Expansion for the critical point of site percolation: the first three terms

We expand the critical point for site percolation on the $d$-dimensional hypercubic lattice in terms of inverse powers of $2d$, and we obtain the first three terms rigorously. This is achieved using the lace expansion.


Introduction
We study site percolation on the hypercubic lattice Z d .To this end, we fix a parameter p ∈ [0, 1] and create a random subgraph of Z d as follows.Each site (or vertex) x ∈ Z d , independently of all other sites, is declared occupied with probability p (and vacant otherwise).A bond (edge) between two nearestneighbor sites in Z d is an edge of the random subgraph if and only if the two sites are occupied.Denote by θ(p) the probability that there is a path starting at the origin 0 ∈ Z d and diverging to infinity that consists only of occupied vertices.This allows us to define the critical point as p c := inf p ∈ [0, 1] : θ(p) > 0 . (1.1) It is standard that 0 < p c < 1 in all dimensions d ≥ 2. In general, it is not possible to write down an explicit value for p c = p c (d) (see Table 1 for numerical values), a notable exception is site percolation on the two-dimensional triangular lattice (when p c = 1/2).However, it is possible to derive an asymptotic expansion for p c (d) when d → ∞.Indeed, it is known in the physics literature that The first four terms were found by Gaunt, Ruskin, and Sykes in 1976 [5] through exact enumeration, the final term has been obtained by Mertens and Moore [16] by exploiting involved numerical methods.When writing this in powers of In this paper, we extent the previously known first term by establishing the second and third term, including a rigorous bound on the error term.Table 1: Critical values for percolation on Z d , rounded to multiples of 10 −4 .The only rigorously obtained value is for bond percolation in dimension 2 (marked with * ).All other values are obtained through numerical simulation; the values for d ≥ 4 are reported in Grassberger [8] and Mertens and Moore [16].
Grimmett and Stacey [10] prove that p site c > p bond c on Z d for all dimensions d ≥ 2. This difference must be reflected in the asymptotic expansion for p c .Indeed, Hara and Slade [12] and van der Hofstad and Slade [15] rigorously obtain a series expansion for bond percolation as which indeed differs from the expansion of p site c in Theorem 1.1.Again, more precise estimates are known by non-rigorous methods [4,16]: We remark that (1.4) was proved in [15] also for the d-dimensional cube.More recently, an asymptotic expansion was also proven for the Hamming graph [3].
Borel summability of the coefficients.It appears that the methods devised in this paper allow to obtain an expansion as in Theorem 1.1 to all orders.Writing s = 1 2d and pc (s) = p c (d), this means that there is a real sequence (α n ) n∈N such that for any M ∈ N, This was proved for bond percolation by Hofstad and Slade [14] (additionally, it was proved that the coefficients α n are rational).However, it is expected that the radius of convergence for this series expansion is zero (even though rigorous evidence is lacking), and this non-convergence is valid in greater generality for series expansions of critical thresholds of various statistical mechanical models.The reason is that the sequence of absolute values |α 1 |, |α 2 |, |α 3 |, . . .grows very rapidly, cf.(1.3), and therefore it is not possible to compute pc (s) from the sequence (α n ).
Instead, we believe that the coefficients are Borel summable.Suppose pc (s) has an analytic extension to the complex disc C = {z ∈ C : Re(z −1 ) > 1}, and suppose there is L > 0 such that for all s ∈ C and all M , we have pc (s) then Sokal [17] proves that the Borel transform B(t) = ∞ n=1 α n t n /n! exists, and pc (s) equals the Borel sum pc (s) = 1 s ∞ 0 e −t/s B(t) dt. (1.7) It is, however, unclear how an analytic extension of pc (s) could be obtained.
A rare example for which we know Borel summability is the exact solution K c (d) of the spherical model.Gerber and Fisher [6] prove that there is an expansion of K c (d) in powers of 1/d, that the radius of convergence is zero, but that we may interpret the expansion as a Borel sum as described above.They also prove that the signs of the coefficients of K n oscillate: the first 12 terms are positive, the next 8 are negative, the next 9 are positive, and so on.For the well-known model of self-avoiding walk, Graham [7] proves bounds for the connective constant as in (1.6).

Strategy of proof, outline of the paper
Theorem 1.1 heavily builds upon the results obtained in [13].We use Section 2 to collect the necessary notation and results from [13] in order to prove our main result.At the heart of these results is an identity for τ p .From this, we almost immediately get an identity for p c in terms of so-called laceexpansion coefficients (see Definition 2.5).It will be clear that sufficient control over the coefficients will result in the expansion of Theorem 1.1.In fact, the results from [13] immediately give the first term of (1.3).
For the other terms in Theorem 1.1, however, we require even better control of these coefficients, which is provided by Lemma 3.1.Section 3 proves Theorem 1.1 assuming Lemma 3.1.The latter is at the heart of this paper and is proved in Section 5.As a preparation for the proof, Section 4 introduces some new notation on connection events and proves bounds on them.Those bounds are in essence an extension of some of the bounds presented in Section 2.

Site percolation: Model and basic definitions
We introduce the model more formally.Given p ∈ [0, 1], we can choose our probability space to be ({0, 1} Z d , F, P p ), where the σ-algebra F is generated by the cylinder sets, and P p = x∈Z d Ber(p).We call ω ∈ {0, 1} Z d a configuration and say that a site x ∈ Z d is open or occupied in ω if ω(x) = 1.If ω(x) = 0, we say that the site x is closed or vacant.We often identify ω with the set {x ∈ Z d : ω(x) = 1}.
For two points x = y ∈ Z d and a configuration ω, we write x ←→ y (and say that x is connected to y) if there are points Here, and throughout the paper, we write |x| We define the cluster of x to be C (x) = {x} ∪ {y ∈ ω : x ←→ y}.Note that apart form x itself, points in C (x) need to be occupied.
For an absolutely summable function f : and k • x = d j=1 k j x j denotes the scalar product.

The lace expansion in high dimension
We use this section to state the definitions and results from [13] needed in the proof of Theorem 1.1.
3. Let {u ←→ x in A} be the event that there is a path from u to x, all of whose internal vertices are elements of ω ∩ A.

5.
We define the modified cluster of x with a designated vertex u as Note that we introduce Ω = 2d.For better readability, we stick to using Ω for the remainder of the paper.We also address the Landau notation f (Ω) ≤ O(g(Ω)) that will appear frequently throughout the paper.It is always to be understood in the sense that there exists some d 0 and a constant C(d 0 ), such that f (Ω) ≤ Cg(Ω) for all Ω ≥ d 0 .The constant C may depend on other appearing parameters.
We remark that {x ←→ y in We state two elementary observations made in [13] involving J that will be important later on.The following, more specific definitions are important to define the lace-expansion coefficients: Definition 2.4 (Extended connection events).Let v, u, x ∈ Z d and A ⊆ Z d .

Define {u
In words, this is the event that u is connected to x, but either any path from u to x has an interior vertex in A , or x itself lies in A .
3. Define the event We remark that {u We can now define the lace-expansion coefficients.To this end, let (ω i ) i∈N0 be a sequence of independent site percolation configurations.For an event E taking place on ω i , we highlight this by writing E i .We also stress the dependence of random variables on the particular configuration they depend on.For example, we write C (u; ω i ) to denote the cluster of u in configuration i.
It is proved in [13] that the functions (Π (n) p (x)) n∈N0 are (absolutely) summable for every x and that Π p is thus well defined.We remark that E (u i−1 , u i ; C i−1 ) i takes place solely on ω i only if C i−1 is regarded as a fixed set; otherwise it takes place on ω i−1 as well as ω i .Proposition 2.6 summarizes the main results of [13] (namely, Theorem 1.1 and Proposition 4.2).
Then there is d 0 ≥ 6 such that, for all d > d 0 , τ p satisfies the Ornstein-Zernike equation (2.1) where we take the right-hand side to be ∞ for k = 0. Thirdly, 2dp ≤ 1 + C/d, and lastly, for n ∈ N 0 , As a consequence, we also have p x Π p (x) ≤ C/d.

Diagrammatic bounds
In the proofs to follow, we need another result from [13].We formulate it in terms of a diagrammatic notation, as we are going to make use of this later as well.To this end, we introduce some quantities related to τ p .
Definition 2.7 (Modified two-point functions and triangles).Let x ∈ Z d and define We need the following bounds obtained in [13].
As part of the proof that bounds the functions Π (i) p in [13], a first bound is formulated in terms of a long sum over products of the modified two-point functions.In a second step, those are decomposed into products of the modified triangles.We need a formulation of this intermediate bound on Π (i) p for i ∈ {1, 2} for Section 5, as well as a pictorial representation.We first state the needed bound on Π  Π (1)  p (x) ≤ w,u,t,z,x∈Z d : The bounds in [13] are formulated only for p < p c , but as the bounds are increasing in p, a limit argument easily extends them to the critical point.We now show how we represent the bound in (2.4) in terms of pictorial diagrams.As the bound on Π (2) p is even longer to write down, Lemma 2.10 is stated only in terms of these pictorial bounds.
The points w, u, t, z, x summed over are represented as squares, factors of τ p are represented as lines, and lines with a '•' ('•') symbol represent factors of τ • p (τ • p ).For example, the factor τ p (w − u) is represented as a line between two squares, which we think of as the points w and u.We interpret the factor τ p (u) as a line between u and the origin.We indicate the position of u and x in the below diagrams.After expanding the two cases in (2.4) according to whether |{t, z, x}| = 1 or |{t, z, x}| = 3, this pictorial representation allows us to rewrite the bound in (2.4) as We now formulate the bound on Π p ; more precisely, we are going to insert a case distinguishing indicator, resulting in two bounds. (2.5) (2.6)

Convolution bounds
The last result from [13] we need to state is going to be important for the proofs of Section 4.
Lemma 2.11 (Bounds on convolutions of J and τ p , [13, Lemma 4 Again, Lemma 4.6 in [13] is stated only for p < p c .A close look at the proof reveals that the bounds are sufficient for the statement to extend to p c .While the former bound is part of Proposition 2.6, the latter follows from the infra-red bound (2.2) by observing that | D(k)| ≤ 1.The bound for k = 0 follows from the continuity of the Fourier transform.

Lemma 3.1 (Expansion of lace-expansion coefficients).
As d → ∞, Lemma 3.1 is the union of Lemmas 5.1, 5.2, 5.3, which are proved in Section 5.As a preparation for these proofs, we need Section 4. These proofs are lengthy considerations of numerous percolation configurations in search for contributions of the right order of magnitude (in terms of powers of Ω −1 ).They are very mechanical in that they boil down to counting exercises and case distinctions.This also means that no new ideas are needed to extend Lemma 3.1 to higher orders of Ω −1 and expand the higher-order coefficients Π (3) , Π (4) , etc.The necessary effort increases exponentially however.
Proof of Theorem 1.1.Let first p < p c .Taking the Fourier transform of (2.1) and solving for τ p at k = 0 gives From here on out, we abbreviate Π = Π pc (0) and Π (m) = Π ) for all m ≥ 0. We can use this to describe Ωp c in more detail as Plugging in the expansion for Π (0) and Π (1) from Lemma 3.
. Using this and the first identity of (3.3) in (3.4) gives

Further bounds on connection events
This section extracts some results that are frequently used in the proofs of Section 5. We start by defining l-step connections.← − → v} as the event that u is connected to v via a path that contains at least l edges, and let τ (l) We define {u (≥l) ← −− → v} as the event that u is connected to v and the shortest path between u and v is of length at least l.Furthermore, let {u ← −− → v} be the event that u and v are connected by a path of length at most l.Lastly, set {u 2. We define {u ← −−− → v} as the event that u and v lie in a cycle of length at least l, where all sites-except possibly u and v-are occupied.

Let {u (≥l)
⇐ == ⇒ v} be the event that {u ⇐⇒ v} and the shortest cycle containing u and v (with all other vertices occupied) is of length at least l.Similarly, let {u ⇐ == ⇒ v} be the event that {u ⇐⇒ v} and the shortest cycle containing u and v is of length at most l, and let {u See Figure 1 for an illustration of (l) .We remark that τ (1) p = τ p .Moreover, note that Z d is bipartite and thus contains no cycles of odd length, which is why {u ⇐=⇒ v} and The bounds stated in Lemma 4.2 provide the core tools in dealing with lower-order terms in the bounds on Π (i) in the proofs of Section 5. Moreover, Proof.We observe that Iterating this yields To prove the first part in (4.2), note that by the BK inequality, (u, x, 0).
To prove the second part of (4.2), we combine (4.4) with Observation 2.3, yielding where the last inequality is due to Lemma 2.11.To prove the bound on τ p , we set m = |x|.note that with the bound (4.4), we can apply Observation 2.3 to obtain The first term, i.e. the term including a convolution with τ p , is bounded using Lemma 2.11.The second term, i.e. the convolutions over J, are bounded using Observation 2.2.
To prove (4.3), we split .First observe that when l 1 = l 2 = 0, When We can rewrite the left-hand side of (4.6) as where we used the same sequence of bounds as in (4.5).
Lastly, we state an observation that appears enough times throughout the arguments of Section 5 for us to extract and state it here.
Proof.Let A = {a}.We know that E (u, v; A) ⊂ {u ←→ v}.If a is vacant, then the shortest possible u-v-path that may be occupied is of length 4 and the claim holds.
On the other hand, if a is occupied, then {u ←→ v} holds.However, {u A ← − → a} also holds, and so for E (u, v; A) to hold, a cannot be a pivotal vertex.But in order for a not to be pivotal, there needs to be a second u-v-path, avoiding a.But either t is vacant, or t = a; in both cases, a second u-v-path must be of length at least 4, proving the claim.
5 Detailed analysis of the first three lace-expansion coefficients 5.1 Analysis of Π (0)   We recall that we write Π (i) = Π (i) pc (0).We will also abbreviate P = P pc and τ = τ pc throughout Section 5. We use (3.3) a lot throughout Section 5, and we recall that it states and follows from Proposition 2.6.Moreover, we will use (4.1) of Lemma 4.2 frequently in the proofs to follow and will not mention every time we do so.Lemma 5.1 (Finer asymptotics of Π (0) ).As d → ∞, Proof.Recall that Π (0) = x P(0 ⇐⇒ x) − J(x).This sum only gets contributions from |x| ≥ 2. Now, where the last identity is due to Lemma 4.2.We first consider 4-cycles.The only points x with |x| ≥ 2 that can form a 4-cycle with the origin are those with |x| = 2, x ∞ = 1.There are 1  2 Ω(Ω − 2) such points.
We are left to consider points |x| ≥ 2 contained in cycles of length 6 that also contain the origin.Note that this is possible for |x| ∈ {2, 3} and x ∞ ∈ {1, 2}.We first claim that x ∞ = 2 gives a contribution of order O(Ω −2 ).Indeed, there are Ω points x with |x| = 2 and x ∞ = 2, and any such point is contained in at most cΩ many origin-including cycles of length 6 (where c is some absolute constant).Any given 6-cycle has probability p 4 c of being present, and so the contribution is at most cΩ 2 p 4 c = O(Ω −2 ).Similarly, there are at most Ω(Ω−2) points x with |x| = 3, x ∞ = 2, and any such point is contained in exactly one origin-including cycle of length 6.Hence, this contributes at most Ω 2 p 4 c = O(Ω −2 ) as well.Let now |x| = 3, x ∞ = 1.There are 1  6 Ω(Ω − 2)(Ω − 4) such points.Such a point spans a (3dimensional) cube with the origin, in which two internally disjoint paths of respective length 3, making up the sought-after 6-cycle, have to be occupied.There are 9 such cycles.By inclusion-exclusion, |x|=3, x ∞=1 P(0 (5.2) Lastly, consider one of the 1 2 Ω(Ω − 2) points x = v 1 + v 2 with |x| = 2, x ∞ = 1, and |v i | = 1.Note that there are precisely two paths of length 2 from 0 to x, namely the ones using v i .To produce a relevant contribution to {0 ⇐ == ⇒ x}, we claim that exactly one of the two vertices must be vacant and the other occupied.Indeed, if both are occupied, then there is a 4-cycle containing 0 and x.If both are vacant, then the shortest possible cycle containing 0 and x is of length 8.
We assume v 1 to be occupied and v 2 to be vacant (the reverse gives the same contribution by symmetry, and we respect it with a factor of 2).It remains to count the number of paths of length 4 from 0 to x that avoid v 1 and v 2 .Avoiding ±v i gives Ω − 4 options for the first step.There are two options for the second step (namely, to a neighbor of v 1 or v 2 ).Steps 3 and 4 are now fixed: Out of the two shortest paths to x, one is via v i , and is not an option.In conclusion, the probability that there is a 0-x-path of length 4 traversing some fixed neighbor of 0 (which is not ±v Summing up (5.1), (5.2), and (5.3) finishes the proof.

Analysis of Π (1)
Lemma 5.2 (Finer asymptotics of Π (1) ).As d → ∞, Proof.Abbreviating C 0 = C u (0; ω 0 ), we recall that While this is a double sum over all points in Z d , we first prove that only small values of u give relevant contributions.To this end, assume that |u| ≥ 3. We use the pictorial representation of the bound in Lemma 2.9 and decompose it in terms of modified triangles introduced in Definition 2.7.In the below pictorial diagrams, points over which the supremum is taken (in particular, those points are not summed over) are represented by colored disks.The indicator that two such points (disks) may not coincide is represented by a disrupted two-sided arrow.Lemma 2.9 together with Proposition 2.8 then gives where the last identity is due to Lemma 4.2.When we encounter similar diagrams to the ones in (5.5) at later stages of this paper, we decompose them in the same way as performed in (5.5), but in less detail.We consider the cases of |u| ∈ {1, 2} separately.For both, we make further case distinctions according to the value of |x|.The contributions are summarized in the following table : Π (1) : Contributions of |u| = 1.By rotational symmetry, we can drop the sum over u, and rewrite (5.4) as P (E (u, x; C 0 ) 1 ) . (5.7) In (5.7) and in the following, we take u to be an arbitrary (but fixed) neighbor of the origin.We recall that ω i is a sequence of independent percolation configurations and an event event with subscript i takes place on ω i .Moreover, E (u, x; C 0 ) is indexed to take place on configuration 1, which is only accurate if C 0 is regarded as a fixed set; otherwise the event takes place on ω 0 and ω 1 .We proceed by splitting the sum over x in (5.7) (respectively, (5.6)) into different cases.The case of |u| = 1, x = 0 contributes Ωp c : The event E (u, v; C 0 ) 1 in (5.7) holds, the sum collapses to 1, and the contribution is Ωp c .
• Let x = ±u and v ∈ ω 1 .Note first that there are Ω − 2 choices for x, and we can treat them equally by symmetry.Now, Note that all three appearing events on the right are independent of each other.Observing  In case (a), the black path is γ 1 , the red and dotted one is γ 2 .In case (b), the two 0-x-paths are marked as black chains of arrows.In case (c), {v 1 , v 2 } ∩ ω 0 = {v 1 } and the only relevant u-x-path is marked in black.
we can replace the sum over x by a factor of (Ω − 2) and write (5.7) as ∈ ω 1 , and 0 / ∈ ω 1 .For E (u, x; C 0 ) 1 to hold, there needs to be a ω 1 -path between u and x.Its pivotal points cannot lie in C 0 however.First, note that any relevant path between u and x is of length 4, as We now investigate the 4-paths from u to x that avoid 0 and v-from Lemma 5.1, we already know that there are 2(Ω − 4) of them.Let z be one of the Ω − 4 unit vectors satisfying dim u, x, z = 3, where we let • denote the span.We denote by γ 1 and γ 2 the two u-x-paths of length 4 that visit y 1 := u + z.W.l.o.g., γ 1 visits y 2 := y 1 + x second and y 3 := y 2 − u third, whereas γ 2 visits z second and y 3 third.Let {γ i ⊆ ω 1 } denote the event that the 3 internal vertices of γ i are ω 1 -occupied.See Figure 2a for an illustration.We now show that only γ 1 produces a relevant term.Assume first that y Resolving the right-hand side of (5.8) by a union bound gives four connection events.The shortest ω 1 -path from u to x of non-vacant vertices is of length 4.Moreover, the shortest ω 1 -path from y 1 to y 3 of non-vacant vertices that avoids z is of length 4 as well, and so (5.7) is bounded by We now show that γ 1 ∈ ω 1 gives a contribution.Note that under ) for all i by Lemma 4.2, and so, by inclusion-exclusion, • Let x = ±u, v / ∈ ω 1 , and 0 ∈ ω 1 .By Observation 4.3, There are 1 2 Ω 2 choices for x.We first consider the Ω − 1 choices neighboring u and, among those, exclude the special case x = 2u first.For x a neighbor of u, we set v := x − u.
←−→ x} 0 , and so the contribution to (5.7) is bounded by Ωp c τ (4) • Let 2u = x ∼ u and v ∈ ω 0 .There are Ω − 2 choices for x.The event E (u, x; C 0 ) 1 holds, and so Ωp c 2u =x∼u • Let 2u = x ∼ u and v / ∈ ω 0 .We partition and treat the second event by observing Ωp c 2u =x∼u As the only 2-paths from 0 to x go through u and v respectively, we can focus on paths of length 4 avoiding v and u.Hence, the status of v is independent of such paths.Let z be one of the Ω − 4 neighbors of 0 with dim u, v, z = 3.For any such z, there are two 0-x-paths of length 4 that first visit z and avoid {v, u}.More precisely, these paths are (0, z, u + z, x + z, x) and (0, z, v + z, x + z, x).Let Q 4 (z) denote the event that at least one of these paths is in ω 0 .See Figure 2b for an illustration.As the events {Q 4 (z)} are pairwise independent, • Let |u − x| = 3 and x ∞ = 2.There are Ω − 1 choices for x.Let 2v = x.Note first that The complementary event is that x / ∈ C 0 and the presence of a u-x-path of length 3. The former implies v / ∈ ω 0 .There are at most four potential sites that can make up internal vertices on a u-x-path of length 3, namely 0, v, u + v, u + 2v.To avoid potential pivotality of 0 and v and still guarantee a path of length 3, we require {v + u, v + 2u} ⊆ ω 1 .But both these vertices are of distance at least 2 from the origin, and at least one of them must be in C 0 .In conclusion, , where |v i | = 1.We first show that contributions arise when precisely one point in {v 1 , v 2 } is ω 0 -occupied.Note that when both v 1 and v 2 are vacant in ω 0 , the contribution to (5.7) is bounded by ∈ ω 0 (the other case is identical and is respected by counting the contribution twice).There are and set y = u + v 1 .We claim that the only u-x-path of length 3 that produces a relevant contribution is (u, z 1 , z 2 , x).See Figure 2c for an illustration.First, assume z 1 / ∈ ω 1 .Note that the only other paths of length 3 from u to x go through either 0 or y.But {0, y} ⊆ C 0 , and so neither 0 nor y can be a pivotal point.Hence, E (u, x; C 0 ) 1 ∩{z 1 / ∈ ω 1 } enforces {0, y} ⊆ ω 1 .To get to x and avoid pivotality of any points in C 0 , at least two points in {v 1 , v 2 , z 1 } must be occupied, and the contribution to (5.7) is at most If z 1 ∈ ω 1 and z 2 / ∈ ω 1 , then the only u-x-path of length 3 through z 1 visits v 2 ∈ C 0 .This gives a contribution of O(Ω −2 ) by the same bound as above.We may turn to the case ). Applying the same bounds, we obtain a contribution to (5.7) of 2Ωp c By assumption, there needs to be some z ∈ γ with z ∈ C 0 .Consequently, z cannot be a pivotal point and so there needs to be another u-x-path γ in ω 1 that contains a point z / ∈ γ with z ∈ C 0 .Assume first that both γ, γ are paths of length 3.If they are disjoint, then the contribution to (5.7) is at most 9Ω 3 p 5 c = O(Ω −2 ).If they share their first vertex, then, in the terminology of Figure 2c, it must be either y or z 1 (otherwise 0 is pivotal).W.l.o.g., γ must then pass through z 2 and so z = z 2 ∈ C 0 needs to hold, and the contribution to (5.7) is at most Assume next that γ is of length 5.As γ and γ share at most one internal vertex (and there are two internal vertices in γ), we count a factor of p c for the unique vertex of γ, and the contribution to (5.7) is at most . Similarly, when both γ and γ are of length at least 5, the contribution is O(Ω −2 ).

The case of |u|
←−→ x} 1 , then the contribution to (5.6) is at most We first claim that only x ∈ C 0 produces a relevant contribution.Indeed, if x / ∈ C 0 , and as there is no u-x-path of length 4, we must have ←−→ z i } 0 , and so (5.7) is at most Turning to x ∈ C 0 , note that when {z 1 , z 2 } ⊆ ω 1 , then (5.7) is at most W.l.o.g., we assume that z 1 ∈ ω 1 (and z 2 / ∈ ω 1 ) and (by symmetry) count the contribution twice.Now, the contribution to (5.7) is equal to If v 1 ∈ ω 0 , then z 1 ∈ C 0 and so z 1 cannot be pivotal, which, in turn, forces {u ←−→ x} 1 .But this was already shown to produce an O(Ω −2 ) contribution.Further, if {0 ←−→ x} 0 , then (5.11) is at most , and so 0 must be ω 0 -connected to x by a path of length 3.
There are precisely two 0-x-paths of length 3 that use neither v 1 nor u, namely and so (5.11) becomes Contributions of |u| = 2.If u is one of the Ω points with |u| = 2 = u ∞ , then Π (1) is bounded by Ωp c x P(0 ⇐⇒ u)τ p (u − x).For fixed j = |u − x|, this is bounded by We now show that we can impose some further restrictions on u and x.Recall the bound in (5.5), and observe that if x / ∈ C 0 , then Similar considerations enforce that |x| ≤ 3 and |x − u| ≤ 2 as well as {0 (≤4) ⇐ == ⇒ u} 0 .Before going into the different cases, we note that there are 1  2 Ω(Ω − 2) choices for u = v 1 + v 2 (where |v i | = 1), and on every choice, {v 1 , v 2 } ⊆ ω 0 need to hold for a relevant contribution to arise.Taking all this into consideration, the contribution to Π (1) becomes where v 1 and v 2 is a pair of arbitrary but fixed independent unit vectors (and The case of |u| = 2, x = 0 contributes O(Ω −2 ): As |u − x| = 2, the contribution to (5.12) is at most The case of |u| = 2, |x| = 1 contributes Ω −1 + O(Ω −2 ): Note that we only need to consider x ∈ {v 1 , v 2 } (otherwise |u − x| = 3).For these choices of x, both x ∈ C 0 and E (u, x; C 0 ) 1 hold and the contribution to (5.12) is as claimed.
5.3 Analysis of Π (2)   Lemma 5.3 (Asymptotics of Π (2) ).As d → ∞, Proof.For the proof, we recall that where C 0 = C u (0; ω 0 ) and C 1 = C v (u; ω 1 ).We first show that when either v / ∈ C 0 or x / ∈ C 1 , then the contribution to Π (2) is O(Ω −2 ).Indeed, by Lemma 2.10 and Proposition 2.8, We expanded the third diagram in (5.14) to get the two diagrams of (5.15).We next show that only |u| = 1 gives a relevant contribution.Indeed, We can thus fix u to be an arbitrary neighbor of the origin and need to investigate Before going into specific cases, we exclude some of them right away: When |x| ∨ |u − x| ≥ 4, then the contribution to (5.16) is by Lemma 4.2.In the above, a line decorated with a '∼' symbol denotes a direct edge.Similarly, when |v| ≥ 3 or |x − v| ≥ 3, the contribution to (5.16) is at most (0, t, x) = O(Ω −2 ).
We now investigate (5.16) by splitting the double sum over v and x.We organize this by considering the three main cases for |v| ∈ {0, 1, 2}.An overview of the contributions is given in the following table: The events E (u, v; C 0 ) 1 and {v ∈ C 0 } hold.The case of |x| = 1 contributes 2Ω −1 + O(Ω −2 ): First, consider the choice of x = u.It is easy to see that the event in (5.16) holds and the contribution is Ωp ←−→ x} 1 and we receive a contribution of order O(Ω −2 ).Consider now one of the Ω − 2 remaining choices for x and set z = u + x.Then Let therefore x be one of the Ω − 2 remaining neighbors of u and note that {x ∈ C 1 } holds.
We set z ←−→ x} 2 by Observation 4.3, and the contribution to (5.16) By a similar argument to the one below (5.9), the contribution to (5.16) becomes We may therefore take v = ±u to be one of the Ω − 2 remaining neighbors of the origin.Set t = v + u.We first claim that t / ∈ ω 1 results in an O(Ω −2 ) contribution.Note that, by Observation 4.3, ←−→ v} 1 .As there is only one choice of x such that u ∼ x ∼ v and at most Ω choices such that |x| = 3 and x ∼ v, we can bound (5.16) by It remains to bound the last probability.There are at most Ω choices for x.If {u ←−→ x}, then the contribution is O(Ω −2 ).Note that the u-v-path in ω 1 cannot use and is independent of the status of 0, as the origin may not be a pivotal point.Hence, if 0 ∈ ω 1 , the contribution is at most Ωp c τ (4) As a consequence, we can focus on t ∈ ω 1 , and (5.16) reduces to But under t ∈ ω 1 , we have E (u, v; C 0 ) 1 = {t / ∈ Piv(u, v)} 1 ∪ {t / ∈ C 0 } 0 .The latter event has probability 1 − O(Ω −1 ), and so we can can instead investigate Ω 2 p 2 c (1 − O(Ω −1 )) where u and v are two arbitrary (but fixed) neighbors of 0 (satisfying (u = ±v).
The contribution of x = 0 is Ω −1 + O(Ω −2 ): Note that x ∈ C 1 holds, and so does E (v, x; C 1 ) 2 .Hence, the contribution to (5.18) is Ω ←−→ x} 1 , and so the contribution is at most The contribution of |x| = 2 is 2Ω −1 + O(Ω −2 ): We can restrict to the choices of x where v ∼ x by the considerations made in the beginning of the proof.
Let now x be one of the Ω − 3 remaining neighbors of v.As v ∼ x, we have E (v, x; C 1 ) 2 = {x ∈ C 1 }.We set z = x + u (see Figure 3b) and assume first that z / ∈ ω 1 .Then ←−→ x off {t} ∪ {t

Observation 2 . 2 (
Convolutions of J, [13, Observation 4.4]).Let m ∈ N and x ∈ Z d with m ≥ |x|.Then there is a constant c = c(m, x) with c ≤ m! such that

Figure 1 :
Figure1: An illustration of the diagrammatic quantity(l) .The '∼' symbol on the line between 0 and u means that |u| = 1.

Observation 4 . 3 .
Let a ∈ Z d .Let further u = v be two neighbors of a, and set t

Figure 2 :
Figure2: An illustration of several appearing cases for |u| = 1.In the first two cases, 0 and v are vacant in ω 1 .In case (a), the black path is γ 1 , the red and dotted one is γ 2 .In case (b), the two 0-x-paths are marked as black chains of arrows.In case (c), {v 1 , v 2 } ∩ ω 0 = {v 1 } and the only relevant u-x-path is marked in black.
then the same bounds with at least one factor of Ω in the choice of x gives a contribution of O(Ω −2 ).The case of |u| = 1, |x| ≥ 4 contributes O(Ω −2 ): The bound is the same as in (5.10).

Figure 3 :
Figure 3: An illustration of several appearing cases for |v| = 1.In (a), the two paths from u to x of length 3 that avoid 0 and t are drawn.In (b), the path along t, z which ensures x ∈ C 1 for a contribution of Ω −1 is drawn.In (c), the scenario |x − u| = 2 = |v − x| is shown, and the path along z ensuring {v ←→ x} 2 is drawn in black.