The Markov gap in the presence of islands

The Markov gap \cite{Hayden:2021gno}, namely the difference between reflected entropy and mutual information, is explicitly computed in the defect extremal surface model, JT gravity, and the generic 2d extremal black holes, in vacuum states. The phases that contain various island contributions are considered, and their existence is carefully checked. Moreover, we show explicitly how the Markov gap originates from the OPE coefficient of the boundary CFT. And, as a generalization of \cite{Hayden:2021gno}, the lower bound of the Markov gap is given by $\frac{c}{3}\log 2$ times the number of EWCS boundaries on minimal surfaces. We propose a boundary way of counting the lower bound for the Markov gap, which states that the lower bound is given by $\frac{c}{3}\log 2$ times the number of gaps between two boundary regions in vacuum states. We discuss the limitation and possible generalization of the boundary counting, and its relation to tripartite entanglement.


Introduction
The von Neumann entropy is an excellent measure of quantum entanglement between two subsystems in a pure state and thus is usually referred to as entanglement entropy.Based on the developments in holographic entanglement entropy [2][3][4][5][6], the island formula for entanglement entropy is proposed as [7][8][9][10][11] where the region I A is known as A's island as it is separate from A, the second term is the quantum entanglement of bulk matter.(1.1) stems from the QES formula for holographic entanglement entropy in AdS/CFT correspondence [5], and is derived via gravitational path integral in a specific JT gravity [8,11].With (1.1), the unitary Page curves for many black holes have been successfully recovered, making significant progress toward the information paradox. 1owever, the von Neumann entropy ceases to be a good measure of entanglement for tripartite systems or mixed states.A measure of entanglement for mixed states is of significant importance, as it can be used to probe the entanglement structure of the state, and, on the other hand, the states we encounter are not always pure.
By subtracting the mutual information, one can define some UV-finite quantities, i.e., g = 2E p − I and h = S R − I [65], where I(A : B) is the mutual information.In particular, non-vanishing g and h imply non-trivial tripartite entanglement [64,65].For g = 0, the state must be in a so-called triangle state |ψ ABC up to local isometries [65].The triangle state is free of non-trivial tripartite entanglement as it is formed by bipartite-entangled states.For h = 0, the state must be in the sum of triangle states (SOTS) [65].In general, g ≥ h, which means some types of tripartite entanglement cannot be seen by h.In a holographic CFT at large-c limit, it is conjectured that the two quantities coincide g = h.Specifically, for a 1D tripartite holographic spin chain on a circle, the authors of [65] found that g = h = c 3 log 2. A similar discovery was also made by Wen with balanced partial entanglement [54].Later on, h is shown to be related to the Markov recovery map, and a non-vanishing h precludes a perfect Markov recovery map [1], because of which h is termed as the Markov gap by Hayden, Parrikar, and Sorce (HPS).Using the geometric approach, they proved that for a static state in pure AdS 3 , the lower bound of the Markov gap of boundary regions is related to the number of boundaries of EWCS: which is nice and neat. is the AdS radius.A BPE version of (1.2) was proposed in [55], and an interface CFT (ICFT) version was studied in [66].While (1.2) is proved to be valid for CFT 2 with a pure AdS 3 dual, it remains to be explored in the other cases, among which the presence of islands is of great interest.Firstly, it is natural to consider the presence of an island as it arises after Page time during black hole evaporation.Secondly, the island formula for reflected entropy has been proposed in [67,68].It is interesting to see if this island formula admits a lower bound for Markov gap like (1.2).We explicitly compute the Markov gaps for various phases in a model based on AdS/BCFT correspondence [69], called defect extremal surface (DES) model [70,71].
In DES model, the RT formula is corrected by the quantum defect theory living on an end-of-the-world (EoW) brane in the bulk.It is very exciting that by combining AdS/BCFT with braneworld holography [72][73][74], the island formula emerges in an effective 2d description of DES model [70].The reflected entropy and entanglement negativity has been studied in this model [71,75,76].Our results favor the HPS inequality even in the presence of islands, if we do not take the boundary of EWCS on brane into account.Even so, the geometric proof of (1.2) in our cases is not a trivial extension of HPS's, as generally the EoW brane in the bulk is neither necessarily along geodesics nor at asymptotic infinity.
The inequalities (1.2) are stated from a bulk point of view, as it incorporates EWCS.We expect that, for a vacuum state, one can also read off some lower bound for the Markov gap from the boundary theory viewpoint.This thought, together with our results, prompts us to conjecture that h(A : B) ≥ c 3 log 2 × (# of gaps between A ∪ I R,A and B ∪ I R,B ), for I(A : B) > 0, (1.3) where c is the central charge, and I R is the reflected island for the corresponding region at the asymptotic boundary [68,70].We test the boundary proposal (1.3) in DES model, JT gravity, and generic 2d extremal black holes for various phases.And the results satisfy (1.3) with the same lower bound.In addition, we show, using an explicit example, how the lower bound of the Markov gap originates from the OPE coefficient, which may kindle the general proof of (1.3) in future work.However, in the most general situations where the boundary region contains multi-intervals, even though the inequality (1.3) is satisfied, the lower bound given by counting gaps could be underestimated.We will discuss this in more detail and provide a generalization for multi-interval regions in Sec. 6.This paper is organized as follows.In Sec.2, we first introduce reflected entropy, the Markov gap, and the HPS inequality.Then we propose a DES version and a boundary version of HPS inequality.In Sec.3, we calculate the Markov gap in the DES model for several phases, both disjoint and adjacent, and compare the results with our proposal.In Sec.4,we do a similar calculation in JT gravity, totally from a boundary point of view, using the island formula.In Sec.5, we calculate the lower bound for the Markov gap in general 2D extremal black hole setups.In Sec.6, we discuss our results and proposal.Throughout this paper, we consider only the ground states of the field, and all the phases are assumed to be time-symmetric.We will use "a b" to indicate that a approaches b but a − b still has a relatively small value. 2 The Markov gap and its bulk and boundary inequalities

The Markov gap
Reflected entropy [63] was proposed as the von Neumann entropy in a canonically purified state | √ ρ AB ABA * B * in the doubled Hilbert space which serves as a measure of entanglement between A and B. The reflected entropy has been widely studied in various systems [77][78][79][80][81][82].
In [1], the difference between reflected entropy S R (A : B) and mutual information I(A : B) = S(A) + S(B) − S(A ∪ B) is called Markov gap h(A : B) ≡ S R (A : B) − I(A : B), as a non-vanishing h precludes a perfect Markov recovery map ρ ABB * = R B→BB * (ρ AB ).Moreover, h is considered as a smoking gun of certain tripartite entanglement, that is, a pure state |ψ ABC is a sum of triangle states iff h(A : B) = 0 [64,65].In [63], the Markov gap is identified with conditional mutual information where the conditional mutual information is defined as I(A : C|B) = I(A : BC) − I(A : B). (2. 3) The Markov gap satisfies the following inequality in information-theoretic language where F is the quantum fidelity (2.5)

HPS inequality and its bulk and boundary version in presence of island
In AdS/CFT correspondence, Hayden, Parrikar and Sorce (HPS) show that the Markov gap satisfies the following inequality h(A : B) ≥ 2G N log 2 × (# of cross-section boundaries) = c 3 log 2 × (# of cross-section boundaries) (2.6) to O(G 0 N ) in pure AdS 3 space by using geometric argument [1]. is the AdS radius.In the second line of Eq. (2.6), we used Brown-Henneaux formula c = 3 /2G N [83].
The island contribution naturally arises in many situations, for example, an evaporating black hole.So it should inevitably be taken into account.Based on AdS/BCFT correspondence, the defect extremal surface (DES) model has been proposed as the holographic counterpart of the island formula [70,71], that is, the island formula emerges when we consider the effective 2D description of DES model by partial dimension reduction.Our observation in Sec.3 will indicate the HPS inequality (2.6) is also obeyed if we do not take the EWCS boundary on brane into account.Or one can instead make a little modification of HPS's statement h(A : B) ≥ 2G N log 2 × (# of cross-section boundaries on the minimal surface of AB) = c 3 log 2 × (# of cross-section boundaries on the minimal surface of AB). (2.7) We also provide a geometric interpretation of our claim (2.7) for DES model in Appendix.D in the case that the brane tension is zero, but in general, this claim remains to be proved.On the other hand, (2.6) and (2.7) are counting the lower bound of the Markov gap from the bulk point of view.In principle, this lower bound can be obtained from boundary theory.Furthermore, we expect that one is also able to read off some lower bounds from the topology of the boundary regions.Therefore, based on our observation, we propose that h(A : B) ≥ c 3 log 2 × (# of gaps between A ∪ I R,A and B ∪ I R,B ), for where I R is the island for reflected entropy.In fact, our inequality (2.8) also works without an island.As shown in Fig. 1, for pure AdS 3 , there are two gaps between two disjoint boundary intervals and one gap between two adjacent boundary intervals and thus according to our inequality (2.8), we have h ≥ 2c 3 log 2 for disjoint intervals and h ≥ c 3 log 2 for adjacent intervals, which are consistent with the explicit calculation [52,65] and HPS inequality (2.6).Mind that on a time slice of vacuum CFT, a region containing infinity ∞ is also regarded as a gap.
In Sec.3 and Sec.4,we will show that (2.8) holds generally for DES model, JT gravity and generic 2D extremal black holes.Phases with disjoint and adjacent intervals will be considered separately.Before going deep into the detailed calculations of the Markov gap for DES models, we will also qualitatively analyze the recovery map of these phases, following the analysis in [1], which will enlighten the physical origin of the Markov gap of these phases.
The holographic dual of BCFT on a half space in vacuum.

Review of DES
In this section, we will review the entanglement entropy and reflected entropy of the DES model [70,71].

Entanglement entropy
DES model is based on AdS/BCFT correspondence.The holographic dual of a 2d BCFT on a half space in a vacuum state is known to be a part of Poincaré AdS 3 + end-of-the-world (EoW) brane Q where the Neumann boundary condition is imposed.The AdS 3 geometry is given by where these coordinates are related via The EoW brane Q lives at y = λx, where λ is related to the brane tension T by The boundary BCFT lives at x ≥ 0. The entanglement entropy for an interval [0, L] on BCFT in the ground state is given by the area of the RT surface Γ, which connects the endpoint L and EoW brane (see Fig. 2), where θ is defined as (cos θ) −1 = cosh(ρ/ ) and the second term c 6 arctanh(sin θ 0 ) is the boundary entropy of BCFT.
If the brane has zero tension or no matter is on the brane, the brane is orthogonal to the BCFT at the boundary of CFT, which is our origin.Now we add CFT matter on the brane and turn on the tension, and the brane will no longer be orthogonal to BCFT [70].This brane can be regarded as a defect in the bulk.Holographically, the matter field on the brane also contributes to the entanglement entropy of a BCFT region and now the entanglement entropy on BCFT is given by defect extremal surface (DES) proposal in which Γ is the defect extremal surface (the corresponding minimal surface).S Defect [D] is the bulk semi-classical entanglement entropy and D is a region on the brane where the bulk matter live on.

Intervals on the brane
Let us first consider S Defect [D].For an brane interval D = [x 1 , 0] with x 1 < 0 and touching the BCFT, we have where log g corresponds to the boundary entropy of the brane [69,84], and c is the central charge for CFT on the brane.In this paper, we take c = c and log g = 0. We see that entanglement entropy does not depend on the length of the interval on the brane.This nice fact makes the calculation in this model tractable with ease.< l a t e x i t s h a 1 _ b a s e 6 4 = " Z + q S L q G G S / n f s For an brane interval D = [x 1 , x 2 ] with x 1 < x 2 < 0, S Defect (D) possesses two phases that correspond to the dominance of the two channels: the bulk operator product expansion (OPE) and the boundary operator product expansion (BOE) [70,85] and where η is given by

Intervals on BCFT
Now let us consider S DES .For an interval [0, L] on BCFT, S Defect is irrelevant to length or position on the brane, thus the extremization and minimization procedures reduce to finding a shortest geodesic distance between brane and a boundary point and the result is where we defined For an interval [M, L] on BCFT, there are two phases for S DES (see Fig. 3).The connected phase and the disconnected phase   where the critical point is Note that there is an extremal value for S DES only when η > 1, which is automatically satisfied as η c > 1.

DES formula v.s. Island formula
One can also seek the effective 2D boundary description for the DES model by taking partial dimension reduction, then the island formula emerges and gives the same result for entanglement entropy as the DES formula [70].To be specific, as shown in Fig. 4, insert an imaginary boundary Q with (t, x, y) = (t, 0, y) to decompose the bulk into two parts W 1 and W 2 .For W 1 , by performing Randall-Sundrum reduction along ρ direction, one could obtain a 2d gravity theory + CFT on the brane Q with the area term For W 2 , we choose its dual BCFT description on the half-space boundary.Ultimately, we arrive at the 2D effective description of DES model.By using boundary QES formula in this 2D boundary, the entanglement entropy for an interval [0, L] agrees with the DES result, that is, where S CFT is the effective entropy of CFT in this 2D boundary and , ds 2 = Ω −2 dwd w. (3.20)

Reflected entropy
In DES model, the reflected entropy can be understood in both boundary theory and bulk theory viewpoints.In boundary island point of view, the reflected entropy is proposed to be [71] where γ = ∂I A ∩ ∂I B is the intersection of entanglement wedge cross-section and brane Q.
In the bulk point of view, the reflected entropy is given by The two proposals are equivalent.
For phase-D1 (Fig. 5) where the entanglement wedge2 of AB is just the disconnected union of entanglement wedges of A and B, the reflected entropy vanishes Throughout this paper, without loss of generality, we assume that region B is large enough so that it always receives the island contribution.
In phase-D2 (Fig. 6), the entanglement wedge of AB is connected (B has an entanglement island while A does not), and the entanglement wedge cross-section is the minimal geodesic with two endpoints on RT surfaces of [0, b 1 ] and [b 2 , b 3 ].Given that the reflected entropy S R is dual to the area of EWCS, then we have . (3.24) (3.24) is just twice the distance between two parallel geodesics in hyperbolic space ds 2 = 2 (dx 2 + dy 2 )/y 2 in unit of 1 4G N .We leave the derivation of (3.24) in Appendix.A. (3.24) can be also obtained by employing the cross-section formula in [52] with the cross-ratio here In phase-D2, on the other hand, one can also compute the reflected entropy using twist operators as was done in Refs.[63,68] S R (A : Again, the cross-ratio x here should be properly chosen, and one easily checks that (3.27) gives the same result as (3.24) and (3.25).
For phase-D4 (Fig. 8) where A and B both have their islands and the entanglement wedge of AB is connected, the reflected entropy is [71] S R (A : where a is the island cross-section.

The Markov gap
In the following, we compute the Markov gap in several phases in DES model.The goal of this section is to show that the Markov gap in DES model satisfies the inequalities (2.7) and (2.8).In some phases, we will present the conditions for the dominance of the phase.These conditions are given by some inequalities that will be useful for later calculation.

Disjoint intervals
We first consider the two regions We assume that b 4 is large enough so that B always admits the island.

Phase-D1
In phase-D1, the entanglement wedge of AB is disconnected, Fig. 5.The reflected entropy and mutual information vanish

Phase-D2
In phase-D2, on one hand, there is no island for On the other hand, we require that the entanglement wedge of Before going deep into the computation of the Markov gap, let us analyze the Markov recovery ρ BAA * = R A→AA * (ρ AB ) first, using the same argument as in [1].In Fig. 6, we stretched the canonical purification of ρ AB and the entanglement wedges of AB, AA * , BAA * .For phase-D2, the entanglement wedge of AB together with AA * cannot cover all the entanglement wedge of BAA * so that there are two jagged lines whose small tubular neighborhoods are completely visible to BAA * , but not to AA * and AB.Thus the Markov recovery ρ BAA * = R A→AA * (ρ AB ) must be precluded and we expect a non-vanishing Markov gap for phase-D2.Now we compute the Markov gap.The entanglement entropy for A is and for B with an island And the entanglement entropy for AB is The areas of the RT and DES surfaces are given by Then The mutual information is then The reflected entropy in this phase is given by (3.24), with which we get the Markov gap Hereafter by "b 3 b 2 " we mean we let b 2 → b 3 but assume the gap [b 2 , b 3 ] still exists so that the phase still makes sense.In the limit b 3 b 2 , we obtain which is saturated at b 1 b 2 b 3 .Note that the reflected entropy S R , the mutual information I and thus h are independent of θ 0 .So one can always tune θ 0 to make this phase happen, that is, to satisfy (3.30) and (3.31).
. Phase-D3.A and B are disjoint.A admits no entanglement island but a reflected island.
There is one boundary of EWCS, denoted by a dashed curve, and one gap between A ∪ I R,A and There is a jagged surface in this phase denoted in blue.
In Appendix.D, we also give a geometric interpretation of this lower bound in the case that the brane tension is zero.For phase-D2, both boundaries of the cross-section are on the minimal surfaces of AB and here both inequalities (2.6) and (2.7) give the same lower bound 2c 3 log 2, which is consistent with our calculation above.On the other hand, from the island boundary viewpoint, there are two gaps (the purple dashed circles in Fig. 6) between A ∪ I R,A and B ∪ I R,B , which also indicates the lower bound 2c 3 log 2 according to our boundary inequality (2.8).

Phase-D3
It is possible that the cross-section of phase-D2 is anchored at the brane, that is phase-D3 (Fig. 7).For this phase to exist, we require .
In this case, the mutual information is the same as (3.41).And the reflected entropy is given by (3.28) where the island cross section Apply (3.45) and (3.44), we have (3.50) We are only interested in the existence of this phase, then it is sufficient to pick a special case.Set As shown in Fig. 7, the entanglement wedge of AB together with AA * does not cover all BAA * and there is a jagged line for phase-D3.Thus a non-vanishing Markov gap is expected.Now let us compute the Markov gap.Use (3.41) and (3.28), and we obtain where we have used (3.44) in the second line.The equality in second line is taken at critical point S(A) RT = S(A) island for A, which is dependent of θ 0 that is in turn related to the brane tension.So (3.51) is saturated when b 2 b 3 and near the critical point of A. To sum up, In fact, it is possible that we cannot take the equality, as if f is negative, phase-D2 takes over the reflected entropy.Then it is necessary to check this.First, at the critical point, (3.49) takes equality.Then it is easy to see that in the limit b 2 b 3 , we have Therefore, we deduce that f ≥ 2 log 2, implying there is no problem taking this limit.On the other hand, (3.50) must be compatible with (3.54), and this can be seen explicitly by the R.H.S of which is divergent as → 0.
In fact, for phase-D3, only one boundary of the cross-section is anchored at the minimal surface of A ∪ B. Then the inequality (2.7) also implies h ≥ c 3 log 2. Besides, one can also obtain the same lower bound using our boundary inequality (2.8).Note that although there is no entanglement island for A, the cross-section for phase-D3 is anchored at the brane so that there is an island of reflected entropy for A and thus only one gap between A ∪ I R,A and B ∪ I R,B .

Phase-D4
For phase-D4 (Fig. 8) where both intervals contain islands, the entanglement wedge of AB together with that of AA * cannot cover all the entanglement wedge of BAA * so that there is a jagged line and thus we also expect the non-vanishing Markov gap for phase-D4.
The entanglement entropies for A and B in phase-D4 are And the entanglement entropy for A ∪ B is Then the mutual information is given by The reflected entropy for phase-D4 is given by (3.28) . Phase-A1.In this phase, the entanglement islands of A and B are identically their reflected island.
where a = √ b 2 b 3 .Then, the Markov gap is given by with the equality taken at b 2 b 3 .
For phase-D4, the analysis of the lower bound in terms of our inequalities (2.7) and(2.8) is similar to phase-D3 and they also give the lower bound c 3 log 2.

Phase-A1
In phase-A1 (Fig. 9), the two intervals A and B are adjacent and both contain island.The entanglement entropies for A and B in phase-A1 read The entanglement entropy for AB is Then the mutual information is given by Now let us derive the reflected entropy.According to proposal (3.22), we have in large-c limit The first term can be computed via correlation functions of twist operators.We refer to [71,85] for details, and just quote the result here . The second term in (3.66) is just twice the length of a geodesic connecting b 2 and −a, which is given by [69,70] The quantities L and θ 0 satisfy Then the reflected entropy is given by For ξ > 1, we find no real solution to ∂ a S R = 0.For ξ < 1, the minimization process reduces to finding the entanglement island for [0, b 2 ], which is a = b 2 .Substitute a = b 2 in (3.69) and (3.70), and we get Now the reflected entropy is given by Thus the correct Markov gap is A vanishing Markov gap implies the existence of a perfect recovery map.
One may notice that (3.65) can be obtained by the following replacement in (3.59) Naively, if we take the same replacement in reflected entropy (3.60), we get which leads to This is owing to the fact that when we evaluate the reflected entropy, or equivalently the entanglement wedge cross-section, we cannot take b 3 = b 2 + , otherwise the cutoff of y coordinate would become y UV ∼ /2.We demonstrate this in Appendix.C.The factor 1/2 in y UV contributes the term c 3 log 2 in (3.76) and (3.77).Recall that all formulae should use the standard cutoff y UV = .In this sense, we should really set b 3 − b 2 = 2 , and this gives us the correct result: Based on above analysis, we conclude that when evaluating reflected entropy in the adjacent limit, the x-axis gap between two intervals should be 2 .For phase-A1, there are no boundaries anchored at the minimal surfaces of A ∪ B and no boundary gaps between A ∪ I R,A and B ∪ I R,B , thus both the inequalities (2.7) and (2.8) imply that the lower bound is zero.

Phase-A2
Phase-A2 (Fig. 10) is just like phase-D3 except that A and B are now adjacent.On one hand, we require A has no island.On the other hand, the area of entanglement wedge cross-section should be less than that of phase-A3, which is given in next phase by (3.93)The above conditions lead to the following inequalities Though, this phase is easy to calculate, we still show how to obtain this phase by taking adjacent limit from phase-D3.First, let b 3 − b 2 = , and we obtain the mutual information from (3.41) . Phase-A2.The region A admits no entanglement island but a reflected island.The entanglement wedges satisfy Then let b 3 − b 2 = 2 , and we have the reflected entropy from (3.28) It is easy to confirm that the mutual information is indeed given by (3.82).The reflected entropy for A : B is given by (3.73), which is exactly (3.83).Then the Markov gap is given by which is just the condition (3.80).Notably, in this case, the Markov gap between A and B is just the difference between two different phases of S(A), and is guaranteed to be nonnegative.The equality in (3.84) is taken when the phase transition between phase-A1 and phase-A2 happens.Away from the phase transition, we have h > 0 and thus an imperfect Markov recovery for phase-A2.We should also apply the second condition (3.81), and this leads to . Phase-A3.In this phase, A and B are adjacent, and A admits no island.
Combine (3.84) and (3.85), and we have That is, the Markov gap is not only lower-bounded but also upper-bounded in this phase.The analysis of Markov recovery for phase-A1 and phase-A2 is as follows.For both phases, as shown in Fig. 9 and Fig. 10, the entanglement wedge of AB together with AA * covers all the entanglement wedge of BAA * .However, this information is not enough to tell us whether there is a perfect Markov recovery or not.In this case, one should resort to the direct calculation of h, which informs us that there is a perfect Markov recovery (h = 0) for phase-A1 while no perfect Markov recovery (h > 0) for phase-A2 away from the phase transition.In fact, as shown in Fig. 10, for another Markov recovery map ρ ABB * = R B→BB * (ρ AB ), the entanglement wedge of AB together with that of BB * cannot cover all the entanglement wedge of ABB * , which obviously signals an imperfect Markov recovery for phase-A2.

Phase-A3
Phase-A3 (Fig. 11) is like phase-D2 except that now A and B are adjacent.Unlike phase-D2, there is only one jagged surface due to the vanishing spacing between A and B. The Markov recovery is precluded and a non-vanishing Markov gap is expected.The condition for this phase to dominate is where the first inequality follows from that A has no entanglement island, while the second inequality follows from that its reflected entropy should be smaller than (3.73).
We can do adjacent limit from phase-D2 to get the result of this phase.First, let b 3 = b 2 + to get the mutual information from (3.41) And then we let b 3 = b 2 + 2 to get the reflected entropy We can also obtain these from direct calculation.In this phase, the mutual information is given by (3.82).And the reflected entropy equals twice the minimum length of geodesics that connect b 2 and the RT surface of [0, b 1 ].The minimum length can be derived with simple geometric relation where L reads We leave the derivation of the above result in Appendix.A. Inserting (3.92) into (3.91),we obtain So we obtain the same result as from the adjacent limit.Now we consider if this phase could exist.Rewrite (3.88) as Obviously, this can be satisfied for b 2 very close to b 1 , and (3.87) can be fulfilled by tuning θ 0 .Subtract the mutual information from reflected entropy, and we have the Markov gap As we have said, if b 1 is close to b 2 , the requirement (3.88) can be satisfied.

The Markov gap in JT gravity
In this section, we consider the JT gravity model in [7,11], where the AdS 2 JT gravity, coupled with CFT matter, is glued with a flat CFT.We do not apply the double holography description [68].Instead, we work in a pure boundary way to test (2.8).

Entanglement entropy for extremal JT black holes coupled to a bath
Consider a system where a 2D Jackiw-Teitelboim (JT) gravity with CFT matter is glued with a flat 2D CFT bath along its boundary, x = 0, at which the transparent boundary condition is imposed.The action is For extremal JT black holes, the metric and the dilaton in the gravity region are given by where x ± = t ± x, x ∈ (−∞, 0] and φ r is the renormalized boundary field [87].x = 0 is the interface of the JT gravity with the 2D flat CFT. According to the quantum extremal surface prescription [5], the entanglement entropy for an interval [0, b] on the flat CFT bath (see Fig. 12) is given by the generalized entropy S gen [0,b] (a) with the island by where the boundary of island a, which is on the gravity side, is given by extremization and minimization of the generalized entropy where ã = a/q, b = b/q and q = 6φ r /c.

The entanglement entropy for an interval
given by the minimum entanglement entropy among possible saddles.And this is the key point to recovering the Page curve [8,11].So we have where the generalized entropy + K S e y 7 z X I F l q o x L u z 6 N u j f 4 x u A g e + e T 6 6 i i 3 R q a 8 9 c u k d q h X 0 2 8 < l a t e x i t s h a 1 _ b a s e 6 4 = " d x 8 j q C V 3 d 8 m e c 9 x u 9 l + K C p z q N j c = " < l a t e x i t s h a 1 _ b a s e 6 4 = " Island < l a t e x i t s h a 1 _ b a s e 6 4 = " y G m g i T r 0 3 where σ and σ are twist operators.Thus at large-c limit, we have (4.8)

The Markov gap
We consider only the phases in which A and B are disjoint.The adjacent cases can be obtained by taking the adjacent limit, that is, set b 3 − b 2 = in mutual information and b 3 − b 2 = 2 in reflected entropy.

Phase-D2
Consider phase-D2 (Fig. 13) where have no entanglement island and reflected island, but B admits both entanglement and reflected island.The conditions are For the entanglement wedge of A ∪ B to be connected, we require the mutual information I(A : B) > 0.
The entanglement entropies for A, B and A ∪ B are S(B) = S gen [0,b 3 ] (a 3 ), (4.12) Then the mutual information is given by In this case, the reflected entropy is given by (3.27).Physically speaking, since we are working in a field theory manner, we should use (3.27), even though it is mathematically equivalent to (3.24) and (3.25).Then the Markov gap is given by Notice that φ 0 does not appear in (4.15) and mutual information (4.14), so we can always tune φ 0 to satisfy (4.9) and (4.10).It is not hard to find that ∂ b 3 h > 0, so that h monotonically increases with b 3 , the minimum is at b 3 → b 2 .In this limit, the reflected entropy reads We write the reflected entropy in a suggestive way, in which the term 2c 3 log 2 is isolated, and we explicitly include the divergent factor −(c/3) log(b 3 − b 2 ).But the divergence in S R is doomed to be cancelled by mutual information.
The Markov gap is then  < l a t e x i t s h a 1 _ b a s e 6 4 = " d x 8 j q C V 3 d 8 m e c 9 x u 9 l + K C p z q N j c = "

Phase-D3
In this phase (Fig. 15), we require A to have no entanglement island, but admits the reflected island, which gives the conditions for this phase S R (A : B) < The entanglement entropies are Then the mutual information is given by The reflected entropy can be derived via replica trick by the correlation functions of twist operators, and the result is [68] S R (A : in which a is island cross-section ∂I R,A ∩ ∂I R,B , given by the following equation We arrive at the Markov gap where we have used (4.19).
If q 1, we get a simple solution to island cross section (4.26), that is This is reminiscent of the DES result of a in (3.28).In the limit q → 0, the second term φ r /a in generalized entropy (4.3) can be ignored, so that the generalized entropy is given by an effective term plus a constant area term φ 0 .This is indeed similar to the boundary QES description of the DES model [70].In addition, we also have Inserting (4.28) and (4.29) into (4.27),we get < l a t e x i t s h a 1 _ b a s e 6 4 = " s 0 H N X U p 4 Z E f x P J s + 8 l L + Y P e b U j s         For general q, it is hard to analytically solve the lower bound of the Markov gap.Numerically we find that the R.H.S of (4.27) grows with b 3 , and clearly, if b 3 = b 2 the second line of (4.27) equals c 3 log 2, as a in this case reduces to a 2 .Therefore, we conclude that

Phase-D4
In phase-D4 (Fig. 16), both A and B have their islands, and the mutual information I(A : B) > 0. The entanglement entropies for A, B and AB are Then the mutual information is given by The reflected entropy for this phase is given by Then the Markov gap is where the second line is just the second line in (4.27).So we conclude that The Markov gap for generic 2D extremal black holes In this section, we derive the Markov gap in a rather generic 2D extremal black hole coupled to CFT at the large central charge limit.The computation is performed using the correlation functions of twist operators.

Setups
The metric of generic 2d extremal black holes can be written in a conformally-flat vacuum coordinate 4 , that is, where x ± = t ± x.At the static time slice, the entanglement entropy for an interval [0, b] on radiation region (see Fig. 17) is given by minimization of the generalized entropy, that is, where after minimization, −a represents the boundary of the island of the interval [0, b].Area(−a) is the area term at the boundary of island −a, and the effective entropy for 2D CFT is

The Markov gap
Without loss of generality, we consider only phase-D2 and phase-D4.

Phase-D2
With the same procedure in Sec.4.2, the Markov gap for phase-D2 (Fig. 13) is given by where the cross-ratio is given by It is easy to find that as b 2 → b 3 , the Markov gap (5.5) decreases.Thus it is sufficient to prove our boundary inequality (2.8) in the limit of b 2 → b 3 .The reflected entropy in this limit is reduced to which can be further written as Using (5.2), (5.3) and (5.4), (5.8) can be written as where math = means that the equal sign should be understood from the mathematical aspect rather than the physical aspect.Notice that we have (5.12) Again, the equal sign here should be understood from the mathematical aspect rather than the physical aspect.With the same argument as in the previous phase, we can arrive at where we used The equality in (5.21) is taken at b 2 b 3 .As expected from (2.8), the lower bound is c 3 log 2 as there is only one gap between A ∪ I R,A and B ∪ I R,B .
From the above analysis for phase-D4, it is insightful to see that the lower bound of Markov gap c 3 log 2 stems from the 3-point structure constant from the boundary viewpoint.

Discussion
We have studied the Markov gap in the DES model, JT gravity and generic 2d extremal black holes in the presence of islands for different phases.Some of these phases are not considered in the literature.For example, phase-D3, where A has no entanglement island but admits a reflected island.In doing this, we correct some little errors in literature as by-products.Then, all the results respect the bulk inequality (2.7) and the boundary inequality (2.8).However, the rigorous proof remains unknown, either from the bulk gravity side or the boundary theory side.We point out the obstacle.In [1], this inequality is proved by using a property of the right-angled pentagon.That is, for a right-angled pentagon in hyperbolic space, the lengths of its three sides satisfy α + β − σ ≥ log 2, where α and β are adjacent, and σ is non-adjacent to α and β.The right-angled pentagon is enclosed by geodesics and degenerate sides at infinity.In the DES model, the EoW brane, which locates along θ 0 in bulk, is neither a geodesic nor asymptotic infinity in Poincaré half-plane 5 .While all the results respect (2.8), there are some points we would like to stress.For two single intervals A and B, whereas AB admits entanglement island I AB , there could be no island cross-section a 6 , like in phase-D2 and phase-A3.Since I R,A ∪ I R,B = I AB , we have either I R,A = I AB , I R,B = ∅, or I R,B = I AB , I R,A = ∅.This can be determined from bulk using the entanglement wedge cross-section, which divides the entanglement wedge of AB into two parts.Nevertheless, from the boundary topology, this is subtle.If only one of them admits an entanglement island, it is natural to assign the reflected island to this one.If both A and B have their entanglement island, there is always an island cross-section that will divide the reflected island into their corresponding parts.To show this in DES model, However, the simple relation (2.8) should be considered as a property of the vacuum state because we did not input much information about the state.For the CFT in a mixed state, there must be further tripartite entanglement between A, B, and a generic purification.In this sense, the HPS inequality (2.6) has more promising validity in general states, as the information about the state is embedded in its gravity dual.But if the lower bound of the Markov gap will change or not requires further investigation.
One final remark.The lower bound of h varies in a discontinuous way as we change the length of an interval and undergo some phase transitions7 .But this does not mean the Markov gap h varies always discontinuously.For example, as we vary the length of [b 1 , b 2 ] in phase-D2 in fig.6, we will encounter a phase transition to phase-D3 in fig. 7. The lower bound changes immediately from 2c  3 log 2 to c 3 log 2. Though the EWCS undergoes a discontinuous change, its area is continuous (so does the reflected entropy S R ), as the phase transition happens when the two possible areas of EWCS coincide.Therefore, h is continuous.

Conclusion
In this paper, we studied the Markov gap h ≡ S R − I, especially its lower bound, in the DES, JT gravity models, and generic 2d extremal black holes.Phases with different island configurations are considered.To get reasonable results, we correct some formulae in the literature.Explicitly, we show how the lower bound of the Markov gap stems from the OPE coefficient.This may shed light on general proof of (2.8).Our results support the HPS inequality (2.6), with a specification that the lower bound only counts the boundaries of EWCS on minimal surfaces.So (2.6) could be a more general statement for holographic CFT.However, the general geometric proof for DES model or for island dominance requires further study.
We proposed a boundary statement (2.8), that the lower bound of the Markov gap h(A : B) is given by c 3 log 2 times the number of gaps between I R,A ∪ A and I R,B ∪ B. This statement is justified in all the phases we considered.An analysis of the relation between a gap and c 3 log 2 is made in Appendix.C, where we find that the different cutoffs for the gap in mutual information and reflected entropy give rise to c 3 log 2. However, (2.8) breaks down in certain situations where the boundary regions contain multi-intervals and EWCS is disconnected, as only topological information is included in (2.8).For multiinterval regions, we provide a possible generalization in Sec.6, and (2.8) is a trivial case.On the other hand, this statement does not work for states other than vacuum states.The entanglement entropy of vacuum states is characterized by the length of a region, which is not true for generic states where other parameters appear.A more generic proof and a physical interpretation of (2.8) from boundary theory are desired, potentially belonging to future exploration.
Apart from reflected entropy, the Markov gap can also be defined by other mixedstate measures claimed to be dual to EWCS.It is interesting to see if there are similar inequalities for other "Markov gaps".For example, in a generic purification instead of the canonical one that corresponds to the definition of reflected entropy, the authors of [55] proposed a generalized Markov gap based on partial entanglement entropy.The holographic entanglement negativity E may also admit a "Markov gap" with a similar HPS inequality.
But the prefactor should be c 4 log 2. In some sense, this problem reduces to checking the dualities between EWCS and these quantities.Nevertheless, they may provide further insights and perspectives, as they have different physical origins.
The target geodesic is shown in green in Fig. 21, and it must be perpendicular to the circle with radius b 1 .Suppose the green half-circle is centered at x = Ω with a radius r = b 2 − Ω.We denote its intersection with the geodesic as P 1 .By the Euclidean Pythagorean theorem, we have the following equation 2 The solution is

C Adjacent limit
We sketched how to obtain the adjacent results from disjoint phases in Sec.3.2.2.Here we present a more concrete example on this point.In Poincaré half-plane, the metric is divergent near the boundary CFT y = 0, corresponding to the IR divergence of the bulk space.Set the y UV = , and the entanglement entropy for an interval with length 2l is given by the area of the RT surface in unit of Setting this cutoff means that we only measure the length of geodesics above y = .Note that the formula works in the limit → 0.
We would like to get mutual information for the case in which A vanishes from that where A is finite.Mutual information is just a combination of entanglement entropies, and these entropies are given by the area of their RT surfaces (C.1).One can achieve this goal by simply setting S(A) = 0, which is effectively equivalent to 2l = , even though (C.1) may not work for 2l ∼ .Now we consider the reflected entropy or entanglement wedge cross-section.Suppose A and B are gapped by two small intervals, as in Fig. 22.Then entanglement wedge crosssection is shown in Fig. 22 with two ends on the RT surfaces of AB.Holographically, the reflected entropy is given by twice the area of entanglement wedge cross-section.When evaluating the reflected entropy, we should also set the cutoff as y UV = to make sure calculations are consistent.We let the two gaps to be [−L − l, −L + l] and [L − l, L + l].The reflected entropy is given by (3.25) We would like to see the vanishing limit of the two gaps.This cannot be obtained from letting the length of the gap to be 2l = , as the corresponding y-cutoff becomes y UV = /2, see Fig. 22.This is not consistent with y UV = we set for entanglement entropy.In this sense, we have to set l = , that is b 3 − b 2 = 2 in Sec.3.2.2, to get the adjacent limit.For there is no gap between A and B, we have In a word, we can effectively take 2l = in mutual information and l = in reflected entropy to get the results in corresponding adjacent phases.It is manifest that this procedure results in an additional term − c 3 log 2 in the Markov gap, as the cutoff is doomed to be canceled there.This partially explains why the lower bound of the Markov gap is related to the number of gaps between A and B.

D Geometric interpretation of the lower bound with no brane tension
In this section, we will give a geometric interpretation of the lower bound of S R − I when the brane is tension free.Without loss of generality, we will only consider phase-D2 and phase-D4.As shown in Fig. 23 In fact, similar to pure AdS 3 , here one may also obtain the lower bound by counting the number of the boundaries of EWCS.However, for AdS/BCFT, only the boundary on the minimal surfaces of A ∪ B contributes to the lower bound while the boundary on the brane does not, as we can see from phase-D4.This is why we generalize the original HPS inequality to (2.7).
. For instance, Area[E W ]/4G N ?BPE for holographic 2D CFT in the ground state, where ?

Figure 1 .
Figure 1.The gap (denoted by purple curves) between two boundary regions A and B of a time slice of AdS 3 vacuum.The lower bound of the Markov gap for the left phase is 2c 3 log 2, and c 3 log 2 for the right as there is no boundary on asymptotic infinity.

Figure 3 .
Figure 3.The entanglement entropy phases for an interval [M, L] in DES model.Left: connected phase.Right: disconnected phase.
Y k O S U I / 5 P d b v 9 p / 2 Z y 4 s B S t g O C l + e 8 k g 7 v S U v O t A m C 3 x 3 v 3 v 0 H D x 8 d P e 4 + e f r s + X G v / + J K F 5 W i M K M F L 9 R N S j R w J m F m m O F w U y o g I u V w n a 4 + N P 3 r N S j N C v n N b E p I B F l I l j N K j E P z X i 8 W x C y V s J 8 U W T O z q e e 9 Q e B P h u F o e I o D f x i E p + H E i W A y O j 8 b 4 9 A

Gravity < l a t e x i t s h a 1 _
b a s e 6 4 = " E 2 2 1 R Y z l o N r n D L y w y R 5 R a b 6 c L Y c = " > A A A C i X i c d Z H f a t s w F M Y V 7 1 + X b m u 7 X f Z G L A y 6 G 2 O n z p b s q i x l 7 L K D p i 3 Y p s j y c S I i y Z 4 k b w T h 1 9 j t 9 l p 9 m 8 q J A 0 v Z D g g + f u e T d P h O V n G m T R D c 9 b x H j 5 8 8 f b b 3 v L / / 4 u W r g 8 O j 1 1 e 6 r B W F G S 1 5 q W 4 y o o E z C T P D D I e b S g E R G Y f r b D l t + 9 c / Q G l W y k u z q i A V Z C 5 Z w S g x D i W J I G a h h J 1 + u W x u D w e B P w y i 4 e g U B / 5 o P D k N A y c m 0 S S K h j j 0 g 3 U N U F c X t 0 e 9 e Z K X t B Y g D e V E 6 z g M K p N a o g y j H J p + U m u o C F 2 S O c R O S i J A p 3 Y 9 d I P f O Z L j o l T u S I P X 9 O 8 b l

+Figure 4 .
Figure 4. 2D effective description of DES model by performing partial reduction for W 1 and AdS/CFT for W 2 .

) where a 1 2 Figure 5 .
Figure 5. Phase-D1.The blue and yellow regions are the entanglement wedges of A and B, respectively.

Figure 6 .
Figure 6.Phase-D2.Left: The blue and yellow regions denote the entanglement wedges of A and B, respectively.The green region denotes the shared entanglement wedge of AB.The black dashed line denotes the entanglement wedge cross-section between A and B. From the 2d boundary viewpoint, there are islands on the brane.I and I R denote the island for entanglement entropy and reflected entropy, respectively.The purple dashed circles are the boundary gap between A ∪ I R,A and B∪I R,B , and there are two gaps for phase-D2.Right: The canonical purification of ρ AB and the entanglement wedges for AB, AA * , BAA * .The red double-headed arrows denote the identification between two spacelike surfaces of the entanglement wedges.The small tubular neighborhoods of two blue jagged lines here are completely visible to BAA * , while not to AB or AA * .The entanglement wedges satisfy W(AA * B) ⊃ W(AB) ∪ W(AA * ).This signals the non-vanishing Markov gap for this phase.
which is equivalent to a non-vanishing mutual information between A and B = [b 3 , b 4 ]: I(A : B) ≥ 0. This condition gives

Figure 8 . 2 .
Figure 8. Phase-D4.In this phase, the entanglement islands for A and B are different from their reflected island.One boundary of EWCS and one gap between A ∪ I R,A and B ∪ I R,B give the lower bound h ≥ c 3 log 2. There is one jagged surface, with W(AA * B) ⊃ W(AA * ) ∪ W(AB).

Figure 12 .
Figure 12.JT gravity plus CFT matters is jointed with a flat CFT bath.
A X P J y E c = < / l a t e x i t > I R,B < l a t e x i t s h a 1 _ b a s e 6 4 = " s d 3 k r P B i m o G B H c d S y K e D c b c 1 0 S g = " > A A A C g X i c b Z F N a 9 t A E I b X S j 9 S 9 y t p j 7 2 I m k J y q J F S k x Z 6 C b S H H l O o k 4 A k z G g 1 s h f v r p T d U Y t Z 9 B d 6 T f 5 a / 0 1 X t g K 1 0 4 G F l 3 e e 2 Z 2 d y W s p L E X R n 0 G w 9 + D h o 8 f 7 T 4

3 <
5 g s 6 b n e w 6 y 0 u u e M y z x V Y p s q 4 t O v T q D u D b w w O s n e + u o 4 q 2 q 2 h O X / t E q k d + t X E u 4 u 4 L y 5 O x v H p e P J 9 M j qb 9 E v a Z 2 / Y W 3 b E Y v a R n b F v 7 J x N G W c L 9 p v d s N t g L z g O o u B k g w a D v u Y 1 2 4 r g 8 1 8O z M S T < / l a t e x i t > a l a t e x i t s h a 1 _ b a s e 6 4 = " l Z I S z Z D H 8 z I M Z t A z 0 f A 6 i 8 a N 9 K Y = " > A A A C g X i c b Z F N a 9 t A E I b X S t u k 7 k c + e u x F 1 B S S Q 4 0 U T F P o J Z A e e k y h T g K S M K P V y F 6 8 u 1 J 3 R y 1 m 0 r 7 o D 7 N s e l 1 x z m e c K L F N l X N r 3 a d S 1 w X c G B z k 4 H 1 1 P F d 3 e 0 J y / d o 3 U j f 1 q 4 s N F 3 B Q X p 7 P 4 z W z + e T 4 5 m w 9 L O m L P 2 Q s 2 Z T F 7 y 8 7 Y J 3 b O F o y z i v 1 k v 9 j v I A i m w e s g 3 q H B a K h 5 x v Y i e P 8 X i i j E X A = = < / l a t e x i t > b 2 H w b n D a T a M d J l v 7 6 3 + / s 8 1 1 e S 2 E p i v 4 M g g c P d x 4 9 3 n 0 y f L r 3 7 P m L / Y O X l 7 Z q D M c p r 2 R l r n O w K I X G K Q m S e F 0 b B J V L v M o X n 7 v 8 1 S 0 a K y r 9 j Z Y 1 Z g p m W p S C A 3 n r 4 t P N / i g a R 6 s I 7 4 u 4 F y P H w b n D a T a M d J l v 7 6 3 + / s 8 1 1 e S 2 E p i v 4 M g g c P d x 4 9 3 n 0 y f L r3 7 P m L / Y O X l 7 Z q D M c p r 2 R l r n O w K I X G K Q m S e F 0 b B J V L v M o X n 7 v 8 1 S 0 a K y r 9 j Z Y 1 Z g p m W p S C A 3 n r 4 t P N / i g a R 6 s I 7 4 u 4 F y P W x / n N w W C W F h V v F G r i E q x N 4 q i m z I E h w S W 2 w 7 S x W A N f w A w T L z U o t J l b d d q G b 7 1 T h G V l / N E U r t x / K x w o a 5 c q 9 6 Q C m t v t X G f + L 5 c 0 V J 5 m T u i 6 I d R 8 / V D Z y J C q s P t 2 W A i D n O T S C + B G + F 5 D P g c D n P x w h q n G H 7 x S C n T h U p s b a J M 4 c 6 n E k g 5 H c W r E b E 5 H 7 R b 2 f Y N L 7 r j M c w W W q T I u 7 f o 0 6 s 7 g a 4 O D 7 J 0 v r q O K d m N o z l + 7 Q G q H f j X x 9 i L u i 8 s P 4 / h k P L m Y j M 4 m / Z J 2 2 W v 2 h h 2 y m H 1 k Z + w r O 2 d T x h m y n + w X + x 2 w 4 F 1 w H L x f o 8 G g r 3 n F N i I 4 / Q v J 9 8 O X < / l a t e x i t > B < l a t e x i ts h a 1 _ b a s e 6 4 = " 1 8 o X A B J M + W / 9 9 f 6 E O d J y D y d F + j Y = " > A A A C g 3 i c b Z H f a 9 s w E M c V b 1 2 7 7 F e 7 P e 7 F L A w 6 B s E O Z e v L o N A 9 b G 8 d L G n B N u U s n x M t k u x J 5 5 U g / D / 0 t f v P 9 t 9 M T l 1 Y 0 h 0 I v n z v c 9 L p L q + l s B R F f w b B g 4 c 7 j 3 b 3 H g 2 H 4 Z x A 8 e P j o 8 d 7 + k + H T Z 8 9 f H B w e v b x 2 Z W 0 F T k W p S j v L w K G S B q c k S e G s s g g 6 U 3 i T L c + 7 / M 1 3 t E 6 W 5 o p W F a Y a 5 k Y W U g B 5 a 3 a h g P j 5 x d X t 4 S g c h + v g 9 0 X U i x H r 4 / L 2 a D B P 8 l L U G g 0 J B c 7 F U V h R 2 o A l K R S 2 w 6 R 2 W I F Y w h x j L w 1 o d G m z b r j l b 7 2 T 8 6 K 0 / h j i a / f f i g a 0 c y u d e V I D L d x u r j P / l 4 t r K j 6 m j

Figure 13 .
Figure 13.Phase-D2 for JT gravity.The purple dashed circles denote the boundary gaps between A ∪ I R,A and B ∪ I R,B

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n E b d G 3 x l c C Y 7 5 4 t r q b x Z W 5 r z b e e A T d + f J t 4 8 x G N x 8 2 E Y f x y O L k e D s 1 F 3 p G 3 y j h y Q I x K T E 3 J G v p I L M i a c V O Q X + U 3 + B L t B H J w G n 1 Z o 0 O t q 3 p K 1 C M 7 / A X P J y E c = < / l a t e x i t > I R,B < l a t e x i t s h a 1 _ b a s e 6 4 = " s d 3 k r P B i m o G B H c d S y K e D c b c 1 0 S g = " > A A A C g X i c b Z F N a 9 t A E I b X S j 9 S 9 y t p j 7 2 I m k J y q J F S k x Z 6 C b S H H l O o k 4 A k z G g 1 s h f v r p T d U Y t Z 9 B d 6 T f 5 a / 0 1 X t g K 1 0 4 G F l 3 e e 2 Z 2 d y W s p L E X R n 0 G w 9 + D h o 8 f 7 T 4 Z P n z 1 / 8 f L g 8 N W F r R r D c c o r WZ m r H C x K o X F K g i R e 1 Q Z B 5 R I v 8 + W X L n / 5 E 4 0 V l f 5 B q x o z B X M t S s G B O u s 9 z D 7 M D k b R O F p H e F / E v R i x P s 5 n h 4 N 5 W l S 8 U a i J S 7 A 2 i a O a M g e G B J f Y D t P G Y g 1 8 C X N M v N S g 0 G Z u 3 W w b v v N O E Z a V 8 U d T u H b / r X C g r F 2 p 3 J M K a G F 3 c 5 3 5 v 1 z S U P k p c 0 L X D a H m m 4 f K R o Z U h d 3 P w 0 I Y 5 C R X X g A 3 w v c a 8 g U Y 4 O T n M 0 w 1 / u K V U q A L l 9 r c Q J v E m U s l l n Q 0 i l M j5 g s 6 b n e w 6 y 0 u u e M y z x V Y p s q 4 t O v T q D u D b w w O s n e + u o 4 q 2 q 2 h O X / t E q k d + t X E u 4 u 4 L y 5 O x v H p e P J 9 M j q b 9 E v a Z 2 / Y W 3 b E Y v a R n b F v 7 J x N G W c L 9 p v d s N t g L z g O o u B k g w a D v u Y 1 2 4 r g 8 1 8 O z M S T < / l a t e x i t > a l a t e x i t s h a 1 _ b a s e 6 4 = " l Z I S z Z D H 8 z I M Z t A z 0 f A 6 i 8 a N 9 K Y = " > A A A C g X i c b Z F N a 9 t A E I b X S t u k 7 k c + e u x F 1 B S S Q 4 0 U T F P o J Z A e e k y h T g K S M K P V y F 6 8 u 1 J 3 R y 1 m 0V / o t f 1 r / T d d 2 Q r U T g c W X t 5 5 Z n d 2 J q + l s B R F f w b B 3 q P H T / Y P n g 6 f P X / x 8 v D o + O T G V o 3 h O O W V r M x d D h a l 0 D g l Q R L v a o O g c o m 3 + f K q y 9 9 + R 2 N F p b / S q s Z M w V y L U n C g z n o H s 3 h 2 N I r G 0 T r C h y L u x Y j 1 c T 0 7 H s z T o u K N Q k 1 c g r V J H N W U O T A k u M R 2 m D Y W a + B Lm G P i p Q a F N n P r Z t v w r X e K s K y M P 5 r C t f t v h Q N l 7 U r l n l R A C 7 u b 6 8 z / 5 Z K G y g + Z E 7 p u C D X f P F Q 2 M q Q q 7 H 4 e F s I g J 7 n y A r g R v t e Q L 8 A A J z + f Y a r x B 6 + U A l 2 4 1 O Y G 2 i T O X C q x p N N R n B o x X 9 B Z u 4 N 9 2 + K S e y 7 z X I F l q o x L u z 6 N u j f 4 x u A g e + e T 6 6 i i 3 R q a 8 9 c u k d q h X 0 2 8 u 4 i H 4 u Z 8 H L 8 f T 7 5 M R p e T f k k H 7 D V 7 w 0 5 Z z C 7 Y J f v M r t m U c b Z g P 9 k v 9 j v Y C 8 6 C K D j f o M G g r 3 n F t i L 4 + B c K m M S R < / l a t e x i t > a 1 H w b n D a T a M d J l v 7 6 3 + / s 8 1 1 e S 2 E p i v 4 M g g c P d x 4 9 3 n 0 y f L r 3 7 P m L / Y O X l 7 Z q D M c p r 2 R l r n O w K I X G K Q m S e F 0 b B J V L v M o X n 7 v 8 1 S 0 a K y r 9 j Z Y 1 Z g p m W p S C A 3 n r 4 t P N / i g a R 6 s I 7 4 u 4 F y PW x / n N w W C W F h V v F G r i E q x N 4 q i m z I E h w S W 2 w 7 S x W A N f w A w T L z U o t J l b d d q G b 7 1 T h G V l / N E U r t x / K x w o a 5 c q 9 6 Q C m t v t X G f + L 5 c 0 V J 5 m T u i 6 I d R 8 / V D Z y J C q s P t 2 W A i D n O T S C + B G + F 5 D P g c D n P x w h q n G H 7 x S C n T h U p s b a J M 4 c 6 n E k g 5 H c W r E b E 5 H 7 R b 2 f Y N L 7 r j M c w W W q T I u 7 f o 0 6 s 7 g a 4 O D 7 J 0 v r q O K d m N o z l + 7 Q G q H f j X x 9 i L u i 8 s P 4 / h k P L m Y j M 4 m / Z J 2 2 W v 2 h h 2 y m H 1 k Z + w r O 2 d T x h m y n + w X + x 2 w 4 F 1 w H L x f o 8 G g r 3 n F N i I 4 / Q v H 3 c O W < / l a t e x i t > A < l a t e x i t s h a 1 _ b a s e 6 4 = " M i V W r 1 f j 9 O y r t O x Y i i P 8 K W u A J y I = " > A A A C f n i c b Z F R b 9 M w E M f d w G C U A R s 8 8 h J R g b a H l Q R V Y 4 8 T 8 M D j J t F t U h J N F + f S W r W d Y F + G K i u f g F f 4 cH w b n D a T a M d J l v 7 6 3 + / s 8 1 1 e S 2 E p i v 4 M g g c P d x 4 9 3 n 0 y f L r 3 7 P m L / Y O X l 7 Z q D M c p r 2 R l r n O w K I X G K Q m S e F 0 b B J V L v M o X n 7 v 8 1 S 0 a K y r 9 j Z Y 1 Z g p m W p S C A 3 n r 4 t P N / i g a R 6 s I 7 4 u 4 F y PW x / n N w W C W F h V v F G r i E q x N 4 q i m z I E h w S W 2 w 7 S x W A N f w A w T L z U o t J l b d d q G b 7 1 T h G V l / N E U r t x / K x w o a 5 c q 9 6 Q C m t v t X G f + L 5 c 0 V J 5 m T u i 6 I d R 8 / V D Z y J C q s P t 2 W A iD n O T S C + B G + F 5 D P g c D n P x w h q n G H 7 x S C n T h U p s b a J M 4 c 6 n E k g 5 H c W r E b E 5 H 7 R b 2 f Y N L 7 r j M c w W W q T I u 7 f o 0 6 s 7 g a 4 O D 7 J 0 v r q O K d m N o z l + 7 Q G q H f j X x 9 i L u i 8 s P 4 / h k P L m Y j M 4 m / Z J 2 2 W v 2 h h 2 y m H 1 k Z + w r O 2 d T x h m y n + w X + x 2 w 4 F 1 w H L x f o 8 G g r 3 n F N i I 4 / Q v J 9 8 O X < / l a t e x i t > B < l a t e x i t s h a 1 _ b a s e 6 4 = " 1 8 o X A B J M + W / 9 9 f 6 E O d J y D y d F + j Y = " > A A A C g 3 i c b Z H f a 9 s w E M c V b 1 2 7 7 F e 7 P e 7 F L A w 6 B s E O Z e v L o N A 9 b G 8 d 1 y f j K P T 8 e T r y e h s 0 i 9 p n 7 1 m b 9 g x i 9 g H d s a + s E s 2 Z Y I p 9 p P 9 Y r + D v e B 9 M A l O N 2 g w 6 G t e s a 0 I P v 0 F m D j G H Q = = < / l a t e x i t > Flat CFT < l a t e x i t s h a 1 _ b a s e 6 4 = " F 6 2 d Z 4 F J W S T 5 R t i g q / 3 P D F B C e G 8 = " > A A A C g H i c b Z F N j 9 M w E I b d 8 L W U j 9 2 F I 5 e I C l E O l A R V g D i t B A e O i 6 C 7 K y X R a u J M U q u 2 E + w J q L L y E 7 j C b + P f 4 L R Z i X Y Z y d K r d 5 6 x x z N 5 I 4 W l K P o z C m 7 c v H X 7 z s H d 8 b 3 7 D x 4 e H h 0 / O r N 1 a z g u e C 1 r c 5 G D R S k 0 L k i Q x I v G I K h c 4 n m + + t D n z 7 + j s a L W X 2 n d Y K a g 0 q I U H M h b X 1 7 C 8 8 u j S T S L N h F e F / E g J m y I 0 8 Y D T U P G Y 7 E b z / C w d 0 x B 4 = < / l a t e x i t > a 0 < l a t e x i t s h a 1 _ b a s e 6 4 = " 7 O P 1 i b + K k d w v x g 9 K 8 O u G e S o O y e I = " > A A A C j 3 i c b Z F d a 9 s w F I Y V 7 6 v N v p L t s j d i Y d D B C P Y I 7 a 5 G x n a x 3 n V j a Q u 2 C b J 8 n I h I s i s d d w T h f 7 L b 7 T / t 3 0 x O H V j S H R C 8 v O e R d H h P V k l h M Q z / 9 I J 7 9 x 8 8 f H R w 2 H / 8 5 O m z 5 4 P h i w t b 1 o b D j J e y N F c Z s y

Figure 15 .
Figure 15.Phase-D3 for JT gravity.The purple dashed circle denotes the boundary gap between A ∪ I R,A and B ∪ I R,B .a denotes the island cross-section.we show the Markov gap against b 2 − b 1 with a specific parameter setting, that is b 1 = 1, φ r = 100 and c = 12000, the same as that used in [68].For fixed b 1 , the minima are located at positions that satisfy b 3 − b 2 b 2 − b 1 .As b 3 − b 2 gets smaller, the minimum approaches h = 2c 3 log 2 at b 2 − b 1 = 0.

1 <
9 B Z u 4 N 9 2 + K S e y 7 z X I F l q o x L u z 6 N u j f 4 x u A g e + e T 6 6 i i 3 R q a 8 9 c u k d q h X 0 2 8 u 4i H 4 u Z 8 H L 8 f T 7 5 M R p e T f k k H 7 D V 7 w 0 5 Z z C 7 Y J f v M r t m U c b Z g P 9 k v 9 j v Y C 8 6 C K D j f o M G g r 3 n F t i L 4 + B c K m M S R < / l a t e x i t >a l a t e x i t s h a 1 _ b a s e 6 4 = " d x 8 j q C V 3 d 8 m e c 9 x u 9 l + K C p z q N j c = " >A A A C g H i c b Z F R b 9 M w E M f d j I 1 R x u j g k Z e I C q m 8 l H i q A P E 0 C R 5 4 H I J u k 5 K o c p x L a t V 2 g n 0 B V V Y + A q / j s / F t c N p M o h 0 n W f r r f 7 + z z 3 d Z L Y X F K P o z C A 4 e H B 4 9 P H 4 0 f H z y 5 P T p 6 O z Z l a 0 a w 2 H O K 1 m Z m 4 x Z k E L D H A V K u K k N M J V J u M 5 W H 7 v 8 9 Q 8 w V l T 6 G 6 5 r S B U r t S g E Z + i t r 9 m C L k b j a B p t I r w v a C / G p I / L x d m g T P K K N w o 0 c s m s j W l U Y + q Y Q c E l t M O k s V A z v m I l x F5 q p s C m b t N r G 7 7 y T h 4 W l f F H Y 7 h x / 6 1 w T F m 7 V p k n F c O l 3 c 9 1 5 v 9 y c Y P

1 < 2 < 3 <
d 4 E d I s G g 7 7 m O d m J 4 M N f i A 7 E W w = = < / l a t e x i t > b l a t e x i t s h a 1 _ b a s e 6 4 = " N 1 z f 5 E F X o A 0 P b H W o e O p Y J J V R P n g = " > A A A C g H i c b Z F N j 9 M w E I b d 8 LW U r 1 0 4 c o m o k M q l J K s K E K e V 4 M B x E X R 3 p S S q J s 4 k t W o 7 w Z 6 A K i s / g S v 8 N v 4 N T p u V a J e R L L 1 6 5 x l 7 P J M 3 U l i K o j + j 4 N b t O 3 f v H d 0 f P 3 j 4 6 P G T 4 5 O n F 7 Z u D c c F r 2 V t r n K w K I X G B Q m S e N U Y B J V L v M z X H / r 8 5 X c 0 V t T 6 K 2 0 a z B R U W p S C A 3 n r S 7 4 8 X R 5 P o l m 0 j f C m i A c x Y U O c L 0 9 G V V r U v F W o i U u w N o m j h j I H h g S X 2 I 3 T 1 m I D f A 0 V J l 5 q U G g z t + 2 1 C 1 9 6 p w j L 2 v i j K d y 6 / 1 Y 4 U N Z u V O 5 J B b S y h 7 n e / F 8 u a a l 8 l z m h m 5 Z Q 8 9 1 D Z S t D q s P + 4 2 E h D H K S G y + A G + F 7 D f k K D H D y 4 x m n G n / w W i n Q h U t t b q B L 4 s y l E k u a T u L U i G p F r 7 o D 7 N s e l 1 x z m e c K L F N l X N r 3 a d S 1 w X c G B z k 4 H 1 1 P F d 3 e 0 J y / d o 3 U j f 1 q 4 s N F 3 B Q X p 7 P 4 z W z + e T 4 5 m w 9 L O m L P 2 Q s 2 Z T F 7 y 8 7 Y J 3 b O F o y z i v 1 k v 9 j v I A im w e s g 3 q H B a K h 5 x v Y i e P 8 X i i j E X A = = < / l a t e x i t > b l a t e x i t s h a 1 _ b a s e 6 4 = " 7 5V 1 Y g a W Q r r U / z h j 8 Q t K Y A i y C F 8 = " > A A A C g H i c b Z F N j 9 M w E I b d 8 L W U r 1 0 4 c o m o k M q l J F A B 4 r Q S H D g u g u 6 u l E T V x J m k V m 0 n 2 B N Q Z e U n c I X f x r / B a b M S 7 T K S p V f v P G O P Z / J G C k t R 9 G c U 3 L h 5 6 / a d o 7 v j e / c f P H x 0 f P L 4 3 N a t 4 b j g t a z N Z Q 4 W p d C 4 I E E S L x u D o H K J F / n 6 Q 5 + / + I 7 G i l p / p U 2 D m Y J K i 1 J w I G 9 9 y Z e v l 8 e T a B Z t I 7 w u 4 k F M 2 B B n y 5 N R l R Y 1 b x V q 4 h K s T e K o o c y B I c E l d u O 0 t d g A X 0 O F i Z c a F N r M b X v t w u f e K c K y N v 5 o C r f u v x U O l L U b l X t S A a 3 s Y a 4 3 / 5 d L W i r f Z U 7 o p i X U f P d Q 2 c q Q 6 r D / e F g I g 5 z k x g v g R v h e Q 7 4 C A 5 z 8 e M a p x h + 8 V g p 0 4 V K b G + i S O H O p x J K m k z g 1 o l r R i + 4 A + 7 b H J V d c 5 r k C y 1 Q Z l / Z 9 G n V l 8 J 3 B Q Q 7 O R 9 d T R b c 3 N O e v X S N 1 Y 7 + a + H A R 1 8 X 5 q 1 n 8 Z j b / P J + c z o c l H b G n 7 Bm b s p i 9 Z a f s E z t j C 8 Z Z x X 6 y X + x 3 E A T T 4 G U Q 7 9 B g N N Q 8 Y X s R v P 8 L j E L E X Q = = < / l a t e x i t > b l a t e x i t s h a 1 _ b a s e 6 4 = " D I 0 k a U b Q M r 2 5 B j H l A 1 y n H w b n D a T a M d J l v 7 6 3 + / s 8 1 1 e S 2 E p i v 4 M g g c P d x 4 9 3 n 0 y f L r 37 P m L / Y O X l 7 Z q D M c p r 2 R l r n O w K I X G K Q m S e F 0 b B J V L v M o X n 7 v 8 1 S 0 a K y r 9 j Z Y 1 Z g p m W p S C A 3 n r 4 t P N / i g a R 6 s I 7 4 u 4 F y P W x / n N w W C W F h V v F G r i E q x N 4 q i m z I E h w S W 2 w 7 S x W A N f w A w T L z U o t J l b d d q G b 7 1 T h G V l / N E U r t x / K x w o a 5 c q 9 6 Q C m t v t X G f + L 5 c 0 V J 5 m T u i 6 I d R 8 / V D Z y J C q s P t 2 W A i D n O T S C + B G + F 5 D P g c D n P x w h q n G H 7 x S C n T h U p s b a J M 4 c 6 n E k g 5 H c W r E b E 5 H 7 R b 2 f Y N L 7 r j M c w W W q T I u 7 f o 0 6 s 7 g a 4 O D 7 J 0 v r q O K d m N o z l + 7 Q G q H f j X x 9 i L u i 8 s P 4 / h k P L m Y j M 4 m / Z J 2 2 W v 2 h h 2 y m H 1 k Z + w r O 2 d T x h m y n + w X + x 2 w 4 F 1 w H L x f o 8 G g r 3 n F N i I 4 / Q v H 3 cO W < / l a t e x i t > A < l a t e x i t s h a 1 _ b a s e 6 4 = " H w b n D a T a M d J l v 7 6 3 + / s 8 1 1 e S 2 E p i v 4 M g g c P d x 4 9 3 n 0 y f L r 3 7 P m L / Y O X l 7 Z q D M c p r 2 R l r n O wK I X G K Q m S e F 0 b B J V L v M o X n 7 v 8 1 S 0 a K y r 9 j Z Y 1 Z g p m W p S C A 3 n r 4 t P N / i g a R 6 s I 7 4 u 4 F y P W x / n N w W C W F h V v F G r i E q x N 4 q i m z I E h w S W 2 w 7 S x W A N f w A w T L z U o t J l b d d q G b 7 1 T h G V l / N E U r t x / K x w o a 5 c q 9 6 Q C m t v t X G f + L 5 c 0 V J 5 m T u i 6 I d R 8 / V D Z y J C q s P t 2 W A i D n O T S C + B G + F 5 D P g c D n P x w h q n G H 7 x S C n T h U p s b a J M 4 c 6 n E k g 5 H c W r E b E 5 H 7 R b 2 f Y N L 7 r j M c w W W q T I u 7 f o 0 6 s 7 g a 4 O D 7 J 0 v r q O K d m N o z l + 7 Q G q H f j X x 9 i L u i 8 s P 4 / h k P L m Y j M 4 m / Z J 2 2 W v 2 h h 2 y m H 1 k Z + w r O 2 d T x h m y n + w X + x 2 w 4 F 1 w H L x f o 8 G g r 3 n F N i I 4 / Q v J 9 8 O X < / l a t e x i t > B < l a t e x i t s h a 1 _ b a s e 6 4 = " 1 8 o X A B J M + W / 9 9 f 6 E O d J y D y d F + j Y = " > A A A C g 3 i c b Z H f a 9 s w E M c V b 1 2 7 7 F e 7 P e 7 F L A w 6 B s E O Z e v L o N A 9 b G 8 d L G n B N u U s n x M t k u x J 5 5 U g / D / 0 t f v P 9 t 9 M T l 1 Y 0 h 0 I v n z v c 9 L p L q + l s B R F f w b B g 4 c 7 j 3 b 3 H g + f P H 3 2 / M X + w c u Z r R r D c c o r W Z m L H C x K o X F K g i R e 1 A Z B 5 R L P 8 + V p l z / / h c a K S n + n V Y 2 Z g r k W p e B A 3 p p 9 t R J 0 c b k / i s b R O s L 7 I u 7 F i P V x d n k w m K d F x R u F mr g E a 5 M 4 q i l z Y E h w i e 0 w b S z W w J c w x 8 R L D Q p t 5 t b t t u F b 7 x R h W R l / N I V r 9 9 8 K B 8 r a l c o 9 q Y d w X s 8 k 4 / j A + + j Y Z n R z 1 S 9 p j r 9 k b d s h i 9 p G d s C / s j E 0 Z Z z / Y N b t h v 4 O d 4 H 0 w C X o 2 G P Q 1 r 9 h G B J / + A v N P x e A = < / l a t e x i t > Island < l a t e x i t s h a 1 _ b a s e 6 4 = " y G m g i T r 0 33 g K E h T b q K X L f D K L 3 Z k = " > A A A C h X i c b Z F N a 9 t A E I b X 6 k d S 9 y N J e + x l q S m k h x o p m L S 3 B l J C j y n E i U E S Y b Q a 2 Y t 3 V + r u q M U I / Y l e 2 z / W f 9 O V r U D t d G D h 5 Z 1 n d m d n s k p J R 2 H 4 Z x A 8 e P j o 8 d 7 + k + H T Z 8 9 f H B w e v b x 2 Z W 0 F T k W p S j v L w K G S B q c k S e G s s g g 6 U 3 i T L c + 7 / M 1 3 t E 6 W 5 o p W F a Y a 5 k Y W U g B 5 a 3 a h g P j 5 x d X t 4 S g c h + v g 9 0 X U i x H r 4 / L 2 a D B P 8 l L U G g 0 J B c 7 F U V h R 2 o A l K R S 2 w 6 R 2 W I F Y w h x j L w 1 o d G m z b r j l b 7 2 T 8 6 K 0 / h j i a / f f i g a 0 c y u d e V I D L d x u r j P / l 4 t r K j 6 m j T R V T W j E 5 q G i V p x K 3 v 2 e 5 9 K i I L X y A o S V v l c u F m B B k J / R M D H 4 Q 5 R a g 8 m b x G U W 2 j h K m 0 R h Q c e j K L F y v q B 3 7 Q 7 2 b Y u L 7 7 j U c z k W i b Z N 0 v V p 9 Z 0 h N o Y A 1 T u f m 4 7 K 2 6 2 h N f 7 a J V I 7 9 K u J d h d x X1 y f j K P T 8 e T r y e h s 0 i 9 p n 7 1 m b 9 g x i 9 g H d s a + s E s 2 Z Y I p 9 p P 9 Y r + D v e B 9 M A l O N 2 g w 6 G t e s a 0 I P v 0 F m D j G H Q = = < / l a t e x i t > Flat CFT < l a t e x i t s h a 1 _ b a s e 6 4 = " 7 O P 1 i b + K k d w v x g 9 K 8 O u G e S o O y e I = " > A A A C j 3 i c b Z F d a 9 s w F I Y V 7 6 v N v p L t s j d i Y d D B C P Y I 7 a 5 G x n a x 3 n V j a Q u 2 C b J 8 n I h I s i s d d w T h f 7 L b 7 T / t 3 0 x O H V j S H R C 8 v O e R d H h P V k l h M Q z / 9 I J 7 9 x 8 8 f H R w 2 H / 8 5 O m z 5 4 P h i w t b 1 o b D j J e y N F c Z s y s C r D p + 9 V E + 4 u 4 K y 7 e j a O T 8 e T r Z D S d d E s 6I E f k F T k m E T k l U / K F n J M Z 4 e S G / C S / y O 9 g G J w G H 4 L p L R r 0 u j s v y U 4 F Z 3 8 B K D 7 J 0 Q = = < / l a t e x i t > I R,A < l a t e x i t s h a 1 _ b a s e 6 4 = " F 6 2 d Z 4 F J W S T 5 R t i g q / 3 P D F B C e G 8 = " > A A A C g H i c b Z F N j 9 M w E I b d 8 L W U j 9 2 F I 5 e I C l E O l A R V g D i t B A e O i 6 C 7 K y X R a u J M U q u 2 E + w J q L L y E 7 j C b + P f 4 L R Z i X Y Z y d Kr d 5 6 x x z N 5 I 4 W l K P o z C m 7 c v H X 7 z s H d 8 b 3 7 D x 4 e H h 0 / O r N 1 a z g u e C 1 r c 5 G D R S k 0 Y D T U P G Y 7 E b z / C w d 0 x B 4 = < / l a t e x i t > a 0 < l a t e x i t s h a 1 _ b a s e 6 4 = " 7 6 h 6 M p d e l F d 3 R q m z u N 3 F / l X 3 b K I = " > A A A C i 3 i c b Z H R b t M w F I b d D N g o D D Z 2 y Y 2 1 C m n c V A l U g B A X m w C J 3 W 0 S 3 S Y l o X K c k 9 a q 7 W T 2 C a i y 8 h 7 c j r f i b X C 6 V F o 7 j m T p 1 3 8 + 2 0 f / y S o p L I b h 3 1 6 w 9 e D h o + 2 d x / 0 n T 3 e f P d / b f 3 F h y 9 p w G P N S l u Y q Y x a k 0 D B G g R K u K g N M Z R I u s / n n t n / 5 E 4 w V p f 6 O i w p S x a Z a F I I z 9 N a P R D G c c S b p 6 c S d N s H L D p + 9 V E m 4 u 4 L y 7 e D K N 3 w 9 H 5 a H A 8 6 p a 0 Q 1 6 S Q 3 J E I v K e H J N v 5 I y M C S e G / C Y 3 5 E + w G 7 w N P g a f b t G g 1 9 0 5 I G s V f P 0 H P + z J D g = = < / l a t e x i t > I A < l a t e x i t s h a 1 _ b a s e 6 4 = " L k 9 n J e P + n m 3

2 Figure 16 .
Figure 16.Phase-D4 for JT gravity.The purple dashed circle denotes the boundary gap between A ∪ I R (A) and B ∪ I R (B). a denotes the island cross-section.

Figure 17 .
Figure 17.An illustration for the interval of 2D extremal black holes.
) where d −ab is the distance between two points (t −a , x −a ) and (t b , x b ) on the time slice t −a = t b of the flat spacetimes ds 2 = −dx + dx − .For an interval [b 1 , b 2 ] which does not admit an entanglement island, the entanglement entropy is S([b 1 , b

. 4 )
This reduces to c 3 log d b 1 b 2 for flat CFT with Ω = 1.

Figure 20 .
Figure 20.The distance between two geodesics, denoted by black circles, is the geodesic length between P 1 and P 2 on the green circle.

3 )
Then the length of the geodesic between b 2 and P 1 is given byD(P 1 , b 2 )

Figure 22 .
Figure 22.A and B are gapped by two small intervals.The green line and red line denote RT surfaces for A and B, respectively.The black dashed line denotes the entanglement wedge cross-section E W between A and B. The cutoff is y UV = , which requires a gap of length 2 .