CLUSTER-SIZE DECAY IN SUPERCRITICAL LONG-RANGE PERCOLATION

. We study the cluster-size distribution of supercritical long-range percolation on Z d , where two vertices x,y ∈ Z d are connected by an edge with probability p( k x − y k ) := p min { 1 , β α k x − y k − αd } for parameters p ∈ (0 , 1) , α > 1 , and β > 0 . We show that when α > 1 + 1 /d , and either β or p is suﬃciently large, the probability that the origin is in a ﬁnite cluster of size at least k decays as exp (cid:0) − Θ( k ( d − 1) /d ) (cid:1) . This corresponds to classical results for nearest-neighbor Bernoulli percolation on Z d , but is in contrast to long-range percolation with α < 1 + 1 /d , when the exponent of the stretched exponential decay changes to 2 − α . This result, together with our accompanying paper establishes the phase diagram of long-range percolation with respect to cluster-size decay. Our proofs rely on combinatorial methods that show that large delocalized components are unlikely to occur. As a side result we determine the asymptotic growth of the second-largest connected component when the graph is restricted to a ﬁnite box.


Introduction
For nearest-neighbor Bernoulli percolation on Z d [7] it is well known [1,3,8,20,26,28] that in the supercritical case, the distribution of the number of vertices in the cluster containing the origin follows subexponential decay.Let us write |C(0)| for the number of vertices in the cluster containing the origin, and assume p > p c (Z d ).Then it holds that (1.1) Figure 1: Simulation of long-range percolation in dimension 2 restricted to a finite box.The largest connected component C (1)   n is colored blue, the second-largest connected component C (2)  n red, and the component containing the origin C(0) green.
The decay rate in (1.1) -stretched exponential decay with exponent (d − 1)/d -can be intuitively explained as follows: a cluster C with at least k vertices has at least Θ(k (d−1)/d ) edges on its (outer) boundary.All these edges need to be absent.In other words, the tail decay in (1.1) is driven by surface tension.Recently, this result was extended to supercritical Bernoulli percolation on certain classes of transitive graphs [9,23].Related works are also [29,34], which determine the size of the second-largest component in a finite box for random geometric graphs, also known as continuum percolation, obtaining the same exponent (d − 1)/d for the cluster-size decay.
In our accompanying papers [24,25], we study the supercritical cluster-size decay in a large class of spatial random graph models where at least one of the degree distribution and the edge-length distribution obey heavy tails: long-range percolation [2,35], scalefree percolation on Z d and in the continuum [12,13]; geometric inhomogeneous random graphs [6], hyperbolic random graphs [27], the ultra-small scale-free geometric network [37]; the scale-free Gilbert model [21], the Poisson Boolean model with random radii [17], the age-and the weight-dependent random connection models [18,19].
In these models the tail in (1.1) stays still stretched exponential, but is at least as light as the right-hand-side of (1.1).The new exponent -say ζ -is at least (d − 1)/d, with its formula depending on the model parameters.Generally speaking, the accompanying papers [24,25] treat cluster-sizes whenever the decay is strictly lighter than the righthand side of (1.1).There, we leave the part of the phase diagram open where the model parameters are such that the conjectured exponent in (1.1) stays (d − 1)/d and the tail decay is driven by surface tension as in nearest-neighbor percolation.
Each edge with length at most β is present with probability p, resembling spread-out percolation in this length-range.The long-range nature is apparent beyond radius β.Our parametrization with dα as the power of the polynomial decay matches the notation of the accompanying paper [24].Other common parametrizations of this power are s [5,10,11] and d + α [22].
In Theorem 1.3 we do not require β or p to be sufficiently large, and we also allow one-dimensional models: when d = 1, LRP is supercritical when α ≤ 2 = 1 + 1/d and when p, β are sufficiently large [15,35].When d = 1 and α > 2, LRP is subcritical for any p, β > 0 such that p(1) < 1 [35], so Theorems 1.2 and 1.3 together give a complete picture for the cluster-size decay for supercritical long-range percolation (under the additional assumption that β or p is sufficiently large when α > 1 + 1/d).In [25], we also study the phase boundary α = 1 + 1/d.In that case the lower bounds (1.6) and (1.7) contain lower-order correction factors, which we conjecture to be sharp.We omit further details here.To the extent of our knowledge, for LRP, the only related results regarding the distribution of smaller clusters in supercritical LRP is an upper bound on the second-largest component with unidentified exponent by Crawford and Sly [11] for α ∈ (1,2) in dimension 1 and α ∈ (1, 1 + 2/d) in dimensions 2 and higher.For (sub)critical LRP with α ∈ (1, 2), a polynomial upper bound on P(|C(0)| ≥ n) is established in [22].
Before proceeding to the technical contributions, we remark that our results could be generalized to a more general class of random graph models on Z d .Theorem 1.2 extends to random graph models on Z d with independent edges for any connectivity function that has a lighter tail than p in Theorem 1.2, provided that the probability of 'short-range' edges is still sufficiently high.For instance, our methods extend to longrange percolation models in which the connection probability decays superpolynomially, and also provide an alternative proof for the cluster size decay in spread out percolation, a special case in [9] (in spread-out percolation two vertices within distance R connect independently by an edge with probability p.We refrain from proving the result in this generality, since it would require many technically involved changes in our already technical companion paper [24].We nevertheless formulate the following comment. Remark 1.4.Consider the percolation model on Z d where each pair of vertices x, y ∈ Z d is connected by an edge with probability p(∥x − y∥) for some function p : [0, ∞) → [0, 1), independently of other vertex pairs.Let J : [0, ∞) → [0, 1) be a function that satisfies sup r>0 J(r) < 1, and Then we have the following two cases: 1) If the connectivity function p is of the form p(∥x∥) = J(∥x∥/β), and there is an ε > 0 such that J(x) > ε whenever x < ε, then (1.3)-(1.5)can be proven for all sufficiently large β depending on ε.
2) If the connectivity function is of the form ) can be proven for all p sufficiently close to 1.
The integral in the condition (1.8) represents the order of the expected number of edges {x, y} for which the line-segment (x, y) crosses a fixed box of volume one.If this number is finite, Theorem 1.2 holds in more generality.The connectivity function p from Definition 1.1 satisfies the integrability condition (1.8) if and only if α > 1 + 1/d.We conjecture that the upper bounds in (1.3)-(1.4)remain valid if (1.8) is violated (but no longer match the lower bounds).However, this would require a different proof technique.
We state a proposition that contains the main technical contribution of this paper.Together with statements from our companion paper [24], where we establish the relation between the second-largest component and the cluster-size decay for spatial random graph models more generally, this proposition will readily imply Theorem 1.2.

Proposition 1.5 (Second-largest component, upper bound). Consider supercritical longrange percolation on
2) is sufficiently large (depending on p, α, d), or if p(1 ∧ β) dα is sufficiently close to 1, then there exists a constant A > 0 such that for all k sufficiently large and for all n satisfying n(log n) −2d/(d−1) ≥ k, We believe that Proposition 1.5, and hence Theorem 1.2, should hold for any values of β, p that lead to a supercritical graph: however, this would require non-trivial adaptations of our proof techniques.

Idea of proof
The proof of Proposition 1.5 relies on a careful first-moment analysis in which we count all possible candidates of isolated components of size at least k.The starting point is the classic isoperimetric inequality which states that any set S of at least k vertices has an edge-boundary of size |∂S| = Ω(k (d−1)/d ).These edges need to be absent when S is a connected component (or simply component below), i.e., detached from the rest of the graph.The combinatorial difficulty arises when we account for all possible candidate components S: the structure of S is more complex than for nearest-neighbor bond percolation in Z d , since S can be "delocalized" in space.The second difficulty arises in the finite box Λ n ⊆ Z d , where we need to take boundary effects into account caused by possibly shared boundaries of ∂S and ∂Λ n .
To resolve these two complications, we distinguish two types of components: the first type consists of several "blocks" connected by long edges: each block is a connected subset of Z d (with respect to nearest-neighbor relation in Z d ).We consider each possible combination of blocks with fixed total outer edge-boundary size m, and give an upper bound on the probability that these blocks form a connected component by counting all possible spanning trees on these blocks.We show that the combinatorial factor arising from counting all potential components with boundary m is at most exponential in m.We then use the large value of β or p in our favor to prove that the probability that such a component is formed and isolated is sufficiently small.The second type of potential component S contains a large block that has a large overlap with the boundary of Λ n , and consequently |∂S| inside Λ n may be small.Again, a simple enumeration of all such blocks would yield too large combinatorial factors.
Instead, we use only certain holes of S and an adapted isoperimetric inequality to still ensure that many edges need to be absent.Then we group potential large components, such that a large number of holes coincide for each potential large component in the same group.This makes combinatorial factors much smaller, and we obtain the right decay.

Organization
In Section 2, we derive an intermediate upper bound for P |C (2)  n | ≥ k , defining the two types of components formally.Then, we state two lemmas and show that they imply Proposition 1.5.We prove the two lemmas in separate sections.In the last section we use the result of Proposition 1.5 to prove Theorem 1.2.

Notation
Let H = (V H , E H ) be a graph.For two sets A, B ⊆ V H , we write A ∼ H B if there exist x ∈ A, y ∈ B such that {x, y} ∈ E H , and A ̸ ∼ H B if no such pair exists.We leave out the subscript H if the graph is clear from the context.For A ⊆ V H , we write H[A] for the induced subgraph of H on vertices in A. Denote by Z d ∞ the graph on the vertex set Z d and an edge between x, y ∈ Z d if and only if ∥x − y∥ ∞ = 1.Similarly, let Z d 1 be the graph on the vertex set Z d and an edge between x, y ∈ Z d if and only if ∥x − y∥ 1 = 1.As already mentioned, we write ∥ • ∥ := ∥ • ∥ 2 .For two sets A, B ⊆ Z d , denote by ∥A − B∥ p = min{∥x − y∥ p | x ∈ A, y ∈ B}.For any graph structure, we say that a path π = (v 1 , v 2 , v 3 , . . . ) is self-avoiding if its vertices are all distinct.For x, y ∈ R, we write x ∧ y := min(x, y), x ∨ y := max(x, y).We sometimes abuse notation and write ∪ i≥1 A i rather than i≥1 A i for the union.We also refer to Definition 2.4 below for the exterior boundary of A ⊆ Z d with respect to either Λ n or Z d , denoted by ∂ ext A, resp.∂ ext A, and the interior boundaries ∂ int A and ∂ int A with respect to Λ n , resp.Z d .

Preliminaries and setup
Throughout the rest of the paper, we assume that d ≥ 2, α > 1 + 1/d, and n 1/d ∈ N.This latter assumption means that the box Λ n contains exactly n vertices and avoids integer parts in our formulas.Adaptation to arbitrary n is straightforward.
We now formalize the concepts from the proof outline in Section 1.1 that eventually lead to two lemmas, one for each of the two described types of components.To ensure that the upcoming definitions naturally follow each other, we will postpone the (sometimes standard) proofs of intermediate claims to the appendix.We start with a definition to describe sets of Λ n that form (subsets of) the second-largest component.

Definition 2.1 (Connected sets and blocks). We call a non-empty
A sequence of at least two sets A 1 , A 2 , . . .
We say that a set A ⊆ Λ n consists of blocks A 1 , . . ., A b if A i is a (non-empty) block for 1 ≤ i ≤ b, if the sequence (A i ) i≤b is 1-disconnected, and their union equals A.
We say that a vertex x is surrounded by A ∈ A if each infinite 1-connected selfavoiding path starting from x contains a vertex of A. We define for A ∈ A its closure Ā as Ā = A ∪ {x ∈ Z d : x surrounded by A}. (2.2) We call the maximal 1-connected subsets of Ā \ A the holes of A, and write H A for the collection of holes.
We make a few comments.The 'vertices on the boundary' of A, that we shall shortly define, are not surrounded by A, but they belong to both A and Ā.The closures of blocks will be used for the first type of components described in Section 1.1.Take now a block A ⊆ Λ n .Then x can only be surrounded by A if x ∈ Λ n , hence A ⊆ Λ n implies that Ā ⊆ Λ n .Due to the presence of long-range edges, a component in long-range percolation may consist of multiple 1-disconnected blocks (some of them possibly consisting of a single vertex).We define the notion of a block graph.

Definition 2.2 (Block graph).
Let A 1 , . . ., A b ∈ 1 A be a sequence of 1-disconnected blocks, and consider a graph In words, the vertices of each block are contracted to a single vertex in the block graph, and two corresponding vertices for the blocks i and j are connected in the block graph if and only if there exists an edge in the original graph G between (some vertices in) the two blocks.We continue with a simple claim, with proof in the appendix on page 30.
Claim 2.3 (Unique block-decomposition of components).Let A be any subset of vertices in Λ n .There exists b ∈ N such that A can be uniquely partitioned into a 1-disconnected sequence (A i ) i≤b of blocks up to permutation of the blocks.Further, if A is the vertex set of a connected component C of G n , then the block graph H Gn ((A i ) i≤b ) is connected.
Later, we will enumerate subsets S ⊆ Λ n of vertices that potentially form a component of LRP in Λ n .To ensure that a subset is isolated from the rest of the graph, there must be no edge from S to its "surrounding" inside Λ n .This motivates the following definition of boundaries with respect to Λ n .Definition 2.4 (Boundaries).Let A ⊆ Λ n .We define the exterior boundary of A with respect to Λ n and Z d , respectively, as We define the interior boundary of A with respect to Λ n and Z d , respectively, as If, in words, we mention the exterior or interior boundary of A then -unless explicitly specified differently -we mean with respect to Λ n .
We mention that ∂ int Λ n is the 'usual' vertex boundary of Λ n .The boundary ∂ ext A may contain vertices outside Λ n , and will be useful in the enumeration of subsets forming isolated components below.It may happen that a block A contains (many) vertices of ∂ int Λ n .On such regions, A may not have exterior boundary vertices, implying that there ∂ int A is also empty.The next claim contains basic properties of blocks, their closures, and their boundaries, with proof in the Appendix on page 30.Claim 2.5 (Blocks, their closures and their boundaries).The following five statements hold: The statements (ii)-(iii) say that if B 1 and B 2 are 1-disconnected, then either B 1 is inside a hole of B 2 or the other way round (ii), or their closures are also 1-disconnected (iii).Part (v) states that a hole H of a 1-connected block B cannot contain further holes.
The fourth statement (iv) of the preceding claim is [14,Lemma 2.1].The next claim shows that the sizes of the boundaries with respect to Λ n and Z d are of the same order, provided that the set has cardinality at most 3n/4.Moreover, the claim contains an isoperimetric inequality that we extensively use below.The proof is given in the appendix on page 31.
The next lemma is due to Peierls [33].We give a proof in the appendix on page 34.More general results also appeared in [4,36].We will use this claim to enumerate blocks that satisfy A = Ā.Lemma 2.7 (Peierls' argument [33]).There exists a constant c pei > 0 such that for all x ∈ Z d and m ∈ N, The proof relies on the fact that ∂ int A is * -connected [ We extend this notation also to subsets of A. For each of the sets and k ∈ N we define an event, namely and (2.9) The following deterministic claim holds for any graph on vertices in Λ n .It shows that the union of these events contains the event {|C (2)  n | ≥ k}.In particular, the proof reveals why we could restrict to sets with A = Ā in the definition of A small in (2.7).Claim 2.8.Consider the graph G n from Definition 1.1, with C (2)   n the second-largest component of G n .Then |C (2)  n The size restriction n/2 is possible since otherwise C (2)   n would be the largest component.Take holds for all i ≤ b, and also b ≤ n/2, and i≤b |A i | ≥ k.
We distinguish two cases.Either (1) there is at least one block that is in A large or (2) all the blocks are in A \ A large .In the first case the event E 2 in (2.9) holds: a block of C satisfying A large (say A) is by definition 1-disconnected from the other blocks of C, and since C is a component of G n , A is G n -disconnected from its exterior boundary, hence E 2 (G n ) defined in (2.9) holds.
Cluster-sizes in long-range percolation In case (2) all the blocks (A i ) i≤b are in A \ A large , and, since they form the component C, the graph G n spanned on ∪ i≤b A i is connected, while their union is disconnected in G n from the rest of the graph.Finally their disjointness and |C| ∈ Formally we describe this event as (changing the sets to be denoted by B i to avoid clash of notation later): Taking a union over the number of blocks and combining the two cases, we arrive at Take now any graph G n on the vertex set Z ∩ Λ n .Think of this as the realization of G n .
We will show the implication that (2.10) Given G n for which in the union.The conditions of Claim 2.5 are satisfied for any pair B i , B j with i ̸ = j, hence for each pair, Bi and Bj are either 1disconnected disjoint sets, or one contains fully the other one.Choose now those sets in { B1 , . . ., Bb } that are not contained in any other set in the same list.We then obtain an integer b ′ ≤ b and a 1- (2.11) Since ( B1 , . . ., Bb ′ ) is a 1-disconnected sequence of blocks that are equal to their own closure, (2.10) since the left-hand side was assumed in E 1 (b, G n ), and the block graph being connected is a less demanding event than the actual spanned graph being connected, and the second containment follows since each set of edges in G n that ensures that H Gn ((B i ) i≤b ) is connected, also ensures that the block graph on the closures of (B i ) i≤b is connected.
The second inclusion (2.13) follows by using ∪ i≤b B i ⊆ ∪ i≤b ′ Bi in (2.11).For the third inclusion (2.14) we have to argue that the set of edges that is excluded on the right-hand side is contained in the set of excluded edges on the left-hand side.Clearly We state two lemmas that together with Claim 2.8 prove Proposition 1.5.We prove the two lemmas in the following sections.Lemma 2.9 (Unlikely block graphs).Let G n be long-range percolation on Λ n as in Definition 1.1 with d ≥ 2, α > 1 + 1/d.There exists a constant c 2.9 = c 2.9 (d, α) > 0 such that for all k, n sufficiently large whenever p and β guarantee that the base of k (d−1)/d is sufficiently small.
We give a brief intuition on the powers occurring on the right-hand side of (2.15).
When β ≥ 1 in (2.15), the factor (1 − p) β comes from the fact that any vertex within β distance from the boundary of total size at least k (d−1)/d of the blocks ∪ i≤b A i in (2.8) must be G n -disconnected from ∪ i≤b A i .An edge between two vertices within distance β is absent with probability 1 − p by (1.2).In this case pβ dα ≥ 1 is possible; this factor arises from counting the number of possible spanning trees on the blocks.When β < 1, we only exclude edges of length 1 around the boundary of ∪ i≤b A i , and the factor pβ dα is at most 1.
We proceed to the lemma dealing with ( The function f (p, β) describes how large p and β should be in the next lemma.It is a technical artifact of the proof that we did not optimize.We write x ∨ y = max(x, y).
To prove (2.17), we use that any set A with size below n/2 but closure Ā with size above 3n/4 must have boundary at least Θ(n (d−1)/d ), and large holes to which it cannot be connected by an edge in G n .The (log n) −2 factor is a technical artifact of our proof based on the pigeon-hole principle.In both Lemma 2.9 and 2.10, the challenge is that a simple union bound would yield too large combinatorial factors arising from the possibilities for the sets (A i ) i≤b and A, respectively.See also Section 1.1.

Spanning trees on block graphs
We work towards proving Lemma 2.9.Recall the event E 1 (b) := E 1 (b, G n ) from (2.8), and A small for the blocks in Λ n from (2.7) on which there should be a connected block graph H Gn ((A i ) i≤b ) (see Definition 2.2).By a union bound over all possible 1disconnected sequences of blocks (A 1 , . . ., A b ) ∈ 1 A small whose total size is at least k, we obtain where the factor 1/b! corrects for the permutations of (A 1 , . . ., A b ) yielding the same blocks, but ordered differently.Using the independence of edges in long-range percolation in Definition 1.1, we obtain The block graph H Gn ((A i ) i≤b ) can only be connected if it contains a spanning tree on its blocks.To count these spanning trees, we introduce the rooted labeled f -tree.In the following definition, we use that each tree on b vertices has b − 1 edges.
. We call f the vector of forward degrees.A rooted labeled tree on b vertices is an f -tree if the root has label 1, and it has an outgoing edge to each of the vertices with labels 2, . . ., f 1 + 1, vertex 2 has an outgoing edge to each of the vertices with labels f 1 + 2, . . ., f 1 + f 2 + 1, and so on, the vertex with label j has an outgoing edge to each of the vertices with labels 2+ ) is a directed edge in an f -tree, then we say that i is the parent of j and j is the child of i.We say that the labeled block graph H Gn ((A i ) i≤b ) is f -connected, if it contains an f -tree on its vertices (1, 2, . . ., b).
Given a forward-degree vector f and a labeled set of vertices, the f -tree is uniquely determined.The construction ensures that the block with label b must be a leaf, i.e., it has forward degree f b = 0, and its parent corresponds to the label of the last nonzero entry of f .Further, given a tree T with labeling f ∈ F b , upon removing the leaf with label b, we obtain a tree T \ {b} on {1, . . ., b − 1} with a labeling in F b−1 .
Further, an f -tree always has vertex 1 as its root, and the forward neighbors of any vertex have consecutive labels.Hence, not all the b! labelings of a tree T are valid labelings, i.e., no vector f ∈ F b can be associated to some labelings.However, for a fixed tree T on a connected block graph H Gn ((A i ) i≤b ), there is at least one permutation σ of (1, 2, . . ., b) with σ(1) = 1 and a vector f In other words, we can relabel the blocks so that the new labeling (1, σ(2), . . .σ(b)) is a proper labeling of T , for some f ∈ F b in Definition 3.1.We denote the set of permutations of (1, 2, . . ., b) with 1 a fixed point by S 1 b .Note that the choice of the spanning tree T may not be unique if H Gn ((A i ) i ) is connected.We obtain on the first factor inside the sum in counting the isomorphisms of rooted trees: namely, the (consecutive) labels of the forward neighbors of any vertex v may be permuted (yielding the factor f v ! for each vertex), resulting in permuting the labels in the forward-subtrees of v accordingly.For any such (f , σ) and (f ′ , σ ′ ), we then also have that Hence, in the above union each rooted tree T (with root fixed) on H Gn (( We substitute this into (3.1), and use that the product b i=1 1 fi! is invariant under label permutations.So we arrive at .
We now argue that the sum over the permutations and the factor 1/(b − 1)! cancel each other.Given A 1 , let (B 2 , . . ., B b ) ∈ 1 A small be arbitrary blocks of total size at least k −|A 1 |, also 1-disconnected from A 1 .Then, for any permutation σ ∈ S 1 b , in the summations over the blocks A 2 , . . ., A b , there is precisely one combination of blocks such that A σ(i) = B i for all i ≤ b.Hence, when summing over all permutations σ ∈ S 1 b , we counted the case that the blocks are A 1 , B 2 , . . ., B b exactly (b − 1)! times.This cancels the factor 1/(b − 1)!, and we arrive at Lastly, we prescribe the sizes of the boundaries of the blocks.We introduce the possible boundary-length vectors: Then, The next two statements will imply Lemma 2.9.

Statement 3.2 (Counting spanning trees)
. Let G n be long-range percolation on Λ n as in Definition 1.
The bound given by the previous statement increases exponentially in While this might look harmful, the next statement will compensate for this, so that the two bounds combined, for appropriate choices of the constants p and β, still gives a bound decaying exponentially in i∈[b] m i .
Cluster-sizes in long-range percolation We show first that Lemma 2.9 follows from these statements, and then prove the statements in the remainder of the section.
Proof of Lemma 2.9, assuming Statements 3.2 and 3.3.Substituting the bounds from Statements 3.2 and 3.3 into the right-hand side of (3.4) yields for β ≥ 1 that In what follows we evaluate the summation over the vectors m ∈ M b (k).We recall from (3.3) that m represents the vector of interior boundary sizes of 1-connected sets (A i ) i≤b ∈ 1 A small with total size at least k, and A i = Āi ≤ 3n/4 for all i ≤ b by the definition of A small in (2.7).In (3.3), the boundary is taken with respect to Z d , i.e., not with respect to Λ n .By the isoperimetric inequality in Claim 2.6, for all blocks /d is concave and increasing, we obtain for all m ∈ M b (k) and any Hence, separating the summation in (3.7) according to the possible values of i m i = ℓ ≥ k (d−1)/d , we arrive at Since b ≤ ⌊n/2⌋, we obtain by a union bound over the number of blocks that (3.9) The two sums are independent of each other.We bound the first sum from above by n log n.We obtain The two constants C 3.2 and c 3.3 depend only on d and α.We may assume that p and β are such that pβ dα (1 − p) β•c3.3 is smaller than (2C 3.2 ) −2 , and under this assumption, we obtain Lemma 2.9 when β ≥ 1.
and conclude in the same way.

Proof of Statement 3.2
We start with a geometric claim.

Claim 3.4.
There exists a constant C 3.4 > 0 such that for each block A ∈ A small in Λ n and all r ∈ N, (3.10) Proof.We start counting line-segments of the right length with endpoints in Z d crossing a single unit square that will be centered later at some vertex in ∂ int A.
Let B 0 := [−1/2, 1/2] d .For two vertices x, y ∈ Z d , let L x,y denote the segment between x and y on the unique line connecting x and y.Define We will show that there exists a constant C 3.4 > 0 such that |Cross(r)| ≤ C 3.4 r d .
(3.11) Indeed, for each pair (x, y) ∈ Cross(r), at least one of the inequalities ∥x∥ ≥ r/2 and ∥y∥ ≥ r/2 is satisfied.Without loss of generality we may assume that ∥x∥ ≥ r/2, and then also ∥x∥ ≤ r + 1. Fix then such a vertex x.Let S x (r 0 ) denote the smallest spherical cone with apex at x that completely contains B 0 , with r 0 being the radius of this cone.Let S x (r) denote a cone with apex x that has the same boundary lines (and the same angle) as S x (r 0 ), but radius exactly r.Then, since ∥x∥ ∈ [r/2, r + 1] by assumption, Further, every y ∈ Z d such that (x, y) ∈ Cross(r) must be contained in S x (r + 1) \ S x (r), since all half-lines emanating from x that cross B 0 are contained in S x (∞), and ∥x − y∥ ∈ (r, r + 1].Since the radius of S x (r + 1) is at most by a factor two larger than the radius r 0 of S x = S x (r 0 ) for all r ≥ 1, by homothety of the cones, |(S x (r + 1) \ S x (r)) ∩ Z d | is bounded from above by a dimension-dependent constant, and so for each x with ∥x∥ ∈ [r/2, r + 1], the number of pairs (x, y) ∈ Cross(r) is bounded from above by a dimension-dependent constant.Summing over all the at most O(r d ) many such x, we obtain (3.11) for some C 3.4 > 0.
To arrive to (3.10), the block A ∈ A small in (2.7) ensures that A = Ā.Its interior boundary ∂ int A is then * -connected by Claim 2.5 Part (iv) ([14, Lemma 2.1]).This implies that there exists a * -connected surface fully contained in ∂ int A separating vertices in A \ ∂ int A from vertices in Z d \ A. Hence, for each pair x ∈ A and y / ∈ A, there exists at least one vertex z ∈ ∂ int A such that L x,y intersects the axis-parallel box z + B 0 .Here x = z may occur.The statement of the claim now follows by (3.11) when summing the at most C 3.4 r d such pairs for each vertex z on the boundary of A.
We continue with a lemma treating the connectedness of the block graphs, i.e., the inner summation on the left-hand side of (3.5) in Statement 3.2.We point out that in this lemma we only bound the event that the block graph H Gn is connected, not the event that the actual graph is connected.Lemma 3.5.Let G n be long-range percolation on Λ n as in Definition 1.
We comment that it is this lemma in the proof that crucially uses that α > 1 + 1/d.
Proof.We will prove the statement by induction on b.We first define the finite constant C 3.5 , using that α > 1 + 1/d as follows: We start with the initialization.Assume first that b = 1, which corresponds to a tree on a single vertex (representing the block A 1 ), so its forward degree is f 1 = 0. We obtain Since A = Ā for all A ∈ A small by definition in (2.7), we can apply Lemma 2.7, which yields, since |Λ n | = n, Since m f1 1 = m 0 1 = 1, this finishes the induction base for (3.12).We now advance the induction.Assume (3.12) holds up to b − 1.Let f ∈ F b and consider the summation over the last block A b ∈ A small on the left-hand side in (3.12).By construction of the f -tree in Definition 3.1, the b-th block is a leaf in the f -tree, and f b = 0. Its parent in the f -tree is the largest vertex-label ℓ in f that is nonzero, and the remaining labeled graph upon removing b is a tree, with a labeling in F b−1 (see the comment below Definition 3.1).Then, the forward degrees of this new tree are given by f Independence of edges in G n by Definition 1.1 yields where in the subscripts of the sums (and also in the remainder of the proof) we implicitly assume that the blocks (A i ) i≤b ∈ 1 A small are 1-disconnected, and that | ∂ int A i | = m i for all i ≤ b.We focus on the summation over A b .Here, A b ∈ A small and A b is 1-disconnected from A ℓ .We decompose the sum according to the length r of an edge (x, y) connecting x ∈ A ℓ and y ∈ A b , and use that ∥x − y∥ > 1 by the 1-disconnectedness of A ℓ , A b .By the connection probability (1.2), p(1 ∧ β/r) dα < pβ dα r −dα holds for all r > 0. So, by a union bound, it follows that Cluster-sizes in long-range percolation Substituting this back into (3.15)yields with the constant C 3.5 from (3.13), We substitute this bound back into (3.14), and use the induction hypothesis: e cpeimi m fi i .
To obtain the fourth row we used that f ′ i = f i for all i ̸ = ℓ, i ≤ b − 1, and f ′ ℓ = f ℓ − 1 by construction, yielding the 1/m ℓ factor.We can rearrange the expression and obtain (3.12), using that f b = 0 (the last block is a leaf).This finishes the proof.
We are ready to prove Statement 3.2.
Proof of Statement 3.2.Using the result of Lemma 3.5 on the left-hand side of (3.5) of Statement 3.2, we arrive at e cpeimi m fi i f i ! . (3.16) We first analyze a single summand, i.e., the value for a fixed f ∈ F b .Since It follows from standard differentiation techniques that for any a, x ≥ 1, the function g a (x) = (ae/x) x is maximized at x = a.Maximizing all factors (m i e/f i ) fi at f i = m i yields that (m i e/f i ) fi ≤ e mi for all i ≤ b.Since by definition of F b in Definition 3.1 we have e cpeimi m fi i Using again that f 1 + . . .+ f b = b − 1 for all f ∈ F b , and using the same combinatorial bounds as for m above (3.8),we obtain

Proof of Statement 3.3
We start with a geometric claim.Recall A small from (2.7), and holes from Def. 2.1.Claim 3.6.Let (A 1 , . . ., A b ) ∈ 1 A be a 1-disconnected sequence of blocks without holes (i.e., A i = Āi for all i ≤ b), with A : As a result of (2.4), x ∈ ∂ int A i , establishing the statement.We turn to prove * -connectedness of (Λ n \ A) ∪ (∪ i≤b ∂ int A i ).Using Claim 2.3, we decompose the set Λ n \A into a 1-disconnected sequence of 1-connected blocks (B j ) j≤b ′ for some b ′ ≥ 1: (3.17) We define now an auxiliary graph.For each set ∂ int A i we associate a vertex a i for i ≤ b, and also for each block B j a vertex t j , for j ≤ b ′ .We define an auxiliary graph H on the vertex set V = {a 1 . . ., a b , t 1 , . . ., t b ′ }.We say that a i ∼ H t j if there is a pair of vertices We will show that H consists of a single connected component.This then implies that To show that H consists of a single connected component, we argue as follows.Let f : (Λ n \ A) ∪ (∪ i≤b ∂ int A i ) → {0, 1} be a function that is constant on * -connected subsets of its domain.Since we assumed that A i = Āi for all i ≤ b, Claim 2.5(iv) is applicable which states that ∂ int A i is * -connected for each i ≤ b.Further, (B j ) j≤b ′ are blocks, i.e., 1-connected and also * -connected.So, f (x) = f (y) =: f H (a i ) for all x, y ∈ ∂ int A i , and also f (x) = f (y) =: f H (t j ) for all x, y ∈ B j .This defines a function f H : V → {0, 1}.Since f is constant on * -connected subsets of its domain, f H is constant on each of the connected components of H. Using again that each ∂ int A i is * -connected and * -connected to all vertices in A i , f (x) = f H (a i ) for all x ∈ ∂ int A i , and f can be uniquely extended to a function g that takes the value g(x) := f H (a i ) on all vertices of A i .If we then set g(y) := f H (t j ) for all y ∈ B j , then g : Λ n → {0, 1} is a function that is constant on *connected components of (Λ n \A)∪(∪ i≤b A i ) = (Λ n \A)∪A = Λ n .Now we may notice that Λ n consists of a single * -connected component, hence g must be constant everywhere.This implies that f and hence f H must have been also a constant everywhere.This implies that H consists of a single connected component and so Proof of Statement 3.3.Let A 1 , . . ., A b ∈ 1 A small , and denote A := ∪ i≤b A i .We first assume β ≥ 1.We define the set of potential edges between the interior boundary of A with respect to Z d and the set of vertices outside A within distance β as Considering the event on the left-hand side of (3.6) in Statement 3.3, there should be no edges in G n between (∪ i≤b ∂ int A i ) and Λ n \ A. In particular, all edges in ∆(A) must be Cluster-sizes in long-range percolation absent.The distance β is chosen so that all edges of this length are present in G n with probability p by (1.2).Hence, Our goal is to show that for some constant c = c(d) > 0, since then (3.19) holds with c = c ′ δ.In order to show (3.20), our first goal is to find enough pairs of vertices in ∆(A) around a linear fraction of vertices in ∪ i≤b ∂ int A i .For this, we claim that a set T := {(x ℓ , y ℓ )} ℓ≥1 with the following properties exists: and ∥x ℓ − y ℓ ∥ = 1 for all ℓ ≥ 1; (ii) each vertex z ∈ Λ n appears at most once in a pair in T ; Note that requirement (i) implies that all (x ℓ , y ℓ ) ∈ T are elements of Λ n × Λ n .We now show that a set T exists.Consider the following greedy algorithm: order the vertices in ∪ i ∂ int A i in an arbitrary order, to obtain the list (v Since each v j is in ∪ i ∂ int A i , for each v j there is at least one vertex y j ∈ ∪ i ∂ ext A i ⊆ (Λ n \A) with ∥v j − y j ∥ = 1 by Definition 2.4 (recall that the sets A 1 , . . ., A b are 1-disconnected).
Starting with T 1 := {(v 1 , y 1 )}, going through the ordering of (v j ) j one-by-one, append the pair (v j , y j ) to the list T j−1 , if and only if y j has not been contained in any pair of T j−1 yet and so obtain T j .Set then T := T M .Since any y ∈ ∪ i ∂ ext A i neighbors at most 2d many interior boundary vertices, adding a certain pair (v j , y j ) only affects at most 2d − 1 other indices where a pair may not be added later.Hence, (since the set on the right-hand side contains ∂ int Λ n , which has cardinality Θ(n (d−1)/d ), the set on the right-hand side has size at least R for n sufficiently large, so such a self-avoiding path of length R then exists).By the triangle inequality, Cluster-sizes in long-range percolation Define now the type typ(z Then define the set of (unordered) pairs representing potential edges in G n ∆(x ℓ , y ℓ ) := {z 1 , typ(z (3.24) The inclusion holds since for each of these pairs, exactly one element is in Λ n \ A and the other one is in ∪ i≤b ∂ int A i , and the distance between the two vertices of each pair is at most β by (3.23).We claim that To see the inequality with ⋄, we show that each potential edge {z, z ′ } ∈ Λ n × Λ n appears at most twice in a set in the union in the middle.Consider {z, z ′ } ∈ Λ n × Λ n .First, assume that there exists ℓ such that (z, z ′ ) = (x ℓ , y ℓ ) ∈ T or (z ′ , z) = (x ℓ , y ℓ ) ∈ T .Without loss of generality, we assume that the pair is ordered such that (z, z ′ ) = (x ℓ , y ℓ ) ∈ T .Then there is no (x j , y j ) ∈ T different from (x ℓ , y ℓ ) such that {z, z ′ } ∈ ∆(x j , y j ), since each element in ∆(x j , y j ) contains either x j or y j , which are different from x ℓ and from y ℓ by requirement (ii) in the construction of T .Moreover, the element {z, z ′ } = {x ℓ , y ℓ } is contained at most once in the set ∆(x ℓ , y ℓ ), since the first coordinates in (3.24) are all different as they form a self-avoiding path, and the first coordinates do no not contain x ℓ = z by (3.22).So, if (z, z ′ ) ∈ T , then this pair of vertices only appears once in ∪ (x ℓ ,y ℓ )∈T ∆(x ℓ , y ℓ ) .Next, assume that (z, z ′ ), (z ′ , z) / ∈ T , but {z, z ′ } is contained in some ∆(x ℓ , y ℓ ).Then, either z or z ′ must be equal to either x ℓ or to y ℓ by (3.24).Assume without loss of generality that z ∈ {x ℓ , y ℓ }, and therefore z ′ / ∈ {x ℓ , y ℓ }.Thus, {z, z ′ } is contained exactly once in ∆(x ℓ , y ℓ ).The only way that {z, z ′ } could be in a set ∆(x ℓ ′ , y ℓ ′ ) for some (x ℓ ′ , y ℓ ′ ) ̸ = (x ℓ , y ℓ ), is when z ′ ∈ {x ℓ ′ , y ℓ ′ } and (x ℓ ′ , y ℓ ′ ) ∈ T for some ℓ ′ ̸ = ℓ.(This can only happen if the self-avoiding path from x ℓ passes through either x ℓ ′ or through y ℓ ′ and the self-avoiding path from x ℓ ′ passes through either x ℓ or y ℓ ).This argument implies that the element {z, z ′ } can be contained at most twice in a set in the union in (3.25), namely in ∆(x ℓ , y ℓ ) and ∆(x ℓ ′ , y ℓ ′ ), and the inequality ⋄ in (3.25) we use that each vertex on the interior boundary is within distance one from a vertex on the exterior boundary, hence Assume now β < 1.Each vertex on the interior boundary of A = ∪ i≤b A i is at distance 1 from at least one vertex in Λ n \ A by definition.We again use that the conditions of the isoperimetric inequality in Claim 2.6 are satisfied, so that |δ int A i | ≥ δm i for all i ≤ b.Each individual edge of length 1 is absent with probability (1 − pβ dα ).Thus, only excluding such edges, we obtain the second case of Statement 3.3 since c 3.3 (d) ≤ δ.

Counting holes
We turn to the proof of Lemma 2.10.We set up a few preliminaries about holes.
Recall that A denotes the family of 1-connected blocks in Λ n from (2.1), and that A large = {A ∈ A : | Ā| > 3n/4, |A| ≤ n/2} from (2.7).Recall also from Definition 2.1 that the holes H A of a 1-connected set A ∈ A are the 1-connected subsets of Ā \ A. By Definition 2.1, each hole H ∈ H A is surrounded by A. This implies that H does not intersect the boundary of the box ∂ int Λ n .Hence, it follows by Definition 2.4 of the boundaries that for all H ∈ H A , The following definition (and claim) of principal holes will ensure that the total size of the boundaries of the holes of A in (4.2) is sufficiently large compared to the combinatorial factor arising from the number of possible sets A there.

Definition 4.1 (Principal holes)
Since |H A (i)| is an integer for all i, the inequality |H A (i)| ≥ ⌈h n (i)⌉ also holds whenever the inequality in (4.3) holds.We define for i ∈ N and observe that A might appear in both A large (i) and A large (j) if both type i and type j are principal for A. We define the following β-dependent constants: A large (i). (

4.6)
There exists a constant c 4.2 = c 4.2 (d) > 0 such that for all A ⊆ Λ n , with any i A being any principal hole-type for A, since the sum converges to π 2 /6 = 1.64... < 2. This contradicts the assumption that the total size is at least n/4, so there must be at least one principal hole-type, say i A .The restriction i ≤ ⌈log 2 n⌉ follows since the number of vertices in Λ n is n, and H > 2 (⌈log 2 n⌉+1)−1 thus can never be satisfied.This shows (4.6).
We turn to (4.7).As argued before (4.1), holes do not intersect the boundary of the box this yields that for some constant c 4.2 = c 4.2 (d) > 0. Lastly, we prove (4.8).Using again that ∂ ext H ≥ |H| (d−1)/d for each hole, we obtain for any hole H with type i ≥ i ⋆ , This finishes the proof of the claim.
We use the inclusion in (4.6) to bound the event on right-hand side in (4.2) (with the convention that the empty sum from 1 to i ⋆ is 0).After a union bound we arrive at We now bound these two sums, the first one corresponding to small principal hole types, the second one corresponding to large principal hole types.

Excluding small principal hole types
The claim shows that the probability that large sets with small principal hole types appear as a component of G n , decays exponentially in n whenever the base of the second exponential factor is strictly smaller than 1/2.Note in particular that this error term is not present when β < 1.
Proof.We may assume that i ⋆ (β) ≥ 2 in (4.5), since otherwise the sum would be empty and the bound holds trivially.By definition of i ⋆ and R 2 (β) in (4.5), we may therefore assume that log 2 R 2 > 0, which is equivalent to R 2 (β) = ⌊β/ √ d⌋ > 1 and β ≥ 2 √ d.We assume R 2 (β) > 1 throughout the remainder of the proof.We start estimating a single summand on the left-hand side of (4.11).Consider some A ∈ A large (i).By Definition 4.1, We now find enough potential edges that all must be absent in order for the event {A ̸ ∼ Gn ∪ H∈H A ∂ int H} in (4.11) to occur.By (4.7) in Claim 4.2, since i is a principal hole-type of A ∈ A large (i) We now obtain a lower bound on |A| using the isoperimetric inequality of Z d in Claim 2.6.By definition of ∂ int A in Definition 2.4, and since ∂ int A ⊇ ∂ int Ā by Claim 2.5(i), it follows from Claim 2.6 applied to Ā that for all A ∈ A large , So, by (4.12), {x, y} : These edges must be all absent in order for {A ̸ ∼ Gn ∪ H∈H A ∂ int H} to occur for A ∈ A large (i) in (4.11).The connection probability in (1.2) ensures that two vertices within ℓ 1 -distance ⌊β⌋ are connected with probability p. Hence using a union bound and then the independence of edges, we obtain where we used that A large (i) counts subsets of Λ n , and the number of subsets of Λ n is at most 2 n .This bounds a single summand in (4.11).To evaluate the sum, recalling that , where for the last inequality we used that for all i We recall that we may assume R 2 (β) = ⌊β/ √ d⌋ > 1 by the reasoning at the beginning of the proof.So, the exponent of β in the numerator becomes 1 − 1/(d − 1) = (d − 2)/(d − 1) in (4.11).The prefactor i ⋆ is a (β-dependent) constant by (4.5).Therefore, the statement in (4.11) follows by adapting the constant in the exponent.

Excluding large principal holes
We turn to the sum in (4.10).Claim 4.4 (Sets with large principal hole-types are unlikely components).Let G n be long-range percolation on Λ n as in Definition 1.1 with d ≥ 2, α > 1, and i ⋆ (β) from (4.5).There exists c 4.4 = c 4.4 (d) > 0 such that for all n sufficiently large Err large := whenever p and β guarantee that the base of c 4.4 (log n) −2 n (d−1)/d is sufficiently small..
Proof.Similarly to the small principal hole-types, we will find enough potential edges that all must be absent in order for the events on the left-hand-side in (4.13) to occur.
Fix an ordering L of vertices in Λ n so that x 1 < L x 2 < L • • • < L x n with respect to this ordering (e.g., the lexicographic ordering).For a block A ∈ A large (i) (which has at least ⌈h n (i)⌉ holes of type i by (4.3)), we order its holes H A in such a way that the holes of type i are H (1) A , . . ., H , and that for all r < s ≤ |H A (i)| the vertices smallest in the ordering within H (r) A and H (s) A -say x r ∈ H (r) A and x s ∈ H (s) A -satisfy x r < L x s .We obtain an upper bound when we exclude edges from A towards only its first ⌈h n (i)⌉ holes of type i: where to get the last row we only look at edges emanating from A that are on the exterior boundaries of the holes.This is an upper bound since ∂ ext H (j) A ⊆ A by (4.1) for all j ≤ ⌈h n (i)⌉.If for two blocks A, A ′ ∈ A large (i), the first ⌈h n (i)⌉ holes coincide, also the exterior boundaries of these first ⌈h n (i)⌉ holes coincide, and the event in (4.14) excludes the exact same edges.So, a simple union bound over A in (4.14) would overcount the non-presence of those edges too many times.Instead, we carry out a union bound over all possible lists of the first ⌈h n (i)⌉ holes.To this end, we consider for all A ∈ A large (i) the following: In words, the set D i (A) has exactly ⌈h n (i)⌉ many holes, all of which are type i, and its holes coincide with the first ⌈h n (i)⌉ type-i holes of A. D(i) collects all sets D that arise in this way for some A ∈ A large (i).Since D i (A) shares the first ⌈h n (i)⌉ type-i holes with A, also the exterior boundaries of those holes agree between A and D i (A).So, A .Hence, in (4.14) we can group the blocks in A large (i) that all map to the same D ∈ D(i), and obtain that Here we used that √ d is the maximal ∥ • ∥ 2 -distance between * -adjacent vertices, which gives ℓ 2 -distance at most β when β ≥ 1.When β < 1, R 2 = 1 and the ℓ 2 -distance between y x and x is 1 by Definition 2.4.Then, the edges {{x, z} : D , z ∈ B x } all need to be absent for the event in (4.16) to occur.When β ≥ 1, the distance bound ∥x − z∥ ≤ β ensures that all these edges are present with probability p by (1.2).When β < 1, the edge {x, y x } is not present with probability 1 − pβ dα .Combining the two cases with (4.14) and (4.16), it follows by the independence of the edges in G n that (4.17)We now encode the holes of D ∈ D(i) similar to the encoding of the blocks in Section 3. We write x D := (x 1 , . . ., x ⌈hn(i)⌉ ) for the vertices with the smallest label in the Lordering within the respective holes H (1)  D , . .Returning to (4.16), we decompose the summation on the right-hand side as follows: D with H (j) D ∋ x j and |∂ int H (j) D | = m j to count the size of D(i, x, m).The lemma can be applied since for each hole H, we have H = H by Claim 2.5(v).Hence, there are at most exp(c pei m j ) possible holes of interior boundary size m j containing x j .So, for all x ∈ Λ .Using these bounds in (4.19), and that R 2 = ⌊β/ √ d⌋ ∨ 1, we obtain The constant c pei depends only on d.In what follows, we assume that p and β are such that the first factor e (cpei+2) is at most the second factor to the power −1/2 (equivalently, the second factor is at most e −2(cpei+2) ).Then, the summands decay in m and the sum is dominated by its first term.This gives for some C > 0 and n sufficiently large that We now bound the binomial coefficient as n h ≤ n h /h!, and then e h = ∞ i=0 h i /i! ≥ h h /h! which implies h! ≥ (h/e) h .This estimate substituted into the bound on the binomial where we also used that 2 < e to obtain the right-hand-side.Using this bound in the right-hand side of (4.20), we may compare the exponents.Let i • = i • (d) be the smallest i ∈ N such that for all n ≥ 1 and all i ≥ i • , Then for i ≥ i • , where the last bound follows from the assumption on the second factor between brackets in the second line being smaller than e −2(cpei+2) above (4.20).Now we treat the case i < i • .Using that i • is a constant that only depends on d, (comparing the coefficients of c 4.2 n in (4.20) to (4.21)) we require that p, β are such that for all i < i • and all n ≥ 1, the surface-order large devation result of [14, Theorem 1.1] applies to C (1)  n (G (nn) n ), and (5.3) immediately follows.
Assume now that p(1 ∧ β) dα is not sufficiently close to 1 for the nearest-neighbor subgraph to ensure the required result.Let where for all n sufficiently large ( independently of other edges.The same is true when both x, y ] vertices and edge probability p.The probability that the graph diameter of G If the graph diameter is at most 2, then G n (Q i ) must be connected.Thus, for any fixed ε > 0, by choosing β (and hence also C(β)) large enough depending on ε, the probability that G n (Q i ) is connected is at least 1 − ε.Further, the graphs (G n (Q i )) i≤m(n) are independent since they are induced subgraphs of long-range percolation on vertices in disjoint boxes, and edges are present independently in G n by Definition 1.1.
We define a deterministic auxiliary graph G. Every box Q i corresponds to a vertex v i , for each i ≤ m(n), and two vertices v i , v j in G are adjacent if the corresponding boxes Q i , Q j are adjacent, i.e., they share a (d − 1)-dimensional face.(Similarly, one can define the 1-distance between any two vertices v i , v j by the length of the shortest path between v i , v j via adjacent vertices.)Hence, the vertices of G then form a box Λ m(n) of volume m(n) of Z d 1 .This we call the re-normalized lattice.Now we define a random subgraph H of G.We declare a vertex v i of G active when G n (Q i ) is connected.Edges of H will be only present between active and adjacent vertices in G. Assuming that two vertices v i , v j ∈ H are adjacent in G and both active, we declare the edge between v 1 and v 2 open, equivalently, present in E(H), if there exist vertices where the last inequality holds for arbitrarily small ε > 0 by making C = C(β) in (5.6) large enough.Different edges of E(G) are present conditionally independently in E(H) given that the end-vertices are active.
Finally, let H be the induced graph obtained from H on active vertices and open edges E(H).By the observation above, the vertices Then H stochastically dominates a site-bond percolation of Z 1 d in Λ m(n) .More precisely, since vertices of G are active independently with probability at least 1 − ε, and edges of G between adjacent vertices are present conditionally independently again with probability at least 1 − ε in H, each edge in the renormalised lattice Λ m(n) is open with probability at least (1 − ε) 3 .The model is 1-dependent, since the state of any edge {v i , v j } of H depends only on edges sharing at least one vertex with {v i , v j }.
Since ε can be chosen arbitrarily small, by [30, Theorem 0.0] or [32, Remark 6.2], the graph H therefore stochastically dominates iid nearest-neighbor bond percolation G ⋆ on Λ m(n) with parameter p ⋆ that can also be made arbitrarily close to 1. Hence for the sizes of the largest connected components, there is a coupling such that |C (1)  m(n) (H)| ≥ |C (1)  m(n) (G ⋆ )| holds.Thus, [14, Theorem 1.1] applies to |C (1)  m(n) (G ⋆ )|, and so for some c(β) > 0 we obtain that using (5.5)P |C (1)  m (5.8) An active v i ∈ G corresponds to a box Q i that an contains at least C(β) vertices and the graphs G(Q i ) are connected, so it holds deterministically that |C (1)  n This, combined with (5.8) implies (5.3).
We turn now to prove (5.4).Consider a smaller box Λ 2 −d n .Define then We argue that {C (1)  n ≥ ℓ} ⊆ {Z ℓ ≥ ℓ}.Indeed, if the largest component is at least of size ℓ then in Z ℓ at least ℓ many indicators are 1.Then, using (5.3) with ρ2 −d for a lower bound, and applying a Markov's inequality with ℓ = ρ2 −d n followed by a union bound yields that If for all x ∈ Λ 2 −d n it would hold that P |C 2 −d n (x)| ≥ ρ2 −d n ≤ ρ/2, then the right-hand side would be at most 1/2, while the left-hand side tends to 1.
Hence, there must exist n be such a vertex.Then, by the translation invariance of the infinite model G ∞ , looking at the component of the origin Hence, we obtain Hence, (5.4) follows by adapting the constant ρ.
Since both prerequisites of Proposition 5.1 are satisfied, this finishes the proof of the upper bounds of Theorem 1.2.

Lower bounds
For the lower bound we adapt the lower bound from [24], rephrased to the model of long-range percolation of Definition 1.1, by setting the vertex set to Z d and all vertex marks to 1 in [24].The lower bound of cluster-size decay and second-largest component that we are about the cite -[24, Proposition 7.1] -requires that in a box of volume ℓ a linear sized (at least ρℓ) giant component on vertices with marks in the interval [1, polylog(ℓ)] exists, with probability at least ρ > 0. Since in long-range percolation all vertex marks are identical to 1, this requirement of [24, Proposition 7.1] turns into the requirement (5.9) below for LRP.Moreover, the number m Z from [24, Proposition 7.1] equals 1 when we restrict to long-range percolation with α > 1 + 1/d: in the setting without vertex marks, m Z counts the number of maximizers in the set {2 − α, (d − 1)/d}.(5.9) Then there exists A > 0 such that for all n ∈ [Ak, ∞], (5.10) Moreover, there exists δ, ε > 0, such that for all (finite) n sufficiently large (5.11) In Lemma 5.2 we has just proved in (5.4) the requirement (5.9).The requirements in Theorem 1.2 are more restrictive then the assumption p ∧ β < 1 here, so the lower bounds in Theorem 1.2 follow from Proposition 5.3.In particular, (5.11) implies the lower bound in (1.3), and after taking logarithm of both sides, (5.10) implies the lower bound in (1.4).
Cluster-sizes in long-range percolation For the first case assume that x ∈ B \ ∂ int B. Then all its Z d 1 -neighboring vertices were also in B by Definition 2.4 of the interior boundary.Thus x is surrounded by B, and hence x ∈ B. Similarly, the neighboring vertices are also in B, contradicting that x ∈ ∂ int B.
For the second case assume that x ∈ (∪ H∈H B H). Then x was surrounded by B, but then also all its Z d 1 -neighboring vertices were either a member of B or surrounded by B, contradicting again that x ∈ ∂ int B.
We move on to part (ii).Assume that B1 ∩ B2 ̸ = ∅, then there exists x ∈ B1 ∩ B2 .Since B 1 and B 2 are 1-disconnected, it is excluded that x ∈ B 1 ∩ B 2 .Assume there exists some x ∈ ( B2 \ B 2 ) ∩ B 1 .Then, since x ∈ ( B2 \ B 2 ), x is surrounded by B 2 .Further, any vertex y ∈ B 1 must be surrounded by B 2 , since ∥B 1 − B 2 ∥ 1 ≥ 2 and B 1 itself is 1-connected.Hence, B 1 ⊆ B2 .Further, vertices surrounded by B 1 are then also surrounded by B 2 , so we obtain B1 ⊆ B2 .The argument when there exists some x ∈ ( B1 \ B 1 ) ∩ B 2 follows analogously yielding that in that case B2 ⊆ B1 .Lastly, assume that there exists x ∈ ( B1 \ B 1 ) ∩ ( B2 \ B 2 ), i.e., x is in the intersection of a hole of B 2 and a hole of B 1 , in particular it is surrounded by both B 1 and B 2 .Then there exists, for some j ≥ 1, a (self-avoiding) path π = (x, x 1 , . . ., x j ) on Z d 1 such that x ℓ ∈ ( B1 \ B 1 ) ∩ ( B2 \ B 2 ) for all ℓ ≤ j − 1 and then x j ∈ ∂ int B 1 ∪ ∂ int B 2 .That is, from x we start a path π of 'hole' vertices until we hit one of the interior boundaries of sets B 1 or B 2 .Assume w.l.o.g. that x j ∈ ∂ int B 1 .Since there were no vertices from B 2 on the path and x is surrounded by B 2 , it must follow that also x j is surrounded by B 2 .Similar to the previous case, it follows that B1 ⊆ B2 .
Part (iii) claims that when B1 ∩ B2 = ∅, and initially B 1 , B 2 are 1-disconnected, then ∥ B1 − B2 ∥ 1 ≥ 2. By definition of ∥ • ∥ 1 between sets, we have Each 1-connected path from x i ∈ Bi \ B i to any y / ∈ Bi must cross a vertex in ∂ int Bi . Consequently, where the second inequality follows since ∂ int B1 ⊆ ∂ int B 1 by part (i), and the third inequality since by 1-disconnectedness of B 1 and B 2 the second term on the right-hand side is at least two.The third and fourth term in the minimum in (A.1) can be bounded similarly.It follows that B1 and B2 are 1-disconnected.
The fourth statement is immediate from [14, Lemma 2.1] which states that the interior and exterior boundaries with respect to Z d 1 of any * -connected set are * -connected.We turn to the last statement, and show that for any hole H of a block B, H = H, i.e., that H does not contain holes.This is true since H was formed as a maximal 1-connected subset of the vertices in Λ n \ B surrounded by B, see below (2.2) in Definition 2.1.So if there were a hole J inside H, then J must intersect B, which would then contradict the 1-connectedness of B, since ∥J − ∂ ext H∥ 1 ≥ 2 as H fully surrounds J.The fact that the interior and exterior boundaries of a hole are * -connected follows now from Part (iv).
Proof of Claim 2.6.The proof is inspired by an argument by Deuschel and Pisztora [14,Proof of (A.3)].The inequalities with (⋆) in (2.5) follow by standard isoperimetric inequalities, but we will also derive them below.
We will first show the bounds for ∂ int and ∂ int .At the end of the proof we adjust it to ∂ ext and ∂ ext .We start by showing that there exists δ ′ > 0 such that ∂ int A ≥ δ ′ ∂ int A for all A ⊆ Λ n of size at most 3n/4.Fix such a set A ⊆ Λ n .We recall an inequality related to the isoperimetric inequality by Loomis and Whitney [31,Theorem 2].For a set A ⊆ Λ n , let S i := π i (A) denote the projection π i of A onto the i-th coordinate hyperplane.That is, for a vertex with coordinates x = (x 1 , . . ., x i−1 , x i , x i+1 , . . ., x d ) we define π i x := (x 1 , . . ., x i−1 , 0, x i+1 , . . ., x d ).Then [31,Theorem 2] proves that Let i ⋆ be the coordinate dimension that contains the largest projected set S i⋆ (ties broken arbitrarily), so that as a result of (A.We abbreviate S ⋆ = S i⋆ , and write π ⋆ for the the i ⋆ -th projection.For s ∈ S ⋆ we define the pre-image of s as π ↑ ⋆ := {y ∈ A : π ⋆ (y) = s} We now describe 'fibers' of A, informally, where A has a full n 1/d -length straight line segment on the i ⋆ th coordinate connecting two opposite faces of the box Λ n .Formally, we call a vertex s ∈ S ⋆ a projection of a fiber or shortly a fiber if there is no vertex in ∂ int A that projects to s via π ⋆ , and define the set F := {s ∈ S ⋆ : ∄y ∈ ∂ int A with π ⋆ (y) = s}. (A.4) The pre-image of any fiber does not contain any vertex of ∂ int A within Λ n , hence, it contains a full length-n 1/d line L s connecting the two opposite faces of Λ n , with π ⋆ (L s ) = s.This is because all vertices that share all coordinates with s except the i ⋆ th coordinate, project to s via π ⋆ , so A must contain all of them (the possibility of A containing none of L s is excluded by assuming s ∈ S ⋆ ), otherwise there would be a boundary vertex of A among them.Then the pre-image of any such fiber intersects the box-boundary ∂ int Λ n in exactly 2 vertices, which then must be also boundary vertices of A with respect to Z d : By the definition in (A.4), the pre-image of vertices in F are disjoint of ∂ int A (the interior boundary of A with respect to Λ n ) by the definition in (2.4), see also the text below (2.4).By the definition (A.4), the pre-image of each vertex z ∈ S ⋆ \ F contains at least one vertex in ∂ int A. A similar argument as the one for fibers shows that the pre-image of each vertex z ∈ S ⋆ \ F contains at least two vertices in ∂ int A. Namely, if z is not a projection of a fiber, then the line segment L z ∩ A must not equal L z and hence it contains at least one vertex pair x, y so that x ∈ A, y ∈ Λ n \ A. In this case y ∈ ∂ ext A and so x ∈ ∂ int A.
Further, either there is a second such vertex pair, or if there is no second such vertex pair then L s ∩ A must contain one vertex of L s ∩ ∂ int Λ n , yielding that the pre-image of z contains at least two vertices of ∂ int A.   Further, for any configuration of A ∩ L z , we see that the difference between the above two intersections is the number of vertices that A ∩ L s ∩ ∂ int Λ n contains.We obtain We characterize z ∈ S \ F according to this difference.For j ∈ {0, 1, 2} we define

Lemma 2 . 10 (
No large isolated component).Let G n be long-range percolation on Λ n as in Definition 1.1 with d ≥ 2, α > 1 + 1/d.There exists a constant c 2.10 = c 2.10 (d) > 0 such that for all n sufficiently large

Statement 3 . 3 (
Isolation).Let G n be long-range percolation on Λ n as in Definition 1.1 with d ≥ 2, α > 1 + 1/d.There exists c 3.3 = c 3.3 (d) > 0 such that for any m ∈ M b (1), and any 1-disconnected blocks since the forward degree of the vertex ℓ decreased by one upon removing the leaf b.With this notation at hand,

(3. 15 )
Using(2.6)  and that the summation in(3.12)requires that A b has boundary size m b , we can bound the last sum over A b from above by exp(c pei m b ).Next, we can apply Claim 3.4 to evaluate the summation over x ∈ A ℓ , y ∈ Z d \ A ℓ , since this sum equals the cardinality described in (3.10) with A = A ℓ .The conditions of the claim are satisfied since A ℓ = Āℓ by assuming A ℓ ∈ A small .Hence x∈A ℓ , y∈Z d \A ℓ 1 {∥x−y∥∈(r,r+1]} A b ∋y 1 ≤ e cpeim b C 3.4 r d | ∂ int A ℓ | = e cpeim b C 3.4 r d m ℓ . for some C(d, α) = C(c, C 3.5 (d, α)), where to obtain the last row we used that b − 1 ≤ b ≤ m 1 + . . .+ m b as each block has at least one interior boundary vertex.Substituting this back into (3.16),we obtain with c ′′ (d, α) := 2 + c pei + c ′ n f ∈F b (C 3.5 pβ dα ) b−1 i∈[b]

Claim 4 . 3 (
Sets with small principal hole-types are unlikely components).Let G n be long-range percolation on Λ n as in Definition 1.1 with d ≥ 2, α > 1, and i ⋆ (β) from (4.5).Then there exists c 4.3 = c 4.3 (d) > 0 such that for all β > 0 and n sufficiently large, Err small := i⋆−1 i=1 . , H ⌈hn (i)⌉ D .Let us write Λ ⌈hn(i)⌉ n,< for the vectors x ∈ Λ ⌈hn(i)⌉ n with x r < L x s for all r < s.By the initial ordering of the holes above (4.14),x D ∈ Λ ⌈hn(i)⌉ n,< .Let then m j := |∂ int H (j) D | for all j ≤ ⌈h n (i)⌉, and write m D := (m 1 , . . ., m ⌈hn(i)⌉ ).Define then for all x ∈ Λ ⌈hn(i)⌉ n,< , and m ∈ N ⌈hn(i)⌉ : D(i, x, m) := {D ∈ D(i) : x D = x, m D = m}.The set D ∈ D(i) has the first ⌈h n (i)⌉ holes of some A ∈ A large (i) where D(A) = D, and for that A, hole-type i was principal in terms of Definition 4.1 and (4.4).Definition 4.1 readily implies that the inequality (4.7) in Claim 4.2 holds for the first ⌈h n (i)⌉ many holes of A, and in turn of D with i A replaced by i in (4.7).Hence, the total interior boundary size m := ⌈hn(i)⌉ j=1 m j satisfies that m ≥ c 4.2 i −2 2 −i/d n.So for m ≥ c 4.2 i −2 2 −i/d n we introduce the possible boundary-length vectors with total size m: M i (m) := m ∈ N ⌈hn(i)⌉ : m 1 + . . .+ m ⌈hn(i)⌉ = m .
For any block B, ∂ int B and ∂ ext B are * -connected.(v) For any hole H of a block B, we have that H = H, so ∂ int H and ∂ ext H are * - connected.
14, Lemma 2.1] (which would not hold if A contained holes, or may not hold if one replaces ∂ int A by ∂ int A).Recall the notation A 1 , . . .A b ∈ 1 A in Definition 2.1 meaning that the blocks A 1 , . . ., A b are 1-disconnected.
) which then immediately yields(3.6)incombination with (3.18) when β ≥ 1.In what follows we estimate |∆(A)|.In order to do so, we will make use of the boundary ∂ int A i , i.e., the interior boundary with respect to the box Λ n .Using that all blocks in A small have size at most 3n/4 by definition in (2.7), the conditions of the isoperimetric inequality in Claim 2.6 are satisfied, and hence δm i ≤ |∂ int A i | ≤ m i for all i ≤ b.Hence, (3.19) is equivalent to showing that that there exists ⋆ has an exterior boundary of size at least R 2 (the exterior boundary of a hole is a subset of ∂ int A) to which the vertices on the interior boundary of the hole (a subset of ∂ ext A) should not be connected in G n .(Large blocks have a principal hole-type).For all A ∈ A large , there exists i A ≤ ⌈log 2 n⌉ such that hole-type i A is principal for the set A, i.e., .5)Here, if two vertices are within ∥ • ∥ ∞ -distance R 2 then they are within ∥ • ∥ 2 -distance β, hence they are connected by an edge with probability p(1 ∧ β) dα in G n by (1.2).The definition of i ⋆ ensures that any hole of type i ≥ i We argue by contradiction for the first part.By definition of A large in (2.7), | Ā| ≥ 3n/4, and also |A| ≤ n/2.Hence, the total size of the holes is at least n/4, i.e., | ∪ H∈H A H| = H∈H A |H| ≥ n/4 hold.Suppose (4.3) holds in the opposite direction for all i ≥ 1.Since the holes are 1-disconnected, it follows from the size requirement in .7) Moreover, for each hole H with type i ≥ i ⋆ in (4.5), |∂ ext H| ≥ R 2 .(4.8) Proof.
and |∂ ext H| ≥ |H| (d−1)/d by Claim 2.6 for each hole.Combined with |H| > 2 i A −1 for all H ∈ H A (i A ) and the lower bound |H A We combine the following three observations to bound a single summand in the last row.First, each vertex x ∈ ∂ int H(j)D is at distance one from at least one vertex yx ∈ ∂ ext H (j) D ⊆ D ⊆ Λ n (by definition, a hole H (j) D does not intersect ∂ int Λ n ).Second, for all j ≤ ⌈h n (i)⌉, |∂ ext H (j)D | ≥ R 2 by (4.8) and the fact that i ≥ i ⋆ .Third, the exterior boundary of a hole∂ ext H (j) D is * -connected by Claim 2.5(v).Recall R 2 = ⌊β/ √ d⌋ ∨ 1 ≥ 1 from (4.5).Hence, for each vertex x ∈ ∂ int H (j) D , starting from y x ∈ ∂ ext H (j) D , one can find a * -connected set of vertices B x ⊆ ∂ ext H (j) D that satisfies |B x | ≥ R 2 , and ∀z ∈ B x : ∥x − z∥ 2 ≤ β ∨ 1.
Now we evaluate the number of terms of the last three summations in(4.19).Each block D ∈ D(i, x, m) is uniquely characterized by its ⌈h n (i)⌉ holes by (4.15) (since these are the only holes of D).Having fixed the vectors x and m, we apply Lemma 2.7 -Peierls' argument-to each hole H(j) .6)Let us say that Q i , Q j are adjacent boxes if they share a (d − 1)-dimensional face.Since the diameter of each box is at most β/2, by (1.2), if Q i , Q j are adjacent boxes,