How fast planar maps get swallowed by a peeling process

The peeling process is an algorithmic procedure that discovers a random planar map step by step. In generic cases such as the UIPT or the UIPQ, it is known [Curien&Le Gall, Scaling limits for the peeling process on random maps, Ann. Inst. Henri Poincar\'e Probab. Stat. 53, 1 (2017), 322-357] that any peeling process will eventually discover the whole map. In this paper we study the probability that the origin is not swallowed by the peeling process until time $n$ and show it decays at least as $n^{-2c/3}$ where \[c \approx 0.12831235141783245423674486573872854933142662048339843...\] is defined via an integral equation derived using the Lamperti representation of the spectrally negative $3/2$-stable L\'evy process conditioned to remain positive [Chaumont, Kyprianou&Pardo, Some explicit identities associated with positive self-similar Markov processes, Stochastic Process. Appl. 119, 3 (2009), 980-1000] which appears as a scaling limit for the perimeter process. As an application we sharpen the upper bound of the sub-diffusivity exponent for random walk of [Benjamini&Curien, Simple random walk on the uniform infinite planar quadrangulation: subdiffusivity via pioneer points, Geom. Funct. Anal. 23, 2 (2013), 501-531].


Introduction
One of the main tool to study random maps is the so-called peeling procedure which is a step-by-step Markovian exploration of the map. Such a procedure was conceived by Watabiki [ ] and then formalized and used in the setting of the Uniform In nite Planar Triangulation (UIPT in short) by Angel [ ]. In this paper we will use Budd's peeling process [ ] which enables to treat many di erent models in a uni ed way. A peeling of a rooted map m is an increasing sequence of sub-maps with holes e 0 ⊂ e 1 ⊂ · · · ⊂ m, where e 0 is the root-edge of m, and where e i+1 is obtained from e i by selecting an edge (the edge to peel) on the boundary of one of its holes and then either gluing one face to it, or identifying it to another edge of the same hole; in the latter case this may split the hole into two new holes or make it disappear. When the map m = m ∞ is in nite and one-ended (a map of the plane) we usually " ll-in" the nite holes of e i and consider instead a sequence of sub-maps with one hole e 0 ⊂ e 1 ⊂ · · · ⊂ m ∞ called a lled-in peeling process. Obviously, the peeling process depends on the algorithm used to choose the next edge to peel and any algorithm is allowed, provided it does not use the information outside e n (it is 'Markovian') see [ ] for details.
In this work we will study the peeling process on random in nite critical Boltzmann planar maps with bounded face-degrees. More precisely, if q = (q i ) i ≥1 is a non-negative sequence with nite support we de ne the q-Boltzmann weight of a nite map m as w(m) = f ∈Faces (m) q deg(f ) . ♠ Département de Mathématiques, Univ. Paris-Sud, Université Paris-Saclay and IUF.
nicolas.curien@gmail.com ♣ Département de Mathématiques, Univ. Paris-Sud, Université Paris-Saclay and FMJH. cyril.marzouk@u-psud.fr We suppose that q is admissible (the above measure has nite total mass) and critical in the sense that we cannot increase any of the q i 's and remain admissible, see [ , , ]

Figure :
Example of the first few steps of a (filled-in) peeling process; the next edge to peel is indicated in fat blue, and when this edge is identified with another one, the la er is indicated in dashed blue. The filled-in holes are indicated in pink (we do not represent the submap they contain). At the last step, the root-edge gets swallowed.
Let (e i ; i ≥ 0) be a ( lled-in Markovian) peeling of M ∞ . The proof of [ , Theorem ] recast below into Proposition shows that (at least in the case of the UIPQ) such an exploration cannot escape too fast to in nity in the sense that whatever the peeling algorithm used e n cannot reach distances more than n 1/3+o(1) from the origin inside M ∞ . On the other hand, it is known (see [ , Corollary ] based on [ ]) that the exploration will eventually reveal the whole map, that is n ≥0 e n = M ∞ , a.s.
Our main result gives a quantitative version of the last display. We let ∂e n denote the boundary of the sub-map e n , de ned as the edges adjacent to its unique hole.
Theorem (The peeling swallows the root). Let M ∞ be an in nite Boltzmann planar map with bounded face-degrees and let (e n ) n ≥0 be any ( lled-in Markovian) peeling process of M ∞ . If R denotes the root-edge of M ∞ then we have as n → ∞: where c ≈ 0.1283123514178324542367448657387285493314266204833984375... is the positive solution to Of course, a given peeling process may swallow the root-edge much faster, but our upper bound holds for any peeling algorithm, and it is sharp for the worst algorithm, in which we always select the edge which is at the opposite of the root-edge on ∂e n . Theorem holds with the same proof for Angel's peeling process on the UIPT or UIPQ. The intriguing constant c ≈ 0.1283... comes from a calculation done on the Lévy process arising in the Lamperti representation of the 3/2-stable spectrally negative Lévy process conditioned to stay positive. Theorem should hold true for any critical generic Boltzmann map. Let us note that the proof of Theorem actually shows that for any sub-map e with a unique hole and any edge E ∈ ∂e, if denotes the length of the hole, then if we start the peeling process with e 0 = e, we have P(E ∈ ∂e n ) ≤ (n/ ) −2c/3+o(1) .

Application.
A particular peeling algorithm was designed in [ ] to study the simple random walk on the UIPQ and show it is sub-di usive with exponent 1/3: the maximum displacement of the walk after n steps is at most of order n 1/3+o(1) . The proof in [ ] actually does not use much of the random walk properties since it is valid for any peeling algorithm (see Proposition below). It shows in fact that the maximal displacement of the walk until the discovery of the nth pioneer point is n 1/3+o(1) . To improve it we give a bound on the number of pioneer points. For convenience we state the result for the walk on the dual of the UIPQ which is more adapted to Budd's peeling process but the proof could likely be adapted to the walk on the dual or primal lattice of any generic in nite Boltzmann map. A pioneer edge is roughly speaking a dual edge crossed by the walk, such that its target is not disconnected from in nity by the past trajectory; we refer to Section and Figure for a formal de nition.
Proposition . Let ( ì E n ) n ≥0 be the oriented dual edges visited by a simple random walk on the faces of the UIPQ started from the face adjacent to the right of the root-edge. There exists γ > 0 such that if π n is the number of pioneer edges of the walk up to time n then we have The basic idea is to use reversibility to convert the discovery of a pioneer edge at time n into the event in which the root-edge is still exposed after n steps of the walk as in [ , Lemma ] and then use Theorem . Our proof gives a rather low value γ = c/(12 + 2c) ≈ 0.01046881621... with c from Theorem ; it could optimised, but it cannot exceed γ = 2c/(3 + 2c) ≈ 0.07880082179... and this would require a longer argument to yield a sub-di usivity exponent slightly larger than 0.3, which is still far from the conjectural value of 1/4 for the simple random walk on generic random planar maps. We therefore restricted ourselves to this weaker exponent which still improves on the very general upper bound on sub-di usivity exponent for unimodular random graphs discovered by Lee [ , Theorem . ]. In the case of the UIPT (type II) Gwynne & Miller [ , Theorem . ] have proven that the exponent is at least 1/4 and an ongoing work of Gwynne & Hutchcroft [ ] shall yield the corresponding upper bound (still in the case of the UIPT). We do not think that our methods can lead to this optimal exponent.
The results of this paper can also be adapted to the case of peeling processes (and random walks on the dual and primal) on Boltzmann maps with large faces [ ].

. Peeling of a Boltzmann planar map
Let us give a brief description of the peeling process of Budd [ ], with a point of view inspired by [ , ] to which we refer for more details. Fix an in nite one-ended map m ∞ ; a sub-map e of m ∞ is a map with a distinguished simple face called the hole, such that one can recover m ∞ by gluing inside the unique hole of e a planar map with a (not necessarily simple) boundary of perimeter matching that of the hole of e (this map is then uniquely de ned). Then a ( lled-in) peeling exploration of m ∞ is an increasing sequence (e i ) i ≥0 of sub-maps of m ∞ containing the root-edge. Such an exploration depends on an algorithm A which associates with each sub-map e an edge on the boundary of its hole; this algorithm can be deterministic or random, but in the latter case the randomness involved must be independent of the rest of the map. Then the peeling exploration of m ∞ with algorithm A is the sequence (e i ) i ≥0 of sub-maps of m ∞ constructed recursively as follows. First e 0 always only consists of two simple faces of degree 2 and an oriented edge, the hole is the face on the left of this root-edge; this corresponds to the root-edge of the map m ∞ that we open up in two. Then for each i ≥ 0, conditional on e i , we choose the edge A(e i ) on the boundary of its hole, and we face two possibilities depicted in Figure : (i) either the face in m ∞ on the other side of A(e i ) is not already present in e i ; in this case e i+1 is obtained by adding this face to e i glued onto A(e i ) and without performing any other identi cation of edges, (ii) or the other side of A(e i ) in m ∞ actually corresponds to a face already discovered in e i . In this case e i+1 is obtained by performing the identi cation of the two edges in the hole of e i . This usually creates two holes, but since m ∞ is one-end, we decide to ll-in the one containing a nite part of m ∞ .

Figure :
Illustration of a (filled-in) peeling step in a one-ended bipartite map. The peeling edge is depicted in blue. In the first case we add a new face adjacent to this edge, and in the second case we identify two edges on the boundary of the hole, this splits the hole into two components and we fill-in the finite one.
We de ne the perimeter of a sub-map e of M ∞ to be the number of edges on the boundary of its simple hole, we denote it by |∂e|. If (e n ) n ≥0 is a peeling exploration of M ∞ , we consider its perimeter process given by (|∂e n |) n ≥0 ; note that |∂e 0 | = 2 from our convention. Budd [ ] de nes in terms of q a probability measure ν ; we shall denote by S = (S n ) n ≥0 a random walk started from 2 and with step distribution ν. When q is admissible, critical and nitely supported, the support of the law ν is bounded above, furthermore this law is centered and belongs to the domain of attraction of a stable law with index 3/2; precisely, . Let (e n ) n ≥0 be any peeling exploration of M ∞ . The perimeter process (|∂e n |) n ≥0 is a Markov chain whose law does not depend on the peeling algorithm A. More precisely, it has the same law as S ↑ , a random walk with step distribution ν conditioned to stay positive (see e.g. [ ]). Moreover, the convergence in distribution holds in the sense of Skorokhod, where ϒ ↑ is a version conditioned to never hit (−∞, 0) of a 3/2-stable spectrally negative Lévy process (see e.g. [ ]).
Let us mention that a jump of this process, say |∂e n | − |∂e n−1 | = k with k ∈ Z, corresponds in the n-th step of the peeling exploration to the discovery of a new face of degree k + 2 if k ≥ −1, that is case (i) in Figure , and it corresponds to case (ii) in Figure if k ≤ −2: the selected edge A(e n ) is identi ed to another one on the boundary, and 'swallows a bubble' of length −(k + 2), either to the left or to the right, equally likely.

. Proof of Theorem
We now prove Theorem , that is: for any peeling process, we have P(R ∈ ∂e n ) ≤ n −2c/3+o(1) and we will compute the value c in the next section. We rst consider the case of the UIPT which corresponds to M ∞ with q = (432 −1/4 1 k=3 ) k ≥1 , for which ν is supported by {. . . , −2, −1, 0, 1}.
Proof of Theorem for the UIPT. First observe that on the event {R ∈ ∂e n }, for each 0 ≤ k ≤ n − 1, if we identify the peel edge A(e k ) to another one on the boundary and swallow a bubble longer than 1 2 |∂e k |, then this identi cation must be on the correct side, so that the root-edge does not belong to this bubble. If we put P k = |∂e k | and ∆P k = P k+1 − P k , then it follows from the Markov property that We rst claim that appart from an event whose probability is op(n) i.e. decays faster than any polynomial, the process P reaches values of order n 2/3+o(1) in the rst n steps: From the convergence of the process (P n ) n ≥0 in Lemma , we see that P n is of order n 2/3 ; in particular, there exists η ∈ (0, 1) such that for every N large enough, for every integer z ∈ (0, N 2/3 ), we have P z (sup 0≤k ≤N P k > N 2/3 ) > η. Then x ε > 0; by splitting the interval [0, n] in n 3ε /2 sub-intervals of length n 1−3ε /2 , we deduce from the Markov property that for every n large enough, P sup 0≤k ≤n We then decompose the process into scales: for every i ≥ 0, let us put τ i = inf {k ≥ 0 : P k ≥ 2 i }; observe that since the jumps of P are bounded by one, then P τ i = 2 i . Using the last display and the strong Markov property we deduce that E 2 −#{0≤k ≤n−1:∆P k <−P k /2} ≤ log 2 (n 2/3−ε ) i=0 E 2 i 2 −#{0≤k <τ i +1 :∆P k <−P k /2} + op(n).
If θ (z) = inf {t ≥ 0 : ϒ ↑ t ≥ z} is the rst passage time above level z > 0, the convergence of the process (P n ) n ≥0 in Lemma then implies that where the last line de nes the constant c. Putting back the pieces together, we deduce by Cesàro summation that for n large enough we have P(R ∈ ∂e n ) ≤ e −c ln(2) log 2 (n 2/3−ε )+o(1) = n −2c/3+o (1) and the proof in the case of the UIPT is complete.
For more general maps, the perimeter process does not increase only by one, so it is not true that |e τ i | = 2 i a.s. and the scales are not independent. Nonetheless, if the degrees are uniformly bounded, say by D < ∞, then |e τ i | ∈ {2 i , . . . , 2 i + D} and the scales do become independent at the limit: one can still use the strong Markov property to get the bound and sup 2 i ≤z ≤2 i +D E z [2 −#{0≤k <τ i +1 :∆P k <−P k /2} ] converges as previously to e −c ln 2 .
Remark: The factor 2/3 coming from the lower bound sup 0≤k ≤n P k ≥ n 2/3−ε with high probability cannot be improved. Indeed it is easy to check in our case of increments bounded above that P(sup 0≤k ≤n P k ≤ n 2/3+ε ) = op(n). This shows that the exponent in Theorem is optimal for the worst algorithm which always peels at the opposite of the root-edge on ∂e n . .

Calculation of c via Lamperti representation
We now characterize the value c de ned in ( ) using the Lamperti representation of ϒ ↑ . We start with a simple calculation on general Lévy processes.
Let ξ = (ξ t ) t ≥0 be a Lévy processes with drift a ∈ R, no Brownian part, and Lévy measure Π(d ) which we assume possesses no atom and is supported on R − so ξ makes only negative jumps (see e.g. Bertoin's book [ , Chapter VII]); assume also that ξ does not drift towards −∞. We can consider the Laplace transform of ξ and de ne its characteristic exponent via the Lévy-Khintchine representation as Recall that θ (z) = inf {t ≥ 0 : ξ t ≥ z} is the rst passage time above level z > 0; since ξ does not drift towards −∞ and does not make positive jumps, we have θ (z) < ∞ and ξ θ (z) = z almost surely.
Lemma . Let F : R − → R − be a function which is identically 0 in a neighborhood of 0. If we note ∆ξ t = ξ t − ξ t − ≤ 0 for the value of the jump at time t, then for every z ≥ 0 we have Proof. For c > 0 we consider the positive càdlàg process Clearly log Z c (t) has stationary and independent increments. We will choose c so that Z c is a martingale and for this it is su cient to tune c so that E[Z c (t)] = 1. Appealing to the Lévy-Itō decomposition and the exponential formula for Poisson random measure (see e.g. [ ], Chapter . and Chapter , Theorem and its proof), and using the Lévy-Khintchine representation of ψ (·), we get Hence if we pick c = c F satisfying the assumption of the lemma then Z c is a positive martingale. Furthermore by our assumptions Z c (t ∧ θ (z)) is bounded by e cz so we can apply the optional sampling theorem and get the statement of the lemma (using also that ξ θ (z) = z).
Let us apply this result to a well-chosen Lévy process. Let (ϒ ↑ t ) t ≥0 be the 3/2-stable Lévy process with no negative jumps conditioned to stay positive and started from ϒ ↑ 0 = 1. The constant c de ned in ( ) satis es e −c ln 2 = E[2 −N ] where Note that, as mentioned in the introduction of this section, this quantity is scale-invariant so it does not depend on the choice of the normalization of ϒ ↑ . The process ϒ ↑ is a so-called positive self-similar Markov process, and by the Lamperti representation it can be represented as the (time-changed) exponential of a Lévy process: for every t ≥ 0, , where the random time change τ t will not be relevant in what follows. In particular, through this representation, for every u ∈ (0, 1), the random variable u N is transformed into the variable exp t ≤θ (ln 2) ln u · 1 {∆ξ t <− ln 2} for the associated Lévy process ξ . This is of the form of the previous lemma with F : x → ln u · 1 {x <− ln 2} . Caballero and Chaumont [ , Corollary ] have computed explicitly the characteristics of ξ ; in particular ξ drifts towards +∞ and has a Lévy measure given by Furthermore in our spectrally negative case, its Laplace exponent has been computed in Chaumont, Kyprianou and Pardo [ , just before Lemma ]: for every λ ≥ 0, To compute the value of E[ξ 1 ] we can use the asymptotic behaviour as λ → ∞: It is easy to see that in the right-hand side of the last display, when λ → ∞, the only contribution which can be of the order of λ 3/2 must appear in the vicinity of 0, hence we can replace the Lévy measure by its equivalent as → 0 and the last display becomes where the last asymptotical equivalence is a standard calculation. Hence we deduce that E[ξ 1 ] = 2π 3 . We can thus apply the last lemma and conclude that for every u ∈ (0, 1), where c u satis es 2π 3 Performing the change of variable x = e in the second integral and using the representation Γ(a)Γ(b) Γ(a+b) = ∫ 1 0 x a−1 (1 − x) b−1 dx gives us the equivalent de nition of c u by an integral equation: The constant c de ned in ( ) and which appears in Theorem corresponds to u = 1/2, and it can easily be evaluated numerically to nd that c = c 1/2 ≈ 0.1283123514178324542...

Pioneer edges and sub-di usivity
In this section we suppose that q = ( 1 12 1 k =4 ) k ≥1 so that M ∞ is the UIPQ although that there is little doubt that everything can be extended to the more general context of generic Boltzmann maps. We rst recast the main result of [ ] in the case of Budd's peeling process. We denote by ρ the origin of the root-edge.
Proposition (Depth of a peeling process). Let (e n ) n ≥0 be any ( lled-in Markovian) peeling process of the UIPQ with origin ρ, then the sequence n −1/3 max{d gr (ρ, x); x ∈ e n } n ≥1 is tight.
Sketch of proof. We follow [ , Section . ]. Throughout the proof, if (Y n ) n ≥0 and (Z n ) n ≥0 are two positive processes, we shall write Y n Z n if (Y n /Z n ) n ≥0 is tight. Also Y n Z n if Z n Y n , and nally Y n Z n if we have both Y n Z n and Y n Z n .
For every n ≥ 0, let D − n and D + n be respectively the minimal and the maximal distance to the origin ρ of a vertex in ∂e n so that where Ball(M ∞ , r ) denotes the hull of the ball in M ∞ of radius r centered at the origin of M ∞ . Using |e n | n 4/3 (see [ ]) and |Ball(M ∞ , r )| r 4 (see [ ]) the two inclusions of ( ) give D − n n 1/3 and D + n n 1/3 . We will rst show that D + n n 1/3 or simply D + n − D − n n 1/3 since we already know D − n n 1/3 . But as in [ , Section . ] we can bound D + n − D − n by the aperture of M ∞ \e n which is the maximal graph distance between points on its (general) boundary. Since conditionally on e n the later is a UIPQ with a boundary of perimeter |∂e n | we can use [ , Section . ] to deduce that its aperture is |∂e n |. Finally since |∂e n | n 2/3 by [ ] we indeed deduce that D + n − D − n n 1/3 . The end of the proof is then the same as [ , Section . ] and use the fact that the hull of the ball of radius r in the UIPQ has no long tentacles of length r .
To prove Proposition we shall use a particular peeling algorithm, introduced in [ ] that we adapt to the peeling of [ ] and the walk on the faces, see also [ , Chapter ]. We let our walk start from the root-face on the right of the root-edge of M ∞ . Every time the walk wants to cross an edge on the boundary of the explored region, we peel this edge. We then de ne the pioneer edges of the walk as the peel edges. We let π n denote the number of pioneer edges amongst the rst n steps of the walk.

Figure :
Le : A simple random walk on the dual map, started at the face on the right-hand side of the root-edge, and about to reach a pioneer edge. Right: the same scenery in reverse time where the root-edge is not swallowed a er n steps of walk.
Proof of Proposition . We use the reversibility trick [ , Lemma ]. Let ( ì E i ) i ≥1 be the edges visited by the walk. Fix n ≥ 1, since the UIPQ is a stationary and reversible random graph (see [ ]) we can reverse the rst n steps of the path and deduce that the probability that ì E n is pioneer equals the probability that (one side of) the root-edge R of M ∞ is still on the boundary of e π n after performing n steps of random walk, see Fig. . We then split this probability as follows: P( ì E n is pioneer) = P R ∈ ∂e π n ≤ P (π n ≤ n α ) + P (R ∈ ∂e n α ) , for some small α > 0 that we tune later on. According to Theorem , the second term is bounded by n −2cα /3+o(1) . Concerning the rst term, if π n ≤ n α then the walk has been con ned in Ball(M ∞ , 4n α ) for the rst n steps. However we claim that the probability that the walk stays con ned in a given nite region G for a long time is bounded as follows: