aainstitutetext: Institute for Theoretical Physics, University of Amsterdam,
1090 GL Amsterdam, The Netherlands

# Island mirages

Andrew Rolph<EMAIL_ADDRESS>
###### Abstract

We point out a loophole in the proof that the Island Finder conditions
Bousso2021  are sufficient for the existence of islands. We explore examples
which satisfy the conditions, but have no islands, which we call island
mirages. We also describe and give resolutions to two new puzzles associated
with islands, including a direct tension with the quantum Bousso bound.

##  1 Introduction

The black hole information problem (BHIP) is a tension between quantum
mechanical unitarity and the insensitivity of Hawking radiation to initial
conditions, and has been a persistent and impelling puzzle in quantum gravity
since its conception Hawking1975 ; Hawking1976 . The Page curve, the von
Neumann entropy of emitted radiation as a function of time, is one sharp
diagnostic of unitarity in black hole evaporation Page1993 . Deriving a Page
curve consistent with unitarity, one that ends at zero von Neumann entropy for
a black hole formed from collapse of matter in a pure state, is an important
step in resolving the BHIP, though it is not the whole story.

Much of the recent progress in deriving a Page curve consistent with unitarity
has been in developing the techniques to calculate the entropy of Hawking
radiation, or more generally of quantum systems coupled to gravity. This was
inspired in part by the Ryu-Takayanagi formula for calculating von Neumann
entropy in holographic CFTs Ryu_2006 ; Freedman2017 , and its generalisations
Hubeny2007 ; Engelhardt2014 ; Harper2018 ; Rolph2021 , and culminated in the
island formula Almheiri2019b  and quantum maximin prescription Akers2019 . The
island formula is a prescription for computing fine-grained entropy in quantum
systems coupled to gravity in terms of semiclassical entropy. See Almheiri2021
for a review of islands, including an explanation of what is meant by fine-
grained, coarse-grained, renormalised and semiclassical entropy in this
context. The island prescription is

$S(R)=\min\text{ext}_{I}\left(\frac{\text{Area}(\partial
I)}{4G_{N}}+S_{ren.}(R\cup I)\right)+S_{c.t.}(R)$ (1)

where $S_{ren.}$ is the renormalised semiclassical von Neumann entropy, and
$S_{c.t.}$ the entropy counterterm that isolates the UV divergences associated
to the region’s boundary. The prescription instructs us to extremise a
functional over all possible islands $I$, including the trivial empty island
$I=\varnothing$, then pick the minimal extremum.

The main technical obstacle to explicitly finding islands is evaluating
$S_{ren.}(R\cup I)$ on the set of all possible islands. The cases where it is
possible, such as in lower dimensional and doubly holographic models
Almheiri2019b ; Almheiri2019 ; Almheiri2019a ; Almheiri2020 , are sparse. This
motivates finding necessary and/or sufficient conditions for the existence of
islands. Sets of necessary and/or sufficient conditions for the existence of
non-trivial islands may be easier to evaluate than the island formula (1)
itself. A set of necessary conditions were proposed in Hartman2020 , and a set
of sufficient conditions in the Island Finder paper Bousso2021 . The
literature which relies on these sets of conditions is growing, so it is
important to know if and when they can be relied upon.

The two conditions given in Bousso2021  for a region $R$ to have a non-empty
island are in terms of an ‘island detector’111Credit to Ahmed Almheiri for the
name island detector. region $I^{\prime}$ and require

$\qquad 1.\quad S_{gen}(I^{\prime}\cup R)<S_{gen}(R)$ (2)

and either

$2.a.\quad\pm\Theta_{\pm}(I^{\prime}\cup R)\geq 0$ (3)

or

$2.b.\quad\pm\Theta_{\pm}(I^{\prime}\cup R)\leq 0.$ (4)

where $\Theta_{\pm}$ is the quantum expansion in the future-directed outward
$(+)$ and inward $(-)$ null directions normal to $\partial I^{\prime}$222See
Bousso2015  for a precise definition of quantum expansion..

In this paper we first point out a loophole in the proof that eqs. 2, 3 and 4
are sufficient conditions for $R$ to have a non-empty island. The proof given
in Bousso2021  is by contradiction: it is assumed that (1) there is an island
detector region $I^{\prime}$ satisfying conditions (2) to (4), and (2) that
the island $I$ is empty. Then it is shown that these two assumptions lead to a
contradiction. The loophole is the implicit assumption that at least one of
the time slices on which the empty island is the quantum maximin surface
intersects the domain of dependence of $I^{\prime}$333Assuming $I^{\prime}\cup
R$ is quantum normal, i.e. (3). We will consider the quantum anti-normal case
separately.. If this assumption is violated then the representative
$\tilde{I}^{\prime}$ \- the intersection of $D(I^{\prime})$ with the time
slice - is itself empty and an intermediate step in the proof is invalid.

We preempt and address counterarguments to this loophole which assert that the
implicit assumption mentioned above is always valid. One counterargument is
based on the claim that the past and future tips of $D(I^{\prime})$ are
truncated by singularities, and the basis of this claim are two puzzling
contradictions that arise when considering small representatives
$\tilde{I}^{\prime}$ near either tip of $D(I^{\prime})$.

The first puzzle is a direct tension with the quantum Bousso bound. Assuming
that either of $\partial D^{\pm}(I^{\prime})$ is a quantum lightsheet, then
the quantum Bousso bound implies that $S_{gen}(\tilde{I}^{\prime}\cup R)$ must
decrease as the representative $\tilde{I}^{\prime}$ is deformed towards the
tip of the $D^{\pm}(I^{\prime})$. A contradiction, $S_{gen}(R)<S_{gen}(R)$, is
found when the tip is reached and $\tilde{I}^{\prime}=\varnothing$. We discuss
several resolutions to this puzzle. The first two are an evasion of the
tension with the quantum Bousso bound: (1) there is no
$\tilde{I}^{\prime}=\varnothing$ because of truncation of $D(I^{\prime})$ by
curvature singularities, or (2) $\partial D^{\pm}(I^{\prime})$ fails to be a
quantum lightsheet. The third resolution is that (3) the quantum Bousso bound
is violated, but that this is not unexpected because to reach
$\tilde{I}^{\prime}=\varnothing$ one has to pass outside the regime of
validity of the semiclassical approximation. There is more than one
resolution, and how the tension is resolved depends on the set-up.

The second puzzle is that the existence of island detectors naively and
incorrectly implies that a ball-shaped region of the Minkowski vacuum violates
the Bekenstein area bound. The authors of Hartman2020  showed, with some mild
assumptions, that any region $\mathcal{I}$ that satisfies
$S_{gen.}(\mathcal{I}\cup R)<S_{gen.}(R)$, which includes island detector
representatives, also violates the Bekenstein area bound. The second puzzle
arises when applying this result to a representative that is smaller than the
local curvature and temperature scales. If such a representative
$\tilde{I}^{\prime}$ exists for a given island detector $I^{\prime}$, then the
state on it is approximately that of the vacuum in flat space, which should
not violate the Bekenstein area bound.

The puzzles are of broader importance than may initially appear, because they
also need to be resolved for islands. As far as the puzzles are concerned
islands are simply special cases of island detectors in the sense that they
satisfy (2) and saturate both conditions (3) and (4) by their definitional
extremality.

We discuss in depth the resolution of these two puzzles for islands in
asymptotically flat evaporating black holes. The past domain of dependence of
the island has no curvature singularities to prevent one from considering
arbitrarily small representatives, so the quantum Bousso bound is apparently
violated. A key part of resolving the Bekenstein bound puzzle is the sub-
Planckian radial separation of the boundary of the island and the past
lightcone of the radiation region, which is an obstacle to specifying ‘nice’
time slices with trace extrinsic curvature $|K|\ll l_{p}^{-1}$.

Outline: In Sec. 2 we review Island Finder and explain the loophole in the
proof. In Sec. 3 we discuss counterarguments to the loophole, which naturally
leads us to the two puzzles, to which we propose resolutions. In Sec. 4 we
give some open questions unresolved by this work and ideas for future
research. In Appendix A we outline two set-ups which may have explicit
examples of island mirages.

##  2 Island Finder

In this section we discuss the Island Finder conditions and reproduce the
proof given that they are sufficient conditions for the existence of a non-
empty island Bousso2021 . We point out loopholes in the proof.

###  2.1 Conditions

Figure 1: Two (asymptotically flat) quantum systems, one with dynamical
gravity and one without. $R$ is a region in the non-gravitating system, and
$I^{\prime}$ a region in the gravitating system. $\Sigma$ is a time slice, and
$\tilde{I}^{\prime}$ the representative of $I^{\prime}$ on that time slice as
defined in (9). If $I^{\prime}$ satisfies conditions (5) and (7) then it is an
island detector for $R$.

Consider two disjoint achronal regions, $I^{\prime}$ and $R$, in a spacetime
$M$ with $R$ in a weakly gravitating subregion. The claim of Bousso2021  is
that if two conditions are satisfied by the generalised entropy of
$I^{\prime}\cup R$ then this is sufficient to guarantee that $R$ has a non-
empty island somewhere in $M$.

We will call an $I^{\prime}$ that satisfies the two conditions an island
detector region for $R$. If an island detector region exists and $R$ does have
a non-empty island $I$, then the conditions are working as intended. If an
island detector region exists and the island $I$ for $R$ is empty, then we
have a faulty island detector, and we call such an $I^{\prime}$ an island
mirage.

The first of the two conditions is

$(1)\quad S_{gen}(I^{\prime}\cup R)<S_{gen}(R)$ (5)

There is an equivalent form of this inequality,

$I(I^{\prime},R)>S_{gen}(I^{\prime})$ (6)

which has the benefit of both sides being UV-finite even if we decouple
gravity, $G_{N}\to 0$, in the region that contains $R$ Casini2007 ;
Bombelli1986 ; Hartman2020 . One quick comment is $S_{gen}(I^{\prime})$ scales
as the surface area of $I^{\prime}$ in Planck units, so there is a manifest
need for a large degree of correlation between $I^{\prime}$ and $R$ in order
to satisfy the condition.

The second of the two sufficient conditions is that $I^{\prime}\cup R$ be
either quantum anti-normal or quantum normal. Quantum normality is the
property that the generalised entropy of a spatial region increases under
deformations of the region’s boundary in outward null directions. This is a
generalisation of the classical notion of a normal spatial region whose
surface area increases under outward deformations. The definition of quantum
normality is

$(2)\quad\pm\Theta_{\pm}(I^{\prime}\cup R)\geq 0$ (7)

where $\Theta_{\pm}$ is the quantum expansion in the outward/inward future
null directions Bousso2015 . Quantum anti-normality is the opposite of quantum
normality in the sense that the generalised entropy decreases under outward
null deformations. Any subregion of a time reflection symmetric slice is
automatically quantum normal or anti-quantum normal, so any $I^{\prime}\cup R$
on such a slice which satisfies condition (5) satisfies both of the Island
Finder conditions. Also, if $I^{\prime}\cup R$ is quantum normal then so is
$I^{\prime}$, which means the detector region $I^{\prime}$ satisfies one of
the necessary conditions to itself be an island Almheiri2019a ; Hartman2020 .
Note that (7) is a property that is required to hold locally at every point on
the boundary of $I^{\prime}\cup R$, and that it depends on the non-local
entanglement entropy.

###  2.2 QFC and quantum maximin

In the proof the quantum focussing conjecture (QFC) and quantum maximin
prescription are assumed Bousso2015 ; Akers2019 . We assume that the spacetime
is stably causal, because that is the causality condition assumed for the
quantum maximin prescription Akers2019 , and because we do not want to exclude
asymptotically AdS spacetimes444Classical maximin was originally introduced as
a reformulation of the HRT holographic entanglement entropy prescription which
applies to asymptotically locally AdS spacetimes Hubeny2007 ; Wall2012 . Since
AdS is not globally hyperbolic it has no Cauchy slices, so the notion of
Cauchy slices had to be enlarged to AdS-Cauchy slices which are geodesically
complete achronal surfaces Wall2012 ; Engelhardt2014 . Stably causal
spacetimes have a global time function, and the function’s level sets are
constant time slices Wald:1984rg .

Quantum maximin is a prescription for calculating entropy of subregions in
non-holographic systems as well as holographic ones Akers2019 . A relevant
class of models are AdS black holes evaporating into a non-gravitating bath,
where in the non-gravitational description we couple a holographic and non-
holographic system together Almheiri2019a . If $R$ is a subregion of the non-
holographic system, then the quantum maximin prescription for calculating its
exact von Neumann entropy is

$S(R)=\max_{\Sigma}\min_{I\subset\Sigma}\left(\frac{\text{Area}(\partial
I)}{4G_{N}}+S_{ren.}(R\cup I)\right)$ (8)

The quantum maximin prescription (8) is equivalent to the island formula,
assuming the QFC and a certain stability condition on $I$ Akers2019 . The
prescription minimises over candidate islands on a given time slice, then
maximises over time slices. Let us call such a maximising time slice $\Sigma$
a maximin slice. The maximin slice need not be unique, particularly if the
island is empty.

###  2.3 Proof

The proof given in Bousso2021  is a proof by contradiction. Start by assuming
that the two conditions (5) and (7) are satisfied, and that the true island
$I$ is the empty set, i.e. there is no quantum extremal region $I$ that lowers
the generalised entropy of $R$. We will treat the case where $I^{\prime}\cup
R$ is quantum normal first, and consider the quantum anti-normal case
separately.

$I^{\prime}\cup R$ is quantum normal so the quantum expansion in the inward
null directions is initially negative. Consider a time slice $\Sigma$ that
contains $R$ and define the representative $\tilde{I}^{\prime}$ of
$I^{\prime}$ on $\Sigma$ as

$\tilde{I}^{\prime}:=D(I^{\prime})\cap\Sigma$ (9)

where $D(I^{\prime})$ is the domain of dependence of $I^{\prime}$. QFC states
that the quantum expansion cannot increase along a null congruence, so that
the quantum expansion must remain negative along the null congruence from
$I^{\prime}$ to $\tilde{I}^{\prime}$. The generalised entropy can only
decrease under such inward null deformations:

$\begin{split}S_{gen}(\tilde{I}^{\prime}\cup R)&\leq S_{gen}(I^{\prime}\cup
R)\\\ &<S_{gen}(R)\end{split}$ (10)

The second line follows from (5).

The contradiction appears if we take $\Sigma$ to be a time slice on which the
codim-2 quantum maximin surface lives. From the quantum maximin prescription
the true island $I$ gives the lowest generalised entropy on its maximin slice,
and since we assumed $I$ to be the empty set we have that

$\begin{split}S_{gen}(\tilde{I}^{\prime}\cup R)&\geq S_{gen}(I\cup R)\\\
&=S_{gen}(R)\end{split}$ (11)

which directly contradicts (10). The conclusion is that the island cannot be
empty if the two conditions (5) and (7) are satisfied.

When $I^{\prime}\cup R$ is quantum anti-normal the essential steps in the
proof are the same, except that the representative $\tilde{I}^{\prime}$ of
$I^{\prime}$ on a time slice $\Sigma$ is defined as

$\tilde{I}^{\prime}:=J(I^{\prime})\cap\Sigma$ (12)

with $J(I^{\prime})$ all points that can be reached from $I^{\prime}$ with
causal curves Bousso2021 .

###  2.4 Loophole

The proof in a nutshell is that the conditions imply that every time slice
that contains $R$ and intersects $D(I^{\prime})$ has a non-empty
$\tilde{I}^{\prime}$ for which $S_{gen}(\tilde{I}^{\prime}\cup R)<S_{gen}(R)$.
This rules out that time slice being the maximin slice for a true island that
is empty, as the true island must give the lowest generalised entropy on the
maximin slice.

One loophole in the proof is the implicit assumption that the maximin slice of
the empty island intersects $D(I^{\prime})$. If the true island is empty, and
its maximin slice does not intersect $D(I^{\prime})$, then by the definition
(9) we have $\tilde{I}^{\prime}=\varnothing$, and the specific step that led
to a contradiction that fails here is (10). No contradiction is reached and
the proof fails.

The QFC does not constrain, in relation to $S_{gen}(I^{\prime}\cup R)$, the
generalised entropy of subregions $X\cup R$, with $X\neq\varnothing$, on time
slices that do not intersect $D(I^{\prime})$. QFC is defined on null
congruences, and by definition the integral curves in a congruence are non-
intersecting, so congruences end at caustics. QFC states that the quantum
expansion cannot increase along a null congruence, but the tips of
$D(I^{\prime})$ are caustics. The QFC does constrain $S_{gen.}(X\cup R)$ for
$X=\varnothing$ if we apply it to a quantum normal $I^{\prime}\cup R$ and
integrate it over the whole of either of $\partial D^{\pm}(I^{\prime})$:

$S_{gen.}(R)\leq S_{gen.}(I^{\prime}\cup R)$ (13)

Thus there is a direct tension between island detectors and the QFC, if there
exists an empty representative $\tilde{I}^{\prime}=\varnothing$, because (13)
contradicts (5). This is a loophole counterargument that we will return to in
Sec. 3.2.

Figure 2: The Penrose diagram of a small AdS black hole that forms from
collapse then evaporates away to nothing. $\Sigma$ is a time slice.
$I^{\prime}$ is a region whose representative $\tilde{I^{\prime}}:=\Sigma\cap
J(I^{\prime})=\varnothing$, for which the step (10) of the proof in Bousso2021
is incorrect. This is an example of the causal structure needed for
$\tilde{I}^{\prime}$ to be empty in the case of $I^{\prime}\cup R$ being
quantum anti-normal.

When $I^{\prime}\cup R$ is quantum normal any time slice that does not
intersect $D(I^{\prime})$ gives an empty representative $\tilde{I}^{\prime}$.
To have an empty representative when $I^{\prime}\cup R$ is quantum anti-normal
there must exist a time slice that does not intersect $J(I^{\prime})$. This is
not possible in globally hyperbolic spacetimes, because all causal curves from
$I^{\prime}$ will intersect all Cauchy slices. Recall however that quantum
maximin is applicable not only to globally hyperbolic spacetimes, but also
spacetimes that are stably causal and not globally hyperbolic. This allows for
time slices on which $\tilde{I}^{\prime}$ is empty. An example of a stably
causal spacetime with an $I^{\prime}$ and a time slice $\Sigma$ on which the
representative $\tilde{I}^{\prime}$ is empty is shown in Fig. 2. Singularities
are barriers to causal curves from $I^{\prime}$. This figure is only an
example of the causal structure needed for an island mirage when
$I^{\prime}\cup R$ is quantum anti-normal - we have not shown nor do we claim
that the Island Finder conditions are satisfied in this case.

Whether $I^{\prime}\cup R$ is quantum normal or anti-normal, there are time
slices on which the representative $\tilde{I}^{\prime}=\varnothing$, and
$S_{gen}(\tilde{I}^{\prime}\cup R)\nleqslant S_{gen}(I^{\prime}\cup R)$. No
contradiction is reached, so a true island that is empty is not ruled out. The
island finder conditions are insufficient to establish existence of islands,
and the conditions may be satisfied without $R$ having a non-empty island. The
conditions are sufficient only to establish nonexistence of empty islands
whose maximin slice intersects the domain of dependence of $I^{\prime}$ (when
$I^{\prime}\cup R$ is quantum normal). The concommitant upper entropy bound
given in Bousso2021  is also not proven if the proof that the conditions are
sufficient for a non-empty island has a loophole.

##  3 Loophole counterarguments and island puzzles

###  3.1 Time slices for empty maximin surfaces

The first objection one may have to the loophole is that if the true island is
empty, then is it not irrelevant which time slice we use when calculating the
entropy? Can we not always choose a time slice with a non-empty intersection
with $D(I^{\prime})$? The loophole is indeed closed if all empty islands have
maximin slices which intersect $D(I^{\prime})$. As a reminder, we defined
maximin slices to be the maximising subset of time slices in the quantum
maximin prescription (8).

Now it is true that when the island is empty the von Neumann entropy of $R\cup
I$ is the same for all time slices that contain $R$, and that in general a
subset of those time slices will intersect $D(I^{\prime})$. What is not true
is that all such time slices are maximin slices for the empty island. If the
true island is empty then by maximin there is certainly at least one time
slice on which the empty-island entropy is the minimal entropy, but there is
no guarantee that the no-island entropy is the minimal entropy on every time
slice containing $R$.

Assuming that the island is empty, a time slice fails to be a maximin slice
when on that slice there is a subregion $\mathcal{X}$ for which

$\frac{\text{Area}(\partial\mathcal{X})}{4G_{N}}+S_{ren.}(R\cup\mathcal{X})<S_{ren.}(R)$
(14)

As an example of this, consider a time slice that contains a subregion
$\mathcal{X}$ whose boundary has zero area, i.e. whose components are null
hypersurfaces, and for which $\mathcal{X}$ is entangled with $R$ such that
$S_{ren.}(\mathcal{X}\cup R)<S_{ren.}(R)$. Similar to the setups considered in
Hartman2020 , we could take a thermofield double state on AdS and a non-
gravitating bath, take $R$ to be the whole of the bath, and $\mathcal{X}$ to
be a ball-shaped region in the AdS whose boundary has been deformed into
zigzag union of many null hypersurfaces. The time slices that contain this
$\mathcal{X}$ (and $R$) fail to be maximin slices for the empty island.

We have argued that for an empty island not every time slice (containing $R$)
is necessarily a maximin slice, but for the loophole to work we need something
stronger: that no time slice that intersects $D(I^{\prime})$ is a maximin
slice when the island is empty. This requires that on all such time slices
with $\Sigma\cap D(I^{\prime})\neq\varnothing$ there exists an $\mathcal{X}$
with $\mathcal{X}\cup R\subseteq\Sigma$ such that (14) is satisfied. Island
Finder Bousso2021  showed that (14) is satisfied by the representative
$\tilde{I}^{\prime}$ as defined in (9). Thus an $I^{\prime}$ that satisfies
conditions (5) and (7) only rules out empty islands whose maximin slices
intersect $D(I^{\prime})$; it is not sufficient for the existence of a non-
empty island.

###  3.2 Quantum Bousso bound puzzle

Another objection to the loophole is the apparent contradiction with the
quantum Bousso bound when there a time slice that does not intersect
$D(I^{\prime})$, with $I^{\prime}\cup R$ satisfying conditions (5) and (7).

The version of the quantum Bousso bound that we use is that which is implied
by the QFC Bousso2015 . There is a different version of the quantum Bousso
bound that has been proven using positivity of relative entropy Casini2008 ;
Bousso2014 . In that version of the bound the entanglement entropy is
regulated by vacuum subtraction, and vacuum-subtracted entanglement entropy is
inequivalent to the renormalised entropy used in $S_{gen.}$ Bousso2014 ;
Bousso2015 . We need to use the version implied by the QFC, because it is that
version that constrains the renormalised and generalised entropies that appear
in the island formula.

The quantum Bousso bound applies to quantum light sheets, which are null
hypersurfaces on which the quantum expansion is every nonpositive. Since
$I^{\prime}\cup R$ is quantum normal by assumption, QFC implies that the past
and future boundaries of the domain of dependence $D(I^{\prime})$ are quantum
light sheets, and the quantum Bousso bound states that Strominger2003

$S_{gen.}(I^{\prime}\cup R)\geq S_{gen.}(\tilde{I}^{\prime}\cup R)$ (15)

where as before, $\tilde{I}^{\prime}:=\Sigma\cap D(I^{\prime})$.

We assume that the spacetime is stably causal, which is the same causality
condition assumed in the quantum maximin prescription, such that there is a
global time function $t$. With a reparametrisation of $t$ we can take the
$t=0$ slice to contain $I^{\prime}$, and the $t=\pm 1$ slices to intersect
$D(I^{\prime})$ at the future and past tips of $D(I^{\prime})$. In the limit
of taking time slices to the tips of the domain of dependence, $|t|\to 1^{-}$,
we have $\text{Area}(\partial\tilde{I}^{\prime})\to 0$. If we assume also that
as we deform towards either tip of $D(I^{\prime})$ that

$\lim_{|t|\to 1^{-}}S_{ren.}(\tilde{I}^{\prime}\cup R)=S_{ren.}(R)$ (16)

then the $|t|\to 1^{-}$ limit of the quantum Bousso bound (15) implies

$S_{gen.}(I^{\prime}\cup R)\geq S_{gen.}(R),$ (17)

which directly contradicts (5). Thus as we deform $\tilde{I}^{\prime}$ towards
either tip of $D(I^{\prime})$ then there is a direct tension with the quantum
Bousso bound even before the exact point of $\tilde{I}^{\prime}=\varnothing$
is reached, if (16) is true.

Figure 3: An evaporating black hole in asymptotically Minkowski spacetime.
Radiation region $R$ has an island $I$. Also shown are the boundary of the
causal past of $R$, the boundary of the past domain of dependence of $I$, and
an earlier time representative $\tilde{I}$ of $I$. This is a counterexample to
the claim that the existence of a region satisfying conditions (5) and (7)
implies the existence of singularities which remove both tips of the domain of
dependence.

Islands, not just island detector regions, are in direct tension with the
quantum Bousso bound by the same chain of logic. Recall that islands are
special cases of island detector regions in the sense that they satisfy (5)
and saturate the bounds (7). As an example, consider an evaporating black hole
in asymptotically flat spacetime, as depicted in Fig. 3. The radiation region
$R$ is taken to be sufficiently large to have an island $I$ in the black hole
interior. The representative $\tilde{I}:=\Sigma\cap D(I)$ on an earlier time
slice satisfies

$S_{gen}(\tilde{I}\cup R)\leq S_{gen}(I\cup R)$ (18)

by the quantum Bousso bound and QFC. Again we reach the apparent contradiction
$S_{gen}(R)<S_{gen}(R)$ when we reach the past tip of $D(I)$ and
$\tilde{I}=\varnothing$.

This contradiction needs to be resolved whenever conditions (5) and (7) are
satisfied, whether or not there is a non-empty island in the spacetime. Here
are several ways the contradiction can be evaded:

1\. $D(I^{\prime})$ is truncated by curvature singularities.

One possible resolution is that when conditions (5) and (7) are satisfied
there are always singularities to the future and past of $I^{\prime}$ that
remove the tips of $D(I^{\prime})$, so that the problematic limit (16) does
not exist. In some spacetimes this would force all time slices to intersect
$D(I^{\prime})$, so there would be no empty representatives
$\tilde{I}^{\prime}$ and the loophole would be closed . If truncation by
singularities were the only resolution to the tension with the quantum Bousso
bound then conditions (5) and (7) would also be sufficient conditions for the
existence of singularities555This was also pointed out in BoussoTalk ..

There is a simple counterexample to the claim that there must be
singularities, which is the evaporating black hole of Fig. 3. While there is a
singularity to the future of $I$ there is none to the past. Curvature
singularities may resolve the tension with the quantum Bousso bound in some
cases, but there must be a different resolution for the past domain of
dependence of an island in an evaporating black hole.

2\. $\partial D^{\pm}(I^{\prime})$ fails to be a quantum lightsheet.

Another way in which the quantum Bousso bound can fail to be applicable is if
only a subset of $\partial D(I^{\prime})$ is a quantum lightsheet. Classical
lightsheets, with everyhere negative classical expansion, can terminate due to
the formation of caustics where $\theta\to-\infty$. Assuming the null energy
condition is satisfied, the Raychaudhuri equation implies that caustics form,
and so lightsheets possibly end after finite affine parameter whenever the
classical expansion is negative Raychaudhuri:1953yv . The truncation of
lightsheets underlied an argument against the critically illuminated black
hole counterexample to the classical Bousso bound Lowe1999 666Angular caustics
also save the classical Bousso bound when there is anisotropy in a collapsing
shell, which increases its entropy but also decreases the extent of the
lightsheet Bousso1999 . The critically illuminated black hole does not violate
the vacuum-subtracted entropy version of the quantum Bousso bound Bousso2014 .
Vacuum-subtracted von Neumann entropy is not classical, in that it is not
generally additive, so the resolution in Bousso2014  is not applicable to the
classical Bousso bound.. This argument against Lowe’s counterexample was later
realised to be flawed; $\partial D(I^{\prime})$ does not fail to be a
classical lightsheet, despite null geodesics in the congruence orthogonal to
$\partial I^{\prime}$ meeting other null geodesics and caustics Akers2018
777Thanks to Raphael Bousso for discussions on this point..

What is true for classical lightsheets is not necessarily true for quantum
lightsheets. If quantum lightsheets can end after finite affine parameter,
perhaps because of a quantum analogue of a classical caustic,
$\Theta\to-\infty$, then the quantum Bousso bound can fail to apply once the
null generator deforming $\partial\tilde{I}^{\prime}$ reaches that point. We
will return to the question of whether $\Theta$ must diverge to $-\infty$ in
finite affine parameter on a quantum lightsheet in the discussion section 4.

It does not seem likely that $\partial D(I^{\prime})$ failing to be a quantum
lightsheet can resolve the tension with the quantum Bousso bound for islands
for all evaporating black hole set ups, as the truncation of quantum
lightsheets seems non-universal in that it depends on the state, and in
particular the stress-energy tensor of the infalling matter. We would like a
more robust resolution for islands in evaporating black holes.

3\. Blueshift effects.

There is a blueshift effect for Hawking modes near the horizon, and it is
tempting to claim that the blueshift of Hawking modes towards the past to
super-Planckian energies is what leads to violations of the quantum Bousso
bound, but we posit that this is not the case.

The renormalised entropy of the island comes from summing over interior
Hawking modes below the renormalisation scale. As we propagate the interior
modes backwards in time onto the representative region $\tilde{I}$ the
relative blueshift - the ratio of frequencies as measured by free-falling
observers that cross the event horizon at different times $v_{1}$ and $v_{2}$
\- grows exponentially as Jacobson2005

$\frac{\omega_{1}}{\omega_{2}}\approx e^{(v_{2}-v_{1})/2r_{s}}$ (19)

Towards earlier infalling times the area of the boundary of the representative
$\tilde{I}$ decreases, yet $\tilde{I}$ continues to be able to contain the
same modes which purify the partner modes in $R$ despite the shrinking volume,
due to the reduction in wavelength from the blueshift. Once the mode
blueshifts to a frequency above the renormalisation scale, whatever that may
be, it switches from contributing to $S_{ren.}$ to contributing to the entropy
counterterm. Naively the blueshift effect leads to modes with super-Planckian
energies that backreact on the geometry, a firewall.

We are assuming no drama for infalling observers which implicitly fixes the
entanglement between interior and exterior outgoing modes such that there is
no backreaction Almheiri2013 . Altering the state by removing a single Hawking
mode from the island would lead to a firewall in the past, which would lead to
a curvature singularity.

To summarise, blueshift effects do not seem to resolve the puzzle because with
the no drama assumption there are no curvature singularities in the
neighbourhood of the past tip of $D(I)$ that could explain the violation of
the quantum Bousso bound.

4\. The naive result (16) was incorrect due to an unavoidable departure from
the semiclassical regime when taking the limit. Violations of the quantum
Bousso bound are not necessarily unexpected when the bound is applied outside
of the semiclassical regime, such as to regions of sub-Planckian size, or when
it is derived from integrating the QFC outside of the semiclassical regime.

The last resolution to the contradiction that we’ll mention is that the naive
result (16) is incorrect, because to take the limit you must consider regions
$\tilde{I}^{\prime}$ of Planckian size. Generalised entropy is a UV finite and
regulator-independent quantity Bianchi2012 ; Cooperman2013 , but it is a
semiclassical quantity. The generalised entropy of a region of Planckian size
is physically meaningless. It may be mathematically well-defined if the
semiclassical approximation is taken, in the sense that the region can at
least be specified if quantum fluctuations of the background geometry are
neglected, but the result has no relation to the physical reality.

A simpler set-up where we can see what goes wrong if we trust formulas for
entropy beyond their physical regime of validity is if we consider an interval
of length $L$ in a 2d CFT in the vacuum state, whose entanglement entropy is
$S(L)=\frac{c}{3}\log(L/\epsilon)$. We can evaluate $S(L)$ for $L<\epsilon$,
but it physically meaningless and gives unphysical negative entropy. Just as
in (16), we can take a naive limit

$\lim_{L\to 0}S(L)=-\infty$ (20)

and get a result which is not the correct entropy of the empty set, which is
zero. There is an important distinction between intervals of infinitesimal and
zero length.

The QFC and the quantum Bousso bound are not physically meaningful when
applied to regions of sub-Planckian size, and violations of these bounds are
not surprising in these applications as we are outside of the semiclassical
regime.

This does not immediately resolve the tension with the quantum Bousso bound
when it is applied to the whole of $\partial D^{\pm}$ as for an island
detector in (13), or for an island as in (18), because in these cases no
subregions of sub-Planckian size are directly involved in the bound.

We may derive the quantum Bousso bound, as applied in (13), by integrating the
QFC over $\partial D^{\pm}(I^{\prime})$. To do so we must however apply the
QFC to sub-Planckian regions, as we integrate towards either tip of
$D(I^{\prime})$, where violations of the QFC are not unexpected. A bound that
can be derived by integrating the QFC along a null congruence like $\partial
D^{\pm}(I^{\prime})$ is not always false - that would rule out the GSL
Bousso2015  \- but violations of the bound are not necessarily unexpected
since we are applying the QFC outside of the semiclassical regime.

###  3.3 Bekenstein bound puzzle

In the previous subsection we described a tension with the quantum Bousso
bound that arises when taking the infinitesimal limit of the representative of
either of an island or a quantum normal island detector, as well as beyond
that when the representative is empty. In this subsection we describe a
different puzzle that arises before the representative reaches Planckian
length scales.

This puzzle is relevant to the loophole described in section 2.4, because just
like the Bousso bound puzzle if the only resolution were that curvature
singularities truncate $D(I^{\prime})$ before the tension is reached then it
is more difficult for the empty island maximin slice to not intersect
$D(I^{\prime})$, though it is still possible in a stably causal spacetime.

Suppose there exists a region $R$ with island detector $I^{\prime}$, with
$I^{\prime}\cup R$ quantum normal. As we follow null generators along
$\partial D(I^{\prime})$, assuming for now that no curvature singularities are
reached, then the QFC implies

$\begin{split}S_{gen}(\tilde{I}^{\prime}\cup R)&\leq S_{gen}(I^{\prime}\cup
R)\\\ &<S_{gen}(R)\end{split}$ (21)

Now in Hartman2020  it was argued that any $\tilde{I}^{\prime}$ which
satisfies $S_{gen}(\tilde{I}^{\prime}\cup R)<S_{gen}(R)$ also satisfies

$S_{ren.}(\tilde{I}^{\prime})>\frac{\text{Area}(\partial\tilde{I}^{\prime})}{4G_{N}}$
(22)

This means that the representative violates of the Bekenstein area bound. The
puzzle arises when $\tilde{I}^{\prime}$ is parametrically smaller than the
local curvature and temperature scale, but still larger than $l_{p}$. When
$\tilde{I}^{\prime}$ is this small the reduced state on it is approximately
the vacuum in flat spacetime, which should not violate the Bekenstein area
bound. Since $\tilde{I}^{\prime}$ is still larger than the Planck scale we can
cannot claim that violations of the QFC in (21) are expected due to the size
of the $\tilde{I}^{\prime}$.. Let us assume that the QFC has not been violated
and look for other resolutions.

This puzzle needs to be resolved for both islands and island detectors. Let us
see how this puzzle manifests itself in a setup with islands. Take an
asymptotically flat black hole formed from collapse, see Fig. 3. By applying
QFC to $S_{gen}(I\cup R)$ we get

$S_{gen}(\tilde{I}\cup R)\leq S_{gen}(I\cup R)<S_{gen.}(R)$ (23)

which again, by the argument given for condition 1 in Hartman2020 , implies
that the representative $\tilde{I}$ violates the Bekenstein area bound. If the
state on $\tilde{I}$ is approximately the vacuum then it should not violate
the Bekenstein area bound.

How small $\tilde{I}$ needs to be before the state on it is approximately the
vacuum depends on how the black hole was formed. For stellar collapse
$\tilde{I}$ will be a ball-shaped region in the centre of the star soon after
the event horizon formed, and it needs to be smaller than the local
temperature scale. For a collapsing shell of radiation, the Vaidya metric, the
geometry inside the shell of collapsing radiation is flat and $\tilde{I}$ can
be just inside. In either case, QFC along with the island’s quantum
extremality property implies that the generalised entropy cannot increase as
we deform along null generators from $I$ to $\tilde{I}$, and we again reach
the incorrect conclusion that a ball-shaped region of the vacuum violates the
Bekenstein area bound.

####  3.3.1 Is Bekenstein’s bound necessarily violated?

To explore this puzzle further we will reproduce the derivation from
Hartman2020  that

$S_{gen.}(\mathcal{I}\cup R)<S_{gen.}(R)\implies
S_{ren.}(\mathcal{I})>\frac{\text{Area}(\partial I)}{4G_{N}}$ (24)

We want to see what approximations were made in the derivation and if those
approximations are ever invalid. In Hartman2020  the region $\mathcal{I}$ was
an island, while for us the region of interest is a representative
$\tilde{I}^{\prime}$ or $\tilde{I}$ of an island detector or an island, but
since they all satisfy the left hand side of (24) the derivation is identical.

Figure 4: Representative cartoon of the subregions of time slice $\Sigma$ used
in the proof of (24).

The first step is to surround $\mathcal{I}$ with a thin region $C$ of width
$\delta\gg l_{p}$. $C$ shares a boundary with $\mathcal{I}$, such that
$\partial\mathcal{I}\backslash\partial C=\varnothing$, and is spacelike
separated from $R$. See Fig. 4. Applying monotonicity of mutual information

$I(X,Y\cup Z)\geq I(X,Y),$ (25)

to the region complement to $\mathcal{I}\cup C$, which is a superset of $R$,
we get

$\begin{split}I(\mathcal{I},(\mathcal{I}\cup C)^{c})&\geq I(\mathcal{I},R)\\\
&>S_{gen}(\mathcal{I})\end{split}$ (26)

We may rewrite (26) in the following way, which makes its relation to the
Bekenstein area bound transparent,

$S_{ren.}(\mathcal{I})-\frac{\text{Area}(\partial\mathcal{I})}{4G_{N}}\geq
S_{ren.}(\mathcal{I})+S_{ren.}(C)-S_{ren.}(\mathcal{I}\cup C).$ (27)

This is done by expanding the mutual information in terms of the renormalised
entropies,

$I(\mathcal{I},(\mathcal{I}\cup
C)^{c})=S_{ren.}(\mathcal{I})+S_{ren.}(\mathcal{I}\cup C)-S_{ren.}(C),$ (28)

and rearranging. Note that no approximations have been made so far.

We reach the conclusion that $\mathcal{I}$ violates the Bekenstein area bound
at leading order in $G_{N}$,

$S_{ren.}(\mathcal{I})-\frac{\text{Area}(\partial\mathcal{I})}{4G_{N}}\geq 0,$
(29)

if and only if the right hand side of (27) is non-negative or $O(G_{N}^{0})$.
Conversely, we need

$S_{ren.}(\mathcal{I})+S_{ren.}(C)-S_{ren.}(\mathcal{I}\cup
C)\lesssim-O(G_{N}^{-1})$ (30)

in order to avoid the conclusion from (27) that $\mathcal{I}$ violates the
Bekenstein bound. $\mathcal{I}$ and $C$ must have super-additive renormalised
entropies. Renormalised entropy can have unusual properties for an entropy,
including superadditivity and negativity Wall2009 , so this is not obviously
impossible. The right hand side of (27) is not the mutual information between
$\mathcal{I}$ and $C$, which would be non-negative by subadditivity, because
the overlap of the boundaries $\mathcal{I}$ and $C$ is nonzero so there is a
leftover piece when the boundary counterterms are added back in.

In Hartman2020  it is assumed that

$S_{ren.}(\mathcal{I}\cup C)=S_{ren.}(\mathcal{I})$ (31)

and

$S_{ren.}(C)=0$ (32)

up to corrections that are subleading in $G_{N}^{-1}$ with respect to the area
term. This is a reasonable assumption if there exists a time slice on which
the state is semiclassical and with a subregion $C$ as defined before, and of
width $\delta\gg l_{p}$ but smaller than the local energy and curvature
scales. Then $S_{ren.}(C)\sim\delta^{-(d-2)}\ll G_{N}^{-1}$. As a first
comment, note that one class of examples where there is no $C$ of width
$\delta\gg l_{p}$ is if $\partial\mathcal{I}$ and $\partial R$ are null-
separated.

Figure 5: An evaporating black hole in asymptotically Minkowski spacetime. $C$
surrounds the island $I$, with $\partial I\backslash\partial C=\varnothing$,
and is spacelike separated from $R$. We show that $C$ cannot have both width
$\delta\gg l_{p}$ and trace extrinsic curvature $|K|\ll l_{p}^{-1}$.

####  3.3.2 Resolution for islands in evaporating black holes

As a first step to resolving this subsection’s Bekenstein bound related puzzle
for an island in an asymptotically flat evaporating black hole, let us re-
examine the assumptions (31) and (32) for this case. We will show that there
is no time slice on which (1) there exists a $C$ of proper width $\delta\gg
l_{p}$ and (2) the curvature scale of the slice, as quantified by its trace
extrinsic curvature $K$, is not Planckian.

The boundaries of $R$ and the island are not null-separated as can be seen in
Fig. 5, so there are time slices with regions $C$ of non-zero width. We are
working in the vacuum with respect to an infalling observer, so a natural set
of coordinates to use are Painlevé-Gullstrand:

$ds^{2}=-dT^{2}+\left(dr+\sqrt{\frac{r_{s}}{r}}dT\right)^{2}+r^{2}d\Omega_{d-2}^{2}$
(33)

where $T$ is the proper time for a radially infalling observer initially at
rest at infinity. We will do an analysis in a region perturbatively smaller
than the length scale set by the black hole evaporation time, so we neglect
the $T$ dependence of $r_{s}$.

The boundaries of $R$ and $I$ are not null-separated, but on a constant
infalling time slice the radial separation of the island and the past
lightcone of $R$ is Planckian Penington2019 , so on that particular time slice
there is no $C$ of spatial width $\delta\gg l_{p}$. Let $r_{I}$ be the radial
position of the boundary of the island on a constant $T$ slice, and $r_{lc}$
the radial position of the intersection of $\partial J^{-}(R)$ with that
slice. The reason why the radial separation $(r_{lc}-r_{I})$ must be Planckian
is because the gradient of the entropy must be order $O(1/G_{N})$ in order to
compete with the $O(1)$ gradient of area in the extremality condition

$\partial_{r_{I}}\left(\frac{\text{Area}(\partial I)}{4G_{N}}+S_{ren.}(I\cup
R)\right)=0$ (34)

The gradient of the area is $\partial_{r_{I}}\text{Area}(\partial I)=8\pi
r_{I}$, while by dimensional analysis the leading divergence of the
renormalised entropy is $\partial_{r_{I}}S_{ren.}(I\cup R)\sim
1/(r_{lc}-r_{I})$ Bousso2021 . To satisfy the extremality condition (34) thus
requires a radial separation $(r_{lc}-r_{I})\sim l_{p}^{2}/r_{I}$.

The representative $\tilde{I}$ at earlier times has an even smaller radial
separation from $\partial J^{-}(R)$ than $I$, as the separation of lightrays
from the event horizon decreases exponentially in advanced time $v:=t+r^{*}$
towards the past. The maximum width an annular region $C$ surrounding
$\tilde{I}$ on a constant $T$ slice can be is exponentially smaller than for
$I$.

Now we discuss other time slicings, as we have only shown that there is no $C$
of width $\delta\gg l_{p}$ on the constant $T$ slice that intersects $\partial
I$. There are certainly other coordinate systems with time slices on which
there is a $C$ of arbitrarily large spatial width, because the null separation
of $\partial R$ and $\partial I$ in the future outward direction is
infinite888It may seem counterintuitive that two nearly-null separated points
can have large spatial separation. To get an intuitition for this, consider
the spatial geodesic connecting two spacelike separated points in 2d
Minkowski. This has proper length $-\Delta x^{+}\Delta x^{-}$ which diverges
when $\Delta x^{+}\to\infty$ if $\Delta x^{-}$ is fixed and finite..

$C$ is annular with inner boundary fixed to be $\partial I$. For a fixed time
slice $\Sigma$ the outer boundary of the maximum width $C$ is
$\Sigma\cap\partial J^{-}(R)$, which is as far as $C$ can extend radially
while staying spacelike separated from $R$, as shown in Fig. 5.

For fixed $I$ we can increase the maximum width $C$ by deforming the time
slice such that $\Sigma\cap\partial J^{-}(R)$ moves towards future null
infinity. The cost of increasing the maximum possible width of $C$ by
deforming the time slice between $\partial I$ and $\partial J^{-}(R)$ towards
the null limit is that the slice in this region approaches an annulus-like
subregion of a null cone, and null cones have a divergent trace extrinsic
curvature.

Let us assume that $C$ stays in the neighbourhood ($\Delta T,\Delta r\ll
r_{s}$) of $\partial I$, and justify that assumption later. The geometry in
this neighbourhood is approximately

$ds^{2}\approx-dT^{2}+d\rho^{2}+r_{s}^{2}d\Omega^{2}_{d-2}$ (35)

where $\rho$ is the Lemaître radial coordinate, which is related to the
Schwarzschild radial coordinate by

$r(T,\rho)=\left(\frac{3}{2}(\rho-T)\right)^{2/3}r_{s}^{1/3}.$ (36)

The radial separations of $\partial I$ and $\partial J^{-}(R)$ on a constant
$T$ slice are approximately the same in the two coordinate systems

$(\rho_{lc}-\rho_{i})\approx(r_{lc}-r_{i})\sim l_{p}^{2}/r_{s}$ (37)

because near $r=r_{s}$ we have $d\rho\approx dr+dT$.

Now consider a spherically symmetric time slice that intersects $\partial I$
and in its neighbourhood has slope $d\rho/dT=\sqrt{1+\alpha^{2}}$ with
$\alpha>0$ to keep the slice spacelike. The trace extrinsic curvature of this
time slice in the neighbourhood of $\partial I$ is the same as the surface of
a cone in Minkowski spacetime, which is

$|K|=\frac{1}{\alpha r_{s}}.$ (38)

This $K$ diverges in the null limit $\alpha\to 0^{+}$. Recall that to maximise
the possible width of $C$ we want to move the intersection $\Sigma\cap\partial
J^{-}(R)$ towards future null infinity, so let us take $1\gg\alpha\gg
l_{p}/r_{s}$, which is as close to the null limit as possible while keeping
the extrinsic curvature sub-Planckian, $|K|\ll l_{p}^{-1}$. The slice
intersects $\partial J^{-}(R)$ after time

$(\sqrt{1+\alpha^{2}}-1)\Delta
T=(\rho_{lc}-\rho_{I})\sim\frac{l_{p}^{2}}{r_{s}}$ (39)

Thus $\Delta T\ll r_{s}$ given the range of $\alpha$, which justifies the
earlier approximation (35). The largest possible $C$ on the time slice defined
has width given by the radial geodesic length between $\partial I$ and the
intersection of the slice with $\partial J^{-}(R)$, which is

$\delta=\alpha\Delta T\ll l_{p}.$ (40)

The conclusion is that there is no time slice on which there exists a $C$ with
both width $\delta\gg l_{p}$ and with $|K|\ll l_{p}^{-1}$.

We have not shown that islands satisfy the Bekenstein bound, nor do we claim
that they do. We have only shown that one particular assumption that is
sufficient to show that islands necessarily violate the bound is invalid for
the black hole set up of Fig. 5. The expectation is that islands violate the
Bekenstein bound, while their early time representatives $\tilde{I}$ of size
below the local curvature scale, as depicted in Fig. 3, do not. In order to
avoid the conclusion that the Bekenstein bound is violated we have seen from
equation (27) that it is necessary for all possible regions $C$ to give
strongly ($\sim G_{N}^{-1}$) superadditive renormalised entropy.

Let us make some preliminary remarks on how $\tilde{I}$ and $C$ can have
superadditive renormalised entropy, but leave a careful analysis for future
work. We have shown that the largest width $C$ can have on a time slice with
$|K|\ll l_{p}^{-1}$ is sub-Planckian for $I$, and it will be exponentially
smaller still for $\tilde{I}$ because of the convergence towards the event
horizon of past-directed ingoing light rays. As mentioned in Sec. 3.2,
semiclassical entropies of regions of sub-Planckian size can be formally
evaluated, even if they are not physically meaningful. If
$S_{ren.}(\tilde{I}\cup C)\approx S_{ren.}(\tilde{I})$, which seems reasonable
as $L_{\tilde{I}}\gg l_{p}\gg L_{C}$, then superadditivity requires
$S_{ren.}(C)<0$. Regions of width smaller than the renormalisation scale can
have negative $S_{ren.}$. For example the renormalised vacuum entropy of an
interval in a 2d CFT is proportional to $\log(L/\mu)$, with $\mu$ the
renormalisation scale, which is negative for $L<\mu$. For $\tilde{I}$ the
maximum width $C$ can be is exponentially smaller than $l_{p}$, so
$S_{ren.}(C)\lesssim-O(G_{N}^{-1})$ may be possible.

##  4 Discussion and future work

In section 2.4 we pointed out a loophole in the proof given in Bousso2021
that their set of conditions are sufficient for the existence of a non-empty
island. There are trivial fixes to their set of conditions to make them
sufficient, such as an additional condition that all time slices in the
spacetime intersect $D(I^{\prime})$ (or $J(I^{\prime})$ if $I^{\prime}\cup R$
is anti-quantum normal rather than quantum normal). This particular additional
condition would be undesirably restrictive though: it is satisfied for the
recollapsing FRW universe example of Hartman2020  but not for evaporating
asymptotically AdS or flat Schwarzschild black holes. The bottom line is that
we do not have a set of conditions that are (1) sufficient to establish
existence of a non-empty island, (2) easier to evaluate than the island
formula, and (3) are not overly restrictive in that they fail to detect
islands for whole classes of examples where islands are expected or have been
explicitly found.

Having pointed out loopholes in the proof, it is natural to look for explicit
counterexamples, i.e. island mirages. In appendix A we outline two promising
setups which we intend to explore further. There may be examples of island
mirages which are very different in character from those suggested as
possibilities in the appendix.

In section 3.2 we mentioned that it is unclear when or whether a quantum
lightsheet will truncate after finite affine parameter. Such a result would be
important as it limits the applicability of the quantum Bousso bound.
Classically, null geodesics in a congruence with negative expansion reach
caustics after finite affine parameter because of the bound

$\theta^{\prime}\leq-\frac{1}{d-2}\theta^{2}$ (41)

which follows from the Raychaudhuri equation for null geodesic congruences and
assuming the null energy condition. Similarly, a negative quantum expansion
leads to the ‘quantum caustic’ $\Theta\to-\infty$ after finite affine
parameter if the following quantum analogue of (41) is true:

$\frac{\delta}{\delta
V(y_{2})}\Theta[V;y_{1}]\leq-\frac{1}{d-2}\Theta(y_{1})\Theta(y_{2})\delta(y_{1}-y_{2})$
(42)

using the notation from Bousso2015 . This conjecture (42) is stronger than
QFC. The on-diagonal components give (41) in the classical limit, and give a
stronger version of QNEC in the weak gravity limit999Which in 4d may not be
satisfied by a ball-shaped region of the CFT vacuum in Minkowski spacetime. We
thank Aron Wall for pointing this out.. We feel that these conjectures and
their implications for the quantum Bousso bound are worth further study.

In section 3.3 we described and gave possible resolutions to the Bekenstein
bound puzzle, and discussed the puzzle in the context of islands in
evaporating black holes in detail. We did not however give a complete argument
for how the early time representative $\tilde{I}$ of the island can avoid the
implication that it violates the Bekenstein bound with superadditive
renormalised entropies, as it seems it must. For future work we believe that
the islands in AdS2 black holes found in Almheiri2019  would be a fruitful
arena in which to explore this puzzle in explicit calculational detail.

###### Acknowledgements.

I would like to thank Tarek Anous, Raphael Bousso, Sean Colin-Ellerin, Ricardo
Espindola, Jackson Fliss, Bahman Najian, Dominik Neuenfeld, Dora
Nikolakopoulou, Edgar Shaghoulian, Arvin Shahbazi-Moghaddam, Shan-Ming Ruan,
Manus Visser, and Aron Wall for useful discussions. Special thanks to Ben
Freivogel for many productive discussions and comments on an earlier draft.
This research was supported by the Stichting Nederlandse Wetenschappelijk
Onderzoek Instituten (NWO-I).

##  Appendix A Possible island mirages

An island mirage is an island detector $I^{\prime}$ that satisfies conditions
(5) and (7), but for which $R$ has no island. Mirages, if they exist, are
counterexamples to the claim that the conditions are sufficient to guarantee
that $R$ has an non-empty island somewhere. In this appendix we outline two
setups which we were not able to explicitly show have island mirages, but that
we feel are candidates worth further study in future work.

###  A.1 Small, critically illuminated, isolated AdS black holes

Figure 6: A small, critically illuminated, isolated AdS black hole, and an
auxiliary non-gravitating flat spacetime. The radiation shown includes the
incoming radiation that keeps the black mass constant, the outgoing Hawking
modes, and their interior partners. The ingoing radiation is maximally mixed
and purified by the degrees of freedom in $R$ in an auxiliary spacetime.

Here we consider a small isolated AdS black hole that is critically
illuminated by mixed state radiation that is purified by an auxiliary non-
gravitational system, see Fig. 6. Unlike typical island setups, isolated AdS
black holes are not coupled to a non-gravitational bath; the asymptotic
boundary conditions are reflecting for all time. Critically illuminated black
holes have an incoming flux of energy that exactly balances the outgoing flux
due to Hawking radiation, such that black hole has constant mass for some
finite period of time Lowe1999 . In our setup what an external observer sees
is the formation of a small AdS black hole from collapse, that is then
critically illuminated for some Schwarzschild time $t_{illum.}$, then allowed
to evaporate away before the earliest Hawking radiation has had time to
reflect and return from the boundary. For AdS black holes to be small enough
to evaporate requires a mass less than

$Ml_{AdS}\ll\left(\frac{l_{AdS}}{l_{p}}\right)^{\frac{d^{2}-1}{2d-1}}$ (43)

and there are no small AdSd black holes in $d\leq 3$ Harlow2016 .

The parametric separation of time scales is taken to be

$t_{p}\ll t_{evap.}\ll t_{illum.}\ll l_{AdS}$ (44)

There are several reasons to look for an island mirage in such a setup:

  1. 1.

Isolated AdS black holes do not have islands. Every point in the black hole
interior lies on a past causal horizon, so GSL rules out any quantum extremal
surface Jacobson2003 ; Engelhardt2021 .

  2. 2.

Critically illuminated black holes are a counterexample to the classical
Bousso bound as an arbitrarily large amount of classical entropy can pass
through the event horizon while the horizon area stays constant Lowe1999 . An
arbitrarily large amount of entropy passing through the horizon is helpful for
satisfying condition (5).

  3. 3.

Small isolated AdS black holes only differ from asymptotically flat black
holes, which do have islands, far from the black hole.

Let us give a few comments about critically illuminated black holes. They are
a counterexample to the classical Bousso bound, though not to GSL as the
Hawking radiation continually adds to the entropy outside the black hole but
it does not cross the lightsheet Lowe1999 . The classical Bousso bound is in
effect violated by quantum effects in the form of Hawking radiation. The
critically illuminated black hole requires fine tuning of the incoming
radiation, and microscopic fluctuations in the incoming radiation generally
leads to the lightsheet truncating in finite affine parameter due to caustics
Flanagan2000 ; Bousso2000 ; Bousso2002 .

Since there can be no islands in isolated AdS black holes, to find an island
mirage we just need to find an $I^{\prime}$ that satisfies the conditions (5)
and (7). In Fig. 6 we depict three such candidate $I^{\prime}$ wich we will
now discuss in turn. $I^{\prime}_{a}$ satisfies (5) because it contains all
the ingoing radiation so $S_{ren.}(I^{\prime}_{a}\cup R)\approx 0$, and the
boundary of $\partial I^{\prime}_{a}$ has small area, but it fails to be
quantum normal because it is deep in the classically trapped region.
$I^{\prime}_{b}\cup R$ is quantum normal by GSL, because $\partial
I_{b}^{\prime}$ lies on both a future and past causal horizon, but it seems
likely to fail to satisfy (5) because it contains all the interior modes.
$I^{\prime}_{c}$ is in a sense in between $I^{\prime}_{a}$ and
$I^{\prime}_{b}$, the Goldilocks candidate. The future inward null quantum
expansion is non-positive by GSL, but since $\partial I^{\prime}_{c}$ is
inside the event horizon it does not lies on a future causal horizon, so GSL
does not apply to the future outward direction. To be quantum normal it must
be outside the quantum trapped region, yet it must also contain enough of the
incoming radiation to satisfy (5), and it is not clear whether both of these
can be simultaneously satisfied.

###  A.2 Large AdS black hole with infalling shell

Figure 7: Left: A large, isolated AdS black hole formed from a collapsing
star. The star is in a mixed state of entropy $S_{star}$ that is purified by
$R$ in an auxiliary non-gravitating spacetime. Spherical symmetry is assumed,
and $I^{\prime}$ are ball regions whose boundaries are points in the figure.
To be an island mirage $I^{\prime}\cup R$ must be both outside the quantum
apparent horizon, depicted in blue, and the inside the red hyperentropic
region. Right: A null shell of pure radiation is added. The null shell has a
large entropy gradient in the $x^{+}$ direction, so as to push the quantum
apparent horizon into the red hyperentropic region.

In this example we consider a large, isolated AdS black hole formed from a
collapsing star, as shown in Fig. 7. The star is in a mixed state, with
entropy $S_{star}$. The star is purified by $R$ in an auxiliary non-
gravitating spacetime. As an isolated AdS black hole, GSL again rules out non-
trivial quantum extremal surfaces, so there are no islands. We assume
spherical symmetry and look for ball-shaped island mirages $I^{\prime}$,
centred on the origin, which satisfy the conditions (5) and (7).

To satisfy (5) and lower the generalised entropy, the boundary of $I^{\prime}$
must be in the red region of Fig. 7. There the surface area of $I^{\prime}$ is
small and $I^{\prime}$ contains a sufficiently large fraction of the
collapsing star, which by construction is entangled with $R$, to lower the
generalised entropy. To satisfy (7), $\partial I^{\prime}$ must be outside the
quantum trapped region, whose boundary is the quantum apparent horizon
depicted in blue in Fig. 7.

Take a point at the radial origin of the star after it has collapsed inside
its Schwarzschild radius. The subset of the future directed null cone that
starts from this point to a cut of the lightcone outside the classical trapped
region is a lightsheet. If we assume the star’s entropy to be extensive, then
the classical Bousso bound places an upper limit on the entropy flux through
the lightsheet

$S(L)\leq\frac{\text{Area}(\partial L)}{4G_{N}}.$ (45)

This rules out an $I^{\prime}$ whose boundary is outside the classically
trapped region, and whose classical entropy violates the Bekenstein bound. If
we assume that this island detector region violates the Bekenstein bound, i.e.
there is a $C$ for which $I^{\prime}$ and $C$ do not have superadditive
renormalised entropies (see section 3.3), then we reach the conclusion that an
island detector cannot exist outside the classically trapped region if the
entropy of the matter forming the black hole is approximately classical, i.e.
extensive.

We need to add matter whose entropy is not classical in order to evade the
classical Bousso bound, in order to have an $I^{\prime}$ that is both quantum
normal and violates the Bekenstein bound Hartman2020 . We would like to ‘push’
the quantum apparent horizon inward, as depicted in the right hand side of
Fig. 7 by creating a large positive entropy gradient in the $x^{+}$ direction,
so making $\Theta_{+}$ positive where it was negative before.

We can try adding a thin shell of radiation of thickness $\Delta x^{+}$. Take
the shell to be composed of two concentric layers that are maximally entangled
with each other, but with the whole shell in a pure state. A ball shaped
region that is concentric with the shell will then have a large entropy
gradient as a function of radius at the shell radius. A positive entropy
gradient in the $x^{+}$ direction has the potential to flip the sign of
$\Theta_{+}$ and so push the quantum apparent horizon in.

We do have to be careful about the backreaction of the null shell though and
the effect on the classical expansion. One promising sign is that the quantum
apparent horizon of the right hand side of Fig. 7 seems not to violate the
QFC. The QFC would forbid $\Theta_{+}(I^{\prime}\cup R)$ from changing signs
twice as we deform $I^{\prime}$ in the $x^{+}$ direction, but here it only
changes sign once. A more careful analysis of the interplay between the
entropy gradient and backreaction of the null shell is needed before this is a
convincing example of an island mirage.

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