Anatomy of a gaussian giant: supercritical level-sets of the free field on random regular graphs
AAnatomy of a gaussian giant: supercritical level-sets of the freefield on random regular graphs
Guillaume Conchon--Kerjan ∗ February 23, 2021
Abstract
In this paper, we study the level-set of the zero-average Gaussian Free Field on a uniformrandom d -regular graph above an arbitrary level h ∈ ( −∞ , h (cid:63) ), where h (cid:63) is the level-setpercolation threshold of the GFF on the d -regular tree T d . We prove that w.h.p as thenumber n of vertices diverges, the GFF has a unique giant connected component C ( n )1 of size η ( h ) n + o ( n ), where η ( h ) is the probability that the root percolates in the correspondingGFF level-set on T d . This gives a positive answer to the conjecture of [2] for most regulargraphs. We also prove that the second largest component has size Θ(log n ).Moreover, we show that C ( n )1 shares the following similarities with the giant component ofthe supercritical Erd˝os-R´enyi random graph. First, the diameter and the typical distancebetween vertices are Θ(log n ). Second, the 2-core and the kernel encompass a given positiveproportion of the vertices. Third, the local structure is a branching process conditioned tosurvive, namely the level-set percolation cluster of the root in T d (in the Erd˝os-R´enyi case,it is known to be a Galton-Watson tree with a Poisson distribution for the offspring). The Gaussian Free Field (GFF) on a transient graph G is a gaussian process indexed by thevertices. Its covariance is given by the Green function, hence it carries a lot of information onthe structure of G and on the behaviour of random walks, giving a base motivation for its study.Level-set percolation of the GFF has been investigated since the 1980s ([9, 18]). Lately, oneimportant incentive has been to gain information on the vacant set of random interlacements([17, 25]), via Dynkin-type isomorphism theorems ([13, 22]). It was subject to much attentionin the last decade on Z d ([11, 17, 20, 24]). On such a lattice where the Green function decayspolynomially with the distance between vertices, it provides a percolation model with long-rangeinteractions. ∗ LPSM UMR 8001, Universit´e de Paris (ex-Diderot), e-mail: gconchon[at]lpsm.paris a r X i v : . [ m a t h . P R ] F e b ore recently, level-set percolation was studied on transient rooted trees ([1, 4, 26]). There is aphase transition at a critical threshold h (cid:63) ∈ R : if h < h (cid:63) , the connected component of the rootin the level-set above h of the GFF has a positive probability to be infinite, and if h > h (cid:63) , thisprobability is zero.One can define an analogous field on a finite connected graph, the zero-average GaussianFree Field , whose covariance is given by the zero-average Green function (see Section 1.2).A natural question is whether some characteristics of the GFF on an infinite graph G can betransferred to a sequence of finite graphs ( G n ) n ≥ whose local limit is G . For instance, one mightask whether a phase transition for the existence of an infinite connected component of the level-set in G corresponds to a phase transition for the emergence of a ”macroscopic” component ofsize Θ( |G n | ) in the level set in G n . For G = Z d , Ab¨acherli [3] studied the zero-average GFF onthe torus.Ab¨acherli and Cerny recently investigated the GFF on the d -regular tree T d [1], and the zero-average GFF on some d -regular graphs (large girth expanders) in a companion paper [2]. Inthis setting, many essential questions (such as the value of h (cid:63) , or the sharpness of the phasetransition at h (cid:63) for the zero-average GFF) remain open. In this paper, we answer some ofthem, and relate the percolation level-sets to other classical random graphs, in particular theErd˝os-R´enyi model (Section 1.4).To do so, we refine some properties of [1] on T d . In a work in progress [16], we study furtherthe GFF on T d and the random walk on the level-sets. In all this work, we fix an integer d ≥
3. We denote T d the infinite d -regular tree rooted at anarbitrary vertex ◦ , and G n a uniform d -regular random graph for n ≥ d is odd, consideronly even n ). Let V n be its vertex set and π n be the uniform measure on V n , i.e. π n ( x ) = 1 /n for every x ∈ V n . Gaussian Free Field on regular trees
The GFF ϕ T d on T d is a centred gaussian field ( ϕ T d ( x )) x ∈ T d indexed by the vertices of T d , andwith covariances given by the Green function G T d : for all x, y ∈ T d , Cov( ϕ T d ( x ) , ϕ T d ( y )) = G T d ( x, y ). Recall that G T d ( x, y ) = E T d x (cid:88) k ≥ { X k = y } where ( X k ) k ≥ is a discrete-time SRW (Simple Random Walk) on T d . In general, we will denote P G µ the law of a SRW on a graph G with initial distribution µ . Gaussian Free Field on finite graphs If G n is connected, the zero-average GFF ψ G n on G n is a centred gaussian field ( ψ G n ( x )) x ∈G n G n , and with covariances given by the zero-average Green function G G n on G n : for all x, y ∈ G n ,Cov( ψ G n ( x ) , ψ G n ( y )) = G G n ( x, y ) := E G n x (cid:20)(cid:90) + ∞ (cid:18) { X t = y } − n (cid:19) dt (cid:21) where ( X t ) t ≥ is a continuous time SRW on G n started at x , defined as follows. Let ( ζ i ) i ≥ be asequence of independent exponential variables of parameter 1. Let ( X k ) k ≥ be a SRW startedat x , independent of ( ξ i ) i ≥ . Then for all t ≥ X t := X k ( t ) , with k ( t ) := sup k ≥ (cid:80) ki =1 ζ i ≤ t . G G n is symmetric, finite and positive semidefinite. This ensures that ψ G n is well-defined (see [3]for details, in particular Remark 1.2). Two layers of randomness
Denote P ann and E ann the annealed law and expectation for the joint realization of G n and of ψ G n on it. For a fixed realization of G n , denote P G n and E G n the quenched law and expectation. Define the level set E ≥ hϕ T d := { x ∈ T d | ϕ T d ( x ) ≥ h } . Let C h ◦ be the connected component of E ≥ hϕ T d containing the root ◦ . Similarly, define the level sets E ≥ hψ G n := { x ∈ G n | ψ G n ( x ) ≥ h } for n ≥ i ≥
1, let C ( n ) i be the i -th largest connected component of E ≥ hψ G n . In [1] (Theorems 4.3 and5.1), Ab¨acherli and ˇCern´y showed that there exists a constant h (cid:63) such thatif h > h (cid:63) , η ( h ) := P T d ( |C h ◦ | = + ∞ ) = 0, and if h < h (cid:63) , η ( h ) >
0. (1)In [2] (Theorems 3.1 and 4.1), they proved that P ann -w.h.p.: if h > h (cid:63) , |C ( n )1 | = O (log n ), andif h < h (cid:63) , at least ξn vertices of E ≥ hψ G n are in components of size at least n δ , for some constants δ, ξ > h . They even found deterministic conditions on G n , satisfied w.h.p., sothat these events hold P G n -w.h.p. (see the discussion in Section 1.4).Thus, in the supercritical case h < h (cid:63) , a positive proportion of the vertices is in at least”mesoscopic” components (there is no explicit lower bound for δ ). In this paper, we prove theexistence of a giant component: Theorem 1.
Let h < h (cid:63) . |C ( n )1 | n P ann −→ η ( h ) , (2) where P ann −→ stands for convergence in P ann -probability as n → + ∞ . Moreover, there exists K > such that P ann (cid:16) K − log n ≤ |C ( n )2 | ≤ K log n (cid:17) −→ n → + ∞ . (3)Note that by Markov’s inequality, for any ε > E n ) n ≥ such that P ann ( E n ) →
1, w.h.p. G n is such that P G n ( E n ) ≥ − ε . Thus, w.h.p. on G n , the conclusions ofTheorem 1 hold with arbitrarily large P G n -probability.3e also establish some structural properties of C ( n )1 . Let C ( n ) be the of C ( n )1 , obtainedby deleting recursively the vertices of degree 1 of C ( n )1 and their edges. Let K ( n ) be the kernel of C ( n )1 , i.e. C ( n ) where simple paths are contracted to a single edge, so that the vertices of K ( n ) are those of C ( n ) with degree at least 3. Theorem 2. Global structure of C ( n )1 Fix h < h (cid:63) . There exist K , K > such that | C ( n ) | n P ann −→ K (4) and | K ( n ) | n P ann −→ K . (5) Moreover, there exists K > such that if D ( n )1 is the diameter of C ( n )1 , then P ann ( D ( n )1 ≤ K log n ) −→ n → + ∞ . (6) Last, there exists λ h > such that for every ε > , π ,n ( { ( x, y ) ∈ ( C ( n )1 ) , (1 − ε ) log λ h n ≤ d C [ n )1 ( x, y ) ≤ (1 + ε ) log λ h n } ) P ann −→ , (7) where π ,n is the uniform measure on ( C ( n )1 ) and d C ( n )1 the usual graph distance on C ( n )1 . Inother words, the typical distance between vertices of C ( n )1 is log λ h n . We will see in Section 3 that λ h is the growth rate of C h ◦ conditioned on being infinite.Say that a random graph G is the local limit of the random graph sequence ( G n ) n ≥ if G n converges to G in distribution w.r.t to the local topology (see for instance the lecture notes ofCurien [19] for a precise definition). Theorem 3. Local limit of C ( n )1 The local limit of C ( n )1 is C h ◦ conditioned to be infinite: for every radius k ≥ , for every rooted tree T of height k , let V ( T ) n := { x ∈ C ( n )1 , B C ( n )1 ( x, k ) = T } and p T := P T d ( B C h ◦ ( ◦ , k ) = T | |C h ◦ | = + ∞ ) .Then | V ( T ) n ||C ( n )1 | P ann −→ p T . GFF percolation versus bond percolation E ≥ hψ G n undergoes the same phase transition as some classical bond percolation models for thesize of the largest connected component. We draw a comparison with the Erd˝os-R´enyi randomgraph (i.e. bond percolation on the complete graph), introduced by Gilbert in [15]: for a con-stant p > n ∈ N , ER( n, p/n ) is the graph on n vertices such that for every pair of vertices4 , y , there is an edge between x and y with probability p/n , independently of all other pairs ofvertices. Erd˝os and R´enyi [14] showed that the supercritical regime corresponds to p > p <
1. Theorems 1, 2 and 3 hold for ER( n, p/n ) as n → + ∞ , for anyfixed p > C h ◦ being replaced by a Galton-Watson tree whose offspring distribution is Poissonwith parameter p , and λ h being replaced by p .As for Bernoulli bond percolation on G n (each edge of G n is deleted with probability 1 − q , in-dependently of the others), the same phase transition holds for the size of the largest connectedcomponent, the critical threshold being q = 1 / ( d −
1) (Theorem 3.2 of [7]).
The structure of C ( n )1 It was shown recently in [10] that the distribution of the giant component of ER( n, p/n ) is con-tinuous w.r.t. to a random graph which can be explicitly described. Its kernel is a configurationmodel whose vertices have i.i.d. degrees with a Poisson distribution (conditioned on being atleast 3). In particular, it is an expander. The lengths of the simple paths in the 2-core are i.i.d.geometric random variables. See Theorem 1 of [10] for details. This clearly implies Theorem 2for ER( n, p/n ).We conjecture that the kernel K ( n ) is an expander for every h < h (cid:63) . The main obstacle togathering information on its global structure is that if ψ G n is revealed on a positive proportionof the vertices of K ( n ) (and hence of G n ), then it could affect substantially ψ G n on the remainingvertices. In particular, if h > ψ G n on thediscovered vertices is positive. But by (18), the average of the GFF on the remaining verticeswould be negative, hence below the threshold h . Deterministic regular graphs
The results of [2] and [7] hold in fact for any deterministic sequence of large-girth expanders(conditions (I) and (II) in Proposition 1), which is w.h.p. the case for G n .Can we find deterministic conditions, holding w.h.p. for G n , so that for any sequence of d -regular graphs ( G n ) n ≥ satisfying them, Theorem 1 holds? In our proofs, averaging on therandomness of G n is a crucial ingredient to control the presence of cycles on large subgraphsof G n , and allows us to extend some arguments of [2], where ψ G n is locally approximated by ϕ T d . Behaviour at criticality
Almost nothing is known about C h (cid:63) ◦ (even the value of h (cid:63) ), though one might conjecture thatit is a.s. finite, as critical Galton-Watson trees. In the Erd˝os-R´enyi model, the critical case isby far the most interesting. When p = 1 + Θ( n − / ), the i -th largest connected component hassize ς i n / , where ς i is an a.s. finite random variable (see the celebrated paper by Aldous [6]).Its structure is similar to a modified Brownian tree [5]. Hence, there is a finite but arbitrarylarge number of components of the biggest order, and some of their main characteristics, such5s their cardinality, are random.It would be interesting to look for the size and shape of |C ( n ) i | for h = h (cid:63) , or for an hypotheticalcritical window ( h (cid:63) − ε n , h (cid:63) + ε n ) for a sequence ( ε n ) n ≥ converging to 0 at an appropriate speed. Our proofs rely on two main arguments:1) An annealed exploration of E ≥ hψ G n (Proposition 5), where the structure of G n is progressivelyrevealed (there is a standard sequential construction of G n , see Section 2). Each newlydiscovered vertex is given an independent standard normal variable. Then ψ G n is builtvia a recursive procedure, using these gaussian variables (Proposition 3).2) A comparison of ψ G n and ϕ T d (Proposition 15) : on a tree-like subgraph T of G n , suchthat there are no cycles in G n at distance κ log log n of T for a large enough constant κ ,there is a bijective map Φ between T and an isomorphic subtree of T d and a coupling of ψ G n and ϕ T d so that sup y ∈ T | ψ G n ( y ) − ϕ T d (Φ( y )) | ≤ log − n. We stress the fact that we reveal ψ G n only after having explored G n : if we reveal ψ G n at a givenvertex, it conditions the structure of G n and thus the pairings of the still unmatched half-edges,so that we cannot use the sequential construction any more to further explore the graph. Hence,during the exploration, we will need to build an approximate version of ψ G n , depending on thegaussian variables of 1). This makes some proofs tedious, in particular that of (3). The base exploration
The exploration that we will perform in all proofs, with some modifications, is as follows: pick x ∈ V n , and reveal its connected component C G n ,hx in E ≥ hψ G n in a breadth-first way, as well as itsneighbourhood up to distance a n = κ log log n . Until we meet a cycle, the explored zone is atree T x , growing at least like C h +log − n ◦ , and at most like C h − log − n ◦ by 2).On one hand, C h +log − n ◦ has a probability (cid:39) η ( h + log − n ) = η ( h ) + o (1) to be infinite, with agrowth rate λ h > o (1)as long as we reveal o ( √ n ) vertices (since we perform o ( √ n ) pairings of half-edges having eacha probability o ( √ n ) /n to involve two already discovered vertices). Thus, with P ann -probability η ( h ) + o (1), T x and ∂T x will reach a size Θ( √ n log − κ (cid:48) n ) for some constant κ (cid:48) > C h − log − n ◦ has a probability 1 − η ( h − log − n ) = 1 − η ( h ) + o (1) to be finite, andwith P ann -probability 1 − η ( h ) + o (1), |C G n ,hx | = o ( √ n ) (Proposition 19). Proof of (2) . 6e show that for any two vertices x, y ∈ V n , there is a P ann -probability η ( h ) + o (1) that theyare connected in E ≥ hψ G n . To do so, we explore the connected components of x and y , that wecouple with independent copies of C h +log − n ◦ , so that with probability η ( h ) + o (1), the explo-rations from x and y are disjoint, and ∂T x and ∂T y have Θ( √ n log − κ (cid:48) n ) vertices. Next, wedraw multiple paths between T x and T y , with an ”envelope” of radius Θ(log log n ) around eachof them, the joining balls (Section 6.1). The probability that E ≥ hψ G n percolates through at leastone joining ball is 1 − o (1).Then, by a second moment argument, we prove that P ann -w.h.p., the number of couples( x, y ) ∈ V n such that y ∈ C G n ,hx is ( η ( h ) + o (1)) n (Lemma 23).In the same way, from the fact that |C G n ,hx | = o ( √ n ) with P ann -probability 1 − η ( h ) + o (1),we deduce that at least (1 − η ( h ) + o (1)) n vertices are in connected components of size o ( √ n )(Lemma 22).Those two facts together force the existence of a connected component of size ( η ( h ) + o (1)) n . Proof of (3).
The most difficult part is the upper bound. We show that for K large enough, for x ∈ V n , P ann ( K log n ≤ |C G n ,hx | ≤ K − n ) = o (1 /n ), and conclude by a union bound on x and a corollaryof the proof of (2), namely that |C ( n )2 | /n P ann −→ o (1 /n ) requires three additional ingredients: • the size of C h ◦ conditioned on being finite has exponential moments (Proposition 11), inparticular, P T d ( |C h ◦ | ≥ c log n, |C h ◦ | < + ∞ ) = o (1 /n ) for a large enough constant c , • when exploring k vertices around x , there is a probability Θ( k /n ) that a cycle arises, sothat we will need to handle at least one cycle to fully explore C G n ,hx , • we need a better approximation of ψ G n than log − n in 2): with probability at leastΘ(1 /n ), we will meet too many vertices with an approximate value of ψ G n that are in[ h − log − n, h + log − n ], so that we can not tell whether they are in C G n ,hx or not beforethe end of the exploration. To remedy this, we replace the ”security radius” a n in 2) bysome r n = Θ(log n ), so that we approximate ψ G n up to a difference n − Θ(1) . Other proofs .The proofs of Theorems 2 and 3 are based on slightly modified explorations, and are muchsimpler.
In Section 2, we review some basic properties of G n (structure, Green function and GFF). InSection 3, we study the GFF on T d . In Section 4, we establish a coupling between recursiveconstructions of the GFF on T d and on a tree-like neighbourhood of G n . In Section 5, we explore7he connected component of a vertex in E ≥ hψ G n . In Section 6, we prove (2). In Section 7, weprove (3). In Section 8, we prove Theorems 2 and 3. In this paper, edges are non-oriented, and thus graphs are undirected. For any graph G , de-note d G the standard graph distance on its vertex set V , and for every vertex x and R ≥ B G ( x, R ) := { y, d G ( x, y ) ≤ R } and ∂B G ( x, R + 1) = B G ( x, R + 1) \ B G ( x, R ). For any S ⊆ V ,let similarly B G ( S, R ) := ∪ x ∈ S B G ( x, R ) and ∂B G ( S, R + 1) = B G ( S, R + 1) \ B G ( S, R ). If x and y are neighbours, we denote B G ( x, y, R ) the subgraph of G obtained by taking all paths oflength R starting at x and not going through y .The tree excess of G is tx( G ) = v − e + s , where v := | V | , e is the number of edges in G and s the number of connected components. Note that tx( G ) = 0 if and only if all components of G are trees.A rooted tree is a tree T with a distinguished vertex ◦ , the root . The height h T ( x ) of avertex x in T is d T ( ◦ , x ). If T is finite, its boundary ∂T is the set of vertices of maximal height.The offspring of x is the set of vertices y such that x is on any path from ◦ to y . For r ≥ r -offspring of x is its offspring at distance r of x , and its offspring up to generation r is its offspring at distance at most r . If y is in the 1-offspring of x , then y is a child of x , and x is its parent . In this case, write x = y .If x, y are neighbours in T , the cone from x out of y is the rooted subtree of T with root x and vertex set { z ∈ T | y is not on the shortest path from x to z } .An isomorphism between two rooted trees T and T (cid:48) is a bijection Φ : T → T (cid:48) preserving theroot and the height, and such that for all vertices x, y ∈ T , there is an edge between x and y ifand only if there is an edge between Φ( x ) and Φ( y ).Unless mention of the contrary, all random walks are in discrete time. We will write T A (resp. H A ) for the first exit (resp. hitting) time of A by a SRW.For two probability distributions µ, µ (cid:48) on R , we write µ ≤ st. µ (cid:48) (or µ (cid:48) ≥ st. µ ) if µ (cid:48) dominatesstochastically µ , i.e. there exist two random variables X ∼ µ and X (cid:48) ∼ µ (cid:48) on the same proba-bility space such that X ≤ X (cid:48) a.s. G n G n can be generated sequentially as follows: attach d half-edges to each vertex of V n . Pick anarbitrary half-edge, and match it to another half-edge chosen uniformly at random. Choose a8emaining half-edge and match it to another unpaired half-edge chosen uniformly and indepen-dently of the previous matching, and so on until all half-edges have been paired. The resultingmulti-graph M n is not necessary simple , i.e. it might have loops and multiple edges. The prob-ability that M n is simple has a positive limit as n → + ∞ , and conditionally on {M n is simple } , M n is distributed as G n (see for instance Section 7 of [27] for a reference).In particular, an event true w.h.p. on M n is also true w.h.p. on G n , so that it is enough toprove all our results on M n . In the rest of the paper, we will even write G n for M n for the sakeof simplicity.This Section is devoted to proving this result: Proposition 1.
There exists K > such that w.h.p. as n → + ∞ , G n satisfies:(I) G n is a K -expander , i.e. the spectral gap λ G n of G n is at least K (the spectral gap is thesmallest eigenvalue of I − P where I is the identity matrix and P the transition matrix ofthe SRW on G n ),(II) for all x ∈ G n , B G n ( x, (cid:98) K log n (cid:99) ) contains at most one cycle.Moreover, there exists K > such that w.h.p. on G n , it holds: for all x ∈ V n such that tx( B G n ( x, (cid:98) K log log n (cid:99) )) = 0 , (cid:12)(cid:12)(cid:12)(cid:12) G G n ( x, x ) − d − d − (cid:12)(cid:12)(cid:12)(cid:12) ≤ log − n. (8) If moreover y is a neighbour of x , (cid:12)(cid:12)(cid:12)(cid:12) G G n ( x, y ) − d − (cid:12)(cid:12)(cid:12)(cid:12) ≤ log − n. (9)Say that a given realization of G n is a good graph when (I), (II), (11), (8) and (9) hold. (8) and(9) illustrate the fact that G G n is close to G T d on a tree-like neighbourhood: it is well-knownthat for all x, y ∈ T d , G T d ( x, y ) = ( d − − d T d ( x,y ) d − . (10)A quick computation can be found in [28], Lemma 1.24.By Proposition 1.1 of [2], (I) and (II) imply that for some K , K > n large enough,for all x, y ∈ V n , | G G n ( x, y ) | ≤ K ( d − d G n ( x,y ) ∨ n − K . (11)Throughout this paper, we will often make binomial estimations, because the number of edgesbetween two sets of vertices in G n is close to a binomial random variable, as highlighted in theLemma below. We will use repeatedly the following classical inequalities: for n ≥ m ≥ p ∈ (0 , Z ∼ Bin( n, p ): P ( Z ≥ m ) ≤ (cid:18) nm (cid:19) p m , P ( Z ≤ m ) ≤ (cid:18) nm (cid:19) (1 − p ) m , (cid:18) nm (cid:19) ≤ n m m ! ≤ n m . (12)The following Lemma is an important consequence of the sequential construction of G n .9 emma 2 ( Binomial number of connections ) . Let m ∈ N , let V , V be disjoint subsets of V n . Write m := | V | and m := | V | . Suppose the only information we have on G n is a set E ofits edges that has been revealed. Let m E := | E | and denote P E the law of G n conditionally on thisinformation. Repeat the following operation m times: pick an arbitrary vertex v ∈ V having atleast one unmatched half-edge, and pair it with an other half-edge. Add its other endpoint v (cid:48) in V , if it was not already in it. Let s be the number of times that v (cid:48) ∈ V . Then s ≤ st. Bin (cid:18) m, m n − ( m E + m ) (cid:19) . (13) In particular,a) for fixed k ∈ N there exists C ( k ) > so that for n large enough, if m E + m + m + m < n/ , P E ( s ≥ k ) ≤ C ( k ) (cid:16) m mn (cid:17) k . (14) b) for k = k ( n ) → + ∞ and n large enough, if we have m E + m + m + m < n/ and kn > m + m + m E ) m , then P E ( s ≥ k ) ≤ . k . (15) Proof.
Pick v ∈ V , such that v has an unmatched half-edge e . There are at most m verticesin V , so that there are at most dm unmatched half-edges that belong to its vertices. And thetotal number of unmatched half-edges is at least dn − | E | + m ) ≥ d ( n − m E − m ). Thus,the probability that e is matched with a half-edge belonging to a vertex of V is not greaterthan dm d ( n − m E − m ) = m n − ( m E + m ) . The successive matchings are performed independently, and (13)follows.Let Z ∼ Bin (cid:16) m, m n − ( m E + m ) (cid:17) . By (12), for k ∈ N : P E ( Z ≥ k ) ≤ (cid:0) mk (cid:1) (cid:16) m n − ( m E + m ) (cid:17) k ≤ (cid:0) mk (cid:1) (cid:16) m n/ (cid:17) k ≤ k k ! m k m k n k .This yields (14). Moreover, if k → + ∞ as n → + ∞ and kn > m + m + m E ) m , by Stirling’sformula, we have that for n large enough: P E ( Z ≥ k ) ≤ (cid:16) (2 e +0 . m mkn (cid:17) k < . k , and (15)follows.It is straightforward to adapt this to the case V = V . m is replaced by m + m in (13) and(14), and (15) does not change. Throughout this paper, we will refer to these equations withoutdistinction for the cases V = V and V ∩ V = ∅ . Proof of Proposition 1.
By Theorem 1 of [8] and the Cheeger bound, G n satisfies (I) w.h.p.As for (II), fix K >
0. For all x ∈ V n , one obtains B G n ( x, (cid:98) K log n (cid:99) ) by proceeding to at most d ( d − (cid:98) K log n (cid:99) pairings of half-edges. If K is small enough, d ( d − (cid:98) K log n (cid:99) < n / − m = 1, m E = 0, m = d ( d − (cid:98) K log n (cid:99) and k = 2, for n large enough: P (tx( B G n ( x, (cid:98) K log n (cid:99) )) ≥ ≤ C (2) (cid:16) n / n (cid:17) ≤ n − / .
10y a union bound on x ∈ V n , w.h.p. G n is such that for all x ∈ V n , tx( B G n ( x, (cid:98) K log n (cid:99) )) ≤ U := B G n ( x, (cid:98) K log log n (cid:99) ) and W := B T d ( ◦ , (cid:98) K log log n (cid:99) )are isomorphic. Then G U G n ( x, x ) = G W T d ( ◦ , ◦ ), where we let G A G n ( y, z ) := E G n y [ (cid:80) T A k =0 { X k = z } ] for every y, z ∈ V n and A (cid:40) V n .Recall that T A is the exit time of A by the SRW ( X k ) k ≥ . Similarly for every B (cid:40) T d , and y, z ∈ T d , G B T d ( y, z ) := E T d y [ T B (cid:88) k =0 { X k = z } ] . (16)On one hand, by the strong Markov property applied to the exit time T W , G W T d ( ◦ , ◦ ) = G T d ( ◦ , ◦ ) − E T d ◦ [ G T d ( ◦ , X T W )] = d − d − − E T d ◦ [ G T d ( ◦ , X T W )].On the other hand, by Lemma 1.4 of [3], for all y, z ∈ V n and A (cid:40) V n : G A G n ( y, z ) = G G n ( y, z ) − E G n y [ G G n ( z, X T A )] + E G n y [ T A ] n , (17)so that G U G n ( x, x ) = G G n ( x, x ) − E G n x [ G G n ( x, X T U )] + E G nx [ T U ] n . Therefore, (cid:12)(cid:12)(cid:12)(cid:12) G G n ( x, x ) − d − d − (cid:12)(cid:12)(cid:12)(cid:12) ≤ | E T d ◦ [ G T d ( ◦ , X T W )] | + | E G n x [ G G n ( x, X T U )] | + E G n x [ T U ] n . By (10) and (11), if K is large enough, then for large enough n : | E T d ◦ [ G T d ( ◦ , X T W )] | + | E G n x [ G G n ( x, X T U )] | ≤ log − n .Note that T U is stochastically dominated by the hitting time H of (cid:98) K log log n (cid:99) by a SRW( Z k ) k ≥ on Z starting at 0, whose transition probabilities from any vertex are d − d towards theright and 1 /d towards the left. By Markov’s exponential inequality, there exists a constant c > n large enough and every k > n / , P ( H ≥ k ) ≤ P ( Z k ≤ (cid:98) K log log n (cid:99) ) ≤ P ( Z k ≤ ( d − d − / k ) ≤ e − ck .Hence for n large enough, E G n x [ T U ] ≤ E [ H ] ≤ n / + (cid:80) k ≥ n / ke − ck ≤ n / . Thus, (cid:12)(cid:12)(cid:12)(cid:12) G G n ( x, x ) − d − d − (cid:12)(cid:12)(cid:12)(cid:12) ≤ log − n + n − / ≤ log − n for large enough n , and this yields (8). One proves (9) by the same reasoning.11 .2 GFF on G n The name ”zero-average” comes from the fact that a.s., (cid:88) x ∈ V n ψ G n ( x ) = 0 (18)since Var (cid:0)(cid:80) x ∈ V n ψ G n ( x ) (cid:1) = (cid:80) x,y ∈ V n G G n ( x, y ) = 0 . This prevents the existence of a domain Markov property. However, there exists a recursiveconstruction of ψ G n : Proposition 3 (Lemma 2.6 in [2]) . Let A (cid:40) V n , x ∈ V n \ A . Write σ ( A ) := σ ( { ψ G n ( y ) , y ∈ A } ) .Let ( X k ) k ≥ be a SRW on G n and let H A be the hitting time of A . Conditionally on σ ( A ) , ψ G n ( x ) is a gaussian variable, E G n [ ψ G n ( x ) | σ ( A )] = E G n x [ ψ G n ( X H A )] − E G n x [ H A ] E G n π n [ H A ] E G n π n [ ψ G n ( X H A )] (19) and Var G n ( ψ G n ( x ) | σ ( A )) = G G n ( x, x ) − E G n x [ G G n ( x, X H A )] + E G n x [ H A ] E G n π n [ H A ] E G n π n [ G G n ( x, X H A )] . (20) Lemma 4.
Suppose that max x ∈ V n G G n ( x, x ) ≤ K . Then for all ∆ > , if n is large enough, P G n (cid:18) max x ∈ V n | ψ G n ( x ) | ≥ log / n (cid:19) ≤ n − ∆ . (21) In particular, by Proposition 1 and (11), w.h.p. G n satisfies (21).Proof. Let N ∼ N (0 , K ). If n is large enough, then for all x ∈ V n , P G n (cid:16) | ψ G n ( x ) | ≥ log / n (cid:17) ≤ P G n (cid:16) | N | ≥ log / n (cid:17) ≤ (cid:32) − log / n K (cid:33) ≤ n − ∆ − by Markov’s inequality applied to the function u (cid:55)→ exp (cid:16) log / n K u (cid:17) . By a union bound on all x ∈ V n , P G n (cid:16) max x ∈G n | ψ G n ( x ) | > log / n (cid:17) ≤ n − ∆ . Proposition 5 ( Joint realization of G n and ψ G n ) . A realization of ( G n , ψ G n ) is given bythe following process. Let ( ξ i ) i ≥ be a sequence of i.i.d. standard normal variables. A move consists in: • choosing an unpaired half-edge e and matching it to another unpaired half-edge chosenuniformly at random (independently of ( ξ i ) i ≥ ), or • choosing x ∈ V n and k ∈ N such that ξ k has not yet been attributed, and attributing ξ k to x . t each move, the choice of e, x or k might depend in an arbitrary way on the previous moves,i.e. on the matchings and on the value of the normal variables attributed before, but not onthe value of the remaining normal variables. Perform moves until all half-edges are paired, andevery vertex x ∈ V n has received a normal variable, that we denote ξ x . Erase loops and replaceeach multiple edge by a single edge.To generate ψ G n , let x , . . . , x n be the vertices of V n , listed in the order in which they receivedtheir normal variable. Let ψ G n ( x ) := (cid:112) G G n ( x , x ) ξ x . For i = 2 , . . . , n successively, define A i := { x , . . . , x i − } . Recall that we write σ ( A i ) for σ ( { ψ G n ( y ) , y ∈ A i } ) . Let ψ G n ( x i ) := E G n [ ψ G n ( x i ) | σ ( A i )] + ξ x i (cid:112) Var( ψ G n ( x i ) | σ ( A i )) .Proof. Clearly, the graph obtained after pairing all the half-edges is distributed as G n . ξ x is astandard normal variable independent of the realization of G n . Finally, remark that for every i ≥ ξ x i is a standard normal variable, independent of the realization of G n and of σ ( A i ), sothat we can conclude by Proposition 3. T d In Section 3.1, we characterize C h ◦ as a branching process, with a recursive construction (Proposi-tion 6). Then, in Section 3.2, we establish its exponential growth, conditionally on {|C h ◦ | = + ∞} .The main results are Propositions 9, 11 and 13. C h ◦ as a branching process There is an alternative definition of ϕ T d , starting from its value at ◦ and expanding recursivelyto its neighbours. It shows that C h ◦ is an infinite-type branching process, the type of a vertex x being ϕ T d ( x ). Proposition 6 ( Recursive construction of the GFF ,[1]) . Define a gaussian field ϕ on T d as follows: let ( ξ y ) y ∈ T d be a family of i.i.d. N (0 , random variables. Let ϕ ( ◦ ) := (cid:113) d − d − ξ ◦ .For every y ∈ T d \ {◦} , define recursively ϕ ( y ) := (cid:113) dd − ξ y + d − ϕ ( y ) , where y is the parent of y . Then ϕ d. = ϕ T d . Proposition 6 is the corollary of a more general domain Markov property (see for instance Lemma1.2 of [21] where it is stated for Z d , but the proof can readily be adapted to any transient graph).Let G be a transient graph, G G the Green function on it and ϕ G the associated GFF. For U (cid:40) G ,and x, y ∈ G , let G U G ( x, y ) := E G x (cid:34) T U (cid:88) k =0 X k = y (cid:35) . (22)Define the field ϕ U G on G by ϕ U G ( x ) := ϕ G ( x ) − E G x [ ϕ G ( X T U )] for all x ∈ G .13 roposition 7 ( Domain Markov property ) . ϕ U G is a gaussian field, independent from ( ϕ G ( x )) x ∈G\ U , with covariances given by Cov ( φ U G ( x ) , ϕ U G ( y )) = G U G ( x, y ) . We apply it with G = T d and U = T y the subtree from y in T d , for every y ∈ T d \ {◦} , to getProposition 6. See [1], (1.4)-(1.9) for details.Write P T d for the law of ϕ T d , and P T d a for P T d ( · | ϕ T d ( ◦ ) = a ), a ∈ R . This constructiongives a monotonicity property for ϕ T d . A set S ⊂ R T d is said to be increasing if for any(Φ (1) z ) z ∈ T d , (Φ (2) z ) z ∈ T d ∈ R T d such that Φ (1) z ≤ Φ (2) z for all z ∈ T d , (Φ (1) z ) z ∈ T d ∈ S only if(Φ (2) z ) z ∈ T d ∈ S . Say that the event { ϕ T d ∈ S } is increasing if S is increasing. Lemma 8 ( Conditional monotonicity ) . If E is an increasing event, then a (cid:55)→ P T d a ( E ) isnon-decreasing on R .Proof. Let a , a ∈ R such that a > a . It suffices to give a coupling between a GFF ϕ (1) T d conditioned on ϕ (1) T d ( ◦ ) = a and a GFF ϕ (2) T d conditioned on ϕ (2) T d ( ◦ ) = a such that a.s., forevery z ∈ T d , ϕ (1) T d ( z ) ≥ ϕ (2) T d ( z ). To do this, let ( ξ y ) y ∈ T d be i.i.d. standard normal variables,and define recursively ϕ (1) T d and ϕ (2) T d as in Proposition 6. Then for every z ∈ T d of height k ≥ ϕ (1) T d ( z ) = ϕ (2) T d ( z ) + ( a − a )( d − − k . Let Z hk := C h ◦ ∩ ∂B T d ( ◦ , k ) be the k -th generation of C h ◦ . We first prove the following: Proposition 9.
There exists λ h > such that lim k → + ∞ P T d ( |Z hk | > λ kh /k ) = η ( h ) and lim k → + ∞ P T d ( |Z hk | < kλ kh ) = 1 . Moreover, h (cid:55)→ λ h is a decreasing homeomorphism from ( −∞ , h (cid:63) ) to (1 , d − . We will need the following Lemma (whose proof is immediate from Propositions 3.1 and 3.3 of[26] and Proposition 2.1 (ii) of [1]). Let ◦ be an arbitrary neighbour of ◦ . Let T + d be the conefrom ◦ out of ◦ . Write C h, + ◦ := C h ◦ ∩ T + d . For k ≥
1, let Z h, + k := C h, + ◦ ∩ ∂B T + d ( ◦ , k ). Lemma 10.
There exists a function χ h that is continuous with a positive minimum χ min on [ h, + ∞ ) , and that vanishes on ( −∞ , h ) . Moreover, M hk := λ − kh (cid:88) x ∈Z h, + k χ h ( ϕ T d ( x )) is a martingale w.r.t. the filtration F k := σ (cid:16) ϕ T d ( x ) , x ∈ B T + d ( ◦ , k ) (cid:17) , k ≥ , and has an a.s.limit M h ∞ . roof of Proposition 9. We first establish that lim k → + ∞ P T d ( |Z hk | > λ kh /k ) = η ( h ). Clearly,lim sup k → + ∞ P T d ( |Z hk | > λ kh /k ) ≤ P T d ( |C h ◦ | = + ∞ ) = η ( h ) . Reciprocally, denote E + = {|C h, + ◦ | = + ∞} and E + k = {|Z h, + k | ≥ λ kh /k } . By Theorem 4.3 of [1],lim k → + ∞ P T d ( E + k ) = P T d ( E + ) > . (23)Hence, for any ε >
0, for k large enough, P T d ( E + k ) ≥ P T d ( E + ) − ε. (24)Let T − d be the cone from ◦ out of ◦ , C h, −◦ := C h ◦ ∩ T − d , and Z h, − k := C h, −◦ ∩ ∂B T − d ( ◦ , k ) for k ≥
1. Let E − = {|C h, −◦ | = + ∞} and E − k = {|Z h, − k | ≥ λ kh /k } . Define E := { ϕ T d ( ◦ ) ≥ h }∩{ C h, + ◦ is finite } and E k := { ϕ T d ( ◦ ) ≥ h } ∩ { Z h, + k = ∅} . We have P T d ( E − k ∩ E k ) ≥ P T d ( E − ∩ E ) − P T d ( E − ∩ E ∩ ( E k ) c ) − P T d ( E − ∩ E ∩ ( E − k ) c ) . Define M h, −∞ on C h, −◦ as M h ∞ on C h, + ◦ . From the proof of Theorem 4.3 in [1], we get that P T d ( { M h, −∞ > } ∩ ( E − k ) c ) →
0. And, by Proposition 4.2 in [1], P T d ( E − ∩ { M h, −∞ = 0 } ) = 0,so that P T d ( E − ∩ ( E − k ) c ) →
0. Moreover, ( E k ) k ≥ is an increasing sequence of events and E = ∪ k ≥ E k , so that P T d ( E ∩ ( E k ) c ) →
0. Hence, for k large enough, P T d ( E − k ∩ E k ) ≥ P T d ( E − ∩ E ) − ε. (25)Note that {|C h ◦ | = + ∞} = E + (cid:116) ( E − ∩ E ), so that P ( E + ) + P ( E − ∩ E ) = η ( h ) . And for all k ≥ E + k (cid:116) ( E − k ∩ E k ) ⊆ {|Z hk | > λ kh /k } , therefore, if k is large enough, by (24) and (25), one has P T d ( |Z hk | > λ kh /k ) ≥ η ( h ) − ε. Since ε > k → + ∞ P T d ( |Z hk | > λ kh /k ) = η ( h ).Now, we show that P T d ( |Z hk | < kλ kh ) →
1. For all k ≥
1, by definition of M hk , M hk ≥ χ h, min λ − kh (cid:12)(cid:12) Z hk (cid:12)(cid:12) .From the proof of Proposition 3.3 in [26], M h ∞ is a.s. finite. Therefore, k − λ − kh (cid:12)(cid:12) Z hk (cid:12)(cid:12) → P T d ( |Z h, + k | ≥ kλ kh / → . In the same way, P T d ( |Z h, − k − | ≥ kλ kh / →
0. Since Z hk ⊆ Z h, + k ∪ Z h, − k − , we are done.The last part of the Proposition comes directly from Propositions 3.1 and 3.3 in [26].Next, we establish finer results on the growth of C h ◦ . |C h ◦ | has exponential moments: Proposition 11.
There exists a constant K > such that as k → + ∞ , max a ≥ h P T d a ( k ≤ |C h ◦ | < + ∞ ) = o (exp( − K k )) . (26)15ince {Z hk (cid:54) = ∅} ⊂ {|C h ◦ | ≥ k } , we have the following straightforward consequence: Corollary 12.
For k large enough, for every a ≥ h , P T d a ( |C h ◦ | = + ∞ ) ≤ P T d a ( Z hk (cid:54) = ∅ ) ≤ P T d a ( |C h ◦ | = + ∞ ) + e − K k . In addition, there are upper and lower large deviations for the growth rate of Z hk : Proposition 13.
For every ε > , there exists C > such that for every k ∈ N large enough, max a ≥ h P T d a ( k − log |Z hk | (cid:54)∈ [log( λ h − ε ) , log( λ h + ε ) + k − log χ h ( a )] | Z hk (cid:54) = ∅ ) ≤ exp( − Ck ) . (27) This holds when replacing C h ◦ by C h, + ◦ , and Z hk by Z h, + k . A crucial idea to prove Propositions 11 and 13 is to make a finite scaling, in order to get abranching process that is uniformly supercritical w.r.t. to the value of ϕ T d ( ◦ ): Lemma 14.
There exists (cid:96) ∈ N such that for every a ≥ h , E T d a [ |Z h(cid:96) | ] ≥ E T d a [ |Z h, + (cid:96) | ] ≥ E T d h [ |Z h, + (cid:96) | ] > . Proof.
Write E k := {|Z h, + k | ≥ λ kh /k } . By (23), there exists ε > k ≥ ε − , P T d ( E k ) ≥ ε . For a large enough, P T d ( ϕ T d ( ◦ ) ≥ a ) < ε/
2. Note that E k is anincreasing event, so that by Lemma 8, the map a (cid:48) (cid:55)→ P T d a (cid:48) ( E k ) is non-decreasing. Therefore, forevery a (cid:48) ≥ a and k ≥ M , if ν ∼ N (0 , d − d − ) denotes the law of ϕ T d ( ◦ ), P T d a (cid:48) ( E k ) ≥ P T d a ( E k ) ≥ (cid:90) a −∞ P T d b ( E k ) ν ( db ) ≥ P T d ( E k ) − P T d ( ϕ T d ( ◦ ) ≥ a ) ≥ ε/ . From the construction of Proposition 6, it is straightforward that p := P T d h ( ◦ has one child z in C h, + ◦ , and ϕ T d ( z ) ≥ a ) > (cid:96) ∈ N large enough, E T d h [ |Z h, + (cid:96) | ] ≥ pε × λ (cid:96)h (cid:96) > . In addition, for every M ∈ R , {|{Z h, + (cid:96) | ≥ M } is an increasing event. By Lemma 8, for every a ≥ h , E T d a [ |Z h, + (cid:96) | ] ≥ E T d h [ |Z h, + (cid:96) | ]. Since Z h, + (cid:96) ⊆ Z h(cid:96) a.s., the conclusion follows. Proof of Proposition 11.
Fix a ≥ h , and let (cid:96) ∈ N be such that the conclusion of Lemma 14holds. Let F := ∂B C h ◦ ( ◦ , (cid:96) ). For j ≥
1, if F j (cid:54) = ∅ , choose an arbitrary vertex z j ∈ F j . Let O j be the (cid:96) -offspring of z j in C h ◦ and let F j +1 := O j ∪ F j \ { z j } . Thus, we explore C h ◦ by revealingsubtrees of height ≤ (cid:96) , so that at each step, we see at most ( d −
1) + . . . + ( d − (cid:96) ≤ d (cid:96) +1 newvertices. Hence, if |C h ◦ | ≥ k for some k ∈ N , there will be at least (cid:98) k/d (cid:96) +1 (cid:99) steps before C h ◦ isfully explored. 16y Lemma 8 (applied to {|Z h, + (cid:96) | ≥ k } for every k ≥ j ≥ | F j | dominatesstochastically a sum S j of j i.i.d. random variables of law ρ (cid:96),h −
1, where ρ (cid:96),h is the law of |Z h, + (cid:96) | conditionally on ϕ T d ( ◦ ) = h . (28)Therefore, P T d a ( k ≤ |C h ◦ | < + ∞ ) ≤ + ∞ (cid:88) j = (cid:98) k/d (cid:96) +1 (cid:99) P ( S j ≤ . But ρ (cid:96),h is bounded and has a positive expectation by Lemma 14, therefore there exist c, c (cid:48) > P ( S j ≤ ≤ ce − c (cid:48) j for all j ≥
1. Hence, P T d a ( k ≤ |C h ◦ | < + ∞ ) ≤ c + ∞ (cid:88) j = (cid:98) k/d (cid:96) +1 (cid:99) e − c (cid:48) j ≤ c exp( − c (cid:48) (cid:98) k/d (cid:96) +1 (cid:99) )1 − e − c (cid:48) and (26) follows. Proof of Proposition 13.
Let ε >
0. By definition of M hk and Lemma 10, {|Z h, + k | ≥ χ h ( a )( λ h + ε ) k } ⊂ { M hk ≥ λ − kh χ h, min χ h ( a )( λ h + ε ) k } so that by Markov’s inequality, for any a ≥ h , P T d a ( |Z h, + k | ≥ χ h ( a )( λ h + ε ) k ) ≤ P T d a (cid:32) M hk ≥ χ h, min χ h ( a ) (cid:18) λ h + ελ h (cid:19) k (cid:33) ≤ χ − h, min (cid:18) λ h λ h + ε (cid:19) k χ h ( a ) − E T d a [ M hk ] ≤ χ − h, min (cid:18) λ h λ h + ε (cid:19) k . This yields the upper large deviation for Z h, + k (and for Z hk , using the facts that Z hk ⊆ Z h, + k ∪Z h, − k − and that |Z h, − k − | and |Z h, + k − | have the same law).It remains to prove that for some C > k large enough,max a ≥ h P T d a ( k − log |Z hk | ≤ log( λ h − ε ) | Z hk (cid:54) = ∅ ) ≤ exp( − Ck ) . (29)We proceed in two steps. We first light the growth of C h ◦ by showing that if Z (cid:96)n (cid:54) = ∅ , theprobability that |Z (cid:96)n | = o ( n ) decays exponentially with n , where (cid:96) is such that Lemma 14holds. Then, if Z (cid:96)n has at least Θ( n ) vertices, each of them has a positive probability to havea Kn -offspring of size ≥ λ Knh / ( Kn ) ≥ ( λ h − ε ) k with k := ( K + (cid:96) ) n and for a large enoughconstant K , independently of the others vertices. Hence the probability that |Z k | ≤ ( λ h − ε ) k decays exponentially with n , and thus with k . First step.
Recall the exploration of C h ◦ of the proof of Proposition 11, but perform it in abreadth-first way: reveal first the (cid:96) -offspring of ◦ , then the (cid:96) -offspring of each vertex of Z h(cid:96) ,then the (cid:96) -offspring of each vertex of Z h (cid:96) , and so on. For n ≥
1, if Z h(cid:96)n (cid:54) = ∅ , let j + 1 be the first17tep at which we explore the offspring of a vertex of Z (cid:96)n . Note that j ≥ n . As in the proof ofProposition 11, there exist (cid:15), c, c (cid:48) > P ( S i ≤ (cid:15)i ) ≤ ce − c (cid:48) i for every i ≥
1. Hence, forevery n ≥
1, min a ≥ h P T d a ( |Z h(cid:96)n | ≥ (cid:15)n |Z h(cid:96)n (cid:54) = ∅ ) ≥ − (cid:88) i ≥ n ce − c (cid:48) i ≥ − c − e − c (cid:48) e − c (cid:48) n . (30) Second step.
Let K be a positive integer constant, and let F be the set of vertices z ∈ Z h(cid:96)n such that the Kn -offspring of z has at least λ Knh /n vertices. This step mainly comes down toshowing that the probability that F is empty decays exponentially with n .Define the events E n := {|Z h, + Kn | ≥ λ Knh /n } and E (cid:48) n := {|Z h, + Kn − | ≥ λ Knh /n } .We first show that p := min a ≥ h P T d a ( E n ) > . (31)By (23), lim inf n → + ∞ P T d ( E (cid:48) n ) =: p (cid:48) >
0. Let a be such that P ( ϕ T d ( ◦ ) ≥ a )
4. For n largeenough, (cid:82) a ≥ h P T d a ( E (cid:48) n ) ν ( da ) > p (cid:48) /
2, hence (cid:82) a h P T d a ( E (cid:48) n ) ν ( da ) > p (cid:48) / , where we recall that ν is the density of ϕ T d ( ◦ ). Since E (cid:48) n is an increasing event, by Lemma 8:min a ≥ a P T d a ( E (cid:48) n ) ≥ p (cid:48) / a ≥ h P T d a ( ∃ z ∈ Z h, +1 , ϕ T d ( z ) ≥ a ) = P T d h ( ∃ z ∈ Z h, +1 , ϕ T d ( z ) ≥ a ) =: p (cid:48)(cid:48) > p ≥ p (cid:48) p (cid:48)(cid:48) / | F | ≥ st. Bin( |Z h(cid:96)n | , p (cid:48)(cid:48) ). Thus by (30), for n large enough,min a ≥ h P T d a ( | F | ≥ |Z h(cid:96)n (cid:54) = ∅ ) ≥ − P T d a ( |Z h(cid:96)n | ≤ (cid:15)n |Z h(cid:96)n (cid:54) = ∅ ) − (1 − p (cid:48)(cid:48) ) (cid:15)j ≥ − ce − c (cid:48) n , up to changing the values of the constants c and c (cid:48) . Therefore,max a ≥ h P T d a (cid:16) |Z h ( K + (cid:96) ) n | < λ Knh /n | Z h(cid:96)n (cid:54) = ∅ (cid:17) ≤ ce − c (cid:48) j ≤ ce − c (cid:48) n . If K is large enough, then for n large enough, λ Knh /n > ( λ h − ε ) ( K + (cid:96) )( n +1) , so thatmax a ≥ h P T d a (cid:16) |Z h ( K + (cid:96) ) n | < ( λ h − ε ) ( K + (cid:96) )( n +1) | Z h(cid:96)n (cid:54) = ∅ (cid:17) ≤ ce − c (cid:48) n . We adjust the conditionning: {Z h ( K + (cid:96) ) n (cid:54) = ∅} ⊂ {Z h(cid:96)n (cid:54) = ∅} , and |Z h ( K + (cid:96) ) n | < ( λ h − ε ) ( K + (cid:96) )( n +1) on {Z h(cid:96)n (cid:54) = ∅} \ {Z h ( K + (cid:96) ) n (cid:54) = ∅} . Therefore, there exists n ≥ n ≥ n ,max a ≥ h P T d a (cid:16) |Z h ( K + (cid:96) ) n | < ( λ h − ε ) ( K + (cid:96) )( n +1) | Z h ( K + (cid:96) ) n (cid:54) = ∅ (cid:17) ≤ c exp( − c (cid:48) ( K + (cid:96) )( n + 1)) , (32)18here the new value of c (cid:48) depends on the constants K and (cid:96) . This yields (29) for large enoughmultiples of K + (cid:96) . One can readily replace each Z hm by Z h, + m in this reasoning (for any m ≥ Z h, +( K + (cid:96) ) n instead of Z h ( K + (cid:96) ) n .It remains to show the result for non multiples of K + (cid:96) . Let k ≥ ( K + (cid:96) ) n . Write k =( K + (cid:96) ) n + m , with 0 ≤ m ≤ ( K + (cid:96) ) −
1. Note that on {Z hk (cid:54) = ∅} , Z hm has at least one vertexwhose ( k − m )-offspring is not empty. Hencemax a ≥ h P T d a (cid:16) |Z hk | < ( λ h − ε ) k | Z hk (cid:54) = ∅ (cid:17) ≤ max a ≥ h P T d a (cid:16) |Z h, + k − m | < ( λ h − ε ) k | Z h, + k − m (cid:54) = ∅ (cid:17) ≤ max a ≥ h P T d a (cid:16) |Z h, + k − m | < ( λ h − ε ) ( k − m )+( K + (cid:96) ) | Z h, + k − m (cid:54) = ∅ (cid:17) ≤ c exp( − c (cid:48) ( K + (cid:96) )( n + 1)) ≤ ce − c (cid:48) k , where the third inequality comes from (32). Adapting this last computation for Z h, + k is imme-diate. This concludes the proof of (27). ψ G n Let κ > a n := (cid:98) κ log d − log n (cid:99) . (33)The following statement is the main result of this section. It shows that a recursive constructionof ψ G n , under some assumptions on the subset A ⊂ V n of vertices where ψ G n is already known,is very close to the construction of ϕ T d in Proposition 6. It is a crucial tool for comparing ψ G n and ϕ T d in the exploration in the next section. It is analogous to Proposition 2.7 of [2],where the assumptions on A are slightly different: they are suited to a deterministic d -regulargraph satisfying (I) and (II), while ours will be adapted to an annealed exploration, where therandomness of G n plays a role. Our proof is very similar, but we feel that the general argumentis sufficiently subtle and interesting to merit a full account. Proposition 15.
If the constant κ from (33) is large enough, then the following holds for n large enough. Assume that G n is a good graph as defined in Proposition 1, and that A ⊂ V n satisfies • | A | ≤ n log − n , • tx( B G n ( A, a n )) = tx( A ) , and • max z ∈ A | ψ G n ( z ) | ≤ log / n . or every y ∈ ∂B G n ( A, , writing y for the unique neighbour of y in A , we have: (cid:12)(cid:12)(cid:12)(cid:12) E G n [ ψ G n ( y ) | σ ( A )] − d − ψ G n ( y ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ log − n (34) and (cid:12)(cid:12)(cid:12)(cid:12) Var G n ( ψ G n ( y ) | σ ( A )) − dd − (cid:12)(cid:12)(cid:12)(cid:12) ≤ log − n. (35) A yya n T y Figure 1.In the proof, we will need the fact below, which is a particular case of Lemma 3.3 in [29].
Lemma 16 ( Geometric repulsion ) . Let s ∈ N , and A ⊂ V n such that tx( B G n ( A, s )) = tx( A ) .Let x ∈ V n \ B G n ( A, s ) , let ( X j ) j ≥ be a SRW started at x and τ its first hitting time of A . Then τ dominates stochastically a geometric random variable of parameter ( d − − s .Proof of Proposition 15. Let us first prove (34). By Proposition 3, E G n [ ψ G n ( y ) | σ ( A )] = E G n y [ ψ G n ( X H A )] − E G n y [ H A ] E G n π n [ H A ] E G n π n [ ψ G n ( X H A )] , where ( X j ) j ≥ is a SRW on G n . Write T y for B G n ( y, y, a n − y byour assumptions on A and y (it consists of the paths of length a n − y and notgoing through y ).Let ∂T y be the ( a n − y in T y . Let τ be the hitting time of ∂T y by ( X j ). Thensplitting the first term of the RHS above into three, we obtain E G n [ ψ G n ( y ) | σ ( A )] = E G n y [ ψ G n ( X H A ) H A ≤ τ ] + E G n y [ ψ G n ( X H A ) H A >τ ] − E G n y [ H A ] E G n π n [ H A ] E π n [ ψ G n ( X H A )] . (36)On { H A ≤ τ } , X H A = y . And, P G n y ( H A ≤ τ ) is the probability that a SRW on Z started at 1with transition probabilities 1 /d towards the left and ( d − /d towards the right hits 0 before20 n . It is classical that its difference with 1 / ( d −
1) decays exponentially with a n (see for instanceTheorem 4.8.9 in [12]). Hence, if κ is large enough, (cid:12)(cid:12)(cid:12)(cid:12) P G n y ( H A ≤ τ ) − d − (cid:12)(cid:12)(cid:12)(cid:12) ≤ log − n, (37)and since max z ∈ A | ψ G n ( z ) | ≤ log / n , (cid:12)(cid:12)(cid:12)(cid:12) E G n y [ ψ G n ( X H A ) H A ≤ τ ] − d − ψ G n ( y ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ log − n. Therefore, to establish (34), it is enough to show that (cid:12)(cid:12)(cid:12)(cid:12) E G n y [ ψ G n ( X H A ) H A >τ ] − d − d − E G n π n [ ψ G n ( X H A )] (cid:12)(cid:12)(cid:12)(cid:12) ≤ − n (38)and (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) d − d − − E G n y [ H A ] E G n π n [ H A ] (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) E G n π n [ ψ G n ( X H A )] ≤ log − n. (39)By the strong Markov property, letting p z := P G n y ( H A > τ, X H ∂Ty = z ) for z ∈ ∂T y : E G n y [ ψ G n ( X H A ) H A >τ ] = (cid:80) z ∈ ∂T y p z E G n z [ ψ G n ( X H A )].For all z ∈ ∂T y , by Lemma 16 (with s = a n ), if κ is large enough, then for large enough n , P G n z ( H A < log n ) ≤ − (1 − ( d − − a n ) log n ≤ n ( d − − a n ≤ log − n .Therefore, (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) E G n y [ ψ G n ( X H A ) H A >τ ] − (cid:88) z ∈ ∂T y p z (cid:88) z (cid:48) ∈ V n P G n z ( X (cid:98) log n (cid:99) = z (cid:48) ) E G n z (cid:48) [ ψ G n ( X H A )] (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ P G n z ( H A < log n ) max z ∈ A | ψ G n ( z ) |≤ log − n. By Corollary 2.1.5 of [23] and (I), (cid:12)(cid:12)(cid:12) P G n z (cid:16) X (cid:98) log n (cid:99) = z (cid:48) (cid:17) − π n ( z (cid:48) ) (cid:12)(cid:12)(cid:12) ≤ e − λ G n log n ≤ n for n large enough and all z, z (cid:48) ∈ V n , so that (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) E G n y [ ψ G n ( X H A ) H A >τ ] − (cid:88) z ∈ ∂T y p z E G n π n [ ψ G n ( X H A )] (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ − n. Finally, note that (cid:12)(cid:12)(cid:12)(cid:80) z ∈ ∂T y p z − d − d − (cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12) P G n y ( H A > τ ) − d − d − (cid:12)(cid:12)(cid:12) ≤ log − n by (37), and this yields(38).Let τ (cid:48) be the hitting time of ∂T y ∪ { y } by ( X j ). By the strong Markov property again, we have E G n y [ H A ] = E G n y (cid:2) τ (cid:48) (cid:3) + (cid:88) z ∈ ∂T y P G n y ( X τ (cid:48) = z ) E G n z [ H A ] . (40)21f ( Z j ) j ≥ is a SRW on Z starting at 1 with transition probabilities 1 /d towards the left and( d − /d towards the right, then τ (cid:48) has the law of the hitting time of { , a n } by ( Z j ). Hence, forall k ≥ P y ( τ (cid:48) ≥ k ) ≤ P ( Z k ≤ a n ) . But Z j is the sum of j i.i.d. increments that are boundeda.s. and with mean ( d − /d >
0. We apply the exponential Markov inequality and obtain thatfor some γ > n large enough, for all k ≥ log n , P ( Z k ≤ a n ) ≤ exp( − γk ) . Thus, E G n y (cid:2) τ (cid:48) (cid:3) ≤ log n P G n y ( τ (cid:48) ≤ log n ) + (cid:88) k ≥ log n k exp( − γk ) ≤ n. (41)By (3.20) of [29], E G n π n [ H A ] ≥ n | A | ≥ log n /
4, so that dividing by E G n π n [ H A ] in (40), we get (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) E G n y [ H A ] E G n π n [ H A ] − (cid:88) z ∈ ∂T y p (cid:48) z E G n z [ H A ] E G n π n [ H A ] (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ n log n ≤ log − n, where p (cid:48) z := P G n y ( X τ (cid:48) = z ). By (37), (cid:12)(cid:12)(cid:12)(cid:80) z ∈ ∂T y p (cid:48) z − d − d − (cid:12)(cid:12)(cid:12) ≤ log − n. Therefore, (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) E G n y [ H A ] E G n π n [ H A ] − d − d − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ − n + max z ∈ ∂T y (cid:12)(cid:12)(cid:12)(cid:12) E G n z [ H A ] E G n π n [ H A ] − (cid:12)(cid:12)(cid:12)(cid:12) . To conclude the proof of (39), and thus of (34), it is enough to show thatmax z ∈ ∂T y (cid:12)(cid:12)(cid:12)(cid:12) E G n z [ H A ] E G n π n [ H A ] − (cid:12)(cid:12)(cid:12)(cid:12) ≤ − n. (42)We adapt for this the proof of Proposition 3.5 in [29]. For all z ∈ ∂T y , we can write E G n z [ H A ] ≤ E G n z (cid:104) H A H A < log n (cid:105) + (cid:80) z (cid:48) ∈ V n P G n z ( X (cid:98) log n (cid:99) = z (cid:48) , H A ≥ log n )( E G n z (cid:48) [ H A ] + log n ),hence using Corollary 2.1.5 of [23] and the fact that π n is uniform on V n , E G n z [ H A ] ≤ log n + (cid:88) z (cid:48) ∈ V n (cid:16) π n ( z (cid:48) ) + e − λ G n log n (cid:17) E G n z (cid:48) [ H A ] ≤ log n + (1 + ne − λ G n log n ) E G n π n [ H A ] . (43)Recall that E G n π n [ H A ] ≥ log n/
4. For n large enough, we have by (43): E G n z [ H A ] E G n π n [ H A ] ≤ n log n + 1 + ne − λ G n log n ≤ − n. (44)Conversely, E G n z [ H A ] ≥ (cid:88) z (cid:48) ∈ V n P G n z ( X (cid:98) log n (cid:99) = z (cid:48) ) E G n z (cid:48) [ H A ] − (cid:88) z (cid:48) ∈ V n P G n z ( X (cid:98) log n (cid:99) = z (cid:48) , H A ≤ log n ) E G n z (cid:48) [ H A ] ≥ (cid:88) z (cid:48) ∈ V n ( π n ( z (cid:48) ) − e − λ G n log n ) E G n z (cid:48) [ H A ] − P G n z ( H A ≤ log n ) sup z (cid:48) ∈ V n E G n z (cid:48) [ H A ] ≥ (1 − ne − λ G n log n ) E G n π n [ H A ] − P G n z ( H A ≤ log n ) sup z (cid:48) ∈ V n E G n z (cid:48) [ H A ] . (43) and (44) are in fact true for all z ∈ V n , so that for large enough n , E G n z [ H A ] E G n π n [ H A ] ≥ − ne − λ G n log n − P G n z ( H A ≤ log n )(1 + 5 log − n ) ≥ − − n, provided that κ is large enough: by Lemma 16 applied as below (39), we have22 G n z ( H A ≤ log n ) ≤ log − n for every z ∈ ∂T y .Here lies the main difference with Proposition 3.5 of [29], where there might be more cycles in B G n ( A, a n ) than in A (i.e. the trees planted on the boundary of A might intersect). One has touse a weaker version of Lemma 16, and the lower bound becomes 1 − | A | log − ∆ n , ∆ being anypositive constant ( κ has to be large enough w.r.t. ∆).Thus, the proof of (34) is complete (note that the required lower bounds on κ given by Lemma 16are uniform in y and A ).We prove (35) in the same fashion. Note in particular that by (20),Var G n ( ψ G n ( y ) | σ ( A )) = G G n ( y, y ) − E G n y [ G G n ( y, X H A )] + E G n y [ H A ] E G n π n [ H A ] E G n π n [ G G n ( y, X H A )]so that we get, noticing that H A ≥ τ (cid:48) (the hitting time of ∂T y ∪ { y } ) a.s., (cid:12)(cid:12)(cid:12)(cid:12) Var G n ( ψ G n ( y ) | σ ( A )) − dd − (cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:12)(cid:12)(cid:12)(cid:12) G G n ( y, y ) − E G n y [ G G n ( y, X H A ) H A = τ (cid:48) ] − dd − (cid:12)(cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) E G n y [ G G n ( y, X H A ) H A >τ (cid:48) ] − E G n y [ H A ] E π n [ H A ] E G n π n [ G G n ( y, X H A )] (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (45)We have E G n y [ G G n ( y, X H A ) H A = τ (cid:48) ] = E G n y [ G G n ( y, X τ (cid:48) ) H A = τ (cid:48) ]= E G n y [ G G n ( y, X τ (cid:48) )] − E G n y [ G G n ( y, X τ (cid:48) ) H A >τ (cid:48) ] . Remark that | E G n y [ G G n ( y, X τ (cid:48) ) H A >τ (cid:48) ] | ≤ max z ∈ ∂T y | G G n ( y, z ) | ≤ log − n if κ is large enough, by (11). Now, by (17) applied to the tree T y (note that T T y = τ (cid:48) ), we get | G G n ( y, y ) − E G n y [ G G n ( y, X τ (cid:48) )] − G T y G n ( y, y ) | ≤ E G ny [ τ (cid:48) ] n By (41), E G ny [ τ (cid:48) ] n = O (log − n ), therefore, the first term of the RHS of (45) is | G T y G n ( y, y ) − dd − | + O (log − n ).But T y is isomorphic to B := B T + d ( ◦ , a n − G T y G n ( y, y ) = G B T d ( ◦ , ◦ ) = G T d ( ◦ , ◦ ) − P T d ◦ ( T B = ◦ ) G T d ( ◦ , ◦ ) − P T d ◦ ( T B (cid:54) = ◦ ) G T d ( ◦ , z )for any z ∈ T + d at distance a n − ◦ , by cylindrical symmetry of T + d . By (10), if κ is largeenough: G T y G n ( y, y ) = d − d − − P T d ◦ ( T B = ◦ ) d − + O (log − n ).By the same reasoning as that leading to (37), P T d ◦ ( T B = ◦ ) = d − + O (log − n ) for κ largeenough, hence 23 G T y G n ( y, y ) − dd − | = O (log − n ).All in all, we get that the first term of the RHS of (45) is O (log − n ).One applies to the second term of the RHS of (45) the same reasoning as that for (38) and (39).Note in particular that since G n is a good graph, (11) implies that max x,y ∈ V n | G G n ( x, y ) | ≤ K ,an upper bound easier to use than max z ∈ A | ψ G n ( z ) | ≤ log / n . ψ G n around a vertex In this section, for x ∈ V n , we explore a tree-like neighborhood T x of x that is isomorphic toa realization of C h +log − n ◦ . We obtain the latter through an array of i.i.d. standard normalvariables, independent of the pairings of the half-edges of G n . As long as we do not meet cycles,we reveal T x generation by generation, until | ∂T x | ≥ n / log − κ − n if C h +log − n ◦ is infinite withan exponential growth (which happens with probability (cid:39) η ( h )). Then by Proposition 5, arealization of ψ G n is given by a recursive construction with the same normal variables as thoseof the realization of ϕ T d . If G n is a good graph, the way T x has been explored allows to applyProposition 15 and to ensure that T x ⊆ C G n ,hx , the connected component of x in E ≥ hψ G n . Thus, weprove that with P ann -probability arbitrarily close to η ( h ), C G n ,hx has size at least n / log − κ − n (Proposition 17). The exploration.
Fix x ∈ V n . Let b n := ( d − − a n log − n, (46)where we recall the definition of a n from (33). Let ( ξ k,i ) k,i ≥ be an array of independentvariables, each of law N (0 , G n . Let T x bea tree rooted at x initially reduced to { x } , and T x its respective counterpart in T d , rooted at ◦ . Let also Φ be at every moment of the exploration an isomorphism from T x to T x . The exploration from x consists of the following steps: • at step 0, reveal the pairings of the half-edges of B G n ( x, a n ). Let ϕ T d ( ◦ ) = (cid:112) G T d ( ◦ , ◦ ) ξ , .Stop the exploration if tx( B G n ( x, a n )) > ϕ T d ( ◦ ) < h + log − n . • at step k ≥
1, reveal the remaining edges of B G n ( T x , a n + 1). Let O k − be the set of thevertices of T x of height k −
1. Denote x k, , x k, , . . . , x k,m the children of O k − (in the tree B G n ( T x , m ∈ N .By Proposition 6, a realization of the GFF on T x is given by the recursive construction ϕ T d (Φ( x k,i )) = d − ϕ T d (Φ( x k,i )) + ξ k,i (cid:113) dd − . Add to T x the vertices x k,i of O k − such that ϕ T d (Φ( x k,i )) ≥ h + log − n . Add to T x the corresponding vertices Φ( x k,i ).24top the exploration at step k ≥ B G n ( T x , a n + 1),C2 no vertex was added to T x during the ( k − | O k − | ≥ n / b n .C4 k > log λ h n .If the exploration is stopped at some step k , at which only C3 is met, say that it is successful .In this case, by Proposition 5, we can sample ψ G n as follows: we reveal the remaining pairingsof half-edges in G n . We set ψ G n ( x ) = ϕ T d ( ◦ ). For all k, i ≥
1, if A k,i = { x } ∪ { x (cid:96),j | ( (cid:96), j ) ≺ ( k, i ) } where ≺ is the lexicographical order on N , ψ G n ( x k,i ) = E G n [ ψ G n ( x k,i ) | σ ( A k,i )] + ξ k,i (cid:113) Var( ψ G n ( x k,i ) | σ ( A k,i )) . (47)Let S ( x ) := { the exploration from x is successful } . We prove the following: Proposition 17. P ann ( S ( x ) ∩ { T x ⊆ C G n ,hx } ) −→ n → + ∞ η ( h ) . (48) Remark 1 ( Exploration size ) . Denote R x the set of vertices seen during the exploration(i.e. such at least one of their half-edges has been paired). Note that for n large enough, forevery x ∈ V n , by C3, C4 and (46), T x contains less than n / ( d − − a n log − n log λ h n ≤ n / ( d − − a n log − n vertices, so that | R x | ≤ n / ( d − − a n log − n × (1 + ( d −
1) + . . . + ( d − a n +1 ) ≤ n / log − n . In order to prove Proposition 17, we first show that C h +log − n ◦ either has an exponential growthwith probability close to η ( h ), or dies out before reaching height log λ h n with probability closeto 1 − η ( h ). Lemma 18.
Let F ( n ) k := { n / b n ≤ |Z h +log − nk | ≤ dn / b n } and F (cid:48) ( n ) k := {Z h +log − nk − = ∅} for k ≥ . Let k := inf { k ≥ , F ( n ) k or F (cid:48) ( n ) k happens } , F ∗ ( n ) := { k ≤ log λ h n } ∩ F ( n ) k and F (cid:48)∗ ( n ) := { k ≤ log λ h n } ∩ F (cid:48) ( n ) k . Then, as n → + ∞ : P ann ( F ∗ ( n ) ) → η ( h ) and P ann ( F (cid:48)∗ ( n ) ) → − η ( h ) . (49) Proof.
Note that for every n ≥ F (cid:48)∗ ( n ) ∩ F ∗ ( n ) = ∅ , implying P ann ( F (cid:48)∗ ( n ) ) + P ann ( F ∗ ( n ) ) ≤ n → + ∞ P ann ( F ∗ ( n ) ) ≥ η ( h ) (50)25nd lim inf n → + ∞ P ann ( F (cid:48)∗ ( n ) ) ≥ − η ( h ) (51)Let ε ∈ (0 , η ( h )), and let δ > | η ( h + δ ) − η ( h ) | ≤ ε . h (cid:48) → η ( h (cid:48) ) is continuous on R \ { h (cid:63) } by Theorem 3.1 of [1], hence such δ exists. By Proposition 9,lim inf n → + ∞ P ann (cid:16) ∃ k ≤ log λ h n, |Z h +log − nk | > n / b n (cid:17) ≥ lim inf n → + ∞ P ann (cid:16) ∃ k ≤ log λ h n, |Z h + δk | > n / b n (cid:17) ≥ η ( h + δ ) ≥ η ( h ) − ε. Since |Z h +log − nk | ≤ d |Z h +log − nk − | , {∃ k ≤ log λ h n, |Z h +log − nk | > n / b n } ⊆ ∪ k ≤ log λh n F ( n ) k .Hence, lim inf n → + ∞ P ann ( F ∗ ( n ) ) ≥ lim inf n → + ∞ P ann (cid:16) ∪ k ≤ log λh n F ( n ) k (cid:17) ≥ η ( h ) − ε ,and this shows (50).For n ≥ e /δ , C h + δ ◦ ⊆ C h +log − n ◦ ⊆ C h ◦ . Note that for n ≥ P ann ( F (cid:48)∗ ( n ) ) ≥ P ann (cid:16) Z h +log − n (cid:98) log λh n (cid:99)− = ∅ (cid:17) − P ann (cid:16) {Z h +log − n (cid:98) log λh n (cid:99)− = ∅} ∩ {∃ k ≥ , |Z h +log − nk | ≥ n / b n } (cid:17) ≥ P ann (cid:16) Z h (cid:98) log λh n (cid:99)− = ∅ (cid:17) − P ann ( {|C h +log − n ◦ | < + ∞} ∩ {∃ k ≥ , |Z h +log − nk | ≥ n / b n } ) ≥ P ann (cid:16) Z h (cid:98) log λh n (cid:99)− = ∅ (cid:17) − P ann ( |C h +log − n ◦ | < + ∞ | ∃ k ≥ , |Z h +log − nk | ≥ n / b n ) . The first term of the RHS converges to 1 − η ( h ) as n → + ∞ . For k ≥ v ∈ Z h +log − nk , if T v isthe subtree from v in C h +log − n ◦ and C ◦ ( h, δ ) the connected component of ◦ in ( {◦}∪ E ≥ h + δϕ T d ) ∩ T + d , P ann ( | T v | < + ∞ ) ≤ P T d h ( |C ◦ ( h, δ ) | < + ∞ )by Lemma 8. Thus, P ann ( |C h ◦ | < + ∞ | ∃ k ≥ , |Z hk | ≥ n / b n ) ≤ P T d h ( |C ◦ ( h, δ ) | < + ∞ ) n / b n . Therefore, it only remains to prove that P T d h ( |C ◦ ( h, δ ) | < + ∞ ) <
1. By Lemma 8, the map a (cid:55)→ P T d a ( |C ◦ ( h, δ ) | = + ∞ ) is non-decreasing. By (23), (cid:82) + ∞ h + δ P T d a ( |C ◦ ( h, δ ) | = + ∞ ) ν ( da ) > a ≥ h + δ , for every a ≥ a , P T d a ( |C ◦ ( h, δ ) | = + ∞ ) ≥ P T d a ( |C ◦ ( h, δ ) | = + ∞ ) > p > P T d h ( ◦ has one child z such that ϕ T d ( z ) ≥ a ) ≥ p .Therefore, P T d h ( |C ◦ ( h, δ ) | = + ∞ ) ≥ p P T d a ( |C ◦ ( h, δ ) | = + ∞ ) >
0, and P T d h ( |C ◦ ( h, δ ) | < + ∞ ) < roof of Proposition 17. We first establish that C1 happens with P ann -probability o (1). Then,if there is no cycle in B G n ( T x , a n ), we can apply Proposition 15, to bound the difference between ϕ T d and ψ G n .By Remark 1, at most dn / log − n matchings of half-edges are performed during the explo-ration. By (14) with k = m = 1, m E = 0 and m ≤ dn / log − n , the probability to create atleast one cycle during these matchings is less than log − n for large enough n . Therefore, P ann (C1 happens) → . (52)Note that if C1 does not happen, then on F ∗ ( n ) , (resp. F (cid:48)∗ ( n ) ), C2 (resp. C3) is satisfied, butnot C4. Moreover, on F ∗ ( n ) , (resp. F (cid:48)∗ ( n ) ), C3 (resp. C2) does not hold, so that P ann ( F ∗ ( n ) ) − P ann (C1 happens) ≤ P ann ( S ( x )) ≤ − P ann ( F (cid:48)∗ ( n ) ) + P ann (C1 happens) . Thus, by (49) and (52), P ann ( S ( x )) → η ( h ) . (53)Now, we compare ψ G n with ϕ T d . Note that by C4 and the triangular inequality, { T x (cid:54)⊆ C G n ,hx } ⊆ {∃ y ∈ T x , | ψ G n ( y ) − ϕ T d (Φ( y )) | ≥ log − n } ⊆ ∪ y ∈ T x E ( y ),where E ( x ) := {| ψ G n ( x ) − ϕ T d (Φ − ( x )) | ≥ log − n } and E ( y ) := {| ψ G n ( y ) − ϕ T d (Φ − ( y )) | ≥ | ψ G n ( y ) − ϕ T d (Φ − ( y )) | + 2 log − n } for y (cid:54) = x .Suppose that G n is a good graph. For x k,i ∈ T x \{ x } , on E ( n ) k,i := { max y (cid:48) ∈ A k,i | ψ G n ( y (cid:48) ) | < log / n } ,we can apply Proposition 15 (note that | A k,i | ≤ n / by Remark 1 and that tx( B G n ( A k,i , a n )) =tx( A k,i ) by C1). Writing y = x k,i and ξ = ξ k,i , we get for n large enough: | ψ G n ( y ) − ϕ T d (Φ − ( y )) | ≤ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ψ G n ( y ) − ϕ T d (Φ − ( y )) d − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + log − n + (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:32)(cid:113) Var G n ( ψ G n ( y ) | σ ( A )) − (cid:114) d − d (cid:33) ξ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) and (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:113) Var G n ( ψ G n ( y ) | σ ( A )) − (cid:114) d − d (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:114) d − d − log − n − (cid:114) d − d (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:114) d − d d d −
1) log − n ≤ log − n. Let E (cid:48) ( y ) := E ( y ) ∩ E ( n ) k,i . We have P G n ( E (cid:48) ( y )) ≤ P ( | ξ | log − n ≥ log − n ) ≤ n − (54)27y the exponential Markov inequality used as in the proof of Lemma 4. Moreover, by (8), if κ is large enough, we obtain in the same manner: P G n ( E ( x )) ≤ P G n ( | ξ , | log − n ≥ log − n ) ≤ n − . (55)By Remark 1, | T x | ≤ n / . By (54), (55) and a union bound on y ∈ T x , P G n ( ∪ y ∈ T x E (cid:48) ( y )) ≤ n − / with E (cid:48) ( x ) := E ( x ). And, by Lemma 4, P G n ( ∪ ( k,i ): x k,i ∈ T x E ( n ) k,i ) ≤ n − for large enough n .Thus for n large enough, if G n is a good graph, P G n ( T x (cid:54)⊆ C G n ,hx ) ≤ P G n ( ∪ y ∈ T x E ( y )) ≤ n − / + n − ≤ n − , so that by Proposition 1 and (53): P ann ( S ( x ) ∩ { T x ⊆ C G n ,hx } ) → η ( h ) . For x ∈ V n , the lower exploration is the exploration of Section 5.1, modified by replacing h + log − n by h − log − n , so that we compare T x and C h − log − n ◦ . If it is stopped at some step k at which only C2 is met, say that it is aborted . Write A ( x ) := { the lower exploration from x is aborted } ∩ {C G n ,hx ⊆ T x } . Proposition 19. P ann ( A ( x )) −→ n → + ∞ − η ( h ) . (56)The proof follows from a direct adaptation of Lemma 18 and Proposition 17. Note in particularthat P ann (C1 happens) = o (1), and P ann ( A ( x )) = P ann ( Z h − log − nk − = ∅ ) + o (1) = 1 − η ( h ) + o (1). In Section 6.1, we prove that two vertices x, y ∈ V n are in the same connected component of E ≥ hψ G n with P ann -probability −→ n → + ∞ η ( h ) (Proposition 21). Then in Section 6.2, we use a secondmoment argument to get concentration and to show (2). Let us describe our strategy to establish Proposition 21.We perform explorations as in Section 5.1 from x and y . If they are both successful and donot meet (which happens with probability (cid:39) η ( h ) ), we develop disjoint balls, denoted ”joiningballs”, from ∂T x to ∂T y (Section 6.1.1). Each of them is rooted at a vertex of ∂T x , hits ∂T y atexactly one vertex, and has a ”security radius” of depth a n around its path from ∂T x to ∂T y (see Figure 2). Then, we realize ψ G n on T x , T y and those balls (Section 6.1.2). If they are all28isjoint and tree-like, once we have revealed ψ G n on T x and T y , this security radius allows us toapply Proposition 15 to approximate ψ G n on the paths from ∂T x to ∂T y by ϕ T d .Let us explain with a back-of-the-enveloppe computation how the joining balls allow to connect T x and T y in E ≥ hψ G n . Since | T y | (cid:39) n / b n by C3, the probability that for a given z ∈ ∂T x , exactlyone of the vertices at distance γ log d − log n from z is in ∂T y is (cid:39) P (Bin(( d − γ log d − log n , n / b n dn ) = 1) (cid:39) log γ n × n − / b n .And there are (cid:39) n / b n vertices in ∂T x , hence we can expect that that the number of joiningballs is at least (cid:39) n / b n × log γ n × n − / b n = b n log γ n , provided that we can control someundesirable events (such as an intersection between balls, or a cycle in a ball). This is thepurpose of Lemma 20.And we know that for large r ∈ N and v ∈ ∂B T d ( ◦ , r ), P T d ( v ∈ C h ◦ ) is of order ( λ h / ( d − r ,by Proposition 13. Taking r = γ log d − log n , the probability that E ≥ hψ G n percolates from ∂T x to ∂T y through a given joining ball is (cid:39) log γ (log d − λ h − n , if we can approximate ψ G n by ϕ T d . For γ large enough w.r.t κ (recall (46) and (33), and recall that λ h > b n log γ n × log γ (log d − λ h − n ≥ log − κ − n × log γ log d − λ h n >> E ≥ hψ G n will percolate through at least one joining ball from ∂T x to ∂T y . For x, y ∈ V n , write x h ↔ y if y ∈ C G n ,hx . Let ( ξ z,k,i ) z ∈{ x,y } ,k,i ≥ be an array of i.i.d. standardnormal variables independent from everything else. Define the joint exploration from x and y as the exploration from x (with the ( ξ x,k,i )’s), then the exploration from y (with the ( ξ y,k,i )’s),as in Section 5.1, with the additional conditionC7 the exploration is stopped as soon as R x ∩ R y (cid:54) = ∅ ,where R x (resp. R y ) is the set of vertices seen during the exploration from x (resp. from y ). Construction of the joining balls.
If both explorations are successful and C7 does not happen (denote S ( x, y ) this event), we addthe following steps: let γ > a (cid:48) n := (cid:98) γ log d − log n (cid:99) . (57)Denote z , . . . , z | ∂T x | the vertices of ∂T x . For j = 1 , , . . . , | ∂T x | successively, build B ∗ ( z j , a (cid:48) n )the subgraph of G n obtained as follows (see Figure 2). Let initially B ∗ ( z j , a (cid:48) n ) := B G n ( z j , z j , a n )be the subtree from z j of height a n in the tree B G n ( T x , a n ). Recall the definition of R x fromRemark 1. Let B ∗ j := ∪ j (cid:48)
1) to B ∗ ( z j , a (cid:48) n ). If | B ∗ ( z j , a (cid:48) n − a n ) ∩ B G n ( T y , a n ) | (cid:54) = 1,the construction of B ∗ ( z j , a (cid:48) n ) stops.Else, let v j (0) be the unique vertex of B ∗ ( z j , a (cid:48) n − a n ) ∩ B G n ( T y , a n ). Iftx(( B ∗ ( z j , a (cid:48) n − a n ) ∪ R j ) \ { v j (0) } ) > tx( R j ),the construction of B ∗ ( z j , a (cid:48) n ) stops.Else, for k = a (cid:48) n − a n , . . . , a (cid:48) n − B ∗ ( z j , k ) ∪ R j ) = tx( B ∗ ( z j , a (cid:48) n − a n ) ∪ R j ),add the neighbours of B ∗ ( z j , k ) to B ∗ ( z j , a (cid:48) n ). Then, the construction of B ∗ ( z j , a (cid:48) n ) is completed.In this case only, and iftx( B ∗ ( z j , a (cid:48) n ) ∪ R j ) = tx( B ∗ ( z j , a (cid:48) n − a n ) ∪ R j ),say that B ∗ ( z j , a (cid:48) n ) is a joining ball . In other words, we obtain a joining ball if, revealing the a (cid:48) n -offspring of z j , we have a unique connection to ∂T y , at distance a (cid:48) n − a n from z j , and no cycleis discovered in the whole construction (except when the joining ball reaches ∂B G n ( T y , a n )).Write J := { j ≤ | ∂T x | , B ∗ ( z j , a (cid:48) n ) is a joining ball } and S (cid:48) ( x, y ) := S ( x, y ) ∩{| J | ≥ log γ − κ − n } ( S ( x, y ) was define above 57)). yT y T x xz v j ( a n ) v j ( a ) a n a n B ∗ ( z j ; a n ) Figure 2. B ∗ ( z j , a (cid:48) n ) is a joining ball.30 emma 20. Fix γ > κ + 18 . P ann ( S (cid:48) ( x, y )) −→ n → + ∞ η ( h ) . Proof.
Denote F ∗ ( n ) ( x ) (resp. F ∗ ( n ) ( y )) the event F ∗ ( n ) for x (resp. y ). As in the proof ofLemma 18, we get that P ann ( F ∗ ( n ) ( x ) ∩ F ∗ ( n ) ( y )) = P ann ( F ∗ ( n ) ( x )) P ann ( F ∗ ( n ) ( y )) → η ( h ) and P ann ( F (cid:48)∗ ( n ) ( x ) ∪ F (cid:48)∗ ( n ) ( y )) → − η ( h ) .Moreover, P ann (C1 or C7 happens) →
0. Note indeed that by Remark 1, less than 2 dn / log − n half-edges are revealed during the explorations from x and y , which allows to control C7 as wedid for C1 in (52). Thus,lim sup n → + ∞ | P ann ( S (cid:48) ( x, y )) − η ( h ) | ≤ lim sup n → + ∞ P ann ( S ( x, y ) ∩ {| J | < log γ − κ − n } )and it remains to prove thatlim sup n → + ∞ P ann ( S ( x, y ) ∩ {| J | < log γ − κ − n } ) = 0 . (59)We proceed in two steps: in step 1, we control the number of spoiled vertices, and the numberof vertices of ∂B G n ( T y , a n ) that are hit when building the B ∗ ( z j , a (cid:48) n )’s (if a large proportion ofthose vertices are in R j , then it significantly affects the probability that B ∗ ( z j , a (cid:48) n ) is a joiningball). In step 2, we estimate the probability that for a given j , B ∗ ( z j , a (cid:48) n ) is a joining ball,provided that the bounds of step 1 hold. This gives a binomial lower bound for | J | . Step 1.
Note that by Remark 1 and (33), | ∂B G n ( T x , a n ) | + | ∂B G n ( T y , a n ) | ≤ n / log − n . Notealso that for every j ≤ | ∂T x | , B ∗ ( z j , a (cid:48) n ) contains less than ( d − a (cid:48) n ≤ log γ n half-edges. Hence:at every moment of the exploration, less than n / log γ n half-edges have been seen. (60)Let B ∗ := ∪ j ≤| ∂T x | B ∗ ( z j , a (cid:48) n ) . (61)To reveal the edges of B ∗ , one proceeds to at most n / log γ n pairings of half-edges by (60).Any pairing that results in an edge e between some B ∗ ( z j , k ) and B G n ( T y , a n ) then leads to atmost 1 + ( d −
1) + . . . + ( d − a n ≤ d − a n ≤ log κ +1 n ≤ log γ − n vertices of B ∗ ( z j , a (cid:48) n ) ∩ B G n ( T y , a n ), since the construction of B ∗ ( z j , a (cid:48) n ) stops if such an edgehappens at distance less than a (cid:48) n − a n of z j (and recall that we choose γ > κ + 18 > κ + 2).Thus, by (15) with k = (cid:98) log γ +1 n (cid:99) , m < n / log γ n and m + m + m E < n / log γ n (due to60), for n large enough: P ann ( S ( x, y ) ∩ {| B G n ( T y , a n ) ∩ B ∗ | ≥ log γ n } ) ≤ . log n ≤ n − . (62)31et N be the total number of spoiled vertices. By (15) with the same parameters, P ann ( S ( x, y ) ∩ { N ≥ log γ n } ) ≤ n − . (63) Step 2.
Recall the definition of B ∗ j from (58). For j ≤ m , denote S j := S ( x, y ) ∩ {| B G n ( T y , a n ) ∩ B ∗ j | ≤ log γ n } ∩ { z j is not spoiled } . (64)Suppose that for every j ≥ P ann ( B ∗ ( z j , a (cid:48) n ) is a joining ball | S j ) ≥ n − / log γ − κ − n. (65)On E := {| B G n ( T y , a n ) ∩ B ∗ | < log γ n } ∩ { N < log γ n } , the number of j ’s such that S j holdsis at least | ∂T x | − log γ n ≥ n / log − κ − n by (C3) and (46). Thus, if Z ∼ Bin (cid:0) (cid:98) n / log − κ − n (cid:99) , n − / log γ − κ − n (cid:1) , P ann (cid:0) S ( x, y ) ∩ {| J | ≤ log γ − κ − n } (cid:1) ≤ P ( Z ≤ log γ − κ − n ) + P ann ( S ( x, y ) ∩ E c ) . For large enough n , P ann ( S ( x, y ) ∩ E c ) = o ( n − ) by (62) and (63). Moreover, one checks easily(using γ > κ + 18 and (12)) that for n large enough:max ≤ k ≤(cid:98) log γ − κ − n (cid:99) P ( Z = k ) ≤ /n .This yields (59). Hence, it only remains to prove (65).Remark that P ann ( B ∗ ( z j , a (cid:48) n ) is a joining ball | S j ) ≥ p p p where: • p := P ann ( E |S j ) and E := S j ∩ { no cycle is created and no connection to R j is madewhen revealing B ∗ ( z j , a (cid:48) n − a n − } , • p := P ann ( E |E ) where E := E ∩ { exactly one edge connects B ∗ ( z j , a (cid:48) n − a n −
1) and D := ∂B G n ( T y , a n ) \{ B G n ( B G n ( T y , a n ) ∩ B ∗ j , a n ) } }∩{ no cycle is created and no connectionto ∂B G n ( T y , a n ) ∪ B ∗ j is made when revealing the other edges of B ∗ ( z j , a (cid:48) n − a n ) } , • p := P ann ( E |E ) where E := E ∩{ no cycle is created and no connection to ∂B G n ( T y , a n ) ∪ B ∗ j is made when revealing the remaining edges of B ∗ ( z j , a (cid:48) n ) } .This definition of D guarantees that B ∗ ( z j , a (cid:48) n ) will not intersect a previously realized joiningball when growing the subtree from v j (0) in B G n ( T y , a n ).(14) with k = 1, m , m ≤ log γ n and m E , m ≤ n / log γ n due to (60) yields for n large enough: p i ≥ − C (1) log γ n max( n / log γ n, γ n ) n ≥ − n − / (66)for i ∈ { , } . Therefore, p p ≥ / n large enough.On E , reveal the pairings of the half-edges of ∂B ∗ ( z j , a (cid:48) n − a n −
1) one by one. E holds if:32 one given half-edge is matched to a half-edge of D , which happens with probability atleast | D | dn , and • each other half-edge is matched to a half-edge that had not been seen before (by (60), foreach half-edge this happens with probability at least 1 − n / log γ ndn − n / log γ n ≥ − n / log γ nn ).Since ∂B ∗ ( z j , a (cid:48) n − a n −
1) has ( d − | ∂B ∗ ( z j , a (cid:48) n − a n − | unpaired half-edges, p ≥ ( d − | ∂B ∗ ( z j , a (cid:48) n − a n − | | D | dn (cid:16) − n / log γ nn (cid:17) | ∂B ∗ ( z j ,a (cid:48) n − a n − |− By (33) and (57), one checks easily that on E ,log γ − κ − n ≤ | ∂B ∗ ( z j , a (cid:48) n − a n − | ≤ log γ n ,and that on S j (defined in (64)), | D | ≥ | ∂B G n ( T y ,a n ) | ≥ n / log − n by (46) and C3. Hence for n large enough, p ≥ ( d −
1) log γ − κ − n n / log − ndn (cid:32) − n / log γ ndn (cid:33) log γ n ≥ n − / log γ − κ − n. With (66), this entails P ann ( B ∗ ( z j , a (cid:48) n ) is a joining ball | S j ) ≥ p p p ≥ n − / log γ − κ − for n large enough (uniformly on the realization of B ∗ j ), and thus (65), so that the proof of theLemma is complete. ψ G n on the joint exploration Suppose that we are on S (cid:48) ( x, y ). By Proposition 5, we can realize ψ G n on T x as in (47) withthe ( ξ x,k,i ) k,i ≥ . Then we can realize it in a similar way on T y with the ( ξ y,k,i ) k,i ≥ , lettingrecursively ψ G n ( y k,i ) = E G n [ ψ G n ( y k,i ) | σ ( A k,i )] + ξ y,k,i (cid:112) Var( ψ G n ( y k,i ) | σ ( A k,i ))with A k,i := T x ∪ { y (cid:96),j | ( (cid:96), j ) ≺ ( k, i ) } and ≺ the lexicographical order on N .Recall that J = { j ≥ , B ∗ ( z j , a (cid:48) n ) is a joining ball } and that for j ∈ J , we denote v j (0) theunique vertex of B ∗ ( z j , a (cid:48) n − a n ) ∩ B G n ( T y , a n ). Since no cycle is discovered when revealing B ∗ ( z j , a (cid:48) n ) \ B ∗ ( z j , a (cid:48) n − a n ), the intersection of B ∗ ( z j , a (cid:48) n − a n ) and T y is a unique vertex v j ( a n ),which is in the a n -offspring of v j (0) in the tree B ∗ ( z j , a (cid:48) n ) rooted at z j . Then we realize ψ G n on B ∗ ( z j , a (cid:48) n − a n ) and on the shortest path P j from v (0) to v ( a n ) as in (47), via a family ofi.i.d. N (0 ,
1) random variables ( ξ j,k,i ) k,i ≥ . In the tree T j := B ∗ ( z j , a (cid:48) n − a n ) ∪ P j with root z j ,denoting z j,k,i the i -th vertex at generation k and33 j,k,i := T x ∪ T y ∪ {∪ j (cid:48) Let γ > κ + 18. By Lemma 20,lim sup n → + ∞ P ann ( S ∗ ( x, y )) ≤ lim n → + ∞ P ann ( S (cid:48) ( x, y )) = η ( h ) . Let E n := {G n is not a good graph } ∪ { max z ∈ V n | ψ G n ( z ) | ≥ log / n } . By Proposition 1 andLemma 4, P ann ( E n ) → 0. Therefore, it is enough to show thatlim sup n → + ∞ P ann ( E cn ∩ ( S (cid:48) ( x, y ) \ S ∗ ( x, y )) ) = 0.By a straightforward adaptation of the reasoning below (53),lim n → + ∞ P ann ( E cn ∩ ( S (cid:48) ( x, y ) \ S (cid:48)(cid:48) ( x, y )) ) = 0,where S (cid:48)(cid:48) ( x, y ) := S (cid:48) ( x, y ) ∩ {∀ z ∈ T x ∪ T y , ψ G n ( z ) ≥ h + (log − n ) / } . Hence, we are left withproving that lim sup n → + ∞ P ann ( E cn ∩ ( S (cid:48)(cid:48) ( x, y ) \ S ∗ ( x, y )) ) = 0 . (68)We use again a binomial argument. For j ∈ J in increasing order, generate the GFF on T j as in(67). Denote E j the event that z j and v j ( a n ) are in the same connected component of E ≥ hψ G n ∩ T j .Note that on S (cid:48)(cid:48) ( x, y ), T x ⊆ C G n ,hx and T y ⊆ C G n ,hy , so that S (cid:48)(cid:48) ( x, y ) ∩ ( ∪ j ∈ J E j ) ⊆ S ∗ ( x, y ).Suppose that for every j ∈ J , P ann ( E j |S (cid:48)(cid:48) ( x, y ) ∩ E cn ) ≥ log γ ( K / − n, (69)where K := log d − ((1 + λ h ) / Z ∼ Bin( (cid:98) log γ − κ − n (cid:99) , log γ ( K / − n ), we have P ann ( E cn ∩ ( S (cid:48)(cid:48) ( x, y ) \ S ∗ ( x, y )) ) ≤ P ( Z = 0) . But if κ and γ/κ are large enough so that γ − κ − 18 + γ ( K / − 1) = γK / − κ − > n → + ∞ P ( Z = 0) = 0 and this yields (68). Thus, we are left with showing (69).We split the proof of (69) in two parts. First, we prove that P ann ( v j (0) ∈ C z j | ( S (cid:48)(cid:48) ( x, y ) ∩ E cn ) ) ≥ log ( γ − κ )( K / − n, (70)where C z j is the connected component of z j in E ≥ hψ G n ∩ B ∗ ( z j , a (cid:48) n − a n ). Second, we show thatfor some constant K > d and h ), P ann ( ∀ v ∈ P j , ψ G n ( v ) ≥ h | ( S (cid:48)(cid:48) ( x, y ) ∩ E cn ∩ { v j (0) ∈ C z j } ) ) ≥ log − K κ n. (71)We prove that both hold for n large enough, uniformly in j ∈ T j and on the realization of the ψ G n on T ∪ . . . ∪ T j − , as long as we are in E cn (so that we can apply Proposition 15).(70) and (71) imply indeed that P ann ( E j |S (cid:48)(cid:48) ( x, y ) ∩ E cn ) ≥ log ( γ − κ )( K / − − K κ n ≥ log γ ( K / − n if γ/κ is large enough. Part 1: proof of (70).Since | A j,k,i | ≤ n / and tx( B G n ( A j,k,i , a n )) = tx( A j,k,i ) for all k, i ≥ 0, we can apply Propo-sition 15 as below (53) to bound the difference between ψ G n on C z j and ϕ T d on an isomorphicsubtree of T d , with the following coupling: ϕ T d ( ◦ ) := ψ G n ( z j ), and then ϕ T d is defined as inProposition 6 via ( ξ j,k,i ) k,i ≥ . Recall that on S (cid:48)(cid:48) ( x, y ), we have that ψ G n ( z j ) ≥ h + (log − n ) / δ > n ,min a ≥ h +(log − n ) / P T d a (cid:16) |Z h +(log − n ) / , + a (cid:48) n − a n | ≥ ( λ h − δ ) a (cid:48) n − a n (cid:17) ≥ p P T d ( E + )2where P T d ( E + ) > p := min a ≥ h +(log − n ) / P T d a ( ∃ v ∈ B T + d ( ◦ , , ϕ T d ( z ) ≥ h + 1) > δ (cid:48) > log − n such that λ h + δ (cid:48) > λ h − δ (such δ (cid:48) exists by Proposition 9,if n is large enough), Z h + δ (cid:48) , + a (cid:48) n − a n ⊆ Z h +(log − n ) / , + a (cid:48) n − a n .Proposition 15 yields then P ann (cid:18) | ∂ C z j | ≥ ( λ h − δ ) a (cid:48) n − a n (cid:12)(cid:12)(cid:12)(cid:12) S (cid:48)(cid:48) ( x, y ) ∩ E cn (cid:19) ≥ p P ( E + )2 + o (1) ≥ p P ( E + )3 . By cylindrical symmetry of B T + d ( ◦ , a (cid:48) n − a n ), we even have P ann (cid:18) v (0) ∈ ∂ C z j (cid:12)(cid:12)(cid:12)(cid:12) S (cid:48)(cid:48) ( x, y ) ∩ E cn (cid:19) ≥ p P ( E + )3 ( λ h − δ ) a (cid:48) n − a n | ∂B T + d ( ◦ , a (cid:48) n − a n ) |≥ p P ( E + )3 (cid:18) λ h − δd − (cid:19) a (cid:48) n − a n . Since K = log d − ((1 + λ h ) / δ small enough yields (70).35 art 2: proof of (71).Denote v j (1) , . . . , v j ( a n − 1) the vertices from v j (0) to v j ( a n ) on the path P j . Remark that itsuffices to prove that there exists a constant K > n large enough, for every k ∈ { , . . . , a n } , P ann (cid:0) ψ G n ( v j ( k )) ≥ h | ( S (cid:48)(cid:48) ( x, y ) ∩ E cn ∩ { ψ G n ( v j ( k − ≥ h } ) (cid:1) ≥ ( d − − K . (72)In the notation of (67), v j ( k ) = y j,k + a (cid:48) n − a n , for 1 ≤ k ≤ a n . Write A k := A j,k + a (cid:48) n − a n , .Suppose that for n large enough and all k ∈ { , . . . , a n } , on S (cid:48)(cid:48) ( x, y ) ∩ E cn ∩ { ψ G n ( v j ( k − ≥ h } : E G n [ ψ G n ( v j ( k )) | σ ( A k )] > −| h | − G n ( ψ G n ( v j ( k )) | σ ( A k )) > d − . (74)Then (72) holds with K := − log d − P ( Y ≥ ( | h | + | h | +1 d − ) / √ d − Y ∼ N (0 , k ≥ 1, note that by construction of B ∗ ( z j , a (cid:48) n ), v j ( k − 1) and v j ( a n ) are the only verticesof ∂A k at distance less than a n of v j ( k ). Let ( X s ) s ≥ be a discrete time SRW started at v j ( k ),and τ := inf { s ≥ , d G n ( v j ( k ) , X s ) ≥ a n } . Write H for the hitting time of A k by ( X s ). Letting a := P G n v j ( k ) ( X H = v j ( k − , H < τ ) and a := P G n v j ( k ) ( X H = v j ( a n ) , H < τ ),we get as in the proof of Proposition 15 that for large enough n , for every realization of G n and ψ G n ( A k ) in S (cid:48)(cid:48) ( x, y ) ∩ E cn ∩ { ψ G n ( v j ( k − ≥ h } : E G n [ ψ G n ( v j ( k )) | σ ( A k )] > a ψ G n ( v j ( k − a ψ G n ( v j ( a n )) − log − n. Since 0 ≤ a + a ≤ ψ G n ( v j ( k − , ψ G n ( v j ( a n ))) ≥ h ≥ −| h | , (73) follows.Split V := Var G n ( ψ G n ( v j ( k )) | σ ( A k )) as follows: V = G G n ( v j ( k ) , v j ( k )) − E G n v j ( k ) (cid:2) G G n ( v j ( k ) , X H ) { H<τ } (cid:3) − E G n v j ( k ) (cid:2) G G n ( v j ( k ) , X H ) { H ≥ τ } (cid:3) + E G n v j ( k ) [ H ] E G n π n [ H ] E G n π n [ G G n ( v j ( k ) , X H )] . By (11), (8) and (9), if κ is large enough, for n large enough, G G n ( v j ( k ) , v j ( k )) − a G G n ( v j ( k ) , v j ( k − − a G G n ( v j ( k ) , v j ( a n )) > d − d − − a + a d − − log − n ≥ d − − log − n. As below (45), we get that (cid:12)(cid:12)(cid:12) E G n v j ( k ) (cid:2) G G n ( v j ( k ) , X H ) { H<τ } (cid:3) − a G G n ( v j ( k ) , v j ( k − − a G G n ( v j ( k ) , v j ( a n )) (cid:12)(cid:12)(cid:12) ≤ log − n (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) E G n v j ( k ) (cid:2) G G n ( v j ( k ) , X H ) { H ≥ τ } (cid:3) − E G n v j ( k ) [ H ] E G n π n [ H ] E G n π n [ G G n ( v j ( k ) , X H )] (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ log − n. These three inequalities imply that for large enough n , for every realization of G n and ψ G n ( A k )in S (cid:48)(cid:48) ( x, y ) ∩ E cn ∩ { ψ G n ( v j ( k − ≥ h } :Var G n ( ψ G n ( v j ( k )) | σ ( A k )) ≥ d − − − n > d − . This shows (74) and the proof is complete. E ≥ hψ G n Write x h ↔ y if x and y are in the same connected component of E ≥ hψ G n , for x, y ∈ V n . In thissection, we prove (2) of Theorem 1 via an argument on the number of pairs of vertices suchthat x h ↔ y . Let S n be the set of pairs of distinct x, y ∈ V n such that x h ↔ y . Let A n ⊂ V n bethe set of vertices x such that |C G n ,hx | ≤ n / .We first suppose that the following two Lemmas hold, and show (2). Then, we derive themfrom Propositions 19 and 21, using a second moment argument. Lemma 22. For every ε > , lim n → + ∞ P ann ( | A n | ≥ (1 − η ( h ) − ε ) n ) = 1 . Lemma 23. For every ε > , lim n → + ∞ P ann ( | S n | ≥ ( η ( h ) / − ε ) n ) = 1 .Proof of (2). Fix ε > 0. Remark that every connected component of E ≥ hψ G n is either includedin A n , or does not intersect A n . Let ( (cid:98) γ ( n ) i ) i ≥ (resp. ( γ ( n ) i ) i ≥ ) be the sizes of the connectedcomponents in A n (resp. not in A n ), listed in decreasing order.Let E n := {| A n | ≥ (1 − η ( h ) − ε ) n } ∩ {| S n | ≥ ( η ( h ) / − ε ) n } . On E n , we have that (cid:80) i ≥ (cid:98) γ ( n ) i ( (cid:98) γ ( n ) i − 1) + (cid:80) i ≥ γ ( n ) i ( γ ( n ) i − 1) = 2 S n ≥ η ( h ) n − εn .Moreover, we have by definition of A n : (cid:80) i ≥ (cid:98) γ ( n ) i ( (cid:98) γ ( n ) i − ≤ (cid:80) i ≥ (cid:98) γ ( n ) i √ n ≤ n / .Thus, for n large enough, γ ( n )1 ( | V n | − | A n | ) ≥ (cid:88) i ≥ γ ( n ) i ( γ ( n ) i − ≥ ( η ( h ) − ε ) n . But | V n | − | A n | ≤ ( η ( h ) + ε ) n , so that γ ( n )1 ≥ η ( h ) − εη ( h ) + ε n ≥ (( η ( h ) − η ( h ) − ε ) n. Since γ ( n )1 is the cardinality of a set included in V n \ A n , one has γ ( n )1 ≤ ( η ( h ) + ε ) n . Note thatfor n large enough, γ ( n )1 > √ n ≥ (cid:98) γ ( n )1 . Therefore, γ ( n )1 = |C ( n )1 | , and we have(( η ( h ) − η ( h ) − ε ) n ≤ |C ( n )1 | ≤ ( η ( h ) + ε ) n E n . By Lemmas 22 and 23, lim n → + ∞ P ann ( E n ) = 1. Since ε was arbitrary, the proof iscomplete. Remark 2. Note that we have |C ( n )2 | = max( γ ( n )2 , (cid:98) γ ( n )1 ) . Since on E n , (cid:98) γ ( n )1 ≤ √ n and γ ( n )2 ≤ | V n | − | A n | − γ ( n )1 ≤ (1 + 4 η ( h ) − ) εn ,we get that |C ( n )2 | /n P ann −→ .Proof of Lemma 22. Let A (cid:48) n be the set of vertices such that their lower exploration (Section 5.2)is aborted. By Remark 1, for n large enough, A (cid:48) n ⊆ A n , and it is enough to prove the Lemmafor A (cid:48) n instead of A n .Let ε ∈ (0 , n large enough and every x ∈ V n , we have | P ann ( x ∈ A (cid:48) n ) − (1 − η ( h )) | ≤ ε. (75)We claim that for n large enough, for all distinct x, y ∈ V n , | Cov ann ( x ∈ A (cid:48) n , y ∈ A (cid:48) n ) | ≤ ε. (76)Indeed, Cov ann ( x ∈ A (cid:48) n , y ∈ A (cid:48) n ) = P ann ( x, y ∈ A (cid:48) n ) − P ann ( x ∈ A (cid:48) n ) P ann ( y ∈ A (cid:48) n ). Then by (75),we have | P ann ( x ∈ A (cid:48) n ) P ann ( y ∈ A (cid:48) n ) − (1 − η ( h )) | ≤ ε + ε ≤ ε .Perform successively the lower explorations from x and then from y as in Section 5.2 (with theadditional condition C7). We get P ann (C7 happens) = o (1) in the same way than (52). Then,revealing ψ G n on R x ∪ R y and comparing it to ϕ T d as below (53), we obtain | P ann ( x, y ∈ A (cid:48) n ) − (1 − η ( h )) | ≤ ε .This shows (76).We apply Bienaym´e-Chebyshev’s inequality: P ann ( | A (cid:48) n | ≤ (1 − η ( h ) − ε / ) n ) ≤ P ann ( | | A (cid:48) n | − E ann [ | A (cid:48) n | ] | ≥ ε / n ) ≤ √ εn (cid:88) x,y ∈ V n Cov ann ( x ∈ A (cid:48) n , y ∈ A (cid:48) n ) ≤ n + n ( n − ε √ εn ≤ √ ε for n large enough. Since ε can be taken arbitrarily small, the proof is complete. Proof of Lemma 23. Let ε ∈ (0 , S ∗ n the set of pairs x, y ∈ V n such that S ∗ ( x, y ) holds.Since S ∗ n ⊆ S n , it is enough to prove the Lemma for S ∗ n instead of S n . First, by Proposition 21,38 ann [ | S ∗ n | ] ≥ ( η ( h ) − ε ) n ( n − ≥ ( η ( h ) / − ε ) n for large enough n . Second, we claim that for n large enough and all distinct x, y, w, t ∈ V n , | Cov ann ( S ∗ ( x,y ) , S ∗ ( w,t ) ) | ≤ ε. (77)Remark that if this holds, then we can conclude by a second moment computation as in theproof of Lemma 22. We haveCov ann ( S ∗ ( x,y ) , S ∗ ( w,t ) ) = P ann ( S ∗ ( x, y ) ∩ S ∗ ( w, t )) − P ann ( S ∗ ( x, y )) P ann ( S ∗ ( w, t )).By Proposition 21, for n large enough, | P ann ( S ∗ ( x, y )) P ann ( S ∗ ( w, t )) − η ( h ) | ≤ ε. (78)Now, perform successively the exploration of Section 5.1 from x , then from y , then from z and finally from t (via an array of i.i.d. standard normal variables ( ξ v,k,i ) v ∈{ x,y,w,t } ,k,i ≥ ). Weadd the following condition: for any v ∈ { x, y, w, t } , the exploration from v is stopped assoon as it meets a vertex seen in a previous exploration. The probability that this happens is o (1) by Remark 1 and (14), since o ( √ n ) vertices and half-edges are revealed during these fourexplorations. Therefore, as for (53), we get that for n large enough, P ann (the explorations from x, y, z, t are all successful) ∈ ( η ( h ) − ε/ , η ( h ) − ε/ . (79)If these explorations are successful, develop balls from ∂T x to ∂T y as described in Section 6.1,with R j := R x ∪ R y ∪ R w ∪ R t ∪ B ∗ j for z j ∈ ∂T x . Then do the same from ∂T w to ∂T t , this timewith R j := R x ∪ R y ∪ ( ∪ z ∈ ∂T x B ∗ ( z, a (cid:48) n )) ∪ R w ∪ R t ∪ B ∗ j for z j ∈ ∂T w . Finally, reveal ψ G n on T x , T y , T w , T t and on the joining balls from T x to T y and from T w to T t , in that order.One can adapt readily the proof of Lemma 20 to show that with P ann -probability 1 − o (1), ifthe four explorations are successful then there are at least log γ − κ − n joining balls from ∂T x (resp. ∂T w ) to ∂T y (resp. ∂T t ). Note in particular that the estimations of (60), (62) and (63)still hold. It is also straightforward to carry the arguments of the proof of Proposition 21, andwe finally have | P ann ( S ∗ ( x, y ) ∩ S ∗ ( w, t )) − P ann (the explorations from x, y, w, t are all successful) | ≤ ε/ In this Section, we prove (3). We start by the lower bound in Section 7.1, showing the existence P ann -w.h.p. of a component (different from C ( n )1 ) having Θ(log n ) vertices.Then, to show that |C ( n )2 | = O (log n ) P ann -w.h.p., we perform an exploration of a new kind,starting from some x ∈ V n . It consists of three phases (Sections 7.2 to 7.4), during which39e assign a pseudo-GFF (cid:100) ψ G n to the vertices that we visit. (cid:100) ψ G n is defined via a recursiveconstruction that mimics Proposition 6, as long as there are no cycles (in Section 5, the analogousof the pseudo-GFF was ϕ T d on T x ). Finally, we reveal the true values of ψ G n one by one onthe set of vertices we have explored via Proposition 5 (Section 7.5), and show that either |C G n ,hx | = O (log n ), or |C G n ,hx | = Θ( n ), in which case C G n ,hx = C ( n )1 by Remark 2. Contrary toSection 5, we need this alternative to hold for every x ∈ V n , P ann -w.h.p. By a union bound, itis enough to prove thatfor x ∈ V n , P ann ( {|C G n ,hx | = O (log n ) } ∪ {|C G n ,hx | = Θ( n ) } ) = o (1 /n ). (80) First phase (Section 7.2). We explore the connected component C of x ∈ V n in the set { y ∈ V n , (cid:100) ψ G n ( y ) ≥ h − n − a } for some constant a > 0. More precisely, we give a mark to eachvertex y such that | (cid:100) ψ G n ( y ) − h | ≤ n − a , and explore each connected component C (cid:48) of C \ M ,where M is the set of marked vertices, until- (i) C (cid:48) is fully explored and has no more than O (log n ) vertices, or- (ii) K log n vertices of C (cid:48) have been seen but not yet explored, for some constant K fixed inthe second phase.We replace a n of (33) by a ”security radius” r n = Θ(log n ). Adapting Proposition 15 (Lemma 30),this allows us in Section 7.5 to bound the difference between (cid:100) ψ G n and ψ G n by n − a , so that forevery connected component C (cid:48) of C \ M , either C (cid:48) ⊆ C G n ,hx or C (cid:48) ∩ C G n ,hx = ∅ .If we kept a n , (cid:100) ψ G n would approximate ψ G n only with precision log − Θ(1) n . With probabilityΘ(1 /n ), there would be too much vertices y such that | (cid:100) ψ G n ( y ) − h | ≤ log − Θ(1) n , hence for whichwe cannot know by anticipation whether they will be in C G n ,hx or not.Moreover, we do not have P (C1 happens) = O (1 /n ) as soon as the number of vertices exploredgoes to infinity with n . We will need to accept the possible occurrence of one cycle. When thishappens, we have to define (cid:100) ψ G n in a slightly different manner. In Section 7.5, we need a variantof Lemma 30 to control the difference between (cid:100) ψ G n and ψ G n (Lemma 31). Second phase (Section 7.3). If (i) happens for every component C (cid:48) , the exploration is over.For each C (cid:48) such that (ii) happens, we explore its K log n remaining vertices, this time in afashion similar to Section 5.1. Each of these explorations has a probability bounded away from0 to be successful. If K is large enough, with probability at least 1 − o (1 /n ), at least one ofthese explorations is successful, and has a boundary of size Θ( n / b n ). Third phase(Section 7.4). For every C (cid:48) such that (ii) happens, we show that the successfulexploration of the second phase is connected to a positive proportion of the vertices of V n , viaan adaptation of the joint exploration in Section 6.1. This yields (80). In this section, we prove the existence of K > P ann ( |C ( n )2 | ≥ K − log n ) → . (81)40o show the existence of x ∈ V n such that the size of its connected component is exactly oforder log n requires an exploration in which we compare C G n ,hx to both C h +log − n ◦ and C h − log − n ◦ .Then, the proof strategy simply consists in performing explorations from different vertices of V n , one after another, until one of them is successful. Proposition 11 hints that the probabilitythat C h ◦ has size c log n should be of order n − f ( c ) , where f is an unknown function of c > f ( c ) −→ c → 0. Hence, for K − = c small enough, performing at most n / explorations willbe enough. The exploration. Let K > 0. For x ∈ V n , we modify the lower exploration of Section 5.2,replacing C3 by | B C h − log − n ◦ ( ◦ , k − | ≥ K log n . If at some step k , the lower explorationis stopped exclusively because of C2, and if | B C h +log − n ◦ ( ◦ , k − | ≥ K log n , say that it issuccessful.Pick x ∈ V n . Denote S the set of vertices seen during its lower exploration (i.e. verticeshaving at least one half-edge paired during the exploration). For 0 ≤ i ≤ (cid:98) n / (cid:99) , if the lowerexploration from x i is not successful, let S i be the set of vertices seen in the lower explorations of x , . . . , x i . Pick x i +1 ∈ V n \ S i and perform its lower exploration, stopping it if a vertex of S i isseen, in which case it is not successful. Let E i,n := { the lower exploration from x i is successful } and E n := ∪ (cid:98) n / (cid:99) i =0 E i,n .Suppose that the following result holds: Lemma 24. If K is small enough, then lim n → + ∞ P ann ( E n ) = 1 . On E n , let i ≥ x i is log-size successful. ApplyingProposition 15 as below (53), we get that P ann ( C h +log − n ◦ ⊆ Φ( C G n ,hx i ) ⊆ C h − log − n ◦ ),where Φ is an isomorphism from C G n ,hx i to T d . In this case, K log n ≤ |C G n x i | ≤ K log n . Withhigh P ann -probability, |C ( n )1 | > K log n by (2), so that C G n ,hx i (cid:54) = C ( n )1 . This yields (81) with K = K − . It remains to establish the Lemma. Proof of Lemma 24. Clearly, it is enough to show that for K small enough, for n large enoughand every 1 ≤ i ≤ n / , P ann ( E i,n | ∩ i − j =0 E cj,n ) ≥ n − / , (82)since it would imply that P ann ( E n ) ≥ − (1 − n − / ) (cid:98) n / (cid:99) −→ n → + ∞ n ≥ ≤ i ≤ (cid:98) n / (cid:99) , we have E i,n ⊇ E line ,n \ ( { C1 happens } ∪ E i,n, meet ),where E i,n, meet := { the lower exploration from x i meets S i − } ,41 line ,n := {C h +log − n ◦ = C h − log − n ◦ = L (cid:98) K log n (cid:99) } and for any k ≥ L k is a ”line” subtree of T d ,i.e. it is rooted at ◦ , has k + 1 vertices and total height k .During each exploration, less than n / vertices and half-edges are seen for n large enough.Thus | S i | ≤ n / . Hence, for i ≥ 1, by (14) with k = 1, m ≤ n / , m = 1 and m E , m ≤ n / ,for n large enough: P ann ( E i,n, meet ) + P ann (C1 happens) ≤ n − / .Thus to establish (82), it only remains to show that for K small enough and n large enough, P ann ( E line ,n ) ≥ n − / . Let v , . . . , v d be the children of ◦ in T d . Remark that for n large enough, P ann ( E line ,n ) ≥ P T d ( ϕ T d ( ◦ ) ∈ [ h + 1 / , h + 1]) pp (cid:48)(cid:98) K log n (cid:99) p (cid:48)(cid:48) ,where p := inf a ∈ [ h +1 / ,h +1] P T d a ( { ϕ T d ( v ) ∈ [ h + 1 / , h + 1] } ∩ {∀ i ∈ { , . . . , d } , ϕ T d ( v i ) < h − } ), p (cid:48) := inf a ∈ [ h +1 / ,h +1] P T d a ( { ϕ T d ( v ) ∈ [ h + 1 / , h + 1] } ∩ {∀ i ∈ { , . . . , d − } , ϕ T d ( v i ) < h − } ), p (cid:48)(cid:48) := inf a ∈ [ h +1 / ,h +1] P T d a ( ∀ i ∈ { , . . . , d − } , ϕ T d ( v i ) < h − p, p (cid:48) , p (cid:48)(cid:48) > 0. Taking K < − (12 log p (cid:48) ) − yields theresult. In this section, we define the first phase of the exploration, and show that it is successful with P ann -probability 1 − n − / (Proposition 25). We will need a variant of Lemma 14, namelyLemma 26. We postpone its statement and proof to the end of this section.Let a, K, K (cid:48) > 0. For every n ∈ N , define r n := (cid:98) . 05 log d − n (cid:99) (83)Let δ ∈ (0 , h (cid:63) − h ) and (cid:96) ∈ N be such that the conclusion of Lemma 26 holds. The exploration. Let x ∈ V n . I - We first assume that we do not meet any cycle throughout the exploration. Let M be theset of marked vertices. Initially, M = ∅ . Let (cid:100) ψ G n ( x ) ∼ N (0 , d − d − ). If ϕ T d ( ◦ ) < h − n − a , stopthe exploration. Else, give a mark to x (hence add it to M ).While M (cid:54) = ∅ , pick y ∈ M and proceed to its subexploration . The subexploration. Let T y be the subexploration tree, through which we will run in a42readth-first way. Initially T y = { y } .While 1 ≤ | ∂T y | ≤ K log n , perform a step : take y ∈ ∂T y of minimal height and if y (cid:54) = x , let y be its only neighbour where (cid:100) ψ G n has already been defined. Note that if y (cid:54) = y , y is the parentof y in T y . Reveal all the edges of B G n ( y , y , r n + (cid:96) ), where we recall that B G n ( y , y , r n + (cid:96) ) isthe graph obtained by taking all paths of length r n + (cid:96) starting at y and not going through y .Since we suppose that no cycle arises, B G n ( y , y , (cid:96) ) is a tree, that we root at y . If y = y = x ,replace B G n ( y , y , r n + (cid:96) ) and B G n ( y , y , (cid:96) ) by B G n ( x, r n + (cid:96) ) and B G n ( x, (cid:96) ) respectively, anddo the same.We construct its subtree T y ( y ) in { z, (cid:100) ψ G n ( z ) ≥ h + n − a } . We start with T y ( y ) = { y } .For k = 1 , , ...(cid:96) − y k, , . . . , y k,m the children of the ( k − T y ( y ). Let ( ξ y ,k,i ) k,i ≥ be an array of i.i.d. variables of law N (0 , (cid:100) ψ G n ( y k,i ) := 1 d − (cid:100) ψ G n ( y k,i ) + (cid:114) dd − ξ y ,k,i . (84)Add y k,i to T y ( y ) if (cid:100) ψ G n ( y k,i ) ≥ h + n − a , and give a mark to y k,i (and thus add it to M ) if h − n − a ≤ (cid:100) ψ G n ( y k,i ) < h + n − a .Finally, include T y ( y ) in T y , add the vertices of ∂T y ( y ) to N y and take y away from N y . Thestep is then over.If | ∂T y | (cid:54)∈ [1 , K log n ], the subexploration is finished. Say that it is fertile if | ∂T y | > K log n ,and infertile else (hence if ∂T y = ∅ ).If M = ∅ , the exploration from y is finished. II - Suppose now that a unique cycle C arises in the subexploration of y , in the step from y ,for some y, y ∈ V n . Let m =: | C | be the number of vertices in C . There are two cases. Case 1: When C is discovered, there are already k consecutive vertices y , . . . , y k of C where (cid:100) ψ G n has been defined, for some 1 ≤ k ≤ m − 1. Reveal B G n ( C, r n ). Denote z , . . . , z m − k theremaining vertices of C , such that z (cid:54) = y is a neighbour of y , and z i is a neighbour of z i − for i ≥ 2. Give a mark to z , . . . , z m − k . Take y away from ∂T y . If y k was in ∂T (cid:101) y for some (cid:101) y whosesubexploration was performed previously, take it away from that set.43 y = y e yy k y k − z m − k z z T y ( y ) T y T e y C Figure 3. Case 1. Marked vertices are in red. C consists of the thick edges. Here, y = y .Remark that we could have y = (cid:101) y , and that Case 1 includes the event that B G n ( y , y , r n + (cid:96) )meets a vertex seen in a previous subexploration or step (as it is the case in this example).We now define (cid:100) ψ G n on the z i ’s. To do so, we mimic a recursive construction of the GFF on G m ,the infinite connected d -regular graph G m having a unique cycle C m of length m (such a con-struction always exists on a transient graph by Proposition 7, Proposition 6 being a particularcase of that on T d ). G m consists of a cycle C m of length m , with d − T + d attached toeach vertex of C m , thus it is clear that the SRW is transient and that the Green function andthe GFF are well-defined.Let u k , . . . , u , v , . . . , v m − k be the vertices of C m , listed consecutively. Let U := G m \{ u , . . . , u k } ,( X j ) j ≥ a SRW on G m and recall that T U is the exit time of U . Define α := P G m v ( X T U = u , T U < + ∞ ), β := P G m v ( X T U = u k , T U < + ∞ ) { k> } , γ := E G m v [ (cid:80) T U − j =0 { X j = v } ],and let ( ξ i ) i ≥ be a family of i.i.d standard normal variables, independent of everything else.Define (cid:100) ψ G n ( z ) := α (cid:100) ψ G n ( y ) + β (cid:100) ψ G n ( y k ) + √ γξ .Then for i ≥ 2, define recursively (cid:100) ψ G n ( z i ) := α m − ( k + i − (cid:100) ψ G n ( z i − ) + β m − ( k + i − (cid:100) ψ G n ( y k ) + (cid:112) γ m − ( k + i − ξ i (85)where we set U i := G m \ { u , . . . , u k , v , . . . , v i − } , α m − ( k + i − := P G m v i ( X T Ui = v i − , T U i < + ∞ ), β m − ( k + i − := P G m v i ( X T Ui = u k , T U i < + ∞ ) and γ m − ( k + i − := E G m v i [ (cid:80) T Ui − j =0 { X j = v i } ]. Case 2: (cid:100) ψ G n has not been defined on any vertex of C . There exists a unique path of consecutivevertices y , . . . , y j for some j ≥ y j ∈ C , and for 2 ≤ i ≤ j − (cid:100) ψ G n ( y i ) has not44een defined and y i (cid:54)∈ C . Reveal B G n ( { y . . . , y j − } ∪ C, r n ). Give a mark to the vertices of { y , . . . , y j − } ∪ C . Take y away from ∂T y . y Cy j y y y j − y Figure 4. Case 2. Marked vertices are in red.We define (cid:100) ψ G n on { y , . . . , y j − } ∪ C in a way similar to Case 1. Let ( ξ i ) i ≥ be a sequence ofi.i.d. standard normal variables, independent of everything else. For i = 1 , , . . . , j , set (cid:100) ψ G n ( y i ) := α (cid:48) j − i (cid:100) ψ G n ( y i − ) + (cid:113) γ (cid:48) j − i ξ i , (86)where α (cid:48) j − i := P G m z ( H { z (cid:48) } < + ∞ ) and γ (cid:48) j − i := E G m z [ (cid:80) H { z (cid:48)} − l =0 { X l = z } ], z, z (cid:48) being two neigh-bours in G m such that z (resp. z (cid:48) ) is at distance j − i (resp. j − i + 1) of C m . Then, define (cid:100) ψ G n on C as in Case 1 with k = 1. After the discovery of C . Resume the subexplorations as before, starting by the subex-ploration interrupted when C was discovered. For any marked vertex y : when proceeding tothe step from y in the subexploration from y , do not reveal the edges of B G n ( y , y , r n + (cid:96) ) inthe direction of already marked neighbours of y . III - If a second cycle arises, the exploration from x is over, and is not successful.If at some point,D1 M = ∅ ,D2 (cid:100) ψ G n has been defined on at most (cid:98) K (cid:48) log n (cid:99) vertices, andD3 at most one cycle has been discovered,say that the first phase of the exploration from x is successful. Denote S ( x ) this event. If allsubexploration trees were infertile, then the exploration from x is over, and said to be successful.Denote S , stop ( x ) this event. Proposition 25. Fix K, a > . If K (cid:48) is large enough, then for large enough n and every x ∈ V n , P ann ( S ( x )) ≥ − n − / . roof. In a nutshell, the argument is as follows: by our choice of (cid:96) and Lemma 26, the samereasoning as that in the beginning of Proposition 11 shows that the size of a subexplorationtree has a bounded expectation (as n → + ∞ ) with exponential moments. Hence if K (cid:48)(cid:48) > ≥ − n − , the first K (cid:48)(cid:48) log n subexplorations encompass less than K (cid:48)(cid:48) log n vertices. The total number of vertices seen is at most K (cid:48)(cid:48) log n × ( d − r n ≤ n / .In that case, Lemma 2 entails that we see at most one cycle, with probability at least 1 − n − / .This allows to control |M| , since a cycle brings less than 2 d (cid:96) (cid:48) r n = Θ(log n ) marked vertices.Without cycles, each vertex where (cid:100) ψ G n is defined has a chance O ( n − a ) to get a mark. Hence for K (cid:48)(cid:48) large enough, with probability at least 1 − n − / , less than K (cid:48)(cid:48) log n vertices get a mark inthe first K (cid:48)(cid:48) log n explorations, ensuring that the exploration from x is finished - and successful.Let E ,n be the event that two cycles are discovered before n / vertices have been seen duringthe exploration from x . By (14) with k = 2, m = 1, m E = 0 and m ≤ n / , P ann ( E ,n ) ≤ n − / . (87)Suppose that we perform the subexploration from some vertex y . Let y ∈ ∂T y , with y (cid:54) = y .If no cycle arises when revealing the r n + (cid:96) offspring of y , then | ∂T y ( y ) | − 1, the increment of | N y | during that step, dominates stochastically ρ (cid:96),h,δ − 1, by definition of ρ (cid:96),h,δ at Lemma 26,as soon as n − a < δ . If a cycle arises (which happens at most once on E c ,n ), at most twovertices are marked and taken away from N y . After j steps (if the subexploration is not over), | N y | ≥ st. S j − − 2, where S j − is the sum of j − ρ (cid:96),h,δ − 1, hencetaking values in the bounded interval [0 , d (cid:96) ], and with a positive expectation by Lemma 26 (the − j − y , and the +2 from thepossibility that two vertices can be taken away from N y if there is a cycle). Hence P ann (the subexploration from y lasts more than j steps) ≤ P ( S j − ≤ K log n + 2)By the exponential Markov inequality, there exist constants c, c (cid:48) > j ≥ K ( E [ ρ (cid:96),h,δ ] − − log n , P ( S j − ≤ K log n + 2) ≤ ce − c (cid:48) j .Now, let K (cid:48)(cid:48) > (cid:98) K (cid:48)(cid:48) log n (cid:99) subexplorations (or on all subexplorationsif there are less than (cid:98) K (cid:48)(cid:48) log n (cid:99) of them). Let N be the total number of steps during thosesubexplorations. On E c ,n , N is stochastically dominated by a sum S of (cid:98) K (cid:48)(cid:48) log n (cid:99) i.i.d. variablesof some law µ (independent of n ) such thatfor every j ≥ K ( E [ ρ (cid:96),h,δ ] − − log n , µ ([ j, + ∞ )) ≤ ce − c (cid:48) j .Hence, letting E ,n := { N ≥ K (cid:48)(cid:48) log n } , by the exponential Markov inequality, P ann ( E ,n ∩ E c ,n ) ≤ P ann ( S ≥ K (cid:48)(cid:48) (cid:98) K (cid:48)(cid:48) log n (cid:99) ) ≤ ( E [ e c (cid:48) Y/ ] e − c (cid:48) K (cid:48)(cid:48) µ/ ) (cid:98) K (cid:48)(cid:48) log n (cid:99) n and for Y ∼ µ . Taking K (cid:48)(cid:48) large enough, we have for n large enough: P ann ( E ,n ∩ E c ,n ) ≤ n − . (88)In a step of a subexploration, less than ( d − r n ≤ n / new vertices are seen. When a cycle C is revealed, there are at most 2 r n new vertices in C (and on the path leading to C , in Case2), so that less than 3 r n ( d − r n ≤ n / new vertices are seen. Therefore, on E c ,n ∩ E c ,n , lessthan n / vertices are seen during the first (cid:98) K (cid:48)(cid:48) log n (cid:99) subexplorations.Moreover, in a step of a subexploration, less than d (cid:96) vertices are added to the subexplorationtree. Hence if K (cid:48) > K (cid:48)(cid:48) d (cid:96) , on E c ,n , D2 holds.We now estimate the total number of vertices that will receive a mark. When a cycle C appears,our construction implies that less than 3 r n vertices receive a mark. When performing m stepsin a subexploration, the number of marked vertices obtained is stochastically dominated by abinomial random variable Bin( m, d (cid:96) n − a ). Indeed, at each step, we reveal (cid:100) ψ G n (and thus ϕ T d )on less than d (cid:96) vertices. And for any vertex y ∈ T d \ {◦} , by Proposition 6,max a (cid:48) ≥ h P T d ( ϕ T d ( y ) ∈ [ h − n − a , h + n − a ] | ϕ T d ( y ) = a (cid:48) ) ≤ n − a (cid:112) πd/ ( d − ≤ n − a . Hence, if E ,n is the event that more than 3 r n + K (cid:48)(cid:48) log n marks are given during the first (cid:98) K (cid:48)(cid:48) log n (cid:99) subexplorations and if Z ∼ Bin( (cid:100) K (cid:48)(cid:48) log n (cid:101) , C d (cid:96) n − a ), P ann ( E c ,n ∩ E c ,n ∩ E ,n ) ≤ P (cid:18) Z ≥ K (cid:48)(cid:48) n (cid:19) ≤ (cid:18) (cid:100) K (cid:48)(cid:48) log n (cid:101) K (cid:48)(cid:48) log n (cid:19) ( C d (cid:96) n − a ) K (cid:48)(cid:48) log n , thus by(12), for large enough n , P ann ( E c ,n ∩ E c ,n ∩ E ,n ) ≤ (cid:16) (cid:100) K (cid:48)(cid:48) log n (cid:101) C d (cid:96) n − a (cid:17) K (cid:48)(cid:48) log n ≤ n − . (89)Taking K (cid:48)(cid:48) > 1, on E c ,n ∩ E c ,n ∩ E c ,n , less than 1 + 3 r n + K (cid:48)(cid:48) log n ≤ (cid:98) K (cid:48)(cid:48) log n (cid:99) vertices receivea mark during the first (cid:98) K (cid:48)(cid:48) log n (cid:99) subexplorations, so that M = ∅ after at most (cid:98) K (cid:48)(cid:48) log n (cid:99) subexplorations. Therefore, on E c ,n ∩ E c ,n ∩ E c ,n , conditions D1, D2 and D3 hold. By (87), (88)and (89), for large enough n P ann ( E c ,n ∩ E c ,n ∩ E c ,n ) ≥ − n − − n − / ≥ − n − / , and this concludes the proof.We state here the variant of Lemma 14. Let δ ∈ [0 , h (cid:63) − h ). For (cid:96) ≥ 1, write Z (cid:96) for the (cid:96) -thgeneration of the connected component of ◦ in ( {◦} ∪ E ≥ h + δϕ T d ) ∩ T + d . Let ρ (cid:96),h,δ be the law of |Z (cid:96) | conditionally on ϕ T d ( ◦ ) = h , so that ρ (cid:96),h = ρ (cid:96),h, (recall (28)). The following result is astraightforward consequence of Lemma 14 (applied to h + δ instead of h ) and from the fact that P T d h ( ∃ v ∈ ∂B ( ◦ , , ϕ T d ( v ) ≥ h + δ ) > Lemma 26. For every δ ∈ [0 , h (cid:63) − h ) , if (cid:96) is large enough, E [ ρ (cid:96),h,δ ] > . .3 Second phase If S , stop ( x ) holds, i.e. all subexploration trees are infertile, the exploration is over. In thisSection, we suppose that S ( x ) \S , stop ( x ) holds. For every fertile tree, we perform an explorationsimilar to that of Section 5.1 from each vertex of its boundary, and show that with probability atleast 1 − n − / , at least one exploration per fertile tree is successful, and hence has a boundaryof size Θ( n / b n ) (Proposition 27). We illustrate this second phase in Figure 5 below.Let T , . . . T m be the fertile subexploration trees for some positive integer m . For every q ∈{ , . . . , m } , denote y q, , y q, , . . . the vertices of ∂T q . For q = 1 , , . . . successively, we perform theexplorations from y q,i , i ≥ ξ y q,i ,k,j ) k,j ≥ ). If y is the j -th vertex of the k -th generation in the explorationtree of y q,i , let recursively (cid:100) ψ G n ( y ) := (cid:100) ψ G n ( y ) d − (cid:114) dd − ξ y q,i ,k,j , so that (cid:100) ψ G n plays the role of ϕ T d on T d in Section 5.1. We implement two modifications: • we do not explore towards y q,i , the parent of y q,i in T q (hence we identify (cid:100) ψ G n to the GFFon a subtree of T + d instead of T d ), and • we do not stop the exploration if (cid:100) ψ G n ( y q,i ) < h + log − n (we only know a priori from thefirst phase that (cid:100) ψ G n ( y q,i ) ≥ h + n − a ), and • we stop the exploration if it meets a vertex already discovered in the first phase or duringthe previous exploration of some y q (cid:48) ,j (thus with q (cid:48) < q , or q (cid:48) = q and j < i ).A vertex y q,i whose exploration is successful is back-spoiled if one vertex of its exploration isseen later during the exploration of y q (cid:48) ,j . Let E ,n := {∃ q ≤ m, all successful explorations of y q, , y q, , . . . , are back-spoiled } On S ( x ) := ( S ( x ) \ S , stop ( x )) ∩ E c ,n ,say that the second phase is successful. Proposition 27. If K of Proposition 25 is large enough, then for n large enough and every x ∈ V n , P ann ( S ( x ) ∪ S , stop ( x )) ≥ − n − / (90) Proof. Say that y q,i is spoiled if it is met during the previous exploration of some y q (cid:48) ,j . Define E ,n := { at least (cid:98) n (cid:99) vertices are spoiled or back-spoiled } . We claim that for n largeenough, P ann ( E ,n ) ≤ n − , (91)48nd for some constant K > d and h ), for each of these vertices y q,i , P ann (the exploration from y q,i is successful | y q,i is not spoiled) ≥ K . (92)On E c ,n , if (92) holds, there are at least ( K − n non-spoiled vertices on each ∂T q (andat most ( K + 1) log n since each step of a subexploration brings less than log n vertices to ∂T q ),where K was defined in the beginning of Section 7.2. And, if more than (cid:98) n (cid:99) vertices ofeach ∂T q are successful, one of them will be successful and not back-spoiled, thus fulfilling therequirement of S ( x ).If K is large enough, for n large enough, the probability that no more than (cid:98) n (cid:99) explo-rations from ∂T q are successful is at most (cid:0) (cid:98) ( K +1) log n (cid:99)(cid:98) n (cid:99) (cid:1) (1 − K ) ( K − n ≤ n (cid:16) ( K +1) K +1 ( K − K − (1 − K ) K − (cid:17) log n = o ( n − )by Stirling’s formula. Hence by a union bound on 1 ≤ q ≤ m , noticing that m ≤ K (cid:48) /K for n large enough by D2, we have: P ann (( S ( x ) ∪ S , stop ( x )) c ) ≤ P ann ( S ( x ) c ) + P ann ( E c ,n ) + mn − ≤ n − / by Proposition 25 and (91), and this concludes the proof. Hence, it remains to establish (91)and (92). x C infertile subtree infertilesubtree T T T T y ; y ; y ; y ; y ; y ; y ; y ; y ; y ; Figure 5. Marked vertices are in red. There are two infertile and four fertile subexplorationtrees. Lightgray areas correspond to the explorations of the y q,i ’s in the second phase. y , isspoiled by y , . y , is back-spoiled by y , . Each of the subxploration trees (delimited by the49arked vertices) will be either included in C G n ,hx or have no common vertex with C G n ,hx ,depending on the value of ψ G n on the marked vertices. Proof of (91). Note that by D2, by Remark 1 and by (83),less than n / log − n vertices and half-edges have been seen in the first two phases. (93)In particular, less than n / log − n edges are built during second phase. Since there are at most K (cid:48) log n y q,i ’s by D2, each new edge has a probability at most K (cid:48) log n/ ( n − n / log − n ) ≤ K (cid:48) n − log n to spoil a vertex. Thus, the number of spoiled vertices is stochastically dominated by a randomvariable Z ∼ Bin( n / log − n, K (cid:48) n − log n ). For n large enough, P ( Z ≥ ≤ (cid:0) n / log − n (cid:1) (cid:16) K (cid:48) log nn (cid:17) ≤ n nn ≤ n − .Moreover, by (15) with k = (cid:98) 999 log n (cid:99) and m , m , m E , m ≤ n / log − n due to (93), P ann (more than (cid:98) 999 log n (cid:99) vertices are back-spoiled) ≤ n − .(91) follows. Proof of (92). By (93) and (14) with k = 1 and m , m , m E , m ≤ n / log − n , for n largeenough, P ann (a cycle is created during the exploration from y q,i ) ≤ log − n. The law of (cid:100) ψ G n on the exploration tree from y q,i is that of ϕ T d on an isomorphical subtree of T + d (and not T d , since we do not explore towards y q,i ), with ϕ T d ( ◦ ) = (cid:100) ψ G n ( y q,i ). Denote C n ◦ theconnected component of ◦ in E ≥ h +log − n, + ϕ T d ∪ {◦} , and Z k its k -th generation for every k ≥ P ann (the exploration from y q,i is successful | y q,i is not spoiled) ≥ p n − log − n ,where p n := min b ≥ h + n − a P T d b ( ∃ k ≤ log λ h n, |Z k | ≥ n / b n ) (recall that (cid:100) ψ G n ( y q,i ) ≥ h + n − a ).Let δ ∈ (0 , h (cid:63) − h ). Clearly, there exists p (cid:48) > n large enough,min b ≥ h + n − a P T d b ( ∃ v ∈ Z , ϕ T d ( v ) ≥ h + δ ) > p (cid:48) .For ε > d − ( λ h + δ − ε ) ≥ (3 log d − λ h ) / ε exists by continuityof h (cid:48) (cid:55)→ λ h (cid:48) , Proposition 9), for n ∈ N , p (cid:48)(cid:48) n := min b ≥ h + δ P T d b ( ∃ k ≤ log λ h n − , |Z h +log − n, + k | ≥ n / b n ) ≥ min b ≥ h + δ P T d b ( |Z h + δ, + (cid:98) log λh n (cid:99) − | ≥ n / b n ) ≥ min b ≥ h + δ P T d b ( |Z h + δ, + (cid:98) log λh n (cid:99) − | ≥ ( λ h + δ − ε ) (cid:98) log λh n (cid:99) − ) . By Proposition 13, lim inf n → + ∞ p (cid:48)(cid:48) n =: p (cid:48)(cid:48) > 0. Since p n ≥ p (cid:48) p (cid:48)(cid:48) n for all n ≥ P ann (the exploration from y q,i is successful | y q,i is not spoiled) ≥ p n − log − n ≥ p (cid:48) p (cid:48)(cid:48) for n large enough, and we can take K = p (cid:48) p (cid:48)(cid:48) . This shows (92).50 .4 Third phase Suppose now that we are on S ( x ). For 1 ≤ q ≤ m , denote y q one vertex of ∂T q whoseexploration was successful and not back-spoiled in the second phase, T y q its exploration tree,and B q the boundary of T y q . In this section, we connect each B q to Θ( n ) vertices in a fashionsimilar to Section 6.1. However, revealing the GFF on a positive proportion of the verticeswould prevent us to use an approximation of the GFF as in Proposition 15.To circumvent this difficulty, denote R , the set of vertices seen in the first two phases, andpartition V n \ R , arbitrarily in sets D , D , . . . , D r for some r ∈ N such that | D | = | D | = . . . = | D r − | = (cid:98) K log n (cid:99) , (94)for some constant K > 0. We will connect each B q to a positive proportion of the verticesof D only, with P ann -probability 1 − n − (Proposition 28), before revealing the GFF on thevertices of the first two phases, and on the connexion from D to B q in Section 7.5. The resultwill follow by symmetry of the D i ’s and a union bound on 1 ≤ i ≤ r − The exploration.1) The w -explorations. Let w , . . . , w (cid:98) K log n (cid:99) be the vertices of D . We proceed successivelyto the w -explorations of w , w , . . . , i.e. for i ≥ 1, we perform the exploration from w i as inSection 5.1, but stop it if we reach a vertex seen in the first two phases or in the w -explorationof some w j , j ≤ i − 1. In particular, if w i was discovered during the exploration of some w j , j ≤ i − 1, say then that w i is w - spoiled and do not proceed to its w -exploration. Denote R w the set of vertices seen during all the w -explorations.For i ≥ 1, if we explore w i and C3 happens, say that the w -exploration from w i is w -successful .Let s be the number of w -successful vertices. Let w i , . . . , w i s be the w -successful vertices with i < . . . < i s . Let T w ij be the exploration tree of w i j , for j ∈ { , . . . , s } . Take away • from each ∂T w ij : the vertices that are seen in the w -exploration of some w i , for i > i j , • from each B q : the vertices z such that B G n ( z, z, a n ) intersects R w .Say that those vertices are w -back-spoiled . 2) The joining balls. For q = 1 , . . . , m successively, we develop balls from B q to the ∂T w ij ’s,1 ≤ j ≤ s , with a few modifications w.r.t the construction of Section 6.1.1: let z ,q , z ,q , . . . bethe vertices of B q . For z i,q ∈ B q , let B ∗ i,q := ∪ ( i (cid:48) ,q (cid:48) ): q (cid:48) Proof of (95). We adapt the proof of Lemma 20. Since there are less than log n w i ’s(and | R , | is controlled by (93)), we can replace (60) byless than n / log γ +2 n vertices are seen during the three phases . (97)Let B ∗ := ∪ mq =1 ∪ z ∈ B q B ∗ ( z, a (cid:48) n ). (62) becomes P ann ( S ( x ) ∩ { | B ∗ ∩ ( ∪ s j =1 B G n ( T z ij , a n )) | ≥ log γ n } ) ≤ n − (98)Let N be the number of vertices of ∪ mq =1 B q that are spoiled, i.e. the vertices z such that B ∗ ( z, a n ) = B G n ( z, z, a n ) is hit by a previously constructed B ∗ ( z (cid:48) , a (cid:48) n ). (63) becomes P ann ( S ( x ) ∩ { N ≥ log γ n } ) ≤ n − . (99)In addition, by (15) with k = log n , m = 1, m , m E , m ≤ n / log − / n , P ann ( E ,n ) ≤ n − , (100)where E ,n := S ( x ) ∩ (cid:8) more than log n vertices are w -back-spoiled (cid:9) . Then, define S i,q := S ( x ) ∩ E c ,n ∩ {| ( ∪ s j =1 B G n ( T w ij , a n )) ∩ B ∗ i,q | ≤ log γ n } ∩ { z i,q is not spoiled } .It is straightforward to adapt the proof of (65) to get that for every 1 ≤ J ≤ s , the probabilitythat B ∗ ( z i,q , a (cid:48) n ) is a J -joining ball, conditionally on S i,q , is at least n − / log γ − κ − n . On S ( x ) ∩ { N ≤ log γ n } ∩ E c ,n , at least n / b n − log n − log γ n ≥ n / (cid:98) log − κ − n (cid:99) z i,q ’s are neither spoiled nor w -back-spoiled by (46). As in the end of the proof of Lemma 20,if γ > κ + 18, we get that for n large enough: for every ( J, q ) ∈ ( { , . . . , s } ∩ { , . . . , m } ) and Z ∼ Bin( n / (cid:98) log − κ − n (cid:99) , n − / log γ − κ − n ): P ann ( S ( x ) ∩ { there are less than log γ − κ − n J -joining balls from B q } ) ≤ P ann (cid:16) S ( x ) ∩ ( { | B ∗ ∩ ( ∪ s j =1 B G n ( T z ij , a n )) | ≥ log γ n } ∪ { N ≥ log γ n } ∪ E ,n ) (cid:17) + P ( Z ≤ log γ − κ − n ) ≤ n − + n max ≤ k ≤ log γ − κ − n P ( Z = k ) by (98), (99) and (100) ≤ n − . s ≤ | D | ≤ log n by (94) and m ≤ K (cid:48) log n by D2, a union bound on ( J, q ) yields (95). Proof of (96). We now estimate the probability that at most log n w -explorations are w -successful. Note that by Remark 1 and (93), | R , | + | R w | ≤ n / log − / n. (101)Hence if C > k = (cid:98) C log n (cid:99) and m , m , m E , m ≤ n / log − / n , P ann ( S ( x ) ∩ { more than C log n w i ’s are w -spoiled } ) ≤ n − . (102)Moreover, for every i ≥ 1, conditionally on the fact that w i is not w -spoiled, the probability thatthe w -exploration from w i is stopped because it reaches a vertex of R , or a vertex seen in theexploration of some w j , j < i is o (1) by (14) with k = 1, m = 1, m , m E , m ≤ n / log − / n .Hence, a straightforward adaptation of the proof of (53) yields P ann (the exploration from w i is w -successful | S ( x ) ∩ { w i is not w -spoiled } ) ≥ η ( h ) / . (103)Take K > C . By (102), (103) and (94), if Z ∼ Bin( (cid:98) ( K log n ) / (cid:99) , η ( h ) / 2) and n is largeenough, P ann ( S ( x ) ∩ { s ≤ log n } ) ≤ n − + P ( Z ≤ log n ) . One checks easily that if K is large enough, then for large enough n , P ( Z ≤ log n ) ≤ n − , and (96) follows. ψ G n on the three phases Let R ψ (resp. R ψ ) be the set of vertices where (cid:100) ψ G n has been defined during the first (resp.second) phase, and R ψ be the set of vertices in the w -successful w -explorations and on the J -joining balls, for all 1 ≤ J ≤ m , on which we will realize ψ G n on the third phase.By Proposition 5, we can realize ψ G n jointly with G n by • proceeding to the three phases of the exploration from x , • revealing the remaining pairings of half-edges of the G n , • defining ψ G n on R ψ ∪ R ψ , in the same order as (cid:100) ψ G n has been defined, using the samestandard normal variables: we let ψ G n ( x ) := (cid:100) ψ G n ( x ) (cid:113) d − d − (cid:112) G G n ( x, x ),53nd for every y , we let A y := { z ∈ V n , (cid:100) ψ G n ( z ) was defined before (cid:100) ψ G n ( y ) } and ψ G n ( y ) := E G n [ ψ G n ( y ) | σ ( A y )] + ξ y (cid:112) Var( ψ G n ( y ) | σ ( A y )) , for every y ∈ R ψ , where ξ y is the normal variable used when defining (cid:100) ψ G n ( y ), • revealing ψ G n on R ψ , and finally on V n \ ( R ψ ∪ R ψ ∪ R ψ ).Let E ,n := {G n is a good graph } ∩ { max z ∈ V n | ψ G n ( z ) | ≤ log / n } , S ψ ( x ) := S ( x ) ∩ { sup y ∈ R ψ | ψ G n ( y ) − (cid:100) ψ G n ( y ) | ≤ n − a / } , S ψ ( x ) := S ( x ) ∩ S ψ ( x ) ∩ { sup y ∈ R ψ | ψ G n ( y ) − (cid:100) ψ G n ( y ) | ≤ (log − n ) / } , and S ψ ,i ( x ) := S ψ ( x ) ∩ {∀ q ∈ { , . . . , m } at least log n vertices of D i are connected to T q in E ≥ hψ G n } for every i ≥ a > n largeenough: P ann (( S ( x ) \ S ψ ( x )) ∩ E ,n ) ≤ n − , (104) P ann (( S ( x ) \ S ψ ( x )) ∩ E ,n ) ≤ n − , (105)and for every 1 ≤ i ≤ r − P ann (( S ( x ) \ S ψ ,i ( x )) ∩ E ,n ) ≤ n − . (106)Letting S ψ , stop ( x ) := S , stop ( x ) ∩ S ψ ( x ), (104), (105), 106) and (90) imply that P ann (( S ψ , stop ( x ) ∪ ( ∩ r − i =1 S ψ ,i ( x )) ) ∩ E ,n ) ≥ − n − / . (107)On S ψ , stop ( x ) ∪ ( ∩ r − i =1 S ψ ,i ( x )), we have the following alternative: • either C G n ,hx contains a subexploration tree T q whose exploration was fertile, the explo-ration from y q is successful and connected to at least log n vertices of every D i , 1 ≤ i ≤ r − E ≥ hψ G n ; • or C G n ,hx contains no such tree, and C G n ,hx ⊆ R ψ , so that |C G n ,hx | ≤ K log n , where we take K ≥ K (cid:48) , and where K (cid:48) is the constant of Proposition 25.Note that the second case comprises S ψ , stop ( x ) but is a priori not included in it: there couldexist fertile subexploration trees not connected to x in E ≥ hψ G n if ψ G n is below h on the appropriatemarked vertices.In the first case, C G n ,hx contains at least log n vertices of each D i , 1 ≤ i ≤ r − 1, so that by (93)and (94) for n large enough: |C G n ,hx | ≥ ( r − 1) log n ≥ log n n − | R , | − | D r | K log n ≥ n K . K := (2 K ) − , for every n large enough, we have on S ψ , stop ( x ) ∪ ( ∩ r − i =1 S ψ ,i ( x )): |C G n ,hx | ≤ K log n or |C G n ,hx | ≥ K n. By (107) and a union bound on all x ∈ V n , P ann ( K log n ≤ |C ( n )2 | ≤ K n ) ≤ n − / + P ann ( E c ,n ).By Remark 2, |C ( n )2 | ≤ K n P ann -w.h.p. By Proposition 1 and Lemma 4, P ann ( E c ,n ) → P ann ( |C ( n )2 | ≤ K log n ) → ψ G n on the first phase: proof of (104)Proposition 29. Let a, K , (cid:96) be such that Proposition 25 holds with K (cid:48) = K , and such thatthe conclusion of Lemmas 30 and 31 hold. Then for n large enough, (104) holds. To prove Proposition 29, we need two variants of Proposition 15. The first consists in replacingthe ”security radius” a n = Θ(log log n ) by r n = Θ(log n ). Lemma 30. If a > is small enough, then the following holds for n large enough. Assumethat G n is a good graph, that A ⊂ V n satisfies • | A | ≤ n / , • tx( B G n ( A, r n )) = tx( A ) , and • max z ∈ A | ψ G n ( z ) | ≤ log / n .For every y ∈ ∂B G n ( A, , writing y for the unique neighbour of y in A , we have: (cid:12)(cid:12)(cid:12)(cid:12) E G n [ ψ G n ( y ) | σ ( A y )] − d − ψ G n ( y ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ n − a (108) and (cid:12)(cid:12)(cid:12)(cid:12) Var G n ( ψ G n ( y ) | σ ( A y )) − dd − (cid:12)(cid:12)(cid:12)(cid:12) ≤ n − a . (109) Proof. We follow the argument of Proposition 15, with a few adjustments.First, the bounds in (37) and (38) are in fact e − ca n for some constant c > 0, and one can replace a n by r n .Second, concerning the proof of (39), note that (41) still holds for large enough n since for k ≥ . n and a constant γ > k nor on n , P ( Z k ≤ r n ) ≤ P ( Z k ≤ . 05 log n ) ≤ P ( Z k ≤ d − d k ) ≤ e − γk .55he condition | A | ≤ n / implies that E π n [ H A ] ≥ n / / 4, so that we can adapt the argumentbelow (40), as well as the proof of (42).The second variant of Proposition 15 corresponds to Case 1 and Case 2 of the first phase of theexploration. Let us introduce some notations before stating the Lemma. Recall that for m ≥ G m is the connected (and infinite) d -regular graph with a unique cycle C m of length m . Recallthe definitions of α k , β k , γ k , α (cid:48) k and γ (cid:48) k from (85) and (86).For k ≥ 0, let z k be a vertex at distance k of the cycle C m in G m , and z k be a neighbourof z k at distance k + 1 of C m . Note that B G m ( z k , z k , r n ) (the subgraph of G m obtained bytaking all paths of length r n starting at z k and not going through z k ) contains C m if and onlyif k ≤ r n − (cid:100) m/ (cid:101) . Lemma 31. If a > is small enough, then the following holds for n large enough (uniformlyin m ≥ ).Assume that G n is a good graph, that A ⊆ V n is such that • | A | ≤ n / , • tx( A ) = 0 , and • max z ∈ A | ψ G n ( z ) | ≤ log / n . Case 1. Let y ∈ V n , and suppose that– y has a neighbour y in A ,– for some ≤ k < m , there exists (cid:98) y in A , a path P of length m − k from y to (cid:98) y whose onlyvertex in A is (cid:98) y , and a path P (cid:48) in A of length k − from (cid:98) y to y , so that C := P ∪ P (cid:48) ∪ ( y, y ) is a cycle of length m (and (cid:98) y = y if k = 1 ), and– tx( B G n ( A ∪ C, r n )) = 1 .Then (cid:12)(cid:12) E G n [ ψ G n ( y ) | σ ( A )] − α k ψ G n ( y ) − β k ψ G n ( (cid:98) y ) (cid:12)(cid:12) ≤ n − a (110) and (cid:12)(cid:12) Var G n ( ψ G n ( y ) | σ ( A )) − γ k (cid:12)(cid:12) ≤ n − a . (111) Case 2. Let y ∈ V n , and suppose that– y has a unique neighbour y in A ,– for some ≤ k ≤ r n − (cid:100) m/ (cid:101) , B G n ( y, y, r n ) is isomorphic to B G m ( z k , z k , r n ) , and– tx( B G n ( A ∪ P ∪ C, r n )) = 1 , where C is the cycle in B G n ( y, y, r n ) and P the path from y to C . hen (cid:12)(cid:12) E G n [ ψ G n ( y ) | σ ( A )] − α (cid:48) k ψ G n ( y ) (cid:12)(cid:12) ≤ n − a (112) and (cid:12)(cid:12) Var G n ( ψ G n ( y ) | σ ( A )) − γ (cid:48) k (cid:12)(cid:12) ≤ n − a . (113) Proof. We will only show (112) and (113). The other proofs are very similar and left to thereader.We start with the proof of (112). We follow the proof scheme of (34). Let τ be the hitting timeof ∂B G n ( y, y, r n ) \ { y } by a SRW ( X k ) k ≥ . Note that { H A ≤ τ } ⊆ { X H A = y } . We write E G n [ ψ G n ( y ) | σ ( A )] = P G n y ( X H A = y, H A ≤ τ ) ψ G n ( y ) + E G n y [ ψ G n ( X H A ) H A >τ ] − E G n y [ H A ] E G n π n [ H A ] E G n π n [ ψ G n ( X H A )] . Since B G n ( y, y, r n ) and B G m ( z k , z k , r n ) are isomorphic, P G n y ( X H A = y, H A ≤ τ ) = P G m z k ( X H ∂BGm ( zk,zk,rn ) = z k ) =: α (cid:48)(cid:48) k .Since { X H ∂BGm ( zk,zk,rn ) = z k } ⊆ { H { z k } < + ∞} , we have α (cid:48)(cid:48) k ≤ α (cid:48) k .Reciprocally, on { H { z k } < + ∞} \ { X H ∂BGm ( zk,zk,rn ) = z k } , a SRW starting at distance r n of C m has to reach C m . As in the proof of (37), a comparison with a biased SRW on Z shows thatthis happens with a probability O ( e − cr n ) for some constant c > d andwe get that if a small enough, then for large enough n , | P G n y ( X H A = y, H A ≤ τ ) − α (cid:48) k | ≤ n − a .It remains to establish (cid:12)(cid:12)(cid:12)(cid:12) E G n y [ ψ G n ( X H A ) H A >τ ] − E G n y [ H A ] E G n π n [ H A ] E G n π n [ ψ G n ( X H A )] (cid:12)(cid:12)(cid:12)(cid:12) ≤ n − a . (114)To do so, one adapts the proofs of (38) and (39) exactly as in the proof of Lemma 30: if H A > τ ,( X k ) k ≥ leaves B G n ( A ∪ C ∪ P, r n ) before hitting A . By Lemma 16 with s = r n , if a is smallenough, ( X k ) does not hit A within the next log n steps with probability at least(1 − ( d − − r n ) log n ≥ − n ( d − − r n ≥ − n − a .Then, we use Corollary 2.1.5 of [23] as six lines above (40): after (cid:98) log n (cid:99) steps, the fact that G n is an expander forces the empirical distribution of X k to be very close to the uniform distribution π n . (112) follows.For (113), we follow the proof scheme of (35). If τ (cid:48) is the exit time of B G n ( y, y, r n ), we have (cid:12)(cid:12) Var G n ( ψ G n ( y ) | σ ( A )) − γ (cid:48) k (cid:12)(cid:12) ≤ (cid:12)(cid:12) G G n ( y, y ) − E G n y [ G G n ( y, X H A ) H A = τ (cid:48) ] − γ (cid:48) k (cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) E G n y [ G G n ( y, X H A ) H A >τ (cid:48) ] − E G n y [ H A ] E π n [ H A ] E G n π n [ G G n ( y, X H A )] (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) . (115)57e deal with the second term of the RHS as (114) to show that it is O ( n − a ). As below (45),we have E G n y [ G G n ( y, X H A ) H A = τ (cid:48) ] = E G n y [ G G n ( y, X τ (cid:48) )] − O ( n − a )if a is small enough, by (11). Now, by (17) applied to D := B G n ( y, y, r n ) (note that T D = τ (cid:48) )and (41) (which still holds, as remarked in the proof of Lemma 30), we get | G G n ( y, y ) − E G n y [ G G n ( y, X τ (cid:48) )] − G D G n ( y, y ) | ≤ E G ny [ τ (cid:48) ] n = O ( n − a ).But D is isomorphic to E := B G m ( z k , z k , r n ), so that G D G n ( y, y ) = G EG m ( z k , z k ) = G G m ( z k , z k ) − P G m z k ( T E = z k ) G G m ( z k , z k ) − P G m z k ( T B (cid:54) = z k ) G G m ( z k , z )for any z ∈ ∂B G m ( E, \ { z k } , by cylindrical symmetry of B G m ( E, a is small enough, then for n large enough, G D G n ( y, y ) = G EG m ( z k , z k ) = G G m ( z k , z k ) − P G m z k ( T E = z k ) G G m ( z k , z k ) + O ( n − a ).One easily adapts the reasoning leading to (37), despite the presence of one cycle, to get P G m z k ( T E = z k ) = P G m z k ( H { z k } < + ∞ ) + O ( n − a ) for a small enough.Note indeed that { T E = z k } ⊆ { H { z k } < + ∞} . Reciprocally, if z ∈ ∂B G m ( E, \ { z k } , a SRWstarting at z has a probability decaying exponentially with r n to reach z k , since there are atmost two injective paths from z to z k , and each contains at least r n − d ) to enter a subtree isomorphic T + d and to neverleave it.Since γ (cid:48) k = G G m ( z k , z k ) − P G m z k ( H { z k } < + ∞ ) G G m ( z k , z k ), we obtain | G D G n ( y, y ) − γ (cid:48) k | = O ( n − a ).All in all, we get that the first term of the RHS of (115) is O ( n − a ), and (113) follows. Proof of Proposition 29. We proceed as below (53) in the proof of Proposition 17. Denote E n := {∃ y ∈ R ψ , | ψ G n ( y ) − (cid:100) ψ G n ( y ) | ≥ n − a } ∩ E ,n ∩ S ( x ).On E n , (cid:100) ψ G n is defined on at most K log n vertices by D2 (and our choice of K (cid:48) = K ), so thatby the triangular inequality, either | (cid:100) ψ G n ( x ) − ψ G n ( x ) | ≥ n − a log − n , or there exists y such that | (cid:100) ψ G n ( y ) − ψ G n ( y ) | ≥ n − a log − n + sup y ∈ R ψy | (cid:100) ψ G n ( y ) − ψ G n ( y ) | ,where R ψy is the set of vertices where (cid:100) ψ G n has been defined before (cid:100) ψ G n ( y ). Let E ( y ) := {| (cid:100) ψ G n ( y ) − ψ G n ( y ) | ≥ n − a log − n + sup y ∈ R ψy | (cid:100) ψ G n ( y ) − ψ G n ( y ) |} ∩ E n y (cid:54) = x , we can apply Lemma 31 (if (cid:100) ψ G n ( y ) is defined in Case 1 or Case 2) or Lemma 30 (inthe other cases), so that P ann ( E ( y )) ≤ P ann ( n − a | ξ y | ≥ n − a log − n − n − a ) ≤ P ann ( | ξ y | ≥ n a/ ) ≤ n − by the exponential Markov inequality, and P ann ( E ( x )) ≤ n − by the same argument, where weset E ( x ) := {| (cid:100) ψ G n ( x ) − ψ G n ( x ) | ≥ n − a log − n } ∩ E n . Hence, we get by a union bound on y ∈ R ψ that for large enough n , P ann ( E n ) ≤ n − K log n ≤ n − . The conclusion follows. ψ G n on the second phase: proof of (105) It is enough to show that for each y q,i whose exploration in the second phase is successful, P ann (cid:16) { sup z ∈ T yq,i \{ y q,i } | (cid:100) ψ G n ( z ) − ψ G n ( z ) | ≥ log − n } ∩ E ,n (cid:17) ≤ n − ,where T y q,i is its exploration tree, and to conclude by a union bound on y q,i . This follows froma straightforward adaptation of the reasoning below (53). Note that the n − in the RHS of (54)can be replaced by any polynomial in n . ψ G n on the third phase: proof of (106) By symmetry, it is enough to consider the case i = 1. Following readily the argument of theproof of (69), we get that the probability that E ≥ hψ G n percolates through a given J -joining ballis at least log γ ( K / − n , for any J . By (95) and (96), and a union bound on every couple( J, q ) ∈ { , . . . , s } × { , . . . m } , P ann (( S ( x ) \ S ψ , ( x )) ∩ E ,n ) ≤ n − + s m P ( Z = 0),where Z ∼ Bin( (cid:98) log γ − κ − n (cid:99) , log γ ( K / − n ). If κ and γ/κ are large enough, then for n largeenough, P ( Z = 0) = (1 − log γ ( K / − n ) (cid:98) log γ − κ − n (cid:99) ≤ n − , and (106) follows. C ( n )1 In this section, we establish Theorem 3. 59 roof of Theorem 3. The proof mimics the reasoning of Lemma 22. Let k ≥ T bea rooted tree of height k , with no vertex of degree more than d . Let x ∈ V n . We perform anexploration as in Section 5.1. Denote S T ( x ) the event that the exploration is successful, that S T ( x ) ⊆ C G n ,hx and that B G n ( x, k ) = T . We claim that P ann ( S T ( x )) −→ n → + ∞ P T d ( B T d ( ◦ , k ) = T, |C h ◦ | = + ∞ ) . (116)The proof goes as those of Lemma 18 and Proposition 17. In the proof of Lemma 18, replace F ( n ) j by F ( n ) T,j := F ( n ) j ∩ { B C h ◦ ( ◦ , k ) = T } and F (cid:48) ( n ) j by F (cid:48) ( n ) T,j := F (cid:48) ( n ) j ∪ { B C h ◦ ( ◦ , k ) (cid:54) = T } , for every j ≥ 1. We also check easily that P T d ( C h +log − n ◦ ∩ B T d ( ◦ , k ) = C h − log − n ◦ ∩ B T d ( ◦ , k )) → B G n ( x, k ) P ann -w.h.p., as Proposition 17 only ensures that C h +log − n ◦ ∩ B T d ( ◦ , k ) ⊆ B G n ( x, k ) w.h.p.Moreover, we get as for (76): sup x,y ∈ V n | Cov ann ( S T ( x ) , S T ( y )) | −→ n → + ∞ . (117)Let ε > 0. Applying Bienaym´e-Chebyshev’s inequality as in Lemma 22, we get P ann ( | |S T | − P T d ( B T d ( ◦ , k ) = T, |C h ◦ | = + ∞ ) n | ≤ εn ) −→ n → + ∞ , (118)where S T is the set of vertices x ∈ V n such that S T ( x ) holds. Let S ⊂ V n be the set of verticessuch that their exploration is successful. By Theorem 1 and a reasoning as in the proof ofLemma 22, with P ann -probability 1 − o (1),(i) |C ( n )2 | ≤ n / , so that S T ⊆ S ∩ C ( n )1 ,(ii) | |C ( n )1 | − η ( h ) n | ≤ εn ,(iii) | |S| − η ( h ) n | ≤ εn .Suppose that these three assumptions and (118) hold. By (i), S T = S ∩ V ( T ) n . Therefore, |S T | ≤ | V ( T ) n | ≤ |S T | + |C ( n )1 ∩ S c | , so that by (ii), (iii) and (118),( P T d ( B T d ( ◦ , k ) = T, |C h ◦ | = + ∞ ) − ε ) n ≤ | V ( T ) n | ≤ ( P T d ( B T d ( ◦ , k ) = T, |C h ◦ | = + ∞ ) + 3 ε ) n. Moreover, | |C ( n )1 | − η ( h ) n | ≤ εn by (ii), so that for ε small enough, P T d ( B T d ( ◦ , k ) = T, |C h ◦ | = + ∞ ) η ( h ) − √ ε ≤ | V ( T ) n ||C ( n )1 | ≤ P T d ( B T d ( ◦ , k ) = T, |C h ◦ | = + ∞ ) η ( h ) + √ ε. But P T d ( B T d ( ◦ , k ) = T, |C h ◦ | = + ∞ ) η ( h ) = P T d ( B T d ( ◦ , k ) = T, |C h ◦ | = + ∞ ) P T d ( |C h ◦ | = + ∞ )= P T d ( B T d ( ◦ , k ) = T | |C h ◦ | = + ∞ ) . And since we can take ε arbitrarily small, the conclusion follows.60 .2 The core and the kernel We now prove (4) and (5) of Theorem 2, starting with (4). Let K (resp. K ) be the probabilitythat ◦ has at least 2 (resp. 3) children with an infinite offspring in C h ◦ , under P T d .Suppose that for x, y ∈ V n , P ann ( x ∈ C ( n ) ) −→ n → + ∞ K (119)and Cov ann ( x ∈ C ( n ) , y ∈ C ( n ) ) −→ n → + ∞ . (120)Then (4) follows by a second moment argument as in Lemma 22. Hence, it is enough to prove(119) and (120). Proof of (119). For x ∈ V n , we perform the exploration in Section 5.1 from x , replacing C3 bythe following condition: for every neighbour v of x , stop exploring the subtree from v at step k + 1 if the k -offspring of v has at least n / b n vertices. Stop the exploration if this happens forat least two neighbours of x . In this case, say that the exploration successful.We adapt easily the proofs of Lemma 18 and Proposition 17 to show that for x ∈ V n , P ann (the exploration from x is successful) −→ n → + ∞ K . (121)Indeed, K is the probability that the realization of ϕ T d to which we couple ψ G n is such that ◦ hasat least two children with an infinite offspring. Then, as in Lemma 18, there is a probability 1 − o (1) that the offspring of these children grows at an exponential rate close to λ h (Proposition 13).Thus, letting F ( n ) v,k := ∪ ≤ j ≤ k { the j -offspring of v has at least n / b n vertices } for every child v of ◦ , and F ( n )core ,k := ∪ v ,v children of ◦ ( F ( n ) v ,k − ∩ F ( n ) v ,k − ), we get that P ann ( ∪ ≤ k ≤(cid:98) log λh n (cid:99) F ( n )core ,k ) −→ n → + ∞ K and P ann ( ∪ ≤ k ≤(cid:98) log λh n (cid:99) { at most one child of ◦ has a non-empty k -offspring } ) −→ n → + ∞ − K .As for (52), we see that we do not meet any cycle with P ann -probability 1 − o (1). This yields(121).If the exploration from x is successful, let x , x be two children of x such that the explorationsubtrees T x and T x from x and x satisfy min( | ∂T x | , | ∂T x | ) ≥ n / b n . Then, let K > w , . . . , w (cid:98) K log n (cid:99) ∈ V n be vertices that have not been met in the exploration from x .Proceed to their w -exploration as described in the first part of the construction in Section 7.4.By Remark 1, o ( √ n ) vertices are seen during the exploration from x and the w -explorations.Applying (14) with k = 1, m , m , m E , m = o ( √ n ), we get that with 1 − o (1) P ann -probability,none of the w -explorations intersects the exploration from x , and no w i is spoiled or back-spoiled. 61s in (103), we get that for each w i , its w -exploration has probability at least η ( h ) / w -successful. Hence with P ann -probability at least 1 − (1 − η ( h ) / (cid:98) K log n (cid:99) = 1 − o (1), thereexists i such that the w -exploration from w i is successful. Denote ∂T w i the boundary of itsexploration tree. Write S core ( x ) := { the exploration from x is successful }∩{∃ i ≥ , the w -exploration from w i is successful } . We have just shown thatlim inf n → + ∞ P ann ( S core ( x )) ≥ lim inf n → + ∞ P ann (the exploration from x is successful) ≥ K . (122)Next, we grow joining balls from ∂T x to ∂T w i , and then from ∂T x to ∂T w i . We proceed asin the second part of the construction in Section 7.4 with m = 2 and s = 1. Similarly to (95),we get that P ann ( S core ( x ) ∩ ( S core,joining,1 ∪ S core,joining,2 )) = o (1),with S core,joining,i := { there are less than log γ − κ − n joining balls from ∂T x i to ∂T w i } for i =1 , ψ G n on the exploration from x , on the w -exploration from w i , on the joiningballs from ∂T x to ∂T w i , and on the joining balls from ∂T x to ∂T w i , in that order. Denote S core,connect the event that there exists a joining ball B from ∂T x to ∂T w i and another B from ∂T x to ∂T w i such that min y ∈ T x ∪ T x ∪ T wi ∪ B ∪ B ψ G n ( y ) ≥ h . As in the proof of Proposition 21,we get that P ann ( S core ( x ) ∩ S c core,connect ) = o (1). Hencelim inf n → + ∞ P ann ( x, x , x , z are in a cycle of C G n x and |C G n x | ≥ n / ) ≥ lim inf n → + ∞ P ann ( S core ( x )).If |C ( n )2 | < n / , this cycle is in C ( n )1 , and thus in C ( n ) so that by (3), for large enough n and any x ∈ V n , one has by (122):lim inf n → + ∞ P ann ( x ∈ C ( n ) ) ≥ lim inf n → + ∞ P ann ( S core ( x )) ≥ K .Reciprocally, for x ∈ V n , turn the exploration into a lower exploration, replacing h + log − n by h − log − n (as in Section 5.2). Say that the lower exploration from x is aborted if forsome k ≤ log log n , at most one child of x has a non-empty ( k − A core ( x ) := { the lower exploration from x is aborted } . We get as in the proof of (121): P ann ( A core ( x )) −→ n → + ∞ − K . (123)Moreover, revealing ψ G n on T x , we can apply Proposition 15 as below (53) to get that P ann ( A core ( x ) ∩ {C G n ,hx ∩ B G n ( x, (cid:98) log log n (cid:99) ) ⊆ T x } ) −→ n → + ∞ − K . (124)62or each neighbour y of x , denote C y its connected component in C G n x \ { x } . If the explorationis aborted and C G n ,hx ∩ B G n ( x, (cid:98) log log n (cid:99) ) ⊆ T x , then x has at most one neighbour y such that C y ∪ { x } is not a tree. Hence x (cid:54)∈ C ( n ) . Thus, lim inf n → + ∞ P ann ( x (cid:54)∈ C ( n ) ) ≥ − K and (119)follows. Proof of (120). By (119), for x, y ∈ V n , P ann ( x ∈ C ( n ) ) P ann ( y ∈ C ( n ) ) −→ n → + ∞ K .It remains to show that P ann ( x, y ∈ C ( n ) ) −→ n → + ∞ K .Perform the exploration from x as in the beginning of the proof of (119), then do the same from y (and stop the latter if it reaches a vertex of the exploration from x ). Since o ( √ n ) verticesare revealed during these explorations (see Remark 1), then by (14), the probability that theexploration from y meets that of x is o (1). Thus by (121), P ann (the explorations from x and y are both successful) −→ n → + ∞ K . Then, let x , x (resp. y , y ) be children of x (resp. y ) whose exploration is successful. Let K > w , . . . , w (cid:98) K log n (cid:99) ∈ V n be vertices that have not been met in the explorations from x and y , and proceed to their w -exploration. If there exists i ≥ w -explorationfrom w i is successful, build joining balls from ∂T x , ∂T x , ∂T y and ∂T y to T w i . Finally, reveal ψ G n on T x , on T y , on T w i on the joining balls from ∂T x , ∂T x , ∂T y and ∂T y , in that order.As in the proof of (122) and below, we get thatlim inf n → + ∞ P ann ( x, y ∈ C ( n ) ) ≥ K . (125)Conversely, if we perform the lower explorations from x and y as defined in the end of the proofof (119), we easily get that P ann ( ∃ z ∈ { x, y } , the lower exploration from z is aborted) −→ n → + ∞ − K .Then, we reveal ψ G n on T z . Following the reasoning below (53), we get that P ann -w.h.p., B C G n,hz ( z, (cid:98) log log n (cid:99) ) ⊆ T z , and thus P ann ( { the lower exploration from z is aborted } ∩ { z ∈ C ( n ) } ) = o (1).This yields lim inf n → + ∞ P ann ( ∃ z ∈ { x, y } , z (cid:54)∈ C ( n ) ) ≥ − K .Together with (125), this concludes the proof.This reasoning can be readily adapted to prove (5), with a modification of the exploration(requiring that at least three children of x have a successful exploration).63 .3 The typical distance The proof of (7) goes as that of (2), with a slight modification of the explorations of Section 5.Fix ε ∈ (0 , k ≥ (1 / ε/ 3) log λ h n . By Proposition 13, as n → + ∞ : P T d (cid:18) |Z h (cid:98) (1 / ε/ 3) log λh n (cid:99) | > n / b n (cid:12)(cid:12)(cid:12)(cid:12) |C h +log − n ◦ | = + ∞ (cid:19) → x, y are connected in E ≥ hψ G n via the successful explorations from x and y and the joining balls from ∂T x to ∂T y , thenfor n large enough: d E ≥ hψ G n ( x, y ) ≤ / ε/ 3) log λ h n + a (cid:48) n ≤ (1 + ε ) log λ h n .Hence, by Lemma 23, (2) and (3), P ann ( E ,n ∩ E ,n ∩ E ,n ) −→ n → + ∞ E ,n := {|{ ( x, y ) ∈ V n , d E ≥ hψ G n ( x, y ) ≤ (1 + ε ) log λ h n }| ≥ ( η ( h ) − ε ) n } , E ,n := {∀ i ≥ , |C ( n ) i | ≤ K log n } and E ,n := { ( η ( h ) − ε ) n ≤ |C ( n )1 | ≤ ( η ( h ) + ε ) n } .On E ,n , |{ ( x, y ) ∈ V n \ ( C ( n )1 ) , d E ≥ hψ G n ( x, y ) ≤ (1 + ε ) log λ h n }| ≤ n / , so that on E ,n ∩ E ,n , |{ ( x, y ) ∈ ( C ( n )1 ) , d C ( n )1 ( x, y ) ≤ (1 + ε ) log λ h n }| ≥ ( η ( h ) − ε ) n .Thus, on E ,n ∩ E ,n ∩ E ,n : π ,n (cid:16) { ( x, y ) ∈ ( C ( n )1 ) , d C ( n )1 ( x, y ) ≤ (1 + ε ) log λ h n } (cid:17) ≥ η ( h ) − ε ( η ( h ) + ε ) ≥ − ε + 2 η ( h ) ε + ε ( η ( h ) + ε ) ≥ − (3 + 2 η ( h )) η ( h ) ε. It remains to show that the typical distance in C ( n )1 is at least (1 − ε ) log λ h n . Modify the lowerexploration of Section 5.2: say that it is aborted if • C1 did not happen, and • it is stopped at some step k ≤ (1 / − ε/ 2) log λ h n , or less than n / − ε/ vertices andhalf-edges have been seen at step (cid:98) (1 / − ε/ 2) log λ h n (cid:99) .Suppose that P ann (the lower exploration from x is aborted) = 1 − o (1) . (126)For x, y ∈ V n , perform the lower exploration from x , then that from y , and stop it if it meets avertex of the exploration from x . This happens with P ann -probability o (1) by (14) with k = 1, m , m , m, m E = o ( √ n ) (recall Remark 1).Hence by (126):64 ann ( E ab ( x, y ))=1 − o (1), with E ab ( x, y ):= { the lower explorations from x and y are aborted } ,Then, reveal ψ G n on the exploration trees T x and T y . Applying Proposition 15 as below (53),we get that: P ann ( E ab ( x, y ) ∩ E x ∩ E y ) = 1 − o (1),with E x := { B C G n,hx ( x, (cid:98) (1 / − ε/ 2) log λ h n (cid:99) ) ⊆ T x } and E y := { B C G n,hy ( y, (cid:98) (1 / − ε/ 2) log λ h n (cid:99) ) ⊆ T y } .On E ab ( x, y ) ∩ E x ∩ E y , if x, y ∈ C ( n )1 , then d C ( n )1 ( x, y ) ≥ (1 − ε ) log λ h n . Therefore, for ev-ery x, y ∈ V n , P ann ( E x,y ) = o (1), with E x,y := { x, y ∈ C ( n )1 , d C ( n )1 ( x, y ) < (1 − ε ) log λ h n } .Similarly, for all distinct x, y, z, t ∈ V n , we get that P ann ( E x,y ∩ E z,t ) = o (1), so thatCov ann ( E x,y , E z,t ) = o (1).Thus by Bienaym´e-Chebyshev’s inequality, P ann ( E ,n ∩ {|{ ( x, y ) ∈ V n , d C ( n )1 ( x, y ) ≤ (1 − ε ) log λ h n }| ≥ εn } ) −→ n → + ∞ . For ε > n large enough, on E ,n ∩ {|{ ( x, y ) ∈ V n , d C ( n )1 ( x, y ) ≤ (1 − ε ) log λ h n }| ≥ εn } , π ,n (cid:16) { ( x, y ) ∈ ( C ( n )1 ) , d C ( n )1 ( x, y ) ≤ (1 − ε ) log λ h n } (cid:17) ≤ εη ( h ) . This concludes the proof of (7).Thus, it remains to establish (126). Note first that P ann (C1 happens) = o (1) by (14) with k = 1, m = 1, m E = 0 and m = o ( √ n ) by Remark 1. Therefore, it is enough to prove that P T d ( | B C h − log − n ◦ ( ◦ , (cid:98) (1 / − ε/ 2) log λ h n (cid:99) + a n ) | < n / − ε/ ) → , since if this event happens, less than n / − ε/ vertices and half-edges have been seen at step (cid:98) (1 / − ε/ 2) log λ h n (cid:99) . By (33), it suffices to show that P T d ( | B C h − log − n ◦ ( ◦ , (cid:98) (1 / − ε/ 2) log λ h n (cid:99) ) | < n / − ε/ ) → , (127)To do so, we first prove that for n large enough and every log log n ≤ k ≤ (1 / − ε/ 2) log λ h n , P T d (cid:16) |Z h − log − nk | ≥ n / − ε/ (cid:17) ≤ e − Ck + n − . (128)Let δ > λ h ≤ λ h − δ ≤ λ h + ε/ 10 (which is possible since h (cid:48) (cid:55)→ λ h (cid:48) is decreasing andcontinuous, Proposition 9). By Proposition 13, there exists C > ε ) such that65or n large enough, for every log log n ≤ k ≤ (1 / − ε/ 2) log λ h n and a ≥ h , P T d a (cid:16) |Z h − log − nk | ≥ n / − ε/ χ h − δ ( a ) (cid:17) ≤ P T d a (cid:16) |Z h − δk | ≥ n / − ε/ χ h − δ ( a ) (cid:17) ≤ P T d a (cid:16) log |Z h − δk | ≥ (1 / − ε/ λ h log λ h n + log χ h − δ ( a ) (cid:17) ≤ P T d a (cid:18) log |Z h − δk | ≥ ( λ h − δ − ε/ 10) 1 / − ε/ / − ε/ k + log χ h − δ ( a ) (cid:19) ≤ P T d a (cid:16) log |Z h − δk | ≥ ( λ h − δ − ε/ ε/ k + log χ h − δ ( a ) (cid:17) ≤ P T d a (cid:16) log |Z h − δk | ≥ λ h − δ (1 − ε/ ε/ k + log χ h − δ ( a ) (cid:17) ≤ P T d a (cid:16) log |Z h − δk | ≥ λ h − δ (1 + ε/ k + log χ h − δ ( a ) (cid:17) ≤ e − Ck . By Proposition 2.1 of [2], there exists c > h (cid:48) ≤ h (cid:63) and a ≥ d − 1, one has χ h (cid:48) ( a ) ≤ ca − log d − λ h (cid:48) ≤ ca . Since χ h (cid:48) is continuous on [ h, + ∞ ) (Lemma 10), we have for n large enough max h ≤ a ≤ log n χ h − δ ( a ) < n ε/ , so that P T d (cid:16) |Z h − log − nk | ≥ n / − ε/ (cid:17) ≤ e − Ck + P T d ( ϕ T d ( ◦ ) ≥ log n )Using the exponential Markov inequality as in Lemma 4, we get P T d ( ϕ T d ( ◦ ) ≥ log n ) ≤ n − .This yields (128). Then, for n large enough, this implies P T d ( | B C h ◦ ( ◦ , (cid:98) (1 / − ε/ 2) log λ h n (cid:99) ) | ≥ n / − ε/ ) ≤ P T d ( ∃ k ∈ [log log n, (1 / − ε/ 2) log λ h n ] , |Z h − log − nk | ≥ n / − ε/ ) ≤ (cid:98) (1 / − ε/ 2) log λh n (cid:99) (cid:88) k = (cid:98) log log n (cid:99) ( e − Ck + n − ) ≤ / log log n. (127) and the conclusion follow. Recall that D ( n )1 is the diameter of C ( n )1 . In Section 7, we have in fact proven that there existsa constant K > E n := {∀ x ∈ C ( n )1 , | B C ( n )1 ( x, (cid:98) K log n (cid:99) ) | ≥ K n } , P ann ( E n ) −→ n → + ∞ . Namely, one can take K = K + 3log − λ h .Suppose that E n holds. Let x ∈ C ( n )1 . If ∂B C ( n )1 ( x , (cid:98) K log n (cid:99) + 1) = ∅ , then D ( n )1 ≤ K log n + 2.Else, let x ∈ ∂B C ( n )1 ( x , (cid:98) K log n (cid:99) + 1). For i = 1 , 2, we have66 B C ( n )1 ( x i , (cid:98) K log n (cid:99) ) | ≥ K n and B C ( n )1 ( x i , (cid:98) K log n (cid:99) ) ⊆ B C ( n )1 ( x , (cid:98) K log n (cid:99) ).Moreover, B C ( n )1 ( x , (cid:98) K log n (cid:99) ) ∩ B C ( n )1 ( x , (cid:98) K log n (cid:99) ) = ∅ . Thus, we have | B C ( n )1 ( x , (cid:98) K log n (cid:99) ) | ≥ K n .For i ≥ 2, if ∂B C ( n )1 ( x , (3 i − (cid:98) K log n (cid:99) + 1) = ∅ , then D ( n )1 ≤ i − K log n + 2. Else,let x i +1 ∈ ∂B C ( n )1 ( x , (3 i − (cid:98) K log n (cid:99) + 1). As for i = 2, we get that | B C ( n )1 ( x , (3 i − (cid:98) K log n (cid:99) ) | − | B C ( n )1 ( x , (3 i − (cid:98) K log n (cid:99) ) | ≥ | B C ( n )1 ( x i , K log n ) | ≥ K n. But there are only n vertices in V n . Therefore, ∂B C ( n )1 ( x , (3 i − (cid:98) K log n (cid:99) + 1) = ∅ for some i ≤ K − , and D ( n )1 ≤ K − K log n. This shows (6). References [1] Angelo Ab¨acherli and Jiˇr´ı ˇCern´y. Level-set percolation of the Gaussian free field on regulargraphs I: Regular trees. arXiv e-prints , page https://arxiv.org/abs/1909.01973, Sep 2019.[2] Angelo Ab¨acherli and Jiˇr´ı ˇCern´y. Level-set percolation of the Gaussian free field on regulargraphs II: Finite expanders. arXiv e-prints , page https://arxiv.org/abs/1909.01972, Sep2019.[3] Angelo Ab¨acherli. Local picture and level-set percolation of the gaussian free field on alarge discrete torus. Stochastic Processes and their Applications , 129(9):3527 – 3546, 2019.[4] Angelo Ab¨acherli and Alain-Sol Sznitman. Level-set percolation for the gaussian free fieldon a transient tree. Ann. Inst. H. Poincar´e Probab. Statist. , 54(1):173–201, 02 2018.[5] Louigi Addario-Berry, Nicolas Broutin, and Christina Goldschmidt. The continuum limitof critical random graphs. Probab. Theory Related Fields , 152(3–4):367–406, 2012.[6] David Aldous. Brownian excursions, critical random graphs and the multiplicative coales-cent. Ann. Probab. , 25(2):812–854, 1997.[7] Noga Alon, Itai Benjamini, and Alan Stacey. Percolation on finite graphs and isoperimetricinequalities. Ann. Probab. , 32(3):1727–1745, 07 2004.[8] B´ela Bollob´as. The isoperimetric number of random regular graphs. European Journal ofCombinatorics , 9(3):241 – 244, 1988. 679] Jean Bricmont, Joel Lebowitz, and Christian Maes. Percolation in strongly correlatedsystems: The massless gaussian field. Journal of Statistical Physics , 48:1249–1268, 011987.[10] Jian Ding, Eyal Lubetzky, and Yuval Peres. Anatomy of the giant component: the strictlysupercritical regime. European J. Combin. , 35:155–168, 2014.[11] Alexander Drewitz, Alexis Pr´evost, and Pierre-Fran¸ccois Rodriguez. The Sign Clusters ofthe Massless Gaussian Free Field Percolate on { Z d , d ≥ } (and more). Communicationsin Mathematical Physics , page 1398, August 2018.[12] Rick Durrett. Probability: Theory and examples (5th edition). https://services.math.duke.edu/~rtd/PTE/PTE5_011119.pdf , 2019.[13] Nathalie Eisenbaum, Haya Kaspi, Michael B. Marcus, Jay Rosen, and Zhan Shi. A ray-knight theorem for symmetric markov processes. Ann. Probab. , 28(4):1781–1796, 10 2000.[14] P. Erd˝os and A. R´enyi. On the evolution of random graphs. Magyar Tud. Akad. Mat.Kutat´o Int. K¨ozl. , 5:17–61, 1960.[15] E. N. Gilbert. Random graphs. Ann. Math. Statist. , 30:1141–1144, 1959.[16] Conchon-Kerjan Guillaume. Supercritical GFF percolation on a regular tree : anatomy ofthe infinite cluster. To appear.[17] Titus Lupu. From loop clusters and random interlacements to the free field. Ann. Probab. ,44(3):2117–2146, 05 2016.[18] S. A. Molchanov and A. K. Stepanov. Percolation in random fields. i. Theoretical andMathematical Physics , 55(2):478–484, 1983.[19] Curien Nicolas. Random graphs - the local convergence point of view. , 2018.[20] Serguei Popov and Balazs Rath. On decoupling inequalities and percolation of excursionsets of the gaussian free field. Journal of Statistical Physics , 159, 07 2013.[21] Pierre-Fran¸cois Rodriguez and Alain-Sol Sznitman. Phase transition and level-set percola-tion for the gaussian free field. Communications in Mathematical Physics , 320(2):571–601,2013.[22] C. Sabot and P. Tarr`es. Inverting ray-knight identity. Probability Theory and RelatedFields , 165(3):559–580, Aug 2016.[23] Laurent Saloff-Coste. Lectures on finite Markov chains , pages 301–413. Springer BerlinHeidelberg, Berlin, Heidelberg, 1997. 6824] Alain-Sol Sznitman. Random interlacements and the gaussian free field. The Annals ofProbability , 40, 02 2011.[25] Alain-Sol Sznitman. An isomorphism theorem for random interlacements. Electron. Com-mun. Probab. , 17:9 pp., 2012.[26] Alain-Sol Sznitman. Coupling and an application to level-set percolation of the gaussianfree field. Electron. J. Probab. , 21:26 pp., 2016.[27] Remco van der Hofstad. Random graphs and complex networks: volume I . Cambridge Se-ries in Statistical and Probabilistic Mathematics. Cambridge University Press, Cambridge,2016.[28] Wolfgang Woess. Random Walks on Infinite Graphs and Groups , volume 138 of CambridgeTracts in Mathematics . Cambridge University Press, United Kingdom, paperback re-edition(with corrections) edition, 2008. 352 S.[29] Jiˇr´ı ˇCern´y, Augusto Teixeira, and David Windisch. Giant vacant component left by arandom walk in a random d -regular graph.