Geometry of the random interlacement

We consider the geometry of random interlacements on the $d$-dimensional lattice. We use ideas from stochastic dimension theory developed in \cite{benjamini2004geometry} to prove the following: Given that two vertices $x,y$ belong to the interlacement set, it is possible to find a path between $x$ and $y$ contained in the trace left by at most $\lceil d/2 \rceil$ trajectories from the underlying Poisson point process. Moreover, this result is sharp in the sense that there are pairs of points in the interlacement set which cannot be connected by a path using the traces of at most $\lceil d/2 \rceil-1$ trajectories.


Introduction
The model of random interlacements was introduced by Sznitman in [Szn10], on the graph Z d , d ≥ 3. Informally, the random interlacement is the trace left by a Poisson point process on the space of doubly infinite trajectories on Z d . The intensity measure of the Poisson process is given by uν, u > 0 and ν is a measure on the space of doubly infinite trajectories, see (2.7) below. This is a site percolation model that exhibits infiniterange dependence, which for example presents serious complications when trying to adapt techniques developed for standard independent site percolation.
In [Szn10], it was proved that the random interlacement on Z d is always a connected set. In this paper we prove a stronger statement (for precise formulation, see Theorem 2.2): Given that two vertices x, y ∈ Z d belong to the interlacement set, it is a.s. possible to find a path between x and y contained in the trace left by at most ⌈d/2⌉ trajectories from the underlying Poisson point process. Moreover, this result is sharp in the sense that a.s. there are pairs of points in the interlacement set which cannot be connected by a path using the traces of at most ⌈d/2⌉ − 1 trajectories.
Our method is based on the concept of stochastic dimension (see Section 2.2 below) introduced by Benjamini, Kesten, Peres and Schramm, [BKPS04]. They studied the geometry of the so called uniform spanning forest, and here we show how their techniques can be adapted to the study of the geometry of the random interlacements.
In Section 2.1 we introduce the model of random interlacements more precisely. In Section 2.2 we give the required background on stochastic dimension and random relations from [BKPS04]. Finally the precise statement and proof of Theorem 2.2 is split in two parts: the lower bound is given in Sections 5 and the upper bound in Section 4.
Throughout the paper, c and c ′ will denote dimension dependant constants, and their values may change from place to place. Dependence of additional parameters will be indicated, for example c(u) will stand for a constant depending on d and u.
During the last stages of this research we have learned that B. Rath and A. Sapozhnikov, see [RS10], have solved this problem independently. Their proof is significantly different from the proof we present in this paper.

Preliminaries
In this section we recall the model of random interlacements from [Szn10] and the concept of stochastic dimension from [BKPS04].

Random interlacements
Let W and W + be the spaces of doubly infinite and infinite trajectories in Z d that spend only a finite amount of time in finite subsets of Z d : The canonical coordinates on W and W + will be denoted by X n , n ∈ Z and X n , n ∈ N respectively. We endow W and W + with the sigma-algebras W and W + , respectively which are generated by the canonical coordinates. For γ ∈ W , let range(γ) = γ(Z). Furthermore, consider the space W * of trajectories in W modulo time shift: Let π * be the canonical projection from W to W * , and let W * be the sigma-algebra on W * given by {A ⊂ W * : (π * ) −1 (A) ∈ W}. Given K ⊂ Z d and γ ∈ W + , letH K (γ) denote the hitting time of K by γ:H (2.1) For x ∈ Z d , let P x be the law on (W + , W + ) corresponding to simple random walk started at x, and for K ⊂ Z d , let P K x be the law of simple random walk, conditioned on not hitting K. Define the equilibrium measure of K: Define the capacity of a set K ⊂ Z d as The normalized equilibrium measure of K is defined as e K (·) = e K (·)/cap(K).
(2.4) For x, y ∈ Z d we let |x − y| = x − y 1 We will repeatedly make use of the following well-known estimates of hitting-probabilities. For any x, y ∈ Z d with |x − y| ≥ 1, see for example Theorem 4.3.1 in [LL10]. Next we define a Poisson point process on W * × R + . The intensity measure of the Poisson point process is given by the product of a certain measure ν and the Lebesque measure on R + . The measure ν was constructed by Sznitman in [Szn10], and now we characterize it. For K ⊂ Z d , let W K denote the set of trajectories in W that enter K. Let W * K = π * (W K ). Define Q K to be the finite measure on W K such that for A, B ∈ W + and x ∈ Z d , The measure ν is the unique σ-finite measure such that (2.7) The existence and uniqueness of the measure was proved in Theorem 1.1 of [Szn10]. Consider the set of point measures in W * × R + : where W * K denotes the set of trajectories in W * that intersect K. Also consider the space of point measures on W * : For u > u ′ ≥ 0, we define the mapping ω u ′ ,u from Ω intoΩ by (2.10) If u ′ = 0, we write ω u . Sometimes we will refer trajectories in ω u , rather than in the support of ω u . On Ω we let P be the law of a Poisson point process with intensity measure given by ν(dw * )dx. Observe that under P, the point process ω u,u ′ is a Poisson point process onΩ with intensity measure (u − u ′ )ν(dw * ). Given σ ∈Ω, we define which we call the random interlacement set between levels u ′ and u. In case u ′ = 0, we write I u .
We introduce a decomposition of ω u as follows. Let ω 0 u be the point measure supported on those w * i ∈ supp(ω u ) for which 0 ∈ w * i (Z). Then proceed inductively: given ω 0 u , ..., ω k−1 u , define ω k u to be the point measure supported on those

Stochastic dimension
In this section, we recall some definitions and results from [BKPS04] and adapt them to our needs. For x, y ∈ Z d , let xy = 1 + |x − y|. Suppose W ⊂ Z d is finite and that τ is a tree on W . Let τ = Π {x,y}∈τ xy . Finally let W = min τ τ where the minimum is taken over all trees on the vertex set W . For example, for n vertices x 1 , ..., x n , x 1 ...x n is the minimum of n n−2 products with n − 1 factors in each.
Definition 2.1. Let R be a random subset of Z d × Z d with distribution P. We will think of R as a relation and for (x, y) and P[xRy, zRv] ≤ c xy −α zv −α + c xyzv −α , (2.14) for all x, y, z, v ∈ Z d . If R has stochastic dimension d − α, then we write dim S (R) = d − α.
Observe that inf x,y∈Z d P[xRy] > 0 if and only if dim S (R) = d.
Definition 2.2. Let R and M be any two random relations. We define the composition RM to be the set of all (x, z) ∈ Z d × Z d such that there exists some y ∈ Z d for which xRy and yMz holds. The n-fold composition of a relation R will be denoted by R (n) .
Next we restate Theorem 2.4 of [BKPS04], which we will use extensively.
Theorem 2.1. Let L, R ⊂ Z d be two independent random relations. Then when dim S (L) and dim S (R) exist.
For each x ∈ Z d , we choose a trajectory w x ∈ W + according to P x . Also assume that w x and w y are independent for x = y and that the collection (w x ) x∈Z is independent of ω.
We will take interest in several particular relations, defined in terms of ω and the collection ( If t 1 = 0, we will write M t 2 as shorthand for M t 1 ,t 2 .
In addition we have For d = 3, 4 the theorem follows easily from the fact that two independent simple random walk trajectories intersect each other almost surely, and we omit the details of these two cases. From now on, we will assume that d ≥ 5. A key step in the proof of our main theorem, is to show that for every x, y ∈ Z d , we have P[xC ⌈d/2⌉ y] = 1.
Proof. Clearly, it is enough to consider the case t 1 = 0 and t 2 = u ∈ (0, ∞). First recall that the trajectories in supp(ω u ) that intersect x ∈ Z d can be sampled in the following way (see for example Theorem 1.1 and Proposition 1.3 of [Szn10]): 1. Sample a Poisson random number N with mean ucap(x) 2. Then sample N independent double sided infinite trajectories, where each such trajectory is given by a simple random walk path started at x, together with a simple random walk path started at x conditioned on never returning to x.
We now establish a lower bound of P[xM u y]. Since any trajectory in supp(ω u ) intersecting x contains a simple random walk trajectory started at x, we obtain that (2.17) Thus the condition (2.13) is established and it remains to establish the more complicated condition (2.14). For this, fix distinct vertices x, y, z, v ∈ Z d and put K = {x, y, z, v}. Our next task is to find an upper bound of P[xM u y, and deal with the two terms on the right hand side of (2.18) separately. For a point measureω ≤ ω u , we write "xM u y inω" if there is a trajectory in supp(ω) whose range contains both x and y. Observe that if w * ∈ supp(ω u −ω u ) and x, y ∈ range(w * ), then at least one of z or v does not belong to range(w * ). Hence, the events {xM u y in ω u −ω u } and {zM u v in ω u −ω u } are defined in terms of disjoint sets of trajectories, and thus they are independent under P. We get that where in the second equality we used the independence that was mentioned above. In addition, we have (2.20) We now find an upper bound on P[ω u = 0]. In view of (2.19), (2.20) and (2.18), in order to establish (2.14) with α = d − 2, it is sufficient to show that Using the method of sampling the trajectories from ω u containing x and the fact that the law of a simple random walk started at x conditioned on never returning to x is dominated by the law of a simple random walk started at x (here we use that the trajectory of a simple random walk after the last time it visits x has the same distribution as a the trajectory of a simple random walk conditioned on not returning to x), we obtain that P[ω u = 0] is bounded from above by the probability that at least one of N independent double sided simple random walks started at x hits each of y, z, v. Here N again is a Poisson random variable with mean ucap(x). We obtain that where we in the last inequality made use of the inequality ] is the probability that a double sided simple random walk starting at x hits each of y, z, v. In order to bound this probability, we first obtain some quite standard hitting estimates. We have where the sums are over all permutations of z, y, v. Similarly, for any choice of x 1 and x 2 from {y, z, v} with x 1 = x 2 , and and (2.28) We have Using the independence between (X n ) n≥0 and (X ′ n ) n≥0 , it readily follows that From (2.28), (2.29) and (2.30) and a union bound, we obtain Proof. We start with the relation L. For x, y ∈ Z d , we have From (2.32) and (2.5), we obtain In addition, for x, y, z, w ∈ Z d , using the independence between the walks w x and w y , we get Recall the definition of the walks (w x ) x∈Z d from above.
Proof. We have Definition 3.2. Let E be a random relation and v ∈ Z d . Define the right tail field corresponding to the vertex v be Definition 3.3. Let E be a random relation. Define the remote tail field to be We say that E is remote tail trivial if F Rem E is trivial.

Left and right tail trivialities
Recall the definition of the walks (w x ) x∈Z d Section 2.2.
Lemma 3.1. The relation L is left tail trivial. The relation R is right tail trivial.
Proof. We start with the relation L. For any x ∈ Z d , we have Since the σ-algebra on the right hand side of (3.4) is trivial ([Dur10] Theorem 6.7.5), F L L (x) is trivial for every x ∈ Z d . Hence, L is left tail trivial. Similarly, we obtain that R is right tail trivial.

Remote tail triviality
We omit the details of the following lemma.
(3.5) Proof. Write η B = (η B − η K ) + η K . Observe that η B − η K and η K are independent random variables with distributions Pois(ucap(B) − ucap(K)) and Pois(ucap(K)) respectively. Consequently , conditioned on never returning to K, on the same probability space with initial starting points are independent and there is a n = n(k, ǫ, K) > 0 large enough for which Lemma 3.5. Let u > 0. The relation M u is remote tail trivial.
Proof. First we show that it is enough to prove that F Rem Mu is independent of F K = σ{xM u y : x, y ∈ K} for every finite K ⊂ Z d . So assume this independence. Let A ∈ F R Mu and let K n be finite sets such that K n ⊂ K n+1 for every n and ∪ n K n = Z d . Let Let 0 < r 1 < r 2 be such that K ⊂ B(0, r 1 ). Later, r 1 and r 2 will be chosen to be large. Fix ǫ > 0. Let N = η B(0,r 1 ) . Let C = C(K) > 0 and D = D(r 1 ) > 0 be so large that B(0, r 1 ))) distributed and conditioned on N, (w i (0)) N i=1 are i.i.d. with distributionẽ B(0,r 1 ) (·), ((w i (k)) k≥0 ) N i=1 are independent simple random walks, and ((w i (k)) k≤0 ) N i=1 are independent simple random walks conditioned on never returning to B(0, r 1 ) (see for example Theorem 1.1 and Proposition 1.3 of [Szn10]). Letting τ i be the last time (w i (k)) k≥0 visits B(0, r 1 ), we have have that ((w i (k)) k≥τ i ) N i=1 are independent simple random walks conditioned on never returning to B(0, r 1 ). We define the vector ξ = (w 1 (0), . . . , w N (0), w 1 (τ 1 ), . . . , w N (τ N )) ∈ ∂(B(0, r 1 )) 2N .
Let κ + i = inf{k > τ i : w i (k) ∈ ∂B(0, r 2 )} and let κ − i = sup{k < 0 : w i (k) ∈ ∂B(0, r 2 )}. Define the vector Observe that since A belongs to F Rem Mu and |κ are conditionally independent. Therefore, conditioned on N andγ, the events A and B are conditionally independent. It follows that for any n ∈ N and anyx ∈ (∂B(0, r 2 )) 2n (3.11) Hence, to obtain (3.7) it will be enough to show that the double sum appearing in the right hand side of (3.11) can be made arbitrarily small by choosing r 1 sufficiently large, and then r 2 sufficiently large. This will be done in several steps.
Using Lemma 3.3 we can choose r 1 big enough such that for every m < C, (3.14) We now estimate the last two lines of (3.14) separately. We have  For the last line of (3.14), we get (3.16) Combining (3.14), (3.15) and (3.16) we obtain that (3.17) By Lemma 3.4, we can choose r 2 > r 1 large enough so that for any n ≤ D and for anyȳ ∈ (∂B(0, r 1 )) 2n ,  We now have what we need to bound (3.11).
(3.20) where the first inequality follows from the triangle inequality. Since ǫ > 0 is arbitrary, we deduce that P[A|B] = P[A] from (3.11) and (3.20). The triviality of the sigma algebra F Rem Mu is therefore established.

Upper bound
In this section, we provide the proof of the upper bound of Theorem 2.2. Throughout this section, fix n = ⌈d/2⌉. Recall the definition of the trajectories (w x ) x∈Z d from Section 2.2. We have proved in Lemma 2.5 that the random relation C n has stochastic dimension d, and therefore inf x,y∈Z d P[xC n y] > 0. Since L is left tail trivial, R is right tail trivial and the relations M u(i−1)/n,ui/n are remote tail trivial for i = 1, ..., n, we obtain from Corollary 3.4 of [BKPS04] that P[xC n y] = 1 for every x, y ∈ Z d . (4.1) Now fix x and y and let A 1 be the event that x ∈ I u/n and A 2 be the event that y ∈ I (n−1)u/n,u . Put A = A 1 ∩ A 2 . We now use (4.1) to argue that M u(i−1)/n,ui/n y A = 1. (4.2) To see this, first observe that A is the event that ω 0,u/n (W * x ) ≥ 1 and ω u(n−1)/n,u (W * y ) ≥ 1. Consequently, on A, I(ω u/n | W * y ) contains at least one trace of a simple random walk started at x and hence stochastically dominates range(w y ). In the same way, I(ω u(n−1)/n,u/n | W * x )stochastically dominates range(w y ). Thus we obtain M u(i−1)/n,ui/n y A ≥ P[xC n y] = 1, (4.3) giving (4.2). Equation (4.2) implies that if x ∈ I u/n and y ∈ I (n−1)u/n,u , then x and y are P-a.s. connected in the ranges of at most ⌈d/2⌉ trajectories from supp(ω u ). Now let I 1 = [t 1 , t 2 ] ⊂ [0, u] and I 2 = [t 3 , t 4 ] ⊂ [0, u] be disjoint intervals. Let A I 1 ,I 2 be the event that x ∈ I t 1 ,t 2 and y ∈ I t 3 ,t 4 . The proof of (4.2) is easily modified to obtain P x n i=1 M u(i−1)/n,ui/n y A I 1 ,I 2 = 1. (4.4) Observe that up to a set of measure 0, we have where the union is over all disjoint intervals I 1 = [t 1 , t 2 ], I 2 = [t 3 , t 4 ] ⊂ [0, u] with rational distinct endpoints. Observe that all the events in the countable union on the right hand side of (4.5) have positive probability. In addition, due to (4.4), conditioned on any of them, we have x n i=1 M u(i−1)/n,ui/n y a.s. Therefore, we finally conclude that P x n i=1 M u(i−1)/n,ui/n y x, y ∈ I u = 1, (4.6) finishing the proof of the upper bound of Theorem 2.2.

Lower bound
In this section, we provide the proof of the lower bound of Theorem 2.2. More precisely, we show that with probability one, there are vertices x and y contained in I u that are not connected by a path using at most ⌈ d 2 ⌉ − 1 trajectories from supp(ω u ). Recall the definition of the decomposition of ω u into ω 0 u , ω 1 u ,... from Section 2.1. For k = 0, 1, ..., In addition, let V −1 = {0} and V −2 = ∅. Observe that with this notation, Here ω u| A denotes ω u restricted to the set of trajectories A ⊂ W * . We also observe that conditioned on ω 0 u , ..., ω k−1 u , under P, ω k u is a Poisson point process on W * with intensity measure u½ W * V k−1 \W * V k−2 ν(dw * ), (5.1) see the Appendix for details. We now construct the vector (ω 0 u , ...,ω k u ) with the same law as the vector (ω 0 u , ..., ω k u ) in the following way. Suppose σ 0 , σ 1 , ... are i.i.d. with the same law as ω u . Letω 0 u = σ 0 | W * {0} and then proceed inductively as follows: Givenω 0 u , ...,ω k u , defineV and letV −1 = {0} andV −2 = ∅. Then let ) .
Using (5.1) one checks that in this procedure, for any k ≥ 0, the vector (ω 0 u , ...,ω k u ) has the same law as (ω 0 u , ..., ω k u ). Let m = ⌈ d 2 ⌉ − 1. We now get that The event 0V m−1 ←→ x is the event that there is some l ≤ m − 1 and trajectories γ i ∈ω i u , i = 0, ..., l, such that γ i (Z) ∩ γ i+1 (Z) = ∅, 0 ∈ γ 0 (Z) and x ∈ γ l (Z). Sinceω i u ≤ σ i , we obtain where we use the notation M u (σ i ) for the random relation defined in the same way as M u , but using σ i instead of ω u . Now use the independence of the σ i 's and the fact that dim S (M u (σ i )) = 2, to obtain that for any l ≤ m − 1, For n ≥ 1, let x n = ne 1 . For n ≥ 1, we define the events A n = {x n ∈ I u (ω m−1 u )} and B n = {x n ∈ I u } Using (5.6) we can find a sequence (n k ) ∞ k=1 such that )} is non-empty. In particular, on the event that 0 ∈ I u , we can a.s. find points belonging to I u that cannot be reached from 0 using the ranges of at most m = ⌈d/2⌉ − 1 trajectories from supp(ω u ).

Open questions
The following question was asked by Itai Benjamini: Given two points x, y ∈ Z d , estimate the probability that x and y are connected by at most d 2 trajectories intersecting a ball of radius r around the origin.
Answering the first question can help solve the question of how one finds the d 2 trajectories connecting two points in an efficient manner.

Appendix
Here we show a technical lemma (Lemma 7.2 below) needed in the proof of the lower bound in Section 5. For the proof of Lemma 7.2, we need the following lemma, which is standard and we state without proof.
Lemma 7.1. Let X be a Poisson point process on W * , with intensity measure ρ. Let A ⊂ W * be chosen at random independently of X. Then, conditioned on A, the point process 1 A X is a Poisson point process on W * with intensity measure 1 A ρ.
Write ω u = ∞ k=0 ω k u where ω k u is defined in the end of Section 2.1. Put V −2 = ∅ and V −1 = {0} and V k = I k i=0 ω i u , k = 0, 1, ... Proof. We will proceed by induction. First consider the case k = 0. We have ω 0 The sets of trajectories W * {0} and W * \ W * {0} are non-random. Therefore we get that, using for example Proposition 3.6 in [Res08], ω 0 u andω 0 u are Poisson point processes with intensity measures that agree with (7.4) and (7.5) respectively. In addition, the sets of trajectories W * {0} and W * \ W * {0} are disjoint, and therefore ω 0 u and ω 0 u are independent. Now fix some k ≥ 0 and assume that the assertion of the lemma is true for k. Observe that we have ω k+1 u =ω k u | W * I(ω k u ) (7.6) andω k+1 u =ω k u | W * \W * I(ω k u ) . (7.7) By the induction assumption, ω k u andω k u are independent Poisson process under P k−1 . In particular, under P k−1 ,ω k u and W * I(ω k u ) are independent. Therefore, using Lemma 7.1 and (7.6), we see that if we further condition on ω k u , the point process ω k+1 u is a Poisson point process on W * with intensity measure given by u½ W * I(ω k u ) ½ (W * \W * (7.9)