turns out that our procedure for finding island is much simpler.

There are some possible generalizations of our analysis.

  * •

Our general analysis should be simply generalized to the asymptotically AdS
black holes in GDT by gluing a flat bath. Since after gluing the flat bath,
the whole spacetime is similar to the asymptotically flat black hole and cut-
off surface $y_{a}$ can be chosen to be boundary of the AdS space.

  * •

In this work we only consider the classical solutions of GDT. It is also
possible to include the quantum effect which comes from the conformal anomaly
following for example Hartman:2020swn .

  * •

It is also possible to generalize our results to single-sided black hole and
consider a truly evaporating black hole. Some examples are Hartman:2020swn ;
Wang:2021mqq .

###### Acknowledgements.

I would like to thank Yi-Jun He, Yang An and many of the members of KITS for
interesting related discussions. JT is supported by the National Youth Fund
No.12105289 and the internal fund of KITS.

##  Appendix A Review of 2D GDT

### Conventions

The local Lorentz metric and the Lorentz transformation invariant tensor are
chosen to be

$\displaystyle\eta_{ab}=\eta^{ab}=\begin{pmatrix}-1&0\\\
0&1\end{pmatrix},\quad\epsilon^{a}_{~{}b}=\begin{pmatrix}0&1\\\
1&0\end{pmatrix}.$ (174)

Thus the Levi-Civita tensors are

$\displaystyle\epsilon^{ac}=\epsilon^{a}_{~{}b}\eta^{bc}=\begin{pmatrix}0&1\\\
-1&0\end{pmatrix},\quad\epsilon_{ab}=\eta_{ac}\epsilon^{c}_{~{}b}=\begin{pmatrix}0&-1\\\
1&0\end{pmatrix}.$ (175)

The volume form is related to the local Lorentz basis $e^{a}$ via

$\displaystyle\epsilon$ $\displaystyle=$
$\displaystyle\frac{1}{2}\epsilon_{ab}e^{a}\wedge
e^{b}=\frac{1}{2}\epsilon_{ab}e^{a}_{~{}\mu}e^{b}_{~{}\nu}dx^{\mu}\wedge
dx^{\nu}=\frac{1}{2}\epsilon_{ab}\left(e^{a}_{~{}1}e^{b}_{~{}0}-e^{a}_{~{}0}e^{b}_{~{}1}\right)\operatorname{d}x^{1}\wedge\operatorname{d}x^{0}$
(176) $\displaystyle=$
$\displaystyle\left(e^{1}_{~{}1}e^{0}_{~{}0}-e^{1}_{~{}0}e^{0}_{~{}1}\right)\operatorname{d}x^{1}\wedge\operatorname{d}x^{0}=\sqrt{-g}\operatorname{d}x^{1}\wedge\operatorname{d}x^{0}\rightarrow\sqrt{-g}\operatorname{d}^{2}x.$
(177)

In 2d the spin connection should be proportional to $\epsilon$:
$\omega^{a}_{~{}b}=\omega\epsilon^{a}_{~{}b}$ and Ricci tensor two-form is
then given by

$\displaystyle R_{ab}$ $\displaystyle=$
$\displaystyle\operatorname{d}\omega\epsilon_{ab},\quad
R_{ab}=\frac{1}{2}(R_{\mu\nu})_{ab}dx^{\mu}\wedge
dx^{\nu},\quad(R_{\mu\nu})_{ab}=\epsilon_{ab}\left(\partial_{\mu}\omega_{\nu}-\partial_{\nu}\omega_{\mu}\right).$
(178)

So the Ricci scalar is

$\displaystyle(R_{\mu\nu})_{ab}e^{a\mu}e^{b\nu}$ $\displaystyle=$
$\displaystyle\epsilon^{\mu\nu}(\partial_{\mu}\omega_{\nu}-\partial_{\nu}\omega_{\mu})=2\epsilon^{\mu\nu}\partial_{\mu}\omega_{\nu}=2|e|^{-1}\tilde{\epsilon}^{\mu\nu}\partial_{\mu}\omega_{\nu}$
(179) $\displaystyle=$ $\displaystyle
2|e|^{-1}(\partial_{0}\omega_{1}-\partial_{1}\omega_{0})$ (180)

On the other hand we have

$\displaystyle\operatorname{d}\omega=\partial_{\mu}\omega_{\nu}dx^{\mu}\wedge
dx^{\nu}=(\partial_{1}\omega_{0}-\partial_{0}\omega_{1})dx^{1}\wedge
dx^{0}=-\frac{1}{2}R\sqrt{-g}\operatorname{d}^{2}x.$ (181)

The torsion two-form is given by

$\displaystyle T^{a}=(D)^{a}_{~{}b}e^{b}$ $\displaystyle=$
$\displaystyle(\delta^{a}_{~{}b}\operatorname{d}+\omega^{a}_{~{}b})e^{b}=\operatorname{d}e^{a}+\omega^{a}_{~{}b}\wedge
e^{b},$ (182)

with its components are

$\displaystyle T^{a}_{\mu\nu}$ $\displaystyle=$
$\displaystyle\partial_{\mu}e^{a}_{\nu}-\partial_{\nu}e^{a}_{\mu}+(\omega_{\mu})^{a}_{~{}b}e^{b}_{\nu}-(\omega_{\nu})^{a}_{~{}b}e^{b}_{\mu}$
(183) $\displaystyle=$ $\displaystyle D_{\mu}e^{a}_{\nu}-D_{\nu}e^{a}_{\mu}.$
(184)

It is convenient to use the Light-cone gauge:

$\displaystyle x^{\pm}=\frac{1}{\sqrt{2}}(x^{0}\pm x^{1}),\quad
x^{0}=\frac{1}{\sqrt{2}}(x^{+}+x^{-}),\quad
x^{1}=\frac{1}{\sqrt{2}}(x^{+}-x^{-}).$ (185)

The Lorentz transformation connecting these gauges is

$\displaystyle\Lambda^{a}_{~{}\bar{a}}=\frac{1}{\sqrt{2}}\begin{pmatrix}1&1\\\
1&-1\end{pmatrix}.$ (186)

Thus we can find that

$\displaystyle\eta_{\bar{a}\bar{b}}=\begin{pmatrix}0&-1\\\
-1&0\end{pmatrix},\quad\epsilon_{\bar{a}\bar{b}}=\begin{pmatrix}0&1\\\
-1&0\end{pmatrix},\quad\epsilon^{\bar{a}}_{~{}\bar{b}}=\begin{pmatrix}1&0\\\
0&-1\end{pmatrix}$ (187)

such that the torsion form (183) can be expressed as

$\displaystyle T^{\pm}=(\operatorname{d}\pm\omega)e^{\pm}.$ (188)

###  A.1 The first order formalism of GDT

The action (2) is equivalent to

$\displaystyle
I_{\text{gen}}[e_{a},\omega,X,X^{a}]=\int\left(X\operatorname{d}\omega+X_{a}(\operatorname{d}e^{a}+\epsilon^{a}_{~{}b}\omega\wedge
e^{b})+\frac{1}{2}\epsilon^{ab}e_{a}\wedge
e_{b}\mathcal{V}(X,X^{c}X_{c})\right).$

We will first solve all its classical solution and then prove the equivalence.
Varying with respect to $\omega$ gives

$\displaystyle
X\operatorname{d}\delta\omega+X^{a}\epsilon_{ab}\delta\omega\wedge
e^{b}=-\operatorname{d}X\wedge\delta\omega-X^{a}\epsilon_{ab}e^{b}\wedge\delta\omega\rightarrow$
$\displaystyle\operatorname{d}X+X^{a}\epsilon_{ab}e^{b}=0.$ (190)

Varying with respect to $e$ we get

$\displaystyle X^{a}\operatorname{d}\delta
e_{a}+X^{a}\epsilon_{a}^{~{}b}\omega\wedge\delta
e_{b}+\frac{1}{2}\epsilon^{ab}\left(\delta e_{a}\wedge e_{b}-e_{a}\wedge\delta
e_{b}\right)\mathcal{V}\rightarrow$
$\displaystyle-\operatorname{d}X^{a}\wedge\delta
e_{a}+X^{b}\epsilon_{b}^{~{}a}\omega\wedge\delta
e_{a}-\epsilon^{ab}e_{b}\wedge\delta e_{a}\mathcal{V}\rightarrow$
$\displaystyle
dX^{a}-X^{b}\epsilon_{b}^{~{}a}\omega+\epsilon^{ab}e_{b}\mathcal{V}=dX^{a}+X^{b}\epsilon_{~{}b}^{a}\omega+\epsilon^{ab}e_{b}\mathcal{V}=0,$
(191)

where in the last line we used
$\epsilon_{b}^{~{}a}=\eta_{bc}\epsilon^{c}_{~{}d}\eta^{da}=-\epsilon_{~{}b}^{a}$.
The other two equations of motion are

$\displaystyle\operatorname{d}\omega+\frac{1}{2}\epsilon^{ab}e_{a}\wedge
e_{b}\frac{\partial\mathcal{V}}{\partial X}=0,$ (192)
$\displaystyle\operatorname{d}e_{a}+\epsilon_{a}^{~{}b}\omega\wedge
e_{b}+\frac{1}{2}\epsilon^{ab}e_{a}\wedge
e_{b}\frac{\partial\mathcal{V}}{\partial X^{a}},.$ (193)

In the light-cone gauge the equations of motion become

$\displaystyle\operatorname{d}X+X^{+}e^{-}-X^{-}e^{+}=0,$ (194)
$\displaystyle(\operatorname{d}\pm\omega)X^{\pm}\pm\mathcal{V}e^{\pm}=0,$
(195) $\displaystyle\operatorname{d}\omega+\epsilon\frac{\partial V}{\partial
X}=0,$ (196)
$\displaystyle(\operatorname{d}\pm\omega)e^{\pm}+\epsilon\frac{\partial\mathcal{V}}{\partial
X_{\pm}}=(\operatorname{d}\pm\omega)e^{\pm}-\epsilon\frac{\partial\mathcal{V}}{\partial
X^{\mp}}=0,$ (197)

where the volume form is $\epsilon=e^{+}\wedge e^{-}$ and in the last line we
have used $X_{\pm}=-X^{\mp}$. From (195) we get

$\displaystyle
X^{-}\operatorname{d}X^{+}+X^{+}\operatorname{d}X^{-}+\mathcal{V}(X^{-}e^{+}-X^{+}e^{-})=0,$
(198)

then using (194) we get

$\displaystyle\operatorname{d}(X^{-}X^{+})+\mathcal{V}(X^{-}X^{+},X)\operatorname{d}X=0.$
(199)

This equation indicates that these exists a conserved quantity defined by
integrating (199).

If $X^{+}\neq 0$, from (195) we can get

$\displaystyle\omega=-\frac{\operatorname{d}X^{+}}{X^{+}}-Z\mathcal{V},\quad
Z\equiv\frac{e^{+}}{X^{+}},$ (200)

and from (194) we can get

$\displaystyle e^{-}=-\frac{\operatorname{d}X}{X^{+}}+X^{-}Z.$ (201)

Substituting the expression of volume form

$\displaystyle\epsilon=\frac{1}{2}\epsilon_{ab}e^{a}\wedge e^{b}=e^{+}\wedge
e^{-}=\operatorname{d}X\wedge Z$ (202)

into (197) gives

$\displaystyle\operatorname{d}e^{+}+\omega\wedge e^{+}-\operatorname{d}X\wedge
Z\frac{\partial\mathcal{V}}{\partial X^{-}}=0$ (203)
$\displaystyle=X^{+}\operatorname{d}Z+\operatorname{d}X^{+}\wedge
Z-\operatorname{d}X^{+}\wedge Z-\operatorname{d}X\wedge
Z\frac{\partial\mathcal{V}}{\partial X^{-}}=0.$ (204)

Therefore we end up with

$\displaystyle\operatorname{d}Z=-\frac{Z\wedge\operatorname{d}X}{X^{+}}\frac{\partial\mathcal{V}}{\partial
X^{-}}.$ (205)

Taking the ansatz of $Z$ as

$\displaystyle
Z=\operatorname{d}ve^{Q(X)},\quad\operatorname{d}Z=e^{Q(X)}\frac{\operatorname{d}Q}{\operatorname{d}X}\operatorname{d}X\wedge
dv\,,$ (206)

and substituting into (205) gives

$\displaystyle\frac{\operatorname{d}Q}{\operatorname{d}X}=\frac{1}{X^{+}}\frac{\partial\mathcal{V}}{\partial
X^{-}},\rightarrow$ (207) $\displaystyle
Q=\int^{X}\frac{1}{X^{+}}\frac{\partial\mathcal{V}}{\partial X^{-}}.$ (208)

Recall that the metric is

$\displaystyle\operatorname{d}s^{2}=\eta_{ab}e^{a}e^{b}=-2e^{+}e^{-}=2(Z\operatorname{d}X-X^{+}X^{-}Z^{2})=2e^{Q}(\operatorname{d}v\operatorname{d}X-e^{Q}Y\operatorname{d}^{2}v),$
(209)

where $Y\equiv X^{+}X^{-}$. So all solutions555in the linear dilaton vacua for
all generalized dilaton gravity models obey a generalized Birkhoff theorem, in
the sense that all solutions exhibit a Killing vector $\partial_{v}$. The
solution space is parameterized by two constants of integration. The one
coming from the integration of (199) is non-trivial, while the one coming from
(208) is trivial and can be fixed by a choice of units.

###  A.2 Back to Second order formalism

First we separate out the torsion-free part of the spin-connection. To do that
we notice

$\displaystyle\star
T_{a}=\star\left(\operatorname{d}e_{a}+\epsilon_{a}^{~{}b}\omega\wedge
e_{b}\right)=\star\operatorname{d}e_{a}+\epsilon_{a}^{~{}b}\omega^{c}\star(e_{c}\wedge
e_{b}).$ (210)

Using

$\displaystyle e_{a}\wedge
e_{b}=e_{a\mu}e_{b\nu}\operatorname{d}x^{\mu}\wedge\operatorname{d}x^{\nu},$
(211) $\displaystyle\star e_{a}\wedge
e_{b}=e_{a\mu}e_{b\nu}\star\operatorname{d}x^{\mu}\wedge\operatorname{d}x^{\nu}=e_{a\mu}e_{b\nu}\epsilon^{\mu\nu}=\epsilon_{ab}$
(212)

we get

$\displaystyle\star
T_{a}=\star\operatorname{d}e_{a}+\epsilon_{a}^{~{}b}\omega^{c}\epsilon_{cb}=\star\operatorname{d}e_{a}-\omega_{a}.$
(213)

So we can rewrite the spin-connection as

$\displaystyle\omega=\omega^{a}e_{a}=\left(\star\operatorname{d}e_{a}-\star
T_{a}\right)e^{a}=e^{a}\star\operatorname{d}e_{a}-e^{a}\star T_{a}.$ (214)

Then $\tilde{\omega}=e^{a}\star\operatorname{d}e_{a}$ is the torsion-free part
which in terms of components is given by

$\displaystyle\star\operatorname{d}e_{a}=\partial_{\mu}(e_{\nu})_{a}\epsilon^{\mu\nu},\quad\tilde{\omega}=e^{a}\partial_{\mu}(e_{\nu})_{a}\epsilon^{\mu\nu}.$
(215)

Recall that the action in the first formalism is

$\displaystyle I_{\text{gen}}\sim\int
X\operatorname{d}\omega+\epsilon\mathcal{V}+X^{a}T_{a}.$ (216)

The first term can be manipulated as

$\displaystyle
X\operatorname{d}\omega=-\operatorname{d}X\wedge\omega=-\operatorname{d}X\wedge(\tilde{\omega}-e^{a}\star
T_{a})=X\operatorname{d}\tilde{\omega}+\operatorname{d}X\wedge e^{a}\star
T_{a}.$ (217)

Note that

$\displaystyle\operatorname{d}\tilde{\omega}=\partial_{\mu}\omega_{\nu}\operatorname{d}x^{\mu}\wedge\operatorname{d}x^{\nu}\rightarrow-\frac{R}{2}\sqrt{-g}\operatorname{d}^{2}x,$
(218)

which is exactly the first term in the action (2). It is obvious that

$\displaystyle\epsilon\mathcal{V}(X,X^{a}X_{a})\rightarrow\sqrt{-g}\mathcal{V}(X,X^{a}X_{a})\operatorname{d}^{2}x.$
(219)

So the last thing to do is to remove $X^{a}$ with the help of equation of
motion (193):

$\displaystyle T_{a}=-\frac{1}{2}\epsilon^{bc}e_{b}\wedge
e_{c}\frac{\partial\mathcal{V}}{\partial X^{a}}\rightarrow$ (220)
$\displaystyle\star
T_{a}=-\frac{1}{2}\epsilon^{bc}\epsilon_{bc}\frac{\partial\mathcal{V}}{\partial
X^{a}}=\frac{\partial\mathcal{V}}{\partial X^{a}}$ (221)

and (190):

$\displaystyle\partial_{\mu}X+X^{a}\epsilon_{a}^{~{}b}e_{b\mu}=0\rightarrow$
(222)
$\displaystyle\partial_{\mu}X+X^{a}e_{a}^{\nu}\epsilon_{\nu\mu}=0\rightarrow$
(223) $\displaystyle X^{a}=-e^{a}_{\nu}\epsilon^{\mu\nu}\partial_{\mu}X.$
(224)

So terms involved with $T^{a}$ are cancelled to each other:

$\displaystyle
X^{a}T_{a}=\frac{1}{2}e^{a}_{\nu}\epsilon^{\mu\nu}\partial_{\mu}X\epsilon^{cb}e_{c}\wedge
e_{b}\frac{\partial\mathcal{V}}{\partial X^{a}}\rightarrow
e^{a}_{\nu}\epsilon^{\mu\nu}\partial_{\mu}X\frac{\partial\mathcal{V}}{\partial
X^{a}}\sqrt{-g}\operatorname{d}^{2}x,$ (225)
$\displaystyle\operatorname{d}X\wedge e^{a}\star
T_{a}=\partial_{\mu}Xe^{a}_{\nu}\frac{\partial\mathcal{V}}{\partial
X^{a}}\operatorname{d}x^{\mu}\wedge\operatorname{d}x^{\nu}\rightarrow-e^{a}_{\nu}\epsilon^{\mu\nu}\partial_{\mu}X\frac{\partial\mathcal{V}}{\partial
X^{a}}\sqrt{-g}\operatorname{d}^{2}x.$ (226)

Then we arrive at the action in second order formalism

$\displaystyle-\frac{1}{2}\int\sqrt{-g}\left(XR-2\mathcal{V}(X,-(\partial
X)^{2})\right),$ (227)

where we have used

$\displaystyle X^{a}X_{a}=-(\partial X)^{2}.$ (228)

Therefore we find that the $Q$ function is given by

$\displaystyle Q=\int^{X}U(y)\operatorname{d}y,$ (229)

and the conserved quantity (199) is given by

$\displaystyle C_{0}=e^{Q}Y+w,\quad Y=X^{+}X^{-},\quad
w=\int^{X}e^{Q}V(y)\operatorname{d}y$ (230) $\displaystyle
dC_{0}=e^{Q}\operatorname{d}Y+Ye^{Q}U(X)\operatorname{d}X+e^{Q}V\operatorname{d}X=0.$
(231)

Using this we can rewrite the metric as

$\displaystyle\operatorname{d}s^{2}$ $\displaystyle=$ $\displaystyle
2e^{Q}\operatorname{d}v(\operatorname{d}X+(w(X)-C_{0})\operatorname{d}v)$
(232) $\displaystyle=$ $\displaystyle
2\operatorname{d}v\operatorname{d}\tilde{X}+\xi(\tilde{X})\operatorname{d}v^{2},$
(233)

where we have introduced

$\displaystyle\operatorname{d}\tilde{X}=\operatorname{d}Xe^{Q},\quad\xi(\tilde{X})=2e^{Q}(w-C_{0}).$
(234)

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